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Loring Tu Introduction to Manifolds

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A graduate-level textbook by Loring W. Tu, second edition, covering manifold theory in seven chapters and twenty-nine sections. The visible contents start with smooth functions, tangent vectors as derivations, the exterior algebra of multicovectors and differential forms on Euclidean space. The prefaces mention de Rham cohomology, the Lie derivative, boundary orientation, and an appendix on quaternions. It sits in Phil's Wedge World / Sjamaar Forms folder, but the text shown contains no annotations by him.

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yLoringW.Tu AnIntroduction toManifolds Second Edition QDSpringer For other titles in this series, go to www.springer.com/series/223Universitext Editorial Board (North America): S. Axler K.A. Ribet An Introduction to ManifoldsLoring W. Tu Second Edition c°Editorial board : Sheldon Axler, San Francisco State University Vincenzo Capasso, Università degli Studi di MilanoCarles Casacuberta, Universitat de BarcelonaAngus MacIntyre, Queen Mary, University of LondonKenneth Ribet, University of California, Berkeley Claude Sabbah, CNRS, École Polytechnique Endre Süli, University of Oxford Wojbor Woyczy ´nski, Case Western Reserve UniversityLoring W. Tu Department of Mathematics Tufts University Medford, MA 02155 [email protected] ISBN 978-1-4419-7399-3 10013, USA), except for brief excerpts in connection with reviews or scholarly analysis. Use in connectionAll rights reserved. This work may not be translated or copied in whole or in part without the written with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use in this publication of trade names, trademarks, service marks, and similar terms, even if they are not identiﬁed as such, is not to be taken as an expression of opinion as to whether or not they are subject to proprietary rights. Printed on acid-free paper Springer is part of Springer Science+Business Media (www.springer.com)e-IS BN 978-1-4419-7400-6 DOI 10.1007/978-1-4419-7400-6 Library of Congress Control Number: Mathematics Subject Classification (2010): 58-01, 58Axx, 58A05, 58A10, 58A12Springer New York Dordrecht Heidelberg London 2010936466 ©Springer Science+ Business Media, LLC 2011 permission of the publisher (Springer Science+Business Media, LLC, 233 Spring Street, New York, NY Dedicated to the memory of Raoul Bott Preface to the Second Edition This is a completely revised edition, with more than ﬁfty pages of new material scattered throughout. In keeping with the conventional meaning of chapters and sections, I have reorganized the book into twenty-nine sections in seven chapters. The main additions are Section 20 on the Lie derivative and interior multiplication,two intrinsic operations on a manifold too important to leave out, new criteria in Section 21 for the boundary orientation, and a new appendix on quaternions and the symplectic group. Apart from correcting errors and misprints, I have thought through every proof again, clariﬁed many passages, and added new examples, exercises, hints, and solu-tions. In the process, every section has been rewritten, sometimes quite drastically. The revisions are so extensive that it is not possible to enumerate them all here. Each chapter now comes with an introductory essay giving an overview of what is to come.To provide a timeline for the development of ideas, I have indicated whenever possi- ble the historical origin of the concepts, and have augmented the bibliography with historical references. Every author needs an audience. In preparing the second edition, I was partic- ularly fortunate to have a loyal and devoted audience of two, George F. Leger andJeffrey D. Carlson, who accompanied me every step of the way. Section by section, they combed through the revision and gave me detailed comments, corrections, and suggestions. In fact, the two hundred pages of feedback that Jeff wrote was in itself amasterpiece of criticism. Whatever clarity this book ﬁnally achieves results in a large measure from their effort. To both George and Jeff, I extend my sincere gratitude. I have also beneﬁted from the comments and feedback of many other readers, includ- ing those of the copyeditor, David Kramer. Finally, it is a pleasure to thank Philippe Courr` ege, Mauricio Gutierrez, and Pierre V ogel for helpful discussions, and the In-stitut de Math´ ematiques de Jussieu and the Universit´ e Paris Diderot for hosting me during the revision. As always, I welcome readers’ feedback. Paris, France Loring W. Tu June 2010 Preface to the First Edition It has been more than two decades since Raoul Bott and I published Differential Forms in Algebraic Topology. While this book has enjoyed a certain success, it does assume some familiarity with manifolds and so is not so readily accessible to the av-erage ﬁrst-year graduate student in mathematics. It has been my goal for quite some time to bridge this gap by writing an elementary introduction to manifolds assuming only one semester of abstract algebra and a year of real analysis. Moreover, given the tremendous interaction in the last twenty years between geometry and topology on the one hand and physics on the other, my intended audience includes not onlybudding mathematicians and advanced undergraduates, but also physicists who want a solid foundation in geometry and topology. With so many excellent books on manifolds on the market, any author who un- dertakes to write another owes to the public, if not to himself, a good rationale. First and foremost is my desire to write a readable but rigorous introduction that gets the reader quickly up to speed, to the point where for example he or she can compute de Rham cohomology of simple spaces. A second consideration stems from the self-imposed absence of point-set topol- ogy in the prerequisites. Most books laboring under the same constraint deﬁne a manifold as a subset of a Euclidean space. This has the disadvantage of making quotient manifolds such as projective spaces difﬁcult to understand. My solutionis to make the ﬁrst four sections of the book independent of point-set topology and to place the necessary point-set topology in an appendix. While reading the ﬁrst four sections, the student should at the same time study Appendix A to acquire the point-set topology that will be assumed starting in Section 5. The book is meant to be read and studied by a novice. It is not meant to be encyclopedic. Therefore, I discuss only the irreducible minimum of manifold theory that I think every mathematician should know. I hope that the modesty of the scope allows the central ideas to emerge more clearly. In order not to interrupt the ﬂow of the exposition, certain proofs of a more routine or computational nature are left as exercises. Other exercises are scattered throughout the exposition, in their natural context. In addition to the exercises em- bedded in the text, there are problems at the end of each section. Hints and solutions x Preface to selected exercises and problems are gathered at the end of the book. I have starred the problems for which complete solutions are provided. This book has been conceived as the ﬁrst volume of a tetralogy on geometry and topology. The second volume is Differential Forms in Algebraic Topology cited above. I hope that V olume 3, Differential Geometry: Connections, Curvature, and Characteristic Classes , will soon see the light of day. V olume 4, Elements of Equiv- ariant Cohomology , a long-running joint project with Raoul Bott before his passing away in 2005, is still under revision. This project has been ten years in gestation. During this time I have bene- ﬁted from the support and hospitality of many institutions in addition to my own;more speciﬁcally, I thank the French Minist` ere de l’Enseignement Sup´ erieur et de la Recherche for a senior fellowship (bourse de haut niveau), the Institut Henri Poincar´ e, the Institut de Math´ ematiques de Jussieu, and the Departments of Mathe- matics at the ´Ecole Normale Sup´ erieure (rue d’Ulm), the Universit´ e Paris 7, and the Universit´ e de Lille, for stays of various length. All of them have contributed in some essential way to the ﬁnished product. I owe a debt of gratitude to my colleagues Fulton Gonzalez, Zbigniew Nitecki, and Montserrat Teixidor i Bigas, who tested the manuscript and provided many use-ful comments and corrections, to my students Cristian Gonzalez-Martinez, Christo- pher Watson, and especially Aaron W. Brown and Jeffrey D. Carlson for their de- tailed errata and suggestions for improvement, to Ann Kostant of Springer and her team John Spiegelman and Elizabeth Loew for editing advice, typesetting, and man- ufacturing, respectively, and to Steve Schnably and Paul G´ erardin for years of un-wavering moral support. I thank Aaron W. Brown also for preparing the List of Notations and the T EX ﬁles for many of the solutions. Special thanks go to George Leger for his devotion to all of my book projects and for his careful reading of manyversions of the manuscripts. His encouragement, feedback, and suggestions have been invaluable to me in this book as well as in several others. Finally, I want to mention Raoul Bott, whose courses on geometry and topology helped to shape my mathematical thinking and whose exemplary life is an inspiration to us all. Medford, Massachusetts Loring W. Tu June 2007 Contents Preface to the Second Edition ...................................... vii Preface to the First Edition ........................................ ix A Brief Introduction .............................................. 1 Chapter 1 Euclidean Spaces §1 Smooth Functions on a Euclidean Space . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.1 C∞Versus Analytic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.2 Taylor’s Theorem with Remainder . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 §2 Tangent Vectors in Rnas Derivations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 2.1 The Directional Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 2.2 Germs of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.3 Derivations at a Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 2.4 Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 2.5 Vector Fields as Derivations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 §3 The Exterior Algebra of Multicovectors . . . . . . . . . . . . . . . . . . . . . . . . . . 18 3.1 Dual Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193.2 Permutations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 3.3 Multilinear Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 3.4 The Permutation Action on Multilinear Functions . . . . . . . . . . . . . 233.5 The Symmetrizing and Alternating Operators . . . . . . . . . . . . . . . . . 243.6 The Tensor Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 3.7 The Wedge Product. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 3.8 Anticommutativity of the Wedge Product . . . . . . . . . . . . . . . . . . . . 273.9 Associativity of the Wedge Product . . . . . . . . . . . . . . . . . . . . . . . . . 28 3.10 A Basis for k-Covectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 xii Contents Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 §4 Differential Forms on Rn. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 4.1 Differential 1-Forms and the Differential of a Function . . . . . . . . . 34 4.2 Differential k-Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 4.3 Differential Forms as Multilinear Functions on Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 4.4 The Exterior Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 4.5 Closed Forms and Exact Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 4.6 Applications to Vector Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 4.7 Convention on Subscripts and Superscripts . . . . . . . . . . . . . . . . . . . 44 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 Chapter 2 Manifolds §5 Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 5.1 Topological Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 5.2 Compatible Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 5.3 Smooth Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 5.4 Examples of Smooth Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 §6 Smooth Maps on a Manifold . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 6.1 Smooth Functions on a Manifold . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 6.2 Smooth Maps Between Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . 61 6.3 Diffeomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 6.4 Smoothness in Terms of Components . . . . . . . . . . . . . . . . . . . . . . . . 63 6.5 Examples of Smooth Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 6.6 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 6.7 The Inverse Function Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 §7 Quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 7.1 The Quotient Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 7.2 Continuity of a Map on a Quotient . . . . . . . . . . . . . . . . . . . . . . . . . . 72 7.3 Identiﬁcation of a Subset to a Point . . . . . . . . . . . . . . . . . . . . . . . . . 73 7.4 A Necessary Condition for a Hausdorff Quotient . . . . . . . . . . . . . . 73 7.5 Open Equivalence Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 7.6 Real Projective Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 7.7 The Standard C∞Atlas on a Real Projective Space . . . . . . . . . . . . . 79 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 Chapter 3 The Tangent Space §8 The Tangent Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 Contents xiii 8.1 The Tangent Space at a Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 8.2 The Differential of a Map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 878.3 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 888.4 Bases for the Tangent Space at a Point . . . . . . . . . . . . . . . . . . . . . . . 898.5 A Local Expression for the Differential . . . . . . . . . . . . . . . . . . . . . . 918.6 Curves in a Manifold . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 928.7 Computing the Differential Using Curves . . . . . . . . . . . . . . . . . . . . 958.8 Immersions and Submersions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 8.9 Rank, and Critical and Regular Points . . . . . . . . . . . . . . . . . . . . . . . 96 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98 §9 Submanifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 9.1 Submanifolds. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1009.2 Level Sets of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1039.3 The Regular Level Set Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 9.4 Examples of Regular Submanifolds . . . . . . . . . . . . . . . . . . . . . . . . . 106 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 §10 Categories and Functors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 10.1 Categories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11010.2 Functors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11110.3 The Dual Functor and the Multicovector Functor . . . . . . . . . . . . . . 113 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 §11 The Rank of a Smooth Map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 11.1 Constant Rank Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11511.2 The Immersion and Submersion Theorems . . . . . . . . . . . . . . . . . . . 11811.3 Images of Smooth Maps. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12011.4 Smooth Maps into a Submanifold . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 11.5 The Tangent Plane to a Surface in R 3. . . . . . . . . . . . . . . . . . . . . . . . 125 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 §12 The Tangent Bundle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 12.1 The Topology of the Tangent Bundle . . . . . . . . . . . . . . . . . . . . . . . . 12912.2 The Manifold Structure on the Tangent Bundle . . . . . . . . . . . . . . . . 132 12.3 Vector Bundles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 12.4 Smooth Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13612.5 Smooth Frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 §13 Bump Functions and Partitions of Unity . . . . . . . . . . . . . . . . . . . . . . . . . 140 13.1 C ∞Bump Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140 13.2 Partitions of Unity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 13.3 Existence of a Partition of Unity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 §14 Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 14.1 Smoothness of a Vector Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 xiv Contents 14.2 Integral Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152 14.3 Local Flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154 14.4 The Lie Bracket. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157 14.5 The Pushforward of Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . 15914.6 Related Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161 Chapter 4 Lie Groups and Lie Algebras §15 Lie Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164 15.1 Examples of Lie Groups. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16415.2 Lie Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16715.3 The Matrix Exponential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169 15.4 The Trace of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 15.5 The Differential of det at the Identity . . . . . . . . . . . . . . . . . . . . . . . . 174 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174 §16 Lie Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178 16.1 Tangent Space at the Identity of a Lie Group . . . . . . . . . . . . . . . . . . 17816.2 Left-Invariant Vector Fields on a Lie Group . . . . . . . . . . . . . . . . . . 18016.3 The Lie Algebra of a Lie Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18216.4 The Lie Bracket on gl(n,R). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 16.5 The Pushforward of Left-Invariant Vector Fields . . . . . . . . . . . . . . 18416.6 The Differential as a Lie Algebra Homomorphism . . . . . . . . . . . . . 185 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187 Chapter 5 Differential Forms §17 Differential 1-Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 17.1 The Differential of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19117.2 Local Expression for a Differential 1-Form . . . . . . . . . . . . . . . . . . . 19117.3 The Cotangent Bundle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19217.4 Characterization of C ∞1-Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193 17.5 Pullback of 1-Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19517.6 Restriction of 1-Forms to an Immersed Submanifold . . . . . . . . . . . 197 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199 §18 Differential k-Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200 18.1 Differential Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20018.2 Local Expression for a k-Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202 18.3 The Bundle Point of View . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20318.4 Smooth k-Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 18.5 Pullback of k-Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204 18.6 The Wedge Product. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20518.7 Differential Forms on a Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206 Contents xv 18.8 Invariant Forms on a Lie Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208 §19 The Exterior Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210 19.1 Exterior Derivative on a Coordinate Chart . . . . . . . . . . . . . . . . . . . . 21119.2 Local Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211 19.3 Existence of an Exterior Derivative on a Manifold . . . . . . . . . . . . . 212 19.4 Uniqueness of the Exterior Derivative . . . . . . . . . . . . . . . . . . . . . . . 21319.5 Exterior Differentiation Under a Pullback . . . . . . . . . . . . . . . . . . . . 214 19.6 Restriction of k-Forms to a Submanifold . . . . . . . . . . . . . . . . . . . . . 216 19.7 A Nowhere-Vanishing 1-Form on the Circle . . . . . . . . . . . . . . . . . . 216Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218 §20 The Lie Derivative and Interior Multiplication . . . . . . . . . . . . . . . . . . . . 221 20.1 Families of Vector Fields and Differential Forms . . . . . . . . . . . . . . 221 20.2 The Lie Derivative of a Vector Field . . . . . . . . . . . . . . . . . . . . . . . . . 223 20.3 The Lie Derivative of a Differential Form . . . . . . . . . . . . . . . . . . . . 226 20.4 Interior Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22720.5 Properties of the Lie Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229 20.6 Global Formulas for the Lie and Exterior Derivatives . . . . . . . . . . 232 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233 Chapter 6 Integration §21 Orientations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236 21.1 Orientations of a Vector Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236 21.2 Orientations and n-Covectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238 21.3 Orientations on a Manifold . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240 21.4 Orientations and Differential Forms . . . . . . . . . . . . . . . . . . . . . . . . . 242 21.5 Orientations and Atlases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246 §22 Manifolds with Boundary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248 22.1 Smooth Invariance of Domain in R n. . . . . . . . . . . . . . . . . . . . . . . . . 248 22.2 Manifolds with Boundary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250 22.3 The Boundary of a Manifold with Boundary . . . . . . . . . . . . . . . . . . 253 22.4 Tangent Vectors, Differential Forms, and Orientations . . . . . . . . . . 253 22.5 Outward-Pointing Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254 22.6 Boundary Orientation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256 §23 Integration on Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 260 23.1 The Riemann Integral of a Function on Rn. . . . . . . . . . . . . . . . . . . 260 23.2 Integrability Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262 23.3 The Integral of an n-Form on Rn. . . . . . . . . . . . . . . . . . . . . . . . . . . . 263 23.4 Integral of a Differential Form over a Manifold . . . . . . . . . . . . . . . 265 xvi Contents 23.5 Stokes’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269 23.6 Line Integrals and Green’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 271Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272 Chapter 7 De Rham Theory §24 De Rham Cohomology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274 24.1 De Rham Cohomology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274 24.2 Examples of de Rham Cohomology . . . . . . . . . . . . . . . . . . . . . . . . . 276 24.3 Diffeomorphism Invariance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27824.4 The Ring Structure on de Rham Cohomology . . . . . . . . . . . . . . . . . 279 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 280 §25 The Long Exact Sequence in Cohomology . . . . . . . . . . . . . . . . . . . . . . . . 281 25.1 Exact Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281 25.2 Cohomology of Cochain Complexes . . . . . . . . . . . . . . . . . . . . . . . . 283 25.3 The Connecting Homomorphism . . . . . . . . . . . . . . . . . . . . . . . . . . . 284 25.4 The Zig-Zag Lemma. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287 §26 The Mayer–Vietoris Sequence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288 26.1 The Mayer–Vietoris Sequence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288 26.2 The Cohomology of the Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29226.3 The Euler Characteristic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295 §27 Homotopy Invariance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296 27.1 Smooth Homotopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296 27.2 Homotopy Type. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29727.3 Deformation Retractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299 27.4 The Homotopy Axiom for de Rham Cohomology . . . . . . . . . . . . . 300 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301 §28 Computation of de Rham Cohomology . . . . . . . . . . . . . . . . . . . . . . . . . . . 302 28.1 Cohomology Vector Space of a Torus. . . . . . . . . . . . . . . . . . . . . . . . 30228.2 The Cohomology Ring of a Torus . . . . . . . . . . . . . . . . . . . . . . . . . . . 303 28.3 The Cohomology of a Surface of Genus g. . . . . . . . . . . . . . . . . . . . 306 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 310 §29 Proof of Homotopy Invariance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311 29.1 Reduction to Two Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31129.2 Cochain Homotopies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312 29.3 Differential Forms on M×R. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312 29.4 A Cochain Homotopy Between i ∗ 0andi∗ 1. . . . . . . . . . . . . . . . . . . . . 314 29.5 Veriﬁcation of Cochain Homotopy . . . . . . . . . . . . . . . . . . . . . . . . . . 315 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 316 Contents xvii Appendices §A Point-Set Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317 A.1 Topological Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317 A.2 Subspace Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 320 A.3 Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321 A.4 First and Second Countability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323 A.5 Separation Axioms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324 A.6 Product Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 326 A.7 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327 A.8 Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329 A.9 Boundedness in Rn. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332 A.10 Connectedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332 A.11 Connected Components . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333 A.12 Closure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 334 A.13 Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337 §B The Inverse Function Theorem on Rnand Related Results . . . . . . . . . 339 B.1 The Inverse Function Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 339 B.2 The Implicit Function Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 339 B.3 Constant Rank Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344 §C Existence of a Partition of Unity in General . . . . . . . . . . . . . . . . . . . . . . . 346 §D Linear Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 349 D.1 Quotient Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 349 D.2 Linear Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 350 D.3 Direct Product and Direct Sum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352 §E Quaternions and the Symplectic Group . . . . . . . . . . . . . . . . . . . . . . . . . . 353 E.1 Representation of Linear Maps by Matrices . . . . . . . . . . . . . . . . . . 354E.2 Quaternionic Conjugation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355 E.3 Quaternionic Inner Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 356 E.4 Representations of Quaternions by Complex Numbers . . . . . . . . . 356 E.5 Quaternionic Inner Product in Terms of Complex Components . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357 E.6H-Linearity in Terms of Complex Numbers . . . . . . . . . . . . . . . . . . 357 E.7 Symplectic Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 358 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359 Solutions to Selected Exercises Within the Text ........................ 361 Hints and Solutions to Selected End-of-Section Problems ............... 367 xviii Contents List of Notations ................................................. 387 References ...................................................... 395 Index........................................................... 397 Chapter 1 Euclidean Spaces The Euclidean space Rnis the prototype of all manifolds. Not only is it the simplest, but locally every manifold looks like Rn. A good understanding of Rnis essential in generalizing differential and integral calculus to a manifold. Euclidean space is special in having a set of standard global coordinates. This is both a blessing and a handicap. It is a blessing because all constructions on Rn can be deﬁned in terms of the standard coordinates and all computations carried out explicitly. It is a handicap because, deﬁned in terms of coordinates, it is often not ob-vious which concepts are intrinsic, i.e., independent of coordinates. Since a manifold in general does not have standard coordinates, only coordinate-independent concepts mension n, it is not possible to integrate functions, because the integral of a function depends on a set of coordinates. The objects that can be integrated are differential forms. It is only because the existence of global coordinates permits an identiﬁcationof functions with differential n-forms on R nthat integration of functions becomes possible on Rn. Our goal in this chapter is to recast calculus on Rnin a coordinate-free way suit- able for generalization to manifolds. To this end, we view a tangent vector not as an arrow or as a column of numbers, but as a derivation on functions. This is followed by an exposition of Hermann Grassmann’s formalism of alternating multilinear func-tions on a vector space, which lays the foundation for the theory of differential forms. Finally, we introduce differential forms on R n, together with two of their basic oper- ations, the wedge product and the exterior derivative, and show how they generalize and simplify vector calculus in R3. §1 Smooth Functions on a Euclidean Space ∞ manifolds. For this reason, we begin with a review of C∞functions on Rn.will make sense on a manifold. For example, it turns out that on a manifold of di- The calculus of Cfunctions will be our primary tool for studying higher-dimensional 3 L.W. Tu, An Introduction to Manifolds, Universitext, DOI 10.1007/978-1-4419-7400-6_1, © Springer Science+Business Media, LLC 2011 4§1 Smooth Functions on a Euclidean Space 1.1C∞Versus Analytic Functions Write the coordinates on Rnasx1,..., xnand let p= (p1,..., pn)be a point in an open set UinRn. In keeping with the conventions of differential geometry, the indices on coordinates are superscripts , not subscripts. An explanation of the rules for superscripts and subscripts is given in Subsection 4.7. Deﬁnition 1.1. Letkbe a nonnegative integer. A real-valued function f:U→Ris said to be Ckat p∈Uif its partial derivatives ∂jf ∂xi1···∂xij of all orders j≤kexist and are continuous at p. The function f:U→RisC∞ at p if it is Ckfor all k≥0; in other words, its partial derivatives ∂jf/∂xi1···∂xij of all orders exist and are continuous at p. A vector-valued function f:U→Rm is said to be Ckat p if all of its component functions f1,..., fmareCkatp. We say that f:U→RmisCkon U if it is Ckat every point in U. A similar deﬁnition holds for a C∞function on an open set U. We treat the terms “ C∞” and “smooth” as synonymous. Example 1.2. (i) A C0function on Uis a continuous function on U. (ii) Let f:R→Rbef(x)=x1/3. Then f′(x)=/braceleftigg 1 3x−2/3forx/ne}a⊔ionslash=0, undeﬁned for x=0. Thus the function fisC0but not C1atx=0. (iii) Let g:R→Rbe deﬁned by g(x)=/integraldisplayx 0f(t)dt=/integraldisplayx 0t1/3dt=3 4x4/3. Then g′(x)=f(x)=x1/3, sog(x)isC1but not C2atx=0. In the same way one can construct a function that is Ckbut not Ck+1at a given point. (iv) The polynomial, sine, cosine, and exponential functions on the real line are all C∞. Aneighborhood of a point in Rnis an open set containing the point. The function fisreal-analytic atpif in some neighborhood of pit is equal to its Taylor series atp: f(x)=f(p)+∑ i∂f ∂xi(p)(xi−pi)+1 2!∑ i,j∂2f ∂xi∂xj(p)(xi−pi)(xj−pj) +···+1 k!∑ i1,...,ik∂kf ∂xi1···∂xik(p)(xi1−pi1)···(xik−pik)+···, 1.2 Taylor’s Theorem with Remainder 5 in which the general term is summed over all 1 ≤i1,..., ik≤n. A real-analytic function is necessarily C∞, because as one learns in real anal- ysis, a convergent power series can be differentiated term by term in its region of convergence. For example, if f(x)=sinx=x−1 3!x3+1 5!x5−···, then term-by-term differentiation gives f′(x)=cosx=1−1 2!x2+1 4!x4−···. The following example shows that a C∞function need not be real-analytic. The idea is to construct a C∞function f(x)onRwhose graph, though not horizontal, is “very ﬂat” near 0 in the sense that all of its derivatives vanish at 0. xy 1 Fig. 1.1. AC∞function all of whose derivatives vanish at 0. Example 1.3 ( A C∞function very ﬂat at 0).Deﬁne f(x)onRby f(x)=/braceleftigg e−1/xforx>0, 0 for x≤0. (See Figure 1.1. )By induction, one can show that fisC∞onRand that the deriva- tives f(k)(0)are equal to 0 for all k≥0 (Problem 1.2). The Taylor series of this function at the origin is identically zero in any neigh- borhood of the origin, since all derivatives f(k)(0)equal 0. Therefore, f(x)cannot be equal to its Taylor series and f(x)is not real-analytic at 0. 1.2 Taylor’s Theorem with Remainder Although a C∞function need not be equal to its Taylor series, there is a Taylor’s theorem with remainder for C∞functions that is often good enough for our purposes. In the lemma below, we prove the very ﬁrst case, in which the Taylor series consists of only the constant term f(p). We say that a subset SofRnisstar-shaped with respect to a point pinSif for every xinS, the line segment from ptoxlies in S(Figure 1.2) . 6§1 Smooth Functions on a Euclidean Space /Bullet/Bullet/Bullet pqx Fig. 1.2. Star-shaped with respect to p, but not with respect to q. Lemma 1.4 (Taylor’s theorem with remainder). Let f be a C∞function on an open subset U of Rnstar-shaped with respect to a point p =(p1,..., pn)in U. Then there are functions g 1(x),..., gn(x)∈C∞(U)such that f(x)=f(p)+n ∑ i=1(xi−pi)gi(x),gi(p)=∂f ∂xi(p). Proof. Since Uis star-shaped with respect to p, for any xinUthe line segment p+t(x−p), 0≤t≤1, lies in U(Figure 1.3 ). So f(p+t(x−p))is deﬁned for 0≤t≤1. /Bullet/Bullet pxU Fig. 1.3. The line segment from ptox. By the chain rule, d dtf(p+t(x−p))=∑(xi−pi)∂f ∂xi(p+t(x−p)). If we integrate both sides with respect to tfrom 0 to 1, we get f(p+t(x−p))/bracketrightbig1 0=∑(xi−pi)/integraldisplay1 0∂f ∂xi(p+t(x−p))dt. (1.1) Let gi(x)=/integraldisplay1 0∂f ∂xi(p+t(x−p))dt. 1.2 Taylor’s Theorem with Remainder 7 Then gi(x)isC∞and (1.1) becomes f(x)−f(p)=∑(xi−pi)gi(x). Moreover, gi(p)=/integraldisplay1 0∂f ∂xi(p)dt=∂f ∂xi(p). ⊓ ⊔ In case n=1 and p=0, this lemma says that f(x)=f(0)+xg1(x) for some C∞function g1(x). Applying the lemma repeatedly gives gi(x)=gi(0)+xgi+1(x), where gi,gi+1areC∞functions. Hence, f(x)=f(0)+x(g1(0)+xg2(x)) =f(0)+xg1(0)+x2(g2(0)+xg3(x)) ... =f(0)+g1(0)x+g2(0)x2+···+gi(0)xi+gi+1(x)xi+1. (1.2) Differentiating (1.2) repeatedly and evaluating at 0, we get gk(0)=1 k!f(k)(0),k=1,2,..., i. So (1.2) is a polynomial expansion of f(x)whose terms up to the last term agree with the Taylor series of f(x)at 0. Remark. Being star-shaped is not such a restrictive condition, since any open ball B(p,ε)={x∈Rn|/bardblx−p/bardbl<ε} is star-shaped with respect to p. If fis aC∞function deﬁned on an open set U containing p, then there is an ε>0 such that p∈B(p,ε)⊂U. When its domain is restricted to B(p,ε), the function fis deﬁned on a star-shaped neighborhood of pand Taylor’s theorem with remainder applies. NOTATION . It is customary to write the standard coordinates on R2asx,y, and the standard coordinates on R3asx,y,z. 8§1 Smooth Functions on a Euclidean Space Problems 1.1. A function that is C2but not C3 Letg:R→Rbe the function in Example 1.2(iii). Show that the function h(x)=/integraltextx 0g(t)dtis C2but not C3atx=0. 1.2.* A C∞function very ﬂat at 0 Letf(x)be the function on Rdeﬁned in Example 1.3. (a) Show by induction that for x>0 and k≥0, the kth derivative f(k)(x)is of the form p2k(1/x)e−1/xfor some polynomial p2k(y)of degree 2 kiny. (b) Prove that fisC∞onRand that f(k)(0)=0 for all k≥0. 1.3. A diffeomorphism of an open interval with R LetU⊂RnandV⊂Rnbe open subsets. A C∞map F:U→Vis called a diffeomorphism if it is bijective and has a C∞inverse F−1:V→U. (a) Show that the function f:]−π/2,π/2[→R,f(x)=tanx, is a diffeomorphism. (b) Let a,bbe real numbers with a<b. Find a linear function h:]a,b[→]−1,1[, thus proving that any two ﬁnite open intervals are diffeomorphic. The composite f◦h:]a,b[→Ris then a diffeomorphism of an open interval with R. (c) The exponential function exp: R→]0,∞[is a diffeomorphism. Use it to show that for any real numbers aandb, the intervals R,]a,∞[, and]−∞,b[are diffeomorphic. 1.4. A diffeomorphism of an open cube with Rn Show that the map f:/bracketrightig −π 2,π 2/bracketleftign →Rn,f(x1,..., xn)=(tan x1,..., tanxn), is a diffeomorphism. 1.5. A diffeomorphism of an open ball with Rn Let0= (0,0)be the origin and B(0,1)the open unit disk in R2. To ﬁnd a diffeomorphism between B(0,1)andR2, we identify R2with the xy-plane in R3and introduce the lower open hemisphere S:x2+y2+(z−1)2=1,z<1, inR3as an intermediate space (Figure 1.4) . First note that the map f:B(0,1)→S,(a,b)/ma√s⊔o→(a,b,1−/radicalbig 1−a2−b2), is a bijection. (a) The stereographic projection g :S→R2from(0,0,1)is the map that sends a point (a,b,c)∈Sto the intersection of the line through (0,0,1)and(a,b,c)with the xy-plane. Show that it is given by (a,b,c)/ma√s⊔o→(u,v)=/parenleftbigga 1−c,b 1−c/parenrightbigg ,c=1−/radicalbig 1−a2−b2, with inverse (u,v)/ma√s⊔o→/parenleftbiggu√ 1+u2+v2,v√ 1+u2+v2,1−1√ 1+u2+v2/parenrightbigg . 1.2 Taylor’s Theorem with Remainder 9 /Bullet/Bullet /Bullet /Bullet/Bullet /Bullet /BulletS S (0,0,1) (a,b,0)(a,b,c) (a,b,c) (u,v,0)B(0,1) R2⊂R3 0 0( ) Fig. 1.4. A diffeomorphism of an open disk with R2. (b) Composing the two maps fandggives the map h=g◦f:B(0,1)→R2, h(a,b)=/parenleftbigga√ 1−a2−b2,b√ 1−a2−b2/parenrightbigg . Find a formula for h−1(u,v)=( f−1◦g−1)(u,v)and conclude that his a diffeomorphism of the open disk B(0,1)withR2. (c) Generalize part (b) to Rn. 1.6.* Taylor’s theorem with remainder to order 2 Prove that if f:R2→RisC∞, then there exist C∞functions g11,g12,g22onR2such that f(x,y)=f(0,0)+∂f ∂x(0,0)x+∂f ∂y(0,0)y +x2g11(x,y)+xyg12(x,y)+y2g22(x,y). 1.7.* A function with a removable singularity Letf:R2→Rbe aC∞function with f(0,0)=∂f/∂x(0,0)=∂f/∂y(0,0)=0. Deﬁne g(t,u)=  f(t,tu) tfort/ne}a⊔ionslash=0, 0 for t=0. Prove that g(t,u)isC∞for(t,u)∈R2. (Hint: Apply Problem 1.6.) 1.8. Bijective C∞maps Deﬁne f:R→Rbyf(x) =x3. Show that fis a bijective C∞map, but that f−1is not C∞. (This example shows that a bijective C∞map need not have a C∞inverse. In complex analysis, the situation is quite different: a bijective holomorphic map f:C→Cnecessarily has a holomorphic inverse.) 10§2 Tangent Vectors in Rnas Derivations §2 Tangent Vectors in Rnas Derivations In elementary calculus we normally represent a vector at a point pinR3algebraically as a column of numbers v= v1 v2 v3  or geometrically as an arrow emanating from p(Figure 2.1 ). /Bullet pv Fig. 2.1. A vector vatp. Recall that a secant plane to a surface in R3is a plane determined by three points of the surface. As the three points approach a point pon the surface, if the corre- sponding secant planes approach a limiting position, then the plane that is the lim-iting position of the secant planes is called the tangent plane to the surface at p. Intuitively, the tangent plane to a surface at pis the plane in R 3that just “touches” the surface at p. A vector at pis tangent to a surface in R3if it lies in the tangent plane at p(Figure 2.2) . /Bulletp v Fig. 2.2. A tangent vector vto a surface at p. Such a deﬁnition of a tangent vector to a surface presupposes that the surface is embedded in a Euclidean space, and so would not apply to the projective plane, for example, which does not sit inside an Rnin any natural way. Our goal in this section is to ﬁnd a characterization of tangent vectors in Rnthat will generalize to manifolds. 2.1 The Directional Derivative In calculus we visualize the tangent space Tp(Rn)atpinRnas the vector space of all arrows emanating from p. By the correspondence between arrows and column 2.2 Germs of Functions 11 vectors, the vector space Rncan be identiﬁed with this column space. To distinguish between points and vectors, we write a point in Rnasp=(p1,..., pn)and a vector in the tangent space Tp(Rn)as v= v1 ... vn or/an}bracke⊔le{⊔v1,..., vn/an}bracke⊔ri}h⊔. We usually denote the standard basis for RnorTp(Rn)bye1,..., en. Then v=∑viei for some vi∈R. Elements of Tp(Rn)are called tangent vectors (or simply vectors ) atpinRn. We sometimes drop the parentheses and write TpRnforTp(Rn). The line through a point p=(p1,..., pn)with direction v=/an}bracke⊔le{⊔v1,..., vn/an}bracke⊔ri}h⊔inRnhas parametrization c(t)=( p1+tv1,..., pn+tvn). Itsith component ci(t)ispi+tvi. IffisC∞in a neighborhood of pinRnandvis a tangent vector at p, the directional derivative offin the direction vatpis deﬁned to be Dvf=lim t→0f(c(t))−f(p) t=d dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0f(c(t)). By the chain rule, Dvf=n ∑ i=1dci dt(0)∂f ∂xi(p)=n ∑ i=1vi∂f ∂xi(p). (2.1) In the notation Dvf, it is understood that the partial derivatives are to be evaluated atp, since vis a vector at p. SoDvfis a number, not a function. We write Dv=∑vi∂ ∂xi/vextendsingle/vextendsingle/vextendsingle/vextendsingle p for the map that sends a function fto the number Dvf. To simplify the notation we often omit the subscript pif it is clear from the context. The association v/ma√s⊔o→Dvof the directional derivative Dvto a tangent vector voffers a way to characterize tangent vectors as certain operators on functions. To makethis precise, in the next two subsections we study in greater detail the directional derivative D vas an operator on functions. 2.2 Germs of Functions Arelation on a set Sis a subset RofS×S. Given x,yinS, we write x∼yif and only if(x,y)∈R. The relation Ris an equivalence relation if it satisﬁes the following three properties for all x,y,z∈S: (i) (reﬂexivity) x∼x, (ii) (symmetry) if x∼y, then y∼x, 12§2 Tangent Vectors in Rnas Derivations (iii) (transitivity) if x∼yandy∼z, then x∼z. As long as two functions agree on some neighborhood of a point p, they will have the same directional derivatives at p. This suggests that we introduce an equivalence relation on the C∞functions deﬁned in some neighborhood of p. Consider the set of all pairs(f,U), where Uis a neighborhood of pandf:U→Ris aC∞function. We say that(f,U)isequivalent to(g,V)if there is an open set W⊂U∩Vcontaining p such that f=gwhen restricted to W. This is clearly an equivalence relation because it is reﬂexive, symmetric, and transitive. The equivalence class of (f,U)is called the germ offatp. We write C∞ p(Rn), or simply C∞ pif there is no possibility of confusion, for the set of all germs of C∞functions on Rnatp. Example. The functions f(x)=1 1−x with domain R−{1}and g(x)=1+x+x2+x3+··· with domain the open interval ]−1,1[have the same germ at any point pin the open interval]−1,1[. Analgebra over a ﬁeld Kis a vector space Aover Kwith a multiplication map µ:A×A→A, usually written µ(a,b)=a·b, such that for all a,b,c∈Aandr∈K, (i) (associativity) (a·b)·c=a·(b·c), (ii) (distributivity) (a+b)·c=a·c+b·canda·(b+c)=a·b+a·c, (iii) (homogeneity) r(a·b)=( ra)·b=a·(rb). Equivalently, an algebra over a ﬁeld Kis a ring A(with or without multiplicative identity) that is also a vector space over Ksuch that the ring multiplication satisﬁes the homogeneity condition (iii). Thus, an algebra has three operations: the addition and multiplication of a ring and the scalar multiplication of a vector space. Usually we omit the multiplication sign and write abinstead of a·b. A map L:V→Wbetween vector spaces over a ﬁeld Kis called a linear map or alinear operator if for any r∈Kandu,v∈V, (i)L(u+v)=L(u)+L(v); (ii)L(rv)=rL(v). To emphasize the fact that the scalars are in the ﬁeld K, such a map is also said to be K-linear . IfAandA′are algebras over a ﬁeld K, then an algebra homomorphism is a linear map L:A→A′that preserves the algebra multiplication: L(ab) =L(a)L(b)for all a,b∈A. The addition and multiplication of functions induce corresponding operations on C∞ p, making it into an algebra over R(Problem 2.2). 2.3 Derivations at a Point 13 2.3 Derivations at a Point For each tangent vector vat a point pinRn, the directional derivative at pgives a map of real vector spaces Dv:C∞ p→R. By (2.1), DvisR-linear and satisﬁes the Leibniz rule Dv(f g)=( Dvf)g(p)+f(p)Dvg, (2.2) precisely because the partial derivatives ∂/∂xi|phave these properties. In general, any linear map D:C∞ p→Rsatisfying the Leibniz rule (2.2) is called aderivation at p or apoint-derivation ofC∞ p. Denote the set of all derivations at p byDp(Rn). This set is in fact a real vector space, since the sum of two derivations at pand a scalar multiple of a derivation at pare again derivations at p(Problem 2.3). Thus far, we know that directional derivatives at pare all derivations at p, so there is a map φ:Tp(Rn)→Dp(Rn), (2.3) v/ma√s⊔o→Dv=∑vi∂ ∂xi/vextendsingle/vextendsingle/vextendsingle/vextendsingle p. Since Dvis clearly linear in v, the map φis a linear map of vector spaces. Lemma 2.1. If D is a point-derivation of C∞ p, then D( c)=0for any constant function c. Proof. Since we do not know whether every derivation at pis a directional derivative, we need to prove this lemma using only the deﬁning properties of a derivation at p. ByR-linearity, D(c) =cD(1). So it sufﬁces to prove that D(1) =0. By the Leibniz rule (2.2), D(1)=D(1·1)=D(1)·1+1·D(1)=2D(1). Subtracting D(1)from both sides gives 0 =D(1). ⊓ ⊔ TheKronecker delta δis a useful notation that we frequently call upon: δi j=/braceleftigg 1 if i=j, 0 if i/ne}a⊔ionslash=j. Theorem 2.2. The linear map φ:Tp(Rn)→Dp(Rn)deﬁned in (2.3) is an isomor- phism of vector spaces. Proof. To prove injectivity, suppose Dv=0 for v∈Tp(Rn). Applying Dvto the coordinate function xjgives 0=Dv(xj)=∑ ivi∂ ∂xi/vextendsingle/vextendsingle/vextendsingle/vextendsingle pxj=∑ iviδj i=vj. 14§2 Tangent Vectors in Rnas Derivations Hence, v=0 and φis injective. To prove surjectivity, let Dbe a derivation at pand let(f,V)be a representative of a germ in C∞ p. Making Vsmaller if necessary, we may assume that Vis an open ball, hence star-shaped. By Taylor’s theorem with remainder (Lemma 1.4) there are C∞functions gi(x)in a neighborhood of psuch that f(x)=f(p)+∑(xi−pi)gi(x),gi(p)=∂f ∂xi(p). Applying Dto both sides and noting that D(f(p))= 0 and D(pi)=0 by Lemma 2.1, we get by the Leibniz rule (2.2) D f(x)=∑(Dxi)gi(p)+∑(pi−pi)Dgi(x)=∑(Dxi)∂f ∂xi(p). This proves that D=Dvforv=/an}bracke⊔le{⊔Dx1,..., Dxn/an}bracke⊔ri}h⊔. ⊓ ⊔ This theorem shows that one may identify the tangent vectors at pwith the deriva- tions at p. Under the vector space isomorphism Tp(Rn)≃Dp(Rn), the standard basis e1,..., enforTp(Rn)corresponds to the set ∂/∂x1|p,..., ∂/∂xn|pof partial deriva- tives. From now on, we will make this identiﬁcation and write a tangent vector v=/an}bracke⊔le{⊔v1,..., vn/an}bracke⊔ri}h⊔=∑vieias v=∑vi∂ ∂xi/vextendsingle/vextendsingle/vextendsingle/vextendsingle p. (2.4) The vector space Dp(Rn)of derivations at p, although not as geometric as ar- rows, turns out to be more suitable for generalization to manifolds. 2.4 Vector Fields Avector ﬁeld X on an open subset UofRnis a function that assigns to each point p inUa tangent vector XpinTp(Rn). Since Tp(Rn)has basis{∂/∂xi|p}, the vector Xp is a linear combination Xp=∑ai(p)∂ ∂xi/vextendsingle/vextendsingle/vextendsingle/vextendsingle p,p∈U,ai(p)∈R. Omitting p, we may write X=∑ai∂/∂xi, where the aiare now functions on U. We say that the vector ﬁeld XisC∞on U if the coefﬁcient functions aiare all C∞onU. Example 2.3.OnR2−{0}, letp=(x,y). Then X=−y/radicalbig x2+y2∂ ∂x+x/radicalbig x2+y2∂ ∂y=/angbracketleftigg −y/radicalbig x2+y2,x/radicalbig x2+y2/angbracketrightigg is the vector ﬁeld in Figure 2.3(a). As is customary, we draw a vector at pas an arrow emanating from p. The vector ﬁeld Y=x∂/∂x−y∂/∂y=/an}bracke⊔le{⊔x,−y/an}bracke⊔ri}h⊔, suitably rescaled, is sketched in Figure 2.3(b). 2.4 Vector Fields 15 202 11 0 -2 -1-1 -2◦ (a) The vector ﬁeld XonR2−{0} (b) The vector ﬁeld /an}bracke⊔le{⊔x,−y/an}bracke⊔ri}h⊔onR2 Fig. 2.3. Vector ﬁelds on open subsets of R2. One can identify vector ﬁelds on Uwith column vectors of C∞functions on U: X=∑ai∂ ∂xi←→ a1 ... an . This is the same identiﬁcation as (2.4), but now we are allowing the point pto move inU. The ring of C∞functions on an open set Uis commonly denoted by C∞(U)or F(U). Multiplication of vector ﬁelds by functions on Uis deﬁned pointwise: (f X)p=f(p)Xp,p∈U. Clearly, if X=∑ai∂/∂xiis aC∞vector ﬁeld and fis aC∞function on U, then f X=∑(f ai)∂/∂xiis aC∞vector ﬁeld on U. Thus, the set of all C∞vector ﬁelds on U, denoted by X(U), is not only a vector space over R, but also a module over the ringC∞(U). We recall the deﬁnition of a module. Deﬁnition 2.4. IfRis a commutative ring with identity, then a (left) R-module is an abelian group Awith a scalar multiplication map µ:R×A→A, usually written µ(r,a)=ra, such that for all r,s∈Randa,b∈A, (i) (associativity) (rs)a=r(sa), (ii) (identity) if 1 is the multiplicative identity in R, then 1 a=a, (iii) (distributivity) (r+s)a=ra+sa,r(a+b)=ra+rb. 16§2 Tangent Vectors in Rnas Derivations IfRis a ﬁeld, then an R-module is precisely a vector space over R. In this sense, a module generalizes a vector space by allowing scalars in a ring rather than a ﬁeld. Deﬁnition 2.5. LetAandA′beR-modules. An R-module homomorphism from A toA′is a map f:A→A′that preserves both addition and scalar multiplication: for alla,b∈Aandr∈R, (i)f(a+b)=f(a)+f(b), (ii)f(ra)=r f(a). 2.5 Vector Fields as Derivations IfXis aC∞vector ﬁeld on an open subset UofRnandfis aC∞function on U, we deﬁne a new function X fonUby (X f)(p)=Xpffor any p∈U. Writing X=∑ai∂/∂xi, we get (X f)(p)=∑ai(p)∂f ∂xi(p), or X f=∑ai∂f ∂xi, which shows that X fis aC∞function on U. Thus, a C∞vector ﬁeld Xgives rise to anR-linear map C∞(U)→C∞(U), f/ma√s⊔o→X f. Proposition 2.6 (Leibniz rule for a vector ﬁeld). If X is a C∞vector ﬁeld and f and g are C∞functions on an open subset U of Rn, then X(f g)satisﬁes the product rule (Leibniz rule): X(f g)=( X f)g+f Xg. Proof. At each point p∈U, the vector Xpsatisﬁes the Leibniz rule: Xp(f g)=( Xpf)g(p)+f(p)Xpg. Aspvaries over U, this becomes an equality of functions: X(f g)=( X f)g+f Xg. ⊓ ⊔ IfAis an algebra over a ﬁeld K, aderivation ofAis aK-linear map D:A→A such that D(ab)=( Da)b+aDb for all a,b∈A. 2.5 Vector Fields as Derivations 17 The set of all derivations of Ais closed under addition and scalar multiplication and forms a vector space, denoted by Der( A). As noted above, a C∞vector ﬁeld on an open set Ugives rise to a derivation of the algebra C∞(U). We therefore have a map ϕ:X(U)→Der(C∞(U)), X/ma√s⊔o→(f/ma√s⊔o→X f). Just as the tangent vectors at a point pcan be identiﬁed with the point-derivations of C∞ p, so the vector ﬁelds on an open set Ucan be identiﬁed with the derivations of the algebra C∞(U); i.e., the map ϕis an isomorphism of vector spaces. The injectivity of ϕis easy to establish, but the surjectivity of ϕtakes some work (see Problem 19.12). Note that a derivation at pis not a derivation of the algebra C∞ p. A derivation at p is a map from C∞ ptoR, while a derivation of the algebra C∞ pis a map from C∞ ptoC∞ p. Problems 2.1. Vector ﬁelds LetXbe the vector ﬁeld x∂/∂x+y∂/∂yandf(x,y,z)the function x2+y2+z2onR3. Com- pute X f. 2.2. Algebra structure on C∞p Deﬁne carefully addition, multiplication, and scalar multiplication in C∞p. Prove that addition inC∞pis commutative. 2.3. Vector space structure on derivations at a point LetDandD′be derivations at pinRn, and c∈R. Prove that (a) the sum D+D′is a derivation at p. (b) the scalar multiple cDis a derivation at p. 2.4. Product of derivations LetAbe an algebra over a ﬁeld K. IfD1andD2are derivations of A, show that D1◦D2is not necessarily a derivation (it is if D1orD2=0), but D1◦D2−D2◦D1is always a derivation of A. 18§3 The Exterior Algebra of Multicovectors §3 The Exterior Algebra of Multicovectors As noted in the introduction, manifolds are higher-dimensional analogues of curves and surfaces. As such, they are usually not linear spaces. Nonetheless, a basic principle in manifold theory is the linearization principle, according to which every manifold can be locally approximated by its tangent space at a point, a linear object. In this way linear algebra enters into manifold theory. Instead of working with tangent vectors, it turns out to be more fruitful to adopt the dual point of view and work with linear functions on a tangent space. After all, there is only so much that one can do with tangent vectors, which are essentially arrows, but functions, far more ﬂexible, can be added, multiplied, scalar-multiplied, and composed with other maps. Once one admits linear functions on a tangent space,it is but a small step to consider functions of several arguments linear in each argu- ment. These are the multilinear functions on a vector space. The determinant of a matrix, viewed as a function of the column vectors of the matrix, is an example ofa multilinear function. Among the multilinear functions, certain ones such as the determinant and the cross product have an antisymmetric oralternating property: they change sign if two arguments are switched. The alternating multilinear func- tions with karguments on a vector space are called multicovectors of degree k , or k-covectors for short. Hermann Grassmann (1809–1877)It took the genius of Hermann Grassmann, a nineteenth-century German mathematician, linguist, and high-school teacher, to recognize the impor- tance of multicovectors. He constructed a vast ed- iﬁce based on multicovectors, now called the exte- rior algebra , that generalizes parts of vector calcu- lus from R3toRn. For example, the wedge prod- uct of two multicovectors on an n-dimensional vec- tor space is a generalization of the cross product in R3(see Problem 4.6). Grassmann’s work was little appreciated in his lifetime. In fact, he was turned down for a university position and his Ph.D. thesis rejected, because the leading mathematicians of hisday such as M¨ obius and Kummer failed to under- stand his work. It was only at the turn of the twenti- eth century, in the hands of the great differential ge-ometer ´Elie Cartan (1869–1951), that Grassmann’s exterior algebra found its just recognition as the algebraic basis of the theory of dif- ferential forms. This section is an exposition, using modern terminology, of some of Grassmann’s ideas. 3.1 Dual Space 19 3.1 Dual Space IfVandWare real vector spaces, we denote by Hom (V,W)the vector space of all linear maps f:V→W. Deﬁne the dual space V∨ofVto be the vector space of all real-valued linear functions on V: V∨=Hom(V,R). The elements of V∨are called covectors or 1-covectors onV. In the rest of this section, assume Vto be a ﬁnite-dimensional vector space. Let e1,..., enbe a basis for V. Then every vinVis uniquely a linear combination v=∑vieiwith vi∈R. Let αi:V→Rbe the linear function that picks out the ith coordinate, αi(v)=vi. Note that αiis characterized by αi(ej)=δi j=/braceleftigg 1 for i=j, 0 for i/ne}a⊔ionslash=j. Proposition 3.1. The functions α1,..., αnform a basis for V∨. Proof. We ﬁrst prove that α1,..., αnspan V∨. Iff∈V∨andv=∑viei∈V, then f(v)=∑vif(ei)=∑f(ei)αi(v). Hence, f=∑f(ei)αi, which shows that α1,..., αnspan V∨. To show linear independence, suppose ∑ciαi=0 for some ci∈R. Applying both sides to the vector ejgives 0=∑ iciαi(ej)=∑ iciδi j=cj,j=1,..., n. Hence, α1,..., αnare linearly independent. ⊓ ⊔ This basis α1,..., αnforV∨is said to be dual to the basis e1,..., enforV. Corollary 3.2. The dual space V∨of a ﬁnite-dimensional vector space V has the same dimension as V. Example 3.3 ( Coordinate functions ).With respect to a basis e1,..., enfor a vector space V, every v∈Vcan be written uniquely as a linear combination v=∑bi(v)ei, where bi(v)∈R. Let α1,..., αnbe the basis of V∨dual to e1,..., en. Then αi(v)=αi/parenleftigg ∑ jbj(v)ej/parenrightigg =∑ jbj(v)αi(ej)=∑ jbj(v)δi j=bi(v). Thus, the dual basis to e1,..., enis precisely the set of coordinate functions b1,..., bn with respect to the basis e1,..., en. 20§3 The Exterior Algebra of Multicovectors 3.2 Permutations Fix a positive integer k. Apermutation of the set A={1,..., k}is a bijection σ:A→ A. More concretely, σmay be thought of as a reordering of the list 1 ,2,..., kfrom its natural increasing order to a new order σ(1),σ(2),..., σ(k). The cyclic permutation , (a1a2···ar)where the aiare distinct, is the permutation σsuch that σ(a1) =a2, σ(a2) =a3,...,σ(ar−1) =( ar),σ(ar) =a1, and σﬁxes all the other elements of A. A cyclic permutation (a1a2···ar)is also called a cycle of length r or an r-cycle. Atransposition is a 2-cycle, that is, a cycle of the form (a b)that interchanges aand b, leaving all other elements of Aﬁxed. Two cycles (a1···ar)and(b1···bs)are said to be disjoint if the sets{a1,..., ar}and{b1,..., bs}have no elements in common. Theproduct τσof two permutations τandσofAis the composition τ◦σ:A→A, in that order; ﬁrst apply σ, then τ. A simple way to describe a permutation σ:A→Ais by its matrix /bracketleftbigg 1 2··· k σ(1)σ(2)···σ(k)/bracketrightbigg . Example 3.4.Suppose the permutation σ:{1,2,3,4,5}→{ 1,2,3,4,5}maps 1,2, 3,4,5 to 2,4,5,1,3 in that order. As a matrix, σ=/bracketleftbigg1 2 3 4 5 2 4 5 1 3/bracketrightbigg . (3.1) To write σas a product of disjoint cycles, start with any element in {1,2,3,4,5}, say 1, and apply σto it repeatedly until we return to the initial element; this gives a cycle: 1/ma√s⊔o→2/ma√s⊔o→4→1. Next, repeat the procedure beginning with any of the remaining elements, say 3, to get a second cycle: 3 /ma√s⊔o→5/ma√s⊔o→3. Since all elements of {1,2,3,4,5}are now accounted for, σequals(1 2 4)(3 5): 1 4 23 5 From this example, it is easy to see that any permutation can be written as a product of disjoint cycles (a1···ar)(b1···bs)···. LetSkbe the group of all permutations of the set {1,..., k}. A permutation is even orodddepending on whether it is the product of an even or an odd number of transpositions. From the theory of permutations we know that this is a well-deﬁned concept: an even permutation can never be written as the product of an odd numberof transpositions and vice versa. The sign of a permutation σ, denoted by sgn (σ)or sgnσ, is deﬁned to be +1 or−1 depending on whether the permutation is even or odd. Clearly, the sign of a permutation satisﬁes sgn(στ)=sgn(σ)sgn(τ) (3.2) forσ,τ∈Sk. 3.2 Permutations 21 Example 3.5.The decomposition (1 2 3 4 5)=( 1 5)(1 4)(1 3)(1 2) shows that the 5-cycle (1 2 3 4 5)is an even permutation. More generally, the decomposition (a1a2···ar)=( a1ar)(a1ar−1)···(a1a3)(a1a2) shows that an r-cycle is an even permutation if and only if ris odd, and an odd per- mutation if and only if ris even. Thus one way to compute the sign of a permutation is to decompose it into a product of cycles and to count the number of cycles of even length. For example, the permutation σ=(1 2 4)(3 5)in Example 3.4 is odd because (1 2 4)is even and (3 5)is odd. Aninversion in a permutation σis an ordered pair (σ(i),σ(j))such that i<j butσ(i)>σ(j). To ﬁnd all the inversions in a permutation σ, it sufﬁces to scan the second row of the matrix of σfrom left to right; the inversions are the pairs (a,b) with a>bandato the left of b. For the permutation σin Example 3.4, from its matrix (3.1) we can read off its ﬁve inversions: (2,1),(4,1),(5,1),(4,3), and(5,3). Exercise 3.6 (Inversions).* Find the inversions in the permutation τ=(1 2 3 4 5 )of Exam- ple 3.5. A second way to compute the sign of a permutation is to count the number of inversions, as we illustrate in the following example. Example 3.7.Letσbe the permutation of Example 3.4. Our goal is to turn σinto the identity permutation 1by multiplying it on the left by transpositions. (i) To move 1 to its natural position at the beginning of the second row of the matrix ofσ, we need to move it across the three elements 2 ,4,5. This can be accom- plished by multiplying σon the left by three transpositions: ﬁrst (5 1), then (4 1), and ﬁnally (2 1): σ=/bracketleftbigg 1 2 3 4 5 2 4 5 1 3/bracketrightbigg (5 1)−−−→/bracketleftbigg 2 4 1 5 3/bracketrightbigg (4 1)−−−→/bracketleftbigg 2 1 4 5 3/bracketrightbigg (2 1)−−−→/bracketleftbigg 1 2 4 5 3/bracketrightbigg . The three transpositions (5 1),(4 1), and(2 1)correspond precisely to the three inversions of σending in 1. (ii) The element 2 is already in its natural position in the second row of the matrix. (iii) To move 3 to its natural position in the second row, we need to move it across two elements 4, 5. This can be accomplished by /bracketleftbigg1 2 3 4 5 1 2 4 5 3/bracketrightbigg (5 3)−−−→/bracketleftbigg 1 2 4 3 5/bracketrightbigg (4 3)−−−→/bracketleftbigg 1 2 3 4 5/bracketrightbigg =1. Thus, (4 3)(5 3)(2 1)(4 1)(5 1)σ=1. (3.3) 22§3 The Exterior Algebra of Multicovectors Note that the two transpositions (5 3)and(4 3)correspond to the two inversions ending in 3. Multiplying both sides of (3.3) on the left by the transpositions(4 3), then(5 3), then(2 1), and so on eventually yields σ=(5 1)(4 1)(2 1)(5 3)(4 3). This shows that σcan be written as a product of as many transpositions as the number of inversions in it. With this example in mind, we prove the following proposition. Proposition 3.8. A permutation is even if and only if it has an even number of inver- sions. Proof. We will obtain the identity permutation 1by multiplying σon the left by a number of transpositions. This can be achieved in ksteps. (i) First, look for the number 1 among σ(1),σ(2),..., σ(k). Every number preced- ing 1 in this list gives rise to an inversion, for if 1 =σ(i), then(σ(1),1),..., (σ(i−1),1)are inversions of σ. Now move 1 to the beginning of the list across thei−1 elements σ(1),..., σ(i−1). This requires multiplying σon the left by i−1 transpositions: σ1=(σ(1)1)···(σ(i−1)1)σ=/bracketleftbigg 1σ(1)···σ(i−1)σ(i+1)···σ(k)/bracketrightbigg . Note that the number of transpositions is the number of inversions ending in 1. (ii) Next look for the number 2 in the list: 1 ,σ(1),..., σ(i−1),σ(i+1),..., σ(k). Every number other than 1 preceding 2 in this list gives rise to an inversion (σ(m),2). Suppose there are i2such numbers. Then there are i2inversions ending in 2. In moving 2 to its natural position 1 ,2,σ(1),σ(2),..., we need to move it across i2numbers. This can be accomplished by multiplying σ1on the left by i2transpositions. Repeating this procedure, we see that for each j=1,..., k, the number of trans- positions required to move jto its natural position is the same as the number of in- versions ending in j. In the end we achieve the identity permutation, i.e, the ordered list 1,2,..., k, from σ(1),σ(2),..., σ(k)by multiplying σby as many transpositions as the total number of inversions in σ. Therefore, sgn (σ)=(−1)# inversions in σ.⊓ ⊔ 3.3 Multilinear Functions Denote by Vk=V×···× Vthe Cartesian product of kcopies of a real vector space V. A function f:Vk→Risk-linear if it is linear in each of its karguments: f(...,av+bw,...)=a f(...,v,...)+b f(...,w,...) for all a,b∈Randv,w∈V. Instead of 2-linear and 3-linear, it is customary to say “bilinear” and “trilinear.” A k-linear function on Vis also called a k-tensor onV. We will denote the vector space of all k-tensors on VbyLk(V). Iffis ak-tensor on V, we also call kthedegree off. 3.4 The Permutation Action on Multilinear Functions 23 Example 3.9 ( Dot product on Rn).With respect to the standard basis e1,..., enfor Rn, the dot product , deﬁned by f(v,w)=v•w=∑ iviwi,where v=∑viei,w=∑wiei, is bilinear. Example. The determinant f(v1,..., vn)=det[v1···vn], viewed as a function of the ncolumn vectors v1,..., vninRn, isn-linear. Deﬁnition 3.10. Ak-linear function f:Vk→Rissymmetric if f/parenleftbig vσ(1),..., vσ(k)/parenrightbig =f(v1,..., vk) for all permutations σ∈Sk; it is alternating if f/parenleftbig vσ(1),..., vσ(k)/parenrightbig =(sgnσ)f(v1,..., vk) for all σ∈Sk. Examples. (i) The dot product f(v,w)=v•wonRnis symmetric. (ii) The determinant f(v1,..., vn)=det[v1···vn]onRnis alternating. (iii) The cross product v×wonR3is alternating. (iv) For any two linear functions f,g:V→Ron a vector space V, the function f∧g:V×V→Rdeﬁned by (f∧g)(u,v)=f(u)g(v)−f(v)g(u) is alternating. This is a special case of the wedge product , which we will soon deﬁne. We are especially interested in the space Ak(V)of all alternating k-linear func- tions on a vector space Vfork>0. These are also called alternating k-tensors , k-covectors , ormulticovectors of degree k onV. For k=0, we deﬁne a 0 -covector to be a constant, so that A0(V)is the vector space R. A 1-covector is simply a covector. 3.4 The Permutation Action on Multilinear Functions Iffis ak-linear function on a vector space Vandσis a permutation in Sk, we deﬁne a new k-linear function σfby (σf)(v1,..., vk)=f/parenleftbig vσ(1),..., vσ(k)/parenrightbig . Thus, fis symmetric if and only if σf=ffor all σ∈Skandfis alternating if and only if σf=(sgnσ)ffor all σ∈Sk. When there is only one argument, the permutation group S1is the identity group and a 1-linear function is both symmetric and alternating. In particular, A1(V)=L1(V)=V∨. 24§3 The Exterior Algebra of Multicovectors Lemma 3.11. Ifσ,τ∈Skand f is a k-linear function on V, then τ(σf)=( τσ)f . Proof. Forv1,..., vk∈V, τ(σf)(v1,..., vk)=( σf)/parenleftbig vτ(1),..., vτ(k)/parenrightbig =(σf)(w1,..., wk) ( letting wi=vτ(i)) =f/parenleftbig wσ(1),..., wσ(k)/parenrightbig =f/parenleftbig vτ(σ(1)),..., vτ(σ(k))/parenrightbig =f/parenleftbig v(τσ)(1),..., v(τσ)(k)/parenrightbig =(τσ)f(v1,..., vk). ⊓ ⊔ In general, if Gis a group and Xis a set, a map G×X→X, (σ,x)/ma√s⊔o→σ·x is called a left action ofGonXif (i)e·x=x, where eis the identity element in Gandxis any element in X, and (ii)τ·(σ·x)=( τσ)·xfor all τ,σ∈Gandx∈X. Theorbit of an element x∈Xis deﬁned to be the set Gx:={σ·x∈X|σ∈G}. In this terminology, we have deﬁned a left action of the permutation group Skon the space Lk(V)ofk-linear functions on V. Note that each permutation acts as a linear function on the vector space Lk(V)since σfisR-linear in f. Aright action ofGonXis deﬁned similarly; it is a map X×G→Xsuch that (i)x·e=x, and (ii)(x·σ)·τ=x·(στ) for all σ,τ∈Gandx∈X. Remark. In some books the notation for σfisfσ. In that notation, (fσ)τ=fτσ, not fστ. 3.5 The Symmetrizing and Alternating Operators Given any k-linear function fon a vector space V, there is a way to make a symmetric k-linear function S ffrom it: (S f)(v1,..., vk)=∑ σ∈Skf/parenleftbig vσ(1),..., vσ(k)/parenrightbig or, in our new shorthand, S f=∑ σ∈Skσf. Similarly, there is a way to make an alternating k-linear function from f. Deﬁne A f=∑ σ∈Sk(sgnσ)σf. 3.6 The Tensor Product 25 Proposition 3.12. If f is a k-linear function on a vector space V, then (i)the k-linear function S f is symmetric, and (ii)the k-linear function A f is alternating. Proof. We prove (ii) only, leaving (i) as an exercise. For τ∈Sk, τ(A f)=∑ σ∈Sk(sgnσ)τ(σf) =∑ σ∈Sk(sgnσ)(τσ)f (by Lemma 3.11 ) =(sgnτ)∑ σ∈Sk(sgnτσ)(τσ)f(by (3.2)) =(sgnτ)A f, since as σruns through all permutations in Sk, so does τσ. ⊓ ⊔ Exercise 3.13 (Symmetrizing operator).* Show that the k-linear function S fis symmetric. Lemma 3.14. If f is an alternating k-linear function on a vector space V, then A f = (k!)f . Proof. Since for alternating fwe have σf= (sgnσ)f, and sgn σis±1, we must have A f=∑ σ∈Sk(sgnσ)σf=∑ σ∈Sk(sgnσ)(sgnσ)f=(k!)f.⊓ ⊔ Exercise 3.15 (Alternating operator).* Iffis a 3-linear function on a vector space Vand v1,v2,v3∈V, what is(A f)(v1,v2,v3)? 3.6 The Tensor Product Letfbe a k-linear function and ganℓ-linear function on a vector space V. Their tensor product is the(k+ℓ)-linear function f⊗gdeﬁned by (f⊗g)(v1,..., vk+ℓ)=f(v1,..., vk)g(vk+1,..., vk+ℓ). Example 3.16 ( Bilinear maps ).Lete1,..., enbe a basis for a vector space V,α1,..., αnthe dual basis in V∨, and/an}bracke⊔le{⊔,/an}bracke⊔ri}h⊔:V×V→Ra bilinear map on V. Set gi j=/an}bracke⊔le{⊔ei,ej/an}bracke⊔ri}h⊔∈ R. Ifv=∑vieiandw=∑wiei, then as we observed in Example 3.3, vi=αi(v)and wj=αj(w). By bilinearity, we can express /an}bracke⊔le{⊔,/an}bracke⊔ri}h⊔in terms of the tensor product: /an}bracke⊔le{⊔v,w/an}bracke⊔ri}h⊔=∑viwj/an}bracke⊔le{⊔ei,ej/an}bracke⊔ri}h⊔=∑αi(v)αj(w)gi j=∑gi j(αi⊗αj)(v,w). Hence,/an}bracke⊔le{⊔,/an}bracke⊔ri}h⊔=∑gi jαi⊗αj. This notation is often used in differential geometry to describe an inner product on a vector space. Exercise 3.17 (Associativity of the tensor product). Check that the tensor product of multi- linear functions is associative: if f,g, and hare multilinear functions on V, then (f⊗g)⊗h=f⊗(g⊗h). 26§3 The Exterior Algebra of Multicovectors 3.7 The Wedge Product If two multilinear functions fandgon a vector space Vare alternating, then we would like to have a product that is alternating as well. This motivates the deﬁnition of the wedge product , also called the exterior product : for f∈Ak(V)andg∈Aℓ(V), f∧g=1 k!ℓ!A(f⊗g); (3.4) or explicitly, (f∧g)(v1,..., vk+ℓ) =1 k!ℓ!∑ σ∈Sk+ℓ(sgnσ)f/parenleftbig vσ(1),..., vσ(k)/parenrightbig g/parenleftbig vσ(k+1),..., vσ(k+ℓ)/parenrightbig . (3.5) By Proposition 3.12, f∧gis alternating. When k=0, the element f∈A0(V)is simply a constant c. In this case, the wedge product c∧gis scalar multiplication, since the right-hand side of (3.5) is 1 ℓ!∑ σ∈Sℓ(sgnσ)cg/parenleftbig vσ(1),..., vσ(ℓ)/parenrightbig =cg(v1,..., vℓ). Thus c∧g=cgforc∈Randg∈Aℓ(V). The coefﬁcient 1 /k!ℓ! in the deﬁnition of the wedge product compensates for repetitions in the sum: for every permutation σ∈Sk+ℓ, there are k! permutations τ inSkthat permute the ﬁrst karguments vσ(1),..., vσ(k)and leave the arguments of g alone; for all τinSk, the resulting permutations στinSk+ℓcontribute the same term to the sum, since (sgnστ)f/parenleftbig vστ(1),..., vστ(k)/parenrightbig =(sgnστ)(sgnτ)f/parenleftbig vσ(1),..., vσ(k)/parenrightbig =(sgnσ)f/parenleftbig vσ(1),..., vσ(k)/parenrightbig , where the ﬁrst equality follows from the fact that (τ(1),..., τ(k))is a permutation of (1,..., k). So we divide by k! to get rid of the k! repeating terms in the sum coming from permutations of the karguments of f; similarly, we divide by ℓ! on account of theℓarguments of g. Example 3.18.Forf∈A2(V)andg∈A1(V), A(f⊗g)(v1,v2,v3)= f(v1,v2)g(v3)−f(v1,v3)g(v2)+f(v2,v3)g(v1) −f(v2,v1)g(v3)+f(v3,v1)g(v2)−f(v3,v2)g(v1). Among these six terms, there are three pairs of equal terms, which we have lined up vertically in the display above: f(v1,v2)g(v3)=−f(v2,v1)g(v3),and so on . Therefore, after dividing by 2, (f∧g)(v1,v2,v3)=f(v1,v2)g(v3)−f(v1,v3)g(v2)+f(v2,v3)g(v1). 3.8 Anticommutativity of the Wedge Product 27 One way to avoid redundancies in the deﬁnition of f∧gis to stipulate that in the sum (3.5), σ(1),..., σ(k)be in ascending order and σ(k+1),..., σ(k+ℓ)also be in ascending order. We call a permutation σ∈Sk+ℓa(k,ℓ)-shufﬂe if σ(1)<···<σ(k)and σ(k+1)<···<σ(k+ℓ). By the paragraph before Example 3.18, one may rewrite (3.5) as (f∧g)(v1,..., vk+ℓ) =∑ (k,ℓ)-shufﬂes σ(sgnσ)f/parenleftbig vσ(1),..., vσ(k)/parenrightbig g/parenleftbig vσ(k+1),..., vσ(k+ℓ)/parenrightbig . (3.6) Written this way, the deﬁnition of (f∧g)(v1,..., vk+ℓ)is a sum of/parenleftbigk+ℓ k/parenrightbig terms, in- stead of(k+ℓ)! terms. Example 3.19 ( Wedge product of two covectors ).*Iffandgare covectors on a vector space Vandv1,v2∈V, then by (3.6), (f∧g)(v1,v2)=f(v1)g(v2)−f(v2)g(v1). Exercise 3.20 (Wedge product of two 2-covectors). Forf,g∈A2(V), write out the deﬁnition off∧gusing(2,2)-shufﬂes. 3.8 Anticommutativity of the Wedge Product It follows directly from the deﬁnition of the wedge product (3.5) that f∧gis bilinear infand in g. Proposition 3.21. The wedge product is anticommutative: if f ∈Ak(V)and g∈ Aℓ(V), then f∧g=(−1)kℓg∧f. Proof. Deﬁne τ∈Sk+ℓto be the permutation τ=/bracketleftbigg1···ℓ ℓ+1···ℓ+k k+1···k+ℓ1··· k/bracketrightbigg . This means that τ(1)=k+1,..., τ(ℓ)=k+ℓ,τ(ℓ+1)=1,..., τ(ℓ+k)=k. Then σ(1)=στ(ℓ+1),..., σ(k)=στ(ℓ+k), σ(k+1)=στ(1),..., σ(k+ℓ)=στ(ℓ). For any v1,..., vk+ℓ∈V, 28§3 The Exterior Algebra of Multicovectors A(f⊗g)(v1,..., vk+ℓ) =∑ σ∈Sk+ℓ(sgnσ)f/parenleftbig vσ(1),..., vσ(k)/parenrightbig g/parenleftbig vσ(k+1),..., vσ(k+ℓ)/parenrightbig =∑ σ∈Sk+ℓ(sgnσ)f/parenleftbig vστ(ℓ+1),..., vστ(ℓ+k)/parenrightbig g/parenleftbig vστ(1),..., vστ(ℓ)/parenrightbig =(sgnτ)∑ σ∈Sk+ℓ(sgnστ)g/parenleftbig vστ(1),..., vστ(ℓ)/parenrightbig f/parenleftbig vστ(ℓ+1),..., vστ(ℓ+k)/parenrightbig =(sgnτ)A(g⊗f)(v1,..., vk+ℓ). The last equality follows from the fact that as σruns through all permutations in Sk+ℓ, so does στ. We have proven A(f⊗g)=( sgnτ)A(g⊗f). Dividing by k!ℓ! gives f∧g=(sgnτ)g∧f. Exercise 3.22 (Sign of a permutation).* Show that sgn τ=(−1)kℓ. ⊓ ⊔ Corollary 3.23. If f is a multicovector of odd degree on V, then f ∧f=0. Proof. Letkbe the degree of f. By anticommutativity, f∧f=(−1)k2f∧f=−f∧f, since kis odd. Hence, 2 f∧f=0. Dividing by 2 gives f∧f=0. ⊓ ⊔ 3.9 Associativity of the Wedge Product The wedge product of a k-covector fand anℓ-covector gon a vector space Vis by deﬁnition the (k+ℓ)-covector f∧g=1 k!ℓ!A(f⊗g). To prove the associativity of the wedge product, we will follow Godbillon [14] by ﬁrst proving a lemma on the alternating operator A. Lemma 3.24. Suppose f is a k-linear function and g an ℓ-linear function on a vector space V. Then (i)A(A(f)⊗g)=k!A(f⊗g), and (ii)A(f⊗A(g))=ℓ!A(f⊗g). Proof. (i) By deﬁnition, A(A(f)⊗g)=∑ σ∈Sk+ℓ(sgnσ)σ/parenleftigg ∑ τ∈Sk(sgnτ)(τf)⊗g/parenrightigg . 3.9 Associativity of the Wedge Product 29 We can view τ∈Skalso as a permutation in Sk+ℓﬁxing k+1,..., k+ℓ. Viewed this way, τsatisﬁes (τf)⊗g=τ(f⊗g). Hence, A(A(f)⊗g)=∑ σ∈Sk+ℓ∑ τ∈Sk(sgnσ)(sgnτ)(στ)(f⊗g). (3.7) For each µ∈Sk+ℓand each τ∈Sk, there is a unique element σ=µτ−1∈Sk+ℓsuch thatµ=στ, so each µ∈Sk+ℓappears once in the double sum (3.7) for each τ∈Sk, and hence k! times in total. So the double sum (3.7) can be rewritten as A(A(f)⊗g)=k!∑ µ∈Sk+ℓ(sgnµ)µ(f⊗g)=k!A(f⊗g). The equality in (ii) is proved in the same way. ⊓ ⊔ Proposition 3.25 (Associativity of the wedge product). Let V be a real vector space and f ,g,h alternating multilinear functions on V of degrees k ,ℓ,m, respec- tively. Then (f∧g)∧h=f∧(g∧h). Proof. By the deﬁnition of the wedge product, (f∧g)∧h=1 (k+ℓ)!m!A((f∧g)⊗h) =1 (k+ℓ)!m!1 k!ℓ!A(A(f⊗g)⊗h) =(k+ℓ)! (k+ℓ)!m!k!ℓ!A((f⊗g)⊗h) (by Lemma 3.24(i) ) =1 k!ℓ!m!A((f⊗g)⊗h). Similarly, f∧(g∧h)=1 k!(ℓ+m)!A/parenleftbigg f⊗1 ℓ!m!A(g⊗h)/parenrightbigg =1 k!ℓ!m!A(f⊗(g⊗h)). Since the tensor product is associative, we conclude that (f∧g)∧h=f∧(g∧h). ⊓ ⊔ By associativity, we can omit the parentheses in a multiple wedge product such as(f∧g)∧hand write simply f∧g∧h. Corollary 3.26. Under the hypotheses of the proposition, f∧g∧h=1 k!ℓ!m!A(f⊗g⊗h). 30§3 The Exterior Algebra of Multicovectors This corollary easily generalizes to an arbitrary number of factors: if fi∈ Adi(V), then f1∧···∧ fr=1 (d1)!···(dr)!A(f1⊗···⊗ fr). (3.8) In particular, we have the following proposition. We use the notation [bi j]to denote the matrix whose (i,j)-entry is bi j. Proposition 3.27 (Wedge product of 1-covectors). Ifα1,..., αkare linear func- tions on a vector space V and v 1,..., vk∈V, then (α1∧···∧ αk)(v1,..., vk)=det[αi(vj)]. Proof. By (3.8), (α1∧···∧ αk)(v1,..., vk)=A(α1⊗···⊗ αk)(v1,..., vk) =∑ σ∈Sk(sgnσ)α1/parenleftbig vσ(1)/parenrightbig ···αk/parenleftbig vσ(k)/parenrightbig =det[αi(vj)]. ⊓ ⊔ An algebra Aover a ﬁeld Kis said to be graded if it can be written as a direct sum A=/circleplustext∞ k=0Akof vector spaces over Ksuch that the multiplication map sends Ak×AℓtoAk+ℓ. The notation A=/circleplustext∞ k=0Akmeans that each nonzero element of Ais uniquely a ﬁnite sum a=ai1+···+aim, where aij/ne}a⊔ionslash=0∈Aij. A graded algebra A=⊕∞ k=0Akis said to be anticommutative or graded commutative if for all a∈Akandb∈Aℓ, ab=(−1)kℓba. Ahomomorphism of graded algebras is an algebra homomorphism that preserves the degree. Example. The polynomial algebra A=R[x,y]is graded by degree; Akconsists of all homogeneous polynomials of total degree kin the variables xandy. For a ﬁnite-dimensional vector space V, say of dimension n, deﬁne A∗(V)=∞/circleplusdisplay k=0Ak(V)=n/circleplusdisplay k=0Ak(V). With the wedge product of multicovectors as multiplication, A∗(V)becomes an an- ticommutative graded algebra, called the exterior algebra or the Grassmann algebra of multicovectors on the vector space V. 3.10 A Basis for k-Covectors 31 3.10 A Basis for k-Covectors Lete1,..., enbe a basis for a real vector space V, and let α1,..., αnbe the dual basis forV∨. Introduce the multi-index notation I=(i1,..., ik) and write eIfor(ei1,..., eik)andαIforαi1∧···∧ αik. Ak-linear function fonVis completely determined by its values on all k-tuples (ei1,..., eik). If fis alternating, then it is completely determined by its values on (ei1,..., eik)with 1≤i1<···<ik≤n; that is, it sufﬁces to consider eIwith Iin strictly ascending order. Lemma 3.28. Let e 1,..., enbe a basis for a vector space V and let α1,..., αnbe its dual basis in V∨. If I=(1≤i1<···<ik≤n)and J=(1≤j1<···<jk≤n)are strictly ascending multi-indices of length k, then αI(eJ)=δI J=/braceleftigg 1for I=J, 0for I/ne}a⊔ionslash=J. Proof. By Proposition 3.27, αI(eJ)=det[αi(ej)]i∈I,j∈J. IfI=J, then[αi(ej)]is the identity matrix and its determinant is 1. IfI/ne}a⊔ionslash=J, we compare them term by term until the terms differ: i1=j1, ..., iℓ−1=jℓ−1,iℓ/ne}a⊔ionslash=jℓ, ... . Without loss of generality, we may assume iℓ<jℓ. Then iℓwill be different from j1,..., jℓ−1(because these are the same as i1,..., iℓ, and Iis strictly ascending), and iℓwill also be different from jℓ,jℓ+1,..., jk(because Jis strictly ascending). Thus, iℓwill be different from j1,..., jk, and the ℓth row of the matrix [ai(ej)]will be all zero. Hence, det [ai(ej)]= 0. ⊓ ⊔ Proposition 3.29. The alternating k-linear functions αI, I=(i1<···<ik), form a basis for the space A k(V)of alternating k-linear functions on V. Proof. First, we show linear independence. Suppose ∑cIαI=0,cI∈R, and Iruns over all strictly ascending multi-indices of length k. Applying both sides to eJ,J= (j1<···<jk), we get by Lemma 3.28, 0=∑ IcIαI(eJ)=∑ IcIδI J=cJ, since among all strictly ascending multi-indices Iof length k, there is only one equal toJ. This proves that the αIare linearly independent. To show that the αIspan Ak(V), let f∈Ak(V). We claim that 32§3 The Exterior Algebra of Multicovectors f=∑f(eI)αI, where Iruns over all strictly ascending multi-indices of length k. Let g=∑f(eI)αI. Byk-linearity and the alternating property, if two k-covectors agree on all eJ, where J=(j1<···<jk), then they are equal. But g(eJ)=∑f(eI)αI(eJ)=∑f(eI)δI J=f(eJ). Therefore, f=g=∑f(eI)αI. ⊓ ⊔ Corollary 3.30. If the vector space V has dimension n, then the vector space A k(V) of k-covectors on V has dimension/parenleftbign k/parenrightbig . Proof. A strictly ascending multi-index I=(i1<···<ik)is obtained by choosing a subset of knumbers from 1 ,..., n. This can be done in/parenleftbign k/parenrightbig ways. ⊓ ⊔ Corollary 3.31. If k>dimV, then A k(V)=0. Proof. Inαi1∧···∧ αik, at least two of the factors must be the same, say αj=αℓ= α. Because αis a 1-covector, α∧α=0 by Corollary 3.23, so αi1∧···∧ αik=0.⊓ ⊔ Problems 3.1. Tensor product of covectors Lete1,..., enbe a basis for a vector space Vand let α1,..., αnbe its dual basis in V∨. Suppose [gi j]∈Rn×nis an n×nmatrix. Deﬁne a bilinear function f:V×V→Rby f(v,w)=∑ 1≤i,j≤ngi jviwj forv=∑vieiandw=∑wjejinV. Describe fin terms of the tensor products of αiandαj, 1≤i,j≤n. 3.2. Hyperplanes (a) Let Vbe a vector space of dimension nandf:V→Ra nonzero linear functional. Show that dimker f=n−1. A linear subspace of Vof dimension n−1 is called a hyperplane inV. (b) Show that a nonzero linear functional on a vector space Vis determined up to a multi- plicative constant by its kernel, a hyperplane in V. In other words, if fandg:V→Rare nonzero linear functionals and ker f=kerg, then g=c ffor some constant c∈R. 3.3. A basis for k-tensors LetVbe a vector space of dimension nwith basis e1,..., en. Let α1,..., αnbe the dual basis forV∨. Show that a basis for the space Lk(V)ofk-linear functions on Vis{αi1⊗···⊗ αik} for all multi-indices (i1,..., ik)(not just the strictly ascending multi-indices as for Ak(L)). In particular, this shows that dim Lk(V)=nk. (This problem generalizes Problem 3.1.) 3.10 A Basis for k-Covectors 33 3.4. A characterization of alternating k-tensors Letfbe a k-tensor on a vector space V. Prove that fis alternating if and only if fchanges sign whenever two successive arguments are interchanged: f(..., vi+1,vi,...)=−f(..., vi,vi+1,...) fori=1,..., k−1. 3.5. Another characterization of alternating k-tensors Letfbe ak-tensor on a vector space V. Prove that fis alternating if and only if f(v1,..., vk)= 0 whenever two of the vectors v1,..., vkare equal. 3.6. Wedge product and scalars LetVbe a vector space. For a,b∈R,f∈Ak(V), and g∈Aℓ(V), show that a f∧bg=(ab)f∧g. 3.7. Transformation rule for a wedge product of covectors Suppose two sets of covectors on a vector space V,β1,..., βkandγ1,..., γk, are related by βi=k ∑ j=1ai jγj,i=1,..., k, for a k×kmatrix A=[ai j]. Show that β1∧···∧ βk=(detA)γ1∧···∧ γk. 3.8. Transformation rule for k-covectors Letfbe ak-covector on a vector space V. Suppose two sets of vectors u1,..., ukandv1,..., vk inVare related by uj=k ∑ i=1ai jvi,j=1,..., k, for a k×kmatrix A=[ai j]. Show that f(u1,..., uk)=(det A)f(v1,..., vk). 3.9. Vanishing of a covector of top degree LetVbe a vector space of dimension n. Prove that if an n-covector ωvanishes on a basis e1,..., enforV, then ωis the zero covector on V. 3.10.* Linear independence of covectors Letα1,..., αkbe 1-covectors on a vector space V. Show that α1∧···∧ αk/ne}a⊔ionslash=0 if and only if α1,..., αkare linearly independent in the dual space V∨. 3.11.* Exterior multiplication Letαbe a nonzero 1-covector and γak-covector on a ﬁnite-dimensional vector space V. Show that α∧γ=0 if and only if γ=α∧βfor some (k−1)-covector βonV. 34§4 Differential Forms on Rn §4 Differential Forms on Rn Just as a vector ﬁeld assigns a tangent vector to each point of an open subset UofRn, so dually a differential k-form assigns a k-covector on the tangent space to each point ofU. The wedge product of differential forms is deﬁned pointwise as the wedge product of multicovectors. Since differential forms exist on an open set, not just at a single point, there is a notion of differentiation for differential forms. In fact, there is a unique one, called the exterior derivative, characterized by three natural properties. Although we deﬁne it using the standard coordinates of Rn, the exterior derivative turns out to be independent of coordinates, as we shall see later, and is therefore intrinsic to a manifold. It is the ultimate abstract extension to a manifold of the gradient, curl, and divergence of vector calculus in R3. Differential forms extend Grassmann’s exterior algebra from the tangent space at a point globally to anentire manifold. Since its creation around the turn of the twentieth century, generally credited to ´E. Cartan [5] and H. Poincar´ e [34], the calculus of differential forms has had far-reaching consequences in geometry, topology, and physics. In fact, certainphysical concepts such as electricity and magnetism are best formulated in terms of differential forms. In this section we will study the simplest case, that of differential forms on an open subset of R n. Even in this setting, differential forms already provide a way to unify the main theorems of vector calculus in R3. 4.1 Differential 1-Forms and the Differential of a Function The cotangent space toRnatp, denoted by T∗ p(Rn)orT∗ pRn, is deﬁned to be the dual space (TpRn)∨of the tangent space Tp(Rn). Thus, an element of the cotangent space T∗ p(Rn)is a covector or a linear functional on the tangent space Tp(Rn). In parallel with the deﬁnition of a vector ﬁeld, a covector ﬁeld or adifferential 1-form on an open subset UofRnis a function ωthat assigns to each point pinUa covector ωp∈T∗ p(Rn), ω:U→/uniondisplay p∈UT∗ p(Rn), p/ma√s⊔o→ωp∈T∗ p(Rn). Note that in the union/uniontext p∈UT∗ p(Rn), the sets T∗ p(Rn)are all disjoint. We call a differential 1-form a 1 -form for short. From any C∞function f:U→R, we can construct a 1-form df, called the dif- ferential off, as follows. For p∈UandXp∈TpU, deﬁne (df)p(Xp)=Xpf. A few words may be in order about the deﬁnition of the differential. The directional derivative of a function in the direction of a tangent vector at a point psets up a bilinear pairing 4.1 Differential 1-Forms and the Differential of a Function 35 Tp(Rn)×C∞ p(Rn)→R, (Xp,f)/ma√s⊔o→/an}bracke⊔le{⊔Xp,f/an}bracke⊔ri}h⊔=Xpf. One may think of a tangent vector as a function on the second argument of this pairing:/an}bracke⊔le{⊔Xp,·/an}bracke⊔ri}h⊔. The differential (df)patpis a function on the ﬁrst argument of the pairing: (df)p=/an}bracke⊔le{⊔·,f/an}bracke⊔ri}h⊔. The value of the differential dfatpis also written df|p. Letx1,..., xnbe the standard coordinates on Rn. We saw in Subsection 2.3 that the set{∂/∂x1|p,..., ∂/∂xn|p}is a basis for the tangent space Tp(Rn). Proposition 4.1. If x1,..., xnare the standard coordinates on Rn, then at each point p∈Rn,{(dx1)p,...,(dxn)p}is the basis for the cotangent space T∗ p(Rn)dual to the basis{∂/∂x1|p,..., ∂/∂xn|p}for the tangent space T p(Rn). Proof. By deﬁnition, (dxi)p/parenleftigg ∂ ∂xj/vextendsingle/vextendsingle/vextendsingle/vextendsingle p/parenrightigg =∂ ∂xj/vextendsingle/vextendsingle/vextendsingle/vextendsingle pxi=δi j. ⊓ ⊔ Ifωis a 1-form on an open subset UofRn, then by Proposition 4.1, at each point pinU,ωcan be written as a linear combination ωp=∑ai(p)(dxi)p, for some ai(p)∈R. As pvaries over U, the coefﬁcients aibecome functions on U, and we may write ω=∑aidxi. The covector ﬁeld ωis said to be C∞on U if the coefﬁcient functions aiare all C∞onU. Ifx,y, and zare the coordinates on R3, then dx,dy, and dzare 1-forms on R3. In this way, we give meaning to what was merely a notation in elementary calculus. Proposition 4.2 (The differential in terms of coordinates). If f:U→Ris a C∞ function on an open set U in Rn, then df=∑∂f ∂xidxi. (4.1) Proof. By Proposition 4.1, at each point pinU, (df)p=∑ai(p)(dxi)p (4.2) for some real numbers ai(p)depending on p. Thus, df=∑aidxifor some real functions aionU. To ﬁnd aj, apply both sides of (4.2) to the coordinate vector ﬁeld ∂/∂xj: df/parenleftbigg∂ ∂xj/parenrightbigg =∑ iaidxi/parenleftbigg∂ ∂xj/parenrightbigg =∑ iaiδi j=aj. On the other hand, by the deﬁnition of the differential, df/parenleftbigg∂ ∂xj/parenrightbigg =∂f ∂xj. ⊓ ⊔ 36§4 Differential Forms on Rn Equation (4.1) shows that if fis aC∞function, then the 1-form dfis also C∞. Example. Differential 1-forms arise naturally even if one is interested only in tangent vectors. Every tangent vector Xp∈Tp(Rn)is a linear combination of the standard basis vectors: Xp=∑ ibi(Xp)∂ ∂xi/vextendsingle/vextendsingle/vextendsingle/vextendsingle p. In Example 3.3 we saw that at each point p∈Rn, we have bi(Xp) = ( dxi)p(Xp). Hence, the coefﬁcient biof a vector at pwith respect to the standard basis ∂/∂x1|p, ...,∂/∂xn|pis none other than the dual covector dxi|ponRn. As pvaries, bi=dxi. 4.2 Differential k-Forms More generally, a differential form ωof degree k or a k-form on an open subset U ofRnis a function that assigns to each point pinUan alternating k-linear function on the tangent space Tp(Rn), i.e., ωp∈Ak(TpRn). Since A1(TpRn) =T∗ p(Rn), the deﬁnition of a k-form generalizes that of a 1-form in Subsection 4.1. By Proposition 3.29, a basis for Ak(TpRn)is dxI p=dxi1p∧···∧ dxikp,1≤i1<···<ik≤n. Therefore, at each point pinU,ωpis a linear combination ωp=∑aI(p)dxI p,1≤i1<···<ik≤n, and a k-form ωonUis a linear combination ω=∑aIdxI, with function coefﬁcients aI:U→R. We say that a k-form ωisC∞onUif all the coefﬁcients aIareC∞functions on U. Denote by Ωk(U)the vector space of C∞k-forms on U. A 0-form on Uassigns to each point pinUan element of A0(TpRn)=R. Thus, a 0-form on Uis simply a function on U, and Ω0(U)=C∞(U). There are no nonzero differential forms of degree >non an open subset of Rn. This is because if deg dxI>n, then in the expression dxIat least two of the 1-forms dxiαmust be the same, forcing dxI=0. Thewedge product of a k-form ωand anℓ-form τon an open set Uis deﬁned pointwise: (ω∧τ)p=ωp∧τp,p∈U. In terms of coordinates, if ω=∑IaIdxIandτ=∑JbJdxJ, then ω∧τ=∑ I,J(aIbJ)dxI∧dxJ. In this sum, if IandJare not disjoint on the right-hand side, then dxI∧dxJ=0. Hence, the sum is actually over disjoint multi-indices: 4.3 Differential Forms as Multilinear Functions on Vector Fields 37 ω∧τ=∑ I,Jdisjoint(aIbJ)dxI∧dxJ, which shows that the wedge product of two C∞forms is C∞. So the wedge product is a bilinear map ∧:Ωk(U)×Ωℓ(U)→Ωk+ℓ(U). By Propositions 3.21 and 3.25, the wedge product of differential forms is anticom- mutative and associative. In case one of the factors has degree 0, say k=0, the wedge product ∧:Ω0(U)×Ωℓ(U)→Ωℓ(U) is the pointwise multiplication of a C∞ℓ-form by a C∞function: (f∧ω)p=f(p)∧ωp=f(p)ωp, since as we noted in Subsection 3.7, the wedge product with a 0-covector is scalar multiplication. Thus, if f∈C∞(U)andω∈Ωℓ(U), then f∧ω=fω. Example. Letx,y,zbe the coordinates on R3. The C∞1-forms on R3are f dx+gdy+hdz, where f,g,hrange over all C∞functions on R3. The C∞2-forms are f dy∧dz+gdx∧dz+hdx∧dy and the C∞3-forms are f dx∧dy∧dz. Exercise 4.3 (A basis for 3-covectors).* Letx1,x2,x3,x4be the coordinates on R4andpa point inR4. Write down a basis for the vector space A3(Tp(R4)). With the wedge product as multiplication and the degree of a form as the grading, the direct sum Ω∗(U) =/circleplustextn k=0Ωk(U)becomes an anticommutative graded algebra overR. Since one can multiply C∞k-forms by C∞functions, the set Ωk(U)ofC∞k- forms on Uis both a vector space over Rand a module over C∞(U), and so the direct sum Ω∗(U)=/circleplustextn k=0Ωk(U)is also a module over the ring C∞(U)ofC∞functions. 4.3 Differential Forms as Multilinear Functions on Vector Fields Ifωis aC∞1-form and Xis aC∞vector ﬁeld on an open set UinRn, we deﬁne a function ω(X)onUby the formula ω(X)p=ωp(Xp),p∈U. Written out in coordinates, 38§4 Differential Forms on Rn ω=∑aidxi, X=∑bj∂ ∂xjfor some ai,bj∈C∞(U), so ω(X)=/parenleftbig∑aidxi/parenrightbig/parenleftbigg ∑bj∂ ∂xj/parenrightbigg =∑aibi, which shows that ω(X)isC∞onU. Thus, a C∞1-form on Ugives rise to a map fromX(U)toC∞(U). This function is actually linear over the ring C∞(U); i.e., if f∈C∞(U), then ω(f X) =fω(X). To show this, it sufﬁces to evaluate ω(f X)at an arbitrary point p∈U: (ω(f X))p=ωp(f(p)Xp)(deﬁnition of ω(f X)) =f(p)ωp(Xp)(ωpisR-linear) =(fω(X))p (deﬁnition of fω(X)). LetF(U) =C∞(U). In this notation, a 1-form ωonUgives rise to an F(U)- linear map X(U)→F(U),X/ma√s⊔o→ω(X). Similarly, a k-form ωonUgives rise to a k-linear map over F(U), X(U)×···×X(U)/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright ktimes→F(U), (X1,..., Xk)/ma√s⊔o→ω(X1,..., Xk). Exercise 4.4 (Wedge product of a 2-form with a 1-form).* Letωbe a 2-form and τa 1- form onR3. IfX,Y,Zare vector ﬁelds on M, ﬁnd an explicit formula for (ω∧τ)(X,Y,Z)in terms of the values of ωandτon the vector ﬁelds X,Y,Z. 4.4 The Exterior Derivative To deﬁne the exterior derivative of aC∞k-form on an open subset UofRn, we ﬁrst deﬁne it on 0-forms: the exterior derivative of a C∞function f∈C∞(U)is deﬁned to be its differential df∈Ω1(U); in terms of coordinates, Proposition 4.2 gives df=∑∂f ∂xidxi. Deﬁnition 4.5. Fork≥1, ifω=∑IaIdxI∈Ωk(U), then dω=∑ IdaI∧dxI=∑ I/parenleftigg ∑ j∂aI ∂xjdxj/parenrightigg ∧dxI∈Ωk+1(U). Example. Letωbe the 1-form f dx+gdyonR2, where fandgareC∞functions on R2. To simplify the notation, write fx=∂f/∂x,fy=∂f/∂y. Then 4.4 The Exterior Derivative 39 dω=df∧dx+dg∧dy =(fxdx+fydy)∧dx+(gxdx+gydy)∧dy =(gx−fy)dx∧dy. In this computation dy∧dx=−dx∧dyanddx∧dx=dy∧dy=0 by the anticom- mutative property of the wedge product (Proposition 3.21 and Corollary 3.23). Deﬁnition 4.6. LetA=⊕∞ k=0Akbe a graded algebra over a ﬁeld K. An antideriva- tionof the graded algebra Ais aK-linear map D:A→Asuch that for a∈Akand b∈Aℓ, D(ab)=( Da)b+(−1)kaDb. (4.3) If there is an integer msuch that the antiderivation Dsends AktoAk+mfor all k, then we say that it is an antiderivation of degree m . By deﬁning Ak=0 for k<0, we can extend the grading of a graded algebra Ato negative integers. With this extension, the degree mof an antiderivation can be negative. (An example of an antiderivation of degree−1 is interior multiplication, to be discussed in Subsection 20.4.) Proposition 4.7. (i)The exterior differentiation d :Ω∗(U)→Ω∗(U)is an antiderivation of degree 1: d(ω∧τ)=( dω)∧τ+(−1)degωω∧dτ. (ii)d2=0. (iii)If f∈C∞(U)and X∈X(U), then(df)(X)=X f . Proof. (i) Since both sides of (4.3) are linear in ωand in τ, it sufﬁces to check the equality forω=f dxIandτ=gdxJ. Then d(ω∧τ)=d(f gdxI∧dxJ) =∑∂(f g) ∂xidxi∧dxI∧dxJ =∑∂f ∂xigdxi∧dxI∧dxJ+∑f∂g ∂xidxi∧dxI∧dxJ. In the second sum, moving the 1-form (∂g/∂xi)dxiacross the k-form dxIresults in the sign(−1)kby anticommutativity. Hence, d(ω∧τ)=∑∂f ∂xidxi∧dxI∧gdxJ+(−1)k∑f dxI∧∂g ∂xidxi∧dxJ =dω∧τ+(−1)kω∧dτ. (ii) Again by the R-linearity of d, it sufﬁces to show that d2ω=0 for ω=f dxI. We compute: 40§4 Differential Forms on Rn d2(f dxI)=d/parenleftbigg ∑∂f ∂xidxi∧dxI/parenrightbigg =∑∂2f ∂xj∂xidxj∧dxi∧dxI. In this sum if i=j, then dxj∧dxi=0; if i/ne}a⊔ionslash=j, then ∂2f/∂xi∂xjis symmetric in i andj, but dxj∧dxiis alternating in iandj, so the terms with i/ne}a⊔ionslash=jpair up and cancel each other. For example, ∂2f ∂x1∂x2dx1∧dx2+∂2f ∂x2∂x1dx2∧dx1 =∂2f ∂x1∂x2dx1∧dx2+∂2f ∂x1∂x2(−dx1∧dx2)=0. Therefore, d2(f dxI)=0. (iii) This is simply the deﬁnition of the exterior derivative of a function as the differ- ential of the function. ⊓ ⊔ Proposition 4.8 (Characterization of the exterior derivative). The three proper- ties of Proposition 4.7 uniquely characterize exterior differentiation on an open set U inRn; that is, if D :Ω∗(U)→Ω∗(U)is(i)an antiderivation of degree 1such that (ii)D2=0and(iii)(D f)(X)=X f for f∈C∞(U)and X∈X(U), then D=d. Proof. Since every k-form on Uis a sum of terms such as f dxi1∧···∧ dxik, by linearity it sufﬁces to show that D=don a k-form of this type. By (iii), D f=df onC∞functions. It follows that Ddxi=DDxi=0 by (ii). A simple induction on k, using the antiderivation property of D, proves that for all kand all multi-indices Iof length k, D(dxI)=D(dxi1∧···∧ dxik)=0. (4.4) Finally, for every k-form f dxI, D(f dxI)=( D f)∧dxI+f D(dxI) (by (i)) =(d f)∧dxI(by (ii) and (4.4) ) =d(f dxI) (deﬁnition of d). Hence, D=donΩ∗(U). ⊓ ⊔ 4.5 Closed Forms and Exact Forms Ak-form ωonUisclosed ifdω=0; it is exact if there is a (k−1)-form τsuch that ω=dτonU. Since d(dτ) =0, every exact form is closed. In the next section we will discuss the meaning of closed and exact forms in the context of vector calculus onR3. Exercise 4.9 (A closed 1-form on the punctured plane). Deﬁne a 1-form ωonR2−{0}by ω=1 x2+y2(−ydx+xdy). Show that ωis closed. 