differentialforms - mathematics...draft: march28,2018 contents preface v introduction v organization...
TRANSCRIPT
Draft: March 28, 2018
Differential Forms
Victor Guillemin & Peter J. Haine
Draft: March 28, 2018
Draft: March 28, 2018
Contents
Preface vIntroduction vOrganization viNotational Conventions xAcknowledgments xi
Chapter 1. Multilinear Algebra 11.1. Background 11.2. Quotient spaces & dual spaces 31.3. Tensors 81.4. Alternating ð-tensors 111.5. The space ð¬ð(ðâ) 171.6. The wedge product 201.7. The interior product 231.8. The pullback operation on ð¬ð(ðâ) 251.9. Orientations 29
Chapter 2. Differential Forms 332.1. Vector fields and one-forms 332.2. Integral Curves for Vector Fields 372.3. Differential ð-forms 442.4. Exterior differentiation 462.5. The interior product operation 512.6. The pullback operation on forms 542.7. Divergence, curl, and gradient 592.8. Symplectic geometry & classical mechanics 63
Chapter 3. Integration of Forms 713.1. Introduction 713.2. The Poincaré lemma for compactly supported forms on rectangles 713.3. The Poincaré lemma for compactly supported forms on open subsets of ðð 763.4. The degree of a differentiable mapping 773.5. The change of variables formula 803.6. Techniques for computing the degree of a mapping 853.7. Appendix: Sardâs theorem 92
Chapter 4. Manifolds & Forms on Manifolds 974.1. Manifolds 974.2. Tangent spaces 1044.3. Vector fields & differential forms on manifolds 109
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iv Contents
4.4. Orientations 1164.5. Integration of forms on manifolds 1244.6. Stokesâ theorem & the divergence theorem 1284.7. Degree theory on manifolds 1334.8. Applications of degree theory 1374.9. The index of a vector field 143
Chapter 5. Cohomology via forms 1495.1. The de Rham cohomology groups of a manifold 1495.2. The MayerâVietoris Sequence 1585.3. Cohomology of Good Covers 1655.4. Poincaré duality 1715.5. Thom classes & intersection theory 1765.6. The Lefschetz Theorem 1835.7. The KÃŒnneth theorem 1915.8. Äech Cohomology 194
Appendix a. Bump Functions & Partitions of Unity 201
Appendix b. The Implicit Function Theorem 205
Appendix c. Good Covers & Convexity Theorems 211
Bibliography 215
Index of Notation 217
Glossary of Terminology 219
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Preface
Introduction
Formostmath undergraduates oneâs first encounterwith differential forms is the changeof variables formula in multivariable calculus, i.e. the formula
(1) â«ððâð| det ðœð|ðð¥ = â«
ððððŠ
In this formula, ð and ð are bounded open subsets of ðð, ðⶠð â ð is a bounded contin-uous function, ðⶠð â ð is a bijective differentiable map, ðâðⶠð â ð is the functionð â ð, and det ðœð(ð¥) is the determinant of the Jacobian matrix.
ðœð(ð¥) â [ððððð¥ð(ð¥)] ,
As for the âðð¥â and âððŠâ, their presence in (1) can be accounted for by the fact that in single-variable calculus, with ð = (ð, ð), ð = (ð, ð), ðⶠ(ð, ð) â (ð, ð), and ðŠ = ð(ð¥) a ð¶1functionwith positive first derivative, the tautological equation ððŠðð¥ =
ðððð¥ can be rewritten in the form
ð(ðâðŠ) = ðâððŠ
and (1) can be written more suggestively as
(2) â«ððâ(ð ððŠ) = â«
ððððŠ
One of the goals of this text on differential forms is to legitimize this interpretation of equa-tion (1) in ð dimensions and in fact, more generally, show that an analogue of this formulais true when ð and ð are ð-dimensional manifolds.
Another related goal is to prove an important topological generalization of the changeof variables formula (1). This formula asserts that if we drop the assumption that ð be abijection and just require ð to be proper (i.e., that pre-images of compact subsets ofð to becompact subsets of ð) then the formula (1) can be replaced by
(3) â«ððâ(ð ððŠ) = deg(ð) â«
ððððŠ
where deg(ð) is a topological invariant of ð that roughly speaking counts, with plus andminus signs, the number of pre-image points of a generically chosen point of ð.1
This degree formula is just one of a host of results which connect the theory of differ-ential forms with topology, and one of the main goals of this book will explore some of theother examples. For instance, for ð an open subset of ð2, we define ðº0(ð) to be the vector
1It is our feeling that this formula should, like formula (1), be part of the standard calculus curriculum,particularly in view of the fact that there now exists a beautiful elementary proof of it by Peter Lax (see [6,8,9]).
v
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vi Preface
space of ð¶â functions on ð. We define the vector space ðº1(ð) to be the space of formalsums(4) ð1 ðð¥1 + ð2 ðð¥2 ,where ð1, ð2 â ð¶â(ð). We define the vector space ðº2(ð) to be the space of expressions ofthe form(5) ððð¥1 ⧠ðð¥2 ,where ð â ð¶â(ð), and for ð > 2 define ðºð(ð) to the zero vector space.
On these vector spaces one can define operators(6) ðⶠðºð(ð) â ðºð+1(ð)by the recipes
(7) ðð â ðððð¥1ðð¥1 +ðððð¥2ðð¥2
for ð = 0,
(8) ð(ð1 ðð¥1 + ð2 ðð¥2) = (ðð2ðð¥1â ðð1ðð¥2) ðð¥1 ⧠ðð¥
for ð = 1, and ð = 0 for ð > 1. It is easy to see that the operator
(9) ð2 ⶠðºð(ð) â ðºð+2(ð)is zero. Hence,
im(ðⶠðºðâ1(ð) â ðºð(ð)) â ker(ðⶠðºð(ð) â ðºð+1(ð)) ,and this enables one to define the de Rham cohomology groups of ð as the quotient vectorspace
(10) ð»ð(ð) â ker(ðⶠðºð(ð) â ðºð+1(ð))im(ðⶠðºðâ1(ð) â ðºð(ð))
.
It turns out that these cohomology groups are topological invariants of ð and are, infact, isomorphic to the cohomology groups ofð defined by the algebraic topologists. More-over, by slightly generalizing the definitions in equations (4), (5) and (7) to (10) one candefine these groups for open subsets of ðð and, with a bit more effort, for arbitrary ð¶âmanifolds (as we will do in Chapter 5); and their existence will enable us to describe inter-esting connections between problems in multivariable calculus and differential geometryon the one hand and problems in topology on the other.
To make the context of this book easier for our readers to access we will devote therest of this introduction to the following annotated table of contents, chapter by chapterdescriptions of the topics that we will be covering.
Organization
Chapter 1: Multilinear algebraAs we mentioned above one of our objectives is to legitimatize the presence of the ðð¥
and ððŠ in formula (1), and translate this formula into a theorem about differential forms.However a rigorous exposition of the theory of differential forms requires a lot of algebraicpreliminaries, and thesewill be the focus of Chapter 1.Weâll begin, in Sections 1.1 and 1.2, byreviewingmaterial that we hopemost of our readers are already familiar with: the definitionof vector space, the notions of basis, of dimension, of linear mapping, of bilinear form, and
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Organization vii
of dual space and quotient space. Then in Section 1.3 we will turn to the main topics ofthis chapter, the concept of ð-tensor and (the future key ingredient in our exposition ofthe theory of differential forms in Chapter 2) the concept of alternating ð-tensor. Those ðtensors come up in fact in two contexts: as alternating ð-tensors, and as exterior forms, i.e.,in the first context as a subspace of the space of ð-tensors and in the second as a quotientspace of the space of ð-tensors. Both descriptions of ð-tensors will be needed in our laterapplications. For this reason the second half of Chapter 1 ismostly concernedwith exploringthe relationships between these two descriptions and making use of these relationships todefine a number of basic operations on exterior forms such as the wedge product operation(see §1.6), the interior product operation (see §1.7) and the pullback operation (see §1.8).We will alsomake use of these results in Section 1.9 to define the notion of an orientation foran ð-dimensional vector space, a notion that will, among other things, enable us to simplifythe change of variables formula (1) by getting rid of the absolute value sign in the term| det ðœð|.
Chapter 2: Differential FormsThe expressions in equations (4), (5), (7) and (8) are typical examples of differential
forms, and if this were intended to be a text for undergraduate physics majors we woulddefine differential forms by simply commenting that theyâre expressions of this type. Weâllbegin this chapter, however, with the following more precise definition: Let ð be an opensubset of ðð. Then a ð-form ð on ð is a âfunctionâ which to each ð â ð assigns an elementof ð¬ð(ðâðð), ððð being the tangent space to ð at ð, ðâðð its vector space dual, and ð¬ð(ðâð )the ðth order exterior power of ðâðð. (It turns out, fortunately, not to be too hard to recon-cile this definition with the physics definition above.) Differential 1-forms are perhaps bestunderstood as the dual objects to vector fields, and in Sections 2.1 and 2.2 we elaborate onthis observation, and recall for future use some standard facts about vector fields and theirintegral curves. Then in Section 2.3 we will turn to the topic of ð-forms and in the exercisesat the end of Section 2.3 discuss a lot of explicit examples (that we strongly urge readers ofthis text to stare at). Then in Sections 2.4 to 2.7 we will discuss in detail three fundamen-tal operations on differential forms of which weâve already gotten preliminary glimpses inthe equation (3) and equations (5) to (9), namely the wedge product operation, the exteriordifferential operation, and the pullback operation. Also, to add to this list in Section 2.5 wewill discuss the interior product operation of vector fields on differential forms, a general-ization of the duality pairing of vector fields with one-forms that we alluded to earlier. (Inorder to get a better feeling for this material we strongly recommend that one take a lookat Section 2.7 where these operations are related to the div, curl, and grad operations infreshman calculus.) In addition this section contains some interesting applications of thematerial above to physics, in Section 2.8 to electrodynamics and Maxwellâs equation, as wellas to classical mechanisms and the HamiltonâJacobi equations.
Chapter 3: Integration of FormsAs we mentioned above, the change of variables formula in integral calculus is a special
case of a more general result: the degree formula; and we also cited a paper of Peter Laxwhich contains an elementary proof of this formula which will hopefully induce future au-thors of elementary text books in multivariate calculus to include it in their treatment of theRiemann integral. In this chapter we will also give a proof of this result but not, regrettably,Laxâs proof. The reason why not is that we want to relate this result to another result which
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viii Preface
will have some important de Rham theoretic applications when we get to Chapter 5. To de-scribe this result letð be a connected open set in ðð and ð = ððð¥1 â§â¯â§ ðð¥ð a compactlysupported ð-form on ð. We will prove:
Theorem 11. The integral â«ð ð = â«ð ððð¥1â¯ðð¥ð of ð over ð is zero if and only if ð = ððwhere ð is a compactly supported (ð â 1)-form on ð.
An easy corollary of this result is the following weak version of the degree theorem:Corollary 12. Letð andð be connected open subsets ofðð andðⶠð â ð a propermapping.Then there exists a constant ð¿(ð) with the property that for every compactly supported ð-formð on ð
â«ððâð = ð¿(ð)â«
ðð .
Thus to prove the degree theorem it suffices to show that this ð¿(ð) is the deg(ð) in formula(3) and this turns out not to be terribly hard.
The degree formula has a lot of interesting applications and at the end of Chapter 3 wewill describe two of them. The first, the Brouwer fixed point theorem, asserts that if ðµð isthe closed unit ball in ðð and ðⶠðµð â ðµð is a continuous map, then ð has a fixed point,i.e. there is a point ð¥0 â ðµð that gets mapped onto itself by ð. (We will show that if this werenot true the degree theorem would lead to an egregious contradiction.)
The second is the fundamental theorem of algebra. Ifð(ð§) = ð0 + ð1ð§ +⯠+ ððâ1ð§ðâ1 + ð§ð
is a polynomial function of the complex variable ð§, then it has to have a complex root ð§0satisfying ð(ð§0) = 0. (Weâll show that both these results are more-or-less one line corolariesof the degree theorem.)
Chapter 4: Forms on ManifoldsIn the previous two chapters the differential forms that we have been considering have
been forms defined on open subsets of ðð. In this chapter weâll make the transition to formsdefined on manifolds; and to prepare our readers for this transition, include at the beginningof this chapter a brief introduction to the theory of manifolds. (It is a tradition in under-graduate courses to define ð-dimensional manifolds as ð-dimensional submanifolds of ðð,i.e., as ð-dimensional generalizations of curves and surfaces in ð3, whereas the tradition ingraduate level courses is to define them as abstract entities. Since this is intended to be a textbook for undergraduates, weâll adopt the first of these approaches, our reluctance to doingso being somewhat tempered by the fact that, thanks to the Whitney embedding theorem,these two approaches are the same.) In section Section 4.1 we will review a few general factsabout manifolds, in section Section 4.2, define the crucial notion of the tangent space ðððto a manifold ð at a point ð and in sections Sections 4.3 and 4.4 show how to define dif-ferential forms on ð by defining a ð-form to be, as in §2.4, a function ð which assigns toeach ð â ð an element ðð of ð¬ð(ðâðð). Then in Section 4.4 we will define what it meansfor a manifold to be oriented and in Section 4.5 show that ifð is an oriented ð-dimensionalmanifold and ð a compactly supported ð-form, the integral of ð is well-defined. Moreover,weâll prove in this section a manifold version of the change of variables formula and in Sec-tion 4.6 prove manifold versions of two standard results in integral calculus: the divergencetheorem and Stokes theorem, setting the stage for the main results of Chapter 4: the mani-fold version of the degree theorem (see §4.7) and a number of applications of this theorem(among them the JordanâBrouwer separation theorem, the GaussâBonnet theorem and theIndex theorem for vector fields (see §§4.8 and 4.9).
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Organization ix
Chapter 5: Cohomology via FormsGiven an ð-dimensional manifold let ðºð(ð) be the space of differential forms on ð of
degree ð, let ðⶠðºð(ð) â ðºð+1(ð) be exterior differentiation and let ð»ð(ð) be the ðth deRham cohomology group ofð: the quotient of the vector space
ðð(ð) â ker(ðⶠðºð(ð) â ðºð+1(ð))by the vector space
ðµð(ð) â im(ðⶠðºðâ1(ð) â ðºð(ð)) .In Chapter 5 we will make a systematic study of these groups and also show that some ofthe results about differential forms that we proved in earlier chapters can be interpreted ascohomological properties of these groups. For instance, in Section 5.1 we will show thatthe wedge product and pullback operations on forms that we discussed in Chapter 2 givesrise to analogous operations in cohomology and that for a compact oriented ð-dimensionalmanifold the integration operator on forms that we discussed in Chapter 4 gives rise to apairing
ð»ð(ð) à ð»ðâð(ð) â ð ,and in fact more generally ifð is non-compact, a pairing
(13) ð»ðð (ð) à ð»ðâð(ð) â ð ,
where the groups ð»ðð(ð) are the DeRham cohomology groups that one gets by replacingthe spaces of differential formsðºð(ð) by the corresponding spacesðºðð(ð) of compactly sup-ported differential forms.
One problem one runs into in the definition of these cohomology groups is that sincethe spacesðºð(ð) andðºðð (ð) are infinite dimensional in general, thereâs no guarantee that thesame wonât be the case for the spacesð»ð(ð) andð»ðð (ð), and weâll address this problem inSection 5.3.More explicitly, we will say thatð has finite topology if it admits a finite coveringby open sets ð1,âŠðð such that for every multi-index ðŒ = (ð1,⊠, ðð), where 1 †ðð †ð,the intersection(14) ððŒ â ðð1 â©â¯ â© ðððis either empty or is diffeomorphic to a convex open subset ofðð. We will show that ifð hasfinite topology, then its cohomology groups are finite dimensional and, secondly, we willshow that many manifolds have finite topology. (For instance, if ð is compact then ð hasfinite topology; for details see §§5.2 and 5.3.)
We mentioned above that if ð is not compact it has two types of cohomology groups:the groupsð»ð(ð) and the groupsð»ðâðð (ð). In Section 5.5 we will show that ifð is orientedthen for ð â ðºð(ð) and ð â ðºðâðð (ð) the integration operation
(ð, ð) ⊠â«ðð ⧠ð
gives rise to a pairingð»ð(ð) à ð»ðâðð (ð) â ð ,
and that ifð is connected this pairing is non-degenerate, i.e., defines a bijective linear trans-formation
ð»ðâðð (ð) ⥲ ð»ð(ð)â .This results in the Poincaré duality theorem and it has some interesting implications whichwe will explore in Sections 5.5 to 5.7. For instance in Section 5.5 we will show that if ð andð are closed oriented submanifolds of ð and ð is compact and of codimension equal to
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x Preface
the dimension of ð, then the intersection number ðŒ(ð, ð) of ð and ð in ð is well-defined(no matter how badly ð and ð intersect). In Section 5.6 we will show that ifð is a compactoriented manifold and ðⶠð â ð a ð¶â map one can define the Lefschetz number of ð asthe intersection number ðŒ(ð¥ð, graph(ð)) where ð¥ð is the diagonal inð à ð, and view thisas a âtopological countâ of the number of fixed points of the map ð.
Finally in Section 5.8 we will show that if ð has finite topology (in the sense we de-scribed above), then one can define an alternative set of cohomology groups forð, the Äechcohomology groups of a cover U, and we will sketch a proof of the association that this Äechcohomology and de Rham cohomology coincide, leaving a lot of details as exercises (how-ever, with some hints supplied).
AppendicesIn Appendix a we review some techniques in multi-variable calculus that enable one to
reduce global problems to local problems and illustrate these techniques by describing sometypical applications. In particular we show how these techniques can be used to make senseof âimproper integralsâ and to extend differentiable maps ðⶠð â ð on arbitrary subsetsð â ðð to open neighborhoods of these sets.
In Appendix b we discuss another calculus issue: how to solve systems of equations
{{{{{
ð1(ð¥) = ðŠ1â®
ðð(ð¥) = ðŠðfor ð¥ in terms of ðŠ (here ð1,âŠ, ðð are differentiable functions on an open subset of ðð). Toanswer this question we prove a somewhat refined version of what in elementary calculustexts is referred to as the implicit function theorem and derive as corollaries of this theoremtwo results that we make extensive use of in Chapter 4: the canonical submersion theoremand the canonical immersion theorem.
Finally, in Appendix c we discuss a concept which plays an important role in the for-mulation of the âfinite topologyâ results that we discussed in Chapter 5, namely the conceptof a âgood coverâ. In more detail letð be an ð-dimensional manifold an let U = {ððŒ}ðŒâðŒ bea collection of open subsets ofð with the property thatâðŒâðŒððŒ = ð.
Then U is a good cover ofð if for every finite subset
{ðŒ1,âŠ, ðŒð} â ðŒ
the intersection ððŒ1 â©â¯ â© ððŒð is either empty or is diffeomorphic to a convex open subsetof ðð. In Appendix c we will prove that good covers always exist.
Notational Conventions
Below we provide a list of a few of our common notational conventions.â used to define a term; the term being defined appears to the left of the colonâ subsetâ denotes an map (of sets, vector spaces, etc.)⪠inclusion mapâ surjective map⥲ a map that is an isomorphismâ difference of sets: if ðŽ â ð, thenð â ðŽ is the complement of ðŽ inð
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Acknowledgments xi
Acknowledgments
To conclude this introduction we would like to thank Cole Graham and Farzan Vafafor helping us correct earlier versions of this text, a task which involved compiling long listof typos and errata. (In addition a lot of pertinent comments of theirs helped us improvethe readability of the text itself.)
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CHAPTER 1
Multilinear Algebra
1.1. Background
We will list below some definitions and theorems that are part of the curriculum of astandard theory-based course in linear algebra.1
Definition 1.1.1. A (real) vector space is a setð the elements of which we refer to as vectors.The set ð is equipped with two vector space operations:(1) Vector space addition: Given vectors ð£1, ð£2 â ð one can add them to get a third vector,ð£1 + ð£2.
(2) Scalar multiplication: Given a vector ð£ â ð and a real number ð, one can multiply ð£ byð to get a vector ðð£.These operations satisfy a number of standard rules: associativity, commutativity, dis-
tributive laws, etc. which we assume youâre familiar with. (See Exercise 1.1.i.) In additionwe assume that the reader is familiar with the following definitions and theorems.(1) The zero vector in ð is the unique vector 0 â ð with the property that for every vectorð£ â ð, we have ð£ + 0 = 0 + ð£ = ð£ and ðð£ = 0 if ð â ð is nonzero.
(2) Linear independence: A (finite) set of vectors, ð£1,âŠ, ð£ð â ð is linearly independent ifthe map
(1.1.2) ðð â ð , (ð1,âŠ, ðð) ⊠ð1ð£1 +⯠+ ððð£ðis injective.
(3) A (finite) set of vectors ð£1,âŠ, ð£ð â ð spans ð if the map (1.1.2) is surjective.(4) Vectors ð£1,âŠ, ð£ð â ð are a basis of ð if they span ð and are linearly independent; in
other words, if themap (1.1.2) is bijective.Thismeans that every vector ð£ can be writtenuniquely as a sum
(1.1.3) ð£ =ðâð=1ððð£ð .
(5) Ifð is a vector space with a basis ð£1,âŠ, ð£ð, thenð is said to be finite dimensional, and ðis the dimension ofð. (It is a theorem that this definition is legitimate: every basis has tohave the same number of vectors.) In this chapter all the vector spaces weâll encounterwill be finite dimensional. We write dim(ð) for the dimension of ð.
(6) A subsetð â ð is a subspace if it is vector space in its own right, i.e., for all ð£, ð£1, ð£2 â ðand ð â ð, both ðð£ and ð£1 + ð£2 are in ð (where the addition and scalar multiplicationis given as vectors of ð).
(7) Letð andð be vector spaces. A mapðŽâ¶ ð â ð is linear if for ð£, ð£1, ð£2 â ð and ð â ðwe have
(1.1.4) ðŽ(ðð£) = ððŽð£1Such a course is a prerequisite for reading this text.
1
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2 Chapter 1: Multilinear Algebra
andðŽ(ð£1 + ð£2) = ðŽð£1 + ðŽð£2 .
(8) Let ðŽâ¶ ð â ð be a linear map of vector spaces. The kernel of ðŽ is the subset of ðdefined by
ker(ðŽ) â { ð£ â ð |ðŽ(ð£) = 0 } ,i.e, is the set of vectors in ð which ðŽ sends to the zero vector inð. By equation (1.1.4)and item 7, the subset ker(ðŽ) is a subspace of ð.
(9) By (1.1.4) and (7) the set-theoretic image im(ðŽ)of ðŽ is a subspace ofð. We call im(ðŽ)the imageðŽ ofðŽ.The following is an important rule for keeping track of the dimensionsof kerðŽ and imðŽ:
(1.1.5) dim(ð) = dim(ker(ðŽ)) + dim(im(ðŽ)) .(10) Linear mappings and matrices: Let ð£1,âŠ, ð£ð be a basis of ð andð€1,âŠ,ð€ð a basis ofð.
Then by (1.1.3) ðŽð£ð can be written uniquely as a sum,
(1.1.6) ðŽð£ð =ðâð=1ðð,ðð€ð , ðð,ð â ð .
The ð à ð matrix of real numbers, [ðð,ð], is the matrix associated with ðŽ. Conversely,given such anð à ðmatrix, there is a unique linear map ðŽ with the property (1.1.6).
(11) An inner product on a vector space is a mapðµâ¶ ð à ð â ð
with the following three properties:(a) Bilinearity: For vectors ð£, ð£1, ð£2, ð€ â ð and ð â ð we have
ðµ(ð£1 + ð£2, ð€) = ðµ(ð£1, ð€) + ðµ(ð£2, ð€)and
ðµ(ðð£, ð€) = ððµ(ð£, ð€) .(b) Symmetry: For vectors ð£, ð€ â ð we have ðµ(ð£, ð€) = ðµ(ð€, ð£).(c) Positivity: For every vector ð£ â ð we have ðµ(ð£, ð£) ⥠0. Moreover, if ð£ â 0 thenðµ(ð£, ð£) > 0.
Remark 1.1.7. Notice that by property (11b), property (11a) is equivalent toðµ(ð€, ðð£) = ððµ(ð€, ð£)
andðµ(ð€, ð£1 + ð£2) = ðµ(ð€, ð£1) + ðµ(ð€, ð£2) .
Example 1.1.8. The map (1.1.2) is a linear map. The vectors ð£1,âŠ, ð£ð spanð if its image ofthis map isð, and the ð£1,âŠ, ð£ð are linearly independent if the kernel of this map is the zerovector in ðð.
The items on the list above are just a few of the topics in linear algebra that weâre assum-ing our readers are familiar with. We have highlighted them because theyâre easy to state.However, understanding them requires a heavy dollop of that indefinable quality âmathe-matical sophisticationâ, a quality which will be in heavy demand in the next few sections ofthis chapter. We will also assume that our readers are familiar with a number of more low-brow linear algebra notions: matrix multiplication, row and column operations on matrices,transposes of matrices, determinants of ð à ðmatrices, inverses of matrices, Cramerâs rule,recipes for solving systems of linear equations, etc. (See [10, §§1.1 & 1.2] for a quick reviewof this material.)
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§1.2 Quotient spaces & dual spaces 3
Exercises for §1.1Exercise 1.1.i. Our basic example of a vector space in this course is ðð equipped with thevector addition operation
(ð1,âŠ, ðð) + (ð1,âŠ, ðð) = (ð1 + ð1,âŠ, ðð + ðð)and the scalar multiplication operation
ð(ð1,âŠ, ðð) = (ðð1,âŠ, ððð) .Check that these operations satisfy the axioms below.(1) Commutativity: ð£ + ð€ = ð€ + ð£.(2) Associativity: ð¢ + (ð£ + ð€) = (ð¢ + ð£) + ð€.(3) For the zero vector 0 â (0,âŠ, 0) we have ð£ + 0 = 0 + ð£.(4) ð£ + (â1)ð£ = 0.(5) 1ð£ = ð£.(6) Associative law for scalar multiplication: (ðð)ð£ = ð(ðð£).(7) Distributive law for scalar addition: (ð + ð)ð£ = ðð£ + ðð£.(8) Distributive law for vector addition: ð(ð£ + ð€) = ðð£ + ðð€.Exercise 1.1.ii. Check that the standard basis vectors ofðð: ð1 = (1, 0,âŠ, 0), ð2 = (0, 1, 0,âŠ, 0),etc. are a basis.
Exercise 1.1.iii. Check that the standard inner product on ðð
ðµ((ð1,âŠ, ðð), (ð1,âŠ, ðð)) =ðâð=1ðððð
is an inner product.
1.2. Quotient spaces & dual spaces
In this section we will discuss a couple of items which are frequently, but not always,covered in linear algebra courses, but which weâll need for our treatment of multilinear al-gebra in §§1.3 to 1.8.
The quotient spaces of a vector spaceDefinition 1.2.1. Let ð be a vector space andð a vector subspace of ð. Að-coset is a setof the form
ð£ +ð â { ð£ + ð€ |ð€ â ð} .It is easy to check that if ð£1 â ð£2 â ð, the cosets ð£1 +ð and ð£2 +ð, coincide while if
ð£1âð£2 â ð, the cosets ð£1+ð and ð£2+ð are disjoint.Thus the distinctð-cosets decomposeð into a disjoint collection of subsets of ð.Notation 1.2.2. Let ð be a vector space andð â ð a subspace. We write ð/ð for the setof distinctð-cosets in ð.Definition 1.2.3. Let ð be a vector space andð â ð a subspace. Define a vector additionoperation on ð/ð by setting(1.2.4) (ð£1 +ð) + (ð£2 +ð) â (ð£1 + ð£2) + ðand define a scalar multiplication operation on ð/ð by setting(1.2.5) ð(ð£ +ð) â (ðð£) + ð .These operations make ð/ð into a vector space, called quotient space of ð byð.
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It is easy to see that the operations on ð/ð from Definition 1.2.3 are well-defined. Forinstance, suppose ð£1 +ð = ð£â²1 +ð and ð£2 +ð = ð£â²2 +ð. Then ð£1 â ð£â²1 and ð£2 â ð£â²2 are inð, so
(ð£1 + ð£2) â (ð£â²1 + ð£â²2) â ð ,and hence (ð£1 + ð£2) + ð = (ð£â²1 + ð£â²2) + ð.
Definition 1.2.6. Let ð be a vector space andð â ð a subspace. Define a map quotientmap(1.2.7) ðⶠð â ð/ðby setting ð(ð£) â ð£ + ð. It is clear from equations (1.2.4) and (1.2.5) that ð is linear, andthat it maps ð surjectively onto ð/ð.
Observation 1.2.8. Note that the zero vector in the vector space ð/ð is the zero coset0 + ð = ð. Hence ð£ â ker(ð) if and only if ð£ + ð = ð, i.e., ð£ â ð. In other words,ker(ð) = ð.
In the Definitions 1.2.3 and 1.2.6, ð andð do not have to be finite dimensional, but ifthey are then by equation (1.1.5) we have
dim(ð/ð) = dim(ð) â dim(ð) .We leave the following easy proposition as an exercise.
Proposition 1.2.9. Let ðŽâ¶ ð â ð a linear map of vector spaces. Ifð â ker(ðŽ) there existsa unique linear map ðŽâ¯ ⶠð/ð â ð with the property that ðŽ = ðŽâ¯ â ð, where ðⶠð â ð/ðis the quotient map.
The dual space of a vector spaceDefinition 1.2.10. Letð be a vector space.Writeðâ the set of all linear functions âⶠð â ð.Define a vector space structure on ðâ as follows: if â1, â2 â ðâ, then define the sum â1 + â2to be the map ð â ð given by
(â1 + â2)(ð£) â â1(ð£) + â2(ð£) ,which is clearly is linear. If â â ðâ is a linear function and ð â ð, define ðâ by
(ðâ)(ð£) â ð â â(ð£) ;then ðâ is clearly linear.
The vector space ðâ is called the dual space of ð.
Suppose ð is ð-dimensional, and let ð1,âŠ, ðð be a basis of ð. Then every vector ð£ â ðcan be written uniquely as a sum
ð£ = ð1ð1 +⯠+ ðððð , ðð â ð .Let(1.2.11) ðâð (ð£) = ðð .If ð£ = ð1ð1 +â¯+ðððð and ð£â² = ðâ²1ð1 +â¯+ðâ²ððð then ð£+ ð£â² = (ð1 + ðâ²1)ð1 +â¯+ (ðð + ðâ²ð)ðð, so
ðâð (ð£ + ð£â²) = ðð + ðâ²ð = ðâð (ð£) + ðâð (ð£â²) .This shows that ðâð (ð£) is a linear function of ð£ and hence ðâð â ðâ.
Claim 1.2.12. If ð is an ð-dimensional vector space with basis ð1,âŠ, ðð, then ðâ1 ,âŠ, ðâð is abasis of ðâ.
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Proof. First of all note that by (1.2.11)
(1.2.13) ðâð (ðð) = {1, ð = ð0, ð â ð .
If â â ðâ let ðð = â(ðð) and let ââ² = âðððâð . Then by (1.2.13)
ââ²(ðð) =ðâð=1ðððâð (ðð) = ðð = â(ðð) ,
i.e., â and ââ² take identical values on the basis vectors, ðð. Hence â = ââ².Suppose next that âðð=1 ðððâð = 0. Then by (1.2.13), ðð = (â
ðð=1 ðððâð )(ðð) = 0 for ð =
1,âŠ, ð. Hence the vectors ðâ1 ,âŠ, ðâð are linearly independent. â¡
Let ð andð be vector spaces and ðŽâ¶ ð â ð, a linear map. Given â â ðâ the com-position, â â ðŽ, of ðŽ with the linear map, âⶠð â ð, is linear, and hence is an element ofðâ. We will denote this element by ðŽââ, and we will denote by
ðŽâ ⶠðâ â ðâ
the map, â ⊠ðŽââ. It is clear from the definition that
ðŽâ(â1 + â2) = ðŽââ1 + ðŽââ2and that
ðŽâ(ðâ) = ððŽââ ,i.e., that ðŽâ is linear.
Definition 1.2.14. Letð andð be vector spaces andðŽâ¶ ð â ð, a linear map. We call themap ðŽâ ⶠðâ â ðâ defined above the transpose of the map ðŽ.
We conclude this section by giving a matrix description of ðŽâ. Let ð1,âŠ, ðð be a basisof ð and ð1,âŠ, ðð a basis ofð; let ðâ1 ,âŠ, ðâð and ðâ1 ,âŠ, ðâð be the dual bases of ðâ andðâ. Suppose ðŽ is defined in terms of ð1,âŠ, ðð and ð1,âŠ, ðð by theðÃðmatrix, [ðð,ð], i.e.,suppose
ðŽðð =ðâð=1ðð,ððð .
Claim 1.2.15. The linear mapðŽâ is defined, in terms ofðâ1 ,âŠ, ðâð and ðâ1 ,âŠ, ðâð by the trans-pose matrix (ðð,ð).
Proof. Let
ðŽâðâð =ðâð=1ðð,ððâð .
Then
ðŽâðâð (ðð) =ðâð=1ðð,ððâð (ðð) = ðð,ð
by (1.2.13). On the other hand
ðŽâðâð (ðð) = ðâð (ðŽðð) = ðâð (ðâð=1ðð,ððð) =
ðâð=1ðð,ððâð (ðð) = ðð,ð
so ðð,ð = ðð,ð. â¡
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Exercises for §1.2Exercise 1.2.i. Let ð be an ð-dimensional vector space andð a ð-dimensional subspace.Show that there exists a basis, ð1,âŠ, ðð of ð with the property that ð1,âŠ, ðð is a basis ofð.
Hint: Induction on ð â ð. To start the induction suppose that ð â ð = 1. Let ð1,âŠ, ððâ1be a basis ofð and ðð any vector in ð âð.
Exercise 1.2.ii. In Exercise 1.2.i show that the vectors ðð â ð(ðð+ð), ð = 1,âŠ, ð â ð are abasis of ð/ð, where ðⶠð â ð/ð is the quotient map.
Exercise 1.2.iii. In Exercise 1.2.i letð be the linear span of the vectors ðð+ð for ð = 1,âŠ, ðâð.Show that the map
ð â ð/ð , ð¢ ⊠ð(ð¢) ,is a vector space isomorphism, i.e., show that it maps ð bijectively onto ð/ð.
Exercise 1.2.iv. Let ð, ð andð be vector spaces and let ðŽâ¶ ð â ð and ðµâ¶ ð â ð belinear mappings. Show that (ðŽðµ)â = ðµâðŽâ.
Exercise 1.2.v. Let ð = ð2 and letð be the ð¥1-axis, i.e., the one-dimensional subspace
{ (ð¥1, 0) | ð¥1 â ð }of ð2.(1) Show that theð-cosets are the lines, ð¥2 = ð, parallel to the ð¥1-axis.(2) Show that the sum of the cosets, âð¥2 = ðâ and âð¥2 = ðâ is the coset âð¥2 = ð + ðâ.(3) Show that the scalar multiple of the coset, âð¥2 = ðâ by the number, ð, is the coset, âð¥2 =ððâ.
Exercise 1.2.vi.(1) Let (ðâ)â be the dual of the vector space, ðâ. For every ð£ â ð, let evð£ ⶠðâ â ð be the
evaluation function evð£(â) = â(ð£). Show that the evð£ is a linear function on ðâ, i.e., anelement of (ðâ)â, and show that the map
(1.2.16) ev = ev(â) ⶠð â (ðâ)â , ð£ ⊠evð£is a linear map of ð into (ðâ)â.
(2) If ð is finite dimensional, show that the map (1.2.16) is bijective. Conclude that there isa natural identification of ð with (ðâ)â, i.e., that ð and (ðâ)â are two descriptions ofthe same object.
Hint: dim(ðâ)â = dimðâ = dimð, so by equation (1.1.5) it suffices to show that(1.2.16) is injective.
Exercise 1.2.vii. Letð be a vector subspace of a finite dimensional vector space ð and let
ðâ = { â â ðâ | â(ð€) = 0 for all ð€ â ð} .ðâ is called the annihilator ofð in ðâ. Show thatðâ is a subspace of ðâ of dimensiondimð â dimð.
Hint: By Exercise 1.2.i we can choose a basis, ð1,âŠ, ðð of ð such that ð1,âŠðð is a basisofð. Show that ðâð+1,âŠ, ðâð is a basis ofðâ.
Exercise 1.2.viii. Let ð and ðâ² be vector spaces and ðŽâ¶ ð â ðâ² a linear map. Show that ifð â ker(ðŽ), then there exists a linear map ðµâ¶ ð/ð â ðâ² with the property thatðŽ = ðµ â ð(where ð is the quotient map (1.2.7)). In addition show that this linear map is injective ifand only if ker(ðŽ) = ð.
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§1.2 Quotient spaces & dual spaces 7
Exercise 1.2.ix. Let ð be a subspace of a finite dimensional vector space, ð. From theinclusion map, ð ⶠðâ â ðâ, one gets a transpose map,
ðâ ⶠ(ðâ)â â (ðâ)â
and, by composing this with (1.2.16), a map
ðâ â ev ⶠð â (ðâ)â .
Show that this map is onto and that its kernel is ð. Conclude from Exercise 1.2.viii thatthere is a natural bijective linear map
ðⶠð/ð â (ðâ)â
with the property ðâð = ðâ âev. In other words,ð/ð and (ðâ)â are two descriptions of thesame object. (This shows that the âquotient spaceâ operation and the âdual spaceâ operationare closely related.)
Exercise 1.2.x. Letð1 andð2 be vector spaces andðŽâ¶ ð1 â ð2 a linear map. Verify that forthe transpose map ðŽâ ⶠðâ2 â ðâ1 we have:
ker(ðŽâ) = im(ðŽ)â
andim(ðŽâ) = ker(ðŽ)â .
Exercise 1.2.xi. Let ð be a vector space.(1) Let ðµâ¶ ð à ð â ð be an inner product on ð. For ð£ â ð let
âð£ ⶠð â ð
be the function: âð£(ð€) = ðµ(ð£, ð€). Show that âð£ is linear and show that the map
(1.2.17) ð¿â¶ ð â ðâ , ð£ ⊠âð£is a linear mapping.
(2) Ifð is finite dimensional, prove that ð¿ bijective. Conclude that ifð has an inner productone gets from it a natural identification of ð with ðâ.
Hint: Since dimð = dimðâ it suffices by equation (1.1.5) to show that ker(ð¿) = 0.Now note that if ð£ â 0 âð£(ð£) = ðµ(ð£, ð£) is a positive number.
Exercise 1.2.xii. Let ð be an ð-dimensional vector space and ðµâ¶ ð à ð â ð an innerproduct on ð. A basis, ð1,âŠ, ðð of ð is orthonormal if
(1.2.18) ðµ(ðð, ðð) = {1, ð = ð0, ð â ð .
