very math oriented 52htext-2015
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A full set of university course lecture notes by Yakov Eliashberg, dated January 2015, found in Phil's Wedge World folder as a math-oriented reference. Part I covers dual spaces, multilinear functions, tensor and exterior products, orientation, volume, dualities and complex vector spaces. Part II covers topology basics, vector fields, differential forms, the exterior derivative and integration on manifolds. Part III covers Stokes' theorem and applications such as homotopy and winding and linking numbers.
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Extracted text (machine-read; may contain errors)
Math 52H: Multilinear algebra, di erential
forms and Stokes' theorem
Yakov Eliashberg
January 2015
2
Contents
I Multilinear Algebra 7
1 Linear and multilinear functions 9
1.1 Dual space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.2 Canonical isomorphism between ( V)andV. . . . . . . . . . . . . . . . . . . . . . 11
1.3 The mapA. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.4 Multilinear functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
1.5 Quotient space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
1.6 Symmetric bilinear functions and quadratic forms . . . . . . . . . . . . . . . . . . . . 16
2 Tensor and exterior products 19
2.1 Tensor product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
2.2 Spaces of multilinear functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2.3 Symmetric and skew-symmetric tensors . . . . . . . . . . . . . . . . . . . . . . . . . 21
2.4 Symmetrization and anti-symmetrization . . . . . . . . . . . . . . . . . . . . . . . . 22
2.5 Exterior product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
2.6 Spaces of symmetric and skew-symmetric tensors . . . . . . . . . . . . . . . . . . . . 26
2.7 OperatorAon spaces of tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
3 Orientation and Volume 31
3.1 Orientation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
3.2 Orthogonal transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
3.3 Determinant and Volume . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
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3.4 Volume and Gram matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
4 Dualities 37
4.1 Duality between k-forms and ( n k)-forms on a n-dimensional Euclidean space V. 37
4.2 Euclidean structure on the space of exterior forms . . . . . . . . . . . . . . . . . . . 43
4.3 Contraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
5 Complex vector spaces 51
5.1 Complex numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
5.2 Complex vector space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
5.3 Complex linear maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
II Calculus of di erential forms 59
6 Topological preliminaries 61
6.1 Elements of topology in a vector space . . . . . . . . . . . . . . . . . . . . . . . . . . 61
6.2 Everywhere and nowhere dense sets . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
6.3 Compactness and connectedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
6.4 Connected and path-connected components . . . . . . . . . . . . . . . . . . . . . . . 66
6.5 Continuous maps and functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
7 Vector elds and di erential forms 71
7.1 Di erential and gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
7.2 Smooth functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
7.3 Gradient vector eld . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
7.4 Vector elds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
7.4.1 Gradient vector eld . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
7.5 Di erential forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
7.6 Coordinate description of di erential forms . . . . . . . . . . . . . . . . . . . . . . . 77
7.7 Smooth maps and their di erentials . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
7.8 Operator f. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
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7.9 Coordinate description of the operator f. . . . . . . . . . . . . . . . . . . . . . . . 81
7.10 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
7.11 Pfaan equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
8 Exterior di erential 85
8.1 Coordinate de nition of the exterior di erential . . . . . . . . . . . . . . . . . . . . . 85
8.2 Properties of the operator d. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
8.3 Curvilinear coordinate systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
8.4 Geometric de nition of the exterior di erential . . . . . . . . . . . . . . . . . . . . . 91
8.5 More about vector elds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
8.6 Casen= 3. Summary of isomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . 95
8.7 Gradient, curl and divergence of a vector eld . . . . . . . . . . . . . . . . . . . . . . 96
8.8 Example: expressing vector analysis operations in spherical coordinates . . . . . . . 97
8.9 Complex-valued di erential k-forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
9 Integration of di erential forms and functions 103
9.1 Useful technical tools: partition of unity and cut-o functions . . . . . . . . . . . . . 103
9.2 One-dimensional Riemann integral for functions and di erential 1-forms . . . . . . . 106
9.3 Integration of di erential 1-forms along curves . . . . . . . . . . . . . . . . . . . . . . 109
9.4 Integrals of closed and exact di erential 1-forms . . . . . . . . . . . . . . . . . . . . . 114
9.5 Integration of functions over domains in high-dimensional spaces . . . . . . . . . . . 115
9.6 Fubini's Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129
9.7 Integration of n-forms over domains in n-dimensional space . . . . . . . . . . . . . . 132
9.8 Manifolds and submanifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
9.8.1 Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
9.8.2 Gluing construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142
9.8.3 Examples of manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149
9.8.4 Submanifolds of an n-dimensional vector space . . . . . . . . . . . . . . . . . 151
9.8.5 Submanifolds with boundary . . . . . . . . . . . . . . . . . . . . . . . . . . . 153
9.9 Tangent spaces and di erential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155
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9.10 Vector bundles and their homomorphisms . . . . . . . . . . . . . . . . . . . . . . . . 158
9.11 Orientation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159
9.12 Integration of di erential k-forms over k-dimensional submanifolds . . . . . . . . . . 159
III Stokes' theorem and its applications 165
10 Stokes' theorem 167
10.1 Statement of Stokes' theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167
10.2 Proof of Stokes' theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171
10.3 Integration of functions over submanifolds . . . . . . . . . . . . . . . . . . . . . . . . 173
10.4 Work and Flux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179
10.5 Integral formulas of vector analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182
10.6 Expressing div and curl in curvilinear coordinates . . . . . . . . . . . . . . . . . . . . 183
11 Applications of Stokes' formula 187
11.1 Integration of closed and exact forms . . . . . . . . . . . . . . . . . . . . . . . . . . . 187
11.2 Approximation of continuous functions by smooth ones . . . . . . . . . . . . . . . . . 188
11.3 Homotopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190
11.4 Winding and linking numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196
11.5 Properties of k-forms onk-dimensional manifolds . . . . . . . . . . . . . . . . . . . . 199
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Part I
Multilinear Algebra
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Chapter 1
Linear and multilinear functions
1.1 Dual space
LetVbe a nite-dimensional real vector space. The set of all linear functions on Vwill be denoted
byV.
Proposition 1.1. Vis a vector space of the same dimension as V.
Proof. One can add linear functions and multiply them by real numbers:
(l1+l2)(x) =l1(x) +l2(x)
(l)(x) =l(x) forl;l1;l22V; x2V; 2R
It is straightforward to check that all axioms of a vector space are satis ed for V. Let us now
check that dim V= dimV.
Choose a basis v1:::vnofV. For anyx2Vlet0
BBB@x1
...
xn1
CCCAbe its coordinates in the basis v1:::vn.
Notice that each coordinate x1;:::;xncan be viewed as a linear function on V. Indeed,
1) the coordinates of the sum of two vectors are equal to the sum of the corresponding coordinates;
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2) when a vector is multiplied by a number, its coordinates are being multiplied by the same
number.
Thusx1;:::;xnare vectors from the space V. Let us show now that they form a basis of V.
Indeed, any linear function l2Vcan be written in the form l(x) =a1x1+:::+anxnwhich means
thatlis a linear combination of x1:::xnwith coecients a1;:::;an. Thusx1;:::;xngenerateV.
On the other hand, if a1x+:::+anxnis the 0-function, then all the coecients must be equal to 0;
i.e. functions x1;:::;xnare linearly independent. Hence x1;:::;xnform a basis of Vand therefore
dimV=n= dimV:
The spaceVis called dual toVand the basis x1;:::;xndual tov1:::vn.1
Exercise 1.2. Prove the converse: given any basis l1;:::;lnofVwe can construct a dual basis
w1;:::;wnofVso that the functions l1;:::;lnserve as coordinate functions for this basis.
Recall that vector spaces of the same dimension are isomorphic. For instance, if we x bases in
both spaces, we can map vectors of the rst basis into the corresponding vectors of the second basis,
and extend this map by linearity to an isomorphism between the spaces. In particular, sending a
basisS=fv1;:::;vngof a space Vinto the dual basis x1;:::;xnof the dual space Vwe can
establish an isomorphism iS:V!V. However, this isomorphism is not canonical , i.e. it depends
on the choice of the basis v1;:::;vn.
IfVis a Euclidean space, i.e. a space with a scalar product hx;yi, then this allows us to de ne
another isomorphism V!V, di erent from the one described above. This isomorphism associates
with a vector v2Va linear function lv(x) =hv;xi. We will denote the corresponding map V!V
byD. Thus we haveD(v) =lvfor any vector v2V.
Exercise 1.3. Prove thatD:V!Vis an isomorphism. Show that D=iSfor any orthonormal
basisS.
The isomorphism Dis independent of a choice of an orthonormal basis, but is still not completely
canonical: it depends on a choice of a scalar product. However, when talking about Euclidean spaces ,
this isomorphism is canonical.
1It is sometimes customary to denote dual bases in VandVby the same letters but using lower indices for V
and upper indices for V, e.g.v1;:::;vnandv1;:::;vn. However, in these notes we do not follow this convention.
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Remark 1.4. The de nition of the dual space Valso works in the in nite-dimensional case.
Exercise 1.5. Show that both maps iSandDare injective in the in nite case as well. However,
neither one is surjective if Vis in nite-dimensional.
1.2 Canonical isomorphism between (V)and V
The space ( V), dual to the dual space V, iscanonically isomorphic in the nite-dimensional case
toV. The word canonically means that the isomorphism is \god-given", i.e. it is independent of
any additional choices.
When we write f(x) we usually mean that the function fis xed but the argument xcan vary.
However, we can also take the opposite point of view, that xis xed but fcan vary. Hence, we can
view the point xas a function on the vector space of functions.
Ifx2Vandf2Vthen the above argument allows us to consider vectors of the space Vas
linear functions on the dual space V. Thus we can de ne a map I:V!Vby the formula
x7!I(x)2(V);whereI(x)(l) =l(x) for any l2V:
Exercise 1.6. Prove that if Vis nite-dimensional then Iis an isomorphism. What can go wrong
in the in nite-dimensional case?
1.3 The mapA
Given a mapA:V!Wone can de ne a dual mapA:W!Vas follows. For any linear
functionl2Wwe de ne the function A(l)2Vby the formulaA(l)(x) =l(A(x)); x2V. In
other words,A(l) =lA.2
2In fact, the above formula makes sense in much more general situation. Given any map : X!Ybetween two
setsXandYthe formula (h) =h de nes a map :F(Y)!F(X) between the spaces of functions on Yand
X. Notice that this map goes in the direction opposite to the direction of the map .
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Given basesBv=fv1;:::;vngandBw=fw1;:::;wkgin the vector spaces VandWone can
associate with the map Aa matrixA=MBvBw(A). Its columns are coordinates of the vectors
A(vj);j= 1;:::;n , in the basisBw. Dual spaces VandWhave dual bases X=fx1;:::xng
andY=fy1;:::;ykgwhich consist of coordinate functions corresponding to the basis BvandBw.
Let us denote by Athe matrix of the dual map Awith respect to the bases YandX, i. e.
A=MYX(A).
Proposition 1.7. The matrices AandAare transpose to each other, i.e. A=AT.
Proof. By the de nition of the matrix of a linear map we should take vectors of the basis Y=
fy1;:::;ykg, apply to them the map A, expand the images in the basis X=fx1;:::xngand write
the components of these vectors as columns of the matrixA. Set ~yi=A(yi); i= 1;:::;k . For
any vector u=nP
j=iujvj2V;we have ~yi(u) =yi(A(u)). The coordinates of the vector A(u) in the
basisw1;:::;wkmay be obtained by multiplying the matrix Aby the column0
BBB@u1
...
un1
CCCA. Hence,
~yi(u) =yi(A(u)) =nX
j=1aijuj:
But we also have
nX
j=1aijxj(u) =nX
j=1aijuj:
Hence, the linear function ~ yi2Vhas an expansionnP
j=1aijxjin the basis X=fx1;:::xngof the
spaceV. Hence the i-th column of the matrix Aequals0
BBB@ai1
...
ain1
CCCA, so that the whole matrix A
has the form
A=0
BBB@a11ak1
:::
a1nakn1
CCCA=AT:
12
Exercise 1.8. Given a linear map A:V!Wwith a matrix A, ndA(yi).
Answer. The mapAsends the coordinate function yionWto the functionnP
j=1aijxjonV, i.e. to
its expression in coordinates xi.
Proposition 1.9. Consider linear maps
UA!VB!W :
Then (BA )=AB.
Proof. For any linear function l2Wwe have
(BA )(l)(x) =l(B(A(x)) =A(B(l))(x)
for anyx2U.
Exercise 1.10. Suppose that Vis a Euclidean space and Ais a linear map V!V. Prove that
for any two vectors X;Y2Vwe have
hA(X);Yi=hX;D 1AD(Y)i: (1.3.1)
Solution. By de nition of the operator Dwe have
hX;D 1(Z)i=Z(X)
for any vector Z2V. Applying this to Z=AD(Y) we see that the right-hand send of (1.3.1) is
equal to toAD(Y)(X). On the other hand, the left-hand side can be rewritten as D(Y)(A(X)).
ButAD(Y)(X) =D(Y)(A(X)).
Let us recall that if Vis an Euclidean space, then operator B:V!Vis called adjoint to
A:V!Vif for any two vectors X;Y2Vone has
hA(X);Yi=hX;B(Y)i:
The adjoint operator always exist and unique. It is denoted by A?. Clearly, (A?)?=A:In any
orthonormal basis the matrices of an operator and its adjoint are transpose to each other. An
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operatorA:V!Vis called self-adjoint ifA?=A, or equivalently, if for any two vectors
X;Y2Vone has
hA(X);Yi=hX;A(Y)i:
The statement of Exercise 1.10 can be otherwise expressed by saying that the adjoint operator
A?is equal toD 1AD:V!V. In particular, an operator A:V!Vis self-adjoint if and
only ifAD=DA .
Remark 1.11. As it follows from Proposition 1.7 and Exercise 1.10 the matrix of a self-adjoint
operator in any orthonormal basis is symmetric. This is not true in an arbitrrary basis.
1.4 Multilinear functions
A function l(X1;X2;:::;Xk) ofkvector arguments X1;:::;Xk2Vis calledk-linear (or multi-
linear) if it is linear with respect to each argument when all other arguments are xed. We say
bilinear instead of 2-linear. Multilinear functions are also called tensors . Sometimes, one may also
say a \k-linear form", or simply k-form instead of a \ k-linear functions". However, we will reserve
the termk-form for a skew-symmetric tensors which will be de ned in Section 2.3 below.
If one xes a basis v1:::vnin the space Vthen with each bilinear function f(X;Y ) one can
associate a square nnmatrix as follows. Set aij=f(vi;vj). ThenA= (aij)i;j=1;:::;nis called the
matrix of the function fin the basis v1;:::;vn.For any 2 vectors
X=nX
1xivi; Y=nX
1yjvj
we have
f(X;Y ) =f0
@nX
i=1xivi;nX
j=1yjvj1
A=nX
i;j=1xiyjf(vi;vj) =nX
i;j=1aijxiyj=XTAY :
Exercise 1.12. How does the matrix of a bilinear function depend on the choice of a basis?
Answer. The matrices Aand ~Aof the bilinear form f(x;y) in the bases v1;:::;vnand ~v1;:::; ~vn
are related by the formula ~A=CTAC, whereCis the matrix of transition from the basis v1:::vn
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to the basis ~ v1:::~vn, i.e the matrix whose columns are the coordinates of the basis ~ v1:::~vnin the
basisv1:::vn.
Similarly, with a k-linear function f(X1;:::;Xk) onVand a basis v1;:::;vnone can associate
a \k-dimensional" matrix
A=fai1i2:::ik; 1i1;:::;ikng;
where
ai1i2:::ik=f(vi1;:::;vik):
IfXi=nP
j=1xijvj; i= 1;:::;k; then
f(X1;:::;Xk) =nX
i1;i2;:::ik=1ai1i2:::ikx1i1x2i2:::xkik;
see Proposition 2.1 below.
1.5 Quotient space
LetVbe a vector space and LVbe its linear subspace. Given a vector a2Vlet us denote byLa
the ane subspace a+L=fa+x;x2Lg. Note that two ane subspaces LaandLb,a;b2V
concide ifa b2Land are disjoint if a b =2L. Consider the set, denoted by V=L of all ane
subspaces parallel to L. The set can be made into a vector space by de ning the operations by the
formulas
La+Lb:=La+b; La:=La;a;b2V; 2R
Ifa0andb0are other vectors in Vsuch thata0 a;b0 b2Lthen (a0+b0) (a+b)2Land
a0 a2L, and hence La0+b0=La+b; La0= La.
The vector space V=L is called the quotient space ofVbyL.
In other words, we can say that V=L is obtained from Vby identifying vectors which di er by
a vector in L. The operations in Vthen naturally descend to the opertions on the quotient space.
IfNVbe any linear subspace such that dim L+ dimN= dimVandL\N= 0 then the
mapN!V=L given by the formula x2N7!Lx2V=L is an isomorphism (why?). In particular,
dim(V=L) = dimV dimL= codimVL.
15
IfVis a Euclidean space then we can choose as Nthe orthogonal complement L?ofLinV,
and thusL=V is isomorphic to L?. The advantage of the quotient construction that it is canonical
whileL?depends on a choice of the Euclidean structure (scalar product).
1.6 Symmetric bilinear functions and quadratic forms
A function Q:V!Ron a vector space Vis called quadratic if there exists a bilinear function
f(X;Y ) such that
Q(X) =f(X;X ); X2V: (1.6.1)
One also uses the term quadratic form . The bilinear function f(X;Y ) is not uniquely determined
by the equation (1.6.1). For instance, all the bilinear functions x1y2; x2y1and1
2(x1y2+x2y1) on
R2de ne the same quadratic form x1x2.
On the other hand, there is a 1-1 corerspondence between quadratic form and symmetric bilinear
functions. A bilinear function f(X;Y ) is called symmetric if f(X;Y ) =f(Y;X) for allX;Y2V.
Lemma 1.13. Given a quadratic form Q:V!Rthere exists a unique symmetric bilinear form
f(X;Y )such thatQ(X) =f(X;X );X2V.
Proof. IfQ(X) =f(X;X ) for a symmetric fthen
Q(X+Y) =f(X+Y;X +Y) =f(X;X ) +f(X;Y ) +f(Y;X) +f(Y;Y)
=Q(X) + 2f(X;Y ) +Q(Y);
and hence
f(X;Y ) =1
2
Q(X+Y) Q(X) Q(Y)
: (1.6.2)
I leave it as an exercise to check that the formula (1.6.2) always de nes a symmetric bilinear
function.
LetS=fv1;:::;vngis a basis of V. The matrix A= (aij) of a symmetric bilinear form f(X;Y )
in the basis Sis called also the matrix of the corresponding quadratic form Q(X) =f(X;X ). This
matrix is symmetric, and
Q(X) =X
ijaijxixj=a11x2
1++annx2
n+ 2X
i<jaijxixj:
16
Thus the matrix Ais diagonal if and only if the quadratic form Qis the sum of squares (with
coecients). Let us recall that if one changes the basis Sto a basiseS=fev1;:::;evngthen the
matrix of a bilinear form fchanges toeC=CTAC.
Exercise 1.14. (Sylvester's inertia law) Prove that there is always exists a basis eS=fev1;:::;evng
in which a quadratic form Qis reduced to a sum of squares. The number of positive, negative and
zero coecients with the squares is independent of the choice of the basis.
Thus, in some coordinate system a quadratic form can always be written as
kX
1x2
i+k+lX
k+1x2
j; k+ln:
The number kof negative squares is called the negative index or simply the index of the quadratic
formQ, the total number of k+lof non-zero squares is called the rank of the form. It coincides
with the rank of the matrix of the form in any basis. A bilinear (and quadratic) form is called
non-degenerate if its rank is maximal possible, i.e. equal to n. For a non-degenerate quadratic for
Qthe di erence l kbetween the number of positive and negative squares is called the signature .
A quadratic form Qis called positive de nite ifQ(X)0 and ifQ(X) = 0 then X= 0.
Equivalently, one can say that a form is positive de nite if it is non-degenerate and its negative
index is equal to 0.
17
18
Chapter 2
Tensor and exterior products
2.1 Tensor product
Given ak-linear function and al-linear function , one can form a ( k+l)-linear function, which
will be denoted by
and called the tensor product of the functions and . By de nition
(X1;:::;Xk;Xk+1;:::;Xk+l) := (X1;:::;Xk) (Xk+1;:::;Xk+l):
For instance, the tensor product two linear functions l1andl2is a bilinear function l1
l2de ned
by the formula
l1
l2(U;V) =l1(U)l2(V):
Letv1:::vnbe a basis in Vandx1;:::;xna dual basis in V, i.e.x1;:::;xnare coordinates of a
vector with respect to the basis v1;:::;vn.
The tensor product xi
xjis a bilinear function xi
xj(Y;Z) =yizj. Thus a bilinear function
fwith a matrix Acan be written as a linear combination of the functions xi
xjas follows:
f=nX
i;j=1aijxi
xj;
whereaijis the matrix of the form fin the basis v1:::vn. Similarly any k-linear function fwith
a \k-dimensional" matrix A=fai1i2:::ikgcan be written (see 2.1 below) as a linear combination of
functions
xi1
xi2
xik;1i1;i2;:::;ikn:
19
Namely, we have
f=nX
i1;i2;:::ik=1ai1i2:::ikxi1
xi2
xik:
2.2 Spaces of multilinear functions
Allk-linear functions, or k-tensors , on a given n-dimensional vector space Vthemselves form a
vector space, which will be denoted by V
k. The space V
1is, of course, just the dual space V.
Proposition 2.1. Letv1;:::vnbe a basis of V, andx1;:::;xkbe the dual basis of Vformed
by coordinate functions with respect to the basis V. Thennkk-linear functions xi1
xik,
1i1;:::;ikn;form a basis of the space V
k.
Proof. Take ak-linear function FfromV
kand evaluate it on vectors vi1;:::;vik:
F(vi1;:::;vik) =ai1:::ik:
We claim that we have
F=X
1i1;:::;iknai1:::ikxi1
xik:
Indeed, the functions on the both sides of this equality being evaluated on any set of kbasic vectors
vi1;:::;vik, give the same value ai1:::ik. The same argument shows that ifP
1i1;:::;iknai1:::ikxi1
xik= 0;then all coecients ai1:::ikshould be equal to 0. Hence the functions xi1
xik,
1i1;:::;ikn;are linearly independent, and therefore form a basis of the space V
k
Similar to the case of spaces of linear functions, a linear map A:V!Winduces a linear map
A:W
k!V
k, which sends a k-linear function F2W
kto ak-linear functionA(F)2V
k,
de ned by the formula
A(F)(X1;:::;Xk) =F(A(X1);:::;A(Xk)) for any vectors X1;:::Xk2V :
Exercise 2.2. SupposeVis provided with a basis v1;:::;vnandxi1
xik,1i1;:::;ikn;is
the corresponding basis of the space V
k. Suppose that the map A:V!Vhas a matrix A= (aij)
in the basis v1;:::;vn. Find the matrix of the map A:V
k!V
kin the basis xi1
xik.
20
2.3 Symmetric and skew-symmetric tensors
A multilinear function (tensor) is called symmetric if it remains unchanged under the transposition
of any two of its arguments:
f(X1;:::;Xi;:::;Xj;:::;Xk) =f(X1;:::;Xj;:::;Xi;:::;Xk)
Equivalently, one can say that a k-tensor fis symmetric if
f(Xi1;:::;Xik) =f(X1;:::;Xk)
for any permutation i1;:::;ikof indices 1 ;:::;k .
Exercise 2.3. Show that a bilinear function f(X;Y )is symmetric if and only if its matrix (in any
basis) is symmetric.
Notice that the tensor product of two symmetric tensors usually is not symmetric.
Example 2.4. Any linear function is (trivially) symmetric. However, the tensor product of two
functionsl1
l2is not a symmetric bilinear function unless l1is proportional to l2. On the other
hand, the function l1
l2+l2
l1is symmetric.
A tensor is called skew-symmetric (or anti-symmetric) if it changes its sign when one transposes
any two of its arguments:
f(X1;:::;Xi;:::;Xj;:::;Xk) = f(X1;:::;Xj;:::;Xi;:::;Xk):
Equivalently, one can say that a k-tensor fis anti-symmetric if
f(Xi1;:::;Xik) = ( 1)inv(i1:::ik)f(X1;:::;Xk)
for any permutation i1;:::;ikof indices 1 ;:::;k , where inv( i1:::ik) is the number of inversions in
the permutation i1;:::;ik. Recall that two indices ik;ilform an inversion ifk<l butik>il.
The matrix Aof a bilinear skew-symmetric function is skew-symmetric, i.e.
AT= A:
21
Example 2.5. The determinant det(X1;:::;Xn)(considered as a function of columns X1;:::;Xn
of a matrix) is a skew-symmetric n-linear function.
Exercise 2.6. Prove that any n-linear skew-symmetric function on Rnis proportional to the de-
terminant.
Linear functions are trivially anti-symmetric (as well as symmetric).
As in the symmetric case, the tensor product of two skew-symmetric functions is not skew-
symmetric. We will de ne below in Section 2.5 a new product, called an exterior product of skew-
symmetric functions, which will again be a skew-symmetric function.
2.4 Symmetrization and anti-symmetrization
The following constructions allow us to create symmetric or anti-symmetric tensors from arbitrary
tensors. Let f(X1;:::;Xk) be ak-tensor. Set
fsym(X1;:::;Xk) :=X
(i1:::ik)f(Xi1;:::;Xik)
and
fasym(X1;:::;Xk) :=X
(i1:::ik)( 1)inv(i1;:::;ik)f(Xi1;:::;Xik)
where the sums are taken over all permutations i1;:::;ikof indices 1 ;:::;k . The tensors fsym
andfasymare called, respectively, symmetrization and anti-symmetrization of the tensor f. It is
now easy to see that
Proposition 2.7. The function fsymis symmetric. The function fasymis skew-symmetric. If f
is symmetric then fsym=k!fandfasym= 0. Similarly, if fis anti-symmetric then fasym=k!f,
fsym= 0.
Exercise 2.8. Letx1;:::;xnbe coordinate function in Rn. Find (x1
x2
:::
xn)asym.
Answer. The determinant.
22
2.5 Exterior product
For our purposes skew-symmetric functions will be more important. Thus we will concentrate on
studying operations on them.
Skew-symmetric k-linear functions are also called exteriork-forms . Let be an exterior k-form
and an exterior l-form. We de ne an exterior ( k+l)-form ^ ,the exterior product of and
, as
^ :=1
k!l!(
)asym:
In other words,
^ (X1;:::;Xk;Xk+1;:::;Xk+l) =1
k!l!X
i1;:::ik+l( 1)inv(i1;:::;ik+l) (Xi1:::;Xik) (Xik+1;:::;Xik+l);
where the sum is taken over all permutations of indices 1 ;:::;k +l.
Note that because the anti-symmetrization of an anti-symmetric k-tensor amounts to its mul-
tiplication by k!, we can also write
^ (X1;:::;Xk+l) =X
i1<:::<ik;ik+1<:::<ik+l( 1)inv(i1;:::;ik+l) (Xi1;:::;Xik) (Xik+1;:::;Xik+l);
where the sum is taken over all permutations i1;:::;ik+lof indices 1 ;:::;k +l.
Exercise 2.9. The exterior product operation has the following properties:
For any exterior k-form and exterior l-form we have ^ = ( 1)kl ^ .
Exterior product is linear with respect to each factor:
( 1+ 2)^ = 1^ + 2^
( )^ =( ^ )
fork-forms ; 1; 2,l-form and a real number .
Exterior product is associative: ( ^ )^!= ^( ^!).
23
First two properties are fairly obvious. To prove associativity one can check that both sides of
the equality ( ^ )^!= ^( ^!) are equal to
1
k!l!m!(
!)asym:
In particular, if ; and!are 1-forms, i.e. if k=l=m= 1 then
^ ^!= (
!)asym:
This formula can be generalized for computing the exterior product of any number of 1-forms:
1^^ k= ( 1
k)asym: (2.5.1)
Example 2.10. x1^x2=x1
x2 x2
x1. For 2 vectors, U=0
BBB@u1
...
un1
CCCA;V=0
BBB@v1
...
vn1
CCCA;we have
x1^x2(U;V) =u1v2 u2v1= u1v1
u2v2 :
For 3 vectors U;V;W we have
x1^x2^x3(U;V;W ) =
=x1^x2(U;V)x3(W) +x1^x2(V;W )x3(U) +x1^x2(W;U )x3(V) =
(u1v2 u2v1)w3+ (v1w2 v2w1)u3+ (w1u2 w2u1)v3= u1v1w1
u2v2w2
u3v3w3 :
The last equality is just the expansion formula of the determinant according to the last row.
Proposition 2.11. Any exterior 2-formfcan be written as
f=X
1i<jnaijxi^xj
24
Proof. We had seen above that any bilinear form can be written as f=P
ijai;jxi
xj. Iffis
skew-symmetric then the matrix A= (aij) is skew-symmetric, i.e. aii= 0; aij= ajifori6=j.
Thus,f=P
1i<jnaij(xi
xj xj
xi) =P
1i<jnaijxi^xj.
Exercise 2.12. Prove that any exterior k-formfcan be written as
f=X
1i;<:::<iknai1:::ikxi1^xi2^:::xik:
The following proposition can be proven by induction over k, similar to what has been done in
Example 2.10 for the case k= 3.
Proposition 2.13. For anyk1-formsl1;:::;lkandkvectorsX1;:::;Xkwe have
l1^^lk(X1;:::;Xk) = l1(X1)::: l 1(Xk)
::: ::: :::
lk(X1)::: lk(Xk) : (2.5.2)
Corollary 2.14. The 1-formsl1;:::;lkare linearly dependent as vectors of Vif and only if
l1^:::^lk= 0. In particular, l1^:::^lk= 0ifk>n = dimV.
Proof. Ifl1;:::;lkare dependent then for any vectors X1;:::;Xk2Vthe rows of the determinant
in the equation (2.13) are linearly dependent. Therefore, this determinant is equal to 0, and hence
l1^:::^lk= 0. In particular, when k > n then the forms l1;:::;lkare dependent (because
dimV= dimV =n).
On the other hand, if l1;:::;lkare linearly independent, then the vectors l1;:::;lk2Vcan be
completed to form a basis l1;:::;lk;lk+1;:::;lnofV. According to Exercise 1.2 there exists a basis
w1;:::;wnofVthat is dual to the basis l1;:::;lnofV. In other words, l1;:::;lncan be viewed as
coordinate functions with respect to the basis w1;:::;wn. In particular, we have li(wj) = 0 ifi6=j
andli(wi) = 1 for all i;j= 1;:::;n . Hence we have
l1^^lk(w1;:::;wk) = l1(w1)::: l 1(wk)
::: ::: :::
lk(w1)::: lk(wk) = 1::: 0
::: 1:::
0::: 1 = 1;
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i.e.l1^^lk6= 0.
Proposition 2.13 can be also deduced from formula (2.5.1).
Corollary 2.14 and Exercise 2.12 imply that there are no non-zero k-forms on an n-dimensional
space fork>n .
2.6 Spaces of symmetric and skew-symmetric tensors
As was mentioned above, k-tensors on a vector space Vform a vector space under the operation of
addition of functions and multiplication by real numbers. We denoted this space by V
k. Symmet-
ric and skew-symmetric tensors form subspaces of this space V
k, which we denote, respectively,
bySk(V) and k(V). In particular, we have
V=S1(V) = 1(V)
.
Exercise 2.15. What is the dimension of the spaces Sk(V)andk(V)?
Answer.
dim k(V) =0
@n
k1
A=n!
k!(n k)!
dimSk(V) =(n+k 1)!
k!(n 1)!:
The basis of k(V) is formed by exterior k-formsxi1^xi2^:::^xik;1i1<i2<:::<i kn.
2.7 OperatorAon spaces of tensors
For any linear operator A:V!Wwe introduced above in Section 1.3 the notion of a dual linear
operatorA:W!V. NamelyA(l) =lAfor any element l2V, which is just a linear
function on V. In this section we extend this construction to k-tensors for k1, i.e. we will de ne
a mapA:W
k!V
k.
26
Given ak-tensor 2W
kandkvectorsX1;:::;Xk2Vwe de ne
A( )(X1;:::;Xk) = (A(X1);:::;A(Xk)):
Note that if is symmetric, or anti-symmetric, so is A( ). Hence, the map Aalso induces the
mapsSk(W)!Sk(V) and k(W)!k(V). We will keep the same notation Afor both of
these maps as well.
Proposition 2.16. LetA:V!Wbe a linear map. Then
1.A(
) =A( )
A( )for any 2W
k; 2W
l;
2.A( asym) = (A( ))asym;A( sym) = (A( ))sym;
3.A( ^ ) =A( )^A( )for any 2k(W); 2l(W).
IfB:W!Uis another linear map then (BA )=AB.
Proof.
1. Take any k+lvectorsX1;:::;Xk+l2V. Then by de nition of the operator Awe have
A(
)(X1;:::;Xk+l) =
(A(X1);:::;A(Xn+k) =
(A(X1);:::;A(Xk) (A(Xk+1);:::;A(Xn+k)) =
A( )(X1;:::;Xk)A( )(Xk+1;:::;Xk+n) =
A( )
A( )(X1;:::;Xk+l):
2. GivenkvectorsX1;:::;Xk2Vwe get
A( asym)(X1;:::;Xk) = asym(A(X1);:::;A(Xk)) =X
(i1:::ik)( 1)inv(i1;:::;ik) (A(Xi1);:::;A(Xik)) =
X
(i1:::ik)( 1)inv(i1;:::;ik)A( )(Xi1;:::;Xik) = (A( ))asym(X1;:::;Xk);
where the sum is taken over all permutations i1;:::;ikof indices 1;:::;k . Similarly one proves that
A( sym) = (A( ))sym.
3.A( ^ ) =1
k!l!A((
)asym)) =1
k!l!(A(
))asym=A( )^A( ).
The last statement of Proposition 2.16 is straightforward and its proof is left to the reader.
27
Let us now discuss how to compute A( ) in coordinates. Let us x bases v1;:::;vmand
w1;:::;wnin spacesVandW. Letx1;:::;xmandy1;:::;ynbe coordinates and
A=0
BBB@a11::: a 1m
.........
an1::: anm1
CCCA
be the matrix of a linear map A:V!Win these bases. Note that the map Ain these coordinates
is given by nlinear coordinate functions:
y1=l1(x1;:::;xm) =a11x1+a12x2+:::+a1mxm
y2=l2(x1;:::;xm) =a21x1+a22x2+:::+a2mxm
:::
yn=ln(x1;:::;xk) =an1x1+an2x2+:::+anmxn
We have already computed in Section 1.3 that A(yk) =lk=mP
j=1akjxj,k= 1;:::;n . Indeed, the
coecients of the function lk=A(yk) form thek-th column of the transpose matrix AT. Hence,
using Proposition 2.16 we compute:
A(yj1
yjk) =lj1
ljk
and
A(yj1^^yjk) =lj1^^ljk:
Consider now the case when V=W,n=m, and we use the same basis v1;:::;vnin the source
and target spaces.
Proposition 2.17.
A(x1^^xn) = detAx1^^xn:
Note that the determinant det Ais independent of the choice of the basis. Indeed, the matrix
of a linear map changes to a similar matrix C 1ACin a di erent basis, and det C 1AC= detA.
Hence, we can write det Ainstead of det A, i.e. attribute the determinant to the linear operator A
rather than to its matrix A.
28
Proof. We have
A(x1^^xn) =l1^^ln=nX
i1=1a1i1xi1^^nX
in=1aninxin=
nX
i1;:::;in=1a1i1:::aninxi1^^xin:
Note that the in the latter sum all terms with repeating indices vanish, and hence we can replace
this sum by a sum over all permutations of indices 1 ;:::;n . Thus, we can continue
A(x1^^xn) =X
i1;:::;ina1i1:::aninxi1^^xin=
0
@X
i1;:::;in( 1)inv(i1;:::;in)a1i1:::anin1
Ax1^^xn= detAx1^^xn:
Exercise 2.18. Apply the equality
A(x1^^xk^xk+1^^xn) =A(x1^^xk)^A(xk+1^^xn)
for a mapA:Rn!Rnto deduce the formula for expansion of a determinant according to its rst
krows:
detA=X
i1<<ik;j1<<jn k;im6=jl( 1)inv(i1;:::;jn k) a1;i1::: a 1;ik
.........
ak;i1::: ak;ik ak+1;j1::: ak+1;jn k
.........
an;j1::: an;jn k :
29
30
Chapter 3
Orientation and Volume
3.1 Orientation
We say that two bases v1;:::;vkandw1;:::;wkof a vector space Vde ne the same orientation of
Vif the matrix of transition from one of these bases to the other has a positive determinant. Clearly,
if we have 3 bases, and the rst and the second de ne the same orientation, and the second and the
third de ne the same orientation then the rst and the third also de ne the same orientation. Thus,
one can subdivide the set of all bases of Vinto the two classes. All bases in each of these classes
de ne the same orientation; two bases chosen from di erent classes de ne opposite orientation of
the space. To choose an orientation of the space simply means to choose one of these two classes of
bases.
There is no way to say which orientation is \positive" or which is \negative"|it is a question
of convention. For instance, the so-called counter-clockwise orientation of the plane depends from
which side we look at the plane. The positive orientation of our physical 3-space is a physical, not
mathematical, notion.
Suppose we are given two oriented spaces V;W of the same dimension. An invertible linear map
(= isomorphism) A:V!Wis called orientation preserving if it maps a basis which de nes the
given orientation of Vto a basis which de nes the given orientation of W.
Any non-zero exterior n-formonVinduces an orientation of the space V. Indeed, the preferred
set of bases is characterized by the property (v1;:::;vn)>0.
31
3.2 Orthogonal transformations
LetVbe a Euclidean vector space. Recall that a linear operator U:V!Vis called orthogonal if
it preserves the scalar product, i.e. if
hU(X);U(Y)i=hX;Yi; (3.2.1)
for any vectors X;Y2V. Recall that we have
hU(X);U(Y)i=hX;U?(U(Y))i;
whereU?:V!Vis the adjoint operator to U, see Section 1.3 above.
Hence, the orthogonality of an operator Uis equivalent to the identity U?U= Id, orU?=U 1.
Here we denoted by Id the identity operator, i.e. Id( X) =Xfor anyX2V.
Let us recall, see Exercise 1.10, that the adjoint operator U?is related to the dual operator
U:V!Vby the formula
U?=D 1UD:
Hence, for an orthogonal operator U, we haveD 1UD=U 1, i.e.