4.6 Applications to Vector Calculus 41 A collection of vector spaces {Vk}∞ k=0with linear maps dk:Vk→Vk+1such that dk+1◦dk=0 is called a differential complex or a cochain complex . For any open subset UofRn, the exterior derivative dmakes the vector space Ω∗(U)ofC∞forms onUinto a cochain complex, called the de Rham complex ofU: 0→Ω0(U)d→Ω1(U)d→Ω2(U)→···. The closed forms are precisely the elements of the kernel of d, and the exact forms are the elements of the image of d. 4.6 Applications to Vector Calculus The theory of differential forms uniﬁes many theorems in vector calculus on R3. We summarize here some results from vector calculus and then show how they ﬁt intothe framework of differential forms. By a vector-valued function on an open subset UofR 3, we mean a function F=/an}bracke⊔le{⊔P,Q,R/an}bracke⊔ri}h⊔:U→R3. Such a function assigns to each point p∈Ua vector Fp∈ R3≃Tp(R3). Hence, a vector-valued function on Uis precisely a vector ﬁeld on U. Recall the three operators gradient, curl, and divergence on scalar- and vector-valuedfunctions on U: {scalar func.}grad−−→{ vector func.}curl−−→{ vector func.}div−→{ scalar func.}, grad f= ∂/∂x ∂/∂y ∂/∂z f= fx fy fz , curl P Q R = ∂/∂x ∂/∂y ∂/∂z × P Q R = Ry−Qz −(Rx−Pz) Qx−Py , div P Q R = ∂/∂x ∂/∂y ∂/∂z · P Q R =Px+Qy+Rz. Since every 1-form on Uis a linear combination with function coefﬁcients of dx, dy, and dz, we can identify 1-forms with vector ﬁelds on Uvia Pdx+Qdy+Rdz←→ P Q R . Similarly, 2-forms on Ucan also be identiﬁed with vector ﬁelds on U: Pdy∧dz+Qdz∧dx+Rdx∧dy←→ P Q R , 42§4 Differential Forms on Rn and 3-forms on Ucan be identiﬁed with functions on U: f dx∧dy∧dz←→ f. In terms of these identiﬁcations, the exterior derivative of a 0-form fis df=∂f ∂xdx+∂f ∂ydy+∂f ∂zdz←→ ∂f/∂x ∂f/∂y ∂f/∂x =grad f; the exterior derivative of a 1-form is d(Pdx+Qdy+Rdz) =(Ry−Qz)dy∧dz−(Rx−Pz)dz∧dx+(Qx−Py)dx∧dy, (4.5) which corresponds to curl P Q R = Ry−Qz −(Rx−Pz) Qx−Py ; the exterior derivative of a 2-form is d(Pdy∧dz+Qdz∧dx+Rdx∧dy) =(Px+Qy+Rz)dx∧dy∧dz, (4.6) which corresponds to div P Q R =Px+Qy+Rz. Thus, after appropriate identiﬁcations, the exterior derivatives don 0-forms, 1- forms, and 2-forms are simply the three operators grad, curl, and div. In summary, on an open subset UofR3, there are identiﬁcations Ω0(U)d−−−−→ Ω1(U)d−−−−→ Ω2(U)d−−−−→ Ω3(U) ≃/arrowbt≃/arrowbt≃/arrowbt≃/arrowbt C ∞(U)−−−−→ gradX(U)−−−−→ curlX(U)−−−−→ divC∞(U). Under these identiﬁcations, a vector ﬁeld /an}bracke⊔le{⊔P,Q,R/an}bracke⊔ri}h⊔onR3is the gradient of a C∞ function fif and only if the corresponding 1-form Pdx+Qdy+Rdz isdf. Next we recall three basic facts from calculus concerning grad, curl, and div. Proposition A. curl(grad f)= 0 00 . 4.6 Applications to Vector Calculus 43 Proposition B. div/parenleftigg curl P Q R /parenrightigg =0. Proposition C. OnR3, a vector ﬁeld Fis the gradient of some scalar function f if and only if curl F=0. Propositions A and B express the property d2=0 of the exterior derivative on open subsets of R3; these are easy computations. Proposition C expresses the fact that a 1-form on R3is exact if and only if it is closed. Proposition C need not be true on a region other than R3, as the following well-known example from calcu- lus shows. Example. IfU=R3−{z-axis}, and Fis the vector ﬁeld F=/angbracketleftbigg−y x2+y2,x x2+y2,0/angbracketrightbigg onR3, then curl F=0, but Fis not the gradient of any C∞function on U. The reason is that if Fwere the gradient of a C∞function fonU, then by the fundamental theorem for line integrals, the line integral /integraldisplay C−y x2+y2dx+x x2+y2dy over any closed curve Cwould be zero. However, on the unit circle Cin the(x,y)- plane, with x=costandy=sintfor 0≤t≤2π, this integral is /integraldisplay C−ydx+xdy=/integraldisplay2π 0−(sint)dcost+(cost)dsint=2π. In terms of differential forms, the 1-form ω=−y x2+y2dx+x x2+y2dy is closed but not exact on U. (This 1-form is deﬁned by the same formula as the 1-form ωin Exercise 4.9, but is deﬁned on a different space.) It turns out that whether Proposition C is true for a region Udepends only on the topology of U. One measure of the failure of a closed k-form to be exact is the quotient vector space Hk(U):={closed k-forms on U} {exact k-forms on U}, called the kthde Rham cohomology ofU. The generalization of Proposition C to any differential form on Rnis called the Poincar ´e lemma : for k≥1, every closed k-form on Rnis exact. This is of course 44§4 Differential Forms on Rn equivalent to the vanishing of the kth de Rham cohomology Hk(Rn)fork≥1. We will prove it in Section 27. The theory of differential forms allows us to generalize vector calculus from R3 toRnand indeed to a manifold of any dimension. The general Stokes theorem for a manifold that we will prove in Subsection 23.5 subsumes and uniﬁes the fundamental theorem for line integrals, Green’s theorem in the plane, the classical Stokes theorem for a surface in R3, and the divergence theorem. As a ﬁrst step in this program, we begin the next chapter with the deﬁnition of a manifold. 4.7 Convention on Subscripts and Superscripts In differential geometry it is customary to index vector ﬁelds with subscripts e1,..., en, and differential forms with superscripts ω1,..., ωn. Being 0-forms, coordinate functions take superscripts: x1,..., xn. Their differentials, being 1-forms, should also have superscripts, and indeed they do: dx1,..., dxn. Coordinate vector ﬁelds ∂/∂x1,...,∂/∂xnare considered to have subscripts because the iin∂/∂xi, although a superscript for xi, is in the lower half of the fraction. Coefﬁcient functions can have superscripts or subscripts depending on whether they are the coefﬁcient functions of a vector ﬁeld or of a differential form. For a vector ﬁeld X=∑aiei, the coefﬁcient functions aihave superscripts; the idea is that the superscript in ai“cancels out” the subscript in ei. For the same reason, the coefﬁcient functions bjin a differential form ω=∑bjdxjhave subscripts. The beauty of this convention is that there is a “conservation of indices” on the two sides of an equality sign. For example, if X=∑ai∂/∂xi, then ai=(dxi)(X). Here both sides have a net superscript i. As another example, if ω=∑bjdxj, then ω(X)=/parenleftbig∑bjdxj/parenrightbig/parenleftbigg ∑ai∂ ∂xi/parenrightbigg =∑biai; after cancellation of superscripts and subscripts, both sides of the equality sign have zero net index. This convention is a useful mnemonic aid in some of the transforma- tion formulas of differential geometry. Problems 4.1. A 1-form on R3 Letωbe the 1-form zdx−dzand let Xbe the vector ﬁeld y∂/∂x+x∂/∂yonR3. Compute ω(X)anddω. 4.2. A 2-form on R3 At each point p∈R3, deﬁne a bilinear function ωponTp(R3)by 4.7 Convention on Subscripts and Superscripts 45 ωp(a,b)=ωp  a1 a2 a3 , b1 b2 b3  =p3det/bracketleftbigga1b1 a2b2/bracketrightbigg , for tangent vectors a,b∈Tp(R3), where p3is the third component of p=(p1,p2,p3). Since ωpis an alternating bilinear function on Tp(R3),ωis a 2-form on R3. Write ωin terms of the standard basis dxi∧dxjat each point. 4.3. Exterior calculus Suppose the standard coordinates on R2are called randθ(thisR2is the(r,θ)-plane, not the (x,y)-plane). If x=rcosθandy=rsinθ, calculate dx,dy, and dx∧dyin terms of dranddθ. 4.4. Exterior calculus Suppose the standard coordinates on R3are called ρ,φ, and θ. If x=ρsinφcosθ,y= ρsinφsinθ, and z=ρcosφ, calculate dx,dy,dz, and dx∧dy∧dzin terms of dρ,dφ, and dθ. 4.5. Wedge product Letαbe a 1-form and βa 2-form on R3. Then α=a1dx1+a2dx2+a3dx3, β=b1dx2∧dx3+b2dx3∧dx1+b3dx1∧dx2. Simplify the expression α∧βas much as possible. 4.6. Wedge product and cross product The correspondence between differential forms and vector ﬁelds on an open subset of R3in Subsection 4.6 also makes sense pointwise. Let Vbe a vector space of dimension 3 with basis e1,e2,e3, and dual basis α1,α2,α3. To a 1-covector α=a1α1+a2α2+a3α3onV, we associate the vector vα=/an}bracke⊔le{⊔a1,a2,a3/an}bracke⊔ri}h⊔∈R3. To the 2-covector γ=c1α2∧α3+c2α3∧α1+c3α1∧α2 onV, we associate the vector vγ=/an}bracke⊔le{⊔c1,c2,c3/an}bracke⊔ri}h⊔∈R3. Show that under this correspondence, the wedge product of 1-covectors corresponds to the cross product of vectors in R3: ifα= a1α1+a2α2+a3α3andβ=b1α1+b2α2+b3α3, then vα∧β=vα×vβ. 4.7. Commutator of derivations and antiderivations LetA=⊕∞ k=−∞Akbe a graded algebra over a ﬁeld Kwith Ak=0 for k<0. Let mbe an integer. A superderivation of A of degree m is aK-linear map D:A→Asuch that for all k, D(Ak)⊂Ak+mand for all a∈Akandb∈Aℓ, D(ab)=( Da)b+(−1)kma(Db). IfD1andD2are two superderivations of Aof respective degrees m1andm2, deﬁne their commutator to be [D1,D2]=D1◦D2−(−1)m1m2D2◦D1. Show that [D1,D2]is a superderivation of degree m1+m2. (A superderivation is said to be even orodddepending on the parity of its degree. An even superderivation is a derivation; an odd superderivation is an antiderivation.) Chapter 2 Manifolds Intuitively, a manifold is a generalization of curves and surfaces to higher dimen- sions. It is locally Euclidean in that every point has a neighborhood, called a chart, homeomorphic to an open subset of Rn. The coordinates on a chart allow one to carry out computations as though in a Euclidean space, so that many concepts fromR n, such as differentiability, point-derivations, tangent spaces, and differential forms, carry over to a manifold. Bernhard Riemann (1826–1866)Like most fundamental mathematical concepts, the idea of a manifold did not originate with a sin- gle person, but is rather the distillation of years ofcollective activity. In his masterpiece Disquisitiones generales circa superﬁcies curvas (“General Inves- tigations of Curved Surfaces”) published in 1827, Carl Friedrich Gauss freely used local coordinates on a surface, and so he already had the idea ofcharts. Moreover, he appeared to be the ﬁrst to con- sider a surface as an abstract space existing in its own right, independent of a particular embedding ina Euclidean space. Bernhard Riemann’s inaugural lecture ¨Uber die Hypothesen, welche der Geometrie zu Grunde liegen (“On the hypotheses that under- lie geometry”) in G¨ ottingen in 1854 laid the foun- dations of higher-dimensional differential geometry.Indeed, the word “manifold” is a direct translation of the German word “Mannigfaltigkeit,” which Riemann used to describe the objects of his inquiry. This was followed by the work of Henri Poincar´ e in the late nineteenthcentury on homology, in which locally Euclidean spaces ﬁgured prominently. The late nineteenth and early twentieth centuries were also a period of feverish develop- ment in point-set topology. It was not until 1931 that one ﬁnds the modern deﬁnition of a manifold based on point-set topology and a group of transition functions [37]. © Springer Science+Business Media, LLC 201147 L.W. Tu, An Introduction to Manifolds, Universitext, DOI 10.1007/978-1-4419-7400-6_2, 48§5 Manifolds In this chapter we give the basic deﬁnitions and properties of a smooth manifold and of smooth maps between manifolds. Initially, the only way we have to verifythat a space is a manifold is to exhibit a collection of C ∞compatible charts covering the space. In Section 7 we describe a set of sufﬁcient conditions under which a quotient topological space becomes a manifold, giving us a second way to construct manifolds. §5 Manifolds While there are many kinds of manifolds—for example, topological manifolds, Ck- manifolds, analytic manifolds, and complex manifolds—in this book we are con- cerned mainly with smooth manifolds. Starting with topological manifolds, which are Hausdorff, second countable, locally Euclidean spaces, we introduce the conceptof a maximal C ∞atlas, which makes a topological manifold into a smooth manifold. This is illustrated with a few simple examples. 5.1 Topological Manifolds We ﬁrst recall a few deﬁnitions from point-set topology. For more details, see Ap- pendix A. A topological space is second countable if it has a countable basis. A neighborhood of a point pin a topological space Mis any open set containing p. An open cover ofMis a collection{Uα}α∈Aof open sets in Mwhose union/uniontext α∈AUα isM. Deﬁnition 5.1. A topological space Mislocally Euclidean of dimension n if every point pinMhas a neighborhood Usuch that there is a homeomorphism φfrom U onto an open subset of Rn. We call the pair (U,φ:U→Rn)achart ,Uacoordinate neighborhood or a coordinate open set , and φacoordinate map or a coordinate system onU. We say that a chart (U,φ)iscentered atp∈Uifφ(p)=0. Deﬁnition 5.2. Atopological manifold is a Hausdorff, second countable, locally Euclidean space. It is said to be of dimension n if it is locally Euclidean of dimen- sion n. For the dimension of a topological manifold to be well deﬁned, we need to know that for n/ne}a⊔ionslash=man open subset of Rnis not homeomorphic to an open subset of Rm. This fact, called invariance of dimension , is indeed true, but is not easy to prove directly. We will not pursue this point, since we are mainly interested in smooth manifolds, for which the analogous result is easy to prove (Corollary 8.7). Of course, if a topological manifold has several connected components, it is possible for each component to have a different dimension. 5.2 Compatible Charts 49 Example. The Euclidean space Rnis covered by a single chart (Rn,1Rn), where 1Rn:Rn→Rnis the identity map. It is the prime example of a topological manifold. Every open subset of Rnis also a topological manifold, with chart (U,1U). Recall that the Hausdorff condition and second countability are “hereditary prop- erties”; that is, they are inherited by subspaces: a subspace of a Hausdorff space is Hausdorff (Proposition A.19) and a subspace of a second-countable space is secondcountable (Proposition A.14). So any subspace of R nis automatically Hausdorff and second countable. Example 5.3 ( A cusp ).The graph of y=x2/3inR2is a topological manifold (Fig- ure5.1(a)) . By virtue of being a subspace of R2, it is Hausdorff and second count- able. It is locally Euclidean, because it is homeomorphic to Rvia(x,x2/3)/ma√s⊔o→x. (a) Cusp (b) Crossp Fig. 5.1. Example 5.4 ( A cross ).Show that the cross in R2inFigure 5.1 with the subspace topology is not locally Euclidean at the intersection p, and so cannot be a topological manifold. Solution. Suppose the cross is locally Euclidean of dimension nat the point p. Then phas a neighborhood Uhomeomorphic to an open ball B:=B(0,ε)⊂Rnwith pmapping to 0. The homeomorphism U→Brestricts to a homeomorphism U− {p}→ B−{0}. Now B−{0}is either connected if n≥2 or has two connected components if n=1. Since U−{p}has four connected components, there can be no homeomorphism from U−{p}toB−{0}. This contradiction proves that the cross is not locally Euclidean at p. ⊓ ⊔ 5.2 Compatible Charts Suppose(U,φ:U→Rn)and(V,ψ:V→Rn)are two charts of a topological man- ifold. Since U∩Vis open in Uandφ:U→Rnis a homeomorphism onto an open subset of Rn, the image φ(U∩V)will also be an open subset of Rn. Similarly, ψ(U∩V)is an open subset of Rn. Deﬁnition 5.5. Two charts (U,φ:U→Rn),(V,ψ:V→Rn)of a topological manifold are C∞-compatible if the two maps 50§5 Manifolds φ◦ψ−1:ψ(U∩V)→φ(U∩V),ψ◦φ−1:φ(U∩V)→ψ(U∩V) areC∞(Figure 5.2) . These two maps are called the transition functions between the charts. If U∩Vis empty, then the two charts are automatically C∞-compatible. To simplify the notation, we will sometimes write UαβforUα∩UβandUαβ γfor Uα∩Uβ∩Uγ. φ ψ U V φ(U∩V) Fig. 5.2. The transition function ψ◦φ−1is deﬁned on φ(U∩V). Since we are interested only in C∞-compatible charts, we often omit mention of “C∞” and speak simply of compatible charts. Deﬁnition 5.6. AC∞atlas or simply an atlas on a locally Euclidean space Mis a collection U={(Uα,φα)}of pairwise C∞-compatible charts that cover M, i.e., such thatM=/uniontext αUα. | | | | −π 0 π 2π φ1(A) φ1(B) φ2(B) φ2(A)φ1(U1) φ2(U2)| |U1U2 )()( AB Fig. 5.3. AC∞atlas on a circle. Example 5.7 ( A C∞atlas on a circle ).The unit circle S1in the complex plane Cmay be described as the set of points {eit∈C|0≤t≤2π}. Let U1andU2be the two open subsets of S1(seeFigure 5.3 ) U1={eit∈C|−π<t<π}, U2={eit∈C|0<t<2π}, 5.2 Compatible Charts 51 and deﬁne φα:Uα→Rforα=1,2 by φ1(eit)=t,−π<t<π, φ2(eit)=t,0<t<2π. Both φ1andφ2are branches of the complex log function (1/i)logzand are home- omorphisms onto their respective images. Thus, (U1,φ1)and(U2,φ2)are charts on S1. The intersection U1∩U2consists of two connected components, A={eit|−π<t<0}, B={eit|0<t<π}, with φ1(U1∩U2)=φ1(A∐B)=φ1(A)∐φ1(B)=]−π,0[∐]0,π[, φ2(U1∩U2)=φ2(A∐B)=φ2(A)∐φ2(B)=] π,2π[∐]0,π[. Here we use the notation A∐Bto indicate a union in which the two subsets Aand Bare disjoint. The transition function φ2◦φ−1 1:φ1(A∐B)→φ2(A∐B)is given by (φ2◦φ−1 1)(t)=/braceleftigg t+2πfort∈]−π,0[, t fort∈]0,π[. Similarly, (φ1◦φ−1 2)(t)=/braceleftigg t−2πfort∈]π,2π[, t fort∈]0,π[. Therefore, (U1,φ1)and(U2,φ2)areC∞-compatible charts and form a C∞atlas on S1. Although the C∞compatibility of charts is clearly reﬂexive and symmetric, it is not transitive. The reason is as follows. Suppose (U1,φ1)isC∞-compatible with (U2,φ2), and(U2,φ2)isC∞-compatible with (U3,φ3). Note that the three coordinate functions are simultaneously deﬁned only on the triple intersection U123. Thus, the composite φ3◦φ−1 1=(φ3◦φ−1 2)◦(φ2◦φ−1 1) isC∞, but only on φ1(U123), not necessarily on φ1(U13)(Figure 5.4) . A priori we know nothing about φ3◦φ−1 1onφ1(U13−U123)and so we cannot conclude that (U1,φ1)and(U3,φ3)areC∞-compatible. We say that a chart (V,ψ)iscompatible with an atlas {(Uα,φα)}if it is compat- ible with all the charts (Uα,φα)of the atlas. Lemma 5.8. Let{(Uα,φα)}be an atlas on a locally Euclidean space. If two charts (V,ψ)and(W,σ)are both compatible with the atlas {(Uα,φα)}, then they are compatible with each other. 52§5 Manifolds φ1(U123)φ1 φ2 φ3U1 U2 U3 Fig. 5.4. The transition function φ3◦φ−1 1isC∞onφ1(U123). /Bullet /Bullet /Bullet/Bullet p ψ(p) φα(p) σ(p) φα◦ψ−1σ◦φ−1αV W Uαψ σ φα Fig. 5.5. Two charts (V,ψ),(W,σ)compatible with an atlas. Proof. (See Figure 5.5. ) Let p∈V∩W. We need to show that σ◦ψ−1isC∞at ψ(p). Since{(Uα,φα)}is an atlas for M,p∈Uαfor some α. Then pis in the triple intersection V∩W∩Uα. By the remark above, σ◦ψ−1=(σ◦φ−1 α)◦(φα◦ψ−1)isC∞onψ(V∩W∩Uα), hence at ψ(p). Since pwas an arbitrary point of V∩W, this proves that σ◦ψ−1is C∞onψ(V∩W). Similarly, ψ◦σ−1isC∞onσ(V∩W). ⊓ ⊔ Note that in an equality such as σ◦ψ−1=(σ◦φ−1 α)◦(φα◦ψ−1)in the proof above, the maps on the two sides of the equality sign have different domains. What the equality means is that the two maps are equal on their common domain. 5.3 Smooth Manifolds An atlasMon a locally Euclidean space is said to be maximal if it is not contained in a larger atlas; in other words, if Uis any other atlas containing M, thenU=M. 5.4 Examples of Smooth Manifolds 53 Deﬁnition 5.9. Asmooth orC∞manifold is a topological manifold Mtogether with a maximal atlas. The maximal atlas is also called a differentiable structure onM. A manifold is said to have dimension nif all of its connected components have dimension n. A 1-dimensional manifold is also called a curve , a 2-dimensional manifold a surface , and an n-dimensional manifold an n-manifold . In Corollary 8.7 we will prove that if an open set U⊂Rnis diffeomorphic to an open set V⊂Rm, then n=m. As a consequence, the dimension of a manifold at a point is well deﬁned. In practice, to check that a topological manifold Mis a smooth manifold, it is not necessary to exhibit a maximal atlas. The existence of anyatlas on Mwill do, because of the following proposition. Proposition 5.10. Any atlas U={(Uα,φα)}on a locally Euclidean space is con- tained in a unique maximal atlas. Proof. Adjoin to the atlas Uall charts (Vi,ψi)that are compatible with U. By Propo- sition 5.8 the charts (Vi,ψi)are compatible with one another. So the enlarged collec- tion of charts is an atlas. Any chart compatible with the new atlas must be compatible with the original atlas Uand so by construction belongs to the new atlas. This proves that the new atlas is maximal. LetMbe the maximal atlas containing Uthat we have just constructed. If M′is another maximal atlas containing U, then all the charts in M′are compatible with U and so by construction must belong to M. This proves that M′⊂M. Since both are maximal, M′=M. Therefore, the maximal atlas containing Uis unique. ⊓ ⊔ In summary, to show that a topological space Mis aC∞manifold, it sufﬁces to check that (i)Mis Hausdorff and second countable, (ii)Mhas a C∞atlas (not necessarily maximal). From now on, a “manifold” will mean a C∞manifold. We use the terms “smooth” and “ C∞” interchangeably. In the context of manifolds, we denote the standard coor- dinates on Rnbyr1,..., rn. If(U,φ:U→Rn)is a chart of a manifold, we let xi= ri◦φbe the ith component of φand write φ=(x1,..., xn)and(U,φ)=(U,x1,..., xn). Thus, for p∈U,(x1(p),..., xn(p))is a point in Rn. The functions x1,..., xnare called coordinates orlocal coordinates onU. By abuse of notation, we sometimes omit the p. So the notation (x1,..., xn)stands alternately for local coordinates on the open set Uand for a point in Rn. By a chart(U,φ)about p in a manifold M, we will mean a chart in the differentiable structure of Msuch that p∈U. 5.4 Examples of Smooth Manifolds Example 5.11 ( Euclidean space) .The Euclidean space Rnis a smooth manifold with a single chart (Rn,r1,...,rn), where r1,...,rnare the standard coordinates on Rn. 54§5 Manifolds Example 5.12 ( Open subset of a manifold ).Any open subset Vof a manifold Mis also a manifold. If {(Uα,φα)}is an atlas for M, then{(Uα∩V,φα|Uα∩V}is an atlas for V, where φα|Uα∩V:Uα∩V→Rndenotes the restriction of φαto the subset Uα∩V. Example 5.13 ( Manifolds of dimension zero ).In a manifold of dimension zero, every singleton subset is homeomorphic to R0and so is open. Thus, a zero-dimensional manifold is a discrete set. By second countability, this discrete set must be countable. Example 5.14 ( Graph of a smooth function ).For a subset of A⊂Rnand a function f:A→Rm, the graph offis deﬁned to be the subset (Figure 5.6 ) Γ(f)={(x,f(x))∈A×Rm}. IfUis an open subset of Rnandf:U→RnisC∞, then the two maps /Bullet/Circle/Bullet/Circle /Bullet/Bullet x(x,f(x)) RnRm ( )Γ(f) U Fig. 5.6. The graph of a smooth function f:Rn⊃U→Rm. φ:Γ(f)→U,(x,f(x))/ma√s⊔o→x, and (1,f):U→Γ(f), x/ma√s⊔o→(x,f(x)), are continuous and inverse to each other, and so are homeomorphisms. The graph Γ(f)of aC∞function f:U→Rmhas an atlas with a single chart (Γ(f),φ), and is therefore a C∞manifold. This shows that many of the familiar surfaces of calculus, for example an elliptic paraboloid or a hyperbolic paraboloid, are manifolds. Example 5.15 ( General linear groups ).For any two positive integers mandnlet Rm×nbe the vector space of all m×nmatrices. Since Rm×nis isomorphic to Rmn, we give it the topology of Rmn. The general linear group GL(n,R)is by deﬁnition GL(n,R):={A∈Rn×n|detA/ne}a⊔ionslash=0}=det−1(R−{0}). Since the determinant function 5.4 Examples of Smooth Manifolds 55 det:Rn×n→R is continuous, GL (n,R)is an open subset of Rn×n≃Rn2and is therefore a manifold. Thecomplex general linear group GL(n,C)is deﬁned to be the group of non- singular n×ncomplex matrices. Since an n×nmatrix Ais nonsingular if and only if det A/ne}a⊔ionslash=0, GL(n,C)is an open subset of Cn×n≃R2n2, the vector space of n×n complex matrices. By the same reasoning as in the real case, GL (n,C)is a manifold of dimension 2 n2. φ1 φ2φ4 φ3U1 U2U3 U4 Fig. 5.7. Charts on the unit circle. Example 5.16 ( Unit circle in the (x,y)-plane ).In Example 5.7 we found a C∞atlas with two charts on the unit circle S1in the complex plane C. It follows that S1is a manifold. We now view S1as the unit circle in the real plane R2with deﬁning equation x2+y2=1, and describe a C∞atlas with four charts on it. We can cover S1with four open sets: the upper and lower semicircles U1,U2, and the right and left semicircles U3,U4(Figure 5.7 ). On U1andU2, the coordinate function xis a homeomorphism onto the open interval ]−1,1[on the x-axis. Thus, φi(x,y) =xfori=1,2. Similarly, on U3andU4,yis a homeomorphism onto the open interval ]−1,1[on the y-axis, and so φi(x,y)=yfori=3,4. It is easy to check that on every nonempty pairwise intersection Uα∩Uβ,φβ◦φ−1 α isC∞. For example, on U1∩U3, (φ3◦φ−1 1)(x)=φ3/parenleftig x,/radicalbig 1−x2/parenrightig =/radicalbig 1−x2, which is C∞. On U2∩U4, (φ4◦φ−1 2)(x)=φ4/parenleftig x,−/radicalbig 1−x2/parenrightig =−/radicalbig 1−x2, which is also C∞. Thus,{(Ui,φi)}4 i=1is aC∞atlas on S1. Example 5.17 ( Product manifold ).IfMandNareC∞manifolds, then M×Nwith its product topology is Hausdorff and second countable (Corollary A.21 and Propo- sition A.22). To show that M×Nis a manifold, it remains to exhibit an atlas on it. Recall that the product of two set maps f:X→X′andg:Y→Y′is f×g:X×Y→X′×Y′,(f×g)(x,y)=( f(x),g(y)). 56§5 Manifolds Proposition 5.18 (An atlas for a product manifold). If{(Uα,φα)}and{(Vi,ψi)} are C∞atlases for the manifolds M and N of dimensions m and n, respectively, then the collection {(Uα×Vi,φα×ψi:Uα×Vi→Rm×Rn)} of charts is a C∞atlas on M×N. Therefore, M×N is a C∞manifold of dimension m+n. Proof. Problem 5.5. ⊓ ⊔ Example. It follows from Proposition 5.18 that the inﬁnite cylinder S1×Rand the torus S1×S1are manifolds (Figure 5.8 ). Inﬁnite cylinder. Torus. Fig. 5.8. Since M×N×P=(M×N)×Pis the successive product of pairs of spaces, if M,N, and Pare manifolds, then so is M×N×P. Thus, the n-dimensional torus S1×···× S1(ntimes) is a manifold. Remark. LetSnbe the unit sphere (x1)2+(x2)2+···+(xn+1)2=1 inRn+1. Using Problem 5.3 as a guide, it is easy to write down a C∞atlas on Sn, showing that Snhas a differentiable structure. The manifold Snwith this differen- tiable structure is called the standard n-sphere . One of the most surprising achievements in topology was John Milnor’s dis- covery [27] in 1956 of exotic 7-spheres, smooth manifolds homeomorphic but not diffeomorphic to the standard 7-sphere. In 1963, Michel Kervaire and John Milnor [24] determined that there are exactly 28 nondiffeomorphic differentiable structuresonS 7. It is known that in dimensions <4 every topological manifold has a unique dif- ferentiable structure and in dimensions >4 every compact topological manifold has a ﬁnite number of differentiable structures. Dimension 4 is a mystery. It is not known 5.4 Examples of Smooth Manifolds 57 whether S4has a ﬁnite or inﬁnite number of differentiable structures. The statement thatS4has a unique differentiable structure is called the smooth Poincar ´e conjecture . As of this writing in 2010, the conjecture is still open. There are topological manifolds with no differentiable structure. Michel Kervaire was the ﬁrst to construct an example [23]. Problems 5.1. The real line with two origins LetAandBbe two points not on the real line R. Consider the set S=(R−{0})∪{A,B}(see Figure 5.9 ). /Bullet/BulletA B Fig. 5.9. Real line with two origins. For any two positive real numbers c,d, deﬁne IA(−c,d)= ]−c,0[∪{A}∪]0,d[ and similarly for IB(−c,d), with Binstead of A. Deﬁne a topology on Sas follows: On (R−{0}), use the subspace topology inherited from R, with open intervals as a basis. A basis of neighborhoods at Ais the set{IA(−c,d)|c,d>0}; similarly, a basis of neighborhoods at Bis{IB(−c,d)|c,d>0}. (a) Prove that the map h:IA(−c,d)→]−c,d[deﬁned by h(x)=xforx∈]−c,0[∪]0,d[, h(A)=0 is a homeomorphism. (b) Show that Sis locally Euclidean and second countable, but not Hausdorff. 5.2. A sphere with a hair A fundamental theorem of topology, the theorem on invariance of dimension, states that if twononempty open sets U⊂R nandV⊂Rmare homeomorphic, then n=m(for a proof, see [18, p. 126]). Use the idea of Example 5.4 as well as the theorem on invariance of dimension to prove that the sphere with a hair in R3(Figure 5.10) is not locally Euclidean at q. Hence it cannot be a topological manifold. 58§5 Manifolds /Bulletq Fig. 5.10. A sphere with a hair. 5.3. Charts on a sphere LetS2be the unit sphere x2+y2+z2=1 inR3. Deﬁne in S2the six charts corresponding to the six hemispheres—the front, rear, right, left, upper, and lower hemispheres (Figure 5.11) : U1={(x,y,z)∈S2|x>0}, φ1(x,y,z)=(y,z), U2={(x,y,z)∈S2|x<0}, φ2(x,y,z)=(y,z), U3={(x,y,z)∈S2|y>0}, φ3(x,y,z)=(x,z), U4={(x,y,z)∈S2|y<0}, φ4(x,y,z)=(x,z), U5={(x,y,z)∈S2|z>0}, φ5(x,y,z)=(x,y), U6={(x,y,z)∈S2|z<0}, φ6(x,y,z)=(x,y). Describe the domain φ4(U14)ofφ1◦φ−1 4and show that φ1◦φ−1 4isC∞onφ4(U14). Do the same for φ6◦φ−1 1. U6U5 U4 U3 U1U2 Fig. 5.11. Charts on the unit sphere. 5.4.* Existence of a coordinate neighborhood Let{(Uα,φα)}be the maximal atlas on a manifold M. For any open set UinMand a point p∈U, prove the existence of a coordinate open set Uαsuch that p∈Uα⊂U. 5.5. An atlas for a product manifold Prove Proposition 5.18. 6.1 Smooth Functions on a Manifold 59 §6 Smooth Maps on a Manifold Now that we have deﬁned smooth manifolds, it is time to consider maps between them. Using coordinate charts, one can transfer the notion of smooth maps fromEuclidean spaces to manifolds. By the C ∞compatibility of charts in an atlas, the smoothness of a map turns out to be independent of the choice of charts and is there- fore well deﬁned. We give various criteria for the smoothness of a map as well asexamples of smooth maps. Next we transfer the notion of partial derivatives from Euclidean space to a co- ordinate chart on a manifold. Partial derivatives relative to coordinate charts allow us to generalize the inverse function theorem to manifolds. Using the inverse func-tion theorem, we formulate a criterion for a set of smooth functions to serve as local coordinates near a point. 6.1 Smooth Functions on a Manifold /Bullet φ(p) φ(U)⊂Rn/BulletpUM Rf φ Fig. 6.1. Checking that a function fisC∞atpby pulling back to Rn. Deﬁnition 6.1. LetMbe a smooth manifold of dimension n. A function f:M→R is said to be C∞orsmooth at a point p inMif there is a chart (U,φ)about pinM such that f◦φ−1, a function deﬁned on the open subset φ(U)ofRn, isC∞atφ(p) (seeFigure 6.1 ). The function fis said to be C∞on M if it is C∞at every point of M. Remark 6.2.The deﬁnition of the smoothness of a function fat a point is indepen- dent of the chart (U,φ), for if f◦φ−1isC∞atφ(p)and(V,ψ)is any other chart about pinM, then on ψ(U∩V), f◦ψ−1=(f◦φ−1)◦(φ◦ψ−1), which is C∞atψ(p)(seeFigure 6.2 ). 60§6 Smooth Maps on a Manifold /Bullet p /Bullet ψ(p)/Bullet φ(p)Rf φ◦ψ−1ψ−1φ−1U V Fig. 6.2. Checking that a function fisC∞atpvia two charts. In Deﬁnition 6.1, f:M→Ris not assumed to be continuous. However, if fis C∞atp∈M, then f◦φ−1:φ(U)→R, being a C∞function at the point φ(p)in an open subset of Rn, is continuous at φ(p). As a composite of continuous functions, f=(f◦φ−1)◦φis continuous at p. Since we are interested only in functions that are smooth on an open set, there is no loss of generality in assuming at the outset that fis continuous. Proposition 6.3 (Smoothness of a real-valued function). Let M be a manifold of dimension n, and f :M→Ra real-valued function on M. The following are equiv- alent: (i)The function f :M→Ris C∞. (ii)The manifold M has an atlas such that for every chart (U,φ)in the atlas, f◦φ−1:Rn⊃φ(U)→Ris C∞. (iii)For every chart (V,ψ)on M, the function f ◦ψ−1:Rn⊃ψ(V)→Ris C∞. Proof. We will prove the proposition as a cyclic chain of implications. (ii)⇒(i): This follows directly from the deﬁnition of a C∞function, since by (ii) every point p∈Mhas a coordinate neighborhood (U,φ)such that f◦φ−1isC∞at φ(p). (i)⇒(iii): Let (V,ψ)be an arbitrary chart on Mand let p∈V. By Remark 6.2, f◦ψ−1isC∞atψ(p). Since pwas an arbitrary point of V,f◦ψ−1isC∞onψ(V). (iii)⇒(ii): Obvious. ⊓ ⊔ The smoothness conditions of Proposition 6.3 will be a recurrent motif through- out the book: to prove the smoothness of an object, it is sufﬁcient that a smoothnesscriterion hold on the charts of some atlas. Once the object is shown to be smooth, it then follows that the same smoothness criterion holds on every chart on the manifold. Deﬁnition 6.4. LetF:N→Mbe a map and ha function on M. The pullback ofh byF, denoted by F ∗h, is the composite function h◦F. In this terminology, a function fonMisC∞on a chart (U,φ)if and only if its pullback(φ−1)∗fbyφ−1isC∞on the subset φ(U)of Euclidean space. 6.2 Smooth Maps Between Manifolds 61 6.2 Smooth Maps Between Manifolds We emphasize again that unless otherwise speciﬁed, by a manifold we always mean aC∞manifold. We use the terms “ C∞” and “smooth” interchangeably. An atlas or a chart on a smooth manifold means an atlas or a chart contained in the differentiable structure of the smooth manifold. We generally denote a manifold by Mand its dimension by n. However, when speaking of two manifolds simultaneously, as in a map f:N→M, we will let the dimension of Nbenand that of Mbem. Deﬁnition 6.5. LetNandMbe manifolds of dimension nandm, respectively. A continuous map F:N→MisC∞at a point p inNif there are charts (V,ψ)about F(p)inMand(U,φ)about pinNsuch that the composition ψ◦F◦φ−1, a map from the open subset φ(F−1(V)∩U)ofRntoRm, isC∞atφ(p)(see Figure 6.3 ). The continuous map F:N→Mis said to be C∞if it is C∞at every point of N. φ(p)V F(p)/Bullet /Bullet/Bullet /BulletU pF φ−1ψ N M Fig. 6.3. Checking that a map F:N→MisC∞atp. In Deﬁnition 6.5, we assume F:N→Mcontinuous to ensure that F−1(V)is an open set in N. Thus, C∞maps between manifolds are by deﬁnition continuous. Remark 6.6 ( Smooth maps into Rm).In case M=Rm, we can take (Rm,1Rm)as a chart about F(p)inRm. According to Deﬁnition 6.5, F:N→RmisC∞atp∈Nif and only if there is a chart (U,φ)about pinNsuch that F◦φ−1:φ(U)→RmisC∞ atφ(p). Letting m=1, we recover the deﬁnition of a function being C∞at a point. We show now that the deﬁnition of the smoothness of a map F:N→Mat a point is independent of the choice of charts. This is analogous to how the smoothness of afunction N→Ratp∈Nis independent of the choice of a chart on Nabout p. Proposition 6.7. Suppose F :N→M is C ∞at p∈N. If(U,φ)is any chart about p in N and (V,ψ)is any chart about F (p)in M, then ψ◦F◦φ−1is C∞atφ(p). Proof. Since FisC∞atp∈N, there are charts (Uα,φα)about pinNand(Vβ,ψβ) about F(p)inMsuch that ψβ◦F◦φ−1 αisC∞atφα(p). By the C∞compatibility 62§6 Smooth Maps on a Manifold of charts in a differentiable structure, both φα◦φ−1andψ◦ψ−1 βareC∞on open subsets of Euclidean spaces. Hence, the composite ψ◦F◦φ−1=(ψ◦ψ−1 β)◦(ψβ◦F◦φ−1 α)◦(φα◦φ−1) isC∞atφ(p). ⊓ ⊔ The next proposition gives a way to check smoothness of a map without specify- ing a point in the domain. Proposition 6.8 (Smoothness of a map in terms of charts). Let N and M be smooth manifolds, and F :N→M a continuous map. The following are equivalent: (i)The map F :N→M is C∞. (ii)There are atlases Ufor N and Vfor M such that for every chart (U,φ)inUand (V,ψ)inV, the map ψ◦F◦φ−1:φ(U∩F−1(V))→Rm is C∞. (iii)For every chart (U,φ)on N and (V,ψ)on M, the map ψ◦F◦φ−1:φ(U∩F−1(V))→Rm is C∞. Proof. (ii)⇒(i): Let p∈N. Suppose (U,φ)is a chart about pinUand(V,ψ)is a chart about F(p)inV. By (ii), ψ◦F◦φ−1isC∞atφ(p). By the deﬁnition of a C∞ map, F:N→MisC∞atp. Since pwas an arbitrary point of N, the map F:N→M isC∞. (i)⇒(iii): Suppose (U,φ)and(V,ψ)are charts on NandMrespectively such that U∩F−1(V)/ne}a⊔ionslash=∅. Let p∈U∩F−1(V). Then(U,φ)is a chart about pand(V,ψ)is a chart about F(p). By Proposition 6.7, ψ◦F◦φ−1isC∞atφ(p). Since φ(p)was an arbitrary point of φ(U∩F−1(V)), the map ψ◦F◦φ−1:φ(U∩F−1(V))→Rm isC∞. (iii)⇒(ii): Clear. ⊓ ⊔ Proposition 6.9 (Composition of C∞maps). If F:N→M and G :M→P are C∞ maps of manifolds, then the composite G ◦F:N→P is C∞. Proof. Let(U,φ),(V,ψ), and(W,σ)be charts on N,M, and Prespectively. Then σ◦(G◦F)◦φ−1=(σ◦G◦ψ−1)◦(ψ◦F◦φ−1). Since FandGareC∞, by Proposition 6.8(i) ⇒(iii), σ◦G◦ψ−1andψ◦F◦φ−1are C∞. As a composite of C∞maps of open subsets of Euclidean spaces, σ◦(G◦F)◦ φ−1isC∞. By Proposition 6.8(iii) ⇒(i), G◦FisC∞. ⊓ ⊔ 6.4 Smoothness in Terms of Components 63 6.3 Diffeomorphisms Adiffeomorphism of manifolds is a bijective C∞map F:N→Mwhose inverse F−1 is also C∞. According to the next two propositions, coordinate maps are diffeomor- phisms, and conversely, every diffeomorphism of an open subset of a manifold with an open subset of a Euclidean space can serve as a coordinate map. Proposition 6.10. If(U,φ)is a chart on a manifold M of dimension n, then the coordinate map φ:U→φ(U)⊂Rnis a diffeomorphism. Proof. By deﬁnition, φis a homeomorphism, so it sufﬁces to check that both φ andφ−1are smooth. To test the smoothness of φ:U→φ(U), we use the atlas {(U,φ)}with a single chart on Uand the atlas{(φ(U),1φ(U))}with a single chart onφ(U). Since1φ(U)◦φ◦φ−1:φ(U)→φ(U)is the identity map, it is C∞. By Proposition 6.8(ii)⇒(i), φisC∞. To test the smoothness of φ−1:φ(U)→U, we use the same atlases as above. Since φ◦φ−1◦1φ(U)=1φ(U):φ(U)→φ(U), the map φ−1is also C∞.⊓ ⊔ Proposition 6.11. Let U be an open subset of a manifold M of dimension n. If F:U→F(U)⊂Rnis a diffeomorphism onto an open subset of Rn, then(U,F) is a chart in the differentiable structure of M. Proof. For any chart (Uα,φα)in the maximal atlas of M, both φαandφ−1 αareC∞ by Proposition 6.10. As composites of C∞maps, both F◦φ−1 αandφα◦F−1areC∞. Hence,(U,F)is compatible with the maximal atlas. By the maximality of the atlas, the chart(U,F)is in the atlas. ⊓ ⊔ 6.4 Smoothness in Terms of Components In this subsection we derive a criterion that reduces the smoothness of a map to the smoothness of real-valued functions on open sets. Proposition 6.12 (Smoothness of a vector-valued function). Let N be a manifold and F :N→Rma continuous map. The following are equivalent: (i)The map F :N→Rmis C∞. (ii)The manifold N has an atlas such that for every chart (U,φ)in the atlas, the map F◦φ−1:φ(U)→Rmis C∞. (iii)For every chart (U,φ)on N, the map F ◦φ−1:φ(U)→Rmis C∞. Proof. (ii)⇒(i): In Proposition 6.8(ii), take Vto be the atlas with the single chart (Rm,1Rm)onM=Rm. (i)⇒(iii): In Proposition 6.8(iii), let (V,ψ)be the chart (Rm,1Rm)onM=Rm. (iii)⇒(ii): Obvious. ⊓ ⊔ Proposition 6.13 (Smoothness in terms of components). Let N be a manifold. A vector-valued function F :N→Rmis C∞if and only if its component functions F1,..., Fm:N→Rare all C∞. 64§6 Smooth Maps on a Manifold Proof. The map F:N→RmisC∞ ⇐⇒ for every chart (U,φ)onN, the map F◦φ−1:φ(U)→RmisC∞(by Proposi- tion 6.12) ⇐⇒ for every chart (U,φ)onN, the functions Fi◦φ−1:φ(U)→Rare all C∞ (deﬁnition of smoothness for maps of Euclidean spaces) ⇐⇒ the functions Fi:N→Rare all C∞(by Proposition 6.3). ⊓ ⊔ Exercise 6.14 (Smoothness of a map to a circle).* Prove that the map F:R→S1,F(t) = (cost,sint)isC∞. Proposition 6.15 (Smoothness of a map in terms of vector-valued functions). Let F:N→M be a continuous map between two manifolds of dimensions n and m respectively. The following are equivalent: (i)The map F :N→M is C∞. (ii)The manifold M has an atlas such that for every chart (V,ψ)=( V,y1,..., ym)in the atlas, the vector-valued function ψ◦F:F−1(V)→Rmis C∞. (iii)For every chart (V,ψ)=(V,y1,..., ym)on M, the vector-valued function ψ◦F: F−1(V)→Rmis C∞. Proof. (ii)⇒(i): Let Vbe the atlas for Min (ii), and let U={(U,φ)}be an arbitrary atlas for N. For each chart (V,ψ)in the atlas V, the collection {(U∩F−1(V),φ|U∩F−1(V))}is an atlas for F−1(V). Since ψ◦F:F−1(V)→Rm isC∞, by Proposition 6.12(i) ⇒(iii), ψ◦F◦φ−1:φ(U∩F−1(V))→Rm isC∞. It then follows from Proposition 6.8(ii) ⇒(i) that F:N→MisC∞. (i)⇒(iii): Being a coordinate map, ψisC∞(Proposition 6.10). As the composite of twoC∞maps, ψ◦FisC∞. (iii)⇒(ii): Obvious. ⊓ ⊔ By Proposition 6.13, this smoothness criterion for a map translates into a smooth- ness criterion in terms of the components of the map. Proposition 6.16 (Smoothness of a map in terms of components). Let F :N→M be a continuous map between two manifolds of dimensions n and m respectively. The following are equivalent: (i)The map F :N→M is C∞. (ii)The manifold M has an atlas such that for every chart (V,ψ)=( V,y1,..., ym)in the atlas, the components yi◦F:F−1(V)→Rof F relative to the chart are all C∞. (iii)For every chart (V,ψ)=(V,y1,..., ym)on M, the components yi◦F:F−1(V)→ Rof F relative to the chart are all C∞. 6.5 Examples of Smooth Maps 65 6.5 Examples of Smooth Maps We have seen that coordinate maps are smooth. In this subsection we look at a few more examples of smooth maps. Example 6.17 ( Smoothness of a projection map ).LetMandNbe manifolds and π:M×N→M,π(p,q)=pthe projection to the ﬁrst factor. Prove that πis aC∞ map. Solution. Let(p,q)be an arbitrary point of M×N. Suppose (U,φ)=( U,x1,..., xm) and(V,ψ) = ( V,y1,..., yn)are coordinate neighborhoods of pandqinMandN respectively. By Proposition 5.18, (U×V,φ×ψ)=( U×V,x1,..., xm,y1,..., yn)is a coordinate neighborhood of (p,q). Then /parenleftbig φ◦π◦(φ×ψ)−1/parenrightbig (a1,..., am,b1,..., bn)=( a1,..., am), which is a C∞map from (φ×ψ)(U×V)inRm+ntoφ(U)inRm, soπisC∞at(p,q). Since(p,q)was an arbitrary point in M×N,πisC∞onM×N. Exercise 6.18 (Smoothness of a map to a Cartesian product).* LetM1,M2, and Nbe manifolds of dimensions m1,m2, and nrespectively. Prove that a map (f1,f2):N→M1×M2 isC∞if and only if fi:N→Mi,i=1,2, are both C∞. Example 6.19.In Examples 5.7 and 5.16 we showed that the unit circle S1deﬁned by x2+y2=1 inR2is aC∞manifold. Prove that a C∞function f(x,y)onR2restricts to aC∞function on S1. Solution. To avoid confusing functions with points, we will denote a point on S1 asp= (a,b)and use x,yto mean the standard coordinate functions on R2. Thus, x(a,b)=aandy(a,b)=b. Suppose we can show that xandyrestrict to C∞functions onS1. By Exercise 6.18, the inclusion map i:S1→R2,i(p)=( x(p),y(p))is then C∞onS1. As the composition of C∞maps, f|S1=f◦iwill be C∞onS1(Proposition 6.9). Consider ﬁrst the function x. We use the atlas (Ui,φi)from Example 5.16. Since xis a coordinate function on U1and on U2, by Proposition 6.10 it is C∞onU1∪U2= S1−{(±1,0)}. To show that xisC∞onU3, it sufﬁces to check the smoothness of x◦φ−1 3:φ3(U3)→R: /parenleftbig x◦φ−1 3/parenrightbig (b)=x/parenleftig/radicalbig 1−b2,b/parenrightig =/radicalbig 1−b2. OnU3, we have b/ne}a⊔ionslash=±1, so that√ 1−b2is aC∞function of b. Hence, xisC∞onU3. OnU4,/parenleftbig x◦φ−1 4/parenrightbig (b)=x/parenleftig −/radicalbig 1−b2,b/parenrightig =−/radicalbig 1−b2, which is C∞because bis not equal to±1. Since xisC∞on the four open sets U1,U2, U3, and U4, which cover S1,xisC∞onS1. The proof that yisC∞onS1is similar. 66§6 Smooth Maps on a Manifold Armed with the deﬁnition of a smooth map between manifolds, we can deﬁne a Lie group. Deﬁnition 6.20. ALie group1is aC∞manifold Ghaving a group structure such that the multiplication map µ:G×G→G and the inverse map ι:G→G,ι(x)=x−1, are both C∞. Similarly, a topological group is a topological space having a group structure such that the multiplication and inverse maps are both continuous. Note that a topo-logical group is required to be a topological space, but not a topological manifold. Examples. (i) The Euclidean space R nis a Lie group under addition. (ii) The set C×of nonzero complex numbers is a Lie group under multiplication. (iii) The unit circle S1inC×is a Lie group under multiplication. (iv) The Cartesian product G1×G2of two Lie groups (G1,µ1)and(G2,µ2)is a Lie group under coordinatewise multiplication µ1×µ2. Example 6.21 ( General linear group ).In Example 5.15 we deﬁned the general linear group GL(n,R)={A=[a i j]∈Rn×n|detA/ne}a⊔ionslash=0}. As an open subset of Rn×n, it is a manifold. Since the (i,j)-entry of the product of two matrices AandBin GL(n,R), (AB)i j=n ∑ k=1aikbk j, is a polynomial in the coordinates of AandB, matrix multiplication µ: GL(n,R)×GL(n,R)→GL(n,R) is aC∞map. Recall that the (i,j)-minor of a matrix Ais the determinant of the submatrix of Aobtained by deleting the ith row and the jth column of A. By Cramer’s rule from linear algebra, the (i,j)-entry of A−1is (A−1)i j=1 detA·(−1)i+j((j,i)-minor of A), which is a C∞function of the ai j’s provided det A/ne}a⊔ionslash=0. Therefore, the inverse map ι: GL(n,R)→GL(n,R)is also C∞. This proves that GL (n,R)is a Lie group. 1Lie groups and Lie algebras are named after the Norwegian mathematician Sophus Lie (1842–1899). In this context, “Lie” is pronounced “lee,” not “lye.” 6.6 Partial Derivatives 67 In Section 15 we will study less obvious examples of Lie groups. NOTATION . The notation for matrices presents a special challenge. An n×nmatrix Acan represent a linear transformation y=Ax, with x,y∈Rn. In this case, yi= ∑jai jxj, soA=[ai j]. An n×nmatrix can also represent a bilinear form /an}bracke⊔le{⊔x,y/an}bracke⊔ri}h⊔=xTAy with x,y∈Rn. In this case,/an}bracke⊔le{⊔x,y/an}bracke⊔ri}h⊔=∑i,jxiai jyj, soA=[a i j]. In the absence of any context, we will write a matrix as A=[a i j], using a lowercase letter ato denote an entry of a matrix Aand using a double subscript ( )i jto denote the (i,j)-entry. 6.6 Partial Derivatives On a manifold Mof dimension n, let(U,φ)be a chart and faC∞function As a function into Rn,φhasncomponents x1,..., xn. This means that if r1,..., rnare the standard coordinates on Rn, then xi=ri◦φ. For p∈U, we deﬁne the partial derivative ∂f/∂xiof f with respect to xiat pto be ∂ ∂xi/vextendsingle/vextendsingle/vextendsingle/vextendsingle pf:=∂f ∂xi(p):=∂/parenleftbig f◦φ−1/parenrightbig ∂ri(φ(p)):=∂ ∂ri/vextendsingle/vextendsingle/vextendsingle/vextendsingle φ(p)/parenleftbig f◦φ−1/parenrightbig . Since p=φ−1(φ(p)), this equation may be rewritten in the form ∂f ∂xi/parenleftbig φ−1(φ(p))/parenrightbig =∂/parenleftbig f◦φ−1/parenrightbig ∂ri(φ(p)). Thus, as functions on φ(U), ∂f ∂xi◦φ−1=∂/parenleftbig f◦φ−1/parenrightbig ∂ri. The partial derivative ∂f/∂xiisC∞onUbecause its pullback (∂f/∂xi)◦φ−1isC∞ onφ(U). In the next proposition we see that partial derivatives on a manifold satisfy the same duality property ∂ri/∂rj=δi jas the coordinate functions rionRn. Proposition 6.22. Suppose(U,x1,..., xn)is a chart on a manifold. Then ∂xi/∂xj=δi j. Proof. At a point p∈U, by the deﬁnition of ∂/∂xj|p, ∂xi ∂xj(p)=∂/parenleftbig xi◦φ−1/parenrightbig ∂rj(φ(p))=∂/parenleftbig ri◦φ◦φ−1/parenrightbig ∂rj(φ(p))=∂ri ∂rj(φ(p))= δi j.⊓ ⊔ Deﬁnition 6.23. LetF:N→Mbe a smooth map, and let (U,φ) =( U,x1,..., xn) and(V,ψ)=( V,y1,..., ym)be charts on NandMrespectively such that F(U)⊂V. Denote by Fi:=yi◦F=ri◦ψ◦F:U→R 68§6 Smooth Maps on a Manifold theith component of Fin the chart (V,ψ). Then the matrix [∂Fi/∂xj]is called theJacobian matrix ofFrelative to the charts (U,φ)and(V,ψ). In case Nand Mhave the same dimension, the determinant det [∂Fi/∂xj]is called the Jacobian determinant ofFrelative to the two charts. The Jacobian determinant is also written as∂(F1,..., Fn)/∂(x1,..., xn). When MandNare open subsets of Euclidean spaces and the charts are (U,r1, ...,rn)and(V,r1,..., rm), the Jacobian matrix [∂Fi/∂rj], where Fi=ri◦F, is the usual Jacobian matrix from calculus. Example 6.24 ( Jacobian matrix of a transition map ).Let(U,φ)=( U,x1,..., xn)and (V,ψ) =( V,y1,..., yn)be overlapping charts on a manifold M. The transition map ψ◦φ−1:φ(U∩V)→ψ(U∩V)is a diffeomorphism of open subsets of Rn. Show that its Jacobian matrix J(ψ◦φ−1)atφ(p)is the matrix [∂yi/∂xj]of partial deriva- tives at p. Solution. By deﬁnition, J(ψ◦φ−1)=[ ∂(ψ◦φ−1)i/∂rj], where ∂/parenleftbig ψ◦φ−1/parenrightbigi ∂rj(φ(p))=∂/parenleftbig ri◦ψ◦φ−1/parenrightbig ∂rj(φ(p))=∂/parenleftbig yi◦φ−1/parenrightbig ∂rj(φ(p))=∂yi ∂xj(p). 6.7 The Inverse Function Theorem By Proposition 6.11, any diffeomorphism F:U→F(U)⊂Rnof an open subset U of a manifold may be thought of as a coordinate system on U. We say that a C∞ map F:N→Mislocally invertible or a local diffeomorphism atp∈Nifphas a neighborhood Uon which F|U:U→F(U)is a diffeomorphism. Given nsmooth functions F1,..., Fnin a neighborhood of a point pin a man- ifold Nof dimension n, one would like to know whether they form a coordinate system, possibly on a smaller neighborhood of p. This is equivalent to whether F= (F1,..., Fn):N→Rnis a local diffeomorphism at p. The inverse function theorem provides an answer. Theorem 6.25 (Inverse function theorem for Rn).Let F :W→Rnbe a C∞map deﬁned on an open subset W of Rn. For any point p in W, the map F is locally invertible at p if and only if the Jacobian determinant det[∂Fi/∂rj(p)]is not zero. This theorem is usually proved in an undergraduate course on real analysis. See Appendix B for a discussion of this and related theorems. Because the inverse func- tion theorem for Rnis a local result, it easily translates to manifolds. Theorem 6.26 (Inverse function theorem for manifolds). Let F :N→M be a C∞ map between two manifolds of the same dimension, and p ∈N. Suppose for some charts(U,φ)=( U,x1,..., xn)about p in N and (V,ψ)=( V,y1,..., yn)about F(p) in M, F(U)⊂V. Set Fi=yi◦F. Then F is locally invertible at p if and only if its Jacobian determinant det[∂Fi/∂xj(p)]is nonzero. 6.7 The Inverse Function Theorem 69 /Bullet/Bullet /Bullet /BulletpF ψ◦F◦φ−1F(p) φ(p) ψ(F(p))U V φ ψ ≃ ≃φ(U) ψ(V) Fig. 6.4. The map Fis locally invertible at pbecause ψ◦F◦φ−1is locally invertible at φ(p). Proof. Since Fi=yi◦F=ri◦ψ◦F, the Jacobian matrix of Frelative to the charts (U,φ)and(V,ψ)is /bracketleftbigg∂Fi ∂xj(p)/bracketrightbigg =/bracketleftbigg∂(ri◦ψ◦F) ∂xj(p)/bracketrightbigg =/bracketleftbigg∂(ri◦ψ◦F◦φ−1) ∂rj(φ(p))/bracketrightbigg , which is precisely the Jacobian matrix at φ(p)of the map ψ◦F◦φ−1:Rn⊃φ(U)→ψ(V)⊂Rn between two open subsets of Rn. By the inverse function theorem for Rn, det/bracketleftbigg∂Fi ∂xj(p)/bracketrightbigg =det/bracketleftbigg∂ri◦(ψ◦F◦φ−1) ∂rj(φ(p))/bracketrightbigg /ne}a⊔ionslash=0 if and only if ψ◦F◦φ−1is locally invertible at φ(p). Since ψandφare diffeomor- phisms (Proposition 6.10), this last statement is equivalent to the local invertibility ofFatp(seeFigure 6.4 ). ⊓ ⊔ We usually apply the inverse function theorem in the following form. Corollary 6.27. Let N be a manifold of dimension n. A set of n smooth func- tions F1,..., Fndeﬁned on a coordinate neighborhood (U,x1,..., xn)of a point p∈N forms a coordinate system about p if and only if the Jacobian determinant det[∂Fi/∂xj(p)]is nonzero. Proof. LetF=(F1,..., Fn):U→Rn. Then det[∂Fi/∂xj(p)]/ne}a⊔ionslash=0 ⇐⇒ F:U→Rnis locally invertible at p(by the inverse function theorem) ⇐⇒ there is a neighborhood WofpinNsuch that F:W→F(W)is a diffeomor- phism (by the deﬁnition of local invertibility) 70§6 Smooth Maps on a Manifold ⇐⇒(W,F1,..., Fn)is a coordinate chart about pin the differentiable structure of N(by Proposition 6.11). ⊓ ⊔ Example. Find all points in R2in a neighborhood of which the functions x2+y2−1,y can serve as a local coordinate system. Solution. Deﬁne F:R2→R2by F(x,y)=/parenleftbig x2+y2−1,y/parenrightbig . The map Fcan serve as a coordinate map in a neighborhood of pif and only if it is a local diffeomorphism at p. The Jacobian determinant of Fis ∂/parenleftbig F1,F2/parenrightbig ∂(x,y)=det/bracketleftbigg 2x2y 0 1/bracketrightbigg =2x. By the inverse function theorem, Fis a local diffeomorphism at p=(x,y)if and only ifx/ne}a⊔ionslash=0. Thus, Fcan serve as a coordinate system at any point pnot on the y-axis. Problems 6.1. Differentiable structures on R LetRbe the real line with the differentiable structure given by the maximal atlas of the chart (R,φ=1:R→R), and let R′be the real line with the differentiable structure given by the maximal atlas of the chart (R,ψ:R→R), where ψ(x)=x1/3. (a) Show that these two differentiable structures are distinct. (b) Show that there is a diffeomorphism between RandR′. (Hint: The identity map R→R is not the desired diffeomorphism; in fact, this map is not smooth.) 6.2. The smoothness of an inclusion map LetMandNbe manifolds and let q0be a point in N. Prove that the inclusion map iq0:M→ M×N,iq0(p)=( p,q0), isC∞. 6.3.* Group of automorphisms of a vector space LetVbe a ﬁnite-dimensional vector space over R, and GL(V)the group of all linear auto- morphisms of V. Relative to an ordered basis e= (e 1,..., en)forV, a linear automorphism L∈GL(V)is represented by a matrix [ai j]deﬁned by L(ej)=∑ iai jei. The map φe: GL(V)→GL(n,R), L/ma√s⊔o→[ai j], is a bijection with an open subset of Rn×nthat makes GL (V)into a C∞manifold, which we denote temporarily by GL (V)e. If GL(V)uis the manifold structure induced from another ordered basis u=(u 1,..., un)forV, show that GL (V)eis the same as GL (V)u. 6.4. Local coordinate systems Find all points in R3in a neighborhood of which the functions x,x2+y2+z2−1,zcan serve as a local coordinate system. 7.1 The Quotient Topology 71 §7 Quotients Gluing the edges of a malleable square is one way to create new surfaces. For ex- ample, gluing together the top and bottom edges of a square gives a cylinder; gluing together the boundaries of the cylinder with matching orientations gives a torus (Fig-ure7.1). This gluing process is called an identiﬁcation or aquotient construction . Fig. 7.1. Gluing the edges of a malleable square. The quotient construction is a process of simpliﬁcation. Starting with an equiv- alence relation on a set, we identify each equivalence class to a point. Mathematicsabounds in quotient constructions, for example, the quotient group, quotient ring, or quotient vector space in algebra. If the original set is a topological space, it is always possible to give the quotient set a topology so that the natural projection mapbecomes continuous. However, even if the original space is a manifold, a quotient space is often not a manifold. The main results of this section give conditions under which a quotient space remains second countable and Hausdorff. We then study real projective space as an example of a quotient manifold. Real projective space can be interpreted as a quotient of a sphere with antipodal points identiﬁed, or as the set of lines through the origin in a vector space. Thesetwo interpretations give rise to two distinct generalizations—covering maps on the one hand and Grassmannians of k-dimensional subspaces of a vector space on the other. In one of the exercises, we carry out an extensive investigation of G(2,4), the Grassmannian of 2-dimensional subspaces of R 4. 7.1 The Quotient Topology Recall that an equivalence relation on a set Sis a reﬂexive, symmetric, and transitive relation. The equivalence class [x]ofx∈Sis the set of all elements in Sequivalent tox. An equivalence relation on Spartitions Sinto disjoint equivalence classes. We denote the set of equivalence classes by S/∼and call this set the quotient ofSby the equivalence relation ∼. There is a natural projection map π:S→S/∼that sends x∈Sto its equivalence class [x]. Assume now that Sis a topological space. We deﬁne a topology on S/∼by declaring a set UinS/∼to be open if and only if π−1(U)is open in S. Clearly, both the empty set ∅and the entire quotient S/∼are open. Further, since 72§7 Quotients π−1/parenleftbigg/uniondisplay αUα/parenrightbigg =/uniondisplay απ−1(Uα) and π−1/parenleftbigg/intersectiondisplay iUi/parenrightbigg =/intersectiondisplay iπ−1(Ui), the collection of open sets in S/∼is closed under arbitrary unions and ﬁnite inter- sections, and is therefore a topology. It is called the quotient topology onS/∼. With this topology, S/∼is called the quotient space ofSby the equivalence relation ∼. With the quotient topology on S/∼, the projection map π:S→S/∼is automatically continuous, because the inverse image of an open set in S/∼is by deﬁnition open inS. 7.2 Continuity of a Map on a Quotient Let∼be an equivalence relation on the topological space Sand give S/∼the quotient topology. Suppose a function f:S→Yfrom Sto another topological space Yis constant on each equivalence class. Then it induces a map ¯f:S/∼→ Yby ¯f([p])= f(p)forp∈S. In other words, there is a commutative diagram Sf/d47/d47 π /d15/d15Y. S/∼¯f/d61/d61/d124/d124/d124/d124/d124/d124/d124/d124 Proposition 7.1. The induced map ¯f:S/∼→ Y is continuous if and only if the map f:S→Y is continuous. Proof. (⇒)If¯fis continuous, then as the composite ¯f◦πof continuous functions, fis also continuous. (⇐)Suppose fis continuous. Let Vbe open in Y. Then f−1(V)=π−1(¯f−1(V))is open in S. By the deﬁnition of quotient topology, ¯f−1(V)is open in S/∼. Since V was arbitrary, ¯f:S/∼→ Yis continuous. ⊓ ⊔ This proposition gives a useful criterion for checking whether a function ¯fon a quotient space S/∼is continuous: simply lift the function ¯ftof:=f◦πonSand check the continuity of the lifted map fonS. For examples of this, see Example 7.2 and Proposition 7.3. 7.4 A Necessary Condition for a Hausdorff Quotient 73 7.3 Identiﬁcation of a Subset to a Point IfAis a subspace of a topological space S, we can deﬁne a relation ∼onSby declaring x∼xfor all x∈S (so the relation is reﬂexive) and x∼yfor all x,y∈A. This is an equivalence relation on S. We say that the quotient space S/∼is obtained from Sbyidentifying A to a point . Example 7.2.LetIbe the unit interval [0,1]andI/∼the quotient space obtained from Iby identifying the two points {0,1}to a point. Denote by S1the unit circle in the complex plane. The function f:I→S1,f(x)=exp(2πix), assumes the same value at 0 and 1 ( Figure 7.2 ), and so induces a function ¯f:I/∼→ S1. /Bullet 0 1f Fig. 7.2. The unit circle as a quotient space of the unit interval. Proposition 7.3. The function ¯f:I/∼→ S1is a homeomorphism. Proof. Since fis continuous, ¯fis also continuous by Proposition 7.1. Clearly, ¯fis a bijection. As the continuous image of the compact set I, the quotient I/∼is compact. Thus, ¯fis a continuous bijection from the compact space I/∼to the Hausdorff space S1. By Corollary A.36, ¯fis a homeomorphism. ⊓ ⊔ 7.4 A Necessary Condition for a Hausdorff Quotient The quotient construction does not in general preserve the Hausdorff property or second countability. Indeed, since every singleton set in a Hausdorff space is closed, ifπ:S→S/∼is the projection and the quotient S/∼is Hausdorff, then for any p∈S, its image{π(p)}is closed in S/∼. By the continuity of π, the inverse image π−1({π(p)}) = [ p]is closed in S. This gives a necessary condition for a quotient space to be Hausdorff. Proposition 7.4. If the quotient space S /∼is Hausdorff, then the equivalence class [p]of any point p in S is closed in S. 74§7 Quotients Example. Deﬁne an equivalence relation ∼onRby identifying the open interval ]0,∞[to a point. Then the quotient space R/∼is not Hausdorff because the equiva- lence class ]0,∞[of∼inRcorresponding to the point ]0,∞[inR/∼is not a closed subset of R. 7.5 Open Equivalence Relations In this section we follow the treatment of Boothby [3] and derive conditions under which a quotient space is Hausdorff or second countable. Recall that a map f:X→Y of topological spaces is open if the image of any open set under fis open. Deﬁnition 7.5. An equivalence relation ∼on a topological space Sis said to be open if the projection map π:S→S/∼is open. In other words, the equivalence relation ∼onSis open if and only if for every open set UinS, the set π−1(π(U))=/uniondisplay x∈U[x] of all points equivalent to some point of Uis open. Example 7.6.The projection map to a quotient space is in general not open. For example, let∼be the equivalence relation on the real line Rthat identiﬁes the two points 1 and−1, and π:R→R/∼the projection map. /Bullet /Bullet/Bullet ( ) −2 0−1 1π () Fig. 7.3. A projection map that is not open. The map πis open if and only if for every open set VinR, its image π(V)is open inR/∼, which by the deﬁnition of the quotient topology means that π−1(π(V))is open inR. Now let Vbe the open interval ]−2,0[inR. Then π−1(π(V))= ]−2,0[∪{1}, which is not open in R(Figure 7.3 ). Therefore, the projection map π:R→R/∼is not an open map. Given an equivalence relation ∼onS, letRbe the subset of S×Sthat deﬁnes the relation R={(x,y)∈S×S|x∼y}. We call Rthegraph of the equivalence relation ∼. 7.5 Open Equivalence Relations 75 /Bullet SSR UV(x,y) Fig. 7.4. The graph Rof an equivalence relation and an open set U×Vdisjoint from R. Theorem 7.7. Suppose∼is an open equivalence relation on a topological space S. Then the quotient space S /∼is Hausdorff if and only if the graph R of ∼is closed in S×S. Proof. There is a sequence of equivalent statements: Ris closed in S×S ⇐⇒(S×S)−Ris open in S×S ⇐⇒ for every (x,y)∈S×S−R, there is a basic open set U×Vcontaining (x,y) such that (U×V)∩R=∅(Figure 7.4 ) ⇐⇒ for every pair x≁yinS, there exist neighborhoods UofxandVofyinSsuch that no element of Uis equivalent to an element of V ⇐⇒ for any two points [x]/ne}a⊔ionslash=[y]inS/∼, there exist neighborhoods UofxandVof yinSsuch that π(U)∩π(V)=∅inS/∼. (∗) We now show that this last statement ( ∗) is equivalent to S/∼being Hausdorff. First assume (∗). Since∼is an open equivalence relation, π(U)andπ(V)are disjoint open sets in S/∼containing [x]and[y]respectively. Therefore, S/∼is Hausdorff. Conversely, suppose S/∼is Hausdorff. Let [x]/ne}a⊔ionslash= [y]inS/∼. Then there exist disjoint open sets AandBinS/∼such that [x]∈Aand[y]∈B. By the surjectivity of π, we have A=π(π−1A)andB=π(π−1B)(see Problem 7.1). Let U=π−1Aand V=π−1B. Then x∈U,y∈V, and A=π(U)andB=π(V)are disjoint open sets in S/∼. ⊓ ⊔ If the equivalence relation ∼is equality, then the quotient space S/∼isSitself and the graph Rof∼is simply the diagonal ∆={(x,x)∈S×S}. In this case, Theorem 7.7 becomes the following well-known characterization of a Hausdorff space by its diagonal (cf. Problem A.6). Corollary 7.8. A topological space S is Hausdorff if and only if the diagonal ∆in S×S is closed. 76§7 Quotients Theorem 7.9. Let∼be an open equivalence relation on a topological space S with projection π:S→S/∼. IfB={Bα}is a basis for S, then its image {π(Bα)}under πis a basis for S /∼. Proof. Since πis an open map,{π(Bα)}is a collection of open sets in S/∼. Let W be an open set in S/∼and[x]∈W,x∈S. Then x∈π−1(W). Since π−1(W)is open, there is a basic open set B∈Bsuch that x∈B⊂π−1(W). Then [x]=π(x)∈π(B)⊂W, which proves that{π(Bα)}is a basis for S/∼. ⊓ ⊔ Corollary 7.10. If∼is an open equivalence relation on a second-countable space S, then the quotient space S /∼is second countable. 7.6 Real Projective Space Deﬁne an equivalence relation on Rn+1−{0}by x∼y⇐⇒ y=txfor some nonzero real number t, where x,y∈Rn+1−{0}. The real projective space RPnis the quotient space of Rn+1−{0}by this equivalence relation. We denote the equivalence class of a point (a0,..., an)∈Rn+1−{0}by[a0,..., an]and let π:Rn+1−{0}→RPnbe the pro- jection. We call [a0,..., an]homogeneous coordinates onRPn. Geometrically, two nonzero points in Rn+1are equivalent if and only if they lie on the same line through the origin, so RPncan be interpreted as the set of all lines through the origin in Rn+1. Each line through the origin in Rn+1meets the unit /Bullet /Bullet/Bullet Fig. 7.5. A line through 0 in R3corresponds to a pair of antipodal points on S2. sphere Snin a pair of antipodal points, and conversely, a pair of antipodal points on Sndetermines a unique line through the origin (Figure 7.5 ). This suggests that we deﬁne an equivalence relation ∼onSnby identifying antipodal points: 7.6 Real Projective Space 77 x∼y⇐⇒ x=±y,x,y∈Sn. We then have a bijection RPn↔Sn/∼. Exercise 7.11 (Real projective space as a quotient of a sphere).* Forx=(x1,..., xn)∈Rn, let/bardblx/bardbl=/radicalbig ∑i(xi)2be the modulus of x. Prove that the map f:Rn+1−{0}→ Sngiven by f(x)=x /bardblx/bardbl induces a homeomorphism ¯f:RPn→Sn/∼. (Hint: Find an inverse map ¯g:Sn/∼→RPn and show that both ¯fand ¯gare continuous.) Example 7.12 ( The real projective line RP1). /Bullet/Bullet /Bullet /Bullet /Bullet 0 −a a Fig. 7.6. The real projective line RP1as the set of lines through 0 in R2. Each line through the origin in R2meets the unit circle in a pair of antipodal points. By Exercise 7.11, RP1is homeomorphic to the quotient S1/∼, which is in turn homeomorphic to the closed upper semicircle with the two endpoints identiﬁed (Figure 7.6 ). Thus,RP1is homeomorphic to S1. Example 7.13 ( The real projective plane RP2).By Exercise 7.11, there is a homeo- morphism RP2≃S2/{antipodal points}=S2/∼. For points not on the equator, each pair of antipodal points contains a unique point in the upper hemisphere. Thus, there is a bijection between S2/∼and the quotient of the closed upper hemisphere in which each pair of antipodal points on the equatoris identiﬁed. It is not difﬁcult to show that this bijection is a homeomorphism (see Problem 7.2). LetH 2be the closed upper hemisphere H2={(x,y,z)∈R3|x2+y2+z2=1,z≥0} and let D2be the closed unit disk 78§7 Quotients D2={(x,y)∈R2|x2+y2≤1}. These two spaces are homeomorphic to each other via the continuous map ϕ:H2→D2, ϕ(x,y,z)=( x,y), and its inverse ψ:D2→H2, ψ(x,y)=/parenleftig x,y,/radicalbig 1−x2−y2/parenrightig . OnH2, deﬁne an equivalence relation ∼by identifying the antipodal points on the equator: (x,y,0)∼(−x,−y,0),x2+y2=1. OnD2, deﬁne an equivalence relation ∼by identifying the antipodal points on the boundary circle: (x,y)∼(−x,−y),x2+y2=1. Then ϕandψinduce homeomorphisms ¯ϕ:H2/∼→ D2/∼,¯ψ:D2/∼→ H2/∼. In summary, there is a sequence of homeomorphisms RP2∼→S2/∼∼→H2/∼∼→D2/∼ that identiﬁes the real projective plane as the quotient of the closed disk D2with the antipodal points on its boundary identiﬁed. This may be the best way to picture RP2 (Figure 7.7 ). /Bullet /Bullet Fig. 7.7. The real projective plane as the quotient of a disk. The real projective plane RP2cannot be embedded as a submanifold of R3. How- ever, if we allow self-intersection, then we can map RP2intoR3as a cross-cap (Fig- ure7.8). This map is not one-to-one. 7.7 The Standard C∞Atlas on a Real Projective Space 79 /Bullet/Bullet /Bullet /BulletA B CD/Bullet/Bullet/Bullet /BulletAB CD /Bullet/Bullet/Bullet /BulletB CD /Bullet/BulletB=D A=C Fig. 7.8. The real projective plane immersed as a cross-cap in R3. Proposition 7.14. The equivalence relation ∼onRn+1−{0}in the deﬁnition of RPn is an open equivalence relation. Proof. For an open set U⊂Rn+1−{0}, the image π(U)is open in RPnif and only ifπ−1(π(U))is open in Rn+1−{0}. But π−1(π(U))consists of all nonzero scalar multiples of points of U; that is, π−1(π(U))=/uniondisplay t∈R×tU=/uniondisplay t∈R×{t p|p∈U}. Since multiplication by t∈R×is a homeomorphism of Rn+1−{0}, the set tUis open for any t. Therefore, their union/uniontext t∈R×tU=π−1(π(U))is also open. ⊓ ⊔ Corollary 7.15. The real projective space RPnis second countable. Proof. Apply Corollary 7.10. ⊓ ⊔ Proposition 7.16. The real projective space RPnis Hausdorff. Proof. LetS=Rn+1−{0}and consider the set R={(x,y)∈S×S|y=txfor some t∈R×}. If we write xandyas column vectors, then [x y]is an(n+1)×2 matrix, and Rmay be characterized as the set of matrices [x y]inS×Sof rank≤1. By a standard fact from linear algebra, rk [x y]≤1 is equivalent to the vanishing of all 2 ×2 minors of [x y](see Problem B.1). As the zero set of ﬁnitely many polynomials, Ris a closed subset of S×S. Since∼is an open equivalence relation on S, and Ris closed in S×S, by Theorem 7.7 the quotient S/∼≃RPnis Hausdorff. ⊓ ⊔ 7.7 The Standard C∞Atlas on a Real Projective Space Let[a0,..., an]be homogeneous coordinates on the projective space RPn. Although a0is not a well-deﬁned function on RPn, the condition a0/ne}a⊔ionslash=0 is independent of the choice of a representative for [a0,..., an]. Hence, the condition a0/ne}a⊔ionslash=0 makes sense onRPn, and we may deﬁne 80§7 Quotients U0:={[a0,..., an]∈RPn|a0/ne}a⊔ionslash=0}. Similarly, for each i=1,..., n, let Ui:={[a0,..., an]∈RPn|ai/ne}a⊔ionslash=0}. Deﬁne φ0:U0→Rn by [a0,..., an]/ma√s⊔o→/parenleftbigga1 a0,...,an a0/parenrightbigg . This map has a continuous inverse (b1,..., bn)/ma√s⊔o→[1,b1,..., bn] and is therefore a homeomorphism. Similarly, there are homeomorphisms for each i=1,..., n: φi:Ui→Rn, [a0,..., an]/ma√s⊔o→/parenleftigg a0 ai,...,/hatwideai ai,...,an ai/parenrightigg , where the caret sign /hatwideover ai/aimeans that that entry is to be omitted. This proves thatRPnis locally Euclidean with the (Ui,φi)as charts. On the intersection U0∩U1, we have a0/ne}a⊔ionslash=0 and a1/ne}a⊔ionslash=0, and there are two coor- dinate systems [a0,a1,a2,..., an] /parenleftbigga0 a1,a2 a1,...,an a1/parenrightbigg ./parenleftbigga1 a0,a2 a0,...,an a0/parenrightbiggφ1 φ0 We will refer to the coordinate functions on U0asx1,..., xn, and the coordinate functions on U1asy1,..., yn. On U0, xi=ai a0,i=1,..., n, and on U1, y1=a0 a1,y2=a2 a1, ..., yn=an a1. Then on U0∩U1, 7.7 The Standard C∞Atlas on a Real Projective Space 81 y1=1 x1,y2=x2 x1,y3=x3 x1, ..., yn=xn x1, so (φ1◦φ−1 0)(x)=/parenleftbigg1 x1,x2 x1,x3 x1,...,xn x1/parenrightbigg . This is a C∞function because x1/ne}a⊔ionslash=0 on φ0(U0∩U1). On any other Ui∩Ujan analogous formula holds. Therefore, the collection {(Ui,φi)}i=0,...,nis aC∞atlas for RPn, called the standard atlas . This concludes the proof that RPnis aC∞manifold. Problems 7.1. Image of the inverse image of a map Letf:X→Ybe a map of sets, and let B⊂Y. Prove that f(f−1(B))= B∩f(X). Therefore, iffis surjective, then f(f−1(B))= B. 7.2. Real projective plane LetH2be the closed upper hemisphere in the unit sphere S2, and let i:H2→S2be the inclusion map. In the notation of Example 7.13, prove that the induced map f:H2/∼→ S2/∼ is a homeomorphism. ( Hint: Imitate Proposition 7.3.) 7.3. Closedness of the diagonal of a Hausdorff space Deduce Theorem 7.7 from Corollary 7.8. (Hint : To prove that if S/∼is Hausdorff, then the graph Rof∼is closed in S×S, use the continuity of the projection map π:S→S/∼. To prove the reverse implication, use the openness of π.) 7.4.* Quotient of a sphere with antipodal points identiﬁed LetSnbe the unit sphere centered at the origin in Rn+1. Deﬁne an equivalence relation ∼on Snby identifying antipodal points: x∼y⇐⇒ x=±y,x,y∈Sn. (a) Show that∼is an open equivalence relation. (b) Apply Theorem 7.7 and Corollary 7.8 to prove that the quotient space Sn/∼is Hausdorff, without making use of the homeomorphism RPn≃Sn/∼. 7.5.* Orbit space of a continuous group action Suppose a right action of a topological group Gon a topological space Sis continuous; this simply means that the map S×G→Sdescribing the action is continuous. Deﬁne two points x,yofSto be equivalent if they are in the same orbit; i.e., there is an element g∈Gsuch that y=xg. Let S/Gbe the quotient space; it is called the orbit space of the action. Prove that the projection map π:S→S/Gis an open map. (This problem generalizes Proposition 7.14, in which G=R×=R−{0}andS=Rn+1−{0}. Because R×is commutative, a left R×-action becomes a right R×-action if scalar multiplication is written on the right.) 7.6. Quotient of Rby2πZ Let the additive group 2 πZact onRon the right by x·2πn=x+2πn, where nis an integer. Show that the orbit space R/2πZis a smooth manifold. 82§7 Quotients 7.7. The circle as a quotient space (a) Let{(Uα,φα)}2 α=1be the atlas of the circle S1in Example 5.7, and let ¯φαbe the map φα followed by the projection R→R/2πZ. On U1∩U2=A∐B, since φ1andφ2differ by an integer multiple of 2 π,¯φ1=¯φ2. Therefore, ¯φ1and ¯φ2piece together to give a well-deﬁned map ¯φ:S1→R/2πZ. Prove that ¯φisC∞. (b) The complex exponential R→S1,t/ma√s⊔o→eit, is constant on each orbit of the action of 2 πZ onR. Therefore, there is an induced map F:R/2πZ→S1,F([t]) = eit. Prove that F isC∞. (c) Prove that F:R/2πZ→S1is a diffeomorphism. 7.8. The Grassmannian G(k,n) The Grassmannian G(k,n)is the set of all k-planes through the origin in Rn. Such a k-plane is a linear subspace of dimension kofRnand has a basis consisting of klinearly independent vectors a1,..., akinRn. It is therefore completely speciﬁed by an n×kmatrix A=[a1···ak] of rank k, where the rank of a matrix A, denoted by rk A, is deﬁned to be the number of linearly independent columns of A. This matrix is called a matrix representative of the k-plane. (For properties of the rank, see the problems in Appendix B.) Two bases a1,..., akandb1,..., bkdetermine the same k-plane if there is a change-of- basis matrix g=[gi j]∈GL(k,R)such that bj=∑ iaigi j,1≤i,j≤k. In matrix notation, B=Ag. LetF(k,n)be the set of all n×kmatrices of rank k, topologized as a subspace of Rn×k, and∼the equivalence relation A∼Biff there is a matrix g∈GL(k,R)such that B=Ag. In the notation of Problem B.3, F(k,n)is the set DmaxinRn×kand is therefore an open subset. There is a bijection between G(k,n)and the quotient space F(k,n)/∼. We give the Grassmannian G(k,n)the quotient topology on F(k,n)/∼. (a) Show that∼is an open equivalence relation. ( Hint: Either mimic the proof of Proposi- tion 7.14 or apply Problem 7.5.) (b) Prove that the Grassmannian G(k,n)is second countable. ( Hint: Apply Corollary 7.10.) (c) Let S=F(k,n). Prove that the graph RinS×Sof the equivalence relation ∼is closed. (Hint: Two matrices A=[a1···ak]andB=[b1···bk]inF(k,n)are equivalent if and only if every column of Bis a linear combination of the columns of Aif and only if rk [A B]≤k if and only if all (k+1)×(k+1)minors of [A B]are zero.) (d) Prove that the Grassmannian G(k,n)is Hausdorff. ( Hint: Mimic the proof of Proposi- tion 7.16.) Next we want to ﬁnd a C∞atlas on the Grassmannian G(k,n). For simplicity, we specialize to G(2,4). For any 4×2 matrix A, letAi jbe the 2×2 submatrix consisting of its ith row and jth row. Deﬁne Vi j={A∈F(2,4)|Ai jis nonsingular}. Because the complement of Vi jinF(2,4)is deﬁned by the vanishing of det Ai j, we conclude thatVi jis an open subset of F(2,4). (e) Prove that if A∈Vi j, then Ag∈Vi jfor any nonsingular matrix g∈GL(2,R). 7.7 The Standard C∞Atlas on a Real Projective Space 83 Deﬁne Ui j=Vi j/∼. Since∼is an open equivalence relation, Ui j=Vi j/∼is an open subset ofG(2,4). ForA∈V12, A∼AA−1 12= 1 0 0 1 ∗∗∗∗ =/bracketleftbiggI A 34A−1 12/bracketrightbigg . This shows that the matrix representatives of a 2-plane in U12have a canonical form Bin which B12is the identity matrix. (f) Show that the map ˜φ12:V12→R2×2, ˜φ12(A)=A34A−1 12, induces a homeomorphism φ12:U12→R2×2. (g) Deﬁne similarly homeomorphisms φi j:Ui j→R2×2. Compute φ12◦φ−1 23, and show that it isC∞. (h) Show that{Ui j|1≤i<j≤4}is an open cover of G(2,4)and that G(2,4)is a smooth manifold. Similar consideration shows that F(k,n)has an open cover {VI}, where Iis a strictly ascending multi-index 1 ≤i1<···<ik≤n. For A∈F(k,n), let AIbe the k×ksubmatrix of Aconsisting of i1th,...,ikth rows of A. Deﬁne VI={A∈G(k,n)|detAI/ne}a⊔ionslash=0}. Next deﬁne ˜φI:VI→R(n−k)×kby ˜φI(A)=(AA−1 I)I′, where( )I′denotes the (n−k)×ksubmatrix obtained from the complement I′of the multi- index I. Let UI=VI/∼. Then ˜φinduces a homeomorphism φ:UI→R(n−k)×k. It is not difﬁcult to show that {(UI,φI)}is aC∞atlas for G(k,n). Therefore the Grassmannian G(k,n) is aC∞manifold of dimension k(n−k). 7.9.* Compactness of real projective space Show that the real projective space RPnis compact. ( Hint: Use Exercise 7.11.) Chapter 3 The Tangent Space By deﬁnition, the tangent space to a manifold at a point is the vector space of deriva- tions at the point. A smooth map of manifolds induces a linear map, called its differ- ential , of tangent spaces at corresponding points. In local coordinates, the differential is represented by the Jacobian matrix of partial derivatives of the map. In this sense, the differential of a map between manifolds is a generalization of the derivative of a map between Euclidean spaces. A basic principle in manifold theory is the linearization principle, according to which a manifold can be approximated near a point by its tangent space at the point,and a smooth map can be approximated by the differential of the map. In this way, one turns a topological problem into a linear problem. A good example of the lin- earization principle is the inverse function theorem, which reduces the local invert-ibility of a smooth map to the invertibility of its differential at a point. Using the differential, we classify maps having maximal rank at a point into immersions and submersions at the point, depending on whether the differential is injective or surjective there. A point where the differential is surjective is a regular point of the map. The regular level set theorem states that a level set all of whose points are regular is a regular submanifold, i.e., a subset that locally looks like a coordinate k-plane in R n. This theorem gives a powerful tool for proving that a topological space is a manifold. We then introduce categories and functors, a framework for comparing structural similarities. After this interlude, we return to the study of maps via their differentials. From the rank of the differential, one obtains three local normal forms for smooth maps—the constant rank theorem, the immersion theorem, and the submersion theo- rem, corresponding to constant-rank differentials, injective differentials, and surjec-tive differentials respectively. We give three proofs of the regular level set theorem, a ﬁrst proof (Theorem 9.9), using the inverse function theorem, that actually produces explicit local coordinates, and two more proofs (p. 119) that are corollaries of theconstant rank theorem and the submersion theorem. The collection of tangent spaces to a manifold can be given the structure of a vector bundle; it is then called the tangent bundle of the manifold. Intuitively, a vector bundle over a manifold is a locally trivial family of vector spaces parametrized © Springer Science+Business Media, LLC 201185 L.W. Tu, An Introduction to Manifolds, Universitext, DOI 10.1007/978-1-4419-7400-6_3, 86§8 The Tangent Space by points of the manifold. A smooth map of manifolds induces, via its differential at each point, a bundle map of the corresponding tangent bundles. In this way weobtain a covariant functor from the category of smooth manifolds and smooth maps to the category of vector bundles and bundle maps. Vector ﬁelds, which manifest themselves in the physical world as velocity, force, electricity, magnetism, and so on, may be viewed as sections of the tangent bundle over a manifold. Smooth C ∞bump functions and partitions of unity are an indispensable technical tool in the theory of smooth manifolds. Using C∞bump functions, we give several criteria for a vector ﬁeld to be smooth. The chapter ends with integral curves, ﬂows, and the Lie bracket of smooth vector ﬁelds. §8 The Tangent Space In Section 2 we saw that for any point pin an open set UinRnthere are two equiv- alent ways to deﬁne a tangent vector at p: (i) as an arrow ( Figure 8.1 ), represented by a column vector; /Bullet p a1 ... an  Fig. 8.1. A tangent vector in Rnas an arrow and as a column vector. (ii) as a point-derivation of C∞ p, the algebra of germs of C∞functions at p. Both deﬁnitions generalize to a manifold. In the arrow approach, one deﬁnes a tangent vector at pin a manifold Mby ﬁrst choosing a chart (U,φ)atpand then decreeing a tangent vector at pto be an arrow at φ(p)inφ(U). This approach, while more visual, is complicated to work with, since a different chart (V,ψ)atpwould give rise to a different set of tangent vectors at pand one would have to decide how to identify the arrows at φ(p)inUwith the arrows at ψ(p)inψ(V). The cleanest and most intrinsic deﬁnition of a tangent vector at pinMis as a point-derivation, and this is the approach we adopt. 8.1 The Tangent Space at a Point Just as for Rn, we deﬁne a germ of a C∞function at pinMto be an equivalence class of C∞functions deﬁned in a neighborhood of pinM, two such functions being equivalent if they agree on some, possibly smaller, neighborhood of p. The set of 8.2 The Differential of a Map 87 germs of C∞real-valued functions at pinMis denoted by C∞ p(M). The addition and multiplication of functions make C∞ p(M)into a ring; with scalar multiplication by real numbers, C∞ p(M)becomes an algebra over R. Generalizing a derivation at a point in Rn, we deﬁne a derivation at a point in a manifold M, or a point-derivation ofC∞ p(M), to be a linear map D:C∞ p(M)→R such that D(f g)=( D f)g(p)+f(p)Dg. Deﬁnition 8.1. Atangent vector at a point pin a manifold Mis a derivation at p. Just as for Rn, the tangent vectors at pform a vector space Tp(M), called the tangent space of M at p . We also write TpMinstead of Tp(M). Remark 8.2 ( Tangent space to an open subset ).IfUis an open set containing pin M, then the algebra C∞ p(U)of germs of C∞functions in Uatpis the same as C∞ p(M). Hence, TpU=TpM. Given a coordinate neighborhood (U,φ) = ( U,x1,..., xn)about a point pin a manifold M, we recall the deﬁnition of the partial derivatives ∂/∂xiﬁrst introduced in Section 6. Let r1,..., rnbe the standard coordinates on Rn. Then xi=ri◦φ:U→R. Iffis a smooth function in a neighborhood of p, we set ∂ ∂xi/vextendsingle/vextendsingle/vextendsingle/vextendsingle pf=∂ ∂ri/vextendsingle/vextendsingle/vextendsingle/vextendsingle φ(p)/parenleftbig f◦φ−1/parenrightbig ∈R. It is easily checked that ∂/∂xi|psatisﬁes the derivation property and so is a tangent vector at p. When Mis one-dimensional and tis a local coordinate, it is customary to write d/dt|pinstead of ∂/∂t|pfor the coordinate vector at the point p. To simplify the notation, we will sometimes write ∂/∂xiinstead of ∂/∂xi|pif it is understood at which point the tangent vector is located. 8.2 The Differential of a Map LetF:N→Mbe a C∞map between two manifolds. At each point p∈N, the map Finduces a linear map of tangent spaces, called its differential at p , F∗:TpN→TF(p)M as follows. If Xp∈TpN, then F∗(Xp)is the tangent vector in TF(p)Mdeﬁned by (F∗(Xp))f=Xp(f◦F)∈Rforf∈C∞ F(p)(M). (8.1) Here fis a germ at F(p), represented by a C∞function in a neighborhood of F(p). Since (8.1) is independent of the representative of the germ, in practice we can be cavalier about the distinction between a germ and a representative function for the germ. 88§8 The Tangent Space Exercise 8.3 (The differential of a map). Check that F∗(Xp)is a derivation at F(p)and that F∗:TpN→TF(p)Mis a linear map. To make the dependence on pexplicit we sometimes write F∗,pinstead of F∗. Example 8.4 ( Differential of a map between Euclidean spaces ).Suppose F:Rn→ Rmis smooth and pis a point in Rn. Let x1,..., xnbe the coordinates on Rnand y1,..., ymthe coordinates on Rm. Then the tangent vectors ∂/∂x1|p,..., ∂/∂xn|p form a basis for the tangent space Tp(Rn)and∂/∂y1|F(p),..., ∂/∂ym|F(p)form a basis for the tangent space TF(p)(Rm). The linear map F∗:Tp(Rn)→TF(p)(Rm)is described by a matrix [ai j]relative to these two bases: F∗/parenleftigg ∂ ∂xj/vextendsingle/vextendsingle/vextendsingle/vextendsingle p/parenrightigg =∑ kak j∂ ∂yk/vextendsingle/vextendsingle/vextendsingle/vextendsingle F(p),ak j∈R. (8.2) LetFi=yi◦Fbe the ith component of F. We can ﬁnd ai jby evaluating the right- hand side (RHS) and left-hand side (LHS) of (8.2) on yi: RHS=∑ kak j∂ ∂yk/vextendsingle/vextendsingle/vextendsingle/vextendsingle F(p)yi=∑ kak jδi k=ai j, LHS=F∗/parenleftigg ∂ ∂xj/vextendsingle/vextendsingle/vextendsingle/vextendsingle p/parenrightigg yi=∂ ∂xj/vextendsingle/vextendsingle/vextendsingle/vextendsingle p(yi◦F)=∂Fi ∂xj(p). So the matrix of F∗relative to the bases {∂/∂xj|p}and{∂/∂yi|F(p)}is[∂Fi/∂xj(p)]. This is precisely the Jacobian matrix of the derivative of Fatp. Thus, the differential of a map between manifolds generalizes the derivative of a map between Euclidean spaces. 8.3 The Chain Rule LetF:N→MandG:M→Pbe smooth maps of manifolds, and p∈N. The differentials of FatpandGatF(p)are linear maps TpNF∗,p−−→ TF(p)MG∗,F(p)−−−−→ TG(F(p))P. Theorem 8.5 (The chain rule). If F:N→M and G :M→P are smooth maps of manifolds and p∈N, then (G◦F)∗,p=G∗,F(p)◦F∗,p. Proof. LetXp∈TpNand let fbe a smooth function at G(F(p))inP. Then ((G◦F)∗Xp)f=Xp(f◦G◦F) and ((G∗◦F∗)Xp)f=(G∗(F∗Xp))f=(F∗Xp)(f◦G)=Xp(f◦G◦F).⊓ ⊔ 8.4 Bases for the Tangent Space at a Point 89 Example 8.13 shows that when written out in terms of matrices, the chain rule of Theorem 8.5 assumes a more familiar form as a sum of products of partial deriva-tives. Remark. The differential of the identity map 1 M:M→Mat any point pinMis the identity map 1TpM:TpM→TpM, because ((1M)∗Xp)f=Xp(f◦1M)=Xpf, for any Xp∈TpMandf∈C∞ p(M). Corollary 8.6. If F:N→M is a diffeomorphism of manifolds and p ∈N, then F∗:TpN→TF(p)M is an isomorphism of vector spaces. Proof. To say that Fis a diffeomorphism means that it has a differentiable inverse G:M→Nsuch that G◦F=1NandF◦G=1M. By the chain rule, (G◦F)∗=G∗◦F∗=(1N)∗=1TpN, (F◦G)∗=F∗◦G∗=(1M)∗=1TF(p)M. Hence, F∗andG∗are isomorphisms. ⊓ ⊔ Corollary 8.7 (Invariance of dimension). If an open set U⊂Rnis diffeomorphic to an open set V⊂Rm, then n=m. Proof. LetF:U→Vbe a diffeomorphism and let p∈U. By Corollary 8.6, F∗,p:TpU→TF(p)Vis an isomorphism of vector spaces. Since there are vector space isomorphisms TpU≃RnandTF(p)≃Rm, we must have that n=m.⊓ ⊔ 8.4 Bases for the Tangent Space at a Point As usual, we denote by r1,..., rnthe standard coordinates on Rn, and if(U,φ)is a chart about a point pin a manifold Mof dimension n, we set xi=ri◦φ. Since φ:U→Rnis a diffeomorphism onto its image (Proposition 6.10), by Corollary 8.6 the differential φ∗:TpM→Tφ(p)Rn is a vector space isomorphism. In particular, the tangent space TpMhas the same dimension nas the manifold M. Proposition 8.8. Let(U,φ)=( U,x1,..., xn)be a chart about a point p in a manifold M. Then φ∗/parenleftigg ∂ ∂xi/vextendsingle/vextendsingle/vextendsingle/vextendsingle p/parenrightigg =∂ ∂ri/vextendsingle/vextendsingle/vextendsingle/vextendsingle φ(p). 90§8 The Tangent Space Proof. For any f∈C∞ φ(p)(Rn), φ∗/parenleftigg ∂ ∂xi/vextendsingle/vextendsingle/vextendsingle/vextendsingle p/parenrightigg f=∂ ∂xi/vextendsingle/vextendsingle/vextendsingle/vextendsingle p(f◦φ) ( deﬁnition of φ∗) =∂ ∂ri/vextendsingle/vextendsingle/vextendsingle/vextendsingle φ(p)(f◦φ◦φ−1) (deﬁnition of ∂/∂xi|p) =∂ ∂ri/vextendsingle/vextendsingle/vextendsingle/vextendsingle φ(p)f. ⊓ ⊔ Proposition 8.9. If(U,φ)=( U,x1,..., xn)is a chart containing p, then the tangent space T pM has basis ∂ ∂x1/vextendsingle/vextendsingle/vextendsingle/vextendsingle p,...,∂ ∂xn/vextendsingle/vextendsingle/vextendsingle/vextendsingle p. Proof. An isomorphism of vector spaces carries a basis to a basis. By Propo- sition 8.8 the isomorphism φ∗:TpM→Tφ(p)(Rn)maps ∂/∂x1|p,..., ∂/∂xn|pto ∂/∂r1|φ(p),..., ∂/∂rn|φ(p), which is a basis for the tangent space Tφ(p)(Rn). There- fore, ∂/∂x1|p,..., ∂/∂xn|pis a basis for TpM. ⊓ ⊔ Proposition 8.10 (Transition matrix for coordinate vectors). Suppose(U,x1,..., xn)and(V,y1,..., yn)are two coordinate charts on a manifold M. Then ∂ ∂xj=∑ i∂yi ∂xj∂ ∂yi on U∩V. Proof. At each point p∈U∩V, the sets{∂/∂xj|p}and{∂/∂yi|p}are both bases for the tangent space TpM, so there is a matrix [ai j(p)]of real numbers such that on U∩V, ∂ ∂xj=∑ kak j∂ ∂yk. Applying both sides of the equation to yi, we get ∂yi ∂xj=∑ kak j∂yi ∂yk =∑ kak jδi k(by Proposition 6.22 ) =ai j. ⊓ ⊔ 8.5 A Local Expression for the Differential 91 8.5 A Local Expression for the Differential Given a smooth map F:N→Mof manifolds and p∈N, let(U,x1,..., xn)be a chart about pinNand let(V,y1,..., ym)be a chart about F(p)inM. We will ﬁnd a local expression for the differential F∗,p:TpN→TF(p)Mrelative to the two charts. By Proposition 8.9, {∂/∂xj|p}n j=1is a basis for TpNand{∂/∂yi|F(p)}m i=1is a basis for TF(p)M. Therefore, the differential F∗=F∗,pis completely determined by the numbers ai jsuch that F∗/parenleftigg ∂ ∂xj/vextendsingle/vextendsingle/vextendsingle/vextendsingle p/parenrightigg =m ∑ k=1ak j∂ ∂yk/vextendsingle/vextendsingle/vextendsingle/vextendsingle F(p),j=1,..., n. Applying both sides to yi, we ﬁnd that ai j=/parenleftigg m ∑ k=1ak j∂ ∂yk/vextendsingle/vextendsingle/vextendsingle/vextendsingle F(p)/parenrightigg yi=F∗/parenleftigg ∂ ∂xj/vextendsingle/vextendsingle/vextendsingle/vextendsingle p/parenrightigg yi=∂ ∂xj/vextendsingle/vextendsingle/vextendsingle/vextendsingle p(yi◦F)=∂Fi ∂xj(p). We state this result as a proposition. Proposition 8.11. Given a smooth map F :N→M of manifolds and a point p ∈N, let(U,x1,..., xn)and(V,y1,..., ym)be coordinate charts about p in N and F (p)in M respectively. Relative to the bases {∂/∂xj|p}for T pN and{∂/∂yi|F(p)}for T F(p)M, the differential F ∗,p:TpN→TF(p)M is represented by the matrix [∂Fi/∂xj(p)], where Fi=yi◦F is the ith component of F. This proposition is in the spirit of the “arrow” approach to tangent vectors. Here each tangent vector in TpNis represented by a column vector relative to the basis {∂/∂xj|p}, and the differential F∗,pis represented by a matrix. Remark 8.12 ( Inverse function theorem ).In terms of the differential, the inverse func- tion theorem for manifolds (Theorem 6.26) has a coordinate-free description: a C∞ map F:N→Mbetween two manifolds of the same dimension is locally invertible at a point p∈Nif and only if its differential F∗,p:TpN→Tf(p)Matpis an isomor- phism. Example 8.13 ( The chain rule in calculus notation ).Suppose w=G(x,y,z)is aC∞ function: R3→Rand(x,y,z)=F(t)is aC∞function: R→R3. Under composition, w=(G◦F)(t)=G(x(t),y(t),z(t)) becomes a C∞function of t∈R. The differentials F∗,G∗, and(G◦F)∗are repre- sented by the matrices  dx/dt dy/dt dz/dt ,/bracketleftbigg∂w ∂x∂w ∂y∂w ∂z/bracketrightbigg ,anddw dt, 92§8 The Tangent Space respectively. Since composition of linear maps is represented by matrix multiplica- tion, in terms of matrices the chain rule (G◦F)∗=G∗◦F∗is equivalent to dw dt=/bracketleftbigg∂w ∂x∂w ∂y∂w ∂z/bracketrightbigg dx/dt dy/dt dz/dt =∂w ∂xdx dt+∂w ∂ydy dt+∂w ∂zdz dt. This is the usual form of the chain rule taught in calculus. 8.6 Curves in a Manifold Asmooth curve in a manifold Mis by deﬁnition a smooth map c:]a,b[→Mfrom some open interval ]a,b[into M. Usually we assume 0 ∈]a,b[and say that cis a curve starting at p ifc(0) =p. The velocity vector c′(t0)of the curve cat time t0∈]a,b[is deﬁned to be c′(t0):=c∗/parenleftigg d dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t0/parenrightigg ∈Tc(t0)M. We also say that c′(t0)is the velocity of cat the point c(t0). Alternative notations for c′(t0)are dc dt(t0)andd dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t0c. NOTATION . When c:]a,b[→Ris a curve with target space R, the notation c′(t) can be a source of confusion. Here tis the standard coordinate on the domain ]a,b[. Letxbe the standard coordinate on the target space R. By our deﬁnition, c′(t)is a tangent vector at c(t), hence a multiple of d/dx|c(t). On the other hand, in calculus notation c′(t)is the derivative of a real-valued function and is therefore a scalar. If it is necessary to distinguish between these two meanings of c′(t)when cmaps into R, we will write ˙ c(t)for the calculus derivative. Exercise 8.14 (Velocity vector versus the calculus derivative). * Letc:]a,b[→Rbe a curve with target space R. Verify that c′(t)=˙c(t)d/dx|c(t). Example. Deﬁne c:R→R2by c(t)=( t2,t3). (See Figure 8.2. ) Then c′(t)is a linear combination of ∂/∂xand∂/∂yatc(t): c′(t)=a∂ ∂x+b∂ ∂y. To compute a, we evaluate both sides on x: 8.6 Curves in a Manifold 93 1 −11 −1xy Fig. 8.2. A cuspidal cubic. a=/parenleftbigg a∂ ∂x+b∂ ∂y/parenrightbigg x=c′(t)x=c∗/parenleftbiggd dt/parenrightbigg x=d dt(x◦c)=d dtt2=2t. Similarly, b=/parenleftbigg a∂ ∂x+b∂ ∂y/parenrightbigg y=c′(t)y=c∗/parenleftbiggd dt/parenrightbigg y=d dt(y◦c)=d dtt3=3t2. Thus, c′(t)=2t∂ ∂x+3t2∂ ∂y. In terms of the basis ∂/∂x|c(t),∂/∂y|c(t)forTc(t)(R2), c′(t)=/bracketleftbigg2t 3t2/bracketrightbigg . More generally, as in this example, to compute the velocity vector of a smooth curve cinRn, one can simply differentiate the components of c. This shows that our deﬁnition of the velocity vector of a curve agrees with the usual deﬁnition in vector calculus. Proposition 8.15 (Velocity of a curve in local coordinates). Let c :]a,b[→M be a smooth curve, and let (U,x1,..., xn)be a coordinate chart about c( t). Write ci=xi◦c for the ith component of c in the chart. Then c′(t)is given by c′(t)=n ∑ i=1˙ci(t)∂ ∂xi/vextendsingle/vextendsingle/vextendsingle/vextendsingle c(t). Thus, relative to the basis {∂/∂xi|p}for T c(t)M, the velocity c′(t)is represented by the column vector ˙c1(t)... ˙cn(t) . Proof. Problem 8.5. ⊓ ⊔ 94§8 The Tangent Space Every smooth curve catpin a manifold Mgives rise to a tangent vector c′(0)in TpM. Conversely, one can show that every tangent vector Xp∈TpMis the velocity vector of some curve at p, as follows. Proposition 8.16 (Existence of a curve with a given initial vector). For any point p in a manifold M and any tangent vector X p∈TpM, there are ε>0and a smooth curve c :]−ε,ε[→M such that c (0)=p and c′(0)=Xp. /Bullet 0/Bulletαφ pXp=∑ai∂ ∂xi/vextendsingle/vextendsingle/vextendsingle/vextendsingle p∑ai∂ ∂ri/vextendsingle/vextendsingle/vextendsingle/vextendsingle 0r2 r1 U Fig. 8.3. Existence of a curve through a point with a given initial vector. Proof. Let(U,φ) = ( U,x1,..., xn)be a chart centered at p; i.e., φ(p) =0∈Rn. Suppose Xp=∑ai∂/∂xi|patp. Let r1,..., rnbe the standard coordinates on Rn. Then xi=ri◦φ. To ﬁnd a curve catpwith c′(0)=Xp, start with a curve αinRn with α(0)=0andα′(0)=∑ai∂/∂ri|0. We then map αtoMviaφ−1(Figure 8.3 ). By Proposition 8.15, the simplest such αis α(t)=( a1t,..., ant),t∈]−ε,ε[, where εis sufﬁciently small that α(t)lies in φ(U). Deﬁne c=φ−1◦α:]−ε,ε[→ M. Then c(0)=φ−1(α(0))= φ−1(0)=p, and by Proposition 8.8, c′(0)=( φ−1)∗α∗/parenleftbiggd dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0/parenrightbigg =(φ−1)∗/parenleftbigg ∑ai∂ ∂ri/vextendsingle/vextendsingle/vextendsingle/vextendsingle 0/parenrightbigg =∑ai∂ ∂xi/vextendsingle/vextendsingle/vextendsingle/vextendsingle p=Xp.