(1) Show that an orthonormal basis exists.Hint: By induction let ð1,âŠ, ðð be vectors with the property (1.2.18) and let ð£ be a
vector which is not a linear combination of these vectors. Show that the vector
ð€ = ð£ âðâð=1ðµ(ðð, ð£)ðð
is nonzero and is orthogonal to the ððâs. Now let ðð+1 = ðð€, where ð = ðµ(ð€,ð€)â 12 .
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8 Chapter 1: Multilinear Algebra
(2) Let ð1,âŠðð and ðâ²1,âŠðâ²ð be two orthonormal bases of ð and let
ðâ²ð =ðâð=1ðð,ððð .
Show that
(1.2.19)ðâð=1ðð,ððð,ð = {
1, ð = ð0, ð â ð .
(3) Let ðŽ be the matrix [ðð,ð]. Show that equation (1.2.19) can be written more compactlyas the matrix identity
(1.2.20) ðŽðŽâº = idðwhere idð is the ð à ð identity matrix.
(4) Let ð1,âŠ, ðð be an orthonormal basis ofð and ðâ1 ,âŠ, ðâð the dual basis ofðâ. Show thatthe mapping (1.2.17) is the mapping, ð¿ðð = ðâð , ð = 1,âŠð.
1.3. Tensors
Definition 1.3.1. Letð be an ð-dimensional vector space and letðð be the set of all ð-tuples(ð£1,âŠ, ð£ð), where ð£1,âŠ, ð£ð â ð, that is, the ð-fold direct sum of ð with itself. A function
ðⶠðð â ðis said to be linear in its ðth variable if, when we fix vectors, ð£1,âŠ, ð£ðâ1, ð£ð+1,âŠ, ð£ð, the mapð â ð defined by(1.3.2) ð£ ⊠ð(ð£1,âŠ, ð£ðâ1, ð£, ð£ð+1,âŠ, ð£ð)is linear.
If ð is linear in its ðth variable for ð = 1,âŠ, ð it is said to be ð-linear, or alternativelyis said to be a ð-tensor. We write Lð(ð) for the set of all ð-tensors in ð. We will agree that0-tensors are just the real numbers, that is L0(ð) â ð.
Let ð1, ð2 ⶠðð â ð be linear functions. It is clear from (1.3.2) that if ð1 and ð2 areð-linear, so is ð1 + ð2. Similarly if ð is ð-linear and ð is a real number, ðð is ð-linear. HenceLð(ð) is a vector space. Note that for ð = 1, âð-linearâ just means âlinearâ, so L1(ð) = ðâ.
Definition 1.3.3. Let ð and ð be positive integers. Amulti-index of ð of length ð is a ð-tupleðŒ = (ð1,âŠ, ðð) of integers with 1 †ðð †ð for ð = 1,âŠ, ð.
Example 1.3.4. Let ð be a positive integer.Themulti-indices of ð of length 2 are in bijectionthe square of pairs of integers integers
(ð, ð) , 1 †ð, ð †ð ,and there are exactly ð2 of them.
We leave the following generalization as an easy exercise.
Lemma 1.3.5. Let ð and ð be positive integers. There are exactly ððmulti-indices of ð of lengthð.
Now fix a basis ð1,âŠ, ðð of ð. For ð â Lð(ð) write(1.3.6) ððŒ â ð(ðð1 ,âŠ, ððð)for every multi-index ðŒ of length ð.
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§1.3 Tensors 9
Proposition 1.3.7. The real numbers ððŒ determine ð, i.e., if ð and ðâ² are ð-tensors and ððŒ =ðâ²ðŒ for all ðŒ, then ð = ðâ².
Proof. By induction on ð. For ð = 1 we proved this result in §1.1. Letâs prove that if thisassertion is true for ð â 1, it is true for ð. For each ðð let ðð be the (ð â 1)-tensor
(ð£1,âŠ, ð£ðâ1) ⊠ð(ð£1,âŠ, ð£ðâ1, ðð) .Then for ð£ = ð1ð1 +â¯ðððð
ð(ð£1,âŠ, ð£ðâ1, ð£) =ðâð=1ðððð(ð£1,âŠ, ð£ðâ1) ,
so the ððâs determine ð. Now apply the inductive hypothesis. â¡
The tensor product operationDefinition 1.3.8. If ð1 is a ð-tensor and ð2 is an â-tensor, one can define a ð + â-tensor,ð1 â ð2, by setting
(ð1 â ð2)(ð£1,âŠ, ð£ð+â) = ð1(ð£1,âŠ, ð£ð)ð2(ð£ð+1,âŠ, ð£ð+â) .This tensor is called the tensor product of ð1 and ð2.
We note that if ð1 is a 0-tensor, i.e., scalar, then tensor product with ð1 is just scalarmultiplication by ð1, and similarly if ð2 is a 0-tensor. That is ð â ð = ð â ð = ðð for ð â ðand ð â Lð(ð).
Properties 1.3.9. Suppose that we are given a ð1-tensor ð1, a ð2-tensor ð2, and a ð3-tensorð3 on a vector space ð.(1) Associativity: One can define a (ð1 + ð2 + ð3)-tensor ð1 â ð2 â ð3 by setting
(ð1 â ð2 â ð3)(ð£1,âŠ, ð£ð+â+ð) â ð1(ð£1,âŠ, ð£ð)ð2(ð£ð+1,âŠ, ð£ð+â)ð3(ð£ð+â+1,âŠ, ð£ð+â+ð) .Alternatively, one can define ð1 â ð2 â ð3 by defining it to be the tensor product of(ð1 â ð2) â ð3 or the tensor product of ð1 â (ð2 â ð3). It is easy to see that both thesetensor products are identical with ð1 â ð2 â ð3:
(ð1 â ð2) â ð3 = ð1 â ð2 â ð3 = ð1 â (ð2 â ð3) .(2) Distributivity of scalar multiplication: We leave it as an easy exercise to check that if ð is
a real number then
ð(ð1 â ð2) = (ðð1) â ð2 = ð1 â (ðð2)(3) Left and right distributive laws: If ð1 = ð2, then
(ð1 + ð2) â ð3 = ð1 â ð3 + ð2 â ð3and if ð2 = ð3, then
ð1 â (ð2 + ð3) = ð1 â ð2 + ð1 â ð3 .
A particularly interesting tensor product is the following. For ð = 1,âŠ, ð let âð â ðâand let
(1.3.10) ð = â1 â⯠â âð .Thus, by definition,
(1.3.11) ð(ð£1,âŠ, ð£ð) = â1(ð£1)â¯âð(ð£ð) .
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A tensor of the form (1.3.11) is called a decomposable ð-tensor. These tensors, as we willsee, play an important role in what follows. In particular, let ð1,âŠ, ðð be a basis of ð andðâ1 ,âŠ, ðâð the dual basis of ðâ. For every multi-index ðŒ of length ð let
ðâðŒ = ðâð1 â⯠â ðâðð .
Then if ðœ is another multi-index of length ð,
(1.3.12) ðâðŒ (ðð1 ,âŠ, ððð) = {1, ðŒ = ðœ0, ðŒ â ðœ
by (1.2.13), (1.3.10) and (1.3.11). From (1.3.12) it is easy to conclude:
Theorem 1.3.13. Let ð be a vector space with basis ð1,âŠ, ðð and 0 †ð †ð be an integer.The ð-tensors ðâðŒ of (1.3.12) are a basis of Lð(ð).
Proof. Given ð â Lð(ð), letðâ² = â
ðŒððŒðâðŒ
where the ððŒâs are defined by (1.3.6). Then
(1.3.14) ðâ²(ðð1 ,âŠ, ððð) = âðŒððŒðâðŒ (ðð1 ,âŠ, ððð) = ððœ
by (1.3.12); however, by Proposition 1.3.7 the ððœâs determine ð, so ðâ² = ð. This proves thatthe ðâðŒ âs are a spanning set of vectors for Lð(ð). To prove theyâre a basis, suppose
âðŒð¶ðŒðâðŒ = 0
for constants, ð¶ðŒ â ð. Then by (1.3.14) with ðâ² = 0, ð¶ðœ = 0, so the ðâðŒ âs are linearly indepen-dent. â¡
As we noted in Lemma 1.3.5, there are exactly ðð multi-indices of length ð and henceðð basis vectors in the set {ðâðŒ }ðŒ, so we have proved
Corollary 1.3.15. Let ð be an ð-dimensional vector space. Then dim(Lð(ð)) = ðð.
The pullback operationDefinition 1.3.16. Let ð andð be finite dimensional vector spaces and let ðŽâ¶ ð â ð bea linear mapping. If ð â Lð(ð), we define
ðŽâðⶠðð â ð
to be the function
(1.3.17) (ðŽâð)(ð£1,âŠ, ð£ð) â ð(ðŽð£1,âŠ,ðŽð£ð) .
It is clear from the linearity of ðŽ that this function is linear in its ðth variable for all ð, andhence is a ð-tensor. We call ðŽâð the pullback of ð by the map ðŽ.
Proposition 1.3.18. The map
ðŽâ ⶠLð(ð) â Lð(ð) , ð ⊠ðŽâð ,
is a linear mapping.
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§1.4 Alternating ð-tensors 11
We leave this as an exercise. We also leave as an exercise the identity
(1.3.19) ðŽâ(ð1 â ð2) = ðŽâ(ð1) â ðŽâ(ð2)
for ð1 â Lð(ð) and ð2 â Lð(ð). Also, if ð is a vector space and ðµâ¶ ð â ð a linearmapping, we leave for you to check that
(1.3.20) (ðŽðµ)âð = ðµâ(ðŽâð)
for all ð â Lð(ð).
Exercises for §1.3Exercise 1.3.i. Verify that there are exactly ðð multi-indices of length ð.
Exercise 1.3.ii. Prove Proposition 1.3.18.
Exercise 1.3.iii. Verify equation (1.3.19).
Exercise 1.3.iv. Verify equation (1.3.20).
Exercise 1.3.v. Let ðŽâ¶ ð â ð be a linear map. Show that if âð, ð = 1,âŠ, ð are elements ofðâ
ðŽâ(â1 â⯠â âð) = ðŽâ(â1) â ⯠â ðŽâ(âð) .Conclude that ðŽâ maps decomposable ð-tensors to decomposable ð-tensors.
Exercise 1.3.vi. Let ð be an ð-dimensional vector space and âð, ð = 1, 2, elements of ðâ.Show that â1 â â2 = â2 â â1 if and only if â1 and â2 are linearly dependent.
Hint: Show that if â1 and â2 are linearly independent there exist vectors, ð£ð, ð =, 1, 2 inð with property
âð(ð£ð) = {1, ð = ð0, ð â ð .
Now compare (â1 â â2)(ð£1, ð£2) and (â2 â â1)(ð£1, ð£2). Conclude that if dimð ⥠2 the tensorproduct operation is not commutative, i.e., it is usually not true that â1 â â2 = â2 â â1.
Exercise 1.3.vii. Let ð be a ð-tensor and ð£ a vector. Define ðð£ ⶠððâ1 â ð to be the map
(1.3.21) ðð£(ð£1,âŠ, ð£ðâ1) â ð(ð£, ð£1,âŠ, ð£ðâ1) .
Show that ðð£ is a (ð â 1)-tensor.
Exercise 1.3.viii. Show that if ð1 is an ð-tensor and ð2 is an ð -tensor, then if ð > 0,
(ð1 â ð2)ð£ = (ð1)ð£ â ð2 .
Exercise 1.3.ix. Let ðŽâ¶ ð â ð be a linear map mapping, and ð£ â ð. Write ð€ â ðŽð£. Showthat for ð â Lð(ð), ðŽâ(ðð€) = (ðŽâð)ð£.
1.4. Alternating ð-tensorsWewill discuss in this section a class of ð-tensors which play an important role inmulti-
variable calculus. In this discussionwewill need some standard facts about the âpermutationgroupâ. For those of you who are already familiar with this object (and I suspect most of youare) you can regard the paragraph below as a chance to re-familiarize yourselves with thesefacts.
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PermutationsDefinition 1.4.1. Let ðŽð be the ð-element set ðŽð â {1, 2,âŠ, ð}. A permutation of orderð is a bijecton ðⶠðŽð ⥲ ðŽð. Given two permutations ð1 and ð2, their product ð1ð2 is thecomposition of ð1 â ð2, i.e., the map,
ð ⊠ð1(ð2(ð)) .For every permutation ð, one denotes by ðâ1 the inverse permutation given by the inversebijection of ð, i.e., defined by
ð(ð) â ð ⺠ðâ1(ð) = ð .Let ðð be the set of all permutations of order ð. One calls ðð the permutation group of ðŽð or,alternatively, the symmetric group on ð letters.
It is easy to check the following.
Lemma 1.4.2. The group ðð has ð! elements.
Definition 1.4.3. Let ð be a positive integer. For every 1 †ð < ð †ð, let ðð,ð be the permu-tation
ðð,ð(â) â{{{{{
ð, â = ðð, â = ðâ, â â ð, ð .
The permuation ðð,ð is called a transposition, and if ð = ð+1, then ðð,ð is called an elementarytransposition.
Theorem 1.4.4. Every permutation in ðð can be written as a product of (a finite number of)transpositions.
Proof. We prove this by induction on ð. The base case when ð = 2 is obvious.For the induction step, suppose that we know the claim for ððâ1. Given ð â ðð, we
have ð(ð) = ð if and only if ðð,ðð(ð) = ð. Thus ðð,ðð is, in effect, a permutation of ðŽðâ1. Byinduction, ðð,ðð can be written as a product of transpositions, so
ð = ðð,ð(ðð,ðð)can be written as a product of transpositions. â¡
Theorem 1.4.5. Every transposition can be written as a product of elementary transpositions.
Proof. Let ð = ððð, ð < ð. With ð fixed, argue by induction on ð. Note that for ð > ð + 1ððð = ððâ1,ððð,ðâ1ððâ1,ð .
Now apply the inductive hypothesis to ðð,ðâ1. â¡
Corollary 1.4.6. Every permutation can be written as a product of elementary transpositions.
The sign of a permutationDefinition 1.4.7. Let ð¥1,âŠ, ð¥ð be the coordinate functions on ðð. For ð â ðð we define
(1.4.8) (â1)ð ââð<ð
ð¥ð(ð) â ð¥ð(ð)ð¥ð â ð¥ð
.
Notice that the numerator and denominator in (1.4.8) are identical up to sign. Indeed,if ð = ð(ð) < ð(ð) = ð, the term, ð¥ð â ð¥ð occurs once and just once in the numerator andonce and just once in the denominator; and if ð = ð(ð) > ð(ð) = ð, the term, ð¥ð âð¥ð, occurs
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§1.4 Alternating ð-tensors 13
once and just once in the numerator and its negative, ð¥ð â ð¥ð, once and just once in thenumerator. Thus
(â1)ð = ±1 .In light of this, we call (â1)ð the sign of ð.
Claim 1.4.9. For ð, ð â ðð we have
(â1)ðð = (â1)ð(â1)ð .
That is, the sign defines a group homomorphism ðð â {±1}.
Proof. By definition,
(â1)ðð = âð<ð
ð¥ðð(ð) â ð¥ðð(ð)ð¥ð â ð¥ð
.
We write the right hand side as a product of
âð<ð
ð¥ð(ð) â ð¥ð(ð)ð¥ð â ð¥ð
= (â1)ð
and
(1.4.10) âð<ð
ð¥ðð(ð) â ð¥ðð(ð)ð¥ð(ð) â ð¥ð(ð)
For ð < ð, let ð = ð(ð) and ð = ð(ð) when ð(ð) < ð(ð) and let ð = ð(ð) and ð = ð(ð) whenð(ð) < ð(ð). Then
ð¥ðð(ð) â ð¥ðð(ð)ð¥ð(ð) â ð¥ð(ð)
=ð¥ð(ð) â ð¥ð(ð)ð¥ð â ð¥ð
(i.e., if ð(ð) < ð(ð), the numerator and denominator on the right equal the numerator anddenominator on the left and, if ð(ð) < ð(ð) are negatives of the numerator and denominatoron the left). Thus equation (1.4.10) becomes
âð<ð
ð¥ð(ð) â ð¥ð(ð)ð¥ð â ð¥ð
= (â1)ð . â¡
Weâll leave for you to check that if ð is a transposition, (â1)ð = â1 and to conclude fromthis:
Proposition 1.4.11. If ð is the product of an odd number of transpositions, (â1)ð = â1 andif ð is the product of an even number of transpositions (â1)ð = +1.
AlternationDefinition 1.4.12. Let ð be an ð-dimensional vector space and ð â Lð(ð£) a ð-tensor. Forð â ðð, define ðð â Lð(ð) to be the ð-tensor
(1.4.13) ðð(ð£1,âŠ, ð£ð) â ð(ð£ðâ1(1),âŠ, ð£ðâ1(ð)) .
Proposition 1.4.14.(1) If ð = â1 â⯠â âð, âð â ðâ, then ðð = âð(1) â⯠â âð(ð).(2) The assignment ð ⊠ðð is a linear map Lð(ð) â Lð(ð).(3) If ð, ð â ðð, we have ððð = (ðð)ð.
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Proof. To (1), we note that by equation (1.4.13)
(â1 â⯠â âð)ð(ð£1,âŠ, ð£ð) = â1(ð£ðâ1(1))â¯âð(ð£ðâ1(ð)) .
Setting ðâ1(ð) = ð, the ðth term in this product is âð(ð)(ð£ð); so the product can be rewrittenas
âð(1)(ð£1)âŠâð(ð)(ð£ð)or
(âð(1) â⯠â âð(ð))(ð£1,âŠ, ð£ð) .We leave the proof of (2) as an exercise.
Now we prove (3). Let ð = â1 â⯠â âð. Then
ðð = âð(1) â⯠â âð(ð) = ââ²1 â⯠â ââ²ðwhere ââ²ð = âð(ð). Thus
(ðð)ð = ââ²ð(1) â⯠â ââ²ð(ð) .But if ð(ð) = ð, ââ²ð(ð) = âð(ð(ð)). Hence
(ðð)ðâðð(1) â⯠â âðð(ð) = ððð . â¡
Definition 1.4.15. Let ð be a vector space and ð ⥠0 an integer. A ð-tensor ð â Lð(ð) isalternating if ðð = (â1)ðð for all ð â ðð.
We denote by Að(ð) the set of all alternating ð-tensors in Lð(ð). By (2) this set is avector subspace of Lð(ð).
It is not easy to write down simple examples of alternating ð-tensors. However, there isa method, called the alternation operation Alt for constructing such tensors.
Definition 1.4.16. Let ð be a vector space and ð a nonnegative integer. The alternationoperation on Lð(ð) is defined as follows: given ð â Lð(ð) let
Alt(ð) â âðâðð(â1)ððð .
The alternation operation enjoys the following properties.
Proposition 1.4.17. For ð â Lð(ð) and ð â ðð,(1) Alt(ð)ð = (â1)ð Altð(2) If ð â Að(ð), then Altð = ð!ð.(3) Alt(ðð) = Alt(ð)ð(4) The map
Alt ⶠLð(ð) â Lð(ð) , ð ⊠Alt(ð)is linear.
Proof. To prove (1) we note that by Proposition 1.4.14:
Alt(ð)ð = âðâðð(â1)ðððð = (â1)ð â
ðâðð(â1)ððððð .
But as ð runs over ðð, ðð runs over ðð, and hence the right hand side is (â1)ð Alt(ð).To prove (2), note that if ð â Að(ð)
Altð = âðâðð(â1)ððð = â
ðâðð(â1)ð(â1)ðð = ð!ð .
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§1.4 Alternating ð-tensors 15
To prove (3), we compute:
Alt(ðð) = âðâðð(â1)ðððð = (â1)ð â
ðâðð(â1)ððððð
= (â1)ð Alt(ð) = Alt(ð)ð .Finally, (4) is an easy corollary of (2) of Proposition 1.4.14. â¡
Wewill use this alternation operation to construct a basis for Að(ð). First, however, werequire some notation.
Definition 1.4.18. Let ðŒ = (ð1,âŠ, ðð) be a multi-index of length ð.(1) ðŒ is repeating if ðð = ðð for some ð â ð .(2) ðŒ is strictly increasing if ð1 < ð2 < ⯠< ðð.(3) For ð â ðð, write ðŒð â (ðð(1),âŠ, ðð(ð)).
Remark 1.4.19. If ðŒ is non-repeating there is a unique ð â ðð so that ðŒð is strictly increasing.
Let ð1,âŠ, ðð be a basis of ð and let
ðâðŒ = ðâð1 â⯠â ðâðð
andððŒ = Alt(ðâðŒ ) .
Proposition 1.4.20.(1) ððŒð = (â1)ðððŒ.(2) If ðŒ is repeating, ððŒ = 0.(3) If ðŒ and ðœ are strictly increasing,
ððŒ(ðð1 ,âŠ, ððð) = {1, ðŒ = ðœ0, ðŒ â ðœ .
Proof. To prove (1) we note that (ðâðŒ )ð = ðâðŒð ; soAlt(ðâðŒð) = Alt(ðâðŒ )ð = (â1)ð Alt(ðâðŒ ) .
To prove (2), suppose ðŒ = (ð1,âŠ, ðð) with ðð = ðð for ð â ð . Then if ð = ððð,ðð , ðâðŒ = ðâðŒð so
ððŒ = ððŒð = (â1)ðððŒ = âððŒ .To prove (3), note that by definition
ððŒ(ðð1 ,âŠ, ððð) = âð(â1)ððâðŒð(ðð1 ,âŠ, ððð) .
But by (1.3.12)
(1.4.21) ðâðŒð(ðð1 ,âŠ, ððð) = {1, ðŒð = ðœ0, ðŒð â ðœ .
Thus if ðŒ and ðœ are strictly increasing, ðŒð is strictly increasing if and only if ðŒð = ðŒ, and(1.4.21) is nonzero if and only if ðŒ = ðœ. â¡
Now let ð be in Að(ð). By Theorem 1.3.13,
ð = âðœððœðâðœ , ððœ â ð .
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16 Chapter 1: Multilinear Algebra
Since ð!ð = Alt(ð),ð = 1ð!âððœ Alt(ðâðœ ) = âððœððœ .
We candiscard all repeating terms in this sum since they are zero; and for every non-repeatingterm, ðœ, we can write ðœ = ðŒð, where ðŒ is strictly increasing, and hence ððœ = (â1)ðððŒ.Conclusion 1.4.22. We can write ð as a sum(1.4.23) ð = â
ðŒððŒððŒ .
with ðŒâs strictly increasing.
Claim 1.4.24. The ððŒâs are unique.
Proof. For ðœ strictly increasing
(1.4.25) ð(ðð1 ,âŠ, ððð) = âðŒððŒððŒ(ðð1 ,âŠ, ððð) = ððœ .
By (1.4.23) the ððŒâs, ðŒ strictly increasing, are a spanning set of vectors for Að(ð), and by(1.4.25) they are linearly independent, so we have proved:
Proposition 1.4.26. The alternating tensors, ððŒ, ðŒ strictly increasing, are a basis for Að(ð).
Thus dimAð(ð) is equal to the number of strictly increasingmulti-indices ðŒ of length ð.We leave for you as an exercise to show that this number is equal to the binomal coefficient
(1.4.27) (ðð) â ð!(ð â ð)!ð!
if 1 †ð †ð. â¡Hint: Show that every strictly increasingmulti-index of length ð determines a ð element
subset of {1,âŠ, ð} and vice-versa.Note also that if ð > ð every multi-index
ðŒ = (ð1,âŠ, ðð)of length ð has to be repeating: ðð = ðð for some ð â ð since the ððâs lie on the interval 1 †ð †ð.Thus by Proposition 1.4.17
ððŒ = 0for all multi-indices of length ð > 0 and
Að(ð) = 0 .Exercises for §1.4
Exercise 1.4.i. Show that there are exactly ð! permutations of order ð.Hint: Induction on ð: Let ð â ðð, and let ð(ð) = ð, 1 †ð †ð. Show that ðð,ðð leaves ð
fixed and hence is, in effect, a permutation of ðŽðâ1.Exercise 1.4.ii. Prove that if ð â ðð is a transposition, (â1)ð = â1 and deduce from thisProposition 1.4.11.
Exercise 1.4.iii. Prove assertion 2 in Proposition 1.4.14.
Exercise 1.4.iv. Prove that dimAð(ð) is given by (1.4.27).
Exercise 1.4.v. Verify that for ð < ð â 1ðð,ð = ððâ1,ððð,ðâ1, ððâ1,ð .
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§1.5 The space ð¬ð(ðâ) 17
Exercise 1.4.vi. For ð = 3 show that every one of the six elements of ð3 is either a transpo-sition or can be written as a product of two transpositions.
Exercise 1.4.vii. Let ð â ðð be the âcyclicâ permutationð(ð) â ð + 1 , ð = 1,âŠ, ð â 1
and ð(ð) â 1. Show explicitly how to write ð as a product of transpositions and compute(â1)ð.
Hint: Same hint as in Exercise 1.4.i.
Exercise 1.4.viii. In Exercise 1.3.vii show that if ð is in Að(ð), ðð£ is in Aðâ1(ð). Show inaddition that for ð£, ð€ â ð and ð â Að(ð) we have (ðð£)ð€ = â(ðð€)ð£.
Exercise 1.4.ix. Let ðŽâ¶ ð â ð be a linear mapping. Show that if ð is in Að(ð), ðŽâð is inAð(ð).Exercise 1.4.x. In Exercise 1.4.ix show that if ð is in Lð(ð) then Alt(ðŽâð) = ðŽâ(Alt(ð)),i.e., show that the Alt operation commutes with the pullback operation.
1.5. The space ð¬ð(ðâ)
In §1.4 we showed that the image of the alternation operation, Alt ⶠLð(ð) â Lð(ð) isAð(ð). In this section we will compute the kernel of Alt.
Definition 1.5.1. A decomposable ð-tensor â1 ââ¯â âð, with â1,âŠ, âð â ðâ, is redundantif for some index ð we have âð = âð+1.
Let Ið(ð) â Lð(ð) be the linear span of the set of redundant ð-tensors.Note that for ð = 1 the notion of redundant does not really make sense; a single vector
â â L1(ðâ) cannot be âredundantâ so we decreeI1(ð) â 0 .
Proposition 1.5.2. If ð â Ið(ð) then Alt(ð) = 0.Proof. Let ð = â1 â⯠â âð with âð = âð+1. Then if ð = ðð,ð+1, ðð = ð and (â1)ð = â1. HenceAlt(ð) = Alt(ðð) = Alt(ð)ð = âAlt(ð); so Alt(ð) = 0. â¡Proposition 1.5.3. If ð â Ið(ð) and ðâ² â Lð (ð) then ð â ðâ² and ðâ² â ð are in Ið+ð (ð).Proof. We can assume that ð and ðâ² are decomposable, i.e., ð = â1 â ⯠â âð and ðâ² =ââ²1 â⯠â ââ²ð and that ð is redundant: âð = âð+1. Then
ð â ðâ² = â1 ââ¯âðâ1 â âð â âð ââ¯âð â ââ²1 â⯠â ââ²ð is redundant and hence in Ið+ð . The argument for ðâ² â ð is similar. â¡
Proposition 1.5.4. If ð â Lð(ð) and ð â ðð, then(1.5.5) ðð = (â1)ðð + ðwhere ð is in Ið(ð).Proof. We can assumeð is decomposable, i.e.,ð = â1ââ¯ââð. Letâs first look at the simplestpossible case: ð = 2 and ð = ð1,2. Then
ðð â (â1)ðð = â1 â â2 + â2 â â1
= 12((â1 + â2) â (â1 + â2) â â1 â â1 â â2 â â2) ,
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and the terms on the right are redundant, and hence in I2(ð). Next let ð be arbitrary andð = ðð,ð+1. If ð1 = â1 â⯠â âðâ2 and ð2 = âð+2 â⯠â âð. Then
ð â (â1)ðð = ð1 â (âð â âð+1 + âð+1 â âð) â ð2is in Ið(ð) by Proposition 1.5.3 and the computation above.The general case: By Theorem 1.4.5, ð can be written as a product ofð elementary transpo-sitions, and weâll prove (1.5.5) by induction onð.
We have just dealt with the caseð = 1.The induction step: theâðâ 1â case implies the âðâ case. Let ð = ððœ where ðœ is a product ofð â 1 elementary transpositions and ð is an elementary transposition. Then
ðð = (ððœ)ð = (â1)ðððœ +â¯= (â1)ð(â1)ðœð +â¯= (â1)ðð +â¯
where the âdotsâ are elements of Ið(ð), and the induction hypothesis was used in the secondline. â¡
Corollary 1.5.6. If ð â Lð(ð), then(1.5.7) Alt(ð) = ð!ð +ð ,
whereð is in Ið(ð).
Proof. By definition Alt(ð) = âðâðð(â1)ððð, and by Proposition 1.5.4, ðð = (â1)ðð +ðð,
withðð â Ið(ð). Thus
Alt(ð) = âðâðð(â1)ð(â1)ðð + â
ðâðð(â1)ððð = ð!ð +ð
whereð = âðâðð(â1)ððð. â¡
Corollary 1.5.8. Let ð be a vector space and ð ⥠1. Then
Ið(ð) = ker(Alt ⶠLð(ð) â Að(ð)) .
Proof. We have already proved that if ð â Ið(ð), then Alt(ð) = 0. To prove the converseassertion we note that if Alt(ð) = 0, then by (1.5.7)
ð = â 1ð!ð .
withð â Ið(ð) . â¡
Putting these results together we conclude:
Theorem 1.5.9. Every element ð â Lð(ð) can be written uniquely as a sum ð = ð1 + ð2,where ð1 â Að(ð) and ð2 â Ið(ð).
Proof. By (1.5.7), ð = ð1 + ð2 with
ð1 =1ð!
Alt(ð)
andð2 = â1ð!ð .
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§1.5 The space ð¬ð(ðâ) 19
To prove that this decomposition is unique, suppose ð1 + ð2 = 0, with ð1 â Að(ð) andð2 â Ið(ð). Then
0 = Alt(ð1 + ð2) = ð!ð1so ð1 = 0, and hence ð2 = 0. â¡
Definition 1.5.10. Let ð be a finite dimensional vector space and ð ⥠0. Define
(1.5.11) ð¬ð(ðâ) â Lð(ð)/Ið(ð) ,i.e., let ð¬ð(ðâ) be the quotient of the vector space Lð(ð) by the subspace Ið(ð). By (1.2.7)one has a linear map:
(1.5.12) ðⶠLð(ð) â ð¬ð(ðâ) , ð ⊠ð + Ið(ð)which is onto and has Ið(ð) as kernel.
Theorem 1.5.13. The map ðⶠLð(ð) â ð¬ð(ðâ)maps Að(ð) bijectively onto ð¬ð(ðâ).
Proof. ByTheorem 1.5.9 every Ið(ð) cosetð+Ið(ð) contains a unique elementð1 ofAð(ð).Hence for every element of ð¬ð(ðâ) there is a unique element of Að(ð) which gets mappedonto it by ð. â¡
Remark 1.5.14. Sinceð¬ð(ðâ) and Að(ð) are isomorphic as vector spaces many treatmentsof multilinear algebra avoid mentioning ð¬ð(ðâ), reasoning that Að(ð) is a perfectly goodsubstitute for it and that one should, if possible, not make two different definitions for whatis essentially the same object.This is a justifiable point of view (and is the point of view takenby in [4,10,12]).There are, however, some advantages to distinguishing betweenAð(ð) andð¬ð(ðâ), as we shall see in §1.6.
Exercises for §1.5Exercise 1.5.i. A ð-tensor ð â Lð(ð) is symmetric if ðð = ð for all ð â ðð. Show that theset ð® ð(ð) of symmetric ð tensors is a vector subspace of Lð(ð).
Exercise 1.5.ii. Let ð1,âŠ, ðð be a basis of ð. Show that every symmetric 2-tensor is of theform
â1â€ð,ðâ€ððð,ððâð â ðâð
where ðð,ð = ðð,ð and ðâ1 ,âŠ, ðâð are the dual basis vectors of ðâ.
Exercise 1.5.iii. Show that if ð is a symmetric ð-tensor, then for ð ⥠2, then ð is in Ið(ð).Hint: Let ð be a transposition and deduce from the identity, ðð = ð, that ð has to be in
the kernel of Alt.
Exercise 1.5.iv (a warning). In general ð® ð(ð) â Ið(ð). Show, however, that if ð = 2 thesetwo spaces are equal.
Exercise 1.5.v. Show that if â â ðâ and ð â Lðâ2(ð), then â â ð â â is in Ið(ð).
Exercise 1.5.vi. Show that if â1 and â2 are inðâ andð â Lðâ2(ð), then â1âðââ2+â2âðââ1is in Ið(ð).
Exercise 1.5.vii. Given a permutation ð â ðð and ð â Ið(ð), show that ðð â Ið(ð).
Exercise 1.5.viii. Let W(ð) be a subspace of Lð(ð) having the following two properties.(1) For ð â ð® 2(ð) and ð â Lðâ2(ð), ð â ð is in W(ð).
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(2) For ð in W(ð) and ð â ðð, ðð is in W(ð).Show that W(ð) has to contain Ið(ð) and conclude that Ið(ð) is the smallest subspace
of Lð(ð) having properties (1) and (2).
Exercise 1.5.ix. Show that there is a bijective linear map
ðŒâ¶ ð¬ð(ðâ) ⥲ Að(ð)with the property
(1.5.15) ðŒð(ð) = 1ð!
Alt(ð)
for allð â Lð(ð), and show that ðŒ is the inverse of themap ofAð(ð) ontoð¬ð(ðâ) describedin Theorem 1.5.13.
Hint: Exercise 1.2.viii.
Exercise 1.5.x. Let ð be an ð-dimensional vector space. Compute the dimension of ð® ð(ð).Hints:
(1) Introduce the following symmetrization operation on tensors ð â Lð(ð):Sym(ð) = â
ðâðððð .
Prove that this operation has properties (2)â(4) of Proposition 1.4.17 and, as a substitutefor (1), has the property: Sym(ð)ð = Sym(ð).
(2) Let ððŒ = Sym(ðâðŒ ), ðâðŒ = ðâð1 ââ¯âðâðð . Prove that { ððŒ | ðŒ is non-decreasing } form a basis
of ðð(ð).(3) Conclude that dim(ð® ð(ð)) is equal to the number of non-decreasing multi-indices of
length ð: 1 †ð1 †ð2 †⯠†âð †ð.(4) Compute this number by noticing that the assignment
(ð1,âŠ, ðð) ⊠(ð1 + 0, ð2 + 1,âŠ, ðð + ð â 1)is a bijection between the set of these non-decreasing multi-indicesand the set of in-creasing multi-indices 1 †ð1 < ⯠< ðð †ð + ð â 1.
1.6. The wedge product
The tensor algebra operations on the spaces Lð(ð)which we discussed in §§1.2 and 1.3,i.e., the âtensor product operationâ and the âpullbackâ operation, give rise to similar oper-ations on the spaces, ð¬ð(ðâ). We will discuss in this section the analogue of the tensorproduct operation.
Definition 1.6.1. Given ðð â ð¬ðð(ðâ), ð = 1, 2 we can, by equation (1.5.12), find a ðð âLðð(ð) with ðð = ð(ðð). Then ð1 â ð2 â Lð1+ð2(ð). The wedge product ð1 ⧠ð2 is defined by(1.6.2) ð1 ⧠ð2 â ð(ð1 â ð2) â ð¬ð1+ð2(ðâ) .Claim 1.6.3. This wedge product is well defined, i.e., does not depend on our choices of ð1 andð2.
Proof. Let ð(ð1) = ð(ðâ²1 ) = ð1. Then ðâ²1 = ð1 +ð1 for someð1 â Ið1(ð), soðâ²1 â ð2 = ð1 â ð2 +ð1 â ð2 .
Butð1 â Ið1(ð) impliesð1 â ð2 â Ið1+ð2(ð) and this implies:ð(ðâ²1 â ð2) = ð(ð1 â ð2) .
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§1.6 The wedge product 21
A symmetric argument shows that ð1 ⧠ð2 is well-defined, independent of the choice ofð2. â¡
More generally let ðð â ð¬ðð(ðâ), for ð = 1, 2, 3, and let ðð = ð(ðð), ðð â Lðð(ð). Define
ð1 ⧠ð2 ⧠ð3 â ð¬ð1+ð2+ð3(ðâ)by setting
ð1 ⧠ð2 ⧠ð3 = ð(ð1 â ð2 â ð3) .As above it is easy to see that this is well-defined independent of the choice of ð1, ð2 and ð3.It is also easy to see that this triple wedge product is just the wedge product of ð1 ⧠ð2 withð3 or, alternatively, the wedge product of ð1 with ð2 ⧠ð3, i.e.,
ð1 ⧠ð2 ⧠ð3 = (ð1 ⧠ð2) ⧠ð3 = ð1 ⧠(ð2 ⧠ð3).We leave for you to check: for ð â ð
(1.6.4) ð(ð1 ⧠ð2) = (ðð1) ⧠ð2 = ð1 ⧠(ðð2)and verify the two distributive laws:
(1.6.5) (ð1 + ð2) ⧠ð3 = ð1 ⧠ð3 + ð2 ⧠ð3and
(1.6.6) ð1 ⧠(ð2 + ð3) = ð1 ⧠ð2 + ð1 ⧠ð3 .As we noted in §1.4, Ið(ð) = 0 for ð = 1, i.e., there are no nonzero âredundantâ ð tensorsin degree ð = 1. Thus
ð¬1(ðâ) = ðâ = L1(ð).A particularly interesting example of a wedge product is the following. Let â1,âŠ, âð â
ðâ = ð¬1(ðâ). Then if ð = â1 â⯠â âð(1.6.7) â1 â§â¯ ⧠âð = ð(ð) â ð¬ð(ðâ) .We will call (1.6.7) a decomposable element of ð¬ð(ðâ).
We will prove that these elements satisfy the following wedge product identity. For ð âðð:(1.6.8) âð(1) â§â¯ ⧠âð(ð) = (â1)ðâ1 â§â¯ ⧠âð .
Proof. For every ð â Lð(ð), ð = (â1)ðð + ð for someð â Ið(ð) by Proposition 1.5.4.Therefore since ð(ð) = 0
ð(ðð) = (â1)ðð(ð) .In particular, if ð = â1 â⯠â âð, ðð = âð(1) â⯠â âð(ð), so
ð(ðð) = âð(1) â§â¯ ⧠âð(ð) = (â1)ðð(ð)= (â1)ðâ1 â§â¯ ⧠âð . â¡
In particular, for â1 and â2 â ðâ
(1.6.9) â1 ⧠â2 = ââ2 ⧠â1and for â1, â2 and â3 â ðâ
â1 ⧠â2 ⧠â3 = ââ2 ⧠â1 ⧠â3 = â2 ⧠â3 ⧠â1 .More generally, it is easy to deduce from equation (1.6.8) the following result (which weâllleave as an exercise).
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22 Chapter 1: Multilinear Algebra
Theorem 1.6.10. If ð1 â ð¬ð(ðâ) and ð2 â ð¬ð (ðâ) then
(1.6.11) ð1 ⧠ð2 = (â1)ðð ð2 ⧠ð1 .
