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Textbook by Michael Spivak (Addison-Wesley, 1965, later printing), here an OCR scan found in the Wedge Stuff folder. It covers functions on Euclidean space, differentiation, integration, integration on chains with differential forms, and integration on manifolds, ending with Stokes' Theorem and the classical Green and Divergence theorems. It is a published book by Spivak, not Phil's own writing; any annotations by Phil were not visible in the text.

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Michael Spivak Brandeis University i'_~ olwll-|nI~ niIi Calculus onManifolds AMODERN APPROACH TO CLASSICAL THEOREMS OFADVANCED CALCULUS W— , ||u—- l ' ‘AN Y ADDISON-WESLEY PUBLISHING COMPANY TheAdvanced Book Program Reading, Massachusetts 'Menlo Park, California °New York Don Mills, Ontario 'Wokingham, England 'Amsterdam 'Bonn Sydney 'Singapore 'Tokyo 'Madrid 'SanIuan 0Paris Seoul 'Milan 'Mexico City 'Taipei Calculus onManifolds AModem Approach toClassical Theorems ofAdvanced Calculus Copyright ©1965 byAddison-Wesley Publishing Company Allrights reserved Library ofCongress Card Catalog Number 66-10910 Manufactured intheUnited States ofAmerica Themanuscript wasputintoproduction onApril 21,1965; thisvolume waspublished onOctober 26,1965 ISBN O-8053-9021-9 2425262728-CRW-9998979695 Twenty-fourth printing, January 1995 Editors’ Foreword Mathematics hasbeen expanding inalldirections atafabulous rateduring thepast halfcentury. New ﬁelds have emerged, thediffusion into other disciplines hasproceeded apace, and ourknowledge oftheclassical areas hasgrown ever more pro- found. Atthesame time, oneofthemost striking trends in modern mathematics istheconstantly increasing interrelation- ship between itsvarious branches. Thus thepresent-day students ofmathematics arefaced with animmense mountain ofmaterial. Inaddition tothetraditional areas ofmathe- matics aspresented inthe traditional manner—and these presentations doabound—there arethenew and often en- lightening ways oflooking atthese traditional areas, andalso thevast new areas teeming with potentialities. Much ofthis newmaterial isscattered indigestibly throughout theresearch journals, and frequently coherently organized only inthe minds orunpublished notes oftheworking mathematicians. And students desperately need tolearn more andmore ofthis material. This series ofbrief topical booklets hasbeen conceived asa possible means totackle and hopefully toalleviate some of ‘U vi Editors’ Foreword these pedagogical problems. They arebeing written byactive research mathematicians, who canlook atthelatest develop- ments, who canusethese developments toclarify and con- dense therequired material, who know what ideas tounder- score and what techniques tostress. Wehope that they will alsoserve topresent totheableundergraduate anintroduction tocontemporary research andproblems inmathematics, and that they willbesufficiently informal that thepersonal tastes andattitudes oftheleaders inmodern mathematics willshine through clearly tothereaders. The area ofdifferential geometry isone inwhich recent developments have effected great changes. That part of differential geometry centered about Stokes’ Theorem, some- times called thefundamental theorem ofmultivariate calculus, istraditionally taught inadvanced calculus courses (second or third year) andisessential inengineering andphysics aswell asinseveral current andimportant branches ofmathematics. However, the teaching ofthis material has been relatively little affected bythese modern developments; sothemathe- maticians must relearn thematerial ingraduate school, and other scientists arefrequently altogether deprived ofit.Dr. Spivak’s book should beahelp tothose who wish tosee Stoke’s Theorem asthemodern Working mathematician sees it.Astudent with agood course incalculus andlinear algebra behind himshould ﬁndthisbook quite accessible. Robert Gunning Hugo Rossi Princeton, New Jersey Waltham, Massachusetts August 1965 Preface This little book isespecially concerned with those portions of “advanced calculus” inwhich thesubtlety oftheconcepts and methods makes rigor diﬂicult toattain atanelementary level. Theapproach taken here uses elementary versions ofmodern methods found insophisticated .mathematics. The formal prerequisites include only aterm oflinear algebra, anodding acquaintance with thenotation ofsettheory, andarespectable ﬁrst-year calculus course (one which atleast mentions the least upper bound (sup) andgreatest lower bound (inf) ofa setofreal numbers). Beyond this acertain (perhaps latent) rapport with abstract mathematics will befound almost essential. The ﬁrst half ofthebook covers that simple part ofad- vanced calculus which generalizes elementary calculus to higher dimensions. Chapter 1contains preliminaries, and Chapters 2and3treat differentiation andintegration. The remainder ofthebook isdevoted tothestudy ofcurves, surfaces, andhigher-dimensional analogues. Here themodern andclassical treatments pursue quite different routes; there are, ofcourse, many points ofcontact, andasigniﬁcant encounter ‘I711 mu Preface occurs inthelastsection. Thevery classical equation repro- duced onthecover appears also asthelasttheorem ofthe book. This theorem (Stokes’ Theorem) has had acurious history andhasundergone astriking metamorphosis. Theﬁrst statement oftheTheorem appears asapostscript toaletter, dated July 2,1850, from SirWilliam Thomson (Lord Kelvin) toStokes. Itappeared publicly asquestion 8 ontheSmith’s Prize Examination for1854. This competitive examination, which was taken annually bythebest mathe- matics students atCambridge University, wassetfrom 1849 to 1882 byProfessor Stokes; bythetime ofhisdeath theresult was known universally asStokes’ Theorem. Atleast three proofs were given byhiscontemporaries: Thomson published one, another appeared inThomson and Tait’s Treatise on Natural Philosophy, and Maxwell provided another inElec- tricity and Magnetism [13]. Since this time thename of Stokes hasbeen applied tomuch more general results, which have ﬁgured soprominently inthe development ofcertain parts ofmathematics that Stokes’ Theorem may becon- sidered acase study inthevalue ofgeneralization. Inthis book there arethree forms ofStokes’ Theorem. Theversion known toStokes appears inthelastsection, along with itsinseparable companions, Green’s Theorem and the Divergence Theorem. These three theorems, theclassical theorems ofthe subtitle, are derived quite easily from a modern Stokes’ Theorem which appears earlier inChapter 5. What theclassical theorems state forcurves andsurfaces, this theorem states forthehigher-dimensional analogues (mani- folds) which arestudied thoroughly intheﬁrstpart ofChapter 5.This study ofmanifolds, which could bejustiﬁed solely on thebasis oftheir importance inmodern mathematics, actually involves nomore effort than acareful study ofcurves andsur- faces alone would require. The reader probably suspects that the modern Stokes’ Theorem isatleast asdiﬂicult astheclassical theorems derived from it.Onthecontrary, itisavery simple con- sequence ofyetanother version ofStokes’ Theorem; thisvery abstract version istheﬁnal and main result ofChapter 4. Preface ix Itisentirely reasonable tosuppose that thedifficulties sofar avoided must behidden here. Yet theproof ofthis theorem is,inthemathematician’s sense, anutter triviality--—a straight- forward computation. Ontheother hand, even thestatement ofthis triviality cannot beunderstood without ahorde of difficult deﬁnitions from Chapter 4.There aregood reasons why thetheorems should allbeeasy andthedeﬁnitions hard. Astheevolution ofStokes’ Theorem revealed, asingle simple principle canmasquerade asseveral difficult results; theproofs ofmany theorems involve merely stripping away thedisguise. The deﬁnitions, ontheother hand, serve atwofold purpose: they are rigorous replacements for vague notions, and machinery forelegant proofs. The ﬁrst two sections of Chapter 4deﬁne precisely, and prove therules formanipulat- ing,what areclassically described as“expressions oftheform” Pdx -1-Qdy +Rdz,orPd:cdy +Qdydz +Rdzdaz. Chains, deﬁned inthethird section, and partitions ofunity (already introduced inChapter 3)freeourproofs from thenecessity of chopping manifolds upintosmall pieces; they reduce questions about manifolds, where everything seems hard, toquestions about Euclidean space, where everything iseasy. Concentrating thedepth ofasubject inthedeﬁnitions is undeniably economical, but itisbound toproduce some difficulties forthestudent. Ihope thereader will beencour- aged tolearn Chapter 4thoroughly bytheassurance that the results willjustify theeffort: theclassical theorems ofthelast section represent only afew, and bynomeans themost im- portant, applications ofChapter 4;many others appear as problems, andfurther developments willbefound byexploring thebibliography. The problems and thebibliography both deserve afew words. Problems appear after every section and arenum- bered (like the theorems) within chapters. Ihave starred those problems whose results areused inthetext, but this precaution should beunnecessary-——the problems arethemost important part ofthebook, and thereader should atleast attempt them all. Itwas necessary tomake thebibliography either very incomplete orunwieldy, since half the major x Preface branches ofmathematics could legitimately berecommended asreasonable continuations ofthematerial inthebook. I have tried tomake itincomplete buttempting. Many criticisms and suggestions were offered during the Writing ofthisbook. Iamparticularly grateful toRichard Palais, Hugo Rossi, Robert Seeley, and Charles Stenard for their many helpful comments. Ihave used thisprinting asanopportunity tocorrect many misprints and minor errors pointed out tomebyindulgent readers. Inaddition, thematerial following Theorem 3-11 hasbeen completely revised andcorrected. Other important changes, which could notbeincorporated inthetext without excessive alteration, arelisted intheAddenda attheendofthe book. Michael Spivak Waltham, Massachusetts March 1968 Editors’ Foreword, Preface, viiContents Functions onEuclidean Space NORM ANDINNER PRODUCT, 1 SUBSETS OFEUCLIDEAN sPAcE,5 FUNCTIONS ANDCONTINUITY, 11 Differentiation BASIC DEFINITIONS, 15 BASIC THEOREMS, 19 PARTIAL DERIVATIVES, 25 DERIVATIVES, 30 mvnnsn FUNCTIONS, 34 IMPLICIT FUNCTIONS, 40 NOTATION, 44 xi xii Contents 3.Integration 46 BASIC DEFINITIONS, 46 MEASURE ZERO ANDCONTENT ZERO, 50 INTEGRABLE FUNCTIONS, 52 FUB1N1’s THEOREM, 56 PARTITIONS orUNITY, 63 GHANGE orVARIABLE, 67 4.Integration onChains 75 ALGEBRAIC PREL1M1NAR1Es, 75 FIELDS ANDFORMS, 86 oEoMETn1o PREL1M1NAn1Es, 97 THEFUNDAMENTAL THEOREM orcALoULUs, 100 5.Integration onManifolds 109 MANIFOLDS, 109 FIELDS ANDFORMS oNMANIFOLDS, 115 sToKEs’ THEOREM ONMANrFoLDs, 122 THEVOLUME ELEMENT, 126 THEoLAssIcAL TuEonEMs, 134 Bibliography, 13.9 Indezv, 141 Calculus onManifolds 1 Functions onEuclidean Space NORM AND INNER PRODUCT Euclidean n-space R"isdeﬁned asthesetofalln-tuples (rel, ...,:v") ofreal numbers rt’:(a“1-tuple ofnumbers” is just anumber and R1=R,thesetofallreal numbers). An element ofR"isoften called apoint inR",and R1,R2,R3are often called theline, theplane, andspace, respectively. Ifat denotes anelement ofR",then atisann-tuple ofnumbers, the ithoneofwhich isdenoted :0";thus wecanwrite :11=(:01, ...,:z:"). Apoint inR"isfrequently also called avector inR", because R”, with .2:-1-y=(zcl+y1, ...an+y") and ax=(axl, ...,aa:"), asoperations, isavector space (over thereal numbers, ofdimension n). Inthisvector space t-here isthenotion ofthelength ofavector x,usually called the norm ofatanddeﬁned by =\/(:z:1)2 +'''-1-(:v")2. Ifn=1,then istheusual absolute value ofanThe rela- tion between thenorm and thevector space structure ofR”is very important. 1 2 Calculus onManifolds 1-1 Theorem. If:c,yER”andaER,then (1) ZO,and =0andonly ifx=0. (2)|E,T‘=1:c"'3y':] § ~Iy;equality holds ifandonly if:1:andy arelinearly dependent. (3)|=v+1/ISlrl+lul- (4)[ax] =|a|' Proof (1)islefttothereader. (2)Ifxandyarelinearly dependent, equality clearly holds. Ifnot, then Ay—as;=40forallAER,so 0<|)\y-xjf=2()\y':—:z:‘)2 ill =A’):(W—Zr2ru‘+Z(W.i=1 i=1 i=1 Therefore theright sideisaquadratic equation inAwith no realsolution, anditsdiscriminant must benegative. Thus Tl, Tl. ‘R. 4(£21 x¢:yt)2 _4£1(x-t)2 .‘Z1 (y.:)2 .<()_ <3)Ix+1/!’=Er=.<=»"+W=El'=1(93i)2 +El=1(?!i)2 'l'22ii=193i?Ji sIr!”+l:»/I”+Zlrl-lrlby<2)=(El+Iii)”. <4>lax!=~/Er=.<ar=>” =~/a”Er=.e*>” =la!~l===l-I The quantity E,T"=1:z:‘y‘ which appears in(2)iscalled the inner product ofasand yand denoted (x,y). The most important properties oftheinner product arethefollowing. 1-2 Theorem. Ifas,2:1,2:2andy,y1,ygarevectors inR" andaER,then (1)(ray)=(r/.=v> (symmetry)- Functions onEuclidean Space 3 (2)(a:c,y) =(:c,ay) =a(:c,y) (bilinearity). (x1 +332; =(why) +(58279) (Iv.:1/1+1/2)=(=v.2/1) +(fave) (3)(:z:,:c) ZO,and(:c,:v) =Oifand (positive deﬁniteness). only if:1:_=l <4)Ix!=~/<=».x>.2_ __2 (5)(a:,y) --ix+yl 4ix l (polarization identity). Proof (xii/> =Ei:-=1x£l/i =Eil=1l/ix‘ = (2)By(1)itsufﬁces toprove (am/> =a(=v.v). ($1 +$22 y)=($12?/> +<x2>l/>-' These follow from theequations I1. Fl- <ax.y>=Z1(war=a =am/>.t= 'Tl. ‘I1. 7! (1131+132.2!) =2(-$1‘+112*)?‘ =ZMilli +2I¢2':y£ i=1 i=1 {=1 =(3311?!)+(way)- (3)and(4)arelefttothereader. lx+@/l2" I-'1’"'1'/|2<5)A 4 =<H(1=+v.r+v>—(=v-v.r—-v>l by(4) =f[(1v.=v) +2(r.v)+(av)—((1%)—2(rv.v)+(2/.v>)l =(1=.v)- I Weconclude this section with some important remarks about notation. The vector (0,...,0)will usually be denoted simply 0.The usual basis ofR"ise1,...,e,,, where e,-=(0,...,1,...,0),with the1intheithplace. IfT:R"-—>R“isalinear transformation, thematrix ofTwith respect totheusual bases ofR"andRmisthernXnmatrix A=(a,-,-), where T(e,-) =Z_?_1a,-.,~e,- —the coeﬁicients ofT(e,-) 4 Calculus onManifolds appear intheithcolumn ofthematrix. IfS:Rm—->RPhas thepXrnmatrix B,then S0Thas thopXnmatrix BA [here SQT(;r) =S(T(:z:)); most books onlinear algebra denote SoTsimply ST]. Toﬁnd T(a:) onecomputes themX1 matrix 1 1y (Z11, ...,(Z1n SE I = I I Q , l/m aml; ---ya/mn xn then T(a:) ==(yl, ...,y"‘). One notational convention greatly simpliﬁes many formulas: if:1:ER"andyERm,then (x,y) denotes (xi, ...,x"',y1,. ..,y”‘) ER"+"‘. Problems. l-].* Prove that lztl_§2;-",_1 lztil. 1-2. When does equality hold inTheorem 1-1(3)? Hint: Re-examine theproof; theanswer isnot“when xandyarelinearly depend- ent.” 1-3.Prove thatIx-ylg[$1+|y].When doesequality hold? 1-4-. Prove that ||x|—|y||§|:z;-— 1-5. The quantity ly—:c|iscalled the distance between :1:and y. Prove and interpret geometrically the “triangle inequality”: lz"-rl Slz—v|+|?/"='>l-1-6. Letfandgbeintegrable on[a,b]. (a)Prove that |f2f- gl_3(_ff,_f2)5 -(fggzﬁ. Hint: Consider separately thecases 0=_jf§,(f —Ag)2 forsome AERand 0< ﬁ’,(f —Ag)?forallAER. (b)Ifequality holds, must f=Agforsome AER? What if fandgarecontinuous? (c)Show that Theorem 1-1(2) isaspecial case of(a). 1-7. Alinear transformation T:R"--> R"isnorm preserving if |T(x)| =|:r:|,and inner product preserving if(T:r:,Ty) =(:r:,y). (a)Prove that Tisnorm preserving ifand only ifTisinner- product preserving. (b)Prove that such alinear transformation Tis1-1andT“1is ofthesame sort. 1-8. Ifx,yER"arenon-zero, theangle between xandy,denoted A(:z:,y), isdeﬁned asarccos ((x,y)/Ix] -lyl),which makes sense by Theorem 1-1(2). The linear transformation Tisangle preserv- ing ifTis1-1,and for:r:,y#50wehave L(Tx,Ty) =£(x,y). Functions onEuclidean Space 5 (a)Prove that ifTisnorm preserving, then Tisangle pre- serving. (b)Ifthere isabasis $1,...,a:..ofR"andnumbers A1,...,A.. such that Tx.-=A.;x.;, prove that Tisangle preserving ifand only ifallIA.-Iareequal. (c)What areallangle preserving T:R"—>R“? cos0,sin01-9. If0s0<1r,letT:R2———> R2have thematrix ( .-s1n 0,cos0 Show that Tisangle preserving andifac#50,then £(a:,Ta;) =6. 1-l0.* IfT:R"‘--> R"isalinear transformation, show that there isa number Msuchthat|T(h)|5M|t\for1.ERm.Hint:Estimate |T(h)| interms of|h|andtheentries inthematrix ofT. l-ll. Ifx,yER"andz,wER",show that ((a:,z),(y,w)) =(a:,y) +(z,w) and |(a',z)l =\/|as|2 +Izli. Note that (x,z) and (y,w) denote points inR"'+"". 1-l2.* Let (R"')"‘ denote thedual space ofthevector space R". If a:ER", deﬁne eaE(R"')"' by<p.,(y) =(x,y). Deﬁne T:R"-—> (R")* byT(x) =tpa. Show that Tisa1-1linear transformation andconclude that every rpE(R")* is<p,,foraunique xER". 1-13."' Ifx,yER",then zrandyarecalled perpendicular (ororthog- onal) if(a;,y) =0.Ifasand yareperpendicular, prove that ls+1/l”=l-"=12+|vl”- SUBSETS OF EUCLIDEAN SPACE Theclosed interval [a,b]hasanatural analogue inR2. This is theclosed rectangle [a,b] X[c,d], deﬁned asthecollection of allpairs (as,y) with a:E[a,b] andyE[c,d]. More generally, ifAERmand BER", then AXBCR"‘+” isdeﬁned as thesetofall(as,y) ER"‘+"' with asEAandyEB.Inpar- ticular, R""+” =RmXR". IfAERm, BER",and CE R1’, then (AXB)XC=AX(BXC),and both ofthese aredenoted simply AXBXC;thisconvention isextended to theproduct ofanynumber ofsets. Theset[a1,b1] X'"'X [a,,,b,,] ER"iscalled aclosed rectangle inR",while theset (01.51) X'''X(a,,,b,.,) ER”iscalled anopen rectangle. More generally asetUER"iscalled open (Figure 1-1) ifforeach asEUthere isanopen rectangle Asuch that a:EACU. Asubset CofR"isclosed ifR"-—Cisopen. Forexam- ple,ifCcontains only ﬁnitely many points, then Cisclosed. Thereader should supply theproof that aclosed rectangle in> Aw’ “< =13-:3-if-Iflfl€3|:fl-{ii131'-if-i::j:':::'> 5"§"3'5§"?I’?if?§3§I5=§3§%§‘§5§§'Ete-<. tn“''A'5:3:3:3:3:3:3:35:3:-:~1-:-:-:-f-1-3':-;-:-Ii-I.':-.1;1;-':':I'.':'r. -I-‘A’ <2;;;;3;;.5.;=:;;-.;.i:2‘-:‘§-&51.€§‘§E5513;555:555 --':=:::1:::21:-1:2:2:2.-e2::=;:r.=;.-;.'1 .<*-..--:;-1;-.1-: . .;;.;;;i§§_Z£5£55-‘:5§i;:§;§§i§§é;5'E{§§_i-153553Fiii=5IiljEtta}?5é;53%;55:5g;;E5z§:;=';';;z_;;;;;;.=;;;;;5:;.§5":=-.'»-':==-'-:1:2-2:;'z:i:.=&:z-E=;%;:'- fie’?}i:=_-'.;I:3;.'2’rzE3551555;:Eg1:i;%_=.=;;:;i-;:,:~Ell:E':32:2-3Z'21'-.'5I:I:.'lI3Z:IiI:':‘: ~' '1: 'L:III-I-I-1-Ii-I-1-'.'.I'-'-I-I-3'1-If-3* E5PiEii;51?;Eiiii?§-:25?El-Ie%§i%':E§;5§§§ii?‘. zi51%Eiiiif-iiEii?ti=5?-=i§i%;%¥.%Es’;eisi&’;s§2;.J35‘?§5§§;f-§'§§§§§§§5_§"§§§E§§f=§§§§;'§5§f55§§§;If5§* :°--tt~’?-.»_.-,'<<¢<a-:1-.§:z< FIGURE 1-1 R"isindeed aclosed set. IfAER"and asER",then oneofthree possibilities must hold (Figure 1-2)Calculus onManifolds -2:5si;i;1'=E;:;i;E33;ia%%J£§i?é%§iI&IzI-2151:’:-I. 1.There isanopen rectangle Bsuch that atEBEA. 2.There isanopen rectangle Bsuch that asEBER"-—~A 3.IfBisany open rectangle with asEB,then Bcontains points ofboth Aand R"-—~A. 20 ...-. 11555IE1:;:I:I'.1:1:1:1;‘-:1:f:I:-;-:-?-.T5;::l;' ag--:.-511.-j.-5;;-3:_=;5§:.=:.-::=.--.-:;;::.-5.-5;; FIGURE 1-2;1;I;'<:::;§;§:‘: :-:-:-:-.-.-.\;3I33:'-:'i3:3 -;1.-5.-_=2;-;.-;;a=";-_:.=;_:.=;.-_:. 51?:§1E3E3{iE=Ei'-5??5?§‘§§‘=Eiiiilifg:F:!:I:I521:1:5:1:11-:-1-'.>:-=:!;1;f:1:=:- ................. ... Functions onEuclidean Space 7 Those points satisfying (1)constitute theinterior ofA,those satisfying (2)theexterior ofA,and those satisfying (3)the boundary ofA. Problems 1-16 to1-18 show that these terms may sometimes have unexpected meanings. Itisnothard toseethat theinterior ofany setAisopen, andthesame istruefortheexterior ofA,which is,infact, the interior ofR"—A. Thus (Problem 1-14) their union isopen, and what remains, theboundary, must beclosed. Acollection (‘Jofopen setsisanopen cover ofA(or,brieﬂy, covers A)ifevery point asEAisinsome open setinthe collection O.For example, if0isthecollection ofallopen intervals (a,a+1)foraER,then 0isacover ofR. Clearly noﬁnite number oftheopen setsin(‘Jwillcover Ror,forthat matter, any unbounded subset ofR. Asimilar situation can also occur forbounded sets. IfL‘)isthecollection ofallopen intefvals (1/n, 1—1/n) forallintegers n>1,then (9isan open cover of(0,1), butagain noﬁnite collection ofsetsin Owillcover (0,1). Although thisphenomenon may notappear particularly scandalous, sets forwhich this state ofaffairs cannot occur areofsuch importance that they have received a special designation: asetAiscalled compact ifevery open cover (9contains aﬁnite subcollection ofopen sets which also covers A. Asetwith only ﬁnitely many points isobviously compact and soistheinﬁnite setAwhich contains 0and thenumbers 1/nforallintegers n(reason: ifOisacover, then 0EUfor some open setUin0;there areonly ﬁnitely many other points ofAnotinU,each requiring atmost onemore open set). Recognizing compact setsisgreatly simpliﬁed bythefollow- ingresults, ofwhich only theﬁrst hasanydepth (i.e., uses any facts about therealnumbers). I-3 Theorem (Heine-Borel). The closed interval [a,b] is compact. Proof. If0isanopen cover of[a,b], let A={a::a§as§band [a,a:j iscovered bysome ﬁnite number ofopen sets inO}. 3 Calculus onManifolds U rt .1‘ or .1" la FIGURE 1-3 Note that aEAand that Aisclearly bounded above (byb). Wewould liketoshow that bEA. This isdone byproving two things about a=least upper bound ofA;namely, (1) aEAand(2)b=a. Since L‘)isacover, aEUforsome UinL9.Then all points insome interval totheleftofaarealsoinU(seeFigure 1-3). Since aistheleast upper bound ofA,there isanasin this interval such that asEA. Thus [a,at] iscovered bysome ﬁnite number ofopen sets ofL‘),while [a:,aj iscovered bythe single setU. Hence [a,aj iscovered byaﬁnite number ofopen sets ofO,and aEA. This proves (1). Toprove that (2)istrue, suppose instead that ct<b. Then there isapoint a:'between aandbsuch that [a,a:’j EU. Since aEA,theinterval [a,aj iscovered byﬁnitely many open sets ofO,while [a,as'] iscovered byU. Hence at’EA, contradicting thefact that orisanupper bound ofA. IfBERmiscompact and atER", itiseasy toseethat lat}XBER”+m iscompact. However, amuch stronger assertion canbemade. 1-4 Theorem. IfBiscompact and L9isanopen cover of {at}XB,then there isanopen setUER"containing atsuch that UXBiscovered byaﬁnite number ofsetsinL9. Proof. Since {as}><Biscompact, wecan assume atthe outset that L9isﬁnite, and weneed only ﬁnd theopen setU such that UXBiscovered byL‘). Foreach yEBthepoint (a:,y) isinsome open setWinL‘). Since Wisopen, wehave (a:,y) EU/yXVyEWforsome open rectangle U/yXV1,. The sets Vycover thecompact set B,soaﬁnite number V.__,,,, ...,l/2,, also cover B. Let U=Uy,(W---(WUyk. Then if(:c’,y’) EUXB,wehave nJI,6EmmﬁmgmgwgmwmummmwﬁwmwmﬁmmmmﬁEﬁﬁgmmﬁﬁH_>AHMm na6__dIE CIvatEn08n0mCnuF \J\)’V\“‘_|‘___vn“l__ém___L_ _IJ'___1+"\k|hIH\"_"_"I____‘I_II‘_______H_H_H__VJ_\__HN"h_ "H_H_VH_H_H___'"_“_U'H“II.-______m_"_H)H_”_“_"_x_IK % _’_JKI \_\\/ I_ 0__\/0'_,|I_:____IVJ‘AFI"(_h\_>__II__L,",__n_H_N_W_H__“H_H_H__ ___III\I,\ ‘\\'Q_JI L__W_\IJ/\___\L____U____H+HlH_A“_n_H________ /_’\J\\\\\__)_\k__I\\__,_I__/v__,___>_,>I/__ 2/\I‘_f_X)\‘f\1r |r___\\\\\\_mmmmWmmmmHmH"_\_MyW“UH"H_H”HmH_HUME“PM“mmmHHHHHw§__\ﬁV_____VI_HHHHH£Hn"""”HhHH"____K_ “___v__£_"‘_Jy___k___"Y__“__I___H__69____r____a___H__“_h__"lb:____h__J“_mmmWWmwm\Q __>\____?ﬂg"lJ_M?“WtV___$u___|__>_____ﬁ_Hggm?I\_,__~_d__J“N____ﬂv___bv_“W_ﬁrﬁwwgpq ____~_¢_hm“uh“H“uniEHhnnumwnnnnnumhmmmmwm___|H_H_“_;\H_%M___?%W_%w)M_$_p_“‘&‘*__WMW‘%¢H_¢__"Hﬁ“H“F“HH_H"HHH_"dH_,_HHUUw/“H1JHUN“?ﬁgNM_m_“__%’_”___~M_m£%_____|_""H“H“HhHNHHuHn‘_ m“H"H_ﬂ“_r_\\“_"“h_“_“_,_{____I“___w__b."_ﬂ_3_w_|$_'H_M_ux_%v%1‘/M \_U»ywdkmwgymmm,_mM_%KW%Uﬁ__¥m__"_H_H_ J__Iipf‘__M’amJ“W” \‘ M__N_M_H_§$nw__wImW_JHg“MW_\\I\__\WH"Hﬁ“_____¢°__$1_'____"_Wk76___“_w"_H_H_______\+__H_“_“_2_H_“_H____H_“_|_|_H_H"_“‘P_v_’_I\|v>_||Hvh>H__|H_____€f_W$_hm,_§_Z__“__?M“'id““Hun?H_“"HH“_H|Hv"_H_"|“_">"_HV_g"___N_“_““N_"____I___“_’_(‘‘_3_\\___€___JVﬁg3_?__*_____6_______ri_‘_.‘_.H.“._HF_¢_Hr__I>\_4x ~mﬁ_ﬁgwmwgm%'§wwWWw§ﬁZNNW“\\IA““"A|““H"H\_W;___‘‘_:_hHHh“_“_H_v$"w\ tI_“_H_HI__~|H_vH_H_H_P.H_"_H_“___n_"_\H“__H_h___v_”__W?Irv__M__@u%__'“___I*__M‘>__ﬂUH__n_"__{|”l"_H|H_J_,1l’I1Mk_WMMEMMWMW/ww"m“"mH“_“MHm“§“m$mv_MNH“MWMHWNHMEHMMMWMMMUm"NW§_g§__uHMHwwuwmnwmumumﬁ_ﬁ”ﬁw_HJ“UH“H_NHHWmM_HmWmHWNHMmHwwmmmmﬁmwmhmg_W%‘HHWMHm"ﬁ@WﬁV*3ﬂ_$__¢_fH__#ﬂ____’v__,$m__anWm“m"Wm%m“"""_mmEWW22IUH.HH.HHHP.nhh_H_HHNh_hHHH_H_H__'alum."UU"H3U_H9?"_“Hnun.“_H”.H_U“HunUnmnnnnhIHHHHHHH_"_HIH_H.Hy“.Haw?"“A."_P_H_U_IVW___in_UV:H_n_n_H,"_"ﬁlo“IU_H.n_H.“.H_“__.H_H_h_"IH_nNIH_“HUN”__HH_____‘ﬁnWP”U“Hun"""\“_"""_H_""P_V%_”__%ﬂw%___w‘_mW“J;~vfdw_“_%%vgmq/’_v”___én_____"§_v\“_m¢___|_y9_u_WANHmwwmmﬁmvmuwmuv/TumHWm“H_WMAHHg?hm"“_HH“H_F_‘H_"_“H_H_"”H__ﬁnumuwxmuwumuhHmuwuudwmwmhwwmwwmnW"NW“mummummgWWE&m_mHmHmHw@mg“WHmHWwMHWWMHMHWm“MMm“WGmHm"M“WmHmm""Hwww"Q_wwW’W¢v“%__W'¢%__W~H"@w%%“_¢%%wJ%&W~.hV_HWHmmwm“m"Lg\mmmMm_HWmWmWmH\4__|_HWH_NWw_M_'lgym’I'I“_£_rA.|éV’‘gmI_AMI__‘_uw_\§Z:___'_'_|M___“w__n__m__“"__w__A__”_Mm“”_“_____AMm___§_I%____¢w_u__H_M’V_ﬂ_vF~¢____m_‘_¥¢Him“\_AIEUJ/Rﬁmmmummmhwmmﬁh_JZ__ﬂ“__H“________mvJ“P’\'_‘)_,‘____“‘'______%__n1,,g___,_V‘v‘__’_wIkg"“HP_'_)_6‘_3%__m"_v_V_'inUsF/V,__fJ‘_;_”\_‘I:/_,,1)“_w_JwWMmWm/(VA{WadhrP_f_HMm___gr:I/%m\_w_&‘WJ_w_W_V¢HWM_A§m\M_m_Ww%_1_ _‘____‘__‘:1,4;,mymigé5”mg”E§_$M_gwmﬁ5’xmg_§_W&_w@_%__€§€_n_$%%WWdJg¢___Mﬂ1d__WA\_&_l$Qaw’Q*__?_w0_A__fI>\/ﬁg’m_#%mgMH_$_(ﬁ_€%EMWZMmgﬁgH__II\\,¢%WWA__u%¢WMé__fMwwgﬁﬂﬁwwmgmnwgtwwI“ ,\_~__‘__|\__‘___\~‘_u6‘z;_\__V_JP,__’_y_k_§________|_____“§_W_,’JAh__vA_\l_____P_‘_’__j$_v_mMW___)_'M‘h”£;_"\g‘kw33_"__g_$$9_____”“ihtww_'’N_'I'6':J“:“_F‘I/J’MW_~_‘%_$’u_h_(“m_'p_www_‘____v__~*__M_ﬂ"r_v@m_;__J___JM__‘_Nfwwﬂﬁ%“Q@_\%_f_%_,(_MM‘3“___|$6”,'3?:____‘um__h_ %m“_%@uW%___£>%“ﬁ)_\_4;“Ft,3W?‘G”___WW_v_M%_Mﬁ“_ﬁ€v¢\_’_u__)__%_v__Wg__fM&%_'*%__ﬂ_W_A%~M$&%vmMg%§~y__{mWm_M_H_wW,l)_“ﬁgQw_,_’_gw1_I\>‘_ fI1Hmg__,__‘N3av__£%I_WM~_“MEI)F_‘A&_W“MMﬂ_%%uﬁM%§~f,2g_:“~m_§_v_g"v__~?€__d__‘_,_A(_¢____QjgQ JHg”_,~W‘$gwv_%W“~Wv__W_WW)ggﬂ_M(_MW_WvAH‘MﬁgI?Mad___;_M WE_,_wgiWgrq%¢__A_;v___P_#3_?,,_'___§__Hg____$m:_§_§_g__MW’T)_Hw_%_w__%__;_'_1*_’%_w_‘_____‘_,$__WM_)’__$___'v“_€w_':f__w__"*Mmm_Hw%M__ﬁ___h___L_‘M’__?