⊓ ⊔ In Deﬁnition 8.1 we deﬁned a tangent vector at a point pof a manifold abstractly as a derivation at p. Using curves, we can now interpret a tangent vector geometri- cally as a directional derivative. Proposition 8.17. Suppose X pis a tangent vector at a point p of a manifold M and f∈C∞ p(M). If c :]−ε,ε[→M is a smooth curve starting at p with c′(0)=Xp, then Xpf=d dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle 0(f◦c). 8.7 Computing the Differential Using Curves 95 Proof. By the deﬁnitions of c′(0)andc∗, Xpf=c′(0)f=c∗/parenleftbiggd dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle 0/parenrightbigg f=d dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle 0(f◦c).⊓ ⊔ 8.7 Computing the Differential Using Curves We have introduced two ways of computing the differential of a smooth map, in terms of derivations at a point (equation (8.1)) and in terms of local coordinates (Proposition 8.11). The next proposition gives still another way of computing the differential F∗,p, this time using curves. Proposition 8.18. Let F :N→M be a smooth map of manifolds, p ∈N, and X p∈ TpN. If c is a smooth curve starting at p in N with velocity X pat p, then F∗,p(Xp)=d dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle 0(F◦c)(t). In other words, F ∗,p(Xp)is the velocity vector of the image curve F ◦c at F(p). Proof. By hypothesis, c(0)=pandc′(0)=Xp. Then F∗,p(Xp)=F∗,p(c′(0)) =(F∗,p◦c∗,0)/parenleftbiggd dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle 0/parenrightbigg =(F◦c)∗,0/parenleftbiggd dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle 0/parenrightbigg (by the chain rule, Theorem 8.5 ) =d dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle 0(F◦c)(t). ⊓ ⊔ Example 8.19 ( Differential of left multiplication ).Ifgis a matrix in the general linear group GL (n,R), letℓg: GL(n,R)→GL(n,R)be left multiplication by g; thus, ℓg(B) =gBfor any B∈GL(n,R). Since GL (n,R)is an open subset of the vector spaceRn×n, the tangent space Tg(GL(n,R))can be identiﬁed with Rn×n. Show that with this identiﬁcation the differential (ℓg)∗,I:TI(GL(n,R))→Tg(GL(n,R))is also left multiplication by g. Solution. LetX∈TI(GL(n,R))=Rn×n. To compute (ℓg)∗,I(X), choose a curve c(t) in GL(n,R)with c(0) =Iandc′(0) =X. Thenℓg(c(t)) = gc(t)is simply matrix multiplication. By Proposition 8.18, (ℓg)∗,I(X)=d dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0ℓg(c(t))=d dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0gc(t)=gc′(0)=gX. In this computation, d/dt|t=0gc(t)=gc′(0)byR-linearity and Proposition 8.15. ⊓ ⊔ 96§8 The Tangent Space 8.8 Immersions and Submersions Just as the derivative of a map between Euclidean spaces is a linear map that best approximates the given map at a point, so the differential at a point serves the same purpose for a C∞map between manifolds. Two cases are especially impor- tant. A C∞map F:N→Mis said to be an immersion at p∈Nif its differential F∗,p:TpN→TF(p)Mis injective, and a submersion at p ifF∗,pis surjective. We callFanimmersion if it is an immersion at every p∈Nand a submersion if it is a submersion at every p∈N. Remark 8.20.Suppose Nand Mare manifolds of dimensions nand mrespec- tively. Then dim TpN=nand dim TF(p)M=m. The injectivity of the differential F∗,p:TpN→TF(p)Mimplies immediately that n≤m. Similarly, the surjectivity of the differential F∗,pimplies that n≥m. Thus, if F:N→Mis an immersion at a point of N, then n≤mand if Fis a submersion a point of N, then n≥m. Example 8.21.The prototype of an immersion is the inclusion of Rnin a higher- dimensional Rm: i(x1,..., xn)=( x1,..., xn,0,..., 0). The prototype of a submersion is the projection of Rnonto a lower-dimensional Rm: π(x1,..., xm,xm+1,..., xn)=( x1,..., xm). Example. IfUis an open subset of a manifold M, then the inclusion i:U→Mis both an immersion and a submersion. This example shows in particular that a submersion need not be onto. In Section 11, we will undertake a more in-depth analysis of immersions and submersions. According to the immersion and submersion theorems to be proven there, every immersion is locally an inclusion and every submersion is locally aprojection. 8.9 Rank, and Critical and Regular Points The rank of a linear transformation L:V→Wbetween ﬁnite-dimensional vector spaces is the dimension of the image L(V)as a subspace of W, while the rank of a matrix Ais the dimension of its column space. If Lis represented by a matrix A relative to a basis for Vand a basis for W, then the rank of Lis the same as the rank ofA, because the image L(V)is simply the column space of A. Now consider a smooth map F:N→Mof manifolds. Its rank at a point pin N, denoted by rk F(p), is deﬁned as the rank of the differential F∗,p:TpN→TF(p)M. Relative to the coordinate neighborhoods (U,x1,..., xn)atpand(V,y1,..., ym)at F(p), the differential is represented by the Jacobian matrix [∂Fi/∂xj(p)](Proposi- tion 8.11), so 8.9 Rank, and Critical and Regular Points 97 rkF(p)=rk/bracketleftbigg∂Fi ∂xj(p)/bracketrightbigg . Since the differential of a map is independent of coordinate charts, so is the rank of a Jacobian matrix. Deﬁnition 8.22. A point pinNis acritical point ofFif the differential F∗,p:TpN→TF(p)M fails to be surjective. It is a regular point ofFif the differential F∗,pis surjective. In other words, pis a regular point of the map Fif and only if Fis a submersion at p. A point in Mis acritical value if it is the image of a critical point; otherwise it is a regular value (Figure 8.4 ). /Bullet/Bullet/Bullet/Bullet /Bullet ×××× ×=critical points =critical valuesf N=torus M=R Fig. 8.4. Critical points and critical values of the function f(x,y,z)=z. Two aspects of this deﬁnition merit elaboration: (i) We do notdeﬁne a regular value to be the image of a regular point. In fact, a regular value need not be in the image of Fat all. Any point of Mnot in the image of Fis automatically a regular value because it is not the image of a critical point. (ii) A point cinMis a critical value if and only if some point in the preimage F−1({c})is a critical point. A point cin the image of Fis a regular value if and only if every point in the preimage F−1({c})is a regular point. Proposition 8.23. For a real-valued function f :M→R, a point p in M is a critical point if and only if relative to some chart (U,x1,..., xn)containing p, all the partial derivatives satisfy ∂f ∂xj(p)=0,j=1,..., n. Proof. By Proposition 8.11 the differential f∗,p:TpM→Tf(p)R≃Ris represented by the matrix 98§8 The Tangent Space /bracketleftbigg ∂f ∂x1(p)···∂f ∂xn(p)/bracketrightbigg . Since the image of f∗,pis a linear subspace of R, it is either zero-dimensional or one-dimensional. In other words, f∗,pis either the zero map or a surjective map. Therefore, f∗,pfails to be surjective if and only if all the partial derivatives ∂f/∂xi(p) are zero. ⊓ ⊔ Problems 8.1.* Differential of a map LetF:R2→R3be the map (u,v,w)=F(x,y)=(x,y,xy). Letp=(x,y)∈R2. Compute F∗(∂/∂x|p)as a linear combination of ∂/∂u,∂/∂v, and ∂/∂w atF(p). 8.2. Differential of a linear map LetL:Rn→Rmbe a linear map. For any p∈Rn, there is a canonical identiﬁcation Tp(Rn)∼→ Rngiven by ∑ai∂ ∂xi/vextendsingle/vextendsingle/vextendsingle/vextendsingle p/ma√s⊔o→a=/an}bracke⊔le{⊔a1,..., an/an}bracke⊔ri}h⊔. Show that the differential L∗,p:Tp(Rn)→TL(p)(Rm)is the map L:Rn→Rmitself, with the identiﬁcation of the tangent spaces as above. 8.3. Differential of a map Fix a real number αand deﬁne F:R2→R2by /bracketleftbiggu v/bracketrightbigg =(u,v)=F(x,y)=/bracketleftbiggcosα−sinα sinα cosα/bracketrightbigg/bracketleftbiggx y/bracketrightbigg . LetX=−y∂/∂x+x∂/∂ybe a vector ﬁeld on R2. Ifp=(x,y)∈R2andF∗(Xp)=(a ∂/∂u+ b∂/∂v)|F(p), ﬁnd aandbin terms of x,y, and α. 8.4. Transition matrix for coordinate vectors Letx,ybe the standard coordinates on R2, and let Ube the open set U=R2−{(x,0)|x≥0}. OnUthe polar coordinates r,θare uniquely deﬁned by x=rcosθ, y=rsinθ,r>0,0<θ<2π. Find ∂/∂rand∂/∂θin terms of ∂/∂xand∂/∂y. 8.5.* Velocity of a curve in local coordinates Prove Proposition 8.15. 8.9 Rank, and Critical and Regular Points 99 8.6. Velocity vector Letp=(x,y)be a point in R2. Then cp(t)=/bracketleftbiggcos2 t−sin2t sin2tcos2 t/bracketrightbigg/bracketleftbiggx y/bracketrightbigg ,t∈R, is a curve with initial point pinR2. Compute the velocity vector c′p(0). 8.7.* Tangent space to a product IfMandNare manifolds, let π1:M×N→Mandπ2:M×N→Nbe the two projections. Prove that for (p,q)∈M×N, (π1∗,π2∗):T(p,q)(M×N)→TpM×TqN is an isomorphism. 8.8. Differentials of multiplication and inverse LetGbe a Lie group with multiplication map µ:G×G→G, inverse map ι:G→G, and identity element e. (a) Show that the differential at the identity of the multiplication map µis addition: µ∗,(e,e):TeG×TeG→TeG, µ∗,(e,e)(Xe,Ye)=Xe+Ye. (Hint: First, compute µ∗,(e,e)(Xe,0)andµ∗,(e,e)(0,Ye)using Proposition 8.18.) (b) Show that the differential at the identity of ιis the negative: ι∗,e:TeG→TeG, ι∗,e(Xe)=−Xe. (Hint: Take the differential of µ(c(t),(ι◦c)(t))= e.) 8.9.* Transforming vectors to coordinate vectors LetX1,..., Xnbenvector ﬁelds on an open subset Uof a manifold of dimension n. Suppose that at p∈U, the vectors (X1)p,...,(Xn)pare linearly independent. Show that there is a chart (V,x1,..., xn)about psuch that (Xi)p=(∂/∂xi)pfori=1,..., n. 8.10. Local maxima A real-valued function f:M→Ron a manifold is said to have a local maximum atp∈Mif there is a neighborhood Uofpsuch that f(p)≥f(q)for all q∈U. (a)*Prove that if a differentiable function f:I→Rdeﬁned on an open interval Ihas a local maximum at p∈I, then f′(p)=0. (b) Prove that a local maximum of a C∞function f:M→Ris a critical point of f. (Hint: LetXpbe a tangent vector in TpMand let c(t)be a curve in Mstarting at pwith initial vector Xp. Then f◦cis a real-valued function with a local maximum at 0. Apply (a).) 100§9 Submanifolds §9 Submanifolds We now have two ways of showing that a given topological space is a manifold: (a) by checking directly that the space is Hausdorff, second countable, and has a C∞ atlas; (b) by exhibiting it as an appropriate quotient space. Section 7 lists some conditions under which a quotient space is a manifold. In this section we introduce the concept of a regular submanifold of a manifold, a subset that is locally deﬁned by the vanishing of some of the coordinate functions. Using the inverse function theorem, we derive a criterion, called the regular level set theorem, that can often be used to show that a level set of a C∞map of manifolds is a regular submanifold and therefore a manifold. Although the regular level set theorem is a simple consequence of the constant rank theorem and the submersion theorem to be discussed in Section 11, deducing it directly from the inverse function theorem has the advantage of producing explicitcoordinate functions on the submanifold. 9.1 Submanifolds The xy-plane in R3is the prototype of a regular submanifold of a manifold. It is deﬁned by the vanishing of the coordinate function z. Deﬁnition 9.1. A subset Sof a manifold Nof dimension nis a regular sub- manifold of dimension kif for every p∈Sthere is a coordinate neighborhood (U,φ) = ( U,x1,..., xn)ofpin the maximal atlas of Nsuch that U∩Sis deﬁned by the vanishing of n−kof the coordinate functions. By renumbering the coordi- nates, we may assume that these n−kcoordinate functions are xk+1,..., xn. We call such a chart (U,φ)inNanadapted chart relative to S. On U∩S,φ= (x1,..., xk,0,..., 0). Let φS:U∩S→Rk be the restriction of the ﬁrst kcomponents of φtoU∩S, that is, φS= (x1,..., xk). Note that (U∩S,φS)is a chart for Sin the subspace topology. Deﬁnition 9.2. IfSis a regular submanifold of dimension kin a manifold Nof dimension n, then n−kis said to be the codimension ofSinN. Remark. As a topological space, a regular submanifold of Nis required to have the subspace topology. Example. In the deﬁnition of a regular submanifold, the dimension kof the subman- ifold may be equal to n, the dimension of the manifold. In this case, U∩Sis deﬁned 9.1 Submanifolds 101 by the vanishing of none of the coordinate functions and so U∩S=U. Therefore, an open subset of a manifold is a regular submanifold of the same dimension. Remark. There are other types of submanifolds, but unless otherwise speciﬁed, by a “submanifold” we will always mean a “regular submanifold.” Example. The interval S:= ]−1,1[on the x-axis is a regular submanifold of the xy-plane (Figure 9.1 ). As an adapted chart, we can take the open square U=]−1,1[ ×]−1,1[with coordinates x,y. Then U∩Sis precisely the zero set of yonU. Vis not an adapted chart. Uis an adapted chart.U V −1 1 −1 1 Fig. 9.1. Note that if V= ]−2,0[×]−1,1[, then(V,x,y)is not an adapted chart relative toS, since V∩Sis the open interval ]−1,0[on the x-axis, while the zero set of yon Vis the open interval ]−2,0[on the x-axis. 0.2 0.4 0.61 −1xy /Bullet /Bullet /Bullet /Bullet /Bullet /Bullet Fig. 9.2. The topologist’s sine curve. Example 9.3.LetΓbe the graph of the function f(x)=sin(1/x)on the interval ]0,1[, and let Sbe the union of Γand the open interval I={(0,y)∈R2|−1<y<1}. The subset SofR2is not a regular submanifold for the following reason: if pis in the interval I, then there is no adapted chart containing p, since any sufﬁciently small neighborhood UofpinR2intersects Sin inﬁnitely many components. (The 102§9 Submanifolds closure of ΓinR2is called the topologist’s sine curve (Figure 9.2 ). It differs from S in including the endpoints (1,sin1),(0,1), and(0,−1).) Proposition 9.4. Let S be a regular submanifold of N and U={(U,φ)}a collection of compatible adapted charts of N that covers S. Then {(U∩S,φS)}is an atlas for S. Therefore, a regular submanifold is itself a manifold. If N has dimension n and S is locally deﬁned by the vanishing of n −k coordinates, then dimS=k. UV S φψ Fig. 9.3. Overlapping adapted charts relative to a regular submanifold S. Proof. Let(U,φ)=( U,x1,..., xn)and(V,ψ)=( V,y1,..., yn)be two adapted charts in the given collection (Figure 9.3 ). Assume that they intersect. As we remarked in Deﬁnition 9.1, in any adapted chart relative to a submanifold Sit is possible to renumber the coordinates so that the last n−kcoordinates vanish on points of S. Then for p∈U∩V∩S, φ(p)=( x1,..., xk,0,..., 0)and ψ(p)=( y1,..., yk,0,..., 0), so φS(p)=( x1,..., xk)and ψS(p)=( y1,..., yk). Therefore,/parenleftbig ψS◦φ−1 S/parenrightbig (x1,..., xk)=( y1,..., yk). Since y1,..., ykareC∞functions of x1,..., xk(because ψ◦φ−1(x1,..., xk,0,..., 0) isC∞), the transition function ψS◦φ−1 SisC∞. Similarly, since x1,..., xkareC∞ functions of y1,..., yk,φS◦ψ−1 Sis also C∞. Hence, any two charts in {(U∩S,φS)} areC∞compatible. Since {U∩S}U∈Ucovers S, the collection{(U∩S,φS)}is aC∞ atlas on S. ⊓ ⊔ 9.2 Level Sets of a Function 103 9.2 Level Sets of a Function Alevel set of a map F:N→Mis a subset F−1({c})={p∈N|F(p)=c} for some c∈M. The usual notation for a level set is F−1(c), rather than the more correct F−1({c}). The value c∈Mis called the level of the level set F−1(c). If F:N→Rm, then Z(F):=F−1(0)is the zero set ofF. Recall that cis a regular value ofFif and only if either cis not in the image of For at every point p∈F−1(c), the differential F∗,p:TpN→TF(p)Mis surjective. The inverse image F−1(c)of a regular value cis called a regular level set . If the zero set F−1(0)is a regular level set of F:N→Rm, it is called a regular zero set . Remark 9.5.If a regular level set F−1(c)is nonempty, say p∈F−1(c), then the map F:N→Mis a submersion at p. By Remark 8.20, dim N≥dimM. Example 9.6 ( The2-sphere in R3).The unit 2-sphere S2={(x,y,z)∈R3|x2+y2+z2=1} is the level set g−1(1)of level 1 of the function g(x,y,z)=x2+y2+z2. We will use the inverse function theorem to ﬁnd adapted charts of R3that cover S2. As the proof will show, the process is easier for a zero set, mainly because a regular submanifold is deﬁned locally as the zero set of coordinate functions. To express S2as a zero set, we rewrite its deﬁning equation as f(x,y,z) =x2+y2+z2−1=0. Then S2=f−1(0). Since ∂f ∂x=2x,∂f ∂y=2y,∂f ∂z=2z, the only critical point of fis(0,0,0), which does not lie on the sphere S2. Thus, all points on the sphere are regular points of fand 0 is a regular value of f. Letpbe a point of S2at which (∂f/∂x)(p) =2x(p)/ne}a⊔ionslash=0. Then the Jacobian matrix of the map (f,y,z):R3→R3is  ∂f ∂x∂f ∂y∂f ∂z ∂y ∂x∂y ∂y∂y ∂z ∂z ∂x∂z ∂y∂z ∂z = ∂f ∂x∂f ∂y∂f ∂z 0 1 0 0 0 1 , and the Jacobian determinant ∂f/∂x(p)is nonzero. By Corollary 6.27 of the inverse function theorem (Theorem 6.26), there is a neighborhood UpofpinR3such that 104§9 Submanifolds (Up,f,y,z)is a chart in the atlas of R3. In this chart, the set Up∩S2is deﬁned by the vanishing of the ﬁrst coordinate f. Thus,(Up,f,y,z)is an adapted chart relative to S2, and(Up∩S2,y,z)is a chart for S2. Similarly, if (∂f/∂y)(p)/ne}a⊔ionslash=0, then there is an adapted chart (Vp,x,f,z)con- taining pin which the set Vp∩S2is the zero set of the second coordinate f. If (∂f/∂z)(p)/ne}a⊔ionslash=0, then there is an adapted chart (Wp,x,y,f)containing p. Since for every p∈S2, at least one of the partial derivatives ∂f/∂x(p),∂f/∂y(p),∂f/∂z(p) is nonzero, as pvaries over all points of the sphere we obtain a collection of adapted charts ofR3covering S2. Therefore, S2is a regular submanifold of R3. By Proposi- tion 9.4, S2is a manifold of dimension 2. This is an important example because one can generalize its proof almost ver- batim to prove that if the zero set of a function f:N→Ris a regular level set, then it is a regular submanifold of N. The idea is that in a coordinate neighborhood (U,x1,..., xn)if a partial derivative ∂f/∂xi(p)is nonzero, then we can replace the coordinate xibyf. First we show that any regular level set g−1(c)of a C∞real function gon a manifold can be expressed as a regular zero set. Lemma 9.7. Let g :N→Rbe a C∞function. A regular level set g−1(c)of level c of the function g is the regular zero set f−1(0)of the function f =g−c. Proof. For any p∈N, g(p)=c⇐⇒ f(p)=g(p)−c=0. Hence, g−1(c)=f−1(0). Call this set S. Because the differential f∗,pequals g∗,pat every point p∈N, the functions fandghave exactly the same critical points. Since ghas no critical points in S, neither does f. ⊓ ⊔ Theorem 9.8. Let g :N→Rbe a C∞function on the manifold N. Then a nonempty regular level set S =g−1(c)is a regular submanifold of N of codimension 1. Proof. Letf=g−c. By the preceding lemma, Sequals f−1(0)and is a regular level set of f. Let p∈S. Since pis a regular point of f, relative to any chart (U,x1,..., xn) about p,(∂f/∂xi)(p)/ne}a⊔ionslash=0 for some i. By renumbering x1,..., xn, we may assume that(∂f/∂x1)(p)/ne}a⊔ionslash=0. The Jacobian matrix of the C∞map(f,x2,..., xn):U→Rnis  ∂f ∂x1∂f ∂x2···∂f ∂xn ∂x2 ∂x1∂x2 ∂x2···∂x2 ∂xn ............ ∂xn ∂x1∂xn ∂x2···∂xn ∂xn = ∂f ∂x1∗···∗ 0 1···0 ............ 0 0···1 . 9.3 The Regular Level Set Theorem 105 So the Jacobian determinant ∂(f,x2,..., xn)/∂(x1,x2,..., xn)atpis∂f/∂x1(p)/ne}a⊔ionslash=0. By the inverse function theorem (Corollary 6.27), there is a neighborhood Upofpon which f,x2,..., xnform a coordinate system. Relative to the chart (Up,f,x2,..., xn) the level set Up∩Sis deﬁned by setting the ﬁrst coordinate fequal to 0, so (Up,f,x2,..., xn)is an adapted chart relative to S. Since pwas arbitrary, Sis a regular submanifold of dimension n−1 inN. ⊓ ⊔ 9.3 The Regular Level Set Theorem The next step is to extend Theorem 9.8 to a regular level set of a map between smooth manifolds. This very useful theorem does not seem to have an agreed-upon name in the literature. It is known variously as the implicit function theorem, the preimage theorem [17], and the regular level set theorem [25], among other nomenclatures. We will follow [25] and call it the regular level set theorem. Theorem 9.9 (Regular level set theorem). Let F :N→M be a C∞map of man- ifolds, with dimN=n and dimM=m. Then a nonempty regular level set F−1(c), where c∈M, is a regular submanifold of N of dimension equal to n −m. Proof. Choose a chart (V,ψ) = ( V,y1,..., ym)ofMcentered at c, i.e., such that ψ(c)=0inRm. Then F−1(V)is an open set in Nthat contains F−1(c). Moreover, inF−1(V),F−1(c)=( ψ◦F)−1(0). So the level set F−1(c)is the zero set of ψ◦F. IfFi=yi◦F=ri◦(ψ◦F), then F−1(c)is also the common zero set of the functions F1,..., FmonF−1(V). pU F−1(c)F−1(V) F ψV N M Rnc 0ψ(V) /Bullet /Bullet /Bullet Fig. 9.4. The level set F−1(c)ofFis the zero set of ψ◦F. Because the regular level set is assumed nonempty, n≥m(Remark 9.5). Fix a point p∈F−1(c)and let(U,φ) = ( U,x1,..., xn)be a coordinate neighborhood of pinNcontained in F−1(V)(Figure 9.4 ). Since F−1(c)is a regular level set, p∈ F−1(c)is a regular point of F. Therefore, the m×nJacobian matrix [∂Fi/∂xj(p)] has rank m. By renumbering the Fiandxj’s, we may assume that the ﬁrst m×m block[∂Fi/∂xj(p)]1≤i,j≤mis nonsingular. Replace the ﬁrst mcoordinates x1,..., xmof the chart (U,φ)byF1,..., Fm. We claim that there is a neighborhood Upofpsuch that (Up,F1,..., Fm,xm+1,..., xn)is a chart in the atlas of N. It sufﬁces to compute its Jacobian matrix at p: 106§9 Submanifolds  ∂Fi ∂xj∂Fi ∂xβ ∂xα ∂xj∂xα ∂xβ = ∂Fi ∂xj∗ 0 I , where 1≤i,j≤mandm+1≤α,β≤n. Since this matrix has determinant det/bracketleftbigg∂Fi ∂xj(p)/bracketrightbigg 1≤i,j≤m/ne}a⊔ionslash=0, the inverse function theorem in the form of Corollary 6.27 implies the claim. In the chart (Up,F1,..., Fm,xm+1,..., xn), the set S:=f−1(c)is obtained by setting the ﬁrst mcoordinate functions F1,..., Fmequal to 0. So (Up,F1,..., Fm, xm+1,...,xn)is an adapted chart for Nrelative to S. Since this is true about every point p∈S,Sis a regular submanifold of Nof dimension n−m.⊓ ⊔ The proof of the regular level set theorem gives the following useful lemma. Lemma 9.10. Let F :N→Rmbe a C∞map on a manifold N of dimension n and let S be the level set F−1(0). If relative to some coordinate chart (U,x1,..., xn)about p∈S, the Jacobian determinant ∂(F1,..., Fm)/∂(xj1,..., xjm)(p)is nonzero, then in some neighborhood of p one may replace xj1,..., xjmby F1,..., Fmto obtain an adapted chart for N relative to S. Remark. The regular level set theorem gives a sufﬁcient but not necessary condition for a level set to be a regular submanifold. For example, if f:R2→Ris the map f(x,y) =y2, then the zero set Z(f) =Z(y2)is the x-axis, a regular submanifold of R2. However, since ∂f/∂x=0 and ∂f/∂y=2y=0 on the x-axis, every point in Z(f)is a critical point of f. Thus, although Z(f)is a regular submanifold of R2, it is not a regular level set of f. 9.4 Examples of Regular Submanifolds Example 9.11 ( Hypersurface ).Show that the solution set Sofx3+y3+z3=1 inR3 is a manifold of dimension 2. Solution. Let f(x,y,z) =x3+y3+z3. Then S=f−1(1). Since ∂f/∂x=3x2, ∂f/∂y=3y2, and ∂f/∂z=3z2, the only critical point of fis(0,0,0), which is not in S. Thus, 1 is a regular value of f:R3→R. By the regular level set theorem (Theorem 9.9), Sis a regular submanifold of R3of dimension 2. So Sis a manifold (Proposition 9.4). ⊓ ⊔ Example 9.12 ( Solution set of two polynomial equations ).Decide whether the subset SofR3deﬁned by the two equations x3+y3+z3=1, x+y+z=0 is a regular submanifold of R3. 9.4 Examples of Regular Submanifolds 107 Solution. Deﬁne F:R3→R2by (u,v)=F(x,y,z)=( x3+y3+z3,x+y+z). Then Sis the level set F−1(1,0). The Jacobian matrix of Fis J(F)=/bracketleftbigguxuyuz vxvyvz/bracketrightbigg =/bracketleftbigg 3x23y23z2 1 1 1/bracketrightbigg , where ux=∂u/∂xand so forth. The critical points of Fare the points (x,y,z)where the matrix J(F)has rank <2. That is precisely where all 2 ×2 minors of J(F) are zero: /vextendsingle/vextendsingle/vextendsingle/vextendsingle3x23y2 1 1/vextendsingle/vextendsingle/vextendsingle/vextendsingle=0,/vextendsingle/vextendsingle/vextendsingle/vextendsingle3x 23z2 1 1/vextendsingle/vextendsingle/vextendsingle/vextendsingle=0. (9.1) (The third condition /vextendsingle/vextendsingle/vextendsingle/vextendsingle3y 23z2 1 1/vextendsingle/vextendsingle/vextendsingle/vextendsingle=0 is a consequence of these two.) Solving (9.1), we get y=±x,z=±x. Since x+y+ z=0 on S, this implies that (x,y,z) = ( 0,0,0). Since(0,0,0)does not satisfy the ﬁrst equation x 3+y3+z3=1, there are no critical points of FonS. Therefore, Sis a regular level set. By the regular level set theorem, Sis a regular submanifold of R3 of dimension 1. ⊓ ⊔ Example 9.13 ( Special linear group ).As a set, the special linear group SL(n,R)is the subset of GL (n,R)consisting of matrices of determinant 1. Since det(AB)=( detA)(detB)and det (A−1)=1 detA, SL(n,R)is a subgroup of GL (n,R). To show that it is a regular submanifold, we let f: GL(n,R)→Rbe the determinant map f(A)=detA, and apply the regular level set theorem to f−1(1)=SL(n,R). We need to check that 1 is a regular value of f. Letai j,1≤i≤n, 1≤j≤n, be the standard coordinates on Rn×n, and let Si j denote the submatrix of A= [a i j]∈Rn×nobtained by deleting its ith row and jth column. Then mi j:=detSi jis the(i,j)-minor ofA. From linear algebra we have a formula for computing the determinant by expanding along any row or any column: if we expand along the ith row, we obtain f(A)=detA=(−1)i+1ai1mi1+(−1)i+2ai2mi2+···+(−1)i+nainmin. (9.2) Therefore ∂f ∂ai j=(−1)i+jmi j. Hence, a matrix A∈GL(n,R)is a critical point of fif and only if all the (n− 1)×(n−1)minors mi jofAare 0. By (9.2) such a matrix Ahas determinant 0. Since every matrix in SL (n,R)has determinant 1, all the matrices in SL (n,R)are regular points of the determinant function. By the regular level set theorem (Theorem 9.9),SL(n,R)is a regular submanifold of GL (n,R)of codimension 1; i.e., dimSL(n,R)=dimGL(n,R)−1=n 2−1. 108§9 Submanifolds Problems 9.1. Regular values Deﬁne f:R2→Rby f(x,y)=x3−6xy+y2. Find all values c∈Rfor which the level set f−1(c)is a regular submanifold of R2. 9.2. Solution set of one equation Letx,y,z,wbe the standard coordinates on R4. Is the solution set of x5+y5+z5+w5=1 inR4a smooth manifold? Explain why or why not. (Assume that the subset is given the subspace topology.) 9.3. Solution set of two equations Is the solution set of the system of equations x3+y3+z3=1,z=xy, inR3a smooth manifold? Prove your answer. 9.4.* Regular submanifolds Suppose that a subset SofR2has the property that locally on Sone of the coordinates is a C∞function of the other coordinate. Show that Sis a regular submanifold of R2. (Note that the unit circle deﬁned by x2+y2=1 has this property. At every point of the circle, there is a neighborhood in which yis aC∞function of xorxis aC∞function of y.) 9.5. Graph of a smooth function Show that the graph Γ(f)of a smooth function f:R2→R, Γ(f)={(x,y,f(x,y))∈R3}, is a regular submanifold of R3. 9.6. Euler’s formula A polynomial F(x0,..., xn)∈R[x0,..., xn]ishomogeneous of degree k if it is a linear com- bination of monomials xi0 0···xinnof degree ∑n j=0ij=k. Let F(x0,..., xn)be a homogeneous polynomial of degree k. Clearly, for any t∈R, F(tx0,..., txn)=tkF(x0,..., xn). (9.3) Show thatn ∑ i=0xi∂F ∂xi=kF. 9.7. Smooth projective hypersurface On the projective space RPna homogeneous polynomial F(x0,..., xn)of degree kis not a function, since its value at a point [a0,..., an]is not unique. However, the zero set in RPnof a homogeneous polynomial F(x0,..., xn)is well deﬁned, since F(a0,..., an)=0 if and only if F(ta0,..., tan)=tkF(a0,..., an)=0 for all t∈R×:=R−{0}. The zero set of ﬁnitely many homogeneous polynomials in RPnis called a real projective variety . A projective variety deﬁned by a single homogeneous polynomial of degree kis called 9.4 Examples of Regular Submanifolds 109 ahypersurface of degree k. Show that the hypersurface Z(F)deﬁned by F(x0,x1,x2) =0 is smooth if ∂F/∂x0,∂F/∂x1, and ∂F/∂x2are not simultaneously zero on Z(F). (Hint: The standard coordinates on U0, which is homeomorphic to R2, are x=x1/x0,y=x2/x0(see Subsection 7.7). In U0,F(x0,x1,x2) =xk 0F(1,x1/x0,x2/x0) =xk 0F(1,x,y). Deﬁne f(x,y) = F(1,x,y). Then fandFhave the same zero set in U0.) 9.8. Product of regular submanifolds IfSiis a regular submanifold of the manifold Mifori=1,2, prove that S1×S2is a regular submanifold of M1×M2. 9.9. Complex special linear group Thecomplex special linear group SL(n,C)is the subgroup of GL (n,C)consisting of n×n complex matrices of determinant 1. Show that SL (n,C)is a regular submanifold of GL (n,C) and determine its dimension. (This problem requires a rudimentary knowledge of complex analysis.) S Sf(N)f(N) ftransversal to SinR2fnot transversal to SinR2 Fig. 9.5. Transversality. 9.10. The transversality theorem AC∞map f:N→Mis said to be transversal to a submanifold S⊂M(Figure 9.5 ) if for every p∈f−1(S), f∗(TpN)+Tf(p)S=Tf(p)M. (9.4) (IfAandBare subspaces of a vector space, their sum A+Bis the subspace consisting of all a+bwith a∈Aandb∈B. The sum need not be a direct sum.) The goal of this exercise is to prove the transversality theorem: if a C∞map f:N→Mis transversal to a regular submanifold Sof codimension kinM, then f−1(S)is a regular submanifold of codimension k inN. When Sconsists of a single point c, transversality of ftoSsimply means that f−1(c) is a regular level set. Thus the transversality theorem is a generalization of the regular level set theorem. It is especially useful in giving conditions under which the intersection of twosubmanifolds is a submanifold. Letp∈f −1(S)and(U,x1,..., xm)be an adapted chart centered at f(p)forMrelative to Ssuch that U∩S=Z(xm−k+1,..., xm), the zero set of the functions xm−k+1,..., xm. Deﬁne g:U→Rkto be the map g=(xm−k+1,..., xm). (a) Show that f−1(U)∩f−1(S)=(g◦f)−1(0). (b) Show that f−1(U)∩f−1(S)is a regular level set of the function g◦f:f−1(U)→Rk. (c) Prove the transversality theorem. 110§10 Categories and Functors §10 Categories and Functors Many of the problems in mathematics share common features. For example, in topol- ogy one is interested in knowing whether two topological spaces are homeomorphic and in group theory one is interested in knowing whether two groups are isomorphic. This has given rise to the theory of categories and functors, which tries to clarify thestructural similarities among different areas of mathematics. A category is essentially a collection of objects and arrows between objects. These arrows, called morphisms, satisfy the abstract properties of maps and are of-ten structure-preserving maps. Smooth manifolds and smooth maps form a category, and so do vector spaces and linear maps. A functor from one category to another preserves the identity morphism and the composition of morphisms. It provides away to simplify problems in the ﬁrst category, for the target category of a functor is usually simpler than the original category. The tangent space construction with the differential of a smooth map is a functor from the category of smooth manifolds with a distinguished point to the category of vector spaces. The existence of the tangent space functor shows that if two manifolds are diffeomorphic, then their tan-gent spaces at corresponding points must be isomorphic, thereby proving the smooth invariance of dimension. Invariance of dimension in the continuous category of topo- logical spaces and continuous maps is more difﬁcult to prove, precisely because thereis no tangent space functor in the continuous category. Much of algebraic topology is the study of functors, for example, the homology, cohomology, and homotopy functors. For a functor to be truly useful, it should be simple enough to be computable, yet complex enough to preserve essential features of the original category. For smooth manifolds, this delicate balance is achieved in the de Rham cohomology functor. In the rest of the book, we will be introducingvarious functors of smooth manifolds, such as the tangent bundle and differential forms, culminating in de Rham cohomology. In this section, after deﬁning categories and functors, we study the dual construc- tion on vector spaces as a nontrivial example of a functor. 10.1 Categories Acategory consists of a collection of elements, called objects , and for any two ob- jects AandB, a set Mor (A,B)of elements, called morphisms from AtoB, such that given any morphism f∈Mor(A,B)and any morphism g∈Mor(B,C), the composite g◦f∈Mor(A,C)is deﬁned. Furthermore, the composition of morphisms is required to satisfy two properties: (i) the identity axiom: for each object A, there is an identity morphism 1A∈ Mor(A,A)such that for any f∈Mor(A,B)andg∈Mor(B,A), f◦1A=fand1A◦g=g; 10.2 Functors 111 (ii) the associative axiom: for f∈Mor(A,B),g∈Mor(B,C), and h∈Mor(C,D), h◦(g◦f)=( h◦g)◦f. Iff∈Mor(A,B), we often write f:A→B. Example. The collection of groups and group homomorphisms forms a category in which the objects are groups and for any two groups AandB, Mor(A,B)is the set of group homomorphisms from AtoB. Example. The collection of all vector spaces over RandR-linear maps forms a category in which the objects are real vector spaces and for any two real vector spaces VandW, Mor(V,W)is the set Hom (V,W)of linear maps from VtoW. Example. The collection of all topological spaces together with continuous maps between them is called the continuous category. Example. The collection of smooth manifolds together with smooth maps between them is called the smooth category. Example. We call a pair (M,q), where Mis a manifold and qa point in M, apointed manifold . Given any two such pairs (N,p)and(M,q), let Mor((N,p),(M,q))be the set of all smooth maps F:N→Msuch that F(p)=q. This gives rise to the category of pointed manifolds . Deﬁnition 10.1. Two objects AandBin a category are said to be isomorphic if there are morphisms f:A→Bandg:B→Asuch that g◦f=1Aand f◦g=1B. In this case both fandgare called isomorphisms . The usual notation for an isomorphism is “ ≃”. Thus, A≃Bcan mean, for ex- ample, a group isomorphism, a vector space isomorphism, a homeomorphism, or a diffeomorphism, depending on the category and the context. 10.2 Functors Deﬁnition 10.2. A(covariant) functor Ffrom one category Cto another category Dis a map that associates to each object AinCan object F(A)inDand to each morphism f:A→Ba morphism F(f):F(A)→F(B)such that (i)F(1A)=1F(A), (ii)F(f◦g)=F(f)◦F(g). 112§10 Categories and Functors Example. The tangent space construction is a functor from the category of pointed manifolds to the category of vector spaces. To each pointed manifold (N,p)we associate the tangent space TpNand to each smooth map f:(N,p)→(M,f(p))we associate the differential f∗,p:TpN→Tf(p)M. The functorial property (i) holds because if 1: N→Nis the identity map, then its differential 1∗,p:TpN→TpNis also the identity map. The functorial property (ii) holds because in this context it is the chain rule (g◦f)∗,p=g∗,f(p)◦f∗,p. Proposition 10.3. LetF:C→Dbe a functor from a category Cto a category D. If f:A→B is an isomorphism in C, thenF(f):F(A)→F(B)is an isomorphism inD. Proof. Problem 10.2. ⊓ ⊔ Note that we can recast Corollaries 8.6 and 8.7 in a more functorial form. Sup- pose f:N→Mis a diffeomorphism. Then (N,p)and(M,f(p))are isomorphic objects in the category of pointed manifolds. By Proposition 10.3, the tangent spacesT pNandTf(p)Mmust be isomorphic as vector spaces and therefore have the same dimension. It follows that the dimension of a manifold is invariant under a diffeo- morphism. If in the deﬁnition of a covariant functor we reverse the direction of the arrow for the morphism F(f), then we obtain a contravariant functor . More precisely, the deﬁnition is as follows. Deﬁnition 10.4. Acontravariant functor Ffrom one category Cto another category Dis a map that associates to each object AinCan object F(A)inDand to each morphism f:A→Ba morphism F(f):F(B)→F(A)such that (i)F(1A)=1F(A); (ii)F(f◦g)=F(g)◦F(f). (Note the reversal of order.) Example. Smooth functions on a manifold give rise to a contravariant functor that as- sociates to each manifold Mthe algebra F(M)=C∞(M)ofC∞functions on Mand to each smooth map F:N→Mof manifolds the pullback map F(F)=F∗:C∞(M)→ C∞(N),F∗(h)=h◦Fforh∈C∞(M). It is easy to verify that the pullback satisﬁes the two functorial properties: (i)(1M)∗=1C∞(M), (ii) if F:N→MandG:M→PareC∞maps, then (G◦F)∗=F∗◦G∗:C∞(P)→ C∞(N). Another example of a contravariant functor is the dual of a vector space, which we review in the next section. 10.3 The Dual Functor and the Multicovector Functor 113 10.3 The Dual Functor and the Multicovector Functor LetVbe a real vector space. Recall that its dual space V∨is the vector space of all linear functionals onV, i.e., linear functions α:V→R. We also write V∨=Hom(V,R). IfVis a ﬁnite-dimensional vector space with basis {e1,..., en}, then by Propo- sition 3.1 its dual space V∨has as a basis the collection of linear functionals {α1,..., αn}deﬁned by αi(ej)=δi j,1≤i,j≤n. Since a linear function on Vis determined by what it does on a basis of V, this set of equations deﬁnes αiuniquely. A linear map L:V→Wof vector spaces induces a linear map L∨, called the dual ofL, as follows. To every linear functional α:W→R, the dual map L∨associates the linear functional VL→Wα→R. Thus, the dual map L∨:W∨→V∨is given by L∨(α)=α◦Lforα∈W∨. Note that the dual of Lreverses the direction of the arrow. Proposition 10.5 (Functorial properties of the dual). Suppose V, W, and S are real vector spaces. (i)If1V:V→V is the identity map on V, then 1∨ V:V∨→V∨is the identity map on V∨. (ii)If f:V→W and g :W→S are linear maps, then (g◦f)∨=f∨◦g∨. Proof. Problem 10.3. ⊓ ⊔ According to this proposition, the dual construction F:()/ma√s⊔o→()∨is a contravari- ant functor from the category of vector spaces to itself: for Va real vector space, F(V) =V∨and for f∈Hom(V,W),F(f) = f∨∈Hom(W∨,V∨). Consequently, iff:V→Wis an isomorphism, then so is its dual f∨:W∨→V∨(cf. Proposi- tion 10.3). Fix a positive integer k. For any linear map L:V→Wof vector spaces, deﬁne thepullback map L∗:Ak(W)→Ak(V)to be (L∗f)(v1,..., vk)=f(L(v1),..., L(vk)) forf∈Ak(W)andv1,..., vk∈V. From the deﬁnition, it is easy to see that L∗is a linear map: L∗(a f+bg)=aL∗f+bL∗gfora,b∈Randf,g∈Ak(W). Proposition 10.6. The pullback of covectors by a linear map satisﬁes the two func- torial properties: 114§10 Categories and Functors (i)If1V:V→V is the identity map on V, then 1∗ V=1Ak(V), the identity map on Ak(V). (ii)If K:U→V and L :V→W are linear maps of vector spaces, then (L◦K)∗=K∗◦L∗:Ak(W)→Ak(U). Proof. Problem 10.6. ⊓ ⊔ To each vector space V, we associate the vector space Ak(V)of all k-covectors onV, and to each linear map L:V→Wof vector spaces, we associate the pullback Ak(L)=L∗:Ak(W)→Ak(V). Then Ak()is a contravariant functor from the category of vector spaces and linear maps to itself. When k=1, for any vector space V, the space A1(V)is the dual space, and for any linear map L:V→W, the pullback map A1(L)=L∗is the dual map L∨:W∨→V∨. Thus, the multicovector functor Ak( )generalizes the dual functor ()∨. Problems 10.1. Differential of the inverse map IfF:N→Mis a diffeomorphism of manifolds and p∈N, prove that (F−1)∗,F(p)=(F∗,p)−1. 10.2. Isomorphism under a functor Prove Proposition 10.3. 10.3. Functorial properties of the dual Prove Proposition 10.5. 10.4. Matrix of the dual map Suppose a linear transformation L:V→¯Vis represented by the matrix A= [ai j]relative to bases e1,..., enforVand ¯e1,..., ¯emfor¯V: L(ej)=∑ iai j¯ei. Letα1,..., αnand ¯α1,..., ¯αmbe the dual bases for V∨and ¯V∨, respectively. Prove that if L∨(¯αi)=∑jbi jαj, then bi j=ai j. 10.5. Injectivity of the dual map (a) Suppose VandWare vector spaces of possibly inﬁnite dimension over a ﬁeld K. Show that if a linear map L:V→Wis surjective, then its dual L∨:W∨→V∨is injective. (b) Suppose VandWare ﬁnite-dimensional vector spaces over a ﬁeld K. Prove the converse of the implication in (a). 10.6. Functorial properties of the pullback Prove Proposition 10.6. 10.7. Pullback in the top dimension Show that if L:V→Vis a linear operator on a vector space Vof dimension n, then the pullback L∗:An(V)→An(V)is multiplication by the determinant of L. 11.1 Constant Rank Theorem 115 §11 The Rank of a Smooth Map In this section we analyze the local structure of a smooth map through its rank. Recall that the rank of a smooth map f:N→Mat a point p∈Nis the rank of its differential atp. Two cases are of special interest: that in which the map fhas maximal rank at a point and that in which it has constant rank in a neighborhood. Let n=dimNand m=dimM. In case f:N→Mhas maximal rank at p, there are three not mutually exclusive possibilities: (i) If n=m, then by the inverse function theorem, fis a local diffeomorphism at p. (ii) If n≤m, then the maximal rank is nandfis an immersion atp. (iii) If n≥m, then the maximal rank is mandfis asubmersion atp. Because manifolds are locally Euclidean, theorems on the rank of a smooth map between Euclidean spaces (Appendix B) translate easily to theorems about mani- folds. This leads to the constant rank theorem for manifolds, which gives a simplenormal form for a smooth map having constant rank on an open set (Theorem 11.1). As an immediate consequence, we obtain a criterion for a level set to be a regu- lar submanifold, which, following [25], we call the constant-rank level set theorem. As we explain in Subsection 11.2, maximal rank at a point implies constant rank in a neighborhood, so immersions and submersions are maps of constant rank. Theconstant rank theorem specializes to the immersion theorem and the submersion the- orem, giving simple normal forms for an immersion and a submersion. The regular level set theorem, which we encountered in Subsection 9.3, is now seen to be a con-sequence of the submersion theorem and a special case of the constant-rank level set theorem. By the regular level set theorem, the preimage of a regular value of a smooth map is a manifold. The image of a smooth map, on the other hand, does not generally have a nice structure. Using the immersion theorem we derive conditions under which theimage of a smooth map is a manifold. 11.1 Constant Rank Theorem Suppose f:N→Mis aC∞map of manifolds and we want to show that the level setf−1(c)is a manifold for some cinM. In order to apply the regular level set theorem, we need the differential f∗to have maximal rank at every point of f−1(c). Sometimes this is not true; even if true, it may be difﬁcult to show. In such cases, the constant-rank level set theorem can be helpful. It has one cardinal virtue: it is notnecessary to know precisely the rank of f; it sufﬁces that the rank be constant. The constant rank theorem for Euclidean spaces (Theorem B.4) has an immediate analogue for manifolds. Theorem 11.1 (Constant rank theorem). Let N and M be manifolds of dimensions n and m respectively. Suppose f :N→M has constant rank k in a neighborhood of 116§11 The Rank of a Smooth Map a point p in N. Then there are charts (U,φ)centered at p in N and (V,ψ)centered at f(p)in M such that for (r1,..., rn)inφ(U), (ψ◦f◦φ−1)(r1,..., rn)=( r1,..., rk,0,..., 0). (11.1) Proof. Choose a chart (¯U,¯φ)about pinNand(¯V,¯ψ)about f(p)inM. Then ¯ψ◦ f◦¯φ−1is a map between open subsets of Euclidean spaces. Because ¯φand ¯ψare diffeomorphisms, ¯ψ◦f◦¯φ−1has the same constant rank kasfin a neighborhood of ¯φ(p)inRn. By the constant rank theorem for Euclidean spaces (Theorem B.4) there are a diffeomorphism Gof a neighborhood of ¯φ(p)inRnand a diffeomorphism F of a neighborhood of (¯ψ◦f)(p)inRmsuch that (F◦¯ψ◦f◦¯φ−1◦G−1)(r1,..., rn)=( r1,..., rk,0,..., 0). Setφ=G◦¯φandψ=F◦¯ψ. ⊓ ⊔ In the constant rank theorem, it is possible that the normal form (11.1) for the function fhas no zeros at all: if the rank kequals m, then (ψ◦f◦φ−1)(r1,..., rn)=( r1,..., rm). From this theorem, the constant-rank level set theorem follows easily. By a neigh- borhood of a subset Aof a manifold Mwe mean an open set containing A. Theorem 11.2 (Constant-rank level set theorem). Let f :N→M be a C∞map of manifolds and c ∈M. If f has constant rank k in a neighborhood of the level set f−1(c)in N, then f−1(c)is a regular submanifold of N of codimension k. Proof. Letpbe an arbitrary point in f−1(c). By the constant rank theorem there are a coordinate chart (U,φ)=( U,x1,..., xn)centered at p∈Nand a coordinate chart (V,ψ)=( V,y1,..., ym)centered at f(p)=c∈Msuch that (ψ◦f◦φ−1)(r1,..., rn)=( r1,..., rk,0,..., 0)∈Rm. This shows that the level set (ψ◦f◦φ−1)−1(0)is deﬁned by the vanishing of the coordinates r1,..., rk. /Bullet /Bulletφ f ψ φ(f−1(c)) f−1(c)c 0 U V Fig. 11.1. Constant-rank level set. The image of the level set f−1(c)under φis the level set (ψ◦f◦φ−1)−1(0) (Figure 11.1 ), since 11.1 Constant Rank Theorem 117 φ(f−1(c))= φ(f−1(ψ−1(0))=( ψ◦f◦φ−1)−1(0). Thus, the level set f−1(c)inUis deﬁned by the vanishing of the coordinate functions x1,..., xk, where xi=ri◦φ. This proves that f−1(c)is a regular submanifold of N of codimension k. ⊓ ⊔ Example 11.3 ( Orthogonal group ).Theorthogonal group O(n)is deﬁned to be the subgroup of GL (n,R)consisting of matrices Asuch that ATA=I, the n×nidentity matrix. Using the constant rank theorem, prove that O (n)is a regular submanifold of GL(n,R). Solution. Deﬁne f: GL(n,R)→GL(n,R)byf(A) =ATA. Then O(n)is the level setf−1(I). For any two matrices A,B∈GL(n,R), there is a unique matrix C∈ GL(n,R)such that B=AC. Denote by ℓCandrC: GL(n,R)→GL(n,R)the left and right multiplication by C, respectively. Since f(AC)=( AC)TAC=CTATAC=CTf(A)C, we have (f◦rC)(A)=(ℓCT◦rC◦f)(A). Since this is true for all A∈GL(n,R), f◦rC=ℓCT◦rC◦f. By the chain rule, f∗,AC◦(rC)∗,A=(ℓCT)∗,ATAC◦(rC)∗,ATA◦f∗,A. (11.2) Since left and right multiplications are diffeomorphisms, their differentials are iso- morphisms. Composition with an isomorphism does not change the rank of a linear map. Hence, in (11.2), rkf∗,AC=rkf∗,A. Since ACandAare two arbitrary points of GL (n,R), this proves that the differential offhas constant rank on GL (n,R). By the constant-rank level set theorem, the orthogonal group O (n)=f−1(I)is a regular submanifold of GL (n,R). NOTATION . Iff:N→Mis a map with constant rank kin a neighborhood of a point p∈N, its local normal form (11.1) relative to the charts (U,φ) = ( U,x1,..., xn) and(V,ψ) = ( V,y1,..., ym)in the constant rank theorem (Theorem 11.1) can be expressed in terms of the local coordinates x1,..., xnandy1,..., ymas follows. First note that for any q∈U, φ(q)=( x1(q),..., xn(q))andψ(f(q))=( y1(f(q)),..., yn(f(q)). Thus, 118§11 The Rank of a Smooth Map (y1(f(q)),..., ym(f(q))= ψ(f(q))=( ψ◦f◦φ−1)(φ(q)) =(ψ◦f◦φ−1)(x1(q),..., xn(q)) =(x1(q),..., xk(q)),0,..., 0)(by (11.1)) . As functions on U, (y1◦f,..., ym◦f)=( x1,..., xk,0,..., 0). (11.3) We can rewrite (11.3) in the following form: relative to the charts (U,x1,..., xn)and (V,y1,..., ym), the map fis given by (x1,..., xn)/ma√s⊔o→(x1,..., xk,0,..., 0). 11.2 The Immersion and Submersion Theorems In this subsection we explain why immersions and submersions have constant rank. The constant rank theorem gives local normal forms for immersions and submer-sions, called the immersion theorem and the submersion theorem respectively. From the submersion theorem and the constant-rank level set theorem, we get two more proofs of the regular level set theorem. Consider a C ∞map f:N→M. Let(U,φ)=( U,x1,..., xn)be a chart about pin Nand(V,ψ)=( V,y1,..., ym)a chart about f(p)inM. Write fi=yi◦ffor the ith component of fin the chart (V,y1,..., ym). Relative to the charts (U,φ)and(V,ψ), the linear map f∗,pis represented by the matrix [∂fi/∂xj(p)](Proposition 8.11). Hence, f∗,pis injective⇐⇒ n≤mand rk[∂fi/∂xj(p)]= n, f∗,pis surjective⇐⇒ n≥mand rk[∂fi/∂xj(p)]= m.(11.4) The rank of a matrix is the number of linearly independent rows of the matrix; it is also the number of linearly independent columns. Thus, the maximum possible rank of an m×nmatrix is the minimum of mandn. It follows from (11.4) that being an immersion or a submersion at pis equivalent to the maximality of rk [∂fi/∂xj(p)]. Having maximal rank at a point is an open condition in the sense that the set Dmax(f)={p∈U|f∗,phas maximal rank at p} is an open subset of U. To see this, suppose kis the maximal rank of f. Then rkf∗,p=k⇐⇒ rk[∂fi/∂xj(p)]= k ⇐⇒ rk[∂fi/∂xj(p)]≥k(since kis maximal ). So the complement U−Dmax(f)is deﬁned by rk[∂fi/∂xj(p)]<k, which is equivalent to the vanishing of all k×kminors of the matrix [∂fi/∂xj(p)]. As the zero set of ﬁnitely many continuous functions, U−Dmax(f)is closed and so Dmax(f)is open. In particular, if fhas maximal rank at p, then it has maximal rank at all points in some neighborhood of p. We have proven the following proposition. 11.2 The Immersion and Submersion Theorems 119 Proposition 11.4. Let N and M be manifolds of dimensions n and m respectively. If a C∞map f :N→M is an immersion at a point p ∈N, then it has constant rank n in a neighborhood of p. If a C∞map f :N→M is a submersion at a point p ∈N, then it has constant rank m in a neighborhood of p. Example. While maximal rank at a point implies constant rank in a neighborhood, the converse is not true. The map f:R2→R3,f(x,y)=( x,0,0), has constant rank 1, but it does not have maximal rank at any point. By Proposition 11.4, the following theorems are simply special cases of the con- stant rank theorem. Theorem 11.5. Let N and M be manifolds of dimensions n and m respectively. (i)(Immersion theorem) Suppose f :N→M is an immersion at p ∈N. Then there are charts (U,φ)centered at p in N and (V,ψ)centered at f (p)in M such that in a neighborhood of φ(p), (ψ◦f◦φ−1)(r1,..., rn)=( r1,..., rn,0,..., 0). (ii)(Submersion theorem) Suppose f :N→M is a submersion at p in N. Then there are charts (U,φ)centered at p in N and (V,ψ)centered at f (p)in M such that in a neighborhood of φ(p), (ψ◦f◦φ−1)(r1,..., rm,rm+1,..., rn)=( r1,..., rm). Corollary 11.6. A submersion f :N→M of manifolds is an open map. Proof. LetWbe an open subset of N. We need to show that its image f(W)is open inM. Choose a point f(p)inf(W), with p∈W. By the submersion theorem, fis locally a projection. Since a projection is an open map (Problem A.7), there is anopen neighborhood UofpinWsuch that f(U)is open in M. Clearly, f(p)∈f(U)⊂f(W). Since f(p)∈f(W)was arbitrary, f(W)is open in M. ⊓ ⊔ The regular level set theorem (Theorem 9.9) is an easy corollary of the submer- sion theorem. Indeed, for a C ∞map f:N→Mof manifolds, a level set f−1(c)is regular if and only if fis a submersion at every point p∈f−1(c). Fix one such point p∈f−1(c)and let(U,φ)and(V,ψ)be the charts in the submersion theorem. Then ψ◦f◦φ−1=π:Rn⊃φ(U)→Rmis the projection to the ﬁrst mcoordinates, π(r1,..., rn)=( r1,..., rm). It follows that on U, ψ◦f=π◦φ=(r1,..., rm)◦φ=(x1,..., xm). Therefore, f−1(c)=f−1(ψ−1(0))=( ψ◦f)−1(0)=Z(ψ◦f)=Z(x1,..., xm), 120§11 The Rank of a Smooth Map showing that in the chart (U,x1,..., xn), the level set f−1(c)is deﬁned by the vanish- ing of the mcoordinate functions x1,..., xm. Therefore, (U,x1,..., xn)is an adapted chart for Nrelative to f−1(c). This gives a second proof that the regular level set f−1(c)is a regular submanifold of N. Since the submersion theorem is a special case of the constant rank theorem, it is not surprising that the regular level set theorem is also a special case of the constant-rank level set theorem. On a regular level set f−1(c), the map f:N→M has maximal rank mat every point. Since the maximality of the rank of fis an open condition, a regular level set f−1(c)has a neighborhood on which fhas constant rank m. By the constant-rank level set theorem (Theorem 11.2), f−1(c)is a regular submanifold of N, giving us a third proof of the regular level set theorem. 11.3 Images of Smooth Maps The following are all examples of C∞maps f:N→M, with N=RandM=R2. Example 11.7. f(t)=( t2,t3). This fis one-to-one, because t/ma√s⊔o→t3is one-to-one. Since f′(0) = ( 0,0), the differential f∗,0:T0R→T(0,0)R2is the zero map and hence not injective; so fis not an immersion at 0. Its image is the cuspidal cubic y2=x3(Figure 11.2) . 1 −11 −1xy Fig. 11.2. A cuspidal cubic, not an immersion. Example 11.8. f(t)=( t2−1,t3−t). Since the equation f′(t)=( 2t,3t2−1)=( 0,0)has no solution in t, this map f is an immersion. It is not one-to-one, because it maps both t=1 and t=−1 to the origin. To ﬁnd an equation for the image f(N), letx=t2−1 and y=t3−t. Then y=t(t2−1)=tx; so y2=t2x2=(x+1)x2. Thus the image of fis the nodal cubic y2=x2(x+1)(Figure 11.3 ). Example 11.9.The map finFigure 11.4 is a one-to-one immersion but its image, with the subspace topology induced from R2, is not homeomorphic to the domain R, because there are points near f(p)in the image that correspond to points in Rfar away from p. More precisely, if Uis an interval about pas shown, there is no neigh- borhood Voff(p)inf(N)such that f−1(V)⊂U; hence, f−1is not continuous. 11.3 Images of Smooth Maps 121 1 −11 −1 xy Fig. 11.3. A nodal cubic, an immersion but not one-to-one. /Bullet /Bulletf f(p) pU ( ) Fig. 11.4. A one-to-one immersion that is not an embedding. Example 11.10 .The manifold MinFigure 11.5 is the union of the graph of y= sin(1/x)on the interval ]0,1[, the open line segment from y=0 to y=1 on the y-axis, and a smooth curve joining (0,0)and(1,sin1). The map fis a one-to-one immersion whose image with the subspace topology is not homeomorphic to R. /Bullet /Bullet /Bullet /Bullet /Bullet /Bullet /Bullet/Bullet f N=R M Fig. 11.5. A one-to-one immersion that is not an embedding. Notice that in these examples the image f(N)is not a regular submanifold of M=R2. We would like conditions on the map fso that its image f(N)would be a regular submanifold of M. Deﬁnition 11.11. AC∞map f:N→Mis called an embedding if (i) it is a one-to-one immersion and 122§11 The Rank of a Smooth Map (ii) the image f(N)with the subspace topology is homeomorphic to Nunder f. (The phrase “one-to-one” in this deﬁnition is redundant, since a homeomor-phism is necessarily one-to-one.) Remark. Unfortunately, there is quite a bit of terminological confusion in the liter- ature concerning the use of the word “submanifold.” Many authors give the image f(N)of a one-to-one immersion f:N→Mnot the subspace topology, but the topol- ogy inherited from f; i.e., a subset f(U)off(N)is said to be open if and only if U is open in N. With this topology, f(N)is by deﬁnition homeomorphic to N. These authors deﬁne a submanifold to be the image of any one-to-one immersion with the topology and differentiable structure inherited from f. Such a set is sometimes called animmersed submanifold ofM.Figures 11.4 and11.5 show two examples of im- mersed submanifolds. If the underlying set of an immersed submanifold is given the subspace topology, then the resulting space need not be a manifold at all! For us, a submanifold without any qualifying adjective is always a regular sub- manifold . To recapitulate, a regular submanifold of a manifold Mis a subset Sof Mwith the subspace topology such that every point of Shas a neighborhood U∩S deﬁned by the vanishing of coordinate functions on U, where Uis a chart in M. ) (/Bullet −π 2π 23π 2f1 −1xy 1 −1( )A B ( ) −π 2/Bullet π 23π 2g1 −1xy 1 −1( )A B Fig. 11.6. The ﬁgure-eight as two distinct immersed submanifolds of R2. Example 11.12 ( The ﬁgure-eight ).The ﬁgure-eight is the image of a one-to-one im- mersion f(t)=( cost,sin2t),−π/2<t<3π/2 (Figure 11.6 ). As such, it is an immersed submanifold of R2, with a topology and manifold structure induced from the open interval ]−π/2,3π/2[byf. Because of the presence of a cross at the origin, it cannot be a regular submanifold of R2. In fact, with the subspace topology of R2, the ﬁgure-eight is not even a manifold. 11.3 Images of Smooth Maps 123 The ﬁgure-eight is also the image of the one-to-one immersion g(t)=( cost,−sin2t),−π/2<t<3π/2 (Figure 11.6 ). The maps fandginduce distinct immersed submanifold structures on the ﬁgure-eight. For example, the open interval from AtoBinFigure 11.6 is an open set in the topology induced from g, but it is not an open set in the topology induced from f, since its inverse image under fcontains an isolated point π/2. We will use the phrase “near p” to mean “in a neighborhood of p.” Theorem 11.13. If f:N→M is an embedding, then its image f (N)is a regular submanifold of M. Proof. Letp∈N. By the immersion theorem (Theorem 11.5), there are local coor- dinates(U,x1,..., xn)near pand(V,y1,..., ym)near f(p)such that f:U→Vhas the form (x1,..., xn)/ma√s⊔o→(x1,..., xn,0,..., 0). /BulletNpU f /Bullet f(p)f(N)V V′ ( ) Fig. 11.7. The image of an embedding is a regular submanifold. Thus, f(U)is deﬁned in Vby the vanishing of the coordinates yn+1,..., ym. This alone does not prove that f(N)is a regular submanifold, since V∩f(N)may be larger than f(U). (Think about Examples 11.9 and 11.10.) We need to show that in some neighborhood of f(p)inV, the set f(N)is deﬁned by the vanishing of m−n coordinates. Since f(N)with the subspace topology is homeomorphic to N, the image f(U) is open in f(N). By the deﬁnition of the subspace topology, there is an open set V′ inMsuch that V′∩f(N)=f(U)(Figure 11.7 ). In V∩V′, V∩V′∩f(N)=V∩f(U)=f(U), andf(U)is deﬁned by the vanishing of yn+1,..., ym. Thus,(V∩V′,y1,..., ym)is an adapted chart containing f(p)forf(N). Since f(p)is an arbitrary point of f(N), this proves that f(N)is a regular submanifold of M. ⊓ ⊔ Theorem 11.14. If N is a regular submanifold of M, then the inclusion i :N→M, i(p)=p, is an embedding. 124§11 The Rank of a Smooth Map Proof. Since a regular submanifold has the subspace topology and i(N)also has the subspace topology, i:N→i(N)is a homeomorphism. It remains to show that i:N→Mis an immersion. Letp∈N. Choose an adapted chart (V,y1,..., yn,yn+1,..., ym)forMabout p such that V∩Nis the zero set of yn+1,..., ym. Relative to the charts (V∩N,y1,..., yn) forNand(V,y1,..., ym)forM, the inclusion iis given by (y1,..., yn)/ma√s⊔o→(y1,..., yn,0,..., 0), which shows that iis an immersion. ⊓ ⊔ In the literature the image of an embedding is often called an embedded subman- ifold. Theorems 11.13 and 11.14 show that an embedded submanifold and a regular submanifold are one and the same thing. 11.4 Smooth Maps into a Submanifold Suppose f:N→Mis aC∞map whose image f(N)lies in a subset S⊂M. IfSis a manifold, is the induced map ˜f:N→SalsoC∞? This question is more subtle than it looks, because the answer depends on whether Sis a regular submanifold or an immersed submanifold of M. Example. Consider the one-to-one immersions fandg:I→R2in Example 11.12, where Iis the open interval ]−π/2,3π/2[inR. Let Sbe the ﬁgure-eight in R2with the immersed submanifold structure induced from g. Because the image of f:I→ R2lies in S, the C∞map finduces a map ˜f:I→S. The open interval from AtoBinFigure 11.6 is an open neighborhood of the origin 0 in S. Its inverse image under ˜fcontains the point π/2 as an isolated point and is therefore not open. This shows that although f:I→R2isC∞, the induced map ˜f:I→Sis not continuous and therefore not C∞. Theorem 11.15. Suppose f :N→M is C∞and the image of f lies in a subset S of M. If S is a regular submanifold of M, then the induced map ˜f:N→S is C∞. Proof. Letp∈N. Denote the dimensions of N,M, and Sbyn,m, and s, respectively. By hypothesis, f(p)∈S⊂M. Since Sis a regular submanifold of M, there is an adapted coordinate chart (V,ψ)=( V,y1,..., ym)forMabout f(p)such that S∩Vis the zero set of ys+1,..., ym, with coordinate map ψS=(y1,..., ys). By the continuity off, it is possible to choose a neighborhood of pwith f(U)⊂V. Then f(U)⊂V∩S, so that for q∈U, (ψ◦f)(q)=( y1(f(q)),..., ys(f(q)),0,..., 0). It follows that on U, ψS◦˜f=(y1◦f,..., ys◦f). Since y1◦f,..., ys◦fareC∞onU, by Proposition 6.16, ˜fisC∞onUand hence at p. Since pwas an arbitrary point of N, the map ˜f:N→SisC∞.⊓ ⊔ 11.5 The Tangent Plane to a Surface in R3125 Example 11.16 ( Multiplication map of SL(n,R)).The multiplication map µ: GL(n,R)×GL(n,R)→GL(n,R), (A,B)/ma√s⊔o→AB, is clearly C∞because (AB)i j=n ∑ k=1aikbk j is a polynomial and hence a C∞function of the coordinates aikandbk j. However, one cannot conclude in the same way that the multiplication map ¯µ: SL(n,R)×SL(n,R)→SL(n,R) isC∞. This is because{ai j}1≤i,j≤nis not a coordinate system on SL (n,R); there is one coordinate too many (See Problem 11.6). Since SL(n,R)×SL(n,R)is a regular submanifold of GL (n,R)×GL(n,R), the inclusion map i: SL(n,R)×SL(n,R)→GL(n,R)×GL(n,R) isC∞by Theorem 11.14; therefore, the composition µ◦i: SL(n,R)×SL(n,R)→GL(n,R) is also C∞. Because the image of µ◦ilies in SL (n,R), and SL(n,R)is a regular submanifold of GL (n,R)(see Example 9.13), by Theorem 11.15 the induced map ¯µ: SL(n,R)×SL(n,R)→SL(n,R) isC∞. 11.5 The Tangent Plane to a Surface in R3 Suppose f(x1,x2,x3)is a real-valued function on R3with no critical points on its zero set N=f−1(0). By the regular level set theorem, Nis a regular submanifold ofR3. By Theorem 11.14 the inclusion i:N→R3is an embedding, so at any point pinN,i∗,p:TpN→TpR3is injective. We may therefore think of the tangent plane TpNas a plane in TpR3≃R3(Figure 11.8 ). We would like to ﬁnd the equation of this plane. Suppose v=∑vi∂/∂xi|pis a vector in TpN. Under the linear isomorphism TpR3≃R3, we identify vwith the vector/an}bracke⊔le{⊔v1,v2,v3/an}bracke⊔ri}h⊔inR3. Let c(t)be a curve lying in Nwith c(0)=pandc′(0)=/an}bracke⊔le{⊔v1,v2,v3/an}bracke⊔ri}h⊔. Since c(t)lies in N,f(c(t))= 0 for allt. By the chain rule, 0=d dtf(c(t))=3 ∑ i=1∂f ∂xi(c(t))(ci)′(t). 126§11 The Rank of a Smooth Map /Bulletv pTpN N Fig. 11.8. Tangent plane to a surface Natp. Att=0, 0=3 ∑ i=1∂f ∂xi(c(0))(ci)′(0)=3 ∑ i=1∂f ∂xi(p)vi. Since the vector v=/an}bracke⊔le{⊔v1,v2,v3/an}bracke⊔ri}h⊔represents the arrow from the point p=(p1,p2,p3) tox=(x1,x2,x3)in the tangent plane, one usually makes the substitution vi=xi−pi. This amounts to translating the tangent plane from the origin to p. Thus the tangent plane to Natpis deﬁned by the equation 3 ∑ i=1∂f ∂xi(p)(xi−pi)=0. (11.5) One interpretation of this equation is that the gradient vector /an}bracke⊔le{⊔∂f/∂x1(p),∂f/∂x2(p), ∂f/∂x3(p)/an}bracke⊔ri}h⊔offatpis normal to any vector in the tangent plane. Example 11.17 ( Tangent plane to a sphere ).Letf(x,y,z) =x2+y2+z2−1. To get the equation of the tangent plane to the unit sphere S2=f−1(0)inR3at(a,b,c)∈S2, we compute ∂f ∂x=2x,∂f ∂y=2y,∂f ∂z=2z. Atp=(a,b,c), ∂f ∂x(p)=2a,∂f ∂y(p)=2b,∂f ∂z(p)=2c. By (11.5) the equation of the tangent plane to the sphere at (a,b,c)is 2a(x−a)+2b(y−b)+2c(z−c)=0, or ax+by+cz=1, since a2+b2+c2=1. 11.5 The Tangent Plane to a Surface in R3127 Problems 11.1. Tangent vectors to a sphere The unit sphere SninRn+1is deﬁned by the equation ∑n+1 i=1(xi)2=1. For p=(p1,..., pn+1)∈ Sn, show that a necessary and sufﬁcient condition for Xp=∑ai∂/∂xi|p∈Tp(Rn+1) to be tangent to Snatpis∑aipi=0. 11.2. Tangent vectors to a plane curve (a) Let i:S1֒→R2be the inclusion map of the unit circle. In this problem, we denote by x,ythe standard coordinates on R2and by ¯ x,¯ytheir restrictions to S1. Thus, ¯ x=i∗xand ¯y=i∗y. On the upper semicircle U={(a,b)∈S1|b>0}, ¯xis a local coordinate, so that ∂/∂¯xis deﬁned. Prove that for p∈U, i∗/parenleftigg ∂ ∂¯x/vextendsingle/vextendsingle/vextendsingle/vextendsingle p/parenrightigg =/parenleftbigg∂ ∂x+∂¯y ∂¯x∂ ∂y/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle p. Thus, although i∗:TpS1→TpR2is injective, ∂/∂¯x|pcannot be identiﬁed with ∂/∂x|p (Figure 11.9 ). /Bullet p∂ ∂x/vextendsingle/vextendsingle/vextendsingle/vextendsingle p ∂ ∂¯x/vextendsingle/vextendsingle/vextendsingle/vextendsingle p Fig. 11.9. Tangent vector ∂/∂¯x|pto a circle. (b) Generalize (a) to a smooth curve CinR2, letting Ube a chart in Con which ¯ x, the restric- tion of xtoC, is a local coordinate. 11.3.* Critical points of a smooth map on a compact manifold Show that a smooth map ffrom a compact manifold NtoRmhas a critical point. (Hint : Let π:Rm→Rbe the projection to the ﬁrst factor. Consider the composite map π◦f:N→R. A second proof uses Corollary 11.6 and the connectedness of Rm.) 11.4. Differential of an inclusion map On the upper hemisphere of the unit sphere S2, we have the coordinate map φ=(u,v), where u(a,b,c)=aand v(a,b,c)=b. So the derivations ∂/∂u|p,∂/∂v|pare tangent vectors of S2at any point p= (a,b,c)on the upper hemisphere. Let i:S2→R3be the inclusion and x,y,zthe standard coordinates on R3. The differential i∗:TpS2→TpR3maps ∂/∂u|p,∂/∂v|pintoTpR3. Thus, 128§11 The Rank of a Smooth Map i∗/parenleftigg ∂ ∂u/vextendsingle/vextendsingle/vextendsingle/vextendsingle p/parenrightigg =α1∂ ∂x/vextendsingle/vextendsingle/vextendsingle/vextendsingle p+β1∂ ∂y/vextendsingle/vextendsingle/vextendsingle/vextendsingle p+γ1∂ ∂z/vextendsingle/vextendsingle/vextendsingle/vextendsingle p, i∗/parenleftigg ∂ ∂v/vextendsingle/vextendsingle/vextendsingle/vextendsingle p/parenrightigg =α2∂ ∂x/vextendsingle/vextendsingle/vextendsingle/vextendsingle p+β2∂ ∂y/vextendsingle/vextendsingle/vextendsingle/vextendsingle p+γ2∂ ∂z/vextendsingle/vextendsingle/vextendsingle/vextendsingle p, for some constants αi,βi,γi. Find(αi,βi,γi)fori=1,2. 11.5. One-to-one immersion of a compact manifold Prove that if Nis a compact manifold, then a one-to-one immersion f:N→Mis an embed- ding. 11.6. Multiplication map in SL(n,R) Letf: GL(n,R)→Rbe the determinant map f(A)=detA=det[ai j]. For A∈SL(n,R), there is at least one (k,ℓ)such that the partial derivative ∂f/∂akℓ(A)is nonzero (Example 9.13). Use Lemma 9.10 and the implicit function theorem to prove that (a) there is a neighborhood of Ain SL(n,R)in which ai j,(i,j)/ne}a⊔ionslash= (k,ℓ), form a coordinate system, and akℓis aC∞function of the other entries ai j,(i,j)/ne}a⊔ionslash=(k,ℓ); (b) the multiplication map ¯µ: SL(n,R)×SL(n,R)→SL(n,R) isC∞. 12.1 The Topology of the Tangent Bundle 129 §12 The Tangent Bundle A smooth vector bundle over a smooth manifold Mis a smoothly varying family of vector spaces, parametrized by M, that locally looks like a product. Vector bundles and bundle maps form a category, and have played a fundamental role in geometryand topology since their appearance in the 1930s [39]. The collection of tangent spaces to a manifold has the structure of a vector bundle over the manifold, called the tangent bundle. A smooth map between two manifolds induces, via its differential at each point, a bundle map of the corresponding tangent bundles. Thus, the tangent bundle construction is a functor from the category ofsmooth manifolds to the category of vector bundles. At ﬁrst glance it might appear that the tangent bundle functor is not a simpliﬁ- cation, since a vector bundle is a manifold plus an additional structure. However,because the tangent bundle is canonically associated to a manifold, invariants of the tangent bundle will give rise to invariants for the manifold. For example, the Chern–Weil theory of characteristic classes, which we treat in another volume, uses differential geometry to construct invariants for vector bundles. Applied to the tan- gent bundle, characteristic classes lead to numerical diffeomorphism invariants fora manifold called characteristic numbers . Characteristic numbers generalize, for instance, the classical Euler characteristic. For us in this book the importance of the vector bundle point of view comes from its role in unifying concepts. A section of a vector bundle π:E→Mis a map from MtoEthat maps each point of Minto the ﬁber of the bundle over the point. As we shall see, both vector ﬁelds and differential forms on a manifold are sections of vector bundles over the manifold. In the following pages we construct the tangent bundle of a manifold and show that it is a smooth vector bundle. We then discuss criteria for a section of a smooth vector bundle to be smooth. 12.1 The Topology of the Tangent Bundle LetMbe a smooth manifold. Recall that at each point p∈M, the tangent space TpM is the vector space of all point-derivations of C∞ p(M), the algebra of germs of C∞ functions at p. The tangent bundle ofMis the union of all the tangent spaces of M: TM=/uniondisplay p∈MTpM. In general, if{Ai}i∈Iis a collection of subsets of a set S, then their disjoint union is deﬁned to be the set/coproductdisplay i∈IAi:=/uniondisplay i∈I({i}×Ai). 130§12 The Tangent Bundle The subsets Aimay overlap, but in the disjoint union they are replaced by nonover- lapping copies. In the deﬁnition of the tangent bundle, the union/uniontext p∈MTpMis (up to notation) the same as the disjoint union/coproducttext p∈MTpM, since for distinct points pandqinM, the tangent spaces TpMandTqMare already disjoint. /Bullet /BulletTpM TqMp qM Fig. 12.1. Tangent spaces to a circle. In a pictorial representation of tangent spaces such as Figure 12.1, where Mis the unit circle, it may look as though the two tangent spaces TpMandTqMintersect. In fact, the intersection point of the two lines in Figure 12.1 represents distinct tangent vectors in TpMandTqM, so that TpMandTqMare disjoint even in the ﬁgure. There is a natural map π:T M→Mgiven by π(v)=pifv∈TpM. (We use the word “natural” to mean that the map does not depend on any choice, for example, the choice of an atlas or of local coordinates for M.) As a matter of notation, we sometimes write a tangent vector v∈TpMas a pair(p,v), to make explicit the point p∈Mat which vis a tangent vector. As deﬁned, T Mis a set, with no topology or manifold structure. We will make it into a smooth manifold and show that it is a C∞vector bundle over M. The ﬁrst step is to give it a topology. If(U,φ)=( U,x1,..., xn)is a coordinate chart on M, let TU=/uniondisplay p∈UTpU=/uniondisplay p∈UTpM. (We saw in Remark 8.2 that TpU=TpM.) At a point p∈U, a basis for TpMis the set of coordinate vectors ∂/∂x1|p,..., ∂/∂xn|p, so a tangent vector v∈TpMis uniquely a linear combination v=n ∑ ici∂ ∂xi/vextendsingle/vextendsingle/vextendsingle/vextendsingle p. In this expression, the coefﬁcients ci=ci(v)depend on vand so are functions on TU. Let ¯ xi=xi◦πand deﬁne the map ˜φ:TU→φ(U)×Rnby v/ma√s⊔o→(x1(p),..., xn(p),c1(v),..., cn(v))=( ¯x1,..., ¯xn,c1,..., cn)(v). (12.1) Then ˜φhas inverse 12.1 The Topology of the Tangent Bundle 131 (φ(p),c1,..., cn)/ma√s⊔o→∑ci∂ ∂xi/vextendsingle/vextendsingle/vextendsingle/vextendsingle p and is therefore a bijection. This means we can use ˜φto transfer the topology of φ(U)×RntoTU: a set AinTUis open if and only if ˜φ(A)is open in φ(U)× Rn, where φ(U)×Rnis given its standard topology as an open subset of R2n. By deﬁnition, TU, with the topology induced by ˜φ, is homeomorphic to φ(U)×Rn. If Vis an open subset of U, then φ(V)×Rnis an open subset of φ(U)×Rn. Hence, the relative topology on TVas a subset of TUis the same as the topology induced from the bijection ˜φ|TV:TV→φ(V)×Rn. Letφ∗:TpU→Tφ(p)(Rn)be the differential of the coordinate map φatp. Since φ∗(v)=∑ci∂/∂ri|φ(p)∈Tφ(p)(Rn)≃Rnby Proposition 8.8, we may identify φ∗(v) with the column vector /an}bracke⊔le{⊔c1,..., cn/an}bracke⊔ri}h⊔inRn. So another way to describe ˜φis˜φ=(φ◦ π,φ∗). LetBbe the collection of all open subsets of T(Uα)asUαruns over all coordi- nate open sets in M: B=/uniondisplay α{A|Aopen in T(Uα),Uαa coordinate open set in M}. Lemma 12.1. (i)For any manifold M, the set TM is the union of all A ∈B. (ii)Let U and V be coordinate open sets in a manifold M. If A is open in TU and B is open in TV, then A ∩B is open in T (U∩V). Proof. (i) Let{(Uα,φα)}be the maximal atlas for M. Then T M=/uniondisplay αT(Uα)⊂/uniondisplay A∈BA⊂T M, so equality holds everywhere. (ii) Since T(U∩V)is a subspace of TU, by the deﬁnition of relative topology, A∩T(U∩V)is open in T(U∩V). Similarly, B∩T(U∩V)is open in T(U∩V). But A∩B⊂TU∩TV=T(U∩V). Hence, A∩B=A∩B∩T(U∩V)=( A∩T(U∩V))∩(B∩T(U∩V)) is open in T(U∩V). ⊓ ⊔ It follows from this lemma that the collection Bsatisﬁes the conditions (i) and (ii) of Proposition A.8 for a collection of subsets to be a basis for some topology onT M. We give the tangent bundle T Mthe topology generated by the basis B. Lemma 12.2. A manifold M has a countable basis consisting of coordinate open sets. 132§12 The Tangent Bundle Proof. Let{(Uα,φα)}be the maximal atlas on MandB={Bi}a countable basis forM. For each coordinate open set Uαand point p∈Uα, choose a basic open set Bp,α∈Bsuch that p∈Bp,α⊂Uα. The collection{Bp,α}, without duplicate elements, is a subcollection of Band is therefore countable. For any open set UinMand a point p∈U, there is a coordinate open set Uα such that p∈Uα⊂U. Hence, p∈Bp,α⊂U, which shows that{Bp,α}is a basis for M. ⊓ ⊔ Proposition 12.3. The tangent bundle T M of a manifold M is second countable. Proof. Let{Ui}∞ i=1be a countable basis for Mconsisting of coordinate open sets. Letφibe the coordinate map on Ui. Since TUiis homeomorphic to the open subset φi(Ui)×RnofR2nand any subset of a Euclidean space is second countable (Ex- ample A.13 and Proposition A.14), TUiis second countable. For each i, choose a countable basis{Bi,j}∞ j=1forTUi. Then{Bi,j}∞ i,j=1is a countable basis for the tan- gent bundle. ⊓ ⊔ Proposition 12.4. The tangent bundle T M of a manifold M is Hausdorff. Proof. Problem 12.1. ⊓ ⊔ 12.2 The Manifold Structure on the Tangent Bundle Next we show that if {(Uα,φα)}is aC∞atlas for M, then{(TUα,˜φα)}is aC∞atlas for the tangent bundle T M, where ˜φαis the map on TUαinduced by φαas in (12.1). It is clear that T M=/uniontext αTUα. It remains to check that on (TUα)∩(TUβ),˜φαand ˜φβareC∞compatible. Recall that if (U,x1,..., xn),(V,y1,..., yn)are two charts on M, then for any p∈U∩Vthere are two bases singled out for the tangent space TpM:{∂/∂xj|p}n j=1 and{∂/∂yi|p}n i=1. So any tangent vector v∈TpMhas two descriptions: v=∑ jaj∂ ∂xj/vextendsingle/vextendsingle/vextendsingle/vextendsingle p=∑ ibi∂ ∂yi/vextendsingle/vextendsingle/vextendsingle/vextendsingle p. (12.2) It is easy to compare them. By applying both sides to xk, we ﬁnd that ak=/parenleftigg ∑ jaj∂ ∂xj/parenrightigg xk=/parenleftigg ∑ ibi∂ ∂yi/parenrightigg xk=∑ ibi∂xk ∂yi. 12.3 Vector Bundles 133 Similarly, applying both sides of (12.2) to ykgives bk=∑ jaj∂yk ∂xj. (12.3) Returning to the atlas {(Uα,φα)}, we write Uαβ=Uα∩Uβ,φα= (x1,..., xn) andφβ=(y1,..., yn). Then ˜φβ◦˜φ−1 α:φα(Uαβ)×Rn→φβ(Uαβ)×Rn is given by (φα(p),a1,..., an)/ma√s⊔o→/parenleftigg p,∑ jaj∂ ∂xj/vextendsingle/vextendsingle/vextendsingle/vextendsingle p/parenrightigg /ma√s⊔o→/parenleftbig (φβ◦φ−1 α)(φα(p)),b1,..., bn/parenrightbig , where by (12.3) and Example 6.24, bi=∑ jaj∂yi ∂xj(p)=∑ jaj∂(φβ◦φ−1 α)i ∂rj(φα(p)). By the deﬁnition of an atlas, φβ◦φ−1 αisC∞. Therefore, ˜φβ◦˜φ−1 αisC∞. This completes the proof that the tangent bundle T Mis aC∞manifold, with{(TUα,˜φα)} as aC∞atlas. 12.3 Vector Bundles On the tangent bundle T M of a smooth manifold M, the natural projection map π:T M→M,π(p,v)=pmakes TMinto a C∞vector bundle over M, which we now deﬁne. Given any map π:E→M, we call the inverse image π−1(p):=π−1({p})of a point p∈Mtheﬁber at p . The ﬁber at pis often written Ep. For any two maps π:E→Mandπ′:E′→Mwith the same target space M, a map φ:E→E′is said to be ﬁber-preserving ifφ(Ep)⊂E′ pfor all p∈M. Exercise 12.5 (Fiber-preserving maps). Given two maps π:E→Mandπ′:E′→M, check that a map φ:E→E′is ﬁber-preserving if and only if the diagram E π/d31/d31/d63/d63/d63/d63/d63/d63/d63/d63φ/d47/d47E′ π′/d126/d126/d126/d126/d126/d126/d126/d126/d126/d126 M commutes. A surjective smooth map π:E→Mof manifolds is said to be locally trivial of rank r if 134§12 The Tangent Bundle (i) each ﬁber π−1(p)has the structure of a vector space of dimension r; (ii) for each p∈M, there are an open neighborhood Uofpand a ﬁber-preserving diffeomorphism φ:π−1(U)→U×Rrsuch that for every q∈Uthe restriction φ|π−1(q):π−1(q)→{q}×Rr is a vector space isomorphism. Such an open set Uis called a trivializing open setforE, and φis called a trivialization ofEover U. The collection{(U,φ)}, with{U}an open cover of M, is called a local trivialization forE, and{U}is called a trivializing open cover ofMforE. AC∞vector bundle of rank r is a triple (E,M,π)consisting of manifolds EandM and a surjective smooth map π:E→Mthat is locally trivial of rank r. The manifold Eis called the total space of the vector bundle and Mthebase space . By abuse of language, we say that Eis avector bundle over M . For any regular submanifold S⊂M, the triple (π−1S,S,π|π−1S)is aC∞vector bundle over S, called the restriction ofEtoS. We will often write the restriction as E|Sinstead of π−1S. Properly speaking, the tangent bundle of a manifold Mis a triple (T M,M,π), and T Mis the total space of the tangent bundle. In common usage, T Mis often referred to as the tangent bundle. π Fig. 12.2. A circular cylinder is a product bundle over a circle. Example 12.6 ( Product bundle ).Given a manifold M, let π:M×Rr→Mbe the projection to the ﬁrst factor. Then M×Rr→Mis a vector bundle of rank r, called the product bundle of rank rover M. The vector space structure on the ﬁber π−1(p)= {(p,v)|v∈Rr}is the obvious one: (p,u)+( p,v)=( p,u+v),b·(p,v)=( p,bv)forb∈R. A local trivialization on M×Ris given by the identity map 1M×R:M×R→M×R. The inﬁnite cylinder S1×Ris the product bundle of rank 1 over the circle ( Fig- ure 12.2 ). 12.3 Vector Bundles 135 Letπ:E→Mbe aC∞vector bundle. Suppose (U,ψ)=( U,x1,..., xn)is a chart onMand φ:E|U∼→U×Rr,φ(e)=/parenleftbig π(e),c1(e),..., cr(e)/parenrightbig , is a trivialization of Eover U. Then (ψ×1)◦φ=(x1,..., xn,c1,..., cr):E|U∼→U×Rn∼→ψ(U)×Rr⊂Rn×Rr is a diffeomorphism of E|Uonto its image and so is a chart on E. We call x1,..., xn thebase coordinates andc1,..., crtheﬁber coordinates of the chart (E|U,(ψ×1)◦ φ)onE. Note that the ﬁber coordinates cidepend only on the trivialization φof the bundle E|Uand not on the trivialization ψof the base U. LetπE:E→M,πF:F→Nbe two vector bundles, possibly of different ranks. Abundle map from EtoFis a pair of maps (f,˜f),f:M→Nand ˜f:E→F, such that (i) the diagram E˜f/d47/d47 πE /d15/d15F πF /d15/d15 Mf/d47/d47N is commutative, meaning πF◦˜f=f◦πE; (ii) ˜fis linear on each ﬁber; i.e., for each p∈M,˜f:Ep→Ff(p)is a linear map of vector spaces. The collection of all vector bundles together with bundle maps between them forms a category. Example. A smooth map f:N→Mof manifolds induces a bundle map (f,˜f), where ˜f:TN→TMis given by ˜f(p,v)=( f(p),f∗(v))∈{f(p)}×Tf(p)M⊂T M for all v∈TpN. This gives rise to a covariant functor Tfrom the category of smooth manifolds and smooth maps to the category of vector bundles and bun-dle maps: to each manifold M, we associate its tangent bundle T(M), and to each C ∞map f:N→Mof manifolds, we associate the bundle map T(f) =/parenleftbig f:N→M,˜f:T(N)→T(M)/parenrightbig . IfEandFare two vector bundles over the same manifold M, then a bundle map from EtoF over M is a bundle map in which the base map is the identity 1M. For a ﬁxed manifold M, we can also consider the category of all C∞vector bundles over MandC∞bundle maps over M. In this category it makes sense to speak of an isomorphism of vector bundles over M . Any vector bundle over Misomorphic over Mto the product bundle M×Rris called a trivial bundle . 136§12 The Tangent Bundle 12.4 Smooth Sections Asection of a vector bundle π:E→Mis a map s:M→Esuch that π◦s=1M, the identity map on M. This condition means precisely that for each pinM,smaps p into the ﬁber Epabove p. Pictorially we visualize a section as a cross-section of the bundle (Figure 12.3 ). We say that a section is smooth if it is smooth as a map from MtoE. πs(M) /Bullet/Bullet pMs(p) Fig. 12.3. A section of a vector bundle. Deﬁnition 12.7. Avector ﬁeld X on a manifold Mis a function that assigns a tangent vector Xp∈TpMto each point p∈M. In terms of the tangent bundle, a vector ﬁeld onMis simply a section of the tangent bundle π:T M→Mand the vector ﬁeld is smooth if it is smooth as a map from MtoT M. Example 12.8.The formula X(x,y)=−y∂ ∂x+x∂ ∂y=/bracketleftbigg −y x/bracketrightbigg deﬁnes a smooth vector ﬁeld on R2(Figure 12.4 , cf. Example 2.3). Fig. 12.4. The vector ﬁeld (−y,x)inR2. 12.5 Smooth Frames 137 Proposition 12.9. Let s and t be C∞sections of a C∞vector bundle π:E→M and let f be a C∞real-valued function on M. Then (i)the sum s +t:M→E deﬁned by (s+t)(p)=s(p)+t(p)∈Ep,p∈M, is a C∞section of E. (ii)the product f s :M→E deﬁned by (f s)(p)=f(p)s(p)∈Ep,p∈M, is a C∞section of E. Proof. (i) It is clear that s+tis a section of E. To show that it is C∞, ﬁx a point p∈Mand letVbe a trivializing open set for Econtaining p, with C∞trivialization φ:π−1(V)→V×Rr. Suppose (φ◦s)(q)=( q,a1(q),..., ar(q)) and (φ◦t)(q)=( q,b1(q),..., br(q)) forq∈V. Because sandtareC∞maps, aiandbiareC∞functions on V(Proposi- tion 6.16). Since φis linear on each ﬁber, (φ◦(s+t))(q)=( q,a1(q)+b1(q),..., ar(q)+br(q)),q∈V. This proves that s+tis aC∞map on Vand hence at p. Since pis an arbitrary point ofM, the section s+tisC∞onM. (ii) We omit the proof, since it is similar to that of (i). ⊓ ⊔ Denote the set of all C∞sections of EbyΓ(E). The proposition shows that Γ(E) is not only a vector space over R, but also a module over the ring C∞(M)ofC∞ functions on M. For any open subset U⊂M, one can also consider the vector space Γ(U,E)ofC∞sections of Eover U. Then Γ(U,E)is both a vector space over Rand aC∞(U)-module. Note that Γ(M,E)=Γ(E). To contrast with sections over a proper subset U, a section over the entire manifold Mis called a global section . 12.5 Smooth Frames Aframe for a vector bundle π:E→Mover an open set Uis a collection of sections s1,..., srofEover Usuch that at each point p∈U, the elements s1(p),..., sr(p) form a basis for the ﬁber Ep:=π−1(p). A frame s1,..., sris said to be smooth orC∞ 138§12 The Tangent Bundle ifs1,..., srareC∞as sections of Eover U. A frame for the tangent bundle TM→M over an open set Uis simply called a frame on U . Example. The collection of vector ﬁelds ∂/∂x,∂/∂y,∂/∂zis a smooth frame on R3. Example. LetMbe a manifold and e1,..., erthe standard basis for Rn. Deﬁne ¯ei:M→M×Rrby ¯ei(p) = ( p,ei). Then ¯ e1,..., ¯eris aC∞frame for the product bundle M×Rr→M. Example 12.10 ( The frame of a trivialization ).Letπ:E→Mbe a smooth vector bundle of rank r. Ifφ:E|U∼→U×Rris a trivialization of Eover an open set U, then φ−1carries the C∞frame ¯ e1,..., ¯erof the product bundle U×Rrto aC∞frame t1,..., trforEover U: ti(p)=φ−1(¯ei(p))= φ−1(p,ei),p∈U. We call t1,..., trtheC∞frame over U of the trivialization φ. Lemma 12.11. Letφ:E|U→U×Rrbe a trivialization over an open set U of a C∞ vector bundle E→M, and t 1,..., trthe C∞frame over U of the trivialization. Then a section s =∑bitiof E over U is C∞if and only if its coefﬁcients birelative to the frame t 1,..., trare C∞. Proof. (⇐)This direction is an immediate consequence of Proposition 12.9. (⇒)Suppose the section s=∑bitiofEover UisC∞. Then φ◦sisC∞. Note that (φ◦s)(p)=∑bi(p)φ(ti(p))=∑bi(p)(p,ei)=/parenleftbig p,∑bi(p)ei/parenrightbig . Thus, the bi(p)are simply the ﬁber coordinates of s(p)relative to the trivialization φ. Since φ◦sisC∞, all the biareC∞. ⊓ ⊔ Proposition 12.12 (Characterization of C∞sections). Letπ:E→M be a C∞ vector bundle and U an open subset of M. Suppose s 1,..., sris a C∞frame for E over U. Then a section s =∑cjsjof E over U is C∞if and only if the coefﬁcients cj are C∞functions on U. Proof. Ifs1,..., sris the frame of a trivialization of Eover U, then the proposition is Lemma 12.11. We prove the proposition in general by reducing it to this case. One direction is quite easy. If the cj’s are C∞functions on U, then s=∑cjsjis aC∞ section on Uby Proposition 12.9. Conversely, suppose s=∑cjsjis aC∞section of Eover U. Fix a point p∈U and choose a trivializing open set V⊂UforEcontaining p, with C∞trivialization φ:π−1(V)→V×Rr. Let t1,..., trbe the C∞frame of the trivialization φ(Example 12.10). If we write sandsjin terms of the frame t1,..., tr, say s=∑bitiandsj= ∑ai jti, the coefﬁcients bi,ai jwill all be C∞functions on Vby Lemma 12.11. Next, express s=∑cjsjin terms of the ti’s: 12.5 Smooth Frames 139 ∑biti=s=∑cjsj=∑ i,jcjai jti. Comparing the coefﬁcients of tigives bi=∑jcjai j. In matrix notation, b= b1 ... br =A c1 ... cr =Ac. At each point of V, being the transition matrix between two bases, the matrix Ais in- vertible. By Cramer’s rule, A−1is a matrix of C∞functions on V(see Example 6.21). Hence, c=A−1bis a column vector of C∞functions on V. This proves that c1,..., cr areC∞functions at p∈U. Since pis an arbitrary point of U, the coefﬁcients cjare C∞functions on U. ⊓ ⊔ Remark 12.13 .If one replaces “smooth” by “continuous” throughout, the discussion in this subsection remains valid in the continuous category. Problems 12.1.* Hausdorff condition on the tangent bundle Prove Proposition 12.4. 12.2. Transition functions for the total space of the tangent bundle Let(U,φ)=( U,x1,..., xn)and(V,ψ)=( V,y1,..., yn)be overlapping coordinate charts on a manifold M. They induce coordinate charts (TU,˜φ)and(TV,˜ψ)on the total space T Mof the tangent bundle (see equation (12.1)), with transition function ˜ψ◦˜φ−1: (x1,..., xn,a1,..., an)/ma√s⊔o→(y1,..., yn,b1,..., bn). (a) Compute the Jacobian matrix of the transition function ˜ψ◦˜φ−1atφ(p). (b) Show that the Jacobian determinant of the transition function ˜ψ◦˜φ−1atφ(p)is (det[∂yi/∂xj])2. 12.3. Smoothness of scalar multiplication Prove Proposition 12.9(ii). 12.4. Coefﬁcients relative to a smooth frame Letπ:E→Mbe a C∞vector bundle and s1,..., sraC∞frame for Eover an open set Uin M. Then every e∈π−1(U)can be written uniquely as a linear combination e=r ∑ j=1cj(e)sj(p),p=π(e)∈U. Prove that cj:π−1U→RisC∞forj=1,..., r. (Hint: First show that the coefﬁcients of e relative to the frame t1,..., trof a trivialization are C∞.) 140§13 Bump Functions and Partitions of Unity §13 Bump Functions and Partitions of Unity A partition of unity on a manifold is a collection of nonnegative functions that sum to 1. Usually one demands in addition that the partition of unity be subordinate to an open cover{Uα}α∈A. What this means is that the partition of unity {ρα}α∈Ais indexed by the same set as the open over {Uα}α∈Aand for each αin the index A, the support of ραis contained in Uα. In particular, ραvanishes outside Uα. The existence of a C∞partition of unity is one of the most important technical tools in the theory of C∞manifolds. It is the single feature that makes the behavior ofC∞manifolds so different from that of real-analytic or complex manifolds. In this section we construct C∞bump functions on any manifold and prove the existence of a C∞partition of unity on a compact manifold. The proof of the existence of a C∞partition of unity on a general manifold is more technical and is postponed to Appendix C. A partition of unity is used in two ways: (1) to decompose a global object on a manifold into a locally ﬁnite sum of local objects on the open sets Uαof an open cover, and (2) to patch together local objects on the open sets Uαinto a global object on the manifold. Thus, a partition of unity serves as a bridge between global andlocal analysis on a manifold. This is useful because while there are always local co- ordinates on a manifold, there may be no global coordinates. In subsequent sections we will see examples of both uses of a C ∞partition of unity. 13.1 C∞Bump Functions Recall that R×denotes the set of nonzero real numbers. The support of a real-valued function fon a manifold Mis deﬁned to be the closure in Mof the subset on which f/ne}a⊔ionslash=0: supp f=clM(f−1(R×))= closure of{q∈M|f(q)/ne}a⊔ionslash=0}inM.1 Letqbe a point in M, and Ua neighborhood of q. By a bump function at q supported in U we mean any continuous nonnegative function ρonMthat is 1 in a neighborhood of qwith supp ρ⊂U. For example, Figure 13.1 is the graph of a bump function at 0 supported in the open interval ]−2,2[. The function is nonzero on the open interval ]−1,1[and is zero otherwise. Its support is the closed interval [−1,1]. Example. The support of the function f:]−1,1[→R,f(x) =tan(πx/2), is the open interval ]−1,1[, not the closed interval [−1,1], because the closure of f−1(R×) is taken in the domain ]−1,1[, not inR. 1In this section a general point is often denoted by q, instead of p, because presembles too much ρ, the notation for a bump function. 13.1 C∞Bump Functions 141 1 2−1−21 Fig. 13.1. A bump function at 0 on R. The only bump functions of interest to us are C∞bump functions. While the continuity of a function can often be seen by inspection, the smoothness of a function always requires a formula. Our goal in this subsection is to ﬁnd a formula for a C∞ bump function as in Figure 13.1. Example. The graph of y=x5/3looks perfectly smooth (Figure 13.2 ), but it is in fact not smooth at x=0, since its second derivative y′′=(10/9)x−1/3is not deﬁned there. 1 −11 −1xy Fig. 13.2. The graph of y=x5/3. In Example 1.3 we introduced the C∞function f(t)=/braceleftigg e−1/tfort>0, 0 for t≤0, with graph as in Figure 13.3. 1 −1t1 f(t) Fig. 13.3. The graph of f(t). 142§13 Bump Functions and Partitions of Unity The main challenge in building a smooth bump function from fis to construct a smooth version of a step function, that is, a C∞function g:R→Rwith graph as in Figure 13.4. Once we have such a C∞step function g, it is then simply a matter of 11g(t) tg(t)=/braceleftigg 0 for t≤0, 1 for t≥1. Fig. 13.4. The graph of g(t). translating, reﬂecting, and scaling the function in order to make its graph look like Figure 13.1. We seek g(t)by dividing f(t)by a positive function ℓ(t), for the quotient f(t)/ℓ(t)will then be zero for t≤0. The denominator ℓ(t)should be a positive function that agrees with f(t)fort≥1, for then f(t)/ℓ(t)will be identically 1 for t≥1. The simplest way to construct such an ℓ(t)is to add to f(t)a nonnegative function that vanishes for t≥1. One such nonnegative function is f(1−t). This suggests that we take ℓ(t)=f(t)+f(1−t)and consider g(t)=f(t) f(t)+f(1−t). (13.1) Let us verify that the denominator f(t)+f(1−t)is never zero. For t>0,f(t)> 0 and therefore f(t)+f(1−t)≥f(t)>0. Fort≤0, 1−t≥1 and therefore f(t)+f(1−t)≥f(1−t)>0. In either case, f(t)+f(1−t)/ne}a⊔ionslash=0. This proves that g(t)is deﬁned for all t. As the quotient of two C∞functions with denominator never zero, g(t)isC∞for all t. As noted above, for t≤0, the numerator f(t)equals 0, so g(t)is identically zero fort≤0. For t≥1, we have 1−t≤0 and f(1−t) =0, so g(t) =f(t)/f(t)is identically 1 for t≥1. Thus, gis aC∞step function with the desired properties. Given two positive real numbers a<b, we make a linear change of variables to map[a2,b2]to[0,1]: x/ma√s⊔o→x−a2 b2−a2. Let h(x)=g/parenleftbiggx−a2 b2−a2/parenrightbigg . Then h:R→[0,1]is aC∞step function such that 13.1 C∞Bump Functions 143 h(x)=/braceleftigg 0 for x≤a2, 1 for x≥b2. (See Figure 13.5. ) 1 h(x) xa2b2 Fig. 13.5. The graph of h(x). Replace xbyx2to make the function symmetric in x:k(x)=h(x2)(Figure 13.6 ). 1 k(x) xa b −a−b Fig. 13.6. The graph of k(x). Finally, set ρ(x)=1−k(x)=1−g/parenleftbiggx2−a2 b2−a2/parenrightbigg . This ρ(x)is aC∞bump function at 0 in Rthat is identically 1 on [−a,a]and has support in [−b,b](Figure 13.7 ). For any q∈R,ρ(x−q)is aC∞bump function at q. ρ(x) xa b −a−b1 Fig. 13.7. A bump function at 0 on R. It is easy to extend the construction of a bump function from RtoRn. To get a C∞bump function at 0inRnthat is 1 on the closed ball B(0,a)and has support in the closed ball B(0,b), set σ(x)=ρ(/bardblx/bardbl)=1−g/parenleftbigg/bardblx/bardblr2−a2 b2−a2/parenrightbigg . (13.2) 144§13 Bump Functions and Partitions of Unity As a composition of C∞functions, σisC∞. To get a C∞bump function at qinRn, take σ(x−q). Exercise 13.1 (Bump function supported in an open set).* Letqbe a point and Uany neighborhood of qin a manifold. Construct a C∞bump function at qsupported in U. In general, a C∞function on an open subset Uof a manifold Mcannot be ex- tended to a C∞function on M; an example is the function sec (x)on the open interval ]−π/2,π/2[inR. However, if we require that the global function on Magree with the given function only on some neighborhood of a point in U, then a C∞extension is possible. Proposition 13.2 ( C∞extension of a function). Suppose f is a C∞function deﬁned on a neighborhood U of a point p in a manifold M. Then there is a C∞function ˜f on M that agrees with f in some possibly smaller neighborhood of p. ( )/Bullet p U| |1 ρf Fig. 13.8. Extending the domain of a function by multiplying by a bump function. Proof. Choose a C∞bump function ρ:M→Rsupported in Uthat is identically 1 in a neighborhood Vofp(Figure 13.8 ). Deﬁne ˜f(q)=/braceleftigg ρ(q)f(q)forqinU, 0 for qnot in U. As the product of two C∞functions on U,˜fisC∞onU. Ifq/∈U, then q/∈supp ρ, and so there is an open set containing qon which ˜fis 0, since supp ρis closed. Therefore, ˜fis also C∞at every point q/∈U. Finally, since ρ≡1 on V, the function ˜fagrees with fonV.⊓ ⊔ 13.2 Partitions of Unity 145 13.2 Partitions of Unity If{Ui}i∈Iis a ﬁnite open cover of M, aC∞partition of unity subordinate to {Ui}i∈I is a collection of nonnegative C∞functions{ρi:M→R}i∈Isuch that supp ρi⊂Ui and ∑ρi=1. (13.3) When Iis an inﬁnite set, for the sum in (13.3) to make sense, we will impose alocal ﬁniteness condition. A collection {Aα}of subsets of a topological space S is said to be locally ﬁnite if every point qinShas a neighborhood that meets only ﬁnitely many of the sets Aα. In particular, every qinSis contained in only ﬁnitely many of the Aα’s. Example 13.3 ( An open cover that is not locally ﬁnite ).LetUr,nbe the open interval/bracketrightbig r−1 n,r+1 n/bracketleftbig on the real line R. The open cover{Ur,n|r∈Q,n∈Z+}ofRis not locally ﬁnite. Deﬁnition 13.4. AC∞partition of unity on a manifold is a collection of nonnegative C∞functions{ρα:M→R}α∈Asuch that (i) the collection of supports, {supp ρα}α∈A, is locally ﬁnite, (ii)∑ρα=1. Given an open cover {Uα}α∈AofM, we say that a partition of unity {ρα}α∈Ais subordinate to the open cover {Uα}if supp ρα⊂Uαfor every α∈A. Since the collection of supports, {supp ρα}α∈A, is locally ﬁnite (condition (i)), every point qlies in only ﬁnitely many of the sets supp ρα. Hence ρα(q)/ne}a⊔ionslash=0 for only ﬁnitely many α. It follows that the sum in (ii) is a ﬁnite sum at every point. Example. LetUandVbe the open intervals ]−∞,2[and]−1,∞[inRrespectively, and let ρVbe a C∞function with graph as in Figure 13.9, for example the function g(t)in (13.1). Deﬁne ρU=1−ρV. Then supp ρV⊂Vand supp ρU⊂U. Thus, {ρU,ρV}is a partition of unity subordinate to the open cover {U,V}. 1 2−1−21 ρV R1 U V) ( Fig. 13.9. A partition of unity {ρU,ρV}subordinate to an open cover {U,V}. 