Hint: It suffices to prove this for decomposable elements i.e., for ð1 = â1 â§â¯ ⧠âð andð2 = ââ²1 â§â¯ ⧠ââ²ð . Now make ðð applications of (1.6.9).
Let ð1,âŠ, ðð be a basis of ð and let ðâ1 ,âŠ, ðâð be the dual basis of ðâ. For every multi-index ðŒ of length ð,
(1.6.12) ðâð1 â§â¯ ⧠ðâðð = ð(ð
âðŒ ) = ð(ðâð1 â⯠â ð
âðð) .
Theorem 1.6.13. The elements (1.6.12), with ðŒ strictly increasing, are basis vectors ofð¬ð(ðâ).
Proof. The elements ððŒ = Alt(ðâðŒ ), for ðŒ strictly increasing, are basis vectors of Að(ð) byProposition 1.4.26; so their images, ð(ððŒ), are a basis of ð¬ð(ðâ). But
ð(ððŒ) = ð (âðâðð(â1)ð(ðâðŒ )ð) = â
ðâðð(â1)ðð(ðâðŒ )ð
= âðâðð(â1)ð(â1)ðð(ðâðŒ ) = ð!ð(ðâðŒ ) . â¡
Exercises for §1.6Exercise 1.6.i. Prove the assertions (1.6.4), (1.6.5), and (1.6.6).
Exercise 1.6.ii. Verify the multiplication law in equation (1.6.11) for the wedge product.
Exercise 1.6.iii. Given ð â ð¬ð(ðâ) let ðð be the ð-fold wedge product of ð with itself, i.e.,let ð2 = ð ⧠ð, ð3 = ð ⧠ð ⧠ð, etc.(1) Show that if ð is odd then for ð > 1, ðð = 0.(2) Show that if ð is decomposable, then for ð > 1, ðð = 0.
Exercise 1.6.iv. If ð and ð are in ð¬ð(ðâ) prove:
(ð + ð)ð =ðââ=0(ðâ)ðâ ⧠ððââ .
Hint:As in freshman calculus, prove this binomial theoremby induction using the iden-tity: (ðâ) = (
ðâ1ââ1) + (
ðâ1â ).
Exercise 1.6.v. Let ð be an element ofð¬2(ðâ). By definition the rank of ð is ð if ðð â 0 andðð+1 = 0. Show that if
ð = ð1 ⧠ð1 +⯠+ ðð ⧠ððwith ðð, ðð â ðâ, then ð is of rank †ð.
Hint: Show thatðð = ð!ð1 ⧠ð1 â§â¯ ⧠ðð ⧠ðð .
Exercise 1.6.vi. Given ðð â ðâ, ð = 1,âŠ, ð show that ð1 ⧠⯠⧠ðð â 0 if and only if the ððâsare linearly independent.
Hint: Induction on ð.
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§1.7 The interior product 23
1.7. The interior product
Weâll describe in this section another basic product operation on the spacesð¬ð(ðâ). Asabove weâll begin by defining this operator on the Lð(ð)âs.
Definition 1.7.1. Letð be a vector space and ð a nonnegative integer. Givenð â Lð(ð) andð£ â ð let ðð£ð be the be the (ð â 1)-tensor which takes the value
(ðð£ð)(ð£1,âŠ, ð£ðâ1) âðâð=1(â1)ðâ1ð(ð£1,âŠ, ð£ðâ1, ð£, ð£ð,âŠ, ð£ðâ1)
on the ð â 1-tuple of vectors, ð£1,âŠ, ð£ðâ1, i.e., in the ðth summand on the right, ð£ gets in-serted between ð£ðâ1 and ð£ð. (In particular the first summand is ð(ð£, ð£1,âŠ, ð£ðâ1) and the lastsummand is (â1)ðâ1ð(ð£1,âŠ, ð£ðâ1, ð£).)
It is clear from the definition that if ð£ = ð£1 + ð£2(1.7.2) ðð£ð = ðð£1ð + ðð£2ð ,and if ð = ð1 + ð2(1.7.3) ðð£ð = ðð£ð1 + ðð£ð2 ,and we will leave for you to verify by inspection the following two lemmas.
Lemma 1.7.4. If ð is the decomposable ð-tensor â1 â⯠â âð then
(1.7.5) ðð£ð =ðâð=1(â1)ðâ1âð(ð£)â1 â⯠â âð â⯠â âð ,
where the âhatâ over âð means that âð is deleted from the tensor product.
Lemma 1.7.6. If ð1 â Lð(ð) and ð2 â Lð(ð)(1.7.7) ðð£(ð1 â ð2) = ðð£ð1 â ð2 + (â1)ðð1 â ðð£ð2 .
We will next prove the important identity.
Lemma 1.7.8. Let ð be a vector space and ð â Lð(ð). Then for all ð£ â ð we have
(1.7.9) ðð£(ðð£ð) = 0 .
Proof. It suffices by linearity to prove this for decomposable tensors and since (1.7.9) istrivially true for ð â L1(ð), we can by induction assume (1.7.9) is true for decomposabletensors of degree ð â 1. Let â1 â ⯠â âð be a decomposable tensor of degree ð. Settingð â â1 â⯠â âðâ1 and â = âð we have
ðð£(â1 â⯠â âð) = ðð£(ð â â) = ðð£ð â â + (â1)ðâ1â(ð£)ðby (1.7.7). Hence
ðð£(ðð£(ð â â)) = ðð£(ðð£ð) â â + (â1)ðâ2â(ð£)ðð£ð + (â1)ðâ1â(ð£)ðð£ð .But by induction the first summand on the right is zero and the two remaining summandscancel each other out. â¡
From (1.7.9) we can deduce a slightly stronger result: For ð£1, ð£2 â ð(1.7.10) ðð£1 ðð£2 = âðð£2 ðð£1 .
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24 Chapter 1: Multilinear Algebra
Proof. Let ð£ = ð£1 + ð£2. Then ðð£ = ðð£1 + ðð£2 so
0 = ðð£ðð£ = (ðð£1 + ðð£2)(ðð£1 + ðð£2)= ðð£1 ðð£1 + ðð£1 ðð£2 + ðð£2 ðð£1 + ðð£2 ðð£2= ðð£1 ðð£2 + ðð£2 ðð£1
since the first and last summands are zero by (1.7.9). â¡
Weâll now show how to define the operation ðð£ on ð¬ð(ðâ). Weâll first prove:
Lemma 1.7.11. If ð â Lð(ð) is redundant then so is ðð£ð.
Proof. Let ð = ð1 â â â â â ð2 where â is in ðâ, ð1 is in Lð(ð) and ð2 is in Lð(ð). Then byequation (1.7.7)
ðð£ð = ðð£ð1 â â â â â ð2 + (â1)ðð1 â ðð£(â â â) â ð2 + (â1)ð+2ð1 â â â â â ðð£ð2 .However, the first and the third terms on the right are redundant and
ðð£(â â â) = â(ð£)â â â(ð£)âby equation (1.7.5). â¡
Now let ð be the projection (1.5.12) of Lð(ð) onto ð¬ð(ðâ) and for ð = ð(ð) â ð¬ð(ðâ)define
(1.7.12) ðð£ð = ð(ðð£ð) .To show that this definition is legitimate we note that if ð = ð(ð1) = ð(ð2), then ð1 â ð2 âIð(ð), so by Lemma 1.7.11 ðð£ð1 â ðð£ð2 â Iðâ1 and hence
ð(ðð£ð1) = ð(ðð£ð2) .Therefore, (1.7.12) does not depend on the choice of ð.
By definition, ðð£ is a linear map ð¬ð(ðâ) â ð¬ðâ1(ðâ). We call this the interior productoperation. From the identities (1.7.2)â(1.7.12) one gets, for ð£, ð£1, ð£2 â ð, ð â ð¬ð(ðâ), ð1 âð¬ð(ðâ), and ð2 â ð¬2(ðâ)
ð(ð£1+ð£2)ð = ðð£1ð + ðð£2ð(1.7.13)ðð£(ð1 ⧠ð2) = ðð£ð1 ⧠ð2 + (â1)ðð1 ⧠ðð£ð2(1.7.14)ðð£(ðð£ð) = 0(1.7.15)
andðð£1 ðð£2ð = âðð£2 ðð£1ð .
Moreover if ð = â1 â§â¯ ⧠âð is a decomposable element of ð¬ð(ðâ) one gets from (1.7.5)
ðð£ð =ðâð=1(â1)ðâ1âð(ð£)â1 â§â¯ ⧠âð â§â¯ ⧠âð .
In particular if ð1,âŠ, ðð is a basis of ð, ðâ1 ,âŠ, ðâð the dual basis of ðâ and ððŒ = ðâð1 â§â¯â§ ðâðð ,
1 †ð1 < ⯠< ðð †ð, then ðððððŒ = 0 if ð â ðŒ and if ð = ðð
(1.7.16) ðððððŒ = (â1)ðâ1ððŒð
where ðŒð = (ð1,âŠ, ð€ð,âŠ, ðð) (i.e., ðŒð is obtained from the multi-index ðŒ by deleting ðð).
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§1.8 The pullback operation on ð¬ð(ðâ) 25
Exercises for §1.7Exercise 1.7.i. Prove Lemma 1.7.4.
Exercise 1.7.ii. Prove Lemma 1.7.6.
Exercise 1.7.iii. Show that ifð â Að, ðð£ð = ððð£ whereðð£ is the tensor (1.3.21). In particularconclude that ðð£ð â Aðâ1(ð). (See Exercise 1.4.viii.)
Exercise 1.7.iv. Assume the dimension of ð is ð and letðº be a nonzero element of the onedimensional vector space ð¬ð(ðâ). Show that the map(1.7.17) ðⶠð â ð¬ðâ1(ðâ) , ð£ ⊠ðð£ðº ,is a bijective linear map. Hint: One can assumeðº = ðâ1 â§â¯â§ ðâð where ð1,âŠ, ðð is a basis ofð. Now use equation (1.7.16) to compute this map on basis elements.
Exercise 1.7.v (cross-product). Let ð be a 3-dimensional vector space, ðµ an inner producton ð and ðº a nonzero element of ð¬3(ðâ). Define a map
â à âⶠð à ð â ðby setting
ð£1 à ð£2 â ðâ1(ð¿ð£1 ⧠ð¿ð£2)where ð is the map (1.7.17) and ð¿â¶ ð â ðâ the map (1.2.17). Show that this map is linearin ð£1, with ð£2 fixed and linear in ð£2 with ð£1 fixed, and show that ð£1 à ð£2 = âð£2 à ð£1.Exercise 1.7.vi. Forð = ð3 let ð1, ð2 and ð3 be the standard basis vectors and ðµ the standardinner product. (See §1.1.) Show that if ðº = ðâ1 ⧠ðâ2 ⧠ðâ3 the cross-product above is thestandard cross-product:
ð1 Ã ð2 = ð3ð2 Ã ð3 = ð1ð3 Ã ð1 = ð2 .
Hint: If ðµ is the standard inner product, then ð¿ðð = ðâð .Remark 1.7.18. One can make this standard cross-product look even more standard byusing the calculus notation: ð1 = ð€, ð2 = ð¥, and ð3 = ᅵᅵ.
1.8. The pullback operation on ð¬ð(ðâ)Let ð andð be vector spaces and let ðŽ be a linear map of ð intoð. Given a ð-tensor
ð â Lð(ð), recall that the pullback ðŽâð is the ð-tensor(ðŽâð)(ð£1,âŠ, ð£ð) = ð(ðŽð£1,âŠ,ðŽð£ð)
in Lð(ð). (See §1.3 and equation (1.3.17).) In this section weâll show how to define a similarpullback operation on ð¬ð(ðâ).
Lemma 1.8.1. If ð â Ið(ð), then ðŽâð â Ið(ð).Proof. It suffices to verify this when ð is a redundant ð-tensor, i.e., a tensor of the form
ð = â1 â⯠â âðwhere âð â ðâ and âð = âð+1 for some index, ð. But by equation (1.3.19),
ðŽâð = ðŽââ1 â⯠â ðŽââðand the tensor on the right is redundant since ðŽââð = ðŽââð+1. â¡
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26 Chapter 1: Multilinear Algebra
Now let ð be an element of ð¬ð(ðâ) and let ð = ð(ð) where ð is in Lð(ð). We define
(1.8.2) ðŽâð = ð(ðŽâð) .
Claim 1.8.3. The left hand side of equation (1.8.2) is well-defined.
Proof. If ð = ð(ð) = ð(ðâ²), then ð = ðâ² + ð for some ð â Ið(ð), and ðŽâð = ðŽâðâ² + ðŽâð.But ðŽâð â Ið(ð), so
ð(ðŽâðâ²) = ð(ðŽâð) . â¡
Proposition 1.8.4. The map ðŽâ ⶠð¬ð(ðâ) â ð¬ð(ðâ) sending ð ⊠ðŽâð is linear. Moreover,(1) If ðð â ð¬ðð(ðâ), ð = 1, 2, then
ðŽâ(ð1 ⧠ð2) = ðŽâ(ð1) ⧠ðŽâ(ð2) .
(2) If ð is a vector space and ðµ ⶠð â ð a linear map, then for ð â ð¬ð(ðâ),
ðµâðŽâð = (ðŽðµ)âð .
Weâll leave the proof of these three assertions as exercises. As a hint, they follow im-mediately from the analogous assertions for the pullback operation on tensors. (See equa-tions (1.3.19) and (1.3.20).)
As an application of the pullback operation weâll show how to use it to define the notionof determinant for a linear mapping.
Definition 1.8.5. Let ð be a ð-dimensional vector space. Then dimð¬ð(ðâ) = (ðð) = 1;i.e., ð¬ð(ðâ) is a one-dimensional vector space. Thus if ðŽâ¶ ð â ð is a linear mapping, theinduced pullback mapping:
ðŽâ ⶠð¬ð(ðâ) â ð¬ð(ðâ) ,is just âmultiplication by a constantâ. We denote this constant by det(ðŽ) and call it the de-terminant of ðŽ, Hence, by definition,
(1.8.6) ðŽâð = det(ðŽ)ð
for all ð â ð¬ð(ðâ).
From equation (1.8.6) it is easy to derive a number of basic facts about determinants.
Proposition 1.8.7. If ðŽ and ðµ are linear mappings of ð into ð, then
det(ðŽðµ) = det(ðŽ) det(ðµ) .
Proof. By (2) and
(ðŽðµ)âð = det(ðŽðµ)ð = ðµâ(ðŽâð)= det(ðµ)ðŽâð = det(ðµ) det(ðŽ)ð ,
so det(ðŽðµ) = det(ðŽ) det(ðµ). â¡
Proposition 1.8.8. Write idð ⶠð â ð for the identity map. Then det(idð) = 1.
Weâll leave the proof as an exercise. As a hint, note that idâð is the identity map onð¬ð(ðâ).
Proposition 1.8.9. If ðŽâ¶ ð â ð is not surjective, then det(ðŽ) = 0.
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§1.8 The pullback operation on ð¬ð(ðâ) 27
Proof. Letð be the image of ðŽ. Then if ðŽ is not onto, the dimension ofð is less than ð,so ð¬ð(ðâ) = 0. Now let ðŽ = ðððµ where ðð is the inclusion map ofð into ð and ðµ is themapping, ðŽ, regarded as a mapping from ð toð. Thus if ð is in ð¬ð(ðâ), then by (2)
ðŽâð = ðµâðâððand since ðâðð is in ð¬ð(ðâ) it is zero. â¡
We will derive by wedge product arguments the familiar âmatrix formulaâ for the deter-minant. Let ð andð be ð-dimensional vector spaces and let ð1,âŠ, ðð be a basis for ð andð1,âŠ, ðð a basis forð. From these bases we get dual bases, ðâ1 ,âŠ, ðâð and ðâ1 ,âŠ, ðâð , forðâandðâ. Moreover, if ðŽ is a linear map of ð intoð and [ðð,ð] the ð à ð matrix describingðŽ in terms of these bases, then the transpose map, ðŽâ ⶠðâ â ðâ, is described in terms ofthese dual bases by the ð à ð transpose matrix, i.e., if
ðŽðð =ðâð=1ðð,ððð ,
then
ðŽâðâð =ðâð=1ðð,ððâð .
(See §1.2.) Consider now ðŽâ(ðâ1 â§â¯ ⧠ðâð ). By (1),
ðŽâ(ðâ1 â§â¯ ⧠ðâð ) = ðŽâðâ1 â§â¯ ⧠ðŽâðâð= â1â€ð1,âŠ,ððâ€ð
(ð1,ð1ðâð1) ⧠⯠⧠(ðð,ððð
âðð) .
Thus,ðŽâ(ðâ1 â§â¯ ⧠ðâð ) = âð1,ð1âŠðð,ðð ð
âð1 â§â¯ ⧠ð
âðð .
If the multi-index, ð1,âŠ, ðð, is repeating, then ðâð1 â§â¯â§ðâðð is zero, and if it is not repeating
then we can writeðð = ð(ð) ð = 1,âŠ, ð
for some permutation, ð, and hence we can rewrite ðŽâ(ðâ1 â§â¯ ⧠ðâð ) as
ðŽâ(ðâ1 â§â¯ ⧠ðâð ) = âðâððð1,ð(1)â¯ðð,ð(ð) (ðâ1 â§â¯ ⧠ðâð )ð .
But(ðâ1 â§â¯ ⧠ðâð )ð = (â1)ððâ1 â§â¯ ⧠ðâð
so we get finally the formula
(1.8.10) ðŽâ(ðâ1 â§â¯ ⧠ðâð ) = det([ðð,ð])ðâ1 â§â¯ ⧠ðâðwhere
(1.8.11) det([ðð,ð]) = âðâðð(â1)ðð1,ð(1)â¯ðð,ð(ð)
summed over ð â ðð. The sum on the right is (as most of you know) the determinant of thematric [ðð,ð].
Notice that if ð = ð and ðð = ðð, ð = 1,âŠ, ð, then ð = ðâ1 ⧠⯠⧠ðâð = ðâ1 ⧠⯠⧠ðâð ,hence by (1.8.6) and (1.8.10),
det(ðŽ) = det[ðð,ð] .
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28 Chapter 1: Multilinear Algebra
Exercises for §1.8Exercise 1.8.i. Verify the three assertions of Proposition 1.8.4.
Exercise 1.8.ii. Deduce from Proposition 1.8.9 a well-known fact about determinants ofð Ã ðmatrices: If two columns are equal, the determinant is zero.
Exercise 1.8.iii. Deduce from Proposition 1.8.7 another well-known fact about determi-nants of ð Ã ðmatrices: If one interchanges two columns, then one changes the sign of thedeterminant.
Hint: Let ð1,âŠ, ðð be a basis of ð and let ðµâ¶ ð â ð be the linear mapping: ðµðð = ðð,ðµðð = ðð and ðµðâ = ðâ, â â ð, ð. What is ðµâ(ðâ1 â§â¯ ⧠ðâð )?
Exercise 1.8.iv. Deduce from Propositions 1.8.3 and 1.8.4 another well-known fact aboutdeterminants of ð Ã ðmatrix: If (ðð,ð) is the inverse of [ðð,ð], its determinant is the inverse ofthe determinant of [ðð,ð].
Exercise 1.8.v. Extract from (1.8.11) a well-known formula for determinants of 2 Ã 2 ma-trices:
det(ð11 ð12ð21, ð22) = ð11ð22 â ð12ð21 .
Exercise 1.8.vi. Show that if ðŽ = [ðð,ð] is an ð à ð matrix and ðŽâº = [ðð,ð] is its transposedetðŽ = detðŽâº.
Hint: You are required to show that the sums
âðâðð(â1)ðð1,ð(1)âŠðð,ð(ð)
andâðâðð(â1)ððð(1),1âŠðð(ð),ð
are the same. Show that the second sum is identical with
âðâðð(â1)ðâ1ððâ1(1),1âŠððâ1(ð),ð .
Exercise 1.8.vii. Let ðŽ be an ð à ðmatrix of the form
ðŽ = (ðµ â0 ð¶) ,
where ðµ is a ð à ð matrix and ð¶ the â à â matrix and the bottom â à ð block is zero. Showthat
det(ðŽ) = det(ðµ) det(ð¶) .Hint: Show that in equation (1.8.11) every nonzero term is of the form
(â1)ððð1,ð(1)âŠðð,ð(ð)ð1,ð(1)âŠðâ,ð(â)where ð â ðð and ð â ðâ.
Exercise 1.8.viii. Let ð andð be vector spaces and let ðŽâ¶ ð â ð be a linear map. Showthat if ðŽð£ = ð€ then for ð â ð¬ð(ðâ),
ðŽâðð€ð = ðð£ðŽâð .Hint:By equation (1.7.14) and Proposition 1.8.4 it suffices to prove this forð â ð¬1(ðâ),
i.e., for ð â ðâ.
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§1.9 Orientations 29
1.9. Orientations
Definition 1.9.1. We recall from freshman calculus that if â â ð2 is a line through theorigin, then â â {0} has two connected components and an orientation of â is a choice ofone of these components (as in the Figure 1.9.1 below).
â
0
Figure 1.9.1. A line in ð2
More generally, if ð¿ is a one-dimensional vector space then ð¿â {0} consists of two com-ponents: namely if ð£ is an element of ð¿ â {0}, then these two components are
ð¿1 = { ðð£ | ð > 0 }and
ð¿2 = { ðð£ | ð < 0 } .
Definition 1.9.2. Let ð¿ be a one-dimensional vector space. An orientation of ð¿ is a choice ofone of a connected component of ð¿â{0}. Usually the component chosen is denoted ð¿+, andcalled the positive component ofð¿â{0} and the other componentð¿â the negative componentof ð¿ â {0}.
Definition 1.9.3. Let (ð¿, ð¿+) be an oriented one-dimensional vector space. A vector ð£ â ð¿is positively oriented if ð£ â ð¿+.
Definition 1.9.4. Let ð be an ð-dimensional vector space. An orientation of ð is an orien-tation of the one-dimensional vector space ð¬ð(ðâ).
One important way of assigning an orientation to ð is to choose a basis, ð1,âŠ, ðð of ð.Then, if ðâ1 ,âŠ, ðâð is the dual basis, we can orient ð¬ð(ðâ) by requiring that ðâ1 ⧠⯠⧠ðâð bein the positive component of ð¬ð(ðâ).
Definition 1.9.5. Let ð be an oriented ð-dimensional vector space. We say that an orderedbasis (ð1,âŠ, ðð) of ð is positively oriented if ðâ1 ⧠⯠⧠ðâð is in the positive component ofð¬ð(ðâ).
Suppose that ð1,âŠ, ðð and ð1,âŠ, ðð are bases of ð and that
(1.9.6) ðð =ðâð=1ðð,ð,ðð .
Then by (1.7.10)ðâ1 â§â¯ ⧠ðâð = det[ðð,ð]ðâ1 â§â¯ ⧠ðâð
so we conclude:
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30 Chapter 1: Multilinear Algebra
Proposition 1.9.7. If ð1,âŠ, ðð is positively oriented, thenð1,âŠ, ðð is positively oriented if andonly if det[ðð,ð] is positive.
Corollary 1.9.8. If ð1,âŠ, ðð is a positively oriented basis of ð, then the basis
ð1,âŠ, ððâ1, âðð, ðð+1,âŠ, ððis negatively oriented.
Now let ð be a vector space of dimension ð > 1 andð a subspace of dimension ð < ð.We will use the result above to prove the following important theorem.
Theorem 1.9.9. Given orientations onð andð/ð, one gets from these orientations a naturalorientation onð.
Remark 1.9.10. What we mean by ânaturalâ will be explained in the course of the proof.
Proof. Let ð = ðâ ð and let ð be the projection ofð ontoð/ð . By Exercises 1.2.i and 1.2.iiwe can choose a basis ð1,âŠ, ðð of ð such that ðð+1,âŠ, ðð is a basis ofð and ð(ð1),âŠ, ð(ðð)a basis ofð/ð. Moreover, replacing ð1 by âð1 if necessary we can assume by Corollary 1.9.8that ð(ð1),âŠ, ð(ðð) is a positively oriented basis of ð/ð and replacing ðð by âðð if neces-sary we can assume that ð1,âŠ, ðð is a positively oriented basis of ð. Now assign toð theorientation associated with the basis ðð+1,âŠ, ðð.
Letâs show that this assignment is ânaturalâ(i.e., does not depend on our choice of basisð1,âŠ, ðð). To see this let ð1,âŠ, ðð be another basis of ð with the properties above and letðŽ = [ðð,ð] be thematrix (1.9.6) expressing the vectors ð1,âŠ, ðð as linear combinations of thevectors ð1,âŠðð. This matrix has to have the form
(1.9.11) ðŽ = (ðµ ð¶0 ð·)
where ðµ is the ðÃðmatrix expressing the basis vectors ð(ð1),âŠ, ð(ðð) ofð/ð as linear com-binations of ð(ð1),âŠ, ð(ðð) and ð· the ð à ðmatrix expressing the basis vectors ðð+1,âŠ, ððofð as linear combinations of ðð+1,âŠ, ðð. Thus
det(ðŽ) = det(ðµ) det(ð·) .However, by Proposition 1.9.7, detðŽ and detðµ are positive, so detð· is positive, and henceif ðð+1,âŠ, ðð is a positively oriented basis ofð so is ðð+1,âŠ, ðð. â¡
As a special case of this theorem suppose dimð = ð â 1. Then the choice of a vectorð£ â ðâð gives one a basis vector ð(ð£) for the one-dimensional spaceð/ð and hence ifðis oriented, the choice of ð£ gives one a natural orientation onð.
Definition 1.9.12. LetðŽâ¶ ð1 â ð2 a bijective linear map of oriented ð-dimensional vectorspaces. We say that ðŽ is orientation-preserving if, for ð â ð¬ð(ðâ2 )+, we have that ðŽâð is inð¬ð(ðâ1 )+.
Example 1.9.13. If ð1 = ð2 then ðŽâð = det(ðŽ)ð so ðŽ is orientation preserving if and onlyif det(ðŽ) > 0.
The following proposition weâll leave as an exercise.
Proposition 1.9.14. Letð1,ð2, andð3 be oriented ð-dimensional vector spaces andðŽð ⶠðð ⥲ðð+1 for ð = 1, 2 be bijective linear maps. Then if ðŽ1 and ðŽ2 are orientation preserving, so isðŽ2 â ðŽ1.
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§1.9 Orientations 31
Exercises for §1.9Exercise 1.9.i. Prove Corollary 1.9.8.
Exercise 1.9.ii. Show that the argument in the proof of Theorem 1.9.9 can be modified toprove that if ð andð are oriented then these orientations induce a natural orientation onð/ð.
Exercise 1.9.iii. Similarly show that ifð and ð/ð are oriented these orientations inducea natural orientation on ð.
Exercise 1.9.iv. Let ð be an ð-dimensional vector space andð â ð a ð-dimensional sub-space. Let ð = ð/ð and let ð ⶠð â ð and ðⶠð â ð be the inclusion and projectionmaps. Suppose ð and ð are oriented. Let ð be in ð¬ðâð(ðâ)+ and let ð be in ð¬ð(ðâ)+. Showthat there exists a ð inð¬ð(ðâ) such that ðâðâ§ð = ð. Moreover show that ðâð is intrinsicallydefined (i.e., does not depend on how we choose ð) and sits in the positive partð¬ð(ðâ)+ ofð¬ð(ðâ).
Exercise 1.9.v. Let ð1,âŠ, ðð be the standard basis vectors ofðð.The standard orientation ofðð is, by definition, the orientation associated with this basis. Show that ifð is the subspaceof ðð defined by the equation ð¥1 = 0, and ð£ = ð1 â ð then the natural orientation ofðassociated with ð£ and the standard orientation of ðð coincide with the orientation given bythe basis vectors ð2,âŠ, ðð ofð.
Exercise 1.9.vi. Letðbe anorientedð-dimensional vector space andð anðâ1-dimensionalsubspace. Show that if ð£ and ð£â² are inðâð then ð£â² = ðð£+ð€, whereð€ is inð andð â ðâ{0}.Show that ð£ and ð£â² give rise to the same orientation ofð if and only if ð is positive.
Exercise 1.9.vii. Prove Proposition 1.9.14.
Exercise 1.9.viii. A key step in the proof of Theorem 1.9.9 was the assertion that the matrixðŽ expressing the vectors, ðð as linear combinations of the vectors ððœ, had to have the form(1.9.11). Why is this the case?
Exercise 1.9.ix.(1) Letð be a vector space,ð a subspace ofð andðŽâ¶ ð â ð a bijective linear map which
mapsð ontoð. Show that one gets from ðŽ a bijective linear mapðµâ¶ ð/ð â ð/ð
with the propertyððŽ = ðµð ,
where ðⶠð â ð/ð is the quotient map.(2) Assume that ð, ð and ð/ð are compatibly oriented. Show that if ðŽ is orientation-
preserving and its restriction toð is orientation preserving then ðµ is orientation pre-serving.
Exercise 1.9.x. Let ð be a oriented ð-dimensional vector space,ð an (ð â 1)-dimensionalsubspace of ð and ð ⶠð â ð the inclusion map. Given ð â ð¬ð(ðâ)+ and ð£ â ð âð showthat for the orientation ofð described in Exercise 1.9.v, ðâ(ðð£ð) â ð¬ðâ1(ðâ)+.
Exercise 1.9.xi. Let ð be an ð-dimensional vector space, ðµâ¶ ð à ð â ð an inner prod-uct and ð1,âŠ, ðð a basis of ð which is positively oriented and orthonormal. Show that thevolume element
vol â ðâ1 â§â¯ ⧠ðâð â ð¬ð(ðâ)
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32 Chapter 1: Multilinear Algebra
is intrinsically defined, independent of the choice of this basis.Hint: Equations (1.2.20) and (1.8.10).
Exercise 1.9.xii.(1) Let ð be an oriented ð-dimensional vector space and ðµ an inner product on ð. Fix an
oriented orthonormal basis, ð1,âŠ, ðð, of ð and let ðŽâ¶ ð â ð be a linear map. Showthat if
ðŽðð = ð£ð =ðâð=1ðð,ððð
and ðð,ð = ðµ(ð£ð, ð£ð), the matrices ððŽ = [ðð,ð] and ððµ = (ðð,ð) are related by: ððµ =ðâºðŽððŽ.
(2) Show that if vol is the volume form ðâ1 â§â¯ ⧠ðâð , and ðŽ is orientation preserving
ðŽâ vol = det(ððµ)12 vol .
(3) ByTheorem 1.5.13 one has a bijective mapð¬ð(ðâ) â Að(ð) .
Show that the element, ðº, of ðŽð(ð) corresponding to the form vol has the property|ðº(ð£1,âŠ, ð£ð)|2 = det((ðð,ð))
where ð£1,âŠ, ð£ð are any ð-tuple of vectors in ð and ðð,ð = ðµ(ð£ð, ð£ð).
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CHAPTER 2
Differential Forms
The goal of this chapter is to generalize to ð dimensions the basic operations of threedimensional vector calculus: divergence, curl, and gradient. The divergence and gradientoperations have fairly straightforward generalizations, but the curl operation is more subtle.For vector fields it does not have any obvious generalization, however, if one replaces vectorfields by a closely related class of objects, differential forms, then not only does it have anatural generalization but it turns out that divergence, curl, and gradient are all special casesof a general operation on differential forms called exterior differentiation.
2.1. Vector fields and one-forms
In this section we will review some basic facts about vector fields in ð variables andintroduce their dual objects: one-forms. We will then take up in §2.3 the theory of ð-formsfor ð > 1. We begin by fixing some notation.
Definition 2.1.1. Let ð â ðð. The tangent space to ðð at ð is the set of pairsðððð â { (ð, ð£) | ð£ â ðð } .
The identification(2.1.2) ðððð ⥲ ðð , (ð, ð£) ⊠ð£makes ðððð into a vector space. More explicitly, for ð£, ð£1, ð£2 â ðð and ð â ð we define theaddition and scalar multiplication operations on ðððð by setting
(ð, ð£1) + (ð, ð£2) â (ð, ð£1 + ð£2)and
ð(ð, ð£) â (ð, ðð£) .
Let ð be an open subset of ðð and ðⶠð â ðð a ð¶1 map. We recall that the derivativeð·ð(ð)ⶠðð â ðð
of ð at ð is the linear map associated with theð Ã ðmatrix
[ ððððð¥ð(ð)] .
It will be useful to have a âbase-pointedâ version of this definition aswell. Namely, if ð = ð(ð)we will define
ððð ⶠðððð â ðððð
to be the mapððð(ð, ð£) â (ð,ð·ð(ð)ð£) .
It is clear from the way we have defined vector space structures on ðððð and ðððð that thismap is linear.
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34 Chapter 2: Differential Forms
Suppose that the image of ð is contained in an open set ð, and suppose ðⶠð â ðð isa ð¶1 map. Then the âbase-pointedâ version of the chain rule asserts that
ððð â ððð = ð(ð â ð)ð .(This is just an alternative way of writingð·ð(ð)ð·ð(ð) = ð·(ð â ð)(ð).)
In 3-dimensional vector calculus a vector field is a functionwhich attaches to each pointð of ð3 a base-pointed arrow (ð, ð£) â ððð3. The ð-dimensional version of this definition isessentially the same.
Definition 2.1.3. Let ð be an open subset of ðð. A vector field ð on ð is a function whichassigns to each point ð â ð a vector ð(ð) â ðððð.
Thus a vector field is a vector-valued function, but its value at ð is an element of a vectorspace ðððð that itself depends on ð.
Examples 2.1.4.(1) Given a fixed vector ð£ â ðð, the function(2.1.5) ð ⊠(ð, ð£)
is a vector field on ðð. Vector fields of this type are called constant vector fields.(2) In particular let ð1,âŠ, ðð be the standard basis vectors ofðð. If ð£ = ðð we will denote the
vector field (2.1.5) by ð/ðð¥ð. (The reason for this âderivation notationâ will be explainedbelow.)
(3) Given a vector field ð onð and a function, ðⶠð â ð we denote by ðð the vector fieldon ð defined by
ð ⊠ð(ð)ð(ð) .(4) Given vector fields ð1 and ð2 on ð, we denote by ð1 + ð2 the vector field on ð defined
byð ⊠ð1(ð) + ð2(ð) .
(5) The vectors, (ð, ðð), ð = 1,âŠ, ð, are a basis of ðððð, so if ð is a vector field on ð, ð(ð)can be written uniquely as a linear combination of these vectors with real numbersð1(ð),âŠ, ðð(ð) as coefficients. In other words, using the notation in example (2) above,ð can be written uniquely as a sum
(2.1.6) ð =ðâð=1ððððð¥ð
where ðð ⶠð â ð is the function ð ⊠ðð(ð).
Definition 2.1.7. We say that ð is a ð¶â vector field if the ððâs are in ð¶â(ð).
Definition 2.1.8. Abasic vector field operation is Lie differentiation. Ifð â ð¶1(ð)we defineð¿ðð to be the function on ð whose value at ð is given by(2.1.9) ð¿ðð(ð) â ð·ð(ð)ð£ ,where ð(ð) = (ð, ð£).
Example 2.1.10. If ð is the vector field (2.1.6) then
ð¿ðð =ðâð=1ðððððð¥ð
,
(motivating our âderivation notationâ for ð).
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§2.1 Vector fields and one-forms 35
We leave the following generalization of the product rule for differentiation as an exer-cise.
Lemma 2.1.11. Letð be an open subset ofðð, ð a vector field onð, andð1, ð2 â ð¶1(ð). Thenð¿ð(ð1 â ð2) = ð¿ð(ð1) â ð2 + ð1 â ð¿ð(ð2) .
We now discuss a class of objects which are in some sense âdual objectsâ to vector fields.
Definition 2.1.12. Let ð â ðð. The cotangent space to ðð at ð is the dual vector spaceðâð ðð â (ððð)â .
An element of ðâð ðð is called a cotangent vector to ðð at ð.
Definition 2.1.13. Let ð be an open subset of ðð. A differential one-form, or simply one-form on ð is a function ð which assigns to each point ð â ð a vector ðð in ðâð ðð.
Examples 2.1.14.(1) Let ðⶠð â ð be a ð¶1 function. Then for ð â ð and ð = ð(ð) one has a linear map(2.1.15) ððð ⶠðððð â ððð
and by making the identification ððð ⥲ ð by sending (ð, ð£) ⊠ð£, the differential ðððcan be regarded as a linear map from ðððð to ð, i.e., as an element of ðâð ðð. Hence theassignment
ð ⊠ðððdefines a one-form on ð which we denote by ðð.
(2) Given a one-formð and a functionðⶠð â ð the pointwise product ofðwithð definesone-form ðð on ð by (ðð)ð â ð(ð)ðð.
(3) Given two one-forms ð1 and ð2 their pointwise sum defines a one-form ð1 + ð2 by(ð1 + ð2)ð â (ð1)ð + (ð2)ð.
(4) The one-forms ðð¥1,âŠ, ðð¥ð play a particularly important role. By equation (2.1.15)
(2.1.16) (ðð¥ð) (ððð¥ð)ð= ð¿ð,ð
i.e., is equal to 1 if ð = ð and zero if ð â ð. Thus (ðð¥1)ð,âŠ, (ðð¥ð)ð are the basis ofðâð ðð dual to the basis (ð/ðð¥ð)ð. Therefore, if ð is any one-form onð, ðð can be writtenuniquely as a sum
ðð =ðâð=1ðð(ð)(ðð¥ð)ð , ðð(ð) â ð ,
and ð can be written uniquely as a sum
(2.1.17) ð =ðâð=1ðððð¥ð
where ðð ⶠð â ð is the function ð ⊠ðð(ð). We say that ð is a ð¶â one-form or smoothone-form if the functions ð1,âŠ, ðð are ð¶â.
We leave the following as an exercise.
Lemma 2.1.18. Let ð be an open subset of ðð. If ðⶠð â ð is a ð¶â function, then
ðð =ðâð=1
ðððð¥ððð¥ð .
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36 Chapter 2: Differential Forms
Suppose that ð is a vector field and ð a one-form on ð â ðð. Then for every ð â ð thevectors ð(ð) â ðððð and ðð â (ðððð)â can be paired to give a number ðð(ð)ðð â ð, andhence, as ð varies, anð-valued function ðððwhich we will call the interior product of ðwithð.Example 2.1.19. For instance if ð is the vector field (2.1.6) and ð the one-form (2.1.17),then
ððð =ðâð=1ðððð .
Thus if ð and ð are ð¶â, so is the function ððð. Also notice that if ð â ð¶â(ð), then as weobserved above
ðð =ðâð=1
ðððð¥ðððð¥ð
so if ð is the vector field (2.1.6) then we have
ðð ðð =ðâð=1ðððððð¥ð= ð¿ðð .