‘ﬁY‘$WH%%3Mw_in°°m§_"ﬁH_H_W___/_‘"_“_mPm‘ﬂm_"mm_“_Hmﬁ____,_TL ____I'_*___“aw/JwagNM*3ﬂy“1:kwL_3.NEWgg_"‘%__my_WM¥g_,%/gm_Qbha_:_____u_“_?_‘_F_a__\‘_§¢__0w__,_,_GxiI _\Hmh“mV_H%_“u__W_“lu_%Hh“£“\H_H_H_/,H_h_H_N”_h_H_U/hm_\__ \N/_/\3\_\\\JI x\VHHH______"‘WHH"_v“_H_"_H_”__“__H__Wm___w_r~H__W__r_wW_H_m___mw"_muHPMh__mMﬂMHum“_H_$__HmMH_r__HW______”mW”"_"H_____"H"_h___ /_‘\"ﬁnNW“W|_H_HW“iH“_w_v“_|“vH_“_“_H_“\Q‘2‘__________‘|_|‘/_TH_h_w_H__|U\/IJI\I_,_gw_“H_mgmhw“H")___,\,\JP___“_’__H|U___w"_H_n|HI//\H_Hh_J/““H_h_H__J_H_“\,J ‘‘__\\Ig\g\__\ \JK_J_>\__‘n‘_\_\ J_,__\\____/_ K\%m____'h_‘__w}_vgkw_an_M$W’“m””mW&MW_M____UMMﬁmuwﬁumumwmhmumvnmmﬂEﬁuwwmumgwwm_M_ ‘NA4t‘Www__ﬁ__\E_mM“mumgwwﬁgW“_“,_\JV\_'\|__1;___H_h_“|HN"Hvm_H_H\|_§_.___%_’‘__“;n_2_u___A___j#_§;_m__§§ﬁg___?D",\\"mHH\_gLg__(@m_i__*%WﬁM_____3_£WWwMW§_“_£{ﬁA__@°“___3%__IwM__|A__,W%N_’_,___N£WI_ 11 .1 ) .1I( .1 _l_______ .1 ‘L) ( H __l BMX‘ItIfll t TN{B p rP HOCxy I .1 yB 6.11 ap 1 I.1 .1 .1yHa1 ‘I .1 l1 1 .1 .1" tll LCaltl l tMm_mcm_@JewU8S_M%i€mCUjrA OUn8ISaa_VhxpSmdmOM nmPS_C6y_1OIJ V__‘MMMwmmwﬁZ tnM®OnUnMmanﬁpp tmnﬂTS.ICnGtmm/FORhS‘_D__D%BOO _dCtmXCC4MSCgmwuogwU________,BAkhlhG_CnyX_n‘HExmmOWf1_$ﬁ€&M TnOm‘Bf_GU,R_POBRQ_gICWT,pmvnFQV%%XLm€% ph_XAmnTUxmyXmBUIOytnsaAwmU_8OabmOO€Mm,w_m_sAy_CSMW.mAhATmmaF,Ca,Oy,R®€u€€nMmffxmCHxSwn"MﬁOuMOTWSCB,m_wVC_hx BBfCgCTC_O__an_UXa€%®FOXOq1bn®€x/Op,yHnLAcmmnmUkm 10 Calculus onManifolds I-7 Corollary. Aclosed bounded subset ofR"iscompact. (The converse isalsotrue (Problem 1-20).) Proof. IfAER"isclosed and bounded, then AEBfor some closed rectangle B. If0isanopen cover ofA,then 0 together with R”--Aisanopen cover ofB.Hence aﬁnite number U1,...,U.,,ofsetsin0,together with R"—Aper- haps, cover B.Then U1,...,U,,cover A.I Problems. 1-14. *Prove that theunion ofany(even inﬁnite) number ofopen setsisopen. Prove that theintersection oftwo(and hence ofﬁnitely many) open sets isopen. Give acounterexample for inﬁnitely many open sets. l-15. Prove that {a2ER":la:—a|<2-}isopen (seealsoProblem 1-27). 1-16. Find theinterior, exterior, andboundary ofthesets {asER“: $1} Ia:ER": =1} Ia:ER”:each as’:isrational}. 1-17. Construct asetAE[0,1] X[0,1] such that Acontains atmost onepoint oneach horizontal andeach vertical linebutboundary A=[0,1] X[0,1]. Hint: Itsuffices toensure that Acontains points ineach quarter ofthesquare [0,1] X[0,1] andalsoineach sixteenth, etc. l-I8. IfAC[0,1] istheunion ofopen intervals (a,-,b,-) such that each rational number in(0,1) iscontained insome (a,;,b,), show that boundary A=[0,1] —A. l-l9.* IfAisaclosed setthat contains every rational number rE[0,1], show that [0,1] EA. 1-20. Prove theconverse ofCorollary 1-7:Acompact subset ofR“is closed andbounded (seealsoProblem 1-28). 1-2l.* (a)IfAisclosed and 2:EA,prove that there isanumber d>Osuchthat ly-113]3dforally EA. (b)IfAisclosed, Biscompact, and AF\B=Z,prove that there iscl>0such that Iy—$13 clforallyEAand ccEB. Hint: Foreach bEBﬁndanopen setUcontaining bsuch that thisrelation holds for:1:EUf\B. (c)Give acounterexample inR2ifAand Bareclosed but neither iscompact. 1-22.* IfUisopen andCEUiscompact, show that there isacompact setDsuch that CEinterior Dand DEU. Functions onEuclidean Space 11 FUNCTIONS AND CONTINUITY Afunction from R”toRm(sometimes called a(vector- valued) function ofnvariables) isarulewhich associates to each point inR"some point inRm; thepoint afunction f associates tozz:isdenoted f(:t). Wewrite f:R”—>Rm(read “f takes R”intoRm” or“f,taking R”intoRm,” depending oncon- text) toindicate that f(a:) ERmisdeﬁned forxER". The notation f:A—>Rmindicates thatf(:c)isdeﬁned only for:1:in thesetA,which iscalled thedomain off.IfBEA,we deﬁne f(B) asthesetofallf(a:) for:1:EB,andifCERmwe deﬁne f*1(C) ={asEA:f(:z:) EC}. The notation f:A—>B indicates that f(A)EB. Aconvenient representation ofafunction f:R2—+Rmay beobtained bydrawing apicture ofitsgraph, thesetofall 3-tuples oftheform (:c,y,f(:c,y)), which isactually aﬁgure in 3-space (see, e.g., Figures 2-1and 2-2ofChapter 2). Iff,g:R”—+R,thefunctions f+g,f—g,f-g,andf/gare deﬁned precisely asintheone-variable case. Iff:A-—>Rm and g:B-—>RP, Where BERm, then the composition gof isdeﬁned bygof(:z:) =g(f(:z:)); thedomain ofgof is Af\f'1(B). Iff:A—>R'""' is1-1, that is,iff(a:) ;¢f(y) when assfy,wedeﬁne f_1:f(A)—>R"bytherequirement that j“1(z) istheunique acEAwith f(a:) =z. Afunction f:A—>Rmdetermines mcomponent functions fl,...,f"‘: A-—>R byf(a:) =(f1(:c), ...,f"‘(x)). Ifcon- versely, mfunctions g1,...,g,,,: A—>Rare given, there isaunique function f:A—>Rmsuch that fl=g,;,namely f(:c) =(g1(a:), ...,g,,,(:z:)). This function fwill bedenoted (g1, ...,g,,,), sothat Wealways have f= (fl, ...,f'"). If1rIR"->R”istheidentity function, 1r(:c) ==av,then 1r':(:U) == xf;thefunction -zr‘iscalled theithprojection function. Thenotation limf(:12)=bmeans, asintheone-variable case, that wecangetf(:1:)asclose tobasdesired, bychoosing atsuf- ﬁciently close to,butnotequal to,a.Inmathematical terms thismeans that forevery number 8>0there isanumber 6>0such that |f(x)—b|<eforalla:inthedomain offwhich 12 Calculus onManifolds satisfy 0<|:c—a|<5.Afunction f:A—>Rmiscalled con- tinuous ataEAiflimf(:v) =f(a),andfissimply called con-37-">0 tinuous ifitiscontinuous ateach aEA.Oneofthepleasant surprises about theconcept ofcontinuity isthat itcanbe deﬁned without using limits. Itfollows from thenext theorem that f:R”—>Rmiscontinuous ifandonly iff'1(U) isopen whenever UERmisopen; ifthedomain offisnotallofR”,a slightly more complicated condition isneeded. 1-8 Theorem. IfAER",afunction f:A—>Rmiscontin- uous ifandonlyifforevery open setUERmthere issome open setVER"such thatf_1(U) =V(WA. Proof. Suppose fiscontinuous. IfaEf_1(U), then f(a) EU.Since Uisopen, there isanopen rectangle Bwith f(a) EBCU.Since fiscontinuous ata,wecanensure that f(:::)EB,provided wechoose :2:insome sufficiently small rectangle Ccontaining a.Dothisforeach aEj_1(U) and letVbetheunion ofallsuch C.Clearly _F1(U) = VHA.The converse issimilar andislefttothereader. I The following consequence ofTheorem 1-8isofgreat importance. 1-9 Theorem. Iff:A—>Rmiscontinuous, where AER”, andAiscompact, thenf(A)ERmiscompact. Proof. Let0beanopen cover off(A). Foreach open set Uin0there isanopen setVUsuch thatj_1(U) =VU(WA. The collection ofallVUisanopen cover ofA. Since Ais compact, aﬁnite number VU,, ...,VU,, cover A.Then U1,...,U,,coverf(A). I Iff:A—>Bisbounded, theextent towhich ffails tobe continuous ataEAcanbemeasured inaprecise way. For 5>0let M(a,f,6) =sup{f(a:)::z: EAandIa:—a|<5}, m(a,f,6) =inf{f(a:)::z: GAandla:—al<6}. Functions onEuclidean Space 13 The oscillation o(f,a) offataisdeﬁned byo(f,a) = lim[M(a,f,5) —m(a,f,5)]. This limit always exists, sinceas->0 M(a,f,5) —m(a,f,5) decreases as5decreases. There aretwo important facts about o(f,a). 1-10 Theorem. Thebounded function fiscontinuous ataif andonly ifo(f,a) =0. Proof. Letfbecontinuous ata.Forevery number 8>0 wecanchoose anumber 5>0sothat |f(:z:) -f(a)| <8for all:2:EAwith |:c—a|<5;thus M(a,f,5) —m(a,f,5) $2E. Since thisistrue forevery 8,wehave o(f,a) =0.The con- verse issimilar andislefttothereader. I I-I1 Theorem. LetAER”beclosed. Iff: A—>Risany bounded function, and 8>0,then {:12EA:o(f,:v) Z5}is closed. Proof. LetB={avEA:o(f,a:) Z6}. Wewish toshow that R"—Bisopen. If:1:ER”—B,then either asEA orelseccEAand o(f,:c) <8.Intheﬁrst case, since Ais closed, there isanopen rectangle Ccontaining :1:such that CER”—AER”—B.Inthe second case there isa 5>0such that M(a,f,5) —m(:v,f,5) <8.LetCbeanopen rectangle containing :1:such that Ia:—y|<5forallyEC. Then ifyECthere isa51such that Ia:—2|<5forallz satisfying |z—y|<51. Thus M(y,f,51) —m(y,_f,51) <8,and consequently o(y,f) <e.Therefore CCR"—B.I Problems. 1-23. Iff:A->R"andaEA,show that limf(:c) =b 3' ' __ CU-‘>0ifandonlyiflimf(:c) —b‘fori- 1,...,m. 1-24. Prove that f: Rmiscontinuous ataifandonly ifeach is. 1-25. Prove that alinear transformation T:R'”—> Rmiscontinuous. Hint: UseProblem 1-10. 1-26. LetA ={(a:,y) ER2: at>0and0 <y<$2}. (a)Show that every straight line through (0,0) contains an interval around (0,0) which isinR2—A. (b)Deﬁne f:R2-> Rbyf(a:) =0ifccEAandf(:c) =1if xEA. ForhER’deﬁne gh:R--> Rbyg;,(t) -f(th), Show that each g;,iscontinuous at0,butfisnotcontinuous at(0,0). 14 1-27 1-28 1-29 1-30Calculus onManifolds Prove that {acER”: Ia:--al<r}isopen byconsidering the function f:R"‘—> Rwith f(a:) =la:—al. IfAER“isnotclosed, show that there isacontinuous function f:A—> Rwhich isunbounded. Hint: IfscER"—Abut acEinterior (R'"—A),letf(y) =1/ly — IfAiscompact, prove that every continuous function f:A——>R takes onamaximum andaminimum value. Letf:[a,b]—> Rbeanincreasing function. Ifx1,...,:c,,E [a,b]aredistinct, Bh0WthatE}‘_10(f,.t,-) <;(t)-1(0). 2 Differentiation BASIC DEFINITIONS Recall that afunction f:R—>Risdifferentiable ataERif there isanumber f'(a) such that (1)£i_ﬁf(a +hZ “'f(a) ___f»(a)_ This equation certainly makes nosense inthegeneral case ofa function f:R"—>Rm,butcanbereformulated inaway that does. IfA:R—+ Risthelinear transformation deﬁned by )\(h) =f’(a) -h,then equation (I)isequivalent to (2,,,mf<<»+h>-to—soI0,h—>0 h Equation (2)isoften interpreted assaying that A+ffa)isa good approximation tofata(seeProblem 2-9). Henceforth wefocus our attention onthelinear transformation )\and reformulate thedeﬁnition ofdilferentiability asfollows. I5 16 Calculus onManifolds Afunction f:R—>Risdifferentiable ataERifthere isa linear transformation A:R—>Rsuch that limf(a+h) _hf(a’) _7‘(h) .___0_ h->0 Inthisform thedeﬁnition hasasimple generalization to higher dimensions: Afunction fIR"—>Rmisdifferentiable ataER"ifthere isalinear transformation A:R”—>Rmsuch that lim,lf(“+ kl("Jf’*<h)|_,0_h->0 Note that hisapoint ofR"and f(a+h)—f(a) —A(h) a point ofRm,sothenorm signs areessential. Thelinear trans- formation Aisdenoted Df(a) andcalled thederivative offat a.Thejustiﬁcation forthephrase “thelinear transformation A”is 2-1 Theorem. Iff:R”—>Rmisdifferentiable ataER" there isaunique linear transformation A:R"—>Rmsuch that f(“+h) -f(a)-W‘) l-.0 lhl I'liml h—>0 Proof. Suppose it:R"—>Rmsatisﬁes hm|r<a+h)-l-h/I<a>-t<h>| =0°h-+0 Ifd(h) =f(a+h)—f(a), then hmI>~<h>—~<h>l_,,,ml>~<h>~d<h>+d<h>—~<h>Ih—+0 lhl h->0 lhl I S1imMh) Td(h)l +lim ld(h) _“(h)l h->0 lhl h—>0 lhl =O. If:1:ER",then ta:—>0ast—>0.Hence for:1:sf0wehave 0=1,ml*<¢e>: Ms-=>l__.l*<1=>-“W.H0 ltxl Therefore A(:v) =u(x). I Differentiation I7 Weshall later discover asimple way ofﬁnding Df(a). For themoment letusconsider thefunction f:R2—+Rdeﬁned by f(:z:,y) =sin1:.Then Df(a,b) =Asatisﬁes A(:c,y) =(cosa)-cc. Toprove this, note that Hmma+hi1»+o—f<<»>b>—>~<hr>|(at)->0 l(h,l¢)l =lim |sin(a +h)—sin a—(cosa)'h|_ 0.1.)->0 l(hi?) l Since sin’(a) =cosa,wehave lim|sin(a +h)— sina—(cosa)-h|_0. h-+0 lhl Since |(h,lt)| Zlhl,itisalsotrue that _|sin(o. +h)—sina—(cosa)-hll1II1- ~ — 2-=0. h—>0 Itisoften convenient toconsider thematrix ofDf(a): R"—> Rmwith respect totheusual bases ofR”and Rm. This mXnmatrix iscalled theJacobian matrix offata, anddenoted f’(a). Iff(:c,y) =sinac,then f'(a,b) =(cosa,0). Iff:R—+ R,then f’(a) isa1X1matrix whose single entry isthenumber which isdenoted f’(a) inelementary calculus. The deﬁnition ofDf(a) could bemade iffwere deﬁned only insome open setcontaining a.Considering only functions deﬁned onR”streamlines thestatement oftheorems and produces noreal lossofgenerality. Itisconvenient todeﬁne afunction f:R”—>Rmtobedifferentiable onAiffisdiffer- entiable ataforeach aEA. Iff:A—>Rm,then fiscalled differentiable iffcanbeextended toadifferentiable function onsome open setcontaining A. Problems. 2-1.* Prove that iff:R"'—> Rmisdifferentiable at aER“,then itiscontinuous ata.Hint: UseProblem 1-10. 2-2. Afunction f:R2—> Risindependent ofthesecond variable if foreach asERwehave f(:c,y1) --=f(a:,y2) forally1,yg ER. Show thatfisindependent ofthesecond variable ifandonly ifthere isa function g:R—> Rsuch that f(:c,y) =g(:c). What isf’(a,b) in terms ofg’? 18 Calculus onManifotds 2-3. Deﬁne when afunction f:R2—>Risindependent oftheﬁrstvaria- bleandﬁndf'(a,b) forsuch f.Which functions areindependent of theﬁrstvariable andalsoofthesecond variable? 2-4-. Let gbeacontinuous real-valued function ontheunit circle {:0ER2: lzcl=1}such that g(0,1) =g(1,0) =0and g(--sc) = —g(a:). Deﬁne f:R2->Rby 37 re)—{M'”(l??l) x#0’0 0 CU". (a)If:cER2andh:R-+Risdeﬁned byh(t) =f(ta:), show that hisdifferentiable. (b)Show that fisnot differentiable at(0,0) unless g=0. Hint: First show that Df(0,0) would have tobe0byconsidering (h,lt) with It=0andthen with h=0. 2-5. Letf:R2->Rbedeﬁned by CUIt/I , 0, f(1W) = ‘\/$2 +yz (xy)I6 0 (:c,y) =0. Show that fisafunction ofthekind considered inProblem 2-4, sothat fisnotdifferentiable at(0,0). 2-6.Let1;R2—>Rbedeﬁned byf(:r,y)-\/tn. Showthatfisnot differentiable at(0,0). 2-7. Letf:R“-> Rbeafunction such that |f(:c)| $ Show that fisdifferentiable at0. 2-8. Letf:R—>R2. Prove thatfisdifferentiable ataERifandonly ifflandf2are,andthat inthiscase ,__o'1>'<a>_I(“)"(o2>'<a>) 2-9. Two functions f,g: R—>Rareequal uptonth order ataif .f(a+h) "9(¢1+h) .11."; he(a)Show that fisdifferentiable ataifand only ifthere isa function goftheform g(:c) =at+a1(;c —a)such that fand gare equal uptoﬁrstorder ata. (b)Iff’(a), ...,f<"'>(a) exist, show that fand thefunction g deﬁned by "<12) _ ye)=Z (e-er.'-0 Diﬂerentiation 19 areequal uptonthorder ata.Hint: Thelimit n-1(i) f(=v)——Z£—.(-‘Q (rv-—-a)‘_ i! lim _--_---_ '_"°.-- ._. :c—>a (97""55)“ may beevaluated byL’Hospital’s rule. BASIC THEOREMS 2-2 Theorem (Chain Rule). Iff:R”—>Rmisdiﬂ'erenti- able ata,and g:R""—> RPisdifferentiable atf(a), then the composition g0f:R"—>RPisdiﬁerentiable ata,and D(9°f)(a) =D9(f(a)) °Df(a)- Remark. This equation canbewritten (9°f)'(a) =9'(f(a)) 'f'(a)- Ifm=n=p=1,weobtain theoldchain rule. Proof. Let b=f(a), letA=Df(a), and letu=Dg(f(a)). Ifwedeﬁne (1)¢>(=v)=f(w)—f(a)—Mr—<1), (2)9(9)=9(9)—9(5)—M9—5). (3)p(rv)=9<=f(w) —9°f(5) —9<=A(¢v—5), then (4)lim—|—‘—p—(£)—l- =0, (5)limw- =0 11->5ll!_bl ’ andwemust show that lim-lg-2-I-— =0.,,_..,lac—cl Now 9(1>)=9(f(1=)) —9(5)—u(>\(Iv -5)) =9(f(1>)) —9(5)—u(f(1>) -f(a)—<e(1>)) by(1) =l9(f(1=)) —9(5)—u(f(1>) -f(5))l +u(<e(1>)) =ll/(f(Iv)) +/»(<9(1v)) by(2)- 20 Calculus onManifolds Thus wemust prove (6)lim-l“’(f("’))l -.0,H,|:c—a] <511*: =°-Equation (7)follows easily from (4)and Problem 1-10. If 8>0itfollows from (5)that forsome 5>0wehave woc»l<eba>—b| ifve>-bl<@. which istrue ifIa:—a|<51,forasuitable 51. Then ltoe»l<eve>—b|=8l<e(Iv) +7\(Iv—<1)l §El<p(5U)| +8M|:c —al forsome M,byProblem 1-10. Equation (6)now follows easily. I 2-3 Theorem (1)Iff:R"->Rmisaconstant function (that is,ifforsome yERmwehave f(a) =yfor all:cER"), then Df(a) =0. (2)Iff:R"->Rmisalinear transformation, then Df(a) =f- (3)Iff:R"'—> Rm, then fisdifferentiable ataER"ifand only ifeach is,and Df(a) =(Df1(<1). ---.Df""(a))- Thus f'(a)isthemXnmatrix whose ithrowis(fl)’(a). (4)Ifs:R2-—>Risdeﬁned bys(:c,y) =cc+y,then Ds(a,b) =s.' (5)Ifp:R2—>Risdeﬁned byp(:c,y) =as-y,then D9(5.5)(¢v.9) =be+99- Thus p'(a,b) =(b,a). Diﬂerentiation 21 Proof lr<a+h>—r<a>—°l...11,19:9—.OI._..,1 (1)lim h->0 lhl Ih-0 lhl fa+h)—f(a)—fool (2)liml h—>0 lhl1Hmlr<a>+f<h>—f<a>-M11__0h—»0 lhl i (3)Ifeach ffisdifferentiable ataand thenA===(Df1(<1). ---,Df'"(a)). f(a+h)—f(a)—A(h) =(f‘(a+5)—f1(5») —Df1(5»)(5). ---. f"‘(o+h)+f"‘(5) -Df"‘(<1)(h)) Therefore me+h>-no->~<h>l lim h—>0 IClhl <1.ilrra +h>—rm)—1>r=<e><h>\ g0 1_ 1m h->01;- If,ontheother hand, fisdifferentiable ata,then ff= 11-‘0fisdifferentiable ataby(2)andTheorem 2-2. (4)follows from (2). (5)LetA(:c,y) =bx+ay. Then be+h,b+Io—pub)—>\<h,t>| 1' _ -________H.__. ., L _I <h.i=iE»o l(h.5)l Now=11m_|f?;’i.a.k>—>0 l(h.7<>)l lhl”iflklslhl.mm5lIltlz iflhl3lit]. Hence |hk|_§lh|2+|lc|2. Therefore Inn<h2+k” --—-—— ---—--ma. =\/h” It’.l(h.5)| “\/hi+it” +’ 22 Calculus onManifolds SO .lhIt|l ?— =0. 1.,}.§iel<h.t>| ' 2-4 Corollary. Iff,g: R”—>Raredifferentiable ata,then D(f+9)(<1»)=Df(a) +D901). D(f'9)(a)=9(5)Df(a) +f(<1)D9(a)- If,moreover, g(a) sé0,then DU/g)(a) =g(a)Df(a) —~l:2(a)Dg(a). l9(a)] Proof. Wewillprove theﬁrstequation andleave theothers tothereader. Sincef +g=s0(f,g), wehave D(f+9)(a)=D8(f(9»),9(<1))<= D(f.9)(5») =8<=(Df(e).D9(a)) =Df(a) +D9(o)- I Wearenowassured ofthedifferentiability ofthose functions f:R"-—> Rm, whose component functions areobtained by addition, multiplication, division, andcomposition, from the functions 1r’:(which arelinear transformations) andthefunc- tions which wecan already differentiate byelementary calculus. Finding Df(:c) orf’(a), however, may beafairly formidable task. Forexample, letf:R2->Rbedeﬁned by f(:c,y) =sin(:cy2). Sincef =sinQ(1r1-[11-212), wehave f'(@.5) =SiI1'(<152) 'l52(1r1)'(a,5) +a(l1r2l2)'(@.5)l =sin’(ab2) -[b2(1r1)'(a,b) +2ab(1r2)'(a,b)] =(cos(ab2)) ~[b2(1,0) +2ab(0,1)] =(b2cos(ab2), 2abcos(ab2)). Fortunately, wewillsoon discover amuch simpler method of computing f’. Problems. 2-10. Usethetheorems ofthissection toﬁndf’forthe following: Diﬂerentiation 23 (a)f(a:/,2) =as”- (b)f(=v.9.z) ==(15%)- (c)f(:c,y) =sin(:c siny). (d)f(:v,y,z) =-=sin(a: sin(y sin2)). (9) .l-(xx?/:3) =37”‘- f(x>l/rz) '7'“37y+z- (S)f(=v.9.z) =(Iv+9)‘- (hlf(r.9) ==siI1(1v9)- (i)f($.19) =[BiI1(1vy)l°°° 8- (.i)f(=v.:9) =(SiI1(=v9). siI1(=vSiI1 9),=v”)- 2-ll. Find f’forthefollowing (where g:R—>Riscontinuous): <t>fa-.9)=--fimt <1»)rat)=-fift-sin(:c sin(y sin z)) <<=>f(rv.9.z) =I... 9.2-12. Afunction f:R"XRm—> R”isbilinear iffor16,181,332 ER”, y,y1,yg ERm, and aERwehave f(e=v.9) =af(w,9) =f(1v,a9). f($1 +372/l/) =.l-(allay) +f(x2/U)! f(xsy1 +I/2) =.l-(xsyl) +.f(:vsl/2)‘ (a)Prove that iffisbilinear, then .Iron)!1---—- =0.(11,530 |(5J¢)l (b)Prove that Df(a,b)(a:,y) =f(a,y) -1-f(:v,b). (c)Show that theformula forDp(a,b) inTheorem 2-3isa special case of(b). 2-13. Deﬁne IP:R“XR“—> RbyIP(:v,y) =(:v,y). (a)Find D(IP)(a,b) and (IP)’(a,b). (b)Iff,g:R—>R“aredifferentiable andh:R—>Risdeﬁned by h(t)=(f(t),g(t)), show that h'(a) =(f'(@)T,9(a)) +(f(@).9'(@)T)- (Note that f’(a)isannX1matrix; itstranspose\f"(a)T isa1Xn matrix, which weconsider asamember ofRm.) (c)Iff:R—>R“isdifferentiable and |f(t)\ ---=1forallt,show that(f'(i)T.f(t)) =0. (d)Exhibit adifferentiable function f:R—> Rsuch that the function |f|deﬁned byIf](t)=lf(t)| isnotdifferentiable. 2-14». LetE1,i=1,...,lcbeEuclidean spaces ofvarious dimensions. Afunction f:E1X---XE;,—> R?’iscalled multilinear if foreach choice of:c,-EE,-,j ;-5ithe function g:E1—>R1’deﬁned by g(a) =f(a-1, ...,:c,-_1,:z:,:c,;+1, ...,:v;,)isalinear transformation. 24 Calculus onManifolds (a)Iffis multilinear and i75j,show that forh=(h1, ...,h;,), with hiEE1,wehave lim h—>0 Hint: Ifg(:c,y) =-"f(a1, ...,:r,...,y,. bilinear. (b)Prove that Itlf(a1, ...,h1;, ...,hj, ...,a1,=)l__c...___ _._--:- A_ _.()_ lhl ..,ap,), then gis Df(a1, ...,a;,)(:v1, ...,a:;,) =Xf(a1, ...,a,;_1,:c,;,a,;_1.1, ...,a;,). =11: 2-15. Regard annXnmatrix asapoint inthen-fold product R"X ---XR“byconsidering each rowasamember ofR“. (a)Prove that det: R"X--'XR“—>Risdifferentiable and Tl _D(det)(a1, ...,a,,)(:c1, ...,:c,1)=Edetlac; . i=1fl 1..(b)Ifa,;,~:R—>Raredifferentiable andf(t)=det(a,-,-(t)), show that a11(t), ...,a1.1(t ) f'(t) =Zdet a,-1’(t), ...,a,~,,'(t) . J,=1 . . a..1(t). ---,om.(t) (0)If(Ii-3t(€l1Ij(l)) as0foralltandb1,...,b,,:R—>Raredif- ferentiable, lets1,...,s.1:R—> Rbethefunctions such that s1(t), ...,s.,(t) arethesolutions oftheequations S1'-1 Show that s1;isdifferentiable andﬁnds,;’(t).a,~-1-(t)s,~(t) =b.-(t) i-1,...,n. Diﬂerentiation 25 2-16. Suppose f:R“—> R“isdifferentiable and has adifferentiable inverse f“1: R"—>R". Show that (f-1)'(a) =[f’(f"1(a))]“1. Hint:f¢>f_1(:v) =:0. PARTIAL DERIVATIVES Webegin theattack ontheproblem ofﬁnding derivatives “one variable atatime.” Iff:R"-—>RandaER",thelimit ,f(a1,... ,a";+h,. ..,a"')—-f(a1,.. .,a"') h—»0 h ifitexists, isdenoted D,-f(a),andcalled theithpartial deriva- tiveoffata.Itisimportant tonote that D,;f(a) istheordi- nary derivative ofacertain function; infact, ifg(a) = f(a1, ...,:c,...,a"), then D,-f(a) =g'(a"). This means that D,—f(a) istheslope ofthetangent lineat(a,f(a)) tothe curve obtained byintersecting thegraph offwith theplane sci=aj,_7' 75i(Figure 2-1). Italsomeans thatcomputation of D,-f(a) isaproblem wecanalready solve. Iff(a‘, ...,a:")is M" l l 4'" I I 11 1 1 / FIGURE 2-1 26 Calculus onManifolds given bysome formula involving 2:1,...,:1:",then weﬁnd D,;f(:1:‘, ...,:1:"') bydifferentiating thefunction whose value atxiisgiven bythe formula when all£137,forj;éi,are thought ofasconstants. For example, iff(:c,y) =sin(:1:y2), then D1f(x,y) =y2cos(:1:y2) and D11f(:1:,y) =2:1:ycos(:1:y2). If, instead, f(:c,y) =:12”,then D1f(:r,y) =y:1:”""1 and D2f(:1:,y) = :11”log:1:. With alittle practice (e.g., theproblems attheendofthis section) you should acquire asgreat afacility forcomputing D,;fasyou already have forcomputing ordinary derivatives. IfD,-f(a) exists forall:1:ER”, weobtain afunction D,;f: R"—>R.Thejthpartial derivative ofthisfunction at:12,that is,D,-(D1 f)(:13),isoften denoted D1,,-f(:1:). Note that thisnota- tion reverses theorder ofiand j.Asamatter offact, the order isusually irrelevant, since most functions (anexception is given intheproblems) satisfy D1,,-f =D1'.t'f- There arevarious delicate theorems ensuring thisequality; thefollowing theorem isquite adequate. Westate ithere butpostpone theproof until later (Problem 3-28). 2-5 Theorem. IfD,-,,-f and D,-,,-f arecontinuous inan open setcontaining a,then Dmf(<1) =D1'.1:f(<1)- The function D,-_,-f iscalled asecond-order (mixed) partial derivative off.Higher-order (mixed) partial derivatives aredeﬁned intheobvious way. Clearly Theorem 2-5can beused toprove theequality ofhigher-order mixed partial derivatives under appropriate conditions. The order ofi1,...,i;,iscompletely immaterial inD,-1, ...,,-1,f iffhascontinuous partial derivatives ofallorders. Afunction with thisproperty iscalled aC”function. Inlater chapters itwillfrequently beconvenient torestrict ourattention toC°° functions. Partial derivatives will beused inthenext section toﬁnd derivatives. They alsohave another important use-—ﬁnding maxima andminima offunctions. Dijferentiation 27 2-6 Theorem. Let AER”. Ifthemaximum (ormini- mum) off: A—>Roccurs atapoint aintheinterior ofAand D,;f(a) exists, then D,;f(a) =O. Proof. Let g,;(x) ==f(a1, ...,x,...,a"). Clearly g,; hasamaximum (orminimum) atat,and g,;isdeﬁned inan open interval containing ai.Hence 0=g,/(at) =D,-f(a). I The reader isreminded that theconverse ofTheorem 2-6 isfalse even ifn=1(iff:R——> Risdeﬁned byf(x) =x3, then f'(0) =0,but 0isnoteven alocal maximum ormini- mum). Ifn>1,the converse ofTheorem 2-6 may fail tobetrue inarather spectacular way. Suppose, forexam- ple,that f:R2—> Risdeﬁned byf(x,y) =x2—y2(Figure 2-2). Then D1f(0,0) =0because g1has aminimum at0, while D2f(0,0) =0because g2hasamaximum at0.Clearly (0,0) isneither arelative maximum norarelative minimum. z 51 l3 U 1 1 Iii I \ 5.5iiitll=l;i§:l.5§.=§.=:7-l-ll;§';i?§it?§'Eil§.5'il§l?iF§?i'??:?\ =£i&iz::5.=z?Ir.1=&=i>isi§i§i.=?;=-:I§=;%i.=£=f1:E \ \ \ \ \ "-1llW ti .1 . if11 FIGURE 2-2 28 Calculus onManifolds IfTheorem 2-6isused toﬁndthemaximum orminimum of fonA,thevalues offatboundary points must beexamined "separately-—a formidable task, since theboundary ofAmay beallofA! Problem 2-27 indicates oneway ofdoing this, andProblem 5-16 states asuperior method which canoften beused. Pmblems. '2-17. Find the partial derivatives ofthe following functions: (3') f’(-‘M/-.1) = (b)f(a:/,-1) =-1- fc).f(-1.9) =sintvsin9)- fd)f‘(:c,y,z) ===sin(x.sin'(y sin2)). (=0).f(-no,-1) =1"‘- (T)f(r.9.-1) =1='+‘- fs)f(==.s.1) =(:1:+1/)‘- (11)f(==.t!) =8il1(=rs)- (i)f(-1=.1/) =[Bin(11!/)l°°" '- 2-18. Find thepartial derivatives ofthefollowing functions (where g:R—>Riscontinuous): (=1)f(r.9) =EH9- rat)-fie. (c)f(a!/) =five- (fir)(d)f(a:/) =f. 9- 2-19. Iff(fv,U) =x"" +(log x)(arctan(arctan(arctan(sin(cos xy)- log(:c +y))))) ﬁnd Dgf(1,y). Hint: There isaneasy way to dothis. 2-20. Find thepartial derivatives offinterms ofthederivatives ofgand hif (H-)f(1v.9) =9(w)h(9)- (blffilhill) =9(1v)h(m- (<1f(rv,9) =9(w)- (df(=v,:l/) =9(9)- Bf(1F1?/) =9(1v+9)- 2-2l.* Letg1,g2I R2—>Rbecontinuous. Deﬁne f:R2—> Rby 37 1/ f(=v,9)=-f91(l.0)dt +ftn-:.1>d1.0 0/""‘\ \|_n/\I1_n/\-—n/ (a)Show that D2f(x,y) =g2(x,y). (b)How should fbedeﬁned sothat D1f(x,y) ==g1(x,y)? (c)Find afunction f:R2—> Rsuch that D1f(x,y) =:1:and D2f(:1:,y) =-y.Find onesuch that D1_f(x,y) =yand D2f(x,y) =x. Differentiation 29 2-22.* Iff:R2-1 Rand Dzf=0,show that fisindependent ofthe second variable. IfD1f=Dgf=0,show that fisconstant. 2-23.‘ LetA={(x,y) ER2::1:<0,orxZ0andyeé0}. (a)Iff:A-—>Rand D1f ==Dgf =0,show that fisconstant. Hint: Note that any two points inAcanbeconnected bya sequence oflines each parallel tooneoftheaxes. (b)Find afunction f:A-1»Rsuch that Dgf=0butfisnot independent ofthesecond variable. 2-24. Deﬁne f:R2-1Rby $2_yz f(a!/) =-={$219 +92 (M)F0’ 0 (x,y) =0. (a)Show that D2f(x,0) =xforallxand D1f(0,y) -=—yfor ally. (b)Show that D1_2f(0,0) iiD2.1f(0,0). 2-25.* Deﬁne f:R--> Rby f<->=1?” iiiShow that fisaC°°function, andfill(0)=0foralli.Hint: .. .6""’ .1/hThe l1m1t f’(0) =hmT =11m—,-5 can beevaluated by h—> h—>0 3 0 L’Hospital’s rule. Itiseasy enough toﬁndf'(x) forxelf0,and f”(0) =limf'(h) /hcanthen befound byL’Hospital’s rule. h—>0 —(:n-*1)". ""'(:r:+l)"* _ 2-26.* Let f(a) -={E 6 itﬁg 1&3’:1: -,. (a)Show that f:R—>RisaC°°function which ispositive on (—1,1) and0elsewhere. (b)Show that there isaC°°function g:R—> [0,1] such that g(x) =0forx$0and g(x) =1for:1:Ze.Hint: IffisaC°° function which ispositive on(0,e) and 0elsewhere, letg(x) = ftf/fit(c)IfaER",deﬁne g:R"—> Rby 9(:v)=f([=v‘ —all/E): ---'f([$" -521/8)- Show that gisaC°°function which ispositive on (a1-e,a1+e)X ---X(a"—e,a"+s) andzero elsewhere. (d)IfAER"isopen andCCAiscompact, show that there is anon-negative C°°function f:A->Rsuch thatf(x) >0forxEC andf=0outside ofsome closed setcontained inA. (e)Show that wecanchoose such anfsothat f:A—>[0,1] and f(x)=1forxEC.Hint: Ifthe function fof(d)satisﬁes f(a) ZsforxEC,consider gQf,where gisthefunction of(b). 