146§13 Bump Functions and Partitions of Unity Remark. Suppose{fα}α∈Ais a collection of C∞functions on a manifold Msuch that the collection of its supports, {supp fα}α∈A, is locally ﬁnite. Then every point qin Mhas a neighborhood Wqthat intersects supp fαfor only ﬁnitely many α. Thus, on Wqthe sum ∑α∈Afαis actually a ﬁnite sum. This shows that the function f=∑fα is well deﬁned and C∞on the manifold M. We call such a sum a locally ﬁnite sum. 13.3 Existence of a Partition of Unity In this subsection we begin a proof of the existence of a C∞partition of unity on a manifold. Because the case of a compact manifold is somewhat easier and already has some of the features of the general case, for pedagogical reasons we give a sep-arate proof for the compact case. Lemma 13.5. If ρ1,..., ρmare real-valued functions on a manifold M, then supp/parenleftbig∑ρi/parenrightbig ⊂/uniondisplay supp ρi. Proof. Problem 13.1. ⊓ ⊔ Proposition 13.6. Let M be a compact manifold and {Uα}α∈Aan open cover of M. There exists a C∞partition of unity{ρα}α∈Asubordinate to{Uα}α∈A. Proof. For each q∈M, ﬁnd an open set Uαcontaining qfrom the given cover and letψqbe aC∞bump function at qsupported in Uα(Exercise 13.1, p. 144). Because ψq(q)>0, there is a neighborhood Wqofqon which ψq>0. By the compactness ofM, the open cover{Wq|q∈M}has a ﬁnite subcover, say {Wq1,..., Wqm}. Let ψq1,..., ψqmbe the corresponding bump functions. Then ψ:=∑ψqiis positive at every point qinMbecause q∈Wqifor some i. Deﬁne ϕi=ψqi ψ,i=1,..., m. Clearly, ∑ϕi=1. Moreover, since ψ>0,ϕi(q)/ne}a⊔ionslash=0 if and only if ψqi(q)/ne}a⊔ionslash=0, so supp ϕi=supp ψqi⊂Uα for some α∈A. This shows that {ϕi}is a partition of unity such that for every i, supp ϕi⊂Uαfor some α∈A. The next step is to make the index set of the partition of unity the same as that of the open cover. For each i=1,..., m, choose τ(i)∈A to be an index such that supp ϕi⊂Uτ(i). We group the collection of functions {ϕi}into subcollections according to τ(i)and deﬁne for each α∈A, ρα=∑ τ(i)=αϕi; 13.3 Existence of a Partition of Unity 147 if there is no ifor which τ(i) =α, the sum above is empty and we deﬁne ρα=0. Then ∑ α∈Aρα=∑ α∈A∑ τ(i)=αϕi=m ∑ i=1ϕi=1. Moreover, by Lemma 13.5, supp ρα⊂/uniondisplay τ(i)=αsupp ϕi⊂Uα. So{ρα}is a partition of unity subordinate to {Uα}. ⊓ ⊔ To generalize the proof of Proposition 13.6 to an arbitrary manifold, it will be necessary to ﬁnd an appropriate substitute for compactness. Since the proof is rathertechnical and is not necessary for the rest of the book, we put it in Appendix C. The statement is as follows. Theorem 13.7 (Existence of a C ∞partition of unity). Let{Uα}α∈Abe an open cover of a manifold M. (i)There is a C∞partition of unity{ϕk}∞ k=1with every ϕkhaving compact support such that for each k, supp ϕk⊂Uαfor some α∈A. (ii)If we do not require compact support, then there is a C∞partition of unity{ρα} subordinate to{Uα}. Problems 13.1.* Support of a ﬁnite sum Prove Lemma 13.5. 13.2.* Locally ﬁnite family and compact set Let{Aα}be a locally ﬁnite family of subsets of a topological space S. Show that every compact set KinShas a neighborhood Wthat intersects only ﬁnitely many of the Aα. 13.3. Smooth Urysohn lemma (a) Let AandBbe two disjoint closed sets in a manifold M. Find a C∞function fonMsuch that fis identically 1 on Aand identically 0 on B. (Hint: Consider a C∞partition of unity {ρM−A,ρM−B}subordinate to the open cover {M−A,M−B}. This lemma is needed in Subsection 29.3.) (b) Let Abe a closed subset and Uan open subset of a manifold M. Show that there is a C∞ function fonMsuch that fis identically 1 on Aand supp f⊂U. 13.4. Support of the pullback of a function LetF:N→Mbe a C∞map of manifolds and h:M→RaC∞real-valued function. Prove that supp F∗h⊂F−1(supp h). (Hint: First show that (F∗h)−1(R×)⊂F−1(supp h).) 148§13 Bump Functions and Partitions of Unity 13.5.* Support of the pullback by a projection Letf:M→Rbe aC∞function on a manifold M. IfNis another manifold and π:M×N→M is the projection onto the ﬁrst factor, prove that supp(π∗f)=(supp f)×N. 13.6. Pullback of a partition of unity Suppose{ρα}is a partition of unity on a manifold Msubordinate to an open cover {Uα}of MandF:N→Mis aC∞map. Prove that (a) the collection of supports {supp F∗ρα}is locally ﬁnite; (b) the collection of functions {F∗ρα}is a partition of unity on Nsubordinate to the open cover{F−1(Uα)}ofN. 13.7.* Closure of a locally ﬁnite union If{Aα}is a locally ﬁnite collection of subsets in a topological space, then /uniondisplay Aα=/uniondisplay Aα, (13.4) where Adenotes the closure of the subset A. Remark. For any collection of subsets Aα, one always has /uniondisplay Aα⊂/uniondisplay Aα. However, the reverse inclusion is in general not true. For example, suppose Anis the closed interval[0,1−(1/n)]inR. Then ∞/uniondisplay n=1An=[0,1)=[ 0,1], but∞/uniondisplay n=1An=∞/uniondisplay n=1/bracketleftbigg 0,1−1 n/bracketrightbigg =[0,1). If{Aα}is a ﬁnite collection, the equality (13.4) is easily shown to be true. 14.1 Smoothness of a Vector Field 149 §14 Vector Fields A vector ﬁeld Xon a manifold Mis the assignment of a tangent vector Xp∈TpM to each point p∈M. More formally, a vector ﬁeld on Mis a section of the tangent bundle T MofM. It is natural to deﬁne a vector ﬁeld as smooth if it is smooth as a section of the tangent bundle. In the ﬁrst subsection we give two other characteriza- tions of smooth vector ﬁelds, in terms of the coefﬁcients relative to coordinate vector ﬁelds and in terms of smooth functions on the manifold. Vector ﬁelds abound in nature, for example the velocity vector ﬁeld of a ﬂuid ﬂow, the electric ﬁeld of a charge, the gravitational ﬁeld of a mass, and so on. The ﬂuid ﬂow model is in fact quite general, for as we will see shortly, every smoothvector ﬁeld may be viewed locally as the velocity vector ﬁeld of a ﬂuid ﬂow. The path traced out by a point under this ﬂow is called an integral curve of the vector ﬁeld. Integral curves are curves whose velocity vector ﬁeld is the restriction of the given vector ﬁeld to the curve. Finding the equation of an integral curve is equivalent to solving a system of ﬁrst-order ordinary differential equations (ODE). Thus, thetheory of ODE guarantees the existence of integral curves. The setX(M)of all C ∞vector ﬁelds on a manifold Mclearly has the structure of a vector space. We introduce a bracket operation [,]that makes it into a Lie algebra. Because vector ﬁelds do not push forward under smooth maps, the Lie algebra X(M) does not give rise to a functor on the category of smooth manifolds. Nonetheless, there is a notion of related vector ﬁelds that allows us to compare vector ﬁelds on two manifolds under a smooth map. 14.1 Smoothness of a Vector Field In Deﬁnition 12.7 we deﬁned a vector ﬁeld Xon a manifold Mto be smooth if the map X:M→T M is smooth as a section of the tangent bundle π:T M→M. In a coordinate chart (U,φ) = ( U,x1,..., xn)onM, the value of the vector ﬁeld Xat p∈Uis a linear combination Xp=∑ai(p)∂ ∂xi/vextendsingle/vextendsingle/vextendsingle/vextendsingle p. Aspvaries in U, the coefﬁcients aibecome functions on U. As we learned in Subsections 12.1 and 12.2, the chart (U,φ)=( U,x1,..., xn)on the manifold Minduces a chart (TU,˜φ)=( TU,¯x1,..., ¯xn,c1,..., cn) on the tangent bundle T M, where ¯ xi=π∗xi=xi◦πand the ciare deﬁned by v=∑ci(v)∂ ∂xi/vextendsingle/vextendsingle/vextendsingle/vextendsingle p,v∈TpM. 150§14 Vector Fields Comparing coefﬁcients in Xp=∑ai(p)∂ ∂xi/vextendsingle/vextendsingle/vextendsingle/vextendsingle p=∑ci(Xp)∂ ∂xi/vextendsingle/vextendsingle/vextendsingle/vextendsingle p,p∈U, we get ai=ci◦Xas functions on U. Being coordinates, the ciare smooth func- tions on TU. Thus, if Xis smooth and (U,x1,..., xn)is any chart on M, then the coefﬁcients aiofX=∑ai∂/∂xirelative to the frame ∂/∂xiare smooth on U. The converse is also true, as indicated in the following lemma. Lemma 14.1 (Smoothness of a vector ﬁeld on a chart). Let(U,φ)=(U,x1,..., xn) be a chart on a manifold M. A vector ﬁeld X =∑ai∂/∂xion U is smooth if and only if the coefﬁcient functions aiare all smooth on U. Proof. This lemma is a special case of Proposition 12.12, with Ethe tangent bundle ofMandsithe coordinate vector ﬁeld ∂/∂xi. Because we have an explicit description of the manifold structure on the tangent bundle T M, a direct proof of the lemma is also possible. Since ˜φ:TU∼→U×Rn is a diffeomorphism, X:U→TUis smooth if and only if ˜φ◦X:U→U×Rnis smooth. For p∈U, (˜φ◦X)(p)=˜φ(Xp)=/parenleftbig x1(p),..., xn(p),c1(Xp),..., cn(Xp)/parenrightbig =/parenleftbig x1(p),..., xn(p),a1(p),..., an(p)/parenrightbig . As coordinate functions, x1,..., xnareC∞onU. Therefore, by Proposition 6.13, ˜φ◦Xis smooth if and only if all the aiare smooth on U. ⊓ ⊔ This lemma leads to a characterization of the smoothness of a vector ﬁeld on a manifold in terms of the coefﬁcients of the vector ﬁeld relative to coordinate frames. Proposition 14.2 (Smoothness of a vector ﬁeld in terms of coefﬁcients). Let X be a vector ﬁeld on a manifold M. The following are equivalent: (i)The vector ﬁeld X is smooth on M. (ii)The manifold M has an atlas such that on any chart (U,φ) =( U,x1,..., xn)of the atlas, the coefﬁcients aiof X=∑ai∂/∂xirelative to the frame ∂/∂xiare all smooth. (iii)On any chart (U,φ) = ( U,x1,..., xn)on the manifold M, the coefﬁcients aiof X=∑ai∂/∂xirelative to the frame ∂/∂xiare all smooth. Proof. (ii)⇒(i): Assume (ii). By the preceding lemma, Xis smooth on every chart (U,φ)of an atlas of M. Thus, Xis smooth on M. (i)⇒(iii): A smooth vector ﬁeld on Mis smooth on every chart (U,φ)onM. The preceding lemma then implies (iii). (iii)⇒(ii): Obvious. ⊓ ⊔ 14.1 Smoothness of a Vector Field 151 Just as in Subsection 2.5, a vector ﬁeld Xon a manifold Minduces a linear map on the algebra C∞(M)ofC∞functions on M: for f∈C∞(M), deﬁne X fto be the function (X f)(p)=Xpf,p∈M. In terms of its action as an operator on C∞functions, there is still another character- ization of a smooth vector ﬁeld. Proposition 14.3 (Smoothness of a vector ﬁeld in terms of functions). A vector ﬁeld X on M is smooth if and only if for every smooth function f on M, the function X f is smooth on M. Proof. (⇒)Suppose Xis smooth and f∈C∞(M). By Proposition 14.2, on any chart (U,x1,..., xn)onM, the coefﬁcients aiof the vector ﬁeld X=∑ai∂/∂xiareC∞. It follows that X f=∑ai∂f/∂xiisC∞onU. Since Mcan be covered by charts, X f isC∞onM. (⇐)Let(U,x1,..., xn)be any chart on M. Suppose X=∑ai∂/∂xionUandp∈U. By Proposition 13.2, for k=1,..., n, each xkcan be extended to a C∞function ˜ xkon Mthat agrees with xkin a neighborhood VofpinU. Therefore, on V, X˜xk=/parenleftbigg ∑ai∂ ∂xi/parenrightbigg ˜xk=/parenleftbigg ∑ai∂ ∂xi/parenrightbigg xk=ak. This proves that akisC∞atp. Since pis an arbitrary point in U, the function akis C∞onU. By the smoothness criterion of Proposition 14.2, Xis smooth. In this proof it is necessary to extend xkto aC∞global function ˜ xkonM, for while it is true that Xxk=ak, the coordinate function xkis deﬁned only on U, not on M, and so the smoothness hypothesis on X fdoes not apply to Xxk.⊓ ⊔ By Proposition 14.3, we may view a C∞vector ﬁeld Xas a linear operator X:C∞(M)→C∞(M)on the algebra of C∞functions on M. As in Proposition 2.6, this linear operator X:C∞(M)→C∞(M)is a derivation: for all f,g∈C∞(M), X(f g)=( X f)g+f(Xg). In the following we think of C∞vector ﬁelds on Malternately as C∞sections of the tangent bundle T M and as derivations on the algebra C∞(M)ofC∞functions. In fact, it can be shown that these two descriptions of C∞vector ﬁelds are equivalent (Problem 19.12). Proposition 13.2 on C∞extensions of functions has an analogue for vector ﬁelds. Proposition 14.4 ( C∞extension of a vector ﬁeld). Suppose X is a C∞vector ﬁeld deﬁned on a neighborhood U of a point p in a manifold M. Then there is a C∞vector ﬁeld ˜X on M that agrees with X on some possibly smaller neighborhood of p. 152§14 Vector Fields Proof. Choose a C∞bump function ρ:M→Rsupported in Uthat is identically 1 in a neighborhood Vofp(Figure 13.8 ). Deﬁne ˜X(q)=/braceleftigg ρ(q)XqforqinU, 0 for qnot in U. The rest of the proof is the same as in Proposition 13.2. ⊓ ⊔ 14.2 Integral Curves In Example 12.8, it appears that through each point in the plane one can draw a circle whose velocity at any point is the given vector ﬁeld at that point. Such a curve is an example of an integral curve of the vector ﬁeld, which we now deﬁne. Deﬁnition 14.5. LetXbe aC∞vector ﬁeld on a manifold M, and p∈M. An integral curve ofXis a smooth curve c:]a,b[→Msuch that c′(t) =Xc(t)for all t∈]a,b[. Usually we assume that the open interval ]a,b[contains 0. In this case, if c(0)=p, then we say that cis an integral curve starting at p and call ptheinitial point ofc. To show the dependence of such an integral curve on the initial point p, we also write ct(p)instead of c(t). Deﬁnition 14.6. An integral curve is maximal if its domain cannot be extended to a larger interval. Example. Recall the vector ﬁeld X(x,y)=/an}bracke⊔le{⊔−y,x/an}bracke⊔ri}h⊔onR2(Figure 12.4 ). We will ﬁnd an integral curve c(t)ofXstarting at the point (1,0)∈R2. The condition for c(t)= (x(t),y(t))to be an integral curve is c′(t)=Xc(t), or /bracketleftbigg˙x(t) ˙y(t)/bracketrightbigg =/bracketleftbigg−y(t) x(t)/bracketrightbigg , so we need to solve the system of ﬁrst-order ordinary differential equations ˙x=−y, (14.1) ˙y=x, (14.2) with initial condition (x(0),y(0))=( 1,0). From (14.1), y=−˙x, so ˙y=−¨x. Substi- tuting into (14.2) gives ¨x=−x. It is well known that the general solution to this equation is x=Acost+Bsint. (14.3) Hence, y=−˙x=Asint−Bcost. (14.4) 14.2 Integral Curves 153 The initial condition forces A=1,B=0, so the integral curve starting at (1,0)is c(t)=( cost,sint), which parametrizes the unit circle. More generally, if the initial point of the integral curve, corresponding to t=0, isp=(x0,y0), then (14.3) and (14.4) give A=x0,B=−y0, and the general solution to (14.1) and (14.2) is x=x0cost−y0sint, y=x0sint+y0cost,t∈R. This can be written in matrix notation as c(t)=/bracketleftbiggx(t) y(t)/bracketrightbigg =/bracketleftbiggcost−sint sintcost/bracketrightbigg/bracketleftbiggx0 y0/bracketrightbigg =/bracketleftbiggcost−sint sintcost/bracketrightbigg p, which shows that the integral curve of Xstarting at pcan be obtained by rotating the point pcounterclockwise about the origin through an angle t. Notice that cs(ct(p))= cs+t(p), since a rotation through an angle tfollowed by a rotation through an angle sis the same as a rotation through the angle s+t. For each t∈R,ct:R2→R2is a diffeo- morphism with inverse c−t. Let Diff(M)be the group of diffeomorphisms of a manifold Mwith itself, the group operation being composition. A homomorphism c:R→Diff(M)is called a one-parameter group of diffeomorphisms ofM. In this example the integral curves of the vector ﬁeld X(x,y)=/an}bracke⊔le{⊔−y,x/an}bracke⊔ri}h⊔onR2give rise to a one-parameter group of dif- feomorphisms of R2. Example. LetXbe the vector ﬁeld x2d/dxon the real line R. Find the maximal integral curve of Xstarting at x=2. Solution. Denote the integral curve by x(t). Then x′(t)=Xx(t)⇐⇒ ˙x(t)d dx=x2d dx, where x′(t)is the velocity vector of the curve x(t), and ˙ x(t)is the calculus derivative of the real-valued function x(t). Thus, x(t)satisﬁes the differential equation dx dt=x2,x(0)=2. (14.5) On can solve (14.5) by separation of variables: dx x2=dt. (14.6) 154§14 Vector Fields Integrating both sides of (14.6) gives −1 x=t+C,orx=−1 t+C, for some constant C. The initial condition x(0)=2 forces C=−1/2. Hence, x(t)= 2/(1−2t). The maximal interval containing 0 on which x(t)is deﬁned is ]−∞,1/2[. From this example we see that it may not be possible to extend the domain of deﬁnition of an integral curve to the entire real line. 14.3 Local Flows The two examples in the preceding section illustrate the fact that locally, ﬁnding anintegral curve of a vector ﬁeld amounts to solving a system of ﬁrst-order ordinary differential equations with initial conditions. In general, if Xis a smooth vector ﬁeld on a manifold M, to ﬁnd an integral curve c(t)ofXstarting at p, we ﬁrst choose a coordinate chart (U, φ)=( U,x1,..., xn)about p. In terms of the local coordinates, Xc(t)=∑ai(c(t))∂ ∂xi/vextendsingle/vextendsingle/vextendsingle/vextendsingle c(t), and by Proposition 8.15, c′(t)=∑˙ci(t)∂ ∂xi/vextendsingle/vextendsingle/vextendsingle/vextendsingle c(t), where ci(t)=xi◦c(t)is the ith component of c(t)in the chart (U,φ). The condition c′(t)=Xc(t)is thus equivalent to ˙ci(t)=ai(c(t))fori=1,..., n. (14.7) This is a system of ordinary differential equations (ODE); the initial condition c(0)= ptranslates to (c1(0),..., cn(0)) = ( p1,..., pn). By an existence and uniqueness theorem from the theory of ODE, such a system always has a unique solution in the following sense. Theorem 14.7. Let V be an open subset of Rn, p0a point in V, and f :V→Rna C∞ function. Then the differential equation dy/dt=f(y),y(0)=p0, has a unique C∞solution y :]a(p0),b(p0)[→V, where ]a(p0),b(p0)[is the maximal open interval containing 0on which y is deﬁned. The uniqueness of the solution means that if z:]δ,ε[→Vsatisﬁes the same differential equation dz/dt=f(z),z(0)=p0, 14.3 Local Flows 155 then the domain of deﬁnition ]δ,ε[ofzis a subset of ]a(p0),b(p0)[andz(t) =y(t) on the interval ]δ,ε[. For a vector ﬁeld Xon a chart Uof a manifold and a point p∈U, this theorem guarantees the existence and uniqueness of a maximal integral curve starting at p. Next we would like to study the dependence of an integral curve on its initial point. Again we study the problem locally on Rn. The function ywill now be a function of two arguments tandq, and the condition for yto be an integral curve starting at the point qis ∂y ∂t(t,q)=f(y(t,q)),y(0,q)=q. (14.8) The following theorem from the theory of ODE guarantees the smooth depen- dence of the solution on the initial point. Theorem 14.8. Let V be an open subset of Rnand f :V→Rna C∞function on V. For each point p 0∈V, there are a neighborhood W of p 0in V, a number ε>0, and a C∞function y:]−ε,ε[×W→V such that ∂y ∂t(t,q)=f(y(t,q)),y(0,q)=q for all(t,q)∈]−ε,ε[×W. For a proof of these two theorems, see [7, Appendix C, pp. 359–366]. It follows from Theorem 14.8 and (14.8) that if Xis any C∞vector ﬁeld on a chart Uandp∈U, then there are a neighborhood WofpinU, anε>0, and a C∞ map F:]−ε,ε[×W→U (14.9) such that for each q∈W, the function F(t,q)is an integral curve of Xstarting at q. In particular, F(0,q)=q. We usually write Ft(q)forF(t,q). /Bullet/Bullet/Bullet qFs(q)Ft(Fs(q))= Ft+s(q) Fig. 14.1. The ﬂow line through qof a local ﬂow. Suppose s,tin the interval ]−ε,ε[are such that both Ft(Fs(q))andFt+s(q)are deﬁned. Then both Ft(Fs(q))andFt+s(q)as functions of tare integral curves of X with initial point Fs(q), which is the point corresponding to t=0. By the uniqueness of the integral curve starting at a point, 156§14 Vector Fields Ft(Fs(q))= Ft+s(q). (14.10) The map Fin (14.9) is called a local ﬂow generated by the vector ﬁeld X . For each q∈U, the function Ft(q)oftis called a ﬂow line of the local ﬂow. Each ﬂow line is an integral curve of X. If a local ﬂow Fis deﬁned on R×M, then it is called a global ﬂow. Every smooth vector ﬁeld has a local ﬂow about any point, but not necessarily a global ﬂow. A vector ﬁeld having a global ﬂow is called a complete vector ﬁeld . If Fis a global ﬂow, then for every t∈R, Ft◦F−t=F−t◦Ft=F0=1M, soFt:M→Mis a diffeomorphism. Thus, a global ﬂow on Mgives rise to a one- parameter group of diffeomorphisms of M. This discussion suggests the following deﬁnition. Deﬁnition 14.9. Alocal ﬂow about a point pin an open set Uof a manifold is a C∞ function F:]−ε,ε[×W→U, where εis a positive real number and Wis a neighborhood of pinU, such that writing Ft(q)=F(t,q), we have (i)F0(q)=qfor all q∈W, (ii)Ft(Fs(q))= Ft+s(q)whenever both sides are deﬁned. IfF(t,q)is a local ﬂow of the vector ﬁeld XonU, then F(0,q)=qand∂F ∂t(0,q)=XF(0,q)=Xq. Thus, one can recover the vector ﬁeld from its ﬂow. Example. The function F:R×R2→R2, F/parenleftbigg t,/bracketleftbigg x y/bracketrightbigg/parenrightbigg =/bracketleftbigg cost−sint sintcost/bracketrightbigg/bracketleftbigg x y/bracketrightbigg , is the global ﬂow on R2generated by the vector ﬁeld X(x,y)=∂F ∂t(t,(x,y))/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0=/bracketleftbigg−sint−cost cost−sint/bracketrightbigg/bracketleftbiggx y/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0 =/bracketleftbigg 0−1 1 0/bracketrightbigg/bracketleftbigg x y/bracketrightbigg =/bracketleftbigg −y x/bracketrightbigg =−y∂ ∂x+x∂ ∂y. This is Example 12.8 again. 14.4 The Lie Bracket 157 14.4 The Lie Bracket Suppose XandYare smooth vector ﬁelds on an open subset Uof a manifold M. We view XandYas derivations on C∞(U). For a C∞function fonU, by Proposi- tion 14.3 the function Y fisC∞onU, and the function (XY)f:=X(Y f)is also C∞ onU. Moreover, because XandYare bothR-linear maps from C∞(U)toC∞(U), the map XY:C∞(U)→C∞(U)isR-linear. However, XYdoes not satisfy the derivation property: if f,g∈C∞(U), then XY(f g)=X((Y f)g+fYg) =(XY f)g+(Y f)(Xg)+(X f)(Y g)+f(XYg). Looking more closely at this formula, we see that the two extra terms (Y f)(Xg) and(X f)(Y g)that make XYnot a derivation are symmetric in XandY. Thus, if we compute Y X(f g)as well and subtract it from XY(f g), the extra terms will disappear, andXY−YXwill be a derivation of C∞(U). Given two smooth vector ﬁelds XandYonUandp∈U, we deﬁne their Lie bracket[X,Y]atpto be [X,Y]pf=(XpY−YpX)f for any germ fof a C∞function at p. By the same calculation as above, but now evaluated at p, it is easy to check that [X,Y]pis a derivation of C∞ p(U)and is therefore a tangent vector at p(Deﬁnition 8.1). As pvaries over U,[X,Y]becomes a vector ﬁeld on U. Proposition 14.10. If X and Y are smooth vector ﬁelds on M, then the vector ﬁeld [X,Y]is also smooth on M. Proof. By Proposition 14.3 it sufﬁces to check that if fis aC∞function on M, then so is[X,Y]f. But [X,Y]f=(XY−YX)f, which is clearly C∞onM, since both XandYare. ⊓ ⊔ From this proposition, we see that the Lie bracket provides a product operation on the vector space X(M)of all smooth vector ﬁelds on M. Clearly, [Y,X]=−[X,Y]. Exercise 14.11 (Jacobi identity). Check the Jacobi identity : ∑ cyclic[X,[Y,Z]]=0. This notation means that one permutes X,Y,Zcyclically and one takes the sum of the resulting terms. Written out, ∑ cyclic[X,[Y,Z]]=[ X,[Y,Z]]+[Y,[Z,X]]+[ Z,[X,Y]]. 158§14 Vector Fields Deﬁnition 14.12. LetKbe a ﬁeld. A Lie algebra over Kis a vector space Vover K together with a product [,]:V×V→V, called the bracket , satisfying the following properties: for all a,b∈KandX,Y,Z∈V, (i) (bilinearity) [aX+bY,Z]=a[X,Z]+b[Y,Z], [Z,aX+bY]=a[Z,X]+b[Z,Y], (ii) (anticommutativity) [Y,X]=−[X,Y], (iii) (Jacobi identity) ∑cyclic[X,[Y,Z]]=0. In practice, we will be concerned only with real Lie algebras , i.e., Lie algebras overR. Unless otherwise speciﬁed, a Lie algebra in this book means a real Lie algebra. Example. On any vector space V, deﬁne[X,Y] =0 for all X,Y∈V. With this bracket, Vbecomes a Lie algebra, called an abelian Lie algebra . Our deﬁnition of an algebra in Subsection 2.2 requires that the product be asso- ciative. An abelian Lie algebra is trivially associative, but in general the bracket of aLie algebra need not be associative. So despite its name, a Lie algebra is in general not an algebra. Example. IfMis a manifold, then the vector space X(M)ofC ∞vector ﬁelds on Mis a real Lie algebra with the Lie bracket [,]as the bracket. Example. LetKn×nbe the vector space of all n×nmatrices over a ﬁeld K. Deﬁne forX,Y∈Kn×n, [X,Y]=XY−YX, where XYis the matrix product of XandY. With this bracket, Kn×nbecomes a Lie algebra. The bilinearity and anticommutativity of [,]are immediate, while the Jacobi identity follows from the same computation as in Exercise 14.11. More generally, if Ais any algebra over a ﬁeld K, then the product [x,y]=xy−yx,x,y∈A, makes Ainto a Lie algebra over K. Deﬁnition 14.13. Aderivation of a Lie algebra Vover a ﬁeld Kis aK-linear map D:V→Vsatisfying the product rule D[Y,Z]=[DY,Z]+[Y,DZ]forY,Z∈V. Example. LetVbe a Lie algebra over a ﬁeld K. For each XinV, deﬁne ad X:V→ Vby adX(Y)=[X,Y]. 14.6 Related Vector Fields 159 We may rewrite the Jacobi identity in the form [X,[Y,Z]]=[[X,Y],Z]+[Y,[X,Z]] or adX[Y,Z]=[ad XY,Z]+[Y,adXZ], which shows that ad X:V→Vis a derivation of V. 14.5 The Pushforward of Vector Fields LetF:N→Mbe a smooth map of manifolds and let F∗:TpN→TF(p)Mbe its differential at a point pinN. IfXp∈TpN, we call F∗(Xp)thepushforward of the vector Xpatp. This notion does not extend in general to vector ﬁelds, since if Xis a vector ﬁeld on Nandz=F(p)=F(q)for two distinct points p,q∈N, then Xpand Xqare both pushed forward to tangent vectors at z∈M, but there is no reason why F∗(Xp)andF∗(Xq)should be equal (see Figure 14.2 ). RR2 pq xF /Bullet/Bullet/BulletX Fig. 14.2. The vector ﬁeld Xcannot be pushed forward under the ﬁrst projection F:R2→R. In one important special case, the pushforward F∗Xof any vector ﬁeld Xon Nalways makes sense, namely, when F:N→Mis a diffeomorphism. In this case, since Fis injective, there is no ambiguity about the meaning of (F∗X)F(p)=F∗,p(Xp), and since Fis surjective, F∗Xis deﬁned everywhere on M. 14.6 Related Vector Fields Under a C∞map F:N→M, although in general a vector ﬁeld on Ncannot be pushed forward to a vector ﬁeld on M, there is nonetheless a useful notion of related vector ﬁelds , which we now deﬁne. 160§14 Vector Fields Deﬁnition 14.14. LetF:N→Mbe a smooth map of manifolds. A vector ﬁeld X onNisF-related to a vector ﬁeld ¯XonMif for all p∈N, F∗,p(Xp)=¯XF(p). (14.11) Example 14.15 ( Pushforward by a diffeomorphism ).IfF:N→Mis a diffeomor- phism and Xis a vector ﬁeld on N, then the pushforward F∗Xis deﬁned. By def- inition, the vector ﬁeld XonNisF-related to the vector ﬁeld F∗XonM. In Sub- section 16.5, we will see examples of vector ﬁelds related by a map Fthat is not a diffeomorphism. We may reformulate condition (14.11) for F-relatedness as follows. Proposition 14.16. Let F :N→M be a smooth map of manifolds. A vector ﬁeld X on N and a vector ﬁeld ¯X on M are F-related if and only if for all g ∈C∞(M), X(g◦F)=( ¯Xg)◦F. Proof. (⇒)Suppose XonNand ¯XonMareF-related. By (14.11), for any g∈C∞(M)and p∈N, F∗,p(Xp)g=¯XF(p)g (deﬁnition of F-relatedness ), Xp(g◦F)=( ¯Xg)(F(p)) ( deﬁnitions of F∗and ¯Xg), (X(g◦F))(p)=( ¯Xg)(F(p)). Since this is true for all p∈N, X(g◦F)=( ¯Xg)◦F. (⇐)Reversing the set of equations above proves the converse. ⊓ ⊔ Proposition 14.17. Let F :N→M be a smooth map of manifolds. If the C∞vector ﬁelds X and Y on N are F-related to the C∞vector ﬁelds ¯X and ¯Y, respectively, on M, then the Lie bracket [X,Y]on N is F-related to the Lie bracket [¯X,¯Y]on M. Proof. For any g∈C∞(M), [X,Y](g◦F)=XY(g◦F)−YX(g◦F) ( deﬁnition of [X,Y]) =X((¯Y g)◦F)−Y((¯Xg)◦F) (Proposition 14.16 ) =(¯X¯Y g)◦F−(¯Y¯Xg)◦F(Proposition 14.16 ) =(( ¯X¯Y−¯Y¯X)g)◦F =([ ¯X,¯Y]g)◦F. By Proposition 14.16 again, this proves that [X,Y]onNand[¯X,¯Y]onMareF- related. ⊓ ⊔ 14.6 Related Vector Fields 161 Problems 14.1.* Equality of vector ﬁelds Show that two C∞vector ﬁelds XandYon a manifold Mare equal if and only if for every C∞ function fonM, we have X f=Y f. 14.2. Vector ﬁeld on an odd sphere Letx1,y1,..., xn,ynbe the standard coordinates on R2n. The unit sphere S2n−1inR2nis deﬁned by the equation ∑n i=1(xi)2+(yi)2=1. Show that X=n ∑ i=1−yi∂ ∂xi+xi∂ ∂yi is a nowhere-vanishing smooth vector ﬁeld on S2n−1. Since all spheres of the same dimen- sion are diffeomorphic, this proves that on every odd-dimensional sphere there is a nowhere- vanishing smooth vector ﬁeld. It is a classical theorem of differential and algebraic topology that on an even-dimensional sphere every continuous vector ﬁeld must vanish somewhere (see[28, Section 5, p. 31] or [16, Theorem 16.5, p. 70]). ( Hint: Use Problem 11.1 to show that X is tangent to S 2n−1.) 14.3. Maximal integral curve on a punctured line LetMbeR−{0}and let Xbe the vector ﬁeld d/dxonM(Figure 14.3 ). Find the maximal integral curve of Xstarting at x=1. /Bullet/Circle /Bullet 0 1 Fig. 14.3. The vector ﬁeld d/dxonR−{0}. 14.4. Integral curves in the plane Find the integral curves of the vector ﬁeld X(x,y)=x∂ ∂x−y∂ ∂y=/bracketleftbiggx −y/bracketrightbigg onR2. 14.5. Maximal integral curve in the plane Find the maximal integral curve c(t)starting at the point (a,b)∈R2of the vector ﬁeld X(x,y)= ∂/∂x+x∂/∂yonR2. 14.6. Integral curve starting at a zero of a vector ﬁeld (a)*Suppose the smooth vector ﬁeld Xon a manifold Mvanishes at a point p∈M. Show that the integral curve of Xwith initial point pis the constant curve c(t)≡p. (b) Show that if Xis the zero vector ﬁeld on a manifold M, and ct(p)is the maximal integral curve of Xstarting at p, then the one-parameter group of diffeomorphisms c:R→Diff(M) is the constant map c(t)≡1M. 162§14 Vector Fields 14.7. Maximal integral curve LetXbe the vector ﬁeld x d/dxonR. For each pinR, ﬁnd the maximal integral curve c(t) ofXstarting at p. 14.8. Maximal integral curve LetXbe the vector ﬁeld x2d/dxon the real line R. For each p>0 inR, ﬁnd the maximal integral curve of Xwith initial point p. 14.9. Reparametrization of an integral curve Suppose c:]a,b[→Mis an integral curve of the smooth vector ﬁeld XonM. Show that for any real number s, the map cs:]a+s,b+s[→M,cs(t)=c(t−s), is also an integral curve of X. 14.10. Lie bracket of vector ﬁelds IffandgareC∞functions and XandYareC∞vector ﬁelds on a manifold M, show that [f X,gY]=f g[X,Y]+f(Xg)Y−g(Y f)X. 14.11. Lie bracket of vector ﬁelds on R2 Compute the Lie bracket/bracketleftbigg −y∂ ∂x+x∂ ∂y,∂ ∂x/bracketrightbigg onR2. 14.12. Lie bracket in local coordinates Consider two C∞vector ﬁelds X,YonRn: X=∑ai∂ ∂xi, Y=∑bj∂ ∂xj, where ai,bjareC∞functions on Rn. Since[X,Y]is also a C∞vector ﬁeld on Rn, [X,Y]=∑ck∂ ∂xk for some C∞functions ck. Find the formula for ckin terms of aiandbj. 14.13. Vector ﬁeld under a diffeomorphism LetF:N→Mbe aC∞diffeomorphism of manifolds. Prove that if gis aC∞function and X aC∞vector ﬁeld on N, then F∗(gX)=(g◦F−1)F∗X. 14.14. Lie bracket under a diffeomorphism LetF:N→Mbe a C∞diffeomorphism of manifolds. Prove that if XandYareC∞vector ﬁelds on N, then F∗[X,Y]=[F∗X,F∗Y]. Chapter 4 Lie Groups and Lie Algebras A Lie group is a manifold that is also a group such that the group operations are smooth. Classical groups such as the general and special linear groups over Rand overC, orthogonal groups, unitary groups, and symplectic groups are all Lie groups. A Lie group is a homogeneous space in the sense that left translation by a group element gis a diffeomorphism of the group onto itself that maps the identity element tog. Therefore, locally the group looks the same around any point. To study the local structure of a Lie group, it is enough to examine a neighborhood of the identity element. It is not surprising that the tangent space at the identity of a Lie groupshould play a key role. The tangent space at the identity of a Lie group Gturns out to have a canonical bracket operation [,]that makes it into a Lie algebra. The tangent space T eGwith the bracket is called the Lie algebra of the Lie group G. The Lie algebra of a Lie group encodes within it much information about the group. Sophus Lie (1842–1899)In a series of papers in the decade from 1874 to 1884, the Norwegian mathematician Sophus Lie ini- tiated the study of Lie groups and Lie algebras. Atﬁrst his work gained little notice, possibly because at the time he wrote mostly in Norwegian. In 1886, Lie became a professor in Leipzig, Germany, and his theory began to attract attention, especially after the publication of the three-volume treatise Theorie der Transformationsgruppen that he wrote in collabora- tion with his assistant Friedrich Engel. Lie’s original motivation was to study the group of transformations of a space as a continuous ana- logue of the group of permutations of a ﬁnite set.Indeed, a diffeomorphism of a manifold Mcan be viewed as a permutation of the points of M. The interplay of group theory, topology, and linear alge- bra makes the theory of Lie groups and Lie algebras © Springer Science+Business Media, LLC 2011L.W. Tu, An Introduction to Manifolds, Universitext, DOI 10.1007/978-1-4419-7400-6_4, 163 164§15 Lie Groups a particularly rich and vibrant branch of mathematics. In this chapter we can but scratch the surface of this vast creation. For us, Lie groups serve mainly as an im-portant class of manifolds, and Lie algebras as examples of tangent spaces. §15 Lie Groups We begin with several examples of matrix groups, subgroups of the general linear group over a ﬁeld. The goal is to exhibit a variety of methods for showing that a group is a Lie group and for computing the dimension of a Lie group. These examples become templates for investigating other matrix groups. A powerful tool, which we state but do not prove, is the closed subgroup theorem. According to this theorem, an abstract subgroup that is a closed subset of a Lie group is itself a Liegroup. In many instances, the closed subgroup theorem is the easiest way to prove that a group is a Lie group. The matrix exponential gives rise to curves in a matrix group with a given initial vector. It is useful in computing the differential of a map on a matrix group. As an example, we compute the differential of the determinant map on the general linear group over R. 15.1 Examples of Lie Groups We recall here the deﬁnition of a Lie group, which ﬁrst appeared in Subsection 6.5. Deﬁnition 15.1. ALie group is aC∞manifold Gthat is also a group such that the two group operations, multiplication µ:G×G→G,µ(a,b)=ab, and inverse ι:G→G,ι(a)=a−1, areC∞. Fora∈G, denote by ℓa:G→G,ℓa(x)=µ(a,x)=ax, the operation of left mul- tiplication bya, and by ra:G→G,ra(x)=xa, the operation of right multiplication bya. We also call left and right multiplications leftandright translations . Exercise 15.2 (Left multiplication).* For an element ain a Lie group G, prove that the left multiplication ℓa:G→Gis a diffeomorphism. Deﬁnition 15.3. A map F:H→Gbetween two Lie groups HandGis aLie group homomorphism if it is a C∞map and a group homomorphism. The group homomorphism condition means that for all h,x∈H, 15.1 Examples of Lie Groups 165 F(hx)=F(h)F(x). (15.1) This may be rewritten in functional notation as F◦ℓh=ℓF(h)◦Ffor all h∈H. (15.2) LeteHandeGbe the identity elements of HandG, respectively. Taking handx in (15.1) to be the identity eH, it follows that F(eH)=eG. So a group homomorphism always maps the identity to the identity. NOTATION . We use capital letters to denote matrices, but generally lowercase letters to denote their entries. Thus, the (i,j)-entry of the matrix ABis(AB)i j=∑kaikbk j. Example 15.4 ( General linear group ).In Example 6.21, we showed that the general linear group GL(n,R)={A∈Rn×n|detA/ne}a⊔ionslash=0} is a Lie group. Example 15.5 ( Special linear group ).The special linear group SL (n,R)is the sub- group of GL (n,R)consisting of matrices of determinant 1. By Example 9.13, SL(n,R)is a regular submanifold of dimension n2−1 of GL(n,R). By Exam- ple 11.16, the multiplication map ¯µ: SL(n,R)×SL(n,R)→SL(n,R) isC∞. To see that the inverse map ¯ι: SL(n,R)→SL(n,R) isC∞, leti: SL(n,R)→GL(n,R)be the inclusion map and ι: GL(n,R)→GL(n,R) the inverse map of GL (n,R). As the composite of two C∞maps, ι◦i: SL(n,R)i→GL(n,R)ι→GL(n,R) is aC∞map. Since its image is contained in the regular submanifold SL (n,R), the induced map ¯ι: SL(n,R)→SL(n,R)isC∞by Theorem 11.15. Thus, SL (n,R)is a Lie group. An entirely analogous argument proves that the complex special linear group SL(n,C)is also a Lie group. Example 15.6 ( Orthogonal group ).Recall that the orthogonal group O (n)is the sub- group of GL (n,R)consisting of all matrices Asatisfying ATA=I. Thus, O (n)is the inverse image of Iunder the map f(A)=ATA. In Example 11.3 we showed that f: GL(n,R)→GL(n,R)has constant rank. By the constant-rank level set theorem, O (n)is a regular submanifold of GL (n,R). 166§15 Lie Groups One drawback of this approach is that it does not tell us what the rank of fis, and so the dimension of O (n)remains unknown. In this example we will apply the regular level set theorem to prove that O( n)is a regular submanifold of GL (n,R). This will at the same time determine the dimension of O(n). To accomplish this, we must ﬁrst redeﬁne the target space of f. Since ATA is a symmetric matrix, the image of flies in Sn, the vector space of all n×nreal symmetric matrices. The space Snis a proper subspace of Rn×nas soon as n≥2. Exercise 15.7 (Space of symmetric matrices).* Show that the vector space Snofn×nreal symmetric matrices has dimension (n2+n)/2. Consider the map f: GL(n,R)→Sn,f(A) =ATA. The tangent space of Snat any point is canonically isomorphic to Snitself, because Snis a vector space. Thus, the image of the differential f∗,A:TA(GL(n,R))→Tf(A)(Sn)≃Sn lies in Sn. While it is true that falso maps GL (n,R)to GL(n,R)orRn×n, if we had taken GL(n,R)orRn×nas the target space of f, the differential f∗,Awould never be surjective for any A∈GL(n,R)when n≥2, since f∗,Afactors through the proper subspace SnofRn×n. This illustrates a general principle: for the differential f∗,Ato be surjective, the target space of fshould be as small as possible. To show that the differential of f: GL(n,R)→Sn,f(A)=ATA, is surjective, we compute explicitly the differential f∗,A. Since GL (n,R)is an open subset of Rn×n, its tangent space at any A∈GL(n,R)is TA(GL(n,R))= TA(Rn×n)=Rn×n. For any matrix X∈Rn×n, there is a curve c(t)in GL(n,R)with c(0)=Aandc′(0)= X(Proposition 8.16). By Proposition 8.18, f∗,A(X)=d dtf(c(t))/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0 =d dtc(t)Tc(t)/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0 =(c′(t)Tc(t)+c(t)Tc′(t))|t=0(by Problem 15.2) =XTA+ATX. The surjectivity of f∗,Abecomes the following question: if A∈O(n)andBis any symmetric matrix in Sn, does there exist an n×nmatrix Xsuch that XTA+ATX=B? Note that since (XTA)T=ATX, it is enough to solve 15.2 Lie Subgroups 167 ATX=1 2B, (15.3) for then XTA+ATX=1 2BT+1 2B=B. Equation (15.3) clearly has a solution: X=1 2(AT)−1B. So f∗,A:TAGL(n,R)→ Snis surjective for all A∈O(n), and O( n)is a regular level set of f. By the regular level set theorem, O (n)is a regular submanifold of GL (n,R)of dimension dimO(n)=n2−dimSn=n2−n2+n 2=n2−n 2. (15.4) 15.2 Lie Subgroups Deﬁnition 15.8. ALie subgroup of a Lie group Gis (i) an abstract subgroup H that is (ii) an immersed submanifold via the inclusion map such that (iii) the group operations on HareC∞. An “abstract subgroup” simply means a subgroup in the algebraic sense, in con- trast to a “Lie subgroup.” The group operations on the subgroup Hare the restrictions of the multiplication map µand the inverse map ιfrom GtoH. For an explanation of why a Lie subgroup is deﬁned to be an immersed submanifold instead of a regular submanifold, see Remark 16.15. Because a Lie subgroup is an immersed subman- ifold, it need not have the relative topology. However, being an immersion, theinclusion map i:H֒→Gof a Lie subgroup His of course C ∞. It follows that the composite µ◦(i×i):H×H→G×G→G isC∞. IfHwere deﬁned to be a regular submanifold of G, then by Theorem 11.15, the multiplication map H×H→Hand similarly the inverse map H→Hwould automatically be C∞, and condition (iii) in the deﬁnition of a Lie subgroup would be redundant. Since a Lie subgroup is deﬁned to be an immersed submanifold, it is necessary to impose condition (iii) on the group operations on H. Example 15.9 ( Lines with irrational slope in a torus ).LetGbe the torus R2/Z2 andLa line through the origin in R2. The torus can also be represented by the unit square with the opposite edges identiﬁed. The image HofLunder the projection π:R2→R2/Z2is a closed curve if and only if the line Lgoes through another lattice point, say (m,n)∈Z2. This is the case if and only if the slope of Lisn/m, a rational number or ∞; then His the image of ﬁnitely many line segments on the unit square. It is a closed curve diffeomorphic to a circle and is a regular submanifold ofR 2/Z2(Figure 15.1 ). If the slope of Lis irrational, then its image Hon the torus will never close up. In this case the restriction to Lof the projection map, f=π|L:L→R2/Z2, is a one-to- one immersion. We give Hthe topology and manifold structure induced from f. It 168§15 Lie Groups (0,0)(3,2) /Bullet/Bullet Fig. 15.1. An embedded Lie subgroup of the torus. can be shown that His a dense subset of the torus [3, Example III.6.15, p. 86]. Thus, His an immersed submanifold but not a regular submanifold of the torus R2/Z2. Whatever the slope of L, its image HinR2/Z2is an abstract subgroup of the torus, an immersed submanifold, and a Lie group. Therefore, His a Lie subgroup of the torus. Exercise 15.10 (Induced topology versus subspace topology).* Suppose H⊂R2/Z2is the image of a line Lwith irrational slope in R2. We call the topology on Hinduced from the bijection f:L∼→Htheinduced topology and the topology on Has a subset of R2/Z2the subspace topology . Compare these two topologies: is one a subset of the other? Proposition 15.11. If H is an abstract subgroup and a regular submanifold of a Lie group G, then it is a Lie subgroup of G. Proof. Since a regular submanifold is the image of an embedding (Theorem 11.14), it is also an immersed submanifold. Letµ:G×G→Gbe the multiplication map on G. Since His an immersed submanifold of G, the inclusion map i:H֒→GisC∞. Hence, the inclusion map i×i:H×H֒→G×GisC∞, and the composition µ◦(i×i):H×H→GisC∞. By Theorem 11.15, because His a regular submanifold of G, the induced map ¯µ:H× H→HisC∞. The smoothness of the inverse map ¯ι:H→Hcan be deduced from the smooth- ness of ι:G→Gjust as in Example 15.5. ⊓ ⊔ A subgroup Has in Proposition 15.11 is called an embedded Lie subgroup , be- cause the inclusion map i:H→Gof a regular submanifold is an embedding (Theo- rem 11.14). Example. We showed in Examples 15.5 and 15.6 that the subgroups SL (n,R)and O(n)of GL(n,R)are both regular submanifolds. By Proposition 15.11 they are embedded Lie subgroups. We state without proof an important theorem about Lie subgroups. If Gis a Lie group, then an abstract subgroup that is a closed subset in the topology of Gis called aclosed subgroup . 15.3 The Matrix Exponential 169 Theorem 15.12 (Closed subgroup theorem). A closed subgroup of a Lie group is an embedded Lie subgroup. For a proof of the closed subgroup theorem, see [38, Theorem 3.42, p. 110]. Examples. (i) A line with irrational slope in the torus R2/Z2is not a closed subgroup, since it is not the whole torus, but being dense, its closure is. (ii) The special linear group SL (n,R)and the orthogonal group O (n)are the zero sets of polynomial equations on GL( n,R). As such, they are closed subsets of GL(n,R). By the closed subgroup theorem, SL (n,R)and O( n)are embedded Lie subgroups of GL (n,R). 15.3 The Matrix Exponential To compute the differential of a map on a subgroup of GL (n,R), we need a curve of nonsingular matrices. Because the matrix exponential is always nonsingular, it is uniquely suited for this purpose. Anorm on a vector space Vis a real-valued function /bardbl·/bardbl:V→Rsatisfying the following three properties: for all r∈Randv,w∈V, (i) (positive-deﬁniteness) /bardblv/bardbl≥0 with equality if and only if v=0, (ii) (positive homogeneity) /bardblrv/bardbl=|r|/bardblv/bardbl, (iii) (subadditivity) /bardblv+w/bardbl≤/bardbl v/bardbl+/bardblw/bardbl. A vector space Vtogether with a norm /bardbl·/bardbl is called a normed vector space . The vector space Rn×n≃Rn2of all n×nreal matrices can be given the Euclidean norm: forX=[x i j]∈Rn×n, /bardblX/bardbl=/parenleftig ∑x2 i j/parenrightig1/2 . Thematrix exponential eXof a matrix X∈Rn×nis deﬁned by the same formula as the exponential of a real number: eX=I+X+1 2!X2+1 3!X3+···, (15.5) where Iis the n×nidentity matrix. For this formula to make sense, we need to show that the series on the right converges in the normed vector space Rn×n≃Rn2. Anormed algebra V is a normed vector space that is also an algebra over R satisfying the submultiplicative property: for all v,w∈V,/bardblvw/bardbl≤/bardbl v/bardbl/bardblw/bardbl. Matrix multiplication makes the normed vector space Rn×ninto a normed algebra. Proposition 15.13. For X,Y∈Rn×n,/bardblXY/bardbl≤/bardbl X/bardbl/bardblY/bardbl. 170§15 Lie Groups Proof. Write X=[x i j]andY=[y i j]and ﬁx a pair of subscripts (i,j). By the Cauchy– Schwarz inequality, (XY)2 i j=/parenleftig ∑ kxikyk j/parenrightig2 ≤/parenleftig ∑ kx2 ik/parenrightig/parenleftig ∑ ky2k j/parenrightig =aibj, where we set ai=∑kx2 ikandbj=∑ky2 k j. Then /bardblXY/bardbl2=∑ i,j(XY)2 i j≤∑ i,jaibj=/parenleftig ∑ iai/parenrightig/parenleftig ∑ jbj/parenrightig =/parenleftig ∑ i,kx2 ik/parenrightig/parenleftig ∑ j,ky2 k j/parenrightig =/bardblX/bardbl2/bardblY/bardbl2.⊓ ⊔ In a normed algebra, multiplication distributes over a ﬁnite sum. When the sum is inﬁnite as in a convergent series, the distributivity of multiplication over the sum requires a proof. Proposition 15.14. Let V be a normed algebra. (i)If a∈V and s mis a sequence in V that converges to s, then as mconverges to as. (ii)If a∈V and ∑∞ k=0bkis a convergent series in V , then a ∑kbk=∑kabk. Exercise 15.15 (Distributivity over a convergent series).* Prove Proposition 15.14. In a normed vector space Va series ∑akis said to converge absolutely if the series ∑/bardblak/bardblof norms converges in R. The normed vector space Vis said to be complete if every Cauchy sequence in Vconverges to a point in V. For example, Rn×nis a complete normed vector space.1It is easy to show that in a complete normed vector space, absolute convergence implies convergence [26, Theorem 2.9.3, p. 126]. Thus, to show that a series ∑Ykof matrices converges, it is enough to show that the series ∑/bardblYk/bardblof real numbers converges. For any X∈Rn×nandk>0, repeated applications of Proposition 15.13 give /bardblXk/bardbl≤/bardbl X/bardblk. So the series ∑∞ k=0/bardblXk/k!/bardblis bounded term by term in absolute value by the convergent series √n+/bardblX/bardbl+1 2!/bardblX/bardbl2+1 3!/bardblX/bardbl3+···=(√n−1)+e/bardblX/bardbl. By the comparison test for series of real numbers, the series ∑∞ k=0/bardblXk/k!/bardblconverges. Therefore, the series (15.5) converges absolutely for any n×nmatrix X. NOTATION . Following standard convention we use the letter eboth for the expo- nential map and for the identity element of a general Lie group. The context should prevent any confusion. We sometimes write exp (X)foreX. 1A complete normed vector space is also called a Banach space , named after the Polish mathematician Stefan Banach, who introduced the concept in 1920–1922. Correspondingly, acomplete normed algebra is called a Banach algebra . 15.4 The Trace of a Matrix 171 Unlike the exponential of real numbers, when AandBaren×nmatrices with n>1, it is not necessarily true that eA+B=eAeB. Exercise 15.16 (Exponentials of commuting matrices). Prove that if AandBare commuting n×nmatrices, then eAeB=eA+B. Proposition 15.17. For X∈Rn×n, d dtetX=XetX=etXX. Proof. Because each (i,j)-entry of the series for the exponential function etXis a power series in t, it is possible to differentiate term by term [35, Theorem 8.1, p. 173]. Hence, d dtetX=d dt/parenleftbigg I+tX+1 2!t2X2+1 3!t3X3+···/parenrightbigg =X+tX2+1 2!t2X3+··· =X/parenleftbigg I+tX+1 2!t2X2+···/parenrightbigg =XetX(Proposition 15.14(ii)) . In the second equality above, one could have factored out Xas the second factor: d dtetX=X+tX2+1 2!t2X3+··· =/parenleftbigg I+tX+1 2!t2X2+···/parenrightbigg X=etXX.⊓ ⊔ The deﬁnition of the matrix exponential eXmakes sense even if Xis a complex matrix. All the arguments so far carry over word for word; one merely has to replace the Euclidean norm /bardblX/bardbl2=∑x2 i jby the Hermitian norm /bardblX/bardbl2=∑|xi j|2, where|xi j| is the modulus of a complex number xi j. 15.4 The Trace of a Matrix Deﬁne the trace of an n×nmatrix Xto be the sum of its diagonal entries: tr(X)=n ∑ i=1xii. Lemma 15.18. (i)For any two matrices X ,Y∈Rn×n,tr(XY)=tr(YX). (ii)For X∈Rn×nand A∈GL(n,R),tr(AXA−1)=tr(X). 172§15 Lie Groups Proof. (i) tr(XY)=∑ i(XY)ii=∑ i∑ kxikyki, tr(Y X)=∑ k(Y X)kk=∑ k∑ iykixik. (ii) Set B=XA−1in (i). ⊓ ⊔ The eigenvalues of an n×nmatrix Xare the roots of the polynomial equation det(λI−X)=0. Over the ﬁeld of complex numbers, which is algebraically closed, such an equation necessarily has nroots, counted with multiplicity. Thus, the advan- tage of allowing complex numbers is that every n×nmatrix, real or complex, has n complex eigenvalues, counted with multiplicity, whereas a real matrix need not haveany real eigenvalue. Example. The real matrix/bracketleftbigg 0−1 1 0/bracketrightbigg has no real eigenvalues. It has two complex eigenvalues, ±i. The following two facts about eigenvalues are immediate from the deﬁnitions: (i) Two similar matrices XandAXA −1have the same eigenvalues, because det(λI−AXA−1)=det/parenleftbig A(λI−X)A−1/parenrightbig =det(λI−X). (ii) The eigenvalues of a triangular matrix are its diagonal entries, because det λI− λ1∗ ... 0 λn  =n ∏ i=1(λ−λi). By a theorem from algebra [19, Th. 6.4.1, p. 286], any complex square matrix Xcan be triangularized; more precisely, there exists a nonsingular complex square matrix Asuch that AXA−1is upper triangular. Since the eigenvalues λ1,..., λnofX are the same as the eigenvalues of AXA−1, the triangular matrix AXA−1must have the eigenvalues of Xalong its diagonal:  λ1∗ ... 0 λn . A real matrix X, viewed as a complex matrix, can also be triangularized, but of course the triangularizing matrix Aand the triangular matrix AXA−1are in general complex. 15.4 The Trace of a Matrix 173 Proposition 15.19. The trace of a matrix, real or complex, is equal to the sum of its complex eigenvalues. Proof. Suppose Xhas complex eigenvalues λ1,..., λn. Then there exists a nonsin- gular matrix A∈GL(n,C)such that AXA−1= λ1∗ ... 0 λn . By Lemma 15.18, tr(X)=tr(AXA−1)=∑λi. ⊓ ⊔ Proposition 15.20. For any X∈Rn×n,det(eX)=etrX. Proof. Case 1. Assume that Xis upper triangular: X= λ1∗ ... 0 λn . Then eX=∑1 k!Xk=∑1 k! λk 1∗ ... 0 λk n = eλ1∗ ... 0 eλn . Hence, det eX=∏eλi=e∑λi=etrX. Case 2. Given a general matrix X, with eigenvalues λ1,..., λn, we can ﬁnd a nonsin- gular complex matrix Asuch that AXA−1= λ1∗ ... 0 λn , an upper triangular matrix. Then eAXA−1=I+AXA−1+1 2!(AXA−1)2+1 3!(AXA−1)3+··· =I+AXA−1+A/parenleftbigg1 2!X2/parenrightbigg A−1+A/parenleftbigg1 3!X3/parenrightbigg A−1+··· =AeXA−1(by Proposition 15.14(ii)) . Hence, 174§15 Lie Groups deteX=det(AeXA−1)=det(eAXA−1) =etr(AXA−1)(by Case 1, since AXA−1is upper triangular ) =etrX(by Lemma 15.18 ). ⊓ ⊔ It follows from this proposition that the matrix exponential eXis always non- singular, because det (eX) =etrXis never 0. This is one reason why the matrix ex- ponential is so useful, for it allows us to write down explicitly a curve in GL (n,R) with a given initial point and a given initial velocity. For example, c(t)=etX:R→ GL(n,R)is a curve in GL (n,R)with initial point Iand initial velocity X, since c(0)=e0X=e0=Iand c′(0)=d dtetX/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0=XetX/vextendsingle/vextendsingle t=0=X. (15.6) Similarly, c(t)=AetX:R→GL(n,R)is a curve in GL (n,R)with initial point Aand initial velocity AX. 15.5 The Differential of detat the Identity Let det: GL (n,R)→Rbe the determinant map. The tangent space TIGL(n,R)to GL(n,R)at the identity matrix Iis the vector space Rn×nand the tangent space T1R toRat 1 isR. So det∗,I:Rn×n→R. Proposition 15.21. For any X∈Rn×n,det∗,I(X)=trX. Proof. We use a curve at Ito compute the differential (Proposition 8.18). As a curve c(t)with c(0)=Iandc′(0)=X, choose the matrix exponential c(t)=etX. Then det∗,I(X)=d dtdet(etX)/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0=d dtettrX/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0 =(trX)ettrX/vextendsingle/vextendsingle t=0=trX. ⊓ ⊔ Problems 15.1. Matrix exponential ForX∈Rn×n, deﬁne the partial sum sm=∑m k=0Xk/k!. (a) Show that for ℓ≥m, /bardblsℓ−sm/bardbl≤ℓ ∑ k=m+1/bardblX/bardblk/k!. (b) Conclude that smis a Cauchy sequence in Rn×nand therefore converges to a matrix, which we denote by eX. This gives another way of showing that ∑∞ k=0Xk/k! is convergent, with- out using the comparison test or the theorem that absolute convergence implies conver-gence in a complete normed vector space. 15.5 The Differential of det at the Identity 175 15.2. Product rule for matrix-valued functions Let]a,b[be an open interval in R. Suppose A:]a,b[→Rm×nandB:]a,b[→Rn×parem×n andn×pmatrices respectively whose entries are differentiable functions of t∈]a,b[. Prove that for t∈]a,b[, d dtA(t)B(t)=A′(t)B(t)+A(t)B′(t), where A′(t)=(dA/dt)(t)andB′(t)=(dB/dt)(t). 15.3. Identity component of a Lie group Theidentity component G 0of a Lie group Gis the connected component of the identity ele- ment einG. Let µandιbe the multiplication map and the inverse map of G. (a) For any x∈G0, show that µ({x}×G0)⊂G0. (Hint: Apply Proposition A.43.) (b) Show that ι(G0)⊂G0. (c) Show that G0is an open subset of G. (Hint: Apply Problem A.16.) (d) Prove that G0is itself a Lie group. 15.4.* Open subgroup of a connected Lie group Prove that an open subgroup Hof a connected Lie group Gis equal to G. 15.5. Differential of the multiplication map LetGbe a Lie group with multiplication map µ:G×G→G, and letℓa:G→Gandrb:G→ Gbe left and right multiplication by aandb∈G, respectively. Show that the differential of µ at(a,b)∈G×Gis µ∗,(a,b)(Xa,Yb)=(r b)∗(Xa)+(ℓ a)∗(Yb)forXa∈Ta(G),Yb∈Tb(G). 15.6. Differential of the inverse map LetGbe a Lie group with multiplication map µ:G×G→G, inverse map ι:G→G, and identity element e. Show that the differential of the inverse map at a∈G, ι∗,a:TaG→Ta−1G, is given by ι∗,a(Ya)=−(ra−1)∗(ℓa−1)∗Ya, where(ra−1)∗=(ra−1)∗,eand(ℓa−1)∗=(ℓa−1)∗,a. (The differential of the inverse at the identity was calculated in Problem 8.8(b).) 15.7.* Differential of the determinant map at A Show that the differential of the determinant map det: GL (n,R)→RatA∈GL(n,R)is given by det∗,A(AX)=( detA)trXforX∈Rn×n. (15.7) 15.8.* Special linear group Use Problem 15.7 to show that 1 is a regular value of the determinant map. This gives a quick proof that the special linear group SL (n,R)is a regular submanifold of GL (n,R). 15.9. Structure of a general linear group 176§15 Lie Groups (a) For r∈R×:=R−{0}, letMrbe the n×nmatrix Mr= r 1 ... 1 =[re 1e2···en], where e1,..., enis the standard basis for Rn. Prove that the map f: GL(n,R)→SL(n,R)×R×, A/ma√s⊔o→/parenleftig AM1/detA,detA/parenrightig , is a diffeomorphism. (b) The center Z(G)of a group Gis the subgroup of elements g∈Gthat commute with all elements of G: Z(G):={g∈G|gx=xgfor all x∈G}. Show that the center of GL (2,R)is isomorphic to R×, corresponding to the subgroup of scalar matrices, and that the center of SL (2,R)×R×is isomorphic to{±1}×R×. The groupR×has two elements of order 2, while the group {±1}×R×has four elements of order 2. Since their centers are not isomorphic, GL (2,R)and SL(2,R)×R×are not isomorphic as groups. (c) Show that h: GL(3,R)→SL(3,R)×R×, A/ma√s⊔o→/parenleftig (detA)1/3A,detA/parenrightig , is a Lie group isomorphism. The same arguments as in (b) and (c) prove that for neven, the two Lie groups GL (n,R) and SL(n,R)×R×are not isomorphic as groups, while for nodd, they are isomorphic as Lie groups. 15.10. Orthogonal group Show that the orthogonal group O (n)is compact by proving the following two statements. (a) O(n)is a closed subset of Rn×n. (b) O(n)is a bounded subset of Rn×n. 15.11. Special orthogonal group SO (2) Thespecial orthogonal group SO(n)is deﬁned to be the subgroup of O (n)consisting of ma- trices of determinant 1. Show that every matrix A∈SO(2)can be written in the form A=/bracketleftbigga c b d/bracketrightbigg =/bracketleftbiggcosθ−sinθ sinθ cosθ/bracketrightbigg for some real number θ. Then prove that SO (2)is diffeomorphic to the circle S1. 15.12. Unitary group Theunitary group U(n)is deﬁned to be U(n)={A∈GL(n,C)|¯ATA=I}, 15.5 The Differential of det at the Identity 177 where ¯Adenotes the complex conjugate of A, the matrix obtained from Aby conjugating every entry of A:(¯A)i j=ai j. Show that U (n)is a regular submanifold of GL (n,C)and that dimU(n)=n2. 15.13. Special unitary group SU(2) Thespecial unitary group SU(n)is deﬁned to be the subgroup of U (n)consisting of matrices of determinant 1. (a) Show that SU (2)can also be described as the set SU(2)=/braceleftbigg/bracketleftbigga−¯b b¯a/bracketrightbigg ∈C2×2/vextendsingle/vextendsingle/vextendsingle/vextendsinglea¯a+b¯b=1/bracerightbigg . (Hint: Write out the condition A−1=¯ATin terms of the entries of A.) (b) Show that SU (2)is diffeomorphic to the three-dimensional sphere S3=/braceleftig (x1,x2,x3,x4)∈R4|x2 1+x2 2+x2 3+x2 4=1/bracerightig . 15.14. A matrix exponential Compute exp/bracketleftbigg0 1 1 0/bracketrightbigg . 15.15. Symplectic group This problem requires a knowledge of quaternions as in Appendix E. Let Hbe the skew ﬁeld of quaternions. The symplectic group Sp(n)is deﬁned to be Sp(n)={A∈GL(n,H)|¯ATA=I}, where ¯Adenotes the quaternionic conjugate of A. Show that Sp (n)is a regular submanifold of GL(n,H)and compute its dimension. 15.16. Complex symplectic group LetJbe the 2 n×2nmatrix J=/bracketleftbigg0In −In0/bracketrightbigg , where Indenotes the n×nidentity matrix. The complex symplectic group Sp(2n,C)is deﬁned to be Sp(2n,C)={A∈GL(2n,C)|ATJA=J}. Show that Sp (2n,C)is a regular submanifold of GL (2n,C)and compute its dimension. ( Hint: Mimic Example 15.6. It is crucial to choose the correct target space for the map f(A)=ATJA.) 178§16 Lie Algebras §16 Lie Algebras In a Lie group G, because left translation by an element g∈Gis a diffeomorphism that maps a neighborhood of the identity to a neighborhood of g, all the local in- formation about the group is concentrated in a neighborhood of the identity, and the tangent space at the identity assumes a special importance. Moreover, one can give the tangent space TeGa Lie bracket [,], so that in addition to being a vector space, it becomes a Lie algebra, called the Lie algebra of the Lie group. This Lie algebra encodes in it much information about the Lie group. The goal of this section is to deﬁne the Lie algebra structure on TeGand to identity the Lie algebras of a few classical groups. The Lie bracket on the tangent space TeGis deﬁned using a canonical isomor- phism between the tangent space at the identity and the vector space of left-invariant vector ﬁelds on G. With respect to this Lie bracket, the differential of a Lie group homomorphism becomes a Lie algebra homomorphism. We thus obtain a functor from the category of Lie groups and Lie group homomorphisms to the category of Lie algebras and Lie algebra homomorphisms. This is the beginning of a reward- ing program, to understand the structure and representations of Lie groups through a study of their Lie algebras. 16.1 Tangent Space at the Identity of a Lie Group Because of the existence of a multiplication, a Lie group is a very special kind of manifold. In Exercise 15.2, we learned that for any g∈G, left translation ℓg:G→G bygis a diffeomorphism with inverse ℓg−1. The diffeomorphism ℓgtakes the identity element eto the element gand induces an isomorphism of tangent spaces ℓg∗=(ℓg)∗,e:Te(G)→Tg(G). Thus, if we can describe the tangent space Te(G)at the identity, then ℓg∗Te(G)will give a description of the tangent space Tg(G)at any point g∈G. Example 16.1 ( The tangent space to GL(n,R)at I).In Example 8.19, we identiﬁed the tangent space GL (n,R)at any point g∈GL(n,R)asRn×n, the vector space of alln×nreal matrices. We also identiﬁed the isomorphism ℓg∗:TI(GL(n,R))→ Tg(GL(n,R))as left multiplication by g:X/ma√s⊔o→gX. Example 16.2 ( The tangent space to SL(n,R)at I).We begin by ﬁnding a condition that a tangent vector XinTI(SL(n,R))must satisfy. By Proposition 8.16 there is a curve c:]−ε,ε[→SL(n,R)with c(0) =Iandc′(0) =X. Being in SL (n,R), this curve satisﬁes detc(t)=1 for all tin the domain ]−ε,ε[. We now differentiate both sides with respect to tand evaluate at t=0. On the left-hand side, we have 16.1 Tangent Space at the Identity of a Lie Group 179 d dtdet(c(t))/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0=(det◦c)∗/parenleftbiggd dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle 0/parenrightbigg =det∗,I/parenleftbigg c∗d dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle 0/parenrightbigg (by the chain rule ) =det∗,I(c′(0)) =det∗,I(X) =tr(X) (by Proposition 15.21 ). Thus, tr(X)=d dt1/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0=0. So the tangent space TI(SL(n,R))is contained in the subspace VofRn×ndeﬁned by V={X∈Rn×n|trX=0}. Since dim V=n2−1=dimTI(SL(n,R)), the two spaces must be equal. Proposition 16.3. The tangent space T I(SL(n,R))at the identity of the special linear group SL(n,R)is the subspace of Rn×nconsisting of all n×n matrices of trace 0. Example 16.4 ( The tangent space to O(n)at I).LetXbe a tangent vector to the orthogonal group O( n)at the identity I. Choose a curve c(t)in O(n)deﬁned on a small interval containing 0 such that c(0)=Iandc′(0)=X. Since c(t)is in O(n), c(t)Tc(t)=I. Differentiating both sides with respect to tusing the matrix product rule (Prob- lem 15.2) gives c′(t)Tc(t)+c(t)Tc′(t)=0. Evaluating at t=0 gives XT+X=0. Thus, Xis a skew-symmetric matrix. LetKnbe the space of all n×nreal skew-symmetric matrices. For example, for n=3, these are matrices of the form  0a b −a0c −b−c0 ,where a,b,c,∈R. The diagonal entries of such a matrix are all 0 and the entries below the diagonal are determined by those above the diagonal. So dimKn=n2−# diagonal entries 2=1 2(n2−n). We have shown that 180§16 Lie Algebras TI(O(n))⊂Kn. (16.1) By an earlier computation (see (15.4)), dimTI(O(n))= dimO(n)=n2−n 2. Since the two vector spaces in (16.1) have the same dimension, equality holds. Proposition 16.5. The tangent space T I(O(n))of the orthogonal group O(n)at the identity is the subspace of Rn×nconsisting of all n×n skew-symmetric matrices. 16.2 Left-Invariant Vector Fields on a Lie Group LetXbe a vector ﬁeld on a Lie group G. We do not assume Xto be C∞. For any g∈G, because left multiplication ℓg:G→Gis a diffeomorphism, the pushforward ℓg∗Xis a well-deﬁned vector ﬁeld on G. We say that the vector ﬁeld Xisleft- invariant if ℓg∗X=X for every g∈G; this means for any h∈G, ℓg∗(Xh)=Xgh. In other words, a vector ﬁeld Xis left-invariant if and only if it is ℓg-related to itself for all g∈G. Clearly, a left-invariant vector ﬁeld Xis completely determined by its value Xeat the identity, since Xg=ℓg∗(Xe). (16.2) Conversely, given a tangent vector A∈Te(G)we can deﬁne a vector ﬁeld ˜AonG by (16.2): (˜A)g=ℓg∗A. So deﬁned, the vector ﬁeld ˜Ais left-invariant, since ℓg∗(˜Ah)=ℓg∗ℓh∗A =(ℓg◦ℓh)∗A(by the chain rule ) =(ℓgh)∗(A) =˜Agh. We call ˜Atheleft-invariant vector ﬁeld on G generated by A ∈TeG. Let L(G)be the vector space of all left-invariant vector ﬁelds on G. Then there is a one-to-one correspondence Te(G)↔L(G), (16.3) Xe←/mapsfromcharX, A/ma√s⊔o→˜A. It is easy to show that this correspondence is in fact a vector space isomorphism. 16.2 Left-Invariant Vector Fields on a Lie Group 181 Example 16.6 ( Left-invariant vector ﬁelds on R).On the Lie group R, the group operation is addition and the identity element is 0. So “left multiplication” ℓgis actually addition: ℓg(x)=g+x. Let us compute ℓg∗(d/dx|0). Sinceℓg∗(d/dx|0)is a tangent vector at g, it is a scalar multiple of d/dx|g: ℓg∗/parenleftbiggd dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle 0/parenrightbigg =ad dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle g. (16.4) To evaluate a, apply both sides of (16.4) to the function f(x)=x: a=ad dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle gf=ℓg∗/parenleftbiggd dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle 0/parenrightbigg f=d dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle 0f◦ℓg=d dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle 0(g+x)=1. Thus, ℓg∗/parenleftbiggd dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle 0/parenrightbigg =d dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle g. This shows that d/dxis a left-invariant vector ﬁeld on R. Therefore, the left-invariant vector ﬁelds on Rare constant multiples of d/dx. Example 16.7 ( Left-invariant vector ﬁelds on GL(n,R)).Since GL (n,R)is an open subset of Rn×n, at any g∈GL(n,R)there is a canonical identiﬁcation of the tangent space Tg(GL(n,R))withRn×n, under which a tangent vector corresponds to an n×n matrix: ∑ai j∂ ∂xi j/vextendsingle/vextendsingle/vextendsingle/vextendsingle g←→[ai j]. (16.5) We use the same letter Bto denote alternately a tangent vector B=∑bi j∂/∂xi j|I∈ TI(G(n,R))at the identity and a matrix B=[b i j]. Let B=∑bi j∂/∂xi j|I∈TI(GL(n,R)) and let ˜Bbe the left-invariant vector ﬁeld on GL (n,R)generated by B. By Exam- ple 8.19, ˜Bg=(ℓg)∗B←→ gB under the identiﬁcation (16.5). In terms of the standard basis ∂/∂xi j|g, ˜Bg=∑ i,j(gB)i j∂ ∂xi j/vextendsingle/vextendsingle/vextendsingle/vextendsingle g=∑ i,j/parenleftigg ∑ kgikbk j/parenrightigg ∂ ∂xi j/vextendsingle/vextendsingle/vextendsingle/vextendsingle g. Proposition 16.8. Any left-invariant vector ﬁeld X on a Lie group G is C∞. Proof. By Proposition 14.3 it sufﬁces to show that for any C∞function fonG, the function X fis also C∞. Choose a C∞curve c:I→Gdeﬁned on some interval I containing 0 such that c(0)=eandc′(0)=Xe. Ifg∈G, then gc(t)is a curve starting atgwith initial vector Xg, since gc(0)=ge=gand (gc)′(0)=ℓg∗c′(0)=ℓg∗Xe=Xg. 182§16 Lie Algebras By Proposition 8.17, (X f)(g)=Xgf=d dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0f(gc(t)). Now the function f(gc(t))is a composition of C∞functions G×I1×c−−→ G×Gµ→ Gf→R, (g,t)/ma√s⊔o−→(g,c(t))/ma√s⊔o→gc(t)/ma√s⊔o→f(gc(t)); as such, it is C∞. Its derivative with respect to t, F(g,t):=d dtf(gc(t)), is therefore also C∞. Since(X f)(g)is a composition of C∞functions, G→G×IF→R, g/ma√s⊔o→(g,0)/ma√s⊔o→F(g,0)=d dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0f(gc(t)), it is a C∞function on G. This proves that Xis aC∞vector ﬁeld on G.⊓ ⊔ It follows from this proposition that the vector space L(G)of left-invariant vector ﬁelds on Gis a subspace of the vector space X(G)of all C∞vector ﬁelds on G. Proposition 16.9. If X and Y are left-invariant vector ﬁelds on G, then so is [X,Y]. Proof. For any ginG,Xisℓg-related to itself, and Yisℓg-related to itself. By Proposition 14.17, [X,Y]isℓg-related to itself. ⊓ ⊔ 16.3 The Lie Algebra of a Lie Group Recall that a Lie algebra is a vector space gtogether with a bracket , i.e., an anticom- mutative bilinear map [,]:g×g→gthat satisﬁes the Jacobi identity (Deﬁnition 14.12). A Lie subalgebra of a Lie algebra gis a vector subspace h⊂gthat is closed under the bracket [,]. By Proposition 16.9, the space L(G)of left-invariant vector ﬁelds on a Lie group Gis closed under the Lie bracket [,]and is therefore a Lie subalgebra of the Lie algebra X(G)of all C∞vector ﬁelds on G. As we will see in the next few subsections, the linear isomorphism ϕ:TeG≃ L(G)in (16.3) is mutually beneﬁcial to the two vector spaces, for each space has something that the other one lacks. The vector space L(G)has a natural Lie algebra structure given by the Lie bracket of vector ﬁelds, while the tangent space at theidentity has a natural notion of pushforward, given by the differential of a Lie group homomorphism. The linear isomorphism ϕ:TeG≃L(G)allows us to deﬁne a Lie bracket on TeGand to push forward left-invariant vector ﬁelds under a Lie group homomorphism. 16.4 The Lie Bracket on gl(n,R) 183 We begin with the Lie bracket on TeG. Given A,B∈TeG, we ﬁrst map them via ϕto the left-invariant vector ﬁelds ˜A,˜B, take the Lie bracket [˜A,˜B]=˜A˜B−˜B˜A, and then map it back to TeGviaϕ−1. Thus, the deﬁnition of the Lie bracket [A,B]∈TeG should be [A,B]=[ ˜A,˜B]e. (16.6) Proposition 16.10. If A,B∈TeG and ˜A,˜B are the left-invariant vector ﬁelds they generate, then [˜A,˜B]=[A,B]˜. Proof. Applying ( )˜to both sides of (16.6) gives [A,B]˜=([ ˜A,˜B]e)˜=[˜A,˜B], since( )˜and( )eare inverse to each other. ⊓ ⊔ With the Lie bracket [,], the tangent space Te(G)becomes a Lie algebra, called theLie algebra of the Lie group G. As a Lie algebra, Te(G)is usually denoted by g. 16.4 The Lie Bracket on gl(n,R) For the general linear group GL (n,R), the tangent space at the identity Ican be identiﬁed with the vector space Rn×nof all n×nreal matrices. We identiﬁed a tangent vector in TI(GL(n,R))with a matrix A∈Rn×nvia ∑ai j∂ ∂xi j/vextendsingle/vextendsingle/vextendsingle/vextendsingle I←→[ai j]. (16.7) The tangent space TIGL(n,R)with its Lie algebra structure is denoted by gl(n,R). Let˜Abe the left-invariant vector ﬁeld on GL (n,R)generated by A. Then on the Lie algebragl(n,R)we have the Lie bracket [A,B]=[ ˜A,˜B]Icoming from the Lie bracket of left-invariant vector ﬁelds. In the next proposition, we identify the Lie bracket in terms of matrices. Proposition 16.11. Let A=∑ai j∂ ∂xi j/vextendsingle/vextendsingle/vextendsingle/vextendsingle I, B=∑bi j∂ ∂xi j/vextendsingle/vextendsingle/vextendsingle/vextendsingle I∈TI(GL(n,R)). If [A,B]=[ ˜A,˜B]I=∑ci j∂ ∂xi j/vextendsingle/vextendsingle/vextendsingle/vextendsingle I, (16.8) then ci j=∑ kaikbk j−bikak j. Thus, if derivations are identiﬁed with matrices via (16.7) , then [A,B]=AB−BA. 184§16 Lie Algebras Proof. Applying both sides of (16.8) to xi j, we get ci j=[˜A,˜B]Ixi j=˜AI˜Bxi j−˜BI˜Axi j =A˜Bxi j−B˜Axi j(because ˜AI=A,˜BI=B), so it is necessary to ﬁnd a formula for the function ˜Bxi j. In Example 16.7 we found that the left-invariant vector ﬁeld ˜Bon GL(n,R)is given by ˜Bg=∑ i,j(gB)i j∂ ∂xi j/vextendsingle/vextendsingle/vextendsingle/vextendsingle gatg∈GL(n,R). Hence, ˜Bgxi j=(gB)i j=∑ kgikbk j=∑ kbk jxik(g). Since this formula holds for all g∈GL(n,R), the function ˜Bxi jis ˜Bxi j=∑ kbk jxik. It follows that A˜Bxi j=∑ p,qapq∂ ∂xpq/vextendsingle/vextendsingle/vextendsingle/vextendsingle I/parenleftbigg ∑ kbk jxik/parenrightbigg =∑ p,q,kapqbk jδipδkq =∑ kaikbk j=(AB)i j. Interchanging AandBgives B˜Axi j=∑ kbikak j=(BA)i j. Therefore, ci j=∑ kaikbk j−bikak j=(AB−BA)i j. ⊓ ⊔ 16.5 The Pushforward of Left-Invariant Vector Fields As we noted in Subsection 14.5, if F:N→Mis aC∞map of manifolds and Xis aC∞ vector ﬁeld on N, the pushforward F∗Xis in general not deﬁned except when Fis a diffeomorphism. In the case of Lie groups, however, because of the correspondence between left-invariant vector ﬁelds and tangent vectors at the identity, it is possibleto push forward left-invariant vector ﬁelds under a Lie group homomorphism. LetF:H→Gbe a Lie group homomorphism. A left-invariant vector ﬁeld X onHis generated by its value A=X e∈TeHat the identity, so that X=˜A. Since a Lie group homomorphism F:H→Gmaps the identity of Hto the identity of G, its differential F∗,eat the identity is a linear map from TeHtoTeG. The diagrams 16.6 The Differential as a Lie Algebra Homomorphism 185 TeHF∗,e/d47/d47 ≃ /d15/d15TeG ≃ /d15/d15 L(H) /d47/d47/d95 /d95 /d95L(G),A/d31 /d47/d47/d95 /d15/d15F∗,eA/d95 /d15/d15 ˜A/d31 /d47/d47/d95 /d95 /d95(F∗,eA)˜ show clearly the existence of an induced linear map F∗:L(H)→L(G)on left- invariant vector ﬁelds as well as a way to deﬁne it. Deﬁnition 16.12. LetF:H→Gbe a Lie group homomorphism. Deﬁne F∗:L(H)→L(G)by F∗(˜A)=( F∗,eA)˜ for all A∈TeH. Proposition 16.13. If F:H→G is a Lie group homomorphism and X is a left- invariant vector ﬁeld on H, then the left-invariant vector ﬁeld F ∗X on G is F-related to the left-invariant vector ﬁeld X. Proof. For each h∈H, we need to verify that F∗,h(Xh)=( F∗X)F(h). (16.9) The left-hand side of (16.9) is F∗,h(Xh)=F∗,h(ℓh∗,eXe)=( F◦ℓh)∗,e(Xe), while the right-hand side of (16.9) is (F∗X)F(h)=(F∗,eXe)˜F(h)(deﬁnition of F∗X) =ℓF(h)∗F∗,e(Xe) ( deﬁnition of left invariance ) =(ℓF(h)◦F)∗,e(Xe) (chain rule ). Since Fis a Lie group homomorphism, we have F◦ℓh=ℓF(h)◦F, so the two sides of (16.9) are equal. ⊓ ⊔ IfF:H→Gis a Lie group homomorphism and Xis a left-invariant vector ﬁeld onH, we will call F∗Xthepushforward of X under F . 16.6 The Differential as a Lie Algebra Homomorphism Proposition 16.14. If F:H→G is a Lie group homomorphism, then its differential at the identity, F∗=F∗,e:TeH→TeG, is aLie algebra homomorphism , i.e., a linear map such that for all A, B∈TeH, F∗[A,B]=[F∗A,F∗B]. 186§16 Lie Algebras Proof. By Proposition 16.13, the vector ﬁeld F∗˜AonGisF-related to the vector ﬁeld ˜AonH, and the vector ﬁeld F∗˜BisF-related to ˜BonH. Hence, the bracket [F∗˜A,F∗˜B] onGisF-related to the bracket [˜A,˜B]onH(Proposition 14.17). This means that F∗/parenleftbig [˜A,˜B]e/parenrightbig =[F∗˜A,F∗˜B]F(e)=[F∗˜A,F∗˜B]e. The left-hand side of this equality is F∗[A,B], while the right-hand side is [F∗˜A,F∗˜B]e=[(F∗A)˜,(F∗B)˜]e(deﬁnition of F∗˜A) =[F∗A,F∗B] ( deﬁnition of [,]onTeG). Equating the two sides gives F∗[A,B]=[F∗A,F∗B]. ⊓ ⊔ Suppose His a Lie subgroup of a Lie group G, with inclusion map i:H→G. Since iis an immersion, its differential i∗:TeH→TeG is injective. To distinguish the Lie bracket on TeHfrom the Lie bracket on TeG, we temporarily attach subscripts TeHandTeGto the two Lie brackets respectively. By Proposition 16.14, for X,Y∈TeH, i∗([X,Y]TeH)=[i∗X,i∗Y]TeG. (16.10) This shows that if TeHis identiﬁed with a subspace of TeGviai∗, then the bracket onTeHis the restriction of the bracket on TeGtoTeH. Thus, the Lie algebra of a Lie subgroup Hmay be identiﬁed with a Lie subalgebra of the Lie algebra of G. In general, the Lie algebras of the classical groups are denoted by gothic letters. For example, the Lie algebras of GL (n,R), SL(n,R), O(n), and U( n)are denoted bygl(n,R),sl(n,R),o(n), andu(n), respectively. By (16.10) and Proposition 16.11, the Lie algebra structures on sl(n,R),o(n), andu(n)are given by [A,B]=AB−BA, as ongl(n,R). Remark 16.15 .A fundamental theorem in Lie group theory asserts the existence of a one-to-one correspondence between the connected Lie subgroups of a Lie group Gand the Lie subalgebras of its Lie algebra g[38, Theorem 3.19, Corollary (a), p. 95]. For the torus R2/Z2, the Lie algebra ghasR2as the underlying vector space and the one-dimensional Lie subalgebras are all the lines through the origin. Each line through the origin in R2is a subgroup of R2under addition. Its image under the quotient map R2→R2/Z2is a subgroup of the torus R2/Z2. If a line has rational slope, then its image is a regular submanifold of the torus. If a line has irrational slope, then its image is only an immersed submanifold of the torus. According to the correspondence theorem just quoted, the one-dimensional connected Lie subgroups 16.6 The Differential as a Lie Algebra Homomorphism 187 of the torus are the images of all the lines through the origin. Note that if a Lie subgroup had been deﬁned as a subgroup that is also a regular submanifold, then one would have to exclude all the lines with irrational slopes as Lie subgroups of the torus, and it would not be possible to have a one-to-one correspondence between the connected subgroups of a Lie group and the Lie subalgebras of its Lie algebra. It is because of our desire for such a correspondence that a Lie subgroup of a Lie group is deﬁned to be a subgroup that is also an immersed submanifold. Problems In the following problems the word “dimension” refers to the dimension as a real vector space or as a manifold. 16.1. Skew-Hermitian matrices A complex matrix X∈Cn×nis said to be skew-Hermitian if its conjugate transpose ¯XTis equal to−X. Let Vbe the vector space of n×nskew-Hermitian matrices. Show that dim V=n2. 16.2. Lie algebra of a unitary group Show that the tangent space at the identity Iof the unitary group U (n)is the vector space of n×nskew-Hermitian matrices. 16.3. Lie algebra of a symplectic group Refer to Problem 15.15 for the deﬁnition and notation concerning the symplectic group Sp (n). Show that the tangent space at the identity Iof the symplectic group Sp (n)⊂GL(n,H)is the vector space of all n×nquaternionic matrices Xsuch that ¯XT=−X. 16.4. Lie algebra of a complex symplectic group (a) Show that the tangent space at the identity Iof Sp(2n,C)⊂GL(2n,C)is the vector space of all 2 n×2ncomplex matrices Xsuch that JXis symmetric. (b) Calculate the dimension of Sp (2n,C). 16.5. Left-invariant vector ﬁelds on Rn Find the left-invariant vector ﬁelds on Rn. 16.6. Left-invariant vector ﬁelds on a circle Find the left-invariant vector ﬁelds on S1. 16.7. Integral curves of a left-invariant vector ﬁeld LetA∈gl(n,R)and let ˜Abe the left-invariant vector ﬁeld on GL (n,R)generated by A. Show thatc(t)=etAis the integral curve of ˜Astarting at the identity matrix I. Find the integral curve of˜Astarting at g∈GL(n,R). 16.8. Parallelizable manifolds A manifold whose tangent bundle is trivial is said to be parallelizable . IfMis a manifold of dimension n, show that parallelizability is equivalent to the existence of a smooth frame X1,..., XnonM. 16.9. Parallelizability of a Lie group Show that every Lie group is parallelizable. 188§16 Lie Algebras 16.10.* The pushforward of left-invariant vector ﬁelds LetF:H→Gbe a Lie group homomorphism and let XandYbe left-invariant vector ﬁelds onH. Prove that F∗[X,Y]=[F∗X,F∗Y]. 16.11. The adjoint representation LetGbe a Lie group of dimension nwith Lie algebra g. (a) For each a∈G, the differential at the identity of the conjugation map ca:=ℓa◦ra−1: G→Gis a linear isomorphism ca∗:g→g. Hence, ca∗∈GL(g). Show that the map Ad: G→GL(g)deﬁned by Ad (a)=ca∗is a group homomorphism. It is called the adjoint representation of the Lie group G. (b) Show that Ad: G→GL(g)isC∞. 16.12. A Lie algebra structure on R3 The Lie algebra o(n)of the orthogonal group O (n)is the Lie algebra of n×nskew-symmetric real matrices, with Lie bracket [A,B]=AB−BA. When n=3, there is a vector space isomor- phism ϕ:o(3)→R3, ϕ(A)=ϕ  0a1a2 −a10a3 −a2−a30  = a1 −a2 a3 =a. Prove that ϕ([A,B])=ϕ(A)×ϕ(B). Thus,R3with the cross product is a Lie algebra. Chapter 5 Differential Forms Differential forms are generalizations of real-valued functions on a manifold. Instead of assigning to each point of the manifold a number, a differential k-form assigns to each point a k-covector on its tangent space. For k=0 and 1, differential k-forms are functions and covector ﬁelds respectively. ´Elie Cartan (1869–1951)Differential forms play a crucial role in manifold theory. First and foremost, they are intrinsic objects associated to any manifold, and so can be used toconstruct diffeomorphism invariants of a manifold. In contrast to vector ﬁelds, which are also intrinsic to a manifold, differential forms have a far richer al- gebraic structure. Due to the existence of the wedge product, a grading, and the exterior derivative, theset of smooth forms on a manifold is both a graded algebra and a differential complex. Such an alge- braic structure is called a differential graded alge- bra. Moreover, the differential complex of smooth forms on a manifold can be pulled back under a smooth map, making the complex into a contravari- ant functor called the de Rham complex of the man- ifold. We will eventually construct the de Rham co-homology of a manifold from the de Rham complex. Because integration of functions on a Euclidean space depends on a choice of coordinates and is not invariant under a change of coordinates, it is not possible to integrate functions on a manifold. The highest possible degree of a differential form is the dimension of the manifold. Among differential forms, those of top degree turn out to transform correctly under a change of coordinates and are precisely the objects that can be integrated. The theory of integration on a manifold would not be possiblewithout differential forms. Very loosely speaking, differential forms are whatever appears under an integral sign. In this sense, differential forms are as old as calculus, and many theorems in © Springer Science+Business Media, LLC 2011189 L.W. Tu, An Introduction to Manifolds, Universitext, DOI 10.1007/978-1-4419-7400-6_5, 190§17 Differential 1-Forms calculus such as Cauchy’s integral theorem or Green’s theorem can be interpreted as statements about differential forms. Although it is difﬁcult to say who ﬁrst gave dif-ferential forms an independent meaning, Henri Poincar´ e [32] and ´Elie Cartan [5] are generally both regarded as pioneers in this regard. In the paper [5] published in 1899, Cartan deﬁned formally the algebra of differential forms on R nas the anticommuta- tive graded algebra over C∞functions generated by dx1,..., dxnin degree 1. In the same paper one ﬁnds for the ﬁrst time the exterior derivative on differential forms.The modern deﬁnition of a differential form as a section of an exterior power of the cotangent bundle appeared in the late forties [6], after the theory of ﬁber bundles came into being. In this chapter we give an introduction to differential forms from the vector bun- dle point of view. For simplicity we start with 1-forms, which already have many ofthe properties of k-forms. We give various characterizations of smooth forms, and show how to multiply, differentiate, and pull back these forms. In addition to the exterior derivative, we also introduce the Lie derivative and interior multiplication, two other intrinsic operations on a manifold. §17 Differential 1-Forms LetMbe a smooth manifold and pa point in M. The cotangent space ofMatp, denoted by T∗ p(M)orT∗ pM, is deﬁned to be the dual space of the tangent space TpM: T∗ pM=(TpM)∨=Hom(TpM,R). An element of the cotangent space T∗ pMis called a covector atp. Thus, a covector ωpatpis a linear function ωp:TpM→R. Acovector ﬁeld , adifferential 1-form , or more simply a 1 -form onM, is a func- tionωthat assigns to each point pinMa covector ωpatp. In this sense it is dual to a vector ﬁeld on M, which assigns to each point in Ma tangent vector at p. There are many reasons for the great utility of differential forms in manifold theory, among which is the fact that they can be pulled back under a map. This is in contrast tovector ﬁelds, which in general cannot be pushed forward under a map. Covector ﬁelds arise naturally even when one is interested only in vector ﬁelds. For example, if Xis a C ∞vector ﬁeld on Rn, then at each point p∈Rn,Xp= ∑ai∂/∂xi|p. The coefﬁcient aidepends on the vector Xp. It is in fact a linear function: TpRn→R, i.e., a covector at p. As pvaries over Rn,aibecomes a cov- ector ﬁeld on Rn. Indeed, it is none other than the 1-form dxithat picks out the ith coefﬁcient of a vector ﬁeld relative to the standard frame ∂/∂x1,...,∂/∂xn. 17.2 Local Expression for a Differential 1-Form 191 17.1 The Differential of a Function Deﬁnition 17.1. Iffis aC∞real-valued function on a manifold M, itsdifferential is deﬁned to be the 1-form dfonMsuch that for any p∈MandXp∈TpM, (df)p(Xp)=Xpf. Instead of (df)p, we also write df|pfor the value of the 1-form dfatp. This is parallel to the two notations for a tangent vector: (d/dt)p=d/dt|p. In Subsection 8.2 we encountered another notion of the differential, denoted by f∗, for a map fbetween manifolds. Let us compare the two notions of the differen- tial. Proposition 17.2. If f:M→Ris a C∞function, then for p ∈M and X p∈TpM, f∗(Xp)=( df)p(Xp)d dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle f(p). Proof. Since f∗(Xp)∈Tf(p)R, there is a real number asuch that f∗(Xp)=ad dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle f(p). (17.1) To evaluate a, apply both sides of (17.1) to x: a=f∗(Xp)(t)=Xp(t◦f)=Xpf=(df)p(Xp).⊓ ⊔ This proposition shows that under the canonical identiﬁcation of the tangent space Tf(p)RwithRvia ad dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle f(p)←→ a, f∗is the same as df. For this reason, we are justiﬁed in calling both of them the differential off. In terms of the differential d f, aC∞function f:M→Rhas a critical point at p∈Mif and only if (d f)p=0. 17.2 Local Expression for a Differential 1-Form Let(U,φ)=( U,x1,..., xn)be a coordinate chart on a manifold M. Then the differ- entials dx1,..., dxnare 1-forms on U. Proposition 17.3. At each point p∈U, the covectors (dx1)p,...,(dxn)pform a basis for the cotangent space T∗ pM dual to the basis ∂/∂x1|p,..., ∂/∂xn|pfor the tangent space T pM. 192§17 Differential 1-Forms Proof. The proof is just like that in the Euclidean case (Proposition 4.1): (dxi)p/parenleftigg ∂ ∂xj/vextendsingle/vextendsingle/vextendsingle/vextendsingle p/parenrightigg =∂ ∂xj/vextendsingle/vextendsingle/vextendsingle/vextendsingle pxi=δi j. ⊓ ⊔ Thus, every 1-form ωonUcan be written as a linear combination ω=∑aidxi, where the coefﬁcients aiare functions on U. In particular, if fis aC∞function on M, then the restriction of the 1-form dftoUmust be a linear combination df=∑aidxi. To ﬁnd aj, we apply the usual trick of evaluating both sides on ∂/∂xj: (df)/parenleftbigg∂ ∂xj/parenrightbigg =∑ iaidxi/parenleftbigg∂ ∂xj/parenrightbigg =⇒∂f ∂xj=∑ iaiδi j=aj. This gives a local expression for df: df=∑∂f ∂xidxi. (17.2) 17.3 The Cotangent Bundle The underlying set of the cotangent bundle T∗Mof a manifold Mis the union of the cotangent spaces at all the points of M: T∗M:=/uniondisplay p∈MT∗ pM. (17.3) Just as in the case of the tangent bundle, the union (17.3) is a disjoint union and there is a natural map π:T∗M→Mgiven by π(α)=pifα∈T∗ pM. Mimicking the construction of the tangent bundle, we give T∗Ma topology as follows. If (U,φ)= (U,x1,..., xn)is a chart on Mandp∈U, then each α∈T∗ pMcan be written uniquely as a linear combination α=∑ci(α)dxi|p. This gives rise to a bijection ˜φ:T∗U→φ(U)×Rn, (17.4) α/ma√s⊔o→(φ(p),c1(α),..., cn(α))=( φ◦π,c1,..., cn)(α). Using this bijection, we can transfer the topology of φ(U)×RntoT∗U. Now for each domain Uof a chart in the maximal atlas of M, letBUbe the collection of all open subsets of T∗U, and let Bbe the union of the BU. As in Subsection 12.1, Bsatisﬁes the conditions for a collection of subsets of T∗Mto be a 17.4 Characterization of C∞1-Forms 193 basis. We give T∗Mthe topology generated by the basis B. As for the tangent bundle, with the maps ˜φ=(x1◦π,..., xn◦π,c1,..., cn)of (17.4) as coordinate maps, T∗M becomes a C∞manifold, and the projection map π:T∗M→Mbecomes a vector bundle of rank nover M, justifying the “bundle” in the name “cotangent bundle.” If x1,..., xnare coordinates on U⊂M, then π∗x1,..., π∗xn,c1,..., cnare coordinates onπ−1U⊂T∗M. Properly speaking, the cotangent bundle of a manifold Mis the triple(T∗M,M,π), while T∗MandMare the total space and the base space of the cotangent bundle respectively, but by abuse of language, it is customary to call T∗M the cotangent bundle of M. In terms of the cotangent bundle, a 1-form on Mis simply a section of the cotan- gent bundle T∗M; i.e., it is a map ω:M→T∗Msuch that π◦ω=1M, the identity map on M. We say that a 1-form ωisC∞if it is C∞as a map M→T∗M. Example 17.4 ( Liouville form on the cotangent bundle ).If a manifold Mhas dimen- sion n, then the total space T∗Mof its cotangent bundle π:T∗M→Mis a manifold of dimension 2 n. Remarkably, on T∗Mthere is a 1-form λ, called the Liouville form (or the Poincar ´e form in some books), deﬁned independently of charts as follows. A point in T∗Mis a covector ωp∈T∗ pMat some point p∈M. IfXωpis a tangent vector to T∗Matωp, then the pushforward π∗/parenleftbig Xωp/parenrightbig is a tangent vector to Matp. Therefore, one can pair up ωpandπ∗/parenleftbig Xωp/parenrightbig to obtain a real number ωp/parenleftbig π∗/parenleftbig Xωp/parenrightbig/parenrightbig . Deﬁne λωp/parenleftbig Xωp/parenrightbig =ωp/parenleftbig π∗/parenleftbig Xωp/parenrightbig/parenrightbig . The cotangent bundle and the Liouville form on it play an important role in the mathematical theory of classical mechanics [1, p. 202]. 17.4 Characterization of C∞1-Forms We deﬁne a 1-form ωon a manifold Mto be smooth ifω:M→T∗Mis smooth as a section of the cotangent bundle π:T∗M→M. The set of all smooth 1-forms on M has the structure of a vector space, denoted by Ω1(M). In a coordinate chart (U,φ)= (U,x1,..., xn)onM, the value of the 1-form ωatp∈Uis a linear combination ωp=∑ai(p)dxi|p. Aspvaries in U, the coefﬁcients aibecome functions on U. We will now derive smoothness criteria for a 1-form in terms of the coefﬁcient functions ai. The devel- opment is parallel to that of smoothness criteria for a vector ﬁeld in Subsection 14.1. By Subsection 17.3, the chart (U,φ)onMinduces a chart (T∗U,˜φ)=( T∗U,¯x1,..., ¯xn,c1,..., cn) onT∗M, where ¯ xi=π∗xi=xi◦πand the ciare deﬁned by α=∑ci(α)dxi|p,α∈T∗ pM. Comparing the coefﬁcients in 194§17 Differential 1-Forms ωp=∑ai(p)dxi|p=∑ci(ωp)dxi|p, we get ai=ci◦ω, where ωis viewed as a map from UtoT∗U. Being coordinate functions, the ciare smooth on T∗U. Thus, if ωis smooth, then the coefﬁcients aiof ω=∑aidxirelative to the frame dxiare smooth on U. The converse is also true, as indicated in the following lemma. Lemma 17.5. Let(U,φ) = ( U,x1,..., xn)be a chart on a manifold M. A 1-form ω=∑aidxion U is smooth if and only if the coefﬁcient functions a iare all smooth. Proof. This lemma is a special case of Proposition 12.12, with Ethe cotangent bun- dleT∗Mandsjthe coordinate 1-forms dxj. However, a direct proof is also possible (cf. Lemma 14.1). Since ˜φ:T∗U→U×Rnis a diffeomorphism, ω:U→T∗Mis smooth if and only if ˜φ◦ω:U→U×Rnis smooth. For p∈U, (˜φ◦ω)(p)=˜φ(ωp)=/parenleftbig x1(p),..., xn(p),c1(ωp),..., cn(ωp)/parenrightbig =/parenleftbig x1(p),..., xn(p),a1(p),..., an(p)/parenrightbig . As coordinate functions, x1,..., xnare smooth on U. Therefore, by Proposition 6.13, ˜φ◦ωis smooth on Uif and only if all aiare smooth on U. ⊓ ⊔ Proposition 17.6 (Smoothness of a 1-form in terms of coefﬁcients). Letωbe a 1-form on a manifold M. The following are equivalent: (i)The1-form ωis smooth on M. (ii)The manifold M has an atlas such that on any chart (U,x1,..., xn)of the atlas, the coefﬁcients a iofω=∑aidxirelative to the frame dxiare all smooth. (iii)On any chart (U,x1,..., xn)on the manifold, the coefﬁcients a iofω=∑aidxi relative to the frame dxiare all smooth. Proof. The proof is omitted, since it is virtually identical to that of Proposition 14.2. ⊓ ⊔ Corollary 17.7. If f is a C∞function on a manifold M, then its differential d f is a C∞1-form on M. Proof. On any chart (U,x1,..., xn)onM, the equality d f=∑(∂f/∂xi)dxiholds. Since the coefﬁcients ∂f/∂xiare all C∞, by Proposition 17.6(iii), the 1-form d f isC∞. ⊓ ⊔ Ifωis a 1-form and Xis a vector ﬁeld on a manifold M, we deﬁne a function ω(X)onMby the formula ω(X)p=ωp(Xp)∈R,p∈M. Proposition 17.8 (Linearity of a 1-form over functions). Letωbe a 1-form on a manifold M. If f is a function and X is a vector ﬁeld on M, then ω(f X)=fω(X). 17.5 Pullback of 1-Forms 195 Proof. At each point p∈M, ω(f X)p=ωp(f(p)Xp)=f(p)ωp(Xp)=( fω(X))p, because ω(X)is deﬁned pointwise, and at each point, ωpisR-linear in its argument. ⊓ ⊔ Proposition 17.9 (Smoothness of a 1-form in terms of vector ﬁelds). A1-form ω on a manifold M is C∞if and only if for every C∞vector ﬁeld X on M, the function ω(X)is C∞on M. Proof. (⇒)Suppose ωis aC∞1-form and Xis aC∞vector ﬁeld on M. On any chart (U,x1,..., xn)onM, by Propositions 14.2 and 17.6, ω=∑aidxiandX=∑bj∂/∂xj forC∞functions ai,bj. By the linearity of 1-forms over functions (Proposition 17.8), ω(X)=/parenleftbig∑aidxi/parenrightbig/parenleftbigg ∑bj∂ ∂xj/parenrightbigg =∑ i,jaibjδi j=∑aibi, aC∞function on U. Since Uis an arbitrary chart on M, the function ω(X)isC∞ onM. (⇐)Suppose ωis a 1-form on Msuch that the function ω(X)isC∞for every C∞ vector ﬁeld XonM. Given p∈M, choose a coordinate neighborhood (U,x1,..., xn) about p. Then ω=∑aidxionUfor some functions ai. Fix an integer j, 1≤j≤n. By Proposition 14.4, we can extend the C∞vector ﬁeld X=∂/∂xjonUto a C∞vector ﬁeld ¯XonMthat agrees with ∂/∂xjin a neighborhood Vj pofpinU. Restricted to the open set Vj p, ω(¯X)=/parenleftbig∑aidxi/parenrightbig/parenleftbigg∂ ∂xj/parenrightbigg =aj. This proves that ajisC∞on the coordinate chart (Vj p,x1,..., xn). On the intersection Vp:=/intersectiontext jVj p, allajareC∞. By Lemma 17.5, the 1-form ωisC∞onVp. So for each p∈M, we have found a coordinate neighborhood Vpon which ωisC∞. It follows thatωis aC∞map from MtoT∗M. ⊓ ⊔ LetF=C∞(M)be the ring of all C∞functions on M. By Proposition 17.9, a 1- form ωonMdeﬁnes a map X(M)→F,X/ma√s⊔o→ω(X). According to Proposition 17.8, this map is both R-linear and F-linear. 