Exercises for §2.1Exercise 2.1.i. Prove Lemma 2.1.18
Exercise 2.1.ii. Prove Lemma 2.1.11
Exercise 2.1.iii. Let ð be an open subset of ðð and ð1 and ð2 vector fields on ð. Show thatthere is a unique vector field ð, on ð with the property
ð¿ðð = ð¿ð1(ð¿ð2ð) â ð¿ð2(ð¿ð1ð)for all ð â ð¶â(ð).Exercise 2.1.iv. The vector field ð in Exercise 2.1.iii is called the Lie bracket of the vectorfields ð1 and ð2 and is denoted by [ð1, ð2]. Verify that the Lie bracket is skew-symmetric, i.e.,
[ð1, ð2] = â[ð2, ð1] ,and satisfies the Jacobi identity
[ð1, [ð2, ð3]] + [ð2, [ð3, ð1]] + [ð3, [ð1, ð2]] = 0 .(And thus defines the structure of a Lie algebra.)
Hint: Prove analogous identities for ð¿ð1 , ð¿ð2 and ð¿ð3 .
Exercise 2.1.v. Let ð1 = ð/ðð¥ð and ð2 = âðð=1 ððð/ðð¥ð. Show that
[ð1, ð2] =ðâð=1
ððððð¥ðððð¥ð
.
Exercise 2.1.vi. Let ð1 and ð2 be vector fields and ð a ð¶â function. Show that[ð1, ðð2] = ð¿ð1ðð2 + ð[ð1, ð2] .
Exercise 2.1.vii. Let ð be an open subset of ðð and let ðŸâ¶ [ð, ð] â ð, ð¡ ⊠(ðŸ1(ð¡),âŠ, ðŸð(ð¡))be a ð¶1 curve. Given a ð¶â one-form ð = âðð=1 ðððð¥ð onð, define the line integral of ð overðŸ to be the integral
â«ðŸð âðâð=1â«ð
ððð(ðŸ(ð¡))ððŸððð¡ðð¡ .
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§2.2 Integral Curves for Vector Fields 37
Show that if ð = ðð for some ð â ð¶â(ð)
â«ðŸð = ð(ðŸ(ð)) â ð(ðŸ(ð)) .
In particular conclude that if ðŸ is a closed curve, i.e., ðŸ(ð) = ðŸ(ð), this integral is zero.
Exercise 2.1.viii. Let ð be the ð¶â one-form on ð2 â {0} defined by
ð = ð¥1ðð¥2 â ð¥2ðð¥1ð¥21 + ð¥22
,
and let ðŸâ¶ [0, 2ð] â ð2â{0} be the closed curve ð¡ ⊠(cos ð¡, sin ð¡). Compute the line integralâ«ðŸ ð, and note that â«ðŸ ð â 0. Conclude that ð is not of the form ðð for ð â ð¶â(ð2 â {0}).
Exercise 2.1.ix. Let ð be the function
ð(ð¥1, ð¥2) ={{{{{{{
arctan ð¥2ð¥1 ð¥1 > 0ð2 , ð¥2 > 0 or ð¥1 = 0arctan ð¥2ð¥1 + ð ð¥1 < 0 .
Recall that âð2 < arctan(ð¡) < ð2 . Show that ð is ð¶â and that ðð is the 1-form ð in Exer-cise 2.1.viii. Why does not this contradict what you proved in Exercise 2.1.viii?
2.2. Integral Curves for Vector Fields
In this section weâll discuss some properties of vector fields which weâll need for themanifold segment of this text. Weâll begin by generalizing to ð-variables the calculus notionof an âintegral curveâ of a vector field.
Definition 2.2.1. Let ð â ðð be open and ð a vector field on ð. A ð¶1 curve ðŸâ¶ (ð, ð) â ðis an integral curve of ð if for all ð¡ â (ð, ð) we have
ð(ðŸ(ð¡)) = (ðŸ(ð¡), ððŸðð¡ (ð¡)) .Remark 2.2.2. If ð is the vector field (2.1.6) and ðⶠð â ðð is the function (ð1,âŠ, ðð),the condition for ðŸ(ð¡) to be an integral curve of ð is that it satisfy the system of differentialequations
(2.2.3) ððŸðð¡(ð¡) = ð(ðŸ(ð¡)) .
We will quote without proof a number of basic facts about systems of ordinary differ-ential equations of the type (2.2.3).1
Theorem 2.2.4 (existence of integral curves). Let ð be an open subset of ðð and ð a vectorfield on ð. Given a point ð0 â ð and ð â ð, there exists an interval ðŒ = (ð â ð, ð + ð), aneighborhoodð0 of ð0 inð, and for every ð â ð0 an integral curve ðŸð ⶠðŒ â ð for ð such thatðŸð(ð) = ð.Theorem 2.2.5 (uniqueness of integral curves). Letð be an open subset of ðð and ð a vectorfield on ð. For ð = 1, 2, let ðŸð ⶠðŒð â ð, ð = 1, 2 be integral curves for ð. If ð â ðŒ1 â© ðŒ2 andðŸ1(ð) = ðŸ2(ð) then ðŸ1|ðŒ1â©ðŒ2 = ðŸ2|ðŒ1â©ðŒ2 and the curve ðŸâ¶ ðŒ1 ⪠ðŒ2 â ð defined by
ðŸ(ð¡) â {ðŸ1(ð¡), ð¡ â ðŒ1ðŸ2(ð¡), ð¡ â ðŒ2
1A source for these results that we highly recommend is [2, Ch. 6].
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38 Chapter 2: Differential Forms
is an integral curve for ð.
Theorem 2.2.6 (smooth dependence on initial data). Let ð â ð â ðð be open subsets. Letð be a ð¶â-vector field on ð, ðŒ â ð an open interval, ð â ðŒ, and âⶠð à ðŒ â ð a map withthe following properties.(1) â(ð, ð) = ð.(2) For all ð â ð the curve
ðŸð ⶠðŒ â ð , ðŸð(ð¡) â â(ð, ð¡)is an integral curve of ð.
Then the map â is ð¶â.
One important feature of the system (2.2.3) is that it is an autonomous system of differ-ential equations: the function ð(ð¥) is a function of ð¥ alone (it does not depend on ð¡). Oneconsequence of this is the following.
Theorem 2.2.7. Let ðŒ = (ð, ð) and for ð â ð let ðŒð = (ð â ð, ð â ð). Then if ðŸâ¶ ðŒ â ð is anintegral curve, the reparametrized curve
(2.2.8) ðŸð ⶠðŒð â ð , ðŸð(ð¡) â ðŸ(ð¡ + ð)is an integral curve.
We recall that ð¶1-function ðⶠð â ð is an integral of the system (2.2.3) if for everyintegral curve ðŸ(ð¡), the function ð¡ ⊠ð(ðŸ(ð¡)) is constant. This is true if and only if for all ð¡
0 = ððð¡ð(ðŸ(ð¡)) = (ð·ð)ðŸ(ð¡) (
ððŸðð¡) = (ð·ð)ðŸ(ð¡)(ð£)
where ð(ð) = (ð, ð£). By equation (2.1.6) the term on the right is ð¿ðð(ð). Hence we conclude:
Theorem 2.2.9. Let ð be an open subset of ðð and ð â ð¶1(ð). Then ð is an integral of thesystem (2.2.3) if and only if ð¿ðð = 0.
Definition 2.2.10. Letð be an open subset of ðð and ð a vector field onð. We say that ð iscomplete if for every ð â ð there exists an integral curve ðŸâ¶ ð â ð with ðŸ(0) = ð, i.e., forevery ð there exists an integral curve that starts at ð and exists for all time.
To see what completeness involves, we recall that an integral curve
ðŸâ¶ [0, ð) â ð ,
with ðŸ(0) = ð is called maximal if it cannot be extended to an interval [0, ðâ²), where ðâ² > ð.(See for instance [2, §6.11].) For such curves it is known that either(1) ð = +â,(2) |ðŸ(ð¡)| â +â as ð¡ â ð,(3) or the limit set of
{ ðŸ(ð¡) | 0 †ð¡ < ð }contains points on the boundary of ð.Hence if we can exclude (2) and (3), we have shown that an integral curve with ðŸ(0) = ð
exists for all positive time. A simple criterion for excluding (2) and (3) is the following.
Lemma 2.2.11. The scenarios (2) and (3) cannot happen if there exists a proper ð¶1-functionðⶠð â ð with ð¿ðð = 0.
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Proof. The identity ð¿ðð = 0 implies that ð is constant on ðŸ(ð¡), but if ð(ð) = ð this impliesthat the curve ðŸ(ð¡) lies on the compact subset ðâ1(ð) â ð, hence it cannot ârun off to infinityâas in scenario (2) or ârun off the boundaryâ as in scenario (3). â¡
Applying a similar argument to the interval (âð, 0] we conclude:
Theorem 2.2.12. Suppose there exists a proper ð¶1-function ðⶠð â ð with the propertyð¿ðð = 0. Then the vector field ð is complete.
Example 2.2.13. Let ð = ð2 and let ð£ be the vector field
ð£ = ð¥3 ðððŠâ ðŠ ððð¥
.
Then ð(ð¥, ðŠ) = 2ðŠ2 + ð¥4 is a proper function with the property above.
Another hypothesis on ð which excludes (2) and (3) is the following.
Definition 2.2.14. The support of ð is set
supp(ð) = { ð â ð | ð(ð) â 0 } .
We say that ð is compactly supported if supp(ð) is compact.
We now show that compactly supported vector fields are complete.
Theorem 2.2.15. If a vector field ð is compactly supported, then ð is complete.
Proof. Notice first that if ð(ð) = 0, the constant curve, ðŸ0(ð¡) = ð, ââ < ð¡ < â, satisfies theequation
ððð¡ðŸ0(ð¡) = 0 = ð(ð) ,
so it is an integral curve of ð. Hence if ðŸ(ð¡), âð < ð¡ < ð, is any integral curve of ð with theproperty ðŸ(ð¡0) = ð for some ð¡0, it has to coincide with ðŸ0 on the interval (âð, ð), and hencehas to be the constant curve ðŸ(ð¡) = ð on (âð, ð).
Now suppose the support supp(ð) of ð is compact. Then either ðŸ(ð¡) is in supp(ð) for allð¡ or is in ð â supp(ð) for some ð¡0. But if this happens, and ð = ðŸ(ð¡0) then ð(ð) = 0, so ðŸ(ð¡)has to coincide with the constant curve ðŸ0(ð¡) = ð, for all ð¡. In neither case can it go off toâor off to the boundary of ð as ð¡ â ð. â¡
One useful application of this result is the following. Suppose ð is a vector field onð â ðð, and one wants to see what its integral curves look like on some compact set ðŽ â ð.Let ð â ð¶â0 (ð) be a bump function which is equal to one on a neighborhood of ðŽ. Thenthe vector field ð = ðð is compactly supported and hence complete, but it is identical withð on ðŽ, so its integral curves on ðŽ coincide with the integral curves of ð.
If ð is complete then for every ð, one has an integral curve ðŸð ⶠð â ð with ðŸð(0) = ð,so one can define a map
ðð¡ ⶠð â ðby settingðð¡(ð) â ðŸð(ð¡). If ð isð¶â, thismapping isð¶â by the smooth dependence on initialdata theorem, and, by definition, ð0 = idð. We claim that the ðð¡âs also have the property
(2.2.16) ðð¡ â ðð = ðð¡+ð .
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40 Chapter 2: Differential Forms
Indeed if ðð(ð) = ð, then by the reparametrization theorem, ðŸð(ð¡) and ðŸð(ð¡ + ð) are bothintegral curves of ð, and since ð = ðŸð(0) = ðŸð(ð) = ðð(ð), they have the same initial point,so
ðŸð(ð¡) = ðð¡(ð) = (ðð¡ â ðð)(ð)= ðŸð(ð¡ + ð) = ðð¡+ð(ð)
for all ð¡. Since ð0 is the identity it follows from (2.2.16) that ðð¡ â ðâð¡ is the identity, i.e.,
ðâð¡ = ðâ1ð¡ ,
so ðð¡ is a ð¶â diffeomorphism. Hence if ð is complete it generates a one-parameter group ðð¡,ââ < ð¡ < â, of ð¶â-diffeomorphisms of ð.
If ð is not complete there is an analogous result, but it is trickier to formulate precisely.Roughly speaking ð generates a one-parameter group of diffeomorphisms ðð¡ but these dif-feomorphisms are not defined on all of ð nor for all values of ð¡. Moreover, the identity(2.2.16) only holds on the open subset of ð where both sides are well-defined.
We devote the remainder of this section to discussing some âfunctorialâ properties ofvector fields and one-forms.
Definition 2.2.17. Letð â ðð andð â ðð be open, and let ðⶠð â ð be a ð¶â map. If ðis a ð¶â-vector field on ð and ð a ð¶â-vector field onð we say that ð and ð are ð-relatedif, for all ð â ð we have
ððð(ð(ð)) = ð(ð(ð)) .
Writing
ð =ðâð=1ð£ðððð¥ð
, ð£ð â ð¶ð(ð)
and
ð =ðâð=1ð€ððððŠð
, ð€ð â ð¶ð(ð)
this equation reduces, in coordinates, to the equation
ð€ð(ð) =ðâð=1
ððððð¥ð(ð)ð£ð(ð) .
In particular, ifð = ð andð is að¶â diffeomorphism, the formula (3.2) defines að¶â-vectorfield onð, i.e.,
ð =ðâð=1ð€ððððŠð
is the vector field defined by the equation
ð€ð =ðâð=1(ððððð¥ðð£ð) â ðâ1 .
Hence we have proved:
Theorem 2.2.18. If ðⶠð ⥲ ð is a ð¶â diffeomorphism and ð a ð¶â-vector field onð, thereexists a unique ð¶â vector field ð onð having the property that ð and ð are ð-related.
Definition 2.2.19. In the setting of Theorem 2.2.18, we denote the vector field ð by ðâð,and call ðâð the pushforward of ð by ð .
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§2.2 Integral Curves for Vector Fields 41
We leave the following assertions as easy exercises (but provide some hints).
Theorem 2.2.20. For ð = 1, 2, let ðð be an open subset of ððð , ðð a vector field on ðð andðⶠð1 â ð2 a ð¶â-map. If ð1 and ð2 are ð-related, every integral curve
ðŸâ¶ ðŒ â ð1of ð1 gets mapped by ð onto an integral curve ð â ðŸâ¶ ðŒ â ð2 of ð2.
Proof sketch. Theorem 2.2.20 follows from the chain rule: if ð = ðŸ(ð¡) and ð = ð(ð)
ððð (ððð¡ðŸ(ð¡)) = ð
ðð¡ð(ðŸ(ð¡)) . â¡
Corollary 2.2.21. In the setting of Theorem 2.2.20, suppose ð1 and ð2 are complete. Let(ðð,ð¡)ð¡âð ⶠðð ⥲ ðð be the one-parameter group of diffeomorphisms generated by ðð. Thenð â ð1,ð¡ = ð2,ð¡ â ð.
Proof sketch. To deduce Corollary 2.2.21 from Theorem 2.2.20 note that for ð â ð, ð1,ð¡(ð)is just the integral curve ðŸð(ð¡) of ð1 with initial point ðŸð(0) = ð.
The notion of ð-relatedness can be very succinctly expressed in terms of the Lie differ-entiation operation. For ð â ð¶â(ð2) let ðâð be the composition, ð â ð, viewed as a ð¶âfunction on ð1, i.e., for ð â ð1 let ðâð(ð) = ð(ð(ð)). Then
ðâð¿ð2ð = ð¿ð1ðâð .
To see this, note that if ð(ð) = ð then at the point ð the right hand side is(ðð)ð â ððð(ð1(ð))
by the chain rule and by definition the left hand side isððð(ð2(ð)) .
Moreover, by definitionð2(ð) = ððð(ð1(ð))
so the two sides are the same. â¡
Another easy consequence of the chain rule is:
Theorem 2.2.22. For ð = 1, 2, 3, let ðð be an open subset of ððð , ðð a vector field on ðð, andfor ð = 1, 2 let ðð ⶠðð â ðð+1 be a ð¶â-map. Suppose that ð1 and ð2 are ð1-related and that ð2and ð3 are ð2-related. Then ð1 and ð3 are (ð2 â ð1)-related.
In particular, if ð1 and ð2 are diffeomorphisms writing ð â ð1 we have(ð2)â(ð1)âð = (ð2 â ð1)âð .
The results we described above have âdualâ analogues for one-forms.
Construction 2.2.23. Let ð â ðð and ð â ðð be open, and let ðⶠð â ð be a ð¶â-map.Given a one-form ð on ð one can define a pullback one-form ðâð on ð by the followingmethod. For ð â ð, by definition ðð(ð) is a linear map
ðð(ð) ⶠðð(ð)ðð â ðand by composing this map with the linear map
ððð ⶠðððð â ðð(ð)ðð
we get a linear mapðð(ð) â ððð ⶠðððð â ð ,
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42 Chapter 2: Differential Forms
i.e., an element ðð(ð) â ððð of ðâð ðð.The pullback one-form ðâð is defined by the assignment ð ⊠(ðð(ð) â ððð).
In particular, if ðⶠð â ð is a ð¶â-function and ð = ðð thenðð â ððð = ððð â ððð = ð(ð â ð)ð
i.e.,ðâð = ðð â ð .
In particular if ð is a one-form of the form ð = ðð, with ð â ð¶â(ð), ðâð is ð¶â. From thisit is easy to deduce:
Theorem 2.2.24. If ð is any ð¶â one-form onð, its pullback ðâð is ð¶â. (See Exercise 2.2.ii.)
Notice also that the pullback operation on one-forms and the pushforward operationon vector fields are somewhat different in character. The former is defined for all ð¶â maps,but the latter is only defined for diffeomorphisms.
Exercises for §2.2Exercise 2.2.i. Prove the reparameterization result Theorem 2.2.20.
Exercise 2.2.ii. Letð be an open subset of ðð,ð an open subset of ðð and ðⶠð â ð a ð¶ðmap.(1) Show that for ð â ð¶â(ð) (2.2) can be rewritten
ðâðð = ððâð .(2) Let ð be the one-form
ð =ðâð=1ðððð¥ð , ðð â ð¶â(ð)
on ð. Show that if ð = (ð1,âŠ, ðð) then
ðâð =ðâð=1ðâðð ððð .
(3) Show that if ð is ð¶â and ð is ð¶â, ðâð is ð¶â.
Exercise 2.2.iii. Let ð be a complete vector field on ð and ðð¡ ⶠð â ð, the one parametergroup of diffeomorphisms generated by ð. Show that if ð â ð¶1(ð)
ð¿ðð =ððð¡ðâð¡ ð |ð¡=0
.
Exercise 2.2.iv.(1) Let ð = ð2 and let ð be the vector field, ð¥1ð/ðð¥2 â ð¥2ð/ðð¥1. Show that the curve
ð¡ ⊠(ð cos(ð¡ + ð), ð sin(ð¡ + ð)) ,for ð¡ â ð, is the unique integral curve of ð passing through the point, (ð cos ð, ð sin ð),at ð¡ = 0.
(2) Let ð = ðð and let ð be the constant vector field: âðð=1 ððð/ðð¥ð. Show that the curve
ð¡ ⊠ð + ð¡(ð1,âŠ, ðð) ,for ð¡ â ð, is the unique integral curve of ð passing through ð â ðð at ð¡ = 0.
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§2.2 Integral Curves for Vector Fields 43
(3) Let ð = ðð and let ð be the vector field, âðð=1 ð¥ðð/ðð¥ð. Show that the curve
ð¡ ⊠ðð¡(ð1,âŠ, ðð) ,for ð¡ â ð, is the unique integral curve of ð passing through ð at ð¡ = 0.
Exercise 2.2.v. Show that the following are one-parameter groups of diffeomorphisms:(1) ðð¡ ⶠð â ð , ðð¡(ð¥) = ð¥ + ð¡(2) ðð¡ ⶠð â ð , ðð¡(ð¥) = ðð¡ð¥(3) ðð¡ ⶠð2 â ð2 , ðð¡(ð¥, ðŠ) = (ð¥ cos(ð¡) â ðŠ sin(ð¡), ð¥ sin(ð¡) + ðŠ cos(ð¡)).
Exercise 2.2.vi. Let ðŽâ¶ ðð â ðð be a linear mapping. Show that the series
exp(ð¡ðŽ) âââð=0
(ð¡ðŽ)ðð!= idð +ð¡ðŽ +
ð¡22!ðŽ2 + ð¡
3
3!ðŽ3 +â¯
converges and defines a one-parameter group of diffeomorphisms of ðð.
Exercise 2.2.vii.(1) What are the infinitesimal generators of the one-parameter groups in Exercise 2.2.v?(2) Show that the infinitesimal generator of the one-parameter group in Exercise 2.2.vi is
the vector fieldâ1â€ð,ðâ€ððð,ðð¥ðððð¥ð
where (ðð,ð) is the defining matrix of ðŽ.
Exercise 2.2.viii. Let ð be the vector field on ð given by ð¥2 ððð¥ . Show that the curve
ð¥(ð¡) = ðð â ðð¡
is an integral curve of ð with initial point ð¥(0) = ð. Conclude that for ð > 0 the curve
ð¥(ð¡) = ð1 â ðð¡
, 0 < ð¡ < 1ð
is a maximal integral curve. (In particular, conclude that ð is not complete.)
Exercise 2.2.ix. Let ð and ð be open subsets of ðð and ðⶠð ⥲ ð a diffeomorphism. Ifðis a vector field on ð, define the pullback of ð to ð to be the vector field
ðâð â (ðâ1â ð) .Show that if ð is a ð¶â function on ð
ðâð¿ðð = ð¿ðâððâð .Hint: Equation (2.2.25).
Exercise 2.2.x. Let ð be an open subset of ðð and ð and ð vector fields on ð. Suppose ð isthe infinitesimal generator of a one-parameter group of diffeomorphisms
ðð¡ ⶠð ⥲ ð , ââ < ð¡ < â .Let ðð¡ = ðâð¡ ð. Show that for ð â ð¶â(ð) we have
ð¿[ð,ð]ð = ð¿ï¿œï¿œð ,where
ᅵᅵ = ððð¡ðâð¡ ð |
ð¡=0.
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44 Chapter 2: Differential Forms
Hint: Differentiate the identityðâð¡ ð¿ðð = ð¿ðð¡ð
âð¡ ð
with respect to ð¡ and show that at ð¡ = 0 the derivative of the left hand side is ð¿ðð¿ðð byExercise 2.2.iii, and the derivative of the right hand side is
ð¿ï¿œï¿œ + ð¿ð(ð¿ðð) .
Exercise 2.2.xi. Conclude from Exercise 2.2.x that
(2.2.25) [ð, ð] = ððð¡ðâð¡ ð |
ð¡=0.
2.3. Differential ð-forms
One-forms are the bottom tier in a pyramid of objects whose ðth tier is the space of ð-forms. More explicitly, given ð â ðð we can, as in Section 1.5, form the ðth exterior powers
ð¬ð(ðâð ðð) , ð = 1, 2, 3,âŠ, ðof the vector space ðâð ðð, and since
(2.3.1) ð¬1(ðâð ðð) = ðâð ðð
one can think of a one-form as a function which takes its value at ð in the space (2.3.1). Thisleads to an obvious generalization.
Definition 2.3.2. Let ð be an open subset of ðð. A ð-form ð on ð is a function whichassigns to each point ð â ð an element ðð â ð¬ð(ðâð ðð).
The wedge product operation gives us a way to construct lots of examples of such ob-jects.
Example 2.3.3. Let ð1,âŠ, ðð be one-forms. Then ð1 â§â¯ ⧠ðð is the ð-form whose valueat ð is the wedge product(2.3.4) (ð1 â§â¯ ⧠ðð)ð â (ð1)ð â§â¯ ⧠(ðð)ð .
Notice that since (ðð)ð is in ð¬1(ðâð ðð) the wedge product (2.3.4) makes sense and is anelement of ð¬ð(ðâð ðð).
Example 2.3.5. Let ð1,âŠ, ðð be a real-valued ð¶â functions on ð. Letting ðð = ððð we getfrom (2.3.4) a ð-form
ðð1 â§â¯ ⧠ðððwhose value at ð is the wedge product(2.3.6) (ðð1)ð â§â¯ ⧠(ððð)ð .
Since (ðð¥1)ð,âŠ, (ðð¥ð)ð are a basis of ðâð ðð, the wedge products
(2.3.7) (ðð¥ð1)ð â§â¯ ⧠(ðð¥1ð)ð , 1 †ð1 < ⯠< ðð †ð
are a basis ofð¬ð(ðâð ). To keep our multi-index notation from getting out of hand, we denotethese basis vectors by (ðð¥ðŒ)ð, where ðŒ = (ð1,âŠ, ðð) and the multi-indices ðŒ range over multi-indices of length ð which are strictly increasing. Since these wedge products are a basis ofð¬ð(ðâð ðð), every element of ð¬ð(ðâð ðð) can be written uniquely as a sum
âðŒððŒ(ðð¥ðŒ)ð , ððŒ â ð
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§2.3 Differential ð-forms 45
and every ð-form ð on ð can be written uniquely as a sum
(2.3.8) ð = âðŒððŒðð¥ðŒ
where ðð¥ðŒ is the ð-form ðð¥ð1 â§â¯ ⧠ðð¥ðð , and ððŒ is a real-valued function ððŒ ⶠð â ð.Definition 2.3.9. The ð-form (2.3.8) is of class ð¶ð if each function ððŒ is in ð¶ð(ð).
Henceforth we assume, unless otherwise stated, that all the ð-forms we consider are ofclass ð¶â. We denote the set of ð-forms of class ð¶â on ð by ðºð(ð).
We will conclude this section by discussing a few simple operations on ð-forms. Let ðbe an open subset of ðð.(1) Given a function ð â ð¶â(ð) and a ð-form ð â ðºð(ð) we define ðð â ðºð(ð) to be theð-form defined by
ð ⊠ð(ð)ðð .(2) Given ð1, ð2 â ðºð(ð), we define ð1 + ð2 â ðºð(ð) to be the ð-form
ð ⊠(ð1)ð + (ð2)ð .
(Notice that this sum makes sense since each summand is in ð¬ð(ðâð ðð).)(3) Givenð1 â ðºð1(ð) andð2 â ðºð2(ð)wedefine theirwedge product,ð1â§ð2 â ðºð1+ð2(ð)
to be the (ð1 + ð2)-formð ⊠(ð1)ð ⧠(ð2)ð .
We recall that ð¬0(ðâð ðð) = ð, so a zero-form is an ð-valued function and a zero formof class ð¶â is a ð¶â function, i.e., ðº0(ð) = ð¶â(ð).The addition and multiplication by functions operations naturally give the sets ðºð(ð)
the structures ofð-vector spacesâwe always regardðºð(ð) as a vector space in thismanner.A fundamental operation on forms is the exterior differention operation which asso-
ciates to a function ð â ð¶â(ð) the 1-form ðð. It is clear from the identity (2.2.3) that ððis a 1-form of class ð¶â, so the ð-operation can be viewed as a map
ðⶠðº0(ð) â ðº1(ð) .In the next section we show that an analogue of this map exists for every ðºð(ð).
Exercises for §2.3Exercise 2.3.i. Let ð â ðº2(ð4) be the 2-form ðð¥1 ⧠ðð¥2 + ðð¥3 ⧠ðð¥4. Compute ð ⧠ð.Exercise 2.3.ii. Let ð1, ð2, ð3 â ðº1(ð3) be the 1-forms
ð1 = ð¥2ðð¥3 â ð¥3ðð¥2 ,ð2 = ð¥3ðð¥1 â ð¥1ðð¥3 ,ð3 = ð¥1ðð¥2 â ð¥2ðð¥1 .
Compute the following.(1) ð1 ⧠ð2.(2) ð2 ⧠ð3.(3) ð3 ⧠ð1.(4) ð1 ⧠ð2 ⧠ð3.Exercise 2.3.iii. Let ð be an open subset of ðð and ð1,âŠ, ðð â ð¶â(ð). Show that
ðð1 â§â¯ ⧠ððð = det[ððððð¥ð]ðð¥1 â§â¯ ⧠ðð¥ð .
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46 Chapter 2: Differential Forms
Exercise 2.3.iv. Let ð be an open subset of ðð. Show that every (ð â 1)-form ð â ðºðâ1(ð)can be written uniquely as a sum
ðâð=1ðððð¥1 â§â¯ ⧠ðð¥ð â§â¯ ⧠ðð¥ð ,
where ðð â ð¶â(ð) and ðð¥ð indicates that ðð¥ð is to be omitted from the wedge productðð¥1 â§â¯ ⧠ðð¥ð.
Exercise 2.3.v. Let ð = âðð=1 ð¥ððð¥ð. Show that there exists an (ðâ1)-form,ð â ðºðâ1(ððâ{0})with the property
ð ⧠ð = ðð¥1 â§â¯ ⧠ðð¥ð .
Exercise 2.3.vi. Let ðœ be the multi-index (ð1,âŠ, ðð) and let ðð¥ðœ = ðð¥ð1 ⧠⯠⧠ðð¥ðð . Showthat ðð¥ðœ = 0 if ðð = ðð for some ð â ð and show that if the numbers ð1,âŠ, ðð are all distinct,then
ðð¥ðœ = (â1)ððð¥ðŒ ,where ðŒ = (ð1,âŠ, ðð) is the strictly increasing rearrangement of (ð1,âŠ, ðð) and ð is the per-mutation
(ð1,âŠ, ðð) ⊠(ð1,âŠ, ðð) .
Exercise 2.3.vii. Let ðŒ be a strictly increasingmulti-index of length ð and ðœ a strictly increas-ing multi-index of length â. What can one say about the wedge product ðð¥ðŒ ⧠ðð¥ðœ?
2.4. Exterior differentiation
Letð be an open subset of ðð. In this section we define an exterior differentiation oper-ation
(2.4.1) ðⶠðºð(ð) â ðºð+1(ð) .
This operation is the fundamental operation in ð-dimensional vector calculus.For ð = 0 we already defined the operation (2.4.1) in §2.1. Before defining it for ð > 0,
we list some properties that we require to this operation to satisfy.
Properties 2.4.2 (desired properties of exterior differentiation). Letð be an open subset ofðð.(1) For ð1 and ð2 in ðºð(ð), we have
ð(ð1 + ð2) = ðð1 + ðð2 .
(2) For ð1 â ðºð(ð) and ð2 â ðºâ(ð) we have
(2.4.3) ð(ð1 ⧠ð2) = ðð1 ⧠ð2 + (â1)ðð1 ⧠ðð2 .
(3) For all ð â ðºð(ð) we have
(2.4.4) ð(ðð) = 0
Let is point out a few consequences of these properties.
Lemma 2.4.5. Let ð be an open subset of ðð. For any functions ð1,âŠ, ðð â ð¶â(ð) we have
ð(ðð1 â§â¯ ⧠ððð) = 0 .
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§2.4 Exterior differentiation 47
Proof. We prove this by induction on ð. For the base case note that by (3)(2.4.6) ð(ðð1) = 0for every function ð1 â ð¶â(ð).
For the induction step, suppose that we know the result for ð â 1 functions and thatwe are given functions ð1,âŠ, ðð â ð¶â(ð). Let ð = ðð2 ⧠⯠⧠ððð. Then by the inductionhypothesis ðð = 0 Combining (2.4.3) with (2.4.6) we see that
ð(ðð1 ⧠ðð2 â§â¯ ⧠ððð) = ð(ðð1 ⧠ð)= ð(ðð1) ⧠ð + (â1) ðð1 ⧠ðð= 0 . â¡
Example 2.4.7. As a special case of Lemma 2.4.5, given a multi-index ðŒ = (ð1,âŠ, ðð) with1 †ðð †ð we have(2.4.8) ð(ðð¥ðŒ) = ð(ðð¥ð1 â§â¯ ⧠ðð¥ðð) = 0 .
Recall that every ð-form ð â ðºð(ð) can be written uniquely as a sum
ð = âðŒððŒðð¥ðŒ , ððŒ â ð¶â(ð)
where the multi-indices ðŒ are strictly increasing. Thus by (2.4.3) and (2.4.8)
(2.4.9) ðð = âðŒðððŒ ⧠ðð¥ðŒ .
This shows that if there exists an operator ðⶠðºâ(ð) â ðºâ+1(ð)with Properties 2.4.2, thenð is necessarily given by the formula (2.4.9). Hence all we have to show is that the operatordefined by this equation (2.4.9) has these properties.
Proposition 2.4.10. Let ð be an open subset of ðð. There is a unique operator ðⶠðºâ(ð) âðºâ+1(ð) satisfying Properties 2.4.2.Proof. The property (1) is obvious.
To verify (2) we first note that for ðŒ strictly increasing equation (2.4.8) is a special caseof equation (2.4.9) (take ððŒ = 1 and ððœ = 0 for ðœ â ðŒ). Moreover, if ðŒ is not strictly increasingit is either repeating, in which case ðð¥ðŒ = 0, or non-repeating in which case there exists apermuation ð â ðð such that ðŒð is strictly increasing. Moreover,(2.4.11) ðð¥ðŒ = (â1)ððð¥ðŒð .Hence (2.4.9) implies (2.4.8) for all multi-indices ðŒ. The same argument shows that for anysum âðŒ ððŒðð¥ðŒ over multi-indices ðŒ of length ð, we have
(2.4.12) ð(âðŒ ððŒðð¥ðŒ) = âðŒðððŒ ⧠ðð¥ðŒ .
(As above we can ignore the repeatingmulti-indices ðð¥ðŒ = 0 if ðŒ is repeating, and by (2.4.11)we can replace the non-repeating multi-indices by strictly increasing multi-indices.)
Suppose now that ð1 â ðºð(ð) and ð2 â ðºâ(ð). Write ð1 â âðŒ ððŒðð¥ðŒ and ð2 ââðœ ððœðð¥ðœ, where ððŒ, ððœ â ð¶â(ð). We see that the wedge product ð1 ⧠ð2 is given by
(2.4.13) ð1 ⧠ð2 = âðŒ,ðœððŒððœðð¥ðŒ ⧠ðð¥ðœ ,
and by equation (2.4.12)
(2.4.14) ð(ð1 ⧠ð2) = âðŒ,ðœð(ððŒððœ) ⧠ðð¥ðŒ ⧠ðð¥ðœ .
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48 Chapter 2: Differential Forms
(Notice that if ðŒ = (ð1,âŠ, ðð) and ðœ = (ðð,âŠ, ðâ), we have ðð¥ðŒ ⧠ðð¥ðœ = ðð¥ðŸ, where ðŸ â(ð1,âŠ, ðð, ð1,âŠ, ðâ). Even if ðŒ and ðœ are strictly increasing, ðŸ is not necessarily strictly in-creasing. However, in deducing (2.4.14) from (2.4.13) we have observed that this does notmatter.)
Now note that by equation (2.2.8)
ð(ððŒððœ) = ððœðððŒ + ððŒðððœ ,
and by the wedge product identities of §1.6,
ðððœ ⧠ðð¥ðŒ = ðððœ ⧠ðð¥ð1 â§â¯ ⧠ðð¥ðð = (â1)ððð¥ðŒ ⧠ðððœ ,
so the sum (2.4.14) can be rewritten:
âðŒ,ðœðððŒ ⧠ðð¥ðŒ ⧠ððœðð¥ðœ + (â1)ðâ
ðŒ,ðœððŒðð¥ðŒ ⧠ðððœ ⧠ðð¥ðœ ,
or
(âðŒðððŒ ⧠ðð¥ðŒ) ⧠(â
ðœððœðð¥ðœ) + (â1)ð(â
ðœðððœ ⧠ðð¥ðœ) ⧠(â
ðŒðððŒ ⧠ðð¥ðŒ) ,
or, finally,ðð1 ⧠ð2 + (â1)ðð1 ⧠ðð2 .
Thus the the operator ð defined by equation (2.4.9) has (2).Let is now check that ð satisfies (3). If ð = âðŒ ððŒðð¥ðŒ, where ððŒ â ð¶â(ð), then by
definition, ðð = âðŒ ðððŒ ⧠ðð¥ðŒ and by (2.4.8) and (2.4.3)
ð(ðð) = âðŒð(ðððŒ) ⧠ðð¥ðŒ ,
so it suffices to check that ð(ðððŒ) = 0, i.e., it suffices to check (2.4.6) for zero forms ð âð¶â(ð). However, by (2.2.3) we have
ðð =ðâð=1
ðððð¥ððð¥ð
so by equation (2.4.9)
ð(ðð) =ðâð=1ð( ðððð¥ð)ðð¥ð =
ðâð=1(ðâð=1
ð2ððð¥ððð¥ððð¥ð) ⧠ðð¥ð
= â1â€ð,ðâ€ð
ð2ððð¥ððð¥ððð¥ð ⧠ðð¥ð .
Notice, however, that in this sum, ðð¥ð ⧠ðð¥ð = âðð¥ð ⧠ðð¥ð and
ð2ððð¥ððð¥ð= ð2ððð¥ððð¥ð
so the (ð, ð) term cancels the (ð, ð) term, and the total sum is zero. â¡
Definition 2.4.15. Let ð be an open subset of ðð. A ð-form ð â ðºð(ð) is closed if ðð = 0and is exact if ð = ðð for some ð â ðºðâ1(ð).
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§2.4 Exterior differentiation 49
By (3) every exact form is closed, but the converse is not true even for 1-forms (seeExercise 2.1.iii). In fact it is a very interesting (and hard) question to determine if an openset ð has the following property: For ð > 0 every closed ð-form is exact.2
Some examples of spaces with this property are described in the exercises at the end of§2.6. We also sketch below a proof of the following result (and ask you to fill in the details).
Lemma 2.4.16 (Poincaré lemma). If ð is a closed form on ð of degree ð > 0, then for everypoint ð â ð, there exists a neighborhood of ð on which ð is exact.
Proof. See Exercises 2.4.v and 2.4.vi. â¡Exercises for §2.4
Exercise 2.4.i. Compute the exterior derivatives of the following differential forms.(1) ð¥1ðð¥2 ⧠ðð¥3(2) ð¥1ðð¥2 â ð¥2ðð¥1(3) ðâððð where ð = âðð=1 ð¥2ð(4) âðð=1 ð¥ððð¥ð(5) âðð=1(â1)ðð¥ððð¥1 â§â¯ ⧠ðð¥ð â§â¯ ⧠ðð¥ðExercise 2.4.ii. Solve the equation ðð = ð for ð â ðº1(ð3), where ð is the 2-form:(1) ðð¥2 ⧠ðð¥3(2) ð¥2ðð¥2 ⧠ðð¥3(3) (ð¥21 + ð¥22)ðð¥1 ⧠ðð¥2(4) cos(ð¥1)ðð¥1 ⧠ðð¥3Exercise 2.4.iii. Let ð be an open subset of ðð.(1) Show that if ð â ðºð(ð) is exact and ð â ðºâ(ð) is closed then ð ⧠ð is exact.
Hint: Equation (2.4.3).(2) In particular, ðð¥1 is exact, so if ð â ðºâ(ð) is closed ðð¥1 ⧠ð = ðð. What is ð?Exercise 2.4.iv. Let ð be the rectangle (ð1, ð1) à ⯠à (ðð, ðð). Show that if ð is in ðºð(ð),then ð is exact.