UD=DU 1: (3.2.2)
Letv1;:::;vnbe an orthonormal basis in VandUbe the matrix ofUin this basis. The matrix
of the adjoint operator in an orthonormal basis is the transpose of the matrix of this operator.
Hence, the equation UU= Id translates into the equation UTU=E, or equivalently UUT=E,
orU 1=UTfor its matrix. Matrices, which satisfy this equation are called orthogonal . If we write
U=0
BBB@u11::: u 1n
::: ::: :::
un1::: unn1
CCCA;
then the equation UTU=Ecan be rewritten as
X
iukiuji=8
><
>:1;ifk=j;
0;ifk6=j;:
32
Similarly, the equation UUT=Ecan be rewritten as
X
iuikuij=8
><
>:1;ifk=j;
0;ifk6=j;:
The above identities mean that columns (and rows) of an orthogonal matrix Uform an or-
thonormal basis of Rnwith respect to the dot-product.
In particular, we have
1 = det(UTU) = det(UT) detU= (detU)2;
and hence det U=1. In other words, the determinant of any orthogonal matrix is equal 1. We
can also say that the determinant of an orthogonal operator is equal to1 because the determinant
of the matrix of an operator is independent of the choice of a basis. Orthogonal transformations
with det = 1 preserve the orientation of the space, while those with det = 1 reverse it.
Composition of two orthogonal transformations, or the inverse of an orthogonal transformation
is again an orthogonal transformation. The set of all orthogonal transformations of an n-dimensional
Euclidean space is denoted by O(n). Orientation preserving orthogonal transformations sometimes
called special , and the set of special orthogonal transformations is denoted by SO(n). For instance
O(1) consists of two elements and SO(1) of one: O(1) =f1; 1g,SO(1) =f1g.SO(2) consists of
rotations of the plane, while O(2) consists of rotations and re
ections with respect to lines.
3.3 Determinant and Volume
We begin by recalling some facts from Linear Algebra. Let Vbe ann-dimensional Euclidean space
with an inner product h;i. Given a linear subspace LVand a point x2V, the projection
projL(x) is a vector y2Lwhich is uniquely characterized by the property x y?L, i.e.
hx y;zi= 0 for any z2L. The lengthjjx projL(x)jjis called the distance fromxtoL; we
denote it by dist( x;L).
LetU1;:::;Uk2Vbe linearly independent vectors. The k-dimensional parallelepiped spanned
by vectorsU1;:::;Ukis, by de nition, the set
P(U1;:::;Uk) =(kX
1jUj; 01;:::;k1)
Span(U1;:::;Uk):
33
Given ak-dimensional parallelepiped P=P(U1;:::;Uk) we will de ne its k-dimensional volume
by the formula
VolP=jjU1jjdist(U2;Span(U1))dist(U3;Span(U1;U2)):::dist(Uk;Span(U1;:::;Uk 1)):(3.3.1)
Of course we can write dist( U1;0) instead ofjjU1jj. This de nition agrees with the de nition of the
area of a parallelogram, or the volume of a 3-dimensional parallelepiped in the elementary geometry.
Proposition 3.1. Letv1;:::;vnbe an orthonormal basis in V. GivennvectorsU1;:::;Unlet us
denote byUthe matrix whose columns are coordinates of these vectors in the basis v1;:::;vn:
U:=0
BBB@u11::: u 1n
...
un1::: unn1
CCCA
Then
VolP(U1;:::;Un) =jdetUj:
Proof. If the vectors U1;:::;Unare linearly dependent then Vol P(U1;:::;Un) = detU= 0.
Suppose now that the vectors U1;:::;Unare linearly independent, i.e. form a basis. Consider rst
the case where this basis is orthonormal. Then the matrix Uis orthogonal. i.e. UUT=E, and
hence detU=1. But in this case Vol P(U1;:::;Un) = 1, and hence Vol P(U1;:::;Un) =jdetUj:
Now let the basis U1;:::;Unbe arbitrary. Let us apply to it the Gram-Schmidt orthonormaliza-
tion process. Recall that this process consists of the following steps. First, we normalize the vector
U1, then subtract from U2its projection to Span( U1), Next, we normalize the new vector U2, then
subtract from U3its projection to Span( U1;U2), and so on. At the end of this process we obtain
an orthonormal basis. It remains to notice that each of these steps a ected Vol P(U1;:::;Un) and
jdetUjin a similar way. Indeed, when we multiplied the vectors by a positive number, both the
volume and the determinant were multiplied by the same number. When we subtracted from a vec-
torUkits projection to Span( U1;:::;Uk 1), this a ected neither the volume nor the determinant.
Corollary 3.2. 1. Letx1;:::;xnbe a Cartesian coordinate system.1Then
VolP(U1;:::;Un) =jx1^:::xn(U1;:::;Un)j:
1i.e. a coordinate system with respect to an orthonormal basis
34
2. LetA:V!Vbe a linear map. Then
VolP(A(U1);:::;A(Un)) =jdetAjVolP(U1;:::;Un):
Proof.
1. According to 2.13, x1^:::xn(U1;:::;Un) = detU.
2.x1^:::xn(A(U1);:::;A(Un)) =A(x1^^xn)(U1;:::;Un) = detAx1^:::xn(U1;:::;Un):
In view of Proposition 3.1 and the rst part of Corollary 3.2 the value
x1^:::xn(U1;:::;Un) = detU
is called sometimes the signed volume of the parallelepiped P(U1;:::;Un). It is positive when the
basisU1;:::;Unde nes the given orientation of the space V, and it is negative otherwise.
Note thatx1^:::xk(U1;:::;Uk) for 0knis the signed k-dimensional volume of the
orthogonal projection of the parallelepiped P(U1;:::;Uk) to the coordinate subspace fxk+1=
=xn= 0g.
For instance, let !be the 2-form x1^x2+x3^x4onR4. Then for any two vectors U1;U22R4
the value!(U1;U2) is the sum of signed areas of projections of the parallelogram P(U1;U2) to the
coordinate planes spanned by the two rst and two last basic vectors.
3.4 Volume and Gram matrix
In this section we will compute the Vol kP(v1;:::;vk) in the case when the number kof vectors is
less than the dimension nof the space.
LetVbe an Euclidean space. Given vectors v1;:::;vk2Vwe can form a kk-matrix
G(v1;:::;vk) =0
BBB@hv1;v1i:::hv1;vki
::: ::: :::
hvk;v1i:::hvk;vki1
CCCA; (3.4.1)
which is called the Gram matrix of vectorsv1;:::;vk.
Suppose we are given Cartesian coordinate system in Vand let us form a matrix Cwhose
columns are coordinates of vectors v1;:::;vk. Thus the matrix Chasnrows andkcolumns. Then
G(v1;:::;vk) =CTC;
35
because in Cartesian coordinates the scalar product looks like the dot-product.
We also point out that if k=nand vectors v1;:::;vnform a basis of V, thenG(v1;:::;vk) is
just the matrix of the bilinear function hX;Yiin the basis v1;:::;vn. It is important to point out
that while the matrix Cdepends on the choice of the basis, the matrix Gdoes not.
Proposition 3.3. Given anykvectorsv1;:::;vkin an Euclidean space Vthe volume VolkP(v1;:::;vk)
can be computed by the formula
VolkP(v1;:::;vk)2= detG(v1;:::;vk) = detCTC; (3.4.2)
whereG(v1;:::;vk)is the Gram matrix and Cis the matrix whose columns are coordinates of
vectorsv1;:::;vkin some orthonormal basis.
Proof. Suppose rst that k=n. Then according to Proposition 3.1 we have Vol kP(v1;:::;vk) =
jdetCj. But detCTC= detC2, and the claim follows.
Let us denote vectors of our orthonormal basis by w1;:::;wn. Consider now the case when
Span(v1;:::;vk)Span(w1;:::;wk): (3.4.3)
In this case the elements in the j-th row of the matrix Care zero if j > k . Hence, if we denote
byeCthe square kkmatrix formed by the rst krows of the matrix C, thenCTC=eCTeCand
thus detCTC= deteCTeC. But deteCTeC= VolkP(v1;:::;vk) in view of our above argument in the
equi-dimensional case applied to the subspace Span( w1;:::;wk)V, and hence
Vol2
kP(v1;:::;vk) = detCTC= detG(v1;:::;vk):
But neither Vol kP(v1;:::;vk), nor the Gram matrix G(v1;:::;vk) depends on the choice of an
orthonormal basis. On the other hand, using Gram-Schmidt process one can always nd an or-
thonormal basis which satis es condition (3.4.3).
Remark 3.4. Note that detG(v1;:::;vk)0anddetG(v1;:::;vk) = 0 if an and only if the
vectorsv1;:::;vkare linearly dependent.
36
Chapter 4
Dualities
4.1 Duality between k-forms and (n k)-forms on a n-dimensional
Euclidean space V
LetVbe ann-dimensional vector space. As we have seen above, the space k(V) ofk-forms, and
the space n k(V) of (n k)-forms have the same dimensionn!
k!(n k)!; these spaces are therefore
isomorphic. Suppose that Vis an oriented Euclidean space, i.e. it is supplied with an orientation
and an inner product h;i. It turns out that in this case there is a canonical way to establish this
isomorphism which will be denoted by
?: k(V)!n k(V):
De nition 4.1. Let be ak-form. Then given any vectors U1;:::;Un k, the value? (U1;:::;Un k)
can be computed as follows. If U1;:::;Un kare linearly dependent then ? (U1;:::;Un k) = 0 . Oth-
erwise, letS?denote the orthogonal complement to the space S= Span(U1;:::;Un k). Choose a
basisZ1;:::;ZkofS?such that
Volk(Z1;:::;Zk) = Voln k(U1;:::;Un k)
and the basis Z1;:::;Zk;U1;:::;Un kde nes the given orientation of the space V. Then
? (U1;:::;Un k) = (Z1;:::;Zk): (4.1.1)
37
Let us rst show that
Lemma 4.2. ? is a(n k)-form, i.e.? is skew-symmetric and multilinear.
Proof. To verify that ? is skew-symmetric we note that for any 1 i<jn qthe bases
Z1;Z2;:::;Zk;U1;:::;Ui;:::;Uj;:::;Un k
and
Z1;Z2;:::;Zk;U1;:::;Uj;:::;Ui;:::;Un k
de ne the same orientation of the space V, and hence
? (U1;:::;Uj;:::;Ui;:::;Un k) = ( Z1;Z2;:::;Zk)
= (Z1;Z2;:::;Zk) = ? (U1;:::;Ui;:::;Uj;:::;Un k):
Hence, in order to check the multi -linearity it is sucient to prove the linearity of with respect
to the rst argument only. It is also clear that
? (U1;:::;Un k) =? (U1;:::;Un k): (4.1.2)
Indeed, multiplication by 6= 0 does not change the span of the vectors U1;:::;Un q, and hence
if? (U1;:::;Un k) = (Z1;:::;Zk) then? (U1;:::;Un k) = (Z1;:::;Zk) = (Z1;:::;Zk).
Thus it remains to check that
? (U1+eU1;U2;:::;Un k) =? (U1;U2;:::;Un k) +? (eU1;U2;:::;Un k)):
Let us denote L:= Span(U2;:::;Un k) and observe that projL(U1+eU1) = projL(U1) +
projL(eU1). DenoteN:=U1 projL(U1) andeN:=eU1 projL(eU1). The vectors NandeNare
normal components of U1andeU1with respect to the subspace L, and the vector N+eNis the
normal component of U1+eU1with respect to L. Hence, we have
? (U1;:::;Un k) =? (N;:::;Un k); ? (eU1;:::;Un k) =? (eN;:::;U n k);
and
? (U1+eU1;:::;Un k) =? (N+eN;:::;U n k):
Indeed, in each of these three cases,
38
- vectors on both side of the equality span the same space;
- the parallelepiped which they generate have the same volume, and
- the orientation which these vectors de ne together with a basis of the complementary space
remains unchanged.
Hence, it is sucient to prove that
? (N+eN;U 2;:::;Un k) =? (N;U 2;:::;Un k) +? (eN;U 2;:::;Un k): (4.1.3)
If the vectors NandeNare linearly dependent, i.e. one of them is a multiple of the other, then
(4.1.3) follows from (4.1.2).
Suppose now that NandeNare linearly independent. Let L?denote the orthogonal complement
ofL= Span(U2;:::;Un k). Then dim L?=k+ 1 and we have N;eN2L?. Let us denote by M
the plane in L?spanned by the vectors NandeN, and byM?its orthogonal complement in L?.
Note that dim M?=k 1.
Choose any orientation of Mso that we can talk about counter-clockwise rotation of this plane.
LetY;eY2Mbe vectors obtained by rotating NandeNinMcounter-clockwise by the angle
2.
ThenY+eYcan be obtained by rotating N+eNinMcounter-clockwise by the same angle
2. Let
us choose in M?a basisZ2;:::;Zksuch that
Volk 1P(Z2;:::;Zk) = Voln k 1P(U2;:::;Un k):
Note that the orthogonal complements to Span( N;U 2;:::;Un k), Span(eN;U 2;:::;Un k), and to
Span(N+eN;U 2;:::;Un k) inVcoincide, respectively, with the orthogonal complements to the the
vectorsN;eNand toN+eNinL?. In other words, we have
(Span(N;U 2;:::;Un k))?
V= Span(Y;Z 2;:::;Zk);
Span(eN;U 2;:::;Un k)?
V= Span(eY;Z 2;:::;Zk) and
Span(N+eN;U 2;:::;Un k)?
V= Span(Y+eY;Z 2;:::;Zk):
Next, we observe that
Voln kP(N;U 2;:::;Un k) = VolkP(Y;Z 2;:::;Zk);
39
Voln kP(eN;U 2;:::;Un k) = VolkP(eY;Z 2;:::;Zk) and
Voln kP(N+eN;U 2;:::;Un k) = VolkP(Y+eY;Z 2;:::;Zk):
Consider the following 3 bases of V:
Y;Z 2;:::;Zk;N;U 2;:::;Un k;
eY;Z 2;:::;Zk;eN;U 2;:::;Un k;
Y+eY;Z 2;:::;Zk;N+eN;U 2;:::;Un k;
and observe that all three of them de ne the same a priori given orientation of V. Thus, by de nition
of the operator ?we have:
? (N+eN;U 2;:::;Un k) = (Y+eY;Z 2;:::;Zk)
= (Y;Z 2;:::;Zk) + (eY;Z 2;:::;Zk) =? (N;U 2;:::;Un k) +? (eN;U 2;:::;Un k):
This completes the proof that ? is an (n k)-form.
Thus the map 7!? de nes a map ?: k(V)!n k(V). Clearly, this map is linear. In
order to check that ?is an isomorphism let us choose an orthonormal basis in Vand consider the
coordinates x1;:::;xn2Vcorresponding to that basis.
Let us recall that the forms xi1^xi2^^xik, 1i1<i2<<ikn, form a basis of the
space k(V).
Lemma 4.3.
?xi1^xi2^^xik= ( 1)inv(i1;:::;ik;j1;:::;jn k)xj1^xj2^^xjn k; (4.1.4)
wherej1<<jn kis the set of indices, complementary to i1;:::;ik. In other words, i1;:::;ik;j1;:::;jn k
is a permutation of indices 1;:::;n .
Proof. Evaluating ?(xi1^^xik) on basic vectors vj1;:::;vjn k, 1j1<<jn qn, we get
0 unless all the indices j1;:::;jn kare all di erent from i1;:::;ik, while in the latter case we get
?(xi1^^xik)(vj1;:::;vjn k) = ( 1)inv(i1;:::;ik;j1;:::;jn k):
40
Hence,
?xi1^xi2^^xik= ( 1)inv(i1;:::;ik;j1;:::;jn k)xj1^xj2^^xjn k:
Thus?establishes a 1 to 1 correspondence between the bases of the spaces k(V) and
the space n k(V), and hence it is an isomorphism. Note that by linearity for any form =
P
1i1<<iknai1:::ikxi1^^xik) we have
? =X
1i1<<iknai1:::ik?(xi1^^xik):
Examples .
1.?C=Cx1^^xn; in other words the isomorphism ?acts on constants (= 0-forms) by
multiplying them by the volume form.
2. InR3we have
?x1=x2^x3;?x2= x1^x3=x3^x1;?x3=x1^x2;
?(x1^x2) =x3;?(x3^x1) =x2;?(x2^x3) =x1:
3. More generally, given a 1-form l=a1x1++anxnwe have
?l=a1x2^^xn a2x1^x3^^xn++ ( 1)n 1anx1^^xn 1:
In particular for n= 3 we have
?(a1x1+a2x2+a3x3) =a1x2^x3+a2x3^x1+a3x1^x2:
Proposition 4.4.
?2= ( 1)k(n k)Id;i.e.?(?!) = ( 1)k(n k)!for anyk-form!:
In particular, if dimension n= dimVis odd then2= Id . Ifnis even and !is ak-form then
?(?!) =!ifkis even, and ?(?!) = !ifkis odd.
41
Proof. It is sucient to verify the equality
?(?!) = ( 1)k(n k)!
for the case when !is a basic form, i.e.
!=xi1^^xik;1i1<<ikn:
We have
?(xi1^^xik) = ( 1)inv(i1;:::;ik;j1;:::;jn k)xj1^xj2^^xjn k
and
?(xj1^xj2^^xjn k) = ( 1)inv(j1;:::;jn k;i1;:::;ik)xi1^^xik:
But the permutations i1:::ikj1:::jn kandj1:::jn ki1:::ikdi er byk(n k) transpositions of
pairs of its elements. Hence, we get
( 1)inv(i1;:::;ik;j1;:::;jn k)= ( 1)k(n k)( 1)inv(j1;:::;jn k;i1;:::;ik);
and, therefore,
?
?(xi1^^xik)
=?
( 1)inv(i1;:::;ik;j1;:::;jn k)xj1^xj2^^xjn k
= ( 1)inv(i1;:::;ik;j1;:::;jn k)?(xj1^xj2^^xjn k)
= ( 1)inv(i1;:::;ik;j1;:::;jn k)+inv(j1;:::;jn k;i1;:::;ik)xi1^^xik
= ( 1)k(n k)xi1^^xik:
Exercise 4.5. (a) For any special orthogonal operator Athe operatorsAand?commute, i.e.
A?=?A:
(b) LetAbe an orthogonal matrix of order nwith detA= 1. Prove that for any k2f1;:::;ng
the absolute value of each k-minorMofAis equal to the absolute value of its complementary
minor of order (n k). (Hint: Apply (a) to the form xi1^^xik).
42
(c) LetVbe an oriented 3-dimensional Euclidean space. Prove that for any two vectors X;Y2V,
their cross-product can be written in the form
XY=D 1(?(D(X)^D(Y))):
4.2 Euclidean structure on the space of exterior forms
Suppose that the space Vis oriented and Euclidean, i.e. it is endowed with an inner product h;i
and an orientation.
Given two forms ; 2k(V),k= 0;:::;n , let us de ne
hh ; ii=?( ^? ):
Note that ^? is ann-form for every k, and hence,hh ; iiis a 0-form, i.e. a real number.
Proposition 4.6. 1. The operation hh;iide nes an inner product on k(V)for eachk=
0;:::;n .
2. IfA:V!Vis a special orthogonal operator then the operator A: k(V)!k(V)is
orthogonal with respect to the inner product hh;ii.
Proof. 1. We need to check that hh ; iiis a symmetric bilinear function on k(V) andhh ; ii>0
unless = 0. Bilinearity is straightforward. Hence, it is sucient to verify the remaining properties
for basic vectors =xi1^^xik; =xj1^^xjk, where 1i1<<ikn, 1j1<<
jkn. Here (x1;:::;xn) is any Cartersian coordinates in Vwhich de ne its given orientation.
Note thathh ; ii= 0 =hh ; iiunlessim=jmfor allm= 1;:::;k , and in the the latter case
we have = . Furthermore, we have
hh ; ii=?( ^? ) =?(x1^^xn) = 1>0:
2. The inner product hh;iiis de ned only in terms of the Euclidean structure and the orientation
ofV. Hence, for any special orthogonal operator A(which preserves these structures) the induced
operatorA: k(V)!k(V) preserves the inner product hh;ii.
Note that we also proved that the basis of k-formsxi1^^xik, 1i1<< ikn, is
orthonormal with respect to the scalar product hh;ii. Hence, we get
43
Corollary 4.7. Suppose that a k-form can be written in Cartesian coordinates as
=X
1i1<<iknai1:::ikxi1^^xik:
Then
jj jj2=hh ; ii=X
1i1<<ikna2
i1:::ik:
Corollary 4.8. For any two exterior k-forms we have
? ^ = ( 1)k(n k) ^? :
Exercise 4.9. 1. Show that for any k-forms we have
hh ; ii=hh? ;? ii:
2. Show that if ; are1-forms on an Euclidean space V. Then
hh ; ii=hD 1( );D 1( )i;
i.e the scalar product hh;iionVis the push-forward by Dof the scalar product h;ionV.
Corollary 4.10. LetVbe a Euclidean n-dimensional space. Choose an orthonormal basis e1;:::;en
inV. Then for any vectors Z1= (z11;:::;zn1);:::;Zk= (z1k;:::;znk)2Vwe have
(VolkP(Z1;:::;Zk))2=X
1i1<<iknZ2
i1;:::;ik; (4.2.1)
where
Zi1;:::;ik= zi11::: zi1k
::: ::: :::
zik1::: zikk :
Proof. Consider linear functions lj=D(Zj) =nP
i=1zijxi2V,j= 1;:::;k . Then
44
l1^^lk=nX
i1=1zi1jxi1^^nX
ik=1zikjxik=
X
i1;:::;ikzi1:::zikxi1^xik=X
1i1<<iknZi1;:::;ikxi1^:::xik: (4.2.2)
In particular, if one has Z1;:::Zk2Span(e1;:::;ek) thenZ1:::k= VolP(Z1;:::;Zk) and hence
l1^^lk=Z1:::kx1^^xk= VolP(Z1;:::;Zk)x1^^xk;
which yields the claim in this case.
In the general case, according to Proposition 4.7 we have
jjl1^^lkjj2=X
1i1<<iknZ2
i1;:::;ik; (4.2.3)
which coincides with the right-hand side of (4.2.1). Thus it remains to check to that
jjl1^^lkjj= VolkP(Z1;:::;Zk):
Given any orthogonal transformation A:V!Vwe have, according to Proposition 4.6, the equality
jjl1^^lkjj=jjAl1^^Alkjj: (4.2.4)
We also note that any orthogonal transformation B:V!Vpreservesk-dimensional volume of all
k-dimensional parallelepipeds:
jVolkP(Z1;:::;Zk)j=jVolkP(B(Z1);:::;B(Zk))j: (4.2.5)
On the other hand, there exists an orthogonal transformation A:V!Vsuch thatA 1(Z1);:::;A 1(Zk)2
Span(e1;:::;ek). DenoteeZj:=A 1(Zj),j= 1;:::;k . Then, according to (3.2.2) we have
elj:=D(eZj) =D(A 1(Zj)) =A(D(Zj)) =Alj:
As was pointed out above we then have
jVolkP(eZ1;:::;eZk)j=jjel1^^elkjj=jjAl1^^Alkjj; (4.2.6)
and hence, the claim follows from (4.2.4) and (4.2.5) applied to B=A 1.
We recall that an alternative formula for computing Vol kP(Z1;:::;Zk) was given earlier in
Proposition 3.3.
45
Remark 4.11. Note that the above proof also shows that for any kvectorsv1;:::;vkwe have
VolkP(v1;:::;vk) =jjl1^^lkjj;
wherelj=D(vi); i= 1;:::;k:
4.3 Contraction
LetVbe a vector space and 2k(V) ak-form. De ne a ( k 1)-form =v by the formula
(X1;:::;Xk 1) = (v;X 1;:::;Xk 1)
for any vectors X1;:::;Xk 12V. We say that the form is obtained by a contraction of with
the vectorv. Sometimes, this operation is called also an interior product of withvand denoted
byi(v) instead ofv . In these notes we will not use this notation.
Proposition 4.12. Contraction is a bilinear operation, i.e.
(v1+v2) =v1 +v2
(v) =(v )
v( 1+ 2) =v 1+v 2
v( ) =(v ):
Herev;v1;v22V; ; 1; 22k(V);2R.
The proof is straightforward.
Let be a non-zero n-form. Then we have
Proposition 4.13. The map :V!n 1(V);de ned by the formula (v) =v is an isomor-
phism between the vector spaces Vandn 1(V).
46
Proof. Take a basis v1;:::;vn. Letx1;:::;xn2Vbe the dual basis, i.e. the corresponding coor-
dinate system. Then =ax1^:::^xn, wherea6= 0. To simplify the notation let us assume that
a= 1, so that
=x1^:::^xn:
Let us compute the images vi ,i= 1;:::;k of the basic vectors. Let us write
vi =nX
1ajx1^^xj 1^xj+1^^xn:
Then
vi (v1;:::;vl 1;vl+1;:::;vn)
=nX
1ajx1^^xj 1^xj+1^^xn(v1;:::;vl 1;vl+1;:::;vn) =al; (4.3.1)
but on the other hand,
vi (v1;:::;vl 1;vl+1;:::;vn) = (vi;v1;:::;vl 1;vl+1;:::;vn)
= ( 1)i 1 (v1;:::;vl 1;vi;vl+1;:::;vn) =8
><
>:( 1)i 1; l=i;
0;otherwise(4.3.2)
Thus,
vi = ( 1)i 1x1^^xi 1^xi+1^^xn:
Hence, the map sends a basis of Vinto a basis of n 1(V), and therefore it is an isomorphism.
Take a vector v=nP
1ajvj. Then we have
v(x1^^xn) =nX
1( 1)i 1aix1^^xi 1^xi+1^^xn: (4.3.3)
This formula can be interpreted as the formula of expansion of a determinant according to the rst
column (or the rst row). Indeed, for any vectors U1;:::;Un 1we have
v (U1;:::;Un 1) = det(v;U 1;:::;Un 1) = a1u1;1::: u 1;n 1
::: ::: ::: :::
anun;1::: un;n 1 ;
47
where0
BBB@u1;i
...
un;i1
CCCAare coordinates of the vector Ui2Vin the basis v1;:::;vn. On the other hand,
v (U1;:::;Un 1) =nX
1( 1)i 1aix1^^xi 1^xi+1^^xn(U1;:::;Un 1)
=a1 u2;1::: u 2;n 1
u3;1::: u 3;n 1
::: ::: :::
un;1::: un;n 1 ++ ( 1)n 1an u1;1::: u 1;n 1
u2;1::: u 2;n 1
::: ::: :::
un 1;1::: un 1;n 1 : (4.3.4)
Suppose that dim V= 3. Then the formula (4.3.3) can be rewritten as
v(x1^x2^x3) =a1x2^x3+a2x3^x1+a3x1^x2;
where0
BBB@a1
a2
a31
CCCAare coordinates of the vector V. Let us describe the geometric meaning of the
operation . Set!=v(x1^x2^x3). Then!(U1;U2) is the volume of the parallelogram de ned
by the vectors U1;U2andv. Letbe the unit normal vector to the plane L(U1;U2)V. Then we
have
!(U1;U2) = AreaP(U1;U2)hv;i:
If we interpret vas the velocity of a
uid
ow in the space Vthen!(U1;U2) is just an amount
of
uid
own through the parallelogram generated by vectors U1andU2for the unit time. It is
called the
ux ofvthrough the parallelogram .
Let us return back to the case dim V=n.
Exercise 4.14. Let
=xi1^^xik;1i1<<ikn
andv= (a1;:::;an). Show that
v =kX
j=1( 1)j+1aijxi1^:::xij 1^xij+1^:::xik:
48
The next proposition establishes a relation between the isomorphisms ?;andD.
Proposition 4.15. LetVbe a Euclidean space, and x1;:::;xnbe coordinates in an orthonormal
basis. Then for any vector v2Vwe have
?Dv=v(x1^^xn):
Proof. Letv= (a1;:::;an). ThenDv=a1x1++anxnand
?Dv=a1x2^^xn a2x1^x3^^xn^x1++ ( 1)n 1anx1^^xn 1:
But according to Proposition 4.13 the ( n 1)-formv(x1^^xnis de ned by the same formula.
We nish this section by the proposition which shows how the contraction operation interacts
with the exterior product.
Proposition 4.16. Let ; be forms of order kandl, respectively, and va vector. Then
v( ^ ) = (v )^ + ( 1)k ^(v ):
Proof. Note that given any indices k1;:::km(not necessarily ordered) we have eixk1^^xkm= 0
ifi =2fk1;:::;kmgandeixk1^^xkm= ( 1)Jxk1^:::i_^xkm;whereJ= inv(i;k1;:::;km)
is the number of variables ahead of xi.
By linearity it is sucient to consider the case when v; ; are basic vector and forms, i.e.
v=ei; =xi1^^xik; =xj1:::xjl:
We havev 6= 0 if and only if the index iis among the indices i1;:::;ik. In that case
v = ( 1)Jxi1^:::i_^xik
and ifv 6= 0 then
v = ( 1)J0xj1^:::i_^xjl;
whereJ= inv(i;i1;:::;ik);J0= inv(i;j1;:::;jl).
If it enters bot not then
v( ^ ) = ( 1)Jxi1^:::i_^xik^xj1^^xjl= (v )^ ;
49
while ^(v ) = 0:
Similarly, if it enters but not then
v( ^ ) = ( 1)J0+mxi1^^xik^xj1^:::i_^xjl= ( 1)k ^(v );
whilev ^ = 0:Hence, in both these cases the formula holds.
Ifxienters both products then ^ = 0, and hence v( ^ ) = 0.
On the other hand,
(v )^ + ( 1)k ^(v ) = ( 1)Jxi1^:::i_^xik^xj1^xjl
+ ( 1)k+J0xi1^^xik^xj1^:::i_^xjl= 0;
because the products xi1^:::i_^xik^xj1^xjlandxi1^^xik^xj1^:::i_^xjldi er
only in the position of xi. In the rst product it is at the ( k+J0)-s position, and in the second
at (J+ 1)-st position. Hence, the di erence in signs is ( 1)J+J0+k+1, which leads to the required
cancellation.
50
Chapter 5
Complex vector spaces
5.1 Complex numbers
The space R2can be endowed with an associative and commutative multiplication operation. This
operation is uniquely determined by three properties:
it is a bilinear operation;
the vector (1 ;0) is the unit;
the vector (0 ;1) satis es (0 ;1)2= (0; 1).
The vector (0 ;1) is usually denoted by i, and we will simply write 1 instead of the vector (1 ;0).
Hence, any point ( a;b)2R2can be written as a+bi, wherea;b2R, and the product of a+biand
c+diis given by the formula
(a+bi)(c+di) =ac bd+ (ad+bc)i:
The plane R2endowed with this multiplication is denoted by Cand called the set of complex
numbers . The real line generated by 1 is called the real axis , the line generated by iis called the
imaginary axis . The set of real numbers Rcan be viewed as embedded into Cas the real axis.
Given a complex number z=x+iy, the numbers xandyare called its realand imaginary parts,
respectively, and denoted by Re zand Imz, so thatz= Rez+iImz.
51
For any non-zero complex number z=a+bithere exists an inverse z 1such thatz 1z= 1.
Indeed, we can set
z 1=a
a2+b2 b
a2+b2i:
The commutativity, associativity and existence of the inverse is easy to check, but it should not
be taken for granted: it is impossible to de ne a similar operation any Rnforn>2.
Givenz=a+bi2Cits conjugate is de ned as z=a bi. The conjugation operation z7!zis
the re
ection of Cwith respect to the real axis RC. Note that
Rez=1
2(z+ z);Imz=1
2i(z z):
Let us introduce the polar coordinates ( r; ) inR2=C. Then a complex number z=x+yi
can be written as rcos +irsin =r(cos +isin ). This form of writing a complex number
is called, sometimes, t trigonometric . The number r=p
x2+y2is called the modulus ofzand
denoted byjzjand is called the argument of and denoted by arg z. Note that the argument is
de ned only mod 2 . The value of the argument in [0 ;2) is sometimes called the principal value
of the argument. When zis real than its modulus jzjis just the absolute value. We also not that
jzj=pzz.
An important role plays the triangle inequality
jz1j jz2j jz1+z2jjz1j+jz2j:
Exponential function of a complex variable
Recall that the exponential function exhas a Taylor expansion
ex=1X
0xn
n!= 1 +x+x2
2+x3
6+::: :
We then de ne for a complex the exponential function by the same formula
ez:= 1 +z+z2
2!++zn
n!+::::
One can check that this power series absolutely converging for allzand satis es the formula
ez1+z2=ez1ez2:
52
Figure 5.1: Leonhard Euler (1707-1783)
In particular, we have
eiy= 1 +iy y2
2! iy3
3!+y4
4!++::: (5.1.1)
=1X
k=0( 1)ky2k
2k!+i1X
k=0( 1)ky2k+1
(2k+ 1)!: (5.1.2)
But1P
k=0( 1)ky2k
2k!= cosyand1P
k=0( 1)ky2k+1
(2k+1)!= siny, and hence we get Euler's formula
eiy= cosy+isiny;
and furthermore,
ex+iy=exeiy=ex(cosy+isiny);
i.e.jex+iyj=ex;arg(ez) =y:In particular, any complex number z=r(cos +isin ) can be
rewritten in the form z=rei . This is called the exponential form of the complex number z.
Note that
ei n
=ein ;
and hence if z=rei thenzn=rnein =rn(cosn +isinn ):
Note that the operation z7!izis the rotation of Ccounterclockwise by the angle
2. More
generally a multiplication operation z7!zw, wherew=eiis the composition of a rotation by
the angleand a radial dilatation (homothety) in times.
53
Exercise 5.1. 1. ComputenP
0coskandnP
1sink.
2. Compute 1 + n
4
+ n
8
+ n
12
+::::
5.2 Complex vector space
In a real vector space one knows how to multiply a vector by a real number. In a complex vector
space there is de ned an operation of multiplication by a complex number. Example is the space
Cnwhose vectors are n-tuplesz= (z1;:::;zn) of complex numbers, and multiplication by any
complex number = +i is de ned component-wise: (z1;:::;zn) = (z1;:::;zn). Complex
vector space can be viewed as an upgrade of a real vector space, or better to say as a real vector
space with an additional structure.
In order to make a real vector space Vinto a complex vector space, one just needs to de ne how
to multiply a vector by i. This operation must be a linear map J:V!Vwhich should satisfy
the conditionJ2= Id, i.eJ(J(v)) =i(iv) = v.
Example 5.2. Consider R2nwith coordinates (x1;y1;:::;xn;yn). Consider a 2n2n-matrix
J=0
BBBBBBBBBBBBBBB@0 1 0 0
1 0 0 0
0 0 0 1
0 0 1 0
:::
0 1
1 01
CCCCCCCCCCCCCCCA
ThenJ2= I. Consider a linear operator J:R2n!R2nwith this matrix, i.e. J(Z) =JZfor
any vector Z2R2nwhich we view as a column-vector. Then J2= Id, and hence we can de ne
onR2na complex structure (i.e.the multiplication by iby the formula
iZ=J(Z); Z2R2n:
This complex vector space is canonically isomorphic to Cn, where we identify the real vector
(x1;y1;:::;xn;yn)2R2nwith a complex vector (z1=x1+iy1;:::;znxn+iyn).
54
On the other hand, any complex vector space can be viewed as a real vector space. In order to
do that we just need to \forget" how to multiply by i. This procedure is called the reali cation of
a complex vector space. For example, the reali cation of CnisR2n. Sometimes to emphasize the
reali cation operation we will denote the reali cation of a complex vector space VbyVR. As the
sets these to objects coincide.
Given a complex vector space Vwe can de ne linear combinationsPivi, wherei2Care
complex numbers, and thus similarly to the real case talk about really dependent, really independent
vectors. Given vectors v1;:::vn2Vwe de ne its complex span SpanC(V1;:::;vn) by the formula
SpanC(v1;:::;vn) =(nX
1ivi; i2C:)
. Abasis of a complex vector space is a system of complex linear independent vectors v1;:::;vn
such that SpanC(v1;:::;vn) =V. The number of vectors in a complex basis is called the complex
dimension ofVand denoted dim CV.
For instance dim Cn=n. On the other hand, its reali cation R2nhas real dimension 2 n. In
particular, Cis a complex vector space of dimension 1, and therefore it is called a complex line
rather than a plane.
Exercise 5.3. Letv1;:::;vnbe a complex basis of a complex vector space V. Find the real basis of
its reali cation VR.
Answer .v1;iv1;v2;iv2;:::;vn;ivn.
There is another important operation which associates with a real vector space Vof real dimen-
sionna complex vector space VCof complex dimension n. It is done in a way similar to how we
made complex numbers out of real numbers. As a real vector space the space VCis just the direct
sumVV=f(v;w);v;w2Vg. This is a real space of dimension 2 n. We then make VVinto
a complex vector space by de ning the multiplication by iby the formula:
i(v;w) = ( w;v):
We will write vector ( v;0) simply by vand (0;v) =i(v;0) byiv. Hence, every vector of VCcan be
written asv+iw, wherev;w2V. Ifv1;:::;vnis a real basis of V, then the same vectors form a
complex basis of VC.
55
Complexi cation of a real vector space
Given a real space Vone can associate with it a complex vector space VC, called the complexi cation
ofV, as follows. It is made of vectors of Vin exactly the same way as complex numbers made of
reals. Namely, VCconsists of expressions X+iY, whereX;Y2V. We de ne a multiplication of a
complex number a+ibby a vector X+iYby a formula
(a+ib)(X+iY) =aX bY+i(aY+bX):
For instance, Cnis canonically isomorphic to ( Rn)C, because any vector Z= (z1=x1+
iy1;:::;zn=xn+iyn)2Cncan be uniquely written as Z=X+iY, whereX= (x1;:::;xn);Y=
(y1;:::;yn)2Rn. Note that the reali cation of the space VC, i.e. the space ( VC)Ris canonically is
justVV=f(X;Y ); X;Y2Rn.
Ifv1;:::;vnis a basis of a VoverR, then the same vectors form a basis of the complexi ed
spaceVCoverC. Thus dim RV= dim CVC.
5.3 Complex linear maps
Complex linear maps and their reali cations
Given two complex vector spaces V;W a mapA:V!Wis called complex linear (orC-linear)
ifA(X+Y) =A(X) +A(Y) andA(X) =A(X) for any vectors X;Y2Vand any complex
number2C. Thus complex linearity is stronger condition than the real linearity. The di erence is
in the additional requirement that A(iX) =iA(X). In other words, the operator Amust commute
with the operation of multiplication by i.
Any linear mapA:C!Cis a multiplication by a complex number a=c+id. If we view C
asR2and right the real matrix of this map in the standard basis 1 ;iwe get the matrix0
@c d
d c1
A.