17.5 Pullback of 1-Forms IfF:N→Mis aC∞map of manifolds, then at each point p∈Nthe differential F∗,p:TpN→TF(p)M 196§17 Differential 1-Forms is a linear map that pushes forward vectors at pfrom NtoM. The codifferential , i.e., the dual of the differential, (F∗,p)∨:T∗ F(p)M→T∗ pN, reverses the arrow and pulls back a covector at F(p)from MtoN. Another notation for the codifferential is F∗=(F∗,p)∨. By the deﬁnition of the dual, if ωF(p)∈T∗ F(p)M is a covector at F(p)andXp∈TpNis a tangent vector at p, then F∗/parenleftbig ωF(p)/parenrightbig (Xp)=/parenleftbig (F∗,p)∨ωF(p)/parenrightbig (Xp)=ωF(p)(F∗,pXp). We call F∗/parenleftbig ωF(p)/parenrightbig thepullback of the covector ωF(p)byF. Thus, the pullback of covectors is simply the codifferential. Unlike vector ﬁelds, which in general cannot be pushed forward under a C∞ map, every covector ﬁeld can be pulled back by a C∞map. If ωis a 1-form on M, its pullback F∗ωis the 1-form on Ndeﬁned pointwise by (F∗ω)p=F∗/parenleftbig ωF(p)/parenrightbig ,p∈N. This means that (F∗ω)p(Xp)=ωF(p)(F∗(Xp)) for all Xp∈TpN. Recall that functions can also be pulled back: if Fis aC∞map from NtoMandg∈C∞(M), then F∗g=g◦F∈C∞(N). This difference in the behavior of vector ﬁelds and forms under a map can be traced to a basic asymmetry in the concept of a function—every point in the domain maps to only one image point in the range, but a point in the range can have several preimage points in the domain. Now that we have deﬁned the pullback of a 1-form under a map, a question naturally suggests itself. Is the pullback of a C∞1-form under a C∞map C∞? To answer this question, we ﬁrst need to establish three commutation properties of the pullback: its commutation with the differential, sum, and product. Proposition 17.10 (Commutation of the pullback with the differential). Let F :N →M be a C∞map of manifolds. For any h ∈C∞(M), F∗(dh)=d(F∗h). Proof. It sufﬁces to check that for any point p∈Nand any tangent vector Xp∈TpN, (F∗dh)p(Xp)=( dF∗h)p(Xp). (17.5) The left-hand side of (17.5) is (F∗dh)p(Xp)=( dh)F(p)(F∗(Xp))(deﬁnition of the pullback of a 1-form) =(F∗(Xp))h (deﬁnition of the differential dh) =Xp(h◦F) (deﬁnition of F∗). The right-hand side of (17.5) is (dF∗h)p(Xp)=Xp(F∗h)(deﬁnition of dof a function) =Xp(h◦F)(deﬁnition of F∗of a function). ⊓ ⊔ 17.6 Restriction of 1-Forms to an Immersed Submanifold 197 Pullback of functions and 1-forms respects addition and scalar multiplication. Proposition 17.11 (Pullback of a sum and a product). Let F :N→M be a C∞map of manifolds. Suppose ω,τ∈Ω1(M)and g∈C∞(M). Then (i)F∗(ω+τ)=F∗ω+F∗τ, (ii)F∗(gω)=( F∗g)(F∗ω). Proof. Problem 17.5. Proposition 17.12 (Pullback of a C∞1-form). The pullback F∗ωof a C∞1-form ω on M under a C∞map F :N→M is C∞1-form on N. Proof. Given p∈N, choose a chart (V,ψ) = ( V,y1,..., yn)inMabout F(p). By the continuity of F, there is a chart (U,φ) = ( U,x1,..., xn)about pinNsuch that F(U)⊂V. On V,ω=∑aidyifor some ai∈C∞(V). On U, F∗ω=∑(F∗ai)F∗(dyi) (Proposition 17.11) =∑(F∗ai)dF∗yi(Proposition 17.10) =∑(ai◦F)d(yi◦F)(deﬁnition of F∗of a function) =∑ i,j(ai◦F)∂Fi ∂xjdxj(equation (17.2)). Since the coefﬁcients (ai◦F)∂Fi/∂xjare all C∞, by Proposition 17.5 the 1-form F∗ωisC∞onUand therefore at p. Since pwas an arbitrary point in N, the pullback F∗ωisC∞onN. ⊓ ⊔ Example 17.13 ( Liouville form on the cotangent bundle ).LetMbe a manifold. In terms of the pullback, the Liouville form λon the cotangent bundle T∗Mintroduced in Example 17.4 can be expressed as λωp=π∗(ωp)at any ωp∈T∗M. 17.6 Restriction of 1-Forms to an Immersed Submanifold LetS⊂Mbe an immersed submanifold and i:S→Mthe inclusion map. At any p∈S, since the differential i∗:TpS→TpMis injective, one may view the tangent space TpSas a subspace of TpM. Ifωis a 1-form on M, then the restriction ofωto Sis the 1-form ω|Sdeﬁned by (ω|S)p(v)=ωp(v)for all p∈Sandv∈TpS. Thus, the restriction ω|Sis the same as ωexcept that its domain has been restricted from MtoSand for each p∈S, the domain of (ω|S)phas been restricted from TpM toTpS. The following proposition shows that the restriction of 1-forms is simply the pullback of the inclusion i. Proposition 17.14. If i:S֒→M is the inclusion map of an immersed submanifold S andωis a1-form on M, then i∗ω=ω|S. 198§17 Differential 1-Forms Proof. Forp∈Sandv∈TpS, (i∗ω)p(v)=ωi(p)(i∗v)(deﬁnition of pullback) =ωp(v) (both iandi∗are inclusions) =(ω|S)p(v)(deﬁnition of ω|S). ⊓ ⊔ To avoid too cumbersome a notation, we sometimes write ωto mean ω|S, relying on the context to make clear that it is the restriction of ωtoS. Example 17.15 ( A1-form on the circle ).The velocity vector ﬁeld of the unit circle c(t)=( x,y)=( cost,sint)inR2is c′(t)=(−sint,cost)=(−y,x). Thus, X=−y∂ ∂x+x∂ ∂y is aC∞vector ﬁeld on the unit circle S1. What this notation means is that if x,yare the standard coordinates on R2andi:S1֒→R2is the inclusion map, then at a point p=(x,y)∈S1, one has i∗Xp=−y∂/∂x|p+x∂/∂y|p, where ∂/∂x|pand∂/∂y|pare tangent vectors at pinR2. Find a 1-form ω=adx+bdyonS1such that ω(X)≡1. Solution. Here ωis viewed as the restriction to S1of the 1-form adx+bdy onR2. We calculate in R2, where dx,dyare dual to ∂/∂x,∂/∂y: ω(X)=( adx+bdy)/parenleftbigg −y∂ ∂x+x∂ ∂y/parenrightbigg =−ay+bx=1. (17.6) Since x2+y2=1 on S1,a=−yandb=xis a solution to (17.6). So ω=−ydx+xdy is one such 1-form. Since ω(X)≡1, the form ωis nowhere vanishing on the circle. Remark. In the notation of Problem 11.2, ωshould be written−¯yd¯x+¯xd¯y, since x,y are functions on R2and ¯x,¯yare their restrictions to S1. However, one generally uses the same notation for a form on Rnand for its restriction to a submanifold. Since i∗x=¯xandi∗dx=d¯x, there is little possibility of confusion in omitting the bar while dealing with the restriction of forms on Rn. This is in contrast to the situation for vector ﬁelds, where i∗(∂/∂¯x|p)/ne}a⊔ionslash=∂/∂x|p. Example 17.16 ( Pullback of a 1-form ).Leth:R→S1⊂R2be given by h(t)=( x,y) =(cost,sint). Ifωis the 1-form−ydx+xdyonS1, compute the pullback h∗ω. Solution. h∗(−ydx+xdy)=−(h∗y)d(h∗x)+(h∗x)d(h∗y) (by Proposition 17.11) =−(sint)d(cost)+(cost)d(sint) =sin2t dt+cos2t dt=dt. 17.6 Restriction of 1-Forms to an Immersed Submanifold 199 Problems 17.1. A 1-form on R2−{(0,0)} Denote the standard coordinates on R2byx,y, and let X=−y∂ ∂x+x∂ ∂yand Y=x∂ ∂x+y∂ ∂y be vector ﬁelds on R2. Find a 1-form ωonR2−{(0,0)}such that ω(X)=1 and ω(Y)=0. 17.2. Transition formula for 1-forms Suppose(U,x1,..., xn)and(V,y1,..., yn)are two charts on Mwith nonempty overlap U∩V. Then a C∞1-form ωonU∩Vhas two different local expressions: ω=∑ajdxj=∑bidyi. Find a formula for ajin terms of bi. 17.3. Pullback of a 1-form on S1 Multiplication in the unit circle S1, viewed as a subset of the complex plane, is given by eit·eiu=ei(t+u),t,u∈R. In terms of real and imaginary parts, (cost+isint)(x+iy)=((cos t)x−(sint)y)+i((sin t)x+(cos t)y). Hence, if g=(cost,sint)∈S1⊂R2, then the left multiplication ℓg:S1→S1is given by ℓg(x,y)=((cos t)x−(sint)y,(sint)x+(cos t)y). Letω=−ydx+xdy be the 1-form found in Example 17.15. Prove that ℓ∗gω=ωfor all g∈S1. 17.4. Liouville form on the cotangent bundle (a) Let(U,φ)=( U,x1,..., xn)be a chart on a manifold M, and let (π−1U,˜φ)=( π−1U,¯x1,..., ¯xn,c1,..., cn) be the induced chart on the cotangent bundle T∗M. Find a formula for the Liouville form λonπ−1Uin terms of the coordinates ¯ x1,..., ¯xn,c1,..., cn. (b) Prove that the Liouville form λonT∗MisC∞. (Hint: Use (a) and Proposition 17.6.) 17.5. Pullback of a sum and a product Prove Proposition 17.11 by verifying both sides of each equality on a tangent vector Xpat a point p. 17.6. Construction of the cotangent bundle LetMbe a manifold of dimension n. Mimicking the construction of the tangent bundle in Section 12, write out a detailed proof that π:T∗M→Mis aC∞vector bundle of rank n. 200§18 Differential k-Forms §18 Differential k-Forms We now generalize the construction of 1-forms on a manifold to k-forms. After deﬁning k-forms on a manifold, we show that locally they look no different from k- forms on Rn. In parallel to the construction of the tangent and cotangent bundles on a manifold, we construct the kth exterior power/logicalandtextk(T∗M)of the cotangent bundle. A differential k-form is seen to be a section of the bundle/logicalandtextk(T∗M). This gives a natural notion of smoothness of differential forms: a differential k-form is smooth if and only if it is smooth as a section of the vector bundle/logicalandtextk(T∗M). The pullback and the wedge product of differential forms are deﬁned pointwise. As examples of differential forms, we consider left-invariant forms on a Lie group. 18.1 Differential Forms Recall that a k-tensor on a vector space Vis ak-linear function f:V×···× V→R. Thek-tensor fisalternating if for any permutation σ∈Sk, f(vσ(1),..., vσ(k))=( sgnσ)f(v1,..., vk). (18.1) When k=1, the only element of the permutation group S1is the identity permutation. So for 1-tensors the condition (18.1) is vacuous and all 1-tensors are alternating (and symmetric too). An alternating k-tensor on Vis also called a k-covector onV. For any vector space V, denote by Ak(V)the vector space of alternating k-tensors onV. Another common notation for the space Ak(V)is/logicalandtextk(V∨). Thus, /logicalandtext0(V∨)=A0(V)=R, /logicalandtext1(V∨)=A1(V)=V∨, /logicalandtext2(V∨)=A2(V),and so on . In fact, there is a purely algebraic construction/logicalandtextk(V), called the kthexterior power of the vector space V, with the property that/logicalandtextk(V∨)is isomorphic to Ak(V). To delve into this construction would lead us too far aﬁeld, so in this book/logicalandtextk(V∨)will simply be an alternative notation for Ak(V). We apply the functor Ak( )to the tangent space TpMof a manifold Mat a point p. The vector space Ak(TpM), usually denoted by/logicalandtextk(T∗ pM), is the space of all alternating k-tensors on the tangent space TpM. Ak-covector ﬁeld onMis a function ωthat assigns to each point p∈Mak-covector ωp∈/logicalandtextk(T∗ pM). Ak-covector ﬁeld is also called a differential k-form, a differential form of degree k , or simply a k-form . Atop form on a manifold is a differential form whose degree is the dimension of the manifold. 18.1 Differential Forms 201 Ifωis ak-form on a manifold MandX1,..., Xkare vector ﬁelds on M, then ω(X1,..., Xk)is the function on Mdeﬁned by (ω(X1,..., Xk))(p)=ωp((X1)p,...,(Xk)p). Proposition 18.1 (Multilinearity of a form over functions). Letωbe a k-form on a manifold M. For any vector ﬁelds X 1,..., Xkand any function h on M, ω(X1,..., hXi,..., Xk)=hω(X1,..., Xi,..., Xk). Proof. The proof is essentially the same as that of Proposition 17.8. ⊓ ⊔ Example 18.2.Let(U,x1,..., xn)be a coordinate chart on a manifold. At each point p∈U, a basis for the tangent space TpUis ∂ ∂x1/vextendsingle/vextendsingle/vextendsingle/vextendsingle p,...,∂ ∂xn/vextendsingle/vextendsingle/vextendsingle/vextendsingle p. As we saw in Proposition 17.3, the dual basis for the cotangent space T∗ pUis (dx1)p,...,(dxn)p. Aspvaries over points in U, we get differential 1-forms dx1,..., dxnonU. By Proposition 3.29 a basis for the alternating k-tensors in/logicalandtextk(T∗ pU)is (dxi1)p∧···∧(dxik)p,1≤i1<···<ik≤n. Ifωis ak-form on Rn, then at each point p∈Rn,ωpis a linear combination ωp=∑ai1···ik(p)(dxi1)p∧···∧(dxik)p. Omitting the point p, we write ω=∑ai1···ikdxi1∧···∧ dxik. In this expression the coefﬁcients ai1···ikare functions on Ubecause they vary with the point p. To simplify the notation, we let Ik,n={I=(i1,..., ik)|1≤i1<···<ik≤n} be the set of all strictly ascending multi-indices between 1 and nof length k, and write ω=∑ I∈Ik,naIdxI, where dxIstands for dxi1∧···∧ dxik. 202§18 Differential k-Forms 18.2 Local Expression for a k-Form By Example 18.2, on a coordinate chart (U,x1,..., xn)of a manifold M, ak-form on Uis a linear combination ω=∑aIdxI, where I∈Ik,nand the aIare functions on U. As a shorthand, we write ∂i=∂/∂xifor the ith coordinate vector ﬁeld. Evaluating pointwise as in Lemma 3.28, we obtain the following equality on UforI,J∈Ik,n: dxI(∂j1,..., ∂jk)=δI J=/braceleftigg 1 for I=J, 0 for I/ne}a⊔ionslash=J.(18.2) Proposition 18.3 (A wedge of differentials in local coordinates). Let(U,x1,..., xn) be a chart on a manifold and f1,..., fksmooth functions on U. Then df1∧···∧ dfk=∑ I∈Ik,n∂(f1,..., fk) ∂(xi1,..., xik)dxi1∧···∧ dxik. Proof. OnU, df1∧···∧ dfk=∑ J∈Ik,ncJdxj1∧···∧ dxjk (18.3) for some functions cJ. By the deﬁnition of the differential, d fi(∂/∂xj)=∂fi/∂xj. Applying both sides of (18.3) to the list of coordinate vectors ∂i1,..., ∂ik, we get LHS=(df1∧···∧ dfk)(∂i1,..., ∂ik)=det/bracketleftbigg∂fi ∂xij/bracketrightbigg by Proposition 3.27 =∂(f1,..., fk) ∂(xi1,..., xik), RHS=∑ JcJdxJ(∂i1,..., ∂ik)=∑ JcJδJ I=cI by Lemma 18.2 . Hence, cI=∂(f1,..., fk)/∂(xi1,..., xik). ⊓ ⊔ If(U,x1,..., xn)and(V,y1,..., yn)are two overlapping charts on a manifold, then on the intersection U∩V, Proposition 18.3 becomes the transition formula for k-forms: dyJ=∑ I∂(yj1,..., yjk) ∂(xi1,..., xik)dxI. Two cases of Proposition 18.3 are of special interest: Corollary 18.4. Let(U,x1,..., xn)be a chart on a manifold, and let f , f1, . . . , fnbe C∞functions on U. Then (i)(1-forms)df=∑(∂f/∂xi)dxi, (ii)(top forms )df1∧···∧ dfn=det[∂fj/∂xi]dx1∧···∧ dxn. Case (i) of the corollary agrees with the formula we derived in (17.2). 18.4 Smooth k-Forms 203 Exercise 18.5 (Transition formula for a 2-form).* If(U,x1,..., xn)and(V,y1,..., yn)are two overlapping coordinate charts on M, then a C∞2-form ωonU∩Vhas two local expres- sions: ω=∑ i<jai jdxi∧dxj=∑ k<ℓbkℓdyk∧dyℓ. Find a formula for ai jin terms of bkℓand the coordinate functions x1,..., xn,y1,..., yn. 18.3 The Bundle Point of View LetMbe a manifold of dimension n. To better understand differential forms, we mimic the construction of the tangent and cotangent bundles and form the set /logicalandtextk(T∗M):=/uniontext p∈M/logicalandtextk(T∗ pM)=/uniontext p∈MAk(TpM) of all alternating k-tensors at all points of the manifold M. This set is called the kth exterior power of the cotangent bundle. There is a projection map π:/logicalandtextk(T∗M)→M given by π(α)=pifα∈/logicalandtextk(T∗ pM). If(U,φ)is a coordinate chart on M, then there is a bijection /logicalandtextk(T∗U)=/uniontext p∈U/logicalandtextk(T∗ pU)≃φ(U)×R(n k), α∈/logicalandtextk(T∗ pU)/ma√s⊔o→(φ(p),{cI(α)}I), where α=∑cI(α)dxI|p∈/logicalandtextk(T∗ pU)andI=(1≤i1<···<ik≤n). In this way we can give/logicalandtextk(T∗U)and hence/logicalandtextk(T∗M)a topology and even a differentiable struc- ture. The details are just like those for the construction of the tangent bundle, so we omit them. The upshot is that the projection map π:/logicalandtextk(T∗M)→Mis aC∞vector bundle of rank/parenleftbign k/parenrightbig and that a differential k-form is simply a section of this bundle. As one might expect, we deﬁne a k-form to be C∞if it is C∞as a section of the bundle π:/logicalandtextk(T∗M)→M. NOTATION . IfE→Mis aC∞vector bundle, then the vector space of C∞sections ofEis denoted by Γ(E)orΓ(M,E). The vector space of all C∞k-forms on Mis usually denoted by Ωk(M). Thus, Ωk(M)=Γ/parenleftig/logicalandtextk(T∗M)/parenrightig =Γ/parenleftig M,/logicalandtextk(T∗M)/parenrightig . 18.4 Smooth k-Forms There are several equivalent characterizations of a smooth k-form. Since the proofs are similar to those for 1-forms (Lemma 17.5 and Propositions 17.6 and 17.9), weomit them. Lemma 18.6 (Smoothness of a k-form on a chart). Let(U,x 1,..., xn)be a chart on a manifold M. A k-form ω=∑aIdxIon U is smooth if and only if the coefﬁcient functions a Iare all smooth on U. 204§18 Differential k-Forms Proposition 18.7 (Characterization of a smooth k-form). Letωbe a k-form on a manifold M. The following are equivalent: (i)The k-form ωis C∞on M. (ii)The manifold M has an atlas such that on every chart (U,φ) = ( U,x1,..., xn) in the atlas, the coefﬁcients a Iofω=∑aIdxIrelative to the coordinate frame {dxI}I∈Ik,nare all C∞. (iii)On every chart (U,φ)=( U,x1,..., xn)on M, the coefﬁcients a Iofω=∑aIdxI relative to the coordinate frame {dxI}I∈Ik,nare all C∞. (iv)For any k smooth vector ﬁelds X 1,..., Xkon M, the function ω(X1,..., Xk)is C∞ on M. We deﬁned the 0-tensors and the 0-covectors to be the constants, that is, L0(V)= A0(V)=R. Therefore, the bundle/logicalandtext0(T∗M)is simply M×Rand a 0-form on Mis a function on M. AC∞0-form on Mis thus the same as a C∞function on M. In our new notation, Ω0(M)=Γ/parenleftbig/logicalandtext0(T∗M)/parenrightbig =Γ(M×R)=C∞(M). Proposition 13.2 on C∞extensions of functions has a generalization to differential forms. Proposition 18.8 ( C∞extension of a form). Suppose τis a C∞differential form deﬁned on a neighborhood U of a point p in a manifold M. Then there is a C∞form ˜τon M that agrees with τon a possibly smaller neighborhood of p. The proof is identical to that of Proposition 13.2. We leave it as an exercise. Of course, the extension ˜τis not unique. In the proof it depends on pand on the choice of a bump function at p. 18.5 Pullback of k-Forms We have deﬁned the pullback of 0-forms and 1-forms under a C∞map F:N→M. For a C∞0-form on M, i.e., a C∞function on M, the pullback F∗fis simply the composition NF→Mf→R,F∗(f)=f◦F∈Ω0(N). To generalize the pullback to k-forms for all k≥1, we ﬁrst recall the pullback of k-covectors from Subsection 10.3. A linear map L:V→Wof vector spaces induces a pullback map L∗:Ak(W)→Ak(V)by (L∗α)(v1,..., vk)=α(L(v1),..., L(vk)) forα∈Ak(W)andv1,..., vk∈V. Now suppose F:N→Mis aC∞map of manifolds. At each point p∈N, the differential F∗,p:TpN→TF(p)M 18.6 The Wedge Product 205 is a linear map of tangent spaces, and so by the preceding paragraph there is a pull- back map (F∗,p)∗:Ak(TF(p)M)→Ak(TpN). This ugly notation is usually simpliﬁed to F∗. Thus, if ωF(p)is ak-covector at F(p) inM, then its pullback F∗/parenleftbig ωF(p)/parenrightbig is the k-covector at pinNgiven by F∗/parenleftbig ωF(p)/parenrightbig (v1,..., vk)=ωF(p)(F∗,pv1,..., F∗,pvk),vi∈TpN. Finally, if ωis ak-form on M, then its pullback F∗ωis the k-form on Ndeﬁned pointwise by (F∗ω)p=F∗/parenleftbig ωF(p)/parenrightbig for all p∈N. Equivalently, (F∗ω)p(v1,..., vk)=ωF(p)(F∗,pv1,..., F∗,pvk),vi∈TpN. (18.4) When k=1, this formula specializes to the deﬁnition of the pullback of a 1-form in Subsection 17.5. The pullback of a k-form (18.4) can be viewed as a composition TpN×···× TpNF∗×···× F∗−−−−−−→ TF(p)M×···× TF(p)MωF(p)−−−→R. Proposition 18.9 (Linearity of the pullback). Let F :N→M be a C∞map. If ω,τ are k-forms on M and a is a real number, then (i)F∗(ω+τ)=F∗ω+F∗τ; (ii)F∗(aω)=aF∗ω. Proof. Problem 18.2. ⊓ ⊔ At this point, we still do not know, other than for k=0,1, whether the pullback of aC∞k-form under a C∞map remains C∞. This very basic question will be answered in Subsection 19.5. 18.6 The Wedge Product We learned in Section 3 that if αandβare alternating tensors of degree kandℓ respectively on a vector space V, then their wedge product α∧βis the alternating (k+ℓ)-tensor on Vdeﬁned by (α∧β)(v1,..., vk+ℓ)=∑(sgnσ)α(vσ(1),..., vσ(k))β(vσ(k+1),..., vσ(k+ℓ)), where vi∈Vandσruns over all (k,ℓ)-shufﬂes of 1 ,..., k+ℓ. For example, if αand βare 1-covectors, then (α∧β)(v1,v2)=α(v1)β(v2)−α(v2)β(v1). The wedge product extends pointwise to differential forms on a manifold: for a k-form ωand anℓ-form τonM, deﬁne their wedge product ω∧τto be the(k+ℓ)- form on Msuch that (ω∧τ)p=ωp∧τp at all p∈M. 206§18 Differential k-Forms Proposition 18.10. Ifωandτare C∞forms on M, then ω∧τis also C∞. Proof. Let(U,x1,..., xn)be a chart on M. On U, ω=∑aIdxI,τ=∑bJdxJ forC∞function aI,bJonU. Their wedge product on Uis ω∧τ=/parenleftbig∑aIdxI/parenrightbig ∧/parenleftbig∑bJdxJ/parenrightbig =∑aIbJdxI∧dxJ. In this sum, dxI∧dxJ=0 ifIandJhave an index in common. If IandJare disjoint, then dxI∧dxJ=±dxK, where K=I∪Jbut reordered as an increasing sequence. Thus, ω∧τ=∑ K/parenleftigg ∑ I∪J=K I,Jdisjoint±aIbJ/parenrightigg dxK. Since the coefﬁcients of dxKareC∞onU, by Proposition 18.7, ω∧τisC∞.⊓ ⊔ Proposition 18.11 (Pullback of a wedge product). If F:N→M is a C∞map of manifolds and ωandτare differential forms on M, then F∗(ω∧τ)=F∗ω∧F∗τ. Proof. Problem 18.3. ⊓ ⊔ Deﬁne the vector space Ω∗(M)ofC∞differential forms on a manifold Mof dimension nto be the direct sum Ω∗(M)=n/circleplusdisplay k=0Ωk(M). What this means is that each element of Ω∗(M)is uniquely a sum ∑n k=0ωk, where ωk∈Ωk(M). With the wedge product, the vector space Ω∗(M)becomes a graded algebra, the grading being the degree of differential forms. 18.7 Differential Forms on a Circle Consider the map h:R→S1,h(t)=( cost,sint). Since the derivative ˙h(t)=(−sint,cost)is nonzero for all t, the map h:R→S1is a submersion. By Problem 18.8, the pullback map h∗:Ω∗(S1)→Ω∗(R)on smooth differential forms is injective. This will allow us to identify the differential forms onS 1with a subspace of differential forms on R. Letω=−ydx+xdybe the nowhere-vanishing form on S1from Example 17.15. In Example 17.16, we showed that h∗ω=dt. Since ωis nowhere vanishing, it is a frame for the cotangent bundle T∗S1over S1, and every C∞1-form αonS1can be 18.8 Invariant Forms on a Lie Group 207 written as α=fωfor some function fonS1. By Proposition 12.12, the function fisC∞. Its pullback ¯f:=h∗fis aC∞function on R. Since pulling back preserves multiplication (Proposition 18.11), h∗α=(h∗f)(h∗ω)= ¯f dt. (18.5) We say that a function gor a 1-form gdtonRisperiodic ofperiod a ifg(t+a)=g(t) for all t∈R. Proposition 18.12. For k=0,1, under the pullback map h∗:Ω∗(S1)→Ω∗(R), smooth k-forms on S1are identiﬁed with smooth periodic k-forms of period 2πonR. Proof. Iff∈Ω0(S1), then since h:R→S1is periodic of period 2 π, the pullback h∗f=f◦h∈Ω0(R)is periodic of period 2 π. Conversely, suppose ¯f∈Ω0(R)is periodic of period 2 π. For p∈S1, let sbe theC∞inverse in a neighborhood Uofpof the local diffeomorphism hand deﬁne f=¯f◦sonU. To show that fis well deﬁned, let s1ands2be two inverses of h over U. By the periodic properties of sine and cosine, s1=s2+2πnfor some n∈Z. Because ¯fis periodic of period 2 π, we have ¯f◦s1=¯f◦s2. This proves that fis well deﬁned on U. Moreover, ¯f=f◦s−1=f◦h=h∗fonh−1(U). Aspvaries over S1, we obtain a well-deﬁned C∞function fonS1such that ¯f= h∗f. Thus, the image of h∗:Ω0(S1)→Ω0(R)consists precisely of the C∞periodic functions of period 2 πonR. As for 1-forms, note that Ω1(S1) =Ω0(S1)ωand Ω1(R) =Ω0(R)dt. The pullback h∗:Ω1(S1)→Ω1(R)is given by h∗(fω) = ( h∗f)dt, so the image of h∗:Ω1(S1)→Ω1(R)consists of C∞periodic 1-forms of period 2 π.⊓ ⊔ 18.8 Invariant Forms on a Lie Group Just as there are left-invariant vector ﬁelds on a Lie group G, so also are there left- invariant differential forms. For g∈G, letℓg:G→Gbe left multiplication by g. A k-form ωonGis said to be left-invariant ifℓ∗ gω=ωfor all g∈G. This means that for all g,x∈G, ℓ∗ g(ωgx)=ωx. Thus, a left-invariant k-form is uniquely determined by its value at the identity, since for any g∈G, ωg=ℓ∗ g−1(ωe). (18.6) Example 18.13 ( A left-invariant 1-form on S1).By Problem 17.3, ω=−ydx+xdy is a left-invariant 1-form on S1. We have the following analogue of Proposition 16.8. Proposition 18.14. Every left-invariant k-form ωon a Lie group G is C∞. 208§18 Differential k-Forms Proof. By Proposition 18.7(iii), it sufﬁces to prove that for any ksmooth vector ﬁelds X1,..., XkonG, the function ω(X1,..., Xk)isC∞onG. Let(Y1)e,...,(Yn)ebe a basis for the tangent space TeGandY1,..., Ynthe left-invariant vector ﬁelds they generate. Then Y1,..., Ynis aC∞frame on G(Proposition 16.8). Each Xjcan be written as a linear combination Xj=∑ai jYi. By Proposition 12.12, the functions ai jareC∞. Hence, to prove that ωisC∞, it sufﬁces to show that ω(Yi1,..., Yik)isC∞for the left-invariant vector ﬁelds Yi1,..., Yik. But (ω(Yi1,..., Yik))(g)=ωg((Yi1)g,...,(Yik)g) =(ℓ∗ g−1(ωe))/parenleftbig ℓg∗(Yi1)e,...,ℓ g∗(Yik)e/parenrightbig =ωe((Yi1)e,...,(Yik)e), which is a constant, independent of g. Being a constant function, ω(Yi1,..., Yik)is C∞onG. ⊓ ⊔ Similarly, a k-form ωonGis said to be right-invariant ifr∗ gω=ωfor all g∈G. The analogue of Proposition 18.14, that every right-invariant form on a Lie group is C∞, is proven in the same way. LetΩk(G)Gdenote the vector space of left-invariant k-forms on G. The linear map Ωk(G)G→/logicalandtextk(g∨),ω/ma√s⊔o→ωe, has an inverse deﬁned by (18.6) and is therefore an isomorphism. It follows that dimΩk(G)G=/parenleftbign k/parenrightbig . Problems 18.1. Characterization of a smooth k-form Write out a proof of Proposition 18.7(i) ⇔(iv). 18.2. Linearity of the pullback Prove Proposition 18.9. 18.3. Pullback of a wedge product Prove Proposition 18.11. 18.4.* Support of a sum or product Generalizing the support of a function, we deﬁne the support of a k-form ω∈Ωk(M)to be supp ω=closure of{p∈M|ωp/ne}a⊔ionslash=0}=Z(ω)c, where Z(ω)cis the complement of the zero set Z(ω)ofωinM. Let ωandτbe differential forms on a manifold M. Prove that (a) supp(ω+τ)⊂supp ω∪supp τ, (b) supp(ω∧τ)⊂supp ω∩supp τ. 18.8 Invariant Forms on a Lie Group 209 18.5. Support of a linear combination Prove that if the k-forms ω1,..., ωr∈Ωk(M)are linearly independent at every point of a manifold Manda1,..., arareC∞functions on M, then suppr ∑ i=1aiωi=r/uniondisplay i=1supp ai. 18.6.* Locally ﬁnite collection of supports Let{ρα}α∈Abe a collection of functions on MandωaC∞k-form with compact support on M. If the collection{supp ρα}α∈Aof supports is locally ﬁnite, prove that ραω≡0 for all but ﬁnitely many α. 18.7. Locally ﬁnite sums We say that a sum ∑ωαof differential k-forms on a manifold Mislocally ﬁnite if the collection {supp ωα}of supports is locally ﬁnite. Suppose ∑ωαand∑ταare locally ﬁnite sums and f is aC∞function on M. (a) Show that every point p∈Mhas a neighborhood Uon which ∑ωαis a ﬁnite sum. (b) Show that ∑ωα+ταis a locally ﬁnite sum and ∑ωα+τα=∑ωα+∑τα. (c) Show that ∑fωαis a locally ﬁnite sum and ∑f·ωα=f·/parenleftbig∑ωα/parenrightbig . 18.8.* Pullback by a surjective submersion In Subsection 19.5, we will show that the pullback of a C∞form is C∞. Assuming this fact for now, prove that if π:˜M→Mis a surjective submersion, then the pullback map π∗:Ω∗(M)→ Ω∗(˜M)is an injective algebra homomorphism. 18.9. Bi-invariant top forms on a compact, connected Lie group Suppose Gis a compact, connected Lie group of dimension nwith Lie algebra g. This exercise proves that every left-invariant n-form on Gis right-invariant. (a) Let ωbe a left-invariant n-form on G. For any a∈G, show that r∗aωis also left-invariant, where ra:G→Gis right multiplication by a. (b) Since dim Ωn(G)G=dim/logicalandtextn(g∨)=1,r∗aω=f(a)ωfor some nonzero real number f(a) depending on a∈G. Show that f:G→R×is a group homomorphism. (c) Show that f:G→R×isC∞. (Hint: Note that f(a)ωe=(r∗aω)e=r∗a(ωa)=r∗aℓ∗ a−1(ωe). Thus, f(a)is the pullback of the map Ad (a−1):g→g. See Problem 16.11.) (d) As the continuous image of a compact connected set G, the set f(G)⊂R×is compact and connected. Prove that f(G)=1. Hence, r∗aω=ωfor all a∈G. 210§19 The Exterior Derivative §19 The Exterior Derivative In contrast to undergraduate calculus, where the basic objects of study are functions, the basic objects in calculus on manifolds are differential forms. Our program now is to learn how to integrate and differentiate differential forms. Recall that an antiderivation on a graded algebra A=/circleplustext∞ k=0Akis anR-linear map D:A→Asuch that D(ω·τ)=( Dω)·τ+(−1)kω·Dτ forω∈Akandτ∈Aℓ. In the graded algebra A, an element of Akis called a homo- geneous element of degree k. The antiderivation is of degree m if degDω=degω+m for all homogeneous elements ω∈A. LetMbe a manifold and Ω∗(M)the graded algebra of C∞differential forms onM. On the graded algebra Ω∗(M)there is a uniquely and intrinsically deﬁned antiderivation called the exterior derivative. The process of applying the exterior derivative is called exterior differentiation . Deﬁnition 19.1. Anexterior derivative on a manifold Mis anR-linear map D:Ω∗(M)→Ω∗(M) such that (i)Dis an antiderivation of degree 1, (ii)D◦D=0, (iii) if fis aC∞function and XaC∞vector ﬁeld on M, then(D f)(X)=X f. Condition (iii) says that on 0-forms an exterior derivative agrees with the differ- ential dfof a function f. Hence, by (17.2), on a coordinate chart (U,x1,..., xn), D f=d f=∑∂f ∂xidxi. In this section we prove the existence and uniqueness of an exterior derivative on a manifold. Using its three deﬁning properties, we then show that the exterior derivative commutes with the pullback. This will ﬁnally allow us to prove that the pullback of a C∞form by a C∞map is C∞. 19.2 Local Operators 211 19.1 Exterior Derivative on a Coordinate Chart We showed in Subsection 4.4 the existence and uniqueness of an exterior derivative on an open subset of Rn. The same proof carries over to any coordinate chart on a manifold. More precisely, suppose (U,x1,..., xn)is a coordinate chart on a manifold M. Then any k-form ωonUis uniquely a linear combination ω=∑aIdxI,aI∈C∞(U). IfDis an exterior derivative on U, then Dω=∑(DaI)∧dxI+∑aIDdxI(by (i)) =∑(DaI)∧dxI(by (iii) and (ii), Dd=D2=0) =∑ I∑ j∂aI ∂xjdxj∧dxI(by (iii)). (19.1) Hence, if an exterior derivative Dexists on U, then it is uniquely deﬁned by (19.1). To show existence, we deﬁne Dby the formula (19.1). The proof that Dsatisﬁes (i), (ii), and (iii) is the same as in the case of Rnin Proposition 4.7. We will denote the unique exterior derivative on a chart (U,φ)bydU. Like the derivative of a function on Rn, an antiderivation DonΩ∗(M)has the property that for a k-form ω, the value of Dωat a point pdepends only on the values of ωin a neighborhood of p. To explain this, we make a digression on local operators. 19.2 Local Operators An endomorphism of a vector space Wis often called an operator onW. For exam- ple, if W=C∞(R)is the vector space of C∞functions on R, then the derivative d/dx is an operator on W: d dxf(x)=f′(x). The derivative has the property that the value of f′(x)at a point pdepends only on the values of fin a small neighborhood of p. More precisely, if f=gon an open set UinR, then f′=g′onU. We say that the derivative is a local operator on C∞(R). Deﬁnition 19.2. An operator D:Ω∗(M)→Ω∗(M)is said to be local if for all k≥0, whenever a k-form ω∈Ωk(M)restricts to 0 on an open set UinM, then Dω≡0 onU. Here by restricting to 0 on U, we mean that ωp=0 at every point pinU, and the symbol “≡0” means “is identically zero”: (Dω)p=0 at every point pinU. An equivalent criterion for an operator Dto be local is that for all k≥0, whenever two k-forms ω,τ∈Ωk(M)agree on an open set U, then Dω≡DτonU. 212§19 The Exterior Derivative Example. Deﬁne the integral operator I:C∞([a,b])→C∞([a,b]) by I(f)=/integraldisplayb af(t)dt. Here I(f)is a number, which we view as a constant function on [a,b]. The integral is not a local operator, since the value of I(f)at any point pdepends on the values offover the entire interval [a,b]. Proposition 19.3. Any antiderivation D on Ω∗(M)is a local operator. Proof. Suppose ω∈Ωk(M)andω≡0 on an open subset U. Let pbe an arbitrary point in U. It sufﬁces to prove that (Dω)p=0. Choose a C∞bump function fatpsupported in U. In particular, f≡1 in a neighborhood of pinU. Then fω≡0 on M, since if a point qis in U, then ωq=0, and if qis not in U, then f(q)=0. Applying the antiderivation property of Dtofω, we get 0=D(0)=D(fω)=( D f)∧ω+(−1)0f∧(Dω). Evaluating the right-hand side at p, noting that ωp=0 and f(p) =1, gives 0 = (Dω)p. ⊓ ⊔ Remark. The same proof shows that a derivation on Ω∗(M)is also a local operator. 19.3 Existence of an Exterior Derivative on a Manifold To deﬁne an exterior derivative on a manifold M, letωbe ak-form on Mandp∈M. Choose a chart (U,x1,..., xn)about p. Suppose ω=∑aIdxIonU. In Subsec- tion 19.1 we showed the existence of an exterior derivative dUonUwith the property dUω=∑daI∧dxIonU. (19.2) Deﬁne(dω)p=(dUω)p. We now show that (dUω)pis independent of the chart U containing p. If(V,y1,..., yn)is another chart about pandω=∑bJdyJonV, then onU∩V, ∑aIdxI=∑bJdyJ. OnU∩Vthere is a unique exterior derivative dU∩V:Ω∗(U∩V)→Ω∗(U∩V). By the properties of the exterior derivative, on U∩V dU∩V/parenleftbig∑aIdxI/parenrightbig =dU∩V/parenleftbig∑bJdyJ/parenrightbig , or ∑daI∧dxI=∑dbJ∧dyJ. 19.4 Uniqueness of the Exterior Derivative 213 In particular,/parenleftbig∑daI∧dxI/parenrightbig p=/parenleftbig∑dbJ∧dyJ/parenrightbig p. Thus,(dω)p=(dUω)pis well deﬁned, independently of the chart (U,x1,..., xn). Aspvaries over all points of M, this deﬁnes an operator d:Ω∗(M)→Ω∗(M). To check properties (i), (ii), and (iii), it sufﬁces to check them at each point p∈M. As in Subsection 19.1, the veriﬁcation reduces to the same calculation as for theexterior derivative on R nin Proposition 4.7. 19.4 Uniqueness of the Exterior Derivative Suppose D:Ω∗(M)→Ω∗(M)is an exterior derivative. We will show that Dcoin- cides with the exterior derivative ddeﬁned in Subsection 19.3. Iffis aC∞function and XaC∞vector ﬁeld on M, then by condition (iii) of Deﬁnition 19.1, (D f)(X)=X f=(df)(X). Therefore, D f=dfon functions f∈Ω0(M). Next consider a wedge product of exact 1-forms df1∧···∧ dfk: D(df1∧···∧ dfk) =D(D f1∧···∧ D fk) (because D fi=dfi) =k ∑ i=1(−1)i−1D f1∧···∧ DD fi∧···∧ D fk(Dis an antiderivation) =0 (D2=0). Finally, we show that Dagrees with don any k-form ω∈Ωk(M). Fix p∈ M. Choose a chart (U,x1,..., xn)about pand suppose ω=∑aIdxIonU. Extend the functions aI,x1,..., xnonUtoC∞functions ˜ aI,˜x1,..., ˜xnonMthat agree with aI,x1,..., xnon a neighborhood of Vofp(by Proposition 18.8). Deﬁne ˜ω=∑˜aId˜xI∈Ωk(M). Then ω≡˜ωonV. Since Dis a local operator, Dω=D˜ωonV. Thus, 214§19 The Exterior Derivative (Dω)p=(D˜ω)p=(D∑˜aId˜xI)p =/parenleftbig∑D˜aI∧d˜xI+∑˜aI∧Dd˜xI/parenrightbig p =/parenleftbig∑d˜aI∧d˜xI/parenrightbig p(because Dd˜xI=DD˜x=0) =/parenleftbig∑daI∧dxI/parenrightbig p(since Dis a local operator ) =(dω)p. We have proven the following theorem. Theorem 19.4. On any manifold M there exists an exterior derivative d :Ω∗(M)→ Ω∗(M)characterized uniquely by the three properties of Deﬁnition 19.1. 19.5 Exterior Differentiation Under a Pullback The pullback of differential forms commutes with the exterior derivative. This fact, together with Proposition 18.11 that the pullback preserves the wedge product, is a cornerstone of calculations involving the pullback. Using these two properties, we will ﬁnally be in a position to prove that the pullback of a C∞form under a C∞map isC∞. Proposition 19.5 (Commutation of the pullback with d).Let F :N→M be a smooth map of manifolds. If ω∈Ωk(M), then dF∗ω=F∗dω. Proof. The case k=0, when ωis aC∞function on M, is Proposition 17.10. Next consider the case k≥1. It sufﬁces to verify dF∗ω=F∗dωat an arbitrary point p∈ N. This reduces the proof to a local computation, i.e., computation in a coordinate chart. If(V,y1,..., ym)is a chart on Mabout F(p), then on V, ω=∑aIdyi1∧···∧ dyik,I=(i1<···<ik), for some C∞functions aIonVand F∗ω=∑(F∗aI)F∗dyi1∧···∧ F∗dyik(Proposition 18.11 ) =∑(aI◦F)dFi1∧···∧ dFik(F∗dyi=dF∗yi=d(yi◦F)=dFi). So dF∗ω=∑d(aI◦F)∧dFi1∧···∧ dFik. On the other hand, F∗dω=F∗/parenleftbig∑daI∧dyi1∧···∧ dyik/parenrightbig =∑F∗daI∧F∗dyi1∧···∧ F∗dyik =∑d(F∗aI)∧dFi1∧···∧ dFik(by the case k=0) =∑d(aI◦F)∧dFi1∧···∧ dFik. Therefore, dF∗ω=F∗dω. ⊓ ⊔ 19.5 Exterior Differentiation Under a Pullback 215 Corollary 19.6. If U is an open subset of a manifold M and ω∈Ωk(M), then (dω)|U=d(ω|U). Proof. Leti:U֒→Mbe the inclusion map. Then ω|U=i∗ω, so the corollary is simply a restatement of the commutativity of dwith i∗. ⊓ ⊔ Example. LetUbe the open set ]0,∞[×]0,2π[in the(r,θ)-planeR2. Deﬁne F:U⊂ R2→R2by F(r,θ)=( rcosθ,rsinθ). Ifx,yare the standard coordinates on the target R2, compute the pullback F∗(dx∧ dy). Solution. We ﬁrst compute F∗dx: F∗dx=dF∗x (Proposition 19.5 ) =d(x◦F) (deﬁnition of the pullback of a function ) =d(rcosθ) =(cosθ)dr−rsinθdθ. Similarly, F∗dy=dF∗y=d(rsinθ)=( sinθ)dr+rcosθdθ. Since the pullback commutes with the wedge product (Proposition 18.11), F∗(dx∧dy)=( F∗dx)∧(F∗dy) =((cosθ)dr−rsinθdθ)∧((sinθ)dr+rcosθdθ) =(rcos2θ+rsin2θ)dr∧dθ(because dθ∧dr=−dr∧dθ) =r dr∧dθ. ⊓ ⊔ Proposition 19.7. If F:N→M is a C∞map of manifolds and ωis a C∞k-form on M, then F∗ωis a C∞k-form on N. Proof. It is enough to show that every point in Nhas a neighborhood on which F∗ω isC∞. Fix p∈Nand choose a chart (V,y1,..., ym)onMabout F(p). Let Fi=yi◦F be the ith coordinate of the map Fin this chart. By the continuity of F, there is a chart(U,x1,..., xn)onNabout psuch that F(U)⊂V. Because ωisC∞, onV, ω=∑ IaIdyi1∧···∧ dyik for some C∞functions aI∈C∞(V)(Proposition 18.7(i) ⇒(ii)). By properties of the pullback, F∗ω=∑(F∗aI)F∗(dyi1)∧··· F∗(dyik)(Propositions 18.9 and 18.11) =∑(F∗aI)dF∗yi1∧···∧ dF∗yik (Proposition 19.5) =∑(aI◦F)dFi1∧···∧ dFik(F∗yi=yi◦F=Fi) =∑ I,J(aI◦F)∂(Fi1,..., Fik) ∂(xj1,..., xjk)dxJ(Proposition 18.3) . 216§19 The Exterior Derivative Since the aI◦Fand∂(Fi1,..., Fik)/∂(xj1,..., xjk)are all C∞,F∗ωisC∞by Propo- sition 18.7(iii)⇒(i). ⊓ ⊔ In summary, if F:N→Mis aC∞map of manifolds, then the pullback map F∗:Ω∗(M)→Ω∗(N)is a morphism of differential graded algebras, i.e., a degree- preserving algebra homomorphism that commutes with the differential. 19.6 Restriction of k-Forms to a Submanifold The restriction of a k-form to an immersed submanifold is just like the restriction of a 1-form, but with karguments. Let Sbe a regular submanifold of a manifold M. If ωis ak-form on M, then the restriction ofωtoSis the k-form ω|SonSdeﬁned by (ω|S)p(v1,..., vk)=ωp(v1,..., vk) forv1,..., vk∈TpS⊂TpM. Thus,(ω|S)pis obtained from ωpby restricting the domain of ωptoTpS×···× TpS(ktimes). As in Proposition 17.14, the restriction of k-forms is the same as the pullback under the inclusion map i:S֒→M. A nonzero form on Mmay restrict to the zero form on a submanifold S. For ex- ample, if Sis a smooth curve in R2deﬁned by the nonconstant function f(x,y), then df=(∂f/∂x)dx+(∂f/∂y)dyis a nonzero 1-form on R2, but since fis identically zero on S, the differential dfis also identically zero on S. Thus,(df)|S≡0. Another example is Problem 19.9. One should distinguish between a nonzero form and a nowhere-zero ornowhere- vanishing form. For example, xdyis a nonzero form on R2, meaning that it is not identically zero. However, it is not nowhere-zero, because it vanishes on the y-axis. On the other hand, dxanddyare nowhere-zero 1-forms on R2. NOTATION . Since pullback and exterior differentiation commute, (d f)|S=d(f|S), so one may write d f|Sto mean either expression. 19.7 A Nowhere-Vanishing 1-Form on the Circle In Example 17.15 we found a nowhere-vanishing 1-form −ydx+x dy on the unit circle. As an application of the exterior derivative, we will construct in a differentway a nowhere-vanishing 1-form on the circle. One advantage of the new method is that it generalizes to the construction of a nowhere-vanishing top form on a smooth hypersurface inR n+1, a regular level set of a smooth function f:Rn+1→R. As we will see in Section 21, the existence of a nowhere-vanishing top form is intimately related to orientations on a manifold. Example 19.8.LetS1be the unit circle deﬁned by x2+y2=1 inR2. The 1-form dx restricts from R2to a 1-form on S1. At each point p∈S1, the domain of (dx|S1)pis Tp(S1)instead of Tp(R2): (dx|S1)p:Tp(S1)→R. 19.7 A Nowhere-Vanishing 1-Form on the Circle 217 Atp=(1,0), a basis for the tangent space Tp(S1)is∂/∂y(Figure 19.1 ). Since (dx)p/parenleftbigg∂ ∂y/parenrightbigg =0, we see that although dxis a nowhere-vanishing 1-form on R2, it vanishes at (1,0) when restricted to S1. p/Bullet∂ ∂y Fig. 19.1. The tangent space to S1atp=(1,0). To ﬁnd a nowhere-vanishing 1-form on S1, we take the exterior derivative of both sides of the equation x2+y2=1. Using the antiderivation property of d, we get 2xdx+2ydy=0. (19.3) Of course, this equation is valid only at a point (x,y)∈S1. Let Ux={(x,y)∈S1|x/ne}a⊔ionslash=0}and Uy={(x,y)∈S1|y/ne}a⊔ionslash=0}. By (19.3), on Ux∩Uy, dy x=−dx y. Deﬁne a 1-form ωonS1by ω=  dy xonUx, −dx yonUy.(19.4) Since these two 1-forms agree on Ux∩Uy,ωis a well-deﬁned 1-form on S1=Ux∪Uy. To show that ωisC∞and nowhere-vanishing, we need charts. Let U+ x={(x,y)∈S1|x>0}. We deﬁne similarly U− x,U+ y,U− y(Figure 19.2 ). On U+ x,yis a local coordinate, and sodyis a basis for the cotangent space T∗ p(S1)at each point p∈U+ x. Since ω=dy/x onU+ x,ωisC∞and nowhere zero on U+ x. A similar argument applies to dy/xonU− x and−dx/yonU+ yandU− y. Hence, ωisC∞and nowhere vanishing on S1. 218§19 The Exterior Derivative xy U+x U−x Fig. 19.2. Two charts on the unit circle. Problems 19.1. Pullback of a differential form LetUbe the open set ]0,∞[×]0,π[×]0,2π[in the(ρ,φ,θ)-spaceR3. Deﬁne F:U→R3 by F(ρ,φ,θ)=( ρsinφcosθ,ρsinφsinθ,ρcosφ). Ifx,y,zare the standard coordinates on the target R3, show that F∗(dx∧dy∧dz)=ρ2sinφdρ∧dφ∧dθ. 19.2. Pullback of a differential form LetF:R2→R2be given by F(x,y)=(x2+y2,xy). Ifu,vare the standard coordinates on the target R2, compute F∗(udu+vdv). 19.3. Pullback of a differential form by a curve Letτbe the 1-form τ= (−ydx +xdy)/(x2+y2)onR2−{0}. Deﬁne γ:R→R2−{0} byγ(t) = (cos t,sint). Compute γ∗τ. (This problem is related to Example 17.16 in that if i:S1֒→R2−{0}is the inclusion, then γ=i◦candω=i∗τ.) 19.4. Pullback of a restriction LetF:N→Mbe aC∞map of manifolds, Uan open subset of M, and F|F−1(U):F−1(U)→U the restriction of FtoF−1(U). Prove that if ω∈Ωk(M), then /parenleftig F|F−1(U)/parenrightig∗ (ω|U)=(F∗ω)|F−1(U). 19.5. Coordinate functions and differential forms Letf1,..., fnbeC∞functions on a neighborhood Uof a point pin a manifold of dimension n. Show that there is a neighborhood Wofpon which f1,..., fnform a coordinate system if and only if (df1∧···∧ dfn)p/ne}a⊔ionslash=0. 19.6. Local operators An operator L:Ω∗(M)→Ω∗(M)issupport-decreasing if supp L(ω)⊂supp ωfor every k- form ω∈Ω∗(M)for all k≥0. Show that an operator on Ω∗(M)is local if and only if it is support-decreasing. 19.7 A Nowhere-Vanishing 1-Form on the Circle 219 19.7. Derivations of C∞functions are local operators LetMbe a smooth manifold. The deﬁnition of a local operator D onC∞(M)is similar to that of a local operator on Ω∗(M):Dislocal if whenever a function f∈C∞(M)vanishes identically on an open subset U, then D f≡0 on U. Prove that a derivation of C∞(M)is a local operator on C∞(M). 19.8. Nondegenerate 2-forms A 2-covector αon a 2 n-dimensional vector space Vis said to be nondegenerate ifαn:= α∧···∧ α(ntimes) is not the zero 2 n-covector. A 2-form ωon a 2 n-dimensional manifold Mis said to be nondegenerate if at every point p∈M, the 2-covector ωpis nondegenerate on the tangent space TpM. (a) Prove that on Cnwith real coordinates x1,y1,..., xn,yn, the 2-form ω=n ∑ j=1dxj∧dyj is nondegenerate. (b) Prove that if λis the Liouville form on the total space T∗Mof the cotangent bundle of an n-dimensional manifold M, then dλis a nondegenerate 2-form on T∗M. 19.9.* Vertical planes Letx,y,zbe the standard coordinates on R3. A plane in R3isvertical if it is deﬁned by ax+by=0 for some (a,b)/ne}a⊔ionslash=(0,0)∈R2. Prove that restricted to a vertical plane, dx∧dy=0. 19.10. Nowhere-vanishing form on S1 Prove that the nowhere-vanishing form ωonS1constructed in Example 19.8 is the form −ydx+xdy of Example 17.15. ( Hint: Consider UxandUyseparately. On Ux, substitute dx=−(y/x)dyinto−ydx+xdy.) 19.11. A C∞nowhere-vanishing form on a smooth hypersurface (a) Let f(x,y)be a C∞function on R2and assume that 0 is a regular value of f. By the regular level set theorem, the zero set Moff(x,y)is a one-dimensional submanifold of R2. Construct a C∞nowhere-vanishing 1-form on M. (b) Let f(x,y,z)be a C∞function on R3and assume that 0 is a regular value of f. By the regular level set theorem, the zero set Moff(x,y,z)is a two-dimensional submanifold of R3. Let fx,fy,fzbe the partial derivatives of fwith respect to x,y,z, respectively. Show that the equalities dx∧dy fz=dy∧dz fx=dz∧dx fy hold on Mwhenever they make sense, and therefore the three 2-forms piece together to give a C∞nowhere-vanishing 2-form on M. (c) Generalize this problem to a regular level set of f(x1,..., xn+1)inRn+1. 19.12. Vector ﬁelds as derivations of C∞functions In Subsection 14.1 we showed that a C∞vector ﬁeld Xon a manifold Mgives rise to a deriva- tion of C∞(M). We will now show that every derivation of C∞(M)arises from one and only one vector ﬁeld, as promised earlier. To distinguish the vector ﬁeld from the derivation, we will temporarily denote the derivation arising from Xbyϕ(X). Thus, for any f∈C∞(M), (ϕ(X)f)(p)=Xpffor all p∈M. 220§19 The Exterior Derivative (a) LetF=C∞(M). Prove that ϕ:X(M)→Der(C∞(M))is anF-linear map. (b) Show that ϕis injective. (c) If Dis a derivation of C∞(M)andp∈M, deﬁne Dp:C∞p(M)→C∞p(M)by Dp[f]=[D˜f]∈C∞ p(M), where[f]is the germ of fatpand ˜fis a global extension of f, such as those given by Proposition 18.8. Show that Dp[f]is well deﬁned. ( Hint: Apply Problem 19.7.) (d) Show that Dpis a derivation of C∞p(M). (e) Prove that ϕ:X(M)→Der(C∞(M))is an isomorphism of F-modules. 19.13. Twentieth-century formulation of Maxwell’s equations In Maxwell’s theory of electricity and magnetism, developed in the late nineteenth century,the electric ﬁeld E=/an}bracke⊔le{⊔E 1,E2,E3/an}bracke⊔ri}h⊔and the magnetic ﬁeld B=/an}bracke⊔le{⊔B1,B2,B3/an}bracke⊔ri}h⊔in a vacuum R3with no charge or current satisfy the following equations: ∇×E=−∂B ∂t,∇×B=∂E ∂t, divE=0, divB=0. By the correspondence in Subsection 4.6, the 1-form EonR3corresponding to the vector ﬁeld Eis E=E1dx+E2dy+E3dz and the 2-form BonR3corresponding to the vector ﬁeld Bis B=B1dy∧dz+B2dz∧dx+B3dx∧dy. LetR4be space-time with coordinates (x,y,z,t). Then both EandBcan be viewed as differential forms on R4. Deﬁne Fto be the 2-form F=E∧dt+B on space-time. Decide which two of Maxwell’s equations are equivalent to the equation dF=0. Prove your answer. (The other two are equivalent to d∗F=0 for a star-operator ∗deﬁned in differential geometry. See [2, Section 19.1, p. 689].) 20.1 Families of Vector Fields and Differential Forms 221 §20 The Lie Derivative and Interior Multiplication The only portion of this section necessary for the remainder of the book is Subsec- tion 20.4 on interior multiplication. The rest may be omitted on ﬁrst reading. The construction of exterior differentiation in Section 19 is local and depends on a choice of coordinates: if ω=∑aIdxI, then dω=∑∂aI ∂xjdxj∧dxI. It turns out, however, that this dis in fact global and intrinsic to the manifold, i.e., independent of the choice of local coordinates. Indeed, for a C∞1-form ωandC∞ vector ﬁelds X,Yon a manifold M, one has the formula (dω)(X,Y)=Xω(Y)−Yω(X)−ω([X,Y]). In this section we will derive a global intrinsic formula like this for the exterior derivative of a k-form. The proof uses the Lie derivative and interior multiplication, two other intrinsic operations on a manifold. The Lie derivative is a way of differentiating a vector ﬁeld or a differential form on a manifold along another vector ﬁeld. For any vector ﬁeld Xon a manifold, the interior multiplication ιXis an antiderivation of degree −1 on differential forms. Being intrinsic operators on a manifold, both the Lie derivative and interior multiplication are important in their own right in differential topology and geometry. 20.1 Families of Vector Fields and Differential Forms A collection{Xt}or{ωt}of vector ﬁelds or differential forms on a manifold is said to be a 1 -parameter family if the parameter truns over some subset of the real line. LetIbe an open interval in Rand let Mbe a manifold. Suppose {Xt}is a 1-parameter family of vector ﬁelds on Mdeﬁned for all t∈Iexcept at t0∈I. We say that the limit limt→t0Xtexists if every point p∈Mhas a coordinate neighborhood (U,x1,..., xn) on which Xt|p=∑ai(t,p)∂/∂xi|pand lim t→t0ai(t,p)exists for all i. In this case, we set lim t→t0Xt|p=n ∑ i=1lim t→t0ai(t,p)∂ ∂xi/vextendsingle/vextendsingle/vextendsingle/vextendsingle p. (20.1) In Problem 20.1 we ask the reader to show that this deﬁnition of the limit of Xtas t→t0is independent of the choice of the coordinate neighborhood (U,x1,..., xn). A 1-parameter family {Xt}t∈Iof smooth vector ﬁelds on Mis said to depend smoothly ontif every point in Mhas a coordinate neighborhood (U,x1,..., xn)on which (Xt)p=∑ai(t,p)∂ ∂xi/vextendsingle/vextendsingle/vextendsingle/vextendsingle p,(t,p)∈I×U, (20.2) 222§20 The Lie Derivative and Interior Multiplication for some C∞functions aionI×U. In this case we also say that {Xt}t∈Iis asmooth family of vector ﬁelds onM. For a smooth family of vector ﬁelds on M, one can deﬁne its derivative with respect to tatt=t0by /parenleftigg d dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=t0Xt/parenrightigg p=∑∂ai ∂t(t0,p)∂ ∂xi/vextendsingle/vextendsingle/vextendsingle/vextendsingle p(20.3) for(t0,p)∈I×U. It is easy to check that this deﬁnition is independent of the chart (U,x1,..., xn)containing p(Problem 20.3). Clearly, the derivative d/dt|t=t0Xtis a smooth vector ﬁeld on M. Similarly, a 1-parameter family {ωt}t∈Iof smooth k-forms on Mis said to depend smoothly ontif every point of Mhas a coordinate neighborhood (U,x1,..., xn)on which (ωt)p=∑bJ(t,p)dxJ|p,(t,p)∈I×U, for some C∞functions bJonI×U. We also call such a family {ωt}t∈Iasmooth family of k-forms onMand deﬁne its derivative with respect to tto be /parenleftigg d dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=t0ωt/parenrightigg p=∑∂bJ ∂t(t0,p)dxJ|p. As for vector ﬁelds, this deﬁnition is independent of the chart and deﬁnes a C∞k- form d/dt|t=t0ωtonM. NOTATION . We write d/dtfor the derivative of a smooth family of vector ﬁelds or differential forms, but ∂/∂tfor the partial derivative of a function of several vari- ables. Proposition 20.1 (Product rule for d/dt).If{ωt}and{τt}are smooth families of k-forms and ℓ-forms respectively on a manifold M, then d dt(ωt∧τt)=/parenleftbiggd dtωt/parenrightbigg ∧τt+ωt∧d dtτt. Proof. Written out in local coordinates, this reduces to the usual product rule in calculus. We leave the details as an exercise (Problem 20.4). ⊓ ⊔ Proposition 20.2 (Commutation of d/dt|t=t0with d).If{ωt}t∈Iis a smooth family of differential forms on a manifold M, then d dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=t0dωt=d/parenleftigg d dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=t0ωt/parenrightigg . Proof. In this proposition, there are three operations—exterior differentiation, dif- ferentiation with respect to t, and evaluation at t=t0. We will ﬁrst show that dand d/dtcommute: 20.2 The Lie Derivative of a Vector Field 223 d dt(dωt)=d/parenleftbiggd dtωt/parenrightbigg . (20.4) It is enough to check the equality at an arbitrary point p∈M. Let(U,x1,..., xn)be a neighborhood of psuch that ω=∑JbJdxJfor some C∞functions bJonI×U. OnU, d dt(dωt)=d dt∑ J,i∂bJ ∂xidxi∧dxJ(note that there is no dtterm) =∑ i,J∂ ∂xi/parenleftbigg∂bJ ∂t/parenrightbigg dxi∧dxJ(since bJisC∞) =d/parenleftigg ∑ J∂bJ ∂tdxJ/parenrightigg =d/parenleftbiggd dtωt/parenrightbigg . Evaluation at t=t0commutes with d, because dinvolves only partial derivatives with respect to the xivariables. Explicitly, /parenleftbigg d/parenleftbiggd dtωt/parenrightbigg/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=t0=/parenleftigg ∑ i,J∂ ∂xi∂ ∂tbJdxi∧dxJ/parenrightigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=t0 =∑ i,J∂ ∂xi/parenleftigg ∂ ∂t/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=t0bJ/parenrightigg dxi∧dxJ=d/parenleftigg ∂ ∂t/vextendsingle/vextendsingle/vextendsingle/vextendsingle t0ωt/parenrightigg . Evaluating both sides of (20.4) at t=t0completes the proof of the proposition. ⊓ ⊔ 20.2 The Lie Derivative of a Vector Field In a ﬁrst course on calculus, one deﬁnes the derivative of a real-valued function fon Rat a point p∈Ras f′(p)=lim t→0f(p+t)−f(p) t. The problem in generalizing this deﬁnition to the derivative of a vector ﬁeld Yon a manifold Mis that at two nearby points pandqinM, the tangent vectors YpandYq are in different vector spaces TpMandTqMand so it is not possible to compare them by subtracting one from the other. One way to get around this difﬁculty is to use the local ﬂow of another vector ﬁeld Xto transport Yqto the tangent space TpMatp. This leads to the deﬁnition of the Lie derivative of a vector ﬁeld. Recall from Subsection 14.3 that for any smooth vector ﬁeld XonMand point pinM, there is a neighborhood Uofpon which the vector ﬁeld Xhas a local ﬂow ; this means that there exist a real number ε>0 and a map ϕ:]−ε,ε[×U→M such that if we set ϕt(q)=ϕ(t,q), then 224§20 The Lie Derivative and Interior Multiplication ∂ ∂tϕt(q)=Xϕt(q),ϕ0(q)=qforq∈U. (20.5) In other words, for each qinU, the curve ϕt(q)is an integral curve of Xwith initial point q. By deﬁnition, ϕ0:U→Uis the identity map. The local ﬂow satisﬁes the property ϕs◦ϕt=ϕs+t whenever both sides are deﬁned (see (14.10)). Consequently, for each tthe map ϕt:U→ϕt(U)is a diffeomorphism onto its image, with a C∞inverse ϕ−t: ϕ−t◦ϕt=ϕ0=1,ϕt◦ϕ−t=ϕ0=1. LetYbe a C∞vector ﬁeld on M. To compare the values of Yatϕt(p)and at p, we use the diffeomorphism ϕ−t:ϕt(U)→Uto push Yϕt(p)intoTpM(Figure 20.1 ). /Bullet/BulletXpYϕt(p)ϕ−t∗(Yϕt(p)) Yp pϕt(p) Fig. 20.1. Comparing the values of Yat nearby points. Deﬁnition 20.3. ForX,Y∈X(M)andp∈M, letϕ:]−ε,ε[×U→Mbe a local ﬂow of Xon a neighborhood Uofpand deﬁne the Lie derivative LXYofYwith respect to Xatpto be the vector (LXY)p=lim t→0ϕ−t∗/parenleftbig Yϕt(p)/parenrightbig −Yp t=lim t→0(ϕ−t∗Y)p−Yp t=d dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0(ϕ−t∗Y)p. In this deﬁnition the limit is taken in the ﬁnite-dimensional vector space TpM. For the derivative to exist, it sufﬁces that {ϕ−t∗Y}be a smooth family of vector ﬁelds on M. To show the smoothness of the family {ϕ−t∗Y}, we write out ϕ−t∗Yin local coordinates x1,..., xnin a chart. Let ϕi tandϕibe the ith components of ϕtand ϕrespectively. Then (ϕt)i(p)=ϕi(t,p)=( xi◦ϕ)(t,p). By Proposition 8.11, relative to the frame {∂/∂xj}, the differential ϕt∗atpis repre- sented by the Jacobian matrix [∂(ϕt)i/∂xj(p)]=[ ∂ϕi/∂xj(t,p)]. This means that 20.2 The Lie Derivative of a Vector Field 225 ϕt∗/parenleftigg ∂ ∂xj/vextendsingle/vextendsingle/vextendsingle/vextendsingle p/parenrightigg =∑ i∂ϕi ∂xj(t,p)∂ ∂xi/vextendsingle/vextendsingle/vextendsingle/vextendsingle ϕt(p). Thus, if Y=∑bj∂/∂xj, then ϕ−t∗/parenleftbig Yϕt(p)/parenrightbig =∑ jbj(ϕ(t,p))ϕ−t∗/parenleftigg ∂ ∂xj/vextendsingle/vextendsingle/vextendsingle/vextendsingle ϕt(p)/parenrightigg =∑ i,jbj(ϕ(t,p))∂ϕi ∂xj(−t,p)∂ ∂xi/vextendsingle/vextendsingle/vextendsingle/vextendsingle p. (20.6) When XandYareC∞vector ﬁelds on M, both ϕiandbjareC∞functions. The formula (20.6) then shows that {ϕ−t∗Y}is a smooth family of vector ﬁelds on M. It follows that the Lie derivative LXYexists and is given in local coordinates by (LXY)p=d dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0ϕ−t∗/parenleftbig Yϕt(p)/parenrightbig =∑ i,j∂ ∂t/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0/parenleftbigg bj(ϕ(t,p))∂ϕi ∂xj(−t,p)/parenrightbigg∂ ∂xi/vextendsingle/vextendsingle/vextendsingle/vextendsingle p. (20.7) It turns out that the Lie derivative of a vector ﬁeld gives nothing new. Theorem 20.4. If X and Y are C∞vector ﬁelds on a manifold M, then the Lie deriva- tiveLXY coincides with the Lie bracket [X,Y]. Proof. It sufﬁces to check the equality LXY= [X,Y]at every point. To do this, we expand both sides in local coordinates. Suppose a local ﬂow for Xisϕ: ]−ε,ε[×U→M, where Uis a coordinate chart with coordinates x1,..., xn. Let X=∑ai∂/∂xiandY=∑bj∂/∂xjonU. The condition (20.5) that ϕt(p)be an integral curve of Xtranslates into the equations ∂ϕi ∂t(t,p)=ai(ϕ(t,p)), i=1,..., n,(t,p)∈]−ε,ε[×U. Att=0,∂ϕi/∂t(0,p)=ai(ϕ(0,p))= ai(p). By Problem 14.12, the Lie bracket in local coordinates is [X,Y]=∑ i,k/parenleftbigg ak∂bi ∂xk−bk∂ai ∂xk/parenrightbigg∂ ∂xi. Expanding (20.7) by the product rule and the chain rule, we get (LXY)p=/bracketleftigg ∑ i,j,k/parenleftbigg∂bj ∂xk(ϕ(t,p))∂ϕk ∂t(t,p)∂ϕi ∂xj(−t,p)/parenrightbigg∂ ∂xi −∑ i,j/parenleftbigg bj(ϕ(t,p))∂ ∂xj∂ϕi ∂t(−t,p)/parenrightbigg∂ ∂xi/bracketrightigg t=0 =∑ i,j,k/parenleftbigg∂bj ∂xk(p)ak(p)∂ϕi ∂xj(0,p)/parenrightbigg∂ ∂xi−∑ i,j/parenleftbigg bj(p)∂ai ∂xj(p)/parenrightbigg∂ ∂xi.(20.8) 226§20 The Lie Derivative and Interior Multiplication Since ϕ(0,p)=p,ϕ0is the identity map and hence its Jacobian matrix is the identity matrix. Thus, ∂ϕi ∂xj(0,p)=δi j,the Kronecker delta . So (20.8) simpliﬁes to LXY=∑ i,k/parenleftbigg ak∂bi ∂xk−bk∂ai ∂xk/parenrightbigg∂ ∂xi=[X,Y]. ⊓ ⊔ Although the Lie derivative of a vector ﬁeld gives us nothing new, in conjunction with the Lie derivative of differential forms it turns out to be a tool of great utility, for example, in the proof of the global formula for the exterior derivative in Theorem20.14. 20.3 The Lie Derivative of a Differential Form LetXbe a smooth vector ﬁeld and ωa smooth k-form on a manifold M. Fix a point p∈Mand let ϕt:U→Mbe a ﬂow of Xin a neighborhood Uofp. The deﬁnition of the Lie derivative of a differential form is similar to that of the Lie derivative of a vector ﬁeld. However, instead of pushing a vector at ϕt(p)topvia(ϕ−t)∗, we now pull the k-covector ωϕt(p)back to pviaϕ∗ t. Deﬁnition 20.5. ForXa smooth vector ﬁeld and ωa smooth k-form on a manifold M, the Lie derivative LXωatp∈Mis (LXω)p=lim t→0ϕ∗ t/parenleftbig ωϕt(p)/parenrightbig −ωp t=lim t→0(ϕ∗ tω)p−ωp t=d dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0(ϕ∗ tω)p. By an argument similar to that for the existence of the Lie derivative LXYin Section 20.2, one shows that {ϕ∗ tω}is a smooth family of k-forms on Mby writing it out in local coordinates. The existence of (LXω)pfollows. Proposition 20.6. If f is a C∞function and X a C∞vector ﬁeld on M, then LXf=X f . Proof. Fix a point pinMand let ϕt:U→Mbe a local ﬂow of Xas above. Then (LXf)p=d dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0(ϕ∗ tf)p (deﬁnition of LXf) =d dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0(f◦ϕt)(p)(deﬁnition of ϕ∗ tf) =Xpf (Proposition 8.17) , since ϕt(p)is a curve through pwith initial vector Xp. ⊓ ⊔ 20.4 Interior Multiplication 227 20.4 Interior Multiplication We ﬁrst deﬁne interior multiplication on a vector space. If βis ak-covector on a vector space Vandv∈V, for k≥2 the interior multiplication orcontraction ofβ with vis the(k−1)-covector ιvβdeﬁned by (ιvβ)(v2,..., vk)=β(v,v2,..., vk),v2,..., vk∈V. We deﬁne ιvβ=β(v)∈Rfor a 1-covector βonVandιvβ=0 for a 0-covector β(a constant) on V. Proposition 20.7. For1-covectors α1,..., αkon a vector space V and v ∈V, ιv(α1∧···∧ αk)=k ∑ i=1(−1)i−1αi(v)α1∧···∧/hatwideαi∧···∧ αk, where the caret /hatwideover αimeans that αiis omitted from the wedge product. Proof. /parenleftig ιv/parenleftig α1∧···∧ αk/parenrightig/parenrightig (v2,..., vk) =/parenleftig α1∧···∧ αk/parenrightig (v,v2,..., vk) =det α1(v)α1(v2)···α1(vk) α2(v)α2(v2)···α2(vk) ............ αk(v)αk(v2)···αk(vk) (Proposition 3.27 ) =k ∑ i=1(−1)i+1αi(v)det[αℓ(vj)]1≤ℓ≤k,ℓ/ne}a⊔ionslash=i 2≤j≤k(expansion along ﬁrst column ) =k ∑ i=1(−1)i+1αi(v)/parenleftig α1∧···∧/hatwideαi∧···∧ αk/parenrightig (v2,..., vk)(Proposition 3.27) . ⊓ ⊔ Proposition 20.8. For v in a vector space V , let ιv:/logicalandtext∗(V∨)→/logicalandtext∗−1(V∨)be interior multiplication by v. Then (i)ιv◦ιv=0, (ii)forβ∈/logicalandtextk(V∨)andγ∈/logicalandtextℓ(V∨), ιv(β∧γ)=( ιvβ)∧γ+(−1)kβ∧ιvγ. In other words, ιvis an antiderivation of degree −1whose square is zero. Proof. (i) Let β∈/logicalandtextk(V∨). By the deﬁnition of interior multiplication, (ιv(ιvβ))(v3,..., vk)=( ιvβ)(v,v3,..., vk)=β(v,v,v3,..., vk)=0, because βis alternating and there is a repeated variable vamong its arguments. 228§20 The Lie Derivative and Interior Multiplication (ii) Since both sides of the equation are linear in βand in γ, we may assume that β=α1∧···∧ αk,γ=αk+1∧···∧ αk+ℓ, where the αiare all 1-covectors. Then ιv(β∧γ) =ιv(α1∧···∧ αk+ℓ) =/parenleftigg k ∑ i=1(−1)i−1αi(v)α1∧···∧/hatwideαi∧···∧ αk/parenrightigg ∧αk+1∧···∧ αk+ℓ +(−1)kα1∧···∧ αk∧k ∑ i=1(−1)i+1αk+i(v)αk+1∧···∧/hatwidestαk+i∧···∧ αk+ℓ (by Proposition 20.7) = (ιvβ)∧γ+(−1)kβ∧ιvγ. ⊓ ⊔ Interior multiplication on a manifold is deﬁned pointwise. If Xis a smooth vector ﬁeld on Mandω∈Ωk(M), then ιXωis the(k−1)-form deﬁned by (ιXω)p=ιXpωp for all p∈M. The form ιXωonMis smooth because for any smooth vector ﬁelds X2,..., XkonM, (ιXω)(X2,..., Xk)=ω(X,X2,..., Xk) is a smooth function on M(Proposition 18.7(iii) ⇒(i)). Of course, ιXω=ω(X) for a 1-form ωandιXf=0 for a function fonM. By the properties of interior multiplication at each point p∈M(Proposition 20.8), the map ιX:Ω∗(M)→Ω∗(M) is an antiderivation of degree −1 such that ιX◦ιX=0. LetFbe the ring C∞(M)ofC∞functions on the manifold M. Because ιXωis a point operator—that is, its value at pdepends only on Xpandωp—it isF-linear in either argument. This means that ιXωis additive in each argument and moreover, for anyf∈F, (i)ιf Xω=fιXω; (ii)ιX(fω)=fιXω. Explicitly, the proof of (i) goes as follows. For any p∈M, (ιf Xω)p=ιf(p)Xpωp=f(p)ιXpωp=(fιXω)p. Hence, ιf Xω=fιXω. The proof of (ii) is similar. Additivity is more or less obvious. Example 20.9 ( Interior multiplication on R2).LetX=x∂/∂x+y∂/∂ybe the ra- dial vector ﬁeld and α=dx∧dythe area 2-form on the plane R2. Compute the contraction ιXα. Solution. We ﬁrst compute ιXdxandιXdy: 20.5 Properties of the Lie Derivative 229 ιXdx=dx(X)=dx/parenleftbigg x∂ ∂x+y∂ ∂y/parenrightbigg =x, ιXdy=dy(X)=dy/parenleftbigg x∂ ∂x+y∂ ∂y/parenrightbigg =y. By the antiderivation property of ιX, ιXα=ιX(dx∧dy)=( ιXdx)dy−dx(ιXdy)=xdy−ydx, which restricts to the nowhere-vanishing 1-form ωon the circle S1in Example 17.15. 20.5 Properties of the Lie Derivative In this section we state and prove several basic properties of the Lie derivative. We also relate the Lie derivative to two other intrinsic operators on differential forms on a manifold: the exterior derivative and interior multiplication. The interplay of these three operators results in some surprising formulas. Theorem 20.10. Assume X to be a C∞vector ﬁeld on a manifold M. (i)The Lie derivative LX:Ω∗(M)→Ω∗(M)is a derivation: it is an R-linear map and if ω∈Ωk(M)andτ∈Ωℓ(M), then LX(ω∧τ)=(LXω)∧τ+ω∧(LXτ). (ii)The Lie derivative LXcommutes with the exterior derivative d. (iii) (Cartan homotopy formula) LX=dιX+ιXd. (iv) (“Product” formula) Forω∈Ωk(M)and Y 1,..., Yk∈X(M), LX(ω(Y1,..., Yk))=(LXω)(Y1,..., Yk)+k ∑ i=1ω(Y1,...,LXYi,..., Yk). Proof. In the proof let p∈Mand let ϕt:U→Mbe a local ﬂow of the vector ﬁeld Xin a neighborhood Uofp. (i) Since the Lie derivative LXisd/dtof a vector-valued function of t, the derivation property of LXis really just the product rule for d/dt(Proposition 20.1). More precisely, (LX(ω∧τ))p=d dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0(ϕ∗ t(ω∧τ))p =d dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0(ϕ∗ tω)p∧(ϕ∗ tτ)p =/parenleftbiggd dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0(ϕ∗ tω)p/parenrightbigg ∧τp+ωp∧d dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0(ϕ∗ tτ)p (product rule for d/dt) =(LXω)p∧τp+ωp∧(LXτ)p. 230§20 The Lie Derivative and Interior Multiplication (ii) LXdω=d dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0ϕ∗ tdω (deﬁnition of LX) =d dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0dϕ∗ tω (dcommutes with pullback) =d/parenleftbiggd dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0ϕ∗ tω/parenrightbigg (by Proposition 20.2) =dLXω. (iii) We make two observations that reduce the problem to a simple case. First, for anyω∈Ωk(M), to prove the equality LXω=(dιX+ιXd)ωit sufﬁces to check it at any point p, which is a local problem. In a coordinate neighborhood (U,x1,..., xn) about p, we may assume by linearity that ωis a wedge product ω=f dxi1∧···∧ dxik. Second, on the left-hand side of the Cartan homotopy formula, by (i) and (ii), LXis a derivation that commutes with d. On the right-hand side, since dandιXare antiderivations, dιX+ιXdis a derivation by Problem 4.7. It clearly commutes with d. Thus, both sides of the Cartan homotopy formula are derivations that commute with d. Consequently, if the formula holds for two differential forms ωandτ, then it holds for the wedge product ω∧τas well as for dω. These observations reduce the veriﬁcation of (iii) to checking LXf=(dιX+ιXd)fforf∈C∞(U). This is quite easy: (dιX+ιXd)f=ιXdf (because ιXf=0) =(df)(X) (deﬁnition of ιX) =X f=LXf(Proposition 20.6) . (iv) We call this the “product” formula, even though there is no product in ω(Y1, ...,Yk), because this formula can be best remembered as though the juxtaposition of symbols were a product. In fact, even its proof resembles that of the product formula in calculus. To illustrate this, consider the case k=2. Let ω∈Ω2(M)and X,Y,Z∈X(M). The proof looks forbidding, but the idea is quite simple. To compare the values of ω(Y,Z)at the two points ϕt(p)andp, we subtract the value at pfrom the value at ϕt(p). The trick is to add and subtract terms so that each time only one of the three variables ω,Y, and Zmoves from one point to the other. By the deﬁnitions of the Lie derivative and the pullback of a function, (LX(ω(Y,Z)))p=lim t→0(ϕ∗ t(ω(Y,Z)))p−(ω(Y,Z))p t =lim t→0ωϕt(p)/parenleftbig Yϕt(p),Zϕt(p)/parenrightbig −ωp(Yp,Zp) t 20.5 Properties of the Lie Derivative 231 =lim t→0ωϕt(p)/parenleftbig Yϕt(p),Zϕt(p)/parenrightbig −ωp/parenleftbig ϕ−t∗/parenleftbig Yϕt(p)/parenrightbig ,ϕ−t∗/parenleftbig Zϕt(p)/parenrightbig/parenrightbig t (20.9) +lim t→0ωp/parenleftbig ϕ−t∗/parenleftbig Yϕt(p)/parenrightbig ,ϕ−t∗/parenleftbig Zϕt(p)/parenrightbig/parenrightbig −ωp/parenleftbig Yp,ϕ−t∗/parenleftbig Zϕt(p)/parenrightbig/parenrightbig t (20.10) +lim t→0ωp/parenleftbig Yp,ϕ−t∗/parenleftbig Zϕt(p)/parenrightbig/parenrightbig −ωp(Yp,Zp) t. (20.11) In this sum the quotient in the ﬁrst limit (20.9) is /parenleftbig ϕ∗ tωϕt(p)/parenrightbig/parenleftbig ϕ−t∗/parenleftbig Yϕt(p)/parenrightbig ,ϕ−t∗/parenleftbig Zϕt(p)/parenrightbig/parenrightbig −ωp/parenleftbig ϕ−t∗/parenleftbig Yϕt(p)/parenrightbig ,ϕ−t∗/parenleftbig Zϕt(p)/parenrightbig/parenrightbig t =ϕ∗ t(ωϕt(p))−ωp t/parenleftbig ϕ−t∗/parenleftbig Yϕt(p)/parenrightbig ,ϕ−t∗/parenleftbig Zϕt(p)/parenrightbig/parenrightbig . On the right-hand side of this equality, the difference quotient has a limit at t=0, namely the Lie derivative (LXω)p, and by (20.6) the two arguments of the difference quotient are C∞functions of t. Therefore, the right-hand side is a continuous function oftand its limit as tgoes to 0 is (LXω)p(Yp,Zp)(by Problem 20.2). By the bilinearity of ωp, the second term (20.10) is lim t→0ωp/parenleftigg ϕ−t∗/parenleftbig Yϕt(p)/parenrightbig −Yp t,ϕ−t∗/parenleftbig Zϕt(p)/parenrightbig/parenrightigg =ωp((LXY)p,Zp). Similarly, the third term (20.11) is ωp(Yp,(LXZ)p). Thus LX(ω(Y,Z))=(LXω)(Y,Z)+ω(LXY,Z)+ω(Y,LXZ). The general case is similar. ⊓ ⊔ Remark. Unlike interior multiplication, the Lie derivative LXωis notF-linear in either argument. By the derivation property of the Lie derivative (Theorem 20.10(i)), LX(fω)=(LXf)ω+fLXω=(X f)ω+fLXω. We leave the problem of expanding Lf Xωas an exercise (Problem 20.7). Theorem 20.10 can be used to calculate the Lie derivative of a differential form. Example 20.11 ( The Lie derivative on a circle) .Letωbe the 1-form−ydx+xdy and let Xbe the tangent vector ﬁeld −y∂/∂x+x∂/∂yon the unit circle S1from Example 17.15. Compute the Lie derivative LXω. 232§20 The Lie Derivative and Interior Multiplication Solution. By Proposition 20.6, LX(x)=Xx=/parenleftbigg −y∂ ∂x+x∂ ∂y/parenrightbigg x=−y, LX(y)=Xy=/parenleftbigg −y∂ ∂x+x∂ ∂y/parenrightbigg y=x. Next we compute LX(−ydx): LX(−ydx)=−(LXy)dx−yLXdx (LXis a derivation) =−(LXy)dx−ydLXx (LXcommutes with d) =−xdx+ydy. Similarly, LX(xdy)=−ydy+xdx. Hence,LXω=LX(−ydx+xdy)=0. 20.6 Global Formulas for the Lie and Exterior Derivatives The deﬁnition of the Lie derivative LXωis local, since it makes sense only in a neighborhood of a point. The product formula in Theorem 20.10(iv), however, givesa global formula for the Lie derivative. Theorem 20.12 (Global formula for the Lie derivative). For a smooth k-form ω and smooth vector ﬁelds X ,Y1,..., Ykon a manifold M, (LXω)(Y1,..., Yk)=X(ω(Y1,..., Yk))−k ∑ i=1ω(Y1,...,[X,Yi],..., Yk). Proof. In Theorem 20.10(iv), LX(ω(Y1,..., Yk))= X(ω(Y1,..., Yk))by Proposition 20.6 andLXYi=[X,Yi]by Theorem 20.4. ⊓ ⊔ The deﬁnition of the exterior derivative dis also local. Using the Lie derivative, we obtain a very useful global formula for the exterior derivative. We ﬁrst derive the formula for the exterior derivative of a 1-form, the case most useful in differentialgeometry. Proposition 20.13. If ωis a C∞1-form and X and Y are C∞vector ﬁelds on a mani- fold M, then dω(X,Y)=Xω(Y)−Yω(X)−ω([X,Y]). Proof. It is enough to check the formula in a chart (U,x1,..., xn), so we may assume ω=∑aidxi. Since both sides of the equation are R-linear in ω, we may further assume that ω=f dg, where f,g∈C∞(U). In this case, dω=d(f dg)=df∧dgand dω(X,Y)=df(X)dg(Y)−df(Y)dg(X)=( X f)Y g−(Y f)Xg, Xω(Y)=X(f dg(Y))= X(f Yg)=( X f)Y g+f XYg, Yω(X)=Y(f dg(X))= Y(f Xg)=( Y f)Xg+fYXg, ω([X,Y])= f dg([X,Y])= f(XY−YX)g. 20.6 Global Formulas for the Lie and Exterior Derivatives 233 It follows that Xω(Y)−Yω(X)−ω([X,Y])=( X f)Y g−(Y f)Xg=dω(X,Y).⊓ ⊔ Theorem 20.14 (Global formula for the exterior derivative). Assume k≥1. For a smooth k-form ωand smooth vector ﬁelds Y 0,Y1,..., Ykon a manifold M, (dω)(Y0,..., Yk)=k ∑ i=0(−1)iYiω(Y0,...,/hatwideYi,..., Yk) +∑ 0≤i<j≤k(−1)i+jω([Yi,Yj],Y0,...,/hatwideYi,...,/hatwideYj,..., Yk). Proof. When k=1, the formula is proven in Proposition 20.13. Assuming the formula for forms of degree k−1, we can prove it by induction for a form ωof degree k. By the deﬁnition of ιY0and Cartan’s homotopy formula (Theorem 20.10(iii)), (dω)(Y0,Y1,..., Yk)=( ιY0dω)(Y1,..., Yk) =(LY0ω)(Y1,..., Yk)−(dιY0ω)(Y1,..., Yk). The ﬁrst term of this expression can be computed using the global formula for the Lie derivative LY0ω, while the second term can be computed using the global formula fordof a form of degree k−1. This kind of veriﬁcation is best done by readers on their own. We leave it as an exercise (Problem 20.6). ⊓ ⊔ Problems 20.1. The limit of a family of vector ﬁelds LetIbe an open interval, Ma manifold, and{Xt}a 1-parameter family of vector ﬁelds on M deﬁned for all t/ne}a⊔ionslash=t0∈I. Show that the deﬁnition of lim t→t0Xtin (20.1), if the limit exists, is independent of coordinate charts. 20.2. Limits of families of vector ﬁelds and differential forms LetIbe an open interval containing 0. Suppose {ωt}t∈Iand{Yt}t∈Iare 1-parameter families of 1-forms and vector ﬁelds respectively on a manifold M. Prove that if lim t→0ωt=ω0and limt→0Yt=Y0, then lim t→0ωt(Yt) =ω0(Y0). (Hint: Expand in local coordinates.) By the same kind of argument, one can show that there is a similar formula for a family {ωt}of 2-forms: lim t→0ωt(Yt,Zt)=ω0(Y0,Z0). 20.3.* Derivative of a smooth family of vector ﬁelds Show that the deﬁnition (20.3) of the derivative of a smooth family of vector ﬁelds on Mis independent of the chart (U,x1,..., xn)containing p. 20.4. Product rule for d/dt Prove that if{ωt}and{τt}are smooth families of k-forms and ℓ-forms respectively on a manifold M, then d dt(ωt∧τt)=/parenleftbiggd dtωt/parenrightbigg ∧τt+ωt∧d dtτt. 234§20 The Lie Derivative and Interior Multiplication 20.5. Smooth families of forms and vector ﬁelds If{ωt}t∈Iis a smooth family of 2-forms and {Yt}t∈Iand{Zt}t∈Iare smooth families of vector ﬁelds on a manifold M, prove that ωt(Xt,Yt)is aC∞function on I×M. 20.6.* Global formula for the exterior derivative Complete the proof of Theorem 20.14. 20.7.F-Linearity and the Lie Derivative Letωbe a differential form, Xa vector ﬁeld, and fa smooth function on a manifold. The Lie derivative LXωis notF-linear in either variable, but prove that it satisﬁes the following identity: Lf Xω=fLXω+df∧ιXω. (Hint: Start with Cartan’s homotopy formula LX=dιX+ιXd.) 20.8. Bracket of the Lie Derivative and Interior Multiplication IfXandYare smooth vector ﬁelds on a manifold M, prove that on differential forms on M LXιY−ιYLX=ι[X,Y]. (Hint: Let ω∈Ωk(M)andY,Y1,..., Yk−1∈X(M). Apply the global formula for LXto (ιYLXω)(Y1,..., Yk−1)=(L Xω)(Y,Y1,..., Yk−1).) 20.9. Interior multiplication on Rn Letω=dx1∧···∧ dxnbe the volume form and X=∑xi∂/∂xithe radial vector ﬁeld on Rn. Compute the contraction ιXω. 20.10. The Lie derivative on the 2-sphere Letω=xdy∧dz−ydx∧dy+zdx∧dyandX=−y∂/∂x+x∂/∂yon the unit 2-sphere S2in R3. Compute the Lie derivative LXω. Chapter 6 Integration On a manifold one integrates not functions as in calculus on Rnbut differential forms. There are actually two theories of integration on manifolds, one in which the inte- gration is over a submanifold and the other in which the integration is over what iscalled a singular chain . Singular chains allow one to integrate over an object such as a closed rectangle in R 2: [a,b]×[c,d]:={(x,y)∈R2|a≤x≤b,c≤y≤d}, which is not a submanifold of R2because of its corners. For simplicity we will discuss only integration of smooth forms over a submani- fold. For integration of noncontinuous forms over more general sets, the reader may consult the many excellent references in the bibliography, for example [3, Section VI.2], [7, Section 8.2], or [25, Chapter 14]. For integration over a manifold to be well deﬁned, the manifold needs to be ori- ented. We begin the chapter with a discussion of orientations on a manifold. We then enlarge the category of manifolds to include manifolds with boundary. Our treatment of integration culminates in Stokes’s theorem for an n-dimensional man- ifold. Stokes’s theorem for a surface with boundary in R3was ﬁrst published as a question in the Smith’s Prize Exam that Stokes set at the University of Cambridge in 1854. It is not known whether any student solved the problem. According to [21, p. 150], the same theorem had appeared four years earlier in a letter of LordKelvin to Stokes, which only goes to conﬁrm that the attribution of credit in mathe- matics is fraught with pitfalls. Stokes’s theorem for a general manifold resulted from the work of many mathematicians, including Vito V olterra (1889), Henri Poincar´ e (1899), Edouard Goursat (1917), and ´Elie Cartan (1899 and 1922). First there were many special cases, then a general statement in terms of coordinates, and ﬁnally ageneral statement in terms of differential forms. Cartan was the master of differen- tial forms par excellence, and it was in his work that the differential form version of Stokes’s theorem found its clearest expression. © Springer Science+Business Media, LLC 2011235 L.W. Tu, An Introduction to Manifolds, Universitext, DOI 10.1007/978-1-4419-7400-6_6, 236§21 Orientations §21 Orientations It is a familiar fact from vector calculus that line and surface integrals depend on the orientation of the curve or surface over which the integration takes place: revers- ing the orientation changes the sign of the integral. The goal of this section is to deﬁne orientation for n-dimensional manifolds and to investigate various equivalent characterizations of orientation. We assume all vector spaces in this section to be ﬁnite-dimensional and real. An orientation of a ﬁnite-dimensional real vector space is simply an equivalence classof ordered bases, two ordered bases being equivalent if and only if their transition matrix has positive determinant. By its alternating nature, a multicovector of top degree turns out to represent perfectly an orientation of a vector space. An orientation on a manifold is a choice of an orientation for each tangent space satisfying a continuity condition. Globalizing n-covectors over a manifold, we ob- tain differential n-forms. An orientation on an n-manifold can also be given by an equivalence class of C ∞nowhere-vanishing n-forms, two such forms being equiva- lent if and only if one is a multiple of the other by a positive function. Finally, athird way to represent an orientation on a manifold is through an oriented atlas , an atlas in which any two overlapping charts are related by a transition function with everywhere positive Jacobian determinant. 21.1 Orientations of a Vector Space OnR1an orientation is one of two directions ( Figure 21.1 ). Fig. 21.1. Orientations of a line. OnR2an orientation is either counterclockwise or clockwise (Figure 21.2) . Fig. 21.2. Orientations of a plane. 21.1 Orientations of a Vector Space 237 e2e3=thumb e1=index ﬁnger Fig. 21.3. Right-handed orientation (e1,e2,e3)ofR3. e2=index ﬁngere3=thumb e1 Fig. 21.4. Left-handed orientation (e2,e1,e3)ofR3. OnR3an orientation is either right-handed (Figure 21.3 ) or left-handed ( Fig- ure 21.4) . The right-handed orientation of R3is the choice of a Cartesian coordinate system such that if you hold out your right hand with the index ﬁnger curling fromthe vector e 1in the x-axis to the vector e2in the y-axis, then your thumb points in the direction of of the vector e3in the z-axis. How should one deﬁne an orientation for R4,R5, and beyond? If we analyze the three examples above, we see that an orientation can be speciﬁed by an orderedbasis forR n. Let e1,..., enbe the standard basis for Rn. ForR1an orientation could be given by either e1or−e1. ForR2the counterclockwise orientation is (e1,e2), while the clockwise orientation is (e2,e1). ForR3the right-handed orientation is (e1,e2,e3), and the left-handed orientation is (e2,e1,e3). For any two ordered bases (u1,u2)and(v1,v2)forR2, there is a unique nonsin- gular 2×2 matrix A=[ai j]such that uj=2 ∑ i=1viai j,j=1,2, 238§21 Orientations called the change-of-basis matrix from(v1,v2)to(u1,u2). In matrix notation, if we write ordered bases as row vectors, for example, [u1u2]for the basis (u1,u2), then [u1u2]=[v 1v2]A. We say that two ordered bases are equivalent if the change-of-basis matrix Ahas pos- itive determinant. It is easy to check that this is indeed an equivalence relation on the set of all ordered bases for R2. It therefore partitions ordered bases into two equiva- lence classes. Each equivalence class is called an orientation ofR2. The equivalence class containing the ordered basis (e1,e2)is the counterclockwise orientation and the equivalence class of (e2,e1)is the clockwise orientation. The general case is similar. We assume all vector spaces in this section to be ﬁnite-dimensional. Two ordered bases u=[u 1···un]andv=[v 1···vn]of a vector space Vare said to be equivalent , written u∼v, ifu=vAfor an n×nmatrix Awith positive determinant. An orientation ofVis an equivalence class of ordered bases. Any ﬁnite-dimensional vector space has two orientations. If µis an orientation of a ﬁnite-dimensional vector space V, we denote the other orientation by −µand call it theopposite of the orientation µ. The zero-dimensional vector space {0}is a special case because it does not have a basis. We deﬁne an orientation on {0}to be one of the two signs +and−. NOTATION . A basis for a vector space is normally written v1,..., vn, without paren- theses, brackets, or braces. If it is an ordered basis, then we enclose it in parenthe- ses:(v1,..., vn). In matrix notation, we also write an ordered basis as a row vector [v1···vn]. An orientation is an equivalence class of ordered bases, so the notation is [(v1,..., vn)], where the brackets now stand for equivalence class. 21.2 Orientations and n-Covectors Instead of using an ordered basis, we can also use an n-covector to specify an orien- tation of an n-dimensional vector space V. This approach to orientations is based on the fact that the space/logicalandtextn(V∨)ofn-covectors on Vis one-dimensional. Lemma 21.1. Let u 1,..., unand v 1,..., vnbe vectors in a vector space V. Suppose uj=n ∑ i=1viai j,j=1,..., n, for a matrix A =[ai j]of real numbers. If βis an n-covector on V, then β(u1,..., un)=( detA)β(v1,..., vn). Proof. By hypothesis, uj=∑ iviai j. Since βisn-linear, 21.2 Orientations and n-Covectors 239 β(u1,..., un)=β/parenleftig ∑vi1ai1 1,...,∑vinainn/parenrightig =∑ai1 1···ainnβ(vi1,..., vin). Forβ(vi1,..., vin)to be nonzero, the subscripts i1,..., inmust be all distinct. An or- dered n-tuple I=(i1,..., in)with distinct components corresponds to a permutation σIof 1,..., nwith σI(j)=ijforj=1,..., n. Since βis an alternating n-tensor, β(vi1,..., vin)=( sgnσI)β(v1,..., vn). Thus, β(u1,..., un)=∑ σI∈Sn(sgnσI)ai1 1···ainnβ(v1,..., vn)=( detA)β(v1,..., vn).⊓ ⊔ As a corollary, if u1,..., unandv1,..., vnare ordered bases of a vector space V, then β(u1,..., un)andβ(v1,..., vn)have the same sign ⇐⇒ detA>0 ⇐⇒ u1,..., unandv1,..., vnare equivalent ordered bases . We say that the n-covector βdetermines orspeciﬁes the orientation (v1,..., vn)if β(v1,..., vn)>0. By the preceding corollary, this is a well-deﬁned notion, indepen- dent of the choice of ordered basis for the orientation. Moreover, two n-covectors β andβ′onVdetermine the same orientation if and only if β=aβ′for some positive real number a. We deﬁne an equivalence relation on the nonzero n-covectors on the n-dimensional vector space Vby setting β∼β′⇐⇒ β=aβ′for some a>0. Thus, in addition to an equivalence class of ordered bases, an orientation of Vis also given by an equivalence class of nonzero n-covectors on V. A linear isomorphism/logicalandtextn(V∨)≃Ridentiﬁes the set of nonzero n-covectors on VwithR−{0}, which has two connected components. Two nonzero n-covectors β andβ′onVare in the same component if and only if β=aβ′for some real number a>0. Thus, each connected component of/logicalandtextn(V∨)−{0}determines an orientation ofV. Example. Lete1,e2be the standard basis for R2andα1,α2its dual basis. Then the 2-covector α1∧α2determines the counterclockwise orientation of R2, since /parenleftbig α1∧α2/parenrightbig (e1,e2)=1>0. Example. Let∂/∂x|p,∂/∂y|pbe the standard basis for the tangent space Tp(R2), and(dx)p,(dy)pits dual basis. Then (dx∧dy)pdetermines the counterclockwise orientation of Tp(R2). 240§21 Orientations 21.3 Orientations on a Manifold Recall that every vector space of dimension nhas two orientations, corresponding to the two equivalence classes of ordered bases or the two equivalence classes of nonzero n-covectors. To orient a manifold M, we orient the tangent space at each point in M, but of course this has to be done in a “coherent” way so that the orienta- tion does not change abruptly anywhere. As we learned in Subsection 12.5, a frame on an open set U⊂Mis an n-tuple (X1,..., Xn)of possibly discontinuous vector ﬁelds on Usuch that at every point p∈U, the n-tuple(X1,p,..., Xn,p)of vectors is an ordered basis for the tangent space TpM. A global frame is a frame deﬁned on the entire manifold M, while a local frame about p∈Mis a frame deﬁned on some neighborhood of p. We introduce an equivalence relation on frames on U: (X1,..., Xn)∼(Y1,..., Yn)⇐⇒(X1,p,..., Xn,p)∼(Y1,p,..., Yn,p)for all p∈U. In other words, if Yj=∑iai jXi, then two frames (X1,..., Xn)and(Y1,..., Yn)are equivalent if and only if the change-of-basis matrix A=[ai j]has positive determinant at every point in U. Apointwise orientation on a manifold Massigns to each p∈Man orientation µpof the tangent space TpM. In terms of frames, a pointwise orientation on Mis simply an equivalence class of possibly discontinuous frames on M. A pointwise orientation µonMis said to be continuous at p∈Mifphas a neighborhood U on which µis represented by a continuous frame ; i.e., there exist continuous vector ﬁelds Y1,..., YnonUsuch that µq=/bracketleftbig/parenleftbig Y1,q,..., Yn,q/parenrightbig/bracketrightbig for all q∈U. The pointwise orientation µiscontinuous on M if it is continuous at every point p∈M. Note that a continuous pointwise orientation need not be represented by a continuous global frame; it sufﬁces that it be locally representable by a continuous local frame. A continuous pointwise orientation on Mis called an orientation onM. A manifold is said to be orientable if it has an orientation. A manifold together with an orientation is said to be oriented . Example. The Euclidean space Rnis orientable with orientation given by the contin- uous global frame (∂/∂r1,..., ∂/∂rn). Example 21.2 ( The open M ¨obius band ).LetRbe the rectangle R={(x,y)∈R2|0≤x≤1,−1<y<1}. The open M¨ obius band M(Figures 21.5 and21.6) is the quotient of the rectangle R by the equivalence relation generated by (0,y)∼(1,−y). (21.1) The interior of Ris the open rectangle U={(x,y)∈R2|0<x<1,−1<y<1}. 21.3 Orientations on a Manifold 241 Fig. 21.5. M¨ obius band. /Bullet /Bullete2 −e2p q 0 1−11 /Bullet /Bulletp q 0 1−11 e1e2 e1e2 e1e2 p=(0,0) q=(1,0) Fig. 21.6. Nonorientability of the M¨ obius band. Suppose the M¨ obius band Mis orientable. An orientation on Mrestricts to an ori- entation on U. To avoid confusion with an ordered pair of numbers, in this example we write an ordered basis without the parentheses. For the sake of deﬁniteness, we ﬁrst assume the orientation on Uto be given by e1,e2. By continuity the orientations at the points (0,0)and(1,0)are also given by e1,e2. But under the identiﬁcation (21.1), the ordered basis e1,e2at(1,0)maps to e1,−e2at(0,0). Thus, at (0,0)the orientation has to be given by both e1,e2ande1,−e2, a contradiction. Assuming the orientation on Uto be given by e2,e1also leads to a contradiction. This proves that the M¨ obius band is not orientable. Proposition 21.3. A connected orientable manifold M has exactly two orientations. Proof. Letµandνbe two orientations on M. At any point p∈M,µpandνpare orientations of TpM. They either are the same or are opposite orientations. Deﬁne a function f:M→{± 1}by f(p)=/braceleftigg 1 if µp=νp, −1 if µp=−νp. Fix a point p∈M. By continuity, there exists a connected neighborhood Uofpon which µ=[(X1,..., Xn)]andν=[(Y1,..., Yn)]for some continuous vector ﬁelds Xi andYjonU. Then there is a matrix-valued function A=[ai j]:U→GL(n,R)such thatYj=∑iai jXi. By Proposition 12.12 and Remark 12.13, the entries ai jare continu- ous, so that the determinant det A:U→R×is continuous also. By the intermediate value theorem, the continuous nowhere-vanishing function det Aon the connected setUis everywhere positive or everywhere negative. Hence, µ=νorµ=−νon 242§21 Orientations U. This proves that the function f:M→{± 1}is locally constant. Since a locally constant function on a connected set is constant (Problem 21.1), µ=νorµ=−ν onM. ⊓ ⊔ 21.4 Orientations and Differential Forms While the deﬁnition of an orientation on a manifold as a continuous pointwise ori- entation is geometrically intuitive, in practice it is easier to manipulate the nowhere- vanishing top forms that specify a pointwise orientation. In this section we show that the continuity condition on pointwise orientations translates to a C∞condition on nowhere-vanishing top forms. Iffis a real-valued function on a set M, we use the notation f>0 to mean that fis everywhere positive on M. Lemma 21.4. A pointwise orientation [(X1,..., Xn)]on a manifold M is continuous if and only if each point p ∈M has a coordinate neighborhood (U,x1,..., xn)on which the function (dx1∧···∧ dxn)(X1,..., Xn)is everywhere positive. Proof. (⇒)Assume that the pointwise orientation µ= [(X1,..., Xn)]onMis continuous. This does not mean that the global frame (X1,..., Xn)is continuous. What it means is that every point p∈Mhas a neighborhood Won which µis represented by a continu- ous frame (Y1,..., Yn). Choose a connected coordinate neighborhood (U,x1,..., xn) ofpcontained in Wand let ∂i=∂/∂xi. Then Yj=∑ibi j∂ifor a continuous ma- trix function [bi j]:U→GL(n,R), the change-of-basis matrix at each point. By Lemma 21.1, /parenleftbig dx1∧···∧ dxn/parenrightbig (Y1,..., Yn)=/parenleftbig det[bi j]/parenrightbig/parenleftbig dx1∧···∧ dxn/parenrightbig (∂1,..., ∂n)=det[bi j], which is never zero, because [bi j]is nonsingular. As a continuous nowhere-vanishing real-valued function on a connected set, (dx1∧···∧ dxn)(Y1,..., Yn)is everywhere positive or everywhere negative on U. If it is negative, then by setting ˜ x1=−x1, we have on the chart (U,˜x1,x2,..., xn)that /parenleftbig d˜x1∧dx2∧···∧ dxn/parenrightbig (Y1,..., Yn)>0. Renaming ˜ x1asx1, we may assume that on the coordinate neighborhood (U,x1,..., xn)ofp, the function (dx1∧···∧ dxn)(Y1,..., Yn)is always positive. Since µ=[(X1,..., Xn)]=[( Y1,..., Yn)]onU, the change-of-basis matrix C=[ci j] such that Xj=∑ici jYihas positive determinant. By Lemma 21.1 again, on U, /parenleftbig dx1∧···∧ dxn/parenrightbig (X1,..., Xn)=( detC)/parenleftbig dx1∧···∧ dxn/parenrightbig (Y1,..., Yn)>0. (⇐)On the chart (U,x1,..., xn), suppose Xj=∑ai j∂i. As before, /parenleftbig dx1∧···∧ dxn/parenrightbig (X1,..., Xn)=/parenleftbig det[ai j]/parenrightbig/parenleftbig dx1∧···∧ dxn/parenrightbig (∂1,..., ∂n)=det[ai j]. 21.4 Orientations and Differential Forms 243 By hypothesis, the left-hand side of the equalities above is positive. Therefore, on U, det[ai j]>0 and[(X1,..., Xn)] = [( ∂1,..., ∂n)], which proves that the pointwise orientation µis continuous at p. Since pwas arbitrary, µis continuous on M.⊓ ⊔ Theorem 21.5. A manifold M of dimension n is orientable if and only if there exists a C∞nowhere-vanishing n-form on M. Proof. (⇒)Suppose[(X1,..., Xn)]is an orientation on M. By Lemma 21.4, each point p has a coordinate neighborhood (U,x1,..., xn)on which /parenleftbig dx1∧···∧ dxn/parenrightbig (X1,..., Xn)>0. (21.2) Let{(Uα,x1 α,..., xn α)}be a collection of these charts that covers M, and let{ρα}be aC∞partition of unity subordinate to the open cover {Uα}. Being a locally ﬁnite sum, the n-form ω=∑αραdx1 α∧···∧ dxn αis well deﬁned and C∞onM. Fix p∈M. Since ρα(p)≥0 for all αandρα(p)>0 for at least one α, by (21.2), ωp(X1,p,..., Xn,p)=∑ αρα(p)/parenleftbig dx1 α∧···∧ dxn α/parenrightbig p(X1,p,..., Xn,p)>0. Therefore, ωis aC∞nowhere-vanishing n-form on M. (⇐)Suppose ωis aC∞nowhere-vanishing n-form on M. At each point p∈M, choose an ordered basis (X1,p,..., Xn,p)forTpMsuch that ωp(X1,p,..., Xn,p)>0. Fixp∈Mand let(U,x1,..., xn)be a connected coordinate neighborhood of p. On U,ω=f dx1∧···∧ dxnfor a C∞nowhere-vanishing function f. Being continuous and nowhere vanishing on a connected set, fis everywhere positive or everywhere negative on U. Iff>0, then on the chart (U,x1,..., xn), /parenleftbig dx1∧···∧ dxn/parenrightbig (X1,..., Xn)>0. Iff<0, then on the chart (U,−x1,x2,..., xn), /parenleftbig d(−x1)∧dx2∧···∧ dxn/parenrightbig (X1,..., Xn)>0. In either case, by Lemma 21.4, µ=[(X1,..., Xn)]is a continuous pointwise orienta- tion on M. ⊓ ⊔ Example 21.6 ( Orientability of a regular zero set ).By the regular level set theorem, if 0 is a regular value of a C∞function f(x,y,z)onR3, then the zero set f−1(0)is a C∞manifold. In Problem 19.11 we constructed a nowhere-vanishing 2-form on the regular zero set of a C∞function. It then follows from Theorem 21.5 that the regular zero set of a C∞function on R3is orientable. As an example, the unit sphere S2inR3is orientable. As another example, since an open M¨ obius band is not orientable (Example 21.2), it cannot be realized as the regular zero set of a C∞function on R3. According to a classical theorem from algebraic topology, a continuous vector ﬁeld on an even-dimensional sphere must vanish somewhere [18, Theorem 2.28, p. 135]. Thus, although the sphere S2has a continuous pointwise orientation, any global frame(X1,X2)that represents the orientation is necessarily discontinuous. 244§21 Orientations Ifωandω′are two nowhere-vanishing C∞n-forms on a manifold Mof dimen- sion n, then ω=fω′for some nowhere-vanishing function fonM. Locally, on a chart(U,x1,..., xn),ω=hdx1∧···∧ dxnandω′=gdx1∧···∧ dxn, where handg areC∞nowhere-vanishing functions on U. Therefore, f=h/gis also a C∞nowhere- vanishing function on U. Since Uis an arbitrary chart, fisC∞and nowhere van- ishing on M. On a connected manifold M, such a function fis either everywhere positive or everywhere negative. In this way the nowhere-vanishing C∞n-forms on a connected orientable manifold Mare partitioned into two equivalence classes by the equivalence relation ω∼ω′⇐⇒ ω=fω′with f>0. To each orientation µ=[(X1,..., Xn)]on a connected orientable manifold M, we associate the equivalence class of a C∞nowhere-vanishing n-form ωonMsuch that ω(X1,..., Xn)>0. (Such an ωexists by the proof of Theorem 21.5.) If µ/ma√s⊔o→[ω], then−µ/ma√s⊔o→[−ω]. On a connected orientable manifold, this sets up a one-to-one correspondence {orientations on M} ←→/braceleftiggequivalence classes of C∞nowhere-vanishing n-forms on M/bracerightigg , (21.3) each side being a set of two elements. By considering one connected component at a time, we see that the bijection (21.3) still holds for an arbitrary orientable manifold,each connected component having two possible orientations and two equivalence classes of C ∞nowhere-vanishing n-forms. If ωis aC∞nowhere-vanishing n-form such that ω(X1,..., Xn)>0, we say that ωdetermines orspeciﬁes the orientation [(X1,..., Xn)]and we call ωanorientation form onM. An oriented manifold can be described by a pair (M,[ω]), where[ω]is the equivalence class of an orientation form on M. We sometimes write M, instead of (M,[ω]), for an oriented manifold if it is clear from the context what the orientation is. For example, unless otherwise speciﬁed, Rnis oriented by dx1∧···∧ dxn. Remark 21.7 ( Orientations on zero-dimensional manifolds ).A connected manifold of dimension 0 is a point. The equivalence class of a nowhere-vanishing 0-form on apoint is either [−1]or[1]. Hence, a connected zero-dimensional manifold is always orientable. Its two orientations are speciﬁed by the two numbers ±1. A general zero-dimensional manifold Mis a countable discrete set of points (Example 5.13), and an orientation on Mis given by a function that assigns to each point of Meither 1 or−1. A diffeomorphism F:(N,[ ωN])→(M,[ωM])of oriented manifolds is said to beorientation-preserving if[F∗ωM]=[ωN]; it is orientation-reversing if[F∗ωM]= [−ωN]. Proposition 21.8. Let U and V be open subsets of Rn, both with the standard orien- tation inherited from Rn. A diffeomorphism F :U→V is orientation-preserving if and only if the Jacobian determinant det[∂Fi/∂xj]is everywhere positive on U. 21.5 Orientations and Atlases 245 Proof. Letx1,..., xnandy1,..., ynbe the standard coordinates on U⊂RnandV⊂ Rn. Then F∗(dy1∧···∧ dyn)=d(F∗y1)∧···∧ d(F∗yn) (Propositions 18.11 and 19.5) =d(y1◦F)∧···∧ d(yn◦F)(deﬁnition of pullback) =dF1∧···∧ dFn =det/bracketleftbigg∂Fi ∂xj/bracketrightbigg dx1∧···∧ dxn(by Corollary 18.4(ii) ). Thus, Fis orientation-preserving if and only if det [∂Fi/∂xj]is everywhere positive onU. ⊓ ⊔ 21.5 Orientations and Atlases Using the characterization of an orientation-preserving diffeomorphism by the sign of its Jacobian determinant, we can describe orientability of manifolds in terms of atlases. Deﬁnition 21.9. An atlas on Mis said to be oriented if for any two overlap- ping charts (U,x1,..., xn)and(V,y1,..., yn)of the atlas, the Jacobian determinant det[∂yi/∂xj]is everywhere positive on U∩V. Theorem 21.10. A manifold M is orientable if and only if it has an oriented atlas. Proof. (⇒)Letµ=[(X1,..., Xn)]be an orientation on the manifold M. By Lemma 21.4, each point p∈Mhas a coordinate neighborhood (U,x1,..., xn)on which /parenleftbig dx1∧···∧ dxn/parenrightbig (X1,..., Xn)>0. We claim that the collection U={(U,x1,..., xn)}of these charts is an oriented atlas. If(U,x1,..., xn)and(V,y1,..., yn)are two overlapping charts from U, then on U∩V, /parenleftbig dx1∧···∧ dxn/parenrightbig (X1,..., Xn)>0 and/parenleftbig dy1∧···∧ dyn/parenrightbig (X1,..., Xn)>0.(21.4) Since dy1∧···∧ dyn=/parenleftbig det[∂yi/∂xj]/parenrightbig dx1∧···∧ dxn, it follows from (21.4) that det[∂yi/∂xj]>0 on U∩V. Therefore, Uis an oriented atlas. (⇒)Suppose{(U,x1,..., xn)}is an oriented atlas. For each p∈(U,x1,..., xn), de- ﬁneµpto be the equivalence class of the ordered basis (∂/∂x1|p,..., ∂/∂xn|p)for TpM. If two charts (U,x1,..., xn)and(V,y1,..., yn)in the oriented atlas contain p, then by the orientability of the atlas, det [∂yi/∂xj]>0, so that(∂/∂x1|p,..., ∂/∂xn|p) is equivalent to (∂/∂y1|p,..., ∂/∂yn|p). This proves that µis a well-deﬁned point- wise orientation on M. It is continuous because every point phas a coordinate neigh- borhood(U,x1,..., xn)on which µ=[(∂/∂x1,..., ∂/∂xn)]is represented by a con- tinuous frame. ⊓ ⊔ 246§21 Orientations Deﬁnition 21.11. Two oriented atlases {(Uα,φα)}and{(Vβ,ψβ)}on a manifold M are said to be equivalent if the transition functions φα◦ψ−1 β:ψβ(Uα∩Vβ)→φα(Uα∩Vβ) have positive Jacobian determinant for all α,β. It is not difﬁcult to show that this is an equivalence relation on the set of oriented atlases on a manifold M(Problem 21.3). In the proof of Theorem 21.10, an oriented atlas {(U,x1,..., xn)}on a mani- foldMdetermines an orientation U∋p/ma√s⊔o→[(∂/∂x1|p,..., ∂/∂xn|p)]onM, and con- versely, an orientation [(X1,..., Xn)]onMgives rise to an oriented atlas {(U,x1,..., xn)} onMsuch that/parenleftbig dx1∧···∧ dxn/parenrightbig (X1,..., Xn)>0 on U. We leave it as an exercise to show that for an orientable manifold M, the two induced maps /braceleftigequivalence classes of oriented atlases on M/bracerightig/d47/d47{orientations on M} /d111/d111 are well deﬁned and inverse to each other. Therefore, one can also specify an orien- tation on an orientable manifold by an equivalence class of oriented atlases. For an oriented manifold M, we denote by−Mthe same manifold but with the opposite orientation. If {(U,φ)}={(U,x1,x2,..., xn)}is an oriented atlas specify- ing the orientation of M, then an oriented atlas specifying the orientation of −Mis {(U,˜φ)}={(U,−x1,x2,..., xn)}. Problems 21.1.