30 Calculus onManifolds 2-27. Deﬁne g,h:{xER2: 51}-1»R2by 9(w,9) =(111.9.\/21—:v2—92). h(=v.9) =(ay.-\/1-$2—92)- Show that themaximum offon{xER2: =1}iseither the maximum offogorthemaximum offo hon {xER2:Ix]$1}. DERIVA TIVES The reader who hascompared Problems 2-10 and 2-17 has probably already guessed thefollowing. 2-7 Theorem. Iff:R"'—-> Rmisdiﬂerentiable ata,then D,-f2(a)existsfor1 §i5m,13j§nandf’(a) isthemXn matrix (D,-ff(a)). Proof. Suppose ﬁrstthat m=1,sothatf:R"->R.Deﬁne h:R-—+R" byh(x) =(al, ...,x,...,a"’), with xinthe jthplace. Then D,-f(a) =(foh)’(aj). Hence, byTheorem 2-2, '1' I<1-h>'<-1')=re)~h'<@-1') Oi O O 1-lSince (foh)’(al) hasthesingle entry D,-f(a), thisshows that D,-f(a) exists andisthejthentry ofthe1Xnmatrix f’(a). Thetheorem now follows forarbitrary insince, byTheorem 2-3, each ffisdifferentiable and the ithrow off’(a) is (f2)’(5)- I=f(a)-lg<~—jth place. There areseveral examples intheproblems toshow that the converse ofTheorem 2-7isfalse. Itistrue, however, ifone hypothesis isadded. Differentiation 31 2-8 Theorem. Iff:R"--> Rm, then Df(a) exists ifall D,-f2(x) exist inanopen setcontaining aand ifeach function D,-ffiscontinuous ata. (Such afunction fiscalled continuously differentiable ata.) Proof. Asintheproof ofTheorem 2-7,itsuffices toconsider thecase m=1,sothatf:R"-—~>R.Then f(a+h)—f(a) =f(a1 -1-71116521 ---1an)—f(a11- --la”) +f(a1 +hl,a2+h2,a3,...,a”) --f(a1 +hl,a2,...,a”) + ~00 —I-f(a1 +h1,. ..,a"'+hm) —f(a1 +hl,...,a"_1' +h'”"1, am). Recall that D1fisthederivative ofthefunction gdeﬁned by g(x) =f(x,a2, ...,a”). Applying themean-value theorem togweobtain f(a1 +h1,a2, ...,a"') --f(a1, ...,a”) =hl-D1f(b1, a2,...,a”) forsome b1between a1anda1+hl. Similarly theithterm inthesum equals hi'Di.f(a1 +hla '''1622-1 "l"hi-1: bi: '''ran) =h2Dif(ci)r forsome c,-. Then Ira+1->—no-om)~1- h—>0 = im -—— 2"? —A" = A "1* lhlTllhl ]Z11>~r<e.~>-D-re->1~1-,.h—>0 - .llril s131,-)Z1|D."f(c.:)-o.f<->1 W slimZ|1>.~f<1-.1)-D-.:f(5)|h—>0 ,-=1 =0, since D,;fiscontinuous ata.I 32 Calculus onManifolds Although thechain rulewasused intheproof ofTheorem 2-7,itcould easily have been eliminated. With Theorem 2-8to provide differentiable functions, andTheorem 2-7toprovide their derivatives, thechain rule may therefore seemyalmost superﬂuous. However, ithasanextremely important corol- lary concerning partial derivatives. 2-9 Theorem. Letg1,...,g,.,,: R"'~—-> Rbecontinuously diﬁerentiable ata,and letf:Rm—> Rbediﬁerentiable at (g1(a), ...,g,,,(a)). Deﬁne thefunction F:R”-—>R by Fe)-ro.<==>, ....t..<->>. rhe- m D,F(a) =2:D,-f(g1(a), ...,g,,,(a)) -D-,;g,-(a). Proof. The function Fisjust thecomposition fog,where g=(g1,...,g,,,). Since g,;iscontinuously differentiable at a,itfollows from Theorem 2-8that gisdifferentiable ata. Hence byTheorem 2-2, F’(5-)=f’(9(5)) '9’(a) = D1g1(a)2 ''2aDng1(a) (D1f(9(a)). ---,D..f(9(a)))' " ' D1gm(a): ---»Dngm-(a) But D,~F(a) istheithentry oftheleftside ofthisequation, while Zl;?"__,_,1D,-f(g1(a), ...,g,,,(a)) -D,;g,-(a) isthe ithentry oftheright side. I Theorem 2-9isoften called thechain rule, but isweaker than Theorem 2-2since gcould bedifferentiable without g,; being continuously differentiable (seeProblem 2-32). Most computations requiring Theorem 2-9arefairly straightforward. Aslight subtlety isrequired forthefunction F:R2-—> R deﬁned by F(11.9)=f(9(1v,9),h(w).5(9)) Differentiation 33 Where h,lt': R-> R. Inorder toapply Theorem 2-9deﬁne h,l<;I R2——> Rby Then h’ 0-D1@(x1l/) D1l¢(~"51?J)0. t’(9). andwecanwrite F(=v.:9) =f(9(-"v,9),5(1=.9)./5(1>.9))- Letting a=(g(x,y),h(x),h(y)), weobtain ItshD1F(1=.9) =D1f(a) -D19(<v.9) +D2f(a) 'h’(Iv), D2F(~"31t!) =D1f(a) 'D29(971y) +Dsf(a) 'l‘3'(Z/)- ould, ofcourse beunnecessary foryoutoactually write 7 down thefunctions handla. Pro 2-29. 2-30. 2-31. 2-32.blems. 2-28. Find expressions forthepartial derivatives ofthe following functions: (5-)F(1=.9) =f(9(1v)l1=(9). 9(=v)+h(9))- (b)F(<v.z/,9) =f(9(:v+9),h(9+z))- (<=)F(:v,1/.9) -'=f(:v”.9”.z”)- (<1)F(:v,9) =f(w,9(1v).h(1v.9))- Letf:R"—>R. ForxER”,thelimit .f(a+tar)-f(a)hm ~~ 1 t—>0 t ifitexists, isdenoted D,,,f(a),andcalled thedirectional deriva- tiveoffata,inthedirection x. (a)Show that D_._,,f(a) =D,;f(a). (b)Show that D1,,f(a) ==tD,,f(a). (c)Iffisdifferentiable ata,show that D,f(a) =Df(a)(x)and therefore D,+1,f(a) =D,,f(a) +D1,f(a). Letfbedeﬁned asinProblem 2-4. Show that D,,f(0,0) exists for allx,butifgre0,then D,,+,,f(0,0) =D,,f(0,0) +D,,f(0,0) isnot true forallxand-y. Letf:R2—>Rbedeﬁned asinProblem 1-26. Show that D,,f(0,0) exists forallx,although fisnoteven continuous at(0,0). (a)Letf:R~—>Rbedeﬁned by 12-__ﬁx) =xsmx xe-*0, 0 x=0. 34 Calculus onManifolds Show that fisdifferentiable at0butf’isnotcontinuous at0. (b)Letf:R2~—>Rbedeﬁned by _ 1 it-.9)-<2”*”” mm2°’0 (x,y) =0. Show that fisdifferentiable at(0,0) but D,;fisnotcontinuous at(0,0). 2-33. Show that thecontinuity ofD1ftatamay beeliminated from the hypothesis ofTheorem 2-8. 2-34. Afunction fIR"-—> Rishomogeneous ofdegree miff(tx) = tmf(x)forallx.Iffisalsodifferentiable, show that ,E_l\4=Iv‘Do‘(rv) =mf(rv)- Hint: Ifg(t)=f(tx), ﬁndg’(1). 2-35. Iff:R"-1Risdifferentiable andf(0)=0,prove that there exist g,-:R"-1Rsuch that Tl- rc-)-2.»-:t.~e>-i=1 1rm¢.-Ifh,(t)-f(a),thenf(x)=[,1,h,’(t)dt. INVERSE FUNCTIONS Suppose that f:R—-> Riscontinuously differentiable inan open setcontaining aandf’(a) 750.Iff’(a)>0,there isan open interval Vcontaining asuch that f'(x) >0forxEV, andasimilar statement holds iff’(a)<0.Thus fisincreas- ing(ordecreasing) onV,andistherefore 1-1with aninverse function f"1deﬁned onsome open interval Wcontaining f(a). Moreover itisnothard toshow thatf‘1isdifferentiable, and foryEWthat 1»._____1_...._. ‘F)2’)r<r‘<9>> Ananalogous discussion inhigher dimensions ismuch more involved, buttheresult (Theorem 2-11) isvery important. Webegin with asimple lemma. Differentiation 35 2-10 Lemma. LetAER"bearectangle andletf:A->R" becontinuously diﬁerentiable. Ifthere isanumber Msuch that ID,-f2(x)l 3Mforallxintheinterior ofA,then Ire)-re)!s1-PM)-=-9! forall17,3!€A. Proof. Wehave re)-rt->-2ml.-1-.91‘,-9+1. --..-"1-1 _.fi(y1: '''if/J--1122)‘: '-'1xn)l- Applying themean-value theorem weobtain fi(l/1: '''ryjaxj-+12 '''awn) _fi(y11 --'2yj_1axj: -'-ax”)'--\. ==(2/i-xi)'D.~'f2(Z='9) forsome z,:,-. The expression ontheright hasabsolute value lessthan orequal toM-Iyj—xi]. Thus Ira-re)!sZ191--="'|-Ms1-M11)—-I1"-1 since each Iyl—xjl$Iy-— Finally Ira)—re)!sZIre)—f‘<-oisr-2M-I9-wtI 2-11 Theorem (InverseFunction Theorem). Suppose that f:R"——>R”iscontinuously diﬁerentiable inanopen setcontain- inga,anddetf’(a)r50.Then there isanopen setVcontaining aandanopen setWcontaining f(a) such thatf:V—->Whasa continuous inverse _F1:W-> Vwhich isdifferentiable andfor allyEWsatisﬁes (F‘)’(v) =[f’(J“‘(9))]"‘- Proof. Let Abethelinear transformation Df(a). Then Aisnon-singular, since detf’(a)rs0.Now D(A_1 Qf)(a)= D(A"'1) (f(a)) oDf(a) =A-10Df(a) isthe identity linear 36 Calculus onManifolds transformation. Ifthetheorem istrueforA"10f,itisclearly trueforf.Therefore wemay assume attheoutset that Aisthe identity. Thus whenever f(a+h)=f(a), wehave lffa+h)""f(a)‘ ’*(5>l_.lhl21 ihl ihi'But .Ire+11>-f(a)_-1-9->11-12 This means that wecannot have f(x) =f(a) forxarbitrarily close to,butunequal to,a.Therefore there isaclosed rec- tangle Ucontaining ainitsinterior such that 1.f(x) ¢f(a) ifxEUand x75a. Since fiscontinuously differentiable inanopen setcontaining a,wecanalso assume that 2.detf’(x) xiOforxEU. 3.lD_,-f2(x) --D,-f’(a)] <1/2-1.2forall1;,1",andx5U. Note that (3)and Lemma 2-10 applied tog(x) =f(x) --x imply forx1,x2 EUthat lf(1?1) '""$1*"(f($2) —¢F2)lSilivl —112l- Since ie.-112i-Ira.)—foal3ha.)-e.—(f(~’IP2)-ed!S"flxl —132i, weobtain 4.lx1—-xel_§2lf(x1) --f(x;1)| forx1,xe EU. Now f(boundary U)isacompact setwhich, by(1),does not contain f(a)(Figure 2-3). Therefore there isanumber d>0 such that |f(a) --f(x)] Zdfor xEboundary U. Let W={y:ly-—f(a)l <d/2}. IfyEWandx Eboundary U, then 5-19-it->1<19-nel- Wewill show that forany yEWthere isaunique xin interior Usuch that f(x) =y.Toprove this consider the MN_m::~U_r_~ __“#‘__r‘‘‘HM‘iii!‘;‘‘1‘QHi‘!iiUiuh mu“HHH____\___\H_____HH_m_HH____Hu\H_‘H:'Hmuﬁmﬂmwmﬁ___"v___m_wH___HHH_>_\HH"__H_vU_“_“_Uw_____H_m"H____H__W“_____H‘H“_Hww”“wmHH_uHH___'|__H|m___|Hmum“”___IUW%H__HH_mH_%%“H__HWvﬂH>HwP%_H_m_‘_um_ﬁm__"“Wwl_'__‘_|Hm“_H_hv"_“'_vd_H_m_WH__H___F“__H"W_"_H_M_h_”___H__H“H_H_HUHF“nu“"“H“H“HHN"“"HHHH“”H__\__hm_HuHh%H“"NHHhUHH“H“HA"Han“__“NH“____HH“_ﬁ____"“____"_uH_HU_hHHHHHH"PHH"“"""‘H"__'.HH__H”"_“___”_“"_HH_|______N/_u._____H 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________h_____x___“_U_H_H_h__“_h_H“I_EH__H“_““__UNHHHHH_“THUH__H__H“_H#_H""|ﬂ_'|ﬂH_H|”_vNH.““pm_H_WI?H‘U?___vH._____’_.“_.v_____\A~t_2E__g_<_'___“_Hm¥_|_H'_______F_vH_”H_H_H__l__mH__"'“'_ _H__mm_I___H___mI_H“_"___H_ _hpU_WNWH__WNWMHWWm"HmHm___H”U"H"EU__II“W"m“mHWm%HwHWmmm“""mH_Wm”NH“Hﬂﬂ“”NH"wHnHH|"n_“|_|______VHHH_h__H__”H_H_“_h_“_“Hi___"_”"_+_ﬂ_"_“'dn_“_“_H_H_H_R_“_N___HN__"__H'H_H‘IdGK__u_"\“"“_In_n_"H__H\_"__ﬂ"_"_“_____"_H__H___V_"___|__v___H_____>“_H_H)“_“_HH_v______"_"'HH___P_H_U_H_H_"_x_hv__P_h3”_“vH_H_W“TH_"_H‘h_"_U_H___H_H_h__hm_"_H‘?__U_ﬂ_"___u_H_“_H_H_H_x_H_"___H_"_H“_“_H_H_H_mu__W_“_H_H'H_"_"_'H_"P‘_H'__H__H__U__v__h_H_"A___H_n_____.H_H_H_v_WN”W_hMm__mM__mH_"hWWE"_H@_____ﬂ“Hm_m~NN_ﬁH__M”ﬂ"w__m“_HWHMH_HmmH_H_H__H_w_HHmH_N__"_hmm__hHWmm_H___wH“wWMHmHP"_K_"_H_”_n|HH___E_U__Ml__q_Hi_H+H_H_H_____N___in_d_h_H_V"_H_"_U_A_H_H_T"_H__P+“_"_..H_H_“_“___v__uH__"‘H_H_"h_H_H_U_H____AH_____v__H_U_H_H_H_H_“___v_“___H__H_"_'H____P_H_H_“_"wm“wm"mm_”__Hgmm“_vHm"_WWmumH3£m"“v“HH“HH_H_H"_HAﬁn"m_“__H___"m"_H“HHmH"|_H__"mumH__m”_"HmHwhmnmﬁuwunuwu___w"_m“m_H__mH__\HH___HHHHU__"_H_"_h”HHHHHHHH""_”_H_m_"/“_H___"|H|_"v__ﬁn"H___|H_HH""‘_vH_ﬁ_>HAHﬁH"_v_>HH"H___""vH>"_HH|“._“+"_">“_“_uu_“_xH__vH“_“.“mHH_“_VH_“__H|||v___|HHIV?_Hl__HGHHHHHEH_5____“vH““"H‘_H“_pI“U_\H_“___E_\ﬂuH_H_H__§_w“wm_H"WWNu“gmHm_m_ﬁgMNEWERmuWWWanHm_H_m_UA_§__Hm__H___“_’H_H‘WMﬁwﬁmm"MMﬁ__Hm_ﬁH1___H_mm_ __II__q|_H_H»_"_"_~“H“H“h|_‘___“_H_NUHHHd>UHF“____I_K“_HHH“hH3___"EU_HH“H“mW“Hh"__“|__H"_H“____)."_ﬁnH_"NM“____H|H"n_”H"___§f“_"_‘__>_>_,__H_H__H"_“____N__““___\>'____‘_'_H__“_U_HHH_P|___J_““_Hh_H__HH__ﬁ|HHH_TH__H_IHI__HHH_H_“_HH"_H_H_|HH|"_P.Hﬁ“+HHHHHH___H_H_HHH_'“_H_H|___NIH“_"_H_HH_H.H_|_|.|.__|bu_P+H_H.__H“_MH___Hﬁ.H_H__HH____"ﬁn“ﬁn“ﬁnHHHHiHHH”_NHH'____HN___H"“‘hH___h__H_HuhH>HI.“ﬂu“H"NHHI“_U"munHUTH.”_UﬁnHU"Huh"HimHUH“H_“VHHu__\N____""_>__“hH“|_HH>H‘_’H__HH"inIR““Hm“H“UHw“HHnN_"”"___“"_‘H_”_uH““H”N__H_H"_"_""MH"_n§H“dH“H_H_____”(____”“_""u“_HHhNHH"h___|nH__Nun“_““HunkH“H_Em_mUH“_”_HmH__NH“H“"H__m"mH"NN"_u__H_HH“H_'|‘HHuH_W_u_HH_ﬂ"_“h“_HUN”HM"HHHHh“HHm“H“HhHH”uH_H”HHH____H_H_H_"_H|"vHml"h_“_H_H_K_h_V_H‘“_"_|_H_"UH“_H_H_"_H|“"_"'"'n_____H.___H____n_H_H|"\__H_P“|“_P_H__'H_"_H'H_”"_K_"|___H_H_H|_____km_"_H_n_h_H_H_v_H_H_>__H___‘____mvuﬂnnnnmﬂnnﬁuH“H"H_m"_Mh“ﬂmmHd“Hm"____“_HHWhm_mHWﬁ““H“HHm“M___"nu___Nﬁ""§U"N“"“_mHm_HJ_>|".H."HH'H_N_\“_Nm_“HNH“““hHHm__"_HHWmﬁ____“”““_H"HN“m_W“m"“"PH"__|H__:|_H:H___v___HHUN”__"__HH__“__W“>H_H#__“hHxhHm___H"M_H'___Hh___H_M__“hH"“u______mm"N__“W_H_"HW___m_m_______W"___'_hhU_______WhNJ_¢____H___“__“_HH_H|H(__u_“|H_H_H_P_vH_vH_h‘H_H|h_'__|_"_H+__“+H_U_v|h____“_H‘H_V+H_N_H|||H__\_d_"_TH|d_HH__U_H|“_H_v|1_:"_”_H_"_“_h__'__”____H_"_h'"_H\‘_“_H_“'_____H_H_V'H_H_H_v“_"_H_”_“_v____H_H_H_vH_“___x___HvH_H__'H|\|“v>'__“_H>H_H_H_n__H_HVH_H+H_"_H|:____H_H_UH__'"_H_"_"_v__"vH______5_"_H_n___HH__’H_“_ﬁ_H_u_|__H_‘H.IE_H"“___.”u_"|_“HHHHuh“H"u__‘"HE/nkhululH_Huh“_“"NIH_"_“”n____‘_H“HHH"H“"__"NH""“_"""““_”"_"_|__u_M__H‘Hun_H_"___H“H'___H_HU_H__H_HHHTH‘H_MyH_H_u_HhH"ﬁnHvH”_"_”___“_H__H___H__Hi"___N_u“__WU_H_HvH_H‘_3'“___H|H_U_H_H_H_H_H_HHU__NU_Hquid“U‘UH_v_V“_U‘HR’_"_K_“"H|H_n_HvH_H|Id_”__>"_H__.H.'.I_P_H_“_'__Y"Un>"h.__UU_Hv“_n_H_H'__"H_"'0?__Vn.“_q_V_n_"+nH"".",_>U.H_H_HHHm___H“_h""“”h_HHHmuMnmmmuunw_HN“U”HP"WWWNun“_H"WMn""“UH_>"_“HmHWmm“H“W"HﬂHHWW"u“F"h“_gm“___“H_"H\H'"_HWwmHwHmQ.“__MunH“HTM“WEHWm“_"N“HHHU"NHM“Wm"mHHmH______H_H“H“H§H_UH"HHHUh“_"_UHHUHHﬂnuhpnwhﬂH“HHH___'HHH_HH”U_"HHWHHH"HhH"""mHH“H__H__"__m'H'"_HHH”HH\_D'H“H"“__HP||_|HH_"_""H"""___H“._"NU“MHHWHN__m_"___NHHHHHﬁ“HH_u_HHH_”_"""_H__uHHPHHh“_HH_‘H__”_"_H”H_HnvHﬂ_Hon$2"___”_“_H_H_U_q_HUNFM‘_w__Uh__'v_H_“_EhP_”H"u“__,_“_“U__“_~_|__|_h‘___h"h.___5*__H_H_J___H_“__'_____;_IIh“.“H_‘“|"_._‘_“H_____m“n“.‘m_h_____o__H_U'H_H_H|__|_Hh|"_H|”"_H_p_"_"__'“_H“ll_HQ”_H_P|H_U_"_“'H_H____|“_'"_H____"__HH_“_|_HH__.P_H_'_H_H_"'.__H‘“_“_VH_H_“+“hH|w_.I_._+H'H_H5.__""7_>|“_nNHH|_"h"“__'__"u||_"”H_““HH_|_""_n“"J./__"”_H(‘"_““___HH‘_§“"""N___v”__‘H_HH"_n"__H””__“Hi_“H_H__H"___“WH__v_§_§_""__x"§_H____£'_:k_HWE¥MHU_|'H_m_HﬁMHmm“WMh_m_M_H_mH_H>_"Hun"_"Wm_H"HE__WM_"“_H'___HmHm_mHmMHWWwWmHw"hmH|H_H_H_H_HHm_H_HH“HHHmmWW""_"“___mH__m__"____m___E_H__‘N_H_H_“H|H_'H"h‘_“_W_HHmHmHH"|Hm_HH'_“_|'Ww__HWmummﬁmﬁn“M‘m_§m%ﬁ_HH__HEum"mﬁH_mh___h____mm_m“mHm__H__m_‘Win"_Hmﬁ__HHmUHmH_HgWuﬁwmﬂ_H“___H“__hMHmH____H\_”m____m__H___H_ww__ﬂm_WH__mhh_“WH__‘_h"mHw_H#“__w_H_H_H_\_|Pw_hNM_”>_|mH"_‘HI?MH_HmHMHHH__H_ﬁ_EBﬁgHm%_"HH_H_H“_HM_\_\_H|_“H VV_\_'_____HUvHHh__v_______“H_____HH_WNm“_HHN“_WH_mhHUPw>___H__"_H__ ll:____F‘H“““;‘““““““‘llA‘<<l‘‘‘_“__:___NHmM"H_“_H“%_m_mKL“_"'h'“wP_HHH_“,“%__"__an_“__”|__"_H_n___"m__H__ \_luvS‘_H."vH_H_ﬂv"_H&“v“>U_"v__.“H_vH_HHh"v‘_____HmW__H_““m"wHMnmHWWmnmHmJDW'_._'”_N_nHH_HHHH:_HH_____|H_U"InnU_P"HH“"H"WWTqWHH_m_HH__ﬁlm“_H”HH“HPH“HH_H”|“'H|m___|H‘UHw”_""NN“_"nHWT_H'||HHv“_|_""_H_”y__|W>_H___""HH___‘_’__NvH_H“_“H“H"____HHh;_HH_v____'UI_‘___.d____Hd.hﬁI.H.____H5"H_“H_HA_“_“+U____H____H____”h_H_H_H_H_H'H|H_z_"h_HH____Mu_HH__|_H"_E"|_”H__H|H___#”HHH____uH___HH_ L 38 Calculus onManifolds function g:U-—>Rdeﬁned by go)=ly--ro>|2=Z<1/'1--f(x))”-1:.-=1 This function iscontinuous andtherefore hasaminimum on U.IfasEboundary U,then, by(5),wehave g(a) <g(x). Therefore theminimum ofgdoes notoccur ontheboundary ofU.ByTheorem 2-6there isapoint ccEinterior Usuch that D,—g(:c) =0forallj,that is 22(y*--f(x))-1),~;*i(.~,;) =0forallj.@:=1 By(2)thematrix (D,~f'7(:c)) hasnon-zero determinant. There- fore wemust have yl--f(x) =0forall11,that isy=f(x). This proves theexistence ofac.Uniqueness follows immedi- ately from (4). IfV=(interior U)f\f-1(W), wehave shown that the function f:V—-—> Whasaninverse f_1: W~—> V.Wecan rewrite (4)as cIrlon --r1<.@/cl 52|;/1*.1/2|for1/1,2/2eW. This shows thatfliscontinuous. Only theproof that fl isdifferentiable remains. Let u=Df(x). Wewillshow thatT1isdifferentiable aty=f(x) with derivative ;f1. Asintheproof ofTheorem 2-2, for x1EV,wehave f($1) =f(x) +M131 "17)+¢>($1 —1?), where lim'¢<“’1"“’>| J0.:z:1—>:v lall """all Therefore u_1(f(1>1) -f($)) =$1-iv+1/—1(<P(971 -93))- Since every ylEWisoftheform _f(x1) forsome x1EV,this canbewritten f‘(y1) =flu) +:f‘(z/1 --y)-rF‘(¢=(F‘(2/1) —f1(y))), Differentiation 39 and ittherefore suﬁlices toshow that lim|#"l(s<F‘<v1>— Fl<y>>>| _0_1/1—>:u l?J1""7-/l Therefore (Problem 1-10) itsuﬂlices toshow that .|<@<1-‘(:1/1) --rlonl1/1—>y lyl —yl Now |<@</“<11/1) -f“<v>>lll/1—yl ..l¢<f1<y1> -r1<y>>l.lr1<y1>--r‘<y>\.|f"l(.1/1) -—f"l(y)| |z/1--yl Since fliscontinuous, fl(y1) ——>fl(y)as3/1~>y.There- foretheﬁrst factor approaches 0.Since, by(6),thesecond factor islessthan 2,theproduct alsoapproaches 0.I Itshould benoted that aninverse function flmay exist even ifdetf’(a)=0.Forexample, iff:R—->Risdeﬁned by f(x) =:03,then f’(0) =Obutfhas the inverse function fl(:c)=<7 Onething iscertain however: ifdetf’(a) =O, then flcannot bedifferentiable atf(a). Toprove thisnote that fofl(:c) =ac.Iffl were differentiable atf(a), the chain rulewould givef’(a) -(fl)’(f(a)) =I,andconsequently detf’(a) -det(fl)’(f(a)) =1,contradicting detf’(a) =0. Problems. 2-36.* Let ACR”beanopen setand f:A—> R" acontinuously differentiable 1-1function such that detf'(a:) vi0 forallav.Show thatf(A) isanopen setandfl:f(A) —>Aisdiffer- entiable. Show alsothat f(B) isopen foranyopen setBCA. 2-37. (a)Let f:R2—+Rbeacontinuously differentiable function. Show that fisnot1-1. Hint: If,forexample, D1f(:v,y) rﬁ0forall (a:,y) insome open setA,consider g:A—+R2deﬁned byg(a:,y) = (f(=v,y),2/). (b)Generalize thisresult tothecase ofacontinuously differen- tiable function f:R"—+Rmwith m<n. 2-38. (a)Iff:R—+ Rsatisﬁes f’(a) ¢0forallaER,show that fis 1-1(onallofR). 40 Calculus onManifolds (b)Deﬁne f:R2--+R2byf(x,y) =(excosy,cl’siny). Show that detf'(Iv,1l/) ¢0forall(a:,y) butf isnot1-1. 2-39. Usethefunction f:R--+Rdeﬁned by x+ lls'n1 #0 - 1-— :2: , f(=v)= 2x sv O x=0, toshow that continuity ofthederivative cannot beeliminated from thehypothesis ofTheorem 2-11. IMPLICIT FUNCTIONS Consider the function f:R2-> Rdeﬁned byf(a:,y) =sol+ yl——1.Ifwechoose (a,b) with f(a,,b) =0andaas1,-1, there are(Figure 2-4) open intervals Acontaining aandB containing bwith thefollowing property: ifasEA,there is aunique yEBwith f(a:,y) =0.Wecantherefore deﬁne ‘ll 1 i!’graph ofg :<==.y>=f(a?/)=01 ; J»'‘l ll’.3 ,6.§ ‘”1 ,, ,1‘ --|n~:——~44-r 1 l | I ‘ 1 Igraph ofgl -1 FIGURE 2-4 Dijferentiation 41 afunction g:A—>Rbythecondition g(x) EBandf(a:,g(:z:)) =O(ifb>O,asindicated inFigure 2-4, then g(x) = \/1-—$2). Forthefunction fweareconsidering there is another number b1such thatf(a,b1) =0.There willalsobe aninterval B1containing b1such that, when asEA,we have f(:c,g1(:z:)) =0foraunique g1(x) EB1(here g1(a:) = -\/1—x2). Both gand g1are differentiable. These functions aresaid tobedeﬁned implicitly bytheequation f(=r,:¢/) =0- Ifwechoose a=1or-1itisimpossible toﬁndanysuch function gdeﬁned inanopen interval containing a.We would like asimple criterion fordeciding when, ingeneral, such afunction canbefound. More generally wemay ask the following: Iff:R"XR-—> Rand f(al, ...,a”',b) =O, when canweﬁnd, foreach (sol,...,:c”)near (al,...,a"'), aunique ynear bsuch that f(xl, ...,:c"',y) =0'? Even more generally, wecanaskabout thepossibility ofsolving anequations, depending upon parameters xl,...,:c",inm unknowns: If f,-:R”"><R’”->R i=1,...,.m and f;(al,...,a"',bl,...,b"l)=O i=l,...,m, when canweﬁnd, foreach (scl,...,:c")near (al,...,a”')a unique (yl,...,y"") near (bl,...,b'"‘) which satisﬁes f,;(:cl, ...,;z:'"',;z/l, ...,y"") ==0‘?Theanswer isprovided by 2-12 Theorem (Implicit Function Theorem). Suppose f:R"XRm~>Rmiscontinuously diﬂerentiable inanopen set containing (a,b) andf(a,b) =0.LetMbetheWtXinmatrix (D»+:'fl(a,b)) 13131'5m- IfdetM;¢‘O,there isanopen setAER”containing aandan open setBERmcontaining b,with thefollowing property: for each 2:EAthere isaunique g(x) EBsuch thatf(:v,g(a:)) =0. Thefunction gisdiﬁerentiable. 42 Calculus onManifolds Proof. Deﬁne F: R”XR"‘—-> R"XRm by F(x,y) = (:2:,f(:v,y)). Then detF’(a,b) =detM;'-40.ByTheorem 2-11 there isanopen setWER”XRmcontaining F(a,b) ==(a,0) andanopen setinR"XRmcontaining (a,b), which wemay take tobeoftheform AXB,such that F:AXB—-> W hasadifferentiable inverse h:W—> AXB.Clearly hisof theform h(:v,y) =(a:,k(x,y)) forsome differentiable function k(since Fisofthisform). Let1r!R"XR"‘—> Rmbedeﬁned by1r(a:,y) =y;then 1roF=f.Therefore f(=v,l<=§<v,?/)) =fOh(=v,2/) =(ir<=F)<=h(r,y) =WQ(FQh)(r,z/) =1r(¢v,y) =2/- Thus f(a:,k(:c,0)) =0;inother words wecandeﬁne g(x) = k(=v,0)- I Since thefunction gisknown tobedifferentiable, itiseasy toﬁnditsderivative. Infact, since fl(a:,g(:c)) =0,taking D,- ofboth sides gives 0=1>.~1"'<1=.g<x>>+'§ 1>.+.r=e.g<=»>> ~D.-re)a=1 i,j=1,...,1n. Since detM#0,these equations canbesolved forD,-g“(a:). Theanswer willdepend onthevarious D,-f'l(a:,g(x)), andthere- foreong(x). This isunavoidable, since thefunction gisnot unique. Reconsidering thefunction f:R2—-> Rdeﬁned by f(a:,y) =:02+g2-1,wenote that two possible functions satisfying f(:c,g(a:)) =0are g(x) =\/1-—x2and g(x) = ——\/1-x2. Differentiating f(:z:,g(x)) =0gives D1.f(1=,9(=v)) +D2.f(¢F,Q($)) 'c'(¢v) =0, OI‘ 2=v+2c(Iv)'c'(=v) =0, 9’(=v)=-—=v/Q(=v), which isindeed thecase foreither g(x) =\/1—-x2org(x) = -\/1—a:2. Diﬂerentiation 43 Ageneralization oftheargument forTheorem 2-12 canbe given, which willbeimportant inChapter 5. 2-13 Theorem. Let f:R”——>R” becontinuously diﬂ'er- entiable inanopen setcontaining a,where p5n.Iff(a)=O andthepXnmatrix (D,-f‘l(a)) hasrank p,then there isan open setACR”containing aandadiﬁerentiable function h: A——>R"with diﬂ'erentiable inverse such that f<>h(:vl, ...,:c")=(:v"_'l"+l, ...,x"). Proof. Wecanconsider fasafunction f:R”""’ XBl’——>R”. IfdetM;£0,then MisthepXpmatrix (D,,__,,+,-f'5(a)), 15i,j5p,then weareprecisely inthesituation considered intheproof ofTheorem 2-12, andasweshowed inthat proof, there ishsuch that foh(a:l, ...,:z:"')=(x"'“l’+l, ...,a:"'). Ingeneral, since (D,-fl(a)) hasrank p,there willbej1< '''<j,, such that thematrix (D,-fl(a)) 15i5p,j= 1'1,...,j,,hasnon-zero determinant. Ifg:R"'——-> R“per- mutes thexisothat g(:vl, ...,:r:”')=(...,xl1, ...,a:"r), then fogisafunction ofthetype already considered, so ((fo g)<=lc)(a:l, ...,:v")=(:v""'l’+l, ...,:v"') forsome k. Leth=g<>lo.I Problems. 2-40. Usetheimplicit function theorem tore-do Prob- lem2-15(0). 2-4-1. Letf:RXR--> Rbedifferentiable. Foreach 2:ERdeﬁne g,,: R—> Rbyg,,(y) =f(a,y). Suppose that foreach 2:there isa unique ywith g,/(y) =0;letc(x)bethisy. (a)IfD2,2f(:v,y) 750forall(:z:,g), show that cisdifferentiable and , H D2,1f(1v,6(¢v))c(st)-- A - D2,2f($,¢($)) Hint: g,/(y) =0canbewritten D2f(a:,y) =O. (b)Show that ifc’(x) =0,then forsome ywehave D2,1f(x!l/) =0! Dzf(=v,z/) =0- (c)Letf(:v,y) =:v(glogy—-y)—-ylogac.Find max (min f(:v,y)). iS==S2 £51151 44 Calculus onManifolds NOTATION This section isabrief andnotentirely unprejudiced discussion ofclassical notation connected with partial derivatives. Thepartial derivative D1f(:0,y,z) isdenoted, among devotees ofclassical notation, by a,, a a alg-lg or-Ior--J-l(r,2/,2) Or"-"f(=v,2/,2)6:0 6:0 6:0 6:0 oranyother convenient similar symbol. This notation forces onetowrite £5<u.v.w> forD1f(u,v,w), although thesymbol ”___'i '““"___ (uavaw)6f(w,z/,2) f or6f(=v,z/,2) ax l(1=.u.z)=(u.v.w) 35'? orsomething similar may beused (and must beused foran expression likeD1f(7,3,2)). Similar notation isused forD2f and D3f.Higher-order derivatives aredenoted bysymbols like 621"(aw-">.la6y6:0D2DLf(xay:z) When f:R—->R,thesymbol 6automatically reverts tod;thus dsin:0 6sin:0i» not —i-d:0 6:0 The mere statement ofTheorem 2-2inclassical notation requires theintroduction ofirrelevant letters. The usual evaluation forD1(f0(g,h)) runs asfollows: Iff(u,v) isafunction and u=g(:0,y) and v=h(:0,y), then ar<g<-->,h<w>> =afrwat+are-»>at6:0 6u 6:0 6v 6:0 [The symbol 6u/6:0 means 6/6:0g(:0,y) and 6/6u f(u,v) means Diﬁerentiation 45 D1f(u,v) =D1f(g(:0,y), h(:0,y)).] This equation isoften written simply ofof6uofat a=aa+aa Note thatfmeans something different onthetwosides ofthe equation! Thenotation df/d:0, always alittle tootempting, hasinspired many\__(usually meaningless) deﬁnitions ofd:0anddfseparately, thesolepurpose ofwhich istomake theequation d= work out. Iff:R2——>Rthen dfisdeﬁned, classically, as 6 3 df=-—-J:d:0+ldy6:0 6y (whatever d:0anddymean). Chapter 4contains rigorous deﬁnitions which enable usto prove theabove equations astheorems. Itisatouchy question whether ornotthese modern deﬁnitions represent a real improvement over classical formalism; this thereader must decide forhimself. 3 Integration BASIC DEFINITIONS The deﬁnition oftheintegral ofafunction f:A—->R,where AER”isaclosed rectangle, issosimilar tothat oftheordi- nary integral that arapid treatment willbegiven. Recall that apartition Pofaclosed interval [a,b] isa sequence t0,...,t;,,where a=to5t15~--5ti,=b. The partition Pdivides theinterval [a,b] into ksubintervals [t,;_1,t,]. Apartition ofarectangle [a1,b1] X---X[a,,,b,,] isacollection P=(P1, ...,P,,), where each P,isapar- tition ofthe interval [a,,b,-]. Suppose, forexample, that P1 =lg, ...,l],;lSELpartition Of[0,1,l)1] and P2 =80, ...,8; isapartition of[a2,b2]. Then thepartition P=(P1,P2) of [a1,b1] X[a2,b2] divides the closed rectangle [a1,b1] X[a2,b2] intoI0-lsubrectangles, atypical onebeing [t,-__1,t,] X[s,-_1,s,-]. Ingeneral, ifP,;divides [a,,b,;] intoN,subintervals, then P= (P1, ...,Pn) dlVld6S [(Z1,l)1] X'‘'X[G,n,l),,] llll/O N= N1- ...-N,, subrectangles. These subrectangles will be called subrectangles ofthe partition P. Suppose now that Aisarectangle, f:A—->Risabounded 46 Integration 47 function, andPisapartition ofA. Foreach subrectangle S ofthepartition let ms(f) =inf{f(<v)= rvES}. Ms(f) =sup{f(rv)= rvES}. andletv(S) bethevolume ofS[thevolume ofarectangle [a1,b1] X''-X[a,,,b,,], andalso of(a1,b1) X---X(a,,,b,,), isdeﬁned as(b1—a1)- ...-(b,, —a,,)]. The lower and upper sums offforParedeﬁned by L(f,P)=Z-18(1) -us)andvo".P>=ZMs(f) -v<-S).S S Clearly L(f,P) 5U(f,P), andaneven stronger assertion (3-2) istrue. 3-I Lemma. Suppose thepartition P’reﬁnes P(that is, each subrectangle ofP’iscontained inasubrectangle ofP). Then L(f,P) 5I/(f.P’) and U(f.P’) SU(f,P)- Proof. Each subrectangle SofPisdivided intoseveral sub- rectangles S1,...,S,,ofP’,sov(S) =v(S1) +---+ v(S.,). Now ’t’)’l,g(f) 5:n,g,(f), since thevalues f(:0) for:0ES include allvalues f(x) for:0ES;(and possibly smaller ones). Thus mslf) 'v(S) =mslfl 'v(S1) +'''+'"1»s(f) ‘"03-) 3'ms1(f) 'v(5'1) +'''+’"?»s..