Indeed,A(1) =a=c+diandA(i) =ai= d+ci, so the rst column of the matrix is equal to0
@c
d1
Aand the second one is equal to0
@ d
c1
A.
If we have bases v1;:::;vnofVandw1;:::wmofWthen one can associate with Aanmn
complex matrixAby the same rule as in the real case.
56
Recall (see Exercise 5.3) that vectors v1;v0
1=iv1;v2;v0
2=iv2;:::;vn;v0
n=ivnandw1;w0
1=
iw1;w2;w0
2=iw2;:::wm;w0
m=iwm) form real bases of the reali cations VRandWRof the spaces
VandW. if
A=0
BBB@a11::: a 1n
:::
am1::: amn1
CCCA
is the complex matrix of Athen the real matrix ARof the mapAis the real basis has order 2 n2n
and is obtained from Aby replacing each complex element akl=ckl+idklby a 22 matrix0
@ckl dkl
dklckl1
A.
Exercise 5.4. Prove that
detAR=jdetAj2:
Complexi cation of real linear maps
Given a real linear map A:V!Wone can de ne a complex linear map AC:VC!WCby the
formula
AC(v+iw) =A(v) +iA(w):
. IfAis the matrix ofAin a basisv1;:::;vnthenAChas the same matrix in the same basis viewed
as a complex basis of VC. The operatorACis called the complexi cation of the operatorA.
In particular, one can consider C-linear functions V!Con a complex vector space V. Complex
coordinates z1;:::;znin a complex basis are examples of C-linear functions, and any other C-linear
function on Vhas a form c1z1+:::cnzn, wherec1;:::;cn2Care complex numbers.
Complex-valued R-linear functions
It is sometimes useful to consider also C-valued R-linear functions on a complex vector space V,
i.e.R-linear maps V!C(i.e. a linear map VR!R2). Such a C-valued function has the form
= +i , where ; are usual real linear functions. For instance the function zonCis a
C-valued R-linear function which is not C-linear.
57
Ifz1;:::;znare complex coordinates on a complex vector space Vthen any R-linear complex-
valued function can be written asnP
1aizi+bizi, whereai;bi2Care complex numbers.
We can furthermore consider complex-valued tensors and, in particular complex-valued exterior
forms. A C-valuedk-formcan be written as +i where and are usual R-valuedk-forms.
For instance, we can consider on Cnthe 2-form !=i
2nP
1zk^zk. It can be rewritten as !=
i
2nP
1(xk+iyk)^(xk iyk) =nP
1xk^yk:
58
Part II
Calculus of di erential forms
59
Chapter 6
Topological preliminaries
6.1 Elements of topology in a vector space
We recall in this section some basic topological notions in a nite-dimensional vector space and
elements of the theory of continuous functions. The proofs of most statements are straightforward
and we omit them.
Let us choose in Va scalar product.
NotationBr(p) :=fx2V;jjx pjj< rg,Dr(p) :=fx2V;jjx pjjrgandSr(p) :=fx2
V;jjx pjj=rgstand for open, closed balls and the sphere of radius rcentered at a point p2V.
Open and closed sets
A setUVis called open if for anyx2Uthere exists >0 such that B(x)U.
A setAVis called closed if its complement VnAis open. Equivalently,
Lemma 6.1. The setAis closed if and only if for any sequence xn2A,n= 1;2;::: which
converges to a point a2V, the limit point abelongs toA.
Remark 6.2. It is important to note that the notion of open and closed sets are independent of
the choice of the auxiliary Euclidean structure in the space V.
Points which appear as limits of sequences of points xn2Aare called limit points ofA.
There are only two subsets of Vwhich are simultaneously open and closed: Vand?.
61
Lemma 6.3. 1. For any family U;2of open sets the unionS
2Uis open.
2. For any family A;2of closed sets the intersectionT
2Ais closed.
3. The unionnS
1Aiof a nite family of closed sets is closed.
4. The intersectionnT
1Uiof a nite family of open sets is open.
By a neighborhood of a pointa2Vwe understand any open set U3p.
Given any subset XVa pointa2Vis called
interior point forAif there is a neighborhood U3psuch thatUA;
boundary point if it is not an interior point neither for Anor for its complement VnA.
We emphasize that a boundary point of Amay or may not belong to A. Equivalently, a point a2V
is a boundary point of Aif it is a limit point both for AandVnA.
The set of all interior points of Ais called the interior of Aand denoted Int A. The set of all
boundary points of Ais called the boundary ofAand denoted @A. The union of all limit points of
Ais called the closure ofAand denoted A.
Lemma 6.4. 1. We have A=A[@A,IntA=An@A.
2.Ais equal to the intersection of all closed sets containg A
3.IntUis the union of all open sets contained in A.
Given a subset XV
- a subset YXis called relatively open inYif there exists an open set UVsuch that
Y=X\U.
- a subsetYXis called relatively closed inYif there exists a closed set AVsuch that
Y=X\A.
One also call relatively and open and closed subsets of Xjust open and closed inX.
Exercise 6.5. Prove that though we de ned open sets using a Euclidean structure on the vector
spaceVthe de nition of open and closed sets is independent of this choice.
62
Figure 6.1: Bernard Bolzano (1781-1848)
6.2 Everywhere and nowhere dense sets
Aclosed setAis called nowhere dense if Int A=?. For instance any nite set is nowhere dense.
Any linear subspace LVis nowhere dense in Vif dimL < dimV. Here is a more interesting
example of a nowhere dense set.
Fix some number <1. For any interval = [ a;b] we denote by theopen interval centered
at the point c=a+b
2, the middle point of , of the total length equal to (b a). We denote by
C() := n. ThusC() consists of two disjoint smaller closed intervals. Let I= [0;1]. Take
C(I) =I1[I2. Take again C(I1)[C(I2) then again apply the operation Cto four new closed
intervals. Continue the process, and take the intersection of all sets arising on all steps of this
construction. The resulted closed set KIis nowhere dense. It is called a Cantor set .
A subsetBAis called everywhere dense inAifBA. For instance the the set Q\Iof
rational points in the interval I= [0;1] is everywhere dense in I.
6.3 Compactness and connectedness
A setAVis called compact if one of the following equivalent conditions is satis ed:
COMP1.Ais closed and bounded.
63
Figure 6.2: Karl Weierstrass (1815-1897)
COMP2. from any in nite sequence of points xn2Aone can choose a subsequence xnkconverging
to a pointa2A.
COMP3. from any family U;2 of open sets covering A, i.e.S
2UA, one can choose
nitely many sets U1;:::;Ukwhich cover A, i.e.kS
1UkA.
The equivalence of these de nitions is a combination of theorems of Bolzano-Weierstrass and
Emile Borel.
A setAis called path-connected if for any two points a0;a12Athere is a continuous path
: [0;1]!Asuch that
(0) =a0and
(1) =a1.
A setAis called connected if one cannot present Aas a unionA=A1[A2such thatA1\A2=?,
A1;A26=?and bothA1andA2are simultaneously relatively closed and open in A.
Lemma 6.6. Any path-connected set is connected.
Proof. Suppose that Ais disconnected. Then it can be presented as a union A=A0[A1of two
non-empty relatively open (and hence relatively closed) subsets. Consider the function :A!R
de ned by the formula
(x) =8
><
>:0; x2A0;
1; x2A1:
64
Figure 6.3: Emile Borel (1871-1956)
We claim that the function is is continuous. Indeed, For each i= 0;1 and any point a2Ai
there exists > 0 such that B(x)\AAi. Hence the function is constant on B(x)\A,
and hence continuous at the point x. Now take points x02A0andx12A1and connect them
by a path
: [0;1]!A(this path exists because Ais path-connected). Consider the function
:=
: [0;1]!R. This function is continuous (as a composition of two continuous maps).
Furthermore, (0) = 0; (1) = 1. Hence, by an intermediate value theorem of Cauchy the function
must take all values in the interval [0 ;1]. But this is a contradiction because by construction the
function takes no other values except 0 and 1.
Lemma 6.7. Any open connected subset URnis path connected.
Proof. Take any point a2U. Denote by Cathe set of all points in Uwhich can be connected with
aby a path. We need to prove that Ca=U.
First, we note that Cais open. Indeed, if b2Cathen using openness of Uwe can nd >0
such the ball B(b)2U. Any point of c2B(b) can be connected by a straight interval IbcB(b)
withb, and hence it can be connected by a path with a, i.e.c2Ca. ThusB(b)Ca, and hence
Cais open. Similarly we prove that the complement UnCais open. Indeed, take b =2Ca. As above,
there exists an open ball B(b)U. ThenB(b)UnCa. Indeed, if it were possible to connect a
pointc2B(b) withaby a path, then the same would be true for b, becausebandcare connected
by the interval Ibc. Thus, we have U=Ca[(UnCa), both sets CaandUnCaare open and Cais
65
non-empty. Hence, UnCais to be empty in view of connectedness of U. Thus,Ca=U, i.e.Uis
path-connected.
Exercise 6.8. In general a connected set need not to be path-connected. A canonical example is
the closure of the graph of the function sin1
x;x2Rn0:. Prove it.
Exercise 6.9. Prove that any non-empty connected ( =path-connected) open subset of Ris equal
to an interval (a;b)(we allow here a= 1 andb=1). If one drops the condition of openness,
then one needs to add a closed and semi-closed intervals and a point.
Remark 6.10. One of the corollaries of this exercise is that in Rany connected set is path-
connected .
Solution. LetARbe a non-empty connected subset. Let a<b be two points of A. Suppose
that a point c2(a;b) does not belong to A. Then we can write A=A0[A1, whereA0=
A\( 1;0);A1=A\(0;1). Both sets A0andA1are relatively open and non-empty, which
contradicts connectedness of A. Hence if two points aandb,a < b , are inA, then the whole
interval [a;b] is also contained in A. Denotem:= infAandM:= supA(we assume that m= 1
ifAis unbounded from below and M= +1ifAis unbounded from above). Then the above
argument shows that the open interval ( m;M ) is contained in A. Thus, there could be 5 cases:
m;M =2A; in this case A= (m;M );
m2A;M =2A; in this case A= [m;M );
m =2A;M2A; in this case A= (m;M ];
m;M2Aandm<M ; in this case A= [m;M ];
m;M2Aandm=M; in this case Aconsists of one point.
6.4 Connected and path-connected components
Lemma 6.11. LetA,2be any family of connected (resp. path-connected) subsets of a vector
spaceV. SupposeT
2A6=?. ThenS
2Ais also connected (resp. path-connected)
66
Proof. Pick a point a2T
2A. Consider rst the case when Aare path connected. Pick a point
a2T
2A. Thenacan be connected by path with all points in Afor any points in 2. Hence,
all points of AandA0can be connected with each other for any ;02.
Suppose now that Aare connected. Denote A:=S
2A. SupposeAcan be presented as a
unionA=U[U0of disjoint relatively open subsets, where we denoted by Uthe set which contains
the pointa2T
2A. Then for each 2 the intersections U:=U\AandU0
:=U0\Aare
relatively open in A. We haveA=U[U0
. By assumption, U3a, and hence U6=?. Hence,
connectedness of Aimplies that U0
=?. But thenU0=S
2U0
=?, and therefore Ais connected.
Given any set AVand a point a2Atheconnected component (resp. path-connected com-
ponentCaAof the point a2Ais the union of all connected (resp. path-connected) subsets
ofAwhich contains the point a. Due to Lemma 6.11 the (path-)connected component Cais itself
(path-)connected, and hence it is the biggest (path-)connected subset of Awhich contains the point
a.The path-connected component of acan be equivalently de ned as the set of all points of Aone
can connect with aby a path in A.
Note that (path-)connected components of di erent points either coincide or do not intersect,
and hence the set Acan be presented as a disjoint union of (path-)-connected components.
Lemma 6.7 shows that for open sets in a vector space Vthe notions of connected and path-
connected components coincide, and due to Exercise 6.9 the same is true for any subsets in R. In
particular, any open set URcan be presented as a union of disjoint open intervals, which are its
connected (= path-connected) components. Note that the number of these intervals can be in nite,
but always countable.
6.5 Continuous maps and functions
LetV;W be two Euclidean spaces and Ais a subset of V. A mapf:A!Wis called continuous
if one of the three equivalent properties hold:
1. For any >0 and any point x2Athere exists >0 such that f(B(x)\A)B(f(x)).
2. If for a sequence xn2Athere exists lim xn=x2Athen the sequence f(xn)2Wconverges
67
tof(x).
3. For any open set UWthe pre-image f 1(U) is relatively open in A.
4. For any closed set BWthe pre-image f 1(B) is relatively closed in A.
Let us verify equivalence of 3 and 4. For any open set UWits complement B=WnUis closed
and we have f 1(U) =Anf 1(B). Hence, if f 1(U) is relatively open, i.e. f 1(U) =U0\Afor
an open set U0V, thenf 1(B) =A\(VnU0), i.e.f 1(B) is relatively closed. The converse is
similar.
Let us deduce 1 from 3. The ball B(f(x)) is open. Hence f 1(B(f(x))) is relatively open in
A. Hence, there exists >0 such that B(x)\Af 1(B(f(x))), i.e.f(B(x)\A)B(f(x)).
We leave the converse and the equivalence of de nition 2 to the reader.
Remark 6.12. Consider a map f:A!Wand denote B:=f(A). Then de nition 3 can be
equivalently stated as follows:
30. For any set UBrelatively open in Bits pre-image f 1(U) is relatively open in A.
De nition 4 can be reformulated in a similar way.
Indeed, we have U=U0\Afor an open set U0W, whilef 1(U) =f 1(U0).
The following theorem summarize properties of continuous maps.
Theorem 6.13. Letf:A!Wbe a continuous map. Then
1. ifAis compact then f(A)is compact;
2. ifAis connected then f(A)is connected;
3. ifAis path connected then f(A)is path-connected.
Proof. 1. Take any in nite sequence yn2f(A). Then there exist points xn2Asuch thatyn=
f(xn);n= 1;:::: Then there exists a converging subsequence xnk!a2A. Then by continuity
limk!1f(xnk) =f(a)2f(A), i.e.f(A) is compact.
2. Suppose that f(A) can be presented as a union B1[B2of simultaneously relatively open
and closed disjoint non-empty sets. Then f 1(B1);f 1(B2)Aare simultaneously relatively open
68
Figure 6.4: George Cantor (1845-1918)
and closed in A, disjoint and non-empty. We also have f 1(B1)[f 1(B2) =f 1(B1[B2) =
f 1(f(A)) =A. HenceAis disconnected which is a contradiction.
3. Take any two points y0;y12f(A). Then there exist x0;x12Asuch thatf(x0) =y0;f(x1) =
y1. ButAis path-connected. Hence the points x0;x1can be connected by a path
: [0;1]!A.
Then the path f
: [0;1]!f(A) connectsy0andy1, i.e.f(A) is path-connected.
Note that in the case W=RTheorem 6.13.1 is just the Weierstrass theorem: a continuos
function on a compact set is bounded and achieves its maximal and minimal values.
We nish this section by a theorem of George Cantor about uniform continuity.
Theorem 6.14. LetAbe compact and f:A!Wa continuous map. Then for any >0there
exists>0such that for any x2Awe havef(B(x))B(f(x)).
Proof. Choose>0. By continuity of ffor every point x2Athere exists (x)>0 such that
f(B(x)(x))B
4(f(x)):
We need to prove that inf
x2A(x)>0. Note that for any point in y2B(x)
2(x) we haveB(x)
2(y)
B(x)(x), and hence f(B(x)
2(y))B(f(y)):By compactness, from the coveringS
x2AB(x)
2(x) we
can choose a nite number of balls B(xj)
2(xj);j= 1;:::;N which still cover A. Then= min
k(xj)
2
satisfy the condition of the theorem, i.e. f(B(x))B(f(x)) for anyx2A.
69
70
Chapter 7
Vector elds and di erential forms
7.1 Di erential and gradient
Given a vector space Vwe will denote by Vxthe vector space Vwith the origin translated to the
pointx2V. One can think of Vxas that tangent space to Vat the point x. Though the parallel
transport allows one to identify spaces VandVxit will be important for us to think about them
as di erent spaces.
Letf:U!Rbe a function on a domain UVin a vector space V. The function fis called
di erentiable at a pointx2Uif there exists a linear function l:Vx!Rsuch that
f(x+h) f(x) =l(h) +o(jjhjj)
for any suciently small vector h, where the notation o(t) stands for any function such that
o(t)
t!
t!00. The linear function lis called the di erential of the function fat the point xand
is denoted by dxf. In other words, fis di erentiable at x2Uif for anyh2Vxthere exists a limit
l(h) = lim
t!0f(x+th) f(x)
t;
and the limit l(h)linearly depends on h. The value l(h) =dxf(h) is called the directional derivative
offat the point xin the direction h. The function fis called di erentiable on the whole domain
Uif it is di erentiable at each point of U.
Simply speaking, the di erentiability of a function means that at a small scale near a point x
the function behaves approximately like a linear function, the di erential of the function at the
71
pointx. However this linear function varies from point to point, and we call the family fdxfgx2Uof
all these linear functions the di erential of the function f, and denote it by df(without a reference
to a particular point x).
Let us summarize the above discussion. Let f:U!Rbe a di erentiable function. Then for
each point x2Uthere exists a linear function dxf:Vx!R, the di erential of fat the point x
de ned by the formula
dxf(h) = lim
t!0f(x+th) f(x)
t;x2U;h2Vx:
We recall that existence of partial derivatives at a point a2Udoes not guarantee the di eren-
tiability of fat the point a. On the other hand if partial derivatives exists in a neighborhood of a
andcontinuous at the point athenfis di erentiable at this point. The functions whose rst partial
derivatives are continuous in uare calledC1-smooth , or sometimes just smooth. Equivalently, we
can say that fis smooth if the di erential dxfcontinuously depends on the point x2U.
Ifv1;:::;vnare vectors of a basis of V, parallel transported to the point x, then we have
dxf(vi) =@f
@xi(x); x2U; i= 1;:::;n;
wherex1;:::;xnare coordinates with respect to the chosen basis v1;:::;vn.
Notice that if fis a linear function,
f(x) =a1x1++anxn;
then for each x2Vwe have
dxf(h) =a1h1++anhn;h= (h1;:::;hn)2Vx:
Thus the di erential of a linear function fat any point x2Vcoincides with this function, parallel
transported to the space Vx. This observation, in particular, can be applied to linear coordinate
functionsx1;:::;xnwith respect to a chosen basis of V.
In Section 7.7 below we will de ne the di erential for maps f:U!W, whereWis a vector
space and not just the real line R.
72
7.2 Smooth functions
We recall that existence of partial derivatives at a point a2Udoes not guarantee the di erentia-
bility offat the point a. On the other hand if partial derivatives exists in a neighborhood of aand
continuous at the point athenfis di erentiable at this point. The functions whose rst partial
derivatives are continuous in uare calledC1-smooth . Equivalently, we can say that fis smooth if
the di erential dxfcontinuously depends on the point x2U.
More generally, for k1 a function f:U!Ris calledCk-smooth all its partial derivatives
up to order kare continuous in U. The space of Ck-smooth functions is denoted by Ck(U). We will
also use the notation C0(U) andC1(U) which stands, respectively, for the spaces of continuous
functions and functions with continuous derivatives of all orders. In this notes we will often speak
of smooth functions without specifying the class of smoothness, assuming that functions have as
many continuous derivatives as necessary to justify our computations.
Remark 7.1. We will often need to consider smooth maps, functions, vectors elds, di erential
forms, etc. de ned on a closed subset Aof a vector space V. We will always mean by that the these
objects are de ned on some open neighborhood UA. It will be not important for us how exactly
these objects are extended to Ubut to make sense of di erentiability we need to assume that they
are extended. In fact, one can de ne what di erentiability means without any extension, but this
would go beyond the goals of these lecture notes.
Moreover, a theorem of Hassler Whitney asserts that any function smooth on a closed subset
AVcan be extended to a smooth function to a neighborhood UA.
7.3 Gradient vector eld
IfVis an Euclidean space, i.e. a vector space with an inner product h;i, then there exists a
canonical isomorphism D:V!V, de ned by the formula D(v)(x) =hv;xiforv;x2V. Of
course,Dde nes an isomorphism Vx!V
xfor eachx2V. Set
rf(x) =D 1(dxf):
73
The vectorrf(x) is called the gradient of the function fat the point x2U. We will also use
the notation grad f(x).
By de nition we have
hrf(x);hi=dxf(h) for any vector h2V :
Ifjjhjj= 1 thendxf(h) =jjrf(x)jjcos', where'is the angle between the vectors rf(x) andh. In
particular, the directional derivative dxf(h) has its maximal value when '= 0. Thus the direction
of the gradient is the direction of the maximal growth of the function and the length of the gradient
equals this maximal value.
As in the case of a di erential, the gradient varies from point to point, and the family of vectors
frf(x)gx2Uis called the gradient vector eld rf.
We discuss the general notion of a vector eld in Section 7.4 below.
7.4 Vector elds
Avector eld von a domain UVis a function which associates to each point x2Ua vector
v(x)2Vx, i.e. a vector originated at the point x.
A gradient vector eld rfof a function fprovides us with an example of a vector eld, but as
we shall see, gradient vector elds form only a small very special class of vector elds.
Letvbe a vector eld on a domain U2V. If we x a basis in V, and parallel transport this
basis to all spaces Vx;x2V, then for any point x2Vthe vectorv(x)2Vxis described by its
coordinates ( v1(x);v2(x);:::;vn(x)). Therefore, to de ne a vector eld on Uis the same as to de ne
nfunctionsv1;:::;vnonU, i.e. to de ne a map ( v1;:::;vn) :U!Rn. We call a vector eld v
Ck-smooth if the functions v1;:::;vnare smooth on U.
Thus, if a basis of Vis xed, then the di erence between the maps U!Rnand vector elds
onUis just a matter of geometric interpretation. When we speak about a vector eld vwe view
v(x) as a vector in Vx, i.e. originated at the point x2U. When we speak about a map v:U!Rn
we viewv(x) as a point of the space V, or as a vector with its origin at 02V.
Vector elds naturally arise in a context of Physics, Mechanics, Hydrodynamics, etc. as force,
velocity and other physical elds.
74
There is another very important interpretation of vector elds as rst order di erential opera-
tors.
LetC1(U) denote the vector space of in nitely di erentiable functions on a domain UV.
Letvbe aC1-smooth vector eld on V. We associate with va linear operator
Dv:C1(U)!C1(U);
given by the formula
Dv(f) =df(v); f2C1(U):
In other words, we compute at any point x2Uthe directional derivative of fin the direction of
the vectorv(x). Clearly, the operator Dvis linear:Dv(af+bg) =aDv(f)+bDv(g) for any functions
f;g2C1(U) and any real numbers a;b2R. It also satis es the Leibniz rule :
Dv(fg) =Dv(f)g+fDv(g):
In view of the above correspondence between vector elds and rst order di erential operators
it is sometimes convenient just to view a vector eld as a di erential operator. Hence, when it will
not be confusing we may drop the notation Dvand just directly apply the vector vto a function
f(i.e. writev(f) instead of Dv(f)).
Letv1;:::;vnbe a basis of V, andx1;:::;xnbe the coordinate functions in this basis. We
would like to introduce the notation for the vector eld obtained from vectors v1;:::;vnby parallel
transporting them to all points of the domain U. To motivate the notation which we are going
to introduce, let us temporarily denote these vector elds by v1;:::;vn. Observe that Dvi(f) =
@f
@xi; i= 1;:::;n . Thus the operator Dviis just the operator@
@xiof takingi-th partial derivative.
Hence, viewing the vector eld vias a di erential operator we will just use the notation@
@xiinstead
ofvi. Given any vector eld vwith coordinate functions a1;a2;:::;an:U!Rwe have
Dv(f)(x) =nX
i=1ai(x)@f
@xi(x);for anyf2C1(U);
and hence we can write v=nP
i=1ai@
@xi. Note that the coecients aihere are functions and not
constants.
75
7.4.1 Gradient vector eld
Suppose that V;h;iis a Euclidean vector space. Choose a (not necessarily orthonormal) basis
v1;:::;vn. Let us nd the coordinate description of the gradient vector eld rf, i.e. nd the
coecients ajin the expansion rf(x) =nP
1ai(x)@
@xi. By de nition we have
hrf(x);hi=dxf(h) =nX
1@f
@xj(x)hj (7.4.1)
for any vector h2Vxwith coordinates ( h1;:::;hn) in the basis v1;:::;vnparallel transported to
Vx. Let us denote gij=hvi;vji. ThusG= (gij) is a symmetric nnmatrix, which is called the
Gram matrix of the basis v1;:::;vn. Then the equation (7.4.1) can be rewritten as
nX
i;j=1gjiaihj=nX
1@f
@xj(x)hj:
Becausehjare arbitrarily numbers it implies that the coecients with hjin the right and left sides
should coincide for all j= 1;:::;n . Hence we get the following system of linear equations:
nX
i=1gijai=@f
@xj(x); j= 1;:::;n; (7.4.2)
or in matrix form
G0
BBB@a1
...
an1
CCCA=0
BBB@@f
@x1(x)
...
@f
@xn(x)1
CCCA;
and thus0
BBB@a1
...
an1
CCCA=G 10
BBB@@f
@x1(x)
...
@f
@xn(x)1
CCCA; (7.4.3)
i.e.
rf=nX
i;j=1gij@f
@xi(x)@
@xj; (7.4.4)
where we denote by gijthe entries of the inverse matrix G 1= (gij) 1
If the basis v1;:::;vnis orthonormal then Gis the unit matrix, and thus in this case
76
rf=nX
1@f
@xj(x)@
@xj; (7.4.5)
i.e.rfhas coordinates (@f
@x1;:::;@f
@xn). However, simple expression (7.4.5) for the gradient holds
only in the orthonormal basis . In the general case one has a more complicated expression
(7.4.4).
7.5 Di erential forms
Similarly to vector elds, we can consider elds of exterior forms , i.e. functions on UVwhich
associate to each point x2Uak-form from k(V
x). These elds of exterior k-forms are called
di erential k-forms .
Thus the relation between k-forms and di erential k-forms is exactly the same as the relation
between vectors and vector- elds. For instance, a di erential 1-form associates with each point
x2Ua linear function (x) on the space Vx. Sometimes we will write xinstead of (x) to leave
space for the arguments of the function (x).
Example 7.2. 1. Letf:V!Rbe a smooth function. Then the di erential dfis a di erential
1-form. Indeed, with each point x2Vit associates a linear function dxfon the space Vx.
As we shall see, most di erential 1-form are not di erentials of functions (just as most vector
elds are not gradient vector elds).
2. A di erential 0-form fonUassociates with each point x2Ua 0-form on Vx, i. e. a number
f(x)2R. Thus di erential 0-forms on Uare just functions U!R.
7.6 Coordinate description of di erential forms
Letx1;:::;xnbe coordinate linear functions on V, which form the basis of Vdual to a chosen
basisv1;:::;vnofV. For each i= 1;:::;n the di erential dxide nes a linear function on each
spaceVx;x2V. Namely, if h= (h1;:::;hn)2Vxthendxi(h) =hi. Indeed
dxxi(h) = lim
t!0xi+thi xi
t=hi;
77
independently of the base point x2V. Thus di erentials dx1;:::;dxnform a basis of the space V
x
for eachx2V. In particular, any di erential 1-form onvcan be written as
=f1dx1+:::+fndxn;
wheref1;:::;fnarefunctions onV. In particular,
df=@f
@x1dx1+:::+@f
@xndxn: (7.6.1)
Let us point out that this simple expression of the di erential of a function holds in an arbitrary
coordinate system , while an analogous simple expression (7.4.5) for the gradient vector eld
is valid only in the case of Cartesian coordinates. This re
ects the fact that while the notion of
di erential is intrinsic and independent of any extra choices, one needs to have a background inner
product to de ne the gradient.
Similarly, any di erential 2-form won a 3-dimensional space can be written as
!=b1(x)dx2^dx3+b2(x)dx3^dx1+b3(x)dx1^dx2
whereb1;b2, andb3are functions on V. Any di erential 3-form
on a 3-dimensional space Vhas
the form
=c(x)dx1^dx2^dx3
for a function conV.
More generally, any di erential k-form can be expressed as
=X
1i1<i2<iknai1:::ikdxi1^^dxik
for some functions ai1:::ikonV.
7.7 Smooth maps and their di erentials
LetV;W be two vector spaces of arbitary (not, necessarily, equal) dimensions and UVbe an
open domain in V.
78
Recall that a map f:U!Wis called di erentiable if for eachx2Uthere exists a linear map
l:Vx!Wf(x)
such that
l(h) = lim
t!0f(x+th) f(x)
t
for anyh2Vx. In other words,
f(x+th) f(x) =tl(h) +o(t);whereo(t)
t!
t!00:
The maplis denoted by dxfand is called the di erential of the map fat the point x2U. Thus,
dxfis a linear map Vx!Wf(x).
The spaceWf(x)can be identi ed with Wvia a parallel transport, and hence sometimes it is
convenient to think about the di erential as a map Vx!W, In particular, in the case of a linear
function . i.e. when W=Rit is customary to do that, and hence we de ned earlier in Section 7.1
the di erential of a function f:U!Rat a point x2Uas a linear function Vx!R, i.e. an
element ofV
x, rather than a linear map Vx!Wf(x).
Let us pick bases in VandWand let (x1;:::;xk) and (y1;:::;yn) be the corresponding coor-
dinate functions. Then each of the spaces VxandWy; x2V; y2Winherits a basis obtained by
parallel transport of the bases of VandW. In terms of these bases, the di erential dxfis given by
the Jacobi matrix
0
BBB@@f1
@x1:::@f1
@xk
::: ::: :::
@fn
@x1:::@fn
@xk1
CCCA
In what follows we will consider only suciently smooth maps, i.e. we assume that all maps
and their coordinate functions are di erentiable as many times as we need it.
7.8 Operator f
LetUbe a domain in a vector space Vandf:U!Wa smooth map. Then the di erential df
de nes a linear map
79
dxf:Vx!Wf(x)
for eachx2V.
Let!be a di erential k-form onW. Thus!de nes an exterior k-form on the space Wyfor
eachy2W.
Let us de ne the di erential k-formf!onUby the formula
(f!)jVx= (dxf)(!jWf(x)):
Here the notation !jWystands for the exterior k-form de ned by the di erential form !on the
spaceWy.
In other words, for any kvectors,H1;:::;Hk2Vxwe have
f!(H1;:::;Hk) =!(dxf(H1);:::;dxf(Hk)):
We say that the di erential form f!isinduced from !by the map f, or thatf!is the pull-back
of!byf.
Example 7.3. Let
=h(x)dx1^^dxn. Then formula (3.2) implies
f
=hfdetDfdx 1^^dxn:
Here
detDf= @f1
@x1:::@f1
@xn
::: ::: :::
@fn
@x1:::@fn
@xn
is the determinant of the Jacobian matrix of f= (f1;:::;fn):
Similarly to Proposition 1.9 we get
Proposition 7.4. Given 2maps
U1f!U2g!U3
and a di erential kform!onU3we have
(gf)(!) =f(g!):
80
An important special case of the pull-back operator fis the restriction operator. Namely Let
LVbe an ane subspace. Let j:L,!Vbe the inclusion map. Then given a di erential k-form
on a domain UVwe can consider the form j on the domain U0:=L\U. This form is called
therestriction of the form toU0and it is usually denoted by j0
U. Thus the restricted form j0
U
is the same form but viewed as function of a point a2U0and vectors T1;:::;Tk2La.
7.9 Coordinate description of the operator f
Consider rst the linear case. Let Abe a linear map V!Wand!2p(W). Let us x coordinate
systemsx1;:::;xkinVandy1;:::;yninW. IfAis the matrix of the map Athen we already have
seen in Section 2.7 that
Ayj=lj(x1;:::;xk) =aj1x1+aj2x2+:::+ajkxk; j= 1;:::;n;
and that for any exterior k-form
!=X
1i1<:::<ipnAi1;:::;ipyi1^:::^yip
we have
A!=X
1i;<:::<ipnAi1:::ipli1^:::^lip:
Now consider the non-linear situation. Let !be a di erential p-form onW. Thus it can be
written in the form
!=X
Ai1:::ip(y)dyi1^:::dyip
for some functions Ai1:::iponW.
LetUbe a domain in Vandf:U!Wa smooth map.
Proposition 7.5. f!=PAi1:::;ip(f(x))dfi1^:::^dfip, wheref1;:::;fnare coordinate functions
of the map f.
81
Proof. For each point x2Uwe have, by de nition,
f!jVx= (dxf)(!jWfx)
But the coordinate functions of the linear map dxfare just the di erentials dxfiof the coordinate
functions of the map f. Hence the desired formula follows from the linear case proven in the previous
proposition.
7.10 Examples
1. Consider the domain U=fr>0;0'<2gon the plane V=R2with cartesian coordinates
(r;'). LetW=R2be another copy of R2with cartesian coordinates ( x;y). Consider a map
P:V!Wgiven by the formula
P(r;') = (rcos';rsin'):
This map introduces ( r;') as polar coordinates on the plane W. Set!=dx^dy. It is called
the area form onW. Then
P!=d(rcos')^d(rsin') = (cos'dr+rd(cos'))^(sin'dr+rd(sin') =
(cos'dr rsin'd')^(sin'dr+rcos'd') =
cos'sin'dr^dr rsin2'd'^dr+rcos2'dr^d' r2sin'cos'd'^d'=
rcos2'dr^d'+rsin2dr^d'=rdr^d':
2.Letf:R2!Rbe a smooth function and the map F:R2!R3be given by the formula
F(x;y) = (x;y;f (x;y))
Let
!=P(x;y;z )dy^dz+Q(x;y;z )dz^dx+R(x;y;z )dx^dy
82
Figure 7.1: Johann Friedrich Pfa (1765{1825)
be a di erential 2-form on R3. Then
F!=P(x;y;f (x;y))dy^df+
+Q(x;y;f (x;y))df^dx+R(x;y;f (x;y))dx^dy
=P(x;y;f (x;y))dy^(fxdx+fydy) +
+Q(x;y;f (x;y))(fxdx+fydy)^dx+
+R(x;y;f (x;y))dx^dy=
= (R(x;y;f (x;y)) P(x;y;f (x;y))fx Q(x;y;f (x;y))fy)dx^dy
wherefx;fyare partial derivatives of f.
3.Ifp>k then the pull-back f!of ap-form!onUto ak-dimensional space Vis equal to 0.
7.11 Pfaan equations
Given a non-zero linear function lon ann-dimensional vector space Vthe equation l= 0 de nes
ahyperplane , i.e. an (n 1)-dimensional subspace of V.
83
y
xzFigure 7.2: Contact structure
Suppose we are given a di erential 1-form on a domain UV. Suppose that x6= 0 for each
x2U.
Then the equation
= 0 (7.11.1)
de nes a hyperplane eld onU, i.e. a family of of hyperplanes x=fx= 0gVx; x2U:
The equation of this type is called Pfaan in honor of a German mathematician Johann
Friedrich Pfa (1765{1825).
Example 7.6. LetV=R3with coordinates ( x;y;z )
1. Let=dz. Then=fdz= 0gis the horizontal plane eld which is equal to Span(@
@x;@
@y).
2. Let=dz ydx: Then the plane eld dz ydxis shown on Fig. 7.11. This plane is non-integrable
in the following sense. There are no surfaces in R3tangent to. This plane eld is called a contact
structure . It plays an important role in symplectic and contact geometry, which is, in turn, the
geometric language for Mechanics and Geometric Optics.
84
Chapter 8
Exterior di erential
8.1 Coordinate de nition of the exterior di erential
Let us denote by by
k(U) the space of all di erential k-forms onU. When we will need to specify
the class of smoothness of coecients of the form we will use the notation
k
l(U) for the space of
di erential k-forms with Cl-smooth coecients, i.e. which depend Cl-smoothly on a point of U.
We will de ne a map
d:
k(U)!
k+1(U);
or more precisely
d:
k
l(U)!
k+1
l 1(U);
(thus assuming that l1), which is called the exterior di erential. In the current form it was
introduced by Elie Cartan, but essentially it was known already to Henri Poincar e.
We rst de ne it in coordinates and then prove that the result is independent of the choice
of the coordinate system. Let us x a coordinate system x1;:::;xninVU. As a reminder, a
di erential k-formw2
k(U) has the form w=P
i1<:::<ikai1:::;ikdxi1^:::^dxikwhereai1:::ikare
functions on the domain U. De ne
dw:=X
i1<:::<ikdai1:::ik^dxi1^:::^dxik:
Examples. 1. Letw2
1(U), i.e.w=nP
i=1aidxi. Then
85
Figure 8.1: Elie Cartan (1869{1951)
dw=nX
i=1dai^dxi=nX
i=10
@nX
j=1@ai
@xjdxj1
A^dxi=X
1i<jn@aj
@xi @ai
@xj
dxi^dxj:
For instance, when n= 2 we have
d(a1dx1+a2dx2) =@a2
@x1 @a1
@x2
dx1^dx2
Forn= 3, we get
d(a1dx1+a2dx2+a3dx3) =@a3
@x2 @a2
@x3
dx2^dx3+@a1
@x3 @a3
@x1
dx3^dx1+@a2
@x1 @a1
@x2
dx1^dx2:
2.Letn= 3 andw2
2(U). Then
w=a1dx2^dx3+a2dx3^dx1+a3dx1^dx2
and
dw =da1^dx2^dx3+da2^dx3^dx1+da3^dx1^dx2
=@a1
@x1+@a2
@x2+@a3
@x3
dx1^dx2^dx3
86
Figure 8.2: Henri Poincar e (1854{1912)
3. For 0-forms, i.e. functions the exterior di erential coincides with the usual di erential of a
function.
8.2 Properties of the operator d
Proposition 8.1. For any two di erential forms, 2
k(U); 2
l(U)we have
d( ^ ) =d ^ + ( 1)k ^d :
Proof. We have
=X
i1<:::<ikai1:::ikdxi1^:::^dxik
=X
j1<:::<jlbji:::jldxj1^:::^dxjl
^ =0
@X
i1<:::<ikai1:::ikdxi1^:::^dxik1
A^0
@X
ji<:::<jlbj1:::jldxj1^:::^dxjl1
A
=X
i1<:::<ikj1<:::<jlai1:::ikbj1:::jldxil^:::^dxik^dxjl^:::^dxjl
87
d( ^ ) =X
i1<:::<ikj1<:::<jl(bj1:::jldai1:::ik+ai1:::ikdbj1:::jl)^dxi1^:::^dxik^dxj1^:::^dxjl
=X
i1<:::<ikj1<:::<jlbj1:::jldai1:::ik^dxi1^:::^dxik^dxjl^:::^dxjl
+X
i1<:::<ikj1<:::<jlai1:::ikdbj1:::jl^dxi1^:::^dxik^dxj1^:::^dxjl
=0
@X
i1<:::<ikdai1:::ik^dxi1:::^dxik1
A^X
bj1:::jldxj1^:::^dxjl
+ ( 1)k0
@X
i1<:::<ikai1:::ikdxi1^:::^dxik1
A^0
@X
ji<:::<jldbj1:::jl^dxj1^:::^dxjl1
A
=d ^ + ( 1)k ^d :
Notice that the sign ( 1)kappeared because we had to make ktransposition to move dbj1:::jlto its
place.