Hint: Let ð = ððð¥1 â§â¯ ⧠ðð¥ð with ð â ð¶â(ð) and let ð be the function
ð(ð¥1,âŠ, ð¥ð) = â«ð¥1
ð1ð(ð¡, ð¥2,âŠ, ð¥ð)ðð¡ .
Show that ð = ð(ððð¥2 â§â¯ ⧠ðð¥ð).Exercise 2.4.v. Let ð be an open subset of ððâ1, ðŽ â ð an open interval and (ð¥, ð¡) productcoordinates on ð à ðŽ. We say that a form ð â ðºâ(ð à ðŽ) is reduced if ð can be written as asum(2.4.17) ð = â
ðŒððŒ(ð¥, ð¡)ðð¥ðŒ ,
(i.e., with no terms involving ðð¡).(1) Show that every form, ð â ðºð(ð à ðŽ) can be written uniquely as a sum:(2.4.18) ð = ðð¡ ⧠ðŒ + ðœ
where ðŒ and ðœ are reduced.
2For ð = 0, ðð = 0 does not imply that ð is exact. In fact âexactnessâ does not make much sense for zeroforms since there are not any â(â1)-formsâ. However, if ð â ð¶â(ð) and ðð = 0 then ð is constant on connectedcomponents of ð (see Exercise 2.2.iii).
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50 Chapter 2: Differential Forms
(2) Let ð be the reduced form (2.4.17) and letðððð¡= â ððð¡ððŒ(ð¥, ð¡)ðð¥ðŒ
and
ððð = âðŒ(ðâð=1
ððð¥ðððŒ(ð¥, ð¡)ðð¥ð) ⧠ðð¥ðŒ .
Show thatðð = ðð¡ ⧠ðð
ðð¡+ ððð .
(3) Let ð be the form (2.4.18). Show that
ðð = âðð¡ ⧠ðððŒ + ðð¡ â§ððœðð¡+ ðððœ
and conclude that ð is closed if and only if
(2.4.19) {ððœðð¡ = ðððŒðððœ = 0 .
(4) Let ðŒ be a reduced (ð â 1)-form. Show that there exists a reduced (ð â 1)-form ð suchthat
(2.4.20) ðððð¡= ðŒ .
Hint: Let ðŒ = âðŒ ððŒ(ð¥, ð¡)ðð¥ðŒ and ð = âððŒ(ð¥, ð¡)ðð¥ðŒ. The equation (2.4.20) reducesto the system of equations
(2.4.21) ððð¡ððŒ(ð¥, ð¡) = ððŒ(ð¥, ð¡) .
Let ð be a point on the interval, ðŽ, and using calculus show that equation (2.4.21) has aunique solution, ððŒ(ð¥, ð¡), with ððŒ(ð¥, ð) = 0.
(5) Show that if ð is the form (2.4.18) and ð a solution of (2.4.20) then the form
(2.4.22) ð â ððis reduced.
(6) LetðŸ = ââðŒ(ð¥, ð¡)ðð¥ðŒ
be a reduced ð-form. Deduce from (2.4.19) that if ðŸ is closed then ððŸðð¡ = 0 and ðððŸ = 0.Conclude that âðŒ(ð¥, ð¡) = âðŒ(ð¥) and that
ðŸ = âðŒâðŒ(ð¥)ðð¥ðŒ
is effectively a closed ð-form onð. Prove that if every closed ð-form onð is exact, thenevery closed ð-form on ð à ðŽ is exact.
Hint: Let ð be a closed ð-form on ð à ðŽ and let ðŸ be the form (2.4.22).
Exercise 2.4.vi. Let ð â ðð be an open rectangle. Show that every closed form on ð ofdegree ð > 0 is exact.
Hint: Let ð = (ð1, ð1) à ⯠à (ðð, ðð). Prove this assertion by induction, at the ðth stageof the induction letting ð = (ð1, ð1) à ⯠à (ððâ1, ððâ1) and ðŽ = (ðð, ðð).
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2.5. The interior product operation
In §2.1 we explained how to pair a one-form ð and a vector field ð to get a function ððð.This pairing operation generalizes.
Definition 2.5.1. Let ð be an open subset of ðð, ð a vector field on ð, and ð â ðºð(ð)Theinterior product of ð with ð is (ð â 1)-form ððð on ð defined by declaring the value of ðððvalue at ð â ð to be the interior product(2.5.2) ðð(ð)ðð .
Note that ð(ð) is inðððð andðð inð¬ð(ðâð ðð), so by definition of interior product (see §1.7),the expression (2.5.2) is an element of ð¬ðâ1(ðâð ðð).
From the properties of interior product on vector spaces which we discussed in §1.7,one gets analogous properties for this interior product on forms. We list these properties,leaving their verification as an easy exercise.
Properties 2.5.3. Letð be an open subset ofðð, ð andð vector fields onð, ð1 and ð2 bothð-forms on ð, and ð a ð-form and ð an â-form on ð.(1) Linearity in the form: we have
ðð(ð1 + ð2) = ððð1 + ððð2 .(2) Linearity in the vector field: we have
ðð+ðð = ððð + ððð .(3) Derivation property: we have
ðð(ð ⧠ð) = ððð ⧠ð + (â1)ðð ⧠ððð(4) The interior product satisfies the identity
ðð(ððð) = âðð(ððð) .(5) As a special case of (4), the interior product satisfies the identity
ðð(ððð) = 0 .(6) Moreover, if ð is decomposable, i.e., is a wedge product of one-forms
ð = ð1 â§â¯ ⧠ðð ,then
ððð =ðâð=1(â1)ðâ1ðð(ðð)ð1 â§â¯ ⧠ᅵᅵð â§â¯ ⧠ðð .
We also leave for you to prove the following two assertions, both of which are specialcases of (6).
Example 2.5.4. If ð = ð/ðð¥ð and ð = ðð¥ðŒ = ðð¥ð1 â§â¯ ⧠ðð¥ðð then
(2.5.5) ððð =ðâð=1(â1)ðâ1ð¿ð,ðððð¥ðŒð
where
ð¿ð,ðð â {1, ð = ðð0, ð â ðð .
and ðŒð = (ð1,âŠ, ð€ð,âŠ, ðð).
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52 Chapter 2: Differential Forms
Example 2.5.6. If ð = âðð=1 ðð ð/ðð¥ð and ð = ðð¥1 â§â¯ ⧠ðð¥ð then
(2.5.7) ððð =ðâð=1(â1)ðâ1ðð ðð¥1 â§â¯ðð¥ðâ¯â§ ðð¥ð .
By combining exterior differentiation with the interior product operation one gets an-other basic operation of vector fields on forms: the Lie differentiation operation. For zero-forms, i.e., forð¶â functions, we defined this operation by the formula (2.1.16). For ð-formswe define it by a slightly more complicated formula.Definition 2.5.8. Let ð be an open subset of ðð, ð a vector field on ð, and ð â ðºð(ð). TheLie derivative of ð with respect to ð is the ð-form(2.5.9) ð¿ðð â ðð(ðð) + ð(ððð) .
Notice that for zero-forms the second summand is zero, so (2.5.9) and (2.1.16) agree.Properties 2.5.10. Let ð be an open subset of ðð, ð a vector field on ð, ð â ðºð(ð), andð â ðºâ(ð). The Lie derivative enjoys the following properties:(1) Commutativity with exterior differentiation: we have(2.5.11) ð(ð¿ðð) = ð¿ð(ðð) .(2) Interaction with wedge products: we have(2.5.12) ð¿ð(ð ⧠ð) = ð¿ðð ⧠ð + ð ⧠ð¿ðð
From Properties 2.5.10 it is fairly easy to get an explicit formula for ð¿ðð. Namely let ðbe the ð-form
ð = âðŒððŒðð¥ðŒ , ððŒ â ð¶â(ð)
and ð the vector field
ð =ðâð=1ððððð¥ð
, ðð â ð¶â(ð) .
By equation (2.5.12)ð¿ð(ððŒðð¥ðŒ) = (ð¿ðððŒ)ðð¥ðŒ + ððŒ(ð¿ððð¥ðŒ)
and
ð¿ððð¥ðŒ =ðâð=1ðð¥ð1 â§â¯ ⧠ð¿ððð¥ðð â§â¯ ⧠ðð¥ðð ,
and by equation (2.5.11)ð¿ððð¥ðð = ðð¿ðð¥ðð
so to compute ð¿ðð one is reduced to computing ð¿ðð¥ðð and ð¿ðððŒ.However by equation (2.5.12)
ð¿ðð¥ðð = ðððand
ð¿ðððŒ =ðâð=1ðððððŒðð¥ð
.
We leave the verification of Properties 2.5.10 as exercises, and also ask you to prove (by themethod of computation that we have just sketched) the following divergence formula.Lemma 2.5.13. Letð â ðð be open, let ð1,âŠ, ðð â ð¶â(ð), and set ð â âðð=1 ðð ð/ðð¥ð. Then
(2.5.14) ð¿ð(ðð¥1 â§â¯ ⧠ðð¥ð) =ðâð=1(ððððð¥ð)ðð¥1 â§â¯ ⧠ðð¥ð .
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§2.5 The interior product operation 53
Exercises for §2.5Exercise 2.5.i. Verify the Properties 2.5.3 (1)â(6).
Exercise 2.5.ii. Show that if ð is the ð-form ðð¥ðŒ and ð the vector field ð/ðð¥ð, then ððð isgiven by equation (2.5.5).
Exercise 2.5.iii. Show that ifð is the ð-formðð¥1â§â¯â§ðð¥ð and ð the vector fieldâðð=1 ððð/ðð¥ð,then ððð is given by equation (2.5.7).
Exercise 2.5.iv. Let ð be an open subset of ðð and ð a ð¶â vector field on ð. Show that forð â ðºð(ð)
ðð¿ðð = ð¿ðððand
ðð(ð¿ðð) = ð¿ð(ððð) .Hint: Deduce the first of these identities from the identity ð(ðð) = 0 and the second
from the identity ðð(ððð) = 0.
Exercise 2.5.v. Given ðð â ðºðð(ð), for ð = 1, 2 show thatð¿ð(ð1 ⧠ð2) = ð¿ðð1 ⧠ð2 + ð1 ⧠ð¿ðð2 .
Hint: Plug ð = ð1 ⧠ð2 into equation (2.5.9) and use equation (2.4.3) and the derivationproperty of the interior product to evaluate the resulting expression.
Exercise 2.5.vi. Let ð1 and ð2 be vector fields onð and letð be their Lie bracket. Show thatfor ð â ðºð(ð)
ð¿ðð = ð¿ð1(ð¿ð2ð) â ð¿ð2(ð¿ð1ð) .Hint: By definition this is true for zero-forms and by equation (2.5.11) for exact one-
forms. Now use the fact that every form is a sum of wedge products of zero-forms andone-forms and the fact that ð¿ð satisfies the product identity (2.5.12).
Exercise 2.5.vii. Prove the divergence formula (2.5.14).
Exercise 2.5.viii.(1) Let ð = ðºð(ðð) be the form
ð = âðŒððŒ(ð¥1,âŠ, ð¥ð)ðð¥ðŒ
and ð the vector field ð/ðð¥ð. Show that
ð¿ðð = âððð¥ðððŒ(ð¥1,âŠ, ð¥ð)ðð¥ðŒ .
(2) Suppose ððð = ð¿ðð = 0. Show that ð only depends on ð¥1,âŠ, ð¥ðâ1 and ðð¥1,âŠ, ðð¥ðâ1,i.e., is effectively a ð-form on ððâ1.
(3) Suppose ððð = ðð = 0. Show that ð is effectively a closed ð-form on ððâ1.(4) Use these results to give another proof of the Poincaré lemma forðð. Prove by induction
on ð that every closed form on ðð is exact.Hints:
⢠Let ð be the form in part (1) and let
ððŒ(ð¥1,âŠ, ð¥ð) = â«ð¥ð
0ððŒ(ð¥1,âŠ, ð¥ðâ1, ð¡)ðð¡ .
Show that if ð = âðŒ ððŒðð¥ðŒ, then ð¿ðð = ð.
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54 Chapter 2: Differential Forms
⢠Conclude that(2.5.15) ð â ðððð = ðð ðð .
⢠Suppose ðð = 0. Conclude from equation (2.5.15) and (5) of Properties 2.5.3 thatthe form ðœ = ðððð satisfies ððœ = ð(ð)ðœ = 0.
⢠By part (3), ðœ is effectively a closed form on ððâ1, and by induction, ðœ = ððŒ. Thusby equation (2.5.15)
ð = ðððð + ððŒ .
2.6. The pullback operation on forms
Letð be an open subset ofðð,ð an open subset ofðð andðⶠð â ð að¶âmap. Thenfor ð â ð and the derivative of ð at ð
ððð ⶠðððð â ðð(ð)ðð
is a linear map, so (as explained in §1.7) we get from a pullback map
(2.6.1) ððâð â (ððð)â ⶠð¬ð(ðâð(ð)ðð) â ð¬ð(ðâð ðð) .In particular, let ð be a ð-form on ð. Then at ð(ð) â ð, ð takes the value
ðð(ð) â ð¬ð(ðâð ðð) ,so we can apply to it the operation (2.6.1), and this gives us an element:
(2.6.2) ððâð ðð(ð) â ð¬ð(ðâð ðð) .In fact we can do this for every point ð â ð, and the assignment
(2.6.3) ð ⊠(ððð)âðð(ð)defines a ð-form on ð, which we denote by ðâð. We call ðâð the pullback of ð along themap ð. A few of its basic properties are described below.
Properties 2.6.4. Let ð be an open subset of ðð, ð an open subset of ðð and ðⶠð â ð að¶â map.(1) Let ð be a zero-form, i.e., a function ð â ð¶â(ð). Since
ð¬0(ðâð ) = ð¬0(ðâð(ð)) = ðthe map (2.6.1) is just the identity map of ð onto ð when ð = 0. Hence for zero-forms
(2.6.5) (ðâð)(ð) = ð(ð(ð)) ,that is, ðâð is just the composite function, ð â ð â ð¶â(ð).
(2) Let ð â ðº0(ð) and let ð â ðº1(ð) be the 1-form ð = ðð. By the chain rule (2.6.2)unwinds to:
(2.6.6) (ððð)âððð = (ðð)ð â ððð = ð(ð â ð)ðand hence by (2.6.5) we have
(2.6.7) ðâ ðð = ððâð .(3) If ð1, ð2 â ðºð(ð) from equation (2.6.2) we see that
(ððð)â(ð1 + ð2)ð = (ððð)â(ð1)ð + (ððð)â(ð2)ð ,and hence by eq:2.5.3 we have
ðâ(ð1 + ð2) = ðâð1 + ðâð2 .
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§2.6 The pullback operation on forms 55
(4) We observed in §1.7 that the operation (2.6.1) commutes with wedge-product, hence ifð1 â ðºð(ð) and ð2 â ðºâ(ð)
ððâð (ð1)ð ⧠(ð2)ð = ððâð (ð1)ð ⧠ððâð (ð2)ð .In other words
(2.6.8) ðâð1 ⧠ð2 = ðâð1 ⧠ðâð2 .(5) Letð be an open subset of ðð and ðⶠð â ð a ð¶â map. Given a point ð â ð, letð = ð(ð) and ð€ = ð(ð). Then the composition of the map
(ððð)â ⶠð¬ð(ðâð ) â ð¬ð(ðâð )and the map
(ððð)â ⶠð¬ð(ðâð€) â ð¬ð(ðâð )is the map
(ððð â ððð)â ⶠð¬ð(ðâð€) â ð¬ð(ðâð )by formula (1.7.5). However, by the chain rule
(ððð) â (ðð)ð = ð(ð â ð)ðso this composition is the map
ð(ð â ð)âð ⶠð¬ð(ðâð€) â ð¬ð(ðâð ) .
Thus if ð â ðºð(ð)ðâ(ðâð) = (ð â ð)âð .
Let us see what the pullback operation looks like in coordinates. Using multi-indexnotation we can express every ð-form ð â ðºð(ð) as a sum over multi-indices of length ð(2.6.9) ð = â
ðŒððŒðð¥ðŒ ,
the coefficient ððŒ of ðð¥ðŒ being in ð¶â(ð). Hence by (2.6.5)
ðâð = âðâððŒðâ(ðð¥ðŒ)where ðâððŒ is the function of ð â ð.
What about ðâðð¥ðŒ? If ðŒ is the multi-index, (ð1,âŠ, ðð), then by definitionðð¥ðŒ = ðð¥ð1 â§â¯ ⧠ðð¥ðð
soðâ(ðð¥ðŒ) = ðâ(ðð¥ð1) ⧠⯠⧠ð
â(ðð¥ðð)by (2.6.8), and by (2.6.7), we have
ðâðð¥ð = ððâð¥ð = ðððwhere ðð is the ðth coordinate function of the map ð. Thus, setting
ðððŒ â ððð1 â§â¯ ⧠ðððð ,we see that for each multi-index ðŒ we have(2.6.10) ðâ(ðð¥ðŒ) = ðððŒ ,and for the pullback of the form (2.6.9)
(2.6.11) ðâð = âðŒðâððŒ ðððŒ .
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56 Chapter 2: Differential Forms
We will use this formula to prove that pullback commutes with exterior differentiation:(2.6.12) ð(ðâð) = ðâ(ðð) .To prove this we recall that by (2.3.6) we have ð(ðððŒ) = 0, hence by (2.3.1) and (2.6.10) wehave
ð(ðâð) = âðŒð(ðâððŒ) ⧠ðððŒ = â
ðŒðâ(ðððŒ) ⧠ðâ(ðð¥ðŒ)
= ðââðŒðððŒ ⧠ðð¥ðŒ = ðâ(ðð) .
A special case of formula (2.6.10) will be needed in Chapter 4: Let ð and ð be opensubsets of ðð and let ð = ðð¥1 â§â¯ ⧠ðð¥ð. Then by (2.6.10) we have
ðâðð = (ðð1)ð â§â¯ ⧠(ððð)ðfor all ð â ð. However,
(ððð)ð = âððððð¥ð(ð)(ðð¥ð)ð
and hence by formula (1.7.10)
ðâðð = det[ððððð¥ð(ð)] (ðð¥1 â§â¯ ⧠ðð¥ð)ð .
In other words
(2.6.13) ðâ(ðð¥1 â§â¯ ⧠ðð¥ð) = det[ððððð¥ð]ðð¥1 â§â¯ ⧠ðð¥ð .
In Exercise 2.6.iv and equation (2.6.6) we outline the proof of an important topologicalproperty of the pullback operation.
Definition 2.6.14. Let ð be an open subset of ðð, ð an open subset of ðð, ðŽ â ð an openinterval containing 0 and 1 and ð0, ð1 ⶠð â ð two ð¶â maps. A ð¶â map ð¹â¶ ðÃðŽ â ð isa ð¶â homotopy between ð0 and ð1 if ð¹(ð¥, 0) = ð0(ð¥) and ð¹(ð¥, 1) = ð1(ð¥).
If there exists a homotopy between ð0 and ð1, we say that ð0 and ð1 are homotopic andwrite ð0 â ð1.
Intuitively, ð0 and ð1 are homotopic if there exists a family of ð¶â maps ðð¡ ⶠð â ð,where ðð¡(ð¥) = ð¹(ð¥, ð¡), which âsmoothly deform ð0 into ð1â. In Exercise 2.6.iv and equa-tion (2.6.6) you will be asked to verify that for ð0 and ð1 to be homotopic they have tosatisfy the following criteria.
Theorem 2.6.15. Letð be an open subset of ðð,ð an open subset of ðð, and ð0, ð1 ⶠð â ðtwo ð¶â maps. If ð0 and ð1 are homotopic then for every closed form ð â ðºð(ð) the formðâ1 ð â ðâ0 ð is exact.
This theorem is closely related to the Poincaré lemma, and, in fact, one gets from ita slightly stronger version of the Poincaré lemma than that described in Exercise 2.3.ivand equation (2.3.7).
Definition 2.6.16. An open subset ð of ðð is contractible if, for some point ð0 â ð, theidentity map idð ⶠð â ð is homotopic to the constant map
ð0 ⶠð â ð , ð0(ð) = ð0at ð0.
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§2.6 The pullback operation on forms 57
From Theorem 2.6.15 it is easy to see that the Poincaré lemma holds for contractableopen subsets ofðð. Ifð is contractable every closed ð-form onð of degree ð > 0 is exact. Tosee this, note that if ð â ðºð(ð), then for the identity map idâð ð = ð and if ð is the constantmap at a point of ð, then ðâð = 0.
Exercises for §2.6Exercise 2.6.i. Let ðⶠð3 â ð3 be the map
ð(ð¥1, ð¥2, ð¥3) = (ð¥1ð¥2, ð¥2ð¥23 , ð¥33) .Compute the pullback, ðâð for the following forms.(1) ð = ð¥2ðð¥3(2) ð = ð¥1ðð¥1 ⧠ðð¥3(3) ð = ð¥1ðð¥1 ⧠ðð¥2 ⧠ðð¥3Exercise 2.6.ii. Let ðⶠð2 â ð3 be the map
ð(ð¥1, ð¥2) = (ð¥21 , ð¥22 , ð¥1ð¥2) .Complete the pullback, ðâð, for the following forms.(1) ð = ð¥2ðð¥2 + ð¥3ðð¥3(2) ð = ð¥1ðð¥2 ⧠ðð¥3(3) ð = ðð¥1 ⧠ðð¥2 ⧠ðð¥3Exercise 2.6.iii. Let ð be an open subset of ðð, ð an open subset of ðð, ðⶠð â ð a ð¶âmap and ðŸâ¶ [ð, ð] â ð a ð¶â curve. Show that for ð â ðº1(ð)
â«ðŸðâð = â«
ðŸ1ð
where ðŸ1 ⶠ[ð, ð] â ð is the curve, ðŸ1(ð¡) = ð(ðŸ(ð¡)). (See Exercise 2.2.viii.)
Exercise 2.6.iv. Let ð be an open subset of ðð, ðŽ â ð an open interval containing thepoints, 0 and 1, and (ð¥, ð¡) product coordinates on ð à ðŽ. Recall from Exercise 2.3.v that aform ð â ðºâ(ð à ðŽ) is reduced if it can be written as a sum
ð = âðŒððŒ(ð¥, ð¡)ðð¥ðŒ
(i.e., none of the summands involve ðð¡). For a reduced form ð, let ðð â ðºâ(ð) be the form
(2.6.17) ðð â (âðŒâ«1
0ððŒ(ð¥, ð¡)ðð¡)ðð¥ðŒ
and let ð1, ð2 â ðºâ(ð), be the forms
ð0 = âðŒððŒ(ð¥, 0)ðð¥ðŒ
andð1 = âððŒ(ð¥, 1)ðð¥ðŒ .
Now recall from Exercise 2.4.v that every form ð â ðºð(ð à ðŽ) can be written uniquely as asum(2.6.18) ð = ðð¡ ⧠ðŒ + ðœwhere ðŒ and ðœ are reduced.(1) Prove:
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58 Chapter 2: Differential Forms
Theorem 2.6.19. If the form (2.6.18) is closed then
(2.6.20) ðœ1 â ðœ0 = ðððŒ .
Hint: Equation (2.4.19).Let ð0 and ð1 be themaps ofð intoðÃðŽ defined by ð0(ð¥) = (ð¥, 0)and ð1(ð¥) = (ð¥, 1). Show that (2.6.20) can be rewritten
(2.6.21) ðâ1ð â ðâ0ð = ðððŒ .
Exercise 2.6.v. Letð be an open subset of ðð and ð0, ð1 ⶠð â ð be ð¶â maps. Suppose ð0and ð1 are homotopic. Show that for every closed form, ð â ðºð(ð), ðâ1 ð â ðâ0 ð is exact.
Hint: Let ð¹â¶ ðÃðŽ â ð be a homotopy between ð0 and ð1 and let ð = ð¹âð. Show thatð is closed and that ðâ0 ð = ðâ0ð and ðâ1 ð = ðâ1ð. Conclude from equation (2.6.21)that
ðâ1 ð â ðâ0 ð = ðððŒ
where ð = ðð¡ ⧠ðŒ + ðœ and ðŒ and ðœ are reduced.
Exercise 2.6.vi. Show that if ð â ðð is a contractable open set, then the Poincaré lemmaholds: every closed form of degree ð > 0 is exact.
Exercise 2.6.vii. An open subset, ð, of ðð is said to be star-shaped if there exists a pointð0 â ð, with the property that for every point ð â ð, the line segment,
{ ð¡ð + (1 â ð¡)ð0 | 0 †ð¡ †1 } ,
joining ð to ð0 is contained in ð. Show that if ð is star-shaped it is contractable.
Exercise 2.6.viii. Show that the following open sets are star-shaped.(1) The open unit ball
{ ð¥ â ðð | |ð¥| < 1 } .(2) The open rectangle, ðŒ1 Ã⯠à ðŒð, where each ðŒð is an open subinterval of ð.(3) ðð itself.(4) Product sets
ð1 Ã ð2 â ðð â ðð1 Ã ðð2
where ðð is a star-shaped open set in ððð .
Exercise 2.6.ix. Let ð be an open subset of ðð, ðð¡ ⶠð â ð, ð¡ â ð, a one-parameter groupof diffeomorphisms and ð£ its infinitesimal generator. Given ð â ðºð(ð) show that at ð¡ = 0
(2.6.22) ððð¡ðâð¡ ð = ð¿ðð .
Here is a sketch of a proof:(1) Let ðŸ(ð¡) be the curve, ðŸ(ð¡) = ðð¡(ð), and let ð be a zero-form, i.e., an element of ð¶â(ð).
Show thatðâð¡ ð(ð) = ð(ðŸ(ð¡))
and by differentiating this identity at ð¡ = 0 conclude that (2.6.22) holds for zero-forms.(2) Show that if (2.6.22) holds for ð it holds for ðð.
Hint: Differentiate the identity
ðâð¡ ðð = ððâð¡ ð
at ð¡ = 0.
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§2.7 Divergence, curl, and gradient 59
(3) Show that if (2.6.22) holds for ð1 and ð2 it holds for ð1 ⧠ð2.Hint: Differentiate the identity
ðâð¡ (ð1 ⧠ð2) = ðâð¡ ð1 ⧠ðâð¡ ð2at ð¡ = 0.
(4) Deduce (2.6.22) from (1)â(3).Hint: Every ð-form is a sum of wedge products of zero-forms and exact one-forms.
Exercise 2.6.x. In Exercise 2.6.ix show that for all ð¡
(2.6.23) ððð¡ðâð¡ ð = ðâð¡ ð¿ðð = ð¿ððâð¡ ð .
Hint: By the definition of a one-parameter group, ðð +ð¡ = ðð â ðð¡ = ðð â ðð , hence:ðâð +ð¡ð = ðâð¡ (ðâð ð) = ðâð (ðâð¡ ð) .
Prove the first assertion by differentiating the first of these identities with respect to ð andthen setting ð = 0, and prove the second assertion by doing the same for the second of theseidentities.
In particular conclude thatðâð¡ ð¿ðð = ð¿ððâð¡ ð .
Exercise 2.6.xi.(1) By massaging the result above show that
ððð¡ðâð¡ ð = ððð¡ð + ðð¡ðð
whereðð¡ð = ðâð¡ ððð .
Hint: Equation (2.5.9).(2) Let
ðð = â«1
0ðâð¡ ððððð¡ .
Prove the homotopy identity
ðâ1 ð â ðâ0 ð = ððð + ððð .
Exercise 2.6.xii. Letð be an open subset of ðð, ð an open subset of ðð, ð a vector field onð, ð a vector field on ð and ðⶠð â ð a ð¶â map. Show that if ð and ð are ð-related
ðððâð = ðâððð .
2.7. Divergence, curl, and gradient
The basic operations in 3-dimensional vector calculus: gradient, curl and divergenceare, by definition, operations on vector fields. As we see below these operations are closelyrelated to the operations
(2.7.1) ðⶠðºð(ð3) â ðºð+1(ð3)in degrees ð = 0, 1, 2. However, only the gradient and divergence generalize to ð dimensions.(They are essentially the ð-operations in degrees zero and ð â 1.) And, unfortunately, thereis no simple description in terms of vector fields for the other ð-operations. This is one ofthe main reasons why an adequate theory of vector calculus in ð-dimensions forces on onethe differential form approach that we have developed in this chapter.
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60 Chapter 2: Differential Forms
Even in three dimensions, however, there is a good reason for replacing the gradient,divergence, and curl by the three operations (2.7.1). A problem that physicists spend a lot oftime worrying about is the problem of general covariance: formulating the laws of physics insuch a way that they admit as large a set of symmetries as possible, and frequently these for-mulations involve differential forms. An example is Maxwellâs equations, the fundamentallaws of electromagnetism. These are usually expressed as identities involving div and curl.However, as we explain below, there is an alternative formulation of Maxwellâs equationsbased on the operations (2.7.1), and from the point of view of general covariance, this for-mulation is much more satisfactory: the only symmetries of ð3 which preserve div and curlare translations and otations, whereas the operations (2.7.1) admit all diffeomorphisms ofð3 as symmetries.
To describe how the gradient, divergence, and curl are related to the operations (2.7.1)we first note that there are two ways of converting vector fields into forms.
The first makes use of the natural inner product ðµ(ð£, ð€) = âð£ðð€ð on ðð. From thisinner product one gets by Exercise 1.2.ix a bijective linear map
ð¿â¶ ðð ⥲ (ðð)â
with the defining property: ð¿(ð£) = â if and only if â(ð€) = ðµ(ð£, ð€). Via the identification(2.1.2) ðµ and ð¿ can be transferred to ðððð, giving one an inner product ðµð on ðððð and abijective linear map
ð¿ð ⶠðððð ⥲ ðâð ðð .Hence if we are given a vector field ð on ð we can convert it into a 1-form ð⯠by setting(2.7.2) ðâ¯(ð) â ð¿ðð(ð)and this sets up a bijective correspondence between vector fields and 1-forms.
For instanceð = ððð¥ð⺠ð⯠= ðð¥ð ,
(see Exercise 2.7.iii) and, more generally,
(2.7.3) ð =ðâð=1ððððð¥ð⺠ð⯠=
ðâð=1ðððð¥ð .
Example 2.7.4. In particular ifð is að¶â function onð â ðð the gradient ofð is the vectorfield
grad(ð) âðâð=1
ðððð¥ð
,
and this gets converted by (2.7.5) into the 1-form ðð. Thus the gradient operation in vectorcalculus is basically just the exeterior derivative operation ðⶠðº0(ð) â ðº1(ð).
The second way of converting vector fields into forms is via the interior product oper-ation. Namely letðº be the ð-form ðð¥1 â§â¯â§ ðð¥ð. Given an open subset ð of ðð and a ð¶âvector field
(2.7.5) ð =ðâð=1ððððð¥ð
on ð the interior product of ð£ with ðº is the (ð â 1)-form
(2.7.6) ðððº =ðâð=1(â1)ðâ1ðððð¥1 â§â¯ ⧠ðð¥ðâ¯â§ ðð¥ð .
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§2.7 Divergence, curl, and gradient 61
Moreover, every (ð â 1)-form can be written uniquely as such a sum, so (2.7.5) and (2.7.6)set up a bijective correspondence between vector fields and (ð â 1)-forms. Under this corre-spondence the ð-operation gets converted into an operation on vector fields
ð ⊠ððððº .Moreover, by (2.5.9)
ððððº = ð¿ððºand by (2.5.14)
ð¿ððº = div(ð)ðºwhere
(2.7.7) div(ð) =ðâð=1
ððððð¥ð
.
In other words, this correspondence between (ð â 1)-forms and vector fields converts theð-operation into the divergence operation (2.7.7) on vector fields.
Notice that divergence and gradient are well-defined as vector calculus operations in ðdimensions, even though one usually thinks of them as operations in 3-dimensional vectorcalculus. The curl operation, however, is intrinsically a 3-dimensional vector calculus oper-ation. To define it we note that by (2.7.6) every 2-form ð on an open subset ð â ð3 can bewritten uniquely as an interior product,(2.7.8) ð = ðððð¥1 ⧠ðð¥2 ⧠ðð¥3 ,for some vector field ð, and the left-hand side of this formula determines ð uniquely.
Definition 2.7.9. Letð be an open subset of ð3 and ð a vector field onð. From ð we get by(2.7.3) a 1-form ðâ¯, and hence by (2.7.8) a vector field ð satisfying
ðð⯠= ðððð¥1 ⧠ðð¥2 ⧠ðð¥3 .The curl of ð is the vector field
curl(ð) â ð .
We leave for you to check that this definition coincides with the definition one finds incalculus books. More explicitly we leave for you to check that if ð is the vector field
ð = ð1ððð¥1+ ð2ððð¥2+ ð3ððð¥3
thencurl(ð) = ð1
ððð¥1+ ð2ððð¥2+ ð3ððð¥3
where
ð1 =ðð2ðð¥3â ðð3ðð¥2
ð2 =ðð3ðð¥1â ðð1ðð¥3
(2.7.10)
ð3 =ðð1ðð¥2â ðð2ðð¥1
.
To summarize: the gradient, curl, and divergence operations in 3-dimensions are basi-cally just the three operations (2.7.1). The gradient operation is the operation (2.7.1) in de-gree zero, curl is the operation (2.7.1) in degree one, and divergence is the operation (2.7.1)
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62 Chapter 2: Differential Forms
in degree two. However, to define gradient we had to assign an inner product ðµð to the tan-gent space ðððð for each ð â ð; to define divergence we had to equip ð with the 3-formðº and to define curl, the most complicated of these three operations, we needed the innerproducts ðµð and the form ðº. This is why diffeomorphisms preserve the three operations(2.7.1) but do not preserve gradient, curl, and divergence. The additional structures whichone needs to define grad, curl and div are only preserved by translations and rotations.
Maxwellâs equations and differential formsWe conclude this section by showing how Maxwellâs equations, which are usually for-
mulated in terms of divergence and curl, can be reset into âformâ language. The discussionbelow is an abbreviated version of [5, §1.20].
Maxwellâs equations assert:div(ððž) = ð(2.7.11)
curl(ððž) = âððð¡ðð(2.7.12)
div(ðð) = 0(2.7.13)
ð2 curl(ðð) = ð +ððð¡ððž(2.7.14)
where ððž and ðð are the electric andmagnetic fields, ð is the scalar charge density,ð is thecurrent density and ð is the velocity of light. (To simplify (2.7.15) slightly we assume thatour units of spaceâtime are chosen so that ð = 1.) As above letðº = ðð¥1 ⧠ðð¥2 ⧠ðð¥3 and let
ððž = ð(ððž)ðºand
ðð = ð(ðð)ðº .We can then rewrite equations (2.7.11) and (2.7.13) in the form(2.7.11â²) ðððž = ððºand(2.7.13â²) ððð = 0 .
What about (2.7.12) and (2.7.14)?We leave the following âformâ versions of these equa-tions as an exercise:
(2.7.12â²) ððâ¯ðž = âððð¡ðð
and
(2.7.14â²) ððâ¯ð = ðððº +ððð¡ððž
where the 1-forms, ðâ¯ðž and ðâ¯ð, are obtained from ððž and ðð by the operation (2.7.2).
These equations can be written more compactly as differential form identities in 3 + 1dimensions. Let ðð and ððž be the 2-forms
ðð = ðð â ðâ¯ðž ⧠ðð¡
and(2.7.15) ððž = ððž â ð
â¯ð ⧠ðð¡
and let ð¬ be the 3-formð¬ â ððº + ðððº ⧠ðð¡ .
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§2.8 Symplectic geometry & classical mechanics 63
We will leave for you to show that the four equations (2.7.11)â(2.7.14) are equivalentto two elegant and compact (3 + 1)-dimensional identities
ððð = 0and
ðððž = ð¬ .
Exercises for §2.7Exercise 2.7.i. Verify that the curl operation is given in coordinates by equation (2.7.10).
Exercise 2.7.ii. Verify thatMaxwellâs equations (2.7.11) and (2.7.12) become the equations (2.7.13)and (2.7.14) when rewritten in differential form notation.
Exercise 2.7.iii. Show that in (3+1)-dimensionsMaxwellâs equations take the form of equa-tions (2.7.10) and (2.7.11).
Exercise 2.7.iv. Let ð be an open subset of ð3 and ð a vector field on ð. Show that if ð isthe gradient of a function, its curl has to be zero.
Exercise 2.7.v. If ð is simply connected prove the converse: If the curl of ð vanishes, ð isthe gradient of a function.
Exercise 2.7.vi. Let ð = curl(ð). Show that the divergence of ð is zero.
Exercise 2.7.vii. Is the converse to Exercise 2.7.vi true? Suppose the divergence ofð is zero.Is ð = curl(ð) for some vector field ð?
2.8. Symplectic geometry & classical mechanics
In this section we describe some other applications of the theory of differential formsto physics. Before describing these applications, however, we say a few words about the geo-metric ideas that are involved.
Definition 2.8.1. Let ð¥1,âŠ, ð¥2ð be the standard coordinate functions on ð2ð and for ð =1,âŠ, ð let ðŠð â ð¥ð+ð. The two-form
ð âðâð=1ðð¥ð ⧠ððŠð =
ðâð=1ðð¥ð ⧠ðð¥ð+ð
is called the Darboux form on ð2ð.From the identity
(2.8.2) ð = âð (âðð=1 ðŠððð¥ð) .it follows that ð is exact. Moreover computing the ð-fold wedge product of ð with itself weget
ðð = (ðâðð=1ðð¥ð1 ⧠ððŠð1) â§â¯ â§(
ðâðð=1ðð¥ðð ⧠ððŠðð)
= âð1,âŠ,ðððð¥ð1 ⧠ððŠð1 â§â¯ ⧠ðð¥ðð ⧠ððŠðð .
We can simplify this sum by noting that if the multi-index ðŒ = (ð1,âŠ, ðð) is repeating, thewedge product(2.8.3) ðð¥ð1 ⧠ððŠð1 â§â¯ ⧠ðð¥ðð ⧠ðð¥ðð
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64 Chapter 2: Differential Forms
involves two repeating ðð¥ðð and hence is zero, and if ðŒ is non-repeating we can permute thefactors and rewrite (2.8.3) in the form
ðð¥1 ⧠ððŠ1 â§â¯ ⧠ðð¥ð ⧠ððŠð .(See Exercise 1.6.v.)
Hence since these are exactly ð! non-repeating multi-indices
ðð = ð!ðð¥1 ⧠ððŠ1 â§â¯ ⧠ðð¥ð ⧠ððŠð ,i.e.,
(2.8.4) 1ð!ðð = ðº
where
(2.8.5) ðº = ðð¥1 ⧠ððŠ1 â§â¯ ⧠ðð¥ð ⧠ððŠðis the symplectic volume form on ð2ð.