(f) '"(S-)- The sum, forallS,oftheterms ontheleftside isL(f,P), while thesum ofalltheterms ontheright side isL(f,P’). Hence L(f,P) 5L(f,P’). The proof forupper sums is similar. I 3-2 Corollary. IfPand P’areany twopartitions, then L(f.P') SU(f,P)- Proof. LetP”beapartition which reﬁnes both Pand P’. (For example, letP”=(Pf, ...,P,',,'), where P§'isapar- / 48 Calculus onManifolds tition of[a,,b,-] which reﬁnes both P,andP;-.) Then L(f,P’) SL(f9P,,) SU(f1P”) SU(f,P)- I Itfollows from Corollary 3-2that theleast upper bound of alllower sums forfislessthan orequal tothegreatest lower bound ofallupper sums forf.Afunction f:A-—>Riscalled integrable ontherectangle Aiffisbounded andsup{L(f,P)} =inf{U(f,P)}. This common number isthen denoted f_.1f, and called theintegral offover A. Often, thenotation f,.1f(:0l, ...,:0")d:0l ---dx”isused. Iff:[a,b]——> R,where a5b,then f=f[,,_1,] f.Asimple butuseful criterion for integrability isprovided by 3-3 Theorem. Abounded function f:A-—>Risintegrable ifand only ifforevery 8>Othere isapartition PofAsuch thatU(f,P) —L(f,P) <8. ;- Proof. Ifthiscondition holds, itisclear that sup{L(f,P)} = inf{U(f,P)} andfisintegrable. Ontheother hand, iffis integrable, sothat sup{L(f,P)} =inf{U(f,P)}, then for any8>0there arepartitions PandP’with U(f,P) —L(f,P’) <6.IfP”reﬁnes both PandP’,itfollows from Lemma 3-1 thatU(f.P”) -"I-(f.P”) SU(f,P) -"L(f,P’) <8-I Inthefollowing sections wewillcharacterize theintegrable functions anddiscover amethod ofcomputing integrals. For thepresent weconsider two functions, oneintegrable and one not. 1.Letf:A——>Rbeaconstant function, f(:0) =c.Then forany partition Pand subrectangle Swehave rn,g(f)= M,g(f) =c,sothat L(f,P) -=U(f,P) =Ego -v(S) =c-v(A). Hence f,1f =c-v(A). 2.Letf:[0,1] X[0,1] ——>Rbedeﬁned by f(a: )_0 if:0isrational, ’y 1 if:0isirrational. IfPisapartition, then every subrectangle Swill contain points (:0,y) with :0rational, and also points (:0,y) with :0 Integration 49 irrational. Hence 1nS(f) =0andMS(f) =1,so L(f,P) =Zeus) =0S and I/o.P>=Z1-us) =»<10.11><10.11)-1.S Therefore fisnotintegrable. Problems. 3-1. Letf:[0,1] X[0,1] —>Rbedeﬁned by __0 if05:r<%, fl“) ll if§5:051. Sl10W that flSintegrable and Jl[0,1><[0,1] f= 3-2. Letf:A—>Rbeintegrable and letg=fexcept atﬁnitely many points. Show that gisintegrable and IA)’ =Lgg. 3-3. Letf,g:A—>Rbeintegrable. (a)For any partition PofAand subrectangle S,show that 'ms(f) +'ms(9) Sms(f +9) and Msff +(J) SMs(f) +Ms(9) andtherefore L(f,P) +L(a.P) 5LU+0.P) and U(f+0.P) 5U(f,P) +U(a.P)- (b)Showthatf+gisintegrable and[A1+e=[A1+J‘,,g. (c)Foranyconstant c,show that Jl,.,cf =cIAf. 3-4. Letf:A—>RandletPbeapartition ofA. Show thatfisintegra- bleifandonly ifforeach subrectangle Sthefunction fIS,which consists offrestricted toS,isintegrable, and that inthis case fAf =2S_[SflS- 3-5. Letf,g: A—> Rbeintegrable and suppose f5g.Show that IAI SIA?- 3-6.If1;A-1Risintegrable, showthatI1|isintegrable and[failg f.ilfl- 3-7. Letf:[0,1] X[0,1]——>Rbedeﬁned by 0 asirrational, f(:0,y) = 0 :0rational, yirrational, 1/q :0rational, y=p/qinlowest terms. Sl10W that fisintegrable and f[(),1])<[0,1] f=O. 50 Calculus onManifolds MEASURE ZERO AND CONTENT ZERO Asubset AofR”has(n-dimensional) measure 0ifforevery 8>0there isacover {U1,Ug,U3, ...}ofAbyclosed rec- tangles such that 21.;-’-‘°=11)(U,~) <8.Itisobvious (but never- theless useful toremember) that ifAhasmeasure 0and BCA,then Bhasmeasure 0.The reader may verify that open rectangles may beused instead ofclosed rectangles in thedeﬁnition ofmeasure O. Asetwith only ﬁnitely many points clearly hasmeasure 0. IfAhasinﬁnitely many points which canbearranged ina sequence a1,a2,(Z3,...,then Aalso hasmeasure 0,forif 8>0,wecanchoose U,;tobeaclosed rectangle containing a,-with v(U,-) <8/2". Then E§°=1v(U,-) <E§°=18/2': =8. Thesetofallrational numbers between 0and1isanimpor- tant and rather surprising example ofaninﬁnite setwhose members canbearranged insuch asequence. Toseethat thisisso,listthefractions inthefollowing array intheorder indicated bythearrows (deleting repetitions and numbers greater than 1): /’/’/’/’0/11/12/13/14/1 ////////0/21/22/23/24/2 /’////0/s1/s2/s3/s1/s ////o/1 // Animportant generalization ofthisidea canbegiven. 3-4 Theorem. IfA==A1\JA2UA;,=U ---and each A,hasmeasure 0,then Ahasmeasure 0. Proof. Let8>0.Since A,hasmeasure 0,there isacover {Ui,1,"U1:,2,U,;_3, ...}ofA,byclosed rectangles such that E;-°=1v(U,-,,-) <8/2l. Then thecollection ofallU,-,,-isacover Integration 51 ofA.Byconsidering thearray /‘ /' /' U1,1 U1,2 U1,3 '' / / / U2,1 U2,2 U2,s " / / U3,1 U3,2 U3,3 '' / weseethat this collection canbearranged inasequence V11 V2: V31 ''''Clearly Ei°=1v(Vi) <Ei°=1e/2?: =2E‘ I Asubset AofR”has(n-dimensional) content 0ifforevery 8>0there isaﬁnite cover {U1, ...,U,,} ofAbyclosed rectangles such that Ef'=1v(U,;) <8.IfAhas content 0, then Aclearly hasmeasure 0.Again, open rectangles could beused instead ofclosed rectangles inthedeﬁnition. 3-5 Theorem. Ifa<b,then [a,b] ERdoes nothave con- tentO.Infact, if{U1, ...,U,,} isaﬁnite cover of[a,b] by closed intervals, thenZ§"=,v( U,~)Zb—a. Proof. Clearly wecanassume that each U,-C[a,b]. Let a=to<t1<...<tk=bbeallendpoints ofallU,. Then each v(U,-) isthesum ofcertain t,-—t,-._1. Moreover, each [t,-__1,t,~] liesinatleast oneU,~(namely, anyonewhich contains aninterior point of[t,-..1,t,-]), soE}‘__1v(U,-) ZZ§_1(t,- —t,-..1) =b—a.I Ifa<b,itisalsotrue that [a,b] does nothave measure 0. This follows from 3-6 Theorem. IfAiscompact andhasmeasure 0,then A hascontent 0. Proof. Let 8>O.Since Ahasmeasure O,there isacover {U1,U2, ...}ofAbyopen rectangles such that E§°=1v(Ui) 52 Calculus onManifolds <8.Since Aiscompact, aﬁnite number U1, ...,U,, of theU1,;alsocover Aandsurely E,’;;1v(Ua) <8.I Theconclusion ofTheorem 3-6isfalse ifAis11otcompact. Forexample, letAbethesetofrational numbers between 0 and 1;then Ahas measure O.Suppose, however, that {[a1,b1], ...,[a,,,b,,]} covers A. Then Aiscontained in theclosed set[a1,b1} U---U[a.,,,b,,], andtherefore [0,1] E [a1,b1l U---U[a,,,b,,]. Itfollows from Theorem 3-5that E,§l:____1(b,; —a,;)Z1forany such cover, and consequently A does nothave content 0. Problems. 3-8. Prove that [a1,b1] X---X[a,,,b,,] does nothave content 0ifa,;<b,;foreach i. 3-9. (a)Show that anunbounded setcannot have content 0. (b)Give anexample ofaclosed setofmeasure 0which does not have content 0. 3-10. (a)IfCisasetofcontent 0,show that theboundary ofChas content 0. (b)Give anexample ofabounded setCofmeasure 0such that theboundary ofCdoes nothave measure 0. 3-11. LetAbethesetofProblem 1-18. IfE,?_';1(b,; —at)<1,show that theboundary ofAdoes nothave measure 0. 3-12. Letf:[a,b]—>Rbeanincreasing function. Show that 1:0:fis discontinuous at:0}hasmeasure 0.Hint: UseProblem 1-30 to show that {:0:o(f,:0) >1/n} isﬁnite, foreach integer n. 3-13.* (a)Show that thecollection ofallrectangles [a1,b1] X---X [a,,,b-,,] with alla,;and b,;rational can bearranged inasequence. (b)IfAER"isany setand 6isanopen cover ofA,show that there isasequence U1,U2,U3,...ofmembers of6which also cover A. Hint: Foreach :0EAthere isarectangle B=[a1,b1] X ---X[a,,,b,,] with alla,-and birational such that :0EBEU forsome UE6. INTEGRABLE FUNCTIONS Recall that o(f,:0) denotes theoscillation offat:0. 3-7 Lemma. LetAbeaclosed rectangle andletf:A—-+Rbe abounded function such thato(f,:0) <8forall:0EA. Then there isapartition PofAwith U(f,P) --L(f,P) <8~v(A). Integration 53 Proof. For each :0EAthere isaclosed rectangle U,,, containing :0initsinterior, such that MU,,(f)—-mU,,( f)<8. Since Aiscompact, aﬁnite number U,,,, ...,U,,,, ofthe sets U,,cover A. LetPbeapartition forAsuch that each subrectangle SofPiscontained insome U,,,. Then MS(f)— m,g(f) <8foreach subrectangle SofP,sothat U(f,P) — L(f,P) =EslMs(f) —ms(f)l 'v(5’) <8"v(A)- 3-8 Theorem. LetAbeaclosed rectangle andf:A——>Ra bounded function. Let B={:0:fisnot continuous at Then fisintegrable ifand only ifBisasetofmeasure O. Proof. Suppose ﬁrst that Bhasmeasure 0.Let8>0and letB8={.0:o(f,.0) Z8}. Then BEEB,sothat Behas measure 0.Since (Theorem 1-11) BEiscompact, Behascon- tent O.Thus there isaﬁnite collection U1, ...,U,, of closed rectangles, whose interiors cover BE,such that Z§"=1v(U1) <8.LetPbeapartition ofAsuch that every subrectangle SofPisinoneoftwo groups (see Figure 3-1): W’ f ' F’ 7 11, i _ 1 ‘ I 1 3 ’ il 1 1;§.§=.iTi§i§.iiliii5EFEE§@§§i‘5§§E2;";z§i;§%§';%§aii31}?§"§§i;=.i;E§EifiE_§';?ii§?i‘:2§i5§i_5.§2ii5iiiiiiiiiiiii§.§5.{{§§§i‘§.i§~§i§§?_; EFifiiziﬁ§5552%?iii-i';§%§{3.';€§§i?'_,1§iiiE?§?ii§§E§§i§':';E{§¥.‘.§§§i=;@ if 1'-'_i§‘:3?:3:18.11%?iii‘,1?:1-.125211l-‘E331115:?Z5-1E153}353412111111EIE5171';£1i'§;1§IE5;-':f§fl?l1:§:E:?t31?:1If'I-SF-I-=Z5}§1:[:§'.§Iiii‘-7J§1§?§i1';I;" .-Iii;E?EIYIl1l.1':§§§I':Y':?§=ill‘-YI§3:l:51E: 51:1E1I'E??:1'.§i:2;?ii:EiiiIE£iI§§'i1§i':.f{-I =13 111:1:1i;-:5>11?I:I;I:1:1:1:1:1:;:1:21EI:1;I:5;E:1;f-.1F:I:511:3E111:'-:f;1:1:1:21-_r;I;.:I:1-.'-:1:I:3.1:?!;I;;}'-11521;‘-;I:i :-:;:;'-,1;I$:1;I;I-1;‘;-.3:-';§". .-;1:i:1-.'-:1:T:1:E:{25.115-;1;'-.-I:I:I;}'.E!1!;:;:;‘-:-:-:~'.;:;115:1I;1;'-.1:I:7$;';.’;r§:;:-_:;:. 1ji_1{:§:§:};;1:;:11}:Q:I_:};}:};:;<;:gi:11}1;:121:11]:[:§:;:1:;1;:-_1-3;I;§;§;§:;1;;;1;1;§g§:§;2_-.';:;:;;;:;;;:;:;:;;;1;:;:-E :§;;;;:;!;:';;:;f:l,-§:-' __.;:;:;::§::;:_:;1.;§:1:1:5.;;1:-5;;:;:_;};;:;:;1;' :5:Q:Q1}:;:-_7;;_:;:;;-1;:;:;.;';-;:;;;".;:;:;;- 1».¢.-.;.-_-_-;»:-.-.-.-.;.;.1;._-:-;._;_-:11;.;.-__;.;.;.;»;-:-:-:1:-;-:;;;.;.;-;-:-;-:-;-.;.;.-.;.;.;-;;-.;:g:1:-'.-:;:;:-_'-:- :-;-:-;-'.;:»:_:;:;.; -.-'.-:-_:-_'.g.:;:1:¢.;.;.;-;_;;:;:;1.;.1.;.;-;.;-;;- ;.;.-_.;.g.g.;-;;1.;.;.;;;;;;:;;.;.-_.;.;.;.;_; 1 l ’F551;3:;g?;i:_igi=,:§:1:;<%:§:;:f-53;E513$};‘5:1-:15:§<?;§r5;5;.f;;:3E§:;:§;5g2gig":323£1;:3:é:5;?;E;:;=;:§;;f'_E;¥;;;:g;;== ,:1;;5=;=1;§=;:;-.3;;:;:3:;r531.;5;5;g-.31;;;:;:~,:.;===_i.1;;;' = 1 i ’.-:;»":»'.~'.-'=1.Z"-'1'!I5:3E21:-'$15-2'-'51EF-"$1.115-'.‘-'£»':1":1'1:Fr?-'£I}r3'.~'.*:5ri'¥I£=lf5»';1.-I§;2-':-'i'.t"?-"1515-‘E-‘Er ‘1;.;-;-;»:¢:;:;.;-;-1-;-'.-1-;;:;;1.;.1.1.-_-:-:-1-1-.;;-;-;-:-:-:;:;:¢.1-;.:-;-:-:»:;-;-;-;-.-:-:-;-:-:;:;.;-:-5;:1.;.;-;-:-: _.;:;.:.<->;-;-:-:-:11;;.;.1-;-;--:-;-:-;-'.-:-L-1-.-.;-_-;-:-;-'.;:;f;.'1~:- 5: . 1I:I:11}!E1E11.151E11:11?:§;'»:EIE1E‘-EI521:I;I:I:i;F11:I:1:-viii1:1:1:1:l:I:'-:l:}:I11%;;I:1:1;I:l:;:§i{:}:1:';§:§:L1}1-' .;:}:j-111Eiiii‘-E1:!:KI1EiEi:1:1:-;I :1;1:f:-'11:1;1:I-.I:I13I';f:I:‘-;1:}:;}:;:;:;'. 11 f1}3-;ZI1]1Z:3§;f:§:§t;:11-,1;§:{:11};§1}:§;{:§;f:115111I;l:§:{:§L§'.}I§IlI-.1':5;¢;§:§: 1:121:1:§1';i;IE11'»;I:ii:{:§_1§iif:{'.f!:;§1'>1:l:1;§:§:§: 111;}; 1.511;I$3»;-1-;-‘:1-I;I£11:-1»1-;-:-2-L-I-:-;-:-1-2-2-;;I;'-1-1-1-:-;-:-L-1-1;i;l:-;-I 3.;.;i;-:-I-1-'.-_'.;:;i;-3-1-I-;-:->I-'.-1-:-1~I-7:l;'-:-:-L~;-I-i;'- w~ 1 1 1l1 i if 11 5 1 1'1 11 ‘1 1 , 31 1 11 1 3 1 ti 1 ‘ 11 1 ‘ 1 ‘ l FIGURE 3-1.The shaded rectangles arein51. 54 Calculus onManifolds (1)51,which consists ofsubrectangles S,such that SEU,- forsome i. (2)52,which consists ofsubrectangles Swith SH Be =g_ Let |f(:0)l <Mfor:0EA. Then M,g(f) —-m,g(f) <2M forevery S. Therefore ft Z[Mm-ms(f)l-»<S><2MZaw.)<21/ItSE§1 i=1 Now, ifSE52,then o(f,:0) <8for:0ES.Lemma 3-7 implies that there isareﬁnement P’ofPsuch that ZlMs'(f)-mao>1-»<s'><E~»<s>S’ S forSE$2. Then v<r.P'>-Lo.P’>-Z[Mam-’ms'(f)l~~<s'>S’(SE31 +ZlMs'(f)-'ms'(f)l~»<s'>S’(SE32 <2Mr-:+ Z8-v(S)SE52 52Me+8-t(A). Since Mandv(A) areﬁxed, thisshows that wecanﬁnda partition P’with U(f,P’) —-L(f,P’) assmall asdesired. Thus fisintegrable. Suppose, conversely, that fisintegrable. Since B= B1UB1UB,.U ---,itsufﬁces (Theorem 3-4) toprove that each B1”, hasmeasure 0.Infact wewill show that each B1,,, hascontent 0(since B1,,,iscompact, thisisactually equivalent). If8>O,letPbeapartition ofAsuch that U(f,P) —- L(f,P) <8/n. Letgbethecollection ofsubrectangles S ofPwhich intersect B1,,,. Then 5isacover ofB1,”. Now if Integration 55 SE5,then M,g(f) --mg(f) Z1/n. Thus 52 11(3) SElMs(f) -'’ms(f)l '"(S) SGS SES SElMs(f)'“' ms(f)l 'v($) S e . <-1 Tl andconsequently Z,g€gv(S) <8.I Wehave thus fardealt only with theintegrals offunctions over rectangles. Integrals over other setsareeasily reduced tothis type. IfCER”, thecharacteristic function X17 ofCisdeﬁned by -<r>=l‘i iii? IfCEAforsome closed rectangle Aand f:A—-> Ris bounded, then fgfisdeﬁned asf,1f- X0,provided f-X0is integrable. This certainly occurs (Problem 3-14) iffand X0areintegrable. 3-9 Theorem. Thefunction )((].'A-—->Risintegrable ifand only iftheboundary ofChasmeasure 0(and hence content O). Proof. Ifasisintheinterior ofC,then there isanopen rectangle Uwith asEUEC.Thus X0=1onUandXgis clearly continuous atas.Similarly, if:0isintheexterior ofC, there isanopen rectangle Uwith asEUER”—-C. Hence X0=0onUand X0iscontinuous atas.Finally, ifasisin theboundary ofC,then forevery open rectangle Ucontaining :0,there is11/1€Uf\ C,sothat )((}(y1) =1and there is ygEUfh (Rn —C),sothat x(;(y2) =O.Hence xgisnot continuous atas.Thus {:0:X61isnotcontinuous atas}= boundary C,andtheresult follows from Theorem 3-8. I 56 Calculus onManifolds Abounded setCwhose boundary hasmeasure 0iscalled Jordan-measurable. The integral fgl iscalled the (n-dimensional) content ofC,orthe(n-dimensional) volume ofC.Naturally one-dimensional volume isoften called length, andtwo-dimensional volume, area. Problem 3-11 shows that even anopen setCmay notbe Jordan-measurable, sothat fgfisnotnecessarily deﬁned even ifCisopen andfiscontinuous. This unhappy state ofaffairs willberectiﬁed soon. Problems. 3-14. Show that if_,".g: A->Rareintegrable, sois f-11- 3-15. Show that ifChascontent 0,then CEAforsome closed rectangle Aand CisJordan-measurable and IAX9=O. 3-16. Give anexample ofabounded setCofmeasure 0such that IAX0 does notexist. 3-17. IfCisabounded setofmeasure 0andL1X0exists, show that IAxg=0.Hint: Show that L(f,P) =Oforallpartitions P. Use Problem 3-8. 3-18. Iff:A-+Risnon-negative and IA)’ =0,show that {:r:f(:0) sf0} hasmeasure 0.Hint: Prove that 1:0:f(:0)>1/n} hascontent 0. 3-19. Let Ubethe open setofProblem 3-11. Show that iff=xv except onasetofmeasure 0,then fisnotintegrable on[0,1]. 3-20. Show that anincreasing function f:[a,b]—> Risintegrable on [a,b]. 3-21. IfAisaclosed rectangle, show that CEAisJordan-measurable ifand only ifforevery 8>0there isapartition PofAsuch that E3Eg,v(S) —E,g€g,v(S) <8,where 51consists ofallsubrectan- glesintersecting Cand52allsubrectangles contained inC. 3-22.* IfAisaJordan-measurable setand8>0,show that there isa compact Jordan-measurable setCEAsuch that f,1_g 1<6. FUBINI’S THEOREM The problem ofcalculating integrals issolved, insome sense, byTheorem 3-10, which reduces thecomputation ofintegrals over aclosed rectangle inR”,n>1,tothecomputation of integrals over closed intervals inR. Ofsufﬁcient importance todeserve aspecial designation, this theorem isusually referred toasFubini’s theorem, although itismore orlessa Integration 57 special caseofatheorem proved byFubini long after Theorem 3-10 was known. Theidea behind thetheorem isbest illustrated (Figure 3-2) forapositive continuous function f:[a,b] X[c,d] —-+R. Let to,...,t,,beapartition of[a,b] and divide [a,b] X[c,d] into nstrips bymeans oftheline segments {ti}X[c,d]. Ifgxisdeﬁned byg,,(y) =f(:0,y), then thearea oftheregion under thegraph offandabove {at}X[c,d]is tl 11 fca=ff(w.v)dv- The volume ofthe region under the graph offand above [ta-1.151;] X[c,d] istherefore approximately equal to (t,;-t1_1) 'fff(:0,y)dy, forany :0E[t¢__1,t,;]. Thus Tl lf=Z ff [a,b]X[c,d] i=1 [ti_.1,tt]XlC.dl isapproximately Z§"=1(t¢ —-t,_1) -f§lf(xi,y)dy, with 1131in graph off ._1;;->;-1.1::1;;-.=;;;.;':€;ls:3-1125.‘-."=::.1-51é,1.'-1:I;r2.11.'I,'5:%:=:-:e‘:.=_-.-,-.111.1-=..-.. ...l ;_'.;Ls:1;§r;;:;;;::::I;l;§:§;;:;:5;_:;i:§1;:;;;r::;!11:';".1:1:;25111;:551112;‘-;-;};§:';:;:;:1-;11;:-_:-_:;:;:;1;L;§1§;;'-- -¢-:-.-.;--:-:c-;-:-.-.-.-.--:-:~.'-1-1-:-:-:->.-.-.-.:-:-:-:~-1"'-:11:1}-Ii-I-7;‘.1};T;71111-1-1:1?‘.5;L‘1-':I;l;1:51'-I5?£-E5I-E§-1-1-51;‘-I311:-I-!-I-5§""12:»:1:;:51‘-:1;11:!1-1»;1:;15:;1;:;t;f13i;l;'-5:1:3:;I;k-:1:1;;?:;:k?:1§:5;l;l-.'-1-: '-13:31?11:1:I;T:Iriiil1'.11’-513:1:i:?;713;55'-1'-’;1:5:11133:1121:l:1?1l7§§:Ei:?-': -5.37:3ll1.11.723lllﬁii»ii:'1lIiii?lI1431:I:3IPI-1EifE1-.'<I11115;1:$111!:i:';1;3l5i?'.T:ll 3'211';I§'.;§:§; <=r?1E1%=;=:r.'-E?=':' 2'E=ifE£=?13=%=:1:==1"'--i=1“ =;;';:'1:§=:1-.'-;3;;;:g~g 1'.Y:i-:E:}:1- 1.;;;1§;§I':I .-:41-i~i~---:-l3171i?§-':3l7:7;113:lI' .575‘..-tltli-151117733;-1-:-1:.‘-L ---1- 112ii5.13781121Er¥l§f?1i:i1¥:E:=I1"",..:‘E1§:ErE1EIii?!§€?l1:1f{IiI':i§?:-1. PI :1:]111:T:l:}:]:1f§?l1li§Z'-:3<i';7:1lii1:lt :Iiiii:1:§1]:§1SEi§E??i:'-:}:§iEIZI€I:_'1 §;§:;_3_'1-.2111§:§:§t§;§§l§;1;:Z§;'_:_§::§;g.;:§;;'g1]‘{1_6.:52]:§1§ﬁ:§:{:?1§1§§I§;EI:1L1i?;1:T;1:Il'.l1l<l*;1;>.T;k7<:EkI:i:-.=.;-'.-'.'.-'.-'.-.-.-',-,-:-'.-:-.-,>.=.¢:-:=.11:1;-:11'l<1*=:5:1.';.1;!-111:3!-1'1-I-7:5:§:;:;'1:E§'4'.:::;:::}:;;;;1'-:3:E':=:1$:§;1:1:E:§!ll;I§I;l.‘-.'-.I_Q.-,I;_T_'$j5;.I_-.-,-.-1:-1-1-:5-I-2.:-.;.~'.-I-2-.-.1. -.- -.-_.;.1.;-;-;-.-.1_--:-:-.-.-.;.;.g.1-:2;:;.;i!1:;:;:;:;.; .;.;1;-1-1»:-!-:;!;.:k»;»:-L;.;»'" '.:l>111-L:1-I:i1-1-L-if->'I:I:1:1:-i1;I' :-1-1-1»1'2/1:I:11};-l;I:l$:-;-:-:-I-F _2:E:Ei1':I:?:i:EI': :E:E>E1Ei1';lE1:1:E;§'-" -;-Vi? E=ii:1:§r1rEiE:E1:111=?:1:}:1>E:E=:=<§:' Iiii?I¥<:1:1r&:?rE:E1-1*ifEfEIEl:115.'~:1 ;1:1;3:§1¥'.'»'.i:i:3 ;l'.T;i:§'Z1?1-f-.- -1f1>.1:l:?-_f:['.T; :'-:?:3:;'-;'»_'a:i1l-§:-:§$,t§I;1:i:1'}1:f:1: .1211?11:1;1:f?7:I:?ki:1:1:11311113...1,-__>(_._-.-,. ..--‘-1.11;->1-L-L‘? 11'-.>.' -I'l']‘-'-' -->1!‘-'-'-‘.-I-1'1‘?-'1 --I'll‘I‘.12-7I'.‘l'l'l'-\'-'-'>Z>}1'\'»‘- 'I-1'I'l'l'-‘-'+I'I'-')-1-1'l'T'i’1'- '';-.'"I-1'-:-:-1-1-1-1:if<f:i:-1-$;T:1:-:-:11’ .-111511:-:-1 :-:-:-'..-11:4-$11:-:<-:-IF:1;1=-' 11:1:-'-1-\>:-1-:-:;1:'.1k3:'-:-:-:;;l:1 1-1-PI:F215:-:-;-1:3:-1-:-1-111:1:.'-'15-:'=.'>'?' i"‘1t'-:35if-'-I-1'-f-5'5?"-if55-1'-3i»i-iii»iii’-15-if-if-I’-:1?'-.:--'-I524-L-1-‘.12 11:1F:-I-1-I-1-1:-1-3-1-1-I-5-1\:*:4’;-2-1-1-:-;-i-I-I-1;>;-:-'.-;-:;-:-: 11-1-1‘-2-'.-:-:-:-:--5-1-:¢~'.»;-1---: ' =1. . . I:1:5:f:11ET§:E5.=?1iiliiifiifi-.i:f:i:YI§ $5115:5IE1?iE$:l:f1§:1'.EI§iEi1:EiiIl‘ EIFFIEIE1}ii!F5.‘-E1¥3':'-ET:l:!:=:=:VV1 -: -1-Li-.-1-.1-1-1->2.;.'.~:-1-I-I-1-1-1',;I.;.!.1-.;.;-L-I-I-I-L-1-i;.;1;-:-I-:-.;-:-::>2-L-1'?‘-;-L-2-1-I--;4-bl-L-I-1-1'-III 1:311:17-';1;!;I:1:iI11:T:?;T:l'.l:-11:15, .-II$5151:$<I:7:':»:;:;:;l-,'-;1:T:1;I=;3:3I'-13:‘-1;:-I;111:i:l;'-:Z3L1:I:?:5:1:};}:+1‘ '-'--11;:;i:!:-!';;I1f:i:kl:-:-_ ;3:;1;-:;;;:-_:;::::-§:I::1-_-;;:§-.-,:- _I:::I;-:-Q1-::f;1;:-I;2;?-_I:K:T:l1:- ‘ .-».-1 -I.-'.=.»".'I=.»'.i~:-:-1‘-‘- -‘."-'-i5;-8:»11*.=.i-:'.~'/:»;-'-1f-7-= ----1-1-;-:1:-. .;-:-;-.;.;.;-1-;-1':;:;.;.;.;4.i:.;.;- :-1-:9.;.;.;.;-11--,-;-:-1-:~1-:-.1-- 1 :-1»;-1;-;-:-:-:-I-1;.1-1-;-1-:->1-1:5-1 -1-1;-1.11-st:-1.:-2;-1.:-:-1:-:;!;.;.- :-: 11 : 1 1 -'-T5-:1:1;:-1-:-I-'-:1:-:-1 =:2‘-:1:?:-:-:-:~1-I:i;~:-I:I:l:i:?;1;»;-= -‘ 1 = 1 -1-1;I:I-.'f:;::§:§' ==.51:;2:115I}:;:§:E:;;;}:§:-;1:;:1;§;§;I ' , :''1311121§§5L11'-;]:§1,1-_1;1;I: : ‘1 in 11 '- '- 1 , 1‘ lll/1‘ 11 __L_____ 11 , /// /1 ///' //// / /1 t,._1 I t. b FIGURE 3-2 58 Calculus onManifolds [t,-.__1,t.,;]'. Ontheother hand, sums similar tothese appear in thedeﬁnition off3(ffff(x,y)dy)dx. Thus, ifhisdeﬁned by h(x) =-ffgx =ff,lf(a:,y)dg, itisreasonable tohope that his integrable on[a,b]andthat f=[bh=[b(fdf(x,y)dy) dx. This willindeed turn outtobetrue when fiscontinuous, but inthegeneral case diﬁiculties arise. Suppose, forexample, that thesetofdiscontinuities offis{:00} X[c,d] forsome xoE[a,b]. Then fisintegrable on[a,b] X[c,d] buth(a:0) = f‘,ff(x0,y)dg may not even bedeﬁned. The statement of Fubini’s theorem therefore looks alittle strange, andwillbe followed byremarks about various special cases where simpler statements arepossible. Wewillneed onebitofterminology. Iff:A——>Risa bounded function onaclosed rectangle, then, whether ornot fisintegrable, theleast upper bound ofalllower sums, and thegreatest lower bound ofallupper sums, both exist. They arecalled thelower and upper integrals offonA,and denoted[a,b]X[Mil Llf and Ulf, respectively. 3-10 Theorem (Fubini’s Theorem). Let ACR"and BCRmbeclosed rectangles, andletf:AXB——>Rbeintegrable. ForasEAletgx:B-->Rbedeﬁned bygx(y) =f(x,g) andlet £(¢v)=LIat=LJf(-'r,y)dy,B as)=U!9.=U!f<x.y>dy. Then .6and‘ILareintegrable onAand r=[s=f(LJf<x.wdy)da=.A A f=]<11=f(U!f(x,y)cly) dx.AAXB AXB A Integration 59 (The integrals ontheright side arecalled iterated integrals forf.) Proof. LetPAbeapartition ofAandPBapartition ofB. Together they give apartition PofAXBforwhich any subrectangle Sisoftheform SAXSB,where SAisasub- rectangle ofthepartition PA,andSBisasubrectangle ofthe partition PB.Thus La,P>=Z'm»s(f) -as)=ZmS..><S.<r> 'v(SA><soS S.4,Se =Z(Zm/S.4><SB(f) ~v<sB>) -v<s..>.S4 SB Now, if:1:ESA, then clearly m,g,,XS,,(f) §m,gB(g,,). Conse- quently, foratESAwehave ZmS..><S.<r> ~v<sB>5Zmm.) ~v<sB>sLfg.=so).SB Se B Therefore Z(ZmS.><S.<r> 'v(SB))~~<s..>sL<aP.o.S4 Sn Wethus obtain 3L(£2PA) gU(°B;PA) SU(clL;PA) S where theproof ofthelastinequality isentirely analogous totheproof oftheﬁrst. Since fisintegrable, sup{L(f,P)} = inf{U(f,P)} =JAXBJ". Hence $\1PlL(=9,P~A)l =i11flU(=9,PA)l =_lA><Bf- Inother words, J3isintegrable onAandIAXBJ" =IA13. The assertion for‘llfollows similarly from theinequalities Remarks. 1.Asimilar proof shows that = (LA[f(a:,y)dx) dg--=3] (Ujf(x,y)dx) dy. £2miM. cs-A 60 Calculus onManifolds These integrals arecalled iterated integrals forfinthereverse order from those ofthetheorem. Asseveral problems show, thepossibility ofinterchanging theorders ofiterated integrals hasmany consequences. 2.Inpractice itisoften thecase that each gxisintegrable, sothat fA><Bf =IA(fBf(:z:,y)dy)d:r. This certainly occurs iffiscontinuous. 3.Theworst irregularity commonly encountered isthat g, isnotintegrable foraﬁnite number ofacEA. Inthiscase £(a:) =fBf(x,y)dy forallbutthese ﬁnitely many as.Since IAJ3 remains unchanged ifJ3isredeﬁned ataﬁnite number of points, wecan still write fAXBf =fA(fBf(x,y)dy)dx, pro- vided that fBf(:c,y)dy isdeﬁned arbitrarily, sayas0,when it does notexist. 4.There arecases when thiswillnotwork andTheorem 3-10 must beused asstated. Letf:[0,1] X[0,1]——>Rbedeﬁned by 1 if:0isirrational, 1 ifccisrational and yisirrational, f(a?/) = .__ . .1-1/q 1f:0-p/q1nlowest terms and y1s rational. Thenfis integrable andf[0,1]><[0,1] f=1.Now f},f(x,y)dy =1 ifa:isirrational, anddoes notexist ifatisrational. There- fore hisnotintegrable ifh(x) =f§,f(x,y)dg issetequal to0 when theintegral does notexist. 5.IfA=[a1,b1] X'''X[a,,,b,,] andf:A—+Rissuf- ﬁciently nice, wecanapply Fubini’s theorem repeatedly to obtain /..f=ti"(~~~(Lire, ---we‘)~~")e="~ 6.IfCCAXB,Fubini’s theorem canbeused toevaluate fgf,since thisisbydeﬁnition IAXB xcf. Suppose, forexam- ple,that 0=[-1,11><[-1.11- {<x.y>=!a.t/>1 <11. for=ff,(ff,fat)'xc($v,y)d1l/)dIv-Then Integration 61 Now 1 if;/>\/1-x2org<-\/1-xi, xC(x’y) = 0 otherwise. Therefore [__11f(37»y)'XC(x;?/)d.7/ =f_',”"""ro,y>dy +/\:,T,,ra.y>dy. Ingeneral, ifCCAXB,themain difficulty inderiving expressions for fgf will bedetermining Cf\({x} XB) foracEA. IfC’f\(AX{g/}) foryEBiseasier todeter- mine, oneshould usetheiterated integral [Cf=/B(/4f(rv,2/) 'xo(w,y)dr¢) dy- Problems. 3-23. Let CCAXBbeasetofcontent 0.Let A’CAbethesetofallscEAsuch that {yEB:(a:,y) EC}is notofcontent 0.Show that A’isaset ofmeasure 0.Hint: )((7iS integrable andIAXB xg=_lA‘l1 =fA.B, so_lA‘11 —£=0. 3-24. LetCC[0,1] X[0,1] betheunion ofall{p/q} X[0,1/q], where p/qisarational number in[0,1] written inlowest terms. UseC toshow that theword “measure” inProblem 3-23 cannot be replaced by“content.” 3-25. Useinduction onntoshow that [a1,b1] X---X[a,,,b,,,] isnota setofmeasure 0(orcontent 0)ifat<b,;foreach i. 3-26. Letf:[a,b]-—> Rbeintegrable and non-negative and letA,--= {(:r,g): a3:1:$band03g$f(:c)}. Show that A;isJordan- measurable andhasarea f. 3-27. Iff:[a,b]X[a,b]-+Riscontinuous, show that Lb/.ayf(rv.y)drv do=jajxf(1=.y)dyd=v- Hint: Compute fgf intwo different ways forasuitable set CC[a,b] X[a,b]. 3-28.* UseFubini’s theorem togive aneasy proof that D1,;-4f =D2,1f ifthese are continuous. Hint: IfD;,2f(a) —D2,1f (a)>0, there isarectangle Acontaining asuch that D1,2f —D2,1f > 0onA. 3-29. UseFubini’s theorem toderive anexpression forthevolume of a.setofR3obtained byrevolving aJordan-measurable setinthe yz-plane about thez-axis. 62 Calculus onManifolds 3-30. LetCbethesetinProblem 1-17. Show that [[04,([[0,1]xc(=m/)d-'0) do=LO,“(Loin xo(y.rv)d:¢/)d1= =0 but that _[[0,1]X[0,1] xgC1068 11015 8XiSt. 3-31. IfA=[a1,b1] X---X[a,,,b,,,] and f:A-> Riscontinuous, deﬁne F:A—->Rby FCC) =ﬁa1.:n‘]>< X[an,:n"] f' What isD.-F(a:), for:1:intheinterior ofA? 3-32."’ Letf:[a,b]X[c,d]-->Rbecontinuous andsuppose Dgfiscon- tinuous. Deﬁne F(y) =ff’,f(a:,y)da:. Prove Leibnitz’s rule: F’(y) =ll1D2f(=m/)dw- Hint: F(z/)=lZf(w.:1/)d~'v =ff-.’.(.l'tD2f(rv.:1/)dy + f(x,c))da:. (The proof will show that continuity ofDgfmay be replaced byconsiderably weaker hypotheses.) 3-33. Iff:[a,b]X[c,d]—>Riscontinuous andDgfiscontinuous, deﬁne F(w.r)-f§f(tu)d¢-(a)Find D1F andDQF. (b)Iran) =fﬂ(“’)f(t,x)dt, ﬁnda'(=.=). 3-34."‘ Letg1,g2: R2-> Rbecontinuously differentiable and suppose D192 ==Dggl. AsinProblem 2-21, let ray)-g1<».0>a +L,”ya-.:>d¢. Show that D1f(a:,g) =g1(:c,y). 3-353" (a)Letg:R"—>R"bealinear transformation ofoneofthefol- lowing types: {g(a) =1% '5#1 g(a)=ﬂea‘ {g(e,-) =at '575.7. g(a)=er+eh g(a)=er 9e"=er- IfUisarectangle, show that thevolume ofg(U) isIdetg|-v(U). (b)Prove that Idetg]-v(U) isthevolume ofg(U) foranylinear transformation g:R"-+ R". Hint: Ifdetgas0,then gisthe composition oflinear transformations ofthetype considered in(a). 3-36. (Cavalieri’s principle). LetAandBbeJordan-measurable sub- setsofR3. LetAc={(:c,y): (x,g,c) EA}anddeﬁne BCsimilarly. Suppose each AcandBCareJordan-measurable andhave thesame area. Show that AandBhave thesame volume.{g(a) =en k¢i,.7' () Integration 63 PAR TITIONS OF UNITY Inthissection weintroduce atoolofextreme importance in thetheory ofintegration. 