* Locally constant map on a connected space A map f:S→Ybetween two topological spaces is locally constant if for every p∈Sthere is a neighborhood Uofpsuch that fis constant on U. Show that a locally constant map f:S→Yon a nonempty connected space Sis constant. ( Hint: Show that for every y∈Y, the inverse image f−1(y)is open. Then S=/uniontext y∈Yf−1(y)exhibits Sas a disjoint union of open subsets.) 21.2. Continuity of pointwise orientations Prove that a pointwise orientation [(X1,..., Xn)]on a manifold Mis continuous if and only if every point p∈Mhas a coordinate neighborhood (U,φ) = ( U,x1,..., xn)such that for allq∈U, the differential φ∗,q:TqM→Tf(q)Rn≃Rncarries the orientation of TqMto the standard orientation of Rnin the following sense:/parenleftbig φ∗X1,q,..., φ∗Xn,q/parenrightbig ∼(∂/∂r1,..., ∂/∂rn). 21.3. Equivalence of oriented atlases Show that the relation in Deﬁnition 21.11 is an equivalence relation. 21.4. Orientation-preserving diffeomorphisms LetF:(N,[ωN])→(M,[ωM])be an orientation-preserving diffeomorphism. If {(Vα,ψα)}= {(Vα,y1α,..., ynα)}is an oriented atlas on Mthat speciﬁes the orientation of M, show that {(F−1Vα,F∗ψα)}={(F−1Vα,F1α,..., Fnα)}is an oriented atlas on Nthat speciﬁes the orien- tation of N, where Fiα=yiα◦F. 21.5 Orientations and Atlases 247 21.5. Orientation-preserving or orientation-reversing diffeomorphisms LetUbe the open set (0,∞)×(0,2π)in the(r,θ)-planeR2. We deﬁne F:U⊂R2→R2by F(r,θ)=(r cosθ,rsinθ). Decide whether Fis orientation-preserving or orientation-reversing as a diffeomorphism onto its image. 21.6. Orientability of a regular level set in Rn+1 Suppose f(x1,..., xn+1)is aC∞function on Rn+1with 0 as a regular value. Show that the zero set of fis an orientable submanifold of Rn+1. In particular, the unit n-sphere SninRn+1 is orientable. 21.7. Orientability of a Lie group Show that every Lie group Gis orientable by constructing a nowhere-vanishing top form on G. 21.8. Orientability of a parallelizable manifold Show that a parallelizable manifold is orientable. (In particular, this shows again that everyLie group is orientable.) 21.9. Orientability of the total space of the tangent bundle LetMbe a smooth manifold and π:T M→Mits tangent bundle. Show that if {(U,φ)}is any atlas on M, then the atlas{(TU,˜φ)}onT M, with ˜φdeﬁned in equation (12.1), is oriented. This proves that the total space T Mof the tangent bundle is always orientable, regardless of whether Mis orientable. 21.10. Oriented atlas on a circle In Example 5.16 we found an atlas U={(Ui,φi)}4 i=1on the unit circle S1. IsUan oriented atlas? If not, alter the coordinate functions φito makeUinto an oriented atlas. 248§22 Manifolds with Boundary §22 Manifolds with Boundary The prototype of a manifold with boundary is the closed upper half-space Hn={(x1,..., xn)∈Rn|xn≥0}, with the subspace topology inherited from Rn. The points (x1,..., xn)inHnwith xn>0 are called the interior points ofHn, and the points with xn=0 are called the boundary points ofHn. These two sets are denoted by (Hn)◦and∂(Hn), respec- tively (Figure 22.1) . xn int(Hn) ∂(Hn) Fig. 22.1. Upper half-space. In the literature the upper half-space often means the open set {(x1,..., xn)∈Rn|xn>0}. We require that Hninclude the boundary in order for it to serve as a model for manifolds with boundary. IfMis a manifold with boundary, then its boundary ∂Mturns out to be a manifold of dimension one less without boundary. Moreover, an orientation on Minduces an orientation on ∂M. The choice of the induced orientation on the boundary is a matter of convention, guided by the desire to make Stokes’s theorem sign-free. Of thevarious ways to describe the boundary orientation, two stand out for their simplicity: (1) contraction of an orientation form on Mwith an outward-pointing vector ﬁeld on ∂Mand (2) “outward vector ﬁrst.” 22.1 Smooth Invariance of Domain in Rn To discuss C∞functions on a manifold with boundary, we need to extend the deﬁni- tion of a C∞function to allow nonopen domains. Deﬁnition 22.1. LetS⊂Rnbe an arbitrary subset. A function f:S→Rmis smooth at a point p inSif there exist a neighborhood UofpinRnand a C∞function ˜f:U→Rmsuch that ˜f=fonU∩S. The function is smooth on S if it is smooth at each point of S. 22.1 Smooth Invariance of Domain in Rn249 With this deﬁnition it now makes sense to speak of an arbitrary subset S⊂Rn being diffeomorphic to an arbitrary subset T⊂Rm; this will be the case if and only if there are smooth maps f:S→T⊂Rmandg:T→S⊂Rnthat are inverse to each other. Exercise 22.2 (Smooth functions on a nonopen set).* Using a partition of unity, show that a function f:S→RmisC∞onS⊂Rnif and only if there exist an open set UinRncontaining Sand a C∞function ˜f:U→Rmsuch that f=˜f|S. The following theorem is the C∞analogue of a classical theorem in the contin- uous category. We will use it to show that interior points and boundary points are invariant under diffeomorphisms of open subsets of Hn. Theorem 22.3 (Smooth invariance of domain). Let U⊂Rnbe an open subset, S⊂Rnan arbitrary subset, and f :U→S a diffeomorphism. Then S is open in Rn. More succinctly, a diffeomorphism between an open subset UofRnand an ar- bitrary subset SofRnforces Sto be open in Rn. The theorem is not automatic. A diffeomorphism f:Rn⊃U→S⊂Rntakes an open subset of Uto an open subset ofS. Thus, a priori we know only that f(U)is open in S, not that f(U), which is S, is open in Rn. It is crucial that the two Euclidean spaces be of the same dimension. For example, there are a diffeomorphism between the open interval ]0,1[inR1and the open segment S= ]0,1[×{0}inR2, but Sis not open in R2. Proof. Letf(p)be an arbitrary point in S, with p∈U. Since f:U→Sis a diffeo- morphism, there are an open set V⊂Rncontaining Sand a C∞map g:V→Rnsuch thatg|S=f−1. Thus, Uf→Vg→Rn satisﬁes g◦f=1U:U→U⊂Rn, the identity map on U. By the chain rule, g∗,f(p)◦f∗,p=1TpU:TpU→TpU≃Tp(Rn), the identity map on the tangent space TpU. Hence, f∗,pis injective. Since UandV have the same dimension, f∗,p:TpU→Tf(p)Vis invertible. By the inverse function theorem, fis locally invertible at p. This means that there are open neighborhoods UpofpinUandVf(p)off(p)inVsuch that f:Up→Vf(p)is a diffeomorphism. It follows that f(p)∈Vf(p)=f(Up)⊂f(U)=S. Since Vis open in RnandVf(p)is open in V, the set Vf(p)is open in Rn. By the local criterion for openness (Lemma A.2), Sis open in Rn. ⊓ ⊔ Proposition 22.4. Let U and V be open subsets of the upper half-space Hnand f:U→V a diffeomorphism. Then f maps interior points to interior points and boundary points to boundary points. 250§22 Manifolds with Boundary Proof. Letp∈Ube an interior point. Then pis contained in an open ball B, which is actually open in Rn(not just in Hn). By smooth invariance of domain, f(B)is open inRn(again not just in Hn). Therefore, f(B)⊂(Hn)◦. Since f(p)∈f(B), f(p)is an interior point of Hn. Ifpis a boundary point in U∩∂Hn, then f−1(f(p)) = pis a boundary point. Since f−1:V→Uis a diffeomorphism, by what has just been proven, f(p)cannot be an interior point. Thus, f(p)is a boundary point. ⊓ ⊔ Remark 22.5.Replacing Euclidean spaces by manifolds throughout this subsection, one can prove in exactly the same way smooth invariance of domain for manifolds: if there is a diffeomorphism between an open subset Uof an n-dimensional manifold Nand an arbitrary subset Sof another n-dimensional manifold M, then Sis open inM. 22.2 Manifolds with Boundary In the upper half-space Hnone may distinguish two kinds of open subsets, depend- ing on whether the set is disjoint from the boundary or intersects the boundary ( Fig- ure 22.2 ). Charts on a manifold are homeomorphic to only the ﬁrst kind of open sets. Fig. 22.2. Two types of open subsets of Hn. A manifold with boundary generalizes the deﬁnition of a manifold by allowing both kinds of open sets. We say that a topological space MislocallyHnif every point p∈Mhas a neighborhood Uhomeomorphic to an open subset of Hn. Deﬁnition 22.6. Atopological n-manifold with boundary is a second countable, Hausdorff topological space that is locally Hn. LetMbe a topological n-manifold with boundary. For n≥2, achart onMis deﬁned to be a pair (U,φ)consisting of an open set UinMand a homeomorphism φ:U→φ(U)⊂Hn ofUwith an open subset φ(U)ofHn. As Example 22.9 (p. 254) will show, a slight modiﬁcation is necessary when n=1: we need to allow two local models, the right half-lineH1and the left half-line 22.2 Manifolds with Boundary 251 L1:={x∈R|x≤0}. A chart(U,φ)in dimension 1 consists of an open set UinMand a homeomorphism φofUwith an open subset of H1orL1. With this convention, if (U,x1,x2,..., xn) is a chart of an n-dimensional manifold with boundary, then so is (U,−x1,x2,..., xn) for any n≥1. A manifold with boundary has dimension at least 1, since a manifold of dimension 0, being a discrete set of points, necessarily has empty boundary. A collection{(U,φ)}of charts is a C∞atlas if for any two charts (U,φ)and (V,ψ), the transition map ψ◦φ−1:φ(U∩V)→ψ(U∩V)⊂Hn is a diffeomorphism. A C∞manifold with boundary is a topological manifold with boundary together with a maximal C∞atlas. A point pofMis called an interior point if in some chart (U,φ), the point φ(p) is an interior point of Hn. Similarly, pis aboundary point ofMifφ(p)is a boundary point ofHn. These concepts are well deﬁned, independent of the charts, because if (V,ψ)is another chart, then the diffeomorphism ψ◦φ−1maps φ(p)toψ(p), and so by Proposition 22.4, φ(p)andψ(p)are either both interior points or both boundary points ( Figure 22.3 ). The set of boundary points of Mis denoted by ∂M. /Bullet /Bullet/Bullet φ ψ p Fig. 22.3. Boundary charts. Most of the concepts introduced for a manifold extend word for word to a man- ifold with boundary, the only difference being that now a chart can be either of twotypes and the local model is H n(orL1). For example, a function f:M→RisC∞ at a boundary point p∈∂Mif there is a chart (U,φ)about psuch that f◦φ−1isC∞ atφ(p)∈Hn. This in turn means that f◦φ−1has a C∞extension to a neighborhood ofφ(p)inRn. In point-set topology there are other notions of interior and boundary, deﬁned for a subset Aof a topological space S. A point p∈Sis said to be an interior point ofA if there exists an open subset UofSsuch that p∈U⊂S. The point p∈Sis an exterior point ofAif there exists an open subset UofSsuch that p∈U⊂S−A. 252§22 Manifolds with Boundary Finally, p∈Sis aboundary point ofAif every neighborhood of pcontains both a point in Aand a point not in A. We denote by int (A), ext(A), and bd(A)the sets of interior, exterior, and boundary points respectively of AinS. Clearly, the topological space Sis the disjoint union S=int(A)∐ext(A)∐bd(A). In case the subset A⊂Sis a manifold with boundary, we call int (A)thetopo- logical interior and bd(A)thetopological boundary ofA, to distinguish them from themanifold interior A◦and the manifold boundary ∂A. Note that the topological interior and the topological boundary of a set depend on an ambient space, while the manifold interior and the manifold boundary are intrinsic. Example 22.7 ( Topological boundary versus manifold boundary ).LetAbe the open unit disk in R2: A={x∈R2|/bardblx/bardbl<1}. Then its topological boundary bd (A)inR2is the unit circle, while its manifold boundary ∂Ais the empty set ( Figure 22.4 ). IfBis the closed unit disk in R2, then its topological boundary bd (B)and its manifold boundary ∂Bcoincide; both are the unit circle. ⊂R2 A⊂R2 B⊂H2 D Fig. 22.4. Interiors and boundaries. Example 22.8 ( Topological interior versus manifold interior ).LetSbe the upper half-plane H2and let Dbe the subset ( Figure 22.4 ) D={(x,y)∈H2|y≤1}. The topological interior of Dis the set int(D)={(x,y)∈H2|0≤y<1}, containing the x-axis, while the manifold interior of Dis the set D◦={(x,y)∈H2|0<y<1}, not containing the x-axis. To indicate the dependence of the topological interior of a set Aon its ambient space S, we might denote it by int S(A)instead of int (A). Then in this example, the topological interior intH2(D)ofDinH2is as above, but the topological interior intR2(D)ofDinR2coincides with D◦. 22.4 Tangent Vectors, Differential Forms, and Orientations 253 22.3 The Boundary of a Manifold with Boundary LetMbe a manifold of dimension nwith boundary ∂M. If(U,φ)is a chart on M, we denote by φ′=φ|U∩∂Mthe restriction of the coordinate map φto the boundary. Since φmaps boundary points to boundary points, φ′:U∩∂M→∂Hn=Rn−1. Moreover, if (U,φ)and(V,ψ)are two charts on M, then ψ′◦(φ′)−1:φ′(U∩V∩∂M)→ψ′(U∩V∩∂M) isC∞. Thus, an atlas{(Uα,φα)}forMinduces an atlas{(Uα∩∂M,φα|Uα∩∂M)}for ∂M, making ∂Minto a manifold of dimension n−1 without boundary. 22.4 Tangent Vectors, Differential Forms, and Orientations LetMbe a manifold with boundary and let p∈∂M. As in Subsection 2.2, two C∞ functions f:U→Randg:V→Rdeﬁned on neighborhoods UandVofpinM are said to be equivalent if they agree on some neighborhood Wofpcontained in U∩V. Agerm ofC∞functions at pis an equivalence class of such functions. With the usual addition, multiplication, and scalar multiplication of germs, the set C∞ p(M) of germs of C∞functions at pis anR-algebra. The tangent space T pMatpis then deﬁned to be the vector space of all point-derivations on the algebra C∞ p(M). For example, for pin the boundary of the upper half-plane H2,∂/∂x|pand ∂/∂y|pare both derivations on C∞ p(H2). The tangent space Tp(H2)is represented by a 2-dimensional vector space with the origin at p. Since ∂/∂y|pis a tangent vector toH2atp, its negative−∂/∂y|pis also a tangent vector at p(Figure 22.5 ), although there is no curve through pinH2with initial velocity −∂/∂y|p. /Bulletp −∂ ∂y/vextendsingle/vextendsingle/vextendsingle p Fig. 22.5. A tangent vector at the boundary. Thecotangent space T∗ pMis deﬁned to be the dual of the tangent space: T∗ pM=Hom(TpM,R). Differential k-forms onMare deﬁned as before, as sections of the vector bundle/logicalandtextk(T∗M). A differential k-form is C∞if it is C∞as a section of the vector bundle 254§22 Manifolds with Boundary /logicalandtextk(T∗M). For example, dx∧dyis aC∞2-form on H2. An orientation on an n- manifold Mwith boundary is again a continuous pointwise orientation on M. The discussion in Section 21 on orientations goes through word for word for manifolds with boundary. Thus, the orientability of a manifold with boundary is equivalent to the existence of a C∞nowhere-vanishing top form and to the existence of an oriented atlas. At one point in the proof of Lemma 21.4, it was necessary to replace the chart (U,x1,x2,..., xn)by(U,−x1,x2,..., xn). This would not have been possible for n=1 if we had not allowed the left half-line L1as a local model in the deﬁnition of a chart on a 1-dimensional manifold with boundary. Example 22.9.The closed interval [0,1]is aC∞manifold with boundary. It has an atlas with two charts (U1,φ1)and(U2,φ2), where U1=[0,1[,φ1(x) =x, and U2= ]0,1],φ2(x) =1−x. With d/dxas a continuous pointwise orientation, [0,1]is an oriented manifold with boundary. However, {(U1,φ1),(U2,φ2)}is not an oriented atlas, because the Jacobian determinant of the transition function (φ2◦φ−1 1)(x)=1− xis negative. If we change the sign of φ2, then{(U1,φ1),(U2,−φ2)}is an oriented atlas. Note that−φ2(x)=x−1 maps]0,1]into the left half-line L1⊂R. If we had allowed only H1as a local model for a 1-dimensional manifold with boundary, the closed interval [0,1]would not have an oriented atlas. 22.5 Outward-Pointing Vector Fields LetMbe a manifold with boundary and p∈∂M. We say that a tangent vector Xp∈Tp(M)isinward-pointing ifXp/∈Tp(∂M)and there are a positive real number εand a curve c:[0,ε[→Msuch that c(0) =p,c((0,ε[)⊂M◦, and c′(0) =Xp. A vector Xp∈Tp(M)isoutward-pointing if−Xpis inward-pointing. For example, on the upper half-plane H2, the vector ∂/∂y|pis inward-pointing and the vector −∂/∂y|pis outward-pointing at a point pon the x-axis. A vector ﬁeld along ∂Mis a function Xthat assigns to each point pin∂Ma vector Xpin the tangent space TpM(as opposed to Tp(∂M)). In a coordinate neigh- borhood(U,x1,..., xn)ofpinM, such a vector ﬁeld Xcan be written as a linear combination Xq=∑ iai(q)∂ ∂xi/vextendsingle/vextendsingle/vextendsingle/vextendsingle q,q∈∂M. The vector ﬁeld Xalong ∂Mis said to be smooth at p∈Mif there exists a coordinate neighborhood of pfor which the functions aion∂MareC∞atp; it is said to be smooth if it is smooth at every point p. In terms of local coordinates, a vector Xpis outward-pointing if and only if an(p)<0 (see Figure 22.5 and Problem 22.3). Proposition 22.10. On a manifold M with boundary ∂M, there is a smooth outward- pointing vector ﬁeld along ∂M. Proof. Cover ∂Mwith coordinate open sets/parenleftbig Uα,x1 α,..., xn α/parenrightbig inM. On each Uαthe vector ﬁeld Xα=−∂/∂xn αalong Uα∩∂Mis smooth and outward-pointing. Choose a partition of unity {ρα}α∈Aon∂Msubordinate to the open cover {Uα∩∂M}α∈A. Then one can check that X:=∑ραXαis a smooth outward-pointing vector ﬁeld along ∂M(Problem 22.4). ⊓ ⊔ 22.6 Boundary Orientation 255 22.6 Boundary Orientation In this section we show that the boundary of an orientable manifold Mwith boundary is an orientable manifold (without boundary, by Subsection 22.3). We will designate one of the orientations on the boundary as the boundary orientation. It is easily described in terms of an orientation form or of a pointwise orientation on ∂M. Proposition 22.11. Let M be an oriented n-manifold with boundary. If ωis an ori- entation form on M and X is a smooth outward-pointing vector ﬁeld on ∂M, then ιXωis a smooth nowhere-vanishing (n−1)-form on ∂M. Hence, ∂M is orientable. Proof. Since ωandXare both smooth on ∂M, so is the contraction ιXω(Subsec- tion 20.4). We will now prove by contradiction that ιXωis nowhere-vanishing on ∂M. Suppose ιXωvanishes at some p∈∂M. This means that (ιXω)p(v1,..., vn−1)= 0 for all v1,..., vn−1∈Tp(∂M). Let e1,..., en−1be a basis for Tp(∂M). Then Xp,e1,..., en−1is a basis for TpM, and ωp(Xp,e1,..., en−1)=( ιXω)p(e1,..., en−1)=0. By Problem 3.9, ωp≡0 on TpM, a contradiction. Therefore, ιXωis nowhere van- ishing on ∂M. By Theorem 21.5, ∂Mis orientable. ⊓ ⊔ In the notation of the preceding proposition, we deﬁne the boundary orientation on∂Mto be the orientation with orientation form ιXω. For the boundary orientation to be well deﬁned, we need to check that it is independent of the choice of the ori- entation form ωand of the outward-pointing vector ﬁeld X. The veriﬁcation is not difﬁcult (see Problem 22.5). Proposition 22.12. Suppose M is an oriented n-manifold with boundary. Let p be a point of the boundary ∂M and let X pbe an outward-pointing vector in T pM. An ordered basis (v1,..., vn−1)for T p(∂M)represents the boundary orientation at p if and only if the ordered basis (Xp,v1,..., vn−1)for T pM represents the orientation on M at p. To make this rule easier to remember, we summarize it under the rubric “outward vector ﬁrst.” Proof. Forpin∂M, let(v1,..., vn−1)be an ordered basis for the tangent space Tp(∂M). Then (v1,..., vn−1)represents the boundary orientation on ∂Matp ⇐⇒(ιXpωp)(v1,..., vn−1)>0 ⇐⇒ ωp(Xp,v1,..., vn−1)>0 ⇐⇒(Xp,v1,..., vn−1)represents the orientation on Matp.⊓ ⊔ 256§22 Manifolds with Boundary Example 22.13 ( The boundary orientation on ∂Hn).An orientation form for the standard orientation on the upper half-space Hnisω=dx1∧···∧ dxn. A smooth outward-pointing vector ﬁeld on ∂Hnis−∂/∂xn. By deﬁnition, an orientation form for the boundary orientation on ∂Hnis given by the contraction ι−∂/∂xn(ω)=−ι∂/∂xn(dx1∧···∧ dxn−1∧dxn) =−(−1)n−1dx1∧···∧ dxn−1∧ι∂/∂xn(dxn) =(−1)ndx1∧···∧ dxn−1. Thus, the boundary orientation on ∂H1={0}is given by−1, the boundary orienta- tion on ∂H2, given by dx1, is the usual orientation on the real line R(Figure 22.6(a) ), and the boundary orientation on ∂H3, given by−dx1∧dx2, is the clockwise orien- tation in the (x1,x2)-planeR2(Figure 22.6(b)) . (a) Boundary orientation on ∂H2=R.x1x2x3 (b) Boundary orientation on ∂H3=R2. Fig. 22.6. Boundary orientations. Example. The closed interval [a,b]in the real line with coordinate xhas a standard orientation given by the vector ﬁeld d/dx, with orientation form dx. At the right endpoint b, an outward vector is d/dx. Hence, the boundary orientation at bis given byιd/dx(dx)=+ 1. Similarly, the boundary orientation at the left endpoint ais given byι−d/dx(dx)=−1. Example. Suppose c:[a,b]→Mis aC∞immersion whose image is a 1-dimensional manifold Cwith boundary. An orientation on [a,b]induces an orientation on Cvia the differential c∗,p:Tp([a,b])→TpCat each point p∈[a,b]. In a situation like this, we give Cthe orientation induced from the standard orientation on [a,b]. The boundary orientation on the boundary of Cis given by +1 at the endpoint c(b)and −1 at the initial point c(a). Problems 22.1. Topological boundary versus manifold boundary LetMbe the subset [0,1[∪{2}of the real line. Find its topological boundary bd (M)and its manifold boundary ∂M. 22.6 Boundary Orientation 257 22.2. Topological boundary of an intersection LetAandBbe two subsets of a topological space S. Prove that bd(A∩B)⊂bd(A)∪bd(B). 22.3.* Inward-pointing vectors at the boundary LetMbe a manifold with boundary and let p∈∂M. Show that Xp∈TpMis inward-pointing if and only if in any coordinate chart (U,x1,..., xn)centered at p, the coefﬁcient of (∂/∂xn)p inXpis positive. 22.4.* Smooth outward-pointing vector ﬁeld along the boundary Show that the vector ﬁeld X=∑ραXαdeﬁned in the proof of Proposition 22.10 is a smooth outward-pointing vector ﬁeld along ∂M. 22.5. Boundary orientation LetMbe an oriented manifold with boundary, ωan orientation form for M, and XaC∞ outward-pointing vector ﬁeld along ∂M. (a) If τis another orientation form on M, then τ=fωfor a C∞everywhere-positive function fonM. Show that ιXτ=fιXωand therefore, ιXτ∼ιXωon∂M. (Here “∼” is the equivalence relation deﬁned in Subsection 21.4.) (b) Prove that if Yis another C∞outward-pointing vector ﬁeld along ∂M, then ιXω∼ιYωon ∂M. 22.6.* Induced atlas on the boundary Assume n≥2 and let (U,φ)and(V,ψ)be two charts in an oriented atlas of an orientable n- manifold Mwith boundary. Prove that if U∩V∩∂M/ne}a⊔ionslash=∅, then the restriction of the transition function ψ◦φ−1to the boundary B:=φ(U∩V)∩∂Hn, (ψ◦φ−1)|B:φ(U∩V)∩∂Hn→ψ(U∩V)∩∂Hn, has positive Jacobian determinant. ( Hint: Let φ=(x1,..., xn)andψ=(y1,..., yn). Show that the Jacobian matrix of ψ◦φ−1in local coordinates is block triangular with J(ψ◦φ−1)|Band ∂yn/∂xnas the diagonal blocks, and that ∂yn/∂xn>0.) Thus, if{(Uα,φα)}is an oriented atlas for a manifold Mwith boundary, then the induced atlas{(Uα∩∂M,φα|Uα∩∂M)}for∂Mis oriented. 22.7.* Boundary orientation of the left half-space LetMbe the left half-space {(x1,..., xn)∈Rn|x1≤0}, with orientation form dx1∧···∧ dxn. Show that an orientation form for the boundary orienta- tion on ∂M={(0,x2,..., xn)∈Rn}isdx2∧···∧ dxn. Unlike the upper half-space Hn, whose boundary orientation takes on a sign (Example 22.13), this exercise shows that the boundary orientation for the left half-space has no sign. For this reason some authors use the left half-space as the model of a manifold with boundary,e.g., [7]. 22.8. Boundary orientation on a cylinder LetMbe the cylinder S 1×[0,1]with the counterclockwise orientation when viewed from the exterior ( Figure 22.7(a) ). Describe the boundary orientation on C0=S1×{0}andC1= S1×{1}. 258§22 Manifolds with Boundary C1 C0 (a) Oriented cylinder.X=∑xi∂ ∂xi (b) Radial vector ﬁeld on a sphere. Fig. 22.7. Boundary orientations. 22.9. Boundary orientation on a sphere Orient the unit sphere SninRn+1as the boundary of the closed unit ball. Show that an orientation form on Snis ω=n+1 ∑ i=1(−1)i−1xidx1∧···∧/hatwiderdxi∧···∧ dxn+1, where the caret /hatwideover dxiindicates that dxiis to be omitted. ( Hint: An outward-pointing vector ﬁeld on Snis the radial vector ﬁeld X=∑xi∂/∂xias in Figure 22.7(b). ) xy yz x Fig. 22.8. Projection of the upper hemisphere to a disk. 22.10. Orientation on the upper hemisphere of a sphere Orient the unit sphere SninRn+1as the boundary of the closed unit ball. Let Ube the upper hemisphere U={x∈Sn|xn+1>0}. It is a coordinate chart on the sphere with coordinates x1,..., xn. (a) Find an orientation form on Uin terms of dx1,..., dxn. (b) Show that the projection map π:U→Rn, π(x1,..., xn,xn+1)=(x1,..., xn), is orientation-preserving if and only if nis even ( Figure 22.8 ). 22.6 Boundary Orientation 259 22.11. Antipodal map on a sphere and the orientability of RPn (a) The antipodal map a:Sn→Snon the n-sphere is deﬁned by a(x1,..., xn+1)=(− x1,...,−xn+1). Show that the antipodal map is orientation-preserving if and only if nis odd. (b) Use part (a) and Problem 21.6 to prove that an odd-dimensional real projective space RPn is orientable. 260§23 Integration on Manifolds §23 Integration on Manifolds In this chapter we ﬁrst recall Riemann integration for a function over a closed rect- angle in Euclidean space. By Lebesgue’s theorem, this theory can be extended to integrals over domains of integration, bounded subsets of Rnwhose boundary has measure zero. The integral of an n-form with compact support in an open set of Rnis deﬁned to be the Riemann integral of the corresponding function. Using a partition of unity,we deﬁne the integral of an n-form with compact support on a manifold by writing the form as a sum of forms each with compact support in a coordinate chart. We then prove the general Stokes theorem for an oriented manifold and show how it generalizes the fundamental theorem for line integrals as well as Green’s theorem from calculus. 23.1 The Riemann Integral of a Function on Rn We assume that the reader is familiar with the theory of Riemann integration in Rn, as in [26] or [35]. What follows is a brief synopsis of the Riemann integral of a bounded function over a bounded set in Rn. Aclosed rectangle inRnis a Cartesian product R= [a1,b1]×···×[an,bn]of closed intervals in R, where ai,bi∈R. Let f:R→Rbe a bounded function deﬁned on a closed rectangle R. The volume vol(R)of the closed rectangle Ris deﬁned to be vol(R):=n ∏ i=1(bi−ai). (23.1) Apartition of the closed interval [a,b]is a set of real numbers {p0,..., pn}such that a=p0<p1<···<pn=b. Apartition of the rectangle Ris a collection P={P1,..., Pn}, where each Piis a partition of [ai,bi]. The partition Pdivides the rectangle Rinto closed subrectangles, which we denote by Rj(Figure 23.1 ). We deﬁne the lower sum and the upper sum offwith respect to the partition P to be L(f,P):=∑(inf Rjf)vol(Rj),U(f,P):=∑(sup Rjf)vol(Rj), where each sum runs over all subrectangles of the partition P. For any partition P, clearly L(f,P)≤U(f,P). In fact, more is true: for any two partitions PandP′of the rectangle R, L(f,P)≤U(f,P′), which we show next. 23.1 The Riemann Integral of a Function on Rn261 /Bullet /Bullet /Bullet /Bullet/Bullet/Bullet/Bullet a1b1a2b2 Fig. 23.1. A partition of a closed rectangle. A partition P′={P′ 1,..., P′ n}is areﬁnement of the partition P={P1,..., Pn}if Pi⊂P′ ifor all i=1,..., n. IfP′is a reﬁnement of P, then each subrectangle RjofP is subdivided into subrectangles R′ jkofP′, and it is easily seen that L(f,P)≤L(f,P′), (23.2) because if R′ jk⊂Rj, then inf Rjf≤infR′ jkf. Similarly, if P′is a reﬁnement of P, then U(f,P′)≤U(f,P). (23.3) Any two partitions PandP′of the rectangle Rhave a common reﬁnement Q= {Q1,..., Qn}with Qi=Pi∪P′ i. By (23.2) and (23.3), L(f,P)≤L(f,Q)≤U(f,Q)≤U(f,P′). It follows that the supremum of the lower sum L(f,P)over all partitions PofRis less than or equal to the inﬁmum of the upper sum U(f,P)over all partitions Pof R. We deﬁne these two numbers to be the lower integral/integraltext Rfand the upper integral /integraltext Rf, respectively: /integraldisplay Rf:=sup PL(f,P),/integraldisplay Rf:=inf PL(f,P). Deﬁnition 23.1. LetRbe a closed rectangle in Rn. A bounded function f:R→R is said to be Riemann integrable if/integraltext Rf=/integraltext Rf; in this case, the Riemann integral offis this common value, denoted by/integraltext Rf(x)dx1···dxn, where x1,..., xnare the standard coordinates on Rn. Remark. When we speak of a rectangle [a1,b1]×···×[an,bn]inRn, we have already tacitly chosen ncoordinates axes, with coordinates x1,..., xn. Thus, the deﬁnition of a Riemann integral depends on the coordinates x1,..., xn. Iff:A⊂Rn→R, then the extension of f by zero is the function ˜f:Rn→R such that 262§23 Integration on Manifolds ˜f(x)=/braceleftigg f(x)forx∈A, 0 for x/∈A. Now suppose f:A→Ris a bounded function on a bounded set AinRn. Enclose A in a closed rectangle Rand deﬁne the Riemann integral of fover Ato be /integraldisplay Af(x)dx1···dxn=/integraldisplay R˜f(x)dx1···dxn if the right-hand side exists. In this way we can deal with the integral of a bounded function whose domain is an arbitrary bounded set in Rn. Thevolume vol(A)of a subset A⊂Rnis deﬁned to be the integral/integraltext A1dx1···dxn if the integral exists. This concept generalizes the volume of a closed rectangle deﬁned in (23.1). 23.2 Integrability Conditions In this section we describe some conditions under which a function deﬁned on an open subset of Rnis Riemann integrable. Deﬁnition 23.2. A set A⊂Rnis said to have measure zero if for every ε>0, there is a countable cover {Ri}∞ i=1ofAby closed rectangles Risuch that ∑∞ i=1vol(Ri)<ε. The most useful integrability criterion is the following theorem of Lebesgue [26, Theorem 8.3.1, p. 455]. Theorem 23.3 (Lebesgue’s theorem). A bounded function f :A→Ron a bounded subset A⊂Rnis Riemann integrable if and only if the set Disc(˜f)of discontinuities of the extended function ˜f has measure zero. Proposition 23.4. If a continuous function f :U→Rdeﬁned on an open subset U ofRnhas compact support, then f is Riemann integrable on U. Proof. Being continuous on a compact set, the function fis bounded. Being com- pact, the set supp fis closed and bounded in Rn. We claim that the extension ˜fis continuous. Since ˜fagrees with fonU, the extended function ˜fis continuous on U. It remains to show that ˜fis continuous on the complement of UinRnas well. If p/∈U, then p/∈supp f. Since supp fis a closed subset of Rn, there is an open ball B containing pand disjoint from supp f. On this open ball, ˜f≡0, which implies that ˜fis continuous at p/∈U. Thus, ˜fis continuous on Rn. By Lebesgue’s theorem, fis Riemann integrable on U. ⊓ ⊔ Example 23.5.The continuous function f:]−1,1[→R,f(x) =tan(πx/2), is de- ﬁned on an open subset of ﬁnite length in R, but is not bounded (Figure 23.2 ). The support of fis the open interval ]−1,1[, which is not compact. Thus, the function fdoes not satisfy the hypotheses of either Lebesgue’s theorem or Proposition 23.4. Note that it is not Riemann integrable. 23.3 The Integral of an n-Form on Rn263 xy 1−1 Fig. 23.2. The function f(x)=tan(πx/2)on]−1,1[. Remark. The support of a real-valued function is the closure in its domain of the subset where the function is not zero. In Example 23.5, the support of fis the open interval]−1,1[, not the closed interval [−1,1], because the domain of fis]−1,1[, notR. Deﬁnition 23.6. A subset A⊂Rnis called a domain of integration if it is bounded and its topological boundary bd (A)is a set of measure zero. Familiar plane ﬁgures such as triangles, rectangles, and circular disks are all domains of integration in R2. Proposition 23.7. Every bounded continuous function f deﬁned on a domain of in- tegration A in Rnis Riemann integrable over A. Proof. Let ˜f:Rn→Rbe the extension of fby zero. Since fis continuous on A, the extension ˜fis necessarily continuous at all interior points of A. Clearly, ˜fis continuous at all exterior points of Aalso, because every exterior point has a neigh- borhood contained entirely in Rn−A, on which ˜fis identically zero. Therefore, the set Disc(˜f)of discontinuities of ˜fis a subset of bd (A), a set of measure zero. By Lebesgue’s theorem, fis Riemann integrable on A. ⊓ ⊔ 23.3 The Integral of an n-Form on Rn Once a set of coordinates x1,..., xnhas been ﬁxed on Rn,n-forms on Rncan be identiﬁed with functions on Rn, since every n-form on Rncan be written as ω= f(x)dx1∧···∧ dxnfor a unique function f(x)onRn. In this way the theory of Riemann integration of functions on Rncarries over to n-forms on Rn. Deﬁnition 23.8. Letω=f(x)dx1∧···∧ dxnbe a C∞n-form on an open subset U⊂Rn, with standard coordinates x1,..., xn. Its integral over a subset A⊂Uis deﬁned to be the Riemann integral of f(x): /integraldisplay Aω=/integraldisplay Af(x)dx1∧···∧ dxn:=/integraldisplay Af(x)dx1···dxn, 264§23 Integration on Manifolds if the Riemann integral exists. In this deﬁnition the n-form must be written in the order dx1∧···∧ dxn. To integrate, for example, τ=f(x)dx2∧dx1over A⊂R2, one would write /integraldisplay Aτ=/integraldisplay A−f(x)dx1∧dx2=−/integraldisplay Af(x)dx1dx2. Example. Iffis a bounded continuous function deﬁned on a domain of integration AinRn, the the integral/integraltext Af dx1∧···∧ dxnexists by Proposition 23.7. Let us see how the integral of an n-form ω=f dx1∧···∧ dxnon an open subset U⊂Rntransforms under a change of variables. A change of variables on Uis given by a diffeomorphism T:Rn⊃V→U⊂Rn. Let x1,..., xnbe the standard coordinates on Uandy1,..., ynthe standard coordinates on V. Then Ti:=xi◦T= T∗(xi)is the ith component of T. We will assume that UandVare connected, and write x=(x1,..., xn)andy=(y1,..., yn). Denote by J(T)the Jacobian matrix [∂Ti/∂yj]. By Corollary 18.4(ii), dT1∧···∧ dTn=det(J(T))dy1∧···∧ dyn. Hence, /integraldisplay VT∗ω=/integraldisplay V(T∗f)T∗dx1∧···∧ T∗dxn(Proposition 18.11) =/integraldisplay V(f◦T)dT1∧···∧ dTn(because T∗d=dT∗) =/integraldisplay V(f◦T)det(J(T))dy1∧···∧ dyn =/integraldisplay V(f◦T)det(J(T))dy1···dyn. (23.4) On the other hand, the change-of-variables formula from advanced calculus gives /integraldisplay Uω=/integraldisplay Uf dx1···dxn=/integraldisplay V(f◦T)|det(J(T))|dy1···dyn, (23.5) with an absolute-value sign around the Jacobian determinant. Equations (23.4) and (23.5) differ by the sign of det (J(T)). Hence, /integraldisplay VT∗ω=±/integraldisplay Uω, (23.6) depending on whether the Jacobian determinant det (J(T))is positive or negative. By Proposition 21.8, a diffeomorphism T:Rn⊃V→U⊂Rnis orientation- preserving if and only if its Jacobian determinant det (J(T))is everywhere positive onV. Equation (23.6) shows that the integral of a differential form is not invari- ant under all diffeomorphisms of Vwith U, but only under orientation-preserving diffeomorphisms. 23.4 Integral of a Differential Form over a Manifold 265 23.4 Integral of a Differential Form over a Manifold Integration of an n-form onRnis not so different from integration of a function. Our approach to integration over a general manifold has several distinguishing features: (i) The manifold must be oriented (in fact, Rnhas a standard orientation). (ii) On a manifold of dimension n, one can integrate only n-forms, not functions. (iii) The n-forms must have compact support. LetMbe an oriented manifold of dimension n, with an oriented atlas {(Uα,φα)} giving the orientation of M. Denote by Ωk c(M)the vector space of C∞k-forms with compact support on M. Suppose{(U,φ)}is a chart in this atlas. If ω∈Ωn c(U)is an n-form with compact support on U, then because φ:U→φ(U)is a diffeomorphism, (φ−1)∗ωis an n-form with compact support on the open subset φ(U)⊂Rn. We deﬁne the integral of ωonUto be /integraldisplay Uω:=/integraldisplay φ(U)(φ−1)∗ω. (23.7) If(U,ψ)is another chart in the oriented atlas with the same U, then φ◦ ψ−1:ψ(U)→φ(U)is an orientation-preserving diffeomorphism, and so /integraldisplay φ(U)(φ−1)∗ω=/integraldisplay ψ(U)(φ◦ψ−1)∗(φ−1)∗ω=/integraldisplay ψ(U)(ψ−1)∗ω. Thus, the integral/integraltext Uωon a chart Uof the atlas is well deﬁned, independent of the choice of coordinates on U. By the linearity of the integral on Rn, ifω,τ∈ Ωn c(U), then/integraldisplay Uω+τ=/integraldisplay Uω+/integraldisplay Uτ. Now let ω∈Ωn c(M). Choose a partition of unity {ρα}subordinate to the open cover{Uα}. Because ωhas compact support and a partition of unity has locally ﬁnite supports, all except ﬁnitely many ραωare identically zero by Problem 18.6. In particular, ω=∑ αραω is aﬁnite sum. Since by Problem 18.4(b), supp(ραω)⊂supp ρα∩supp ω, supp(ραω)is a closed subset of the compact set supp ω. Hence, supp (ραω)is com- pact. Since ραωis an n-form with compact support in the chart Uα, its integral/integraltext Uαραωis deﬁned. Therefore, we can deﬁne the integral of ωover Mto be the ﬁnite sum /integraldisplay Mω:=∑ α/integraldisplay Uαραω. (23.8) For this integral to be well deﬁned, we must show that it is independent of the choices of oriented atlas and partition of unity. Let {Vβ}be another oriented atlas 266§23 Integration on Manifolds ofMspecifying the orientation of M, and{χβ}a partition of unity subordinate to {Vβ}. Then{(Uα∩Vβ,φα|Uα∩Uβ)}and{(Uα∩Vβ,ψβ|Uα∩Uβ)}are two new atlases ofMspecifying the orientation of M, and ∑ α/integraldisplay Uαραω=∑ α/integraldisplay Uαρα∑ βχβω(because ∑ βχβ=1) =∑ α∑ β/integraldisplay Uαραχβω(these are ﬁnite sums) =∑ α∑ β/integraldisplay Uα∩Vβραχβω, where the last line follows from the fact that the support of ραχβis contained in Uα∩Vβ. By symmetry, ∑β/integraltext Vβχβωis equal to the same sum. Hence, ∑ α/integraldisplay Uαραω=∑ β/integraldisplay Vβχβω, proving that the integral (23.8) is well deﬁned. Proposition 23.9. Letωbe an n-form with compact support on an oriented man- ifold M of dimension n. If −M denotes the same manifold but with the opposite orientation, then/integraltext −Mω=−/integraltext Mω. Thus, reversing the orientation of Mreverses the sign of an integral over M. Proof. By the deﬁnition of an integral ((23.7) and (23.8)), it is enough to show that for every chart (U,φ)=( U,x1,..., xn)and differential form τ∈Ωn c(U), if(U,¯φ)= (U,−x1,x2,..., xn)is the chart with the opposite orientation, then /integraldisplay ¯φ(U)/parenleftbig¯φ−1/parenrightbig∗τ=−/integraldisplay φ(U)/parenleftbig φ−1/parenrightbig∗τ. Letr1,..., rnbe the standard coordinates on Rn. Then xi=ri◦φandri=xi◦φ−1. With ¯φ, the only difference is that when i=1, −x1=r1◦¯φand r1=−x1◦¯φ−1. Suppose τ=f dx1∧···∧ dxnonU. Then (¯φ−1)∗τ=(f◦¯φ−1)d(x1◦¯φ−1)∧d(x2◦¯φ−1)∧···∧ d(xn◦¯φ−1) =−(f◦¯φ−1)dr1∧dr2∧···∧ drn. (23.9) Similarly,/parenleftbig φ−1/parenrightbig∗τ=/parenleftbig f◦φ−1/parenrightbig dr1∧dr2∧···∧ drn. Since φ◦¯φ−1:¯φ(U)→φ(U)is given by 23.4 Integral of a Differential Form over a Manifold 267 (φ◦¯φ−1)(a1,a2,..., an)=(−a1,a2,..., an), the absolute value of its Jacobian determinant is |J(φ◦¯φ−1)|=|−1|=1. (23.10) Therefore, /integraldisplay ¯φ(U)/parenleftbig¯φ−1/parenrightbig∗τ=−/integraldisplay ¯φ(U)/parenleftbig f◦¯φ−1/parenrightbig dr1···drn(by (23.9)) =−/integraldisplay ¯φ(U)(f◦φ−1)◦(φ◦¯φ−1)|J(φ◦¯φ−1)|dr1···drn(by (23.10)) =−/integraldisplay φ(U)(f◦φ−1)dr1···drn(by the change-of-variables formula) =−/integraldisplay φ(U)/parenleftbig φ−1/parenrightbig∗τ. ⊓ ⊔ The treatment of integration above can be extended almost word for word to oriented manifolds with boundary. It has the virtue of simplicity and is of great utility in proving theorems. However, it is not practical for actual computation of integrals; an n-form multiplied by a partition of unity can rarely be integrated as a closed expression. To calculate explicitly integrals over an oriented n-manifold M, it is best to consider integrals over a parametrized set. Deﬁnition 23.10. Aparametrized set in an oriented n-manifold Mis a subset A together with a C∞map F:D→Mfrom a compact domain of integration D⊂Rnto Msuch that A=F(D)andFrestricts to an orientation-preserving diffeomorphism from int(D)toF(int(D)). Note that by smooth invariance of domain for manifolds (Remark 22.5), F(int(D))is an open subset of M. The C∞map F:D→Ais called aparametrization ofA. IfAis a parametrized set in Mwith parametrization F:D→Aandωis aC∞n- form on M, not necessarily with compact support, then we deﬁne/integraltext Aωto be/integraltext DF∗ω. It can be shown that the deﬁnition of/integraltext Aωis independent of the parametrization and that in case Ais a manifold, it agrees with the earlier deﬁnition of integration over a manifold. Subdividing an oriented manifold into a union of parametrized sets can be an effective method of calculating an integral over the manifold. We will not delve into this theory of integration (see [31, Theorem 25.4, p. 213] or [25,Proposition 14.7, p. 356]), but will content ourselves with an example. Example 23.11 ( Integral over a sphere) .In spherical coordinates, ρis the distance/radicalbig x2+y2+z2of the point (x,y,z)∈R3to the origin, ϕis the angle that the vector /an}bracke⊔le{⊔x,y,z/an}bracke⊔ri}h⊔makes with the positive z-axis, and θis the angle that the vector /an}bracke⊔le{⊔x,y/an}bracke⊔ri}h⊔in the (x,y)-plane makes with the positive x-axis (Figure 23.3(a) ). Let ωbe the 2-form on the unit sphere S2inR3given by 268§23 Integration on Manifolds θϕρ xyz (a) Spherical coordinates in R3θϕ 2ππF (b) A parametrization by spherical coordinates Fig. 23.3. The sphere as a parametrized set. ω=  dy∧dz xforx/ne}a⊔ionslash=0, dz∧dx yfory/ne}a⊔ionslash=0, dx∧dy zforz/ne}a⊔ionslash=0. Calculate/integraldisplay S2ω. Up to a factor of 2, the form ωis the 2-form on S2from Problem 19.11(b). In Riemannian geometry, it is shown that ωis the area form of the sphere S2with respect to the Euclidean metric. Therefore, the integral/integraldisplay S2ωis the surface area of the sphere. Solution. The sphere S2has a parametrization by spherical coordinates ( Figure 23.3(b) ): F(ϕ,θ)=( sinϕcosθ,sinϕsinθ,cosϕ) onD={(ϕ,θ)∈R2|0≤ϕ≤π,0≤θ≤2π}. Since F∗x=sinϕcosθ,F∗y=sinϕsinθ,and F∗z=cosϕ, we have F∗dy=dF∗y=cosϕsinθdϕ+sinϕcosθdθ and F∗dz=−sinϕdϕ, so for x/ne}a⊔ionslash=0, F∗ω=F∗dy∧F∗dz F∗x=sinϕdϕ∧dθ. 23.5 Stokes’s Theorem 269 Fory/ne}a⊔ionslash=0 and z/ne}a⊔ionslash=0, similar calculations show that F∗ωis given by the same formula. Therefore, F∗ω=sinϕdϕ∧dθeverywhere on D, and /integraldisplay S2ω=/integraldisplay DF∗ω=/integraldisplay2π 0/integraldisplayπ 0sinϕdϕdθ=2π/bracketleftbig −cosϕ/bracketrightbigπ 0=4π.⊓ ⊔ Integration over a zero-dimensional manifold The discussion of integration so far assumes implicitly that the manifold Mhas di- mension n≥1. We now treat integration over a zero-dimensional manifold. A com- pact oriented manifold Mof dimension 0 is a ﬁnite collection of points, each point oriented by +1 or−1. We write this as M=∑pi−∑qj. The integral of a 0-form f:M→Ris deﬁned to be the sum /integraldisplay Mf=∑f(pi)−∑f(qj). 23.5 Stokes’s Theorem LetMbe an oriented manifold of dimension nwith boundary. We give its boundary ∂Mthe boundary orientation and let i:∂M֒→Mbe the inclusion map. If ωis an (n−1)-form on M, it is customary to write/integraltext ∂Mωinstead of/integraltext ∂Mi∗ω. Theorem 23.12 (Stokes’s theorem). For any smooth (n−1)-form ωwith compact support on the oriented n-dimensional manifold M, /integraldisplay Mdω=/integraldisplay ∂Mω. Proof. Choose an atlas{(Uα,φα)}forMin which each Uαis diffeomorphic to either RnorHnvia an orientation-preserving diffeomorphism. This is possible since any open disk is diffeomorphic to Rnand any half-disk containing its boundary diameter is diffeomorphic to Hn(see Problem 1.5). Let {ρα}be a C∞partition of unity subordinate to{Uα}. As we showed in the preceding section, the (n−1)-form ραω has compact support in Uα. Suppose Stokes’s theorem holds for Rnand forHn. Then it holds for all the charts Uαin our atlas, which are diffeomorphic to RnorHn. Also, note that (∂M)∩Uα=∂Uα. Therefore, 270§23 Integration on Manifolds /integraldisplay ∂Mω=/integraldisplay ∂M∑ αραω/parenleftbigg ∑ αρα=1/parenrightbigg =∑ α/integraldisplay ∂Mραω/parenleftbigg ∑ αραωis a ﬁnite sum by Problem 18.6/parenrightbigg =∑ α/integraldisplay ∂Uαραω(supp ραω⊂Uα) =∑ α/integraldisplay Uαd(ραω) (Stokes’s theorem for Uα) =∑ α/integraldisplay Md(ραω) ( supp d(ραω)⊂Uα) =/integraldisplay Md/parenleftbig∑ραω/parenrightbig/parenleftbigg ∑ αραωis a ﬁnite sum/parenrightbigg =/integraldisplay Mdω. Thus, it sufﬁces to prove Stokes’s theorem for Rnand forHn. We will give a proof only for H2, since the general case is similar (see Problem 23.4). Proof of Stokes’s theorem for the upper half-plane H2.Letx,ybe the coordinates onH2. Then the standard orientation on H2is given by dx∧dy, and the bound- ary orientation on ∂H2is given by ι−∂/∂y(dx∧dy)=dx. The form ωis a linear combination ω=f(x,y)dx+g(x,y)dy (23.11) forC∞functions f,gwith compact support in H2. Since the supports of fandgare compact, we may choose a real number a>0 large enough that the supports of fand gare contained in the interior of the square [−a,a]×[0,a]. We will use the notation fx,fyto denote the partial derivatives of fwith respect to xandy, respectively. Then dω=/parenleftbigg∂g ∂x−∂f ∂y/parenrightbigg dx∧dy=(gx−fy)dx∧dy, and /integraldisplay H2dω=/integraldisplay H2gxdxdy−/integraldisplay H2fydxdy =/integraldisplay∞ 0/integraldisplay∞ −∞gxdxdy−/integraldisplay∞ −∞/integraldisplay∞ 0fydydx =/integraldisplaya 0/integraldisplaya −agxdxdy−/integraldisplaya −a/integraldisplaya 0fydydx. (23.12) In this expression,/integraldisplaya −agx(x,y)dx=g(x,y)/bracketrightbiga x=−a=0 because supp glies in the interior of [−a,a]×[0,a]. Similarly, 23.6 Line Integrals and Green’s Theorem 271 /integraldisplaya 0fy(x,y)dy=f(x,y)/bracketrightbiga y=0=−f(x,0) because f(x,a)=0. Thus, (23.12) becomes /integraldisplay H2dω=/integraldisplaya −af(x,0)dx. On the other hand, ∂H2is the x-axis and dy=0 on ∂H2. It follows from (23.11) thatω=f(x,0)dxwhen restricted to ∂H2and /integraldisplay ∂H2ω=/integraldisplaya −af(x,0)dx. This proves Stokes’s theorem for the upper half-plane. ⊓ ⊔ 23.6 Line Integrals and Green’s Theorem We will now show how Stokes’s theorem for a manifold uniﬁes some of the theorems of vector calculus on R2andR3. Recall the calculus notation F·dr=Pdx+Qdy+ Rdz forF=/an}bracke⊔le{⊔P,Q,R/an}bracke⊔ri}h⊔andr= (x,y,z). As in calculus, we assume in this section that functions, vector ﬁelds, and regions of integration have sufﬁcient smoothness orregularity properties so that all the integrals are deﬁned. Theorem 23.13 (Fundamental theorem for line integrals). Let C be a curve in R 3, parametrized by r(t)=( x(t),y(t),z(t)), a≤t≤b, and let Fbe a vector ﬁeld on R3. IfF=grad f for some scalar function f , then /integraldisplay CF·dr=f(r(b))−f(r(a)). Suppose in Stokes’s theorem we take Mto be a curve Cwith parametrization r(t),a≤t≤b, and ωto be the function fonC. Then /integraldisplay Cdω=/integraldisplay Cdf=/integraldisplay C∂f ∂xdx+∂f ∂ydy+∂f ∂zdz=/integraldisplay Cgrad f·dr and /integraldisplay ∂Cω=f/bracketrightbigr(b) r(a)=f(r(b))−f(r(a)). In this case Stokes’s theorem specializes to the fundamental theorem for line inte- grals. Theorem 23.14 (Green’s theorem). If D is a plane region with boundary ∂D, and P and Q are C∞functions on D, then /integraldisplay ∂DPdx+Qdy=/integraldisplay D/parenleftbigg∂Q ∂x−∂P ∂y/parenrightbigg dA. 272§23 Integration on Manifolds In this statement, dAis the usual calculus notation for dxdy . To obtain Green’s theorem, let Mbe a plane region Dwith boundary ∂Dand let ωbe the 1-form Pdx+Qdy onD. Then/integraldisplay ∂Dω=/integraldisplay ∂DPdx+Qdy and /integraldisplay Ddω=/integraldisplay DPydy∧dx+Qxdx∧dy=/integraldisplay D(Qx−Py)dx∧dy =/integraldisplay D(Qx−Py)dxdy=/integraldisplay D(Qx−Py)dA. In this case Stokes’s theorem is Green’s theorem in the plane. Problems 23.1. Area of an ellipse Use the change-of-variables formula to compute the area enclosed by the ellipse x2/a2+y2/b2=1 inR2. 23.2. Characterization of boundedness in Rn Prove that a subset A⊂Rnis bounded if and only if its closure ¯AinRnis compact. 23.3.* Integral under a diffeomorphism Suppose NandMare connected, oriented n-manifolds and F:N→Mis a diffeomorphism. Prove that for any ω∈Ωkc(M),/integraldisplay NF∗ω=±/integraldisplay Mω, where the sign depends on whether Fis orientation-preserving or orientation-reversing. 23.4.* Stokes’s theorem Prove Stokes’s theorem for Rnand forHn. 23.5. Area form on the sphere S2 Prove that the area form ωonS2in Example 23.11 is equal to the orientation form xdy∧dz−ydx∧dz+zdx∧dy ofS2in Problem 22.9. Chapter 7 De Rham Theory Henri Poincar´ e (1854–1912)By the fundamental theorem for line integrals (Theorem 23.13), if a smooth vector ﬁeld Fis the gradient of a scalar function f, then for any two points pandqinR3, the line integral/integraltext CF·drover a curve Cfrom ptoqis independent of the curve. In this case, the line integral/integraltext CF·drcan be com- puted in terms of its values at the two endpoints as f(q)−f(p). Similarly, by the classical Stokes the- orem for a surface, the surface integral of smooth avector ﬁeld Fover an oriented surface Swith bound- aryCinR 3can be evaluated as an integral over the curve CifFis the curl of another vector ﬁeld. It is thus of interest to know whether a vector ﬁeld R3 is the gradient of a function or is the curl of an- other vector ﬁeld. By the correspondence of Sec-tion 4.6 between vector ﬁelds and differential forms, this translates into whether a differential form ωon R3is exact. Considerations such as these led Henri Poincar´ e to look for conditions under which a differential form is exact on Rn. Of course, a necessary condition is that the form ωbe closed. Poincar´ e proved in 1887 that for k=1,2,3, ak-form on Rn is exact if and only if it is closed, a lemma that now bears his name. Vito V olterra published in 1889 the ﬁrst complete proof of the Poincar´ e lemma for all k. It turns out that whether every closed form on a manifold is exact depends on the topology of the manifold. For example, on R2every closed k-form is exact for k>0, but on the punctured plane R2−{(0,0)}there are closed 1-forms that are not exact. The extent to which closed forms are not exact is measured by the de Rham cohomology, possibly the most important diffeomorphism invariant of a manifold. © Springer Science+Business Media, LLC 2011L.W. Tu, An Introduction to Manifolds, Universitext, DOI 10.1007/978-1-4419-7400-6_7, 273 274§24 De Rham Cohomology Georges de Rham (1903–1990)In a series of groundbreaking papers, starting with “Analysis situs” [33] in 1895, Poincar´ e intro-duced the concept of homology and laid the founda- tions of modern algebraic topology. Roughly speak- ing, a compact submanifold with no boundary isacycle , and a cycle is homologous to zero if it is the boundary of another manifold. The equivalence classes of cycles under the homology relation are called homology classes . In his doctoral thesis [8] in 1931, Georges de Rham showed that differentialforms satisfy the same axioms as cycles and bound- aries, in effect proving a duality between what are now called de Rham cohomology and singular ho-mology with real coefﬁcients. Although he did not deﬁne explicitly de Rham cohomology in this paper, it was implicit in his work. A formal deﬁnition of de Rham cohomology appeared in 1938 [9]. §24 De Rham Cohomology In this section we deﬁne de Rham cohomology, prove some of its basic properties, and compute two elementary examples: the de Rham cohomology vector spaces ofthe real line and of the unit circle. 24.1 De Rham Cohomology Suppose F(x,y)=/an}bracke⊔le{⊔P(x,y),Q(x,y)/an}bracke⊔ri}h⊔is a smooth vector ﬁeld representing a force on an open subset UofR2, and Cis a parametrized curve c(t)=( x(t),y(t))inUfrom a point pto a point q, with a≤t≤b. Then the work done by the force in moving a particle from ptoqalong Cis given by the line integral/integraltext CPdx+Qdy. Such a line integral is easy to compute if the vector ﬁeld Fis the gradient of a scalar function f(x,y): F=grad f=/an}bracke⊔le{⊔fx,fy/an}bracke⊔ri}h⊔, where fx=∂f/∂xandfy=∂f/∂y. By Stokes’s theorem, the line integral is simply /integraldisplay Cfxdx+fydy=/integraldisplay Cdf=f(q)−f(p). A necessary condition for the vector ﬁeld F=/an}bracke⊔le{⊔P,Q/an}bracke⊔ri}h⊔to be a gradient is that Py=fxy=fyx=Qx. The question is now the following: if Py−Qx=0, is the vector ﬁeld F=/an}bracke⊔le{⊔P,Q/an}bracke⊔ri}h⊔on Uthe gradient of some scalar function f(x,y)onU? 24.1 De Rham Cohomology 275 In Section 4.6 we established a one-to-one correspondence between vector ﬁelds and differential 1-forms on an open subset of R3. There is a similar correspondence on an open subset of any Rn. ForR2, it is as follows: vector ﬁelds←→ differential 1-forms , F=/an}bracke⊔le{⊔P,Q/an}bracke⊔ri}h⊔←→ ω=Pdx+Qdy, grad f=/an}bracke⊔le{⊔fx,fy/an}bracke⊔ri}h⊔←→ df=fxdx+fydy, Qx−Py=0←→ dω=(Qx−Py)dx∧dy=0. In terms of differential forms the question above becomes the following: if the 1- form ω=Pdx+Qdy is closed on U, is it exact? The answer to this question is sometimes yes and sometimes no, depending on the topology of U. Just as for an open subset of Rn, a differential form ωon a manifold Mis said to beclosed ifdω=0, and exact ifω=dτfor some form τof degree one less. Since d2=0, every exact form is closed. In general, not every closed form is exact. LetZk(M)be the vector space of all closed k-forms and Bk(M)the vector space of all exact k-forms on the manifold M. Because every exact form is closed, Bk(M)is a subspace of Zk(M). The quotient vector space Hk(M):=Zk(M)/Bk(M)measures the extent to which closed k-forms fail to be exact, and is called the de Rham co- homology ofMin degree k. As explained in Appendix D, the quotient vector space construction introduces an equivalence relation on Zk(M): ω′∼ωinZk(M)iff ω′−ω∈Bk(M). The equivalence class of a closed form ωis called its cohomology class and denoted by[ω]. Two closed forms ωandω′determine the same cohomology class if and only if they differ by an exact form: ω′=ω+dτ. In this case we say that the two closed forms ωandω′arecohomologous . Proposition 24.1. If the manifold M has r connected components, then its de Rham cohomology in degree 0is H0(M) =Rr. An element of H0(M)is speciﬁed by an ordered r-tuple of real numbers, each real number representing a constant function on a connected component of M. Proof. Since there are no nonzero exact 0-forms, H0(M)=Z0(M)={closed 0-forms}. Supposed fis a closed 0-form on M; i.e., fis aC∞function on Msuch that df=0. On any chart (U,x1,..., xn), df=∑∂f ∂xidxi. 276§24 De Rham Cohomology Thus, df=0 on Uif and only if all the partial derivatives ∂f/∂xivanish identically onU. This in turn is equivalent to fbeing locally constant on U. Hence, the closed 0-forms on Mare precisely the locally constant functions on M. Such a function must be constant on each connected component of M. IfMhasrconnected components, then a locally constant function on Mcan be speciﬁed by an ordered set of rreal numbers. Thus, Z0(M)=Rr. ⊓ ⊔ Proposition 24.2. On a manifold M of dimension n, the de Rham cohomology Hk(M) vanishes for k >n. Proof. At any point p∈M, the tangent space TpMis a vector space of dimension n. Ifωis ak-form on M, then ωp∈Ak(TpM), the space of alternating k-linear functions onTpM. By Corollary 3.31, if k>n, then Ak(TpM)=0. Hence, for k>n, the only k-form on Mis the zero form. ⊓ ⊔ 24.2 Examples of de Rham Cohomology Example 24.3 ( De Rham cohomology of the real line ).Since the real line R1is con- nected, by Proposition 24.1, H0(R1)=R. For dimensional reasons, on R1there are no nonzero 2-forms. This implies that every 1-form on R1is closed. A 1-form f(x)dxonR1is exact if and only if there is aC∞function g(x)onR1such that f(x)dx=dg=g′(x)dx, where g′(x)is the calculus derivative of gwith respect to x. Such a function g(x)is simply an antiderivative of f(x), for example g(x)=/integraldisplayx 0f(t)dt. This proves that every 1-form on R1is exact. Therefore, H1(R1)=0. In combination with Proposition 24.2, we have Hk(R1)=/braceleftigg Rfork=0, 0 for k≥1. Example 24.4 ( De Rham cohomology of a circle ).LetS1be the unit circle in the xy-plane. By Proposition 24.1, because S1is connected, H0(S1) =R, and because S1is one-dimensional, Hk(S1)=0 for all k≥2. It remains to compute H1(S1). Recall from Subsection 18.7 the map h:R→S1,h(t) = ( cost,sint). Let i:[0,2π]→Rbe the inclusion map. Restricting the domain of hto[0,2π]gives a parametrization F:=h◦i:[0,2π]→S1of the circle. In Examples 17.15 and 17.16, we found a nowhere-vanishing 1-form ω=−ydx+xdy onS1and showed thatF∗ω=i∗h∗ω=i∗dt=dt. Thus, 24.2 Examples of de Rham Cohomology 277 /integraldisplay S1ω=/integraldisplay F([0,2π])ω=/integraldisplay [0,2π]F∗ω=/integraldisplay2π 0dt=2π. Since the circle has dimension 1, all 1-forms on S1are closed, so Ω1(S1) = Z1(S1). The integration of 1-forms on S1deﬁnes a linear map ϕ:Z1(S1)=Ω1(S1)→R,ϕ(α)=/integraldisplay S1α. Because ϕ(ω)=2π/ne}a⊔ionslash=0, the linear map ϕ:Ω1(S1)→Ris onto. By Stokes’s theorem, the exact 1-forms on S1are in ker ϕ. Conversely, we will show that all 1-forms in ker ϕare exact. Suppose α=fωis a smooth 1-form on S1 such that ϕ(α)=0. Let ¯f=h∗f=f◦h∈Ω0(R). Then ¯fis periodic of period 2 π and 0=/integraldisplay S1α=/integraldisplay F([0,2π])α=/integraldisplay [0,2π]F∗α=/integraldisplay [0,2π](i∗h∗f)(t)·F∗ω=/integraldisplay2π 0¯f(t)dt. Lemma 24.5. Suppose ¯f is a C∞periodic function of period 2πonRand/integraltext2π 0¯f(u)du= 0. Then ¯f dt=d¯g for a C∞periodic function ¯g of period 2πonR. Proof. Deﬁne ¯ g∈Ω0(R)by ¯g(t)=/integraldisplayt 0¯f(u)du. Since/integraltext2π 0¯f(u)du=0 and ¯fis periodic of period 2 π, ¯g(t+2π)=/integraldisplay2π 0¯f(u)du+/integraldisplayt+2π 2π¯f(u)du =0+/integraldisplayt+2π 2π¯f(u)du=/integraldisplayt 0¯f(u)du=¯g(t). Hence, ¯ g(t)is also periodic of period 2 πonR. Moreover, d¯g=¯g′(t)dt=¯f(t)dt. ⊓ ⊔ Let ¯gbe the periodic function of period 2 πonRfrom Lemma 24.5. By Proposi- tion 18.12, ¯ g=h∗gfor some C∞function gonS1. It follows that d¯g=dh∗g=h∗(dg). On the other hand, ¯f(t)dt=(h∗f)(h∗ω)=h∗(fω)=h∗α. Since h∗:Ω1(S1)→Ω1(R)is injective, α=dg. This proves that the kernel of ϕ consists of exact forms. Therefore, integration induces an isomorphism H1(S1)=Z1(S1) B1(S1)∼→R. In the next section we will develop a tool, the Mayer–Vietoris sequence, using which the computation of the cohomology of the circle becomes more or less routine. 278§24 De Rham Cohomology 24.3 Diffeomorphism Invariance For any smooth map F:N→Mof manifolds, there is a pullback map F∗:Ω∗(M)→ Ω∗(N)of differential forms. Moreover, the pullback F∗commutes with the exterior derivative d(Proposition 19.5). Lemma 24.6. The pullback map F∗sends closed forms to closed forms, and sends exact forms to exact forms. Proof. Suppose ωis closed. By the commutativity of F∗with d, dF∗ω=F∗dω=0. Hence, F∗ωis also closed. Next suppose ω=dτis exact. Then F∗ω=F∗dτ=dF∗τ. Hence, F∗ωis exact. ⊓ ⊔ It follows that F∗induces a linear map of quotient spaces, denoted by F#: F#:Zk(M) Bk(M)→Zk(N) Bk(N),F#([ω])=[F∗(ω)]. This is a map in cohomology, F#:Hk(M)→Hk(N), called the pullback map in cohomology . Remark 24.7.The functorial properties of the pullback map F∗on differential forms easily yield the same functorial properties for the induced map in cohomology: (i) If1M:M→Mis the identity map, then 1# M:Hk(M)→Hk(M)is also the iden- tity map. (ii) If F:N→MandG:M→Pare smooth maps, then (G◦F)#=F#◦G#. It follows from (i) and (ii) that (Hk( ),F#)is a contravariant functor from the category of C∞manifolds and C∞maps to the category of vector spaces and linear maps. By Proposition 10.3, if F:N→Mis a diffeomorphism of manifolds, then F#:Hk(M)→Hk(N)is an isomorphism of vector spaces. In fact, the usual notation for the induced map in cohomology is F∗, the same as for the pullback map on differential forms. Unless there is a possibility of confu- sion, henceforth we will follow this convention. It is usually clear from the context whether F∗is a map in cohomology or on forms. 24.4 The Ring Structure on de Rham Cohomology 279 24.4 The Ring Structure on de Rham Cohomology The wedge product of differential forms on a manifold Mgives the vector space Ω∗(M)of differential forms a product structure. This product structure induces a product structure in cohomology: if [ω]∈Hk(M)and[τ]∈Hℓ(M), deﬁne [ω]∧[τ]=[ω∧τ]∈Hk+ℓ(M). (24.1) For the product to be well deﬁned, we need to check three things about closed forms ωandτ: (i) The wedge product ω∧τis a closed form. (ii) The class [ω∧τ]is independent of the choice of representative for [τ]. In other words, if τis replaced by a cohomologous form τ′=τ+dσ, then in the equa- tion ω∧τ′=ω∧τ+ω∧dσ, we need to show that ω∧dσis exact. (iii) The class [ω∧τ]is independent of the choice of representative for [ω]. These all follow from the antiderivation property of d. For example, in (i), since ωandτare closed, d(ω∧τ)=( dω)∧τ+(−1)kω∧dτ=0. In (ii), d(ω∧σ)=( dω)∧σ+(−1)kω∧dσ=(−1)kω∧dσ(since dω=0), which shows that ω∧dσis exact. Item (iii) is analogous to (ii), with the roles of ω andτreversed. IfMis a manifold of dimension n, we set H∗(M)=n/circleplusdisplay k=0Hk(M). What this means is that an element αofH∗(M)is uniquely a ﬁnite sum of cohomol- ogy classes in Hk(M)for various k’s: α=α0+···+αn,αk∈Hk(M). Elements of H∗(M)can be added and multiplied in the same way that one would add or multiply polynomials, except here multiplication is the wedge product. It is easy to check that under addition and multiplication, H∗(M)satisﬁes all the properties of a ring, called the cohomology ring ofM. The ring H∗(M)has a natural grading by the degree of a closed form. Recall that a ring Aisgraded if it can be written as a direct sum A=/circleplustext∞ k=0Akso that the ring multiplication sends Ak×AℓtoAk×ℓ. A graded ring A=/circleplustext∞ k=0Akis said to be anticommutative if for all a∈Akandb∈Aℓ, 280§24 De Rham Cohomology a·b=(−1)kℓb·a. In this terminology, H∗(M)is an anticommutative graded ring. Since H∗(M)is also a real vector space, it is in fact an anticommutative graded algebra over R. Suppose F:N→Mis aC∞map of manifolds. Because F∗(ω∧τ) =F∗ω∧ F∗τfor differential forms ωand τonM(Proposition 18.11), the linear map F∗:H∗(M)→H∗(N)is a ring homomorphism. By Remark 24.7, if F:N→M is a diffeomorphism, then the pullback F∗:H∗(M)→H∗(N)is a ring isomorphism. To sum up, de Rham cohomology gives a contravariant functor from the cate- gory of C∞manifolds to the category of anticommutative graded rings. If Mand Nare diffeomorphic manifolds, then H∗(M)andH∗(N)are isomorphic as anticom- mutative graded rings. In this way the de Rham cohomology becomes a powerful diffeomorphism invariant of C∞manifolds. Problems 24.1. Nowhere-vanishing 1-forms Prove that a nowhere-vanishing 1-form on a compact manifold cannot be exact. 24.2. Cohomology in degree zero Suppose a manifold Mhas inﬁnitely many connected components. Compute its de Rham cohomology vector space H0(M)in degree 0. ( Hint: By second countability, the number of connected components of a manifold is countable.) 25.1 Exact Sequences 281 §25 The Long Exact Sequence in Cohomology Acochain complex Cis a collection of vector spaces {Ck}k∈Ztogether with a se- quence of linear maps dk:Ck→Ck+1, ···→ C−1d−1−→C0d0−→C1d1−→C2d2−→···, such that dk◦dk−1=0 (25.1) for all k. We will call the collection of linear maps {dk}thedifferential of the cochain complexC. The vector space Ω∗(M)of differential forms on a manifold Mtogether with the exterior derivative dis a cochain complex, the de Rham complex ofM: 0→Ω0(M)d→Ω1(M)d→Ω2(M)d→···,d◦d=0. It turns out that many of the results on the de Rham cohomology of a manifold de- pend not on the topological properties of the manifold, but on the algebraic properties of the de Rham complex. To better understand de Rham cohomology, it is useful toisolate these algebraic properties. In this section we investigate the properties of a cochain complex that constitute the beginning of a subject known as homological algebra . 25.1 Exact Sequences This subsection is a compendium of a few basic properties of exactness that will be used over and over again. Deﬁnition 25.1. A sequence of homomorphisms of vector spaces Af→Bg→C is said to be exact at B if im f=kerg. A sequence of homomorphisms A0f0−→A1f1−→A2f2−→···fn−1−→An that is exact at every term except the ﬁrst and the last is simply said to be an exact sequence . A ﬁve-term exact sequence of the form 0→A→B→C→0 is said to be short exact . The same deﬁnition applies to homomorphisms of groups or modules, but we are mainly concerned with vector spaces. 282§25 The Long Exact Sequence in Cohomology Remark. (i) When A=0, the sequence 0f→Bg→C is exact if and only if kerg=imf=0, so that gis injective. (ii) Similarly, when C=0, the sequence Af→Bg→0 is exact if and only if imf=kerg=B, so that fis surjective. The following two propositions are very useful for dealing with exact sequences. Proposition 25.2 (A three-term exact sequence). Suppose Af→Bg→C is an exact sequence. Then (i)the map f is surjective if and only if g is the zero map; (ii)the map g is injective if and only if f is the zero map. Proof. Problem 25.1. ⊓ ⊔ Proposition 25.3 (A four-term exact sequence). (i)The four-term sequence 0→Af→B→0of vector spaces is exact if and only if f:A→B is an isomorphism. (ii)If Af→B→C→0 is an exact sequence of vector spaces, then there is a linear isomorphism C≃coker f:=B imf. Proof. Problem 25.2. ⊓ ⊔ 25.2 Cohomology of Cochain Complexes 283 25.2 Cohomology of Cochain Complexes IfCis a cochain complex, then by (25.1), imdk−1⊂kerdk. We can therefore form the quotient vector space Hk(C):=kerdk imdk−1, which is called the kth cohomology vector space of the cochain complex C. It is a measure of the extent to which the cochain complex Cfails to be exact at Ck. Elements of the vector space Ckare called cochains of degree k ork-cochains for short. A k-cochain in ker dkis called a k-cocycle and a k-cochain in im dk−1is called ak-coboundary . The equivalence class [c]∈Hk(C)of ak-cocycle c∈kerdkis called itscohomology class . We denote the subspaces of k-cocycles and k-coboundaries of CbyZk(C)andBk(C)respectively. The letter Zfor cocycles comes from Zyklen , the German word for cycles. To simplify the notation we will usually omit the subscript in dk, and write d◦ d=0 instead of dk◦dk−1=0. Example. In the de Rham complex, a cocycle is a closed form and a coboundary is an exact form. IfAandBare two cochain complexes with differentials dandd′respectively, a cochain map ϕ:A→Bis a collection of linear maps ϕk:Ak→Bk, one for each k, that commute with dandd′: d′◦ϕk=ϕk+1◦d. In other words, the following diagram is commutative: ··· /d47/d47Ak−1d/d47/d47 ϕk−1 /d15/d15Akd/d47/d47 ϕk /d15/d15Ak+1 /d47/d47 ϕk+1 /d15/d15··· ··· /d47/d47Bk−1 d′/d47/d47Bk d′/d47/d47Bk+1 /d47/d47···. We will usually omit the subscript kinϕk. A cochain map ϕ:A→Bnaturally induces a linear map in cohomology ϕ∗:Hk(A)→Hk(B) by ϕ∗[a]=[ϕ(a)]. (25.2) To show that this is well deﬁned, we need to check that a cochain map takes cocycles to cocycles, and coboundaries to coboundaries: 284§25 The Long Exact Sequence in Cohomology (i) for a∈Zk(A),d′(ϕ(a))= ϕ(da)=0; (ii) for a′∈Ak−1,ϕ(da′)=d′(ϕ(a′)). Example 25.4. (i) For a smooth map F:N→Mof manifolds, the pullback map F∗:Ω∗(M)→ Ω∗(N)on differential forms is a cochain map, because F∗commutes with d (Proposition 19.5). By the discussion above, there is an induced map F∗:H∗(M)→ H∗(N)in cohomology, as we saw once before, after Lemma 24.6. (ii) If Xis aC∞vector ﬁeld on a manifold M, then the Lie derivative LX:Ω∗(M)→ Ω∗(M)commutes with d(Theorem 20.10(ii)). By (25.2), LXinduces a linear mapL∗ X:H∗(M)→H∗(M)in cohomology. 25.3 The Connecting Homomorphism A sequence of cochain complexes 0→Ai→Bj→C→0 isshort exact ifiandjare cochain maps and for each k, 0→Akik→Bkjk→Ck→0 is a short exact sequence of vector spaces. Since we usually omit subscripts on cochain maps, we will write i,jinstead of ik,jk. Given a short exact sequence as above, we can construct a linear map d∗:Hk(C) →Hk+1(A), called the connecting homomorphism , as follows. Consider the short exact sequences in dimensions kandk+1: 0/d47/d47Ak+1i/d47/d47Bk+1j/d47/d47Ck+1 /d47/d470 0/d47/d47Ak i/d47/d47d/d79/d79 Bk j/d47/d47d/d79/d79 Ck /d47/d47d/d79/d79 0. To keep the notation simple, we use the same symbol dto denote the a priori distinct differentials dA,dB,dCof the three cochain complexes. Start with [c]∈Hk(C). Since j:Bk→Ckis onto, there is an element b∈Bksuch that j(b)=c. Then db∈Bk+1is in ker jbecause jdb=djb(by the commutativity of the diagram ) =dc=0(because cis a cocycle ). By the exactness of the sequence in degree k+1, ker j=imi. This implies that db=i(a)for some ainAk+1. Once bis chosen, this ais unique because iis injective. The injectivity of ialso implies that da=0, since 25.4 The Zig-Zag Lemma 285 i(da)=d(ia)=ddb=0. (25.3) Therefore, ais a cocycle and deﬁnes a cohomology class [a]. We set d∗[c]=[a]∈Hk+1(A). In deﬁning d∗[c]we made two choices: a cocycle cto represent the cohomology class[c]∈Hk(C)and then an element b∈Bkthat maps to cunder j. For d∗to be well deﬁned, one must show that the cohomology class [a]∈Hk+1(A)does not depend on these choices. Exercise 25.5 (Connecting homomorphism).* Show that the connecting homomorphism d∗:Hk(C)→Hk+1(A) is a well-deﬁned linear map. The recipe for deﬁning the connecting homomorphism d∗is best remembered as a zig-zag diagram, a/d47/d47i/d47/d47db b/d31 j/d47/d47/d47/d47/d95d/d79/d79 c, where a/d47/d47/d47/d47dbmeans that amaps to dbunder an injection and b/d31/d47/d47/d47/d47cmeans thatbmaps to cunder a surjection. 25.4 The Zig-Zag Lemma The zig-zag lemma produces a long exact sequence in cohomology from a short exact sequence of cochain complexes. It is most useful when some of the terms in the long exact sequence are known to be zero, for then by exactness, the adjacent maps will be injections, surjections, or even isomorphisms. For example, if the cohomology of one of the three cochain complexes is zero, then the cohomology vector spaces ofthe other two cochain complexes will be isomorphic. Theorem 25.6 (The zig-zag lemma). A short exact sequence of cochain complexes 0→Ai→Bj→C→0 gives rise to a long exact sequence in cohomology: Hk+1(A)i∗/d47/d47···, Hk(A)i∗/d47/d47Hk(B)j∗/d47/d47Hk(C)/d66/d67/d69/d68/d56/d57 d∗/d63/d62/d47/d47 ···j∗/d47/d47Hk−1(C)/d58/d59/d61/d60/d56/d57 d∗/d63/d62/d47/d47 (25.4) 286§25 The Long Exact Sequence in Cohomology where i∗and j∗are the maps in cohomology induced from the cochain maps i and j, and d∗is the connecting homomorphism. To prove the theorem one needs to check exactness at Hk(A),Hk(B), and Hk(C) for each k. The proof is a sequence of trivialities involving what is commonly called diagram-chasing . As an example, we prove exactness at Hk(C). Claim. imj∗⊂kerd∗. Proof. Let[b]∈Hk(B). Then d∗j∗[b]=d∗[j(b)]. In the recipe above for d∗, we can choose the element in Bkthat maps to j(b)to be b. Then db∈Bk+1. Because bis a cocycle, db=0. Following the zig-zag diagram 0 db/d47/d47i/d47/d47 b/d95d/d79/d79 j(b),/d31 j/d47/d47/d47/d47=0 we see that since i(0)=0=db, we must have d∗[j(b)]=[0]. So j∗[b]∈kerd∗.⊓ ⊔ Claim. kerd∗⊂imj∗. Proof. Suppose d∗[c] = [a] =0, where [c]∈Hk(C). This means that a=da′for some a′∈Ak. The calculation of d∗[c]can be represented by the zig-zag diagram a/d47/d47i/d47/d47db a′/d95d/d79/d79 b/d95d/d79/d79 /d31j/d47/d47/d47/d47c, where bis an element in Bkwith j(b)=candi(a)=db. Then b−i(a′)is a cocycle inBkthat maps to cunder j: d(b−i(a′))= db−di(a′)=db−id(a′)=db−ia=0, j(b−i(a′))= j(b)−ji(a′)=j(b)=c. Therefore, j∗[b−i(a′)]=[c]. So[c]∈imj∗. ⊓ ⊔ These two claims together imply the exactness of (25.4) at Hk(C). As for the exactness of the cohomology sequence (25.4) at Hk(A)and at Hk(B), we will leave it as an exercise (Problem 25.3). 25.4 The Zig-Zag Lemma 287 Problems 25.1. A three-term exact sequence Prove Proposition 25.1. 25.2. A four-term exact sequence Prove Proposition 25.2. 25.3. Long exact cohomology sequence Prove the exactness of the cohomology sequence (25.4) at Hk(A)andHk(B). 25.4.* The snake lemma1 Use the zig-zag lemma to prove the following: The snake lemma. A commutative diagram with exact rows 0/d47/d47A1 /d47/d47B1 /d47/d47C1 /d47/d470 0/d47/d47A0 /d47/d47α/d79/d79 B0 /d47/d47β/d79/d79 C0 /d47/d47γ/d79/d79 0 induces a long exact sequence coker α/d47/d47coker β /d47/d47coker γ /d47/d470. 0 /d47/d47kerα/d47/d47kerβ /d47/d47kerγ/d66/d67/d69/d68/d56/d57/d63/d62/d47/d47 1The snake lemma, also called the serpent lemma, derives its name from the shape of the long exact sequence in it, usually drawn as an S. It may be the only result from homological algebra that has made its way into popular culture. In the 1980 ﬁlm It’s My Turn there is a scene in which the actress Jill Clayburgh, who plays a mathematics professor, explains theproof of the snake lemma. 288§26 The Mayer–Vietoris Sequence §26 The Mayer–Vietoris Sequence As the example of the cohomology of the real line R1illustrates, calculating the de Rham cohomology of a manifold amounts to solving a canonically given system of differential equations on the manifold and, in case it is not solvable, to ﬁnding obstructions to its solvability. This is usually quite difﬁcult to do directly. We in-troduce in this section one of the most useful tools in the calculation of de Rham cohomology, the Mayer–Vietoris sequence. Another tool, the homotopy axiom, will come in the next section. 26.1 The Mayer–Vietoris Sequence Let{U,V}be an open cover of a manifold M, and let iU:U→M,iU(p)=p, be the inclusion map. Then the pullback i∗ U:Ωk(M)→Ωk(U) is the restriction map that restricts the domain of a k-form on MtoU:i∗ Uω=ω|U. In fact, there are four inclusion maps that form a commutative diagram: U/d21/d117iU /d40/d40/d81/d81/d81/d81/d81/d81 U∩V/d40/d8jU/d53/d53/d107/d107/d107/d107/d107/d107/d22/d118 jV/d41/d41/d83/d83/d83/d83/d83/d83 M. V/d41/d9 iV/d54/d54/d109/d109/d109/d109/d109/d109 By restricting a k-form from MtoUand to V, we get a homomorphism of vector spaces i:Ωk(M)→Ωk(U)⊕Ωk(V), σ/ma√s⊔o→(i∗ Uσ,i∗ Vσ)=( σ|U,σ|V). Deﬁne the map j:Ωk(U)⊕Ωk(V)→Ωk(U∩V) by j(ω,τ)=j∗ Vτ−j∗ Uω=τ|U∩V−ω|U∩V. (26.1) IfU∩Vis empty, we deﬁne Ωk(U∩V) =0. In this case, jis simply the zero map. We call itherestriction map and jthedifference map . Since the direct sum Ω∗(U)⊕Ω∗(V)is the de Rham complex Ω∗(U∐V)of the disjoint union U∐V, the exterior derivative donΩ∗(U)⊕Ω∗(V)is given by d(ω,τ)=( dω,dτ). Proposition 26.1. Both the restriction map i and the difference map j commute with the exterior derivative d. 26.1 The Mayer–Vietoris Sequence 289 Proof. This is a consequence of the commutativity of dwith the pullback (Proposi- tion 19.5). For σ∈Ωk(M), diσ=d(i∗ Uσ,i∗ Vσ)=( di∗ Uσ,di∗ Vσ)=( i∗ Udσ,i∗ Vdσ)=idσ. For(ω,τ)∈Ωk(U)⊕Ωk(V), d j(ω,τ)=d(j∗ Vτ−j∗ Uω)=j∗ Vdτ−j∗ Udω=jd(ω,τ).⊓ ⊔ Thus, iandjare cochain maps. Proposition 26.2. For each integer k≥0, the sequence 0→Ωk(M)i→Ωk(U)⊕Ωk(V)j→Ωk(U∩V)→0 (26.2) is exact. Proof. Exactness at the ﬁrst two terms Ωk(M)andΩk(U)⊕Ωk(V)is straightfor- ward. We leave it as an exercise (Problem 26.1). We will prove exactness at Ωk(U∩V). To prove the surjectivity of the difference map j:Ωk(U)⊕Ωk(V)→Ωk(U∩V), it is best to consider ﬁrst the case of functions on M=R1. Let fbe aC∞function on U∩Vas in Figure 26.1. We have to write fas the difference of a C∞function on V and a C∞function on U. ( ) ()ρU ρVU Vf ρUf Fig. 26.1. Writing fas the difference of a C∞function on Vand a C∞function on U. Let{ρU,ρV}be a partition of unity subordinate to the open cover {U,V}. Deﬁne fV:V→Rby 290§26 The Mayer–Vietoris Sequence fV(x)=/braceleftigg ρU(x)f(x)forx∈U∩V, 0 for x∈V−(U∩V). Exercise 26.3 (Smooth extension of a function). Prove that fVis aC∞function on V. The function fVis called the extension by zero ofρUffrom U∩VtoV. Similarly, we deﬁne fUto be the extension by zero of ρVffrom U∩VtoU. Note that to “extend” the domain of ffrom U∩Vto one of the two open sets, we multiply by the partition function of the other open set. Since j(−fU,fV)=fV|U∩V+fU|U∩V=ρUf+ρVf=fonU∩V, jis surjective. For differential k-forms on a general manifold M, the formula is similar. For ω∈Ωk(U∩V), deﬁne ωUto be the extension by zero of ρVωfrom U∩VtoU, andωVto be the extension by zero of ρUωfrom U∩VtoV. On U∩V,(−ωU,ωV) restricts to (−ρVω,ρUω). Hence, jmaps(−ωU,ωV)∈Ωk(U)⊕Ωk(V)to ρVω−(−ρUω)=ω∈Ωk(U∩V). This shows that jis surjective and the sequence (26.2) is exact at Ωk(U∩V).⊓ ⊔ It follows from Proposition 26.2 that the sequence of cochain complexes 0→Ω∗(M)i→Ω∗(U)⊕Ω∗(V)j→Ω∗(U∩V)→0 is short exact. By the zig-zag lemma (Theorem 25.6), this short exact sequence of cochain complexes gives rise to a long exact sequence in cohomology, called the Mayer–Vietoris sequence: Hk+1(M)i∗/d47/d47···. Hk(M)i∗/d47/d47Hk(U)⊕Hk(V)j∗/d47/d47Hk(U∩V)/d66/d67/d69/d68/d56/d57 d∗/d63/d62/d47/d47 ···j∗/d47/d47Hk−1(U∩V)/d58/d59/d61/d60/d56/d57 d∗/d63/d62/d47/d47 (26.3) In this sequence i∗andj∗are induced from iandj: i∗[σ]=[i(σ)]=([ σ|U],[σ|V])∈Hk(U)⊕Hk(V), j∗([ω],[τ])=[ j(ω,τ)]=[ τ|U∩V−ω|U∩V]∈Hk(U∩V). By the recipe of Section 25.3, the connecting homomorphism d∗:Hk(U∩V)→ Hk+1(M)is obtained in three steps as in the diagrams below: 26.1 The Mayer–Vietoris Sequence 291 Ωk+1(M)/d47/d47i/d47/d47Ωk+1(U)⊕Ωk+1(V) Ωk(U)⊕Ωk(V)d/d79/d79 j/d47/d47/d47/d47Ωk(U∩V),α/d47/d47i (3)/d47/d47(−dζU,dζV)/d31j/d47/d47/d47/d470 (−ζU,ζV)/d95d(2)/d79/d79 /d31j (1)/d47/d47/d47/d47ζ./d95d/d79/d79 (1) Starting with a closed k-form ζ∈Ωk(U∩V)and using a partition of unity {ρU,ρV}subordinate to{U,V}, one can extend ρUζby zero from U∩Vto ak-form ζVonVand extend ρVζby zero from U∩Vto ak-form ζUonU(see the proof of Proposition 26.2). Then j(−ζU,ζV)=ζV|U∩V+ζU|U∩V=(ρU+ρV)ζ=ζ. (2) The commutativity of the square for dand jshows that the pair (−dζU,dζV) maps to 0 under j. More formally, since jd=djand since ζis a cocycle, j(−dζU,dζV)=jd(−ζU,ζV)=dj(−ζU,ζV)=dζ=0. It follows that the (k+1)-forms−dζUonUanddζVonVagree on U∩V. (3) Therefore,−dζUonUanddζVpatch together to give a global (k+1)-form α onM. Diagram-chasing shows that αis closed (see (25.3)). By Section 25.3, d∗[ζ]=[α]∈Hk+1(M). Because Ωk(M)=0 for k≤−1, the Mayer–Vietoris sequence starts with 0→H0(M)→H0(U)⊕H0(V)→H0(U∩V)→···. Proposition 26.4. In the Mayer–Vietoris sequence, if U, V , and U ∩V are connected and nonempty, then (i)M is connected and 0→H0(M)→H0(U)⊕H0(V)→H0(U∩V)→0 is exact; (ii)we may start the Mayer–Vietoris sequence with 0→H1(M)i∗ →H1(U)⊕H1(V)j∗ →H1(U∩V)→···. Proof. (i) The connectedness of Mfollows from a lemma in point-set topology (Proposi- tion A.44). It is also a consequence of the Mayer–Vietoris sequence. On a nonempty, connected open set, the de Rham cohomology in dimension 0 is simply the vectorspace of constant functions (Proposition 24.1). By (26.1), the map j ∗:H0(U)⊕H0(V)→H0(U∩V) is given by 292§26 The Mayer–Vietoris Sequence (u,v)/ma√s⊔o→v−u,u,v∈R. This map is clearly surjective. The surjectivity of j∗implies that imj∗=H0(U∩V)=kerd∗, from which we conclude that d∗:H0(U∩V)→H1(M)is the zero map. Thus the Mayer–Vietoris sequence starts with 0→H0(M)i∗ →R⊕Rj∗ →Rd∗ →0. (26.4) This short exact sequence shows that H0(M)≃imi∗=kerj∗. Since kerj∗={(u,v)|v−u=0}={(u,u)∈R⊕R}≃R, H0(M)≃R, which proves that Mis connected. (ii) From (i) we know that d∗:H0(U∩V)→H1(M)is the zero map. Thus, in the Mayer–Vietoris sequence, the sequence of two maps H0(U∩V)d∗ →H1(M)i∗ →H1(U)⊕H1(V) may be replaced by 0→H1(M)i∗ →H1(U)⊕H1(V) without affecting exactness. ⊓ ⊔ 26.2 The Cohomology of the Circle In Example 24.4 we showed that integration of 1-forms induces an isomorphism of H1(S1)withR. In this section we apply the Mayer–Vietoris sequence to give an alternative computation of the cohomology of the circle. Cover the circle with two open arcs UandVas in Figure 26.2. The intersection U∩Vis the disjoint union of two open arcs, which we call AandB. Since an open arc is diffeomorphic to an open interval and hence to the real line R1, the cohomology rings of UandVare isomorphic to that of R1, and the cohomology ring of U∩V to that of the disjoint union R1∐R1. They ﬁt into the Mayer–Vietoris sequence, which we arrange in tabular form: S1U∐V U∩V H2→ 0→ 0→ 0 H1 d∗ →H1(S1)→ 0→ 0 H00→Ri∗ →R⊕Rj∗ →R⊕R 26.2 The Cohomology of the Circle 293 U V) ( ) (A B Fig. 26.2. An open cover of the circle. From the exact sequence 0→Ri∗ −→R⊕Rj∗ −→R⊕Rd∗ −→H1(S1)→0 and Problem 26.2, we conclude that dim H1(S1)=1. Hence, the cohomology of the circle is given by Hk(S1)=/braceleftigg Rfork=0,1, 0 otherwise . By analyzing the maps in the Mayer–Vietoris sequence, it is possible to write down an explicit generator for H1(S1). First, according to Proposition 24.1, an element of H0(U)⊕H0(V)is an ordered pair (u,v)∈R⊕R, representing a con- stant function uonUand a constant function vonV. An element of H0(U∩V) = H0(A)⊕H0(B)is an ordered pair (a,b)∈R⊕R, representing a constant function a onAand a constant function bonB. The restriction map j∗ U:Z0(U)→Z0(U∩V)is the restriction of a constant function on Uto the two connected components AandB of the intersection U∩V: j∗ U(u)=u|U∩V=(u,u)∈Z0(A)⊕Z0(B). Similarly, j∗ V(v)=v|U∩V=(v,v)∈Z0(A)⊕Z0(B). By (26.1), j:Z0(U)⊕Z0(V)→Z0(U∩V)is given by j(u,v)=v|U∩V−u|U∩V=(v,v)−(u,u)=( v−u,v−u). Hence, in the Mayer–Vietoris sequence, the induced map j∗:H0(U)⊕H0(V)→ H0(U∩V)is given by j∗(u,v)=( v−u,v−u). The image of j∗is therefore the diagonal ∆inR2: ∆={(a,a)∈R2}. 294§26 The Mayer–Vietoris Sequence Since H1(S1)is isomorphic to R, a generator of H1(S1)is simply a nonzero element. Moreover, because d∗:H0(U∩V)→H1(S1)is surjective and kerd∗=imj∗=∆, such a nonzero element in H1(S1)is the image under d∗of an element (a,b)∈ H0(U∩V)≃R2with a/ne}a⊔ionslash=b. 1 0f VfV VdfVU−fU U−dfU Fig. 26.3. A generator of H1of the circle. So we may start with (a,b)=( 1,0)∈H0(U∩V). This corresponds to a function fwith value 1 on Aand 0 on B. Let{ρU,ρV}be a partition of unity subordinate to the open cover{U,V}, and let fU,fVbe the extensions by zero of ρVf,ρUffrom U∩VtoUand to V, respectively. By the proof of Proposition 26.2, j(−fU,fV)=f onU∩V. From Section 25.3, d∗(1,0)is represented by a 1-form on S1whose restriction to Uis−dfUand whose restriction to VisdfV. Now fVis the function on Vthat is ρUonAand 0 on V−A, sodfVis a 1-form on Vwhose support is contained entirely in A. A similar analysis shows that −dfUrestricts to the same 1-form on A, because ρU+ρV=1. The extension of either dfVor−dfUby zero to a 1-form onS1represents a generator of H1(S1). It is a bump 1-form on S1supported in A (Figure 26.3 ). The explicit description of the map j∗gives another way to compute H1(S1), for by the exactness of the Mayer–Vietoris sequence and the ﬁrst isomorphism theorem of linear algebra, there is a sequence of vector-space isomorphisms H1(S1)=imd∗≃R⊕R kerd∗=R⊕R imj∗≃R⊕R imj∗≃R2 R≃R. 26.3 The Euler Characteristic 295 26.3 The Euler Characteristic If the cohomology vector space Hk(M)of an n-manifold Mis ﬁnite-dimensional for every k, we deﬁne its Euler characteristic to be the alternating sum χ(M)=n ∑ k=0(−1)kdimHk(M). As a corollary of the Mayer–Vietoris sequence, the Euler characteristic of U∪Vis computable from those of U,V, and U∩V, as follows. Exercise 26.5 (Euler characteristics in terms of an open cover). Suppose a manifold M has an open cover {U,V}and the spaces M,U,V, and U∩Vall have ﬁnite-dimensional cohomology. By applying Problem 26.2 to the Mayer–Vietoris sequence, prove that χ(M)−(χ(U)+χ(V))+χ(U∩V)=0. Problems 26.1. Short exact Mayer–Vietoris sequence Prove the exactness of (26.2) at Ωk(M)and at Ωk(U)⊕Ωk(V). 26.2. Alternating sum of dimensions Let 0→A0d0−→A1d1−→A2d2−→···→ Am→0 be an exact sequence of ﬁnite-dimensional vector spaces. Show that m ∑ k=0(−1)kdimAk=0. (Hint: By the rank–nullity theorem from linear algebra, dimAk=dimker dk+dimim dk. Take the alternating sum of these equations over kand use the fact that dimker dk=dimim dk−1 to simplify it.) 296§27 Homotopy Invariance §27 Homotopy Invariance The homotopy axiom is a powerful tool for computing de Rham cohomology. While homotopy is normally deﬁned in the continuous category, since we are primarily interested in smooth manifolds and smooth maps, our notion of homotopy will be smooth homotopy. It differs from the usual homotopy in topology only in that all themaps are assumed to be smooth. In this section we deﬁne smooth homotopy, state the homotopy axiom for de Rham cohomology, and compute a few examples. We postpone the proof of the homotopy axiom to Section 29. 27.1 Smooth Homotopy LetMandNbe manifolds. Two C∞maps f,g:M→Nare(smoothly) homotopic if there is a C∞map F:M×R→N such that F(x,0)=f(x)and F(x,1)=g(x) for all x∈M; the map Fis called a homotopy from ftog. A homotopy Ffrom fto gcan be viewed as a smoothly varying family of maps {ft:M→N|t∈R}, where ft(x)=F(x,t),x∈M, such that f0=fandf1=g. We can think of the parameter tas time and a homotopy as an evolution through time of the map f0:M→N. Iffandgare homotopic, we write f∼g. Since any open interval is diffeomorphic to R(Problem 1.3), in the deﬁnition of homotopy we could have used any open interval containing 0 and 1, instead ofR. The advantage of an open interval over the closed interval [0,1]is that an open interval is a manifold without boundary. /Bullet /Bullet f(x)g(x) Fig. 27.1. Straight-line homotopies. 27.2 Homotopy Type 297 Example 27.1 ( Straight-line homotopy) .LetfandgbeC∞maps from a manifold MtoRn. Deﬁne F:M×R→Rnby F(x,t)=f(x)+t(g(x)−f(x))=( 1−t)f(x)+tg(x). Then Fis a homotopy from ftog, called the straight-line homotopy from ftog (Figure 27.1 ). Exercise 27.2 (Homotopy). LetMandNbe manifolds. Prove that homotopy is an equiva- lence relation on the set of all C∞maps from MtoN. 27.2 Homotopy Type As usual,1Mdenotes the identity map on a manifold M. Deﬁnition 27.3. A map f:M→Nis ahomotopy equivalence if it has a homotopy inverse , i.e., a map g:N→Msuch that g◦fis homotopic to the identity 1MonM andf◦gis homotopic to the identity 1NonN: g◦f∼1Mand f◦g∼1N. In this case we say that Mishomotopy equivalent toN, or that MandNhave the same homotopy type. Example. A diffeomorphism is a homotopy equivalence. /Bullet/Circle/Bullet /Bulletx x /bardblx/bardbl Fig. 27.2. The punctured plane retracts to the unit circle. Example 27.4 ( Homotopy type of the punctured plane ).Leti:S1→R2−{0}be the inclusion map and let r:R2−{0}→ S1be the map 298§27 Homotopy Invariance r(x)=x /bardblx/bardbl. Then r◦iis the identity map on S1. We claim that i◦r:R2−{0}→R2−{0} is homotopic to the identity map. Note that in the deﬁnition of a smooth homo- topy F(x,t), the domain of tis required to be the entire real line. The straight-line homotopy H(x,t)=( 1−t)x+tx /bardblx/bardbl,(x,t)∈(R2−{0})×R, will be ﬁne if tis restricted to the closed interval [0,1]. However, if tis allowed to be any real number, then H(x,t)may be equal to 0. Indeed, for t=/bardblx/bardbl/(/bardblx/bardbl−1), H(x,t) =0, and so Hdoes not map into R2−{0}. To correct this problem, we modify the straight-line homotopy so that for all tthe modiﬁed map F(x,t)is always a positive multiple of xand hence never zero. Set F(x,t)=( 1−t)2x+t2x /bardblx/bardbl=/parenleftbigg (1−t)2+t2 /bardblx/bardbl/parenrightbigg x. Then F(x,t)=0⇐⇒(1−t)2=0 andt2 /bardblx/bardbl=0 ⇐⇒ t=1=0,a contradiction . Therefore, F:(R2−{0})×R→R2−{0}provides a homotopy between the identity map onR2−{0}andi◦r(Figure 27.2 ). It follows that randiare homotopy inverse to each other, and R2−{0}andS1have the same homotopy type. Deﬁnition 27.5. A manifold is contractible if it has the homotopy type of a point. In this deﬁnition, by “the homotopy type of a point” we mean the homotopy type of a set{p}whose single element is a point. Such a set is called a singleton set or just a singleton . Example 27.6 ( The Euclidean space Rnis contractible ).Letpbe a point in Rn, i:{p}→Rnthe inclusion map, and r:Rn→{p}the constant map. Then r◦i= 1{p}, the identity map on {p}. The straight-line homotopy provides a homotopy between the constant map i◦r:Rn→Rnand the identity map on Rn: F(x,t)=( 1−t)x+t r(x)=( 1−t)x+t p. Hence, the Euclidean space Rnand the set{p}have the same homotopy type. 27.3 Deformation Retractions 299 27.3 Deformation Retractions LetSbe a submanifold of a manifold M, with i:S→Mthe inclusion map. Deﬁnition 27.7. Aretraction from MtoSis a map r:M→Sthat restricts to the identity map on S; in other words, r◦i=1S. If there is a retraction from MtoS, we say that Sis aretract ofM. Deﬁnition 27.8. Adeformation retraction from MtoSis a map F:M×R→M such that for all x∈M, (i)F(x,0)=x, (ii) there is a retraction r:M→Ssuch that F(x,1)=r(x), (iii) for all s∈Sandt∈R,F(s,t)=s. If there is a deformation retraction from MtoS, we say that Sis adeformation retract ofM. Setting ft(x)=F(x,t), we can think of a deformation retraction F:M×R→M as a family of maps ft:M→Msuch that (i)f0is the identity map on M, (ii)f1(x)=r(x)for some retraction r:M→S, (iii) for every tthe map ft:M→Mrestricts to the identity on S. We may rephrase condition (ii) in the deﬁnition as follows: there is a retraction r:M→Ssuch that f1=i◦r. Thus, a deformation retraction is a homotopy between the identity map 1Mandi◦rfor a retraction r:M→Ssuch that this homotopy leaves Sﬁxed for all time t. Example. Any point pin a manifold Mis a retract of M; simply take a retraction to be the constant map r:M→{p}. Example. The map Fin Example 27.4 is a deformation retraction from the punctured planeR2−{0}to the unit circle S1. The map Fin Example 27.6 is a deformation retraction from Rnto a singleton{p}. Generalizing Example 27.4, we prove the following theorem. Proposition 27.9. If S⊂M is a deformation retract of M, then S and M have the same homotopy type. Proof. LetF:M×R→Mbe a deformation retraction and let r(x)=f1(x)=F(x,1) be the retraction. Because ris a retraction, the composite Si→Mr→S,r◦i=1S, is the identity map on S. By the deﬁnition of a deformation retraction, the composite 300§27 Homotopy Invariance Mr→Si→M isf1and the deformation retraction provides a homotopy f1=i◦r∼f0=1M. Therefore, r:M→Sis a homotopy equivalence, with homotopy inverse i:S→M. ⊓ ⊔ 27.4 The Homotopy Axiom for de Rham Cohomology We state here the homotopy axiom and derive a few consequences. The proof willbe given in Section 29. Theorem 27.10 (Homotopy axiom for de Rham cohomology). Homotopic maps f 0,f1:M→N induce the same map f∗ 0=f∗ 1:H∗(N)→H∗(M)in cohomology. Corollary 27.11. If f:M→N is a homotopy equivalence, then the induced map in cohomology f∗:H∗(N)→H∗(M) is an isomorphism. Proof (of Corollary ).Letg:N→Mbe a homotopy inverse to f. Then g◦f∼1M,f◦g∼1N. By the homotopy axiom, (g◦f)∗=1H∗(M),(f◦g)∗=1H∗(N). By functoriality, f∗◦g∗=1H∗(M),g∗◦f∗=1H∗(N). Therefore, f∗is an isomorphism in cohomology. ⊓ ⊔ Corollary 27.12. Suppose S is a submanifold of a manifold M and F is a deformation retraction from M to S. Let r :M→S be the retraction r (x)=F(x,1). Then r induces an isomorphism in cohomology r∗:H∗(S)∼→H∗(M). Proof. The proof of Proposition 27.9 shows that a retraction r:M→Sis a homotopy equivalence. Apply Corollary 27.11. ⊓ ⊔ Corollary 27.13 (Poincar ´e lemma). SinceRnhas the homotopy type of a point, the cohomology of Rnis Hk(Rn)=/braceleftigg Rfor k=0, 0for k>0. 27.4 The Homotopy Axiom for de Rham Cohomology 301 More generally, any contractible manifold will have the same cohomology as a point. Example 27.14 ( Cohomology of a punctured plane ).For any p∈R2, the translation x/ma√s⊔o→x−pis a diffeomorphism of R2−{p}withR2−{0}. Because the punctured planeR2−{0}and the circle S1have the same homotopy type (Example 27.4), they have isomorphic cohomology. Hence, Hk(R2−{p})≃Hk(S1)for all k≥0. Example. The central circle of an open M¨ obius band Mis a deformation retract of M(Figure 27.3 ). Thus, the open M¨ obius band has the homotopy type of a circle. By the homotopy axiom, Hk(M)=Hk(S1)=/braceleftigg Rfork=0,1, 0 for k>1. Fig. 27.3. The M¨ obius band deformation retracts to its central circle. Problems 27.1. Homotopy equivalence LetM,N, and Pbe manifolds. Prove that if MandNare homotopy equivalent and NandP are homotopy equivalent, then MandPare homotopy equivalent. 27.2. Contractibility and path-connectedness Show that a contractible manifold is path-connected. 27.3. Deformation retraction from a cylinder to a circle Show that the circle S1×{0}is a deformation retract of the cylinder S1×R. 302§28 Computation of de Rham Cohomology §28 Computation of de Rham Cohomology With the tools developed so far, we can compute the cohomology of many manifolds. This section is a compendium of some examples. 28.1 Cohomology Vector Space of a Torus Cover a torus Mwith two open subsets UandVas shown in Figure 28.1. A B U∩V∼S1∐S1 M U∐V Fig. 28.1. An open cover{U,V}of a torus. Both UandVare diffeomorphic to a cylinder and therefore have the homotopy type of a circle (Problem 27.3). Similarly, the intersection U∩Vis the disjoint union of two cylinders AandBand has the homotopy type of a disjoint union of two circles. Our knowledge of the cohomology of a circle allows us to ﬁll in many terms in the Mayer–Vietoris sequence: M U∐V U∩V H2d∗ 1→H2(M)→ 0 H1d∗ 0→H1(M)i∗ →R⊕Rβ→R⊕R H00→R→R⊕Rα→R⊕R(28.1) LetjU:U∩V→Uand jV:U∩V→Vbe the inclusion maps. Recall that H0 of a connected manifold is the vector space of constant functions on the manifold (Proposition 24.1). If a∈H0(U)is the constant function with value aonU, then j∗ Ua=a|U∩V∈H0(U∩V)is the constant function with the value aon each compo- nent of U∩V, that is, j∗ Ua=(a,a). 