Proposition 8.2. For any di erential k-formwwe have
ddw= 0:
Proof. Letw=P
i1<:::<ikai1:::ikdxi1^:::^dxik. Then we have
dw=X
i1<:::<ikdai1:::ik^dxi1^:::^dxik:
Applying Proposition 8.1 we get
ddw=X
i1<:::<ikddai1:::ik^dxi1^:::^dxik dai1:::ik^ddxi1^:::^dxik+:::
+ ( 1)kdai1:::ik^dxi1^:::^ddxik:
88
Butddf= 0 for any function fas was shown above. Hence all terms in this sum are equal to 0, i.e.
ddw= 0.
De nition. Ak-form!is called closed ifd!= 0. It is called exact if there exists a ( k 1)-form
such thatd=!. The form is called the primitive of the form !. The previous theorem can be
reformulated as follows:
Corollary 8.3. Every exact form is closed.
The converse is not true in general. For instance, take a di erential 1-form
!=xdy ydx
x2+y2
on the punctured plane U=R2n0 (i.e the plane R2with the deleted origin). It is easy to calculate
thatd!= 0, i.e!is closed. On the other hand it is not exact. Indeed, let us write down this form
in polar coordinates ( r;'). We have
x=rcos'; y=rsin':
Hence,
!=1
r2(rcos'(sin'dr+rcos'd') rsin'(cos'dr rsin'd')) =d':
If there were a function HonUsuch thatdH=!, then we would have to have H='+ const,
but this is impossible because the polar coordinate 'is not a continuous univalent function on U.
Hence!is not exact.
However, as we will see later, a closed form is exact if it is de ned on the whole vector space V.
Proposition 8.4. Operatorsfanddcommute, i.e. for any di erential k-formw2
k(W), and
a smooth map f:U!Wwe have
dfw=fdw
.
Proof. Suppose rst that k= 0, i.e.wis a function ':W!R. Thenf'='f. Then
d('f) =fd'. Indeed, for any point x2Uand a vector X2Vxwe have
89
d('f)(X) =d'(dxf(X))
(chain rule)
Butd'(dxf(X)) =f(d'(X)).
Consider now the case of arbitrary k-formw,
w=X
i1<:::<ikai1:::ikdxi1^:::^dxik:
Then
fw=X
i1<:::<ikai1:::ikfdfi1^:::^dfik
wheref1;:::;fnare coordinate functions of the map f. Using the previous theorem and taking into
account that d(dfi) = 0, we get
d(fw) =X
i1<:::<ikd(ai1:::ikf)^dfi1^:::^dfik:
On the other hand
dw=X
i1<:::<ikdai1:::ik^dxi1^:::^dxik
and therefore
fdw=X
i1<:::<ikf(dai1:::ik)^dfi1^:::^dfik:
But according to what is proven above, we have
fdai1:::ik=d(ai1:::ikf)
Thus,
fdw=X
i1<:::<ikd(ai1:::ikf)^dfi1^:::^dfik=dfw
The above theorem shows, in particular, that the de nition of the exterior di erential is in-
dependent of the choice of the coordinate. Moreover, one can even use non-linear (curvilinear)
coordinate systems, like polar coordinates on the plane.
90
8.3 Curvilinear coordinate systems
A (non-linear) coordinate system on a domain Uin ann-dimensional space Vis a smooth map
f= (f1;:::;fn) :U!Rnsuch that
1. For each point x2Uthe di erentials dxf1;:::;dxfn2(Vx)are linearly independent.
2.fis injective, i.e. f(x)6=f(y) forx6=y.
Thus a coordinate map fassociatesncoordinates y1=f1(x);:::,yn=fn(x) with each point
x2U. The inverse map f 1:U0!Uis called the parameterization . HereU0=f(U)Rn
is the image of Uunder the map f. If one already has another set of coordinates x1:::xnon
U, then the coordinate map fexpresses new coordinates y1:::ynthrough the old one, while the
parametrization map expresses the old coordinate through the new one. Thus the statement
gdw=dgw
applied to the parametrization map gjust tells us that the formula for the exterior di erential is
the same in the new coordinates and in the old one.
Consider a space Rnwith coordinates ( u1;:::;un). Thej-th coordinate line is given by equations
ui=ci;i= 1;:::;n ;i6=j:Given a domain U0Rnconsider a parameterization map g:U0!
UV. The images gfui=ci;i6=jg)Uof coordinates lines fui=ci;i6=jgU0are called
coordinate lines in Uwith respect to the curvilinear coordinate system ( u1;:::;un). For instance,
coordinate lines for polar coordinates in R2are concentric circles and rays, while coordinate lines
for spherical coordinates in R3are rays from the origin, and latitudes and meridians on concentric
spheres.
8.4 Geometric de nition of the exterior di erential
We will show later (see Lemma 10.2) that one can give another equivalent de nition of the operator
d) without using any coordinates at all. But as a rst step we give below an equivalent de nition
of the exterior di erential which is manifestly invariant of a choice of ane coordinates.
91
Given a point a2Vand vectors h1;:::;hk2Vawe will denote by Pa(h1;:::;hk) thek-
dimensional parallelepiped
(
a+1
2kX
1ujhj;ju1j;:::;jukj1)
centered at the point awith its sides parallel to the vectors h1;:::;hk, see Fig. ??. For instance,
fork= 1 the parallelepiped Pa(h) is an interval centered at aof lengthjaj.
As it was explained above in Section 8.1 for a 0-form f2
0(U), i.e. a function f:U!Rits
exterior di erential is just its usual di erential, and hence it can be de ned by the formula
(df)a(h) =daf(h) = lim
t!01
t
f
a+th
2
f
a th
2
;t2R;h2Va;a2U
Proposition 8.5. Let 2
k 1
1(U)be a di erential (k 1)-form on a domain URn. Then for
any pointa2Uand any vectors h1;:::;hk2Vawe have
(d )a(h1;:::;hk) = lim
t!01
tkkX
j=1( 1)k 1
a+thj
2(th1;:::;j
_;:::;thk) a thj
2(th1;:::;j
_;:::;thk)
= lim
t!01
tkX
j=1( 1)k 1
a+thj
2(h1;:::;j
_;:::;hk) a thj
2(h1;:::;j
_;:::;hk)
(8.4.1)
For instance, for a 1-form 2
1(U), we have
(d )a(h1;h2) := lim
t!01
t2
a+th1
2(th2) + a th1
2( th2) + a+th2
2( th1) + a th2
2(th1)
= lim
t!01
t
a+th1
2(h2) a th1
2(h2) a+th2
2(h1) + a th2
2(h1)
Proof. First we observe that for a C1-smooth form the limit in the right-hand side of formula
(8.4.1) exists and de ne a k-form (i.e. for each ait is a multilinear skewsymmetric function of
vectorsh1;:::;hk2Rn
a). We can also assume that the vectors h1;:::;hkare linearly independent.
Otherwise, both de nitions give 0. Denote La= Span(h1;:::;hk)Vaand consider the restriction
0:= jL\U. Let (y1;:::;yk) be coordinates in Lacorresponding to the basis h1;:::;hk. Then we
92
can write 0:=PPidy1^:::_i^dykand thus
lim
t!01
t0
@kX
j=1 a+thj
2(h1;:::;i_;:::;hk) a+thj
2(h1;:::;i_;:::;hk)1
A
=kX
j=1kX
i=1lim
t!0Pj(a+thj
2) Pj(a thj
2)
tdy1^:::i_^dyk(h1;:::;j
_;:::;hk)
=kX
j=1kX
i=1@Pj
@yj(a)dy1^:::i_^dyk(h1;:::;i_;:::;hk):(8.4.2)
But
dy1^:::i_^dyk(h1;:::;j
_;:::;hk) =8
><
>:( 1)i 1;ifi=j
0; ifi6=j:
Hence, the expression in (8.4.2) is equal tokP
1( 1)i 1@Pi
@yi(a). But
d 0= kX
1( 1)i 1@Pi
@yi!
dy1^^dyk;
and therefore ( d 0)a(h1;:::;hk) =kP
1( 1)i 1@Pi
@yi(a), i.e. the two de nitions of the exterior di eren-
tial coincide.
8.5 More about vector elds
Similarly to the case of linear coordinates, given any curvilinear coordinate system ( u1;:::;un) in
U, one denotes by
@
@u1;:::;@
@un
the vector elds which correspond to the partial derivatives with respect to the coordinates u1;:::;un.
In other words, the vector eld@
@uiis tangent to the ui-coordinate lines and represents the the ve-
locity vector of the curves u1= const 1;:::;ui 1= consti 1;ui+1= consti+1;:::;un= constn;
parameterized by the coordinate ui.
For instance, suppose we are given spherical coordinates ( r;;' ),r0; 2[0;];2[0;2) in
93
R3. The spherical coordinates are related to the cartesian coordinates ( x;y;z ) by the formulas
x=rsin'cos; (8.5.1)
y=rsin'sin; (8.5.2)
z=rcos': (8.5.3)
Then the vector elds
@
@r;@
@';and@
@
are mutually orthogonal and de ne the same orientation of the space as the standard Cartesian
coordinates in R3. We also havejj@
@rjj= 1. However the length of vector elds@
@'and@
@vary. When
randare xed and 'varies, then the corresponding point ( r;;' ) is moving along a meridian of
radiusrwith a constant angular speed 1. Hence,
jj@
@'jj=r:
Whenrand'are xed and varies, then the point ( r;'; ) is moving along a latitude of radius
rsin'with a constant angular speed 1. Hence,
jj@
@jj=rsin':
Note that it is customary to introduce unit vector elds in the direction of@
@r,@
@and@
@':
er:=@
@r;e':=1
r@
@';e=1
rsin'@
@;
which form an orthonormal basis at every point. The vector elds er;eande'are not de ned at
the origin and at the poles = 0;.
The chain rule allows us to express the vector elds@
@u1;:::;@
@unthrough the vector elds
@
@x1;:::;@
@xn. Indeed, for any function f:U!Rwe have
@f
@ui=nX
j=1@f
@xj@xj
@ui;
and, therefore,
@
@ui=nX
j=1@xj
@ui@
@xj;
94
For instance, spherical coordinates ( r;; ) are related to the cartesian coordinates ( x;y;z ) by
the formulas (8.5.1), and hence we derive the following expression of the vector elds@
@r;@
@;@
@'
through the vector elds@
@x;@
@y;@
@z:
@
@r= sin'cos@
@x+ sin'sin@
@y+ cos'@
@z;
@
@= rsin'sin@
@x+rsin'cos@
@y;
@
@'=rcos'cos@
@x+rcos'sin@
@y rsin'@
@z:
8.6 Case n= 3. Summary of isomorphisms
LetUbe a domain in the 3-dimensional space V. We will consider 5 spaces associated with U.
0(U) =C1(U)|the space of 0-forms, i.e. the space of smooth functions;
k(U) fork= 1;2;3|the spaces of di erential k-forms onU;
Vect(U)|the space of vector elds on U.
Let us x a volume form w2
3(U) that is any nowhere vanishing di erential 3-form. In coordinates
wcan be written as
w=f(x)dx1^dx2^dx3
where the function f:U!Ris never equal to 0. The choice of the form wallows us to de ne the
following isomorphisms.
1. w:C1(U)!
3(U), w(h) =hwfor any function h2C1(U).
2.w: Vect(U)!
2(U)w(v) =vw:
Sometimes we will omit the subscript wan write just and .
Our third isomorphism depends on a choice of a scalar product <;> inV. Let us x a scalar
product. This enables us to de ne an isomorphism
D=D<;>: Vect(U)!
1(U)
95
which associates with a vector eld vonUa di erential 1-form D(v) =hv;:i. Let us write down the
coordinate expressions for all these isomorphisms. Fix a cartesian coordinate system ( x1;x2;x3) in
Vso that the scalar product hx;yiin these coordinates equals x1y1+x2y2+x3y3. Suppose also
thatw=dx1^dx2^dx3. Then (h) =hdx 1^dx2^dx3.
(v) =v1dx2^dx3+v2dx3^dx1+v3dx1^dx2
wherev1;v2;v3are coordinate functions of the vector eld v.
D(v) =v1dx1+v2dx2+v3dx3:
IfVis an oriented Euclidean space then one also has isomorphisms
?:
k(V)!
3 k(V); k= 0;1;2;3:
Ifwis the volume form on Vfor which the unit cube has volume 1 and which de ne the given
orientation of V(equivalently, if w=x1^x2^x3for any Cartesian positive coordinate system on
V), then
w(v) =?D(v);and w=?:
0(V)!
3(V):
8.7 Gradient, curl and divergence of a vector eld
The above isomorphism, combined with the operation of exterior di erentiation, allows us to de ne
the following operations on the vector elds. First recall that for a function f2C1(U);
gradf=D 1(df):
Now letv2Vect (U) be a vector eld. Then its divergence div vis the function de ned by the
formula
divv= 1(d(v))
In other words, we take the 2-form vw(wis the volume form) and compute its exterior di erential
d(vw). The result is a 3-form, and, therefore is proportional to the volume form w, i.e.d(vw) =hw.
96
This proportionality coecient (which is a function; it varies from point to point) is simply the
divergence: div v=h.
Given a vector eld v, its curlis as another vector eld curl vde ned by the formula
curlv:= 1d(Dv) =D 1d(Dv):
If one xes a cartesian coordinate system in Vsuch thatw=dx1^dx2^dx3andhx;yi=
x1y1+x2y2+x3y3then we get the following formulas
gradf=@f
@x1;@f
@x2;@f
@x3
divv=@v1
@x1+@v2
@x2+@v3
@x3
curlv=@v3
@x2 @v2
@x3;@v1
@x3 @v3
@x1;@v2
@x1 @v1
@x2
wherev= (v1;v2;v3).
We will discuss the geometric meaning of these operations later in Section ??.
8.8 Example: expressing vector analysis operations in spherical
coordinates
Given spherical coordinates ( r;; ) inR3, consider orthonormal vector elds
er=@
@r;e'=1
r@
@'ande=1
rsin'@
@
tangent to the coordinate lines and which form orthonormal basis at every point.
OperatorD.
Given a vector eld v=a1er+a2e'+a3ewe computeDv. For any vector eld h=h1er+
h2e+h3e'we havehv;hi=Dv(h). Let us writeDv=c1dr+c2d'+c3d. Then
hv;hi=a1h1+a2h2+a3h3
97
and
Dv(h) = (c1dr+c2d'+c3d)(h1er+h2e'+h3e)
=c1h1dr(er) +c2h2d'(e') +c3h3d(e) =c1h1+c2
rh2+c3
rsin'h3:
Hence,c1=a1; c2=ra2; c3=ra3sin', i.e.
Dv=a1dr+ra2d'+ra3sin'd: (8.8.1)
Operator.First we express the volume form
= dx^dy^dzin spherical coordinates. One
way to do that is just to plug into the form the expression of x;y;z through spherical coordinates.
But we can also argue as follows. We have
= cdr^d'^d. Let us evaluate both sides of this
equality on vectors er;e';e. Then
( er;e';e) = 1 because these vectors elds are at every point
orthonormal and de ne the standard orientation of the space. On the other hand,
(dr^d'^d)(er;e';e)
=1
r2sin'(dr^d'^d)(@
@r;@
@';@
@) =1
r2sin';
and therefore c=r2sin', i.e.
=r2sin'dr^d'^d:
In particular, given a 3-form =adr^d'^dwe get=a
r2sin'.
Let us now compute dr;d'andd. We argue that
dr=Ad'^d;d=Bdr^d';d'=Cd^dr:
Indeed, suppose that dr=Ad'^d+A0d^dr+A00dr^d'and compute the value of both
sides on vectors e;er:
dr(e;er) =dr(e ) = 0;
because the 2-dimensional volume (i.e the area) of the 2-dimensional parallelepiped P(e;er) and
the 1-dimensional volume (i.e. the length) of the 1-dimensional parallelepiped P(e') are both equal
to 1. On the other hand,
(Ad'^d+A0d^dr+A00dr^d')(e;er) =1
rsin'A0;
98
and therefore A0= 0. Similarly, A00= 0. The argument for dandd'is also similar.
It remains to compute the coecients A;B;C . We have
dr(e';e) =dr(er) = 1 =Ad'^d(e';e) =A
r2sin';
d(er;e') =d(e) =1
rsin'=Bdr^d'(er;e') =B
r;
d'(e;er) =d'(e') =1
r=Cd'^dr(e;er) =C
rsin':
Thus, i.e.
A=r2sin'; B =1
sin'; C= sin';
dr=r2sin'd'^d;d=1
sin'dr^d';d'= sin'd^dr: (8.8.2)
Hence we also have
d'^d=1
r2dr;
dr^d'= sin'd;
d^dr=1
sin'd':(8.8.3)
Now we are ready to express the vector analysis operations in spherical coordinates.
Exercise 8.6. (Gradient) Given a function f(r;;' ) expressed in cylindrical coordinates, com-
puterf.
Solution. We have
rf=D 1(frdr+f'd'+fd) =frer+f'
re'+f'
rsin'e:
Here we denote by fr;f'andfthe respective partial derivatives of the function f.
Exercise 8.7. Given a vector eld v=aer+be'+cecompute div vand curl v.
Solution.
99
a)Divergence . We have
v
= (aer+be'+ce)r2sin'dr^d'^d
=ar2sin'd'^d+crdr^d'+brsin'd^dr
and
d(v
) =
(r2ar+ 2ra) sin'+rc+rb'sin'+rbcos'
dr^d'^d
= (ar+2a
r+c
rsin'+b'
r+b
rcot')
:
Hence,
divv=ar+2a
r+c
rsin'+b'
r+b
rcot': (8.8.4)
b)Curl . We have
curlv=D 1dDv=D 1d(adr+rbd' +rcsin'd)
=D 1
(a' b rbr)dr^d'+ (csin'+rcrsin' a)d^dr
+ (rb rc'sin' rccos')d'^d
=D 1 rb rc'sin' rccos'
r2dr+csin'+rcrsin' a
sin'd'
+ (a' b rbr) sin'd
=rb rc'sin' rccos'
r2er+csin'+rcrsin' a
rsin'e'
+a' b rbr
re:(8.8.5)
8.9 Complex-valued di erential k-forms
One can consider complex-valued di erential k-forms. A C-valued di erential 1-form is a eld of
C-valuedk-forms, or simply it is an expression +i , where ; are usual real-valued k-forms. All
operations on complex valued k-forms ( exterior multiplication, pull-back and exterior di erential)
are de ned in a natural way:
( 1+i 1)^( 2+i 2) = 1^ 2 1^ 2+i( 1^ 2+ 1^ 2);
100
f( +i ) =f +if ; d( +i ) =d +id :
We will be in particular interested in complex valued on C. Note that a complex-valued function
(or 0-form) is on a domain UCis just a map f=u+iv:U!C. Its di erential dfis the same as
the di erntial of this map, but it also can be viewed as a C-valued di erential 1-form df=du+idv.
Example 8.8.
dz=dx+idy;d z=dx idy;zdz = (x+iy)(dx+idy) =xdx ydy+i(xdy+ydx);
dz^dz= (dx+idy)^(dx idy) = 2idx^dy:
Exercise 8.9. Prove that d(zn) =nzn 1dzforanyintegern6= 0.
Solution. Let us do the computation in polar coordinates. Then zn=rnein and assuming
thatn6= 0 we have
d(zn) =nrn 1ein dr+inrnein d =nrn 1ein (dr+id ):
On the other hand,
nzn 1dz=nrn 1ei(n 1) d(rei ) =nrn 1ei(n 1)
ei dr+iei d
=nrn 1ein (dr+id ):
Comparing the two expressions we conclude that d(zn) =nzn 1dz.
It follows that the 1-formdz
znis exact on Cn0 forn>1. Indeed,
dz
zn=d1
(1 n)zn 1
:
On the other hand the formdz
zis closed on Cn0 but not exact. Indeed,
dz
z=d(rei )
rei =ei dr+irei d
rei =dr
r+id ;
and henced dz
z
= 0:
On the other hand, we already had seen that the form d , and hencedz
z, is not exact.
101
102
Chapter 9
Integration of di erential forms and
functions
9.1 Useful technical tools: partition of unity and cut-o functions
Let us recall that the support of a function is the closure of the set of points where it is not
equal to 0. We denote the support by Supp( ). We say that issupported in an open set Uif
Supp()U.
Lemma 9.1. There exists a C1function:R![0;1)with the following properties:
(x)0,jxj1;
(x) =( x);
(x)>0forjxj<1.
Proof. There are a lot of functions with this property. For instance, one can be constructed as
follows. Take the function
h(x) =8
><
>:e 1
x2; x> 0
0; x0:(9.1.1)
The function e 1
x2has the property that all its derivatives at 0 are equal to 0, and hence the
functionhisC1-smooth. Then the function (x) :=h(1 +x)h(1 x) has the required properties.
103
Lemma 9.2. Existence of cut-off functions LetCVbe compact set and UCits open
neighborhood. Then there exists a C1-smooth function C;U:V![0;1)with its support in U
which is equal to 1onC
Proof. Let us x a Euclidean structure in Vand a Cartesian coordinate system. Thus we can
identifyVwithRnwith the standard dot-product. Given a point a2Vand >0 let us denote
by a;thebump function on Vde ned by
a;(x) :=jjx ajj2
2
; (9.1.2)
where:R![0;1) is the function constructed in Lemma 9.1. Note that a;(x) is aC1-function
with Supp ( a;) =D:=B(a) and such that a;(x)>0 forx2B(a).
Let us denote by U(C) the-neighborhood of C, i.e.
U(C) =fx2V;9y2C;jjy xjj<:g
There exists >0 such that U(C)U. Using compactness of Cwe can nd nitely many points
z1;:::;zN2Csuch that the balls B(z1);:::;B(zN)UcoverU
2(C), i.e.U
2(C)NS
1B(zj).
Consider a function
1:=NX
1 zi;
2:V!R:
The function 1is positive on U
2(C) and has Supp( 1)U.
The complement E=VnU
2(C) is a closed but unbounded set. Take a large R> 0 such that
BR(0)U. ThenER=DR(0)nU
2(C) is compact. Choose nitely many points x1;:::;xM2ER
such thatMS
1B
4(xi)ER. Notice thatMS
1B
4(xi)\C=?. Denote
2:=MX
1 xi;
4:
Then the function 2is positive on VRand vanishes on C. Note that the function 1+2is positive
onBR(0) and it coincides with 1onC. Finally, de ne the function C;Uby the formula
C;U:=1
1+2
104
onB(R)(0) and extend it to the whole space Vas equal to 0 outside the ball BR(0). ThenC;U= 1
onCand Supp(C;U)U, as required. .
LetCVbe a compact set. Consider its nite covering by open sets U1;:::;UN, i.e.
N[
1UjC:
We say that a nite sequence 1;:::;KofC1-functions de ned on some open neighborhood Uof
CinVforms a partition of unity overCsubordinated to the covering fUjgj=1;:::;N if
KP
1j(x) = 1 for all x2C;
Each function j,j= 1;:::;K is supported in one of the sets Ui,i= 1;:::;K .
Lemma 9.3. For any compact set Cand its open covering fUjgj=1;:::;N there exists a partition of
unity overCsubordinated to this covering.
Proof. In view of compactness of there exists >0 and nitely many balls B(zj) centered at
pointszj2C,j= 1;:::;K , such thatKS
1B(zj)Cand each of these balls is contained in one of
the open sets Uj,j= 1;:::;N . Consider the functions zj;de ned in (9.1.2). We haveKP
1 zj;>0
on some neighborhood UC. LetC;Ube the cut-o function constructed in Lemma 9.2. For
j= 1;:::;K we de ne
j(x) =8
>>><
>>>: zj;(x)C;U(x)
KP
1 zj;(x);ifx2U;
0; otherwise:
Each of the functions is supported in one of the open sets Uj,j= 1;:::;N , and we have for every
x2C
KX
1j(x) =KP
1 zj;(x)
KP
1 zj;(x)= 1:
105
Figure 9.1: Bernhard Riemann (1826-1866)
9.2 One-dimensional Riemann integral for functions and di eren-
tial1-forms
ApartitionPof an interval [ a;b] is a nite sequence a=t0< t 1<< tN=b. We will
denote byTj;j= 0;:::;N the vectortj+1 tj2Rtjand by jthe interval [ tj;tj+1]. The length
tj+1 tj=jjTjjjof the interval jwill be denoted by j. The number max
j=1;:::;Njis called the
neness or the sizeof the partitionPand will be denoted by (P). Let us rst recall the de nition
of (Riemann) integral of a function of one variable. Given a function f: [a;b]!Rwe will form a
lower and upper integral sums corresponding to the partition P:
L(f;P) =N 1X
0( inf
[tj;tj+1]f)(tj+1 tj);
U(f;P) =N 1X
0( sup
[tj;tj+1]f)(tj+1 tj); (9.2.1)
The function is called Riemann integrable if
sup
PL(f;P) = inf
PU(f;P);
and in this case this number is called the (Riemann) integral of the function fover the interval
[a;b].The integrability of fcan be equivalently reformulated as follows. Let us choose a set C=
106
fc1;:::;cN 1g; cj2j, and consider an integral sum
I(f;P;C) =N 1X
0f(cj)(tj+1 tj); cj2j: (9.2.2)
Then the function fis integrable if there exists a limit lim
(P)!0I(f;P;C). In this case this limit is
equal to the integral of fover the interval [ a;b]. Let us emphasize that if we already know that the
function is integrable, then to compute the integral one can choose anysequence of integral sum,
provided that their neness goes to 0. In particular, sometimes it is convenient to choose cj=tj,
and in this case we will write I(f;P) instead of I(f;P;C).
The integral has di erent notations. It can be denoted sometimes byR
[a;b]f, but the most common
notation for this integral isbR
af(x)dx. This notation hints that we are integrating here the di erential
formf(x)dxrather than a function f. Indeed, given a di erential form =f(x)dxwe have
f(cj)(tj+1 tj) = cj(Tj),1and hence
I( ;P;C) =I(f;P;C) =N 1X
0 cj(Tj); cj2j: (9.2.3)
We say that a di erential 1-form isintegrable if there exists a limit lim
(P)!0I( ;P;C), which is
called in this case the integral of the di erential 1-form over the oriented interval [a;b] and will
be denoted byR
!
[a;b] , or simplybR
a . By de nition, we say thatR
[a;b] = R
!
[a;b] . This agrees with the
de nitionR
[a;b] = lim
(P)!0N 1P
1 cj( Tj), and with the standard calculus ruleaR
bf(x)dx= bR
af(x)dx:
Let us recall that a map : [a;b]![c;d] is called a di eomorphism if it is smooth and has a
smooth inverse map 1: [c;d]![a;b]. This is equivalent to one of the following:
(a) =c; (b) =dand 0>0 everywhere on [ a;b]. In this case we say that preserves
orientation.
(a) =d; (b) =cand 0<0 everywhere on [ a;b]. In this case we say that reverses
orientation.
1Here we parallel transported the vector Tjfrom the point tjto the point cj2[tj;tj+1].
107
Theorem 9.4. Let : [a;b]![c;d]be a di eomorphism. Then if a 1-form =f(x)dxis integrable
over [c;d]then its pull-back f is integrable over [a;b], and we have
Z
!
[a;b] =Z
!
[c;d] ; (9.2.4)
if preserves the orientation and
Z
!
[a;b] =Z
[c;d] = Z
!
[c;d] ;
if reverses the orientation.
Remark 9.5. We will show later a stronger result:
Z
!
[a;b] =Z
!
[c;d]
forany : [a;b]![c;d] with (a) =c; (b) =d, which is not necessarily a di eomorphism.
Proof. We consider only the orientation preserving case, and leave the orientation reversing one
to the reader. Choose any partition P=fa=t0<<tN 1<tN=bgof the interval [ a;b] and
choose any set C=fc0;:::;cN 1gsuch thatcj2j. Then the points etj= (tj)2[c;d],j= 0;:::N
form a partition of [ c;d]. Denote this partition by eP, and denoteej:= [etj;etj+1][c;d],ecj= (cj),
eC= (C) =fec0;:::;ecN 1g. Then we have
I( ;P;C) =N 1X
0 cj(Tj) =N 1X
0 ecj(d (Tj)) =
N 1X
0 ecj( 0(cj)j): (9.2.5)
Recall that according to the mean value theorem there exists a point dj2j, such that
eTj=etj+1 etj= (tj+1) (tj) = 0(dj)(tj+1 tj):
Note also that the function 0isuniformly continuous , i.e. for any >0 there exists >0 such
that for any t;t02[a;b] such thatjt t0j<we havej 0(t) 0(t0)j<. Besides, the function 0
108
is bounded above and below by some positive constants: m< 0<M . Hencemj<ej<Mjfor
allj= 1;:::;N 1. Hence, if (P)<then we have
I( ;P;C) I( ;eP;eC) = N 1X
0 ecj
( 0(cj) 0(dj))j =
m N 1X
1f(ecj)ej =
m I( ;eP;eC) : (9.2.6)
When(P)!0 we havee(P) = 0, and hence by assumption I( ;eP;eC)!dR
c , but this implies
thatI( ;P;C) I( ;eP;eC)!0, and thus is integrable over [ a;b] and
bZ
a = lim
(P)!0I( ;P;C) = lim
(eP)!0I( ;eP;eC) =dZ
c :
If we write =f(x)dx, then =f( (t)) 0(t)dtand the formula (9.4) takes a familiar form
of the change of variables formula from the 1-variable calculus:
dZ
cf(x)dx=bZ
af( (t)) 0(t)dt:
9.3 Integration of di erential 1-forms along curves
Curves as paths
Apath, orparametrically given curve in a domain Uin a vector space Vis a map
: [a;b]!U. We
will assume in what follows that all considered paths are di erentiable. Given a di erential 1-form
inUwe de ne the integral of over
by the formula
Z
=Z
[a;b]
:
Example 9.6. Consider the form =dz ydx+xdyonR3. Let
: [0;2]!R3be a helix given
by parametric equations x=Rcost;y=Rsint;z=Ct. Then
Z
=2Z
0(Cdt+R2(sin2tdt+ cos2tdt)) =2Z
0(C+R2)dt= 2(C+R2):
109
Note thatR
= 0 whenC= R2. One can observe that in this case the curve
is tangent to the
plane eldgiven by the Pfaan equation = 0.
Proposition 9.7. Let a pathe
be obtained from
: [a;b]!Uby a reparameterization, i.e.
e
=
, where : [c;d]![a;b]is an orientation preserving di eomorphism. ThenR
e
=R
.
Indeed, applying Theorem 9.4 we get
Z
e
=dZ
ce
=dZ
c (
)bZ
a
=Z
:
A vector
0(t)2V
(t)is called the velocity vector of the path
.
Curves as 1-dimensional submanifolds
A subset Uis called a 1-dimensional submanifold ofUif for any point x2 there is a
neighborhood UxUand a di eomorphism x:Ux!
xRn, such that x(x) = 02Rnand
x( \Ux) either coincides with fx2=:::xn= 0g\
x, or withfx2=:::xn= 0; x10g\
x. In
the latter case the point xis called a boundary point of . In the former case it is called an interior
point of .
A 1-dimensional submanifold is called closed if it is compact and has no boundary. An example
of a closed 1-dimensional manifold is the circle S1=fx2+y2= 1gR2.
WARNING. The word closed is used here in a di erent sense than when one speaks about closed
subsets . For instance, a circle in R2is both, a closed subset and a closed 1-dimensional submanifold,
while a closed interval is a closed subset but not a closed submanifold: it has 2 boundary points.
An open interval in R(or any Rn) is a submanifold without boundary but it is not closed because
it is not compact. A line in a vector space is a 1-dimensional submanifold which is a closed subset
of the ambient vector space. However, it is not compact, and hence not a closed submanifold.
Proposition 9.8. 1. Suppose that a path
: [a;b]!Uis an embedding. This means that
0(t)6= 0 for allt2[a;b]and
(t)6=
(t0)ift6=t0.2Then =
([a;b])is 1-dimensional
compact submanifold with boundary.
2If only the former property is satis ed that
is called an immersion .
110
2. Suppose Uis given by equations F1= 0;:::;Fn 1= 0 whereF1;:::;Fn 1:U!Rare
smooth functions such that for each point x2 the di erential dxF1;:::;dxFn 1are linearly
independent. Then is a1-dimensional submanifold of U.
3. Any compact connected 1-dimensional submanifold Ucan be parameterized either by an
embedding
: [a;b]! ,!Uif it has non-empty boundary, or by an embedding
:S1!
,!Uif it is closed.
Proof. 1. Take a point c2[a;b]. By assumption
0(c)6= 0. Let us choose an ane coordinate
system (y1;:::;yn) inVcentered at the point C=
(c) such that the vector
0(c)2VCcoincide
with the rst basic vector. In these coordinates the map gamma can be written as (
1;:::;
n)
where
0
1(c) = 1,
0
j(c) = 0 forj >1 and
j(c) = 0 for all j= 1;:::;n . By the inverse function
theorem the function
1is a di eomorphism of a neighborhood of conto a neighborhood of 0 in
R(ifcis one of the end points of the interval, then it is a di eomorphism onto the corresponding
one-sided neighborhood of 0). Let be the inverse function de ned on the interval equal to
( ;), [0;) and ( ;0], respectively, depending on whether cis an interior point, c=aorc=b].
so that
1((u)) =ufor anyu2. Denotee =()[a;b]. Then
(e)Ucan be given by
the equations:
y2=e
2(u) :=
2((y1)); :::; yn=e
n(u) :=
n((y1));y12:
Let us denote
=() := max
j=2;:::;nmax
u2je
j(u)j:
Denote
P:=fjy1j;jyjj()g:
We have
()P:
We will show now that for a suciently small we have
([a;b])\P=
(). For every
pointt2[a;b]nInt denote d(t) =jj
(t)
(c)jj:Recall now the condition that
(t)6=
(t0)
fort6=t0. Henced(t)>0 for allt2[a;b]nInt . The function d(t) is continuous and hence
achieve the minimum value on the compact set [ a;b]nInt . Denote d:= min
t2[a;b]nInt d(t)>0:Chose
0<min(d;) and such that (0) = max
j=2;:::;nmax
ju cj<0je
j(u)j<d. Let 0= \fjuj0gj. Then
([a;b])\P0=fy2=e
2(u); :::; yn=e
n(u);y120g:
111
2. Take a point c2 . The linear independent 1-forms dcF1;:::;dcFn 12V
ccan be completed
by a 1-form l2V
cto a basis of V
c. We can choose an ane coordinate system in Vwithcas its
origin and such that the function xncoincides with l. Then the Jacobian matrix of the functions
F1;:::;Fn 1;xnis non-degenerate at c= 0, and hence by the inverse function theorem the map
F= (F1;:::;Fn 1;xn) :V!Rnis invertible in the neighborhood of c= 0, and hence these
functions can be chosen as new curvilinear coordinates y1=F1;:::;yn 1=Fn 1;yn=xnnear the
pointc= 0. In these coordinates the curve is given near cby the equations y1==yn 1= 0:
3. See Exercise ??.
In the case where is closed we will usually parameterize it by a path
: [a;b]! U
with
(a) =
(b). For instance, we parameterize the circle S1=fx2+y2= 1gR2by a path
[0;2]7!(cost;sint). Such
, of course, cannot be an embedding, but we will require that
0(t)6= 0
and that for t6=t0we have
(t)6=
(t0)unless one of these points is aand the other one is b. We
will refer to 1-dimensional submanifolds simply as curves , respectively closed, with boundary etc.
Given a curve its tangent line at a pointx2 is a subspace of Vxgenerated by the velocity
vector
0(t) for any local parameterization
: [a;b]! with
(t) =x. If is given implicitly, as in
9.8.2, then the tangent line is de ned in Vxby the system of linear equations dxF1= 0;:::;dxFn 1=
0.
Orientation of a curve is the continuously depending on points orientation of all its tangent
lines. If the curve is given as a path
: [a;b]! Usuch that
0(t)6= 0 for all t2[a;b] than
it is canonically oriented. Indeed, the orientation of its tangent line lxat a pointx=
(t)2 is
de ned by the velocity vector
0(t)2lx.
It turns out that one can de ne an integral of a di erential form over an oriented compact
curve directly without referring to its parameterization. For simplicity we will restrict our discussion
to the case when the form is continuous.
Let be a compact connected oriented curve. A partition of is a sequence of points P=
fz0;z1;:::;zNgordered according to the orientation of the curve and such that the boundary
points of the curve (if they exist) are included into this sequence. If is closed we assume that
zN=z0. The neness(P) ofPis by de nition is max
j=0;:::;N 1dist(zj;zj+1) (we assume here that V
a Euclidean space).
112
De nition 9.9. Let be a di erential 1-form and a compact connected oriented curve. Let
P=fz0;:::;zNgbe its partition. Then we de ne
Z
= lim
(P)!0I( ;P);
whereI( ;P) =N 1P
0 zj(Zj); Zj=zj+1 zj2Vzj.
When is a closed submanifold then one sometimes uses the notationH
instead ofR
.
Proposition 9.10. If one chooses a parameterization
: [a;b]! which respects the given
orientation of then
Z
=bZ
a
=Z
:
Proof. Indeed, leteP=ft0;:::;tNgbe a partition of [ a;b] such that
(tj) =zj; j= 0;:::;N .
I(
;eP) =N 1X
1
tj(Tj) =N 1X
1 zj(Uj);
whereUj=dtj
(Tj))2Vzjis a tangent vector to at the point zj. Let us evaluate the di erence
Uj Zj. Choosing some Cartesian coordinates in Vwe denote by
1;:::;
nthe coordinate functions
of the path
. Then using the mean value theorem for each of the coordinate functions we get
i(tj+1)
i(tj) =
0
i(ci
j)jfor someci
j2j,i= 1;:::;n ;j= 0;:::;N 1:Thus
Zj=
(tj+1)
(tj) = (
0
1(c1
j);:::;
0
i(cn
j))j:
On the other hand, Uj=dtj
(Tj))2Vzj=
0(tj)j. Hence,
jjZj Ujjj=jvuutnX
1(
0
i(ci
j)
0
i(tj))2:
Note that if (P)!0 then we also have (eP)!0, and hence using smoothness of the path
we
conclude that for any >0 there exists >0 such thatjjZj Ujjj<jfor allj= 1;:::;N . Thus
N 1X
1 zj(eTj) N 1X
1 zj(Zj)!