3-11 Theorem. LetACR"andlet0beanopen cover ofA. Then there isacollection ofC°°functions <pdeﬁned inanopen setcontaining A,with thefollowing properties: (1)Foreach scEAwehave 0§<p(x) §1. (2)ForeachccEAthere isanopen setVcontaining scsuch that allbutﬁnitely many toEare0onV. (3)Foreach asEAwehave Z,,€.1,<p(x) =1(by(2)foreach :1; this sum isﬁnite insome open setcontaining ac). (4)Foreach 10E<11there isanopen setUinOsuch that1,0=0 outside ofsome closed setcontained inU. (Acollection satisfying (1)to(3)iscalled aC°°partition of unity forA. Ifalso satisﬁes (4), itissaid tobesub- ordinate tothecover 0.Inthis chapter wewillonly use continuity ofthefunctions 10.) Proof. Case 1.Aiscompact. Thenaﬁnite number U1,...,U.,ofopen setsinOcover A. Itclearly suﬁices toconstruct apartition ofunity subordinate tothecover (U1, ...,U,,}. Wewill ﬁrst ﬁnd compact sets D1CU1;whose interiors cover A. The sets D1;arecon- structed inductively asfollows. Suppose that D1,...,D1, have been chosen sothat {interior D1, ...,interior D1,, U1.+1, ...,U.,} covers A. Let 0114.1: A-"' ''' Uk+2U ''‘ Then C1.+1 CU1,+1 iscompact. Hence (Problem 1-22) wecan ﬁndacompact setD1,+1 such that Ck+1 C interior Dk+1 and D];+1 C Uk+1. Having constructed thesetsD1,...,D,,, let111,;beanon- negative C°°function which ispositive onD1and0outside of some closed setcontained inU1(Problem 2-26). Since 64 Calculus onManifolds {D1, ...,D,,} covers A,we have ¢1(x) +---+¢,,(:c) >0 forall:1:insome open setUcontaining A. OnUwecandeﬁne .(x)_--___‘h(x) _. “"'mo+--~+1!/..(rv) Iff:U——>[0,1] isaC°°function which is1onAand0outside ofsome closed setinU,then ={f'101,...,f-<p..}isthe desired partition ofunity. Case 2.A=A1UA1UA3U ---,where each A.is compact and A,Cinterior A.~+1. Foreach ilet0,;consist ofallUf)(interior A,;+1 -—A,-_2) forUin0.Then 0.isanopen cover ofthecompact set B.=A.—-interior A.;-1. Bycase 1there isapartition ofunity .forB1,subordinate to0..Foreach :1:EAthesum <10?)=Z ¢(Iv)¢,oEt,a.lli isaﬁnite suminsome open setcontaining x,since ifccEA,;we have ¢>(:t) =0for,0E,-with jZi+2.For each ,0in each ,;,deﬁne ¢>'(:c) =<p(17)/0'(12). The collection ofall<p'is thedesired partition ofunity. Case 8.Aisopen. L613 A; = {atEA: _§iand distance from :ctoboundary AZ1/i}, and apply case 2. Case 4.Aisarbitrary. LetBbetheunion ofallUin0.Bycase 3there isapar- tition ofunity forB;thisisalsoapartition ofunity forA.I Animportant consequence ofcondition (2)ofthetheorem should benoted. LetCCAbecompact. Foreach scEC there isanopen setV,containing scsuch that only ﬁnitely many goEarenot0onVx. Since Ciscompact, ﬁnitely many such V1,cover C. Thus only ﬁnitely many <pE<1>are notOonC. One important application ofpartitions ofunity willillus- trate their main role—--piecing together results obtained locally. Integration 65 Anopen cover 0ofanopen setACR"isadmissible if each UEOiscontained inA.Ifissubordinate to0, f:A-—>Risbounded insome open setaround each point ofA, and lac:fisdiscontinuous atx}has measure 0,then each fAgo-lflexists. Wedeﬁnef tobeintegrable (intheextended sense) if2,5,1, fAgo-lflconverges (theproof ofTheorem 3-11 shows that theqo’smay bearranged inasequence). This implies convergence ofZ,,,E.1.lf4 to'fl,andhence absolute con- vergence of2,51. fAto-f,which wedeﬁne tobeIAf.These deﬁnitions donotdepend on(9or(but seeProblem 3-38). 3-12 Theorem. (1)If\I/isanother partition ofunity, subordinate toanadmis- sible cover 0'ofA,then 21,51, IA1,0'lflalso converges, and ¢E-4P4i.‘G'-'1. (2)IfAandfarebounded, thenfisintegrable intheextended sense. ' (3)IfAisJordan-measurable andfisbounded, thenthisdeﬁni- tion ofIAfagrees with theoldone. Proof (1)Since go-f=0except onsome compact setC,andthere areonly ﬁnitely many 1/1which arenon-zero onC,wecan write gbft-r-Z] Ex!/we-f=Z 2j¢'r'f-r A ¢E‘1> ¢/G1’ ¢E'I>~I»E‘I' This result, applied tolfl,shows theconvergence of2,51. E¢6q,IA #1'<19' and T161106 OfZ¢E¢E¢Eq,l_lA IL'go This absolute convergence justiﬁes interchanging theorder ofsummation intheabove equation; theresulting double sum clearly equals Z,,,G.1,_lA glx-f.Finally, this result applied tolflproves convergence ofZ151, fA11/-lfl. 66 Calculus onManifolds (2)IfAiscontained intheclosedrectangle Bandlf(:c)l5M for:1:EA,andFCisﬁnite, then gp/<p'lfl.€ ZM/e=M[ Z¢SM'"(B).v A ¢€F ¢€F since 2,511 to§_1onA. (3)If6>0there is(Problem 3-22) acompact Jordan-meas- urable CCAsuch that fA__g1 <E.There areonly ﬁnitely many <pE¢I>which arenon-zero onC.IfFC isanyﬁnite collection which includes these, andfAfhas itsoldmeaning, then f- ¢'fl£jlf"¢;F¢'fl es/<1-21¢)¢€F =MJ¢€Z_F¢_§MA!C1_§Me.|lg? ‘Gfl)$4INC. Problems. 3-37. (a)Suppose thatf:(0,1) -->Risanon-negative continuous function. Show that f(t1,1)f exists ifand only if limo_l'§“"’f exists.E-9 (b)LetAn=[1-1/2", 1—1/2"+1]. Suppose thatf: (0,1)-> R satisﬁes _lA,f =(--1)"/n andf(x) =0fora:EanyAn. Show that f(0,1)f does notexist, butlimf(,1,1_e) f=log2. e——>0 3-38. LetAnbeaclosed setcontained in(n,n+1). Suppose that 1;R->Rsatisﬁes f,.1,,f-=(—1)”/n and1=0for:1:EanyA... Find twopartitions ofunity and\I1such that E,,e.1.fR to-fand E,1,E.1.fR 4/-fconverge absolutely todifferent values. CHANGE OF VARIABLE Ifg:[a,b]-—> Riscontinuously differentiable andf:R-—> R iscontinuous, then, asiswell known, a(b) b [f=/(fee)-eh0(0) 0 Integration 67 The proof isvery simple: ifF’=f,then (FOg)’=(f0g)-g’; thus theleftside isF(g(b)) -F(g(a)), while theright side is F=>9(5)-FOg(a)=F(g(b)) -~F(g(a))- Weleave ittothereader toshow that ifgis1-1,then the above formula canbewritten /f= /f°o'|e'l-o((ﬂ.b)) (a,b) (Consider separately thecases where gisincreasing andwhere gisdecreasing.) Thegeneralization ofthisformula tohigher dimensions isbynomeans sotrivial. 3-13 Theorem. LetACR"beanopen setandg:A-—>R" a1-1, continuously differentiable function such that detg’(x) 750forall:1:EA. Iff:g(A) ~—>Risintegrable, then 53.E25M.-lasoldett'|- Proof. Webegin with some important reductions. 1.Suppose there isanadmissible cover 0forAsuch that foreach UE0andanyintegrable fwehave [1-f<r~t>Idett'l~e(U) U Then thetheorem istrue forallofA. (Since gisauto- matically 1-1inanopen setaround each point, itisnotsur- prising that thisistheonly part oftheproof using thefact that gis1-1onallofA.) Proof of(1). The collection ofallg(U) isanopen cover of g(A). Letbeapartition ofunity subordinate tothiscover. Ifgo=Ooutside ofg(U), then, since gis1-1,wehave (to-f)Og 68 Calculus. onManifolds =0outside ofU.Therefore theequation f¢'f=lie-r>=»t1l<:1stt'l.o(U) canbewritten /¢'f=[to-r>»t1Idett'l»9A) A Hence 9-.5*'-. -Z[(¢°9)(f°9)|d@t9'|¢E<l*A =[<r~olderg’l-A Remark. The theorem also follows from theassumption that if=f(f°o)lde1> o’l0"‘(V) forVinsome admissible cover ofg(A). This follows from (1) applied tog‘“1. 2.Itsufﬁces toprove thetheorem forthefunction f=1. Proof of(2). Ifthetheorem holds forf=1,itholds for constant functions. LetVbearectangle ing(A) andPapar- tition ofV.Foreach subrectangle SofPletfsbethecon- stant function 7Tt3(f). Then L<r.P>=2ms(f) us)-Z[fsS SintS =2 f(fs°9)ldel59'SZ /(f°o)|dete’|Sg"1(int S) Sg'1(int S) 5I(feo)|d@1'» g’l-a‘1(V) Since fvfistheleast upper bound ofallL(f,P), this proves that fvf3fa-11v, (fOg)ldet g’l. Asimilar argument, letting 1,,==M,(f), shows thatfer3f,-.,V,(fs g)ldetg|.The result now follows from theabove Remark. Integration 69 3.Ifthetheorem istrue forg:A—>R"andforh:B—>R”, where g(A) CB,then itistrue forh<>g:A—>R". Proof of(3). [1-[1-([o<=h>|de1=h'l) h=o(A) h(0(A)) 0A -f[coh>Qti-[Idsh'|Oti-lastt'|A =[fa(h<=o)|d<->11 (h<=o)'|-A 4.The theorem istrue ifgisalinear transformation. Proof of(4). By(1)and(2)itsuffices toshow foranyopen rectangle Uthat [1=Jldet g’l. 0(U) This isProblem 3-35. Observations (3)and(4)together show that wemay assume foranyparticular aEAthat g'(a) istheidentity matrix: in fact, ifTisthelinear transformation Dg(a), then (T"1 Og)’(a) =I;since thetheorem istrue forT,ifitistrue forT_10git willbetrue forg. Wearenow prepared togive theproof, which preceeds by induction onn.The remarks before thestatement ofthe theorem, together with (1)and (2),prove thecase n=1. Assuming thetheorem indimension n-—1,weprove itin dimension n.Foreach aEAweneed only ﬁndanopen set Uwith aEUCAforwhich thetheorem istrue. Moreover wemay assume that g'(a) =I. Deﬁne h:A—> R" byh(x) =(g1(:c), ...,g"“1(:c),a:”). Then h’(a)=I.Hence insome open U’with aEU’CA, the function his1-1 and deth'(x) #5O.We can thus deﬁne lc:h(U’)—> R”byh(x) =(x1,...,:v"_1,g"(h_1(:c))) andg=It<>h.Wehave thus expressed gasthecomposition .;._,-.-;.,..|._-I-,.,\. .-.-.-.-:,-.\ ..-1.->5?:1?-3':-;-rt-r< '-:‘-.’-:1. . 1}l\_L-I\;.- .. .\ I.:::;_;l,v,%,:,E,E:52:::L,:::‘:._._;_;.;.¢.:_>._._._._ .. 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'\=;T:=.*.~.:';:,:-_.~_;; :4»;-1.-1.‘:-1_..'<.:-.;.~.£»:=.a-'.<. =.I.-.<-2st».-::-‘.-:>§::-:-:--; -'-.-.->1-E"-.-'~\' 1*..-:..-:',:-:-:-r-:.;.:--.-1-:-.:-:» -/-- 70TA]::;:;:-;;:55;:E:1IE!3I-.'§i5:§:»':§i:-§?:?,1-' -/3-3 I Integration 7I oftwo maps, each ofwhich changes fewer than ncoordinates (Figure 3-3). Wemust attend toafewdetails toensure that Irisafunction oftheproper sort. Since (9"’°h'1)’(h(<1)) =(9”')'(a) "[h’(<1)l”1 ==(9")’(@). wehave D,,(g'"’¢>h”])(h(a)) =D,,g'”’(a) =1,sothat h’(h(a)) =I.Thus insome open setVwith h(a) EVCh(U'), the function his1-1and deth’(:c) #O.Letting U=h_1(V) wenow have g=hoh, where h:U—+ R”and It:V—+ R" and h(U) CV. By(3)itsuffices toprove thetheorem forh and h.VVegive theproof forh;theproof forIrissimilar and easier. Let WCUbearectangle oftheform DX[a,,,b,,|, where Disarectangle inR"_1. ByFubini’s theorem /1- 1ds1-~-d.s"*1)ds". h(W) lambﬂl h(D>< lwI) Let h_.,-»»: D—-> R'”"'1 bedeﬁned byh,,»(;c1, ...,:c'”"1) =- (g1(;c1, ...,x"), ...,g'”""1(:c1, ...,;c'”')). Then each h,,n isclearly 1-1and det(h,,n)’(;c1, ...,;c'”’_1) =deth’(:c1, ...,:v"') #0. Moreover I 1d;c1- -'d;c'”"'l= [1d;c1~--d."c"'_1. 11(1)><lx"1) hs"(D) Applying thetheorem intheease n-1therefore gives f1= ids‘---d;c'”’_1)d;c" h(W) [an,b11] = et(hxs)’(.r1, ...,:v"’_1)ld:c'- --da:'”’_1)d.i:”' Z‘"s"/'5/'3C.Q$$§_/"\3 = ldeth'(;c1, ...,;c")ld;c] ---d;c”_1)d:c"’ 1) = deth'l. I The condition detg'(:c) ;é0may beeliminated from the 72 Calculus onManifolds hypotheses ofTheorem 3-13 byusing thefollowing theorem, which often plays anunexpected role. 3-14. Theorem (Sard’s Theorem). Letg:A——>R"becon- tinuously diﬂerentiable, where ACR”isopen, and letB= {asEA:detg'(:c) =0}. Then g(B) hasmeasure 0. Proof. LetUCAbeaclosed rectangle such that allsides ofUhave length l,say. LetE>0.IfNissufficiently large andUisdivided intoN"rectangles, with sides oflength l/N, then foreach ofthese rectangles S,if:cESwehave |1>to=>o -so-to)-to)!<sis-tl5e~/E<1/N) forallyES.IfSintersects Bwecanchoose :1:ESF)B; since detg’(a:) =0,theset{Dg(a:) (y—x):yES}liesinan (n—1)-dimensional subspace VofR”. Therefore theset {h(t)-g(x):yes}lieswithin e\/6(t/N)ofV,sothat {g(y):11es}lieswithin e\/Z(l/N)ofthe(11.-1)-plane V+9($v). Ontheother hand, byLemma 2-10 there isa number Msuch that Ito)-ta)!<Mix-slsMvio/N). Thus, ifSintersects B,theset{g(y): yES}iscontained in acylinder whose height is<28\/h(l/N) andwhose base isan (n-1)-dimensional sphere ofradius <M \/h(l/N). This cylinder hasvolume <C(l/N)"e forsome constant C.There areatmost N"such rectangles S,sog(Uf\B)liesinasetof volume <C(l/N)" -6-N" =Cl"'8.Since this istrue for alle>0,thesetg(Uf\B)hasmeasure 0.Since (Problem 3-13) wecancover allofAwith asequence ofsuch rectangles U,thedesired result follows from Theorem 3-4. I Theorem 3-14. isactually only theeasy part ofSard’s Theorem. The statement andproof ofthedeeper result will befound in[17], page 47. Problems. 3-39. UseTheorem 3-14toprove Theorem 3-13without theassumption detg'(a) #50. Integration 73 3-40 3-41Ifg:R"—> R"and detg'(a) as0,prove that insome open set containing :2:wecanwrite g=T0g,,0---<>g1,where g,;isof theformg.,;(:r)=(s1,...,f,;(:i:), ...,:c"),sheT'isalinear transformation. Show that wecan write g=g,,=> ---=>g1 if andonly ifg'(a) isadiagonal matrix. Deﬁne f:{r:r>0}X(0,2rr) —>R2byf(r,6) =(rcos6,rsin6). (a)Show that fis1-1, compute f’(r,6), and show that detf’(r,6) ;-60forall(r,6). Show that f({r:r>0}X(0,211-)) is thesetAofProblem 2-23. (b)IfP=f_1, show that P(:c,y) =(r(:c,y),6(:z:,y)), where mt:/>=\/e2+t’.arctan y/:1: 2:>0,y>0, rr+arctan y/:c :1:<0, 6(:c,y) = Zrr+arctan y/:1: :1:>0,y<0, rr/2 :1:=0,y>O, 3rr/2 :1;=0,y<O. (Here arctan denotes theinverse ofthefunction tan: (-11-/2,1r/2) —>R.) Find P’(:r,y). The function Piscalled thepolar coor- dinate system onA. (c)LetCCAbetheregion between thecircles ofradii r1and T2andthehalf-lines through 0which make angles of61and01with the:1:-axis. Ifh:C—>Risintegrable andh(:c,y) =g(r(:c,y),6(:r:,y)), show that 1'292 (/_[h =Ifrg(r,6)d0 dr. T 31 IfB,={(:c,y): 1:2+y23r2}, show that 1'21' [It =I[rg(r,0)d0dr. Br 00 (d)IfC,=—r,r] X[—r,r], show that -fax,-1Q_(”'+”’) dz:dy=rr(1—e_”) and T Ie—("”’+”’) dz:dy= e—”'d:c)2. Cr -1’ ’ (e)Prove that lim Ie""“’+”’) dxdy=limfe"'(“”+"’) dz:dy T-YQ Br T-Pﬁ Cr 74 Calculus onManifolds andconclude that i Q [e"""dz:= “Amathematician isonetowhom thatisasobvious asthat twice wemakes four istoyou. Liouville wasamathematician.” -—-Loan KELVIN 4 Integration onChains ALGEBRAIC PRELIMINARIES IfVisavector space (over R),wewilldenote theIt-fold product VX---XVbyVk. Afunction T:V"-—> Ris called multilinear ifforeach iwith 13i3Itwehave T(o1, ...,0,-+ 0,-’,...,o;,)=T(o1, ...,0,-,...,v1,) +T(o1, ...,o.;’,...,o1,), T(o1, ...,ao,, ...,o;,)=aT(o1, ...,2),-,...,o;,). Amultilinear function T:Vk—->Riscalled ah-tensor onV and thesetofallIt-tensors, denoted 5"’(V),becomes avector space (over R)ifforS,TE5"(V) andaERwedeﬁne T)(2)1, ...,7/'11,) =S(?/*1, ...,2/111;) +T(t/'1, ...,7);,;), (aS)(o1, ...,o;,)=a-S(o1, ...,v1,). There isalso anoperation connecting thevarious spaces 5"(V). IfSE5"(V) and TE’Jl(V), wedeﬁne thetensor product ssT6:>'*=+‘(v) by S®T(?/*1, ...,b';1,,?);¢_l_1, ...,b‘;,_l_l) mS(t/'1, ...,2/'15) °T(f/‘k_l_1, ...,t)k_l_[). 75 76 Calculus onManifolds Note that theorder ofthefactors SandTiscrucial here since S®Tand T®Sarefarfrom equal. The following prop- erties of®areleftaseasy exercises forthereader. (S1-I-S2)®T=S1®T-I-S1®T, (aS)®T=S®(aT)=a(S®T), (S®T)®U=S®(T®U). Both (S®T)®Uand S®(T®U)areusually denoted simply S®T®U;higher-order products T1®---®T, aredeﬁned similarly. Thereader hasprobably already noticed that fJ1(V) isjust thedual space V*. Theoperation ®allows ustoexpress the other vector spaces 5'“(V)interms of51(V). 4-1 Theorem. Letv1,...,v,,beabasis for V,and let <p1,...14%bethedual basis, <p,~(v,-) --=6,-,-. Then thesetofall It-fold tensor products <pt'1®"'®<p,',, 1§t1,...,i;,§n isabasis forfl"(V), which therefore hasdimension n'°. Proof. Note that <Pi1 ® '''®i9ix(v.7'1: '''avit) =51;“,-,' ...'5,-,,,,-,, _l1 1fj1='li1,..-1,Ilc=’l;¢, _ 0 otherwise. Ifw1,...,w1,areItvectors with w,-=Z;-"=1a,~,-v,- and Tisin 5"(V), then ‘R T(w1, ...,w1,) = Z a1,,~,' ...-a1,,,-,,T(v,-,, ...,0,-,) J1 1 '1---1.7-ll: ‘R = 2 T(1),;,, ...,1),j,,) '<p,', ®'''®<,01;,,(’w1, ...,w1,). 1'1,...,'ib=1 Thus T=Z}';____,,,_.1T(v,~,, ...,2)g,,)'<p,', ®---®<p,-,. Consequently theor,®'''®<p1,,span 5'°(V). Integration onChains 77 Suppose now that there arenumbers a.',,.....;,, such that 2 6h',.....t.'¢h:.®-'-®<m1.=0- ii,...,i:t==1 Applying both sides ofthisequation to(o,-,, ...,2),-,)yields a,-,,__,,,-,, =0.Thus the <p.;,®'''®<p.-, are linearly independent. I One important construction, familiar forthecase ofdual spaces, canalsobemade fortensors. Iff:V-->Wisalinear transformation, alinear transformation f*:5"(W) ->5"(V) isdeﬁned by f*T(v1: ''°rvlﬁ) mT(f(v1): ''':f(vk)) forTE5'°(W) and v1,...,v1,EV. Itiseasy toverify thatf*(S ®T)=f*S®f*T. The reader isalready familiar with certain tensors, aside from members ofV*. Theﬁrstexample istheinner product (,)E32(R"). Onthegrounds that anygood mathematical commodity isworth generalizing, wedeﬁne aninner product onVtobea2-tensor Tsuch that Tissymmetric, that is T(v,w) ==T(w,o) forv,wEVand such that Tispositive- deﬁnite, that is,T(v,v) >0ifo750.Wedistinguish (,)as the usual inner product onR". The following theorem shows that ourgeneralization isnottoogeneral. 4-2 Theorem. IfTisaninner product onV,there isa basis v1,...,v,,forVsuch that T(v,-,2),-) =6,-,-. (Such a basis iscalled orthonormal with respect toT.) Consequently there isanisomorphism f:R"~—>Vsuch that‘ T(f(:i:),f(y)) =- (:c,y) for:2:,yER". Inother words f*T =(,). Proof. Letw1,...,w,,beanybasis forV.Deﬁne wt’=wt, I T(w1',w2) I, ‘L02 =wg "-'"mi '‘L01, T(’w1 {wi) w,_w T('w1'1'ws) w,T('w2','wa) w2, __ . _. _. . 3 T(w1’,w1') I T(w1’,w2') ’ etc. 78 Calculus onManifolds Itiseasy tocheck that T(w,',w,-') =0ifi5'5jandw.-'rf0so- thatT(w,-',w,-') >0.Nowdeﬁnet,-=w,-'/\/T(w,~',w.;'). The isomorphism fmay bedeﬁned byf(a,-) =v.,~.I Despite itsimportance, theinner product plays afarlesser role than another familiar, seemingly ubiquitous function, thetensor detE5"(R”). Inattempting togeneralize this function, werecall that interchanging tworows ofamatrix changes thesign ofitsdeterminant. This suggests thefol- lowing deﬁnition. AIt-tensor wE5"(V)iscalled alternating if w(o1, ...,o,~,...,1),-,...,v1,) =--—w(v1, ...,1),-,...,v,;,...,v1,) forallv1, ...,v1,EV. (Inthisequation v,-andv,-areinterchanged andallother v’s areleftﬁxed.) The setofallalternating h-tensors isclearly asubspace A'°(V) of5"(V). Since itrequires considerable work toproduce thedeterminant, itisnotsurprising that alternating lo-tensors aredifficult towrite down. There is, however, auniform way ofexpressing allofthem. Recall that thesign ofapermutation 0',denoted sgn0',is+1ifais even and -1ifaisodd. IfTE5'°(V), wedeﬁne Alt(T) by 1Alt(T)(v1, ...,v1,)=F,2sgnc- T(o,,(1,, ...,v,,(,,)), ‘test where S1,isthesetofallpermutations ofthenumbers 1to1:. 4-3 Theorem (1)IfTE:s’“(V), thenAlt(T) EA'°(V). (2) Ifw EAk(V), then Alt(w) =cc. (3)IfTE5'°(V), thenAlt(Alt(T)) =Alt(T). Proof (1)Let(i,j) bethepermutation that interchanges iandjand leaves allother numbers ﬁxed. IfoES1,,let0'= a'(i,j). Then Integration onChains 79 Alt(T)(v1, ...,2),-,...,v,,...,v1,) 1 =“I; 2 sgn“ 'T(v¢(1): ''‘:v°'(j): ''':vfI(?:): ''':v0(k)) HES]: 1=El sgna- T(v,,»(1), ...,v,,1(,-,, ...,v,,1(,-,, ...,v,,»(1,)) sES1. 1 =-ll?‘-l 2 _Sgn 0"'T(v¢'(1): ''':v¢I'(lv‘)) west. =—Alt(T)(v1, ...,v1,). (2) w€Ak(V), ﬂ.1'ld 0'=(’l,j'), 13l'l6I1 w(l),,(1), ...,l),,(k)) = sgn0'-w(v1, ...,v;,). Since every 0-isaproduct ofper- mutations oftheform (i,j),this equation holds ofall0'. Therefore 1 Alt(C0)(U1, ...,1)k) = 2 Sgn0'°OJ(l),(1), ...,?),(k)) sES1. 1=H2Sgna'Sgno"w(v1, ...,v1,) oES:e =oJ(l)1, ...,1);,;). (3)follows immediately from (1)and (2). I Todetermine thedimensions ofA"(V), wewould like a theorem analogous toTheorem 4-1. Ofcourse, iftoEA"’(V) and 11EAl(V), then w®11isusually not inA"’+Z(V). We will therefore deﬁne anew product, thewedge product (U/\17EA'°+z(V) by (ll+l)!w/\1)=-~'i:~Tl—!—~Alt(w ®'l7). (The reason forthestrange coefficient willappear later.) The following properties of/\areleftasanexercise forthereader: (w1+w2)/\17=w1/\n+w2/\r7, w/\(r71+172)=w/\r71+w/\172, aw/\v=w/\an=a(w/\n). w/\'7=(*1)Mﬂ/\w1 f*(w/\v)=f*(w) /\f*(n)- 80 Calculus onManifolds The equation (w/\1))/\6=w/\(:7/\6)istrue but requires more work. 4-4Theorem (1)IfSecs’°(V)sheTEs‘(V)theAlt(S) =0,then AMS®D=AMT®&=o @LMMMw®W®@=wM@®n®® =Alt(weAlt(11s0)). (3)Ifw EA"’(V), -qEAl(V), and 6EA"‘(V), then (cc/\n)/\9=w/\(n/\6) Proof (1)Itz 2=(++mlAlt(we1s>e).k!l!m! (I6+ ®T)(l)1, ...,Uk+;) v€=Zsgn<1'S(vt<1). ~--.v=»<k)) 'T(vea=+1). ---."-<k+l))-k+l IfGCS;,+1 consists ofall0'which leave It+1,.., It+lﬁxed, then 2 sgno'.S(v°’(1): ''':v¢(l5)) 'T(v°'(l¢+1): '~':v°'(l$+l)) o'EG = l:Z Sgn O" 'S(Uar(1), ...,1)o,r(k)):l °T(Uk+1, ...,1)k_l_l) a'ESk =0. Suppose now that 0'0EG. Let G'011=lo"'01120'EG} and letv,,,,(1), ...,v,,,,(1,+1, =w1,...,w1,+1. Then 0E2110@sI1<>"3(ve<1). ---.v-<t))'T(”~<k+1). ---.%<k+1>) =-'l:Sgl’l 0'0‘ E Sgll 0''S(’l0,,r(1), ...,w,.»(1,))l =0.o'EG 'T(w1,+1, ...,’t0k_|_1) Integration onChains 81 Notice that Gm G-<10 =Q’. Infact, if0'EG(WG-<10, then 0-=0''00forsome 0-’EGand 00=0-(a’)"1 EG, acontradiction. We can then continue inthis Way, breaking SH; upintodisjoint subsets; thesum over each subset is0,sothat thesum over S;,,+; isO.The relation Alt(T ®S)=Oisproved similarly. (2)We have Alt(Alt(n ®6)—-n®6)=Alt(17 ®6)—Alt(n ®6)=0. Hence by(1)wehave 0=Alt(w ®[Alt(n ®6)-17®6]) =Alt(w ®Alt(17 ®6))-—Alt(w ®17®6). Theother equality isproved similarly. (3)(w/\17)/\6-;UEk++ll;i_! ;n!)!Alt((w /\17)®6) __(k+Z+m)!(k+Z)! _(lc—l-Z)!'ml hill!A“’(‘°®”®9)' The other equality isproved similarly. I Naturally w/\(1;/\6)and (co/\17)/\6areboth denoted simply w/\n/\6,andhigher-order products 0:1/\'''/\CUT aredeﬁned similarly. Ifv1,...,v,,,isabasis forVand <p1,...,¢>,,isthedual basis, abasis forA"’(V) cannow be constructed quite easily. 4-5 Theorem. Thesetofall sen/\"'/\¢a 1S’i1<’i2<"'<’i1iS’ﬂ isabasis forA'°(V), which therefore hasdimension it“'h!(n-I43)! Proof. IfwEAk(V) C5'°(V), then WecanWrite w= Z ai1,...,i;¢‘Pi1® i''®¢7ik' i1,...,1')¢ 82 Calculus onManifolds Thus w=AM“) = 2 <1a,...,aA1’°(¢t1 ®'''®sea)- 1:1’. IQ‘ix Since each Alt(¢>,;1 ®---®<p.;,c)isaconstant times oneofthe <p-,;1/\'''/\¢,;,c,these elements span A'°(V). Linear inde- pendence isproved asinTheorem 4-1(cf.Problem 4-1). I IfVhasdimension n,itfollows from Theorem 4-5that A'"'(V) hasdimension 1.Thus allalternating n-tensors onV aremultiples ofany non-zero one. Since thedeterminant is anexample ofsuch amember ofA”(R"'), itisnotsurprising toﬁnditinthefollowing theorem. 4-6 Theorem. Letv1,...,v.,,beabasis for V,and let coEA"(V). If'w,;=Z;?=1a¢,-v_,- arenvectors inV,then w(w1, ...,'w,,,) =det(a.,-,-) -w(o1, ...,v,,). Proof. Deﬁne 17E5”'(R'"') by 7l((a11: ''':a1n)2 ''':(an1a '''aann)) =..(2a.,t,-, ...,Za,,,»v_,-). Clearly 11EA”(R'”') so1;=A-detforsome AERand A= 1;(e1, ...,e.,,,)=w(v1, ...,v.,,). I Theorem 4-6shows that anon-zero wEA”(V) splits the bases ofVinto two disjoint groups, those with w(v1, ...,0”) >0and those forwhich w(v1, ...,0”)<0;ifv1,...,v,. and "w1,...,w,,aretwo bases and A=(a,;,-) isdeﬁned by w,-=Ea),-v,-, then o1,...,v,,and w1,...,w.,,areinthe same group ifandonly ifdetA>0.This criterion isinde- pendent ofcoandcanalways beused todivide thebases ofV into two disjoint groups. Either ofthese two groups is called anorientation forV. The orientation towhich a basis v1,...,v.,,belongs isdenoted [v1,...,v,,,]and the Integration onChains 83 other orientation isdenoted -[v1, ...,v.,,,]. InR”wedeﬁne theusual orientation as[e1,...,e,,]. The factthat dimA”(R"') =1isprobably notnew toyou, since detisoften deﬁned astheunique element coEA""(R'"') such that w(e1, ...,e,,)=1.Forageneral vector space V there isnoextra criterion ofthissort todistinguish aparticular wEA"(V). Suppose, however, that aninner product Tfor Visgiven. Ifv1,...,v,,and w1,...,w,,aretwo bases which areorthonormal with respect toT,and thematrix A=(a,-,-) isdeﬁned by'w,;=Zl;"'=1a¢,-v,-, then Tl 5e"=T(w¢,'wj) =2‘%'ka.1'lT(vk,vz)lc,l=1 =2aika,-;,,. k=1 Inother words, ifATdenotes thetranspose ofthematrix A, then wehave A-AT=I,sodetA=il. Itfollows from Theorem 4-6that ifwEA"(V) satisﬁes w(v1, ...,v,,)=i1, then w(w1, ...,w,,) =i1. Ifanorientation p.forVhas also been given, itfollows that there isaunique wEA”'(V) such that w(v1, ...,v».».)=1whenever v1,...,v,,isan orthonormal basis such that [v1,...,v,,]=p..This unique wiscalled the volume element ofV,determined bythe inner product Tand orientation ju.Note that detisthe volume element ofR"determined bytheusual inner product andusual orientation, andthat jdet(v1, ...,v.,,)| isthevol- ume oftheparallelipiped spanned bytheline segments from 0toeach ofv1,...,v,,. Weconclude thissection with aconstruction which wewill restrict toR". Ifv1,...,v,,__1 ER"and tpisdeﬁned by 111 (p('w) =Cl€l] _ 1 vn--1 w 84 Calculus onManifolds then toEA1(R'"'); therefore there isaunique zER"such that v1 (we) =¢>('w) =det , vn-1 U) This zisdenoted v1X'--Xv,,__1 and called thecross product ofv1,...,v.,,_._1. The following properties are immediate from thedeﬁnition: v0'(1)>< '''><%(n-1)=5g11<Y'?J1X '''><vn--11 v1X"'X¢wi'X "'><v-n,_1=f1‘(’v1><"'Xvn_-1), v1X"'X(vt+vt')X"'Xvn-1 =v1><"-Xvi><"'Xv,,__1 +v1><---><v.~’>< '--><vn__1- Itisuncommon inmathematics tohave a“product” that depends onmore than twofactors. Inthecaseoftwovectors v,wER3,weobtain amore conventional looking product, vXwER3. For this reason itissometimes maintained that thecross product’ canbedeﬁned only inR3. Problems. 4-1.* Lete1,...,cnbetheusual basis ofR"and let e1,...,¢,,,bethedual basis. (3.) Sl‘lOW that epgl /\ --'/\qogk (6-5,, ...,e¢,,) =1. What would theright sidebeifthefactor (la+Z)l/kill didnotappear in thedeﬁnition of/\? (b)Show that ipg,/\---/\,0,-,,(v1, ...,v;,)isthedeterminant '11 ofthe kXItminor of - obtained byselecting columns vk i1,...,’l)¢. 4-2. Iff:V—-> Visalinear transformation and dim V=n,then f*:A"(V) —+A"(V) must bemultiplication bysome constant c. Show that c=detf. Integration onChains 85 4-3 4-4 4-5. 4-6 4-7 4-8IfwEA"(V) isthevolume element determined byTandju,and w1,...,w,,EV,show that lw(w1: ---ywn)l =‘\/det (§'ij)’ where g,;,-=T(w,;,w,-). Hint: Ifv1,...,v,,isanorthonormal basis andw,;=2,11 a,;;v,~, show that g.;_,*=2,1,1 a,;;.a;~,-. Ifwisthevolume element ofVdetermined byTand it,and f:R"-> Visanisomorphism such that f*T =(,)andsuch that [f(e1), ...,f(e,.,)] =pt,show thatf*w=det. Ifc:[0,1]-> (R")" iscontinuous and each (c1(t), ...,c"(t)) is abasis forR”,show that [c1(0), ...,c"(0)] =[c1(1), ...,c"(1)]. Hint: Consider detOc. (a)IfvER2,what isvX? (b)Ifv1,...,v,.,....1 ER”are linearly independent, show that [#11,...,v,,,_1, v1X---Xvn_1] istheusual orientation of R". Show that every non-zero coEA”(V) isthevolume element determined bysome inner product Tandorientation itforV. IfcoEA”(V) isavolume element, deﬁne a“cross product” v1X---Xvn_1 interms ofw. 4-9.* Deduce thefollowing properties ofthecross product inR3: 4-10. 