28.2 The Cohomology Ring of a Torus 303 Therefore, for (a,b)∈H0(U)⊕H0(V), α(a,b)=b|U∩V−a|U∩V=(b,b)−(a,a)=( b−a,b−a). Similarly, let us now describe the map β:H1(U)⊕H1(V)→H1(U∩V)=H1(A)⊕H1(B). Since Ais a deformation retract of U, the restriction H∗(U)→H∗(A)is an isomor- phism, so if ωUgenerates H1(U), then j∗ UωUis a generator of H1onAand on B. Identifying H1(U∩V)withR⊕R, we write j∗ UωU=(1,1). Let ωVbe a generator ofH1(V). The pair of real numbers (a,b)∈H1(U)⊕H1(V)≃R⊕R stands for (aωU,bωV). Then β(a,b)=j∗ V(bωV)−j∗ U(aωU)=( b,b)−(a,a)=( b−a,b−a). By the exactness of the Mayer–Vietoris sequence, H2(M)=imd∗ 1 (because H2(U)⊕H2(V)=0) ≃H1(U∩V)/kerd∗ 1(by the ﬁrst isomorphism theorem ) ≃(R⊕R)/imβ ≃(R⊕R)/R≃R. Applying Problem 26.2 to the Mayer–Vietoris sequence (28.1), we get 1−2+2−dimH1(M)+2−2+dimH2(M)=0. Since dim H2(M)=1, this gives dim H1(M)=2. As a check, we can also compute H1(M)from the Mayer–Vietoris sequence using our knowledge of the maps αandβ: H1(M)≃keri∗⊕imi∗(by the ﬁrst isomorphism theorem ) ≃imd∗ 0⊕kerβ (exactness of the M–V sequence ) ≃(H0(U∩V)/kerd∗ 0)⊕kerβ(ﬁrst isomorphism theorem for d∗ 0) ≃((R⊕R)/imα)⊕R ≃R⊕R. 28.2 The Cohomology Ring of a Torus A torus is the quotient of R2by the integer lattice Λ=Z2. The quotient map π:R2→R2/Λ 304§28 Computation of de Rham Cohomology induces a pullback map on differential forms, π∗:Ω∗(R2/Λ)→Ω∗(R2). Since π:R2→R2/Λis a local diffeomorphism, its differential π∗:Tq(R2)→ Tπ(q)(R2/Λ)is an isomorphism at each point q∈R2. In particular, πis a submersion. By Problem 18.8, π∗:Ω∗(R2/Λ)→Ω∗(R2)is an injection. Forλ∈Λ, deﬁne ℓλ:R2→R2to be translation by λ, ℓλ(q)=q+λ,q∈R2. A differential form ¯ωonR2is said to be invariant under translation by λ∈Λif ℓ∗ λ¯ω=¯ω. The following proposition generalizes the description of differential forms on a circle given in Proposition 18.12, where Λwas the lattice 2 πZ. Proposition 28.1. The image of the injection π∗:Ω∗(R2/Λ)→Ω∗(R2)is the sub- space of differential forms on R2invariant under translations by elements of Λ. Proof. For all q∈R2, (π◦ℓλ)(q)=π(q+λ)=π(q). Hence, π◦ℓλ=π. By the functoriality of the pullback, π∗=ℓ∗ λ◦π∗. Thus, for any ω∈Ωk(R2/Λ),π∗ω=ℓ∗ λπ∗ω. This proves that π∗ωis invariant under all translations ℓλ,λ∈Λ. Conversely, suppose ¯ω∈Ωk(R2)is invariant under translations ℓλfor all λ∈Λ. Forp∈R2/Λandv1,..., vk∈Tp(R2/Λ), deﬁne ωp(v1,..., vk)= ¯ω¯p(¯v1,..., ¯vk) (28.2) for any ¯ p∈π−1(p)and ¯v1,..., ¯vk∈T¯pR2such that π∗¯vi=vi. Note that once ¯ pis chosen, ¯ v1,..., ¯vkare unique, since π∗:T¯p(R2)→Tp(R2/Λ)is an isomorphism. For ωto be well deﬁned, we need to show that it is independent of the choice of ¯ p. Now any other point in π−1(p)may be written as ¯ p+λfor some λ∈Λ. By invariance, ¯ω¯p=(ℓ∗ λ¯ω)¯p=ℓ∗ λ(¯ω¯p+λ). So ¯ω¯p(¯v1,..., ¯vk)=ℓ∗ λ(¯ω¯p+λ)(¯v1,..., ¯vk)= ¯ω¯p+λ(ℓλ∗¯v1,...,ℓ λ∗¯vk). (28.3) Since π◦ℓλ=π, we have π∗(ℓλ∗¯vi) =π∗¯vi=vi. Thus, (28.3) shows that ωpis independent of the choice of ¯ p, and ω∈Ωk(R2/Λ)is well deﬁned. Moreover, by (28.2), for any ¯ p∈R2and ¯v1,..., ¯vk∈T¯p(R2), ¯ω¯p(¯v1,..., ¯vk)=ωπ(¯p)(π∗¯v1,..., π∗¯vk)=( π∗ω)¯p(¯v1,..., ¯vk). Hence, ¯ω=π∗ω. ⊓ ⊔ 28.2 The Cohomology Ring of a Torus 305 Letx,ybe the standard coordinates on R2. Since for any λ∈Λ, ℓ∗ λ(dx)=d(ℓ∗ λx)=d(x+λ)=dx, by Proposition 28.1 the 1-form dxonR2isπ∗of a 1-form αon the torus R2/Λ. Similarly, dyisπ∗of a 1-form βon the torus. Note that π∗(dα)=d(π∗α)=d(dx)=0. Since π∗:Ω∗(R2/Z2)→Ω∗(R2)is injective, dα=0. Similarly, dβ=0. Thus, both αandβare closed 1-forms on the torus. Proposition 28.2. Let M be the torus R2/Z2. A basis for the cohomology vector space H∗(M)is represented by the forms 1,α,β,α∧β. Proof. LetIbe the closed interval [0,1], and i:I2֒→R2the inclusion map of the closed square I2intoR2. The composite map F=π◦i:I2֒→R2→R2/Z2repre- sents the torus M=R2/Z2as a parametrized set. Then F∗α=i∗(π∗α)=i∗dx, the restriction of dxto the square I2. Similarly, F∗β=i∗dy. As an integral over a parametrized set, /integraldisplay Mα∧β=/integraldisplay F(I2)α∧β=/integraldisplay I2F∗(α∧β)=/integraldisplay I2dx∧dy=/integraldisplay1 0/integraldisplay1 0dxdy=1. Thus, the closed 2-form α∧βrepresents a nonzero cohomology class on M. Since H2(M)=Rby the computation of Subsection 28.1, the cohomology class [α∧β]is a basis for H2(M). Next we show that the cohomology classes of the closed 1-forms α,βonM constitute a basis for H1(M). Let i1,i2:I→R2be given by i1(t) = ( t,0),i2(t) = (0,t). Deﬁne two closed curves C1,C2inM=R2/Z2as the images of the maps (Figure 28.2 ) ck:Iik−→R2π−→M=R2/Z2,k=1,2, c1(t)=[( t,0)], c2(t)=[( 0,t)]. Each curve Ciis a smooth manifold and a parametrized set with parametrization ci. C1C2 C1 C2 Fig. 28.2. Two closed curves on a torus. 306§28 Computation of de Rham Cohomology Moreover, c∗ 1α=(π◦i1)∗α=i∗ 1π∗α=i∗ 1dx=di∗ 1x=dt, c∗ 1β=(π◦i1)∗β=i∗ 1π∗β=i∗ 1dy=di∗ 1y=0. Similarly, c∗ 2α=0 and c∗ 2β=dt. Therefore, /integraldisplay C1α=/integraldisplay c1(I)α=/integraldisplay Ic∗ 1α=/integraldisplay1 0dt=1 and/integraldisplay C1β=/integraldisplay c1(I)β=/integraldisplay Ic∗ 1β=/integraldisplay1 00=0. In the same way,/integraltext C2α=0 and/integraltext C2β=1. Because/integraltext C1α/ne}a⊔ionslash=0 and/integraltext C2β/ne}a⊔ionslash=0, neither αnorβis exact on M. Furthermore, the cohomology classes [α]and[β]are linearly independent, for if [α]were a multiple of [β], then/integraltext C1αwould have to be a nonzero multiple of/integraltext C1β=0. By Subsection 28.1, H1(M)is two-dimensional. Hence, [α],[β]is a basis for H1(M). In degree 0, H0(M)has basis [1], as is true for any connected manifold M.⊓ ⊔ The ring structure of H∗(M)is clear from this proposition. Abstractly it is the algebra /logicalandtext(a,b):=R[a,b]/(a2,b2,ab+ba),dega=1,degb=1, called the exterior algebra on two generators aandbof degree 1. 28.3 The Cohomology of a Surface of Genus g Using the Mayer–Vietoris sequence to compute the cohomology of a manifold often leads to ambiguities, because there may be several unknown terms in the sequence. We can resolve these ambiguities if we can describe explicitly the maps occurring in the sequence. Here is an example of how this might be done. Lemma 28.3. Suppose p is a point in a compact oriented surface M without bound- ary, and i :C→M−{p}is the inclusion of a small circle around the puncture (Figure 28.3). Then the restriction map i∗:H1(M−{p})→H1(C) is the zero map. Proof. An element [ω]∈H1(M−{p})is represented by a closed 1-form ωonM− {p}. Because the linear isomorphism H1(C)≃H1(S1)≃Ris given by integration over C, to identify i∗[ω]inH1(C), it sufﬁces to compute the integral/integraltext Ci∗ω. 28.3 The Cohomology of a Surface of Genus g 307 p CM Fig. 28.3. Punctured surface. IfDis the open disk in Mbounded by the curve C, then M−Dis a compact oriented surface with boundary C. By Stokes’s theorem, /integraldisplay Ci∗ω=/integraldisplay ∂(M−D)i∗ω=/integraldisplay M−Ddω=0, because dω=0. Hence, i∗:H1(M−{p})→H1(C)is the zero map. ⊓ ⊔ Proposition 28.4. Let M be a torus, p a point in M, and A the punctured torus M − {p}. The cohomology of A is Hk(A)=  R for k=0, R2for k=1, 0 for k>1. Proof. Cover Mwith two open sets, Aand a disk Ucontaining p. Since A,U, and A∩ Uare all connected, we may start the Mayer–Vietoris sequence with the H1(M)term (Proposition 26.4(ii)). With H∗(M)known from Section 28.1, the Mayer–Vietoris sequence becomes M U∐A U∩A∼S1 H2d∗ 1→R→H2(A)→ 0 H10→R⊕Rβ→H1(A)α→ H1(S1) Because H1(U)=0, the map α:H1(A)→H1(S1)is simply the restriction map i∗. By Lemma 28.3, α=i∗=0. Hence, H1(A)=kerα=imβ≃H1(M)≃R⊕R and there is an exact sequence of linear maps 0→H1(S1)d∗ 1→R→H2(A)→0. Since H1(S1)≃R, it follows that H2(A)=0. ⊓ ⊔ 308§28 Computation of de Rham Cohomology Proposition 28.5. The cohomology of a compact orientable surface Σ2of genus 2is Hk(Σ2)=  R for k=0,2, R4for k=1, 0 for k>2. Σ2 U∐V U∩V∼S1 Fig. 28.4. An open cover{U,V}of a surface of genus 2. Proof. Cover Σ2with two open sets UandVas in Figure 28.4. Since U,V, and U∩Vare all connected, the Mayer–Vietoris sequence begins with M U∐V U∩V∼S1 H2→H2(Σ2)→ 0 H10→H1(Σ2)→R2⊕R2α−→R The map α:H1(U)⊕H1(V)→H1(S1)is the difference map α(ωU,ωV)=j∗ VωV−j∗ UωU, where jUand jVare inclusions of an S1inU∩VintoUandV, respectively. By Lemma 28.3, j∗ U=j∗ V=0, so α=0. It then follows from the exactness of the Mayer–Vietoris sequence that H1(Σ2)≃H1(U)⊕H1(V)≃R4 and H2(Σ2)≃H1(S1)≃R. ⊓ ⊔ A genus-2 surface Σ2can be obtained as the quotient space of an octagon with its edges identiﬁed following the scheme of Figure 28.5. To see this, ﬁrst cut Σ2along the circle eas in Figure 28.6. Then the two halves AandBare each a torus minus an open disk ( Figure 28.7 ), so that each half can be represented as a pentagon, before identiﬁcation ( Figure 28.8 ). When AandBare glued together along e, we obtain the octagon in Figure 28.5. 28.3 The Cohomology of a Surface of Genus g 309 abc d c d ab Fig. 28.5. A surface of genus 2 as a quotient space of an octagon. A B e Fig. 28.6. A surface of genus 2 cut along a curve e. d bc ad b c ae e Fig. 28.7. Two halves of a surface of genus 2. c d c deA abab e B Fig. 28.8. Two halves of a surface of genus 2. 310§28 Computation of de Rham Cohomology By Lemma 28.3, if p∈Σ2andi:C→Σ2−{p}is a small circle around pinΣ2, then the restriction map i∗:H1(Σ2−{p})→H1(C) is the zero map. This allows us to compute inductively the cohomology of a compact orientable surface Σgof genus g. Exercise 28.6 (Surface of genus 3).Compute the cohomology vector space of Σ2−{p}and then compute the cohomology vector space of a compact orientable surface Σ3of genus 3. Problems 28.1. Real projective plane Compute the cohomology of the real projective plane RP2(Figure 28.9) . /Bullet /Bullet aa Fig. 28.9. The real projective plane. 28.2. The n-sphere Compute the cohomology of the sphere Sn. 28.3. Cohomology of a multiply punctured plane (a) Let p,qbe distinct points in R2. Compute the de Rham cohomology of R2−{p,q}. (b) Let p1,..., pnbe distinct points in R2. Compute the de Rham cohomology of R2− {p1,..., pn}. 28.4. Cohomology of a surface of genus g Compute the cohomology vector space of a compact orientable surface Σgof genus g. 28.5. Cohomology of a 3-dimensional torus Compute the cohomology ring of R3/Z3. 29.1 Reduction to Two Sections 311 §29 Proof of Homotopy Invariance In this section we prove the homotopy invariance of de Rham cohomology. Iff:M→Nis aC∞map, the pullback maps on differential forms and on coho- mology classes are normally both denoted by f∗. Since this might cause confusion in the proof of homotopy invariance, in this section we revert to our original convention of denoting the pullback of forms by f∗:Ωk(N)→Ωk(M) and the induced map in cohomology by f#:Hk(N)→Hk(M). The relation between these two maps is f#[ω]=[ f∗ω] for[ω]∈Hk(N). Theorem 29.1 (Homotopy axiom for de Rham cohomology). Two smoothly ho- motopic maps f ,g:M→N of manifolds induce the same map in cohomology: f#=g#:Hk(N)→Hk(M). We ﬁrst reduce the problem to two special maps i0andi1:M→M×R, which are the 0-section and the 1-section, respectively, of the product line bundle M×R→M: i0(x)=( x,0),i1(x)=( x,1). Then we introduce the all-important technique of cochain homotopy. By ﬁnding a cochain homotopy between i∗ 0andi∗ 1, we prove that they induce the same map in cohomology. 29.1 Reduction to Two Sections Suppose fandg:M→Nare smoothly homotopic maps. Let F:M×R→Nbe a smooth homotopy from ftog. This means that F(x,0)=f(x),F(x,1)=g(x) (29.1) for all x∈M. For each t∈R, deﬁne it:M→M×Rto be the section it(x)=( x,t). We can restate (29.1) as F◦i0=f,F◦i1=g. By the functoriality of the pullback (Remark 24.7), 312§29 Proof of Homotopy Invariance f#=i# 0◦F#,g#=i# 1◦F#. This reduces proving homotopy invariance to the special case i# 0=i# 1. The two maps i0,i1:M→M×Rare obviously smoothly homotopic via the identity map 1M×R:M×R→M×R. 29.2 Cochain Homotopies The usual method for showing that two cochain maps ϕ,ψ:A→Binduce the same map in cohomology is to ﬁnd a linear map K:A→Bof degree−1 such that ϕ−ψ=d◦K+K◦d. Such a map Kis called a cochain homotopy from ϕtoψ. Note that Kis not assumed to be a cochain map. If ais a cocycle in A, then ϕ(a)−ψ(a)=dKa+Kda=dKa is a coboundary, so that in cohomology ϕ#[a]=[ϕ(a)]=[ ψ(a)]= ψ#[a]. Thus, the existence of a cochain homotopy between ϕandψimplies that the induced maps ϕ#andψ#in cohomology are equal. Remark. Given two cochain maps ϕ,ψ:A→B, if one could ﬁnd a linear map K:A→Bof degree−1 such that ϕ−ψ=d◦KonA, then ϕ#would be equal to ψ#in cohomology. However, such a map almost never exists; it is necessary to have the term K◦das well. The cylinder construction in homology theory [30, p. 65] shows why it is natural to consider d◦K+K◦d. 29.3 Differential Forms on M×R Recall that a sum ∑αωαofC∞differential forms on a manifold Mis said to be locally ﬁnite if the collection{supp ωα}of supports is locally ﬁnite. This means that every point pinMhas a neighborhood Vpsuch that Vpintersects only ﬁnitely many of the sets supp ωα. If supp ωαis disjoint from Vp, then ωα≡0 on Vp. Thus, on Vp the locally ﬁnite sum ∑αωαis actually a ﬁnite sum. As an example, if {ρα}is a partition of unity, then the sum ∑ραis locally ﬁnite. Letπ:M×R→Mbe the projection to the ﬁrst factor. In this subsection we will show that every C∞differential form on M×Ris a locally ﬁnite sum of the following two types of forms: 29.3 Differential Forms on M×R 313 (I)f(x,t)π∗η, (II) f(x,t)dt∧π∗η, where f(x,t)is aC∞function on M×Randηis aC∞form on M. In general, a decomposition of a differential form on M×Rinto a locally ﬁnite sum of type-I and type-II forms is far from unique. However, once we ﬁx an atlas {(Uα,φα)}onM, aC∞partition of unity{ρα}subordinate to{Uα}, and a collection {gα}ofC∞functions on Msuch that gα≡1 on supp ραand supp gα⊂Uα, then there is a well-deﬁned procedure to produce uniquely such a locally ﬁnite sum. The existence of the functions gαfollows from the smooth Urysohn lemma (Prob- lem 13.3). In the proof of the decomposition procedure, we will need the following simple but useful lemma on the extension of a C∞form by zero. Lemma 29.2. Let U be an open subset of a manifold M. If a smooth k-form τ∈ Ωk(U)deﬁned on U has support in a closed subset of M contained in U, then τcan be extended by zero to a smooth k-form on M. Proof. Problem 29.1. ⊓ ⊔ Fix an atlas{(Uα,φα)}, a partition of unity {ρα}, and a collection{gα}ofC∞ functions as above. Then {π−1Uα}is an open cover of M×R, and{π∗ρα}is a partition of unity subordinate to {π−1Uα}(Problem 13.6). Letωbe any C∞k-form on M×Rand let ωα=(π∗ρα)ω. Since ∑π∗ρα=1, ω=∑ α(π∗ρα)ω=∑ αωα. (29.2) Because{supp π∗ρα}is locally ﬁnite, (29.2) is a locally ﬁnite sum. By Problem 18.4, supp ωα⊂supp π∗ρα∩supp ω⊂supp π∗ρα⊂π−1Uα. Letφα=(x1,..., xn). Then on π−1Uα, which is homeomorphic to Uα×R, we have coordinates π∗x1,..., π∗xn,t. For the sake of simplicity, we sometimes write xiin- stead of π∗xi. On π−1Uαthek-form ωαmay be written uniquely as a linear combi- nation ωα=∑ IaIdxI+∑ JbJdt∧dxJ, (29.3) where aIandbJareC∞functions on π−1Uα. This decomposition shows that ωαis a ﬁnite sum of type-I and type-II forms on π−1Uα. By Problem 18.5, the supports ofaIandbIare contained in supp ωα, hence in supp π∗ρα, a closed set in M×R. Therefore, by the lemma above, aIandbJcan be extended by zero to C∞functions on M×R. Unfortunately, dxIanddxJmake sense only on Uαand cannot be extended toM, at least not directly. 314§29 Proof of Homotopy Invariance To extend the decomposition (29.3) to M×R, the trick is to multiply ωαby π∗gα. Since supp ωα⊂supp π∗ραandπ∗gα≡1 on supp π∗ρα, we have the equality ωα=(π∗gα)ωα. Therefore, ωα=(π∗gα)ωα=∑ IaI(π∗gα)dxI+∑ JbJdt∧(π∗gα)dxJ, =∑ IaIπ∗(gαdxI)+∑ JbJdt∧π∗(gαdxJ). (29.4) Now supp gαis a closed subset of Mcontained in Uα, so by Lemma 29.2 again, gαdxIcan be extended by zero to M. Equations (29.2) and (29.4) prove that ωis a locally ﬁnite sum of type-I and type-II forms on M×R. Moreover, given{(Uα,φα)}, {ρα}, and{gα}, the decomposition in (29.4) is unique. 29.4 A Cochain Homotopy Between i∗ 0andi∗ 1 In the rest of the proof, ﬁx an atlas {(Uα,φα)}forM, aC∞partition of unity{ρα} subordinate to{Uα}, and a collection{gα}ofC∞functions on Mas in Section 29.3. Letω∈Ωk(M×R). Using (29.2) and (29.4), we decompose ωinto a locally ﬁnite sum ω=∑ αωα=∑ α,Iaα Iπ∗(gαdxI α)+∑ α,Jbα Jdt∧π∗(gαdxJ α), where we now attach an index αtoaI,bJ,xI, and xJto indicate their dependence onα. Deﬁne K:Ω∗(M×R)→Ω∗−1(M) by the following rules: (i) on type-I forms, K(fπ∗η)=0; (ii) on type-II forms, K(f dt∧π∗η)=/parenleftbigg/integraldisplay1 0f(x,t)dt/parenrightbigg η; (iii) Kis linear over locally ﬁnite sums. Thus, K(ω)=K/parenleftbigg ∑ αωα/parenrightbigg =∑ α,J/parenleftbigg/integraldisplay1 0bα J(x,t)dt/parenrightbigg gαdxJ α. (29.5) Given the data{(Uα,φα)},{ρα},{gα}, the decomposition ω=ωαwith ωαas in (29.4) is unique. Therefore, Kis well deﬁned. It is not difﬁcult to show that so deﬁned, Kis the unique linear operator Ω∗(M×R)→Ω∗−1(M)satisfying (i), (ii), and (iii) (Problem 29.3), so it is in fact independent of the data {(Uα,φα)},{ρα}, and{gα}. 29.5 Veriﬁcation of Cochain Homotopy 315 29.5 Veriﬁcation of Cochain Homotopy We check in this subsection that d◦K+K◦d=i∗ 1−i∗ 0. (29.6) Lemma 29.3. (i)The exterior derivative d is R-linear over locally ﬁnite sums. (ii)Pullback by a C∞map isR-linear over locally ﬁnite sums. Proof. (i) Suppose ∑ωαis a locally ﬁnite sum of C∞k-forms. This implies that every point phas a neighborhood on which the sum is ﬁnite. Let Ube such a neighborhood. Then /parenleftbig d∑ωα/parenrightbig |U=d/parenleftbig/parenleftbig∑ωα/parenrightbig |U/parenrightbig (Corollary 19.6 ) =d/parenleftbig∑ωα|U/parenrightbig =∑d(ωα|U)/parenleftbig∑ωα|Uis a ﬁnite sum/parenrightbig =∑(dωα)|U(Corollary 19.6 ) =/parenleftbig∑dωα/parenrightbig |U. Since Mcan be covered by such neighborhoods, d(∑ωα) =∑dωαonM. The homogeneity property d(rω)=rd(ω)forr∈Randω∈Ωk(M)is trivial. (ii) The proof is similar to (i) and is relegated to Problem 29.2. ⊓ ⊔ By linearity of K,d,i∗ 0, and i∗ 1over locally ﬁnite sums, it sufﬁces to check the equality (29.6) on any coordinate open set. Fix a coordinate open set (U×R, π∗x1,..., π∗xn,t)onM×R. On type-I forms, Kd(fπ∗η)=K/parenleftigg ∂f ∂tdt∧π∗η+∑ i∂f ∂xiπ∗dxi∧π∗η+fπ∗dη/parenrightigg . In the sum on the right-hand side, the second and third terms are type-I forms; they map to 0 under K. Thus, Kd(fπ∗η)=K/parenleftbigg∂f ∂tdt∧π∗η/parenrightbigg =/parenleftbigg/integraldisplay1 0∂f ∂tdt/parenrightbigg η =(f(x,1)−f(x,0))η=(i∗ 1−i∗ 0)(f(x,t)π∗η). Since dK(fπ∗η)=d(0)=0, on type-I forms, d◦K+K◦d=i∗ 1−i∗ 0. On type-II forms, by the antiderivation property of d, dK(f dt∧π∗η)=d/parenleftbigg/parenleftbigg/integraldisplay1 0f(x,t)dt/parenrightbigg η/parenrightbigg =∑/parenleftbigg∂ ∂xi/integraldisplay1 0f(x,t)dt/parenrightbigg dxi∧η+/parenleftbigg/integraldisplay1 0f(x,t)dt/parenrightbigg dη =∑/parenleftbigg/integraldisplay1 0∂f ∂xi(x,t)dt/parenrightbigg dxi∧η+/parenleftbigg/integraldisplay1 0f(x,t)dt/parenrightbigg dη. 316§29 Proof of Homotopy Invariance In the last equality, differentiation under the integral sign is permissible because f(x,t)isC∞. Furthermore, Kd(f dt∧π∗η)=K(d(f dt)∧π∗η−f dt∧dπ∗η) =K/parenleftbigg ∑ i∂f ∂xidxi∧dt∧π∗η/parenrightbigg −K(f dt∧π∗dη) =−∑ i/parenleftbigg/integraldisplay1 0∂f ∂xi(x,t)dt/parenrightbigg dxi∧η−/parenleftbigg/integraldisplay1 0f(x,t)dt/parenrightbigg dη. Thus, on type-II forms, d◦K+K◦d=0. On the other hand, i∗ 1(f(x,t)dt∧π∗η)=0 because i∗ 1dt=di∗ 1t=d(1)=0. Similarly, i∗ 0also vanishes on type-II forms. There- fore, d◦K+K◦d=0=i∗ 1−i∗ 0 on type-II forms. This completes the proof that Kis a cochain homotopy between i∗ 0andi∗ 1. The existence of the cochain homotopy Kproves that the induced maps in cohomology i# 0andi# 1are equal. As we pointed out in Section 29.1, f#=i# 0◦F#=i# 1◦F#=g#. Problems 29.1. Extension by zero of a smooth k-form Prove Lemma 29.2. 29.2. Linearity of pullback over locally ﬁnite sums Leth:N→Mbe a C∞map, and ∑ωαa locally ﬁnite sum of C∞k-forms on M. Prove that h∗(∑ωα)=∑h∗ωα. 29.3. The cochain homotopy K (a) Check that deﬁned by (29.5), the linear map Ksatisﬁes the three rules in Section 29.4. (b) Prove that a linear operator satisfying the three rules in Section 29.4 is unique if it exists. Appendices §A Point-Set Topology Point-set topology, also called “general topology,” is concerned with properties that remain invariant under homeomorphisms (continuous maps having continuous in-verses). The basic development in the subject took place in the late nineteenth and early twentieth centuries. This appendix is a collection of basic results from point-set topology that are used throughout the book. A.1 Topological Spaces The prototype of a topological space is the Euclidean space Rn. However, Euclidean space comes with many additional structures, such as a metric, coordinates, an innerproduct, and an orientation, that are extraneous to its topology. The idea behind the deﬁnition of a topological space is to discard all those properties of R nthat have nothing to do with continuous maps, thereby distilling the notion of continuity to its very essence. In advanced calculus one learns several characterizations of a continuous map, among which is the following: a map ffrom an open subset of RntoRmis contin- uous if and only if the inverse image f−1(V)of any open set VinRmis open in Rn. This shows that continuity can be deﬁned solely in terms of open sets. To deﬁne open sets axiomatically, we look at properties of open sets in Rn. Recall that inRnthedistance between two points pandqis given by d(p,q)=/bracketleftigg n ∑ i=1(pi−qi)2/bracketrightigg1/2 , and the open ball B (p,r)with center p∈Rnand radius r>0 is the set B(p,r)={x∈Rn|d(x,p)<r}. 318§A Point-Set Topology A set UinRnis said to be open if for every pinU, there is an open ball B(p,r)with center pand radius rsuch that B(p,r)⊂U(Figure A.1 ). It is clear that the union of an arbitrary collection {Uα}of open sets is open, but the same need not be true of the intersection of inﬁnitely many open sets. /Bullet pUB(p,r) Fig. A.1. An open set in Rn. Example. The intervals ]−1/n,1/n[,n=1,2,3,..., are all open in R1, but their intersection/intersectiontext∞ n=1]−1/n,1/n[is the singleton set {0}, which is not open. What is true is that the intersection of a ﬁnite collection of open sets in Rnis open. This leads to the deﬁnition of a topology on a set. Deﬁnition A.1. Atopology on a set Sis a collection Tof subsets containing both the empty set ∅and the set Ssuch that Tis closed under arbitrary unions and ﬁnite intersections; i.e., if Uα∈Tfor all αin an index set A, then/uniontext α∈AUα∈Tand if U1,..., Un∈T, then/intersectiontextn i=1Ui∈T. The elements of Tare called open sets and the pair (S,T)is called a topological space . To simplify the notation, we sometimes simply refer to a pair (S,T)as “the topological space S” when there is no chance of confusion. A neighborhood of a point pinSis an open set Ucontaining p. IfT1andT2are two topologies on a set SandT1⊂T2, then we say that T1iscoarser thanT1, or thatT2isﬁner thanT1. A coarser topology has fewer open sets; conversely, a ﬁner topology has more open sets. Example. The open subsets of Rnas we understand them in advanced calculus form a topology on Rn, the standard topology ofRn. In this topology a set Uis open in Rn if and only if for every p∈U, there is an open ball B(p,ε)with center pand radius ε contained in U. Unless stated otherwise, Rnwill always have its standard topology. The criterion for openness in Rnhas a useful generalization to a topological space. Lemma A.2 (Local criterion for openness). Let S be a topological space. A subset A is open in S if and only if for every p ∈A, there is an open set V such that p ∈V⊂A. A.1 Topological Spaces 319 Proof. (⇒)IfAis open, we can take V=A. (⇐)Suppose for every p∈Athere is an open set Vpsuch that p∈Vp⊂A. Then A⊂/uniondisplay p∈AVp⊂A, so that equality A=/uniontext p∈AVpholds. As a union of open sets, Ais open. ⊓ ⊔ Example. For any set S, the collection T={∅,S}consisting of the empty set ∅and the entire set Sis a topology on S, sometimes called the trivial orindiscrete topology . It is the coarsest topology on a set. Example. For any set S, letTbe the collection of all subsets of S. ThenTis a topology on S, called the discrete topology . A singleton set is a set with a single element. The discrete topology can also be characterized as the topology in whichevery singleton subset {p}is open. A topological space having the discrete topology is called a discrete space . The discrete topology is the ﬁnest topology on a set. The complement of an open set is called a closed set . By de Morgan’s laws from set theory, arbitrary intersections and ﬁnite unions of closed sets are closed (Problem A.3). One may also specify a topology by describing all the closed sets. Remark. When we say that a topology is closed under arbitrary union and ﬁnite intersection, the word “closed” has a different meaning from that of a “closed subset.” Example A.3 ( Finite-complement topology on R 1).LetTbe the collection of subsets ofR1consisting of the empty set ∅, the lineR1itself, and the complements of ﬁnite sets. Suppose FαandFiare ﬁnite subsets of R1forα∈some index set A and i=1,..., n. By de Morgan’s laws, /uniondisplay α/parenleftbig R1−Fα/parenrightbig =R1−/intersectiondisplay αFαandn/intersectiondisplay i=1/parenleftbig R1−Fi/parenrightbig =R1−n/uniondisplay i=1Fi. Since the arbitrary intersection/intersectiontext α∈AFαand the ﬁnite union/uniontextn i=1Fiare both ﬁnite, Tis closed under arbitrary unions and ﬁnite intersections. Thus, Tdeﬁnes a topology onR1, called the ﬁnite-complement topology . For the sake of deﬁniteness, we have deﬁned the ﬁnite-complement topology on R1, but of course, there is nothing speciﬁc about R1here. One can deﬁne in exactly the same way the ﬁnite-complement topology on any set. Example A.4 ( Zariski topology ).One well-known topology is the Zariski topology from algebraic geometry. Let Kbe a ﬁeld and let Sbe the vector space Kn. Deﬁne a subset of Knto be Zariski closed if it is the zero set Z(f1,..., fr)of ﬁnitely many polynomials f1,..., fronKn. To show that these are indeed the closed subsets of 320§A Point-Set Topology a topology, we need to check that they are closed under arbitrary intersections and ﬁnite unions. LetI= (f1,..., fr)be the ideal generated by f1,..., frin the polynomial ring K[x1,..., xn]. Then Z(f1,..., fr) =Z(I), the zero set of allthe polynomials in the ideal I. Conversely, by the Hilbert basis theorem [11, §9.6, Th. 21], any ideal in K[x1,..., xn]has a ﬁnite set of generators. Hence, the zero set of ﬁnitely many poly- nomials is the same as the zero set of an ideal in K[x1,..., xn]. IfI=(f1,..., fr)and J=(g1,..., gs)are two ideals, then the product ideal IJ is the ideal in K[x1,..., xn] generated by all products figj, 1≤i≤r, 1≤j≤s. If{Iα}α∈Ais a family of ideals inK[x1,..., xn], then their sum ∑αIαis the smallest ideal in K[x1,..., xn]containing all the ideals Iα. Exercise A.5 (Intersection and union of zero sets). LetIα,I, and Jbe ideals in the polyno- mial ring K[x1,..., xn]. Show that (i)/intersectiondisplay αZ(Iα)=Z/parenleftbigg ∑ αIα/parenrightbigg and (ii) Z(I)∪Z(J)=Z(IJ). The complement of a Zariski-closed subset of Knis said to be Zariski open . If I=(0)is the zero ideal, then Z(I) =Kn, and if I=(1)=K[x1,..., xn]is the entire ring, then Z(I)is the empty set ∅. Hence, both the empty set and Knare Zariski open. It now follows from Exercise A.5 that the Zariski-open subsets of Knform a topology on Kn, called the Zariski topology onKn. Since the zero set of a polynomial onR1is a ﬁnite set, the Zariski topology on R1is precisely the ﬁnite-complement topology of Example A.3. A.2 Subspace Topology Let(S,T)be a topological space and Aa subset of S. DeﬁneTAto be the collection of subsets TA={U∩A|U∈T}. By the distributive property of union and intersection, /uniondisplay α(Uα∩A)=/parenleftbigg/uniondisplay αUα/parenrightbigg ∩A and /intersectiondisplay i(Ui∩A)=/parenleftbigg/intersectiondisplay iUi/parenrightbigg ∩A, which shows that TAis closed under arbitrary unions and ﬁnite intersections. More- over,∅,A∈TA. SoTAis a topology on A, called the subspace topology or the A.3 Bases 321 relative topology ofAinS, and elements of TAare said to be open in A . To empha- size the fact that an open set UinAneed not be open in S, we also say that Uisopen relative to A orrelatively open in A . The subset AofSwith the subspace topology TAis called a subspace ofS. IfAis an open subset of a topological space S, then a subset of Ais relatively open in Aif and only if it is open in S. Example. Consider the subset A=[0,1]ofR1. In the subspace topology, the half- open interval [0,1/2[is open relative to A, because /bracketleftbig 0,1 2/bracketleftbig =/bracketrightbig −1 2,1 2/bracketleftbig ∩A. (See Figure A.2. ) ( [ ) ] 0 1 2-1 21 Fig. A.2. A relatively open subset of [0,1]. A.3 Bases It is generally difﬁcult to describe directly all the open sets in a topology T. What one can usually do is to describe a subcollection BofTsuch that any open set is expressible as a union of open sets in B. Deﬁnition A.6. A subcollection Bof a topology Ton a topological space Sis a basis for the topology Tif given an open set Uand point pinU, there is an open set B∈Bsuch that p∈B⊂U. We also say that Bgenerates the topology Tor thatB is abasis for the topological space S . Example. The collection of all open balls B(p,r)inRn, with p∈Rnandra positive real number, is a basis for the standard topology of Rn. Proposition A.7. A collection Bof open sets of S is a basis if and only if every open set in S is a union of sets in B. Proof. (⇒)SupposeBis a basis and Uis an open set in S. For every p∈U, there is a basic open set Bp∈Bsuch that p∈Bp⊂U. Therefore, U=/uniontext p∈UBp. (⇐)Suppose every open set in Sis a union of open sets in B. Given an open set U and a point pinU, since U=/uniontext Bα∈BBα, there is a Bα∈Bsuch that p∈Bα⊂U. Hence,Bis a basis. ⊓ ⊔ 322§A Point-Set Topology The following proposition gives a useful criterion for deciding whether a collec- tionBof subsets is a basis for some topology. Proposition A.8. A collection Bof subsets of a set S is a basis for some topology T on S if and only if (i)S is the union of all the sets in B, and (ii)given any two sets B 1and B 2∈Band a point p∈B1∩B2, there is a set B∈B such that p∈B⊂B1∩B2(Figure A.3 ). /Bullet pB B1B2 Fig. A.3. Criterion for a basis. Proof. (⇒)(i) follows from Proposition A.7. (ii) IfBis a basis, then B1andB2are open sets and hence so is B1∩B2. By the deﬁnition of a basis, there is a B∈Bsuch that p∈B⊂B1∩B2. (⇐)DeﬁneTto be the collection consisting of all sets that are unions of sets in B. Then the empty set ∅and the set Sare inTandTis clearly closed under arbitrary union. To show that Tis closed under ﬁnite intersection, let U=/uniontext µBµandV=/uniontext νBνbe inT, where Bµ,Bν∈B. Then U∩V=/parenleftbigg/uniondisplay µBµ/parenrightbigg ∩/parenleftbigg/uniondisplay νBν/parenrightbigg =/uniondisplay µ,ν(Bµ∩Bν). Thus, any pinU∩Vis in Bµ∩Bνfor some µ,ν. By (ii) there is a set BpinB such that p∈Bp⊂Bµ∩Bν. Therefore, U∩V=/uniondisplay p∈U∩VBp∈T. ⊓ ⊔ Proposition A.9. LetB={Bα}be a basis for a topological space S, and A a sub- space of S. Then{Bα∩A}is a basis for A. Proof. LetU′be any open set in Aandp∈U′. By the deﬁnition of subspace topol- ogy,U′=U∩Afor some open set UinS. Since p∈U∩A⊂U, there is a basic open setBαsuch that p∈Bα⊂U. Then p∈Bα∩A⊂U∩A=U′, which proves that the collection {Bα∩A|Bα∈B}is a basis for A.⊓ ⊔ A.4 First and Second Countability 323 A.4 First and Second Countability First and second countability of a topological space have to do with the countability of a basis. Before taking up these notions, we begin with an example. We say that a point inRnisrational if all of its coordinates are rational numbers. Let Qbe the set of rational numbers and Q+the set of positive rational numbers. From real analysis, it is well known that every open interval in Rcontains a rational number. Lemma A.10. Every open set in Rncontains a rational point. Proof. An open set UinRncontains an open ball B(p,r), which in turn contains an open cube ∏n i=1Ii, where Iiis the open interval ]pi−(r/√n),pi+(r/√n)[(see Problem A.4). For each i, let qibe a rational number in Ii. Then(q1,..., qn)is a rational point in ∏n i=1Ii⊂B(p,r)⊂U. ⊓ ⊔ Proposition A.11. The collection Bratof all open balls in Rnwith rational centers and rational radii is a basis for Rn. /Bullet /Bulletpqr Fig. A.4. A ball with rational center qand rational radius r/2. Proof. Given an open set UinRnand point pinU, there is an open ball B(p,r′)with positive real radius r′such that p∈B(p,r′)⊂U. Take a rational number rin]0,r′[. Then p∈B(p,r)⊂U. By Lemma A.10, there is a rational point qin the smaller ball B(p,r/2). We claim that p∈B/parenleftig q,r 2/parenrightig ⊂B(p,r). (A.1) (See Figure A.4. ) Since d(p,q)<r/2, we have p∈B(q,r/2). Next, if x∈B(q,r/2), then by the triangle inequality, d(x,p)≤d(x,q)+d(q,p)<r 2+r 2=r. Sox∈B(p,r). This proves the claim (A.1). Because p∈B(q,r/2)⊂U, the collec- tionBratof open balls with rational centers and rational radii is a basis for Rn.⊓ ⊔ 324§A Point-Set Topology Both of the sets QandQ+are countable. Since the centers of the balls in Brat are indexed by Qn, a countable set, and the radii are indexed by Q+, also a countable set, the collection Bratis countable. Deﬁnition A.12. A topological space is said to be second countable if it has a countable basis. Example A.13 .Proposition A.11 shows that Rnwith its standard topology is second countable. With the discrete topology, Rnwould not be second countable. More generally, any uncountable set with the discrete topology is not second countable. Proposition A.14. A subspace A of a second-countable space S is second countable. Proof. By Proposition A.9, if B={Bi}is a countable basis for S, thenBA:={Bi∩ A}is a countable basis for A. ⊓ ⊔ Deﬁnition A.15. LetSbe a topological space and pa point in S. Abasis of neighbor- hoods at p or aneighborhood basis at p is a collection B={Bα}of neighborhoods ofpsuch that for any neighborhood Uofp, there is a Bα∈Bsuch that p∈Bα⊂U. A topological space Sisﬁrst countable if it has a countable basis of neighborhoods at every point p∈S. Example. Forp∈Rn, let B(p,1/n)be the open ball of center pand radius 1 /nin Rn. Then{B(p,1/n)}∞ n=1is a neighborhood basis at p. Thus,Rnis ﬁrst countable. Example. An uncountable discrete space is ﬁrst countable but not second countable. Every second-countable space is ﬁrst countable (the proof is left to Problem A.18). Suppose pis a point in a ﬁrst-countable topological space and {Vi}∞ i=1is a count- able neighborhood basis at p. By taking Ui=V1∩···∩ Vi, we obtain a countable descending sequence U1⊃U2⊃U3⊃··· that is also a neighborhood basis at p. Thus, in the deﬁnition of ﬁrst countability, we may assume that at every point the countable neighborhood basis at the point is a descending sequence of open sets. A.5 Separation Axioms There are various separation axioms for a topological space. The only ones we will need are the Hausdorff condition and normality. Deﬁnition A.16. A topological space SisHausdorff if given any two distinct points x,yinS, there exist disjoint open sets U,Vsuch that x∈Uandy∈V. A Hausdorff space is normal if given any two disjoint closed sets F,GinS, there exist disjoint open sets U,Vsuch that F⊂UandG⊂V(Figure A.5 ). A.5 Separation Axioms 325 /Bullet /Bullet x y UVFG U V Fig. A.5. The Hausdorff condition and normality. Proposition A.17. Every singleton set (a one-point set )in a Hausdorff space S is closed. Proof. Letx∈S. For any y∈S−{x}, by the Hausdorff condition there exist an open setU∋xand an open set V∋ysuch that UandVare disjoint. In particular, y∈V⊂S−U⊂S−{x}. By the local criterion for openness (Lemma A.2), S−{x}is open. Therefore, {x}is closed. ⊓ ⊔ Example. The Euclidean space Rnis Hausdorff, for given distinct points x,yinRn, ifε=1 2d(x,y), then the open balls B(x,ε)andB(y,ε)will be disjoint (Figure A.6) . /Bullet/Bullet xy Fig. A.6. Two disjoint neighborhoods in Rn. Example A.18 ( Zariski topology ).LetS=Knbe a vector space of dimension nover a ﬁeld K, endowed with the Zariski topology. Every open set UinSis of the form S−Z(I), where Iis an ideal in K[x1,..., xn]. The open set Uis nonempty if and only if Iis not the zero ideal. In the Zariski topology any two nonempty open sets intersect: if U=S−Z(I)andV=S−Z(J)are nonempty, then IandJare nonzero ideals and U∩V=(S−Z(I))∩(S−Z(J)) =S−(Z(I)∪Z(J)) (de Morgan’s law) =S−Z(IJ), (Exercise A.5) 326§A Point-Set Topology which is nonempty because IJis not the zero ideal. Therefore, Knwith the Zariski topology is not Hausdorff. Proposition A.19. Any subspace A of a Hausdorff space S is Hausdorff. Proof. Letxandybe distinct points in A. Since Sis Hausdorff, there exist disjoint neighborhoods UandVofxandyrespectively in S. Then U∩AandV∩Aare disjoint neighborhoods of xandyrespectively in A. ⊓ ⊔ A.6 Product Topology TheCartesian product of two sets AandBis the set A×Bof all ordered pairs (a,b) with a∈Aandb∈B. Given two topological spaces XandY, consider the collection Bof subsets of X×Yof the form U×V, with Uopen in XandVopen in Y. We will call elements of Bbasic open sets inX×Y. IfU1×V1andU2×V2are inB, then (U1×V1)∩(U2×V2)=( U1∩U2)×(V1∩V2), which is also in B(Figure A.7). From this, it follows easily that Bsatisﬁes the conditions of Proposition A.8 for a basis and generates a topology on X×Y, called theproduct topology. Unless noted otherwise, this will always be the topology we assign to the product of two topological spaces. U1 U2V1V2 | | | |−− −− XY Fig. A.7. Intersection of two basic open subsets in X×Y. Proposition A.20. Let{Ui}and{Vj}be bases for the topological spaces X and Y , respectively. Then{Ui×Vj}is a basis for X×Y. Proof. Given an open set WinX×Yand point (x,y)∈W, we can ﬁnd a basic open setU×VinX×Ysuch that (x,y)∈U×V⊂W. Since Uis open in Xand{Ui}is a basis for X, A.7 Continuity 327 x∈Ui⊂U for some Ui. Similarly, y∈Vj⊂V for some Vj. Therefore, (x,y)∈Ui×Vj⊂U×V⊂W. By the deﬁnition of a basis, {Ui×Vj}is a basis for X×Y. ⊓ ⊔ Corollary A.21. The product of two second-countable spaces is second countable. Proposition A.22. The product of two Hausdorff spaces X and Y is Hausdorff. Proof. Given two distinct points (x1,y1),(x2,y2)inX×Y, without loss of generality we may assume that x1/ne}a⊔ionslash=x2. Since Xis Hausdorff, there exist disjoint open sets U1,U2inXsuch that x1∈U1andx2∈U2. Then U1×YandU2×Yare disjoint neighborhoods of (x1,y1)and(x2,y2)(Figure A.8), so X×Yis Hausdorff. ⊓ ⊔ (x1,y1)(x2,y2) x1 x2 XY U1 U2 | | | | Fig. A.8. Two disjoint neighborhoods in X×Y. The product topology can be generalized to the product of an arbitrary collection {Xα}α∈Aof topological spaces. Whatever the deﬁnition of the product topology, the projection maps παi:∏αXα→Xαi,παi(∏xα)=xαishould all be continuous. Thus, for each open set UαiinXαi, the inverse image π−1 αi(Uαi)should be open in ∏αXα. By the properties of open sets, a ﬁnite intersection/intersectiontextr i=1π−1 αi(Uαi)should also be open. Such a ﬁnite intersection is a set of the form ∏α∈AUα, where Uαis open in XαandUα=Xαfor all but ﬁnitely many α∈A. We deﬁne the product topology on the Cartesian product ∏α∈AXαto be the topology with basis consisting of sets of this form. The product topology is the coarsest topology on ∏αXαsuch that all the projection maps παi:∏αXα→Xαiare continuous. A.7 Continuity Letf:X→Ybe a function of topological spaces. Mimicking the deﬁnition from advanced calculus, we say that fiscontinuous at a point p inXif for every neigh- borhood Voff(p)inY, there is a neighborhood UofpinXsuch that f(U)⊂V. We say that fiscontinuous on X if it is continuous at every point of X. 328§A Point-Set Topology Proposition A.23 (Continuity in terms of open sets). A function f :X→Y is con- tinuous if and only if the inverse image of any open set is open. Proof. (⇒)Suppose Vis open in Y. To show that f−1(V)is open in X, let p∈f−1(V). Then f(p)∈V. Since fis assumed to be continuous at p, there is a neighborhood Uofpsuch that f(U)⊂V. Therefore, p∈U⊂f−1(V). By the local criterion for openness (Lemma A.2), f−1(V)is open in X. (⇐)Letpbe a point in X, and Va neighborhood of f(p)inY. By hypothesis, f−1(V)is open in X. Since f(p)∈V,p∈f−1(V). Then U=f−1(V)is a neighbor- hood of psuch that f(U)=f(f−1(V))⊂V, sofis continuous at p.⊓ ⊔ Example A.24 ( Continuity of an inclusion map ).IfAis a subspace of X, then the inclusion map i:A→X,i(a)=ais continuous. Proof. IfUis open in X, then i−1(U)=U∩A, which is open in the subspace topol- ogy of A. ⊓ ⊔ Example A.25 ( Continuity of a projection map ).The projection π:X×Y→X, π(x,y)=x, is continuous. Proof. LetUbe open in X. Then π−1(U) =U×Y, which is open in the product topology on X×Y. ⊓ ⊔ Proposition A.26. The composition of continuous maps is continuous: if f :X→Y and g :Y→Z are continuous, then g ◦f:X→Z is continuous. Proof. LetVbe an open subset of Z. Then (g◦f)−1(V)=f−1(g−1(V)), because for any x∈X, x∈(g◦f)−1(V)iffg(f(x))∈Vifff(x)∈g−1(V)iffx∈f−1(g−1(V)). By Proposition A.23, since gis continuous, g−1(V)is open in Y. Similarly, since f is continuous, f−1(g−1(V))is open in X. By Proposition A.23 again, g◦f:X→Z is continuous. ⊓ ⊔ IfAis a subspace of Xandf:X→Yis a function, the restriction offtoA, f|A:A→Y, is deﬁned by (f|A)(a)=f(a). With i:A→Xbeing the inclusion map, the restriction f|Ais the composite f◦i. Since both fandiare continuous (Example A.24) and the composition of continuous functions is continuous (Proposition A.26), we have the following corollary. A.8 Compactness 329 Corollary A.27. The restriction f|Aof a continuous function f :X→Y to a sub- space A is continuous. Continuity may also be phrased in terms of closed sets. Proposition A.28 (Continuity in terms of closed sets). A function f :X→Y is continuous if and only if the inverse image of any closed set is closed. Proof. Problem A.9. ⊓ ⊔ A map f:X→Yis said to be open if the image of every open set in Xis open inY; similarly, f:X→Yis said to be closed if the image of every closed set in Xis closed in Y. Iff:X→Yis a bijection, then its inverse map f−1:Y→Xis deﬁned. In this context, for any subset V⊂Y, the notation f−1(V)a priori has two meanings. It can mean either the inverse image of Vunder the map f, f−1(V)={x∈X|f(x)∈V}, or the image of Vunder the map f−1, f−1(V)={f−1(y)∈X|y∈V}. Fortunately, because y=f(x)if and only if x=f−1(y), these two meanings coincide. A.8 Compactness While its deﬁnition may not be intuitive, the notion of compactness is of central importance in topology. Let Sbe a topological space. A collection {Uα}of open subsets of Sis said to cover S or to be an open cover ofSifS⊂/uniontext αUα. Of course, because Sis the ambient space, this condition is equivalent to S=/uniontext αUα. Asubcover of an open cover is a subcollection whose union still contains S. The topological space Sis said to be compact if every open cover of Shas a ﬁnite subcover. With the subspace topology, a subset Aof a topological space Sis a topological space in its own right. The subspace Acan be covered by open sets in Aor by open sets in S. An open cover of A in S is a collection{Uα}of open sets in Sthat covers A. In this terminology, Ais compact if and only if every open cover of AinAhas a ﬁnite subcover. Proposition A.29. A subspace A of a topological space S is compact if and only if every open cover of A in S has a ﬁnite subcover. Proof. (⇒)Assume Acompact and let{Uα}be an open cover of AinS. This means that A⊂/uniontext αUα. Hence, A⊂/parenleftbigg/uniondisplay αUα/parenrightbigg ∩A=/uniondisplay α(Uα∩A). 330§A Point-Set Topology A Fig. A.9. An open cover of AinS. Since Ais compact, the open cover {Uα∩A}has a ﬁnite subcover {Uαi∩A}r i=1. Thus, A⊂r/uniondisplay i=1(Uαi∩A)⊂r/uniondisplay i=1Uαi, which means that{Uαi}r i=1is a ﬁnite subcover of {Uα}. (⇐)Suppose every open cover of AinShas a ﬁnite subcover, and let {Vα}be an open cover of AinA. Then each Vαis equal to Uα∩Afor some open set UαinS. Since A⊂/uniondisplay αVα⊂/uniondisplay αUα, by hypothesis there are ﬁnitely many sets Uαisuch that A⊂/uniontext iUαi. Hence, A⊂/parenleftbigg/uniondisplay iUαi/parenrightbigg ∩A=/uniondisplay i(Uαi∩A)=/uniondisplay iVαi. So{Vαi}is a ﬁnite subcover of {Vα}that covers A. Therefore, Ais compact.⊓ ⊔ Proposition A.30. A closed subset F of a compact topological space S is compact. Proof. Let{Uα}be an open cover of FinS. The collection{Uα,S−F}is then an open cover of S. By the compactness of S, there is a ﬁnite subcover {Uαi,S−F}that covers S, soF⊂/uniontext iUαi. This proves that Fis compact. ⊓ ⊔ Proposition A.31. In a Hausdorff space S, it is possible to separate a compact subset K and a point p not in K by disjoint open sets; i.e., there exist an open set U ⊃K and an open set V∋p such that U∩V=∅. Proof. By the Hausdorff property, for every x∈K, there are disjoint open sets Ux∋x andVx∋p. The collection{Ux}x∈Kis a cover of Kby open subsets of S. Since Kis compact, it has a ﬁnite subcover {Uxi}. LetU=/uniontext iUxiandV=/intersectiontext iVxi. Then Uis an open set of Scontaining K. Being the intersection of ﬁnitely many open sets containing p,Vis an open set containing p. Moreover, the set U∩V=/uniondisplay i(Uxi∩V) is empty, since each Uxi∩Vis contained in Uxi∩Vxi, which is empty. ⊓ ⊔ A.8 Compactness 331 Proposition A.32. Every compact subset K of a Hausdorff space S is closed. Proof. By the preceding proposition, for every point pinS−K, there is an open set Vsuch that p∈V⊂S−K. This proves that S−Kis open. Hence, Kis closed.⊓ ⊔ Exercise A.33 (Compact Hausdorff space).* Prove that a compact Hausdorff space is nor- mal. (Normality was deﬁned in Deﬁnition A.16.) Proposition A.34. The image of a compact set under a continuous map is compact. Proof. Letf:X→Ybe a continuous map and Ka compact subset of X. Suppose {Uα}is a cover of f(K)by open subsets of Y. Since fis continuous, the inverse images f−1(Uα)are all open. Moreover, K⊂f−1(f(K))⊂f−1/parenleftbigg/uniondisplay αUα/parenrightbigg =/uniondisplay αf−1(Uα). So{f−1(Uα)}is an open cover of KinX. By the compactness of K, there is a ﬁnite subcollection{f−1(Uαi)}such that K⊂/uniondisplay if−1(Uαi)=f−1/parenleftbigg/uniondisplay iUαi/parenrightbigg . Then f(K)⊂/uniontext iUαi. Thus, f(K)is compact. ⊓ ⊔ Proposition A.35. A continuous map f :X→Y from a compact space X to a Haus- dorff space Y is a closed map. Proof. LetFbe a closed subset of the compact space X. By Proposition A.30, Fis compact. As the image of a compact set under a continuous map, f(F)is compact in Y(Proposition A.34). As a compact subset of the Hausdorff space Y,f(F)is closed (Proposition A.32). ⊓ ⊔ A continuous bijection f:X→Ywhose inverse is also continuous is called a homeomorphism. Corollary A.36. A continuous bijection f :X→Y from a compact space X to a Hausdorff space Y is a homeomorphism. Proof. By Proposition A.28, to show that f−1:Y→Xis continuous, it sufﬁces to prove that for every closed set FinX, the set(f−1)−1(F)=f(F)is closed in Y, i.e., that fis a closed map. The corollary then follows from Proposition A.35. ⊓ ⊔ Exercise A.37 (Finite union of compact sets). Prove that a ﬁnite union of compact subsets of a topological space is compact. We mention without proof an important result. For a proof, see [29, Theo- rem 26.7, p. 167, and Theorem 37.3, p. 234]. Theorem A.38 (The Tychonoff theorem). The product of any collection of compact spaces is compact in the product topology. 332§A Point-Set Topology A.9 Boundedness in Rn A subset AofRnis said to be bounded if it is contained in some open ball B(p,r); otherwise, it is unbounded . Proposition A.39. A compact subset of Rnis bounded. Proof. IfAwere an unbounded subset of Rn, then the collection {B(0,i)}∞ i=1of open balls with radius increasing to inﬁnity would be an open cover of AinRnthat does not have a ﬁnite subcover. ⊓ ⊔ By Propositions A.39 and A.32, a compact subset of Rnis closed and bounded. The converse is also true. Theorem A.40 (The Heine–Borel theorem). A subset of Rnis compact if and only if it is closed and bounded. For a proof, see for example [29]. A.10 Connectedness Deﬁnition A.41. A topological space Sisdisconnected if it is the union S=U∪V of two disjoint nonempty open subsets UandV(Figure A.10 ). It is connected if it is not disconnected. A subset AofSisdisconnected if it is disconnected in the subspace topology. U V Fig. A.10. A disconnected space. Proposition A.42. A subset A of a topological space S is disconnected if and only if there are open sets U and V in S such that (i)U∩A/ne}a⊔ionslash=∅, V∩A/ne}a⊔ionslash=∅, (ii)U∩V∩A=∅, (iii)A⊂U∪V. A pair of open sets in S with these properties is called a separation of A(Figure A.11). Proof. Problem A.15. ⊓ ⊔ A.11 Connected Components 333 U V A Fig. A.11. A separation of A. Proposition A.43. The image of a connected space X under a continuous map f:X→Y is connected. Proof. Suppose f(X)is not connected. Then there is a separation {U,V}off(X)in Y. By the continuity of f, both f−1(U)and f−1(V)are open in X. We claim that {f−1(U),f−1(V)}is a separation of X. (i) Since U∩f(X)/ne}a⊔ionslash=∅, the open set f−1(U)is nonempty. (ii) If x∈f−1(U)∩f−1(V), then f(x)∈U∩V∩f(X)=∅, a contradiction. Hence, f−1(U)∩f−1(V)is empty. (iii) Since f(X)⊂U∪V, we have X⊂f−1(U∪V)=f−1(U)∪f−1(V). The existence of a separation of Xcontradicts the connectedness of X. This contra- diction proves that f(X)is connected. ⊓ ⊔ Proposition A.44. In a topological space S, the union of a collection of connected subsets A αhaving a point p in common is connected. Proof. Suppose/uniontext αAα=U∪V, where UandVare disjoint open subsets of/uniontext αAα. The point p∈/uniontext αAαbelongs to UorV. Assume without loss of generality that p∈U. For each α, Aα=Aα∩(U∪V)=( Aα∩U)∪(Aα∩V). The two open sets Aα∩UandAα∩VofAαare clearly disjoint. Since p∈Aα∩U, Aα∩Uis nonempty. By the connectedness of Aα,Aα∩Vmust be empty for all α. Hence, V=/parenleftig/uniondisplay αAα/parenrightig ∩V=/uniondisplay α(Aα∩V) is empty. So/uniontext αAαmust be connected. ⊓ ⊔ A.11 Connected Components Letxbe a point in a topological space S. By Proposition A.44, the union Cxof all connected subsets of Scontaining xis connected. It is called the connected compo- nent ofScontaining x. 334§A Point-Set Topology Proposition A.45. Let C xbe a connected component of a topological space S. Then a connected subset A of S is either disjoint from C xor is contained entirely in C x. Proof. IfAandCxhave a point in common, then by Proposition A.44, A∪Cxis a connected set containing x. Hence, A∪Cx⊂Cx, which implies that A⊂Cx.⊓ ⊔ Accordingly, the connected component Cxis the largest connected subset of S containing xin the sense that it contains every connected subset of Scontaining x. Corollary A.46. For any two points x ,y in a topological space S, the connected com- ponents C xand C yeither are disjoint or coincide. Proof. IfCxandCyare not disjoint, then by Proposition A.45, they are contained in each other. In this case, Cx=Cy. ⊓ ⊔ As a consequence of Corollary A.46, the connected components of Spartition S into disjoint subsets. A.12 Closure LetSbe a topological space and Aa subset of S. Deﬁnition A.47. Theclosure ofAinS, denoted by A, cl(A), or cl S(A), is deﬁned to be the intersection of all the closed sets containing A. The advantage of the bar notation Ais its simplicity, while the advantage of the clS(A)notation is its indication of the ambient space S. IfA⊂B⊂S, then the closure ofAinBand the closure of AinSneed not be the same. In this case, it is useful to have the notations cl B(A)and cl M(A)for the two closures. As an intersection of closed sets, Ais a closed set. It is the smallest closed set containing Ain the sense that any closed set containing Acontains A. Proposition A.48 (Local characterization of closure). Let A be a subset of a topo- logical space S. A point p ∈S is in the closure cl(A)if and only if every neighborhood of p contains a point of A (Figure A.12 ). Here by “local,” we mean a property satisﬁed by a basis of neighborhoods at a point. Proof. We will prove the proposition in the form of its contrapositive: p/∈cl(A)⇐⇒ there is a neighborhood of pdisjoint from A. (⇒) Suppose p/∈cl(A)=/intersectiondisplay {Fclosed in S|F⊃A}. Then p/∈some closed set Fcontaining A. It follows that p∈S−F, an open set disjoint from A. (⇐) Suppose p∈an open set Udisjoint from A. Then the complement F:=S−U is a closed set containing Aand not containing p. Therefore, p/∈cl(A).⊓ ⊔ A.12 Closure 335 /BulletpA Fig. A.12. Every neighborhood of pcontains a point of A. Example. The closure of the open disk B(0,r)inR2is the closed disk B(0,r)={p∈R2|d(p,0)≤r}. Deﬁnition A.49. A point pinSis an accumulation point ofAif every neighborhood ofpinScontains a point of Aother than p. The set of all accumulation points of A is denoted by ac (A). IfUis a neighborhood of pinS, we call U−{p}adeleted neighborhood ofp. An equivalent condition for pto be an accumulation point of Ais to require that every deleted neighborhood of pinScontain a point of A. In some books an accumulation point is called a limit point . Example. IfA=[0,1[∪{2}inR1, then the closure of Ais[0,1]∪{2}, but the set of accumulation points of Ais only the closed interval [0,1]. Proposition A.50. Let A be a subset of a topological space S. Then cl(A)=A∪ac(A). Proof. (⊃)By deﬁnition, A⊂cl(A). By the local characterization of closure (Proposition A.48), ac(A)⊂cl(A). Hence, A∪ac(A)⊂cl(A). (⊂)Suppose p∈cl(A). Either p∈Aorp/∈A. Ifp∈A, then p∈A∪ac(A). Suppose p/∈A. By Proposition A.48, every neighborhood of pcontains a point of A, which cannot be p, since p/∈A. Therefore, every deleted neighborhood of pcontains a point of A. In this case, p∈ac(A)⊂A∪ac(A). So cl(A)⊂A∪ac(A). ⊓ ⊔ Proposition A.51. A set A is closed if and only if A =A. Proof. (⇐)IfA=A, then Ais closed because Ais closed. (⇒)Suppose Ais closed. Then Ais a closed set containing A, so that A⊂A. Because A⊂A, equality holds. ⊓ ⊔ 336§A Point-Set Topology Proposition A.52. If A⊂B in a topological space S, then A⊂B. Proof. Since Bcontains B, it also contains A. As a closed subset of Scontaining A, it contains Aby deﬁnition. ⊓ ⊔ Exercise A.53 (Closure of a ﬁnite union or ﬁnite intersection). LetAandBbe subsets of a topological space S. Prove the following: (a)A∪B=A∪B, (b)A∩B⊂A∩B. The example of A=]a,0[andB=]0,b[in the real line shows that in general, A∩B/ne}a⊔ionslash=A∩B. A.13 Convergence LetSbe a topological space. A sequence inSis a map from the set Z+of positive integers to S. We write a sequence as /an}bracke⊔le{⊔xi/an}bracke⊔ri}h⊔orx1,x2,x3,.... Deﬁnition A.54. The sequence/an}bracke⊔le{⊔xi/an}bracke⊔ri}h⊔converges topif for every neighborhood Uof p, there is a positive integer Nsuch that for all i≥N,xi∈U. In this case we say that pis alimit of the sequence/an}bracke⊔le{⊔xi/an}bracke⊔ri}h⊔and write xi→por lim i→∞xi=p. Proposition A.55 (Uniqueness of the limit). In a Hausdorff space S, if a sequence /an}bracke⊔le{⊔xi/an}bracke⊔ri}h⊔converges to p and to q, then p =q. Proof. Problem A.19. ⊓ ⊔ Thus, in a Hausdorff space we may speak of thelimit of a convergent sequence. Proposition A.56 (The sequence lemma). Let S be a topological space and A a subset of S. If there is a sequence /an}bracke⊔le{⊔ai/an}bracke⊔ri}h⊔in A that converges to p, then p ∈cl(A). The converse is true if S is ﬁrst countable. Proof. (⇒)Suppose ai→p, where ai∈Afor all i. By the deﬁnition of convergence, every neighborhood Uofpcontains all but ﬁnitely many of the points ai. In particular, U contains a point in A. By the local characterization of closure (Proposition A.48), p∈cl(A). (⇐)Suppose p∈cl(A). Since Sis ﬁrst countable, we can ﬁnd a countable basis of neighborhoods{Un}atpsuch that U1⊃U2⊃···. By the local characterization of closure, in each Uithere is a point ai∈A. We claim that the sequence/an}bracke⊔le{⊔ai/an}bracke⊔ri}h⊔converges to p. IfUis any neighborhood of p, then by the deﬁnition of a basis of neighborhoods at p, there is a UNsuch that p∈UN⊂U. For alli≥N, we then have Ui⊂UN⊂U. Therefore, for all i≥N, ai∈Ui⊂U. This proves that/an}bracke⊔le{⊔ai/an}bracke⊔ri}h⊔converges to p. ⊓ ⊔ A.13 Convergence 337 Problems A.1. Set theory IfU1andU2are subsets of a set X, and V1andV2are subsets of a set Y, prove that (U1×V1)∩(U2×V2)=( U1∩U2)×(V1∩V2). A.2. Union and intersection Suppose U1∩V1=U2∩V2=∅in a topological space S. Show that the intersection U1∩U2 is disjoint from the union V1∪V2. (Hint: Use the distributive property of an intersection over a union.) A.3. Closed sets LetSbe a topological space. Prove the following two statements. (a) If{Fi}n i=1is a ﬁnite collection of closed sets in S, then/uniontextn i=1Fiis closed. (b) If{Fα}α∈Ais an arbitrary collection of closed sets in S, then/intersectiontext αFαis closed. A.4. Cubes versus balls Prove that the open cube ]−a,a[nis contained in the open ball B(0,√na), which in turn is contained in the open cube ]−√na,√na[n. Therefore, open cubes with arbitrary centers in Rnform a basis for the standard topology on Rn. A.5. Product of closed sets Prove that if Ais closed in XandBis closed in Y, then A×Bis closed in X×Y. A.6. Characterization of a Hausdorff space by its diagonal LetSbe a topological space. The diagonal ∆inS×Sis the set ∆={(x,x)∈S×S}. Prove that Sis Hausdorff if and only if the diagonal ∆is closed in S×S. (Hint: Prove that Sis Hausdorff if and only if S×S−∆is open in S×S.) A.7. Projection Prove that if XandYare topological spaces, then the projection π:X×Y→X,π(x,y) =x, is an open map. A.8. The ε-δcriterion for continuity Prove that a function f:A→Rmis continuous at p∈Aif and only if for every ε>0, there exists a δ>0 such that for all x∈Asatisfying d(x,p)<δ, one has d(f(x),f(p))<ε. A.9. Continuity in terms of closed sets Prove Proposition A.28. A.10. Continuity of a map into a product LetX,Y1, and Y2be topological spaces. Prove that a map f= (f1,f2):X→Y1×Y2is continuous if and only if both components fi:X→Yiare continuous. A.11. Continuity of the product map Given two maps f:X→X′andg:Y→Y′of topological spaces, we deﬁne their product to be f×g:X×Y→X′×Y′,(f×g)(x,y)=( f(x),g(y)). Note that if π1:X×Y→Xandπ2:X×Y→Yare the two projections, then f×g= (f◦π1,f◦π2). Prove that f×gis continuous if and only if both fandgare continuous. 338§A Point-Set Topology A.12. Homeomorphism Prove that if a continuous bijection f:X→Yis a closed map, then it is a homeomorphism (cf. Corollary A.36). A.13.* The Lindel ¨of condition Show that if a topological space is second countable, then it is Lindel¨ of; i.e., every open cover has a countable subcover. A.14. Compactness Prove that a ﬁnite union of compact sets in a topological space Sis compact. A.15.* Disconnected subset in terms of a separation Prove Proposition A.42. A.16. Local connectedness A topological space Sis said to be locally connected at p ∈Sif for every neighborhood Uof p, there is a connected neighborhood Vofpsuch that V⊂U. The space Sislocally connected if it is locally connected at every point. Prove that if Sis locally connected, then the connected components of Sare open. A.17. Closure LetUbe an open subset and Aan arbitrary subset of a topological space S. Prove that U∩¯A/ne}a⊔ionslash= ∅if and only if U∩A/ne}a⊔ionslash=∅. A.18. Countability Prove that every second-countable space is ﬁrst countable. A.19.* Uniqueness of the limit Prove Proposition A.55. A.20.* Closure in a product LetSandYbe topological spaces and A⊂S. Prove that clS×Y(A×Y)=clS(A)×Y in the product space S×Y. A.21. Dense subsets A subset Aof a topological space Sis said to be dense inSif its closure cl S(A)equals S. (a) Prove that Ais dense in Sif and only if for every p∈S, every neighborhood Uofpcontains a point of A. (b) Let Kbe a ﬁeld. Prove that a Zariski-open subset UofKnis dense in Kn. (Hint: Example A.18.) B.2 The Implicit Function Theorem 339 §B The Inverse Function Theorem on Rnand Related Results This appendix reviews three logically equivalent theorems from real analysis: the inverse function theorem, the implicit function theorem, and the constant rank theo- rem, which describe the local behavior of a C∞map from RntoRm. We will assume the inverse function theorem and from it deduce the other two in the simplest cases. In Section 11 these theorems are applied to manifolds in order to clarify the localbehavior of a C ∞map when the map has maximal rank at a point or constant rank in a neighborhood. B.1 The Inverse Function Theorem AC∞map f:U→Rndeﬁned on an open subset UofRnislocally invertible or a local diffeomorphism at a point pinUiffhas a C∞inverse in some neighborhood of p. The inverse function theorem gives a criterion for a map to be locally invertible. We call the matrix J f=[∂fi/∂xj]of partial derivatives of ftheJacobian matrix of fand its determinant det [∂fi/∂xj]theJacobian determinant off. Theorem B.1 (Inverse function theorem). Let f :U→Rnbe a C∞map deﬁned on an open subset U of Rn. At any point p in U, the map f is invertible in some neigh- borhood of p if and only if the Jacobian determinant det[∂fi/∂xj(p)]is not zero. For a proof, see for example [35, Theorem 9.24, p. 221]. Although the inverse function theorem apparently reduces the invertibility of fon an open set to a single number at p, because the Jacobian determinant is a continuous function, the non- vanishing of the Jacobian determinant at pis equivalent to its nonvanishing in a neighborhood of p. Since the linear map represented by the Jacobian matrix J f(p)is the best linear approximation to fatp, it is plausible that fis invertible in a neighborhood of pif and only if J f(p)is also, i.e., if and only if det (J f(p))/ne}a⊔ionslash=0. B.2 The Implicit Function Theorem In an equation such as f(x,y) =0, it is often impossible to solve explicitly for one of the variables in terms of the other. If we can show the existence of a function y=h(x), which we may or may not be able to write down explicitly, such that f(x,h(x)) = 0, then we say that f(x,y) =0 can be solved implicitly foryin terms ofx. The implicit function theorem provides a sufﬁcient condition on a system of equations fi(x1,..., xn)=0,i=1,..., m, under which locally a set of variables can be solved implicitly as C∞functions of the other variables. Example. Consider the equation 340§B The Inverse Function Theorem on Rnand Related Results f(x,y)=x2+y2−1=0. The solution set is the unit circle in the xy-plane. xy 1 −1 Fig. B.1. The unit circle. From the picture we see that in a neighborhood of any point other than (±1,0), yis a function of x. Indeed, y=±/radicalbig 1−x2, and either function is C∞as long as x/ne}a⊔ionslash=±1. At(±1,0), there is no neighborhood on which yis a function of x. On a smooth curve f(x,y)=0 inR2, ycan be expressed as a function of xin a neighborhood of a point (a,b) ⇐⇒ the tangent line to f(x,y)=0 at(a,b)is not vertical ⇐⇒ the normal vector grad f:=/an}bracke⊔le{⊔fx,fy/an}bracke⊔ri}h⊔tof(x,y)=0 at(a,b) is not horizontal ⇐⇒ fy(a,b)/ne}a⊔ionslash=0. The implicit function theorem generalizes this condition to higher dimensions. We will deduce the implicit function theorem from the inverse function theorem. Theorem B.2 (Implicit function theorem). Let U be an open subset in Rn×Rmand f:U→Rma C∞map. Write (x,y) =( x1,..., xn,y1,..., ym)for a point in U. At a point(a,b)∈U where f (a,b)=0and the determinant det[∂fi/∂yj(a,b)]is nonzero, there exist a neighborhood A ×B of(a,b)in U and a unique function h :A→B such that in A×B⊂U⊂Rn×Rm, f(x,y)=0⇐⇒ y=h(x). Moreover, h is C∞. Proof. To solve f(x,y) =0 for yin terms of xusing the inverse function theorem, we ﬁrst turn it into an inverse problem. For this, we need a map between two open B.2 The Implicit Function Theorem 341 xy /Bullet (a,b)U1 f(x,y)=0F=(x,f) uv V1 /Bullet (a,0) Fig. B.2. F−1maps the u-axis to the zero set of f. sets of the same dimension. Since f(x,y)is a map from an open set UinRn+mto Rm, it is natural to extend fto a map F:U→Rn+mby adjoining xto it as the ﬁrst n components: F(x,y)=( u,v)=( x,f(x,y)). To simplify the exposition, we will assume in the rest of the proof that n=m=1. Then the Jacobian matrix of Fis JF=/bracketleftbigg 1 0 ∂f/∂x∂f/∂y/bracketrightbigg . At the point (a,b), detJF(a,b)=∂f ∂y(a,b)/ne}a⊔ionslash=0. By the inverse function theorem, there are neighborhoods U1of(a,b)andV1of F(a,b) = ( a,0)inR2such that F:U1→V1is a diffeomorphism with C∞inverse F−1(Figure B.2 ). Since F:U1→V1is deﬁned by u=x, v=f(x,y), the inverse map F−1:V1→U1must be of the form x=u, y=g(u,v) for some C∞function g:V1→R. Thus, F−1(u,v)=( u,g(u,v)). The two compositions F−1◦FandF◦F−1give (x,y)=( F−1◦F)(x,y)=F−1(x,f(x,y))=( x,g(x,f(x,y))), (u,v)=( F◦F−1)(u,v)=F(u,g(u,v))=( u,f(u,g(u,v))). Hence, 342§B The Inverse Function Theorem on Rnand Related Results y=g(x,f(x,y)) for all(x,y)∈U1, (B.1) v=f(u,g(u,v))for all(u,v)∈V1. (B.2) Iff(x,y) =0, then (B.1) gives y=g(x,0). This suggests that we deﬁne h(x)= g(x,0)for all x∈R1for which (x,0)∈V1. The set of all such xis homeomorphic to V1∩(R1×{0})and is an open subset of R1. Since gisC∞by the inverse function theorem, his also C∞. Claim. For(x,y)∈U1such that (x,0)∈V1, f(x,y)=0⇐⇒ y=h(x). Proof (of Claim ). (⇒)As we saw already, from (B.1), if f(x,y)=0, then y=g(x,f(x,y))= g(x,0)=h(x). (B.3) (⇐)Ify=h(x)and in (B.2) we set (u,v)=( x,0), then 0=f(x,g(x,0))= f(x,h(x))= f(x,y). ⊓ ⊔ By the claim, in some neighborhood of (a,b)∈U1, the zero set of f(x,y)is precisely the graph of h. To ﬁnd a product neighborhood of (a,b)as in the statement of the theorem, let A1×Bbe a neighborhood of (a,b)contained in U1and let A= h−1(B)∩A1. Since his continuous, Ais open in the domain of hand hence in R1. Then h(A)⊂B, A×B⊂A1×B⊂U1,and A×{0}⊂V1. By the claim, for (x,y)∈A×B, f(x,y)=0⇐⇒ y=h(x). Equation (B.3) proves the uniqueness of h. ⊓ ⊔ Replacing a partial derivative such as ∂f/∂ywith a Jacobian matrix [∂fi/∂yj], we can prove the general case of the implicit function theorem in exactly the same way. Of course, in the theorem y1,..., ymneed not be the last mcoordinates in Rn+m; they can be any set of mcoordinates in Rn+m. Theorem B.3. The implicit function theorem is equivalent to the inverse function theorem. Proof. We have already shown, at least for one typical case, that the inverse function theorem implies the implicit function theorem. We now prove the reverse implica- tion. So assume the implicit function theorem, and let f:U→Rnbe aC∞map deﬁned on an open subset UofRnsuch that at some point p∈U, the Jacobian determinant B.3 Constant Rank Theorem 343 det[∂fi/∂xj(p)]is nonzero. Finding a local inverse for y=f(x)near pamounts to solving the equation g(x,y)=f(x)−y=0 forxin terms of ynear(p,f(p)). Note that ∂gi/∂xj=∂fi/∂xj. Hence, det/bracketleftbigg∂gi ∂xj(p,f(p))/bracketrightbigg =det/bracketleftbigg∂fi ∂xj(p)/bracketrightbigg /ne}a⊔ionslash=0. By the implicit function theorem, xcan be expressed in terms of ylocally near (p,f(p)); i.e., there is a C∞function x=h(y)deﬁned in a neighborhood of f(p) inRnsuch that g(x,y)=f(x)−y=f(h(y))−y=0. Thus, y=f(h(y)). Since y=f(x), x=h(y)=h(f(x)). Therefore, fandhare inverse functions deﬁned near pandf(p)respectively.⊓ ⊔ B.3 Constant Rank Theorem Every C∞map f:U→Rmon an open set UofRnhas a rank at each point pinU, namely the rank of its Jacobian matrix [∂fi/∂xj(p)]. Theorem B.4 (Constant rank theorem). If f:Rn⊃U→Rmhas constant rank k in a neighborhood of a point p ∈U, then after a suitable change of coordinates near p in U and f (p)inRm, the map f assumes the form (x1,..., xn)/ma√s⊔o→(x1,..., xk,0,..., 0). More precisely, there are a diffeomorphism G of a neighborhood of p in U sending p to the origin in Rnand a diffeomorphism F of a neighborhood of f (p)inRmsending f(p)to the origin in Rmsuch that (F◦f◦G)−1(x1,..., xn)=( x1,..., xk,0,..., 0). Proof (for n=m=2, k=1).Suppose f=(f1,f2):R2⊃U→R2has constant rank 1 in a neighborhood of p∈U. By reordering the functions f1,f2or the variables x, y, we may assume that ∂f1/∂x(p)/ne}a⊔ionslash=0. (Here we are using the fact that fhas rank ≥1 atp.) Deﬁne G:U→R2by G(x,y)=( u,v)=( f1(x,y),y). The Jacobian matrix of Gis JG=/bracketleftbigg ∂f1/∂x∂f1/∂y 0 1/bracketrightbigg . 344§B The Inverse Function Theorem on Rnand Related Results Since det JG(p)=∂f1/∂x(p)/ne}a⊔ionslash=0, by the inverse function theorem there are neigh- borhoods U1ofp∈R2andV1ofG(p)∈R2such that G:U1→V1is a diffeomor- phism. By making U1a sufﬁciently small neighborhood of p, we may assume that f has constant rank 1 on U1. OnV1, (u,v)=( G◦G−1)(u,v)=( f1◦G−1,y◦G−1)(u,v). Comparing the ﬁrst components gives u=(f1◦G−1)(u,v). Hence, (f◦G−1)(u,v)=( f1◦G−1,f2◦G−1)(u,v) =(u,f2◦G−1(u,v)) =(u,h(u,v)), where we set h=f2◦G−1. Because G−1:V1→U1is a diffeomorphism and fhas constant rank 1 on U1, the composite f◦G−1has constant rank 1 on V1. Its Jacobian matrix is J(f◦G−1)=/bracketleftbigg1 0 ∂h/∂u∂h/∂v/bracketrightbigg . For this matrix to have constant rank 1, ∂h/∂vmust be identically zero on V1. (Here we are using the fact that fhas rank≤1 in a neighborhood of p.) Thus, his a function of ualone and we may write (f◦G−1)(u,v)=( u,h(u)). Finally, let F:R2→R2be the change of coordinates F(x,y) = ( x,y−h(x)). Then (F◦f◦G−1)(u,v)=F(u,h(u))=( u,h(u)−h(u))=( u,0).⊓ ⊔ Example B.5.If aC∞map f:Rn⊃U→Rndeﬁned on an open subset UofRnhas nonzero Jacobian determinant det (J f(p))at a point p∈U, then by continuity it has nonzero Jacobian determinant in a neighborhood of p. Therefore, it has constant rank nin a neighborhood of p. Problems B.1.* The rank of a matrix Therank of a matrix A, denoted by rk A, is deﬁned to be the number of linearly independent columns of A. By a theorem in linear algebra, it is also the number of linearly independent rows of A. Prove the following lemma. Lemma. Let A be an m×n matrix(not necessarily square ),and k a positive integer. Then rkA≥k if and only if A has a nonsingular k ×k submatrix. Equivalently, rkA≤k−1if and only if all k×k minors of A vanish. (A k×kminor of a matrix A is the determinant of a k ×k submatrix of A .) B.3 Constant Rank Theorem 345 B.2.* Matrices of rank at most r For an integer r≥0, deﬁne Drto be the subset of Rm×nconsisting of all m×nreal matrices of rank at most r. Show that Dris a closed subset of Rm×n. (Hint: Use Problem B.1.) B.3.* Maximal rank We say that the rank of an m×nmatrix Aismaximal if rkA=min(m,n). Deﬁne Dmaxto be the subset of Rm×nconsisting of all m×nmatrices of maximal rank r. Show that Dmaxis an open subset of Rm×n. (Hint: Suppose n≤m. Then Dmax=Rm×n−Dn−1. Apply Problem B.2.) B.4.* Degeneracy loci and maximal-rank locus of a map LetF:S→Rm×nbe a continuous map from a topological space Sto the space Rm×n. The degeneracy locus of rank r ofFis deﬁned to be Dr(F):={x∈S|rkF(x)≤r}. (a) Show that the degeneracy locus Dr(F)is a closed subset of S. (Hint:Dr(F)=F−1(Dr), where Drwas deﬁned in Problem B.2.) (b) Show that the maximal-rank locus ofF, Dmax(F):={x∈S|rkF(x)is maximal}, is an open subset of S. B.5. Rank of a composition of linear maps Suppose V,W,V′,W′are ﬁnite-dimensional vector spaces. (a) Prove that if the linear map L:V→Wis surjective, then for any linear map f:W→W′, rk(f◦L)=rkf. (b) Prove that if the linear map L:V→Wis injective, then for any linear map g:V′→V, rk(L◦g)=rkg. B.6. Constant rank theorem Generalize the proof of the constant rank theorem (Theorem B.4) in the text to arbitrary n,m, andk. B.7. Equivalence of the constant rank theorem and the inverse function theorem Use the constant rank theorem (Theorem B.4) to prove the inverse function theorem (Theo- rem B.1). Hence, the two theorems are equivalent. 346§C Existence of a Partition of Unity in General §C Existence of a Partition of Unity in General This appendix contains a proof of Theorem 13.7 on the existence of a C∞partition of unity on a general manifold. Lemma C.1. Every manifold M has a countable basis all of whose elements have compact closure. Recall that if Ais a subset of a topological space X, the notation Adenotes the closure of AinX. Proof (of Lemma C.1 ).Start with a countable basis BforMand consider the sub- collection Sof elements in Bthat have compact closure. We claim that Sis again a basis. Given an open subset U⊂Mand point p∈U, choose a neighborhood Vof psuch that V⊂UandVhas compact closure. This is always possible since Mis locally Euclidean. SinceBis a basis, there is an open set B∈Bsuch that p∈B⊂V⊂U. Then B⊂V. Because Vis compact, so is the closed subset B. Hence, B∈S. Since for any open set Uand any p∈U, we have found a set B∈Ssuch that p∈B⊂U, the collection Sof open sets is a basis. ⊓ ⊔ Proposition C.2. Every manifold M has a countable increasing sequence of subsets V1⊂V1⊂V2⊂V2⊂···, with each V iopen and Vicompact, such that M is the union of the V i’s(Figure C.1). Proof. By Lemma C.1, Mhas a countable basis {Bi}∞ i=1with each Bicompact. Any basis of Mof course covers M. Set V1=B1. By compactness, V1is covered by ﬁnitely many of the Bi’s. Deﬁne i1to be the smallest integer ≥2 such that V1⊂B1∪B2∪···∪ Bi1. Suppose open sets V1,..., Vmhave been deﬁned, each with compact closure. As before, by compactness, Vmis covered by ﬁnitely many of the Bi’s. If imis the smallest integer≥m+1 and≥im−1such that Vm⊂B1∪B2∪···∪ Bim, then we set Vm+1=B1∪B2∪···∪ Bim. Since a ﬁnite union of compact sets is compact and §C Existence of a Partition of Unity in General 347 Vm+1⊂B1∪B2∪···∪ Bim is a closed subset of a compact set, Vm+1is compact. Since im≥m+1,Bm+1⊂Vm+1. Thus, M=/uniondisplay Bi⊂/uniondisplay Vi⊂M. This proves that M=/uniontext∞ i=1Vi. ⊓ ⊔ ) ) ) ) ) ) ) V1 V2 V3 Vi−1Vi Vi+1 Vi+2 ... [ ] ( )compact open Fig. C.1. A nested open cover. Deﬁne V0to be the empty set. For each i≥1, because Vi+1−Viis a closed subset of the compact Vi+1, it is compact. Moreover, it is contained in the open set Vi+2−Vi−1. Theorem 13.7 (Existence of a C∞partition of unity). Let{Uα}α∈Abe an open cover of a manifold M. (i)There is a C∞partition of unity{ϕk}∞ k=1with every ϕkhaving compact support such that for each k, supp ϕk⊂Uαfor some α∈A. (ii)If we do not require compact support, then there is a C∞partition of unity{ρα} subordinate to{Uα}. Proof. (i) Let{Vi}∞ i=0be an open cover of Mas in Proposition C.2, with V0the empty set. The idea of the proof is quite simple. For each i, we ﬁnd ﬁnitely many smooth bump functions ψi jonM, each with compact support in the open set Vi+2−Vi−1as well as in some Uα, such that their sum ∑jψi jis positive on the compact set Vi+1−Vi. The collection{supp ψi j}of supports over all i,jwill be locally ﬁnite. Since the compact setsVi+1−Vicover M, the locally ﬁnite sum ψ=∑i,jψi jwill be positive on M. Then {ψi j/ψ}is aC∞partition of unity satisfying the conditions in (i). We now ﬁll in the details. Fix an integer i≥1. For each pin the compact set Vi+1−Vi, choose an open set Uαcontaining pfrom the open cover {Uα}. Then p is in the open set Uα∩(Vi+2−Vi−1). Let ψpbe a C∞bump function on Mthat is positive on a neighborhood Wpofpand has support in Uα∩(Vi+2−Vi−1). Since supp ψpis a closed set contained in the compact set Vi+2, it is compact. The collection{Wp|p∈Vi+1−Vi}is an open cover of the compact set Vi+1−Vi, and so there is a ﬁnite subcover {Wp1,..., Wpm}, with associated bump functions 348§C Existence of a Partition of Unity in General ψp1,..., ψpm. Since m,Wpj, and ψpjall depend on i, we relabel them as m(i), Wi 1,..., Wi m(i), and ψi 1,..., ψi m(i). In summary, for each i≥1, we have found ﬁnitely many open sets Wi 1,..., Wi m(i) and ﬁnitely many C∞bump functions ψi 1,..., ψi m(i)such that (1)ψi j>0 on Wi jforj=1,..., m(i); (2)Wi 1,..., Wi m(i)cover the compact set Vi+1−Vi; (3) supp ψi j⊂Uαi j∩(Vi+2−Vi−1)for some αi j∈A; (4) supp ψi jis compact. Asiruns from 1 to ∞, we obtain countably many bump functions {ψi j}. The collection of their supports, {supp ψi j}, is locally ﬁnite, since only ﬁnitely many of these sets intersect any Vi. Indeed, since supp ψℓ j⊂Vℓ+2−Vℓ−1 for allℓ, as soon as ℓ≥i+1, /parenleftig supp ψℓ j/parenrightig ∩Vi=the empty set ∅. Any point p∈Mis contained in the compact set Vi+1−Vifor some i, and there- fore p∈Wi jfor some (i,j). For this (i,j),ψi j(p)>0. Hence, the sum ψ:=∑i,jψi j is locally ﬁnite and everywhere positive on M. To simplify the notation, we now relabel the countable set {ψi j}as{ψ1,ψ2,ψ3,...}. Deﬁne ϕk=ψk ψ. Then ∑ϕk=1 and supp ϕk=supp ψk⊂Uα for some α∈A. So{ϕk}is a partition of unity with compact support such that for each k, supp ϕk⊂Uαfor some α∈A. (ii) For each k=1,2,..., letτ(k)be an index in A such that supp ϕk⊂Uτ(k) as in the preceding paragraph. Group the collection {ϕk}according to τ(k)and deﬁne ρα=∑ τ(k)=αϕk if there is a kwith τ(k)=α; otherwise, set ρα=0. Then ∑ α∈Aρα=∑ α∈A∑ τ(k)=αϕk=∞ ∑ k=1ϕk=1. By Problem 13.7, supp ρα⊂/uniondisplay τ(k)=αsupp ϕk⊂Uα. Hence,{ρα}is aC∞partition of unity subordinate to {Uα}. ⊓ ⊔ D.1 Quotient Vector Spaces 349 §D Linear Algebra This appendix gathers together a few facts from linear algebra used throughout the book, especially in Sections 24 and 25. The quotient vector space is a construction in which one reduces a vector space to a smaller space by identifying a subspace to zero. It represents a simpliﬁcation, much like the formation of a quotient group or of a quotient ring. For a linear map f:V→Wof vector spaces, the ﬁrst isomorphism theorem of linear algebra gives an isomorphism between the quotient space V/kerfand the image of f. It is one of the most useful results in linear algebra. We also discuss the direct sum and the direct product of a family of vector spaces, as well as the distinction between an internal and an external direct sum. D.1 Quotient Vector Spaces IfVis a vector space and Wis a subspace of V, acoset ofWinVis a subset of the form v+W={v+w|w∈W} for some v∈V. Two cosets v+Wandv′+Ware equal if and only if v′=v+wfor some w∈W, or equivalently, if and only if v′−v∈W. This introduces an equivalence relation on V: v∼v′⇐⇒ v′−v∈W⇐⇒ v+W=v′+W. A coset of WinVis simply an equivalence class under this equivalence relation. Any element of v+Wis called a representative of the coset v+W. The set V/Wof all cosets of WinVis again a vector space, with addition and scalar multiplication deﬁned by (u+W)+(v+W)=( u+v)+W, r(v+W)=rv+W foru,v∈Vandr∈R. We call V/Wthequotient vector space orthe quotient space of V by W . Example D.1.ForV=R2andWa line through the origin in R2, a coset of Win R2is a line in R2parallel to W. (For the purpose of this discussion, two lines in R2 areparallel if and only if they coincide or fail to intersect. This deﬁnition differs from the usual one in plane geometry in allowing a line to be parallel to itself.) Thequotient space R 2/Wis the collection of lines in R2parallel to W(Figure D.1 ). 350§D Linear Algebra /Bulletv v+WWL Fig. D.1. Quotient vector space of R2byW. D.2 Linear Transformations LetVandWbe vector spaces over R. A map f:V→Wis called a linear transfor- mation , avector space homomorphism , alinear operator , or a linear map overRif for all u,v∈Vandr∈R, f(u+v)=f(u)+f(v), f(ru)=r f(u). Example D.2.LetV=R2andWa line through the origin in R2as in Example D.1. IfLis a line through the origin not parallel to W, then Lwill intersect each line in R2 parallel to Win one and only one point. This one-to-one correspondence L→R2/W, v/ma√s⊔o→v+W, preserves addition and scalar multiplication, and so is an isomorphism of vector spaces. Thus, in this example the quotient space R2/Wcan be identiﬁed with the lineL. Iff:V→Wis a linear transformation, the kernel offis the set kerf={v∈V|f(v)=0} and the image offis the set imf={f(v)∈W|v∈V}. The kernel of fis a subspace of Vand the image of fis a subspace of W. Hence, one can form the quotient spaces V/kerfandW/imf. This latter space, W/imf, denoted by coker f, is called the cokernel of the linear map f:V→W. For now, denote by Kthe kernel of f. The linear map f:V→Winduces a linear map ¯f:V/K→imf, by ¯f(v+K)=f(v). It is easy to check that ¯fis linear and bijective. This gives the following fundamental result of linear algebra. D.3 Direct Product and Direct Sum 351 Theorem D.3 (The ﬁrst isomorphism theorem). Let f :V→W be a homomor- phism of vector spaces. Then f induces an isomorphism ¯f:V kerf∼−→imf. D.3 Direct Product and Direct Sum Let{Vα}α∈Ibe a family of real vector spaces. The direct product ∏αVαis the set of all sequences (vα)with vα∈Vαfor all α∈I, and the direct sum/circleplustext αVαis the subset of the direct product ∏αVαconsisting of sequences (vα)such that vα=0 for all but ﬁnitely many α∈I. Under componentwise addition and scalar multiplication, (vα)+(wα)=( vα+wα), r(vα)=( rvα),r∈R, both the direct product ∏αVαand the direct sum⊕αVαare real vector spaces. When the index set Iis ﬁnite, the direct sum coincides with the direct product. In particular, for two vector spaces AandB, A⊕B=A×B={(a,b)|a∈Aandb∈B}. Thesumof two subspaces AandBof a vector space Vis the subspace A+B={a+b∈V|a∈A,b∈B}. IfA∩B={0}, this sum is called an internal direct sum and written A⊕iB. In an internal direct sum A⊕iB, every element has a representation as a+bfor a unique a∈Aand a unique b∈B. Indeed, if a+b=a′+b′∈A⊕iB, then a−a′=b′−b∈A∩B={0}. Hence, a=a′andb=b′. In contrast to the internal direct sum A⊕iB, the direct sum A⊕Bis called the external direct sum . In fact, the two notions are isomorphic: the natural map ϕ:A⊕B→A⊕iB, (a,b)/ma√s⊔o→a+b is easily seen to be a linear isomorphism. For this reason, in the literature the internal direct sum is normally denoted by A⊕B, just like the external direct sum. IfV=A⊕iB, then Ais called a complementary subspace to B inV. In Ex- ample D.2, the line Lis a complementary subspace to W, and we may identify the quotient vector space R2/Wwith any complementary subspace to W. In general, if Wis a subspace of a vector space VandW′is a complementary subspace to W, then there is a linear map ϕ:W′→V/W, w′/ma√s⊔o→w′+W. 352§D Linear Algebra Exercise D.4. Show that ϕ:W′→V/Wis an isomorphism of vector spaces. Thus, the quotient space V/Wmay be identiﬁed with any complementary sub- space to WinV. This identiﬁcation is not canonical, for there are many complemen- tary subspaces to a given subspace Wand there is no reason to single out any one of them. However, when Vhas an inner product /an}bracke⊔le{⊔,/an}bracke⊔ri}h⊔, one can single out a canonical complementary subspace, the orthogonal complement ofW: W⊥={v∈V|/an}bracke⊔le{⊔v,w/an}bracke⊔ri}h⊔=0 for all w∈W}. Exercise D.5. Check that W⊥is a complementary subspace to W. In this case, there is a canonical identiﬁcation W⊥∼→V/W. Letf:V→Wbe a linear map of ﬁnite-dimensional vector spaces. It follows from the ﬁrst isomorphism theorem and Problem D.1 that dimV−dim(kerf)=dim(imf). Since the dimension is the only isomorphism invariant of a vector space, we therefore have the following corollary of the ﬁrst isomorphism theorem. Corollary D.6. If f:V→W is a linear map of ﬁnite-dimensional vector spaces, then there is a vector space isomorphism V≃kerf⊕imf. (The right-hand side is an external direct sum because ker fand im fare not sub- spaces of the same vector space.) Problems D.1. Dimension of a quotient vector space Prove that if w1,..., wmis a basis for Wthat extends to a basis w1,..., wm,v1,..., vnforV, then v1+W,..., vn+Wis a basis for V/W. Therefore, dimV/W=dimV−dimW. D.2. Dimension of a direct sum Prove that if a1,..., amis a basis for a vector space Aandb1,..., bnis a basis for a vector space B, then(ai,0),(0,bj),i=1,..., m,j=1,..., n, is a basis for the direct sum A⊕B. Therefore, dimA⊕B=dimA+dimB. 353 §E Quaternions and the Symplectic Group First described by William Rowan Hamilton in 1843, quaternions are elements of the form q=a+ib+jc+kd,a,b,c,d∈R, that add componentwise and multiply according to the distributive property and the rules i2=j2=k2=−1, ij=k,jk=i,ki=j, ij=−ji,jk=−kj,ki=−ik. A mnemonic for the three rules ij=k,jk=i,ki=jis that in going clockwise around the circle i k j , the product of two successive elements is the next one. Under addition and multi- plication, the quaternions satisfy all the properties of a ﬁeld except the commutative property for multiplication. Such an algebraic structure is called a skew ﬁeld or adi- vision ring . In honor of Hamilton, the usual notation for the skew ﬁeld of quaternions isH. A division ring that is also an algebra over a ﬁeld Kis called a division algebra over K. The real and complex ﬁelds RandCare commutative division algebras over R. By a theorem of Ferdinand Georg Frobenius [13] from 1878, the skew ﬁeld Hof quaternions has the distinction of being the only (associative) division algebra over Rother than RandC.1 In this appendix we will derive the basic properties of quaternions and deﬁne the symplectic group in terms of quaternions. Because of the familiarity of complex matrices, quaternions are often represented by 2 ×2 complex matrices. Correspond- ingly, the symplectic group also has a description as a group of complex matrices. One can deﬁne vector spaces and formulate linear algebra over a skew ﬁeld, just as one would for vector spaces over a ﬁeld. The only difference is that over a skewﬁeld it is essential to keep careful track of the order of multiplication. A vector space overHis called a quaternionic vector space. We denote by H nthe quaternionic vector space of n-tuples of quaternions. There are many potential pitfalls stemming from a choice of left and right, for example: 1If one allows an algebra to be nonassociative, then there are other division algebras over R, for example Cayley’s octonians. 354§E Quaternions and the Symplectic Group (1) Should i,j,kbe written on the left or on the right of a scalar? (2) Should scalars multiply on the left or on the right of Hn? (3) Should elements of Hnbe represented as column vectors or as row vectors? (4) Should a linear transformation be represented by multiplication by a matrix on the left or on the right? (5) In the deﬁnition of the quaternion inner product, should one conjugate the ﬁrst or the second argument? (6) Should a sesquilinear form on Hnbe conjugate-linear in the ﬁrst or the second argument? The answers to these questions are not arbitrary, since the choice for one question may determine the correct choices for all the others. A wrong choice will lead toinconsistencies. E.1 Representation of Linear Maps by Matrices Relative to given bases, a linear map of vector spaces over a skew ﬁeld will also be represented by a matrix. Since maps are written on the left of their arguments as inf(x), we will choose our convention so that a linear map fcorresponds to left multiplication by a matrix. In order for a vector in Hnto be multiplied on the left by a matrix, the elements of Hnmust be column vectors, and for left multiplication by a matrix to be a linear map, scalar multiplication on Hnshould be on the right. In this way, we have answered (1), (2), (3), and (4) above. LetKbe a skew ﬁeld and let VandWbe vector spaces over K, with scalar multiplication on the right. A map f:V→Wislinear over KorK-linear if for all x,y∈Vandq∈K, f(x+y)=f(x)+f(y), f(xq)=f(x)q. Anendomorphism or a linear transformation of a vector space Vover Kis aK- linear map from Vto itself. The endomorphisms of Vover Kform an algebra over K, denoted by End K(V). An endomorphism f:V→Visinvertible if it has a two- sided inverse, i.e., a linear map g:V→Vsuch that f◦g=g◦f=1V. An invertible endomorphism of Vis also called an automorphism ofV. The general linear group GL(V)is by deﬁnition the group of all automorphisms of the vector space V. When V=Kn, we also write GL (n,K)for GL(V). To simplify the presentation, we will discuss matrix representation only for en- domorphisms of the vector space Kn. Let eibe the column vector with 1 in the ith row and 0 everywhere else. The set e1,..., enis called the standard basis forKn. If f:Kn→KnisK-linear, then f(ej)=∑ ieiai j for some matrix A= [ai j]∈Kn×n, called the matrix off(relative to the standard basis). Here ai jis the entry in the ith row and jth column of the matrix A. For x=∑jejxj∈Kn, E.2 Quaternionic Conjugation 355 f(x)=∑ jf(ej)xj=∑ i,jeiai jxj. Hence, the ith component of the column vector f(x)is (f(x))i=∑ jai jxj. In matrix notation, f(x)=Ax. Ifg:Kn→Knis another linear map and g(ej)=∑ieibi j, then (f◦g)(ej)=f/parenleftbigg ∑ kekbk j/parenrightbigg =∑ kf(ek)bk j=∑ i,keiai kbk j. Thus, if A=[ai j]andB=[bi j]are the matrices representing fandgrespectively, then the matrix product ABis the matrix representing the composite f◦g. Therefore, there is an algebra isomorphism End K(Kn)∼→Kn×n between endomorphisms of Knandn×nmatrices over K. Under this isomorphism, the group GL (n,K)corresponds to the group of all invertible n×nmatrices over K. E.2 Quaternionic Conjugation Theconjugate of a quaternion q=a+ib+jc+kdis deﬁned to be ¯q=a−ib−jc−kd. It is easily shown that conjugation is an antihomomorphism from the ring Hto itself: it preserves addition, but under multiplication, pq=¯q¯pforp,q∈H. Theconjugate of a matrix A=[ai j]∈Hm×nis¯A=/bracketleftig ai j/bracketrightig , obtained by conjugating each entry of A. The transpose ATof the matrix Ais the matrix whose (i,j)-entry is the(j,i)-entry of A. In contrast to the case for complex matrices, when AandBare quaternion matrices, in general AB/ne}a⊔ionslash=¯A¯B,AB/ne}a⊔ionslash=¯B¯A,and(AB)T/ne}a⊔ionslash=BTAT. However, it is true that ABT=¯BT¯AT, as one sees by a direct computation. 356§E Quaternions and the Symplectic Group E.3 Quaternionic Inner Product Thequaternionic inner product onHnis deﬁned to be /an}bracke⊔le{⊔x,y/an}bracke⊔ri}h⊔=∑ ixiyi=¯xTy,x,y∈Hn, with conjugation on the ﬁrst argument x=/an}bracke⊔le{⊔x1,..., xn/an}bracke⊔ri}h⊔. For any q∈H, /an}bracke⊔le{⊔xq,y/an}bracke⊔ri}h⊔=¯q/an}bracke⊔le{⊔x,y/an}bracke⊔ri}h⊔and/an}bracke⊔le{⊔x,yq/an}bracke⊔ri}h⊔=/an}bracke⊔le{⊔x,y/an}bracke⊔ri}h⊔q. If conjugation were on the second argument, then the inner product would not have the correct linearity property with respect to scalar multiplication on the right. For quaternion vector spaces VandW, we say that a map f:V×W→His sesquilinear overHif it is conjugate-linear on the left in the ﬁrst argument and linear on the right in the second argument: for all v∈V,w∈W, and q∈H, f(vq,w)=¯q f(v,w), f(v,wq)=f(v,w)q. In this terminology, the quaternionic inner product is sesquilinear over H. E.4 Representations of Quaternions by Complex Numbers A quaternion can be identiﬁed with a pair of complex numbers: q=a+ib+jc+kd=(a+ib)+j(c−id)=u+jv←→(u,v). Thus,His a vector space over Cwith basis 1, j, andHnis a vector space over Cwith basis e1,..., en,je1, . . . , jen. Proposition E.1. Let q be a quaternion and let u, v be complex numbers. (i)If q=u+jv, then ¯q=¯u−jv. (ii)juj−1=¯u. Proof. Problem E.1. ⊓ ⊔ By Proposition E.1(ii), for any complex vector v∈Cn, one has jv=¯vj. Although jei=eij, elements of Hnshould be written as u+jv, not as u+vj, so that the map Hn→C2n,u+jv/ma√s⊔o→(u,v), will be a complex vector space isomorphism. For any quaternion q=u+jv, left multiplication ℓq:H→HbyqisH-linear and a fortiori C-linear. Since ℓq(1)=u+jv, ℓq(j)=( u+jv)j=−¯v+j¯u, the matrix of ℓqas aC-linear map relative to the basis 1, jforHoverCis the 2×2 complex matrix/bracketleftbigu−¯v v¯u/bracketrightbig . The map H→EndC(C2),q/ma√s⊔o→ℓqis an injective algebra homomorphism over R, giving rise to a representation of the quaternions by 2 ×2 complex matrices. E.6H-Linearity in Terms of Complex Numbers 357 E.5 Quaternionic Inner Product in Terms of Complex Components Letx=x1+jx2andy=y1+jy2be inHn, with x1,x2,y1,y2∈Cn. We will express the quaternionic inner product /an}bracke⊔le{⊔x,y/an}bracke⊔ri}h⊔in terms of the complex vectors x1,x2,y1,y2∈Cn. By Proposition E.1, /an}bracke⊔le{⊔x,y/an}bracke⊔ri}h⊔=¯xTy=/parenleftbig ¯xT 1−jxT 2/parenrightbig (y1+jy2) (since ¯ x=¯x1−jx2) =/parenleftbig ¯xT 1y1+¯xT 2y2/parenrightbig +j/parenleftbig xT 1y2−xT 2y1/parenrightbig (since xT 2j=j¯xT 2and ¯xT 1j=jxT 1). Let /an}bracke⊔le{⊔x,y/an}bracke⊔ri}h⊔1=¯xT 1y1+¯xT 2y2=n ∑ i=1¯xi 1yi1+¯xi 2yi2 and /an}bracke⊔le{⊔x,y/an}bracke⊔ri}h⊔2=xT 1y2−xT 2y1=n ∑ i=1xi 1yi2−xi 2yi1. So the quaternionic inner product /an}bracke⊔le{⊔,/an}bracke⊔ri}h⊔is the sum of a Hermitian inner product and j times a skew-symmetric bilinear form on C2n: /an}bracke⊔le{⊔,/an}bracke⊔ri}h⊔=/an}bracke⊔le{⊔,/an}bracke⊔ri}h⊔1+j/an}bracke⊔le{⊔,/an}bracke⊔ri}h⊔2. Letx=x1+jx2∈Hn. By skew-symmetry, /an}bracke⊔le{⊔x,x/an}bracke⊔ri}h⊔2=0, so that /an}bracke⊔le{⊔x,x/an}bracke⊔ri}h⊔=/an}bracke⊔le{⊔x,x/an}bracke⊔ri}h⊔1=/bardblx1/bardbl2+/bardblx2/bardbl2≥0. Thenorm of a quaternionic vector x=x1+jx2is deﬁned to be /bardblx/bardbl=/radicalbig /an}bracke⊔le{⊔x,x/an}bracke⊔ri}h⊔=/radicalig /bardblx1/bardbl2+/bardblx2/bardbl2. In particular, the norm of a quaternion q=a+ib+jc+kdis /bardblq/bardbl=/radicalbig a2+b2+c2+d2. E.6H-Linearity in Terms of Complex Numbers Recall that an H-linear map of quaternionic vector spaces is a map that is additive and commutes with right multiplication rqfor any quaternion q. Proposition E.2. Let V be a quaternionic vector space. A map f :V→V isH-linear if and only if it is C-linear and f◦rj=rj◦f . Proof.