(P)0;
113
and therefore
bZ
a
= lim
(eP)!0I(
;eP) = lim
(P)!0I( ;P) =Z
:
9.4 Integrals of closed and exact di erential 1-forms
Theorem 9.11. Let =dfbe an exact 1-form in a domain UV. Then for any path
: [a;b]!U
which connects points A=
(a)andB=
(b)we have
Z
=f(B) f(A):
In particular, if
is a loop thenH
= 0.
Similarly for an oriented curve Uwith boundary @ =B Awe have
Z
=f(B) f(A):
Proof. We haveR
df=bR
a
df=bR
ad(f
) =f(
(b)) f(
(a)) =f(B) f(A).
It turns out that closed forms are locally exact . A domain UVis called star-shaped with
respect to a point a2Vif with any point x2Uit contains the whole interval Ia;xconnecting a
andx, i.e.Ia;x=fa+t(x a);t2[0;1]g. In particular, any convex domain is star-shaped.
Proposition 9.12. Let be a closed 1-form in a star-shaped domain UV. Then it is exact.
Proof. De ne a function F:U!Rby the formula
F(x) =Z
!
Ia;x ; x2U;
where the intervals Ia;xare oriented from 0 to x.
We claim that dF= . Let us identify Vwith the Rnchoosingaas the origin a= 0. Then
can be written as =nP
1Pk(x)dxk;andI0;xcan be parameterized by
t7!tx; t2[0;1]:
114
Hence,
F(x) =Z
!
I0;x =1Z
0nX
1Pk(tx)xkdt: (9.4.1)
Di erentiating the integral over xjas parameters, we get
@F
@xj=1Z
0nX
k=1txk@Pk
@xj(tx)dt+1Z
0Pj(tx)dt:
Butd = 0 implies that@Pk
@xj=@Pj
@xk, and using this we can further write
@F
@xj=1Z
0nX
k=1txk@Pj
@xk(tx)dt+1Z
0Pj(tx)dt=1Z
0tdPj(tx)
dtdt+1Z
0Pj(tx)dt
= (tPj(tx))j1
0 1Z
0Pj(tx)dt+1Z
0Pj(tx)dt=Pj(tx)
Thus
dF=nX
j=1@F
@xjdxj=nX
j=1Pj(x)dx=
9.5 Integration of functions over domains in high-dimensional spaces
Riemann integral over a domain in Rn.
In this section we will discuss integration of bounded functions over bounded sets in a vector space
V. We will x a basis e1;:::;enand the corresponding coordinate system x1;:::;xnin the space
and thus will identify VwithRn. Letdenote the volume form x1^:::xn. As it will be clear
below, the de nition of an integral will not depend on the choice of a coordinate system but only
on the background volume form, or rather its absolute value because the orientation of Vwill be
irrelevant.
We will need a special class of parallelepipeds in V, namely those which are generated by vectors
proportional to basic vectors, or in other words, parallelepipeds with edges parallel to the coordinate
axes. We will also allow these parallelepipeds to be parallel transported anywhere in the space. Let
us denote
P(a1;b1;a2;b2;:::;an;bn) :=faixibi;i= 1;:::;ngRn:
115
We will refer to P(a1;b1;a2;b2;:::;an;bn) as a special parallelepiped , orrectangle .
Let us x one rectangle P:=P(a1;b1;a2;b2;:::;an;bn). Following the same scheme as we used
in the 1-dimensional case, we de ne a partitionPofPas a product of partitions a1=t1
0<<
t1
N1=b1; :::;an=tn
0<<tn
Nn=bn, of intervals [ a1;b1];:::; [an;bn]. For simplicity of notation
we will always assume that each of the coordinate intervals is partitioned into the same number
of intervals, i.e. N1==Nn=N. This de nes a partition of PintoNnsmaller rectangles
Pj=ft1
j1x1t1
j1+1; :::; tn
jnxntn
jn+1g, where j= (j1;:::;jn) and each index jktakes
values between 0 and N 1. Let us de ne
Vol(Pj) :=nY
k=1(tk
jk+1 tk
jk): (9.5.1)
This agrees with the de nition of the volume of a parallelepiped which we introduced earlier (see
formula (3.3.1) in Section 3.3). We will also denote j:= max
k=1;:::;n(tk
jk+1 tk
jk) and(P) := max
j(j).
Let us x a point cj2Pjand denote by Cthe set of all such cj. Given a function f:P!Rwe
form an integral sum
I(f;P;C) =X
jf(cj)Vol(Pj) (9.5.2)
where the sum is taken over all elements of the partition. If there exists a limit lim
(P)!0I(f;P;C)
then the function f:P!Ris called integrable (in the sense of Riemann) over P, and this limit is
called the integral offoverP. There exist several di erent notations for this integral:R
Pf,R
PfdV,
R
PfdVol, etc. In the particular case of n= 2 one often uses notationR
PfdA, orRR
PfdA. Sometime, the
functions we integrate may depend on a parameter, and in these cases it is important to indicate with
respect to which variable we integrate. Hence, one also uses the notation likeR
Pf(x;y)dxn, where
the indexnrefers to the dimension of the space over which we integrate. One also use the notationZ
:::Z
|{z}
Pf(x1;:::xn)dx1:::dxn, which is reminiscent both of the integralR
Pf(x1;:::xn)dx1^^
dxnwhich will be de ned later in Section 9.7 and the notationbnR
an:::b1R
a1f(x1;:::xn)dx1:::dxnfor
n interated integral which will be discussed in Section 9.6.
Alternatively and equivalently the integrability can be de ned via upper and lower integral sum,
116
similar to the 1-dimensional case. Namely, we de ne
U(f;P) =X
jMj(f)Vol(Pj); L(f;P) =X
jmj(f)Vol(Pj);
whereMj(f) = sup
Pjf;m j(f) = inf
Pjf;and say that the function fis integrable over Pif inf
PU(f;P) =
sup
PL(f;P).
Note that inf
PU(f;P) and sup
PL(f;P) are sometimes called upper and lower integrals, respec-
tively, and denoted byR
PfandR
Pf. Thus a function f:P!Ris integrable i R
Pf=R
Pf.
Let us list some properties of Riemann integrable functions and integrals.
Proposition 9.13. Letf;g:P!Rbe integrable functions. Then
1.af+bg, wherea;b2R, is integrable andR
Paf+bg=aR
Pf+bR
Pg;
2. IffgthenR
PfR
Pg;
3.h= max(f;g)is integrable; in particular the functions f+:= max(f;0)andf := max( f;0)
andjfj=f++f are integrable;
4.fgis integrable.
Proof. Parts 1 and 2 are straightforward and we leave them to the reader as an exercise. Let us
check properties 3 and 4.
3. Take any partition PofP. Note that
Mj(h) mj(h)max(Mj(f) mj(f);Mj(g) mj(g)): (9.5.3)
Indeed, we have Mj(h) = max(Mj(f);Mj(g)) andmj(h)max(mj(f);mj(g)). Suppose for deter-
minacy that max( Mj(f);Mj(g)) =Mj(f). We also have mj(h)mj(f). Thus
Mj(h) mj(h)Mj(f) mj(f)max(Mj(f) mj(f);Mj(g) mj(g)):
Then using (9.5.3) we have
U(h;P) L(h;P) =X
j(Mj(h) mj(h))Vol(Pj)
X
jmax (Mj(f) mj(f);Mj(g) mj(g)) Vol(Pj) =
max (U(f;P) L(f;P);U(f;P) L(f;P)):
117
By assumption the right-hand side can be made arbitrarily small for an appropriate choice of the
partitionP, and hence his integrable.
4. We have f=f+ f ,g=g+ g andfg=f+g++f g f+g f g+:Hence, using 1
and 3 we can assume that the functions f;gare non-negative. Let us recall that the functions f;g
are by assumption bounded, i.e. there exists a constant C > 0 such that f;gC. We also have
Mj(fg)Mj(f)Mj(g) andmj(fg)mj(f)mj(g). Hence
U(fg;P) L(fg;P) =X
j(Mj(fg) mj(fg)) Vol(Pj)
X
j(Mj(f)Mj(g) mj(f)mj(g))Vol(Pj) =
X
j(Mj(f)Mj(g) mj(f)Mj(g) +mj(f)Mj(g) mj(f)mj(g)) Vol(Pj)
X
j((Mj(f) mj(f))Mj(g) +mj(f)(Mj(g) mj(g))) Vol(Pj)
C(U(f;P) L(f;P) +U(g;P) L(g;P)):
By assumption the right-hand side can be made arbitrarily small for an appropriate choice of the
partitionP, and hence fgis integrable.
Consider now a bounded subset KRnand choose a rectangle PK. Given any function
f:K!Rone can always extend it to Pas equal to 0. A function f:K!Ris called integrable
overKif this trivial extension fis integrable over P, and we de neR
KfdV :=R
fdV. When this
will not be confusing we will usually keep the notation ffor the above extension.
Volume
We further de ne the volume
Vol(K) =Z
K1dV=Z
P KdV;
provided that this integral exists. In this case we call the set Kmeasurable in the sense of Riemann ,
or just measurable.3Here Kis the characteristic orindicator function of K, i.e. the function which
3There exists a more general and more common notion of measurability in the sense of Lebesgue. Any Riemann
measurable set is also measurable in the sense of Lebesgue, but not the other way around. Historically an attribution
118
is equal to 1 on Kand 0 elsewhere. In the 2-dimensional case the volume is called the area, and in
the 1-dimensional case the length.
Remark 9.14. For any bounded set Athere is de ned a lower and upper volumes,
Vol(A) =Z
AdVVol(A) =Z
AdV:
The set is measurable i Vol (A) = Vol(A):IfVol(A) = 0 then Vol (A) = 0, and hence Ais
measurable and Vol( A) = 0.
Exercise 9.15. Prove that for the rectangles this de nition of the volume coincides with the one
given by the formula (9.5.1) .
The next proposition lists some properties of the volume.
Proposition 9.16. 1. Volume is monotone, i.e. if A;BPare measurable and ABthen
Vol(A)Vol(B).
2. If setsA;BPare measurable then A\B,AnBandA[Bare measurable as well and we
have
Vol(A[B) = Vol(A) + Vol(B) Vol(A\B):
3. IfAcan be covered by a measurable set of arbitrarily small total volume then Vol(A) = 0 .
Conversely, if Vol(A) = 0 then for any >0there exists a >0such that for any partition
Pwith(P)< the elements of the partition which intersect Ahave arbitrarily small total
volume.
4.Ais measurable i Vol (@A) = 0 .
Proof. The rst statement is obvious. To prove the second one, we observe that A[B= max( A; B),
A\B= A B, max( A; B) = A+ B A B, AnB= A A\Band then apply Proposition
9.13. To prove 9.16.3 we rst observe that if a set Bis measurable and Vol B < then then for a
suciently ne partition Pwe haveU( B;P)<VolB+<2. SinceABthen A B, and
of this notion to Riemann is incorrect. It was de ned by Camille Jordan and Giuseppe Peano before Riemann integral
was introduced. What we call in these notes volume is also known by the name Jordan content.
119
thereforeU( A;P)U( B;P)<2. Thus, infPU( A;P) = 0 and therefore Ais measurable and
Vol(A) = 0. Conversely, if Vol( A) = 0 then for any >0 for a suciently ne partition Pwe have
U( A;P)<. ButU( A;P) is equal to the sum of volumes of elements of the partition which have
non-empty intersection with A.
Finally, let us prove 9.16.4. Consider any partition PofPand form lower and upper integral
sums for A. DenoteMj:=Mj( A) andmj=mj( A). Then all numbers Mj;mjare equal to either
0 or 1. We have Mj=mj= 1 ifPjA;Mj=mj= 0 ifPj\A=?andMj= 1;mj= 0 ifPjhas
non-empty intersection with both AandPnA. In particular,
B(P) :=[
j;Mj mj=1Pj@A:
Hence, we have
U( A;P) L( A;P) =X
j(Mj mj)Vol(Pj) = VolB(P):
Suppose that Ais measurable. Then there exists a partition such that U( A;P) L( A;P)< ,
and hence@Ais can be covered by the set B(P) of volume <. Thus applying part 3 we conclude
that Vol(@A) = 0. Conversely, we had seen below that if Vol( @A) = 0 then there exists a partition
such that the total volume of the elements intersecting @Ais<. Hence, for this partition we have
L( A;P)U( A;P)<, which implies the integrability of A, and hence measurability of A.
Corollary 9.17. If a bounded set AVis measurable then its interior IntAand its closure A
are also measurable and we have in this case
VolA= Vol IntA= VolA:
Proof. 1. We have @A@Aand@(IntA)@A. Therefore, Vol @A= Vol@IntA= 0, and
therefore the sets Aand IntAare measurable. Also Int A[@A=Aand IntA\@A=?. Hence, the
additivity of the volume implies that Vol A= Vol@IntA+Vol@A= Vol@IntA:On the other hand,
IntAAA. and hence the monotonicity of the volume implies that Vol Int AVolAVolA:
Hence, VolA= Vol IntA= VolA:
Exercise 9.18. If IntAorAare measurable then this does not imply that Ais measurable. For
instance, if Ais the set of rational points in interval I= [0;1]Rthen IntA=?andA=I.
However, show that Ais not Riemann measurable.
120
2. A setAis called nowhere dense if IntA=?. Prove that if Ais nowhere dense then either
VolA= 0, orAis not measurable in the sense of Riemann. Find an example of a non-measurable
nowhere dense set.
Lipshitz maps
LetV;W be two Euclidean spaces. Recall the de nition of the norm of a linear operator A:V!W:
jjAjj = max
jjxjj=1jjA(x)jj
jjxjj:
Thus we haveA(BR(0))BaR(0)W, where we denoted a:=jjAjj:
Given a subset AVa mapf:A!Wis called Lipshitz if there exists a constant C > 0 such
that for any x;y2Awe have
jjf(y) f(x)jjCjjy xjj:
Lemma 9.19. LetAVbe a compact set. Then any C1-smooth map A!Wis Lipshitz.
Let us recall that given a compact set CV, we say that a map f:C!Wis smooth if it
extends to a smooth map de ned on an open neighborhood UC.
Proof. LetK:A!Rbe the function de ned by K(x) =jjdxfjj;x2A. The function Kis
continuous because fisC1-smooth. Hence it is bounded: there exists a constant E > 0 such that
K(x)Efor allx2A.
Let us rst consider the case when Ais a convex set, i.e. with any two points x;y2Athe interval
connecting them is also contained in A. Given two points x;y2Aat a distance d=jjx yjj>0
consider a path (s) =x+s
d(y x); s2[0;d] which connects them. Note that the velocity vector
0(s) =y x
dhas the unit length. Denote ef(s) =f( (s));s2[0;d]. Note thatef0(s) =df (s)( 0(s))
by the chain rule. In particular,
jjef0(s)jjjjdf (s)jjjj 0(s)jjE:
We also have f(y) f(x) =ef(d) ef(0). Letef= (ef1;:::;efm) be the coordinate functions of ef
in some Cartesian coordinates in W. By the intermediate value theorem, for each k= 1;:::;n we
121
haveefk(d) efk(0) =ef0
k(ck)dfor someck2[0;d]. Hence,jefk(d) efk(0)jCdand therefore
jjf(y) f(x)jj=jjef(d) ef(0)jj=vuutnX
1(efk(d) efk(0))2Epnd=eEjjy xjj; (9.5.4)
where we denoted eE:=Epn.
For a general bounded set Achoose an open neighborhood UAto which the map fextends
C1-smoothly. Let U0Ube a smaller open neighborhood of Asuch thatU0U. We will also
assume that U0is bounded, and hence compact.
For every point x2Athere is(x)>0 such that the closed ball B(x)(x) is contained in U0. We
haveS
x2AB(x)
2(x)A, and hence by compactness of Athere are nitely many balls Bi:=B(xi)
2(xi),
i= 1;:::;N , such thatNS
1BiA. Denote:= min
i=1;:::;N(xi)
2:Then for any two points x;y2Awith
jjy xjjbelong to one of the balls bBi:=B(xi)(xi) which is convex, and hence according to
(9.5.4) we havejjf(y) f(x)jjeEjjy xjj.
Denote byDthe diameter of the compact set f(A), i.e.D:= max
x;y2Ajjf(y) f(x)jj. Then if for
the distance between two points x;y2Aiswe havejjf(y) f(x)jjD
jjy xjj. Finally, if we
denoteC:= max(eE;D
) we get
jjf(y) f(x)jjCjjy xjjfor anyx;y2A:
Volume and smooth maps
Lemma 9.20. LetAVbe a compact set of volume 0andf:V!Wa Lipshitz map, where
dimWdimV. Then Volf(A) = 0 .
Proof. According to Lemma 9.19 there is a constant C > 0 such thatjjf(y) f(x)jjCjjy xjj
for anyx;y2A. In particular, the image f(P) of a cube Pof sizeinVx,x2A, centered at x, is
contained in a cube of size KinWf(x)centered at f(x).
The volume 0 assumption implies that for any >0 there exists a partition of some size >0
of a larger cube containing Asuch that the total volume Nnof cubesP1;:::;PNintersecting A
122
is<. But thenNS
1f(Pj)f(A) while Volm(f(Pj))KMm, and hence
Volmf(A)NnKmm nKmm nto
!0!0:
Corollary 9.21. LetAVbe any compact set and f:A!WaC1-smooth map. Suppose that
n= dimV <m = dimW. Then Vol(f(A)) = 0 .
Indeed,fcan be extended to a smooth map de ned on a neighborhood of A0 inVR(e.g.
as independent of the new coordinate t2R). But Vol n+1(A0) = 0 and mn+ 1. Hence, the
required statement follows from Lemma 9.20.
Remark 9.22. The statement of Corollary 9.21 is wrong for continuous maps. For instance, there
exists a continuous map h: [0;1]!R2such thath([0;1]) is the squaref0x1;x11g. (This is
a famous Peano curve passing through every point of the square.)
Corollary 9.21 is a simplest special case of Sard's theorem which asserts that the set of critical
values of a suciently smooth map has volume 0. More precisely,
Proposition 9.23. (A. Sard, 1942) Given aCk-smooth map f:A!W(whereAis a compact
subset ofV,dimV=n;dimW=m) let us denote by
(f) :fx2A; rankdxf <mg:
Then ifkmax(n m+ 1;1)then Volm(f((f)) = 0 .
Ifm>n then (f) =A, and hence the statement is equivalent to Corollary 9.21.
Proof. We prove the proposition only for the case m=n.
To clarify the main idea we rst consider the case when n= 1 andA= [0;1], i.e.fis a
functionf:A!Rwith a continuous derivative f0. According to Cantor's theorem f0is uniformly
continuous and hence for any there exists >0 such that
jf0(x) f0(y)j<whenjx yj<: (9.5.5)
Let us take a partition of the interval of the size <. LetI1;:::;INbe the interval of the partition
which contain critical points, i.e. points where the derivative is 0. Then for any point cin on
123
of these intervals jf0(c)j< , and hence by the intermidiate value theorem for any two points
x;y2Ij; j= 1;:::;N we havejf(x) f(y)j=jf0(c)jjx yj<, i.e. the image f(Ij) is contained
in an interval of length . But total length of theintrvals Ijis1, and thus f()(f) is covered by
the union of intervals of the total length <. Hence Vol 1(f((f))) = 0.
Consider now the case of a general n. Again the C1-smoothness of fimplies that dxfis uniformly
continuous, i.e. for every >0 there exists >0 such that
jjdxf dyfjj<whenjjx yjj<: (9.5.6)
The inequalityjjdxf dyfjj<means that for any unit vector hinV
jjdxf(h) dyf(h)jj<; (9.5.7)
where we parallel transport the vector hto pointsxandy.
Consider a partition of a cube Qcontaining Aby cubes of size <
2pn, so that the ball sur-
rounding each of the cubes and centered at any point of the cube has radius <.
LetQ1;:::;QNbe the cubes of the partition intersecting ( f). The total volume 2nn n
2Nn
of these cubes is bounded by Vol P, wherePis a xed cube containing A.
Choose a point cj2Qj\(f) for eachj= 1;:::;N LetB1;:::;BNbe balls of radius centered
atcj. As we already pointed out, BjQjfor eachj= 1;:::;N .
The di erential dcjfis degenerate, and hence the image dcjf(Vcj)Wf(cj)is contained in a
codimension 1 subspace LjWf(cj). Let us choose a Cartesian coordinate system ( y1;:::;yn) in
Wf(cj)such thatLj=fyn= 0gwe can view yjas coordinates in Vwith the origin shifted to
f(cj). Let (f1;:::;fn) be the coordinate functions of fwith respect to these coordinates. Then
dcjfn= 0, i.e. the directional derivatives of the function fatcjat every direction are equal to 0.
But then according to inequality (9.5.7) the (absolute value of the) directional derivatives of fnat
any point of Bjare<. Hence, using our above 1-dimensional argument along each radius of Bj,
we conclude that jfn(x)j<, i.e. the image f(Bj) is contained in an -neighborhood Uof the
hyperplane Ljviewed as an ane hyperplane in V. We also recall that the map fis Lipshitz, and
hence the image f(Bj) is contained in a ball BK(f(cj))Wof radiusKcentered at f(cj) for
some constant K > 0. Hence,f(Bj)U\BK(f(cj)), sof(Bj) is contained in a rectangular Pj
with all sides equal to 2 Kand one side of size 2 . In particular, Vol( Pj) =Kn 12nn.
124
Therefore,f((f) can be covered by Nsuch rectangular of total volume
Kn 12nNn= (Kn 1nn
2VolP)!
!00:
Hence, Volf((f)) = 0.
Corollary 9.24. LetARnbe a measurable set and f:A!Rq, whereqn, be aC1-smooth
map. Then the image f(A)Rqis also measurable.
Proof. Ifq > n then Volqf(A) = 0 according to Corollary 9.21, and hence f(A) is measurable.
Suppose that q=n. Then for any interior point a2Asuch that rank daf=nthe imagef(a) is an
interior point of f(A) according to the implicit function theorem. Hence, the boundary @(f(A))
f(@A)[f((f)), where ( f) is the set critical points of f(i.e. points a2Awhere rankdaf <n ).
But Vol@A= 0 and hence, according to Lemma 9.20 Vol( f(@A)) = 0. On the other hand, Sard's
theorem 9.23 implies that Vol f((f)) = 0, and therefore Vol( @f(A)) = 0, which means that f(A)
is measurable.
Properties which hold almost everywhere
We say that some property holds almost everywhere (we will abbreviate a.e.) if it holds in the
complement of a set of volume 0. For instance, we say that a bounded function f:P!Ris almost
everywhere continuous (or a.e. continuous) if it is continuous in the complement of a set APof
volume 0. For instance, a characteristic function of any measurable set is a.e. continuous . Indeed,
it is constant away from the set @Awhich according to Proposition 9.16.4 has volume 0.
Proposition 9.25. Suppose that the bounded functions f;g:P!Rcoincide a.e. Then if fis
integrable, then so is gand we haveR
Pf=R
Pg.
Proof. DenoteA=fx2P:f(x)6=g(x)g. By our assumption, Vol A= 0. Hence, for any there
exists a > 0 such that for every partition Pwith(P)the unionBof all rectangles of
the partition which have non-empty intersection with Ahas volume < . The functions f;gare
bounded, i.e. there exists C > 0 such Cjf(x)j;jg(x)jCfor allx2P. Due to integrability of
fwe can choose small enough so that jU(f;P) L(f;P)jwhen(P). Then we have
jU(g;P) U(f;P)j= X
J:PJBsup
PJg sup
PJf) 2CVolB2C:
125
Similarly,jL(g;P) L(f;P)j2C, and hence
jU(g;P) L(g;P)jjU(g;P) U(f;P)j+jU(f;P) L(f;P)j+jL(f;P) L(g;P)j
+ 4C !
!00;
and hencegis integrable and
Z
Pg= lim
(P)!0U(g;P) = lim
(P)!0U(f;P) =Z
Pf:
Proposition 9.26. 1. Suppose that a function f:P!Ris a.e. continuous. Then fis inte-
grable.
2. LetAVbe compact and measurable, f:U!WaC1-smooth map de ned on a neighbor-
hoodUA. Suppose that dimW= dimV. Thenf(A)is measurable.
Proof. 1. Let us begin with a
Warning . One could think that in view of Proposition 9.25 it is sucient to consider only the case
when the function fis continuous. However, this is not the case, because for a given a.e. continuos
function one cannot, in general, nd a continuos function gwhich coincides with fa.e.
Let us proceed with the proof. Given a partition Pwe denote by JAthe set of multi-indices jsuch
that IntPj\A6=?, and byJAthe complementary set of multi-indices, i.e. for each j2JAwe have
Pj\A=?. Let us denote C:=S
j2JAPj. According to Proposition 9.16.3 for any >0 there exists
a partitionPsuch that Vol( C) =P
j2JAVol(Pj)< . By assumption the function fis continuous
over a compact set B=S
j2JAPj, and hence it is uniformly continuous over it. Thus there exists
>0 such thatjf(x) f(x0)j<provided that x;x02Bandjjx x0jj<. Thus we can further
subdivide our partition, so that for the new ner partition P0we have(P0)< . By assumption
the function fis bounded, i.e. there exists a constant K > 0 such that Mj(f) mj(f)<K for all
126
indices j. Then we have
U(f;P0) L(f;P0) =X
j(Mj(f) mj(f))Vol(Pj) =
X
j;PjB(Mj(f) mj(f))Vol(Pj) +X
j;PjC(Mj(f) mj(f))Vol(Pj)<
VolB+KVolC < (VolP+K):
Hence inf
PU(f;P) = sup
PL(f;P), i.e. the function fis integrable.
2. Ifxis an interior point of AanddetDf (x)6= 0 then the inverse function theorem implies
thatf(x)2Intf(A). DenoteC=fx2A; detDf(x) = 0g. Hence,@f(A)f(@A)[f(C).
But Vol(@A) = 0 because Ais measurable and Vol f(C) = 0 by Sard's theorem 9.23. Therefore,
Vol@f(A) = 0 and thus f(A) is measurable.
Orthogonal invariance of the volume and volume of a parallelepiped
The following lemma provides a way of computing the volume via packing by balls rather then
cubes. An admissible set of balls in Ais any nite set of disjoint balls B1;:::;BKA
Lemma 9.27. LetAbe a measurable set. Then VolAis the supremum of the total volume of
admissible sets of balls in A. Here the supremum is taken over all admissible sets of balls in A.
Proof. Let us denote this supremum by . The monotonicity of volume implies that VolA.
Suppose that <VolA . Let us denote by nthe volume of an n-dimensional ball of radius 1 (we
will compute this number later on). This ball is contained in a cube of volume 2n. It follows then
that the ratio of the volume of any ball to the volume of the cube to which it is inscribed is equal
ton
2n. Choose an <n
2n(VolA ). Then there exists a nite set of disjoint balls B1;:::;BKA
such that Vol(KS
1Bj)> . The volume of the complement C=AnKS
1Bjsatis es
VolC= VolA Vol K[
1Bj!
>VolA :
Hence there exists a partition PofPby cubes such that the total volume of cubes Q1;:::;QLcon-
tained inCis>VolA . Let us inscribe in each of the cubes Qja balleBj. ThenB1;:::;BK;eB1;:::;eBL
127
is an admissible set of balls in A. Indeed, all these balls are disjoint and contained in A. The total
volume of this admissible set is equal to
KX
1VolBj+LX
1VoleBi +n
2n(VolA )> ;
in view of our choice of , but this contradicts to our assumption < VolA. Hence, we have
= VolA.
Lemma 9.28. LetAVbe any measurable set in a Euclidean space V. Then for any linear
orthogonal transformation F:V!Vthe setF(A)is also measurable and we have Vol(F(A)) =
Vol(A).
Proof. First note that if Vol A= 0 then the claim follows from Lemma 9.20. Indeed, an orthogonal
transformation is, of course a smooth map.
Let nowAbe an arbitrary measurable set. Note that @F(A) =F(@A). Measurability of A
implies Vol( @A) = 0. Hence, as we just have explained, Vol( @F(A)) = Vol(F(@A)) = 0, and hence
F(A) is measurable. According to Lemma 9.27 the volume of a measurable set can be computed
as a supremum of the total volume of disjoint inscribed balls. But the orthogonal transformation
Fmoves disjoint balls to disjoint balls of the same size, and hence Vol A= VolF(A):
Next proposition shows that the volume of a parallelepiped can be computed by formula (3.3.1)
from Section 3.3.
Proposition 9.29. Letv1;:::;vn2Vbe linearly independent vectors. Then
VolP(v1;:::;vn) =jx1^^xn(v1:::;vn)j: (9.5.8)
Proof. The formula (9.5.8) holds for rectangles, i.e. when vj=cjejfor some non-zero numbers cj,
j= 1;:::n . Using Lemma 9.28 we conclude that it also holds for any orthogonal basis. Indeed, any
such basis can be moved by an orthogonal transformation to a basis of the above form cjej; j=
1;:::n . Lemma 9.28 ensures that the volume does not change under the orthogonal transformation,
while Proposition 2.17 implies the same about jx1^^xn(v1:::;vn)j.
The Gram-Schmidt orthogonalization process shows that one can pass from any basis to an
orthogonal basis by a sequence of following elementary operations:
128
- reordering of basic vectors, and
shears , i.e. an addition to the last vector a linear combination of the other ones:
v1;:::;vn 1;vn7!v1;:::;vn 1;vn+n 1X
1jvj:
Note that the reordering of vectors v1;:::;vnchanges neither Vol P(v1;:::;vn), nor the absolute
value
jx1^^xn(v1:::;vn)j. On the other hand, a shear does not change
x1^^xn(v1:::;vn):
It remains to be shown that a shear does not change the volume of a parallelepiped. We will consider
here only the case n= 2 and will leave to the reader the extension of the argument to the general
case.
Letv1;v2be two orthogonal vectors in R2. We can assume that v1= (a;0),v2= (0;b) for
a;b > 0, because we already proved the invariance of volume under orthogonal transformations.
Letv0
2=v2+v1= (a0;b), wherea0=a+b. Let us partition the rectangle P=P(v1;v2) intoN2
smaller rectangles Pi;j,i;j= 0;:::;N 1, of equal size. We number the rectangles in such a way
that the rst index corresponds to the rst coordinate, so that the rectangles P00;:::;PN 1;0form
the lower layer, P01;:::;PN 1;1the second layer, etc. Let us now shift the rectangles in k-th layer
horizontally by the vector (kb
N;0). Then the total volume of the rectangles, denoted ePijremains
the same, while when N!1 the volume of part of the parallelogram P(v1;v0
2) that is not covered
by rectangles ePi;j,i;j= 0;:::;N 1 converges to 0.
9.6 Fubini's Theorem
Let us consider Rnas a direct product of RkandRn kfor somek= 1;:::;n 1. We will denote
coordinates in Rkbyx= (x1;:::;xk) and coordinates in Rn kbyy= (y1;:::;yn k), so the
coordinates in Rnare denoted by ( x1;:::;xk;y1;:::;yn k):Given rectangles P1RkandP2
Rn ktheir product P=P1P2is a rectangle in Rn.
129
Figure 9.2: Guido Fubini (1879-1943)
The following theorem provides us with a basic tool for computing multiple integrals.
Theorem 9.30 (Guido Fubini) .Suppose that a function f:P!Ris integrable over P. Given a
pointx2P1let us de ne a function fx:P2!Rby the formula fx(y) =f(x;y); y2P2. Then
Z
PfdVn=Z
P10
B@Z
P2fxdVn k1
CAdVk=Z
P10
@Z
P2fxdVn k1
AdVk:
In particular, if the function fxis integrable for all (or almost all) x2P1then one has
Z
PfdVn=Z
P10
@Z
P2fxdVn k1
AdVk:
Here by writing dVk;dVn kanddVnwe emphasize the integration with respect to the k-, (n k)-
andn-dimensional volumes, respectively.
Proof. Choose any partition P1ofP1andP2ofP2. We will denote elements of the partition P1
byPj
1and elements of the partition P2byPi
2. Then products of Pj;i=Pj
1Pi
2form a partitionP
ofP=P1P2. Let us denote
I(x) :=Z
P2fx; I(x) :=Z
P2fx; x2P1:
Let us show that
L(f;P)L(I;P1)U(I;P1)U(f;P): (9.6.1)
130
Indeed, we have
L(f;P) =X
jX
imj;i(f)VolnPj;i:
Here the rst sum is taken over all multi-indices jof the partitionP1, and the second sum is taken
over all multi-indices iof the partitionP2. On the other hand,
L(I;P1) =X
jinf
x2Pj
10
B@Z
P2fxdVn k1
CAVolkPj
1:
Note that for every x2Pj
1we have
Z
P2fxdVn kL(fx;P2) =X
imi(fx)Voln k(Pi
2)X
imi;j(f)Voln k(Pi
2);
and hence
inf
x2Pj
1Z
P2fxdVn kX
imi;j(f)Voln k(Pi
2):
Therefore,
L(I;P1)X
jX
imi;j(f)Voln k(Pi
2)Volk(Pj
1) =X
jX
imj;i(f)Voln(Pj;i) =L(f;P):
Similarly, one can check that U(I;P1)U(f;P). Together with an obvious inequality L(I;P1)
U(I;P1) this completes the proof of (9.6.1). Thus we have
max(U(I;P1) L(I;P1);U(I;P1) L(I;P1))U(I;P1) L(I;P1)U(f;P) L(f;P):
By assumption for appropriate choices of partitions, the right-hand side can be made <for any
a priori given >0. This implies the integrability of the function I(x) andI(x) overP1. But then
we can writeZ
P1I(x)dVn k= lim
(P1)!0L(I;P1)
andZ
P1I(x)dVn k= lim
(P1)!0U(I;P1):
131
We also have
lim
(P)!0L(f;P) = lim
(P)!0U(f;P) =Z
PfdVn:
Hence, the inequality (9.6.1) implies that
Z
PfdVn=Z
P10
B@Z
P2fxdVn k1
CAdVk=Z
P10
@Z
P2fxdVn k1
AdVk:
Corollary 9.31. Supposef:P!Ris a continuous function. Then
Z
Pf=Z
P1Z
P2fx=Z
P2Z
P1fy:
Thus if we switch back to the notation x1;:::;xnfor coordinates in Rn, and ifP=fa1x1
b1;:::;anxnbngthen we can write
Z
Pf=bnZ
an0
@:::0
@b1Z
a1f(x1;:::;xn)dx11
A:::1
Adxn: (9.6.2)
The integral in the right-hand side of (9.6.2) is called an iterated integral . Note that the order
of integration is irrelevant there. In particular, for continuous functions one can change the order
of integration in the iterated integrals.
9.7 Integration of n-forms over domains in n-dimensional space
Di erential forms are much better suited to be integrated than functions. For integrating a function,
one needs a measure. To integrate a di erential form, one needs nothing except an orientation of
the domain of integration.
Let us start with the integration of a n-form over a domain in a n-dimensional space. Let !be
an-form on a domain UV;dimV=n.
Let us x now an orientation of the space V. Pick any coordinate system ( x1:::xn) that agrees
with the chosen orientation.
132
We proceed similar to the way we de ned an integral of a function. Let us x a rectangle
P=P(a1;b1;a2;b2;:::;an;bn) =faixibi;i= 1;:::;ng. Choose its partition PbyNn
smaller rectangles Pj=ft1
jnx1t1
jn+1; :::; tn
j1x1tn
j1+1g, where j= (j1;:::;jn) and each
indexjktakes values between 0 and N 1. Let us x a point cj2Pjand denote by Cthe set of all
suchcj. We also denote by tjthe point with coordinates t1
j1;:::;tn
jnand byTj;m2Vcj,m= 1;:::;n
the vectortj+1m tj, parallel-transported to the point cj. Here we use the notation j+ 1mfor the
multi-index j1;:::;jm 1;jm+ 1;jm+1;:::;jn. Thus the vector Tj;mis parallel to the m-th basic
vector and has the length jtjm+1 tjmj.
Given a di erential n-form onPwe form an integral sum
I( ;P;C) =X
j (Tj
1;Tj
2;:::;T j;n); (9.7.1)
where the sum is taken over all elements of the partition. We call an n-form integrable if there
exists a limit lim
(P)!0I( ;P;C) which we denote byR
P and call the integral of overP.Note
that if =f(x)dx1^^dxnthen the integral sum I( ;P;C) from (9.7.1) coincides with the
integral sum I(f;P;C) from (9.5.2) for the function f. Thus the integrability of is the same as
integrability of fand we have
Z
Pf(x)dx1^^dxn=Z
PfdV: (9.7.2)
Note, however, that the equality (9.7.2) holds only if the coordinate system (x1;:::;xn)de nes the
given orientation of the space V. The integralR
Pf(x)dx1^^dxnchanges its sign with a change
of the orientation while the integralR
PfdV is not sensitive to the orientation of the space V.
It is not clear from the above de nition whether the integral of a di erential form depends on
our choice of the coordinate system. It turns out that it does not, as the following theorem, which
is the main result of this section, shows. Moreover, we will see that one even can use arbitrarty
curvilinear coordinates.
In what follows we use the convention introduced at the end of Section 6.1. Namely by a
di eomorphism between two closed subsets of vector spaces we mean a di eomorphism between
their neighborhoods.
133
Theorem 9.32. LetA;BRnbe two measurable compact subsets. Let f:A!Bbe an orientation
preserving di eomorphism. Let be a di erential n-form de ned on B. Then ifis integrable over
Bthenf is integrable over Aand we have
Z
Af=Z
B: (9.7.3)
For an orientation reversing di eomorphism fwe haveR
Af= R
B:
Let =g(x)dx1^^dxn. Thenf =gfdetDfdx 1^^dxn, and hence the formula
(9.7.3) can be rewritten as
Z
Pg(x1;:::;xn)dx1^^dxn=Z
PgfdetDfdx 1^^dxn:
Here
detDf= @f1
@x1:::@f1
@xn
::: ::: :::
@fn
@x1:::@fn
@xn
is the determinant of the Jacobian matrix of f= (f1;:::;fn):
Hence, in view of formula (9.7.2) we get the following change of variables formula for multiple
integrals of functions.
Corollary 9.33. [Change of variables in a multiple integral] Letg:B!Rbe an integrable
function and f:A!Ba di eomorphism. Then the function gfis also integrable and
Z
BgdV =Z
AgfjdetDfjdV : (9.7.4)
We begin the proof with the following special case of Theorem 9.32.