Definition 2.8.6. Let ð and ð be open subsets of ð2ð. A diffeomorphism ðⶠð ⥲ ð is asymplectic diffeomorphism or symplectomorphism if ðâð = ð, where ð denotes the Dar-boux form on ð2ð.
Definition 2.8.7. Let ð be an open subset of ð2ð, let(2.8.8) ðð¡ ⶠð ⥲ ð , ââ < ð¡ < âbe a one-parameter group of diffeomorphisms of ð, and ð be the vector field generating(2.8.8) We say that ð is a symplectic vector field if the diffeomorphisms (2.8.8) are symplec-tomorphisms, i.e., for all ð¡ we have ðâð¡ ð = ð.
Let us see what such vector fields have to look like. Note that by (2.6.23)
(2.8.9) ððð¡ðâð¡ ð = ðâð¡ ð¿ðð ,
hence if ðâð¡ ð = ð for all ð¡, the left hand side of (2.8.9) is zero, so
ðâð¡ ð¿ðð = 0 .In particular, for ð¡ = 0, ðð¡ is the identity map so ðâð¡ ð¿ðð = ð¿ðð = 0. Conversely, if ð¿ðð = 0,then ðâð¡ ð¿ðð = 0 so by (2.8.9) ðâð¡ ð does not depend on ð¡. However, since ðâð¡ ð = ð for ð¡ = 0we conclude that ðâð¡ ð = ð for all ð¡. To summarize, we have proved:
Theorem 2.8.10. Let ðð¡ ⶠð â ð be a one-parameter group of diffeomorphisms and ð theinfinitesmal generator of this group. Then ð is symplectic of and only if ð¿ðð = 0.
There is an equivalent formulation of Theorem 2.8.10 in terms of the interior productððð. By (2.5.9)
ð¿ðð = ðððð + ðð ðð .But by (2.8.2) we have ðð = 0, so
ð¿ðð = ðððð .Thus we have shown:
Theorem 2.8.11. A vector field ð on an open subset of ð2ð is symplectic if and only if the formððð is closed.
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§2.8 Symplectic geometry & classical mechanics 65
Letð be an open subset of ð2ð and ð a vector field onð. If ððð is not only closed but isexact we say that ð is a Hamiltonian vector field. In other words, ð is Hamiltonian if(2.8.12) ððð = ðð»for some ð¶â functionð» â ð¶â(ð).
Let is see what this condition looks like in coordinates. Let
(2.8.13) ð =ðâð=1(ððððð¥ð+ ðððððŠð) .
Thenððð = â
1â€ð,ðâ€ððð ðð/ðð¥ð(ðð¥ð ⧠ððŠð) + â
1â€ð,ðâ€ððð ðð/ððŠð(ðð¥ð ⧠ððŠð) .
But
ðð/ðð¥ððð¥ð = {1, ð = ð0, ð â ð
andðð/ðð¥ðððŠð = 0
so the first summand above is âðð=1 ðð ððŠð, and a similar argument shows that the secondsummand is ââðð=1 ðððð¥ð.
Hence if ð is the vector field (2.8.13), then
(2.8.14) ððð =ðâð=1(ðð ððŠð â ðððð¥ð) .
Thus since
ðð» =ðâð=1(ðð»ðð¥ððð¥ð +ðð»ððŠðððŠð)
we get from (2.8.12)â(2.8.14)
ðð =ðð»ððŠð
and ðð = âðð»ðð¥ð
so ð has the form:
(2.8.15) ð =ðâð=1(ðð»ððŠðððð¥ðâ ðð»ðð¥ððððŠð) .
In particular if ðŸ(ð¡) = (ð¥(ð¡), ðŠ(ð¡)) is an integral curve of ð it has to satisfy the system ofdifferential equations
(2.8.16){{{{{{{
ðð¥ððð¡= ðð»ððŠð(ð¥(ð¡), ðŠ(ð¡))
ððŠððð¡= âðð»ðð¥ð(ð¥(ð¡), ðŠ(ð¡)) .
The formulas (2.8.13) and (2.8.14) exhibit an important property of the Darboux form ð.Every one-form on ð can be written uniquely as a sum
ðâð=1(ðð ððŠð â ðððð¥ð)
with ðð and ðð in ð¶â(ð) and hence (2.8.13) and (2.8.14) imply:
Theorem 2.8.17. The map ð ⊠ððð defines a bijective between vector field and one-forms.
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66 Chapter 2: Differential Forms
In particular, for every ð¶â functionð», we get by correspondence a unique vector fieldð = ðð» with the property (2.8.12). We next note that by (1.7.9)
ð¿ðð» = ðð ðð» = ðð(ððð) = 0 .Thus(2.8.18) ð¿ðð» = 0i.e.,ð» is an integral ofmotion of the vector field ð. In particular, if the functionð»â¶ ð â ð isproper, then by Theorem 2.2.12 the vector field, ð, is complete and hence by Theorem 2.8.10generates a one-parameter group of symplectomorphisms.
Remark 2.8.19. If the one-parameter group (2.8.8) is a group of symplectomorphisms, thenðâð¡ ðð = ðâð¡ ð â§â¯ ⧠ðâð¡ ð = ðð, so by (2.8.4)(2.8.20) ðâð¡ ðº = ðº ,where ðº is the symplectic volume form (2.8.5).
Application 2.8.21. The application we want to make of these ideas concerns the descrip-tion, inNewtonianmechanics, of a physical system consisting ofð interacting point-masses.The configuration space of such a system is
ðð = ð3 Ã⯠à ð3 ,where there are ð copies of ð3 in the product, with position coordinates ð¥1,âŠ, ð¥ð andthe phase space is ð2ð with position coordinates ð¥1,âŠ, ð¥ð and momentum coordinates,ðŠ1,âŠ, ðŠð. The kinetic energy of this system is a quadratic function of the momentum co-ordinates
12
ðâð=1
1ðððŠ2ð ,
and for simplicity we assume that the potential energy is a function ð(ð¥1,âŠ, ð¥ð) of theposition coordinates alone, i.e., it does not depend on themomenta and is time-independentas well. Let
(2.8.22) ð» = 12
ðâð=1
1ðððŠ2ð + ð(ð¥1,âŠ, ð¥ð)
be the total energy of the system.
We now show that Newtonâs second law of motion in classical mechanics reduces to theassertion.
Proposition 2.8.23. The trajectories in phase space of the system above are just the integralcurves of the Hamiltonian vector field ðð».Proof. For the function (2.8.22) the equations (2.8.16) become
{{{{{{{{{
ðð¥ððð¡= 1ðððŠð
ððŠððð¡= â ðððð¥ð
.
The first set of equation are essentially just the definitions of momentum, however, if weplug them into the second set of equations we get
(2.8.24) ððð2ð¥ððð¡2= âðððð¥ð
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§2.8 Symplectic geometry & classical mechanics 67
and interpreting the term on the right as the force exerted on the ðth point-mass and theterm on the left as mass times acceleration this equation becomes Newtonâs second law. â¡
In classical mechanics the equations (2.8.16) are known as the HamiltonâJacobi equa-tions. For a more detailed account of their role in classical mechanics we highly recommend[1]. Historically these equations came up for the first time, not in Newtonian mechanics,but in geometric optics and a brief description of their origins there and of their relation toMaxwellâs equations can be found in [5].
We conclude this chapter bymentioning a few implications of the Hamiltonian descrip-tion (2.8.16) of Newtonâs equations (2.8.24).(1) Conservation of energy: By (2.8.18) the energy function (2.8.22) is constant along the
integral curves of ð, hence the energy of the system (2.8.16) does notchange in time.(2) Noetherâs principle: Let ðŸð¡ ⶠð2ð â ð2ð be a one-parameter group of diffeomorphisms of
phase space and ð its infinitesimal generator. Then (ðŸð¡)ð¡âð is a symmetry of the systemabove if each ðŸð¡ preserves the function (2.8.22) and the vector field ð is Hamiltonian.The Hamiltonian condition means that
ððð = ððºfor some ð¶â function ðº, and what Noetherâs principle asserts is that this function is anintegral of motion of the system (2.8.16), i.e., satisfies ð¿ððº = 0. In other words statedmore succinctly: symmetries of the system (2.8.16) give rise to integrals of motion.
(3) Poincaré recurrence: An important theorem of Poincaré asserts that if the function
ð»â¶ ð2ð â ðdefined by (2.8.22) is proper then every trajectory of the system (2.8.16) returns arbi-trarily close to its initial position at some positive time ð¡0, and, in fact, does this not justonce but does so infinitely often. We sketch a proof of this theorem, using (2.8.20), inthe next chapter.
Exercises for §2.8Exercise 2.8.i. Let ðð» be the vector field (2.8.15). Prove that div(ðð») = 0.
Exercise 2.8.ii. Let ð be an open subset of ðð, ðð¡ ⶠð â ð a one-parameter group ofdiffeomorphisms of ð and ð the infinitesmal generator of this group. Show that if ðŒ is a ð-form on ð then ðâð¡ ðŒ = ðŒ for all ð¡ if and only if ð¿ððŒ = 0 (i.e., generalize to arbitrary ð-formsthe result we proved above for the Darboux form).
Exercise 2.8.iii (the harmonic oscillator). Letð» be the functionâðð=1ðð(ð¥2ð +ðŠ2ð )where theððâs are positive constants.(1) Compute the integral curves of ðð».(2) Poincaré recurrence: Show that if (ð¥(ð¡), ðŠ(ð¡)) is an integral curvewith initial point (ð¥0, ðŠ0) â(ð¥(0), ðŠ(0)) and ð an arbitrarily small neighborhood of (ð¥0, ðŠ0), then for every ð > 0there exists a ð¡ > ð such that (ð¥(ð¡), ðŠ(ð¡)) â ð.
Exercise 2.8.iv. Let ð be an open subset of ð2ð and letð»1, ð»2 â ð¶â(ð). Show that
[ðð»1 , ðð»2] = ðð»where
(2.8.25) ð» =ðâð=1
ðð»1ðð¥ððð»2ððŠðâ ðð»2ðð¥ððð»1ððŠð
.
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68 Chapter 2: Differential Forms
Exercise 2.8.v. The expression (2.8.25) is known as the Poisson bracket ofð»1 andð»2, de-noted by {ð»1, ð»2}. Show that it is anti-symmetric
{ð»1, ð»2} = â{ð»2, ð»1}and satisfies the Jacobi identity
0 = {ð»1, {ð»2, ð»3}} + {ð»2, {ð»3, ð»1}} + {ð»3, {ð»1, ð»2}} .Exercise 2.8.vi. Show that
{ð»1, ð»2} = ð¿ðð»1ð»2 = âð¿ðð»2ð»1 .
Exercise 2.8.vii. Prove that the following three properties are equivalent.(1) {ð»1, ð»2} = 0.(2) ð»1 is an integral of motion of ð2.(3) ð»2 is an integral of motion of ð1.Exercise 2.8.viii. Verify Noetherâs principle.
Exercise 2.8.ix (conservation of linearmomentum). Suppose the potentialð in equation (2.8.22)is invariant under the one-parameter group of translations
ðð¡(ð¥1,âŠ, ð¥ð) = (ð¥1 + ð¡,âŠ, ð¥ð + ð¡) .(1) Show that the function (2.8.22) is invariant under the group of diffeomorphisms
ðŸð¡(ð¥, ðŠ) = (ðð¡ð¥, ðŠ) .(2) Show that the infinitesmal generator of this group is the Hamiltonian vector field ððº
where ðº = âðð=1 ðŠð.(3) Conclude from Noetherâs principle that this function is an integral of the vector fieldðð», i.e., that âtotal linear momentâ is conserved.
(4) Show that âtotal linear momentumâ is conserved if ð is the Coulomb potential
âðâ ð
ðð|ð¥ð â ð¥ð|
.
Exercise 2.8.x. Let ð ðð¡ ⶠð2ð â ð2ð be the rotation which fixes the variables, (ð¥ð, ðŠð), ð â ðand rotates (ð¥ð, ðŠð) by the angle, ð¡:
ð ðð¡(ð¥ð, ðŠð) = (cos ð¡ ð¥ð + sin ð¡ ðŠð ,â sin ð¡ ð¥ð + cos ð¡ ðŠð) .(1) Show that ð ðð¡, ââ < ð¡ < â, is a one-parameter group of symplectomorphisms.(2) Show that its generator is the Hamiltonian vector field, ðð»ð , whereð»ð = (ð¥2ð + ðŠ2ð )/2.(3) Letð» be the harmonic oscillator Hamiltonian from Exercise 2.8.iii. Show that the ð ðð¡ âs
preserveð».(4) What does Noetherâs principle tell one about the classical mechanical system with en-
ergy functionð»?Exercise 2.8.xi. Show that ifð is an open subset ofð2ð and ð is a symplectic vector field onðthen for every point ð0 â ð there exists a neighborhoodð0 of ð0 onwhich ð is Hamiltonian.
Exercise 2.8.xii. Deduce from Exercises 2.8.iv and 2.8.xi that if ð1 and ð2 are symplecticvector fields on an open subset ð of ð2ð their Lie bracket [ð1, ð2] is a Hamiltonian vectorfield.
Exercise 2.8.xiii. Let ðŒ = âðð=1 ðŠððð¥ð.(1) Show that ð = âððŒ.
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§2.8 Symplectic geometry & classical mechanics 69
(2) Show that if ðŒ1 is any one-form on ð2ð with the property ð = âððŒ1, thenðŒ = ðŒ1 + ðð¹
for some ð¶â function ð¹.(3) Show that ðŒ = ððð where ð is the vector field
ð â âðâð=1ðŠððððŠð
.
Exercise 2.8.xiv. Let ð be an open subset of ð2ð and ð a vector field on ð. Show that ð hasthe property, ð¿ððŒ = 0, if and only if
ððð = ððððŒ .In particular conclude that if ð¿ððŒ = 0 then ð£ is Hamiltonian.
Hint: Equation (2.8.2).
Exercise 2.8.xv. Letð» be the function
(2.8.26) ð»(ð¥, ðŠ) =ðâð=1ðð(ð¥)ðŠð ,
where the ððâs are ð¶â functions on ðð. Show that(2.8.27) ð¿ðð»ðŒ = 0 .
Exercise 2.8.xvi. Conversely show that if ð» is any ð¶â function on ð2ð satisfying equa-tion (2.8.27) it has to be a function of the form (2.8.26).Hints:(1) Let ð be a vector field on ð2ð satisfying ð¿ððŒ = 0. By equation (2.8.16) we have ð = ðð»,
whereð» = ðððŒ.(2) Show thatð» has to satisfy the equation
ðâð=1ðŠððð»ððŠð= ð» .
(3) Conclude that ifð»ð = ðð»ððŠð thenð»ð has to satisfy the equationðâð=1ðŠððððŠðð»ð = 0 .
(4) Conclude thatð»ð has to be constant along the rays (ð¥, ð¡ðŠ), for 0 †ð¡ < â.(5) Conclude finally thatð»ð has to be a function of ð¥ alone, i.e., does not depend on ðŠ.Exercise 2.8.xvii. Show that if ððð is a vector field
ðâð=1ðð(ð¥)ððð¥ð
on configuration space there is a unique lift of ððð to phase space
ð =ðâð=1ðð(ð¥)ððð¥ð+ ðð(ð¥, ðŠ)
ðððŠð
satisfying ð¿ððŒ = 0.
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CHAPTER 3
Integration of Forms
3.1. Introduction
The change of variables formula asserts that if ð and ð are open subsets of ðð andðⶠð â ð að¶1 diffeomorphism then, for every continuous functionðⶠð â ð the integral
â«ðð(ðŠ)ððŠ
exists if and only if the integral
â«ð(ð â ð)(ð¥) |detð·ð(ð¥)|ðð¥
exists, and if these integrals exist they are equal. Proofs of this can be found in [10] or [12].This chapter contains an alternative proof of this result. This proof is due to Peter Lax. Ourversion of his proof in § 3.5 below makes use of the theory of differential forms; but, asLax shows in the article [8] (which we strongly recommend as collateral reading for thiscourse), references to differential forms can be avoided, and the proof described in §3.5 canbe couched entirely in the language of elementary multivariable calculus.
The virtue of Laxâs proof is that is allows one to prove a version of the change of variablestheorem for other mappings besides diffeomorphisms, and involves a topological invariant,the degree of a map, which is itself quite interesting. Some properties of this invariant, andsome topological applications of the change of variables formula will be discussed in §3.6of these notes.
Remark 3.1.1. The proof we are about to describe is somewhat simpler and more transpar-ent if we assume that ð is a ð¶â diffeomorphism. Weâll henceforth make this assumption.
3.2. The Poincaré lemma for compactly supported forms on rectangles
Definition 3.2.1. Let ð be a ð-form on ðð. We define the support of ð bysupp(ð) â { ð¥ â ðð | ðð¥ â 0 } ,
and we say that ð is compactly supported if supp(ð) is compact.We will denote by ðºðð (ðð) the set of all ð¶â ð-forms which are compactly supported,
and ifð is an open subset of ðð, we will denote byðºðð (ð) the set of all compactly supportedð-forms whose support is contained in ð.
Let ð = ððð¥1 â§â¯â§ ðð¥ð be a compactly supported ð-form with ð â ð¶â0 (ðð). We willdefine the integral of ð over ðð:
â«ððð
to be the usual integral of ð over ðð
â«ððððð¥ .
71
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72 Chapter 3: Integration of Forms
(Since ð is ð¶â and compactly supported this integral is well-defined.)Now let ð be the rectangle
[ð1, ð1] à ⯠à [ðð, ðð] .The Poincaré lemma for rectangles asserts:
Theorem3.2.2 (Poincaré lemma for rectangles). Letð be a compactly supported ð-formwithsupp(ð) â int(ð). Then the following assertions are equivalent.(1) â«ð = 0.(2) There exists a compactly supported (ð â 1)-form ð with supp(ð) â int(ð) satisfying ðð =ð.
We will first prove that (2)â(1).
Proof that (2)â(1). Let
ð =ðâð=1ðð ðð¥1 â§â¯ ⧠ðð¥ð â§â¯ ⧠ðð¥ð ,
(the âhatâ over the ðð¥ð meaning that ðð¥ð has to be omitted from the wedge product). Then
ðð =ðâð=1(â1)ðâ1 ððððð¥ððð¥1 â§â¯ ⧠ðð¥ð ,
and to show that the integral of ðð is zero it suffices to show that each of the integrals
(3.2.2)ð â«ððððððð¥ððð¥
is zero. By Fubini we can compute (3.2.2)ð by first integrating with respect to the variable,ð¥ð, and then with respect to the remaining variables. But
â« ððððð¥ððð¥ð = ð(ð¥)|
ð¥ð=ðð
ð¥ð=ðð= 0
since ðð is supported on ð. â¡
We will prove that (1)â(2) by proving a somewhat stronger result. Let ð be an opensubset of ðð. Weâll say that ð has property ð if every form ð â ðºðð (ð) such that â«ð ð = 0we have ð â ððºðâ1ð (ð). We will prove:
Theorem 3.2.3. Let ð be an open subset of ððâ1 and ðŽ â ð an open interval. Then if ð hasproperty ð, ð à ðŽ does as well.
Remark 3.2.4. It is very easy to see that the open interval ðŽ itself has property ð. (SeeExercise 3.2.i below.) Hence it follows by induction from Theorem 3.2.3 that
intð = ðŽ1 Ã⯠à ðŽð , ðŽð = (ðð, ðð)has property ð, and this proves â(1)â(2)â.
Proof of Theorem 3.2.3. Let (ð¥, ð¡) = (ð¥1,âŠ, ð¥ðâ1, ð¡) be product coordinates onðÃðŽ. Givenð â ðºðð (ðÃðŽ)we can expressð as a wedge product, ðð¡â§ðŒwith ðŒ = ð(ð¥, ð¡) ðð¥1â§â¯â§ðð¥ðâ1and ð â ð¶â0 (ð à ðŽ). Let ð â ðºðâ1ð (ð) be the form
(3.2.5) ð = (â«ðŽð(ð¥, ð¡)ðð¡) ðð¥1 â§â¯ ⧠ðð¥ðâ1 .
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§3.2 The Poincaré lemma for compactly supported forms on rectangles 73
Thenâ«ððâ1ð = â«ððð(ð¥, ð¡)ðð¥ðð¡ = â«
ððð
so if the integral of ð is zero, the integral of ð is zero. Hence since ð has property ð, ð = ððfor some ð â ðºðâ2ð (ð). Let ð â ð¶â(ð) be a bump function which is supported on ðŽ andwhose integral over ðŽ is 1. Setting
ð = âð(ð¡)ðð¡ ⧠ðwe have
ðð = ð(ð¡)ðð¡ ⧠ðð = ð(ð¡)ðð¡ ⧠ð ,and hence
ð â ðð = ðð¡ ⧠(ðŒ â ð(ð¡)ð)= ðð¡ ⧠ð¢(ð¥, ð¡)ðð¥1 â§â¯ ⧠ðð¥ðâ1
whereð¢(ð¥, ð¡) = ð(ð¥, ð¡) â ð(ð¡) â«
ðŽð(ð¥, ð¡)ðð¡
by (3.2.5). Thus
(3.2.6) â«ð¢(ð¥, ð¡) ðð¡ = 0 .
Let ð and ð be the end points of ðŽ and let
(3.2.7) ð£(ð¥, ð¡) = â«ð¡
ðð¢(ð¥, ð ) ðð .
By equation (3.2.6) ð£(ð, ð¥) = ð£(ð, ð¥) = 0, so ð£ is in ð¶â0 (ð à ðŽ) and by (3.2.7), ðð£/ðð¡ = ð¢.Hence if we let ðŸ be the form, ð£(ð¥, ð¡) ðð¥1 â§â¯ ⧠ðð¥ðâ1, we have:
ððŸ = ðð¡ ⧠ð¢(ð¥, ð¡) ðð¥1 â§â¯ ⧠ðð¥ðâ1 = ð â ðð and
ð = ð(ðŸ + ð ) .Since ðŸ and ð are both inðºðâ1ð (ð ÃðŽ) this proves that ð is in ððºðâ1ð (ð ÃðŽ) and hence
that ð à ðŽ has property ð. â¡
Exercises for §3.2Exercise 3.2.i. Let ðⶠð â ð be a compactly supported function of class ð¶ð with supporton the interval (ð, ð). Show that the following are equivalent.
(1) â«ðð ð(ð¥)ðð¥ = 0.(2) There exists a function ðⶠð â ð of class ð¶ð+1 with support on (ð, ð) with ðððð¥ = ð.
Hint: Show that the function
ð(ð¥) = â«ð¥
ðð(ð )ðð
is compactly supported.
Exercise 3.2.ii. Let ð = ð(ð¥, ðŠ) be a compactly supported function on ðð à ðâ with theproperty that the partial derivatives
ðððð¥ð(ð¥, ðŠ) , for ð = 1,âŠ, ð ,
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74 Chapter 3: Integration of Forms
exist and are continuous as functions of ð¥ and ðŠ. Prove the following âdifferentiation underthe integral signâ theorem (which we implicitly used in our proof of Theorem 3.2.3).
Theorem 3.2.8. The function ð(ð¥) â â«ðâ ð(ð¥, ðŠ)ððŠ is of class ð¶1 andðððð¥ð(ð¥) = â« ðð
ðð¥ð(ð¥, ðŠ)ððŠ .
Hints: For ðŠ fixed and â â ðð,ð(ð¥ + â, ðŠ) â ð(ð¥, ðŠ) = ð·ð¥ð(ð, ðŠ)â
for some point ð on the line segment joining ð¥ to ð¥+â. Using the fact thatð·ð¥ð is continuousas a function of ð¥ and ðŠ and compactly supported, conclude the following.
Lemma 3.2.9. Given ð > 0 there exists a ð¿ > 0 such that for |â| †ð¿|ð(ð¥ + â, ðŠ) â ð(ð¥, ðŠ) â ð·ð¥ð(ð¥, ðŠ)â| †ð|â| .
Now let ð â ðâ be a rectangle with supp(ð) â ðð Ã ð and show that
|ð(ð¥ + â) â ð(ð¥) â (â«ð·ð¥ð(ð¥, ðŠ) ððŠ) â| †ð vol(ð)|â| .
Conclude that ð is differentiable at ð¥ and that its derivative is
â«ð·ð¥ð(ð¥, ðŠ)ððŠ .
Exercise 3.2.iii. Let ðⶠððÃðâ â ð be a compactly supported continuous function. Provethe following theorem.
Theorem 3.2.10. If all the partial derivatives of ð(ð¥, ðŠ) with respect to ð¥ of order †ð existand are continuous as functions of ð¥ and ðŠ the function
ð(ð¥) = â«ð(ð¥, ðŠ)ððŠ
is of class ð¶ð.Exercise 3.2.iv. Letð be an open subset of ððâ1,ðŽ â ð an open interval and (ð¥, ð¡) productcoordinates on ð à ðŽ. Recall from Exercise 2.3.v that every form ð â ðºð(ð à ðŽ) can bewritten uniquely as a sum ð = ðð¡ ⧠ðŒ + ðœ where ðŒ and ðœ are reduced, i.e., do not contain afactor of ðð¡.(1) Show that if ð is compactly supported on ð à ðŽ then so are ðŒ and ðœ.(2) Let ðŒ = âðŒ ððŒ(ð¥, ð¡)ðð¥ðŒ. Show that the form
(3.2.11) ð = âðŒ(â«ðŽððŒ(ð¥, ð¡) ðð¡) ðð¥ðŒ
is in ðºðâ1ð (ð).(3) Show that if ðð = 0, then ðð = 0.
Hint: By equation (3.2.11),
ðð = âðŒ,ð(â«ðŽ
ðððŒðð¥ð(ð¥, ð¡) ðð¡) ðð¥ð ⧠ðð¥ðŒ = â«
ðŽ(ðððŒ) ðð¡
and by equation (2.4.19) we have ðððŒ =ððœðð¡ .
Exercise 3.2.v. In Exercise 3.2.iv, show that if ð is in ððºðâ2ð (ð) then ð is in ððºðâ1ð (ð à ðŽ)as follows.
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§3.2 The Poincaré lemma for compactly supported forms on rectangles 75
(1) Let ð = ðð, with ð â ðºðâ2ð (ð) and letð â ð¶â(ð) be a bump functionwhich is supportedon ðŽ and whose integral over ðŽ is one. Setting ð = âð(ð¡)ðð¡ ⧠ð show that
ð â ðð = ðð¡ ⧠(ðŒ â ð(ð¡)ð) + ðœ= ðð¡ ⧠(âðŒ ð¢ðŒ(ð¥, ð¡)ðð¥ðŒ) + ðœ ,
where
ð¢ðŒ(ð¥, ð¡) = ððŒ(ð¥, ð¡) â ð(ð¡) â«ðŽððŒ(ð¥, ð¡)ðð¡ .
(2) Let ð and ð be the end points of ðŽ and let
ð£ðŒ(ð¥, ð¡) = â«ð¡
ðð¢ðŒ(ð¥, ð¡) ðð¡ .
Show that the form âðŒ ð£ðŒ(ð¥, ð¡) ðð¥ðŒ is in ðºðâ1ð (ð à ðŽ) and that
ððŸ = ð â ðð â ðœ + ðððŸ .
(3) Conclude that the form ð â ð(ð + ðŸ) is reduced.(4) Prove that if ð â ðºðð (ð à ðŽ) is reduced and ðð = 0 then ð = 0.
Hint: Let ð = âðŒ ððŒ(ð¥, ð¡) ðð¥ðŒ. Show that ðð = 0 â ððð¡ððŒ(ð¥, ð¡) = 0 and exploit thefact that for fixed ð¥, ððŒ(ð¥, ð¡) is compactly supported in ð¡.
Exercise 3.2.vi. Letð be an open subset ofðð.We say thatð has propertyðð, for 1 †ð < ð,if every closed ð-form ð â ðºðð (ð) is in ððºðâ1ð (ð). Prove that if the open set ð â ððâ1 inExercise 3.2.iv has property ððâ1 then ð à ðŽ has property ðð.
Exercise 3.2.vii. Show that if ð is the rectangle [ð1, ð1] à ⯠à [ðð, ðð] and ð = intð thenð has property ðð.
Exercise 3.2.viii. Letðð be the half-space
(3.2.12) ðð â { (ð¥1,âŠ, ð¥ð) â ðð | ð¥1 †0 }
and let ð â ðºðð (ðð) be the ð-form ð â ððð¥1 â§â¯ ⧠ðð¥ð with ð â ð¶â0 (ðð). Define
(3.2.13) â«ððð â â«
ððð(ð¥1,âŠ, ð¥ð)ðð¥1â¯ðð¥ð
where the right hand side is the usual Riemann integral of ð over ðð. (This integral makessense since ð is compactly supported.) Show that if ð = ðð for some ð â ðºðâ1ð (ðð) then
(3.2.14) â«ððð = â«ððâ1ðâð
where ð ⶠððâ1 â ðð is the inclusion map
(ð¥2,âŠ, ð¥ð) ⊠(0, ð¥2,âŠ, ð¥ð) .
Hint: Let ð = âð ðð ðð¥1 ⧠â¯ðð¥ð⯠⧠ðð¥ð. Mimicking the â(2)â(1)â part of the proofof Theorem 3.2.2 show that the integral (3.2.13) is the integral over ððâ1 of the function
â«0
ââ
ðð1ðð¥1(ð¥1, ð¥2,âŠ, ð¥ð)ðð¥1 .
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3.3. The Poincaré lemma for compactly supported forms on open subsets of ðð
In this section we will generalize Theorem 3.2.2 to arbitrary connected open subsets ofðð.Theorem3.3.1 (Poincaré lemma for compactly supported forms). Letð be a connected opensubset of ðð and let ð be a compactly supported ð-form with supp(ð) â ð. The the followingassertions are equivalent:(1) â«ðð ð = 0.(2) There exists a compactly supported (ð â 1)-form ð with supp ð â ð and ð = ðð.Proof that (2)â(1). The support of ð is contained in a large rectangle, so the integral of ððis zero by Theorem 3.2.2. â¡Proof that (1)â(2). Let ð1 and ð2 be compactly supported ð-forms with support in ð. Wewill write
ð1 ⌠ð2as shorthand notation for the statement: There exists a compactly supported (ð â 1)-form,ð, with support in ð and with ð1 â ð2 = ðð. We will prove that (1)â(2) by proving anequivalent statement: Fix a rectangle, ð0 â ð and an ð-form, ð0, with suppð0 â ð0 andintegral equal to one.
Theorem 3.3.2. If ð is a compactly supported ð-form with supp(ð) â ð and ð = â«ð thenð ⌠ðð0.
Thus in particular if ð = 0, Theorem 3.3.2 says that ð ⌠0 proving that (1)â(2). â¡To prove Theorem 3.3.2 let ðð â ð, ð = 1, 2, 3,âŠ, be a collection of rectangles with
ð = ââð=1 int(ðð) and let ðð be a partition of unity with supp(ðð) â int(ðð). Replacing ðby the finite sum âðð=1 ððð for ð large, it suffices to prove Theorem 3.3.2 for each of thesummands ððð. In other words we can assume that supp(ð) is contained in one of the openrectangles int(ðð). Denote this rectangle by ð. We claim that one can join ð0 to ð by asequence of rectangles as in the figure below.
ᅵ
ᅵ0
Figure 3.3.1. A sequence of rectangles joining the rectangles ð0 and ð
Lemma3.3.3. There exists a sequence of rectanglesð 0,âŠ, ð ð+1 such thatð 0 = ð0,ð ð+1 = ðand int(ð ð) â© int(ð ð+1) is nonempty.Proof. Denote by ðŽ the set of points, ð¥ â ð, for which there exists a sequence of rectangles,ð ð, ð = 0,âŠ,ð + 1 with ð 0 = ð0, with ð¥ â intð ð+1 and with intð ð â© intð ð+1 nonempty.It is clear that this set is open and that its complement is open; so, by the connectivity of ð,ð = ðŽ. â¡
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To prove Theorem 3.3.2 with suppð â ð, select, for each ð, a compactly supportedð-form ðð with supp(ðð) â int(ð ð) â© int(ð ð+1) and with â« ðð = 1. The difference, ðð â ðð+1is supported in intð ð+1, and its integral is zero. So by Theorem 3.2.2, ðð ⌠ðð+1. Similarly,ð0 ⌠ð0 and, setting ð â â«ð, we have ð ⌠ððð. Thus
ðð0 ⌠ðð0 ⌠⯠⌠ððð = ðproving the theorem.
3.4. The degree of a differentiable mapping
Definition 3.4.1. Letð andð be open subsets ofðð andðð. A continuous map ðⶠð â ð,is proper if for every compact subset ðŸ â ð, the preimage ðâ1(ðŸ) is compact in ð.
Proper mappings have a number of nice properties which will be investigated in theexercises below. One obvious property is that if ð is a ð¶â mapping and ð is a compactlysupported ð-form with support on ð, ðâð is a compactly supported ð-form with supporton ð. Our goal in this section is to show that if ð and ð are connected open subsets of ððandðⶠð â ð is a properð¶âmapping then there exists a topological invariant ofð, whichwe will call its degree (and denote by deg(ð)), such that the âchange of variablesâ formula:
(3.4.2) â«ððâð = deg(ð) â«
ðð
holds for all ð â ðºðð (ð).Before we prove this assertion letâs see what this formula says in coordinates. If
ð = ð(ðŠ)ððŠ1 â§â¯ ⧠ððŠðthen at ð¥ â ð
ðâð = (ð â ð)(ð¥) det(ð·ð(ð¥))ðð¥1 â§â¯ ⧠ðð¥ð .Hence, in coordinates, equation (3.4.2) takes the form
(3.4.3) â«ðð(ðŠ)ððŠ = deg(ð) â«
ðð â ð(ð¥) det(ð·ð(ð¥))ðð¥ .
Proof of equation (3.4.2). Let ð0 be a compactly-supported ð-form with supp(ð0) â ð andwith â«ð0 = 1. If we set degð â â«ð ð
âð0 then (3.4.2) clearly holds for ð0. We will provethat (3.4.2) holds for every compactly supported ð-form ð with supp(ð) â ð. Let ð â â«ð ð.Then by Theorem 3.3.1 ð â ðð0 = ðð, where ð is a compactly supported (ð â 1)-form withsupp ð â ð. Hence
ðâð â ððâð0 = ðâðð = ððâð ,and by part (1) of Theorem 3.3.1
â«ððâð = ðâ«ðâð0 = deg(ð) â«
ðð . â¡
We will show in §3.6 that the degree of ð is always an integer and explain why it is aâtopologicalâ invariant of ð.
Proposition 3.4.4. For the moment, however, weâll content ourselves with pointing out a sim-ple but useful property of this invariant. Let ð, ð andð be connected open subsets of ðð andðⶠð â ð and ðⶠð â ð proper ð¶â maps. Then
(3.4.5) deg(ð â ð) = deg(ð) deg(ð) .
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78 Chapter 3: Integration of Forms
Proof. Let ð be a compactly supported ð-form with support onð. Then
(ð â ð)âð = ðâðâð ;so
â«ð(ð â ð)âð = â«
ððâ(ðâð) = deg(ð) â«
ððâð
= deg(ð) deg(ð) â«ðð . â¡
From this multiplicative property it is easy to deduce the following result (which wewill need in the next section).
Theorem 3.4.6. Let ðŽ be a non-singular ð à ðmatrix and ððŽ ⶠðð â ðð the linear mappingassociated with ðŽ. Then deg(ððŽ) = +1 if detðŽ is positive and â1 if detðŽ is negative.
A proof of this result is outlined in Exercises 3.4.v to 3.4.ix below.
Exercises for §3.4Exercise 3.4.i. Letð be an open subset of ðð and (ðð)ðâ¥1 be a partition of unity onð. Showthat the mapping ðⶠð â ð defined by
ð âââð=1ððð
is a proper ð¶â mapping.
Exercise 3.4.ii. Let ð and ð be open subsets of ðð and ðð and let ðⶠð â ð be a propercontinuous mapping. Prove:
Theorem 3.4.7. If ðµ is a compact subset of ð and ðŽ = ðâ1(ðµ) then for every open subset ð0with ðŽ â ð0 â ð, there exists an open subset ð0 with ðµ â ð0 â ð and ðâ1(ð0) â ð0.
Hint: Let ð¶ be a compact subset of ð with ðµ â intð¶. Then the setð = ðâ1(ð¶) â ð0 iscompact; so its image ð(ð) is compact. Show that ð(ð) and ðµ are disjoint and let
ð0 = intð¶ â ð(ð) .
Exercise 3.4.iii. Show that if ðⶠð â ð is a proper continuous mapping andð is a closedsubset of ð, then ð(ð) is closed.
Hint: Letð0 = ðâð. Show that if ð is inðâð(ð), then ðâ1(ð) is contained inð0 andconclude from Exercise 3.4.ii that there exists a neighborhood ð0 of ð such that ðâ1(ð0) iscontained in ð0. Conclude that ð0 and ð(ð) are disjoint.
Exercise 3.4.iv. Let ðⶠðð â ðð be the translation ð(ð¥) = ð¥ + ð. Show that deg(ð) = 1.Hint: Let ðⶠð â ð be a compactly supported ð¶â function. For ð â ð, the identity
(3.4.8) â«ðð(ð¡) ðð¡ = â«
ðð(ð¡ â ð) ðð¡
is easy to prove by elementary calculus, and this identity proves the assertion above in di-mension one. Now let
(3.4.9) ð(ð¥) = ð(ð¥1)â¯ð(ð¥ð)and compute the right and left sides of equation (3.4.3) by Fubiniâs theorem.
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Exercise 3.4.v. Let ð be a permutation of the numbers, 1,âŠ, ð and let ðð ⶠðð â ðð be thediffeomorphism, ðð(ð¥1,âŠ, ð¥ð) = (ð¥ð(1),âŠ, ð¥ð(ð)). Prove that degðð = (â1)ð.
Hint: Let ð be the function (3.4.9). Show that if ð = ð(ð¥) ðð¥1 â§â¯â§ ðð¥ð, then we haveðâð = (â1)ðð.Exercise 3.4.vi. Let ðⶠðð â ðð be the mapping
ð(ð¥1,âŠ, ð¥ð) = (ð¥1 + ðð¥2, ð¥2,âŠ, ð¥ð).Prove that deg(ð) = 1.
Hint: Let ð = ð(ð¥1,âŠ, ð¥ð) ðð¥1 ⧠⯠⧠ðð¥ð where ðⶠðð â ð is compactly supportedand of class ð¶â. Show that
â«ðâð = â«ð(ð¥1 + ðð¥2, ð¥2,âŠ, ð¥ð) ðð¥1â¯ðð¥ðand evaluate the integral on the right by Fubiniâs theorem; i.e., by first integrating with re-spect to the ð¥1 variable and then with respect to the remaining variables. Note that by equa-tion (3.4.8)
â«ð(ð¥1 + ðð¥2, ð¥2,âŠ, ð¥ð) ðð¥1 = â«ð(ð¥1, ð¥2,âŠ, ð¥ð) ðð¥1 .