4-11 4-12(a)e1><e1==0 e2><e1=—ea eaX61= 61Xe2==6s 62X62=0 e:~.Xe2= e1Xe3=—e2 e¢Xe3=e1 e3Xe3=0. (b)vXw=(vzwa -v3w2)e1 +(vswl -—v1w3)e2 +(vlwz —v2w1)e3. -|w|-Isin6],where 6=£(v,w). (vXw,w)=0. (d)(v,wXz) =(w,zXv) =(z,vXw) vX(wXz)=(v,z)w -(v,w)z (vXw)Xz=(v,z)w —(w,z)v. (e) ivXwl =V<v:v> '(w/wl ""<v:w)2' Ifw1,...,w,,,_1 ER”,show that6'2 ..._e1 <<=>I»><wl=lvl(UXwr U)= lw1><---><w.._1|=\/det(gs). where g,;,-=(w,-,w,-). Hint: Apply Problem 4-3 toacertain (n—-1)-dimensional subspace ofR". IfTisaninner product onV,alinear transformation f:V—->V iscalled self-adjoint (with respect toT)ifT(:z:,f(y))=T(f(:v),y) for:v,yEV.Ifv1,...,v,,isanorthonormal basis andA=(at,-) isthematrix offwith respect tothisbasis, show that at-5=ay.-. Iff1, ...,fn_12 Rm-—> R”, deﬁne f1X---Xf,,_12 Rm—+ R" byI1><~~-><f.._1(z>) =f1(z>) ><---><f.._1(r)- UseProb- lem 2-14 toderive aformula forD(f1 X--\-Xf,._1) when f1, ...,f,,_1 aredifferentiable. 86 Calculus onManifolds FIELDS AND FORMS IfpER",thesetofallpairs (p,v), forvER",isdenoted R-np, and called thetangent space ofR”atp.This setis made into avector space inthemost obvious way, bydeﬁning (av)+(aw)=(P,v+w), a'(pap) =(pram)- Avector vER"isoften pictured asanarrow from 0tov;the vector (p,v) ER”,,may bepictured (Figure 4-1) asanarrow with thesame direction and length, butwith initial point p. This arrow goes from ptothepoint p+o,andwetherefore P+v U Ur P FIGURE 4-I Integration onChains 87 deﬁne p+vtobetheend point of(p,v). Wewillusually write (p,v) asUp(read: thevector vatp). The vector space R-np issoclosely allied toR”that many ofthestructures onR”have analogues onR",,. Inparticular theusual inner product (,),,forR"? isdeﬁned by(v,,,wp),, = (v,w), andtheusual orientation forR-npis[(e1)p, ...,(e.,,),,]. Any operation which ispossible inavector space may be performed ineach R'”,,, andmost ofthissection ismerely an elaboration ofthis theme. About thesimplest operation ina vector space istheselection ofavector from it.Ifsuch a selection ismade ineach R”',,, weobtain avector ﬁeld (Figure 4-2). Tobeprecise, avector ﬁeld isafunction Fsuch that F'(p) ER",,foreach pER". Foreach pthere arenumbers F1(p), ...,F”(p) such that F0»)=F1(P)'(@1)p +'''+F"(r) -(6.)... Wethus obtain ncomponent functions Fi:R"-——>R. The vector ﬁeld Fiscalled continuous, differentiable, etc., ifthe functions Flare. Similar deﬁnitions canbemade foravector ﬁeld deﬁned only onanopen subset ofR”. Operations on vectors yield operations onvector ﬁelds when applied ateach point separately. Forexample, ifFandGarevector ﬁelds /\ /--——\\\'//--*' \\ [,\\ //-*"\\\.E///"' \ \""‘//' 41\\~;\‘\\ -’/ --/_-—-____/__.__////\\l ll] _h_~:_.-"""/"'1‘\~l\-_.."'""— /,-1“ ,_.'“:*'____-l-.'.I.'---.s:*-_":--\/ |____/ A ___. \/1/_,_,-'Ii\ \\ \]'/ F- /(\\ ...?’:\\" \ / \\//4“ //4 \\:lili//// FIGURE 4-2 88 Calculus onManifolds andfisafunction, wedeﬁne (F+G)(z>)=F'(r>)+Gov). (F,G>(2>) =(F(r).G(r)>, (f'F')(r)=f(2>)F(r>)- IfF1,...,F',,_1 arevector ﬁelds onR”,then wecan simi- larly deﬁne (F1X'''><Fn-1)(P) =F1(P) X'''XFn_1(P)~ Certain other deﬁnitions arestandard and useful. Wedeﬁne thedivergence, divF ofF,asZ,T‘=1D,;Fi. Ifweintroduce theformal symbolism V=2D5'65, i=1 wecan write, symbolically, divF =(V,F). Ifn=3we write, inconformity with this symbolism, (VXF)(P) =(D2F3 -DsF2)(@1);» +(D3171 "-D1F3)(@2)p +(D1F2 """D2F1)(@s)p- The vector ﬁeld VXFiscalled curlF.The names “diverg- ence” and “curl” arederived from physical considerations which areexplained attheendofthisbook. Many similar considerations may beapplied toafunction cowith w(p) EA'°(R",,); such afunction iscalled ak-form on R", orsimply adifferential form. If<p1(p), ...,<p,,,(p) isthedual basis to(e1),,, ...,(e,,),,, then w(r>)=Zwt.,...,t.(2>) -[¢».:.(2>) /\'''/\¢>t.(2>)]i1<'--<it forcertain functions cog,’_,_,,-,,;theform coiscalled continuous, differentiable, etc., ifthese functions are. Weshall usually assume tacitly that forms andvector ﬁelds aredifferentiable, and “differentiable” will henceforth mean “Cw”; this isa simplifying assumption that eliminates theneed forcounting how many times afunction isdifferentiated inaproof. The sum co+17,product f~w,andwedge product w/\1;aredeﬁned Integration onChains 89 intheobvious way. Afunction fisconsidered tobea0-form andf-wisalso written f/\w. Iff:R"—>Risdifferentiable, then Df(p) EA1(R"'). Bya minor modiﬁcation wetherefore obtain a1-form df,deﬁned by df(r)(vp) =Df(p)(v)- Letusconsider inparticular the1-forms dirl. Itiscustomary toletscidenote thefunction 1|-5. (On R3weoften denote ac‘,:c2,and:03bysc,y,andz.) This standard notation has obvious disadvantages but itallows many classical results tobeexpressed byformulas ofequally classical appearance. Since d:v'i(p)(v,,) =d1r'(p)(v,,) =D1r':(p)(v) =vi,weseethat 6l:B1(p), ...,d:c"(p) isjustthedual basis to(e1),,, ...,(e,,),,. Thus every h-form wcanbewritten w= 2 w,;,_,,__,-,,d:t':1 /\'''/\6l5l3':'°. <.1 <1». Theexpression fordfisofparticular interest. 4-7 Theorem. If R"—>Risdiﬂerentiable, then df= D1f-d:c1—|— ---—|—D,,f-dsc". Inclassical notation, of ofd=-—-—d1 ---—~d "'.f 6x1 in+ +6:c"' in Prvof- df(r)(vp) =Df(P)(v) =3?=1v" '_Dt-f(a») =3i"=1d~’v"'(P)(%) 'Dif(P)- I Ifweconsider now adifferentiable function f:R"—>Rmwe have alinear transformation Df(p): R”-—>Rm. Another minor modiﬁcation therefore produces alinear transformation f=|Rmf(p) deﬁned by f*(Up) "L(Df(P)(v))f(p)- This linear transformation induces alinear transformation f*:A"’(R’"';(,,,)) —>A'°(R"',,). IfwisaIt-form onRmwecan therefore deﬁne ak-form f*wonR”by(f*w)(p)=f*(w(f(p))). 90 Calculus onManifolds Recall this means that ifv1,...,v;.ER"',,, then wehave f*w(z>)(v1. ---mt)=w(f(r))(f*(v1), ---,f*(vt))- Asan antidote totheabstractness ofthese deﬁnitions wepresent atheorem, summarizing theimportant properties off*,which allows explicit calculations off*w. 4-8 Theorem. Iff:R"—>Rmisdiﬂerentiable, then . . . 6° . (1)f*(da:') =E;?=1D,-fl 'dsc’=Z;3'=1 ggdx’. (3)f*(w1 '|'012)=f*(w1) +f*(w2)- (3)f*(9''w)=(9°f) 'f*w- (4)f"'(w/\11)=f*w/\f*v- Proof <1>r*<dr=>,<v.> =dej<r><m.> _-dx'<r><E;;.»’ -D.-r1.- ~».I>;*=.r~1>.-r'"<t>>>.»<.>=Eii=1”J 'Djfi(P) _ =E§'=1D1f‘(P) 'd1>’(P)("p)~ Theproofs of(2),(3),and(4)arelefttothereader. I Byrepeatedly applying Theorem 4-8wehave, forexample, f*(P dscl/\d:c2+Qd:c2/\da:3) =(Pof)[f*(d:c1) /\f"‘(da:2)] -|-(Q°f)lf*(d$2) /\f*(d$3)l- Theexpression obtained byexpanding outeach f*(d:v‘) isquite complicated. (Itishelpful toremember, however, that we have dsvl/\dsvl=(—1)d:c£ /\dxi=0.) Inonespecial case it willbeworth ourwhile tomake anexplicit evaluation. 4-9 Theorem. Iff:R"-—>R"isdiﬂerentiable, then f*(hda:1 /\---/\dx")=(h<>f)(detf’) dzcl/\---/\dx“. Proof. Since f*(hd:c1/\ ~~-/\dzc")=(hof)f*(d:z:1/\ ---/\am"), Integration onChains 91 itsuffices toshow that f*(d:c1 /\'''/\dx") =(detf')d:c1 /\'''/\dsv". LetpER"and letA=(a,-,-) bethematrix off’(p). Here, and Whenever convenient and notconfusing, weshall omit “p” indxl/\'''/\dx"(p), etc. Then f*(d:1:1 /\'''/'\d:z:")(e1, ...,e,,) =dxl/\---/\d:c"'(f.|.e1, ...,f*en) =d:c1/\'--/\ (Ea,-1e,-,...,2a,-nei) r-1 r-1 =det(a,-_,-) -dxl /\---/\da:"'(e1, ...,e,,), byTheorem 4-6. I Animportant construction associated with forms isagen- eralization oftheoperator dwhich changes 0-forms into 1-forms. If wT 2 °-‘t1.....ad$“ /\'''/\dxiki¢1<---<a wedeﬁne a(lc+1)-form dw,thedifferential ofw,by Clo: '7: 2 dwi1,...,’i;c AClilill A '''Adiltil‘ i1<"' <ic n =2 21).,(¢..,;,,,,,_.;,,) 'dSU"/\dscll/\---/\dxlk.1:.<---<1:t a=1 4-10 Theorem (1)d(w+11)=dw+do- (2)IfcoisaIc-form and17isanl-form, then d(co/\17)=dw/\17-|-(—1)ko.v/\dn. (3)d(dw) =O.Brieﬂy, d2=0. (4)IfwisaIt-form onRmandf:R”-——>Rmisdiﬂerentiable, then f*(dw) =d(f*w). 92 Calculus onManifolds Proof (1)Left tothereader. (2)The formula istrue ifw=dash/\'''/\dxlk and 1;=dzvjl /\'''/\dscjl, since allterms vanish. The formula iseasily checked when wisa0-form. The gen- eral formula may bederived from (1)and these two observations. (3)Since at=Z2o.(t.,,_,,,..)d$~ /\dxfl/\---/\det,i1<"'<ita=1 wehave Tl Tl» d<d<»>-Z2Zo.,tu.-.,...,..>de /\as1 1 e.<---<".s=1a= _ _ /\d:t'1/\---/\<i;t'1¢. Inthis sum theterms D,,._,5(¢-.1,-,,_,_,,-,,)d:zt‘9 /\dx“/\dscil /\'''/\dscle and D,3_,.(w,-,,___,;,,)da:“ /\dzrf/\d:z:'i1 /\---/\d:v"'= cancel inpairs. (4)This isclear ifwisa0-form. Suppose, inductively, that (4)istrue when wisaIt-form. Itsuffices toprove (4)for a(h+1)-form ofthetype w/\dx‘. Wehave f*(d(..» /\am)=f*(d.../\as+(-1)'*=t.» /\d(d:c"‘)) =f*(d<»/\dc‘)=f*(dw) /\f*(d1=") =d(f*w/\f*(dw‘)) by(2)and(8)=d(f*(t.» /\da:"?)). | Aform wiscalled closed ifdo:=0and exact ifw=dn,for some 1;.Theorem 4-10 shows that every exact form isclosed, anditisnatural toaskwhether, conversely, every closed form isexact. Ifwisthe1-form Pda:+QdyonR2,then dw=(D1P da:+DQP dy)/\dx+(D1Q da:+DQQ dy)/\dy =(D1Q —D2P)d:c /\dy. Integration onChains 93 Thus, ifdw=0,then D1Q =DQP. Problems 2-21 and3-34 show that there isa0-form fsuch that w=df=D1fda:+ Dgfdg. Ifwisdeﬁned only onasubset ofR2,however, such afunction may notexist. Theclassical example istheform — :1:no='a:*2":-|_Ly§ dd?+a dy deﬁned onR2—0.This form isusually denoted d6(where 6isdeﬁned inProblem 3-41), since (Problem 4-21) itequals d6 ontheset{(:c,y): :1:<0,orasZ0and ysf0},where 6is deﬁned. Note, however, that6cannot bedeﬁned continuously onallofR2—0.Ifco=dfforsome function f:R2—0—> R, then D1f=D16andDgf=D26, sof=6+constant, show- ingthat such anfcannot exist. Suppose thatw=E}‘=1w,; dx‘isa.1-form onR”andtohappens toequal df=E.}‘=1D,-_f -dx‘. We can clearly assume that f(0) =0.AsinProblem 2-35, wehave °‘§\'__<=~*__Ho§.H::l\-43lM=&9*f(x)=—f(iw) di = D,-f(t:c) -ti‘dt = 0J,'(l$) '58':(ll. This suggests that inorder toﬁndf,given w,weconsider the function Iw,deﬁned by 1n Iw(:v) =bfZw,-(tar) ':c':dt. ‘I I-d Note that thedeﬁnition ofIwmakes sense ifwisdeﬁned only onanopen setAER"with the property that whenever :2:EA,thelinesegment from 0to:1:iscontained inA;such anopen setiscalled star-shaped with respect to0(Figure 4-3). Asomewhat involved calculation shows that (ona star-shaped open set)wehave co=d(Iw)provided that wsatis- ﬁesthenecessary condition do=0.The calculation, aswell asthedeﬁnition ofIw,may begeneralized considerably: 94 Calculus onManifolds \ ..~-‘-.-/C\_--..:-;.:>1-_'.;» -‘Iii’:1"'*<:I:¢»1 -.i;‘-'-L\-1.‘; .;'...' f. _._-.‘- -;.‘ '~.:.1-_ Y.'1 -.5;-> __- ' _.; T_'"-'"’-=:-. _ i ,- i" i";':'5'"I;':"5I_.I...-1.11 . , _.,,,-;..’.-_-_.-_» ..'.;>;-_-..-_.»;;'._*.._,__.,\,_~ ' '- "' " ',-*_;'3¢"~“T1';= --_'-"-’-'11-f-5'; ..-- "' J‘ . .-,-‘.._'.-;§:;i__¢'=.-1-, '--.-- it-5i'- / '1 '1“" I '~.-.--. -/-.:.-. '-;;-7-.. “-,_.4, -.--5; -'.;\= »_._ ' »'.;i-'1.11..-I -1: "---1:... '-'-\1;I:;: _.g." }31' "1';-.'1}, '-1:;.= -'15'Tf*'-'4 '-E:-.-1;i:’.>. '-T.=I‘.l';1._*11';f:-. '-i;lu.-. 4;.-<1.1‘=-'5 2-;=9/._--:<;¢._ 1";'i'.'€?‘i Z1 =1."~':¥i1:'11?E-‘ 'i..£'=' .'.-'-EI3.?':'3:f-1 .-1:-'r"'-3' _;:§:-,I-_'-;2gI-'(:- .-;‘.>l;!:-'5'-:i - , .3:-,2-'sa;.;.;,.;.;..- _;.;.-,.¢:;.;.- __;.; FIGURE 4-3 4-11 Theorem (Poincare Lemma). IfAER"isanopen setstar-shaped with respect to0,then every closed form onA isexact. Proof. Wewill deﬁne afunction Ifrom l-forms to(l—1)- forms (foreach l),such that 1(0) =0andcc=-~1(dw) +d(Iw) foranyform co.Itfollows that co==d(Iw) ifdo=0.Let £0 ___: 2 .¢~pit A . T A i1< '''<it Since Aisstar-shaped wecandeﬁne z 1 Ito)-Z):<-1>""1([ c—1w..,...,..<e>dt)1=ei1<*-*<iza:=1 0 /\ d:1:’:*/\"'/\da:"°*/\"' /\d:z:“3-1. (The symbol -~over dad“ indicates that itisomitted.) The Integration onChains 95 proof that 0:=I(do) —l-d(1w) isanelaborate computation: Wehave, using Problem 3-32, 1 d<I<»>-1-Z(f6-1w......,..<e>dt)i1<"'<iz 0 . _ e<e»1/\--- Ada"l n 1 +1:.<<e..Z1,-Z1 (—1)a_1 (OItzDj(wiI' ' Wadi) mid da:"/\d:z:’31/\-"/\da:‘°'/\ ---/\d:r"3¢. (Explain why wehave thefactor ti,instead oftZ“1.) Wealso have Tl do== 2 2D,-(w,;,,____,;,) -dsrj /\dscll /\'''/\d:c':*. n<---<ej=1 Applying Itothe(l-l—1)-form dw,weobtain n 1 dzvll /\'''Adm‘! nz 1 -Z2Z<-1>""1( f61>.-<<».».,...,..><a>d:)r’~=i1<"'<izj=1a=1 0 //\. die"/\de‘1/\-~Adei-A---Adan. Adding, thetriple sums cancel, and weobtain 1 e(1t.)+ 1(e..,)=2z-(fti-1t,,__,__.,(a)ei)i1< '''<iz U ee1/\-~- /\dacil +M dz1.51(Oftz37jDj(¢°i1, ...,a)(?5$)d5) d:v'i1 /\'''/\d:z:":‘ 1 -Z(f~§,[t*e..,...,..<tx>1d:)i1<"'<it O eel/\-~~/\d:c" 2 Z wi1.....itd$i:1 A'''Adz,“i1< <il =0.». 96 Calculus onManifolds Problems. 4-13. (a)Iff:R"——> Rmand g:R"‘-—> RP,show that 4-14 4-15 4-16 4-17 4-18 4-19 4-20(6°f)-t =9-°f-&I1d(9°f)"‘ =f"‘°9"'- (b)Iff,g:R" —->R,show that d(f-g)==f-dg+g-df. Letcbeadifferentiable curve inR",that is,adifferentiable func- tion c:[0,1]-—> R". Deﬁne thetangent vector vofcattas c.,.((e1);) =((c')’(t), ...,(c")’(t)),,(;). Iff:R"-—> R”,show that thetangent vector tof0cattisf...(v). Letf:R——> Rand deﬁne c:R-—> R2byc(t)=(t,f(t)). Show that the end point ofthetangent vector ofcattliesonthe tangent linetothegraph offat(t,f(t)). Letc:[0,1]-—>R"beacurve such that|c(t)|= 1forallt.Show that c(t),,(¢) andthetangent vector tocattareperpendicular. Iff:R"——>R",deﬁne avector ﬁeld fbyf(p) =f(p),, ER"‘,,. (a)Show that every vector ﬁeld FonR"isoftheform ffor some f. (b)Show that divf==trace f’. Iff:R"->R,deﬁne avector ﬁeld grad fby (EI'&df)(P) =D1f(P) '(6119 +'''+Dnf(P) '(enliv- Forobvious reasons wealso write grad f=Vf. IfVf(p) =w,,, prove that D,,f(p) =(v,w) andconclude that Vf(p) isthedirection inwhich fischanging fastest atp. IfFisavector ﬁeld onR3,deﬁne theforms ...}.=F1da:+F2dg+F3dz, cu}=F1dy/\dz+F2dz/\d:c+F3d:r/\dy. (a)Prove that ___ 1 df"'“grad fa 1_2d(°’1~") -°’curlFr d(<..»§.-)-(divF)dxAatAdz. (b)Use(a)toprove that curlgrad f=0, divcurlF=0. (c)IfFisavector ﬁeld onastar-shaped open setAand curlF=0,show that F=grad fforsome function f:A—>R. Similarly, ifdivF=0,show that FmcurlGforsome vectdi ﬁeld GonA. Letf:U-—>R"beadifferentiable function with adifferentiable inverse f'1:f(U) —>R". Ifevery closed form onUisexact, show that thesame istrue forf(U). Hint: Ifdw=0andf"'w=dn, consider (f'1)"‘n. Integration onChains 97 4-21."‘ Prove that onthesetwhere 6isdeﬁned wehave ee=-it ---—e.:z:2+,1/2x+a:2+y2 y GEOMETRIC PRELIMINARIES Asingular n-cube inAER”isacontinuous function c: [0,1]"' -—->A(here [0,1]”' denotes then-fold product [0,1] X--' X[0,1]). WeletR0and[0,1]° both denote {0}. Asingular 0-cube inAisthen afunction f:{0}—->Aor,what amounts to thesame thing, apoint inA.Asingular 1-cube isoften called acurve. Aparticularly simple, but particularly important example ofasingular n-cube inR"isthestandard n-cube I":[0,1]"' —~—>R"deﬁned byI"(:z:) =:1:for:1:E[0,1]"'. Weshall need toconsider formal sums ofsingular n-cubes in Amultiplied byintegers, that is,expressions like 201+302-~403, where c1,02,c3aresingular n-cubes inA.Such aﬁnite sum ofsingular n-cubes with integer coeﬂicients iscalled an n-chain inA. Inparticular asingular n-cube cisalsocon- sidered asann-chain 1-c. Itisclear how n-chains canbe added, andmultiplied byintegers. Forexample 3(¢1+364)+(—2)(¢1 +63+62)=-262 —203+604- (Arigorous exposition ofthis formalism ispresented inProb- lem4-22.) Foreach singular n-chain cinAweshall deﬁne an(n—1)- chain inAcalled theboundary ofcanddenoted dc. The boundary ofI2,forexample, might bedeﬁned asthesum of four singular 1-cubes arranged counterclockwise around the boundary of[0,1]2, asindicated inFigure 4-4(a). Itis actually much more convenient todeﬁne 612asthesum, with theindicated coefficients, ofthefour singular 1-cubes shown inFigure 4-4(b). The precise deﬁnition of61”requires some preliminary notions. For each iwith 1§i§nwedeﬁne two singular (n-1)-cubes Iflm, and If‘,-_,) asfollows. If 98 Calculus onManifolds -1-<———-—--—- —-————> —l +4 +1 (a) ' (b) FIGURE 4-4 :1:E[0,1]"'_1, then II‘,-_0)(:1:) =I"'(:1:1, ...,:1:'i"'1,0,:1:":, ...,:1:""‘1) =(:1:1, ...,:1:"1,0,:1:", ...,:1:"'1), If‘,-_1,(:1:) =I"(:1:1, ...,:1:':"1,1,:1:"i, ...,:1:"'_1) =(ml, ...,:1:""1,1,:1:", ...,:r:"_1). Wecall I[‘,_0, the(i,0)-face ofI"and I{‘,;_1) the(i,1)-face (Figure 4-5). Wethen deﬁne 11. er"=ZZ(-1)*+“1r,,,,.i=1 oz=0,1 Forageneral singular n-cube c:[0,1]"' —>Aweﬁrst deﬁne the (i,a)-face, c(i.a) =C°(I?i,a)) and then deﬁne 66: 2 Z (—l)£+aC(i,,,,). i=1 a=0,1 Finally wedeﬁne theboundary ofann-chain Ea,-c, by 6(Ea,-c,;) =-Ea,~6(c,;). Although these fewdeﬁnitions sufﬁce forallapplications in this book, weinclude here theonestandard property of6. Integration onChains 99 If.» 2 I%1.0) I(1.1) -€ Il1.o) lim) Ii?-0) (a) (b) FIG URE 4-5 4-12 Theorem. Ifcisann-chain inA,then 6(6c) =0. Brieﬂy, 62=0. Proof. Let i_§jand consider (If‘,_,,,))(,-49). If.1:E[0,l]"_2, then, remembering thedeﬁnition ofthe(j,j6)-face ofasingular n-cube, wehave (Iii1:,a))(.1'.6)($) =Iiii,a)(Ili;T61)(x)) _ _ =It-,.><:1‘. »~_.1=:*‘.e.1’. __..1>""2>=I”'(:1:1, ...,.iU"_l,a,:I;", ...,:1:"_1,B,:1:’, ...,:1:”'_'2). Similarly (I?j+1.a>)<i.a> =Iii¢'+1,a>(IfEI1=1>(“’)) _ _ 2 =1I'?',._,_,_,,,(:c.1,l. .._,a:"1,a,a:"',1. ...,:1:"'“) 2 =I”'(:1: ,...,x‘_' ,a,:1:", ...,:1:-'_' ,B,:1:’, ...,:1:”'" ). Thus (I"f',,;,,,,,)(,-,,,, =(If‘,-+1,,_.,,)(,;_,,,, for135;‘. (Itmay help to verify this inFigure 4-5.) Itfollows easily forany singular n-cube Cthﬂll (C(,j,a))(j,,5) =(C(_7'+1,5))(g,a) when S NOW T1 soc)-<1():1Zl<—1>‘+"c<...>)‘I. a00 11- -1;1- -1. ' n n—1 :2 222("r1)':+“+"+5(@<t.-1))(1.5)-1:=1...=0,1j=1 e=o,1 100 Calculus onManifolds Inthis sum (0(,~,.,))(,-,5) and (0(,-+1_,9))(,;,,,, occur with opposite signs. Therefore allterms cancel outinpairs and6(60) =0. Since thetheorem istrue foranysingular n-cube, itisalso true forsingular n-chains. I Itisnatural toaskwhether Theorem 4-12hasaconverse: If 60=0,isthere achain dinAsuch that0=6d?Theanswer depends onAand isgenerally “no.” For example, deﬁne 0:[0,1]——> R2-0byc(t)=(sin21rnt, cos21rnt), where nis anon-zero integer. Then 0(1) =0(0), so60=0.But (Problem 4-26) there isno2-chain 0'inR2—-0,with 60'=c. Problems. 4-22. Letgbethesetofallsingular n-cubes, andZthe integers. Ann-chain isafunction f:5-1Zsuch that f(c) =0 forallbutﬁnitely many 0.Deﬁne f+gandnfby(f+g)(c) = f(c) +g(a) and nf(c) =n-f(c). Show that f+gand nfare n-chains iffandgare. If0ES.let0also denote thefunction f such that f(c) =1andf(a’) =0for0'#0.Show that every n-chain fcanbewritten a1c1 +---+aka), forsome integers a1,...,a;,andsingular n-cubes 01,...,c;,. 4-23. ForR>0andnaninteger, deﬁne thesingular 1-cube 03,": [0,1]-1 R2—0byca,-1(t) =(Rcos21rnt, Rsin2mt). Show that there isasingular 2-cube 0:[0,1]2 -1R2-0such that 03,," -011,,” =6c. 4-24. If0isasingular 1-cube inR2-0with 0(0) =0(1), show that there isaninteger nsuch that 0~—01,,=602forsome 2-chain 02. Hint: First partition [0,1] sothat each 0([t1..1,t1]) iscontained on onesideofsome linethrough O. THE FUNDAMENTAL THEOREM OF CALCULUS The fact that d2-=0and62=0,nottomention thetypo- graphical similarity ofdand 6,suggests some connection between chains andforms. This connection isestablished by integrating forms over chains. Henceforth only differentiable singular n-cubes willbeconsidered. If0.»isa10-form on[0,1]2, then co=fdasl /\'''/\dashfor aunique function f.Wedeﬁne '52*8 "5sm-$5 Integration onChains 101 Wecould alsowrite thisas ffd:1:1/\'-'/\d:1:"’= ff(:1:1,...,:1:'°)d:1:1-'-d:1:"’, [0,11" l0.ll" oneofthereasons forintroducing thefunctions :0‘. IfwisaI0-form onAand0isasingular It-cube inA,wedeﬁne /.=,/stt 10,11» Note, inparticular, that [feel A~~-/\dx"’=f(1'~=)*(fe1=1 /\---/\dxl’)1» [o,11~ =ff(:1:1, ...,x'°)d:1:1 ---d:z:". l0.1l" Aspecial deﬁnition must bemade forI0=0.A0-form 0.1is afunction; if0:{0}——>Aisasingular 0-cube inAwedeﬁne f(.0-1411(0)). Theintegral of0.1over aI0-chain c=Ea,-0; isdeﬁned by !.=§tlt The integral ofa1-form over a1-chain isoften called aline integral. IfPda:+Qdyisa1-form onR2and0:[0,1] ——>R2 isasingular 1-cube (acurve), then onecan(but wewillnot) prove that 1’!- [Pat+Q<11-limZW.)-<=1<1._.>1~P<<=<r>>0 i-1 +[c’(1.>—¢2(11....1)1 -Q(<=(i‘)) where to,...,t,,isapartition of[0,1], thechoice oftiin [t,~...1,t,] isarbitrary, andthelimit istaken over allpartitions 102 Calculus onManifolds asthemaximum of|t,--t,-_1l goes to0.The right side is often taken asadeﬁnition ofLPdx+Qdy. This isanatural deﬁnition tomake, since these sums arevery much likethe sums appearing inthedeﬁnition ofordinary integrals. How- ever such anexpression isalmost impossible towork with and isquickly equated with anintegral equivalent tof[0,1]0*(P da: +Qdy). Analogous deﬁnitions forsurface integrals, that is,integrals of2-forms over singular 2-cubes, areeven more complicated and difficult touse. This isonereason why we have avoided such anapproach. Theother reason isthat the deﬁnition given here istheonethat makes sense inthemore general situations considered inChapter 5. Therelationship between forms, chains, d,and6issummed upintheneatest possible way byStokes’ theorem, sometimes called thefundamental theorem ofcalculus inhigher dimen- sions (ifh=1and0=I1,itreally isthefundamental theorem ofcalculus). 4-13 Theorem (Stokes’ Theorem). If00isa(lo-1)- form onanopen setACR"and0isaI0-chain inA,then C[dw=a[w. Proof. Suppose ﬁrst that 0=I'°and wisa(/0—1)-form on [0,1]". Then wisthesum of(lo-1)-forms ofthetype /\. 1 ' I0fda: /\"'/\dx"/\"'/\d:1:, and itsuffices toprove thetheorem foreach ofthese. This simply involves acomputation: Note that /If‘,-,,,,*(j'd:1:1 /\---/\61¢"A---A66'“)[0, 11¢-I 0 ifjsfi, = [f(:r1,. ..,a,. ..,a:'°)da:1- '-d:1:'° ifj=i.l0.1l" Integration onChains 103 Therefore ffdasl/\-"/\1:t/ail/\"'/\d:1:'°6I'°k 5:.:*~=2 Z (._1)J'+a Il(c]_,a)*(j-dxl A...A5:271? j=1a=0,1 ' /\'''/\dick) =(-1)'1+1 ff(:1:1, ...,1,...,a:'°)da:1 ---611'"l0.1l" +(-1)'1 [A111, ...,0,...,:1:'°)d:1:1 --~d:0".l0,11'= Ontheother hand, /-. l ' I6fd(fd:1: A--- /\d:v‘/\"'/\d:1:)110 --=fD,fd:1:"3/\d:1:1/\"'/\d:li’3/\"'/\d:0" 10.11- =<-1):-1 fD6.l0.1l" ByFubini’s theorem andthefundamental theorem ofcalculus (inonedimension) wehave fe(fee1A ~--/\di:2/\ /\d:1:"),. =(__1)i-16,} ... Dif(x1’ _._,xlc)dxi) dxl .. cTi:2"-d:1:"’ Q1-I Q\H=<-1)"-1/ ~~~[f(:v‘,._-.1.....15 /'\ —f(:1:1,... ,0,... ,:1:")]d:1:1--'d:1:2"'d:1:'° =(—1)'l'"1 ff(:1:1, ...,1,...,a:'°)d:1:1 ---d:1:"’ l0.1l" +(-1)'1 ffal, ...,0,...,:1:'°)d:1:1 ~--d:1:".10.112 Thus Hfdo:=allco. 104 Calculus onManifolds If0isanarbitrary singular I0-cube, working through the deﬁnitions willshow that /....;.=-...60 61* Therefore do:= c*(dw) = d(c*w) = c*w = co. II.1.1.[Finally, if0isaI0-chain Ea,-c,;, wehave fdw=Za,-[dw=2a,;fw=fw.I 1: 1;. 601 60 Stokes’ theorem shares three important attributes with many fully evolved major theorems: 1.Itistrivial. 2.Itistrivial because theterms appearing inithave been properly deﬁned. 3.Ithassigniﬁcant consequences. Since this entire chapter was little more than aseries of deﬁnitions which made thestatement and proof ofStokes’ theorem possible, thereader should bewilling togrant the ﬁrst twoofthese attributes toStokes’ theorem. The restof thebook isdevoted tojustifying thethird. Problems. 4-25. (Independence ofparameterization). Let cbea singular I0-cube and p:[0,1]"-1 [0,1]" a1-1function such that p([0,1]k) =[0,1]'“ anddetp'(1;)20for1;5[0,1]*. If...isa I0-form, show that f...=f...0 co1) 4-26. Show that f_,,,,,, d6=21rn, and useStokes’ theorem toconclude that cR,.. re60forany2-chain 0inR2—0(recall thedeﬁnition of 03,,inProblem 4-23). 4-27. Show that theinteger nofProblem 4-24 isunique. This integer iscalled the winding number ofcaround O. 4-28. Recall that thesetofcomplex numbers Cissimply R2with (a,b) =a+bi. Ifa1,...,a,,EC letf:C—> Cbe = 2”+a1z"'“1 +---+an. Deﬁne the singular 1-cube 03,,-: Integration onChains 105 4-29. 4-30 4-31 4-32 4-33[0,1]—> C-0by03,; =f<>0R_1, and thesingular 2-cube 0by c(8.t)=t-011.2(8) +(1—t)c11.:(s). (a)Show that 60=03,;—011,2, and that 0([0,1] X[0,1]) E C-0ifRislarge enough. (b)Using Problem 4-26, prove theFundamental Theorem of Algebra: Every polynomial 2"’+a1z""'1 +---+anwith a1EC hasaroot inC. Ifoisa1-form fdxon[0,1] with f(0) =f(1), show that there is aunique number Asuch that o—Adx=dgforsome function g with g(0) =g(1). Hint: Integrate o-Ada =dgon[0,1] to ﬁndA. Ifoisa1-form onR2-0such that do=0,prove that co=Ad6 -|-dg forsome AERandg:R2-0-1 R.Hint: If 6R.1"'(w) =ARdiv"l"d(6R): show that allnumbers ARhave thesame value A. Ifoas0,show that there isachain 0such that fco;é0.Usethis fact, Stokes’ theorem and62=0toprove d2=0. (a)Let01,02besingular 1-cubes inR2with 01(0) =02(0) and01(1) =02(1). Show that there isasingular 2-cube 0such that 60= 01—02+03-04,where 03and04aredegenerate, that is,03([0,1]) and04([0,1]) arepoints. Conclude that f,,,o =f,,,o ifoisexact. Give acounterexample onR2-0ifoismerely closed. (b)Ifoisa1-form onasubset ofR2andf.,,o =fc,oforall01, 02with 01(0) =02(0) and 01(1) =02(1), show that oisexact. Hint: Consider Problems 2-21 and3-34. (Afirst course incomplex variables.) Iff:C—>C,deﬁne ftobe differentiable atZ0ECifthelimit f(z)—f(20)re.)=lim--—-z—>z2 Z_Z0 exists. (This quotient involves two complex numbers and this deﬁnition iscompletely different from theone inChapter 2.) Iffisdifferentiable atevery point zinanopen setAandf’is continuous onA,then fiscalled analytic onA. (a)Show that f(z) =2isanalytic andf(z)=2isnot(where :1:-1-iy=:1:—iy). Show that thesum, product, and quotient ofanalytic functions areanalytic. (b)Iff=u+ivisanalytic onA,show that uand vsatisfy theCauchy-Riemann equations: 6u 6v 6u —6v-—-=-— and --=—-6:0 6y 6y 6:1: Calculus onMamlfclds Hint: Use thefact that lim[f(z) —f(z0)]/(2 —20)must bethe Z—§Z0 Same f0? -'==Z0-|-(iv-l-1?-0) and Z=Z0+ (0+i-y) with a',y->0.