Proposition 9.34. The statement of 9.32 holds when =dx1^^dxnand the set Ais the unit
cubeI=In. In other words,
Volf(I) = Z
Af :
134
Figure 9.3: Image of a cube under a di eomorphism and its linearization
We will use below the following notation.
For any set AVand any positive number >0 we denote by Athe setfx;x2Ag. For
any linear operator F:V!Wbetween two Euclidean spaces VandWwe de ne its normjjFjj
by the formula
jjFjj= max
jjvjj=1jjF(v)jj= max
v2V;v6=0jjF(v)jj
jjvjj:
Equivalently, we can de ne jjFjjas follows. The linear map Fmaps the unit sphere in the space V
onto an ellipsoid in the space W. ThenjjFjjis the biggest semi-axis of this ellipsoid.
Let us begin by observing the following geometric fact:
Lemma 9.35. LetI=fjxij1
2;i= 1;:::;ngRnbe the unit cube centered at 0andF:Rn!Rn
a non-degenerate linear map. Take any 2(0;1)and set=
jjF 1jj. Then for any boundary point
z2@Iwe have
B(F(z))(1 +)F(I)n(1 )F(I); (9.7.5)
135
see Fig. 9.7
Proof. Inclusion (9.7.5) can be rewritten as
F 1(B(F(z))(1 +)In(1 )I:
But the set F 1(B(F(z)) is an ellipsoid centered at zwhose greatest semi-axis is equal to jjF 1jj:
Hence, ifjjF 1jjthenF 1(B(F(z))(1 +)In(1 )I.
Recall that we denote by Ithe cubeIscaled with the coecient , i.e.I=fjxij
2;i=
1;:::;ngRn. We will also need
Lemma 9.36. LetURnis an open set, f:U!Rnsuch thatf(0) = 0 . Suppose that fis
di erentiable at 0and its di erential F=d0f:Rn!Rnat0is non-degenerate. Then for any
2(0;1)there exists >0such that
(1 )F(I)f(I)(1 +)F(I); (9.7.6)
see Fig. 9.7.
Proof. First, we note that inclusion (9.7.5) implies, using linearity of F, that for any >0 we
have
B(F(z))(1 +)F(I)n(1 )F(I); (9.7.7)
wherez2@(I) and, as in Lemma 9.35, we assume that =jjF 1jj.
According to the de nition of di erentiability we have
f(h) =F(h) +o(jjhjj):
Denotee:=pn=pnjjF 1jj. There exists >0 such that ifjjhjjthen
jjf(h) F(h)jjejjhjje:
Denote:=pn. ThenIB(0), and hencejjf(z) F(z)jjefor anyz2I. In particular,
for any point z2@(I) we have
f(z)2Be(F(z)) =Bpne(F(z)) =B(F(z));
136
and therefore in view of (9.7.7)
f(@(I))(1 +)F(I)n(1 )F(I):
But this is equivalent to inclusion (9.7.6).
Lemma 9.37. LetF:Rn!Rnbe a non-degenerate orientation preserving linear map, P=
P(v1;:::;vn)a parallelepiped, and =dx1^^dxn. ThenR
F(P)=R
PF. Here we assume that
the orientation of PandF(P)are given by the orientation of Rn.
Proof. We haveR
F(P)=R
F(P)dx1^:::dxn= VolF(P) = (detF)VolP. On the other hand, F=
detF, and henceR
PF= detFR
P= (detF)VolP:
Proof of Proposition 9.34. We havef= (detDf)dx1^^dxn, and hence the form fis
integrable because fisC1-smooth, and hence det Dfis continuous.
Choose a partition Pof the cube IbyNnsmall cubes IK,K= 1;:::;Nn, of the same size1
N.
LetcK2IKbe the center of the cube IK. Then
Z
If=NnX
K=1Z
IKf:
Note that in view of the uniform continuity of the function det Df, for any >0 the number N
can be chosen so large that jdetDf(x) detDf(x0)j< for any two points x;x02IKand any
K= 1;:::;Nn. LetKbe the form det Df(ck)dx1^^dxnonIK. Then
Z
IKf Z
IKK Z
IKjdetDf(x) detDf(cK)jdx1^^dxnVol(IK) =
Nn:
Thus Z
If NnX
K=1Z
IKK : (9.7.8)
Next, let us analyze the integralR
IKK. DenoteFK:=dcK(f). We can assume without loss of
generality that cK= 0, andf(cK) = 0, and hence FKcan be viewed just as a linear map Rn!Rn.
Using Lemma 9.36 we have for a suciently large N
(1 )FK(IK)f(IK)!(1 +)FK(IK):
137
Again in view of compactness of Ithe number can be chosen the same for all cubes IK. Hence
(1 )nVol(FK(IK))Volf(IK)(1 +)nVol(FK(IK)): (9.7.9)
Note that Vol f(IK) =R
f(IK);and hence summing up inequality (9.7.9) over Kwe get
(1 )nNnX
K=1Vol(FK(IK))NnX
K=1Volf(IK) =NnX
K=1Z
f(IK)=Z
f(I)(1 +)nNnX
K=1Vol(FK(IK)):
(9.7.10)
Note thatK=F
Kand by Lemma 9.37 we haveR
IKK=R
IKF
K=R
FK(IK)= Vol(FK(IK)):
Hence, it follows from (9.7.10) that
(1 )nNnX
K=1Z
IKKZ
f(I)(1 +)nNnX
K=1Z
IKK: (9.7.11)
Recall that from (9.7.8) we have
Z
If NnX
K=1Z
IKZ
If+:
Combining with (9.7.11) we get
(1 )n0
@Z
If 1
AZ
f(I)(1 +)n0
@Z
If+1
A: (9.7.12)
Passing to the limit when !0 we get
Z
IfZ
f(I)Z
If; (9.7.13)
i.e.R
If=R
f(I):
Corollary 9.38. The statement of Theorem 9.32 holds for =dx1^^dxnand an arbitrary
measurable A.
Proof. Suppose for determinacy that fpreserves the orientation. By assumption the di eomor-
phismAextends to an open neighborhood UA. Consider a cube containing PA. Choose
138
a >0 and consider a partition PofPby small cube of size < . Denote by A+
the union of
elements of the partition which intersect A, and byA
the union of elements which are completely
insideA. Ifis small enough then A+
U. We have
Z
A fZ
AfZ
A+
f
and
Z
A+
f Z
A f!
!00:
On the other hand, each of the sets A+
is a union of cubes. Hence, Proposition 9.34 implies that
Z
A
f=Z
f(A
)= Vol(A
):
Butf(A
)f(A)f(A+
), and hence Vol( f(A
))Volf(A)Vol(f(A+
)) which implies when
!0 that Volf(A) =R
f(A)=R
Af:
Proof of Theorem 9.32. Let us recall that the di eomorphism fis de ned as a di eomorphism
between open neighborhoods UAandU0B. We also assume that the form is extended to U0
as equal to 0 outside B. The form can be written as hdx 1^^dxn. Let us take a partition Pof
a rectangular containg U0by cubesIjof the same size . Consider forms +
j:=Mj(h)dx1^^dxn
and
j:=mj(h)dx1^^dxnonIj, wheremj(h) = infIjh; M j(h) = supIj(h):Letbe the
form onU0equal to
jon each cube Ij. The assumption of integrability of overBguarantees
that for any >0 ifis chosen small enough we haveR
B+ R
B. The forms fare a.e.
continuous, and hence integrable over Aand we haveR
Af R
AfR
Af+. Hence, if we prove
thatR
Af=R
Bthen this will imply that is integrable andR
A=R
B.
On the other hand,R
B=P
jR
Ij
jandR
Af=P
jR
Bjf
j, whereBj=f 1(Ij). But according
to Corollary ??we haveR
Bjf
j=R
Ij
j, and henceR
Af=R
B.
139
9.8 Manifolds and submanifolds
9.8.1 Manifolds
Manifolds of dimension nare spaces which are locally look like open subsets of Rnbut globally
could be much more complicated. We give a precise de nition below.
LetU;U0Rnbe open sets. A map f:U!U0is called a homeomorpism if it is continuous
one-to-one map which has a continuous inverse f 1:U0!U.
A mapf:U!U0is called aCk-di eomorpism ,k= 1;:::;1;if it isCk-smooth, one-to-one
map which has a Ck-smooth inverse f 1:U0!U. Usually we will omit the reference to the class
of smoothness, and just call fadi eomorphism , unless it will be important to emphasize the class
of smoothness.
A setMis called an n-dimensional Ck-smooth (resp. topological) manifold if there exist subsets
UX,2, where is a nite or countable set of indices, and for every 2 a map
:U!Rnsuch that
M1.M=S
2U.
M2. The image G= (U) is an open set in Rn.
M3. The map viewed as a map U!Gis one-to-one.
M4. For any two sets U;U;;2 the images (U\U); (U\U)Rnare open and
the map
h:= 1
: (U\U)!(U\U)Rn
is aCk-di eomorphism (resp. homeomorphism).
SetsUare called coordinate neighborhoods and maps :U!Rnare called coordinate maps .
The pairs ( U;) are also called local coordinate charts . The maps hare called transiton maps
between di erent coordinate charts. The inverse maps = 1
:G!Uare called (local)
parameterization maps . An atlas is a collection A=fU;g2of all coordinate charts.
One says that two atlases A=fU;g2andA0=fU0
;0
g
2 on the same manifold
Xareequivalent , or that they de ne the same smooth structure on Xif their union A[A0=
140
f(U;);(U0
;0
)g2;
2 is again an atlas on X. In other words, two atlases de ne the same
smooth structure if transition maps from local coordinates in one of the atlases to the local coor-
dinates in the other one are given by smooth functions.
A subsetGMis called open if for every 2 the image (G\U)Rnis open. In
particular, coordinate charts Uthemselves are open, and we can equivalently say that a set Gis
open if its intersection with every coordinate chart is open. By a neighborhood of a pointa2Mwe
will mean any open subset UMsuch thata2U.
Given two smooth manifolds MandfMof dimension mandnthen a map f:M!fMis
called continuous if if for every point a2Mthere exist local coordinate charts ( U;) inMand
(eU;e) infM, such that a2U;f(U)eUand the composition map
G= (U) !Uf!eUe!Rn
is continuous.
Similarly, for k= 1;:::;1a mapf:M!fMis calledCk-smooth if for every point a2M
there exist local coordinate charts ( U;) inMand (eU;e) infM, such thata2U;f(U)eU
and the composition map
G= (U) !Uf!eUe!Rn
isCk-smooth. In other words, a map is continuous or smooth, if it is continuous or smooth when
expressed in local coordinates.
A mapf:M!Nis called a di eomorphism if it is smooth, one-to-one, and the inverse map
is also smooth. One-to-one continuous maps with continuous inverses are called homeomorphisms .
Note that in view of the chain rule the Ck-smoothness is independent of the choice of local
coordinate charts ( U;) and (eU;e). Note that for Ck-smooth manifolds one can talk only
aboutCl-smooth maps for lk. For topological manifolds one can talk only about continuous
maps.
If one replaces condition M2 in the de nition of a manifold by
M2b. The image G= (U) is either an open set in Rnor an intersection of an open set in Rn
withRn
+=fx10g
141
then one gets a de nition of a manifold with boundary .
A slightly awkward nuance in the above de nition is that a manifold with boundary is not a
manifold! It would be, probably, less confusing to write this as a 1 word manifold-with-boundary ,
but of course nobody does that.
The points of a manifold Mwith boundary which are mapped by coordinate maps to points
inRn 1=@Rn
+are called the boundary points ofM. The set of boundary points is called the
boundary ofMand denoted by @M. It is itself a manifold of dimension n 1.
Note that any (interior) point aof ann-dimensional manifold Mhas a neighborhood Bdi eo-
morphic to an open ball B1(0)Rn, while any boundary point has a neighborhood di eomorphic
to a semi-ball B1(0)\fx10gRn.
Exercise 9.39. Prove that a boundary point does not have a neighborhood di eomorphic to an
open ball. In other words, the notion of boundary and interior point of a manifold with boundary
are well de ned.
Next we want to introduce a notion of compactness for subsets in a manifold. Let us recall that
for subsets in a Euclidean vector space we introduced three equivalent de nition of compactness,
see Section 6.1. The rst de nition, COMP1 is unapplicable because we cannot talk about bounded
sets in a manifold. However, de nitions COMP2 and COMP3 make perfect sense in an arbitrary
manifold. For instance, we can say that a subset AMis compact if from any in nite sequence
of points in Aone can choose a subsequence converging to a point in A.
A compact manifold (without boundary) is called closed . Note that the word closed is used
here in a di erent sense than a closed set. For instance, a closed interval is not a closed manifold
because it has a boundary. An open interval or a real line Ris not a closed manifold because it is
not compact. On the other hand, a circle, or a sphere Snof any dimension nis a closed manifold.
The notions of connected and path connected subsets of a manifold are de ned in the same way
as in an Euclidean space.
9.8.2 Gluing construction
The construction which is described in this section is called gluing orquotient construction. It
provides a rich source of examples of manifolds. We discuss here only very special cases of this
142
construction.
a) LetMbe a manifold and U;U0its two open disjoint subsets. Let us moreover assume that
each pointx2Mhas a neighborhood which does not intersect at least one of the sets UandU0.4
Consider a di eomorphism f:U!U0.
Let us denote by M=ff(x)xgthe set obtained from Mby identifying each point x2Uwith
its imagef(x)2U0. In other words, a point of M=ff(x)xgis either a point from x2Mn(U[U0),
or a pair of points ( x;f(x)), wherex2U. Note that there exists a canonical projection :M!
M=ff(x)xg. Namely(x) =xifx =2U[U0,(x) = (x;f(x)) ifx2Uand(x) = (f 1(x);x)
ifx2U0. By our assumption each point x2Mhas a coordinate neighborhood Gx3xsuch that
f(Gx\U)\Gx=?. In particular, the projection jGx:Gx!eGx=(Gx) is one-to-one. We will
declare by de nition that eGxis a coordinate neighborhood of (x)2M=ff(x)xgand de ne a
coordinate map e :eGx!Rnby the formula e = 1. It is not dicult to check that this
construction de ne a structure of an n-dimensional manifold on the set M=ff(x)xg. We will
call the resulted manifold the quotient manifold ofM, or say that M=ff(x)xgis obtained from
Mby gluingUwithU0with the di eomorphism f.
Though the described above gluing construction always produce a manifold, the result could be
quite pathological, if no additional care is taken. Here is an example of such pathology.
Example 9.40. LetM=I[I0be the union of two disjoint open intervals I= (0;2) andI0= (3;5).
ThenMis a 1-dimensional manifold. Denote U:= (0;1)I;U0:= (3;4)I0. Consider a
di eomorphism f:U!U0given by the formula f(t) =t+ 3; t2U. LetfM=M=ff(x)xgbe
the corresponding quotient manifold. In other words, fMis the result of gluing the intervals Iand
I0along their open sub-intervals UandU0. Note that the points 1 2Iand 42I0are not identi ed,
but 1 ;4 are identi ed for an arbitrary small >0. This means that any neighborhood of 1
and any neighborhood of 4 have non-empty intersection.
In order to avoid such pathological examples one usually (but not always) requires that manifolds
satisfy an additional axiom, called Hausdor property :
M5. Any two distinct points x;y2Mhave non-intersecting neighborhoods U3x;G3y:
4Here is an example when this condition is not satis ed :M= (0;2);U= (0;1);U0= (1;2). In this case any
neighborhood of the point 1 intersect both sets, UandU0.
143
Figure 9.4: Felix Hausdor (1868-1942)
In what follows we always assume that the manifolds satisfy the Hausdor property M5.
Let us make the following general remark about di eomorphisms f: (a;b)!(c;d) between
two open intervals. Such di eomorphism is simply a di erentiable function whose derivative never
vanishes and whose range is equal to the interval ( c;d). If derivative is positive then the di eomor-
phism is orientation preserving, and it is orientation reversing otherwise. The function falways
extends to a continuous (but necessarily di erentiable function f: [a;b]![c;d] such that f(a) =c
andf(b) =din the orientation preserving case, and f(a) =df(b) =cin the orientation reversing
case.
Lemma 9.41. Givena;b;a0b02(0;1)such thata<b anda0<b0consider an orientation preserving
di eomorphisms f: (0;a)!(0;a0)and(b;1)!(b01). Then for any ea2(0;a)andeb2(b;1)there
exists a di eomorphism F: (0;1)!(0;1)which coincides with fon(0;ea)and coincides with gon
(eb;1).
Proof. Choose real numbers c;ec;ed;dsuch thata < c <ec <ed < d < b . Consider a cut-o C1-
function: (0;1)!(0;1) which is equal to 1 on (0 ;ea][[ec;ed][[eb;1) and equal to 0 on [ a;c][[d;b].
For positive numbers >0 andC > 0 (which we will choose later) consider a function h;Con
144
(0;1) de ned by the formula
h;C(x) =8
>>>>>>>>><
>>>>>>>>>:(x)f0(x) + (1 (x)); x2(0;a);
; x 2[a;c][[d;b];
C(x) + (1 (x)); x2(c;d);
(x)g0(x) + (1 (x)); x2(b;1):
Note thath;C(x) =f0(x) on (0;ea],h;C(x) =g0(x) on [eb;1) and equal to Con [ec;ed]. De ne the
functionF;C: (0;1)!(0;1) by the formula
F;C(x) =xZ
0h(u)du:
Note that the derivative F0
;Cis positive, and hence the function F;Cis strictly increasing. It
coincides with fon (0;ea] and coincides up to a constant withgon (eb;1). Note that when and
Care small we have F;C(eb)< b0< g(eb), and lim
C!1F;C(eb) =1. Hence, by continuity one can
choose;C > 0 in such a way that F;C(eb) =g(eb). Then the function F=F;Cis a di eomorphism
(0;1)!(0;1) with the required properties.
Lemma 9.42. Suppose that a 1-dimensional manifold M(which satis es the Hausdor axiom M5)
is covered by two coordinate charts, M=U[U0, with coordinate maps :U!(0;1);0:U0!
(0;1)such that (U\U0) = (0;a)[(b;1);0(U\U0) = (0;a0)[(b01)for somea;a0;b;b02(0;1)
witha<b;a0<b0. ThenMis di eomorphic to the circle S1.
Proof. Denote by and 0the parameterization maps 1and (0) 1, and setG:= (U\U0)
andG0:= 0(U\U0). Leth= 0 :G!G0be the transition di eomorphism. There could
be two cases: h((0;a)) = (0;a0);h((b;1)) = (b0;1) andh((0;a)) = (b0;1);h((b;1)) = (0;a). We will
analyze the rst case. The second one is similar.
Lethbe the continuous extension of hto [0;a][[b;1]. We claim that h(0) =a0,h(a) = 0,
h(b) = 1 andh(1) =b0. Indeed, assuming otherwise we come to a contradiction with the Hausdor
property M5. Indeed, suppose h(a) =a0. Note the points A:= (a);A0:= 0(a0)2Mare disjoint.
On the other hand, for any neighborhood
3Aits image (
)Icontains an interval ( a ;a),
145
and similarly for any neighborhood
03A0its image 0(
)Icontains an interval ( a0 ;a0). for
a suciently small . Buth((a ;a)) = (a0 0;a0) for some0>0 and hence
0((a0;a0 0)) = 0h((a;a )) = 00((a;a )) = ((a;a ))
;
i.e.
\
06=?. In other words, any neighborhoods of the distinct points A;A02Mintersect , which
violates axiom M5. Similarly we can check that h(b) =b0.
Now take the unit circle S1R2and consider the polar coordinate onS1. Let us de ne a
mapg0:U0!S1by the formula = 0(x). Thusg0is a di eomorphism of U0onto an arc
ofS1given in polar coordinates by < < 0. The points A0= 0(a0) andB0= 0(b0) are
mapped to points with polar coordinates = a0and = b0. On the intersection U\U0
we can describe the map g0in terms of the coordinate in U. Thus we get a map f:=g0 :
(0;a)[(b;1)!S1. We have f(0) =g0(A0);f(a) =g0(0);f(1) =g0(B0);f(b) =g0(1). Here we
denoted by fthe continuous extension of fto [0;a][[b;1]. Thusf((0;a)) =f a0< < 0gand
f((b;1) =f< < 3 b0g. Note that the di eomorphism fis orientation preserving assuming
that the circle is oriented counter-clockwise. Using Lemma 9.41 we can nd a di eomorphism F
from (0;1) to the arcf a0< < 3 b0gS1which coincides with fon (0;ea)[(eb;1) for
anyea2(0;a) andeb2(b;1). Denoteea0:=h(ea);eb0=h(eb). Notice that the neighborhoods Uand
eU0= 0((ea;eb)) coverM. Hence, the required di eomorphism eF:M!S1we can de ne by the
formula
eF(x) =8
><
>:g(x); x2eU0;
F(x); x2U:
Similarly (and even simpler), one can prove
Lemma 9.43. Suppose that a 1-dimensional manifold M(which satis es the Hausdor axiom M5)
is covered by two coordinate charts, M=U[U0, with coordinate maps :U!(0;1);0:U0!
(0;1)such thatU\U0is connected. Then Mis di eomorphic to the open interval (0;1).
Theorem 9.44. Any (Hausdor ) connected closed 1-dimensional manifold is di eomorphic to the
circleS1.
146
Exercise 9.45. Show that the statement of the above theorem is not true without the axiom M5,
i.e. the assumption that the manifold has the Hausdor property.
Proof. LetMbe a connected closed 1-dimensional manifold. Each point x2Mhas a coordinate
neighborhood Uxdi eomorphic to an open interval. All open intervals are di eomorphic, and hence
we can assume that each neighborhood Gxis parameterized by the interval I= (0;1). Let x:I!
Gxbe the corresponding parameterization map. We haveS
x2MUx=M, and due to compactness of
Mwe can choose nitely many Ux1;:::;Uxksuch thatkS
i=1Uxi=M. We can further assume that
none of these neighborhoods is completrely contained inside another one. Denote U1:=Ux1; 1:=
x1. Note that U1\kS
2Uxk6=?. Indeed, if this were the case then due to connectedness of Mwe
would havekS
2Uxi=?and henceM=U1, but this is impossible because Mis compact. Thus,
there exists i= 2;:::;k such thatUxi\U16=?. We setU2:=Uxi; 2= xi. Consider open sets
G1;2:= 1
1(U1\U2);G2;1= 1
2(U1\U2)I:The transition map h1;2:= 1
2 1jG1;2:G1;2!
G2;1is a di eomorphism.
Let us show that the set G1;2(and hence G2;1) cannot contain more that two connected com-
ponents. Indeed, in that case one of the components of G1;2has to be a subinterval I0= (a;b)
I= (0;1) where 0 < a < b < 1. DenoteI00:=h1;2(I0). Then at least of of the boundary values
of the transition di eomorphism h1;2jI0, sayh1;2(a), which is one of the end points of I00, has to
be an interior point c2I= (0;1). We will assume for determinacy that I00= (c;d)I. But this
contradicts the Hausdor property M5. The argument repeats a similar argument in the proof of
Lemma 9.42.
Indeed, note that 1(a)6= 2(c). Indeed, 1(a) belongs to U1nU2and 2(c) is inU2nU1.
Take any neighborhood
3 1(a) inM. Then 1(
) is an open subset of Iwhich contains the
pointa. Hence 1((a;a+))
, and similarly, for any neighborhood
03 2(c) inMwe have
2((c;c+))
0for a suciently small >0. But 1((a;a+)) = 2(h1;2((a;a))) = 2(c;c+0);
wherec+0=h1;2(a+). Hence
\
06=?, i.e. any two neighborhoods of two distict points 1(a)
and 2(c) have a non-empty intersection, which violates the Hausdor axiom M5.
IfG1;2(0;1) consists of two components then the above argument shows that each of these
components must be adjacent to one of the ends of the interval I, and the same is true about the
147
components of the set G2;1I. Hence, we can apply Lemma 9.42 to conclude that the union U1[U2
is di eomorphic to S1. We also notice that in this case all the remaining neighborhoods Uxjmust
contain inU1[U2. Indeed, each Uxiwhich intersects the circle U1[U2must be completely contained
in it, because otherwise we would again get a contradiction with the Hausdor property. Hence,
we can eleiminate all neighborhoods which intersect U1[U2. But then no other neighborhoods
could be left because otherwise we would have M= (U1[U2)[S
Uxj\(U1[U2)=?Uxj, i.e. the manifold
Mcould be presented as a union of two disjoint non-empty open sets which is impossible due to
connectedness of M. Thus we conclude that in this case M=U1[U2is di eomorphic to S1.
Finally in the case when G1;2consists of 1 component, i.e. when it is connected, one can Use
Lemma 9.43 to show that U1[U2is di eomorphic to an open interval. Hence, we get a covering of
Mbyk 1 neighborhood di eomorphic to S1. Continuing inductively this process we will either
nd at some step two neighborhoods which intersect each other along two components, or continue
to reduce the number of neighborhoods. However, at some moment the rst situation should occur
because otherwise we would get that Mis di eomorphic to an interval which is impossible because
by assumption Mis compact.
b) LetMbe a manifold, f:M!Mbe a di eomorphism. Suppose that fsatis es the fol-
lowing property: There exists a positive integer psuch that for any point x2Mwe havefp(x) =
fff|{z}
p(x) =x, but the points x;f(x):::;fp 1(x)are all disjoint. The setfx;f(x):::;fp 1(x)g
Mis called the trajectory of the point xunder the action of f. It is clear that trajectory of two
di erent points either coincide or disjoint. Then one can consider the quotient space X=f, whose
points are trajectories of points of Munder the action of f. Similarly to how it was done in a) one
can de ne on M=f a structure of an n-dimensional manifold.
c) Here is a version of the construction in a) for the case when trajectory of points are in nite.
Letf:M!Mbe a di eomorphism which satis es the following property: for each point x2M
there exists a neighborhood Ux3xsuch that all sets
:::;f 2(Ux);f 1(Ux);Ux;f(Ux);f2(Ux);:::
are mutually disjoint. In this case the trajectoryf:::;f 2(x);f 1(x);x;f(x);f2(x);:::gof each
point is in nite. As in the case b), the trajectories of two di erent points either coincide or disjoint.
148
The setM=f of all trajectories can again be endowed with a structure of a manifold of the same
dimension as M.
9.8.3 Examples of manifolds
1.n-dimensional sphere Sn.Consider the unit sphere Sn=fjjxjj=s
n+1P
1x2
j= 1gRn+1. Let
introduce on Snthe structure of an n-dimensional manifold. Let N= (0;:::; 1) andS= (0;:::; 1)
be the North and South poles of Sn, respectively.
DenoteU =SnnS;U +=SnnNand consider the maps p:U!Rngiven by the formula
p(x1;:::;xn+1) =1
1xn+1(x1;:::;xn): (9.8.1)
The mapsp+:U+!Rnandp :U !Rnare called stereographic projections from the North
and South poles, respectively. It is easy to see that stereographic projections are one-to one maps.
Note thatU+\U =SnnfS;Ngand both images, p+(U+\U ) andp (U+\U ) coincide with
Rnn0. The map p p 1
+:Rnn0!Rnn0 is given by the formula
p p 1
+(x) =x
jjxjj2; (9.8.2)
and therefore it is a di eomorphism Rnn0!Rnn0.
Thus, the atlas which consists of two coordinate charts ( U+;p+) and (U ;p ) de nes on Sna
structure of an n-dimensional manifold. One can equivalently de nes the manifold Snas follows.
Take two disjoint copies of Rn, let call them Rn
1andRn
2. DenoteM=Rn
1[Rn
2,U=Rn
1n0 and
U0=Rn
2n0. Letf:U!U0be a difeomorphism de ned by the formula f(x) =x
jjxjj2, as in (9.8.2).
ThenSncan be equivalently described as the quotient manifold M=f .
Note that the 1-dimensional sphere is the circle S1. It can be d as follows. Consider the map
T:R!Rgiven by the formula T(x) =x+ 1,x2R. It satis es the condition from 9.8.2 and
hence, one can de ne the manifold R=T. This manifold is di eomorphic to S1.
2.Real projective space. The real projective space RPnis the set of all lines in Rn+1passing
through the origin. One introduces on RPna structure of an n-dimensional manifold as follows.
For eachj= 1;:::;n + 1 let us denote by Ujthe set of lines which are not parallel to the ane
subspace j=fxj= 1g. Clearly,n+1S
1Uj=RPn. There is a natural one-to one map j:Uj!j
149
which associates with each line 2Ujthe unique intersection point of with j. Furthermore,
each jcan be identi ed with Rn, and hence pairs ( Uj;j),j= 1;:::;n + 1 can be chosen as an
atlas of coordinate charts. We leave it to the reader to check that this atlas indeed de ne on RPn
a structure of a smooth manifold, i.e. that the transition maps between di erent coordinate charts
are smooth.
Exercise 9.46. Let us view Snas the unit sphere in Rn+1. Consider a map p:Sn!RPnwhich
associates to a point of Snthe line passing through this point and the origin. Prove that this
two-to-one map is smooth, and moreover a local di eomorphism , i.e. that the restriction of pto
a suciently small neighborhood of each point is a di eomorphism. Use it to show that RPnis
di eomorphic to the quotient space Sn=f, wheref:Sn!Snis the antipodal map f(x) = x.
3.Products of manifolds and n-dimensional tori. Given two manifolds, MandNof
dimensionmandn, respectively, one can naturally endow the direct product
MN=f(x;y);x2M;y2Ng
with a structure of a manifold of dimension m+n. Letf(U;)g2andf(V
;
)g
2 be atlases
forMandN, so that :U!U0
Rm; :V!V0
Rnare di eomorphisms on open
subsets of RmandRn. Then the smooth structure on MNcan be de ned by an atlas
f(UV
;
)g2;
2 ;
where we denote by
:UV
!U0
V0
RmRn=Rm+nare di eomorphisms de ned
by the formula ( x;y)7!((x) (y)) forx2Uandy2V.
One can similarly de ne the direct product of any nite number of smooth manifolds. In par-
ticular the n-dimensional torus Tnis de ned as the product of ncircles:Tn=S1S1
|{z}
n. Let
us recall, that the circle S1is di eomorphic to R=T, i.e. a point of S1is a real number up to adding
any integer. Hence, the points of the torus Tncan viewed as the points of Rnup to adding any
vector with all integer coordinates.
150
9.8.4 Submanifolds of an n-dimensional vector space
LetVbe ann-dimensional vector space. A subset AVis called ak-dimensional submanifold
ofV, or simply a k-submanifold ofV, 0kn, if for any points a2Athere exists a local
coordinate chart ( Ua;u= (u1;:::;un)!Rn) such that u(a) = 0 (i.e. the point ais the origin in
this coordinate system) and
A\Ua=fu= (u1;:::;un)2Ua;uk+1==un= 0g: (9.8.3)
We will always assume the local coordinates at least as smooth as necessary for our purposes (but
at leastC1-smooth), but more precisely, one can talk of Cm- submanifolds if the implied coordinate
systems are at least Cm-smooth.
Note that in the above we can replace the vector space Vby anyn-dimensional manifold, and
thus will get a notion of a k-dimesional submanifold of an n-dimensional manifold V.
Example 9.47. Suppose a subset AUVis given by equations F1==Fn k= 0 for
someCm-smooth functions F1;:::;Fn konU. Suppose that for any point a2Athe di erentials
daF1;:::;daFn k2V
aare linearly independent. Then AUis aCm-smooth submanifold.
Indeed, for each a2Aone can choose a linear functions l1;:::;lk2V
asuch that together
withdaF1;:::;daFn k2V
athey form a basis of V. Consider functions L1;:::;Lk:V!R,
de ned byLj(x) =lj(x a) so thatda(Lj) =lj,j= 1;:::;k . Then the Jacobian det DaFof the
mapF: (L1;:::;Lk;F1;:::;Fn k) :U!Rndoes not vanish at a, and hence the inverse function
theorem implies that this map is invertible in a smaller neighborhood UaUof the point a2A.
Hence, the functions u1=L1;:::;uk=Lk;uk+1=F1;:::;un=Fn kcan be chosen as a local
coordinate system in Ua, and thusA\Ua=fuk+1==un= 0g.
Note that the map u0= (u1;:::;uk) mapsUA
a=Ua\Aonto an open neighborhood eU=
u(Ua)\Rkof the origin in RkRn, and therefore u0= (u1;:::uk) de nes a local coordinates,
so that the pair ( UA
a;u0) is a coordinate chart . The restriction e = jeUof the parameterization
map =u 1mapseUontoUA
a. Thuse aparameterization map for the neighborhood UA
a. The
atlasf(UA
a;u0)ga2Ade nes onaa structure of a k-dimensional manifold. The complementary
dimensionn kis called the codimension of the submanifold A. We will denote dimension and
codimension of Aby dimAand codimA, respectively.
151
As we already mentioned above in Section 9.3 1-dimensional submanifolds are usually called
curves. We will also call 2-dimensional submanifolds surfaces and codimension 1 submanifolds hyper-
surfaces . Sometimes k-dimensional submanifolds are called k-surfaces. Submanifolds of codimension
0 are open domains in V.
An important class form graphicalk-submanifolds. Let us recall that given a map f:B!Rn k,
whereBis a subset BRk, then graph is the set
f=f(x;y)2RkRn k=Rn;x2B;y=f(x)g:
A (Cm)-submanifold AVis called graphical with respect to a splitting : RkRn k!V,
if there exist an open set URkand a (Cm)-smooth map f:U!Rn ksuch that
A= ( f):
In other words, Ais graphical if there exists a coordinate system in Vsuch that
A=fx= (x1;:::;xn); (x1;:::xk)2U;xj=fj(x1;:::;xk); j=k+ 1;:::;ng:
for some open set URkand smooth functions, fk+1;:::;fn:U!R.
For a graphical submanifold there is a global coordinate system given by the projection of the
submanifold to Rk.
It turns out that that any submanifold locally is graphical .
Proposition 9.48. LetAVbe a submanifold. Then for any point a2Athere is a neighborhood
Ua3asuch thatUa\Ais graphical with respect to a splitting of V. (The splitting may depend on
the pointa2A).
We leave it to the reader to prove this proposition using the implicit function theorem.
One can generalize the discussion in this section and de ne submanifolds of any manifold M,
and not just the vector space V. In fact, the de nition (9.8.3) can be used without any changes to
de ne submanifolds of an arbitrary smooth manifold.
A mapf:M!Qis called an embedding of a manifold Minto another manifold Qif it is a
di eomorphism of Monto a submanifold AQ. In other words, fis an embedding if the image
A=f(M) is a submanifold of Qand the map fviewed as a map M!Ais a di eomorphism.
152
One can prove that any n-dimensional manifold can be embedded into RNwith a suciently large
N(in factN= 2n+ 1 is always sucient).
Hence, one can think of manifold assubmanifold of some Rngiven up to a di eomorphism, i.e.
ignoring how this submanifold is embedded in the ambient space.
In the exposition below we mostly restrict our discussion to submanifolds of Rnrather than
general abstract manifolds.
9.8.5 Submanifolds with boundary
A slightly di erent notion is of a submanifold with boundary . A subset AVis called a k-
dimensional submanifold with boundary , or simply a k-submanifold of Vwith boundary , 0k<n ,
if for any points a2Athere is a neighborhood Ua3ainVand local (curvi-linear) coordinates
(u1;:::;un) inUawith the origin at aifone of two conditions is satis ed: condition (9.8.3), or the
following condition
A\Ua=fu= (u1;:::;un)2Ua;u10;uk+1==un= 0g: (9.8.4)
In the latter case the point ais called a boundary point ofA, and the set of all boundary points is
called the boundary ofAand is denoted by @A.
As in the case of submanifolds without boundary, any submanifold with boundary has a struc-
ture of a manifold with boundary.
Exercise 9.49. Prove that if Aisk-submanifold with boundary then @Ais a(k 1)-dimensional
submanifold (without boundary).
Remark 9.50. 1. As we already pointed out when we discussed manifolds with boundary, a
submanifold with boundary is not a submanifold!
2. As it was already pointed out when we discussed 1-dimensional submanifolds with boundary,
the boundary of a k-submanifold with boundary is not the same as its set-theoretic boundary,
though traditionally the same notation @Ais used. Usually this should be clear from the
context, what the notation @Astands for in each concrete case. We will explicitly point this
di erence out when it could be confusing.
153
Figure 9.5: The parameterization introducing local coordinates near an interior point aand a
boundary point b.
A compact manifold (without boundary) is called closed . The boundary of any compact manifold
with boundary is closed, i.e. @(@A) =?.
Example 9.51. An open ball Bn
r=Bn
R(0) =fnP
1x2
j<1gRnis a codimension 0 submanifold,
A closed ball Dn
r=Dn
R(0) =fnP
1x2
j1gRnia codimension 0 submanifold with boundary.
Its boundary @Dn
Ris an (n 1)-dimensional sphere Sn 1
R=fnP
1x2
j= 1gRn. It is a closed
hypersurface. For k= 0;1:::n 1 let us denote by Lkthe subspace Lk=fxk+1==xn= 0g
Rn. Then the intersections
Bk
R=Bn
R\Lk;Dk
R=Dn
R\Lk;andSk 1
R=Sn 1
R\LkRn
are, respectively a k-dimensional submanifold, a k-dimensional submanifold with boundary and a
closed (k 1)-dimensional submanifold of Rn. Among all above examples there is only one (which
one?) for which the manifold boundary is the same as the set-theoretic boundary.
A neighborhood of a boundary point a2@Acan be always locally parameterized by the semi-
154
open upper-half ball
B+(0) =fx= (x1;:::;xk)2Rk;x10;nX
1x2
j<1g:
We will nish this section by de ning submanifolds with piece-wise smooth boundary . A subset
AVis called ak-dimensional submanifold of Vwith piecewise smooth boundary or with boundary
with corners , 0k < n , if for any points a2Athere is a neighborhood Ua3ainVand local
(curvi-linear) coordinates ( u1;:::;un) inUawith the origin at aif one of three condittions satis ed:
conditions (9.8.3), (9.8.4) or the following condition
A\Ua=fu= (u1;:::;un)2Ua;l1(u)0;:::;lm(u)0;uk+1==un= 0g; (9.8.5)
wherem > 1 andl1;:::;lm2(Rk)are linear functions In the latter case the point ais called a
corner point of@A.
Note that the system of linear inequalities l1(u)0;:::;lm(u)0 de nes a convex cone in
Rk. Hence, near a corner point of its boundary the manifold is di eomorphic to a convex cone.
Thus convex polyhedra and their di eomorphic images are important examples of submanifolds
with boundary with corners.