Exercise 3.4.vii. Let ðⶠðð â ðð be the mappingð(ð¥1,âŠ, ð¥ð) = (ðð¥1, ð¥2,âŠ, ð¥ð)
with ð â 0. Show that degð = +1 if ð is positive and â1 if ð is negative.Hint: In dimension 1 this is easy to prove by elementary calculus techniques. Prove it
in ð-dimensions by the same trick as in the previous exercise.
Exercise 3.4.viii.(1) Let ð1,âŠ, ðð be the standard basis vectors of ðð and ðŽ, ðµ and ð¶ the linear mappings
defined by
ðŽðð = {ð1, ð = 1âðð=1 ðð,ððð, ð â 1 ,
ðµðð = {âðð=1 ðððð, ð = 1ðð, ð â 1 ,
(3.4.10)
ð¶ðð = {ð1, ð = 1ðð + ððð1, ð â 1 .
Show that
ðµðŽð¶ðð = {âðð=1 ðððð, ð = 1âðð=1(ðð,ð + ðððð)ðð + ððð1ð1, ð â 1 .
for ð > 1.(2) Let ð¿â¶ ðð â ðð be the linear mapping
(3.4.11) ð¿ðð =ðâð=1âð,ððð , ð = 1,âŠ, ð .
Show that if â1,1 â 0 one can write ð¿ as a product, ð¿ = ðµðŽð¶, whereðŽ, ðµ andð¶ are linearmappings of the form (3.4.10).
Hint: First solve the equationsâð,1 = ðð
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80 Chapter 3: Integration of Forms
for ð = 1,âŠ, ð. Next solve the equations
â1,ð = ð1ððfor ð > 1. Finally, solve the equations
âð,ð = ðð,ð + ððððfor ð, ð > 1.
(3) Suppose ð¿ is invertible. Conclude that ðŽ, ðµ and ð¶ are invertible and verify that Theo-rem 3.4.6 holds for ðµ and ð¶ using the previous exercises in this section.
(4) Show by an inductive argument that Theorem 3.4.6 holds for ðŽ and conclude from(3.4.5) that it holds for ð¿.
Exercise 3.4.ix. To show that Theorem 3.4.6 holds for an arbitrary linear mapping ð¿ of theform (3.4.11) weâll need to eliminate the assumption: â1,1 â 0. Show that for some ð, âð,1 isnonzero, and show how to eliminate this assumption by considering ðð1,ð â ð¿ where ð1,ð isthe transposition 1 â ð.
Exercise 3.4.x. Here is an alternative proof ofTheorem 3.4.6 which is shorter than the proofoutlined in Exercise 3.4.ix but uses some slightly more sophisticated linear algebra.(1) Prove Theorem 3.4.6 for linear mappings which are orthogonal, i.e., satisfy ð¿âºð¿ = idð.
Hints:⢠Show that ð¿â(ð¥21 +⯠+ ð¥2ð) = ð¥21 +⯠+ ð¥2ð.⢠Show that ð¿â(ðð¥1 ⧠⯠⧠ðð¥ð) is equal to ðð¥1 ⧠⯠⧠ðð¥ð or âðð¥1 ⧠⯠⧠ðð¥ð
depending on whether ð¿ is orientation-preserving or orinetation reversing. (SeeExercise 1.2.x.)
⢠Let ð be as in Exercise 3.4.iv and let ð be the form
ð = ð(ð¥21 +⯠+ ð¥2ð) ðð¥1 â§â¯ ⧠ðð¥ð .Show that ð¿âð = ð if ð¿ is orientation-preserving and ð¿âð = âð if ð¿ is orientationreversing.
(2) Prove Theorem 3.4.6 for linear mappings which are self-adjoint (satisfy ð¿âº = ð¿).Hint: A self-adjoint linear mapping is diagonizable: there exists an intervertible lin-
ear mapping,ðⶠðð â ðð such that
(3.4.12) ðâ1ð¿ððð = ðððð , ð = 1,âŠ, ð .(3) Prove that every invertible linearmapping, ð¿, can bewritten as a product, ð¿ = ðµð¶whereðµ is orthogonal and ð¶ is self-adjoint.
Hints:⢠Show that the mapping, ðŽ = ð¿âºð¿, is self-adjoint and that it is eigenvalues (the ððâs
in equation (3.4.12)) are positive.⢠Show that there exists an invertible self-adjoint linear mapping, ð¶, such that ðŽ =ð¶2 and ðŽð¶ = ð¶ðŽ.
⢠Show that the mapping ðµ = ð¿ð¶â1 is orthogonal.
3.5. The change of variables formula
Let ð and ð be connected open subsets of ðð. If ðð ⥲ ð is a diffeomorphism, thedeterminant ofð·ð(ð¥) at ð¥ â ð is nonzero, and hence, sinceð·ð(ð¥) is a continuous functionof ð¥, its sign is the same at every point. We will say that ð is orientation-preserving if thissign is positive and orientation reversing if it is negative. We will prove below:
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§3.5 The change of variables formula 81
Theorem 3.5.1. The degree of ð is +1 if ð is orientation-preserving and â1 if ð is orientationreversing.
We will then use this result to prove the following change of variables formula for dif-feomorphisms.
Theorem 3.5.2. Let ðⶠð â ð be a compactly supported continuous function. Then
(3.5.3) â«ð(ð â ð)(ð¥)|det(ð·ð(ð¥))| = â«
ðð(ðŠ) ððŠ .
Proof of Theorem 3.5.1. Given a point ð1 â ð, let ð2 = âð(ð1) and for ð = 1, 2 let ðð ⶠðð âðð be the translation, ðð(ð¥) = ð¥ + ðð. By equation (3.4.2) and Exercise 3.4.iv the compositediffeomorphism(3.5.4) ð2 â ð â ð1has the samedegree asð, so it suffices to prove the theorem for thismapping.Notice howeverthat this mapping maps the origin onto the origin. Hence, replacing ð by this mapping, wecan, without loss of generality, assume that 0 is in the domain of ð and that ð(0) = 0.
Next notice that if ðŽâ¶ ðð ⥲ ðð is a bijective linear mapping the theorem is true for ðŽ(by Exercise 3.4.ix), and hence if we can prove the theorem forðŽâ1 â ð, equation (3.4.2) willtell us that the theorem is true for ð. In particular, letting ðŽ = ð·ð(0), we have
ð·(ðŽâ1 â ð)(0) = ðŽâ1ð·ð(0) = idðwhere idð is the identity mapping. Therefore, replacing ð by ðŽâ1 â ð, we can assume thatthe mapping ð (for which we are attempting to prove Theorem 3.5.1) has the properties:ð(0) = 0 and ð·ð(0) = idð. Let ð(ð¥) = ð(ð¥) â ð¥. Then these properties imply that ð(0) = 0andð·ð(0) = 0. â¡
Lemma 3.5.5. There exists a ð¿ > 0 such that |ð(ð¥)| †12 |ð¥| for |ð¥| †ð¿.
Proof. Let ð(ð¥) = (ð1(ð¥),âŠ, ðð(ð¥)). Thenððððð¥ð(0) = 0 ;
so there exists a ð¿ > 0 such that
| ððððð¥ð(ð¥)| †12
for |ð¥| †ð¿. However, by the mean value theorem,
ðð(ð¥) =ðâð=1
ððððð¥ð(ð)ð¥ð
for ð = ð¡0ð¥, 0 < ð¡0 < 1. Thus, for |ð¥| < ð¿,
|ðð(ð¥)| â€12
sup |ð¥ð| =12|ð¥| ,
so|ð(ð¥)| = sup |ðð(ð¥)| â€
12|ð¥| . â¡
Let ð be a compactly supported ð¶â function with 0 †ð †1 and with ð(ð¥) = 0 for|ð¥| ⥠ð¿ and ð(ð¥) = 1 for |ð¥| †ð¿2 and let ð ⶠðð â ðð be the mapping
ð(ð¥) â ð¥ + ð(ð¥)ð(ð¥) .
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It is clear that(3.5.6) ð(ð¥) = ð¥ for |ð¥| ⥠ð¿and, since ð(ð¥) = ð¥ + ð(ð¥),
(3.5.7) ð(ð¥) = ð(ð¥) for |ð¥| †ð¿2
.
In addition, for all ð¥ â ðð:
(3.5.8) | ð(ð¥)| ⥠12|ð¥| .
Indeed, by (3.5.6), | ð(ð¥)| ⥠|ð¥| for |ð¥| ⥠ð¿, and for |ð¥| †ð¿| ð(ð¥)| ⥠|ð¥| â ð(ð¥)|ð(ð¥)|
⥠|ð¥| â |ð(ð¥)| ⥠|ð¥| â 12|ð¥|
= 12|ð¥|
by Lemma 3.5.5.Now let ðð be the cube ðð â {ð¥ â ðð | |ð¥| †ð }, and let ððð â ðð â ðð. From (3.5.8)
we easily deduce that
(3.5.9) ðâ1(ðð) â ð2ðfor all ð, and hence that ð is proper. Also notice that for ð¥ â ðð¿,
| ð(ð¥)| †|ð¥| + |ð(ð¥)| †32|ð¥|
by Lemma 3.5.5 and hence
(3.5.10) ðâ1(ðð32 ð¿) â ððð¿ .
We will now prove Theorem 3.5.1.
Proof of Theorem 3.5.1. Since ð is a diffeomorphism mapping 0 to 0, it maps a neighbor-hood ð0 of 0 in ð diffeomorphically onto a neighborhood ð0 of 0 in ð, and, by shrinkingð0 if necessary, we can assume that ð0 is contained in ðð¿/2 and ð0 contained in ðð¿/4. Let ðbe an ð-form with support in ð0 whose integral over ðð is equal to one. Then ðâð is sup-ported inð0 and hence inðð¿/2. Also by (3.5.9) ðâð is supported inðð¿/2.Thus both of theseforms are zero outside ðð¿/2. However, on ðð¿/2, ð = ð by (3.5.7), so these forms are equaleverywhere, and hence
deg(ð) = â«ðâð = â« ðâð = deg( ð) .
Next let ð be a compactly supported ð-form with support in ðð3ð¿/2 and with integralequal to one. Then ðâð is supported inððð¿ by (3.5.10), and hence since ð(ð¥) = ð¥ onððð¿, wehave ðâð = ð. Thus
deg( ð) = â« ðâð = â«ð = 1 .
Putting these two identities together we conclude that deg(ð) = 1. â¡
If the function, ð, in equation (3.5.4) is a ð¶â function, the identity (3.5.3) is an im-mediate consequence of the result above and the identity (3.4.3). If ð is not ð¶â, but is justcontinuous, we will deduce equation (3.5.4) from the following result.
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§3.5 The change of variables formula 83
Theorem 3.5.11. Let ð be an open subset of ðð. If ðⶠðð â ð is a continuous function ofcompact support with suppð â ð; then for every ð > 0 there exists a ð¶â function of compactsupport, ðⶠðð â ð with suppð â ð and
sup |ð(ð¥) â ð(ð¥)| < ð .Proof. Let ðŽ be the support of ð and let ð be the distance in the sup norm from ðŽ to thecomplement of ð. Since ð is continuous and compactly supported it is uniformly continu-ous; so for every ð > 0 there exists a ð¿ > 0 with ð¿ < ð2 such that |ð(ð¥) â ð(ðŠ)| < ð when|ð¥ â ðŠ| †ð¿. Now let ð be the cube: |ð¥| < ð¿ and let ðⶠðð â ð be a non-negative ð¶âfunction with supp ð â ð and
(3.5.12) â«ð(ðŠ)ððŠ = 1 .
Setð(ð¥) = â«ð(ðŠ â ð¥)ð(ðŠ)ððŠ .
By Theorem 3.2.10 ð is a ð¶â function. Moreover, if ðŽð¿ is the set of points in ðð whosedistance in the sup norm from ðŽ is †ð¿ then for ð¥ â ðŽð¿ and ðŠ â ðŽ , |ð¥ â ðŠ| > ð¿ and henceð(ðŠ â ð¥) = 0. Thus for ð¥ â ðŽð¿
â«ð(ðŠ â ð¥)ð(ðŠ)ððŠ = â«ðŽð(ðŠ â ð¥)ð(ðŠ)ððŠ = 0 ,
so ð is supported on the compact set ðŽð¿. Moreover, since ð¿ < ð2 , suppð is contained in ð.Finally note that by (3.5.12) and Exercise 3.4.iv
(3.5.13) â«ð(ðŠ â ð¥) ððŠ = â«ð(ðŠ) ððŠ = 1
and henceð(ð¥) = â«ð(ð¥)ð(ðŠ â ð¥)ððŠ
soð(ð¥) â ð(ð¥) = â«(ð(ð¥) â ð(ðŠ))ð(ðŠ â ð¥)ððŠ
and|ð(ð¥) â ð(ð¥)| †⫠|ð(ð¥) â ð(ðŠ)|ð(ðŠ â ð¥)ððŠ .
But ð(ðŠ â ð¥) = 0 for |ð¥ â ðŠ| ⥠ð¿; and |ð(ð¥) â ð(ðŠ)| < ð for |ð¥ â ðŠ| †ð¿, so the integrand onthe right is less than
ð â« ð(ðŠ â ð¥) ððŠ ,
and hence by equation (3.5.13) we have|ð(ð¥) â ð(ð¥)| †ð . â¡
To prove the identity (3.5.3), let ðŸâ¶ ðð â ð be a ð¶â cut-off function which is one ona neighborhood ð1 of the support of ð is non-negative, and is compactly supported withsupp ðŸ â ð, and let
ð = â« ðŸ(ðŠ) ððŠ .
By Theorem 3.5.11 there exists, for every ð > 0, a ð¶â function ð, with support on ð1 satis-fying
(3.5.14) |ð â ð| †ð2ð
.
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84 Chapter 3: Integration of Forms
Thus
|â«ð(ð â ð)(ðŠ)ððŠ| †â«
ð|ð â ð|(ðŠ)ððŠ
†â«ððŸ|ð â ð|(ð¥ðŠ)ððŠ
†ð2ðâ« ðŸ(ðŠ)ððŠ †ð
2so
(3.5.15) |â«ðð(ðŠ) ððŠ â â«
ðð(ðŠ) ððŠ| †ð
2.
Similarly, the expression
|â«ð(ð â ð) â ð(ð¥)|detð·ð(ð¥)| ðð¥|
is less than or equal to the integral
â«ð(ðŸ â ð)(ð¥)|(ð â ð) â ð(ð¥)| |detð·ð(ð¥)| ðð¥
and by (3.5.14), |(ð â ð) â ð(ð¥)| †ð2ð , so this integral is less than or equal toð2ðâ«(ðŸ â ð)(ð¥)|detð·ð(ð¥)| ðð¥
and hence by (3.5.3) is less than or equal to ð2 . Thus
(3.5.16) |â«ð(ð â ð)(ð¥) | detð·ð(ð¥)|ðð¥ â â«
ðð â ð(ð¥)| detð·ð(ð¥)| ðð¥| †ð
2.
Combining (3.5.15), (3.5.16) and the identity
â«ðð(ðŠ) ððŠ = â«ð â ð(ð¥)|detð·ð(ð¥)| ðð¥
we get, for all ð > 0,
|â«ðð(ðŠ) ððŠ â â«
ð(ð â ð)(ð¥)|detð·ð(ð¥)| ðð¥| †ð
and henceâ«ð(ðŠ) ððŠ = â«(ð â ð)(ð¥)|detð·ð(ð¥)| ðð¥ .
Exercises for §3.5Exercise 3.5.i. Let âⶠð â ð be a non-negative continuous function. Show that if theimproper integral
â«ðâ(ðŠ)ððŠ
is well-defined, then the improper integral
â«ð(â â ð)(ð¥)|detð·ð(ð¥)|ðð¥
is well-defined and these two integrals are equal.Hint: If (ðð)ðâ¥1 is a partition of unity on ð then ðð = ðð â ð is a partition of unity on ð
andâ«ððâððŠ = â«ðð(â â ð(ð¥))|detð·ð(ð¥)|ðð¥ .
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§3.6 Techniques for computing the degree of a mapping 85
Now sum both sides of this identity over ð.Exercise 3.5.ii. Show that the result above is true without the assumption that â is non-negative.
Hint: â = â+ â ââ, where â+ = max(â, 0) and ââ = max(ââ, 0).Exercise 3.5.iii. Show that in equation (3.4.3) one can allow the function ð to be a continu-ous compactly supported function rather than a ð¶â compactly supported function.
Exercise 3.5.iv. Letðð be the half-space (3.2.12) andð andð open subsets of ðð. Supposeðⶠð â ð is an orientation-preserving diffeomorphism mapping ð â© ðð onto ð â© ðð.Show that for ð â ðºðð (ð)
(3.5.17) â«ðâ©ðððâð = â«
ðâ©ððð .
Hint: Interpret the left and right hand sides of this formula as improper integrals overð â© int(ðð) and ð â© int(ðð).Exercise 3.5.v. The boundary ofðð is the set
ððð â { (0, ð¥2,âŠ, ð¥ð) | (ð¥2,âŠ, ð¥ð) â ððâ1 }so the map
ð ⶠððâ1 â ðð , (ð¥2,âŠ, ð¥ð) ⊠(0, ð¥2,âŠ, ð¥ð)in Exercise 3.2.viii maps ððâ1 bijectively onto ððð.(1) Show that the map ðⶠð â ð in Exercise 3.5.iv maps ð â© ððð onto ð â© ððð.(2) Letðâ² = ðâ1(ð) andðâ² = ðâ1(ð). Conclude from (1) that the restriction ofð toðâ©ððð
gives one a diffeomorphismðⶠðâ² â ðâ²
satisfying:ð â ð = ð â ð .
(3) Let ð be in ðºðâ1ð (ð). Conclude from equations (3.2.14) and (3.5.17):
â«ðâ²ðâðâð = â«
ðâ²ðâð
and in particular show that the diffeomorphism ðⶠðâ² â ðâ² is orientation-preserving.
3.6. Techniques for computing the degree of a mapping
Let ð and ð be open subsets of ðð and ðⶠð â ð a proper ð¶â mapping. In thissectionwewill show how to compute the degree ofð and, in particular, show that it is alwaysan integer. From this fact we will be able to conclude that the degree of ð is a topologicalinvariant of ð: if we deform ð smoothly, its degree doesnât change.
Definition 3.6.1. A point ð¥ â ð is a critical point of ð if the derivativeð·ð(ð¥)ⶠðð â ðð
fails to be bijective, i.e., if det(ð·ð(ð¥)) = 0.We will denote the set of critical points of ð by ð¶ð. It is clear from the definition that
this set is a closed subset ofð and hence, by Exercise 3.4.iii,ð(ð¶ð) is a closed subset ofð.Wewill call this image the set of critical values of ð and the complement of this image the set ofregular values of ð. Notice that ð â ð(ð) is contained in ð(ð) â ð(ð¶ð), so if a point ð â ðis not in the image of ð, it is a regular value of ð âby defaultâ, i.e., it contains no points of
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86 Chapter 3: Integration of Forms
ð in the pre-image and hence, a fortiori, contains no critical points in its pre-image. Noticealso that ð¶ð can be quite large. For instance, if ð â ð and ðⶠð â ð is the constant mapwhich maps all of ð onto ð, then ð¶ð = ð. However, in this example, ð(ð¶ð) = {ð}, so the setof regular values of ð is ð â {ð}, and hence (in this example) is an open dense subset of ð.We will show that this is true in general.Theorem 3.6.2 (Sard). If ð and ð are open subsets of ðð and ðⶠð â ð a proper ð¶â map,the set of regular values of ð is an open dense subset of ð.
We will defer the proof of this to Section 3.7 and in this section explore some of itsimplications. Picking a regular value ð of ð we will prove:Theorem 3.6.3. The set ðâ1(ð) is a finite set. Moreover, if ðâ1(ð) = {ð1,âŠ, ðð} there existconnected open neighborhoodsðð of ðð in ð and an open neighborhoodð of ð inð such that:(1) for ð â ð the sets ðð and ðð are disjoint;(2) ðâ1(ð) = ð1 âªâ¯ ⪠ðð,(3) ðmaps ðð diffeomorphically ontoð.Proof. If ð â ðâ1(ð) then, since ð is a regular value, ð â ð¶ð; so
ð·ð(ð)ⶠðð â ðð
is bijective. Hence by the inverse function theorem, ðmaps a neighborhood,ðð of ð diffeo-morphically onto a neighborhood of ð. The open sets
{ ðð | ð â ðâ1(ð) }are a covering of ðâ1(ð); and, since ð is proper, ðâ1(ð) is compact. Thus we can extract afinite subcovering
{ðð1 ,âŠ,ððð}and since ðð is the only point in ððð which maps onto ð, we have that ðâ1(ð) = {ð1,âŠ, ðð}.
Without loss of generality we can assume that theððð âs are disjoint from each other; for,if not, we can replace them by smaller neighborhoods of the ððâs which have this property.By Theorem 3.4.7 there exists a connected open neighborhoodð of ð in ð for which
ðâ1(ð) â ðð1 âªâ¯ ⪠ððð .
To conclude the proof let ðð â ðâ1(ð) â© ððð . â¡
The main result of this section is a recipe for computing the degree of ð by countingthe number of ððâs above, keeping track of orientation.Theorem 3.6.4. For each ðð â ðâ1(ð) let ððð = +1 if ðⶠðð â ð is orientation-preservingand â1 if ðⶠðð âð is orientation reversing. Then
(3.6.5) deg(ð) =ðâð=1ððð .
Proof. Let ð be a compactly supported ð-form onð whose integral is one. Then
deg(ð) = â«ððâð =
ðâð=1â«ðððâð .
Since ðⶠðð âð is a diffeomorphism
â«ðððâð = ±â«
ðð = {1, ð is orientation-preservingâ1, ð is not orientation-preserving .
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§3.6 Techniques for computing the degree of a mapping 87
Thus deg(ð) is equal to the sum (3.6.5). â¡As we pointed out above, a point ð â ð can qualify as a regular value of ð âby defaultâ,
i.e., by not being in the image of ð. In this case the recipe (3.6.5) for computing the degreegives âby defaultâ the answer zero. Letâs corroborate this directly.Theorem 3.6.6. If ðⶠð â ð is not surjective, then deg(ð) = 0.Proof. By Exercise 3.4.iii, ð â ð(ð) is open; so if it is nonempty, there exists a compactlysupported ð-form ð with support in ð â ð(ð) and with integral equal to one. Since ð = 0on the image of ð, ðâð = 0; so
0 = â«ððâð = deg(ð) â«
ðð = deg(ð) . â¡
Remark 3.6.7. In applications the contrapositive of Theorem 3.6.6 is much more usefulthan the theorem itself.Theorem 3.6.8. If deg(ð) â 0, then ðmaps ð surjectively onto ð.
In other words if deg(ð) â 0 the equationð(ð¥) = ðŠ
has a solution, ð¥ â ð for every ðŠ â ð.We will now show that the degree of ð is a topological invariant of ð: if we deform ð by
a âhomotopyâ we do not change its degree. To make this assertion precise, letâs recall whatwe mean by a homotopy between a pair of ð¶â maps. Let ð be an open subset of ðð, ðan open subset of ðð, ðŽ an open subinterval of ð containing 0 and 1, and ð1, ð2 ⶠð â ða pair of ð¶â maps. Then a ð¶â map ð¹â¶ ð à ðŽ â ð is a homotopy between ð0 and ð1 ifð¹(ð¥, 0) = ð0(ð¥) and ð¹(ð¥, 1) = ð1(ð¥). (See Definition 2.6.14.) Suppose now that ð0 and ð1 areproper.Definition 3.6.9. A homotopy ð¹ between ð0 and ð1 is a proper homotopy if the map
ð¹â¯ ⶠð à ðŽ â ð à ðŽdefined by (ð¥, ð¡) ⊠(ð¹(ð¥, ð¡), ð¡) is proper.
Note that if ð¹ is a proper homotopy between ð0 and ð1, then for every ð¡ between 0 and1, the map
ðð¡ ⶠð â ð , ðð¡(ð¥) â ð¹(ð¥, ð¡)is proper.
Now let ð and ð be open subsets of ðð.Theorem 3.6.10. If ð0 and ð1 are properly homotopic, then deg(ð0) = deg(ð1).Proof. Let
ð = ð(ðŠ)ððŠ1 â§â¯ ⧠ððŠðbe a compactly supported ð-form on ð whose integral over ð is 1. The the degree of ðð¡ isequal to
(3.6.11) â«ðð(ð¹1(ð¥, ð¡),âŠ, ð¹ð(ð¥, ð¡)) detð·ð¥ð¹(ð¥, ð¡)ðð¥ .
The integrand in (3.6.11) is continuous and for 0 †ð¡ †1 is supported on a compact subsetof ð à [0, 1], hence (3.6.11) is continuous as a function of ð¡. However, as weâve just proved,deg(ðð¡) is integer valued so this function is a constant. â¡
(For an alternative proof of this result see Exercise 3.6.ix below.)
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88 Chapter 3: Integration of Forms
ApplicationsWeâll conclude this account of degree theory by describing a couple applications.
Application 3.6.12 (The Brouwer fixed point theorem). Let ðµð be the closed unit ball inðð:ðµð â { ð¥ â ðð | âð¥â †1 } .
Theorem 3.6.13. If ðⶠðµð â ðµð is a continuous mapping then ð has a fixed point, i.e., mapssome point, ð¥0 â ðµð onto itself.
The idea of the proof will be to assume that there isnât a fixed point and show that thisleads to a contradiction. Suppose that for every point ð¥ â ðµð we have ð(ð¥) â ð¥. Considerthe ray through ð(ð¥) in the direction of ð¥:
ð(ð¥) + ð (ð¥ â ð(ð¥)) , ð â [0,â) .This ray intersects the boundary ððâ1 â ððµð in a unique point ðŸ(ð¥) (see Figure 3.6.1 below);and one of the exercises at the end of this section will be to show that the mapping ðŸâ¶ ðµð âððâ1 given by ð¥ ⊠ðŸ(ð¥), is a continuous mapping. Also it is clear from Figure 3.6.1 thatðŸ(ð¥) = ð¥ if ð¥ â ððâ1, so we can extend ðŸ to a continuous mapping of ðð into ðð by letting ðŸbe the identity for âð¥â ⥠1. Note that this extended mapping has the property
âðŸ(ð¥)â ⥠1for all ð¥ â ðð and(3.6.14) ðŸ(ð¥) = ð¥for all âð¥â ⥠1. To get a contradiction weâll show that ðŸ can be approximated by a ð¶â mapwhich has similar properties. For this wewill need the following corollary ofTheorem 3.5.11.
Lemma 3.6.15. Let ð be an open subset of ðð, ð¶ a compact subset of ð and ðⶠð â ð acontinuous function which is ð¶â on the complement of ð¶. Then for every ð > 0, there exists að¶â function ðⶠð â ð, such that ð â ð has compact support and |ð â ð| < ð.Proof. Let ð be a bump function which is in ð¶â0 (ð) and is equal to 1 on a neighborhoodof ð¶. By Theorem 3.5.11 there exists a function ð0 â ð¶â0 (ð) such that |ðð â ð0| < ð. Tocomplete the proof, let ð(1 â ð)ð + ð0, and note that
ð â ð = (1 â ð)ð + ðð â (1 â ð)ð â ð0= ðð â ð0 . â¡
By applying Lemma 3.6.15 to each of the coordinates of the map ðŸ, one obtains a ð¶âmap ðⶠðð â ðð such that
âð â ðŸâ < ð < 1and such that ð = ðŸ on the complement of a compact set. However, by (3.6.14), this meansthat ð is equal to the identity on the complement of a compact set and hence (see Exer-cise 3.6.ix) that ð is proper and deg(ð) = 1. On the other hand by (3.6.19) and (3.6.14) wehave âð(ð¥)â > 1 â ð for all ð¥ â ðð, so 0 â im(ð) and hence by Theorem 3.6.4 we havedeg(ð) = 0, which provides the desired contradiction.
Application 3.6.16 (The fundamental theorem of algebra). Letð(ð§) = ð§ð + ððâ1ð§ðâ1 +⯠+ ð1ð§ + ð0
be a polynomial of degree ð with complex coefficients. If we identify the complex planeð â { ð§ = ð¥ + ððŠ | ð¥, ðŠ â ð }
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§3.6 Techniques for computing the degree of a mapping 89
x
ᅵ(ᅵ)
ᅵ(ᅵ)
Figure 3.6.1. Brouwer fixed point theorem
with ð2 via the map ð2 â ð given by (ð¥, ðŠ) ⊠ð§ = ð¥ + ððŠ, we can think of ð as defining amapping
ðⶠð2 â ð2 , ð§ ⊠ð(ð§) .We will prove:
Theorem 3.6.17. The mapping ðⶠð2 â ð2 is proper and deg(ð) = ð.
Proof. For ð¡ â ð let
ðð¡(ð§) â (1 â ð¡)ð§ð + ð¡ð(ð§) = ð§ð + ð¡ðâ1âð=0ððð§ð .
We will show that the mapping
ðⶠð2 à ð â ð2 , (ð§, ð¡) ⊠ðð¡(ð§)is a proper homotopy. Let
ð¶ = sup0â€ðâ€ðâ1|ðð| .
Then for |ð§| ⥠1 we have
|ð0 +⯠+ ððâ1ð§ðâ1| †|ð0| + |ð1||ð§| + ⯠+ |ððâ1| |ð§|ðâ1
†ð¶ð|ð§|ðâ1 ,and hence, for |ð¡| †ð and |ð§| ⥠2ðð¶ð,
|ðð¡(ð§)| ⥠|ð§|ð â ðð¶ð|ð§|ðâ1
⥠ðð¶ð|ð§|ðâ1 .If ðŽ â ð is compact, then for some ð > 0, ðŽ is contained in the disk defined by |ð€| †ð ,and hence the set
{ ð§ â ð | (ð¡, ðð¡(ð§)) â [âð, ð] à ðŽ }is contained in the compact set
{ ð§ â ð | ðð¶|ð§|ðâ1 †ð } .This shows that ð is a proper homotopy.
Thus for each ð¡ â ð, the map ðð¡ ⶠð â ð is proper and
deg(ðð¡) = deg(ð1) = deg(ð) = deg(ð0)However, ð0 ⶠð â ð is just the mapping ð§ ⊠ð§ð and an elementary computation (seeExercises 3.6.v and 3.6.vi below) shows that the degree of this mapping is ð. â¡
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90 Chapter 3: Integration of Forms
In particular for ð > 0 the degree of ð is nonzero; so byTheorem 3.6.4 we conclude thatðⶠð â ð is surjective and hence has zero in its image.
Theorem 3.6.18 (Fundamental theorem of algebra). Every positive-degree polynomial
ð(ð§) = ð§ð + ððâ1ð§ðâ1 +⯠+ ð0with complex coefficients has a complex root: ð(ð§0) = 0 for some ð§0 â ð.
Exercises for §3.6Exercise 3.6.i. Letð be a subset of ðð and let ð(ð¥), ð(ð¥) and ð(ð¥) be real-valued functionsonð of class ð¶ð. Suppose that for every ð¥ â ð the quadratic polynomial
ð(ð¥)ð 2 + ð(ð¥)ð + ð(ð¥)
has two distinct real roots, ð +(ð¥) and ð â(ð¥), with ð +(ð¥) > ð â(ð¥). Prove that ð + and ð â arefunctions of class ð¶ð.
Hint: What are the roots of the quadratic polynomial: ðð 2 + ðð + ð?
Exercise 3.6.ii. Show that the function ðŸ(ð¥) defined in Figure 3.6.1 is a continuous surjec-tion ðµð â ð2ðâ1.
Hint: ðŸ(ð¥) lies on the ray,
ð(ð¥) + ð (ð¥ â ð(ð¥)) , ð â [0,â)
and satisfies âðŸ(ð¥)â = 1. Thus
ðŸ(ð¥) = ð(ð¥) + ð 0(ð¥ â ð(ð¥)) ,
where ð 0 is a non-negative root of the quadratic polynomial
âð(ð¥) + ð (ð¥ â ð(ð¥))â2 â 1 .
Argue from Figure 3.6.1 that this polynomial has to have two distinct real roots.
Exercise 3.6.iii. Show that the Brouwer fixed point theorem isnât true if one replaces theclosed unit ball by the open unit ball.
Hint: Let ð be the open unit ball (i.e., the interior of ðµð). Show that the map
âⶠð â ðð , â(ð¥) â ð¥1 â âð¥â2
is a diffeomorphism of ð onto ðð, and show that there are lots of mappings of ðð onto ððwhich do not have fixed points.
Exercise 3.6.iv. Show that the fixed point in the Brouwer theorem doesnât have to be aninterior point of ðµð, i.e., show that it can lie on the boundary.
Exercise 3.6.v. If we identify ð with ð2 via the mapping (ð¥, ðŠ) ⊠ð¥ + ððŠ, we can think of að-linear mapping of ð into itself, i.e., a mapping of the form
ð§ ⊠ðð§
for a fixed ð â ð, as an ð-linear mapping of ð2 into itself. Show that the determinant of thismapping is |ð|2.
Exercise 3.6.vi.
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§3.6 Techniques for computing the degree of a mapping 91
(1) Let ðⶠð â ð be the mapping ð(ð§) â ð§ð. Show thatð·ð(ð§) is the linear map
ð·ð(ð§) = ðð§ðâ1
given by multiplication by ðð§ðâ1.Hint: Argue from first principles. Show that for â â ð = ð2
(ð§ + â)ð â ð§ð â ðð§ðâ1â|â|
tends to zero as |â| â 0.(2) Conclude from Exercise 3.6.v that
det(ð·ð(ð§)) = ð2|ð§|2ðâ2 .(3) Show that at every point ð§ â ð â {0}, ð is orientation preserving.(4) Show that every point, ð€ â ð â {0} is a regular value of ð and that
ðâ1(ð€) = {ð§1,âŠ, ð§ð}with ðð§ð = +1.
(5) Conclude that the degree of ð is ð.
Exercise 3.6.vii. Prove that the map ð from Exercise 3.6.vi has degree ð by deducing thisdirectly from the definition of degree.
Hints:⢠Show that in polar coordinates, ð is the map (ð, ð) ⊠(ðð, ðð).⢠Let ð be the 2-form ð â ð(ð¥2 +ðŠ2)ðð¥ ⧠ððŠ, where ð(ð¡) is a compactly supportedð¶â function of ð¡. Show that in polar coordinates ð = ð(ð2)ðððâ§ðð, and computethe degree of ð by computing the integrals of ð and ðâð in polar coordinates andcomparing them.
Exercise 3.6.viii. Let ð be an open subset of ðð, ð an open subset of ðð, ðŽ an open subin-terval of ð containing 0 and 1, ð0, ð1 ⶠð â ð a pair of ð¶â mappings, and ð¹â¶ ð à ðŽ â ða homotopy between ð0 and ð1.(1) In Exercise 2.4.iv you proved that if ð â ðºð(ð) and ðð = 0, then(3.6.19) ðâ0 ð â ðâ1 ð = ðð
where ð is the (ð â 1)-form ððŒ in equation (2.6.17). Show (by careful inspection of thedefinition of ððŒ) that if ð¹ is a proper homotopy and ð â ðºðð (ð) then ð â ðºðâ1ð (ð).
(2) Suppose in particular that ð and ð are open subsets of ðð and ð is in ðºðð (ð). Deducefrom equation (3.6.19) that
â«ðâ0 ð = â«ðâ1 ð
and deduce directly from the definition of degree that degree is a proper homotopyinvariant.
Exercise 3.6.ix. Let ð be an open connected subset of ðð and ðⶠð â ð a proper ð¶âmap. Prove that if ð is equal to the identity on the complement of a compact set ð¶, then ðis proper and deg(ð) = 1.
Hints:⢠Show that for every subset ðŽ â ð, we have ðâ1(ðŽ) â ðŽ ⪠ð¶, and conclude from
this that ð is proper.⢠Use the recipe (1.6.2) to compute deg(ð) with ð â ð â ð(ð¶).
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92 Chapter 3: Integration of Forms
Exercise 3.6.x. Let (ðð,ð) be an ðÃðmatrix andðŽâ¶ ðð â ðð the linear mapping associatedwith this matrix. Frobeniusâ Theorem asserts: If the ðð,ð are non-negative then ðŽ has a non-negative eigenvalue. In other words there exists a ð£ â ðð and a ð â ð, ð ⥠0, such thatðŽð£ = ðð£. Deduce this linear algebra result from the Brouwer fixed point theorem.
Hints:⢠We can assume that ðŽ is bijective, otherwise 0 is an eigenvalue. Let ððâ1 be the(ð â 1)-sphere, defined by |ð¥| = 1, and ðⶠððâ1 â ððâ1 the map,
ð(ð¥) = ðŽð¥âðŽð¥â
.
Show that ðmaps the set
ð â { (ð¥1,âŠ, ð¥ð) â ððâ1 | ð¥1,âŠ, ð¥ð ⥠0 }into itself.
⢠It is easy to prove that ð is homeomorphic to the unit ball ðµðâ1, i.e., that thereexists a continuous map ðⶠð â ðµðâ1 which is invertible and has a continuousinverse. Without bothering to prove this fact deduce from it Frobeniusâ theorem.
3.7. Appendix: Sardâs theorem
The version of Sardâs theorem stated in §3.5 is a corollary of the following more generalresult.
Theorem 3.7.1. Letð be an open subset of ðð and ðⶠð â ðð a ð¶â map. Then ðð âð(ð¶ð)is dense in ðð.
Before undertaking to prove this wewill make a few general comments about this result.
Remark 3.7.2. If (ðð)ðâ¥1 are open dense subsets of ðð, the intersectionâðâ¥1ðð is densein ðð. (This follows from the Baire category theorem; see, for instance, [7, Ch. 6 Thm. 34;11, Thm. 48.2] or Exercise 3.7.iv.)
Remark 3.7.3. If (ðŽð)ðâ¥1 is a covering ofð by compact sets,ðð â ðð âð(ð¶ð â©ðŽð) is open,so if we can prove that it is dense then by Remark 3.7.2 we will have proved Sardâs theorem.Hence since we can always cover ð by a countable collection of closed cubes, it suffices toprove: for every closed cube ðŽ â ð, the subspace ðð â ð(ð¶ð â© ðŽ) is dense in ðð.
Remark 3.7.4. Let ðⶠð â ð be a diffeomorphism and let â = ð â ð. Then
(3.7.5) ð(ð¶ð) = â(ð¶â)so Sardâs theorem for â implies Sardâs theorem for ð.
We will first prove Sardâs theorem for the set of super-critical points of ð, the set:
ð¶â¯ð â {ð â ð |ð·ð(ð) = 0 } .
Proposition 3.7.6. Let ðŽ â ð be a closed cube. Then the open set ðð â ð(ðŽ â© ð¶â¯ð) is a densesubset of ðð.
Weâll deduce this from the lemma below.
Lemma 3.7.7. Given ð > 0 one can cover ð(ðŽ â© ð¶â¯ð) by a finite number of cubes of totalvolume less than ð.