(⇒)Clear. (⇐)Suppose fisC-linear and fcommutes with rj. ByC-linearity, fis additive and commutes with rufor any complex number u. Any q∈Hcan be written as q=u+jvfor some u,v∈C; moreover, rq=ru+jv=ru+rv◦rj(note the order reversal in rjv=rv◦rj). Since fis additive and commutes with ru,rv, and rj, it commutes with rqfor any q∈H. Therefore, fisH-linear. ⊓ ⊔ 358§E Quaternions and the Symplectic Group Because the map rj:Hn→Hnis neither H-linear nor C-linear, it cannot be rep- resented by left multiplication by a complex matrix. If q=u+jv∈Hn, where u,v∈Cn, then rj(q)=qj=(u+jv)j=−¯v+j¯u. In matrix notation, rj/parenleftbigg/bracketleftbigg u v/bracketrightbigg/parenrightbigg =/bracketleftbigg −¯v ¯u/bracketrightbigg =c/parenleftbigg/bracketleftbigg 0−1 1 0/bracketrightbigg/bracketleftbigg u v/bracketrightbigg/parenrightbigg =−c/parenleftbigg J/bracketleftbigg u v/bracketrightbigg/parenrightbigg , (E.1) where cdenotes complex conjugation and Jis the 2×2 matrix/bracketleftbig0 1 −1 0/bracketrightbig . E.7 Symplectic Group LetVbe a vector space over a skew ﬁeld Kwith conjugation, and let B:V×V→K be a bilinear or sesquilinear function over K. Such a function is often called a bilinear or sesquilinear form over K. AK-linear automorphism f:V→Vis said to preserve the form Bif B(f(x),f(y)) = B(x,y)for all x,y∈V. The set of these automorphisms is a subgroup of the general linear group GL (V). When Kis the skew ﬁeld R,C, orH, and Bis the Euclidean, Hermitian, or quaternionic inner product respectively on Kn, the subgroup of GL (n,K)consisting of automorphisms of Knpreserving each of these inner products is called the orthog- onal,unitary , orsymplectic group and denoted by O (n), U(n), or Sp(n)respectively. Naturally, the automorphisms in these three groups are called orthogonal ,unitary , orsymplectic automorphisms. In particular, the symplectic group is the group of automorphisms fofHnsuch that /an}bracke⊔le{⊔f(x),f(y)/an}bracke⊔ri}h⊔=/an}bracke⊔le{⊔x,y/an}bracke⊔ri}h⊔for all x,y∈Hn. In terms of matrices, if Ais the quaternionic matrix of such an f, then /an}bracke⊔le{⊔f(x),f(y)/an}bracke⊔ri}h⊔=AxTAy=¯xT¯ATAy=¯xTyfor all x,y∈Hn. Therefore, f∈Sp(n)if and only if its matrix Asatisﬁes ¯ATA=I. Because Hn= Cn⊕jCnis isomorphic to C2nas a complex vector space and an H-linear map is necessarily C-linear, the group GL (n,H)is isomorphic to a subgroup of GL (2n,C) (see Problem E.2). Example. Under the algebra isomorphisms End H(H)≃H, elements of Sp (1)corre- spond to quaternions q=a+ib+jc+kdsuch that ¯qq=a2+b2+c2+d2=1. These are precisely quaternions of norm 1. Therefore, under the chain of real vector space isomorphisms End H(H)≃H≃R4, the group Sp (1)maps to S3, the unit 3- sphere in R4. E.7 Symplectic Group 359 Thecomplex symplectic group Sp(2n,C)is the subgroup of GL (2n,C)consisting of automorphisms of C2npreserving the skew-symmetric bilinear form B:C2n× C2n→C, B(x,y)=n ∑ i=1xiyn+i−xn+iyi=xTJy,J=/bracketleftbigg 0In −In0/bracketrightbigg , where Inis the n×nidentity matrix. If f:C2n×C2n→Cis given by f(x) =Ax, then f∈Sp(2n,C)⇐⇒ B(f(x),f(y)) = B(x,y)for all x,y∈C2n ⇐⇒(Ax)TJAy=xT(ATJA)y=xTJyfor all x,y∈C2n ⇐⇒ ATJA=J. Theorem E.3. Under the injection GL(n,H)֒→GL(2n,C), the symplectic group Sp(n)maps isomorphically to the intersection U(2n)∩Sp(2n,C). Proof. f∈Sp(n) ⇐⇒ f:Hn→HnisH-linear and preserves the quaternionic inner product ⇐⇒ f:C2n→C2nisC-linear, f◦rj=rj◦f, and fpreserves the Hermitian inner product and the standard skew-symmetric bilinear form on C2n(by Proposition E.2 and Section E.5) ⇐⇒ f◦rj=rj◦fandf∈U(2n)∩Sp(2n,C). We will now show that if f∈U(2n), then the condition f◦rj=rj◦fis equivalent tof∈Sp(2n,C). Let f∈U(2n)and let Abe the matrix of frelative to the standard basis inC2n. Then (f◦rj)(x)=( rj◦f)(x)for all x∈C2n ⇐⇒ − Ac(Jx)=−c(JAx)for all x∈C2n(by (E.1)) ⇐⇒ c(¯AJx)=c(JAx)for all x∈C2n ⇐⇒ ¯AJx=JAxfor all x∈C2n ⇐⇒ J=¯A−1JA ⇐⇒ J=ATJA (since A∈U(2n)) ⇐⇒ f∈Sp(2n,C). Therefore, the condition f◦rj=rj◦fis redundant if f∈U(2n)∩Sp(2n,C). By the ﬁrst paragraph of this proof, there is a group isomorphism Sp (n)≃U(2n)∩ Sp(2n,C). ⊓ ⊔ Problems E.1. Quaternionic conjugation Prove Proposition E.1. 360§E Quaternions and the Symplectic Group E.2. Complex representation of an H-linear map Suppose an H-linear map f:Hn→Hnis represented relative to the standard basis e1,..., enby the matrix A=u+jv∈Hn×n, where u,v∈Cn×n. Show that as a C-linear map, f:Hn→Hn is represented relative to the basis e1,..., en,je1,..., jenby the matrix/bracketleftbigu−¯v v¯u/bracketrightbig . E.3. Symplectic and unitary groups of small dimension For a ﬁeld K, the special linear group SL(n,K)is the subgroup of GL (n,K)consisting of all automorphisms of Knof determinant 1, and the special unitary group SU(n)is the subgroup of U(n)consisting of unitary automorphisms of Cnof determinant 1. Prove the following identiﬁcations or group isomorphisms. (a) Sp(2,C)=SL(2,C). (b) Sp(1)≃SU(2). (Hint: Use Theorem E.3 and part (a).) (c) SU(2)≃/braceleftbigg/bracketleftbiggu−¯v v¯u/bracketrightbigg ∈C2×2/vextendsingle/vextendsingle/vextendsingle/vextendsingleu¯u+v¯v=1/bracerightbigg . (Hint: Use part (b) and the representation of quaternions by 2 ×2 complex matrices in Subsection E.4.) Solutions to Selected Exercises Within the Text 3.6 Inversions As a matrix, τ=/bracketleftbig1 2 3 4 5 2 3 4 5 1/bracketrightbig . Scanning the second row, we see that τhas four inversions: (2,1), (3,1),(4,1),(5,1). ♦ ♦ 3.13 Symmetrizing operator Ak-linear function h:V→Ris symmetric if and only if τh=hfor all τ∈Sk. Now τ(S f)=τ∑ σ∈Skσf=∑ σ∈Sk(τσ)f. Asσruns over all elements of the permutation groups Sk, so does τσ. Hence, ∑ σ∈Sk(τσ)f=∑ τσ∈Sk(τσ)f=S f. This proves that τ(S f)=S f. ♦ ♦ 3.15 Alternating operator f(v1,v2,v3)−f(v1,v3,v2)+f(v2,v3,v1)−f(v2,v1,v3)+f(v3,v1,v2)−f(v3,v2,v1).♦ ♦ 3.20 Wedge product of two 2-covectors (f∧g)(v1,v2,v3,v4) =f(v1,v2)g(v3,v4)−f(v1,v3)g(v2,v4)+f(v1,v4)g(v2,v3) +f(v2,v3)g(v1,v4)−f(v2,v4)g(v1,v3)+f(v3,v4)g(v1,v2). ♦ ♦ 3.22 Sign of a permutation We can achieve the permutation τfrom the initial conﬁguration 1 ,2,..., k+ℓinksteps. (1) First, move the element kto the very end across the ℓelements k+1,..., k+ℓ. This requiresℓtranspositions. (2) Next, move the element k−1 across the ℓelements k+1,..., k+ℓ. (3) Then move the element k−2 across the same ℓelements, and so on. Each of the ksteps requires ℓtranspositions. In the end we achieve τfrom the identity using ℓktranspositions. 362 Solutions to Selected Exercises Within the Text Alternatively, one can count the number of inversions in the permutation τ. There are k inversions starting with k+1, namely, (k+1,1),...,(k+1,k). Indeed, for each i=1,...,ℓ, there are kinversions starting with k+i. Hence, the total number of inversions in τiskℓ. By Proposition 3.8, sgn (τ)=(−1)kℓ. ♦ ♦ 4.3 A basis for 3-covectors By Proposition 3.29, a basis for A3(Tp(R4))is/parenleftbig dx1∧dx2∧dx3/parenrightbig p,/parenleftbig dx1∧dx2∧dx4/parenrightbig p,/parenleftbig dx1∧dx3∧dx4/parenrightbig p,/parenleftbig dx2∧dx3∧dx4/parenrightbig p. ♦ ♦ 4.4 Wedge product of a 2-form with a 1-form The(2,1)-shufﬂes are (1<2,3),(1<3,2),(2<3,1), with respective signs +,−,+. By equation (3.6), (ω∧τ)(X,Y,Z)=ω(X,Y)τ(Z)−ω(X,Z)τ(Y)+ω(Y,Z)τ(X). ♦ ♦ 6.14 Smoothness of a map to a circle Without further justiﬁcation, the fact that both cos tand sin tareC∞proves only the smooth- ness of(cost,sint)as a map from RtoR2. To show that F:R→S1isC∞, we need to cover S1 with charts (Ui,φi)and examine in turn each φi◦F:F−1(Ui)→R. Let{(Ui,φi)|i=1,..., 4} be the atlas of Example 5.16. On F−1(U1),φ1◦F(t)=(x◦F)(t)=costisC∞. On F−1(U3), φ3◦F(t)=sintisC∞. Similar computations on F−1(U2)andF−1(U4)prove the smoothness ofF. ♦ ♦ 6.18 Smoothness of a map to a Cartesian product Fix p∈N, let(U,φ)be a chart about p, and let (V1×V2,ψ1×ψ2)be a chart about (f1(p),f2(p)). We will be assuming either (f1,f2)smooth or both fismooth. In either case,(f1,f2)is continuous. Hence, by choosing Usufﬁciently small, we may assume (f1,f2)(U)⊂V1×V2. Then (ψ1×ψ2)◦(f1,f2)◦φ−1=(ψ1◦f1◦φ−1,ψ2◦f2◦φ−1) maps an open subset of Rnto an open subset of Rm1+m2. It follows that (f1,f2)isC∞atpif and only if both f1andf2areC∞atp. ♦ ♦ 7.11 Real projective space as a quotient of a sphere Deﬁne ¯f:RPn→Sn/∼by¯f([x])=[x /bardblx/bardbl]∈Sn/∼. This map is well deﬁned because ¯f([tx])= [tx |tx|] = [±x /bardblx/bardbl] = [x /bardblx/bardbl]. Note that if π1:Rn+1−{0}→RPnandπ2:Sn→Sn/∼are the projection maps, then there is a commutative diagram Rn−{0}f/d47/d47 π1 /d15/d15Sn π2 /d15/d15 RPn ¯f/d47/d47Sn/∼. By Proposition 7.1, ¯fis continuous because π2◦fis continuous. Next deﬁne g:Sn→Rn+1−{0}byg(x)=x. This map induces a map ¯ g:Sn/∼→RPn, ¯g([x])=[ x]. By the same argument as above, ¯ gis well deﬁned and continuous. Moreover, ¯g◦¯f([x])=/bracketleftbiggx /bardblx/bardbl/bracketrightbigg =[x], ¯f◦¯g([x])=[ x], Solutions to Selected Exercises Within the Text 363 so¯fand ¯gare inverses to each other. ♦ ♦ 8.14 Velocity vector versus the calculus derivative As a vector at the point c(t)in the real line, c′(t)equals ad/dx|c(t)for some scalar a. Applying both sides of the equality to x, we get c′(t)x=adx/dx|c(t)=a. By the deﬁnition of c′(t), a=c′(t)x=c∗/parenleftigg d dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle c(t)/parenrightigg x=d dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle c(t)x◦c=d dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle c(t)c=˙c(t). Hence, c′(t)=˙c(t)d/dx|c(t). ♦ ♦ 13.1 Bump function supported in an open set Let(V,φ)be a chart centered at qsuch that Vis diffeomorphic to an open ball B(0,r). Choose real numbers aandbsuch that B(0,a)⊂B(0,b)⊂B(0,b)⊂B(0,r). With the σgiven in (13.2), the function σ◦φ, extended by zero to M, gives the desired bump function. ♦ ♦ 15.2 Left multiplicationLeti a:G→G×Gbe the inclusion map ia(x)= (a,x). It is clearly C∞. Thenℓa(x) =ax= (µ◦ia)(x). Sinceℓa=µ◦iais the composition of two C∞maps, it is C∞. Moreover, because it has a two-sided C∞inverseℓa−1, it is a diffeomorphism. ♦ ♦ 15.7 Space of symmetric matrices Let A=[ai j]= a11a12···a1n ∗a22···a2n ............ ∗ ∗ ··· ann  be a symmetric matrix. The symmetry condition aji=ai jimplies that the entries below the diagonal are determined by the entries above the diagonal, and that there are no further conditions on the the entries above or on the diagonal. Thus, the dimension of Snis equal to the number of entries above or on the diagonal. Since there are nsuch entries in the ﬁrst row, n−1 in the second row, and so on, dimSn=n+(n−1)+(n−2)+···+1=n(n+1) 2. ♦ ♦ 15.10 Induced topology versus the subspace topology A basic open set in the induced topology on His the image under fof an open interval in L. Such a set is not open in the subspace topology. A basic open set in the subspace topology onHis the intersection of Hwith the image of an open ball in R2under the projection π:R2→R2/Z2; it is a union of inﬁnitely many open intervals. Thus, the subspace topology is a subset of the induced topology, but not vice versa. ♦ ♦ 15.15 Distributivity over a convergent series (i) We may assume a/ne}a⊔ionslash=0, for otherwise there is nothing to prove. Let ε>0. Since sm→s, there exists an integer Nsuch that for all m≥N,/bardbls−sm/bardbl<ε//bardbla/bardbl. Then for m≥N, /bardblas−asm/bardbl≤/bardbl a/bardbl/bardbls−sm/bardbl</bardbla/bardbl/parenleftbiggε /bardbla/bardbl/parenrightbigg =ε. 364 Solutions to Selected Exercises Within the Text Hence, asm→as. (ii) Set sm=∑m k=0bkands=∑∞ k=0bk. The convergence of the series ∑∞ k=0bkmeans that sm→s. By (i), asm→as, which means that the sequence asm=∑m k=0abkconverges to a∑∞ k=0bk. Hence, ∑∞ k=0abk=a∑∞ k=0bk. ♦ ♦ 18.5 Transition formula for a 2-form ai j=ω(∂/∂xi,∂/∂xj)=∑ k<ℓbkℓdyk∧dyℓ(∂/∂xi,∂/∂xj) =∑ k<ℓbkℓ/parenleftig dyk(∂/∂xi)dyℓ(∂/∂xj)−dyk(∂/∂xj)dyℓ(∂/∂xi)/parenrightig =∑ k<ℓbkℓ/parenleftbigg∂yk ∂xi∂yℓ ∂xj−∂yk ∂xj∂yℓ ∂xi/parenrightbigg =∑ k<ℓbkℓ∂(yk,yℓ) ∂(xi,xj). ♦ ♦ Alternative solution: by Proposition 18.3, dyk∧dyℓ=∑ i<j∂(yk,yℓ) ∂(xi,xj)dxi∧dxj. Hence, ∑ i<jai jdxi∧dxj=∑ k<ℓbkℓdyk∧dyℓ=∑ i<j∑ k<ℓbkℓ∂(yk,yℓ) ∂(xi,xj)dxi∧dxj. Comparing the coefﬁcients of dxi∧dxjgives ai j=∑ k<ℓbkℓ∂(yk,yℓ) ∂(xi,xj). ♦ ♦ 22.2 Smooth functions on a nonopen set By deﬁnition, for each pinSthere are an open set UpinRnand a C∞function ˜fp:Up→Rm such that f=˜fponUp∩S. Extend the domain of ˜fptoRnby deﬁning it to be zero on Rn−Up. LetU=/uniontext p∈SUp. Choose a partition of unity {σp}p∈SonUsubordinate to the open cover {Up}p∈SofUand deﬁne the function ˜f:U→Rmby ˜f=∑ p∈Sσp˜fp. (∗) Because supp σp⊂Up, the product σp˜fpis zero and hence smooth outside Up; as a product of two C∞functions on Up,σp˜fpisC∞onUp. Therefore, σp˜fpisC∞onRn. Since the sum (∗) is locally ﬁnite, ˜fis well deﬁned and C∞onRnfor the usual reason. (Every point q∈U has a neighborhood Wqthat intersects only ﬁnitely many of of the sets supp σp,p∈S. Hence, the sum (∗) is a ﬁnite sum on Wq.) Letq∈S. Ifq∈Up, then ˜fp(q) =f(q), and if q/∈Up, then σp(q)=0. Thus, for q∈S, one has ˜f(q)=∑ p∈Sσp(q)˜fp(q)=∑ p∈Sσp(q)f(q)=f(q). ♦ ♦ 25.5 Connecting homomorphism The proof that the cohomology class of ais independent of the choice of bas preimage of c can be summarized in the commutative diagram Solutions to Selected Exercises Within the Text 365 a′′da′′ /d95d/d79/d79 b−b′ /d47/d47 i/d47/d47db−db′ /d95d/d79/d79 0./d31 j/d47/d47/d47/d47=a−a′/d47/d47i/d47/d47 Suppose b,b′∈Bkboth map to cunder j. Then j(b−b′) =jb−jb′=c−c=0.By the exactness at Bk,b−b′=i(a′′)for some a′′∈Ak. With the choice of bas preimage, the element d∗[c]is represented by a cocycle a∈Ak+1 such that i(a) =db. Similarly, with the choice of b′as preimage, the element d∗[c]is repre- sented by a cocycle a′∈Ak+1such that i(a′) =db′. Then i(a−a′) =d(b−b′) =di(a′′) = id(a′′). Since iis injective, a−a′=da′′, and thus [a]=[a′]. This proves that d∗[c]is indepen- dent of the choice of b. The proof that the cohomology class of ais independent of the choice of cin the coho- mology class [c]can be summarized by the commutative diagram =0 a−a′db−db′ /d47/d47i/d47/d47 b−b′c−c′ /d31 j/d47/d47/d47/d47 b′′c′′./d31 j/d47/d47/d47/d47/d95d/d79/d79 /d95d/d79/d79 /d95d/d79/d79 Suppose[c] = [c′]∈Hk(C). Then c−c′=dc′′for some c′′∈Ck−1. By the surjectivity of j:Bk−1→Ck−1, there is a b′′∈Bk−1that maps to c′′under j. Choose b∈Bksuch that j(b)=c and let b′=b−db′′∈Bk. Then j(b′)=j(b)−jdb′′=c−d j(b′′)=c−dc′′=c′. With the choice of bas preimage, d∗[c]is represented by a cocycle a∈Ak+1such that i(a)=db. With the choice of b′as preimage, d∗[c]is represented by a cocycle a′∈Ak+1such that i(a′)=db′. Then i(a−a′)=d(b−b′)=ddb′′=0. By the injectivity of i,a=a′, so[a]=[a′]. This shows that d∗[c]is independent of the choice ofcin the cohomology class [c]. ♦ ♦ A.33 Compact Hausdorff space LetSbe a compact Hausdorff space, and A,Btwo closed subsets of S. By Proposition A.30, A andBare compact. By Proposition A.31, for any a∈Athere are disjoint open sets Ua∋aand Va⊃B. Since Ais compact, the open cover {Ua}a∈AforAhas a ﬁnite subcover {Uai}n i=1. Let U=/uniontextn i=1UaiandV=/intersectiontextn i=1Vai. Then A⊂UandB⊂V. The open sets UandVare disjoint because if x∈U∩V, then x∈Uaifor some iandx∈Vaifor the same i, contradicting the fact thatUai∩Vai=∅. ♦ ♦ Hints and Solutions to Selected End-of-Section Problems Problems with complete solutions are starred (*). Equations are numbered consecutively within each problem. 1.2* A C∞function very ﬂat at 0 (a) Assume x>0. For k=1,f′(x)=(1/x2)e−1/x. With p2(y)=y2, this veriﬁes the claim. Now suppose f(k)(x)=p2k(1/x)e−1/x. By the product rule and the chain rule, f(k+1)(x)=p2k−1/parenleftbigg1 x/parenrightbigg ·/parenleftbigg −1 x2/parenrightbigg e−1/x+p2k/parenleftbigg1 x/parenrightbigg ·1 x2e−1/x =/parenleftbigg q2k+1/parenleftbigg1 x/parenrightbigg +q2k+2/parenleftbigg1 x/parenrightbigg/parenrightbigg e−1/x =p2k+2/parenleftbigg1 x/parenrightbigg e−1/x, where qn(y)andpn(y)are polynomials of degree niny. By induction, the claim is true for all k≥1. It is trivially true for k=0 also. (b) For x>0, the formula in (a) shows that f(x)isC∞. For x<0,f(x)≡0, which is trivially C∞. It remains to show that f(k)(x)is deﬁned and continuous at x=0 for all k. Suppose f(k)(0)=0. By the deﬁnition of the derivative, f(k+1)(0)=lim x→0f(k)(x)−f(k)(0) x=lim x→0f(k)(x) x. The limit from the left is clearly 0. So it sufﬁces to compute the limit from the right: lim x→0+f(k)(x) x=lim x→0+p2k/parenleftbig1 x/parenrightbig e−1/x x=lim x→0+p2k+1/parenleftbigg1 x/parenrightbigg e−1/x(1.2.1) =limy→∞p2k+1(y) ey/parenleftbigg replacing1 xbyy/parenrightbigg . Applying l’Hˆ opital’s rule 2 k+1 times, we reduce this limit to 0. Hence, f(k+1)(0) =0. By induction, f(k)(0)=0 for all k≥0. A similar computation as (1.2.1) for lim x→0f(k)(x)=0 proves that f(k)(x)is continuous atx=0. ♦ ♦ 368 Hints and Solutions to Selected End-of-Section Problems 1.3(b)h(t)=( π/(b−a))(t−a)−(π/2). 1.5 (a) The line passing through (0,0,1)and(a,b,c)has a parametrization x=at,y=bt,z=(c−1)t+1. This line intersects the xy-plane when z=0⇔t=1 1−c⇔(x,y)=/parenleftbigga 1−c,b 1−c/parenrightbigg . To ﬁnd the inverse of g, write down a parametrization of the line through (u,v,0)and (0,0,1)and solve for the intersection of this line with S. 1.6* Taylor’s theorem with remainder to order 2 To simplify the notation, we write 0for(0,0). By Taylor’s theorem with remainder, there exist C∞functions g1,g2such that f(x,y)=f(0)+xg1(x,y)+yg2(x,y). (1.6.1) Applying the theorem again, but to g1andg2, we obtain g1(x,y)=g1(0)+xg11(x,y)+yg12(x,y), (1.6.2) g2(x,y)=g2(0)+xg21(x,y)+yg22(x,y). (1.6.3) Since g1(0) =∂f/∂x(0)andg2(0) =∂f/∂y(0), substituting (1.6.2) and (1.6.3) into (1.6.1) gives the result. ♦ ♦ 1.7* A function with a removable singularityIn Problem 1.6, set x=tandy=tu. We obtain f(t,tu) =f(0)+t∂f ∂x(0)+tu∂f ∂y(0)+t2(···), where (···)=g11(t,tu)+ug12(t,tu)+u2g22(t,tu) is aC∞function of tandu. Since f(0)=∂f/∂x(0)=∂f/∂y(0)=0, f(t,tu) t=t(···), which is clearly C∞int,uand agrees with gwhen t=0. ♦ ♦ 1.8See Example 1.2(ii). 3.1 f=∑gi jαi⊗αj. 3.2 (a) Use the formula dimker f+dimim f=dimV. (b) Choose a basis e1,..., en−1for ker f, and extend it to a basis e1,..., en−1,enforV. Let α1,..., αnbe the dual basis for V∨. Write both fandgin terms of this dual basis. 3.3We write temporarily αIforαi1⊗···⊗ αikandeJfor(ej1,..., ejk). Hints and Solutions to Selected End-of-Section Problems 369 (a) Prove that f=∑f(eI)αIby showing that both sides agree on all (eJ). This proves that the set{αI}spans. (b) Suppose ∑cIαI=0. Applying both sides to eJgives cJ=∑cIαI(eJ)=0. This proves that the set{αI}is linearly independent. 3.9To compute ω(v1,..., vn)for any v1,..., vn∈V, write vj=∑ieiai jand use the fact that ωis multilinear and alternating. 3.10* Linear independence of covectors (⇒)Ifα1,..., αkare linearly dependent, then one of them is a linear combination of the others. Without loss of generality, we may assume that αk=k−1 ∑ i=1ciαi. In the wedge product α1∧···∧ αk−1∧(∑k−1 i=1ciαi), every term has a repeated αi. Hence, α1∧···∧ αk=0. (⇐)Suppose α1,..., αkare linearly independent. Then they can be extended to a basis α1,..., αk,..., αnforV∨. Let v1,..., vnbe the dual basis for V. By Proposition 3.27, (α1∧···∧ αk)(v1,..., vk)=det[αi(vj)]= det[δi j]=1. Hence, α1∧···∧ αk/ne}a⊔ionslash=0. ♦ ♦ 3.11* Exterior multiplication (⇐)Clear because α∧α=0. (⇒)Suppose α∧γ=0. Extend αto a basis α1,..., αnforV∨, with α1=α. Write γ= ∑cJαJ, where Jruns over all strictly ascending multi-indices 1 ≤j1<···<jk≤n. In the sum α∧γ=∑cJα∧αJ, all the terms α∧αJwith j1=1 vanish, since α=α1. Hence, 0=α∧γ=∑ j1/ne}a⊔ionslash=1cJα∧αJ. Since{α∧αJ}j1/ne}a⊔ionslash=1is a subset of a basis for Ak+1(V), it is linearly independent, and so all cJ are 0 if j1/ne}a⊔ionslash=1. Thus, γ=∑ j1=1cJαJ=α∧/parenleftbigg ∑ j1=1cJαj2∧···∧ αjk/parenrightbigg . ♦ ♦ 4.1ω(X)=yz,dω=−dx∧dz. 4.2 Write ω=∑i<jci jdxi∧dxj. Then ci j(p) =ωp(ei,ej), where ei=∂/∂xi. Calculate c12(p),c13(p), and c23(p). The answer is ωp=p3dx1∧dx2. 4.3dx=cosθdr−rsinθdθ,dy=sinθ,dr+rcosθdθ,dx∧dy=r dr∧dθ. 4.4dx∧dy∧dz=ρ2sinφdρ∧dφ∧dθ. 4.5α∧β=(a 1b1+a2b2+a3b3)dx1∧dx2∧dx3. 5.3The image φ4(U14)=/braceleftig (x,z)|−1<z<1,0<x<√ 1−z2/bracerightig . The transition function (φ1◦φ−1 4)(x,z) =φ1(x,y,z) = ( y,z) =/parenleftig −√ 1−x2−z2,z/parenrightig is aC∞ function of x,z. 370 Hints and Solutions to Selected End-of-Section Problems 5.4* Existence of a coordinate neighborhood LetUβbe any coordinate neighborhood of pin the maximal atlas. Any open subset of Uβis again in the maximal atlas, because it is C∞compatible with all the open sets in the maximal atlas. Thus Uα:=Uβ∩Uis a coordinate neighborhood such that p∈Uα⊂U. 6.3* Group of automorphisms of a vector space The manifold structure GL (V)eis the maximal atlas on GL (V)containing the coordinate chart (GL(V),φe). The manifold structure GL (V)uis the maximal atlas on GL (V)containing the coordinate chart (GL(V),φu). The two maps φe: GL(V)→Rn×nandφu: GL(V)→Rn×n areC∞compatible, because φe◦φ−1u: GL(n,R)→GL(n,R)is conjugation by the change-of- basis matrix from utoe. Therefore, the two maximal atlases are in fact the same. ♦ ♦ 7.4* Quotient space of a sphere with antipodal points identiﬁed (a) Let Ube an open subset of Sn. Then π−1(π(U))=U∪a(U), where a:Sn→Sn,a(x)=−x, is the antipodal map. Since the antipodal map is a homeomorphism, a(U)is open, and hence π−1(π(U))is open. By the deﬁnition of quotient topology, π(U)is open. This proves that π is an open map. (b) The graph Rof the equivalence relation ∼is R={(x,x)∈Sn×Sn}∪{(x,−x)∈Sn×Sn}=∆∪(1×a)(∆). By Corollary 7.8, because Snis Hausdorff, the diagonal ∆inSn×Snis closed. Since 1× a:Sn×Sn→Sn×Sn,(x,y)/ma√s⊔o→(x,−y)is a homeomorphism, (1×a)(∆)is also closed. As a union of the two closed sets ∆and(1×a)(∆),Ris closed in Sn×Sn. By Theorem 7.7, Sn/∼ is Hausdorff. ♦ ♦ 7.5* Orbit space of a continuous group action LetUbe an open subset of S. For each g∈G, since right multiplication by gis a homeomor- phism S→S, the set Ugis open. But π−1(π(U))=∪g∈GUg, which is a union of open sets, hence is open. By the deﬁnition of the quotient topology, π(U) is open. ♦ ♦ 7.9* Compactness of real projective spaceBy Exercise 7.11 there is a continuous surjective map π:Sn→RPn. Since the sphere Snis compact, and the continuous image of a compact set is compact (Proposition A.34), RPnis compact. ♦ ♦ 8.1* Differential of a map To determine the coefﬁcient ainF∗(∂/∂x)=a∂/∂u+b∂/∂v+c∂/∂w, we apply both sides touto get F∗/parenleftbigg∂ ∂x/parenrightbigg u=/parenleftbigg a∂ ∂u+b∂ ∂v+c∂ ∂w/parenrightbigg u=a. Hence, a=F∗/parenleftbigg∂ ∂x/parenrightbigg u=∂ ∂x(u◦F)=∂ ∂x(x)=1. Similarly, b=F∗/parenleftbigg∂ ∂x/parenrightbigg v=∂ ∂x(v◦F)=∂ ∂x(y)=0 and Hints and Solutions to Selected End-of-Section Problems 371 c=F∗/parenleftbigg∂ ∂x/parenrightbigg w=∂ ∂x(w◦F)=∂ ∂x(xy)=y. SoF∗(∂/∂x)=∂/∂u+y∂/∂w. ♦ ♦ 8.3 One can directly calculate a=F∗(X)uandb=F∗(X)vor more simply, one can apply Problem 8.2. The answer is a=−(sinα)x−(cosα)y,b=(cos α)x−(sinα)y. 8.5* Velocity of a curve in local coordinates We know that c′(t)=∑aj∂/∂xj. To compute ai, evaluate both sides on xi: ai=/parenleftbigg ∑aj∂ ∂xj/parenrightbigg xi=c′(t)xi=c∗/parenleftbiggd dt/parenrightbigg xi=d dt(xi◦c)=d dtci=˙ci(t).♦ ♦ 8.6c′(0)=−2y∂/∂x+2x∂/∂y. 8.7* Tangent space to a product If(U,φ) = ( U,x1,..., xm)and(V,ψ) = ( V,y1,..., yn)are charts about pinMandqinN respectively, then by Proposition 5.18, a chart about (p,q)inM×Nis (U×V,φ×ψ)=( U×V,(π∗ 1φ,π∗ 2ψ))=( U×V,¯x1,..., ¯xn,¯y1,..., ¯yn), where ¯ xi=π∗ 1xiand ¯yi=π∗ 2yi. Let π1∗/parenleftbig ∂/∂¯xj/parenrightbig =∑ai j∂/∂xi. Then ai j=π1∗/parenleftbigg∂ ∂¯xj/parenrightbigg xi=∂ ∂¯xj/parenleftig xi◦π1/parenrightig =∂¯xi ∂¯xj=δi j. Hence, π1∗/parenleftbigg∂ ∂¯xj/parenrightbigg =∑ iδi j∂ ∂xi=∂ ∂xj. This really means that π1∗/parenleftigg ∂ ∂¯xj/vextendsingle/vextendsingle/vextendsingle/vextendsingle (p,q)/parenrightigg =∂ ∂xj/vextendsingle/vextendsingle/vextendsingle/vextendsingle p. (8.7.1) Similarly, π1∗/parenleftbigg∂ ∂¯yj/parenrightbigg =0,π2∗/parenleftbigg∂ ∂¯xj/parenrightbigg =0,π2∗/parenleftbigg∂ ∂¯yj/parenrightbigg =∂ ∂yj. (8.7.2) A basis for T(p,q)(M×N)is ∂ ∂¯x1/vextendsingle/vextendsingle/vextendsingle/vextendsingle (p,q),...,∂ ∂¯xm/vextendsingle/vextendsingle/vextendsingle/vextendsingle (p,q),∂ ∂¯y1/vextendsingle/vextendsingle/vextendsingle/vextendsingle (p,q),...,∂ ∂¯yn/vextendsingle/vextendsingle/vextendsingle/vextendsingle (p,q). A basis for TpM×TqNis /parenleftigg ∂ ∂x1/vextendsingle/vextendsingle/vextendsingle/vextendsingle p,0/parenrightigg ,...,/parenleftigg ∂ ∂xm/vextendsingle/vextendsingle/vextendsingle/vextendsingle p,0/parenrightigg ,/parenleftigg 0,∂ ∂y1/vextendsingle/vextendsingle/vextendsingle/vextendsingle q/parenrightigg ,...,/parenleftigg 0,∂ ∂yn/vextendsingle/vextendsingle/vextendsingle/vextendsingle q/parenrightigg . By (8.7.1) and (8.7.2), the linear map (π1∗,π2∗)maps a basis of T(p,q)(M×N)to a basis of TpM×TqN. It is therefore an isomorphism. ♦ ♦ 372 Hints and Solutions to Selected End-of-Section Problems 8.8(a) Let c(t)be a curve starting at einGwith c′(0)=Xe. Then α(t)=(c(t),e)is a curve starting at (e,e)inG×Gwith α′(0)=(X e,0). Compute µ∗,(e,e)using α(t). 8.9* Transforming vectors to coordinate vectors Let(V,y1,..., yn)be a chart about p. Suppose (Xj)p=∑iai j∂/∂yi|p. Since(X1)p,...,(Xn)p are linearly independent, the matrix A=[ai j]is nonsingular. Deﬁne a new coordinate system x1,..., xnby yi=n ∑ j=1ai jxjfori=1,..., n. (8.9.1) By the chain rule, ∂ ∂xj=∑ i∂yi ∂xj∂ ∂yi=∑ai j∂ ∂yi. At the point p, ∂ ∂xj/vextendsingle/vextendsingle/vextendsingle/vextendsingle p=∑ai j∂ ∂yi/vextendsingle/vextendsingle/vextendsingle/vextendsingle p=(X j)p. In matrix notation,  y1 ... yn =A x1 ... xn ,so x1 ... xn =A−1 y1 ... yn . This means that (8.9.1) is equivalent to xj=∑n i=1/parenleftbig A−1/parenrightbigj iyi. ♦ ♦ 8.10 (a) For all x≤p,f(x)≤f(p). Hence, f′(p)= lim x→p−f(x)−f(p) x−p≥0. (8.10.1) Similarly, for all x≥p,f(x)≤f(p), so that f′(p)= lim x→p+f(x)−f(p) x−p≤0. (8.10.2) The two inequalities (8.10.1) and (8.10.2) together imply that f′(p)=0. 9.1c∈R−{0,−108}. 9.2Yes, because it is a regular level set of the function f(x,y,z,w)=x5+y5+z5+w5. 9.3Yes; see Example 9.12. 9.4* Regular submanifolds Letp∈S. By hypothesis there is an open set UinR2such that on U∩Sone of the coordinates is aC∞function of the other. Without loss of generality, we assume that y=f(x)for some C∞function f:A⊂R→B⊂R, where AandBare open sets in RandV:=A×B⊂U. Let F:V→R2be given by F(x,y)=(x,y−f(x)). Since Fis a diffeomorphism onto its image, it can be used as a coordinate map. In the chart (V,x,y−f(x)),V∩Sis deﬁned by the vanishing of the coordinate y−f(x). This proves that Sis a regular submanifold of R2. ♦ ♦ 9.5(R3,x,y,z−f(x,y))is an adapted chart for R3relative to Γ(f). 9.6Differentiate (9.3) with respect to t. 9.10* The transversality theorem Hints and Solutions to Selected End-of-Section Problems 373 (a)f−1(U)∩f−1(S)=f−1(U∩S)=f−1(g−1(0))=(g◦f)−1(0). (b) Let p∈f−1(U)∩f−1(S) =f−1(U∩S). Then f(p)∈U∩S. Because Sis a ﬁber of g, the pushforward g∗(Tf(p)S)equals 0. Because g:U→Rkis a projection, g∗(Tf(p)M)= T0(Rk). Applying g∗to the transversality equation (9.4), we get g∗f∗(TpN)=g∗(Tf(p)M)=T0(Rk). Hence, g◦f:f−1(U)→Rkis a submersion at p. Since pis an arbitrary point of f−1(U)∩f−1(S)=(g◦f)−1(0), this set is a regular level set of g◦f. (c) By the regular level set theorem, f−1(U)∩f−1(S)is a regular submanifold of f−1(U)⊂ N. Thus every point p∈f−1(S)has an adapted chart relative to f−1(S)inN. ♦ ♦ 10.7 Lete1,..., enbe a basis for Vandα1,..., αnthe dual basis for V∨. Then a basis for An(V)isα1∧···∧ αnandL∗(α1∧···∧ αn)=cα1∧···∧ αnfor some constant c. Suppose L(ej)=∑iai jei. Compute cin terms of ai j. 11.1 Letc(t)=(x1(t),..., xn+1(t))be a curve in Snwith c(0)=pandc′(0)=Xp. Differenti- ate∑i(xi)2(t)=1 with respect to t. Let Hbe the plane{(a1,a2,a3)∈R3|∑aipi=0}. Show thatTp(S2)⊂H. Because both sets are linear spaces and have the same dimension, equality holds. 11.3* Critical points of a smooth map on a compact manifold First Proof . Suppose f:N→Rmhas no critical point. Then it is a submersion. The projection to the ﬁrst factor, π:Rm→R, is also a submersion. It follows that the composite π◦f:N→ Ris a submersion. This contradicts the fact that as a continuous function from a compact manifold to R, the function π◦fhas a maximum and hence a critical point (see Problem 8.10). Second Proof . Suppose f:N→Rmhas no critical point. Then it is a submersion. Since a submersion is an open map (Corollary 11.6), the image f(N)is open in Rm. But the contin- uous image of a compact set is compact and a compact subset of Rmis closed and bounded. Hence, f(N)is a nonempty proper closed subset of Rm. This is a contradiction, because being connected, Rmcannot have a nonempty proper subset that is both open and closed. ♦ ♦ 11.4 Atp=(a,b,c),i∗(∂/∂u|p)=∂/∂x−(a/c)∂/∂z, and i∗(∂/∂v|p)=∂/∂y−(b/c)∂/∂z. 11.5 Use Proposition A.35 to show that fis a closed map. Then apply Problem A.12. 12.1* The Hausdorff condition on the tangent bundle Let(p,v)and(q,w)be distinct points of the tangent bundle T M. Case 1:p/ne}a⊔ionslash=q. Because Mis Hausdorff, pandqcan be separated by disjoint neighborhoods UandV. Then TUandTVare disjoint open subsets of T M containing (p,v)and(q,w), respectively. Case 2:p=q. Let(U,φ)be a coordinate neighborhood of p. Then(p,v)and(p,w)are distinct points in the open set TU. Since TUis homeomorphic to the open subset φ(U)×Rn ofR2n, and any subspace of a Hausdorff space is Hausdorff, TUis Hausdorff. Therefore, (p,v)and(p,w)can be separated by disjoint open sets in TU. ♦ ♦ 13.1* Support of a ﬁnite sum LetAbe the set where ∑ρiis not zero and Aithe set where ρiis not zero: A=/braceleftbig x∈M|∑ρi(x)/ne}a⊔ionslash=0/bracerightbig ,Ai={x∈M|ρi(x)/ne}a⊔ionslash=0}. 374 Hints and Solutions to Selected End-of-Section Problems If∑ρi(x)/ne}a⊔ionslash=0, then at least one ρi(x)must be nonzero. This implies that A⊂/uniontextAi. Taking the closure of both sides gives cl (A)⊂/uniontextAi. For a ﬁnite union,/uniontextAi=/uniontextAi(Exercise A.53). Hence, supp/parenleftbig∑ρi/parenrightbig =cl(A)⊂/uniondisplay Ai=/uniondisplay Ai=/uniondisplay supp ρi. ♦ ♦ 13.2* Locally ﬁnite family and compact set For each p∈K, letWpbe a neighborhood of pthat intersects only ﬁnitely many of the sets Aα. The collection{Wp}p∈Kis an open cover of K. By compactness, Khas a ﬁnite subcover {Wpi}r i=1. Since each Wpiintersects only ﬁnitely many of the Aα, the ﬁnite union W:=/uniontextr i=1Wpiintersects only ﬁnitely many of the Aα. ♦ ♦ 13.3 (a) Take f=ρM−B. 13.5* Support of the pullback by a projection LetA={p∈M|f(p)/ne}a⊔ionslash=0}. Then supp f=clM(A). Observe that (π∗f)(p,q)=f(p)/ne}a⊔ionslash=0 iff p∈A. Hence, {(p,q)∈M×N|(π∗f)(p,q)/ne}a⊔ionslash=0}=A×N. So supp(π∗f)=clM×N(A×N)=clM(A)×N=(supp f)×N by Problem A.20. ♦ ♦ 13.7* Closure of a locally ﬁnite union (⊃)Since Aα⊂/uniontextAα, taking the closure of both sides gives Aα⊂/uniondisplay Aα. Hence,/uniontextAα⊂/uniontextAα. (⊂)Instead of proving/uniontextAα⊂/uniontextAα, we will prove the contrapositive: if p/∈/uniontextAα, then p/∈/uniontextAα. Suppose p/∈/uniontextAα. By local ﬁniteness, phas a neighborhood Wthat meets only ﬁnitely many of the Aα’s, say Aα1,..., Aαm(see the ﬁgure below). /BulletW Aα2Aα1 Aα3p Since p/ne}a⊔ionslash∈Aαfor any α,p/ne}a⊔ionslash∈/uniontextm i=1Aαi. Note that Wis disjoint from Aαfor all α/ne}a⊔ionslash=αi, so W−/uniontextm i=1Aαiis disjoint from Aαfor all α. Because/uniontextm i=1Aαiis closed, W−/uniontextm i=1Aαiis an open set containing pdisjoint from/uniontextAα. By the local characterization of closure (Proposition A.48), p/ne}a⊔ionslash∈/uniontextAα. Hence,/uniontextAα⊂/uniontextAα. ♦ ♦ Hints and Solutions to Selected End-of-Section Problems 375 14.1* Equality of vector ﬁelds The implication in the direction (⇒)is obvious. For the converse, let p∈M. To show that Xp=Yp, it sufﬁces to show that Xp[h] =Yp[h]for any germ [h]ofC∞functions in C∞p(M). Suppose h:U→Ris aC∞function that represents the germ [h]. We can extend it to a C∞ function ˜h:M→Rby multiplying it by a C∞bump function supported in Uthat is identically 1 in a neighborhood of p. By hypothesis, X˜h=Y˜h. Hence, Xp˜h=(X ˜h)p=(Y˜h)p=Yp˜h. (14.1.1) Because ˜h=hin a neighborhood of p, we have Xph=Xp˜handYph=Yp˜h. It follows from (14.1.1) that Xph=Yph. Thus, Xp=Yp. Since pis an arbitrary point of M, the two vector ﬁelds XandYare equal. ♦ ♦ 14.6 Integral curve starting at a zero of a vector ﬁeld (a)*Suppose c(t)=pfor all t∈R. Then c′(t)=0=Xp=Xc(t) for all t∈R. Thus, the constant curve c(t)=pis an integral curve of Xwith initial point p. By the uniqueness of an integral curve with a given initial point, this is the maximal integral curve of Xstarting at p. 14.8 c(t)=1/((1/p)−t)on(−∞,1/p). 14.10 Show that both sides applied to a C∞function honMare equal. Then use Problem 14.1. 14.11−∂/∂y. 14.12 ck=∑i/parenleftbigg ai∂bk ∂xi−bi∂ak ∂xi/parenrightbigg . 14.14 Use Example 14.15 and Proposition 14.17. 15.3 (a) Apply Proposition A.43. (b) Apply Proposition A.43. (c) Apply Problem A.16. (d) By (a) and (b), the subset G0is a subgroup of G. By (c), it is an open submanifold. 15.4* Open subgroup of a connected Lie group For any g∈G, left multiplication ℓg:G→Gbygmaps the subgroup Hto the left coset gH. Since His open and ℓgis a homeomorphism, the coset gHis open. Thus, the set of cosets gH, g∈G, partitions Ginto a disjoint union of open subsets. Since Gis connected, there can be only one coset. Therefore, H=G. ♦ ♦ 15.5 Letc(t)be a curve in Gwith c(0) =a,c′(0) =Xa. Then(c(t),b)is a curve through (a,b)with initial velocity (Xa,0). Compute µ∗,(a,b)(Xa,0)using this curve (Proposition 8.18). Compute similarly µ∗,(a,b)(0,Yb). 15.7* Differential of the determinant map Letc(t)=AetX. Then c(0)=Aandc′(0)=AX. Using the curve c(t)to calculate the differ- ential yields 376 Hints and Solutions to Selected End-of-Section Problems detA,∗(AX)=d dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0det(c(t))=d dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0(detA)detetX =(det A)d dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0ettrX=(det A)trX. ♦ ♦ 15.8* Special linear group If det A=1, then Exercise 15.7 gives det∗,A(AX)=trX. Since tr Xcan assume any real value, det ∗,A:TAGL(n,R)→Ris surjective for all A∈ det−1(1). Hence, 1 is a regular value of det. ♦ ♦ 15.10 (a) O(n)is deﬁned by polynomial equations. (b) If A∈O(n), then each column of Ahas length 1. 15.11 Write out the conditions ATA=I, det A=1. If a2+b2=1, then(a,b)is a point on the unit circle, and so a=cosθ,b=sinθfor some θ∈[0,2π]. 15.14 /bracketleftbigg cosh1 sinh1 sinh1 cosh1/bracketrightbigg , where cosh t=(et+e−t)/2 and cosh t=(et−e−t)/2 are hyperbolic cosine and sine, respec- tively. 15.16 The correct target space for fis the vector space K2n(C)of 2n×2nskew-symmetric complex matrices. 16.4 Letc(t)be a curve in Sp (n)with c(0) =Iandc′(0) =X. Differentiate c(t)TJc(t) =J with respect to t. 16.5 Mimic Example 16.6. The left-invariant vector ﬁelds on Rnare the constant vector ﬁelds ∑n i=1ai∂/∂xi, where ai∈R. 16.9 A basis X1,e,..., Xn,efor the tangent space Te(G)at the identity gives rise to a frame consisting of left-invariant vector ﬁelds X1,..., Xn. 16.10* The pushforward of left-invariant vector ﬁelds Under the isomorphisms ϕH:TeH∼→L(H)andϕG:TeG∼→L(G), the Lie brackets correspond and the pushforward maps correspond. Thus, this problem follows from Proposition 16.14 bythe correspondence. A more formal proof goes as follows. Since XandYare left-invariant vector ﬁelds, X=˜A andY=˜BforA=X eandB=Ye∈TeH. Then F∗[X,Y]=F∗[˜A,˜B]=F∗([A,B]˜)(Proposition 16.10) =(F∗[A,B])˜ (deﬁnition of F∗onL(H)) =[F∗A,F∗B]˜ (Proposition 16.14) =[(F∗A)˜,(F∗B)˜] (Proposition 16.10) =[F∗˜A,F∗˜B] (deﬁnition of F∗onL(H)) =[F∗X,F∗Y]. ♦ ♦ Hints and Solutions to Selected End-of-Section Problems 377 16.11 (b) Let(U,x1,..., xn)be a chart about einG. Relative to this chart, the differential ca∗ateis represented by the Jacobian matrix [∂(xi◦ca)/∂xj|e]. Since ca(x)=axa−1is aC∞ function of xanda, all the partial derivatives ∂(xi◦ca)/∂xj|eareC∞and therefore Ad (a)is a C∞function of a. 17.1 ω=(xdx+ydy)/(x2+y2). 17.2 aj=∑ibi∂yi/∂xj. 17.4 (a) Suppose ωp=∑cidxi|p. Then λωp=π∗(ωp)=∑ciπ∗(dxi|p)=∑ci(π∗dxi)ωp=∑ci(dπ∗xi)ωp=∑ci(d¯xi)ωp. Hence, λ=∑cid¯xi. 18.4* Support of a sum or product (a) If(ω+τ)(p)/ne}a⊔ionslash=0, then ω(p)/ne}a⊔ionslash=0 orτ(p)/ne}a⊔ionslash=0. Hence, Z(ω+τ)c⊂Z(ω)c∪Z(τ)c. Taking the closure of both sides and using the fact that A∪B=A∪B, we get supp(ω+τ)⊂supp ω∪supp τ. (b) Suppose (ω∧τ)p/ne}a⊔ionslash=0. Then ωp/ne}a⊔ionslash=0 and τp/ne}a⊔ionslash=0. Hence, Z(ω∧τ)c⊂Z(ω)c∩Z(τ)c. Taking the closure of both sides and using the fact that A∩B⊂A∩B, we get supp(ω∧τ)⊂Z(ω)c∩Z(τ)c⊂supp ω∩supp τ. ♦ ♦ 18.6* Locally ﬁnite supports Letp∈supp ω. Since{supp ρα}is locally ﬁnite, there is a neighborhood WpofpinMthat intersects only ﬁnitely many of the sets supp ρα. The collection{Wp|p∈supp ω}covers supp ω. By the compactness of supp ω, there is a ﬁnite subcover {Wp1,..., Wpm}. Since each Wpiintersects only ﬁnitely many supp ρα, supp ωintersects only ﬁnitely many supp ρα. By Problem 18.4, supp(ραω)⊂supp ρα∩supp ω. Thus, for all but ﬁnitely many α, supp(ραω)is empty; i.e., ραω≡0. ♦ ♦ 18.8* Pullback by a surjective submersion The fact that π∗:Ω∗(M)→Ω∗(˜M)is an algebra homomorphism follows from Proposi- tions 18.9 and 18.11. Suppose ω∈Ωk(M)is ak-form on Mfor which π∗ω=0 inΩk(˜M). To show that ω=0, pick an arbitrary point p∈M, and arbitrary vectors v1,..., vk∈TpM. Since πis surjective, there is a point ˜ p∈˜Mthat maps to p. Since πis a submersion at ˜ p, there exist ˜ v1,..., ˜vk∈T˜p˜M such that π∗,˜p˜vi=vi. Then 0=(π∗ω)˜p(˜v1,..., ˜vk) (because π∗ω=0) =ωπ(˜p)(π∗˜v1,..., π∗˜vk) (deﬁnition of π∗ω) =ωp(v1,..., vk). 378 Hints and Solutions to Selected End-of-Section Problems Since p∈Mandv1,..., vk∈TpMare arbitrary, this proves that ω=0. Therefore, π∗:Ω∗(M)→ Ω∗(˜M)is injective. ♦ ♦ 18.9 (c) Because f(a)is the pullback by Ad (a−1), we have f(a) =det(Ad(a−1))by Prob- lem 10.7. According to Problem 16.11, Ad (a−1)is aC∞function of a. 19.1 F∗(dx∧dy∧dz)=d(x◦F)∧d(y◦F)∧d(z◦F). Apply Corollary 18.4(ii). 19.2 F∗(udu+vdv) =(2 x3+3xy2)dx+(3x2y+2y3)dy. 19.3 c∗ω=dt. 19.5* Coordinates and differential forms Let(V,x1,..., xn)be a chart about p. By Corollary 18.4(ii), df1∧···∧ dfn=det/bracketleftbigg∂fi ∂xj/bracketrightbigg dx1∧···∧ dxn. So(df1∧···∧ dfn)p/ne}a⊔ionslash=0 if and only if det [∂fi/∂xj(p)]/ne}a⊔ionslash=0. By the inverse function theorem, this condition is equivalent to the existence of a neighborhood Won which the map F:= (f1,..., fn):W→Rnis aC∞diffeomorphism onto its image. In other words, (W,f1,..., fn) is a chart. ♦ ♦ 19.7 Mimic the proof of Proposition 19.3. 19.9* Vertical plane Since ax+byis the zero function on the vertical plane, its differential is identically zero: adx+bdy=0. Thus, at each point of the plane, dxis a multiple of dyor vice versa. In either case, dx∧dy=0. ♦ ♦ 19.11 (a) Mimic Example 19.8. Deﬁne Ux={(x,y)∈M|fx/ne}a⊔ionslash=0}and Uy={(x,y)∈M|fy/ne}a⊔ionslash=0}, where fx,fyare the partial derivatives ∂f/∂x,∂f/∂yrespectively. Because 0 is a regular value of f, every point in Msatisﬁes fx/ne}a⊔ionslash=0 or fy/ne}a⊔ionslash=0. Hence,{Ux,Uy}is an open cover ofM. Deﬁne ω=dy/fxonUxand−dx/fyonUy. Show that ωis globally deﬁned on M. By the implicit function theorem, in a neighborhood of a point (a,b)∈Ux,xis aC∞ function of y. It follows that ycan be used as a local coordinate and the 1-form dy/fxis C∞at(a,b). Thus, ωisC∞onUx. A similar argument shows that ωisC∞onUy. (b) On M,df=fxdx+fydy+fzdz≡0. (c) Deﬁne Ui={p∈Rn+1|∂f/∂xi(p)/ne}a⊔ionslash=0}and ω=(− 1)i−1dx1∧···∧/hatwiderdxi∧···∧ dxn+1 ∂f/∂xionUi. 19.13∇×E=−∂B/∂tand div B=0. 20.3* Derivative of a smooth family of vector ﬁelds Let(V,y1,..., yn)be another coordinate neighborhood of psuch that Hints and Solutions to Selected End-of-Section Problems 379 Xt=∑ jbj(t,q)∂ ∂yjonV. OnU∩V, ∂ ∂xi=∑ j∂yj ∂xi∂ ∂yj. Substituting this into (20.2) in the text and comparing coefﬁcients with the expression for Xt above, we get bj(t,q)=∑ iai(t,q)∂yj ∂xi. Since ∂yj/∂xidoes not depend on t, differentiating both sides of this equation with respect to tgives ∂bj ∂t=∑ i∂ai ∂t∂yj ∂xi. Hence, ∑ j∂bj ∂t∂ ∂yj=∑ i,j∂ai ∂t∂yj ∂xi∂ ∂yj=∑ i∂ai ∂t∂ ∂xi. ♦ ♦ 20.6* Global formula for the exterior derivative By Theorem 20.12, (LY0ω)(Y1,..., Yk)=Y0(ω(Y1,..., Yk))−k ∑ j=1ω(Y1,...,[Y0,Yj],..., Yk) =Y0(ω(Y1,..., Yk))+k ∑ j=1(−1)jω([Y0,Yj],Y1,...,/hatwideYj,..., Yk).(20.6.1) By the induction hypothesis, Theorem 20.14 is true for (k−1)-forms. Hence, −(dιY0ω)(Y1,..., Yk)=−k ∑ i=1(−1)i−1Yi/parenleftig (ιY0ω)(Y1,...,/hatwideYi,..., Yk)/parenrightig −∑ 1≤i<j≤k(−1)i+j(ιY0ω)([Yi,Yj],Y1,...,/hatwideYi,...,/hatwideYj,..., Yk) =k ∑ i=1(−1)iYi/parenleftig ω(Y0,Y1,...,/hatwideYi,..., Yk)/parenrightig +∑ 1≤i<j≤k(−1)i+jω([Yi,Yj],Y0,Y1,...,/hatwideYi,...,/hatwideYj,..., Yk).(20.6.2) Adding (20.6.1) and (20.6.2) gives k ∑ i=0(−1)iYi/parenleftig ω(Y0,...,/hatwideYi,..., Yk)/parenrightig +k ∑ j=1(−1)jω([Y0,Yj],/hatwideY0,Y1,...,/hatwideYj,..., Yk) +∑ 1≤i<j≤k(−1)i+jω([Yi,Yj],Y0,...,/hatwideYi,...,/hatwideYj,..., Yk), which simpliﬁes to the right-hand side of Theorem 20.14. ♦ ♦ 380 Hints and Solutions to Selected End-of-Section Problems 21.1* Locally constant map on a connected space We ﬁrst show that for every y∈Y, the inverse f−1(y)is an open set. Suppose p∈f−1(y). Then f(p)=y. Since fis locally constant, there is a neighborhood Uofpsuch that f(U)={y}. Thus, U⊂f−1(y). This proves that f−1(y)is open. The equality S=/uniontext y∈Yf−1(y)exhibits Sas a disjoint union of open sets. Since Sis connected, this is possible only if there is just one such open set S=f−1(y0). Hence, f assumes the constant value y0onS. ♦ ♦ 21.5 The map Fis orientation-preserving. 21.6 Use Problem 19.11(c) and Theorem 21.5. 21.9 See Problem 12.2. 22.1 The topological boundary bd (M)is{0,1,2}; the manifold boundary ∂Mis{0}. 22.3* Inward-pointing vectors at the boundary (⇐) Suppose (U,φ)=(U,x1,..., xn)is a chart for Mcentered at psuch that Xp=∑ai∂/∂xi|p with an>0. Then the curve c(t)=φ−1(a1t,..., ant)inMsatisﬁes c(0)=p,c(]0,ε[)⊂M◦,andc′(0)=Xp. (22.3.1) SoXpis inward-pointing. (⇒) Suppose Xpis inward-pointing. Then Xp/∈Tp(∂M)and there is a curve c:[0,ε[→M such that (22.3.1) holds. Let (U,φ) = ( U,x1,..., xn)be a chart centered at p. On U∩M, we have xn≥0. If(φ◦c)(t) = (c1(t),..., cn(t)), then cn(0) =0 and cn(t)>0 for t>0. Therefore, the derivative of cnatt=0 is ˙cn(0)= lim t→0+cn(t)−cn(0) t=lim t→0+cn(t) t≥0. Since Xp=∑n i=1˙ci(0)∂/∂xi|p, the coefﬁcient of ∂/∂xn|pinXpis ˙cn(0). In fact, ˙ cn(0)>0 because if ˙ cn(0)were 0, then Xpwould be tangent to ∂Matp. ♦ ♦ 22.4* Smooth outward-pointing vector ﬁeld along the boundary Letp∈∂Mand let(U,x1,..., xn)be a coordinate neighborhood of p. Write Xα,p=n ∑ i=1ai(Xα,p)∂ ∂xi/vextendsingle/vextendsingle/vextendsingle/vextendsingle p. Then Xp=∑ αρα(p)Xα,p=n ∑ i=1∑ αρα(p)ai(Xα,p)∂ ∂xi/vextendsingle/vextendsingle/vextendsingle/vextendsingle p. Since Xα,pis outward-pointing, the coefﬁcient an(Xα,p)is negative by Problem 22.3. Because ρα(p)≥0 for all αwith ρα(p)positive for at least one α, the coefﬁcient ∑αρα(p)ai(Xα,p) of∂/∂xn|pinXpis negative. Again by Problem 22.3, this proves that Xpis outward-pointing. The smoothness of the vector ﬁeld Xfollows from the smoothness of the partition of unity ραand of the coefﬁcient functions ai(Xα,p)as functions of p. ♦ ♦ 22.6* Induced atlas on the boundary Letr1,..., rnbe the standard coordinates on the upper half-space Hn. As a shorthand, we write a=(a1,..., an−1)for the ﬁrst n−1 coordinates of a point in Hn. Since the transition function Hints and Solutions to Selected End-of-Section Problems 381 ψ◦φ−1:φ(U∩V)→ψ(U∩V)⊂Hn takes boundary points to boundary points and interior points to interior points, (i)/parenleftbig rn◦ψ◦φ−1/parenrightbig (a,0)=0, and (ii)/parenleftbig rn◦ψ◦φ−1/parenrightbig (a,t)>0 for t>0, where(a,0)and(a,t)are points in φ(U∩V)⊂Hn. Letxj=rj◦φandyi=ri◦ψbe the local coordinates on the charts (U,φ)and(V,ψ) respectively. In particular, yn◦φ−1=rn◦ψ◦φ−1. Differentiating (i) with respect to rjgives ∂yn ∂xj/vextendsingle/vextendsingle/vextendsingle/vextendsingle φ−1(a,0)=∂(yn◦φ−1) ∂rj/vextendsingle/vextendsingle/vextendsingle/vextendsingle (a,0)=∂(rn◦ψ◦φ−1) ∂rj/vextendsingle/vextendsingle/vextendsingle/vextendsingle (a,0)=0 for j=1,..., n−1. From (i) and (ii), ∂yn ∂xn/vextendsingle/vextendsingle/vextendsingle/vextendsingle φ−1(a,0)=∂/parenleftbig yn◦φ−1/parenrightbig ∂rn/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle (a,0) =lim t→0+/parenleftbig yn◦φ−1/parenrightbig (a,t)−/parenleftbig yn◦φ−1/parenrightbig (a,0) t =lim t→0+/parenleftbig yn◦φ−1/parenrightbig (a,t) t≥0, since both tand/parenleftbig yn◦φ−1/parenrightbig (a,t)are positive. The Jacobian matrix of J= [∂yi/∂xj]of the overlapping charts UandVat a point p= φ−1(a,0)inU∩V∩∂Mtherefore has the form J= ∂y1 ∂x1···∂y1 ∂xn−1∂y1 ∂xn ............ ∂yn−1 ∂x1···∂yn−1 ∂xn−1∂yn−1 ∂xn 0··· 0∂yn ∂xn = A∗ 0 ∂yn ∂xn , where the upper left (n−1)×(n−1)block A=[∂yi/∂xj]1≤i,j≤n−1is the Jacobian matrix of the induced charts U∩∂MandV∩∂Mon the boundary. Since det J(p)>0 and ∂yn/∂xn(p)> 0, we have det A(p)>0. ♦ ♦ 22.7* Boundary orientation of the left half-space Because a smooth outward-pointing vector ﬁeld along ∂Mis∂/∂x1, by deﬁnition an orienta- tion form of the boundary orientation on ∂Mis the contraction ι∂/∂x1(dx1∧dx2∧···∧ dxn)=dx2∧···∧ dxn. ♦ ♦ 22.8 Viewed from the top, C1is clockwise and C0is counterclockwise. 22.9 Compute ιX(dx1∧···∧ dxn+1). 22.10 (a) An orientation form on the closed unit ball is dx1∧···∧ dxn+1and a smooth outward-pointing vector ﬁeld on Uis∂/∂xn+1. By deﬁnition, an orientation form on Uis the contraction 382 Hints and Solutions to Selected End-of-Section Problems ι∂/∂xn+1(dx1∧···∧ dxn+1)=(− 1)ndx1∧···∧ dxn. 22.11 (a) Let ωbe the orientation form on the sphere in Problem 22.9. Show that a∗ω= (−1)n+1ω. 23.1 Letx=auandy=bv. 23.2 Use the Heine–Borel theorem (Theorem A.40). 23.3* Integral under a diffeomorphism Let{(Uα,φα)}be an oriented atlas for Mthat speciﬁes the orientation of M, and{ρα}a partition of unity on Msubordinate to the open cover {Uα}. Assume that F:N→Mis orientation-preserving. By Problem 21.4, {(F−1(Uα),φα◦F)}is an oriented atlas for N that speciﬁes the orientation of N. By Problem 13.6, {F∗ρα}is a partition of unity on N subordinate to the open cover {F−1(Uα)}. By the deﬁnition of the integral, /integraldisplay NF∗ω=∑ α/integraldisplay F−1(Uα)(F∗ρα) (F∗ω) =∑ α/integraldisplay F−1(Uα)F∗(ραω) =∑ α/integraldisplay (φα◦F)(F−1(Uα))(φα◦F)−1∗F∗(ραω) =∑ α/integraldisplay φα(Uα)(φ−1 α)∗(ραω) =∑ α/integraldisplay Uαραω=/integraldisplay Mω. IfF:N→Mis orientation-reversing, then {(F−1(Uα),φα◦F)}is an oriented atlas for Nthat gives the opposite orientation of N. Using this atlas to calculate the integral as above gives−/integraltext NF∗ω. Hence in this case/integraltext Mω=−/integraltext NF∗ω. ♦ ♦ 23.4* Stokes’s theorem for Rnand forHn An(n−1)-form ωwith compact support on RnorHnis a linear combination ω=n ∑ i=1fidx1∧···∧/hatwiderdxi∧···∧ dxn. (23.4.1) Since both sides of Stokes’s theorem are R-linear in ω, it sufﬁces to check the theorem for just one term of the sum (23.4.1). So we may assume ω=f dx1∧···∧/hatwiderdxi∧···∧ dxn, where fis aC∞function with compact support in RnorHn. Then dω=∂f ∂xidxi∧dx1∧···∧ dxi−1∧/hatwiderdxi∧···∧ dxn =(− 1)i−1∂f ∂xidx1∧···∧ dxi∧···∧ dxn. Since fhas compact support in RnorHn, we may choose a>0 large enough that supp flies in the interior of the cube [−a,a]n. Hints and Solutions to Selected End-of-Section Problems 383 Stokes’s theorem for Rn By Fubini’s theorem, one can ﬁrst integrate with respect to xi: /integraldisplay Rndω=/integraldisplay Rn(−1)i−1∂f ∂xidx1···dxn =(− 1)i−1/integraldisplay Rn−1/parenleftbigg/integraldisplay∞ −∞∂f ∂xidxi/parenrightbigg dx1···/hatwiderdxi···dxn =(− 1)i−1/integraldisplay Rn−1/parenleftbigg/integraldisplaya −a∂f ∂xidxi/parenrightbigg dx1···/hatwiderdxi···dxn. But /integraldisplaya −a∂f ∂xidxi=f(..., xi−1,a,xi+1,...)−f(..., xi−1,−a,xi+1,...) =0−0=0, because the support of flies in the interior of [−a,a]n. Hence,/integraltext Rndω=0. The right-hand side of Stokes’s theorem is/integraltext ∂Rnω=/integraltext ∅ω=0, because Rnhas empty boundary. This veriﬁes Stokes’s theorem for Rn. Stokes’s theorem for Hn Case 1:i/ne}a⊔ionslash=n. /integraldisplay Hndω=(− 1)i−1/integraldisplay Hn∂f ∂xidx1···dxn =(− 1)i−1/integraldisplay Hn−1/parenleftbigg/integraldisplay∞ −∞∂f ∂xidxi/parenrightbigg dx1···/hatwiderdxi···dxn =(− 1)i−1/integraldisplay Hn−1/parenleftbigg/integraldisplaya −a∂f ∂xidxi/parenrightbigg dx1···/hatwiderdxi···dxn =0 for the same reason as the case of Rn. As for/integraltext ∂Hnω, note that ∂Hnis deﬁned by the equation xn=0. Hence, on ∂Hn, the 1- form dxnis identically zero. Since i/ne}a⊔ionslash=n,ω=f dx1∧···∧/hatwiderdxi∧···∧ dxn≡0 on ∂Hn, so/integraltext ∂Hnω=0. Thus, Stokes’s theorem holds in this case. Case 2:i=n. /integraldisplay Hndω=(− 1)n−1/integraldisplay Hn∂f ∂xndx1···dxn =(− 1)n−1/integraldisplay Rn−1/parenleftbigg/integraldisplay∞ 0∂f ∂xndxn/parenrightbigg dx1···dxn−1. In this integral /integraldisplay∞ 0∂f ∂xndxn=/integraldisplaya 0∂f ∂xndxn =f(x1,..., xn−1,a)−f(x1,..., xn−1,0) =−f(x1,..., xn−1,0). Hence, 384 Hints and Solutions to Selected End-of-Section Problems /integraldisplay Hndω=(− 1)n/integraldisplay Rn−1f(x1,..., xn−1,0)dx1···dxn−1=/integraldisplay ∂Hnω because(−1)nRn−1is precisely ∂Hnwith its boundary orientation. So Stokes’s theorem also holds in this case. ♦ ♦ 23.5 Take the exterior derivative of x2+y2+z2=1 to obtain a relation among the 1-forms dx,dy, and dzonS2. Then show for example that for x/ne}a⊔ionslash=0, one has dx∧dy=(z/x)dy∧dz. 24.1 Assume ω=d f. Derive a contradiction using Problem 8.10(b) and Proposition 17.2. 25.4* The snake lemma If we view each column of the given commutative diagram as a cochain complex, then thediagram is a short exact sequence of cochain complexes 0→A→B→C→0. By the zig-zag lemma, it gives rise to a long exact sequence in cohomology. In the long exact sequence, H 0(A)=kerα,H1(A)=A1/imα=coker α, and similarly for BandC.♦ ♦ 26.2 Deﬁne d−1=0. Then the given exact sequence is equivalent to a collection of short exact sequences 0→imdk−1→Akdk−→imdk→0, k=0,..., m−1. By the rank–nullity theorem, dimAk=dim(imdk−1)+dim(imdk). When we compute the alternating sum of the left-hand side, the right-hand side will can- cel to 0. ♦ ♦ 28.1 LetUbe the punctured projective plane RP2−{p}andVa small disk containing p. Because Ucan be deformation retracted to the boundary circle, which after identiﬁcation is in factRP1,Uhas the homotopy type of RP1. SinceRP1is homeomorphic to S1,H∗(U)≃ H∗(S1). Apply the Mayer–Vietoris sequence. The answer is H0(RP2)=R,Hk(RP2)=0 for k>0. 28.2 Hk(Sn)=Rfork=0,n, and Hk(Sn)=0 otherwise. 28.3 One way is to apply the Mayer–Vietoris sequence to U=R2−{p},V=R2−{q}. A.13* The Lindel ¨of condition Let{Bi}i∈Ibe a countable basis and {Uα}α∈Aan open cover of the topological space S. For every p∈Uα, there exists a Bisuch that p∈Bi⊂Uα. Since this Bidepends on pandα, we write i=i(p,α). Thus, p∈Bi(p,α)⊂Uα. Now let Jbe the set of all indices j∈Isuch that j=i(p,α)for some pand some α. Then/uniontext j∈JBj=Sbecause every pinSis contained in some Bi(p,α)=Bj. For each j∈J, choose an α(j)such that Bj⊂Uα(j). Then S=/uniontext jBj⊂/uniontext jUα(j). So {Uα(j)}j∈Jis a countable subcover of {Uα}α∈A. ♦ ♦ Hints and Solutions to Selected End-of-Section Problems 385 A.15* Disconnected subset in terms of a separation (⇒)By (iii), A=(U∩V)∩A=(U∩A)∪(V∩A). By (i) and (ii), U∩AandV∩Aare disjoint nonempty open subsets of A. Hence, Ais discon- nected. (⇐)Suppose Ais disconnected in the subspace topology. Then A=U′∪V′, where U′and V′are two disjoint nonempty open subsets of A. By the deﬁnition of the subspace topology, U′=U∩AandV′=V∩Afor some open sets U,VinS. (i) holds because U′andV′are nonempty. (ii) holds because U′andV′are disjoint. (iii) holds because A=U′∪V′⊂U∪V. ♦ ♦ A.19* Uniqueness of the limit Suppose p/ne}a⊔ionslash=q. Since Sis Hausdorff, there exist disjoint open sets UpandUqsuch that p∈Up andq∈Uq. By the deﬁnition of convergence, there are integers NpandNqsuch that for all i≥Np,xi∈Upand for all i≥Nq,xi∈Uq. This is a contradiction, since Up∩Uqis the empty set. ♦ ♦ A.20* Closure in a product (⊂)By Problem A.5, cl (A)×Yis a closed set containing A×Y. By the deﬁnition of closure, cl(A×Y)⊂cl(A)×Y. (⊃)Conversely, suppose (p,y)∈cl(A)×Y. Ifp∈A, then(p,y)∈A×Y⊂cl(A×Y). Suppose p/∈A. By Proposition A.50, pis an accumulation point of A. Let U×Vbe any basis open set in S×Ycontaining (p,y). Because p∈ac(A), the open set Ucontains a point a∈Awith a/ne}a⊔ionslash=p. SoU×Vcontains the point (a,y)∈A×Ywith(a,y)/ne}a⊔ionslash=(p,y). This proves that (p,y) is an accumulation point of A×Y. By Proposition A.50 again, (p,y)∈ac(A×Y)⊂cl(A×Y). This proves that cl (A)×Y⊂cl(A×Y). ♦ ♦ B.1* The rank of a matrix (⇒)Suppose rk A≥k. Then one can ﬁnd klinearly independent columns, which we call a1, ...,ak. Since the m×kmatrix[a1···ak]has rank k, it has klinearly independent rows b1,..., bk. The matrix Bwhose rows are b1,...,bkis ak×ksubmatrix of A, and rk B=k. In other words, Bis a nonsingular k×ksubmatrix of A. (⇐)Suppose Ahas a nonsingular k×ksubmatrix B. Let a1,...,akbe the columns of Asuch that the submatrix [a1···ak]contains B. Since[a1···ak]hasklinearly independent rows, it also has klinearly independent columns. Thus, rk A≥k. ♦ ♦ B.2* Matrices of rank at most r LetAbe an m×nmatrix. By Problem B.1, rk A≤rif and only if all (r+1)×(r+1)minors m1(A),..., ms(A)ofAvanish. As the common zero set of a collection of continuous functions, Dris closed in Rm×n. ♦ ♦ B.3* Maximal rank For deﬁniteness, suppose n≤m. Then the maximal rank is nand every matrix A∈Rm×nhas rank≤n. Thus, Dmax={A∈Rm×n|rkA=n}=Rm×n−Dn−1. Since Dn−1is a closed subset of Rm×n(Problem B.2), Dmaxis open in Rm×n. ♦ ♦ B.4* Degeneracy loci and maximal-rank locus of a map 386 Hints and Solutions to Selected End-of-Section Problems (a) Let Drbe the subset of Rm×nconsisting of matrices of rank at most r. The degeneracy locus of rank rof the map F:S→Rm×nmay be described as Dr(F)={x∈S|F(x)∈Dr}=F−1(Dr). Since Dris a closed subset of Rm×n(Problem B.2) and Fis continuous, F−1(Dr)is a closed subset of S. (b) Let Dmaxbe the subset of Rm×nconsisting of all matrices of maximal rank. Then Dmax(F) =F−1(Dmax). Since Dmaxis open in Rm×n(Problem B.3) and Fis continu- ous, F−1(Dmax)is open in S. ♦ ♦ B.7 Use Example B.5. List of Notations RnEuclidean space of dimension n(p. 4) p=(p1,..., pn) point inRn(p. 4) C∞smooth or inﬁnitely differentiable (p. 4) ∂f/∂xipartial derivative with respect to xi(pp. 4, 67) f(k)(x) thekth derivative of f(x)(p. 5) B(p,r) open ball in Rnwith center pand radius r(pp. 7, 317) ]a,b[ open interval in R1(p. 8) Tp(Rn)orTpRntangent space to Rnatp(p. 10) v= v1 v2 v3 =/an}bracke⊔le{⊔v1,..., vn/an}bracke⊔ri}h⊔column vector (p. 11) {e1,..., en} standard basis for Rn(p. 11) Dvf directional derivative of fin the direction of vatp(p. 11) x∼y equivalence relation (p. 11) C∞ p algebra of germs of C∞functions at pinRn(p. 12) Dp(Rn) vector space of derivations at pinRn(p. 13) X(U) vector space of C∞vector ﬁelds on U(p. 15) Der(A) vector space of derivations of an algebra A(p. 17) δi j Kronecker delta (p. 13) Hom(V,W) vector space of linear maps f:V→W(p. 19) V∨=Hom(V,R) dual of a vector space (p. 19) VkCartesian product V×···× Vofkcopies of V(p. 22) Lk(V) vector space of k-linear functions on V(p. 22) 388 List of Notations (a1a2···ar) cyclic permutation, r-cycle (p. 20) (ab) transposition (p. 20) Sk group of permutations of kobjects (p. 20) sgn(σ)or sgn σ sign of a permutation (p. 20) Ak(V) vector space of alternating k-linear functions on V(p. 23) σf a function facted on by a permutation σ(p. 23) e identity element of a group (p. 24) σ·x left action of σonx(p. 24) x·σ right action of σonx(p. 24) S f symmetrizing operator applied to f(p. 24) A f alternating operator applied to f(p. 24) f⊗g tensor product of multilinear functions fandg(p. 25) f∧g wedge product of multicovectors fandg(p. 26) B=[bi j]or[bi j] matrix whose (i,j)-entry is bi jorbi j(p. 30) det[bi j]or det[bi j] determinant of the matrix [bi j]or[bi j](p. 30) /logicalandtext(V) exterior algebra of a vector space (p. 30) I=(i1,..., ik) multi-index (p. 31) eI k-tuple(ei1,..., eik)(p. 31) αIk-covector αi1∧···∧ αik(p. 31) T∗ p(Rn)orT∗ pRncotangent space to Rn(p. 34) df differential of a function (pp. 34, 191) dxIdxi1∧···∧ dxik(p. 36) Ωk(U) vector space of C∞k-forms on U(pp. 36, 203) Ω∗(U) direct sum/circleplustextn k=0Ωk(U)(p. 37, 206) ω(X) the function p/ma√s⊔o→ωp(Xp)(p. 37) F(U)orC∞(U) ring of C∞functions on U(p. 38) dω exterior derivative of ω(p. 38) fx ∂f/∂x, partial derivative of fwith respect to x(p. 38) /circleplustext∞ k=0Akdirect sum of A0,A1,...(p. 30) grad f gradient of a function f(p. 41) curlF curl of a vector ﬁeld F(p. 41) divF divergence of a vector ﬁeld F(p. 41) Hk(U) kth de Rham cohomology of U(p. 43) List of Notations 389 {Uα}α∈A open cover (p. 48) (U,φ),(U,φ:U→Rn)chart or coordinate open set (p. 48) 1U identity map on U(p. 49) Uαβ Uα∩Uβ(p. 50) Uαβ γ Uα∩Uβ∩Uγ(p. 50) U={(Uα,φα)} atlas (p. 50) C complex plane (p. 50) /coproducttextdisjoint union (pp. 51, 129) φα|Uα∩V restriction of φαtoUα∩V(p. 54) Γ(f) graph of f(p. 54) Km×nvector space of m×nmatrices with entries in K(p. 54) GL(n,K) general linear group over a ﬁeld K(p. 54) M×N product manifold (p. 55) f×g Cartesian product of two maps (p. 55) Snunit sphere in Rn+1(p. 58) F∗h pullback of a function hby a map F(p. 60) J(f)=[ ∂Fi/∂xj] Jacobian matrix (p. 68) det[∂Fi/∂xj] Jacobian determinant (p. 68) ∂(F1,..., Fn) ∂(x1,..., xn)Jacobian determinant (p. 68) µ:G×G→G multiplication on a Lie group (p. 66) ι:G→G inverse map of a Lie group (p. 66) K×nonzero elements of a ﬁeld K(p. 66) S1unit circle in C×(p. 66) A=[a i j],[ai j] matrix whose (i,j)-entry is ai jorai j(p. 67) S/∼ quotient (p. 71) [x] equivalence class of x(p. 71) π−1(U) inverse image of Uunder π(p. 71) RPnreal projective space of dimension n(p. 76) /bardblx/bardbl modulus of x(p. 77) a1∧···∧/hatwideai∧···∧ anthe caret/hatwidemeans to omit ai(p. 80) G(k,n) Grassmannian of k-planes in Rn(p. 82) rkA rank of a matrix A(p. 82 (p. 344) C∞ p(M) germs of C∞functions at pinM(p. 87) 390 List of Notations Tp(M)orTpM tangent space to Matp(p. 87) ∂/∂xi|p coordinate tangent vector at p(p. 87) d/dt|p coordinate tangent vector of a 1-dimensional manifold (p. 87) F∗,porF∗ differential of Fatp(p. 87) c(t) curve in a manifold (p. 92) c′(t):=c∗/parenleftigd dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t0/parenrightig velocity vector of a curve (p. 92) ˙c(t) derivative of a real-valued function (p. 92) φS coordinate map on a submanifold S(p. 100) f−1({c})orf−1(c) level set (p. 103) Z(f)=f−1(0) zero set (p. 103) SL(n,K) special linear group over a ﬁeld K(pp. 107, 109) mi jormi j(A) ( i,j)-minor of a matrix A(p. 107) Mor(A,B) the set of morphisms from AtoB(p. 110) 1A identity map on A(p. 110) (M,q) pointed manifold (p. 111) ≃ isomorphism (p. 111) F,G functors (p. 111) C,D categories (p. 111) {e1,..., en} basis for a vector space V(p. 113) {α1,..., αn} dual basis for V∨(p. 113) L∨dual of linear map L(p. 113) O(n) orthogonal group (p. 117) ATtranspose of a matrix A(p. 117) ℓg left multiplication by g(p. 117) rg right multiplication by g(p. 117) Dmax(F) maximal rank locus of F:S→Rm×n(pp. 118, 345) i:N→M inclusion map (p. 123) T M tangent bundle (p. 129) ˜φ coordinate map on the tangent bundle (p. 130) Ep:=π−1(p) ﬁber at pof a vector bundle (p. 133) X vector ﬁeld (p. 136) Xp tangent vector at p(p. 136) Γ(U,E) vector space of C∞sections of Eover U(p. 137) List of Notations 391 Γ(E):=Γ(M,E) vector space of C∞sections of Eover M(p. 137) supp f support of a function f(p. 140) B(p,r) closed ball in Rnwith center pand radius r(p. 143) A, cl(A), or cl S(A) closure of a set AinS(pp. 148, 334) ct(p) integral curve through p(p. 152) Diff(M) group of diffeomorphisms of M(p. 153) Ft(q)=F(t,q) local ﬂow (p. 156) [X,Y] Lie bracket of vector ﬁelds, bracket in a Lie algebra (pp. 157, 158) X(M) Lie algebra of C∞vector ﬁelds on M(p. 158) Sn vector space of n×nreal symmetric matrices (p. 166) R2/Z2torus (p. 167) /bardblX/bardbl norm of a matrix (p. 169 exp(X)oreXexponential of a matrix X(p. 170) tr(X) trace (p. 171) Z(G) center of a group G(p. 176) SO(n) special orthogonal group (p. 176) U(n) unitary group (p. 176) SU(n) special unitary group (p. 177) Sp(n) compact symplectic group (p. 177) J the matrix/bracketleftig 0In −In0/bracketrightig (p. 177) In n×nidentity matrix (p. 177) Sp(2n,C) complex symplectic group (p. 177) Kn space of n×nreal skew-symmetric matrices (p. 179) ˜A left-invariant vector ﬁeld generated by A∈TeG(p. 180) L(G) Lie algebra of left-invariant vector ﬁelds on G(p. 180) g Lie algebra (p. 182) h⊂g Lie subalgebra (p. 182) gl(n,R) Lie algebra of GL (n,R)(p. 183) sl(n,R) Lie algebra of SL (n,R)(p. 186) o(n) Lie algebra of O (n)(p. 186) u(n) Lie algebra of U (n)(p. 186) (df)p,df|p value of a 1-form at p(p. 191) 392 List of Notations T∗ p(M)orT∗ pM cotangent space at p(p. 190) T∗M cotangent bundle (p. 192) F∗:T∗ F(p)M→T∗ pN codifferential (p. 196) F∗ω pullback of a differential form ωbyF(pp. 196, 205) /logicalandtextk(V∨)=Ak(V) k-covectors on a vector space V(p. 200) ωp value of a differential form ωatp(p. 200) Ik,n the set of strictly ascending multi-indices 1≤i1<···<ik≤n(p. 201) /logicalandtextk(T∗M) kth exterior power of the cotangent bundle (p. 203) Ωk(G)Gleft-invariant k-forms on a Lie group G(p. 208) supp ω support of a k-form (p. 208) dω exterior derivative of a differential form ω(p. 213) ω|S restriction of a differential from ωto a submanifold S (p. 216) ֒→ inclusion map (p. 216) LXY the Lie derivative of a vector ﬁeld Yalong X(p. 224) LXω the Lie derivative of a differential form ωalong X(p. 226) ιvω interior multiplication of ωbyv(p. 227) (v1,..., vn) ordered basis (p. 237) [v1,..., vn] ordered basis as a matrix (p. 238) (M,[ω]) oriented manifold with orientation [ω](p. 244) −M the oriented manifold having the opposite orientation as M (p. 246) Hnclosed upper half-space (p. 248) M◦interior of a manifold with boundary (pp. 248, 252) ∂M boundary of a manifold with boundary (pp. 248, 251) L1left half-line (p. 251) int(A) topological interior of a subset A(p. 252) ext(A) exterior of a subset A(p. 252) bd(A) topological boundary of a subset A(p. 252) {p0,..., pn} partition of a closed interval (p. 260) P={P1,..., Pn} partition of a closed rectangle (p. 260) L(f,P) lower sum of fwith respect to a partition P(p. 260) U(f,P) upper sum of fwith respect to a partition P(p. 260) List of Notations 393 /integraltext Rf upper integral of fover a closed rectangle R(p. 261) /integraltext Rf lower integral of fover a closed rectangle R(p. 261) /integraltext Rf(x)|dx1···dxn| Riemann integral of fover a closed rectangle R(p. 261) /integraltext Uω Riemann integral of a differential form ωover U(p. 263) vol(A) volume of a subset AofRn(p. 262) Disc(f) set of discontinuities of a function f(p. 262) Ωk c(M) vector space of C∞k-forms with compact support on M (p. 265) Zk(M) vector space of closed k-forms on M(p. 275) Bk(M) vector space of exact k-forms on M(p. 275) Hk(M) de Rham cohomology of Min degree k(p. 275) [ω] cohomology class of ω(p. 275) F#orF∗induced map in cohomology (p. 278) H∗(M) the cohomology ring ⊕n k=0Hk(M)(p. 279) C=({Ck}k∈Z,d) cochain complex (p. 281) (Ω∗(M),d) de Rham complex (p. 281) Hk(C) kth cohomology of C(p. 283) Zk(C) subspace of k-cocycles (p. 283) Bk(C) subspace of k-coboundaries (p. 283 d∗:Hk(C)→Hk+1(A)connecting homomorphism (p. 284) /d47/d47/d47/d47 injection or maps to under an injection (p. 285) /d31/d47/d47/d47/d47 maps to under a surjection (p. 285) iU:U→M inclusion map of UinM(p. 288) jU:U∩V→U inclusion map of U∩VinU(p. 288) /d47/d47/d47/d47 surjection (p. 291) χ(M) Euler characteristic of M(p. 295) f∼g fis homotopic to g(p. 296) Σg compact orientable surface of genus g(p. 310) d(p,q) distance between pandq(p. 317) (a,b) open interval (p. 318) (S,T) a set Swith a topology T(p. 318) Z(f1,..., fr) zero set of f1,..., fr(p. 319) Z(I) zero set of all the polynomials in an ideal I(p. 320) 394 List of Notations IJ the product ideal (p. 320) ∑αIα sum of ideals (p. 320) TA subspace topology or relative topology of A(p. 320) Q the set of rational numbers (p. 323) Q+the set of positive rational numbers (p. 323) A×B Cartesian product of two sets AandB(p. 326) Cx connected component of a point x(p. 333) ac(A) the set of accumulation points of A(p. 335) Z+the set of positive integers (p. 336) Dr the set of matrices of rank ≤rinRm×n(p. 345) Dmax the set of matrices of maximal rank in Rm×n(p. 345) Dr(F) degeneracy locus of rank rof a map F:S→Rm×n(p. 345) kerf kernel of a homomorphism f(p. 350) imf image of a map f(p. 350) coker f cokernel of a homomorphism f(p. 350) v+W coset of a subspace W(p. 349) V/W quotient vector space of VbyW(p. 349) ∏αVα,A×B direct product (p. 351) /circleplustext αVα,A⊕B direct sum (p. 351) A+B sum of two vector subspaces (p. 351) A⊕iB internal direct sum (p. 351) W⊥W“perp,” orthogonal complement of W(p. 352) H skew ﬁeld of quaternions (p. 353) End K(V) algebra of endomorphisms of Vover K(p. 354) References [1] V . I. Arnold, Mathematical Methods of Classical Mechanics, 2nd ed., Springer, New York, 1989. [2] P. Bamberg and S. Sternberg, A Course in Mathematics for Students of Physics , V ol. 2, Cambridge University Press, Cambridge, UK, 1990. [3] W. Boothby, An Introduction to Differentiable Manifolds and Riemannian Geometry , 2nd ed., Academic Press, Boston, 1986. [4] R. Bott and L. W. Tu, Differential Forms in Algebraic Topology , 3rd corrected printing, Graduate Texts in Mathematics, V ol. 82, Springer, New York, 1995. [5]´E. Cartan, Sur certaines expressions diff´ erentielles et le probl` eme de Pfaff, Ann. E.N.S. (3), vol. XVI (1899), pp. 239–332 (= Oeuvres compl` etes , vol. II, Gauthier- Villars, Paris, 1953, pp. 303–396). [6] C. Chevalley, Theory of Lie Groups, Princeton University Press, Princeton, 1946. [7] L. Conlon, Differentiable Manifolds, 2nd ed., Birkh¨ auser Boston, Cambridge, MA, 2001. [8] G. de Rham, Sur l’analysis situs des vari´ et´ es ` a n-dimension, Journal Math. 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Kervaire, A manifold which does not admit any differentiable structure, Commen- tarii Mathematici Helvetici 34 (1960), pp. 257–270. [24] M. Kervaire and J. Milnor, Groups of homotopy spheres: I, Annals of Mathematics 77 (1963), pp. 504–537. [25] J. M. Lee, Introduction to Smooth Manifolds , Graduate Texts in Mathematics, V ol. 218, Springer, New York, 2003. [26] J. E. Marsden and M. J. Hoffman, Elementary Classical Analysis , 2nd ed., W. H. Free- man, New York, 1993. [27] J. Milnor, On manifolds homeomorphic to the 7-sphere, Annals of Mathematics 64 (1956), pp. 399–405. [28] J. Milnor, Topology from the Differentiable Viewpoint , University Press of Virginia, Charlottesville, V A, 1965. [29] J. Munkres, Topology , 2nd ed., Prentice–Hall, Upper Saddle River, NJ, 2000. [30] J. Munkres, Elements of Algebraic Topology , Perseus Publishing, Cambridge, MA, 1984. [31] J. Munkres, Analysis on Manifolds, Addison–Wesley, Menlo Park, CA, 1991. [32] H. Poincar´ e, Sur les r´ esidus des int´ egrales doubles, Acta Mathematica 9 (1887), pp. 321–380. Oeuvres , Tome III, pp. 440–489. [33] H. Poincar´ e, Analysis situs, Journal de l’ ´Ecole Polytechnique 1 (1895), pp. 1–121; Oeuvres 6, pp. 193–288. [34] H. Poincar´ e, Les m´ ethodes nouvelles de la m´ ecanique c´ eleste, vol. III, Gauthier-Villars, Paris, 1899, Chapter XXII. [35] W. Rudin, Principles of Mathematical Analysis , 3rd ed., McGraw–Hill, New York, 1976. [36] M. Spivak, A Comprehensive Introduction to Differential Geometry , V ol. 1, 3rd ed., Publish or Perish, Houston, 2005. [37] O. Veblen and J. H. C. Whitehead, A set of axioms for differential geometry, Pro- ceedings of National Academy of Sciences 17 (1931), pp. 551–561. The Mathematical Works of J. H. C. Whitehead , V ol. 1, Pergamon Press, 1962, pp. 93–104. [38] F. Warner, Foundations of Differentiable Manifolds and Lie Groups , Springer, New York, 1983. [39] M. Zisman, Fibre bundles, ﬁbre maps, in History of Topology, edited by I. M. James, Elsevier, Amsterdam, 1999, pp. 605–629. Index abelian Lie algebra, 158 absolute convergence, 170 accumulation point, 335 action, 24 of the permutation group on k-linear functions, 23 adapted chart, 100 adjoint representation, 188 A f, 24 algebra, 12 graded, 30 algebra homomorphism, 12 alternating k-linear function, 23 alternating k-tensor, 23 alternating operator, 24 analytic, 4 anticommutative, 30, 279anticommutativity of the wedge product, 27 antiderivation, 39, 210 degree of, 39, 210 is a local operator, 212 antihomomorphism, 355 ascending multi-index, 31 associative axiom in a category, 111 associativity of the tensor product, 25 of the wedge product, 29 atlas, 50 equivalent oriented atlases, 246 for a regular submanifold, 102maximal, 52oriented, 245 automorphism of a vector space, 354 Banach algebra, 170 Banach space, 170base coordinates on a vector bundle, 135 base space of a vector bundle, 134 basic open set, 326basis, 321 forA k(V), 31 fork-tensors, 32 for the cotangent space, 35, 191for the dual space, 19for the product topology, 326for the tangent space, 89of neighborhoods at a point, 324 bi-invariant top form on a compact connected Lie group, 209 bilinear, 22bilinear form, 358bilinear map as a tensor product of covectors, 25 boundary manifold boundary, 252of an n-manifold with boundary is an (n−1)-manifold without boundary, 253 topological boundary, 252 boundary orientation, 255 boundary point, 251, 252 398 Index ofHn, 248 bounded inRn, 332 bracket of a Lie algebra, 158 bump function, 140bundle map, 135 over a manifold, 135 C ∞extension of a function, 144 C∞function need not be analytic, 5onR n, 4 on a manifold, 59 C∞invariance of domain, 249 C∞manifold, 53 C∞manifold with boundary, 251 C∞map between manifolds, 61 C∞-compatible charts, 49 Ckfunction onRn, 4 Cartan, ´Elie, 18, 189 Cartan homotopy formula, 229 Cartesian product, 326 category, 110center of a group, 176 chain rule for maps of manifolds, 88in calculus notation, 91 change-of-basis matrix, 238 change-of-variables formula, 264characterization of smooth sections, 138 chart, 48 about a point, 53adapted, 100 C ∞-compatible, 49 centered at a point, 48compatible with an atlas, 51on a manifold with boundary, 250 circle a nowhere-vanishing 1-form, 216cohomology of, 292 is a manifold, 55 same homotopy type as the punctured plane, 297closed form, 40, 275 closed map, 329closed set, 319 Zariski closed, 319 closed subgroup, 168closed subgroup theorem, 169closure, 334 characterization of a point in, 334 of a ﬁnite union or ﬁnite intersection, 336of a locally ﬁnite union, 148 coarser topology, 318 coboundary, 283 cochain, 283 of degree k, 283 cochain homotopy, 312 cochain complex, 41, 281cochain map, 283 cocycle, 283 codifferential, 196 codimension, 100cohomologous closed forms, 275cohomology, seede Rham cohomology in degree zero, 280 cohomology class, 275, 283cohomology ring, 279 of a torus, 303 cohomology vector space, 283 of a torus, 302 cokernel, 350 commutator of superderivations, 45 compact, 329 closed subset of a compact space is compact, 330 compact subset of a Hausdorff space is closed, 331 continuous bijection from a compact space to a Hausdorff space is ahomeomorphism, 331 continuous image of a compact set is compact, 331 ﬁnite union of compact sets is compact, 331 product of compact spaces is compact, 331 compatible charts, 49 complementary subspace, 351complete vector ﬁeld, 156 complete normed algebra, 170 Index 399 complete normed vector space, 170 complex general linear group, 55complex special linear group, 109complex symplectic group, 177, 359 Lie algebra of, 187 composite in a category, 110of smooth maps is smooth, 62 conjugate of a matrix, 355of a quaternion, 355 connected, 332 continuous image of a connected set is connected, 333 connected component containing a point, 333 connected space a locally constant map on a connected space is constant, 246 connectedness union of connected sets having a point in common is connected, 333 connecting homomorphism, 284 constant rank theorem, 115, 343constant-rank level set theorem, 116continuity of a map on a quotient space, 72 continuous at a point, 327continuous bijection from a compact space to a Hausdorff space is ahomeomorphism, 331 continuous image of a compact set is compact, 331 continuous image of a connected set is connected, 333 frame, 240 iff the inverse image of any closed set is closed, 329 iff the inverse image of any open set is open, 328 on a set, 327pointwise orientation, 240the projection is continuous, 328 continuous category, 111 contractible, 298 Euclidean space is, 298 contraction, seeinterior multiplication contravariant functor, 112convention on subscripts and superscripts, 44 convergence, 336 absolute, 170 coordinate map, 48coordinate neighborhood, 48 coordinate system, 48 coordinates base, 135 ﬁber, 135 on a manifold, 53 coordinates on a projective space homogeneous, 76 coset, 349coset representative, 349 cotangent bundle, 192 topology on, 192 cotangent space, 34, 190 basis for, 35, 191 of a manifold with boundary, 253 covariant functor, 111 covector, 19, 23 at a point of a manifold, 190of degree 0, 23 on a vector space, 23 covector ﬁeld, 34, 190cover, 329 critical point of a map of manifolds, 97of a smooth map from a compact manifold toR n, 127 critical value of a map of manifolds, 97 cross is not locally Euclidean, 49 cross product relation to wedge product, 45 curl, 41 curve, 53 existence with a given initial vector, 94in a manifold, 92 starting at a point, 92 cuspidal cupic, 120cycle, 274 of length r, 20 cycles disjoint, 20 cyclic permutation, 20 400 Index de Rham, Georges, 274 de Rham cohomology, 43, 275 homotopy invariance, 311 in degree greater than the dimension of the manifold, 276 in degree zero, 275 of a circle, 276, 292of a M¨ obius band, 301 of a multiply punctured plane, 310 of a punctured plane, 301of a punctured torus, 307 of a sphere, 310 of a surface of genus 3, 310of a surface of genus 2, 308 of the real line, 276 of the real projective plane, 310ring structure, 279 de Rham complex, 41, 281 deformation retract, 299 implies the same homotopy type, 299 deformation retraction, 299 degeneracy locus, 345degree of a cochain, 283 of a differential form, 36, 200 of a tensor, 22of an antiderivation, 39, 210 deleted neighborhood, 335 dense, 338derivation at a point, 13, 87 of a constant function is zero, 13 ofC ∞functions is a local operator, 219 of a Lie algebra, 158 of an algebra, 16 derivative of a matrix exponential, 171 determinant differential of, 175 diagram-chasing, 286 diffeomorphism, 63 local, 68 of an open ball with Rn, 8 of an open cube with Rn, 8 of an open interval with R, 8 of open subsets of Rn, 8 orientation-preserving, 244orientation-reversing, 244 difference mapin the Mayer–Vietoris sequence, 288 differentiable structure, 53differential, 281 agrees with exterior derivative on 0-forms, 210 compute using curves, 95 matrix of, 88of a function, 34, 191 in terms of coordinates, 35 relation with differential of a map, 191 of a map, 87 local expression, 91 of left multiplication, 95 of the inverse of a diffeomorphism, 114of the determinant, 174, 175 of the inverse map in a Lie group, 99, 175 of the multiplication map in a Lie group, 99, 175 differential complex, 41differential form, 36, 200 as a multilinear function on vector ﬁelds, 37 closed, 275 degree of, 36exact, 275 local expression, 202 onM×R, 312 on a manifold with boundary, 253pullback, 204 smooth 1-form, 193 smoothness characterizations, 203support of, 208 type-I, 313 type-II, 313wedge product of differential forms, 205 with compact support, 209 differential forms one-parameter family of, 221 differential 1-form, 190 local expression, 191 dimension invariance of, 48, 89 ofA k(V), 32 of the orthogonal group, 167 direct product, 351 direct sum, 351 external, 351internal, 351 directional derivative, 11 Index 401 disconnected, 332 discrete space, 319discrete topology, 319 disjoint cycles, 20 disjoint union, 129distance inR n, 317 div, 41 divergence, 41division algebra, 353 division ring, 353 domain of integration, 263dot product, 23dual functorial properties, 113 of a linear map, 113 dual basis, 19 dual map matrix of, 114 dual space, 19, 113 basis, 19 has the same dimension as the vector space, 19 embedded submanifold, 124 embedding, 121 image is a regular submanifold, 123 endomorphism, 354 invertible, 354matrix of, 354 equivalence class, 71 equivalence of functions, 12equivalence relation, 11 open, 74 equivalent ordered bases, 238 equivalent oriented atlas, 246Euclidean space is contractible, 298 is Hausdorff, 325 is second countable, 324 Euler characteristic, 295 Euler’s formula, 108 even permutation, 20even superderivation, 45exact form, 40, 275 exact sequence, 281 long, 285short, 281, 284 exponentialof a matrix, 169 extension by zero of a form, 316 of a function, 261, 290 exterior algebra, 306 of multicovectors, 30 exterior derivative, 38, 210 characterization, 40 existence, 212global formula, 233 on a coordinate chart, 211 uniqueness, 213 exterior differentiation, seeexterior derivative exterior point, 251exterior power, 200 of the cotangent bundle, 203 exterior product, 26external direct sum, 351 ﬁber of a map, 133of a vector bundle, 133 ﬁber coordinates on a vector bundle, 135 ﬁber-preserving, 133ﬁner topology, 318 ﬁnite-complement topology, 319 ﬁrst countable, 324ﬁrst isomorphism theorem, 351 ﬂow global, 156local, 156 ﬂow line, 156 form, seedifferential form 1-form on an open set, 34a basis for the space of k-covectors, 31 bilinear or sesquilinear, 358 closed, 40 dimension of the space of k-forms, 32 exact, 40 frame, 137, 240 global, 240local, 240of a trivialization, 138 functor contravariant, 112covariant, 111 functorial properties 402 Index of the pullback map in cohomology, 278 fundamental theorem for line integrals, 271 Gauss, Carl Friedrich, 47 general linear group, 54, 354 bracket on the Lie algebra of, 183is a Lie group, 66 tangent space at the identity, 178 generate a topology, 321 germ, 12 of a function on a manifold, 86 global ﬂow, 156global frame, 240global section of a vector bundle, 137 grad, 41graded algebra, 30 homomorphism, 30 graded commutative, 30 graded ring, 279 gradient, 41graph of a function, 54 of a smooth function, 108 of a smooth function is a manifold, 54of an equivalence relation, 74 Grassmann, Hermann, 18Grassmann algebra of multicovectors, 30 Grassmannian, 82Green’s theorem in the plane, 271 half-space, 248 Hausdorff, 324 compact subset of a Hausdorff space is closed, 331 continuous bijection from a compact space to a Hausdorff space is a homeomorphism, 331 product of two Hausdorff spaces is Hausdorff, 327 singleton subset of a Hausdorff space is closed, 325 subspace of a Hausdorff space is Hausdorff, 326 Hausdorff quotient necessary and sufﬁcient condition, 75necessary condition, 73 Hom, 19homeomorphism, 331 homogeneous coordinates, 76 homogeneous element, 210homological algebra, 281homologous, 274 homology class, 274 homomorphism ofR-modules, 16 of algebras, 12 of graded algebras, 30 of Lie groups, 164 homotopic maps, 296 induce the same map in cohomology, 300, 311 homotopy from one map to another, 296 straight-line homotopy, 297 homotopy axiom for de Rham cohomology, 300 homotopy equivalence, 297homotopy invariance of de Rham cohomology, 311 homotopy inverse, 297 homotopy type, 297hypersurface, 106, 109, 216 nowhere-vanishing form on a smooth hypersurface, 219 orientability, 247 ideals product of, 320sum of, 320 identiﬁcation, 71 of a subspace to a point, 73 identity axiom in a category, 110 identity component of a Lie group is a Lie group, 175 image of a linear map, 350 of a smooth map, 120 immersed submanifold, 122immersion, 96, 115 at a point, 96 immersion theorem, 119implicit function theorem, 340 indiscrete topology, 319 Index 403 induced topology, 168 initial point, 152integrable, 261integral of a form on a manifold, 265 invariant under orientation-preserving diffeomorphisms, 264 of an nform onR n, 263 over a zero-dimensional manifold, 269under a diffeomorphism, 272 under reversal of orientation, 266 integral curve, 152 maximal, 152of a left-invariant vector ﬁeld, 187 starting at a point, 152 interior manifold interior, 252 topological interior, 252 interior multiplication, 227 is an antiderivation of degree −1, 227 bracket with Lie derivative, 234contraction, 227 isF-linear, 228 interior point, 251 ofH n, 248 internal direct sum, 351 invariance of domain, 249 invariance of dimension, 48, 89invariant under translation, 304inverse function theorem, 342 for manifolds, 68 forR n, 68, 339 inversion, 21invertible endomorphism, 354 inward-pointing vector, 254 isomorphism of objects in a category, 111 Jacobi identity, 157 Jacobian determinant, 68, 339 Jacobian matrix, 68, 339 k-covector ﬁeld, 200 k-form on an open set, 36 k-linear function, 22 alternating, 23symmetric, 23 k-tensorsa basis for, 32 kernel of a linear map, 350 Kronecker delta, 13 Lebesgue’s theorem, 262 left action, 24left half-line, 250 left multiplication, 164 differential of, 95 left translation, 164 left-invariant form on a compact connected Lie group is right-invariant, 209 on a Lie group, 207 isC ∞, 207 left-invariant vector ﬁeld, 180 bracket of left-invariant vector ﬁelds is left-invariant, 182 integral curves, 187 isC∞, 181 onR, 181 on a circle, 187 on GL(n,R), 181 onRn, 187 Leibniz rule for a vector ﬁeld, 16 length of a cycle, 20 level, 103level set, 103 regular, 103 Lie, Sophus, 66, 163Lie algebra, 182 abelian, 158 of a symplectic group, 187 of a complex symplectic group, 187of a Lie group, 183of a unitary group, 187 over a ﬁeld, 158 Lie algebra homomorphism, 185Lie bracket, 157 Jacobi identity, 157 ongl(n,R), 183 on the tangent space at the identiy of a Lie group, 183 Lie derivative bracket with interior multiplication, 234 Cartan homotopy formula, 229 404 Index commutes with exterior differentiation, 229 global formula, 232 is a derivation on forms, 229notF-linear, 231, 234 of a differential form, 226 of a vector ﬁeld, 224product formula, 229 Lie group, 66, 164 adjoint representation, 188differential of the inverse map, 175differential of the multiplication map, 175 is orientable, 247 parallelizability, 187 Lie group homomorphism, 164 differential is a Lie algebra homomor- phism, 185 Lie subalgebra, 182 Lie subgroup, 167limit of a 1-parameter of family of vector ﬁelds, 221 of a sequence, 336 unique in a Hausdorff space, 336 Lindel¨ of condition, 338 line integrals fundamental theorem, 271 linear algebra, 349linear functional, 113 linear map, 12, 350 over a skew ﬁeld, 354 linear operator, 12, 350 linear transformation, 350, 354 lines with irrational slope in a torus, 167 Liouville form, 193, 199 local coordinate, 53local diffeomorphism, 68, 339local expression for a 1-form, 191 for a k-form, 202 for a differential, 91 local ﬂow, 156, 223 generated by a vector ﬁeld, 156 local frame, 240 local maximum, 99 local operator, 211, 218 is support-decreasing, 218 onC ∞(M), 219local trivialization, 134 locally connected, 338 at a point, 338 locally constant map on a connected space, 246 locally Euclidean, 48locally ﬁnite, 145 collection of supports, 209 sum, 146, 209, 312union closure of, 148 locallyH n, 250 locally invertible, 68, 339locally trivial, 133 long exact sequence in cohomology, 285 lower integral, 261lower sum, 260 manifold has a countable basis consisting of coordinate open sets, 131 n-manifold, 53 open subset is a manifold, 54 open subset is a regular submanifold, 101 pointed, 111smooth, 53topological, 48 manifold boundary, 252 manifold interior, 252manifold with boundary cotangent space of, 253 differential forms, 253orientation, 254smooth, 251 smoothness of a function on, 251 tangent space, 253topological, 250 map closed, 329 open, 329 matrix of a differential, 88 of an endomorphism, 354 matrix exponential, 169 derivative of, 171 maximal atlas, 52 maximal integral curve, 152, 162maximal rank open condition, 118 Index 405 maximal-rank locus, 345 maximum local, 99 Maxwell’s equations, 220 Mayer–Vietoris sequence, 288 measure zero, 262minor (i,j)-minor of a matrix, 66, 107 k×kminor of a matrix, 344 M¨ obius band, 243 has the homotopy type of a circle, 301not orientable, 240 module, 15 module homomorphism, 16modulus of a point in R n, 77 morphism in a category, 110 multi-index, 201 strictly ascending, 31 multicovector, 23multilinear function, 22 alternating, 23 symmetric, 23 near a point, 123 neighborhood, 4, 48, 318neighborhood basis at a point, 324 neighborhood of a set, 116 nondegenerate 2-covector, 219nondegenerate 2-form, 219norm, 169 of a quaternionic vector, 357 normal, 324normed algebra, 169 complete, 170 normed vector space, 169 complete, 170 object in a category, 110 odd permutation, 20 odd superderivation, 451-form a nowhere-vanishing 1-form on the circle, 216 linearity over functions, 194 smooth, 193smoothness characterization, 194 transition formula, 199 one-parameter family, 221 of differential forms, 221 of vector ﬁelds, 221 one-parameter group of diffeomorphisms, 153 open ball, 317open condition, 118open cover, 48, 329 of a subset in a topological space, 329 open equivalence relation, 74open map, 74, 329 open set, 318 in quotient topology, 71inR n, 318 in the subspace topology, 321 relative to a subspace, 321Zariski open, 320 open subgroup of a connected Lie group is the Lie group, 175 open subset of a manifold is a manifold, 54of a manifold is a regular submanifold, 101 operator, 211 is local iff support-decreasing, 218 linear, 12local, 211 opposite orientation, 238 orbit, 24orbit space, 81 ordered bases equivalent, 238 orientable manifold, 240 orientation boundary orientation, 255determined by a top form, 239on a manifold, 240 on a manifold with boundary, 253, 254 on a vector space, 238opposite, 238 oriented atlases, 245 pointwise, 240 orientation-preserving diffeomorphism, 244 iff Jacobian determinant always positive, 244 orientation-reversing diffeomorphism, 244 406 Index oriented atlas, 245 and orientability of a manifold, 245equivalent oriented atlases, 246 oriented manifold, 240 orthogonal complement, 352orthogonal group, 117, 165, 358 dimension, 167 tangent space at the identity, 179 outward-pointing vector, 254 parallelizable manifold, 187 is orientable, 247 parametrization, 267parametrized set, 267 partial derivative on a manifold, 67 partition, 260 partition of unity, 140, 145, 265, 289 existence in general, 147–348existence on a compact manifold, 146pullback of, 148 subordinate to an open cover, 145 under a pullback, 148 period, 207periodic, 207 permutation, 20 cyclic, 20even, 20 is even iff it has an even number of inversions, 22 odd, 20 product of permutations, 20sign of, 20 permutation action onk-linear functions, 23 Poincar´ e, Henri, 190, 273Poincar´ e conjecture smooth, 57 Poincar´ e form, 193 Poincar´ e lemma, 43, 300point operator, 228 point-derivation ofC ∞p, 13 ofC∞p(M), 87 pointed manifold, 111 pointwise orientation, 240 continuous, 240 preserving a form, 358 productof compact spaces is compact, 331 of permutations, 20of two Hausdorff spaces is Hausdorff, 327 of two second-countable spaces is second countable, 327 product bundle, 134 product ideal, 320product manifold, 55 atlas, 56 product map, 337 product rule for matrix-valued functions, 175 product topology, 326 basis, 326 projection map, 71 is continuous, 328 projective line real, 77 projective plane real, 77 projective space as a quotient of a sphere, 77real, 76 projective variety, 108 pullback by a surjective submersion, 209by a projection support of, 148 commutes with the exterior derivative, 214 in cohomology, 278linearity, 205ofk-covectors, 113 of a 1-form, 196 of a covector, 196of a differential form, 204, 205, 278 of a function, 60 support of, 147 of a multicovector, 204 of a partition of unity, 148 of a wedge product, 206 punctured plane same homotopy type as the circle, 297 punctured torus cohomology of, 307 pushforward of a left-invariant vector ﬁeld, 185 of a vector, 159 Index 407 quaternion, 353 quaternionic inner product, 356quaternionic vector space, 353 quotient construction, 71 quotient space, 72 basis, 76necessary and sufﬁcient condition to be Hausdorff, 75 second countable, 76 quotient topology, 72 open set, 71 quotient vector space, 349 rank of a composition of linear maps, 345 of a linear transformation, 96of a matrix, 82, 96, 344 of a smooth map, 96, 115, 343 rational point, 323real Lie algebra, 158real line with two origins is locally Euclidean, second countable, but not Hausdorff, 57 real projective line, 77real projective plane, 77 cohomology of, 310 real projective space, 76 as a quotient of a sphere, 77 Hausdorff, 79 is compact, 83locally Euclidean, 80second countable, 79 standard atlas, 81 real-analytic, 4rectangle, 260reﬁnement, 261 reﬂexive relation, 11 regular zero set, 103regular level set, 103regular level set theorem for a map between manifolds, 105 regular point of a map of manifolds, 97 regular submanifold, 100, 122 atlas, 102 is itself a manifold, 102 regular value of a map of manifolds, 97related vector ﬁelds, 160 relation, 11 equivalence, 11 relative topology, 321 relatively open set, 321restriction of a 1-form to a submanifold, 197 of a form to a submanifold, 216 of a vector bundle to a submanifold, 134 restriction map in the Mayer–Vietoris sequence, 288 retract, 299 retraction, 299Riemann, Bernhard, 47 Riemann integrable, 261 right action, 24right half-line, 250 right multiplication, 164 right translation, 164 right-invariant form on a Lie group, 208 second countability, 48, 324 a subspace of a second-countable space is second countable, 324 of a quotient space, 76product of two second-countable spaces is second countable, 327 section global, 137 of a vector bundle, 136 smooth, 136 separation, 332separation axioms, 324 sequence, 336 sequence lemma, 336sesquilinear, 356 sesquilinear form, 358 S f, 24 short exact sequence of cochain complexes, 284 of vector spaces, 281 shufﬂe, 27sign of a permutation, 20singleton set, 298, 319 in a Hausdorff space is closed, 325 singular chain, 235skew ﬁeld, 353 smooth, 4 408 Index smooth homotopy, 296 smooth hypersurface, 216smooth category, 111 smooth dependence of vector ﬁelds on a parameter, 221 smooth differential form, 203 smooth family of vector ﬁelds, 222 smooth function on a manifold, 59 on a manifold with boundary, 251on an arbitrary subset of R n, 248 onRn, 4 smooth invariance of domain, 249smooth manifold, 53smooth map between manifolds, 61 from a compact manifold to R nhas a critical point, 127 into a submanifold, 124rank of, 96 smooth 1-form, 193 smooth Poincar´ e conjecture, 57smooth section, 136 characterization of, 138 smooth vector ﬁeld, 136 on an open set in R n, 14 smooth vector-valued function, 4 smoothness of a vector ﬁeld as a smooth section of the tangent bundle, 149 in terms of coefﬁcients, 150 in terms of smooth functions, 151 solution set of two equations, 108 solving an equation implicitly, 339 special linear group, 107, 165, 360 is a Lie group, 165 is a manifold, 107 overC, 109 tangent space at the identity, 178 special orthogonal group, 176 special unitary group, 177, 360sphere charts on, 58 cohomology of, 310standard sphere, 56 tangent plane, 126standard n-sphere, 56 standard basis for the vector space K n, 354 standard topology of Rn, 318 star-shaped, 5stereographic projection, 8 Stokes’s theorem, 269 specializes to Green’s theorem in the plane, 272 specializes to the fundamental theorem for line integrals, 271 Stokes, George, 235straight-line homotopy, 297strictly ascending multi-index, 31 subalgebra, 182 subcover, 329submanifold embedded, 124 immersed, 122regular, 100, 122 submersion, 96, 115, 209 at a point, 96is an open map, 119 submersion theorem, 119 subordinate to an open cover, 145subscripts convention on, 44 subspace, 321 of a Hausdorff space is Hausdorff, 326of a second-countable space is second countable, 324 subspace topology, 320 sum of ideals, 320 sum of two subspaces, 351superderivation, 45 even, 45 odd, 45 superscripts convention on, 44 support of a differential form, 208of a function, 140 of a product, 208 of a sum, 208of the pullback by a projection, 148 of the pullback of a function, 147 support-decreasing, 218surface, 53 surface of genus 3 Index 409 cohomology of, 310 surface of genus 2 as the quotient of an octagon, 308 cohomology of, 308 symmetric k-linear function, 23 symmetric relation, 11symmetrizing operator, 24 symplectic group, 177, 358 complex, 177Lie algebra of, 187 tangent bundle, 129 manifold structure, 132 topology of, 129 total space is orientable, 247 tangent plane to a sphere, 126tangent space, 253 at a point of a manifold, 87 basis, 89of a manifold with boundary, 253 toR n, 10 to a general linear group, 178to a special linear group, 178to a unitary group, 187 to an open subset, 87 to an orthogonal group, 179 tangent vector at a boundary point, 253, 254 inR n, 11 on a manifold, 87on a manifold with boundary, 253 Taylor’s theorem with remainder, 5with remainder to order two, 9 tensor, 22 degree of, 22on a vector space, 200 tensor product is associative, 25 of multilinear functions, 25 top form, 200 topological interior, 252 topological boundary, 252topological group, 66topological manifold, 48 with boundary, 250 topological space, 318topologist’s sine curve, 102 topology, 318coarser, 318 discrete, 319ﬁner, 318 ﬁnite-complement, 319 generated by a basis, 321indiscrete, 319relative, 321 standard topology of R n, 318 subspace, 320trivial, 319 Zariski, 319 torus cohomology ring, 303cohomology vector space, 302 lines with irrational slope, 167 total space of a vector bundle, 134 trace of a matrix, 171 transition formula for a 2-form, 203 transition function, 50 transition matrix for coordinate vectors, 90 transitive relation, 12 transpose, 355 transposition, 20transversal map to a submanifold, 109 transversality theorem, 109trilinear, 22trivial bundle, 135 trivial topology, 319 trivialization of a vector bundle over an open set, 134 trivializing open set for a vector bundle, 134 trivializing open cover, 1342-form transition formula, 203 Tychonoff theorem, 331type-I form, 313type-II form, 313 unbounded subset of R n, 332 uniqueness of the limit in a Hausdorff space, 336 unitary group, 176, 358 410 Index tangent space at the identity, 187 upper half-space, 248 upper integral, 261upper sum, 260 Urysohn lemma, 147 vector bundle, 133, 134 locally trivial, 133product bundle, 134 trivial bundle, 135 vector ﬁeld F-related vector ﬁelds, 160 as a derivation of the algebra of C ∞ functions, 16, 219 complete, 156 integral curve, 152 left-invariant, 180Leibniz rule, 16 on a manifold, 136 on an open subset of R n, 14 smoothness condition in Rn, 14 smoothness condition on a manifold, 136 smoothness in terms of coefﬁcients, 150 smoothness in terms of functions, 151 vector ﬁelds one-parameter family of, 221 vector space orientation, 238 vector space homomorphism, 350vector-valued function, 41 smoothness of, 4 velocity of a curve in local coordinates, 93 velocity vector, 92vertical, 219volume of a closed rectangle, 260 of a subset of R n, 262 wedge product is anticommutative, 27is associative, 28, 29of differential forms, 36, 205of forms on a vector space, 26relation to cross product, 45under a pullback, 206 Zariski topology, 319 Zariski closed, 319Zariski open, 320 zero set, 103 intersection and union of zero sets, 320of two equations, 106regular, 103 0-covector, 23zig-zag diagram, 286zig-zag lemma, 285 Loring W. Tu was born in Taipei, Taiwan, and grew up in Taiwan, Canada, and the United States. He attended McGill University and Princeton University as an undergraduate, and obtained his Ph.D. from Harvard University under the supervi- sion of Phillip A. Grifﬁths. He has taught at the University of Michigan, Ann Arbor, and at Johns Hopkins University, and iscurrently Professor of Mathematics at Tufts University in Mas- sachusetts. An algebraic geometer by training, he has done research at the interface of algebraic geometry, topology, and differential geometry, including Hodge theory, degeneracy loci, moduli spaces of vector bundles, and equivariant cohomology. He is the coau- thor with Raoul Bott of Differential Forms in Algebraic Topology .