9.9 Tangent spaces and di erential
Suppose we are given two local parameterizations :G!Aande :eG!A. Suppose that
02G\eGand (0) =e (0) =a2A. Then there exists a neighborhood U3ainAsuch that
U (G)\e (eG).
DenoteG1:= 1(U);eG1:=e 1(eU). Then one has two coordinate charts on U:u= (u1;:::;uk) =
( jG1) 1:U!G1;andeu= (eu1;:::;euk) =
e jeG1 1
:U!eG1:
Denoteh:=ue jeG1=eG1!G1. We have
e = ue = h;
and hence the di erentials d 0andde 0of parameterizations ande at the origin map Rk
0isomor-
phically onto the same k-dimensional linear subspace TVa. Indeed,d0e =d0 d0h. Thus the
spaceT=d0 (Rk
0)Vais independent of the choice of parameterization. It is called the tangent
155
space to the submanifold Aat the point a2Aand will be denoted by TaA. IfAis a submanifold
with boundary and a2@Athen there are de ned both the k-dimensional tangent space TaAand
its (k 1)-dimensional subspace Ta(@A)TaAtangent to the boundary.
Example 9.52. 1. Suppose a submanifold AVis globally parameterized by an embedding
:G!A,!V,GRk. Suppose the coordinates in Rkare denoted by ( u1;:::;uk). Then
the tangent space TaAat a pointa= (b);b2Gis equal to the span
Span@
@u1(a);:::;@
@uk(a)
:
2. In particular, suppose a submanifold Ais graphical and given by equations
xk+1=g1(x1;:::;xk);:::;xn=gn k(x1;:::;xk);(x1;:::;xk)2GRk:
Take points b= (b1;:::bk)2Ganda= (b1;:::;bk;g1(b);:::;gn k(b))2A. ThenTaA=
Span (T1;:::Tk);where
T1=0
@1;0;:::; 0|{z}
k;@g1
@x1(b);:::;@gn k
@x1(b)1
A;
T1=0
@0;1;:::; 0|{z}
k;@g1
@x2(b);:::;@gn k
@x2(b)1
A;
:::
T1=0
@0;0;:::; 1|{z}
k;@g1
@xk(b);:::;@gn k
@xk(b)1
A:
3. Suppose a hypersurface Rnis given by an equation = fF= 0gfor a smooth function
Fde ned on an neighborhood of and such that daF6= 0 for any a2. In other words,
the function Fhas no critical points on . Take a point a2. ThenTaRn
ais given by a
linear equation
nX
1@F
@xj(a)hj= 0; h= (h1;:::;hn)2Rn
a:
Note that sometimes one is interested to de ne Ta as an ane subspace of Rn=Rn
0and
not as a linear subspace of Rn
a. We get the required equation by shifting the origin:
Ta =fx= (x1;:::;xn)Rn;nX
1@F
@xj(a)(xj aj) = 0g:
156
If for some parameterization :G!Awith (0) =athe composition f is di erentiable at
0, and the linear map
d0(f )(d0 ) 1:TaA!Wf(a)
is called the di erential offat the point aand denoted, as usual, by daf. Similarly one can de ne
Cm-smooth maps A!W.
Exercise 9.53. Show that a map f:A!Wis di erentiable at a point a2Ai for some
neighborhood UofainVthere exists a map F:U!Wthat is di erentiable at aand such
thatFjU\A=fjU\A, and we have dFjTaA=daf. As it follows from the above discussion the map
dFjTaAis independent of this extension. Similarly, any Cm-smooth map of Alocally extends to a
Cm-smooth map of a neighborhood of AinV.
Suppose that the image f(A) of a smooth map f:A!Wis contained in a submanifold BW.
In this case the image daf(TaA) is contained in Tf(a)B. Hence, given a smooth map f:A!B
between two submanifolds AVandBWits di erential at a point acan be viewed as a linear
mapdaf:TaA!Tf(a)B.
Let us recall, that given two submanifolds AVandBW(with or without boundary), a
smooth map f:A!Bis called a di eomorphism if there exists a smooth inverse map : B!A,
i.e.fg:B!Bandgf:A!Aare both identity map. The submanifolds AandBare called
in this case di eomorphic.
Exercise 9.54. 1. LetA;B be two di eomorphic submanifolds. Prove that
(a) ifAis path-connected then so is B;
(b) ifAis compact then so is B;
(c) if@A=?then@B=?;
(d) dimA= dimB;5
2. Give an example of two di eomorphic submanifolds, such that one is bounded and the other
is not.
5In fact, we will prove later that even homeomorphic manifolds should have the same dimension.
157
3. Prove that any closed connected 1-dimensional submanifold is di eomorphic to the unit circle
S1=fx2
1+x2
2= 1gR2.
9.10 Vector bundles and their homomorphisms
Let us put the above discussion in a bit more global and general setup.
A collection of all tangent spaces fTaAga2Ato a submanifold Ais called its tangent bundle
and denoted by TAorT(A). This is an example of a more general notion of a vector bundle of
rankrover a set AV. One understands by this a family of r-dimensional vector subspaces
LaVa, parameterized by points of Aand continuously (orCm-smoothly) depending on a. More
precisely one requires that each point a2Ahas a neighborhood UAsuch that there exist linear
independent vector elds v1(a);:::;vr(a)2Lawhich continuously (smoothly, etc.) depend on a.
Besides the tangent bundle T(A) over ak-submanifold Aan important example of a vector
bundle over a submanifold Ais its normal bundle NA=N(A), which is a vector bundle of rank
n kformed by orthogonal complements NaA=T?
aAVaof the tangent spaces TaAofA. We
assume here that Vis Euclidean space.
A vector bundle Lof rankkoverAis called trivial if one can nd kcontinuous linearly
independent vector elds v1(a);:::;vk(a)2Lade ned for alla2A. The setAis called the base of
the bundle L.
An important example of a trivial bundle is the bundle TV=fVaga2V.
Exercise 9.55. Prove that the tangent bundle to the unit circle S1R2is trivial. Prove that the
tangent bundle to S2R3is not trivial, but the tangent bundle to the unit sphere S3R4is
trivial. (The case of S1is easy, ofS3is a bit more dicult, and of S2even more dicult. It turns
out that the tangent bundle TSn 1to the unit sphere Sn 1Rnis trivial if and only if n= 2;4
and 8. The only if part is a very deep topological fact which was proved by F. Adams in 1960.
Suppose we are given two vector bundles, LoverAandeLovereAand a continuous (resp. smooth)
map :A!eA. By a continuous (resp. smooth) homomorphism :L!eLwhich covers the map
:A!eAwe understand a continuous (resp. smooth) family of linear maps a:La!eL (a). For
instance, a Cm-smooth map f:A!Bde nes aCm 1-smooth homomorphism df:TA!TB
158
which covers f:A!B. Heredf=fdafga2Ais the family of linear maps daf:TaA!Tf(a)B,
a2A.
9.11 Orientation
By an orientation of a vector bundle L=fLaga2AoverAwe understand continuously depending
onaorientation of all vector spaces La. An orientation of a submanifold kis the same as an
orientation of its tangent bundle T(A). A co-orientation of ak- submanifold Ais an orientation of
its normal bundle N(A) =T?AinV. Note that not all bundles are orientable , i.e. some bundles
admit no orientation. But if Lis orientable and the base Ais connected, then Ladmits exactly two
orientations. Here is a simplest example of a non-orientable rank 1 bundle of the circle S1R2.
Let us identify a point a2S1with a complex number a=ei , and consider a line la2R2
adirected
by the vector ei
2. Hence, when the point completes a turn around S1the linelarotates by the
angle. We leave it to the reader to make a precise argument why this bundle is not orientable .
In fact, rank 1 bundles are orientable if and only if they are trivial.
If the ambient space Vis oriented then co-orientation and orientation of a submanifold Adeter-
mine each other according to the following rule. For each point a, let us choose any basis v1;:::;vk
ofTa(A) and any basis w1;:::;wn kofNa(A). Thenw1;:::;wn k;v1;:::;vkis a basis of Va=V.
Suppose one of the bundles, say N(A), is oriented. Let us assume that the basis w1;:::;wn kde-
nes this orientation. Then we orient TaAby the basis v1;:::;vkif the basisw1;:::;wn k;v1;:::;vk
de nes the given orientation of V, and we pick the opposite orientation of TaAotherwise.
Example 9.56. (Induced orientation of the boundary of a submanifold.) SupposeAis an
oriented manifold with boundary. Let us co-orient the boundary @Aby orienting the rank 1 normal
bundle toT(@A) inT(A) by the unit ourtward normal to T(@A) inT(A) vector eld. Then the
above rule determine an orientation of T(@A), and hence of @A.
9.12 Integration of di erential k-forms over k-dimensional sub-
manifolds
Let be a di erential k-form de ned on an open set UV.
159
Figure 9.6: The orientation of the surface is induced by its co-orientation by the normal vector n.
The orientation of the boundary us induced by the orientation of the surface.
160
Consider rst a k-dimensional compact submanifold with boundary AUde ned parametri-
cally by an embedding :G!A,!U, whereGRkis possibly with boundary. Suppose that A
is oriented by this embedding. Then we de ne
Z
A :=Z
G :
Note that if we de ne Aby a di erent embedding e :eG!A, then we have e = h, where
h= 1e :eG!Gis a di eomorphism. Hence, using Theorem 9.32 we get
Z
eGe =Z
eGh( ) =Z
G ;
and henceR
A is independent of a choice of parameterization, provided that the orientation is
preserved.
Let nowAbe any compact oriented submanifold with boundary. Let us choose a partition of
unity 1 =KP
1jin a neighborhood of Asuch that each function is supported in some coordinate
neighborhood of A. Denote j=j . Then =KP
1 j, where each form jis supported in one
of coordinate neighborhoods. Hence there exist orientation preserving embeddings j:Gj!A
of domains with boundary GjRk, such that j(Gj)Supp( j),j= 1;:::;K . Hence, we can
de neZ
A j:=Z
Gj
j jandZ
A :=KX
1Z
A j:
Lemma 9.57. The above de nition ofR
A is independent of a choice of a partition of unity.
Proof. Consider two di erent partitions of unity 1 =KP
1jand 1 =eKP
1ejsubordinated to coverings
U1;:::;UKandeU1;:::;eUeK, respectively. Taking the product of two partitions we get another
partition 1 =KP
i=1eKP
j=1ij, whereij:=ij, which is subordinated to the covering by intersections
Ui\Uj,i= 1;:::;K; j = 1;:::;eK. Denote i:=i ;e j:=ej and ij=ij . ThenKP
i=1 ij=e j;
161
eKP
j=1 ij= iand =KP
1 i=eKP
1e j:Then, using the linearity of the integral we get
KX
1Z
A i=KX
i=1eKX
j=1Z
A ij=eKX
1Z
Ae j:
Whenk= 1 the above de nition of the integral coincides with the de nition of the integral of
a 1-form over an oriented curve which was given above in Section 9.2.
Let us extend the de nition of integration of di erential forms to an important case of integration
of 0-form over oriented 0-dimensional submanifolds. Let us recall a compact oriented 0-dimensional
submanifold of Vis just a nite set of points a1;:::;am2Vwith assigned signs to every point.
So in view of the additivity of the integral it is sucient to de ne integration over 1 point with a
sign. On the other hand, a 0-form is just a function f:V!R. So we de ne
Z
af:=f(a):
A partition of unity is a convenient tool for studying integrals, but not so convenient for practical
computations. The following proposition provides a more practical method for computations.
Proposition 9.58. LetAbe a compact oriented submanifold of Vand a di erential k-form given
on a neighborhood of A. Suppose that Apresented as a union A=NS
1Aj, whereAjare codimension
0submanifolds of Awith boundary with corners. Suppose that AiandAjfor anyi6=jintersect
only along pieces of their boundaries. Then
Z
A =NX
1Z
Aj :
In particular, if each Ajis parameterized by an orientation preserving embedding j:Gj!A,
whereGjRkis a domain with boundary with corners. Then
Z
A =NX
1Z
Gj
j :
162
We leave the proof to the reader as an exercise.
Exercise 9.59. Compute the integral
Z
S1=3(x1dx2^dx3+x2dx3^dx1+x3dx1^dx2);
whereSis the sphere
fx2
1+x2
2+x2
3= 1g;
cooriented by its exterior normal vector.
Solution. Let us present the sphere as the union of northern and southern hemispheres:
S=S [S+;whereS =S\fx30g; S+=S\fx30g:
ThenR
S!=R
S+!+R
S !. Let us rst computeR
S+!.
We can parametrize S+by the map ( u;v)!(u;v;p
R2 u2 v2), (u;v)2fu2+v2R2g=
DR. One can check that this parametrization agrees with the prescribed coorientation of S. Thus,
we have
Z
S+!= 1=3Z
DR
udv^dp
R2 u2 v2+vdp
R2 u2 v2^du+p
R2 u2 v2du^dv
:
Passing to polar coordinates ( r;') in the plane ( u;v) we get
Z
S+!= 1=3Z
Prcos'd(rsin')^dp
R2 r2+rsin'dp
R2 r2^d(rcos')
+p
R2 r2d(rcos')^d(rsin');
whereP=f0rR;0'2g. Computing this integral we get
163
Z
S+!=1
3Z
P r3cos2'd'^drp
R2 r2+r3sin2'dr^d'p
R2 r2+p
R2 r2dr^d'
=1
3Z
Pr3
p
R2 r2+rp
R2 r2
dr^d'
=2
3ZR
0rR2
p
R2 r2dr= 2R2
3p
R2 r2ZR
0=2R3
3
Similarly, one can compute thatZ
S !=2R3
3:
Computing this last integral, one should notice the fact that the parametrization
(u;v)7!(u;v; p
R2 u2 v2)
de nes the wrong orientation of S . Thus one should use instead the parametrization
(u;v)7!(v;u; p
R2 u2 v2);
and we get the answerZ
S!=4R3
3:
This is just the volume of the ball bounded by the sphere. The reason for such an answer will be
clear below from Stokes' theorem.
164
Part III
Stokes' theorem and its applications
165
Chapter 10
Stokes' theorem
10.1 Statement of Stokes' theorem
Theorem 10.1. LetAVbe a compact oriented submanifold with boundary (and possibly with
corners). Let !be aC2-smooth di erential form de ned on a neighborhood UA. Then
Z
@A!=Z
Ad!:
Hered!is the exterior di erential of the form !and@Ais the oriented boundary of A.
We will discuss below what exactly Stokes' theorem means for the case k3 andn= dimV3.
Let us begin with the case k= 1; n= 2. ThusV=R2. Letx1;x2be coordinates in R2andU
a domain in R2bounded by a smooth curve = @U. Let us co-orient with the outward normal
to the boundary of U. This de nes a counter-clockwise orientation of .
Let!=P1(x1;x2)dx1+P2(x1;x2)dx2be a di erential 1-form. Then the above Stokes' formula
asserts
Z
Ud!=Z
!;
orZ
U@P2
@x1 @P1
@x2
dx1^dx2=Z
P1dx1+P2dx2:
167
Figure 10.1: George Stokes (1819-1903)
This is called Green's formula. In particular, when d!=dx1^dx2, e.g.!=xdy or!=
1
2(xdy ydx), the integralR
!computes the area of the domain U.
Consider now the case n= 3; k= 2. Thus
V=R3; !=P1dx2^dx3+P2dx3^dx1+P3dx1^dx2:
LetUR3be a domain bounded by a smooth surface S. We co-orient Swith the exterior normal
. Then
d!=@P1
@x1+@P2
@x2+@P3
@x3
dx1^dx2^dx3:
Thus, Stokes' formulaZ
S!=Z
Ud!
gives in this case
Z
SP1dx2^dx3+P2dx3^dx1+P3dx1^dx2=Z
U@P1
@x1+@P2
@x2+@P3
@x3
dx1^dx2^dx3
This is called the divergence theorem or Gauss-Ostrogradski's formula.
168
Figure 10.2: George Green (1793-1841)
Consider the case k= 0; n= 1. Thus !is just a function fon an interval I= [a;b]. The
boundary@Iconsists of 2 points: @I=fa;bg. One should orient the point awith the sign and
the pointbwith the sign +.
Thus, Stokes' formula in this case gives
Z
[a;b]df=Z
f a;+bgf;
orZb
af0(x)dx=f(b) f(a):
This is Newton-Leibnitz' formula. More generally, for a 1-dimensional oriented connected curve
R3with boundary @ =B[( A) and any smooth function fwe get the formula
Z
df=Z
B[( A)f=f(B) f(A);
which we already proved earlier, see Theorem 9.11 .
Consider now the case n= 3; k= 1.
ThusV=R3and!=P1dx1+P2dx2+P3dx3. LetSR3be an oriented surface with boundary
169
Figure 10.3: Carl Friedrich Gauss (1777-1855) Mikhail Ostrogradski (1801-1862)
. We orient in the same way, as in Green's theorem. Then Stokes' formula
Z
Sd!=Z
!
gives in this case
Z
S@P3
@x2 @P2
@x3
dx2^dx3+@P1
@x3 @P3
@x1
dx3^dx1+@P2
@x1 @P1
@x2
dx1^dx2
=Z
P1dx1+P2dx2+P3dx3:
This is the original Stokes' theorem.
Stokes' theorem allows one to clarify the geometric meaning of the exterior di erential.
Lemma 10.2. Let be a di erential k-form in a domain UV. Take any point a2Uand
vectorsX1;:::;Xk+12Va. Given >0let us consider the parallelepiped P(X1;:::;Xk+1)as a
subset ofVwith vertices at points ai1:::ik+1=a+k+1P
1ijXj, where each index ijtakes values 0;1.
Then
d a(X1;:::Xk+1) = lim
!01
k+1Z
@P(X1;:::;Xk+1) :
170
Proof. First, it follows from the de nition of integral of a di erential form that
d a(X1;:::;Xk+1) = lim
!01
k+1Z
P(X1;:::;Xk+1)d : (10.1.1)
Then we can continue using Stokes' formula
d a(X1;:::;Xk+1) = lim
!01
k+1Z
P(X1;:::;Xk+1)d = lim
!01
k+1Z
@P(X1;:::;Xk+1) : (10.1.2)
10.2 Proof of Stokes' theorem
We prove in this section Theorem 10.1. We will consider only the case when Ais a manifold with
boundary without corners and leave the corner case as an exercise to the reader.
Let us cover Aby coordinate neighborhoods such that in each neighborhood Ais given either
by (9.8.3) or (9.8.4). First we observe that it is sucient to prove the theorem for the case of a
form supported in one of these coordinate neighborhoods. Indeed, let us choose nitely many such
neighborhoods covering A. Let 1 =NP
1jbe a partition of unity subordinated to this covering. We
set!j=j!, so that!=NP
1!j, and each of !jis supported in one of coordinate neighborhoods.
Hence, if formula 10.1 holds for each !jit also holds for !.
Let us now assume that !is supported in one of coordinate neighborhoods. Consider the
corresponding parameterization :G!UV,GRn, introducing coordinates u1;:::;un. Then
A\U= (G\L), whereLis equal to the subspace Rk=fuk+1=:::un= 0gin the case (9.8.3) and
the upper-half space Rk\fu10g. By de nition, we haveR
Ad!=R
Ud!=R
G\L d!=R
G\Ld !:1
Though the form e!= !jG\Lis de ned only on G\L, it is supported in this neighborhood,
and hence we can extend it to a smooth form on the whole Lby setting it equal to 0 outside the
neighborhood. With this extension we haveR
G\Lde!=R
Lde!:The (k 1)-forme!can be written in
coordinates u1;:::;ukas
e!=jX
1fj(u)du1^:::j
^duk:
1We assume here that the coordinates u1;:::;ukde ne the given orientation of A.
171
ThenZ
G\Lde!=Z
L kX
1@fj
@uj!
du1^^duk:
Let us choose a suciently R> 0 so that the cube I=fjuijR;i= 1;:::;kgcontains Supp( e!).
Thus in the case (9.8.3) we have
Z
G\Lde!=kX
1Z
Rk@fj
@ujdV=kX
1RZ
R:::RZ
R@fj
@ujdu1:::dun=
kX
1RZ
R:::0
@RZ
R@fj
@ujduj1
Adu1:::duj 1duj+1:::dun= 0 (10.2.1)
because
RZ
R@fj
@ujduj=fj(u1;:::;ui 1;R;ui;:::;un) fj(u1;:::;ui 1; R;ui;:::;un)) = 0:
On the other hand, in this caseR
@A!= 0, because the support of !does not intersect the boundary
ofA. Hence, Stokes' formula holds in this case. In case (9.8.4) we similarly get
Z
G\Lde!=kX
1Z
fu10g@fj
@ujdV=
kX
1RZ
00
@RZ
R:::RZ
R@fj
@ujdun:::du 21
Adu1=RZ
R0
@RZ
R:::RZ
0@f1
@u1du1:::dun 11
Adun=
RZ
R:::RZ
Rf1(0;u2;:::;un)du2:::dun: (10.2.2)
because all terms in the sum with j >1 are equal to 0 by the same argument as in (10.2.1). On
the other hand, in this case
Z
@A!=Z
fu1=0g !=Z Z
fu1=0gf1(0;u2;:::;un)du2^^dun=
RZ
R:::RZ
Rf1(0;u2;:::;un)du2:::dun: (10.2.3)
172
The sign minus appears in the last equality in front of the integral because the induced orien-
tation on the space fu1= 0gas the boundary of the upper-half space fu10gis opposite to the
orientation de ned by the volume form du2^^dun:Comparing the expressions (10.2.2) and
(10.2.3) we conclude thatR
Ad!=R
@A!, as required.
10.3 Integration of functions over submanifolds
In order to integrate functions over a submanifold we need a notion of volume for subsets of the
submanifold.
LetAVbe an oriented k-dimensional submanifold, 0 kn. By de nition, the volume
form=AofA(or the area form ifk= 2, or the length form ifk= 1) is a di erential k-form
onAwhose value on any ktangent vectors v1;:::;vk2TxAequals the oriented volume of the
parallelepiped generated by these vectors.
Given a function f:A!Rwe de ne its integral over Aby the formula
Z
AfdV =Z
AfA; (10.3.1)
and, in particular,
VolA=Z
AA:
Notice that the integralR
AfdV is independent of the orientation of A. Indeed, changing the orienta-
tion we also change the sign of the form A, and hence the integral remains unchanged. This allows
us to de ne the integralR
AfdV even for a non-orientable A. Indeed, we can cover Aby coordinate
charts, nd a subordinated partition of unity and split correspondingly the function f=NP
1fj
in such a way that each function fjis supported in a coordinate neighborhood. By orienting in
arbitrary ways each of the coordinate neighborhoods we can compute each of the integralsR
AfjdV,
j= 1;:::;N . It is straightforward to see that the integralR
AfdV =P
jR
AfjdVis independent of the
choice of the partition of unity.
Let us study in some examples how the form Acan be e ectively computed.
173
Example 10.3. Volume form of a hypersurface. Let us x a Cartesian coordiantes in V. LetAV
is given by the equation
A=fF= 0g
for some function F:V!Rwhich has no critical points on A. The vector eld rFis orthogonal
toA, and
n=rF
jjrFjj
is the unit normal vector eld to A. Assuming Ato be co-oriented by nwe can write down the
volume form of Aas the contraction of nwith the volume form
= dx1^^dxnofRn, i.e.
A=n
=1
jjrFjjnX
1( 1)i 1@F
@xidx1^i_:::^dxn:
In particular, if n= 3 we get the following formula for the area form of an implicitely given
2-dimensional surface A=fF= 0gR3:
A=1r
@F
@x12
+
@F
@x22
+
@F
@x32@F
@x1dx2^dx3+@F
@x2dx3^dx1+@F
@x3dx1^dx2
:(10.3.2)
Example 10.4. Length form of a curve.
Let Rnbe an oriented curve given parametrically by a map
: [a;b]!Rn. Let= be
the length form. Let us compute the form
. Denoting the coordinate in [ a;b] bytand the unit
vector eld on [ a;b] byewe have
=f(t)dt;
where
f(t) =
(e) =
0(t)
=jj
0(t)jj:
In particular the length of is equal to
Z
=bZ
ajj
0(t)jjdt=bZ
avuutnX
i=1(x0
i(t))2dt;
where
(t) = (x1(t);:::;xn(t)):
174
Similarly, given any function f: !Rwe have
Z
fds=bZ
af(
(t))jj
0(t)jjdt:
Example 10.5. Area form of a surface given parametrically.
Suppose a surface SRnis given parametrically by a map : U!RnwhereUin the plane R2
with coordinates ( u;v).
Let us compute the pull-back form S. In other words, we want to express Sin coordinates
u;v. We have
S=f(u;v)du^dv:
To determine f(u;v) take a point z= (u;v)2R2and the standard basis e1;e22R2
z. Then
(S)z(e1;e2) =f(u;v)du^dv(e1;e2): (10.3.3)
On the other hand, by the de nition of the pull-back form we have
(S)z(e1;e2) = (S)(z)(dz(e1);dz(e2)): (10.3.4)
Butdz(e1) =@
@u(z) = u(z) anddz(e2) =@
@v(z) = v(z). Hence from (10.3.3) and (10.3.4) we
get
f(u;v) =S(u;v): (10.3.5)
The value of the form Son the vectors u;vis equal to the area of the parallelogram gener-
ated by these vectors, because the surface is assumed to be oriented by these vectors, and hence
S(u;v)>0. Denoting the angle between uand vby we get2S(u;v) =jjujjjjvjjsin .
Hence
S(u;v)2=jjujj2jjvjj2sin2 =jjujj2jjvjj2(1 cos2 ) =jjujj2jjvjj2 (uv)2;
and therefore,
f(u;v) =S(u;v) =p
jjujj2jjvjj2 (uv)2:
2See a computation in a more general case below in Example 10.6.
175
It is traditional to introduce the notation
E=jjujj2; F= uv; G=jjvjj2;
so that we get
S=p
EG F2du^dv;
and hence we get for any function f:S!R
Z
SfdS=Z
SfS=Z
Uf((u;v))p
EG F2du^dv=Z Z
Uf((u;v))p
EG F2dudv: (10.3.6)
Consider a special case when the surface Sde ned as a graph of a function over a domain
DR2. Namely, suppose
S=fz= (x;y);(x;y)2DR2g:
The surface Sas parametrized by the map
(x;y)7!(x;y; (x;y)):
Then
E=jjxjj2= 1 + 2
x; G=jjyjj2= 1 + 2
y; F= xy= x y;
and hence
EG F2= (1 + 2
x)(1 + 2
y) 2
x 2
y= 1 + 2
x+ 2
y:
Therefore, the formula (10.3.6) takes the form
Z
SfdS=Z Z
Df((x;y))p
EG F2dx^dy=
Z Z
Df(x;y; (x;y))q
1 + 2x+ 2ydxdy: (10.3.7)
Note that the formula (10.3.7) can be also deduced from (10.3.2). Indeed, the surface
S=fz= (x;y);(x;y)2DR2g;
can also be de ned implicitly by the equation
F(x;y;z ) =z (x;y) = 0;(x;y)2D:
176
We have
rF= ( @
@x; @
@y;1);
and, therefore,
Z
SfdS =Z
Sf(x;y;z )
jjrFjj@F
@xdy^dz+@F
@ydz^dx+@F
@zdx^dy
=Z Z
Df(x;y; (x;y))r
1 +
@
@x2
+
@
@y2
@
@xdy^d @
@yd ^dx+dx^dy
=Z Z
Df(x;y; (x;y))r
1 +
@
@x2
+
@
@y2
1 +@
@x2
+@
@y2!
dxdy
=Z Z
Df(x;y; (x;y))s
1 +@
@x2
+@
@y2
dxdy:
Example 10.6. Integration over a parametrically given k-dimensional submanfold.
Consider now a more general case of a parametrically given k-submanifold Ain ann-dimensional
Euclidean space V. We x a Cartersian coordinate system in Vand thus identify VwithRnwith
the standard dot-product.
LetURkbe a compact domain with boundary and :U!Rnbe an embedding. We assume
that the submanifold with boundary A= (U) is oriented by this parameterization. Let Abe the
volume form of A. We will nd an explicit expression for A. Namely. denoting coordinates in
Rkby (u1;:::;uk) we have A=f(u)du1^^duk, and our goal is to compute the function f.
By de nition, we have
f(u) = (A)u(e1;:::;ek) = (A)u(du (e1);:::;du (ek)) =
(A)u@
@u1(u);:::;@
@uk(u)
= Volk
P@
@u1(u);:::;@
@uk(u)
(10.3.8)
In Section 4.2 we proved two formulas for the volume of a parallelepiped. Using formula (4.2.1)
we get
Volk
P@
@u1(u);:::;@
@uk(u)
=sX
1i1<<iknZ2
i1:::ik: (10.3.9)
177
where
Zi1:::ik= @ i1
@u1(u):::@ i1
@uk(u)
::: ::: :::
@ ik
@u1(u):::@ ik
@uk(u) : (10.3.10)
ThusZ
AfdV =Z
AfA=Z
Uf( (u))sX
1i1<<iknZ2
i1:::ikdu1^^duk:
Rewriting this formula for k= 2 we get
Z
AfdV =Z
Uf( (u))vuuutX
1i<jn @ i
@u1@ i
@u2
@ j
@u1@ j
@u2 2
du1^du2: (10.3.11)
Alternatively we can use formula ( ??). Then we get
Volk
P@
@u1(u);:::;@
@uk(u)
=q
det
(D )TD
; (10.3.12)
where
D =0
BBB@@ 1
@u1:::@ 1
@uk
::: ::: :::
@ n
@u1:::@ n
@uk1
CCCA
is the Jacobi matrix of .3Thus using this expression we get
Z
AfdV =Z
AfA=Z
Uf( (u))q
det
(D )TD
dV: (10.3.13)
Exercise 10.7. In casen= 3,k= 2 show explicitely equivalence of formulas (10.3.6) ,(10.3.11)
and(10.3.13) .
Exercise 10.8. Integration over an implicitly de ned k-dimensional submanfold. Suppose that
A=fF1==Fn k= 0gand the di erentials of de ning functions are linearly independent at
points ofA. Show that
A=(dF1^^dFn k)
jjdF1^^dFn kjj:
3The symmetric matrix ( D )TD is the Gram matrix of vectors@
@u1(u);:::;@
@uk(u), and its entrees are pairwise
scalar products of these vectors, see Remark ??.
178
Example 10.9. Let us compute the volume of the unit 3-sphere S3=fx2
1+x2
2+x2
3+x2
4= 1g.
By de nition, Vol( S3) =R
S3n
, where
= dx1^dx2^dx3^dx4, and nis the outward unit
normal vector to the unit ball B4. HereS3should be co-oriented by the vector eld n. Then using
Stokes' theorem we have
Z
S3n
=Z
S3(x1dx2^dx3^dx4 x2dx1^dx3^dx4+x3dx1^dx2^dx4 x4dx1^dx2^dx3) =
4Z
B4dx1^dx2^dx3^dx4= 4Z
B4dV:
(10.3.14)
Introducing polar coordinates ( r; ) and (;) in the coordinate planes ( x1;x2) and (x3;x4) and
using Fubini's theorem we get
Z
B4=2Z
02Z
01Z
0p
1 r2Z
0rddrdd =
221Z
0(r r3)dr=2
2: (10.3.15)
Hence, Vol( S3) = 22.
Exercise 10.10. Find the ratioVoln(Bn
R)
Voln 1(Sn 1
R):
10.4 Work and Flux
We introduce in this section two fundamental notions of vector analysis: a work of a vector eld
along a curve , and a
ux of a vector eld through a surface. Let be an oriented smooth curve in
a Euclidean space VandTthe unit tangent vector eld to . Let vbe another vector eld, de ned
along . The function hv;Tiequals the projection of the vector eld vto the tangent directions
to the curve. If the vector eld vis viewed as a force eld, then the integralR
hv;Tidshas the
meaning of a work Work (v) performed by the eld vto transport a particle of mass 1 along the
curve in the direction determined by the orientation. If the curve is closed then this integral is
sometimes called the circulation of the vector eld valong and denoted byH
hv;Tids. As we
179
already indicated earlier, the signH
in this case has precisely the same meaning asR
, and it is used
only to stress the point that we are integrating along a closed curve.
Consider now a co-oriented hypersurface Vand denote by nthe unit normal vector eld
to which determines the given co-orientation of . Given a vector eld valong we will view
it as the velocity vector eld of a
ow of a
uid in the space. Then we can interpret the integral
Z
hv;nidV
as the
uxFlux (v) ofvthrough , i.e. the volume of
uid passing through in the direction of
nin time 1.
Lemma 10.11. 1. For any co-oriented hypersurface and a vector eld vgiven in its neigh-
borhood we have
hv;ni= (v
);
where
is the volume form in V.
2. For any oriented curve and a vector eld vnear we have
hv;Ti =D(v)j :
Proof. 1. For anyn 1 vectorsT1;:::;Tn 12Tx we have
v
(T1;:::;Tn 1) =
( v;T1;:::;Tn 1) = VolP(v;T1;:::;Tn 1):
Using (3.3.1) we get
VolP(v;T1;:::;Tn 1) =hv;niVoln 1P(T1;:::;Tn 1) =
hv;niVolP(n;T1;:::;Tn 1) =hv;ni(T1;:::;Tn 1): (10.4.1)
2. The tangent space Tx is generated by the vector T, and hence we just need to check that
hv;Ti (T) =D(v)(T). But (T) = Vol( n;T) = 1, and hence
D(v)(T) =hv;Ti=hv;Ti (T):
180
Note that if we are given a Cartesian coordinate system in Vandv=nP
1aj@
@xj, then
v
=n 1X
1( 1)i 1aidx1i_:::dxn;D(v) =nX
1aidxi:
Thus, we have
Corollary 10.12.
Flux (v) =Z
hv;nidV=Z
(v
) =Z
nX
1( 1)i 1aidx1i_:::dxn;
Work (v) =Z
hv;Ti=Z
D(v) =Z
nX
1aidxi:
In particular if n= 3 we have
Flux (v) =Z
a1dx2^dx3+a2dx3^dx1+a3dx1^dx2:
Let us also recall that in a Euclidean space Vwe havev
=D(v). Hence, the equation
!=v
is equivalent to the equation
v=D1
1!
= ( 1)n 1D 1(!):
In particular, when n= 3 we get v=D 1(!). Thus we get
Corollary 10.13. For any di erential (n 1)-form!and an oriented compact hypersurface we
haveZ
!= Flux v;
where v= ( 1)n 1D 1(!):
Integration of functions along curves and surfaces can be interpreted as the work and the
ux
of appropriate vector elds. Indeed, suppose we need to compute an integralR
fds. Consider the
tangent vector eld v(x) =f(x)T(x); x2 , along . Then hv;Ti=fand hence the integral
R
fdscan be interpreted as the work Work (v). Therefore, we have
Z
fds= Work (v) =Z
D(v)
181
.
Note that we can also express vthrough!by the formula v=D 1?!, see Section 8.7.
Similarly, to compute an integralR
fdS let us co-orient the surface with a unit normal to
vector eld n(x); x2 and set v(x) =f(x)n(x). Thenhv;ni=f, and hence
Z
fdS=Z
hv;nidS= Flux (v) =Z
!;
where!=v
.
10.5 Integral formulas of vector analysis
We interpret in this section Stokes' formula in terms of integrals of functions and operations on
vector elds. Let us consider again di erential forms, which one can associate with a vector eld
vin an Euclidean 3-space. Namely, this is a di erential 1-form =D(v) and a di erential 2-form
!=v
, where
= dx^dy^dzis the volume form.
Using Corollary 10.12 we can reformulate Stokes' theorem for domains in a R3as follows.
Theorem 10.14. Letvbe a smooth vector eld in a domain UR3with a smooth (or piece-wise)
smooth boundary . Suppose that is co-oriented by an outward normal vector eld. Then we have
Flux v=Z Z Z
Udivvdxdydz:
Indeed, div v=d!. Hence we have
Z
UdivvdV=Z
U(?d!)dx^dy^dz=Z
Ud!=Z
!=Z
v
= Flux v:
This theorem clari es the meaning of div v:
LetBr(x) be the ball of radius rcentered at a point x2R3, andSr(x) =@Br(x) be its
boundary sphere co-oriented by the outward normal vector eld. Then
divv(x) = lim
r!0FluxSr(x)v
Vol(Br(x)):
182
Theorem 10.15. Letbe a piece-wise smooth compact oriented surface in R3with a piece-wise
smooth boundary =@oriented respectively. Let vbe a smooth vector eld de ned near . Then
Flux (curlv) =Z
hcurlv;nidV=I
vTds= Work v:
To prove the theorem we again use Stokes' theorem and the connection between integrals of
functions and di erential forms. Set =D(v). Then curl v=D 1?d . We have
I
vTds=Z
=Z
d = Flux (D 1?(d )) = Flux (curlv):
Again, similar to the previous case, this theorem clari es the meaning of curl. Indeed, let us
denote by Dr(x;w) the 2-dimensional disc of radius rinR3centered at a point x2R3and
orthogonal to a unit vector w2R3
x. Set
c(x;w) = lim
r!0Work@Dr(x;w)v
r2:
Then
c(x;w) = lim
r!0FluxDr(x;w)(curlv)
r2= lim
r!0R
Dr(x;w)hcurlv;wi
r2=hcurlv;wi:
Hence,jjcurlv(x)jj= max
wc(x;w) and direction of curl v(x) coincides with the direction of the
vector wfor which the maximum value of c(x;w) is achieved.
10.6 Expressing divand curlin curvilinear coordinates
Let us show how to compute div vand curl vof a vector eld vinR3given in a curvilinear
coordinates u1;u2;u3, i.e. expressed through the coordinate vector elds@
@u1;@
@u2and@
@u3. Let
=f(u1;u2;u3)du1^du2^du3
be the volume form dx1^dx2^dx3expressed in coordinates u1;u2;u3.
Let us rst compute ?(du1^du2^du3). We have
?(du1^du2^du3) =?1
fdx1^dx2^dx3
=1
f:
183
Let
v=a1@
@u1+a2@
@u2+a3@
@u3:
Then we have
divv=?d(v
)
=?d 3X
1ai@
@ui!
fdu 1^du2^du3!
=?d(fa1du2^du3+fa2du3^du1+fa3du1^du2)
=?@(fa1)
@u1+@(fa2)
@u2+@(fa3)
@u3
du1^du2^du3
=1
f@(fa1)
@u1+@(fa2)
@u2+@(fa3)
@u3
=@a1
@u1+@a2
@u2+@a3
@u3+1
f@f
@u1a1+@f
@u2a2+@f
@u3a3
:
In particular, we see that the divergence of a vector eld is expressed by the same formulas as in
the cartesian case if and only if the volume form is proportional to the form du1^du2^du3with
aconstant coecient.