(The converse isalso true, ifuandvarecontinuously differentiable; thisismore difficult toprove.) (c)LetT:C—>Cbealinear transformation (where Ciscon- sidered asavector space over R). Ifthematrix ofTwith respect tothebasis (1,2I) is($3) show that Tismultiplication byacom- lexnumber ifandonly ifa=dandb=—-c. Part (b)shows that ananalytic function f:C—+C,considered asafunction f:R2—> R2,hasaderivative Df(z0) which ismultiplication byacomplex number. What complex number isthis? (d)Deﬁne d(w +'l17)= dw+'ld11, [co-I-’l17=‘[w+'l-[17, (w+'l1;)/$9+’l7$=w/\9 --17/\7\+i('q/\9+ co/\)\), and dz=da:+idy. Show that d(f-dz)=0ifand only iffsatisﬁes theCauchy- Riemann equations. (e)Prove theCauchy Integral Theorem: Iffisanalytic onA, then fcfdz=0forevery closed curve c(singular 1-cube with c(0) =c(1)) such that c=60’forsome 2-chain c’inA. (f)Show that ifg(z) =1/z,then g-dz[or(1/z)dz inclassical notation] equals 2Id0+dh forsome function h:C—0--+ R. Conclude that f,,R_n (1/z)dz =21rin. (g)Iffisanalytic on{zzI2]<1},usethefact that g(z) = f(a)/z isanalytic in{zz0<I2]<1}toshow that if0<R1,R2<1.Use (f)toevaluate limfcR_nf(z)/z dzandR—>0 conclude: Cauchy Integral Formula: Iffisanalytic on{zzI2}<1}and cisaclosed curve in{zz0<|z|<1}with winding number n around 0,then n-f(0) =-£37-Jfﬂgzdz. Integraticn onChains 1-:av 2 \ IL‘T’ (=1) 'Z \ (h) Z 1 I 1 5.+ §*@UJ} (0) FIGURE 4-6r '1 1 ll 108 4--34Calculus onManifolds IfF:[0,1]’--> R3and sE[0,1] deﬁne F,,:[0,1]—>R3byF,(t) = F(s,t). IfeachF,isaclosed curve, Fiscalled ahomotopy between theclosed curve F0andtheclosed curve F1.Suppose FandGare homotopies ofclosed curves; ifforeach stheclosed curves F,and G’,donotintersect, thepair (F,G) iscalled ahomotopy between the nonintersecting closed curves F0,G0andF1,G1. Itisintuitively obvious that there isnosuch homotopy with F0,G0thepair of curves shown inFigure 4-6(a),andF1,G1thepair of(b)or(c). The present problem, andProblem 5-33 prove thisfor(b)butthe proof for(c)requires different techniques. (a)Iff,g:[0,1]-->R3arenonintersecting closed curves deﬁne c;,q: [0,1]2—-> R3~—0by ¢!.a('w,v) =f(u)-11(11)- If(F,G) isahomotopy ofnonintersecting closed curves deﬁne Cpggi [0,1]3 —>R3—-0by CF,o(8ﬂ1»v) =6F,,o,(W1) =F(81%) *-"G(8,v)- Show that 6C'F,o =6F,,o, -¢1r,,o,- (b)Ifwisaclosed 2-form onR3—0show that for '-= [02. cn,,,o,, expo, 5 Integration onMamfolds MANIFOLDS IfUand Vareopen sets inR",adifferentiable function h:U—>Vwith adifferentiable inverse h'1: V—> Uwill be called adiffeomorphism. (“Differentiable” henceforth means “C°°”.) Asubset MofR”iscalled als-dimensional manifold (in R”) ifforevery point :2:EMthefollowing condition is satisﬁed: (M) There isanopen setUcontaining :z:,anopen setVCR”, andadiffeomorphism h:U—>Vsuch that h(Uf\M) =vr\(n'">< {0}) ={2/€V=z/"+‘= "'=2/"=0}- Inother words, Uf\Mis,“up todiffeomorphism,” simply BkX{0}(see Figure 5-1). The two extreme cases ofour deﬁnition should benoted: apoint inR”isa0-dimensional manifold, and anopen subset ofR”isann-dimensional manifold. One common example ofann-dimensional manifold isthe 109 110 U ._:;:;;;'(- (E1) (b)_-...;5;:_,.,;-;‘"==1:;;.=;2§;§&§.=i:%;P:%§-H‘..__.;_,_1.x__g..__,,: _..;;.»1,3,-,2-._1-»;..,_ 7---».--.91» '-.;=_~;-:-:-;-.--:-;-;.;.;.- *""‘".:2",->- -=-12.-¢=::::.====:= ‘+1--:-f-fl:l:1:'-.'3:I:f.-?-.- ---:-5,“~‘:1-.-ff!-:-I-l:I;I-1-;§.§:£';.. ":12;:21;:a.=.=2;:;:;.';2;I::;:==a..as=ea:a=::;:.1%;z::::;:‘a-----Calculus onManifolds V $31Eifiiiiiffii5151555EZEEFPFEZEEITIEIEI g h V ;:;f;'-‘E15If;j:j:§:f.-I:1:1:5:1;1:1:T.-Q13:;:;.-j:§:§;§;I:I:-11:1;I;?;;;:;:;.-j:f;§:§:§:I:;':l,'I:";-I:-';'-3:;:;:§:{.-ffr}:§:I:55;I;-';:;:;:;:j:§:§:§_'§:§:§:§.-1:11;‘.-§;§; “"352.'-~"" 5.'.1:}.-ft'F’ "‘~t**‘5'%- -.ii-1:’ 2-1:.‘-5.;1 Q:6,1_( 0x -¢.,-fwg ..3?1% 1,-1;:;:;:1;;:;:;:j:§:_I:f;1,-I-_‘. _', _,'-_...*'_iI-§§_‘-1;53‘;‘fl -~?1;:§:§:j:§EI§-'.i;§I§:§.'-Ff: -:»:-:-'I;I;>;:;I;:;:;:»:-:-;;'-§\:-is 1. ,....-. 1.-;_ F5.-'-.3¢’. Y‘-.1 ’°i;1;:_»r-:-:-:»-‘-!:I;I;I_I:;::- 51Zriré:55555555;-'§5;.‘=E=:=§a ~-"iv".>'/'1-: .;l‘2"‘¥ '.'.'=2.‘31"‘ .1:5;.1Er:1;~E1E151£:€r5::_=;;:;;;-=E=;2. “xi ‘P“$333., Q ,3;W. _. :1:;.=.:.%.-s=:-e:&rs=are -as- .-'...»;s1;.-.1-.=;1=r:1;:?=E1i15:Ys%"fi, 3“?3 -er-.1-,2 ‘.~-;.-;:;I:2:;:;:;.;;!:;.g;=;:s;;;;;;.-1;=:=:1.¥:1:§§: T‘ =; '- It-_,'3"".e__ _ ', 1-_‘I'21",_‘.5.;:fijiiiif-iI:§if§?§1f»'§fI5 " ' -f-EIFS-‘E1EIEIEIZIE15I-'55I§i:'1:=§1ZI?1.i1§1§f§IE11r§:§-;I:§5=£11 '1I!1FIEIII}IEiii}?§5Ei§§EIE=i»'§‘E!:'I{§IEI§5.5E5E:»':1:I:f:1:IEIEI}€IE!E=?' 17*"?51"‘-'3'3'1""'i" _»:;;;I;;;1;I;.";:;153;;I;1§:§;[:l£;!;1_».;:;;» .__;:;:§:5; ;:;:;:;1-1-1»?-:1:-;I;I;Ig.';:-:;:-:»:-:-I-f- FIGURE 5-1. Aone-dimensional manifold inR2and atwo-dimen- sional manifold inR3. Integration onZ11anifolds 1I1 n-sphere S”,deﬁned as{acER"'+1: =1}. Weleave it asanexercise forthereader toprove that condition (M) is satisﬁed. Ifyou areunwilling totrouble yourself with the details, you may instead usethe following theorem, which provides many examples ofmanifolds (note that S"=g—1(O), Where g:R"'+1-> Risdeﬁned byg(x) =}a:}2--1). 5-1 Theorem. Let AER”beopen and letg:A-->R1’ beadiﬁferentiable function such thatg'(a) hasrank pwhenever g(x) =O.Then g—1(O) isan(n-~p)-dimensional manifold in R”. Proof. This follows immediately from Theorem 2-13. I There isanalternative characterization ofmanifolds which isvery important. 5-2 Theorem. Asubset MofR”isais-dimensional mani- fold ifandonly ifforeach point atEElfthefollowing “coordinate condition” issatisﬁed: (C) There isanopen setUcontaining as,anopen setWER7‘, anda1-1dijferentiable function f:W—>R”such that (1)f(W) =Mm U, (2)f’(y) hasrank lsforeach yEW, (3)f'1: f(W) —>Wiscontinuous. [Such afunction fiscalled acoordinate system around :1: (see Figure 5-2).] Proof. Ifillisals-dimensional manifold inR", choose h:U—>Vsatisfying (M). LetW={aERf:(a,0) Eh(M)} and deﬁne f:W—> R"byf(a) =h_1(a,O). Clearly f(W) = MT) Uand f'1 iscontinuous. IfH:U—> B,‘isH(z) = (h1(z), ...,h'°(z)), then H(f(y)) =yforallyEW;there- foreH'(f(y)) -f’(y) =Iandf'(y) must have rank lc. Suppose, conversely, that f:W—>R"satisﬁes condition (C). Leta:=f(y). Assume that thematrix (D,~ff(y)), 1_§i,j_§k hasanon-zero determinant. Deﬁne g:WXR"“'° —>R"by 112 Calculus onManifolds W IE1513!i?:'1ii§i§1§1§1E=E" '1'1E15-'52E121111131545I’I3II5:35;=:f:I:IiI§'?IE=§i§I;-:-:>l-:-i‘;171’i'i;?'I'l 5-L;J;f;;f£1I31{.5111573;3‘1;2§;;;I;;1I-I51-J-i‘i7:3:l;7;If-I-Z»I-1-)-fl;-:l;ZI-01-1-I-I-I-I-_'-1'1-I-I-j-;.;.'Jgl-I.;.;.;.;.:._.;;._.;.;.;:.;»:-1;.;.;.;._.;,,.;.;,;_;_::;.;;; ;;1;?;:_-.';:;:<§:;:f:§E:§:§:?:_I;;;;;;;:;§;:;:;:;:;:;:}:;:f:,_:;:_-;;1;:;;-‘-:i:?;:;:;I;I$5;r;:;:;:;I-:-:-:-:l;i:i:i.’3:3I;I;1;I_-:-r;I-:-:-:~::-::':—:7;1:!;5;?:I:;;:;:-: :;:—.*:';_-.;.;.;.;:;.-:-:-:-:-:-:-r-1+:-:»;-;-;.;1.-.-::-.-:-:-: /2 _:§;]:E13:513:357,5-jIfiE:§:f:f:PS3]I5:335:7:3§1j'I§f?§'§.'{:f7f:_ 5:?7;1:1:i:3§:I:1:-':J:I»§15:3:1:?;i:-:f:1.'i;1:?;I:1:f:1§l£IZ-:-:-;1.-:3:311;7:1,-T;I;?;J;;;f1:_-:11;;;:;','§;{;I:{§ 52;);:;:;:::;' FIGURE 5-2 g(a,b) =f(a) +(O,b). Then detg’(a,b) =det(D,-ff(a)), so detg’(y,0) rf0.ByTheorem 2-11 there isanopen setV1’ containing (y,0) andanopenset V2’containing g(y,0) =:2:such that g:V1’—> V2’hasadifferentiable inverse h:V2’—>V1’. Since F1iscontinuous, {f(a): (a,0) EV1’} =Uf'\ f(W) for some open set U. Let V2=V2’f\ Uand V1=g"1(V2). Then V2f\Misexactly {f(a): (a,0) EV1}={g(a,0): (a,0) EV1}: so h(V2F)M)=e'"‘(T/‘2 F)M)=e"‘({e(a,0)= (61.0)ET/1}) =V17)(BkX{0l)- I One consequence oftheproof ofTheorem 5-2should be noted. Iff1:W1—>R"andf2:W2—>R"aretwocoordinate Integration onManifolds 113 systems, then fil°f1IfY1(f2(W2))""’ Bk isdifferentiable with non-singular Jacobian. Iffact, f{1(y) consists oftheﬁrst hcomponents ofh(y). The half-space H"°ERfisdeﬁned as{avEBk:a:’°Z0}. Asubset MofR”isak-dimensional manif0ld-with- boundary (Figure 5-3) ifforevery point asEMeither condi- tion (M)orthefollowing condition issatisﬁed: (M')There isanopen setUcontaining as,anopen set VER”, and adiffeomorphism h:U-> Vsuch that h(Uf\ M)=Vf\(H"’>< {0}) ={z/€V=z/"2.0and2/’°+‘= =2/”=0} and h(x) haslcthcomponent =O. Itisimportant tonote that conditions (M) and (M’) cannot both hold forthesame as.Infact, ifh1:U1—>V1and h2: U2——>V2satisﬁed (M) and (M’),respectively, then h20h1"1 would beadifferentiable map that takes anopen set inBk,containing h(x), intoasubset ofH’°which isnotopen in R". Since det(h2<>h1'1)’ ¢O,this contradicts Problem 2-36. The setofallpoints :1:EMforwhich condition M’is satisﬁed iscalled theboundary ofManddenoted 6M. This (I1) (b) FIGURE 5-3. Aone-dimensional and atwo-dimensional manifold- with-boundary inR3. 1I4 Calculus onManifolds must notbeconfused with theboundary ofaset,asdeﬁned in Chapter 1(seeProblems 5-3and5-8). Problems. 5-1. IfMisais-dimensional manifold-with-boundary, prove that 6Misa(lc—1)-dimensional manifold andM—-6Mis aIt-dimensional manifold. 5-2. Find acounterexample toTheorem 5-2ifcondition (3)isomitted. Hint: Wrap anopen interval into aﬁgure six. 5-3. (a)LetAER"beanopen setsuch that boundary Aisan(n—1)- dimensional manifold. Show that N=AUboundary Aisan n-dimensional manifold-with-boundary. (Itiswell tobear inmind thefollowing example: ifA={stER": Ix}<1or1<Ix}<2} then N=AUboundary Aisamanifold-with-boundary, but 6N$4boundary A.) (b)Prove asimilar assertion foranopen subset ofann-dimen- sional manifold. 5-4.-. Prove apartial converse ofTheorem 5-1:IfMCR"isak-dimen- sional manifold and2:EM,then there isanopen setACR"con- taining scandadifferentiable function g:A——>R""""‘ such that AF)M =g"'1(O) and g’(g) hasrank n—kwhen g(g) =O. 5-5. Prove that ais-dimensional (vector) subspace ofR"isak-dimen- sional manifold. 1 1 /\....- FIGURE 5-4 Integration onManifolds 115 5-6. Iff:R"-> Rm, the graph offisl(a:,g): g=f(x)}. Show that thegraph offisann-dimensional manifold ifand only iffis differentiable. 5-7. Let K"={asER":a:1= Oand $2,...,a:"'"1> O}. IfM EK" isalc-dimensional manifold and Nisobtained byrevolving M around theaxisas‘----=a:""" -0,show thatNisa(Ia+1)- dimensional manifold. Example: thetoms (Figure 5-4). 5-8. (a)IfMisalo-dimensional manifold inR"and It<n,show that Mhasmeasure 0. (b)IfMisaclosed n-dimensional manifold-with-boundary in R",show that theboundary ofMis6M. Give acounterexample if Misnotclosed. (c)IfMisacompact n-dimensional manifold-with-boundary inR",show that MisJordan-measurable. FIELDS AND FORMS ON MANIFOLDS LetMbeais-dimensional manifold inB”andletf:W—>R” beacoordinate system around as=f(a). Since f’(a)hasrank lo,thelinear transformation f.1.:Rka-—->R”;is1-1,andf.1.(R"’,,) isalo-dimensional subspace ofR”,,. Ifg:V—>R”isanother coordinate system, with as=g(b), then 9*(Rkb) =f*(f—1 °9)*(Rkb) =f*(Bka)- Thus thels-dimensional subspace f*(R"°.1) does notdepend on thecoordinate system f.This subspace isdenoted M1,,and iscalled thetangent space ofMatas(seeFigure 5-5). In later sections wewillusethefactthat there isanatural inner product T1,onM1,,induced bythat onR",,: ifv,wEM1deﬁne T,,(v,w) =(v,w),,. Suppose that Aisanopen setcontaining M,andFisadiffer- entiable vector ﬁeld onAsuch that F(as) EM1,foreach asEM. Iff:~W—> R”isacoordinate system, there isa unique (differentiable) vector ﬁeldG onWsuch thatf.1.(G(a))= F(f(a)) foreach aEW. Wecanalso consider afunction F which merely assigns avector F(as) EM1,foreach asEM; such afunction iscalled a’vector ﬁeld onM. There isstill aunique vector ﬁeld GonWsuch that f*(G(a)) =F(f(a)) for aEW;wedeﬁne Ftobedifferentiable ifGisdifferentiable. Note that ourdeﬁnition does notdepend onthecoordinate I16 Calculus onManifolds ff 3/‘ -/.\v *1 /\€, \ v’ ‘;,>w<' /'//*}’\i~§, 2/ / — /‘\ /\ <‘ \‘>’Q1.\ V m ‘H / \. f .;.$2=<e‘$:¢$;>..»-q>.~;:\-a.~;;:-‘4~ \l;;;I-2)'-"1-.1-:-oz; . 1.-r&'§:¢;lr'\\*?"E1¥3=i:'7¢'2lE; -.‘.¢-:-:1:-rffr-5-.¢:1:1'=~">" --1-'.-.-'l:-Fir.-o':f:> .6‘-’:-"-’»:3:v I l:I<”" .-/'1‘>. -.\{-.-.-PH ..-13;}./<. .0-$;;.T_.i§c.-.1.'F'5'-.1<¢’ "1-I-.'-: .;.-5:1:7:I:i'-zf:-.-1-$xk;$i:h:~.1:i:-:r :-:l:?;§:T:;-, .'1-I7'‘< ""1;-17'.‘I"»J?:33'i‘z”’E1i"Fii1§?i‘$"'§?""€ii?§1'Y7'""'.-.-.-.-.-142;:-.;ﬂ-.o_,:<-:-;g.‘{‘-4t? ~§,. ,1.--'..-.>'~;->:;:-71'». .->.;r;:5>. :1;-.-;.;'\-_;.~:;‘~ ..~:;.;.;.§_.-,g:;.-.;:~_.Q .3;/_:;.'}_§Ng.>;4-:-' ''4:551::1:-:-:-f~:?.1".i-:-:-.<:>-Fiig-1;:/3.‘-. '..¢~:~:5Q;;:-7-11>‘-@1211->1-‘Qt,-:~:-1-1-:-" NI1 '-I-'1-‘v~-‘<6?-t~\§2<t-‘~-‘ ’\<‘ -..~'*‘-}-.1- .. +1-1:1-r "-t3.’=I4".1;'**Y:‘ ‘/““"*'/ 1.€:'A.<¢/-9 .'<=;H"’ -4;.-.;,~ ‘-.-'!‘<‘|'J.'.'K} ;..,;.- {>24 FIGURE 5-5 system chosen: ifg:V-+ R”andg...(H(b))=F(g(b)) forall bEV,then thecomponent functions ofH(b)must equal the component functions ofG(f"1(g(b))), soHisdifferentiable ifGis. Precisely thesame considerations hold forforms. Afunc- tion nowhich assigns w(as) EAP(M 1,)foreach asEMiscalled ap-form OnM. Iff:W—>R"isacoordinate system, then f*wisap-form onW;wedeﬁne wtobedifferentiable iff*wis. Ap-form toonMcanbewritten as to= 2 w1,,___,,;,,da;f‘ /\'''/\dash’. i1<"'<i1. Here thefunctions w,;,,___,1,, aredeﬁned only onM. The deﬁnition ofdwgiven previously would make nosense here, since D,-(w,;,, ___,1,)hasnomeaning. Nevertheless, there isa reasonable way ofdeﬁning (lw. Integration onManifolds 117 5-3 Theorem. There isaunique (p+1)-form clwonM such thatforevery coordinate system f:W—>R"wehave f*(dw) =d(f*w)- Proof. Iff:W—-> R"isacoordinate system with as=f(a) andv1,...,o1,_,_1 EM1,, there areunique w1,...,w1,+1 in R"°1, such that f*(w,;) --121;.Deﬁne do.>(as)(o1, ...,v1,+1) = d(f*w)(a)(w1, ...,w1,+1). One cancheck that thisdeﬁnition ofdw(as) does notdepend onthecoordinate system f,sothat do:iswell-deﬁned. Moreover, itisclear that doshas tobe deﬁned this way, sodo:isunique. Itisoften necessary tochoose anorientation g1,foreach tangent space M1,ofamanifold M. Such choices arecalled consistent (Figure 5-6) provided that forevery coordinate \~/'\~/\~/ ~</\~/ .\_,. \~/\7'< hi (b) FIGURE 5-6. (a)Consistent and (b)inconsistent choices oforien- tations. 118 Calculus onManifolds system f:W-—>R”anda,bEWtherelation lf*((e1)a): ''':f*((els)a)l =I-‘f(c) holds ifand only if [f*((@1)r>), --->f*((@k)b)l =um»)- Suppose orientations 111,have been chosen consistently. If f:W—>R"isacoordinate system such that lf*((e1)a)a ''':f*((elc)a)l :l1f(a) forone, andhence forevery aEW,then fiscalled orien- tation-preserving. Iffisnotorientation-preserving and T:Bk——>R"isalinear transformation with detT=--1,then f0Tisorientation-preserving. Therefore there isanorienta- tion-preserving coordinate system around each point. Iffand gareorientation-preserving and as=f(a) =g(b), then the relation [f*((@1)a). --->f*((eh)a)l =14:1:=[9*((@1)i), ---.9*((@i)z>)] -- 1--:1 .--._._ .' ..;r;=5>E1i=i=i’=3E§i1' .‘ -as‘ "sq:.- : ‘'1;¢:1;;:§:;:;1;1;l:I:1:i:1i:§;§l;._ ,.;-11;1;-:-:7:§:1:5:;:§:[ﬁ:;:;;;-;-§:$§1§E.§§:<g§f€c'é$1 ¢,§§}'ﬁ:§‘»::;:‘f» 3‘?/.~;;.__V--\\n ' '1-I-er-'-'.'.-a.--_-.--Tm-1-,-.111.-;:1;-1-1:2!121:1;I:;:;:{:14:11;:-1:1:I$1111;I:;:;:§:1:5-:11;1-1:!:I:1;I:I:1:§:;1:5;:s;r5i>-:-_ ~. ;,-‘e:>§:'-; , ‘gs-1,,../.<, _....» ....-" €~j¢,;,;=;;‘<=:}_!;:=1:;=‘;5;1;1;_;§;:=;;;;;=-;:;:s; ',-5-i.-R31:a:Iilli:i;?;7:1I5»'- =_-'21.3'35?5{ﬁg$2'7$4t§§Z5::2375i'i-i-'-‘I3='-.'i.-.-.-».- '.'.,-/@5;.,1¢-:-..,-:-.~.-.»;-:4-:Z-2-1-:-:»:-;.;.<.;q1R¥.-.-‘-;”.;.;.;q-:4A-:-.-;-.-.'.;.;.:.3-9;-:-.-.\.-Q5;-;._._"f-11:»-,74;:;:;:;:t:iq1;.f;1:;!1-1-;:-:;1-:;:;1§$;I>:-:1!-:4-I,. 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Ig\% " "\ 2'--> F‘1‘! 1.-ﬁ¢'-~.;.'¢.<!-:'-'-F'+4*- :I;' .-:-li:=E':-‘.3:-5 " 1,11:1;:;:;:;:;-::1¢:;;;1:1 1:;:§:;$:-;§: _-t.-2'"-2'-:-: ;:;"~:;i;-f'%:=:<:;:_';> r'"->_. .-€tI;::.i;3;;;:;:-;L;15:.E*i:i::-22:2;-1 _-:’r_-.'-" ‘I»J.-:'-;-:-1-1-'I»I-€>i-1‘P4:- - 1», ...,-.-,_saw‘?:?Ri=;},€1:<:¥Ii;i1':‘.'-:-:4»2'-:=:-:»;~(.}\§;7;l$)§'$$f<iQ>_i;i M‘:=:;=:=:=:EE:=:=:E===s$:=121$EiliiiiiiiiiiiifiiiiiE1€=E=31E:5;i'g=;=5.';s=§:3s1~:¥:¢17-‘F155-‘2-§'}>.»1I'.._._.,,A} "ll.'11&5.""141..-._.§-. -pi,“ :-;:;:;: -"Q, 31?£1 . ‘F\'-I-S~:-2-s;i:iS:I151???LE:1:?:1ti-21E.5:'l:I3:l->12175:2:-:-s;r;=s:l:1:?:%:¢:1:=re=5; FIGURE 5-7. The Mobius strip, anon-orientable manifold. A basis begins atP,moves totheright and around, and comes back toPwith thewrong orientation. Integration onManifolds 119 implies that l(g_1°.f)*((e1)¢1): ''':(g_1°.f)*((6k)a)l =l(e1)b: '''2(elc)bla --1 rsothat det(g 0f)>0,animportant fact toremember. Amanifold forwhich orientations n1,can bechosen con- sistently iscalled orientable, andaparticular choice ofthe #1iscalled anorientation nofM. Amanifold together with anorientation giscalled anoriented manifold. The classical example ofanon-orientable manifold istheMobius strip. Amodel canbemade bygluing together theends ofastrip of paper which hasbeen given ahalf twist (Figure 5-7). Ourdeﬁnitions ofvector ﬁelds, forms, andorientations can bemade formanifolds-with-boundary also. IfMisah-dimen- sional manifold-with-boundary and asE6M, then (6M)1, is a(ls—1)-dimensional subspace oftheIt-dimensional vector space M1,. Thus there areexactly two unit vectors inM1, which areperpendicular to(6M)1; they canbedistinguished asfollows (Figure 5-8). Iff:W——>R”isacoordinate system with WEH’°andf(0) =at,then only oneofthese unitvectors isfs(co)forsome nowith of<O.This unit vector iscalled the outward unit normal n(a:); itisnothard tocheck that this deﬁnition does notdepend onthecoordinate system f. Suppose that 11isanorientation ofaIt-dimensional manifold- with-boundary M. IfxE6M, choose v1,...,v;1._._1 E(6M)1, sothat [n(a:), o1,...,o;,__1] =111,. Ifitisalso true that [n(a:), w1,...,w;,__.1] =1.11,,then both [v1,...,v;,_1] and [w1, ...,w;,._1] are the same orientation for(6M)1,. This orientation isdenoted (6g)1,. Itiseasy toseethat theorienta- tions (6u)1, forasE6M, areconsistent on6M. Thus ifMis orientable, 6Misalso orientable, and anorientation ,1forM determines anorientation 6nfor6M,called the induced orientation. IfWeapply these deﬁnitions toH"with the usual orientation, weﬁnd that the induced orientation on R’°"1 ={asEHf: as"=0}is(—-1)" times theusual orienta- tion. The reason forsuch achoice willbecome clear inthe next section. IfMisanoriented (n—1)-dimensional manifold inR",a substitute foroutward unit normal vectors canbedeﬁned, I20 Calculus onManifolds .-.-.\~;->;-.-- ,._._-.;:?-91092-:-it1";-.-. . ;i;-1=..~;~:i;1:€1:l:;§4:521%14>>;4_%?'2'.-. 7' -.»1;':~._-,\;.;.;-;2,>;-;g.1<¢§;.;:<;.;.;.,r-1, -:4-, :-:-:-:§:>_~;-_\1-.>:'6:l1-"17:Yrbﬂ:$<->1-I-1-1.‘?-I45-3"/\> .""r 'Y:7§.<§:3:1'>.J:i;1---2-1-2:14:+.-“=1-:1-itr5s\$¢>st:1a;»I~ -2- E-f-1-:11-1-'-Tn-:1-. _;;;:_-:;(;.g;:;;;;-‘,j:3§§;_;1;‘:;;5§;;:-5;-3155.;:1.1;:;.;;;:;._ ~:1:;:;.;:;3;:5_('rc¢:;:-__ -_ .-:b:1:1:1‘<f'¢:'-Mia-;1;1¢z<g$;¢;:;1;.§:<>:-.<:~:-.l:Zs:1;5;I$;; ':I:-‘:I:'¢;;2;:;>.g:;:;:E1; 1-j¢;:_:-.-*';"l"'?:-‘Z'1'I}2*P?‘-7-'-1+D/2-:-2*->1-;¥<;3fZ'I'?~{-2=1;X§*-l-lg-. '1-I-use-1-:-6: ~"1"-"A-l)F\‘i'.-?-'~‘.1i55fl.$I5‘€#7521"iii:!:5‘;I-'i41iF7i3<5'-:3‘1:-:>.]:-. '55$:i:?:*.i:l:f:i:&§:-'1'-. 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Some outward unit normal vectors ofmanifolds-with- boundary inR3. even though Misnotnecessarily theboundary ofann-dimen- sional manifold. If[v1,...,v,,_1] =g1,,wechoose n(as) in R"1, sothat n(a:) isaunit vector perpendicular toM1,and [n(a:), v1,...,v,,._1] istheusual orientation ofR"'1,. Westill call n(a:) theoutward unit normal toM(determined byn). Thevectors n(a:) vary continuously onM,inanobvious sense. Conversely, ifacontinuous family ofunit normal vectors n(a:) isdeﬁned onallofM,then wecandetermine anorientation of M. This shows that such acontinuous choice ofnormal vectors isimpossible ontheMobius strip. Inthepaper model oftheMobius strip thetwo sides ofthepaper (which has thickness) may bethought ofastheend points oftheunit Integration onManifolds I21 normal vectors inboth directions. The impossibility of choosing normal vectors continuously isreﬂected bythe famous property ofthe paper model. The paper model is one-sided (ifyou start topaint itonone side you end up painting itallover); inother words, choosing n(a:) arbitrarily atonepoint, and then bythecontinuity requirement atother points, eventually forces theopposite choice forn(a:) atthe initial point. Problems. 5-9. Show that M1,consists ofthetangent vectors att ofcurves cinMwith c(t)=as. 5-10. Suppose (3isacollection ofcoordinate systems forMsuch that (1)Foreach asEMthere isfE(3which isacoordinate system around as;(2)iff,gEe,then det(f_1Qg)’>0.Show thatthere isaunique orientation ofMsuch that fisorientation-preserving iffE8. 5-11. IfMisann-dimensional manifold-with-boundary inR”,deﬁne n1,astheusual orientation ofM1,=R”1, (the orientation nso deﬁned istheusual orientation ofM). IfatE6M,show that thetwodeﬁnitions ofn(a:) given above agree. 5-12. (a)IfFisadifferentiable vector ﬁeld onMER”,show that there isanopen setADMand adifferentiable vector ﬁeld F onAwith F(:v) =F(a:) foratEM. Hint: Dothis locally and usepartitions ofunity. (b)IfMisclosed, show that wecanchoose A=R”. 5-13. Letg:A—->RPbeasinTheorem 5-1. (a)IfatEM=g'"1(0), leth:U——>R"betheessentially unique diﬁeomorphism such that g<>h(y) ==(y""*"+1, ...,y") and h(O) ===:23.Deﬁne f:R"“P —->R"byf(a) =h(O,a). Show that f1. is1-1sothat then—-pvectors f1.((e1)2), ...,f.,.((e1._1,)2) are linearly independent. (b)Show that orientations g1,canbedeﬁned consistently, so that Misorientable. (c)Ifp=1,show that thecomponents oftheoutward normal atataresome multiple ofD1g(a:), ...,D1g(a:). 5-14. IfMCR”isanorientable (n—1)-dimensional manifold, show that there isanopen setACR"andadifferentiable g:A—->R1so that M=g“1(0) andg’(a:) hasrank 1forasEM. Hint: Prob- lem5-4does thislocally. Usetheorientation tochoose consistent local solutions andusepartitions ofunity. 5-15. LetMbean(n—1)-dimensional manifold inR”. LetM(e)be thesetofend points ofnormal vectors (inboth directions) of length eand suppose eissmall enough sothat M(e)isalso an 122 Calculus onManifolds (n—1)-dimensional manifold. Show that M(e)isorientable (even ifMisnot). What isM(e)ifMistheMobius strip? 5-16. Letg:A—->RPbeasinTheorem 5-1. Iff:R”—->Risdifferentiable andthemaximum (orminimum) offong“1(0) occurs ata,show that there areA1,...,>\,,ER,such that ‘R <1)1>,~r<a>=2M-D,-g*<a> j=1,...,n.i=1 Hint: This equation canbewritten df(a) =Z§‘_=1>\,;dg':(a) andis obvious ifg(x) =(:c"“p+1, ...,x"). Themaximum offong""(0) issometimes called themaximum offsubject totheconstraints gl=0.One can attempt to ﬁnd abysolving thesystem ofequations (1). Inparticular, if g:A—+R,wemust solve n+1equations D1'f(a) =>\Dj9(a), g(a)=0, inn+1 unknowns a1,...,a",)\, which isoften very simple ifweleave theequation g(a) =0forlast. This isLagrange’s method, and theuseful but irrelevant )\iscalled aLagrangian multiplier. The following problem gives anice theoretical use forLagrangian multipliers. 5-17. (a)Let T:R”-> R”beself-adjoint with matrix A=(ctr,-), so that a,-5=a,--;. Iff(x) =(T:z:,:c) =Ea,;,-:vi:z:", show that Dk_f(x) = 2E}”_1a;,,-xi. Byconsidering themaximum of(T:z:,:z:) onS""1 show that there is:1:ES""1 andAERwith Ta:=Xx. (b)IfV={yER”: (:z',y) =O}, show that T(V) CVand T:V—-> Visself-adjoint. (c)Show that Thasabasis ofeigenvectors. STOKES’ THEOREM ON MANIFOLDS Ifwisap-form onalc-dimensional manifold-with-boundary Mandcisasingular p-cube inM,wedeﬁne /¢.,=/at,c [O,1]P precisely asbefore; integrals over p-chains arealsodeﬁned as before. Inthecase p=lcitmay happen that there isan open setWD[0,1]"’ and acoordinate system f:W-—>R"such that c(a:) =f(x) forxE[0,1]'°; aIc-cube inMwillalways be Integration onManifolds 123 understood tobeofthistype. IfMisoriented, thesingular it-cube ciscalled orientation-preserving iffis. 5-4 Theorem. Ifc1,c2: [0,1]" -—~>Maretwoorientation- preserving singular It-cubes intheoriented h-dimensional mani- fold Mand wisalc-form onMsuch that w=Ooutside of v1([9,1]’°) Qc2([0,1]'°), W" c[..=c[... Proof. Wehave [w=fc.*<<»>= [<c;1o¢1>*c2*<w>.l9»1l'° l0,1l" (Here of0c1isdeﬁned only onasubset of[0,1]"’ and the second equality depends onthefact that w=0outside of c1([0,1]"’) ('\c2([0,1]").) Ittherefore suffices toshow that I(cf OC1)*Cg*((.0) = Ic2*(w) ==Iw. [o,111= [o,111= C2 Ifc2*(w) =fdxl /\---/\dxkandcflOc1isdenoted byg, then byTheorem 4-9wehave (C2_1 °C1)*Cg*(w) =g*(fd131 /\'''/\dﬁllk) =(fog)-detg'-dxl /\-'-/\dank =(fog)'ld6l3g'l'd$1 /\---/\dash, since detg’=det(c2_1 0c1)’>0.The result now follows from Theorem 3-13. I Thelastequation inthisproof should help explain why we have had tobesocareful about orientations. Letcobeait-form onanoriented k-dimensional manifold M. Ifthere isanorientation-preserving singular lc-cube cinMsuch that w=0outside ofc([0,1]"), wedeﬁne I(.0= f(.0. M c Theorem 5-4shows IMwdoes notdepend onthechoice ofc. 124 Calculus onManifolds Suppose now that wisanarbitrary lc-form onM. There isan open cover 0ofMsuch that foreach UE0there isanorienta- tion-preserving singular is-cube cwith UCc([0,1]"). Letbe apartition ofunity forMsubordinate tothis cover. We deﬁne .z~=2werp Q provided thesumconverges asdescribed inthediscussion pre- ceding Theorem 3-12 (this iscertainly true ifMiscompact). Anargument similar tothatinTheorem 3-12shows thatIMw doesnotdepend onthecover 0oron. Allourdeﬁnitions could have been given forait-dimensional manifold-with-boundary Mwith orientation u.Let6Mhave theinduced orientation 6n. Letcbeanorientation-preserv- ingit-cube inMsuch that c(;,_0, liesin6Mandistheonly face which hasanyinterior points in6M. Astheremarks after thedeﬁnition of6;.»show, c(;,,0, isorientation-preserving ifItis even, butnotifitisodd. Thus, ifcoisa(it—1)-form onM which is0outside ofc([0,1]"), wehave fa=(-1)’*J w. c(7¢.9) Ontheother hand, c(;,,,0, appears with coefficient (—1)”in6c. Therefore fw== [w=(--1),‘ fw=)£w. 6c (—1)"c(k,o) C(k,0) 6 Ourchoice of6awasmade toeliminate anyminus signs inthis equation, andinthefollowing theorem. 5-5 Theorem (Stokes’ Theorem). IfMisacompact oriented It-dimensional manifold-with-boundary and wisa (lc-1)-form onM,then Jdw=a1£w. (Here 6Misgiven theinduced orientation.) Proof. Suppose ﬁrst that there isanorientation-preserving singular It-cube inM--6M such that w=0outside of Integration onManifolds 125 c([O,1]"). ByTheorem 4-13 and thedeﬁnition ofdo:wehave cfdw =[01, c*(dw) =[0'!];d(c*w) =a}£ c*w =66] w. lJdw=Cjdw=a!w=O, since w=0on6c. Ontheother hand, fag;w=0since co=0 on6M. Suppose next that there isanorientation-preserving singular It-cube inMsuch that c(;,_0, istheonly facein6M,andcc=0 outside ofc([0,1])". Then 1J’dw=6[dw=acfw=6j,/gm. Now consider thegeneral case. There isanopen cover 0 ofMandapartition ofunity forMsubordinate to0such that foreach cpEtheform o-wisofoneofthetwosorts already considered. Wehave 0=d<1>=d(Z ¢>)=2dr.¢E ¢E*I> 2do/\co=O. rp Q Since Miscompact, thisisaﬁnite sum andwehave ¢gq)]Jdq0/\w=0.Then sothat Therefore ‘Jdw=¢g@lJqo'dw=¢g¢ﬂ[d<p/\w+<p'dw =¢g6£(p.w =[.,.|BM Problems. 5-18. IfMisann-dimensional manifold (ormanifold- with-boundary) inR", with theusual orientation, show that 126 Calculus onManifolds _lMfdxl/\---/\dx”, asdeﬁned inthissection, isthesame as IMf,asdeﬁned inChapter 3. 5-19. (a)Show that Theorem 5-5isfalse ifMisnotcompact. Hint: If Misamanifold-with-boundary forwhich 5-5holds, then M—6M isalsoamanifold-with-boundary (with empty boundary). (b)Show that Theorem 5-5holds fornoncompact Mprovided that cuvanishes outside ofacompact subset ofM. 5-20. Ifcoisa(lo-—1)-form onacompact Ic-dimensional manifold M, prove that IMdo.»=0.Give acounterexample ifMisnot compact. 5-21. Anabsolute ls-tensor onVisafunction 11:Vk—-+Roftheform lwlfor0:EAk(V). Anabsolute k-form onMisafunction 17 such that 11(:l7) isanabsolute Ir:-tensor onM,,.Show that {M17 canbedeﬁned, even ifMisnotorientable. 5-22. IfM1ER"isann-dimensional manifold-with-boundary and M2EM1—6M1 isann-dimensional manifold-with-boundary, andM1,M2 arecompact, prove that /@-/Q»BM1 BM: where coisan(n-1)-form onM1,and6M1 and6M2 have theori- entations induced bytheusual orientations ofM1andM2.Hint: Find amanifold-with-boundary Msuch that6M=6M1 U6M1 and such that theinduced orientation on6Magrees with that for 6M1 on6M1 andisthenegative ofthat for6M2on6M2. THE VOLUME ELEMENT LetMbeaIt-dimensional manifold (ormanifold-with-bound- ary) inR",with anorientation u.IfasEM,then 11,,and the inner product Ta,wedeﬁned previously determine avolume element w(:U) EA"’(M,,). Wetherefore obtain anowhere-zero h-form toonM,which iscalled thevolume element onM (determined byu)and denoted dV,even though itisnotgen- erally thedifferential ofa(ls-—1)-form. The volume ofM isdeﬁned as[MdV,provided this integral exists, which is certainly thecase ifMiscompact. “Volume” isusually called length orsurface area forone- and two-dimensional manifolds, andclVisdenoted ds(the “element oflength”) or dA[ordS](the “element of[surface] area”). Aconcrete caseofinterest tousisthevolume element ofan Integration onManifolds 127 oriented surface (two-dimensional manifold) MinR3. Let n(a:) betheunit outward normal atasEM. IfwEA2(M,,) isdeﬁned by v w(v,w) =det(iv ), "(1v) then w(v,w) =1ifvandwareanorthonormal basis ofM,,with [v,w] =ax. Thus dA=w.Ontheother hand, w(v,w) = (vXw,n(:c)) bydeﬁnition ofvXw. Thus wehave dA(v,w) =(22Xw,n(x)). Since vXwisamultiple ofn(a_:) forv,wEM,,, weconclude that dA(v,w) =IvXw] if[v,w] =ax. Ifwewish tocompute thearea ofM,wemust evaluate f[0,1]= c*(dA) fororientation-preserving singular 2-cubes c.Deﬁne E(a)=[D1c‘(a)l2 +[D1c2(@)]2 +[D1c3(a)]2, F(a) =D1c1(a) -D2c1(a) "l"D162(a) 'D202(a) +D1¢3(a) 'D2c3(a). G(a) =lD261(a)l2 "l"lD262(a)l2 "l"lD263(a)l2- Then 0*(dA)((61)a.(@2)a) =dA(¢*((@1)a)1¢*((62)a)) =1(D1c1<a>.D1c*<a>.1>1c3<a>> ><<1>2c‘<<»>.1>2c*<<»>.1>2c3<a>>I=\/E(a)G(a) -F(a)é byProblem 4-9. Thus fc*(dA) -f\/EG—F2.[0.1l’ [9.1l' Calculating surface area isclearly afoolhardy enterprise; fortunately oneseldom needs toknow thearea ofasurface. Moreover, there isasimple expression fordAwhich sufﬁces for theoretical considerations. 128 Calculus onManifolds 5-6 Theorem. LetMbeanoriented two-dimensional man- ifold (ormanifold-with-boundary) inR3andletnbetheunit outward normal. Then (1) dA=n1dy/\dz+n2dz/\da:+n3d:v/\dy. Moreover, onMwehave (2) n1dA=dy/\dz. (3) n2dA=dz/\dx. (4) n3dA=da:/\dy. Proof. Equation (1)isequivalent totheequation v dA(v,w) =det(w '"»(~"v) This isseen byexpanding thedeterminant byminors along thebottom row. Toprove theother equations, letzER3,. Since vXw=an(a:) forsome ozER,wehave (Z,"»(fv)) '(vX‘w,"»(1v)) =<2,"/(1v))<1 =(2,011?/(111)) =<2,"Xw)- Choosing 2=e1,e2,and e3weobtain (2),(3),and (4). I AWord ofcaution: ifwEA2(R3,,) isdeﬁned by w=n1(a) -dy(a) /\dz(a) —|—n2(a) -dz(a) /\da:(a) +'"»3(a) '6111(0) /\d?/(0), itisnottrue, forexample, that n1(a) -w=dg(a) /\dz(a). The two sides give thesame result only when applied to v,wEMa. Afewremarks should bemade tojustify thedeﬁnition of length and surface area wehave given. Ifc:[0,1]-—>R"is differentiable andc([0,1]) isaone-dimensional manifold-with- boundary, itcanbeshown, buttheproof ismessy, that the length ofc([0,1]) isindeed theleast upper bound ofthelengths Integration onManifolds 129 / “‘\\ ‘Inn -___‘- _3......_................_ -._—---I-P -I""" ; “K I-um _-—im-._ ___.-__-III‘ -_-|-I-' ﬂ j W-ii-hi ‘In- - . \ r ‘—— _ J/,.-”"’_k I m I ... _..__._._.11____, 7 1 1 I‘;-T g i ..- *‘*n_m E 4|-IIIﬁ-l- 4_r Am’ -1.5-—-5 1;’___--r§""'- W ﬁ__-"-"*¥i|--___ g W "-1 II I-"*1-1 -- 19; 1 Jib‘___- E-_ﬁn- FIG URE 5-9. Asurface containing 20triangles inscribed inapor- tionofacylinder. Ifthenumber oftriangles isincreased suﬂiciently, by making thebases oftriangles 3,4,7,8,etc., suﬁiciently small, thetotal area oftheinscribed surface canbemade aslarge asdesired. 130 Calculus onManifolds ofinscribed broken lines. Ifc:[0,1]2——> R”, one naturally hopes that thearea ofc([0,1]2) willbetheleast upper bound of theareas ofsurfaces made upoftriangles whose vertices liein c([0,1]2). Amazingly enough, such aleast upper bound is usually nonexistentwone canﬁndinscribed polygonal surfaces arbitrarily close toc([0,1]2) with arbitrarily large area! This isindicated foracylinder inFigure 5-9. Many deﬁnitions ofsurface area have been proposed, disagreeing with each other, butallagreeing with ourdeﬁnition fordifferentiable surfaces. For adiscussion ofthese difficult questions the reader isreferred toReferences [3]or[15]. Problems. 5-23. IfMisanoriented one-dimensional manifold in R"andc:[0,1]->Misorientation-preserving, show that ¢*ds>= [<>12+ +[(c>12.2*/"'-"K s\Ohi 3 5-24. IfMisann-dimensional manifold inR”,with theusual orienta- tion, show that dV=d:r1/\---/\dx”, sothat thevolume of M,asdeﬁned inthissection, isthevolume asdeﬁned inChapter 3. (Note that thisdepends onthenumerical factor inthedeﬁnition of cu/\17.) 5-25. Generalize Theorem 5-6tothecase ofanoriented (n-—1)-dimen- sional manifold inR”. 5-26. (a)Iff:[a,b]——>Risnon-negative and thegraph offinthe my-plane isrevolved around thea:-axis inR3toyield asurface M, show that thearea ofMis b f2111‘\/1+(1')? (b)Compute thearea ofS2. 5-27.IfT:R"-——> R"isanorm preserving linear transformation andM isalc-dimensional manifold inR",show that Mhasthesame volume asT(M). 5-28. (a)IfMisalc-dimensional manifold, show that anabsolute It-tensor [dV[ canbedeﬁned, even ifMisnotorientable, sothat thevolume ofMcanbedeﬁned asfM[dV[. (b)Ifc:[0,21r] X(—1,1)-—> R3isdeﬁned byc(u,v) = (2’cos u+vsin(u/2)cos u,2sinu+vsin(u/2) sinu,vcosu/2), show that C([O,21r] X(—1,1)) isaMobius strip andﬁnd itsarea. Integration onManifolds I31 5-29 5-30 5-31Ifthere isanowhere-zero k-form onalc-dimensional manifold M, show that Misorientable. (a)Iff:[0,1]——>Risdifferentiable andc:[0,1]-—>R2isdeﬁned by c(:z:)=(a:,f(:t)), Showthat¢([o,11) haslength ft,\/1+(102. (b)Show that thislength istheleast upper bound oflengths of inscribed broken lines. Hint: If0=tn§t15---3tn=1, then We-Cs--1>l=\/<r-*1--1>”+<f<*~">-M--1>>”=\/(:1.-t.--1)?+r'<s.-)2<t.- -t.-_.1>2 forsome s,-E[t,;_1,t,]. Consider the2-form wdeﬁned onR3—0by 3xdy /\dz+ydz /\dz+zdzv/\dy w_ ml (;1;2+y2+ z2)f 3 (a)Show that cuisclosed. (b)Show that <»(r)(vp.wp) =Ir] Forr>()letS2(r) ={:0ER3: =r}. Show that corestricted tothetangent space ofS2(r) is1/r2 times thevolume element, andthat I320.) cu=41r. Conclude thatcoisnotexact. Neverthe- lesswedenote tobyd9since, asweshall see,d9istheanalogue of the1-form d6onR2—0. (c)Ifopisatangent vector such that v=hpforsome AER show that d9(p)(v,,,w,,) =0forallwp. Ifatwo-dimensional manifold MinR3ispart ofageneralized cone, that is,M istheunion ofsegments ofrays through theorigin, show that [[11/1 d9=O. (d)LetMER3-—0beacompact two-dimensional manifold- with-boundary such that every raythrough 0intersects Matmost once (Figure 5-10). The union ofthose rays through 0which intersect. M,isasolid cone C(M). Thesolid angle subtended byM isdeﬁned asthearea ofC(M)(WS2,orequivalently as1/r2times thearea ofC(M) KNS2(r) forr>0.Prove that thesolid angle subtended byMis[[111d(-3|. Hint: Choose rsmall enough so that there isathree-dimensional manifold-with-boundary N(asin Figure 5-10) such that 6Nistheunion ofMand C(M) KNS2(r), andapart ofageneralized cone. (Actually, Nwillbeamanifold- with corners‘ seetheremarks attheend ofthenext section.) " 2 132 4"’ ___ —- / 4. l. 11'1 -,._,-..-..=.»-,.1 1;:1-..,1a;_=a§;_;.;3§,§;_3;;;§ -.-.;-;.j-,-.-»_-,-.¢.-.-;, 1-._»,.: .., "- -_\-1.-,,~_ .-.-/--.;'..,--.;>..=;1=-,'. "‘...._...‘.'.-,'.-‘_L»_1.-\;‘,‘.} '»'--,\-'--.1 -.<.-.».:»-_'.=':- '.-1.-.-..-_-:.='-;-‘-' "'--. ""--;:»-\-._-.-.:, -.-.'.-._r.-. -:1-.\_-;-.>_=2;1._,,,._.;,1..- ._\.1-,;~.-..-2-.:a '.';=;'=_=.'='-2'5 _.-_,\|;.-_\-,"_ -‘.'-'.'->L'- "- -.-..--;-=-.--...=--..-.=3='--.=2:=>-.'i-.=1==-\=-».,\.-.--.. .._._,_.__, ,,_ =.==.:a-.;=.-.-rz=:,&:;=1';=g =:_=,':'-1'=':r,'t'=_E':<_2€_'=';gE_-' '"'"- 1I-.I1E-'-?I:'éI'-_=?.-F.EI(€%" " .-.__..__\'.'-I113:-.-.'-.1 =1-.-3;-E '-‘-.1.=3%'=-:-.1‘-.11,-:5-.-.'1\1.1%.-. . §._-_-_-_\.;\_-_'.' '':"?J-K‘-‘Eli‘5-.:'§'.=?.-.31‘-E.::_-.1‘:'51-526--.‘-E'»En ~:~a-.:~1=-='-.-1.=I-'41-+1=:-532-'-.=.»a=~;-. \=-‘-:1‘1'--.1-:.'-.'-V-:-'.'.~.:'-.-=.-.'-¢\'- --1--_:-‘.= -.-.--,111----'9.-;:-.='--. =:-an-.-.:r».~.->1=z'= "‘"' \-;--.-.1».-.21:. .\_. ..._“#F_'\"'».¢-¢.\.=-.a=:.~.'--.=a-.-;->=_=-.x-.==- '-;-:-.';-,=;-.'a+.».=.-.=Calculus onManifolds C(M) ., "'~_ l 1 1 1 1<1:1R3benonintersecting closed curves. l)eﬁne thelinking number l(f,g) offand gby(cf.Problem 4-34) lag)=All‘ll’ 5130 (a)Show that if(F,G) isahomotopy ofnonintersecting closed curves, then l(F0,G0) =l(F1,G1). Integration onManifolds 133 5-33(b)Ifr(u,v) =[f(u) —g(v)[ show that 11 l(f,g)=€;ff -A(u,v)dudv 0 0 where (f1)'(u) (f2)’('11) (f3)'(v») A(u,v) =det( (o‘)’(v) (o2)'(v) (93)’(v) f1(u) —91(1)) f2(u) "".q2(v) f3(u) "".q3(v) (c)Show that l(f,g) =0iffand gboth lieinthemy-plane. The curves ofFigure 4-5(b)aregiven byf(u)=(cosu,sinu,0) and g(o)=(1+cosv,0,sinv). You may easily convince yourself that calculating l(f,g)bytheabove integral ishopeless in thiscase. Thefollowing problem shows how toﬁndl(f,g)without explicit calculations. (a)If(a,b,c) ER3deﬁne (:2:-—a)dy/\dz"["(Z/-"b)dz/\d:1:-[-(z-——c)d:v/\dy d9‘“"”"” [(11—a>2+ (y—W+<z-oat 03' IfMisacompact two-dimensional manifold-with-boundary in R3and (a,b,c) EMdeﬁne Q(a,b,c) = ‘/‘d9(a,b,c). M Let(a,b,c) beapoint onthesame sideofMastheoutward normal and(a’,b’,c’) apoint ontheopposite side. Show that bychoosing (a,b,c) suﬂiciently close to(a’,b',c') wecan make S2(a,b,c) — Q(a",b’,c’) asclose to-"411" asdesired. Hint: First show that if M=6Nthen S2(a,b,c) ='"'41r for(a,b,c) EN——Mand S2(a,b,c) = 0for(a,b,c) EN. (b)Suppose f([0,1]) =6M forsome compact oriented two- dimensional manifold-with-boundary M. (Iffdoes notintersect itself such anMalways exists, even iffisknotted, see[6],page 138.) Suppose that whenever gintersects Mat:1:thetangent vector vof gisnotinMx. Letn+bethenumber ofintersections where v points inthesame direction astheoutward normal and n_the number ofother intersections. Ifn=n+—n“show that ——1n== -—~ fdfl.4-.-r o 134 Calculus onManifolds (c)Prove that _.b _. ... 1>..1<..,t,.>= ldz‘Z‘M1! %\“--.\=-.\'13‘'15‘“Q*2CA3 -—d- —bDsQ(a,b,c) = G)” r3(y as)dx, where r(:z:,y,z) =[(:c,g,z)[. (d)Show that theinteger nof(b)equals theintegral ofProb- lem5-32(b), andusethisresult toshow that l(f,g) =1iffandg arethecurves ofFigure 4-6(b),while l(f,g) =0iffandgarethe curves ofFigure 4-6(c). (These results were known toGauss [7]. The proofs outlined here arefrom [4]pp.409»-411; seealso [13], Volume 2,pp.41-43.) THE CLASSICAL THEOREMS Wehave nowprepared allthemachinery necessary tostate and prove theclassical “Stokes’ type” oftheorems. Wewill indulge inalittle bitofself-explanatory classical notation. 5-7 Theorem (Green’s Theorem). LetMER2beacom- pact two-dimensional manifold-with-boundary. Suppose that a,B.' M-+Rarediﬂerentiable. Then /adx+[6dy =/(D16-— D2a)d:z: /\dy BM M -/1e~a6:2: 6g y- M (Here Misgiven theusual orientation, and6Mtheinduced orientation, alsoknown asthecounterclockwise orientation.) Proof. This isavery special case ofTheorem 5-5, since d(adx+Bdg) =(D16 ~—D2a)d:v /\dy. I Integration onManifolds 135 *5-8 Theorem (Divergence Theorem). Let MER3bea compact three-dimensional manifold-with-boundary and nthe unit outward normal on6M.LetFbeadifferentiable vector ﬁeld onM. Then ﬂjearev =a!Ii(F,n)d/1. This equation isalsowritten interms ofthree differentiable func- tions a,B,'y.' M—->R: aaafff(-5:-+5§—+-ég-)dV= f/i(n1a—[—n2B —[—n3'y)dS. M y 6M Proof. Deﬁne cconMbyw=Fldy /\dz+F2 dz/\dsc+ F3dx/\dy. Then dw=divF dV. According toTheorem 5-6, on6Mwehave n1dA =dy Adz, n2dA =dz /\d:c, n3dA=dx/\dy. Therefore on6Mwehave (F,n)e/1=FlnldA+PWat+ruedA =Fldy /\dz+F2dz /\d:c+F3d:c /\dy =OJ. Thus, byTheorem 5-5wehave [divFdV= [ea=[a=, [(F,n)dA. |M M aM 6M 5-9 Theorem (Stokes’ Theorem). LetMER3beacom- pact oriented two-dimensional manifold-with-boundary andnthe unit outward normal onMdetermined bytheorientation ofM. Let6Mhavetheinduced orientation. LetTbethevector ﬁeld on 6Mwith ds(T) =1andletFbeadiﬁerentiable vector ﬁeld in anopen setcontaining M. Then [<(v><F),n)dA =[<r,T>ds. 136 Calculus onManifolds This equation issometimes written /ad:c+,8dy+'ydz= 6M 1£".'>_’_§.(f ?_€_,"_'Y .<.?E__.‘1sIfin(av 6z)+n2(6z 6w)+n3 (611 62/lldS'M Proof. Deﬁne wonMbycc=F1dx+F2dy+F3dz. Since VXFhascomponents DZF3 -D3F2, D3F1 —D1F3, D1F2 --D2F1, itfollows, asintheproof ofTheorem 5-8, that onMwehave ((VXF),n)dA=(DQF3 -—-D3F2)dy /\dz +(D3F1 —D1F3)dz /\dx +(D1F2 -D2F1)da: /\dy =dw. Ontheother hand, since ds(T) =1,on6Mwehave T1ds=dx, T2ds=dy, T3ds=dz. (These equations may bechecked byapplying both sides to T,,,forasE6M, since T,isabasis for(6M),,,.) Therefore on6Mwehave (F,T) ds=FIT1 ds+F2T2 ds+F3T3 ds =Fldx —[-F2dy+F3dz = CO. Thus, byTheorem 5-5, wehave ((VXF),n) dA= do= to= (F,T) ds. I .1 .1...!...!Theorems 5-8and5-9arethebasis forthenames divFand curlF. IfF(:c) isthevelocity vector ofaﬂuid atas(atsome time) then f,1,;(F,n) dAistheamount ofﬂuid “diverging” from M. Consequently thecondition divF=0expresses Integration onManifolds 137 thefactthat theﬂuid isincompressible. IfMisadisc, then I611;(F,T)dsmeasures theamount that theﬂuid curls around thecenter ofthedisc. Ifthisiszero foralldiscs, then VXF =0,and theﬂuid iscalled irrotational. These interpretations ofdivFandcurlFareduetoMaxwell [13]. Maxwell actually worked with thenegative ofdivF, which heaccordingly called the convergence. For VXF Maxwell proposed “with great difﬁdence” theterminology rotation ofF;thisunfortunate term suggested theabbreviation rotFwhich oneoccasionally stillsees. The classical theorems ofthissection areusually stated in somewhat greater generality than they arehere. Forexam- ple,Green’s Theorem istrueforasquare, andtheDivergence Theorem istrue foracube. These twoparticular facts can beproved byapproximating thesquare orcube bymanifolds- with-boundary. Athorough generalization ofthetheorems of this section requires theconcept ofmanifolds-with-corners; these aresubsets ofR”which are, uptodiffeomorphism, locally aportion ofR3which isbounded bypieces of(lc-—1)- planes. The ambitious reader willﬁnditachallenging exer- cise todeﬁne manifolds-with-corners rigorously and to investigate how theresults ofthis entire chapter may be generalized. Problems. 5-34. Generalize thedivergence theorem tothecase of ann-manifold with boundary inR". 5-35. Applying thegeneralized divergence theorem tothesetM= {:0ER": [:c[3a}and F(:c) =xx,ﬁnd thevolume of8”“! = {:cER":[av]=1}interms ofthen-dimensional volume ofB,,= {:cER":[:c[51}. (This volume is1r""3/(n/2)! ifniseven and 2(""'1)(31r("“1)(3/1 -3-5-...-nifnisodd.) 5-36. Deﬁne FonR3byF(:c),= (0,0,c:z:3), and letMbeacompact three-dimensional manifold-with-boundary with ME{:c:x33 0}. Thevector ﬁeld Fmay bethought ofasthedownward pres- sure ofaﬂuid ofdensity cin{:c:11:330}. Since aﬂuid exerts equal pressures inalldirections, wedeﬁne thebuoyant force onM, duetotheﬂuid, as-fay (F,n) dA. Prove thefollowing theorem. Theorem (Archimedes). The buoyant force onMisequal tothe weight oftheﬂuid displaced byM. Bibliography Ahlfors, Complex Analysis, McGraw-Hill, New York, 1953. Auslander andMacKenzie, Introduction toDiﬁerentiable Manifolds, McGraw-Hill, New York, 1963. Cesari, Surface Area, Princeton University Press, Princeton, New Jersey, 1956. Courant, Differential andIntegral Calculus, Volume II,Interscience, New York, 1937. Dieudonné, Foundations ofModern Analysis, Academic Press, New York, 1960. Fort, Topology of3-Manifolds, Prentice-Hall, Englewood Cliffs, New Jersey, 1962. Gauss, Zurmathematischen Theorie derelectrodynamischen Wirkungen, [4](Nachlass) Werke V,605. Helgason, Differential Geometry and Symmetric Spaces, Academic Press, New York, 1962. Hilton and Wylie, Homology Theory, Cambridge University Press, New York, 1960. Hu, Homotopy Theory, Academic Press, New York, 1959. Kelley, General Topology, Van Nostrand, Princeton, New Jersey, 1955. 139 140 Bibliography 12.Kobayashi and Nomizu, Foundations ofDifferential Geometry, Interscience, New York, 1963. 13.Maxwell, Electricity and Magnetism, Dover, New York, 1954. 14-.Natanson, Theory ofFunctions ofaReal Variable, Frederick Ungar, New York, 1955. 15.Rado, Length and Area, Volume XXX, American Mathematical Society, Colloquium Publications, New York, 1948. 16.deRham, Variétés Diﬂerentiables, Hermann, Paris, 1955. 17.Sternberg, Lectures onDifferential Geometry, Prentice-Hall, Engla- wood Cliffs, New Jersey, 1964. Indezr Absolute differential form, 126 Absolute tensor, 126 Absolute value, 1 Algebra, Fundamental Theorem of, 105 Alternating tensor, 78 Analytic function, 105 Angle, 4 preserving, 4 solid, 131 Approximation, 15 Archimedes, 137 Area, 56 element of,126 surface, 126, 127 Basis, usual forR”,3 Bilinear function, 3,23 Boundary ofachain, 97,98 ofamanifold-with-boundary 113Boundary, ofaset,7 Buoyant force, 137 Cauchy Integral Formula, 106 Cauchy Integral Theorem, 106 Cauchy-Riemann equations, 105 Cavalieri’s principle, 62 Chain, 97,100 Chain rule, 19,32 Change ofvariable, 67-72 Characteristic function, 55 Closed curve, 106 Closed differential form, 92 Closed rectangle, 5 Closed set,5 Compact, 7 Complex numbers, 104 Complex variables, 105 Component function, 11,87 Composition, 11 Cone, generalized, 131 142 Consistent choices oforientation, 117 . Constant function, 20 Constraints, 122 Content, 56 Content zero, 51 Continuous differential form, 88 Continuous function, 12 Continuous vector ﬁeld, 87 Continuously differentiable, 31 Convergence, 137 Coordinate condition, 111 Coordinate system, 111 polar, 73 Counterclockwise orientation, 134 Cover, 7 Cross product, 84 Cube singular, 97 standard n-cube, 97 Curl, 88,137 Curve, 97 closed, 106 differentiable, 96 C°°,26 Degenerate singular cube, 1()5 Derivative, 16 directional, 33 partial, 25 higher-order (mixed), 26 second-order (mixed), 26 Diffeomorphism, 109 Differentiable function, 15,16, 105 continuously, 31 Differentiable curve, 96Index Differential form, onamanifold, 117 differentiable, 117 Dimension ofamanifold, 109 ofamanifold-with-boundary, 113 Directional derivative, 33 Distance, 4 Divergence ofaﬁeld, 88,137 Divergence Theorem, 135 Domain, 11 Dual space, 5 Element ofarea, 126 Element oflength, 126 Element ofvolume, seeVolume element End point, 87 Equal uptonthorder, 18 Euclidean space, 1 Exact differential form, 92 Exterior ofaset,7 Faces ofasingular cube, 98 Field, seeVector ﬁeld Form, seeDifferential form Fubini’s Theorem, 58 Function, 11 analytic, 105 characteristic, 55 component, 11,87 composition of,11 constant, 20 continuous, 12 continuously differentiable, 31 Differentiable differential form, 88 Cw,25 onamanifold, 117 Differentiable vector ﬁeld, 87 onamanifold, 115 Differentiable =Cw,88 Differential, 91 Differential form, 88 absolute, 126 closed, 92 continuous, 88 differentiable, 88 exact, 92differentiable, 15,16,105 homogeneous, 34 identity, 11 implicitly deﬁned, 41 seealsoImplicit Function Theorem integrable, 48 inverse, 11,34-39 seealsoInverse Function Theorem projection, 11 Indea: Fundamental Theorem ofAlgebra, 105 Fundamental Theorem ofCalcu- lus,100-104 Gauss, 134 Generalized cone, 131 Grad f,96 Graph, 11,115 Green’s Theorem, 134 Half-space, 113 Heine-Borel Theorem, 7 Homogeneous function, 34 Homotopy, 108 Identity function, 11 Implicit Function Theorem, 41 Implicitly deﬁned function, 41 Incompressible ﬂuid, 137 Independence ofparameteriza- tion, 104 Induced orientation, 119 Inequality, seeTriangle inequality Inner product, 2,77 preserving, 4 usual, 77,87 Integrable function, 48 Integral, 48 iterated, 59,60 Hne,10l lower, 58 ofaform onamanifold, 123-124 ofaform over achain, 101 over aset,55 over anopen set,65 surface, 102 upper, 58 Integral Formula, Cauchy, 106 Integral Theorem, Cauchy, 106 Interior ofaset,7 Inverse function, 11,34-39 Inverse Function Theorem, 35 Irrotational ﬂuid, 137 Iterated integral, 59,60 Jacobian matrix, 17 Jordan-measurable, 56143 Kelvin, 74 Laclocus, 106 Lagrange’s method, 122 Lagrangian multiplier, 122 Leibnitz’s Rule, 62 Length, 56,126 element of,126 Length =norm, 1 Limit, 11 Line, 1 Line integral, 101 Linking number, 132 Liouville, 74 Lower integral, 58 Lower sum, 47 Manifold, 109 Manifold-with-boundary, 113 Manifold-with-corners, 131, 137 Mathematician (oldstyle), 74 Matrix, 1 Jacobian, 17 transpose of,23,83 Maxima, 26-27 Measure zero, 50 Minima, 26-27 Mobius strip, 119, 120, 130 Multilinear function, 23,75 Multiplier, seeLagrangian multi- plier Norm, 1 Norm preserving, 4 Normal, seeOutward unit normal Notation, 3,44,89 One-one (1-1) function, 11 One-sided surface, 121 Open cover, 7 Open rectangle, 5 Open set,5 Orientable manifold, 119 Orientation, 82,119 consistent choices of,117 counterclockwise, 134 induced, 119 usual, 83,87,121 I44 Orientation-preserving, 118, 123 Oriented manifold, 119 Orthogonal vectors, 5 Orthonormal basis, 77 Oscillation, 13 Outward unit normal, 119, 120 Parameterization, independence of 104 Partial derivative, 25 higher-order (mixed), 26 second-order (mixed), 26 Partition ofaclosed interval, 46 ofaclosed rectangle, 46 ofunity, 63 Perpendicular, 5 Plane, 1 Poincare Lemma, 94 Point, 1 Polar coordinate system, 73 Polarization identity, 5 Positive deﬁniteness, 3,77 Product, seeCross product, Inner product, Tensor product, Wedge product Projection function, 11 Rectangle (closed oropen), 5 Reﬁne apartition, 47 Rotation ofF,137 Sard’s Theorem, 72 Self-adjoint, 85 Sign ofapermutation, 78 Singular n-cube, 97 Solid angle, 131 Space, 1 seealsoDual space, Euclidean space, Half-space, Tangent space Sphere, 111 Standard n-cube, 97 Star-shaped, 93Indea: Stokes’ Theorem, 102, 124, 135 Subordinate, 63 Subrectangles ofapartition, 46 Surface, 127 Surface area, 126, 127 Surface integral, 102 Symmetric, 2,77 Tangent space, 86,115 Tangent vector, 96 Tensor, 75 absolute, 126 alternating, 78 Tensor product, 75 Torus, 115 Transpose ofamatrix, 23,83 Triangle inequality, 4 Unit outward normal, 119, 120 Upper integral, 58 Upper sum, 47 Usual, seeBasis, Inner product, Orientation Variable change of,67-72 complex, seeComplex variables function ofn,11 independent oftheﬁrst, 18 independent ofthesecond, 17 Vector, 1 tangent, 96 Vector ﬁeld, 87 continuous, 87 differentiable, 87 onamanifold, 115 continuous, 87 differentiable, 115 Vector-valued function, 11 Volume, 47,56,126 Volume element, 83,126 Wedge product, 79 Winding number, 104 Addenda 1.Itshould beremarked after Theorem 2-11 (the Inverse Function Theorem) that theformula forflallows ustocon- clude thatF1isactually continuously differentiable (and that itisC”iffis). Indeed, itsuﬂices tonote that theentries of theinverse ofamatrix AareC”functions oftheentries ofA. This follows from “Cramer’s Rule”: (A"1),-,- = (detA35)/(det A),where Allisthematrix obtained from A bydeleting rowiandcolumn j. 2.Theproof oftheﬁrstpart ofTheorem 3-8canbesimpli- ﬁed considerably, rendering Lemma 3-7unnecessary. It suffices tocover Bbytheinteriors ofclosed rectangles U,-with Z,‘1°_1v(U,-) <E,and tochoose foreach :1:EA—Baclosed rectangle V1,, containing a:initsinterior, with MV,(f)— mv,(f) <E.Ifevery subrectangle ofapartition Piscon- tained inoneofsome ﬁnite collection ofU,-’sandV,’s which cover A,and[f(x)] 5MforallxinA,then U(f, P)—L(f,P) <ev(A)+2Me. The proof oftheconverse part contains anerror, since M,(f)—m,(f)Z1/nisguaranteed only iftheinterior ofS intersects B1,,,.Tocompensate forthisitsuﬁices tocover the boundaries ofallsubrectangles ofPwith a.ﬁnite collection of rectangles with total volume <8.These, together with 5, cover B1,,,,andhave total volume <28. 145 I46 Addenda 3.The argument intheﬁrst part ofTheorem 3-14 (Sard’s Theorem) requires alittle ampliﬁcation. IfUEAisaclosed rectangle with sides oflength l,then, because Uiscompact, there isaninteger Nwith thefollowing property: ifUis divided into N"rectangles, with sides oflength l/N, then [D,-g3(w) —D,-g3(z)[ <8/n3 whenever wandzareboth inone such rectangle S.Given :1:ES,letf(z) =Dg(x) (z)—g(z). Then, ifzES, |1>.~re>l =|1>.~r<==>-1>.~r<el<e/In”. SobyLemma 2-10, if:r,yE5,then |Dv(=v)(v —rv)—g(v)+g(x)]=[f(v)-—f(=v)|<Elw"-vl 5e\/It(l/N). 4.Finally, thenotation A"(V) appearing inthis book is incorrect, since itconﬂicts with thestandard deﬁnition of A3(V) (asacertain quotient ofthetensor algebra ofV). For thevector space inquestion (which isnaturally isomorphic to A"(V*) forﬁnite dimensional vector spaces V)thenotation Q"(V) isprobably ontheway tobecoming standard. This substitution should bemade onpages 78-85, 88-89, 116, and 126-128.