For instance, in the spherical coordinates the volume form can be written as
=r2sin'dr^d'^d;4
and hence the divergence of a vector eld
v=a@
@r+b@
@+c@
@'
can be computed by the formula
divv=@a
@r+@b
@+@c
@'+2a
r+ccot':
The general formula for curl vin curvilinear coordinates looks pretty complicated. So instead
of deriving the formula we will explain here how it can be obtained in the general case, and then
illustrate this procedure for the spherical coordinates.
By the de nition we have
4Note that the spherical coordinates ordered as ( r; ; ) determine the same orientation of R3as the cartesian
coordinates ( x;y;z ).
184
curlv=D 1(?(d(D(v)))):
Hence we rst need to compute D(v).
To do this we need to introduce a symmetric matrix
G=0
BBB@g11g12g13
g21g22g23
g31g32g331
CCCA;
where
gij=h@
@ui;@
@uji; i;j = 1;2;3:
The matrix Gis called the Gram matrix .
Notice, that if D(v) =A1du1+Bdu 2+Cdu 3then for any vector h=h1@
@u1+h2@
@u2+h3@
@u3:
we have
D(v)(h) =A1h1+A2h2+A3h3=
A1A2A30
BBB@h1
h2
h31
CCCA=hv;hi:
But
hv;hi=
a1a2a3
G0
BBB@h1
h2
h31
CCCA:
Hence
A1A2A3
=
a1a2a3
G;
or, equivalently, because the Gram matrix Gis symmetric we can write
0
BBB@A1
A2
A31
CCCA=G0
BBB@a1
a2
a31
CCCA;
and therefore,
Ai=nX
i=1gijaj; i= 1;2;3:
185
After computing
!=d(D(v)) =B1du2^du3+B2du3^du1+B3du1^du2
we compute curl vby the formula = curl v=D 1(?!). Let us recall (see Proposition 4.15 above)
that for any vector eld wthe equality Dw=?!is equivalent to the equality w
=!, where
=fdu 1^du2^du3is the volume form. Hence, if
curlv=c1@
@u1+c2@
@u2+c3@
@u3
then we have
w
=fc1du2^du3+fc2du3^du1+fc3du1^du2;
and therefore,
curlv=B1
f@
@u1+B2
f@
@u2+B3
f@
@u3:
Let us use the above procedure to compute curl vof the vector eld
v=a@
@r+b@
@ +c@
@
given in the spherical coordinates. The Gram matrix in this case is the diagonal matrix
G=0
BBB@1 0 0
0r20
0 0r2sin2'1
CCCA:
Hence,
Dv=adr+br2d'+cr2sin2'd;
and
!=d(Dv) =da^dr+d(br2)^d'+d(cr2sin2')^d
=
r@b
@+r2csin 2'+r2sin2'@c
@'
d'^d
+
@a
@'+r2@b
@r+ 2br
dr^d'+
2rcsin2' r2sin2'@c
@r+@a
@
d^dr:
Finally, we get the following expression for curl v:
curlv= r@b
@+r2csin 2'+r2sin2'@c
@'
r2cos @
@r+ @a
@'+r2@b
@r+ 2br
r2cos'@
@'+ 2rcsin2' r2sin2'@c
@r+@a
@
r2cos'@
@:
186
Chapter 11
Applications of Stokes' formula
11.1 Integration of closed and exact forms
Let us recall that a di erential k-form!is called closed ifd!= 0, and that it is called exact if
there exists a ( k 1)-form , called primitive of!, such that !=d .
Any exact form is closed, because d(d ) = 0. Any n-form in an-dimensional space is closed.
Proposition 11.1. a) For a closed k-form!de ned near a (k+ 1)-dimensional submanifold
with boundary @we haveZ
@!= 0:
b) If!is exactk-form de ned near a closed k-dimensional submanifold Sthen
Z
S!= 0:
The proof immediately follows from Stokes' formula. Indeed, in case a) we have
Z
@!=Z
d!= 0:
In case b) we have !=d and@S=?. Thus
Z
Sd =Z
; = 0:
Proposition 11.1b) gives a necessary condition for a closed form to be exact.
187
Example 11.2. The di erential 1-form =1
x2+y2(xdy ydx) de ned on the punctured plane
R2n0 is closed but not exact.
Indeed, it is straightforward to check that is exact (one can simplify computations by passing
to polar coordinates and computing that =d'). To check that it is not exact we compute the
integralR
S , whereSin the unit circle fx2+y2= 1g. We have
Z
S =2Z
0d'= 26= 0:
More generally, an ( n 1)-form
n=nX
i=1( 1)i 1xi
rndx1^i
:::^dxn(dxiis missing)
is closed in Rnn0. However, it is not exact. Indeed, let us show thatR
Sn 1n6= 0, where Sn 1is the
unit sphere oriented as the boundary of the unit ball. Let us recall that the volume form Sn 1on
the unit sphere is de ned as
Sn 1=n
=nX
i=1( 1)i 1xi
rdx1^i
:::^dxn:
Notice that njSn 1=Sn 1, and hence
Z
Sn 1n=Z
Sn 1Sn 1=Z
Sn 1dV= Vol(Sn 1)>0:
11.2 Approximation of continuous functions by smooth ones
Theorem 11.3. LetCVbe a compact domain with smooth boundary. Then any continuous
functionf:C!Rcan beC0-approximated by C1- smooth functions, i.e. for any >0there
exists aC1-smooth function g:C!Rsuch thatjf(x) g(x)j< for anyx2C. Moreover, if
the function fis alreadyC1-smooth in a neighborhood of a closed subset BIntC, then one can
arrange that the function gcoincides with foverB.
Lemma 11.4. There is a continuous extension of ftoV.
188
Sketch of the proof. Letnbe the outward normal vector eld to the boundary @C. If the
boundary is C1-smooth then so is the vector eld n. Consider a map :@C[0;1]!Vgiven
by the formula (x;t) =x+tn;x2@C;t2[0;1]. The di erential of at the points of @C0 has
rankn. (Exercise: prove this. ) Hence by the inverse function theorem for a suciently small >0
the mapis a di eomorphism of @C[0;) ontoUnIntCfor some open neighborhood UC.
Consider a function F:@C[0;)!R, de ned by the formula
F(x;t) =
1 2t
f(x)
ift2[0;
2] andf(x;t) = 0 ift2(
2;). Now we can extend ftoUby the formula f(y) =F 1(y)
ify2UnC, and setting it to 0 outside U.
Consider the function
=1R
D(0) 0;dV 0;;
where 0;is a bump function de ned above in (9.1.2). It is supported in the disc D(0), non-
negative, and satis esZ
D(0) dV= 1:
. Given a continuous function f:V!Rwe de ne a function f:V!Rby the formula
f(x) =Z
f(x y) (y)dny: (11.2.1)
Then
Lemma 11.5. 1. The function fisC1-smooth.
2. For any >0there exists >0such that for all x2Cwe havejf(x) f(x)j<provided
that< .
Proof. 1. By the change of variable formula we have, replacing the variable ybyu=y x:
f(x) =Z
D(0)f(x y) (y)dny=Z
D( x)f( u) (x+u)dnu=Z
Vf( u) (x+u)dnu:
189
But the expression under the latter integral depends on xC1-smoothly as a parameter. Hence, by
the theorem about di erentiating integral over a parameter, we conclude that the function fin
C1-smooth.
2. Fix some 0>0. The function fis uniformly continuous in U0(C). Hence there exists >0
such thatx;x02U0(C) andjjx x0jj< we havejf(x) f(x0)j< . Hence, for <min(0;)
and forx2Cwe have
jf(x) f(x)j=jZ
D(0)f(x y) (y)dny Z
D(0)f(x) (y)dnyj
Z
D(0)jf(x y) f(x)j (y)dnyZ
D(0) (y)dny=: (11.2.2)
Proof of Theorem 11.3. Lemma 11.5 implies that for a suciently small the function g=f
is the required C1-smooth-approximation of the continuous function f. To prove the second part
of the theorem let us assume that fis alreadyC1-smooth on a neighborhood U,BUC. Let
us choose a cut-o function B;Uconstructed in Lemma 9.2 and de ne the required approximation
gby the formula f+ (f f)B;U.
Theorem 11.3 implies a similar theorem form continuous maps C!Rnby applying it to all
coordinate functions.
11.3 Homotopy
LetA;B be any 2 subsets of vector spaces VandW, respectively. Two continuous maps f0;f1:A!
Bare called homotopic if there exists a continuous map F:A[0;1]!Bsuch thatF(x;0) =f0(x)
andF(x;1) =f1(x) for allt2[0;1]. Notice that the family ft:A!B,t2[0;1], de ned by the
formulaft(x) =F(x;t) is a continuous deformation connecting f0andf1. Conversely, any such
continuous deformation fftgt2[0;1]provides a homotopy between f0andf1.
Given a subset CA, we say that a homotopy fftgt2[0;1]is xed over Cifft(x) =f0(x) for
allx2Cand allt2[0;1].
A setAis called contractible if there exists a point a2Aand a homotopy ft:A!A,t2[0;1],
such thatf1= Id andf0is a constant map, i.e. f1(x) =xfor allx2Aandf0(x) =a2Afor all
x2A.
190
Example 11.6. Any star-shaped domain AinVis contractible. Indeed, assuming that it is star-
shaped with respect to the origin, the required homotopy ft:A!A,t2[0;1], can be de ned by the
formulaft(x) =tx; x2A.
Remark 11.7. In what follows we will always assume all homotopies to be smooth. According to
Theorem 11.3 this is not a serious constraint. Indeed, any continuous map can be C0-approximated
by smooth ones, and any homotopy between smooth maps can be C0-approximated by a smooth
homotopy between the same maps.
Lemma 11.8. LetUVbe an open set, Aa compact oriented manifold (possibly with boundary)
and a smooth closed di erential k-form onU. Letf0;f1:A!Ube two maps which are homotopic
relative to the boundary @A. ThenZ
Af
0 =Z
Af
1 :
Proof. LetF:A[0;1]!Ube the homotopy map between f0andf1. By assumption d = 0,
and henceR
A[0;1]Fd = 0:Then, using Stokes' theorem we have
0 =Z
A[0;1]Fd =Z
A[0;1]dF =Z
@(A[0;1])F =Z
@A[0;1]F +Z
A1F +Z
A0F
where the boundary @(A[0;1]) = (A1)\(A0)\(@A[0;1]) is oriented by an outward
normal vector eld n. Note that n=@
@tonA1 and n= @
@tonA0, where we denote by t
the coordinate corresponding to the factor [0 ;1]. First, we notice that F j@A[0;1]= 0 because the
mapFis independent of the coordinate t, when restricted to @A[0;1]. HenceR
@A[0;1]F = 0.
Consider the inclusion maps A!A[0;1] de ned by the formulas j0(x) = (x;0) andj1(x) = (x;1).
Note thatj0;j1are di eomorphisms A!A0 andA!A1, respectively. Note that the map
j1preserves the orientation while j0reverses it. We also have Fj0=f0,Fj1=f1. Hence,
R
A1F =R
Af
1 andR
A0F = R
Af
0 . Thus,
0 =Z
@(A[0;1])F =Z
@A[0;1]F +Z
A1F +Z
A0F =Z
Af
1 Z
Af
0 :
191
Lemma 11.9. LetAbe an oriented m-dimensional manifold, possibly with boundary. Let
(A)
denote the space of di erential forms on Aand
(A[0;1])denote the space of di erential forms
on the product A[0;1]. Letj0;j1:A!A[0;1]be the inclusion maps j0(x) = (x;0)2A[0;1]
andj1(x) = (x;1)2A[0;1]. Then there exists a linear map K:
(A[0;1])!
(A)such that
If is ak-form,k= 1;:::;m thenK( )is a(k 1)-form;
dK+Kd=j
1 j
0;i.e. for each di erential k-form 2
k(A[0;1]one hasdK( ) +
K(d ) =j
1 j
0 :
Remark 11.10. Note that the rst din the above formula denotes the exterior di erential
k(A)!
k(A), while the second one is the exterior di erential
k(A[0;1])!
k(A[0;1]).
Proof. Let us write a point in A[0;1] as (x;t),x2A;t2[0;1]. To construct K( ) for a given
2
k(A[0;1] we rst contract with the vector eld@
@tand then integrate the resultant form
with respect to the t-coordinate. More precisely, note that any k-form onA[0;1] can be written
as = (t) +dt^
(t),t2[0;1], where for each t2[0;1]
(t)2
k(A);
(t)2
k 1(A):
Then@
@t =
(t) and we de ne K( ) =1R
0
(t)dt:
If we choose a local coordinate system ( u1;:::;um) onAthen
(t) can be written as
(t) =
P
1i1<<imhi1:::ik(t)dui1^^duik, and hence
K( ) =1Z
0
(t)dt=X
1i1<<im0
@1Z
0hi1:::ik(t)dt1
Adui1^^duik:
Clearly,Kis a linear operator
k(AI)!
k 1(A):
Note that if = (t) +dt^
(t)2
k(A[0;1]) then
j
0 = (0); j
1 = (1):
We further have
K( ) =1Z
0
(t)dt;
d =dAI =dU (t) +dt^_ (t) dt^dU
(t) =dA (t) +dt^(_ (t) dA
(t));
192
where we denoted _ (t) :=@ (t)
@tandI= [0;1]:Here the notation dAIstands for exterior di erential
on
(AI) anddAdenotes the exterior di erential on
( A). In other words, when we write dA (t)
we view (t) as a form on Adepending on tas a parameter. We do not write any subscript for d
when there could not be any misunderstanding.
Hence,
K(d ) =1Z
0(_ (t) dA
(t))dt= (1) (0) 1Z
0dA
(t)dt;
d(K( )) =1Z
0dA
(t)dt:
Therefore,
d(K( )) +K(d( )) = (1) (0) 1Z
0dA
(t)dt+1Z
0dA
(t)dt
= (1) (0) =j
1( ) j
0( ):
Theorem 11.11. (Poincar e's lemma) LetUbe a contractible domain in V. Then any closed form
inUis exact. More precisely, let F:U[0;1]!Ube the contraction homotopy to a point a2U,
i.e.F(x;1) =x; F(x;0) =afor allx2U. Then if!a closedk-form inUthen
!=dK(F!);
whereK:
k+1(U[0;1])!
k(U)is an operator constructed in Lemma 11.9.
Proof. Consider a contraction homotopy F:U[0;1]!U. ThenFj0(x) =a2Uand
Fj1(x) =xfor allx2U. Consider an operator K:
U)!
(U) constructed above. Thus
Kd+dK=j
1 j
0:
Let!be a closed k-form onU. Denote :=F!. Thus is ak-form onU[0;1]. Note that
d =dF!=Fd!= 0,j
1 = (Fj1)!=!andj
0 = (Fj0)!= 0. Then, using Lemma 11.9
we have
K(d ) +dK( ) =dK( ) =j
1 j
0 =!; (11.3.1)
193
i.e.!=dK(F!).
In particular, any closed form is locally exact .
Example 11.12. Let us work out explicitly the formula for a primitive of a closed 1-form in a
star-shaped domain URn. We can assume that Uis star-shaped with respect to the origin.
Let =nP
1fidxibe a closed 1-form. Then according to Theorem 11.11 we have =dF, where
F=K( ), where : U[0;1]!Uis a contraction homotopy, i.e. ( x;1) =x;(x;1) = 0 for
x2U. can be de ned by the formula ( x;t) =tx; x2U;t2[0;1]. Then
=nX
1fi(tx)d(txi) =nX
1tfi(tx)dxi+nX
1xifi(tx)dt:
Hence,
K( ) =1Z
0@
@t =1Z
0 nX
1xifi(tx)!
dt:
Note that this expression coincides with the expression in formula (9.4.1) in Section 9.4.
Exercise 11.13. Work out an explicit expression for a primitive of a closed 2-form =Pdy^dz+
Qdz^dx+Rdx^dyon a star-shaped domain UR3.
Example 11.14. 1.Rn
+=fx10gRnis not di eomorphic to Rn.Indeed, suppose there exists
such a di eomorphism f:Rn
+!Rn. Denotea:=f(0). Without loss of generality we can assume
thata= 0. Thenef=fjRn
+n0is a di eomorphism Rn
+n0!Rnn0. But Rn
+n0 is star-shaped with
respect to any point with positive coordinate x1, and hence it is contractible. In particular any
closed form on Rn
+n0 is exact. On the other hand, we exhibited above in 11.2 a closed ( n 1)-form
onRnn0 which is not exact.
2.Borsuk's theorem :There is no continuous map Dn!@Dnwhich is the identity on @Dn.We
denote here by Dnthe unit disc in Rnand by@Dnits boundary ( n 1)-sphere.
Proof. Suppose that there is such a map f:Dn!@Dn. One can assume that fis smooth.
Indeed, according to Theorem 11.3 one can approximate fby a smooth map, keeping it xed on
the boundary where it is the identity map, and hence smooth. Take the closed non-exact form n
from Example 11.2 on @Dn. Then n=fnis a closed ( n 1)-form on Dnwhich coincides with
194
[email protected] star-shaped, and therefore nis exact, n=d!. But thenn=d(!j@Dn) which
is a contradiction.
3.Brouwer's fixed point theorem :Any continuous map f:Dn!Dnhas at least 1 xed point.
Proof. Supposef:Dn!Dnhas no xed points. Let us de ne a map F:Dn!@Dnas follows.
For eachx2Dntake a rayrxfrom the point f(x) which goes through xtill it intersects @Dnat a
point which we will denote F(x). The map is well de ned because for any xthe pointsxandf(x)
are distinct. Note also that if x2@Dnthen the ray rxintersects@Dnat the point x, and hence
F(x) =xin this case. But existence of such Fis ruled out by Borsuk' theorem.
k-connected manifolds
A subsetAVis calledk-connected ,k= 0;1;:::; if for anymkany two continuous maps
of discsf0;f1:Dm!Awhich coincide along @Dmare homotopic relative to @Dm. Thus, 0-
connectedness is equivalent to path-connectedness. 1-connected submanifolds are also called simply
connected.
Exercise 11.15. Prove thatk-connectedness can be equivalently de ned as follows: Aisk-connected
if any map f:Sm!A,mkis homotopic to a constant map.
Example 11.16. 1.IfAis contractible then it is k-connected for any k.For some classes of
subsets, e.g. submanifolds, the converse is also true (J.H.C Whitehead's theorem) but this is
a quite deep and non-trivial fact.
2.Then-sphereSnis(n 1)-connected but not n-connected. Indeed, to prove that Sn 1simply
connected we will use the second de nition. Consider a map f:Sk!Sn. We rst notice
that according to Theorem 11.3 we can assume that the map fis smooth. Hence, according
to Corollary 9.21 Vol nf(Sk) = 0 provided that k < n . In particular, there exists a point
p2Snnf(Sk). But the complement of a point pinSnis di eomorphic to Snvis the
stereographic projection from the point p. ButRnis contractible, and hence fis homotopic
to a constant map. On the other hand, the identity map Id : Sn!Snis not homotopic to
a constant map. Indeed, we know that there exists a closed n-form onSn, say the form n
195
from Example 11.2, such thatR
Snn6= 0. Hence,R
SnIdn6= 0. On the other hand if Id were
homotopic to a constant map this integral would vanish.
Exercise 11.17. Prove that Rn+1n0 is (n 1) connected but notn-connected.
Proposition 11.18. LetUVbe am-connected domain. Then for any kmany closed
di erential k-form inUis exact.
Proof. We will prove here only the case m= 1. Though the general case is not dicult, it requires
developing certain additional tools. Let be a closed di erential 1-form. Choose a reference point
b2U. By assumption, Uis path-connected. Hence, any other point xcan be connected to bby
a path
x: [0;1]!U, i.e.
x(0) =b;
x(1) =x. Let us de ne the function F:U!Rby the
formulaF(x) =R
x :Note that due to the simply-connectedness of the domain U, any: [0;1]!U
connecting bandxis homotopic to
xrelative its ends, and hence according to Lemma 11.8 we
haveR
x =R
. Thus the above de nition of the function Fis independent of the choice of paths
x. We claim that the function Fis di erentiable and dF= . Note that if the primitive of exists
than it has to be equal to Fup to an additive constant. But we know that in a suciently small
ballB(a) centered at any point a2Uthere exists a primitive Gof . Hence,G(x) =F(x)+const,
and the the di erentiability of Gimplies di erentiablity of Fand we have dF=dG= .
11.4 Winding and linking numbers
Given a loop
:S1!R2n0 we de ne its winding number around 0 as the integral
w(
) =1
21Z
S2=1
2Z
S1xdy ydx
x2+y2;
where we orient S1as the boundary of the unit disc in R2. For instance, if j:S1,!R2is the
inclusion map then w(j) = 1. For the loop
nparameterized by the map t7!(cosnt;sinnt);t2
[0;2] we havew(
n) =n.
Proposition 11.19. 1. For any loop
the number w(
)is an integer.
2. If loops
0;
1:S1!R2n0are homotopic in R2n0thenw(
0) =w(
1).
196
Figure 11.1: w( ) = 2:
3.w(
) =nthen the loop
is homotopic (as a loop in R2n0) to the loop n: [0;1]!R2n0
given by the formula n(t) = (cos 2nt;sin 2nt).
Proof. 1. Let us de ne the loop
parametrically in polar coordinates:
r=r(s); = (s); s2[0;1];
wherer(0) =r(1) and (1) = (0) + 2n. The form 2in polar coordinates is equal to d , and
hence
Z
=1
21Z
0 0(s)ds= (1) (0)
2=n:
2. This is an immediate corollary of Proposition 11.8.
3. Let us write both loops
andnin polar coordinates. Respectively,we have r=r(t); = (s)
for
andr= 1; = 2nsforn,s2[0;1]. The condition w(
) =nimplies, in view of part 1, that
(1) = (0) + 2n. Then the required homotopy
t,t2[0;1], connecting the loops
0=
and
1=ncan be de ned by the parametric equations r= (1 t)r(s)+t, = t(s) = (1 t) (s)+2nst.
Note that for all t2[0;1] we have t(1) = t(0) + 2n. Therefore,
tis a loop for all t2[0;1].
197
Figure 11.2: l(
1;
2) = 1:
Given two disjoint loops
;:S1!R3(i.e.
(s)6=(t) for anys;t2S1) consider a map
F
;:T2!R3n0, whereT2=S1S1is the 2-torus, de ned by the formula
F
;(s;t) =
(s) (t):
Then the number
l(
;) :=1
42Z
TF
;3=1
42Z
TF
;
xdy^dz+ydz^dx+zdx^dy
(x2+y2+z2)3
2!
is called the linking number of loops
;.1
Exercise 11.20. Prove that
1. The number l(
;) remains unchanged if one continuously deforms the loops
; keeping
them disjoint ;
2. The number l(
;) is an integer for any disjoint loops
;;
1This de nition of the linking number is due to Carl Friedrich Gauss.
198
3.l(
;) =l(;
);
4. Let
(s) = (coss;sins;0);s2[0;2] and(t) = ( 1 +1
2cost;0;1
2sint),t2[0;2]. Then
l(
;) = 1:
11.5 Properties of k-forms on k-dimensional manifolds
Ak-form onk-dimensional submanifold is always closed. Indeed, d is a (k+ 1)-form and hence
it is identically 0 on a k-dimensional manifold.
Remark 11.21. Given ak-dimensional submanifold AV, and ak-form onV, the di erential
dx does not need to vanish at a point x2A. However, d xjTx(A)does vanish.
The following theorem is the main result of this section.
Theorem 11.22. LetAVbe an orientable compact connected k-dimensional submanifold,
possibly with boundary, and a di erential k-form onA.
1. Suppose that @A6=?. Then is exact, i.e. there exists a (k 1)-form onAsuch that
d = .
2. Suppose that Ais closed, i.e. @A=?. Then is exact if and only ifR
A = 0.
To prove Theorem 11.22 we will need a few lemmas.
Lemma 11.23. LetIkbe thek-dimensional cube f 1xj1; j= 1;:::;kg.
1. Let be a di erential k-form onIksuch that
Supp( )\
0Ik 1[[0;1]@Ik 1
=?:
Then there exists a (k 1)-form such thatd = and such that Supp( )\
1Ik 1[[ 1;1]@Ik 1
=
?:
2. Let be a di erential k-form onIksuch that Supp( )IntIkandR
Ik= 0. Then there exists
a(k 1)-form such thatd = andSupp( )IntIk.
199
Proof. We have
=f(x1;:::;xk)dx1^^dxk;
In the rst case of the lemma the function fvanishes on 0Ik 1[[ 1;1]@Ik 1. We will look
for in the form
=g(x1;:::;xk)dx2^^dxk:
Then
d =@g
@x1(x1;:::;xk)dx1^dx2^^dxk:
and hence the equation d = is equivalent to
@g
@x1(x1;:::;xk) =f(x1;:::;xk):
Hence, if we de ne
g(x1;:::;xk) :=x1Z
1f(u;x2;:::;xk)du;
then the form =g(x1;:::;xk)dx2^^dxkhas the required properties.
The second part of the lemma we will prove here only for the case k= 2. The general case can
be handled similarly by induction over k. We have in this case Supp( f)IntI2andR
I2fdS = 0.
Let us denote h(x2) :=1R
1f(x1;x2)dx1:Note thath(u) = 0 ifuis suciently close to 1 or 1.
According to Fubini's theorem,1R
1h(x2)dx2= 0. We can assume that f(x1;x2) = 0 forx11 ,
and henceuR
1f(x1;:::;xk 1;t)dt=h(x1;:::;xk 1) foru2[1 ;1]. Consider any non-negative
C1-function: [1 ;1]!Rsuch that(u) = 1 foru2[1 ;1 2
3] and(u) = 0 foru2[1
3;1].
De ne a function g1:I2!Rby the formula
g1(x1x2) =8
>><
>>:x1R
1f(u;x2)du; x 12[ 1;1 ];
h(x2)(x1); x 12(1 ;1]:
Denote 1=g1(x1;x2)dx2:Thend = on [ 1;1 ][0;1] andd 1=h(x2)0(x1)dx1^dx2on
[1 ;1][0;1]. Note that Supp( 1)IntI2.
200
Let us de ne
g2(x1;x2) :=8
>><
>>:0; x 12[ 1;1 ];
0(x1)x2R
1h(u)du; x 12(1 ;1]
and denote 2=g2(x1;x2)dx1. Thend 2= 0 on [ 1;1 ][ 1;1] and
d 2= h(x2)0(x1)dx1^dx2
on [1 ;1][ 1;1]. Note that g2(x1;1) = 0(x1)1R
1h(u)du= 0. Taking into account that h(u) = 0
whenuis suciently close to 1 or 1 we conclude that h(x1;x2) = 0 near@I2, i.e. Supp( 2)IntI2.
Finally, if we de ne = 1+ 2then we have d = and Supp( )IntI2.
The following lemma is a special case of the, so-called, tubular neighborhood theorem .
Lemma 11.24. LetAVbe a compact k-dimensional submanifold with boundary. Let :
[ 1;1]!Abe an embedding such that (1)2@A, 0(1)?T (1)(@A)and ([0;1))IntA.
Then the embedding can be extended to an embedding : [ 1;1]Ik 1!Asuch that
(t;0) = (t);fort2[ 1;1],02Ik 1;
(1Ik 1)@A;([ 1;1)Ik 1)IntA;
@
@t(1;x)=2T(@A)for allx2Ik 1.
There are many ways to prove this lemma. We will explain below one of the arguments.
Proof. Step 1. We rst construct k 1 ortonormal vector elds 1;:::;kalong = ([ 1;1])
which are tangent to Aand normal to . To do that let us denote by Nuthe normal ( k 1)-
dimensional space NutoTu inTuA. Let us observe that in view of compactness of there is an
>0 with the following property: for any two points u= (t);u0= (t0)2 ,t;t02[ 1;1], such
thatjt t0jthe orthogonal projection Nu!Nu0is non-degenerate (i.e. is an isomorphism).
ChooseN <1
2and consider points uj= (tj), wheretj= 1 +2j
N;j= 1;:::N . Choose any
orthonormal basis 1(0);:::;k(0)2Nu0, parallel transport these vectors to all points of the arc
1= ([ 1;t1]), project them orthogonally to the normal spaces Nuin these points, and then
orthonormalize the resulted bases via the Gram-Schmidt process. Thus we constructed orthonormal
vector elds 1(t);:::k(t)2N (t),t2[ 1;t1]. Now we repeat this procedure beginning with the
201
basis1(t1);:::k(t1)2N (t1)=Nu1and extend the vector elds 1;:::;kto 2= ([t1;t2]).
Continuing this process we will construct the orhonormal vector elds 1;:::;kalong the whole
curve .2
Step 2. Consider a map : [ 1;1]Ik 1!Vgiven by the formula
(t;x1;:::;xk 1) = (t) +k 1X
1xjj(t); t;x 1;:::;xk 12[ 1;1];
where a small positive number will be chosen later. The map is an embedding if is chosen
small enough.3Unfortunately the image ([ 1;1]Ik 1) is not contained in A. We will correct
this in the next step.
Step 3. Take any point a2Aand denote by athe orthogonal projection V!TaA. Let us make
the following additional assumption (in the next step we will show how to get rid of it): there exists
a neighborhood U3a= (1) in@Asuch thata(U)NaTaA. Given >0 let us denote by
B(a) the (k 1)-dimensional ball of radius in the space N TaA. In view of compactness of A
one can choose an >0 such that for all points a2 there exists an embedding ea:B(a)!A
such thataea= Id. Then for a suciently small <p
k 1the mape : [ 1;1]Ik 1!A
de ned by the formula
e (t;x) =e (t) (t;x); t2[ 1;1];x2Ik 1
is an embedding with the required properties.
Step 4 It remains to show how to satisfy the additional condition at the boundary point (1)2
\@Awhich were imposed above in Step 3. Take the point a= (1)2 \@A. Without loss of
generality we can assume that a= 02V. Choose an orthonormal basis v1;:::;vnofVsuch that
v1;:::;vk2Naandvkis tangent to and pointing inward . Let ( y1;:::;yn) be the corresponding
cartesian coordinates in V. Then there exists a neighborhood U3ainAwhich is graphical in
these coordinates and can be given by
yj=j(y1;:::;yk); j=k+ 1;:::;n; ykk(y1;:::;yk 1);jyij;i= 1;:::;k 1;
2Strictly speaking, the constructed vector elds only piece-wise smooth, because we did not make any special
precautions to ensure smoothness at the points uj;j= 1;:::;N 1. This could be corrected via a standard smoothing
procedure.
3Exercise: prove it!
202
where all the rst partial derivatives of the functions k;:::;nvanish at the origin. Take a C1
cut-o function : [0;1)]!Rwhich is equal to 1 on [0 ;1
2] and which is supported in [0 ;1] (see
Lemma 9.2). Consider a map Fgiven by the formula
F(y1;:::;yn) = (y1;:::;yk 1;yk k(y1;:::;yk 1)jjyjj
;yk+1;:::;yn):
For a suciently small >0 this is a di eomorphism supported in an -ball inVcentered in the
origin. On the other hand, the manifold eA=F(A) satis es the extra condition of Step 3.
Lemma 11.25. LetAVbe a (path)-connected submanifold with a non-empty boundary. Then
for any point a2Athere exists an embedding a: [ 1;1]!Asuch that a(0) =a, a(1)2@A
and 0
a(1)?T a(1)(@A):
Sketch of the proof. BecauseAis path-connected with non-empty boundary, any interior point
can be connected by a path with a boundary point. However, this path need not be an embedding.
First, we perturb this path to make it an immersion : [ 1;1]!A, i.e. a map with non-vanising
derivative. This can be done as follows. As in the proof of the previous lemma we consider a su-
ciently small partition of the path, so that two neighboring subdivision points lie in a coordinate
neighborhood. Then we can connect these points by a straight segment in these coordinate neigh-
borhoods. Finally we can smooth the corners via the standard smoothing procedure. Unfortunately
the constructed immersed path may have self-intersection points. First, one can arrange that
there are only nitely many intersections, and then \cut-out the loops", i.e. if (t1) = (t2) for
t1< t2we can consider a new piece-wise smooth path which consists of j[ 1;t1]and j[t2;1]The
new path has less self-intersection points, and thus continuing by induction we will end with a
piece-wise smooth embedding. It remains to smooth again the corners.
Proof of Theorem 11.22. 1. For every point a2Achoose an embedding a: [ 1;1]!A, as in
Lemma 11.25, and using Lemma 11.24 extend ato an embedding a: [ 1;1]Ik 1!Asuch
that
- a(t;0) = (t);fort2[ 1;1], 02Ik 1;
- a(1Ik 1)@A;a([ 1;1)Ik 1)IntA;
-@a
@t(1;x)=2T(@A) for allx2Ik 1.
203
Due to compactness of Awe can choose nitely many such embeddings j= aj,j= 1;:::;N ,
such thatNS
1j(( 1;1]Int(Ik 1) =A. Choose a partition of unity subordinated to this covering
and split the k-form as a sum =KP
1 j, where each iis supported in j(( 1;1]Int(Ik 1))
for somej= 1;:::;N . To simplify the notation we will assume that N=Kand each jis
supported in j(( 1;1]Int(Ik 1)),j= 1;:::;N . Consider the pull-back form e j=
j jon
Ik= [ 1;1]Ik 1. According to Lemma 11.23.1 there exists a ( k 1)-forme jsuch that Supp( e j)
( 1;1]Int(Ik 1) andde j=e j. Let us transport the form e jback toA. Namely, set jequal to
( 1
j)e jon j(( 1;1]Int(Ik 1))Aand extend it as 0 elsewhere on A. Thend j= j, and
henced(NP
1 j) =NP
1 j= .
2. Choose a point a2Aand parameterize a coordinate neighborhood UAby an embedding
:Ik!Asuch that (0) = a. Take a small closed ball D(0)IkRkand denoteeD= (D(0)).
TheneA=AnInteDis a submanifold with non-empty boundary, and @eA=@eD. Let us use part
1 of the theorem to construct a form e oneAsuch thatde = jeA. Let us extent the form e in
any way to a form, still denoted by e on the whole submanifold A. Thende = +where
Supp()eDInt (Ik). Note that
Z
(Ik)=Z
A=Z
A Z
Ade = 0
becauseR
A = 0 by our assumption, andR
Ade = 0 by Stokes' theorem. Thus,kR
I= 0, and
hence, we can apply Lemma 11.23.2 to the form onIkand construct a ( k 1)-formon
Ik 1such thatd= and Supp()IntIk:Now we push-forward the form toA, i.e. take
the formeonAwhich is equal to ( 1)on (Ik) and equal to 0 elsewhere. Finally, we have
d(e +e) =de += ;and hence =e +eis the required primitive of onA.
Corollary 11.26. LetAbe an oriented compact connected k-dimensional submanifold with non-
empty boundary and a di erential k-form onAfrom Theorem 11.22. Then for any smooth map
f:A!Asuch thatfj@A= Id we have
Z
Af =Z
A :
204
Proof. According to Theorem 11.22.1 there exists a form such that =d . Then
Z
Af =Z
Afd =Z
Adf =Z
@Af =Z
@A =Z
A :
Degree of a map
Consider two closed connected oriented submanifolds AV,BWof the same dimension k. Let
!be ann-form onBsuch thatR
B!= 1. Given a smooth map f:A!Bthe integer deg( f) :=R
Af!
is called the degree of the map f.
Proposition 11.27. 1. Given any two k-forms onBsuchR
B!=R
Be!we haveR
Af!=R
Afe!;
for any smooth map f:A!B, and thus deg(f)is independent of the choice of the form !
onBwith the propertyR
A!= 1:
2. If the maps f;g:A!Bare homotopic then deg(f) = deg(g).
3. Letb2Bbe a regular value of the map f. Letf 1(b) =fa1;:::;adg. Then
deg(f) =dX
1sign(detDf(aj)):
In particular, deg(f)is an integer number.
Proof. The second part follows from Lemma 11.8. To prove the rst part, let us write e!=!+,
whereR
B= 0. Using Theorem 11.22.2 we conclude that =d for some (k 1)-form onB.
ThenZ
Afe!=Z
Af!+Z
Af=Z
Af!+Z
Adf =Z
Af!:
Let us prove the last statement of the theorem. By the inverse function theorem there exists a
neighborhood U3binBand neighborhoods U13a1;:::;Ud3adinAsuch that the restrictions
of the map fto the neighborhoods U1;:::;Udare di eomorphisms fjUj:Uj!U,j= 1;:::;d . Let
us consider a form !onBsuch that Supp !UandR
B!=R
U!= 1. Then
deg(f) =Z
Af!=dX
1Z
Ujf!=dX
1sign(detDf(aj));
205
because according to Theorem ??we have
Z
Ujf!= sign(detDf(aj))Z
U!= sign(detDf(aj)):
for eachj= 1;:::;d .
Remark 11.28. Any continuous map f:A!Bcan be approximated by a homotopic to fsmooth
mapA!B, and any two such smooth approximations of fare homotopic. Hence this allows us
to de ne the degree of any continuous mapf:A!B.
Exercise 11.29. 1. Let us view R2asC. In particular, we view the unit sphere S1=S1
1(0) as the
set of complex numbers of modulus 1:
S1=fz2C;jzj= 1g:
Consider a map hn:S1!S1given by the formula hn(z) =zn;z2S1. Then deg( hn) =n.
2. Letf:Sn 1!Sn 1be a map of degree d. Letpbe the north and south poles of Sn+1, i.e.
p= (0;:::; 0;1). Given any point x= (x1;:::;xn+1)2Snnfp+;p gwe denote by (x) the
point
1s
nP
1x2
j(x1;:::;xn)2Sn 1
and de ne a map f:Sn!Snby the formula
f(x) =8
>>><
>>>:p; ifx=p; s
nP
1x2
jf((x));xn+1!
;ifx6=p:
Prove that deg(( f)) =d.4
3. Prove that two maps f;g:Sn!Snare homotopic if and only if they have the same degree.
In particular, any map of degree nis homotopic to the map hn. (Hint: For n=1 this follows from
Proposition 11.19. For n>1 rst prove that any map is homotopic to a suspension. )
4. Give an example of two non-homotopic orientation preserving di eomorphisms T2!T2. Note
that the degree of both these maps is 1. Hence, for manifolds, other than spheres, having the same
degree is not sucient for their homotopy.
4The map fis called the suspension of the map f.
206
5. Let
;:S1!R3be two disjoint loops in R3. Consider a map eF
;:T2!S2de ned by the
formula
eF
;(s;t) =
(s) (t)
jj
(s) (t)jj; s;t2S1:
Prove that l(
;) = deg(eF
;):Use this to solve Exercise 11.20.4 above.
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