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§3.7 Appendix: Sardâs theorem 93
Proof. Let the length of each of the sides ofðŽ be â. Given ð¿ > 0 one can subdivideðŽ intoððcubes, each of volume (â/ð)ð, such that if ð¥ and ðŠ are points of any one of these subcubes
(3.7.8) | ððððð¥ð(ð¥) â ððððð¥ð(ðŠ)| < ð¿ .
Let ðŽ1,âŠ,ðŽð be the cubes in this collection which intersect ð¶â¯ð. Then for ð§0 â ðŽð â© ð¶â¯ð,
ððððð¥ð(ð§0) = 0, so for ð§ â ðŽð we have
(3.7.9) | ððððð¥ð(ð§)| < ð¿
by equation (3.7.8). If ð¥ and ðŠ are points of ðŽð then by the mean value theorem there existsa point ð§ on the line segment joining ð¥ to ðŠ such that
ðð(ð¥) â ðð(ðŠ) =ðâð=1
ððððð¥ð(ð§)(ð¥ð â ðŠð)
and hence by (3.7.9)
|ðð(ð¥) â ðð(ðŠ)| †ð¿ðâð=1|ð¥ð â ðŠð| †ðð¿
âð
.
Thus ð(ð¶ð â©ðŽð) is contained in a cube ðµð of volume (ð ð¿âð )ð, and ð(ð¶ð â©ðŽ) is contained in
a union of cubes ðµð of total volume less than
ðððð ð¿ðâððð= ððð¿ðâð
so if we choose ð¿ such that ððð¿ðâð < ð, weâre done. â¡
Proof of Proposition 3.7.6. To prove Proposition 3.7.6 we have to show that for every pointð â ðð and neighborhood,ð of ð, the setðâ ð(ð¶â¯ð â© ðŽ) is nonempty. Suppose
(3.7.10) ð â ð(ð¶â¯ð â© ðŽ) .Without loss of generality we can assumeð is a cube of volume ð, but the lemma tells usthat ð(ð¶â¯ð â©ðŽ) can be covered by a finite number of cubes whose total volume is less than ð,and hence by (3.7.10)ð can be covered by a finite number of cubes of total volume less thanð, so its volume is less than ð. This contradiction proves that the inclusion (3.7.10) cannothold. â¡
Now we prove Theorem 3.7.1.
Proof of Theorem 3.7.1. Let ðð,ð be the subset of ð where ððððð¥ð â 0. Then
ð = ð¶â¯ð ⪠â1â€ð,ðâ€ððð,ð ,
so to prove the theorem it suffices to show that ðð â ð(ðð,ð â© ð¶ð) is dense in ðð, i.e., itsuffices to prove the theorem with ð replaced by ðð,ð. Let ðð ⶠðð ⥲ ðð be the involutionwhich interchanges ð¥1 and ð¥ð and leaves the remaining ð¥ðâs fixed. Letting ðnew = ðððoldððand ðnew = ðððold, we have, for ð = ðnew and ð = ðnew
(3.7.11) ðð1ðð¥1(ð) â 0
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for all ð â ð so weâre reduced to proving Theorem 3.7.1 for maps ðⶠð â ðð having theproperty (3.7.11). Let ðⶠð â ðð be defined by
(3.7.12) ð(ð¥1,âŠ, ð¥ð) = (ð1(ð¥), ð¥2,âŠ, ð¥ð) .Then
ðâð¥1 = ðâð¥1 = ð1(ð¥1,âŠ, ð¥ð)and
det(ð·ð) = ðð1ðð¥1â 0 .
Thus, by the inverse function theorem, ð is locally a diffeomorphism at every point ð â ð.This means that if ðŽ is a compact subset of ð we can cover ðŽ by a finite number of opensubsets ðð â ð such that ð maps ðð diffeomorphically onto an open subset ðð in ðð. Toconclude the proof of the theorem weâll show that ðð â ð(ð¶ð â© ðð â© ðŽ) is a dense subset ofðð. Let âⶠðð â ðð be themapâ = ðâðâ1. To prove this assertion it suffices byRemark 3.7.4to prove that the set ðð â â(ð¶â) is dense in ðð. This we will do by induction on ð. First notethat for ð = 1, we have ð¶ð = ð¶
â¯ð, so weâve already proved Theorem 3.7.1 in dimension one.
Now note that by (3.7.12) we have ââð¥1 = ð¥1, i.e., â is a mapping of the form
â(ð¥1,âŠ, ð¥ð) = (ð¥1, â2(ð¥),âŠ, âð(ð¥)) .Thus if we let
ðð â { (ð¥2,âŠ, ð¥ð) â ððâ1 | (ð, ð¥2,âŠ, ð¥ð) â ðð }and let âð ⶠðð â ððâ1 be the map
âð(ð¥2,âŠ, ð¥ð) = (â2(ð, ð¥2,âŠ, ð¥ð),âŠ, âð(ð, ð¥2,âŠ, ð¥ð)) .Then
det(ð·âð)(ð¥2,âŠ, ð¥ð) = det(ð·â)(ð, ð¥2,âŠ, ð¥ð) ,and hence
(3.7.13) (ð, ð¥) â ðð â© ð¶â ⺠ð¥ â ð¶âð .Now let ð0 = (ð, ð¥0) be a point in ðð. We have to show that every neighborhood ð of ð0contains a point ð â ððââ(ð¶â). Letðð â ððâ1 be the set of points ð¥ for which (ð, ð¥) â ð. Byinduction ðð contains a point ð¥ â ððâ1 â âð(ð¶âð) and hence ð = (ð, ð¥) is in ð by definitionand in ðð â â(ð¶â) by (3.7.13). â¡
Exercises for §3.7Exercise 3.7.i. What are the set of critical points and the image of the set of critical pointsfor the following maps ð â ð?(1) The map ð1(ð¥) = (ð¥2 â 1)2.(2) The map ð2(ð¥) = sin(ð¥) + ð¥.(3) The map
ð3(ð¥) = {0, 𥠆0ðâ 1ð¥ , ð¥ > 0 .
Exercise 3.7.ii (Sardâs theorem for affine maps). Let ðⶠðð â ðð be an affine map, i.e., amap of the form
ð(ð¥) = ðŽ(ð¥) + ð¥0where ðŽâ¶ ðð â ðð is a linear map and ð¥0 â ðð. Prove Sardâs theorem for ð.
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§3.7 Appendix: Sardâs theorem 95
Exercise 3.7.iii. Letðⶠð â ðbe að¶â functionwhich is supported in the interval (â1/2, 1/2)and has amaximum at the origin. Let (ðð)ðâ¥1 be an enumeration of the rational numbers, andlet ðⶠð â ð be the map
ð(ð¥) =ââð=1ððð(ð¥ â ð) .
Show that ð is a ð¶â map and show that the image of ð¶ð is dense in ð.The moral of this example: Sardâs theorem says that the complement of ð¶ð is dense in ð,
but ð¶ð can be dense as well.
Exercise 3.7.iv. Prove the assertion made in Remark 3.7.4.Hint: You need to show that for every point ð â ðð and every neighborhood ð of ð,
ðâ©âðâ¥1ðð is nonempty. Construct, by induction, a family of closed balls (ðµð)ðâ¥1 such that⢠ðµð â ð,⢠ðµð+1 â ðµð,⢠ðµð â âðâ€ððð,⢠The radius of ðµð is less than 1ð ,
and show thatâðâ¥1 ðµð â â .Exercise 3.7.v. Verify equation (3.7.5).
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CHAPTER 4
Manifolds & Forms on Manifolds
4.1. Manifolds
Our agenda in this chapter is to extend to manifolds the results of Chapters 2 and 3 andto formulate and prove manifold versions of two of the fundamental theorems of integralcalculus: Stokesâ theorem and the divergence theorem. In this section weâll define what wemean by the term manifold, however, before we do so, a word of encouragement. Havinghad a course in multivariable calculus, you are already familiar with manifolds, at least intheir one and two-dimensional emanations, as curves and surfaces in ð3, i.e., a manifold isbasically just an ð-dimensional surface in some high dimensional Euclidean space. Tomakethis definition precise letð be a subset of ðð, ð a subset of ðð and ðⶠð â ð a continuousmap. We recall:
Definition 4.1.1. The map ð is a ð¶â map if for every ð â ð, there exists a neighborhoodðð of ð in ðð and a ð¶â map ðð ⶠðð â ðð which coincides with ð on ðð â© ð.
We also recall:
Theorem 4.1.2. If ðⶠð â ð is a ð¶â map, there exists a neighborhoodð, ofð in ðð and að¶â map ðⶠð â ðð such that ð coincides with ð onð.
(A proof of this can be found in Appendix a.)We will say that ð is a diffeomorphism if ð is a bijection and ð and ðâ1 are both ð¶â
maps. In particular if ð is an open subset of ðð,ð is an example of an object which we willcall a manifold. More generally:
Definition 4.1.3. Letð and ð be nonnegative integers with ð †ð. A subset ð â ðð is anð-manifold if for every ð â ð there exists a neighborhood ð of ð in ðð, an open subset ðin ðð, and a diffeomorphism ðⶠð ⥲ ð â© ð.
Thusð is an ð-manifold if, locally near every point ð,ð âlooks likeâ an open subset ofðð.
Examples 4.1.4.(1) Graphs of functions: Let ð be an open subset of ðð and ðⶠð â ð a ð¶â function. Its
graphð€ð = { (ð¥, ð(ð¥)) â ðð+1 | ð¥ â ð }
is an ð-manifold in ðð+1. In fact the map
ðⶠð â ðð+1 , ð¥ ⊠(ð¥, ð(ð¥))is a diffeomorphism of ð onto ð€ð. (Itâs clear that ð is a ð¶â map, and it is a diffeomor-phism since its inverse is the map ðⶠð€ð â ð given by ð(ð¥, ð¡) â ð¥, which is also clearlyð¶â.)
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(2) Graphs of mappings: More generally if ðⶠð â ðð is a ð¶â map, its graph ð€ð is anð-manifold in ðð+ð.
(3) Vector spaces: Letð be an ð- dimensional vector subspace of ðð, and (ð1,âŠ, ðð) a basisof ð. Then the linear map
(4.1.5) ðⶠðð â ð , (ð¥1,âŠ, ð¥ð) âŠðâð=1ð¥ððð
is a diffeomorphism of ðð onto ð. Hence every ð-dimensional vector subspace of ððis automatically an ð-dimensional submanifold of ðð. Note, by the way, that ifð is anyð-dimensional vector space, not necessarily a subspace of ðð, the map (4.1.5) givesus an identification of ð with ðð. This means that we can speak of subsets of ð asbeing ð-dimensional submanifolds if, via this identification, they get mapped onto ð-dimensional submanifolds of ðð. (This is a trivial, but useful, observation since a lot ofinteresting manifolds occur âin natureâ as subsets of some abstract vector space ratherthan explicitly as subsets of some ðð. An example is the manifold, ð(ð), of orthogonalð Ã ð matrices. This manifold occurs in nature as a submanifold of the vector space ofð by ðmatrices.)
(4) Affine subspaces of ðð: These are manifolds of the form ð + ð, where ð is a vector sub-space of ðð and ð is a specified point in ðð. In other words, they are diffeomorphiccopies of vector subspaces with respect to the diffeomorphism
ðð ⶠðð ⥲ ðð , ð¥ ⊠ð + ð¥ .
Ifð is an arbitrary submanifold of ðð its tangent space a point ð â ð, is an example ofa manifold of this type. (Weâll have more to say about tangent spaces in Section 4.2.)
(5) Product manifolds: For ð = 1, 2, let ðð be an ðð-dimensional submanifold of ððð . Thenthe Cartesian product ofð1 andð2
ð1 à ð2 = { (ð¥1, ð¥2) | ð¥ð â ðð }is an (ð1 + ð2)-dimensional submanifold of ðð1+ð2 = ðð1 à ðð2 .
We will leave for you to verify this fact as an exercise.Hint: For ðð â ðð, ð = 1, 2, there exists a neighborhood,ðð, of ðð in ððð , an open set,
ðð in ððð , and a diffeomorphism ðⶠðð â ðð â© ðð. Let ð = ð1 à ð2, ð = ð1 à ð2 andð = ð1 à ð2, and let ðⶠð â ð â© ð be the product diffeomorphism, (ð(ð1), ð2(ð2)).
(6) The unit ð-sphere: This is the set of unit vectors in ðð+1:ðð = { ð¥ â ðð+1 | ð¥21 +⯠+ ð¥2ð+1 = 1 } .
To show that ðð is an ð-manifold, letð be the open subset of ðð+1 on which ð¥ð+1 ispositive. Ifð is the open unit ball inðð andðⶠð â ð is the function,ð(ð¥) = (1â(ð¥21 +⯠+ ð¥2ð))1/2, then ðð â© ð is just the graph ð€ð of ð. Hence we have a diffeomorphism
ðⶠð â ðð â© ð .More generally, if ð = (ð¥1,âŠ, ð¥ð+1) is any point on the unit sphere, then ð¥ð is nonzerofor some ð. If ð¥ð is positive, then letting ð be the transposition, ð â ð+1 andðð ⶠðð+1 âðð+1, the map
ðð(ð¥1,âŠ, ð¥ð) = (ð¥ð(1),âŠ, ð¥ð(ð))one gets a diffeomorphismððâð ofð onto a neighborhood ofð in ðð and ifð¥ð is negativeone gets such a diffeomorphism by replacing ðð by âðð. In either case we have shownthat for every point ð in ðð, there is a neighborhood of ð in ðð which is diffeomorphicto ð.
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(7) The 2-torus: In calculus books this is usually described as the surface of rotation in ð3obtained by taking the unit circle centered at the point, (2, 0), in the (ð¥1, ð¥3) plane androtating it about the ð¥3-axis. However, a slightly nicer description of it is as the productmanifold ð1 à ð1 in ð4. (As an exercise, reconcile these two descriptions.)
Weâll now turn to an alternative way of looking at manifolds: as solutions of systems ofequations. Let ð be an open subset of ðð and ðⶠð â ðð a ð¶â map.
Definition 4.1.6. A point ð â ðð is a regular value of ð if for every point ð â ðâ1(ð), themap ð is a submersion at ð.
Note that for ð to be a submersion at ð, the differential ð·ð(ð)ⶠðð â ðð has to besurjective, and hence ð has to be less than or equal to ð. Therefore this notion of regularvalue is interesting only ifð ⥠ð.
Theorem4.1.7. Let ð â ðâð. If ð is a regular value ofð, the setð â ðâ1(ð) is an ð-manifold.
Proof. Replacing ð by ðâð â ð we can assume without loss of generality that ð = 0. Let ð âðâ1(0). Since ð is a submersion at ð, the canonical submersion theorem (see Theorem b.16)tells us that there exists a neighborhood ð of 0 in ðð, a neighborhood ð0, of ð in ð and adiffeomorphism ð â ð ⥲ ð0 such that
(4.1.8) ð â ð = ð
where ð is the projection map
ðð = ðð à ðð â ðð , (ð¥, ðŠ) ⊠ð¥ .
Henceðâ1(0) = {0}Ãðð = ðð and by equation (4.1.5)), ðmapsðâ©ðâ1(0) diffeomorphicallyonto ð0 â© ðâ1(0). However, ð â© ðâ1(0) is a neighborhood of 0 in ðð and ð0 â© ðâ1(0) is aneighborhood ofð inð, and, as remarked, these two neighborhoods are diffeomorphic. â¡
Some examples:
Examples 4.1.9.(1) The ð-sphere: Let
ðⶠðð+1 â ðbe the map,
(ð¥1,âŠ, ð¥ð+1) ⊠ð¥21 +⯠+ ð¥2ð+1 â 1 .Then
ð·ð(ð¥) = 2(ð¥1,âŠ, ð¥ð+1)so, if ð¥ â 0, then ð is a submersion at ð¥. In particular, ð is a submersion at all points ð¥on the ð-sphere
ðð = ðâ1(0)so the ð-sphere is an ð-dimensional submanifold of ðð+1.
(2) Graphs: Let ðⶠðð â ðð be a ð¶â map and, as in (2), let
ð€ð â { (ð¥, ðŠ) â ðð à ðð | ðŠ = ð(ð¥) } .
We claim that ð€ð is an ð-dimensional submanifold of ðð+ð = ðð à ðð.
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100 Chapter 4: Manifolds & Forms on Manifolds
Proof. Letðⶠðð à ðð â ðð
be the map ð(ð¥, ðŠ) â ðŠ â ð(ð¥). Then
ð·ð(ð¥, ðŠ) = (âð·ð(ð¥) idð)
where idð is the identity map of ðð onto itself. This map is always of rank ð. Henceð€ð = ðâ1(0) is an ð-dimensional submanifold of ðð+ð. â¡
(3) Let M ð be the set of all ð à ð matrices and let ð®ð be the set of all symmetric ð à ðmatrices, i.e., the set
ð®ð = {ðŽ âM ð | ðŽ = ðŽâº } .The map
ðŽ = (ðð,ð) ⊠(ð1,1, ð1,2,âŠ, ð1,ð, ð2,1,âŠ, ð2,ð,âŠ)gives us an identification
M ð ⥲ ðð2
and the map
ðŽ = (ðð,ð) ⊠(ð1,1,âŠ, ð1,ð, ð2,2,âŠ, ð2,ð, ð3,3,âŠ, ð3,ð,âŠ)gives us an identification
ð®ð ⥲ ðð(ð+1)2 .
(Note that ifðŽ is a symmetric matrix, then ðð,ð = ðð,ð, so this map avoids redundancies.)Let
ð(ð) = {ðŽ âM ð | ðŽâºðŽ = idð } .This is the set of orthogonal ð à ðmatrices, and we will leave for you as an exercise toshow that itâs an ð(ð â 1)/2-manifold.
Hint: Let ðⶠM ð â ð®ð be the map ð(ðŽ) = ðŽâºðŽ â idð. Then
ð(ð) = ðâ1(0) .
These examples show that lots of interestingmanifolds arise as zero sets of submersionsðⶠð â ðð. This is, in fact, not just an accident. We will show that locally every manifoldarises this way. More explicitly let ð â ðð be an ð-manifold, ð a point of ð, ð a neighbor-hood of 0 in ðð, ð a neighborhood of ð in ðð and ðⶠ(ð, 0) ⥲ (ð â© ð, ð) a diffeomor-phism. We will for the moment think of ð as a ð¶â map ðⶠð â ðð whose image happensto lie inð.
Lemma 4.1.10. The linear mapð·ð(0)ⶠðð â ðð is injective.
Proof. Sinceðâ1 ⶠð â© ð ⥲ ð is a diffeomorphism, shrinkingð if necessary, we can assumethat there exists a ð¶â map ðⶠð â ð which coincides with ðâ1on ð â© ð. Since ðmaps ðonto ð â© ð, ð â ð = ðâ1 â ð is the identity map on ð. Therefore,
ð·(ð â ð)(0) = (ð·ð)(ð)ð·ð(0) = idðby the chain rule, and hence ifð·ð(0)ð£ = 0, it follows from this identity that ð£ = 0. â¡
Lemma 4.1.10 says that ð is an immersion at 0, so by the canonical immersion theorem(see Theorem b.18) there exists a neighborhood ð0 of 0 in ð, a neighborhood ðð of ð in ð,and a diffeomorphism
(4.1.11) ðⶠ(ðð, ð) ⥲ (ð0 à ððâð, 0)
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§4.1 Manifolds 101
such that ð â ð = ð, where ð is, as in Appendix b, the canonical immersion
(4.1.12) ð ⶠð0 â ð0 à ððâð , ð¥ ⊠(ð¥, 0) .By (4.1.11) ð maps ð(ð0) diffeomorphically onto ð(ð0). However, by (4.1.8) and (4.1.11)ð(ð0) is defined by the equations, ð¥ð = 0, ð = ð + 1,âŠ,ð. Hence if ð = (ð1,âŠ, ðð) the set,ð(ð0) = ðð â© ð is defined by the equations
(4.1.13) ðð = 0 , ð = ð + 1,âŠ,ð .
Let â = ð â ð, letðⶠðð = ðð à ðâ â ðâ
be the canonical submersion,
ð(ð¥1,âŠ, ð¥ð) = (ð¥ð+1,âŠ, ð¥ð)and let ð = ð â ð. Since ð is a diffeomorphism, ð is a submersion and (4.1.12) can beinterpreted as saying that
(4.1.14) ðð â© ð = ðâ1(0) .Thus to summarize we have proved:
Theorem 4.1.15. Letð be an ð-dimensional submanifold of ðð and let â â ðâ ð. Then forevery ð â ð there exists a neighborhood ðð of ð in ðð and a submersion
ðⶠ(ðð, ð) â (ðâ, 0)such thatð â© ðð is defined by equation (4.1.13).
A nice way of thinking about Theorem 4.1.2 is in terms of the coordinates of the map-ping, ð. More specifically if ð = (ð1,âŠ, ðð) we can think of ðâ1(ð) as being the set ofsolutions of the system of equations
ðð(ð¥) = ðð , ð = 1,âŠ, ðand the condition that ð be a regular value of ð can be interpreted as saying that for everysolution ð of this system of equations the vectors
(ððð)ð =ðâð=1
ððððð¥ð(0)ðð¥ð
in ðâð ðð are linearly independent, i.e., the system (4.1.14) is an âindependent system ofdefining equationsâ forð.
Exercises for §4.1Exercise 4.1.i. Show that the set of solutions of the system of equations
{ð¥21 +⯠+ ð¥2ð = 1ð¥1 +⯠+ ð¥ð = 0
is an (ð â 2)-dimensional submanifold of ðð.
Exercise 4.1.ii. Let ððâ1 â ðð be the (ð â 1)-sphere and let
ðð = { ð¥ â ððâ1 | ð¥1 +⯠+ ð¥ð = ð } .For what values of ð isðð an (ð â 2)-dimensional submanifold of ððâ1?
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Exercise 4.1.iii. Show that ifðð is an ðð-dimensional submanifold of ððð , for ð = 1, 2, thenð1 Ã ð2 â ðð1 Ã ðð2
is an (ð1 + ð2)-dimensional submanifold of ðð1 Ã ðð2 .Exercise 4.1.iv. Show that the set
ð = { (ð¥, ð£) â ððâ1 à ðð | ð¥ â ð£ = 0 }is a (2ðâ2)-dimensional submanifold ofððÃðð. (Here ð¥â ð£ â âðð=1 ð¥ðð£ð is the dot product.)
Exercise 4.1.v. Let ðⶠðð â ðð be a ð¶â map and let ð = ð€ð be the graph of ð. Provedirectly thatð is an ð-manifold by proving that the map
ðŸð ⶠðð â ð , ð¥ ⊠(ð¥, ð(ð¥))is a diffeomorphism.
Exercise 4.1.vi. Prove that the orthogonal group ð(ð) is an ð(ð â 1)/2-manifold.Hints:
⢠Let ðⶠM ð â ð®ð be the mapð(ðŽ) = ðŽâºðŽ â idð .
Show that ð(ð) = ðâ1(0).⢠Show that
ð(ðŽ + ððµ) = ðŽâºðŽ + ð(ðŽâºðµ + ðµâºðŽ) + ð2ðµâºðµ â idð .⢠Conclude that the derivative of ð at ðŽ is the map given by
(4.1.16) ðµ ⊠ðŽâºðµ + ðµâºðŽ .⢠Let ðŽ be in ð(ð). Show that if ð¶ is in ð®ð and ðµ = ðŽð¶/2 then the map (4.1.16)
maps ðµ onto ð¶.⢠Conclude that the derivative of ð is surjective at ðŽ.⢠Conclude that 0 is a regular value of the mapping ð.
The next five exercises, which are somewhat more demanding than the exercises above,are an introduction to Grassmannian geometry.
Exercise 4.1.vii.(1) Let ð1,âŠ, ðð be the standard basis of ðð and letð = span(ðð+1,âŠ, ðð). Prove that if ð
is a ð-dimensional subspace of ðð and(4.1.17) ð â©ð = 0 ,
then one can find a unique basis of ð of the form
(4.1.18) ð£ð = ðð +ââð=1ðð,ððð+ð , ð = 1,âŠ, ð ,
where â = ð â ð.(2) Let ðºð be the set of ð-dimensional subspaces of ðð having the property (4.1.17) and
let M ð,â be the vector space of ð Ã â matrices. Show that one gets from the identities(4.1.18) a bijective map:
(4.1.19) ðŸâ¶ M ð,â â ðºð .Exercise 4.1.viii. Let ðð be the vector space of linear mappings of ðð into itself which areself-adjoint, i.e., have the property ðŽ = ðŽâº.
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§4.1 Manifolds 103
(1) Given a ð-dimensional subspace,ð ofðð letðð ⶠðð â ðð be the orthogonal projectionof ðð onto ð. Show that ðð is in ðð and is of rank ð, and show that (ðð)2 = ðð.
(2) Conversely supposeðŽ is an element of ðð which is of rank ð and has the property,ðŽ2 =ðŽ. Show that if ð is the image of ðŽ in ðð, then ðŽ = ðð.
Definition 4.1.20. We call anðŽ â ðð of the formðŽ = ðð above a rank ð projection operator.
Exercise 4.1.ix. Composing the map
ðⶠðºð â ðð , ð ⊠ððwith the map (4.1.19) we get a map
ðⶠM ð,â â ðð , ð = ð â ðŸ .Prove that ð is ð¶â.
Hints:⢠By GramâSchmidt one can convert (4.1.18) into an orthonormal basis
ð1,ðµ,âŠ, ðð,ðµof ð. Show that the ðð,ðµâs are ð¶â functions of the matrix ðµ = (ðð,ð).
⢠Show that ðð is the linear mapping
ð£ âŠðâð=1(ð£ â ðð,ðµ)ðð,ðµ .
Exercise 4.1.x. Let ð0 = span(ð1,âŠ, ðð) and let ðºð = ð(ðºð). Show that ðmaps a neighbor-hood of ð0 in M ð,â diffeomorphically onto a neighborhood of ðð0 in ðºð.
Hints: ðð is in ðºð if and only if ð satisfies (4.1.17). For 1 †ð †ð let
ð€ð = ðð(ðð) =ðâð=1ðð,ððð +
ââð=1ðð,ððð+ð .
⢠Show that if the matrix ðŽ = (ðð,ð) is invertible, ðð is in ðºð.⢠Let ð â ðºð be the set of all ððâs for which ðŽ is invertible. Show that ðâ1 ⶠð â
M ð,â is the mapðâ1(ðð) = ðµ = ðŽâ1ð¶
where ð¶ = [ðð,ð].
Exercise 4.1.xi. Letðº(ð, ð) â ðð be the set of rank ð projection operators. Prove thatðº(ð, ð)is a ðâ-dimensional submanifold of the Euclidean space ðð â ð
ð(ð+1)2 .
Hints:⢠Show that if ð is any ð-dimensional subspace of ðð there exists a linear mapping,ðŽ â ð(ð)mapping ð0 to ð.
⢠Show that ðð = ðŽðð0ðŽâ1.
⢠Let ðŸðŽ ⶠðð â ðð be the linear mapping,
ðŸðŽ(ðµ) = ðŽðµðŽâ1 .Show that
ðŸðŽ â ðⶠM ð,â â ððmaps a neighborhood of ð0 in M ð,â diffeomorphically onto a neighborhood ofðð in ðº(ð, ð).
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104 Chapter 4: Manifolds & Forms on Manifolds
Remark 4.1.21. Let Gr(ð, ð) be the set of all ð-dimensional subspaces of ðð. The identifi-cation of Gr(ð, ð) with ðº(ð, ð) given by ð ⊠ðð allows us to restate the result above in theform.
Theorem4.1.22. TheGrassmannianGr(ð, ð) of ð-dimensional subspaces ofðð is a ðâ-dimens-ional submanifold of ðð â ð
ð(ð+1)2 .
Exercise 4.1.xii. Show that Gr(ð, ð) is a compact submanifold of ðð.Hint: Show that itâs closed and bounded.
4.2. Tangent spaces
We recall that a subset ð of ðð is an ð-dimensional manifold if for every ð â ð thereexists an open setð â ðð, a neighborhoodð ofð inðð, and að¶â-diffeomorphismðⶠð ⥲ð â© ð.
Definition 4.2.1. We will call ð a parametrization ofð at ð.
Our goal in this section is to define the notion of the tangent space ððð to ð at ð anddescribe some of its properties. Before giving our official definitionweâll discuss some simpleexamples.
Example 4.2.2. Let ðⶠð â ð be a ð¶â function and letð = ð€ð be the graph of ð.
ᅵᅵ
ᅵ
ᅵ
â
ᅵ0
Figure 4.2.1. The tangent line â to the graph of ð at ð0
Then in this figure above the tangent line â to ð at ð0 = (ð¥0, ðŠ0) is defined by theequation
ðŠ â ðŠ0 = ð(ð¥ â ð¥0)where ð = ðâ²(ð¥0) In other words if ð is a point on â then ð = ð0 + ðð£0 where ð£0 = (1, ð)and ð â ð. We would, however, like the tangent space to ð at ð0 to be a subspace of thetangent space to ð2 at ð0, i.e., to be the subspace of the space: ðð0ð
2 = {ð0} Ã ð2, and thisweâll achieve by defining
ðð0ð â { (ð0, ðð£0) | ð â ð } .
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§4.2 Tangent spaces 105
Example 4.2.3. Let ð2 be the unit 2-sphere in ð3. The tangent plane to ð2 at ð0 is usuallydefined to be the plane
{ ð0 + ð£ | ð£ â ð3 and ð£ â ð0 } .However, this tangent plane is easily converted into a subspace ofððð3 via themapð0+ð£ âŠ(ð0, ð£) and the image of this map
{ (ð0, ð£) | ð£ â ð3 and ð£ â ð0 } .will be our definition of ðð0ð
2.
Letâs now turn to the general definition. As above let ð be an ð-dimensional submani-fold of ðð, ð a point ofð, ð a neighborhood of ð in ðð, ð an open set in ðð and
ðⶠ(ð, ð) ⥲ (ð â© ð, ð)a parameterization ofð. We can think of ð as a ð¶â map
ðⶠ(ð, ð) â (ð, ð)whose image happens to lie inð â© ð and we proved in §4.1 that its derivative at ð
(ðð)ð ⶠðððð â ðððð
is injective.
Definition 4.2.4. The tangent space ððð toð at ð is the image of the linear map
(ðð)ð ⶠðððð â ðððð .
In other words, ð€ â ðððð is in ððð if and only if ð€ = ððð(ð£) for some ð£ â ðððð. Moresuccinctly,
(4.2.5) ððð â (ððð)(ðððð) .
(Since ððð is injective, ððð is an ð-dimensional vector subspace of ðððð.)
One problem with this definition is that it appears to depend on the choice of ð. Toget around this problem, weâll give an alternative definition of ððð. In §4.1 we showed thatthere exists a neighborhood ð of ð in ðð (which we can without loss of generality take tobe the same as ð above) and a ð¶â map
(4.2.6) ðⶠ(ð, ð) â (ðð, 0) , where ð = ð â ð ,such that ð â© ð = ðâ1(0) and such that ð is a submersion at all points of ð â© ð, and inparticular at ð. Thus
ððð ⶠðððð â ð0ðð
is surjective, and hence the kernel of ððð has dimension ð. Our alternative definition ofðððis
(4.2.7) ððð â ker(ððð) .
Lemma 4.2.8. The spaces (4.2.5) and (4.2.7) are both ð-dimensional subspaces of ðððð, andwe claim that these spaces are the same.
(Notice that the definition (4.2.7) of ððð does not depend on ð, so if we can show thatthese spaces are the same, the definitions (4.2.5) and (4.2.7)) will depend neither on ð noron ð.)
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106 Chapter 4: Manifolds & Forms on Manifolds
Proof. Since ð(ð) is contained inð â© ð andð â© ð is contained in ðâ1(0), ð â ð = 0, so bythe chain rule(4.2.9) ððð â ððð = ð(ð â ð)ð = 0 .Hence if ð£ â ðððð and ð€ = ððð(ð£), ððð(ð€) = 0. This shows that the space (4.2.5) iscontained in the space (4.2.7). However, these two spaces are ð-dimensional so they co-incide. â¡
From the proof above one can extract a slightly stronger result.
Theorem 4.2.10. Letð be an open subset of ðâ and âⶠ(ð, ð) â (ðð, ð) a ð¶â map. Sup-pose â(ð) is contained inð. Then the image of the map
ðâð ⶠðððâ â ðððð
is contained in ððð.Proof. Let ð be the map (4.2.6). We can assume without loss of generality that â(ð) is con-tained inð, and so, by assumption, â(ð) â ðâ©ð. Therefore, as above, ðââ = 0, and henceðâð(ðððâ) is contained in the kernel of ððð. â¡
This result will enable us to define the derivative of a mapping between manifolds. Ex-plicitly: Let ð be a submanifold of ðð, ð a submanifold of ðð, and ðⶠ(ð, ð) â (ð, ðŠ0)a ð¶â map. By Definition 4.1.1 there exists a neighborhood ð of ð in ðð and a ð¶â mapðⶠð â ðð extending to ð. We will define(4.2.11) (ðð)ð ⶠððð â ððŠ0ðto be the restriction of the map
(ðð)ð ⶠðððð â ððŠ0ðð
to ððð. There are two obvious problems with this definition:(1) Is the space (ðð)ð(ððð) contained in ððŠ0ð?(2) Does the definition depend on ð?
To show that the answer to (1) is yes and the answer to (2) is no, letðⶠ(ð, ð¥0) ⥲ (ð â© ð, ð)
be a parametrization of ð, and let â = ð â ð. Since ð(ð) â ð, â(ð) â ð and hence byTheorem 4.2.10
ðâð¥0(ðð¥0ðð) â ððŠ0ð .
But by the chain ruleðâð¥0 = ððð â ððð¥0 ,
so by (4.2.5)(ððð)(ððð) â ððð
and(ððð)(ððð) = (ðâ)ð¥0(ðð¥0ð
ð)Thus the answer to (1) is yes, and since â = ð â ð = ð â ð, the answer to (2) is no.
From equations (4.2.9) and (4.2.11) one easily deduces:
Theorem 4.2.12 (Chain rule for mappings between manifolds). Let ð be a submanifold ofðâ and ðⶠ(ð, ðŠ0) â (ð, ð§0) a ð¶â map. Then ðððŠ0 â ððð = ð(ð â ð)ð.
Wewill next provemanifold versions of the inverse function theorem and the canonicalimmersion and submersion theorems.
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§4.2 Tangent spaces 107
Theorem 4.2.13 (Inverse function theorem for manifolds). Letð andð be ð-manifolds andðⶠð â ð a ð¶â map. Suppose that at ð â ð the map
ððð ⶠððð â ððð , ð â ð(ð) ,
is bijective. Then ðmaps a neighborhood ð of ð in ð diffeomorphically onto a neighborhoodð of ð in ð.
Proof. Letð andð be open neighborhoods ofð inð and ð inð and letð0 ⶠ(ð0, 0) â (ð, ð)and ð0 ⶠ(ð0, 0) â (ð, ð) be parametrizations of these neighborhoods. Shrinking ð0 and ðwe can assume that ð(ð) â ð. Let
ðⶠ(ð0, ð0) â (ð0, ð0)
be the map ðâ10 â ð â ð0. Then ð0 â ð = ð â ð, so by the chain rule
(ðð0)ð0 â (ðð)ð0 = (ðð)ð â (ðð0)ð0 .
Since (ðð0)ð0 and (ðð0)ð0 are bijective itâs clear from this identity that if ððð is bijectivethe same is true for (ðð)ð0 . Hence by the inverse function theorem for open subsets of ðð,ð maps a neighborhood of ð0 in ð0 diffeomorphically onto a neighborhood of ð0 in ð0.Shrinkingð0 andð0 we assume that these neighborhoods areð0 andð0 and hence that ð isa diffeomorphism. Thus since ðⶠð â ð is the map ð0 â ð â ðâ10 , it is a diffeomorphism aswell. â¡
Theorem 4.2.14 (The canonical submersion theorem for manifolds). Let ð and ð be man-ifolds of dimension ð andð, whereð < ð, and let ðⶠð â ð be a ð¶â map. Suppose that atð â ð the map
ððð ⶠððð â ððð , ð â ð(ð) ,is surjective. Then there exists an open neighborhood ð of ð in ð, and open neighborhood, ðof ð(ð) in ð and parametrizations ð0 ⶠ(ð0, 0) â (ð, ð) and ð0 ⶠ(ð0, 0) â (ð, ð) such thatin the square
ð0 ð0
ð ð
ð0
ðâ10 ðð0
ð0
ð
the map ðâ10 â ð â ð0 is the canonical submersion ð.
Proof. Let ð and ð be open neighborhoods of ð and ð and ð0 ⶠ(ð0, 0) â (ð, ð) andð0 ⶠ(ð0, 0) â (ð, ð) be parametrizations of these neighborhoods. Composing ð0 and ð0with the translations we can assume that ð0 is the origin in ðð and ð0 the origin in ðð,and shrinking ð we can assume ð(ð) â ð. As above let ðⶠ(ð0, 0) â (ð0, 0) be the mapðâ10 â ð â ð0. By the chain rule
(ðð0)0 â (ðð)0 = ððð â (ðð0)0 ,
therefore, since (ðð0)0 and (ðð0)0 are bijective it follows that (ðð)0 is surjective. Hence, byTheorem 4.1.2, we can find an open neighborhood ð of the origin in ðð and a diffeomor-phism ð1 ⶠ(ð1, 0) ⥲ (ð0, 0) such that ð â ð1 is the canonical submersion. Now replace ð0by ð1 and ð0 by ð0 â ð1. â¡
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108 Chapter 4: Manifolds & Forms on Manifolds
Theorem 4.2.15 (The canonical immersion theorem for manifolds). Let ð and ð be mani-folds of dimension ð andð, where ð < ð, and ðⶠð â ð a ð¶â map. Suppose that at ð â ðthe map
ððð ⶠððð â ððð , ð = ð(ð)
is injective. Then there exists an open neighborhood ð of ð in ð, an open neighborhood ð ofð(ð) in ð and parametrizations ð0 ⶠ(ð0, 0) â (ð, ð) and ð0 ⶠ(ð0, 0) â (ð, ð) such that inthe square
ð0 ð0
ð ð
ð0
ðâ10 ðð0
ð0
ð
the map ðâ10 â ð â ð0 is the canonical immersion ð.
Proof. The proof is identical with the proof of Theorem 4.2.14 except for the last step. In thelast step one converts ð into the canonical immersion via a map ð1 ⶠ(ð1, 0) â (ð0, 0) withthe property ð â ð1 = ð and then replaces ð0 by ð0 â ð1. â¡
Exercises for §4.2Exercise 4.2.i. What is the tangent space to the quadric
ð = { (ð¥1,âŠ, ð¥ð) â ðð | ð¥ð = ð¥21 +⯠+ ð¥2ðâ1 }
at the point (1, 0,âŠ, 0, 1)?
Exercise 4.2.ii. Show that the tangent space to the (ð â 1)-sphere ððâ1 at ð is the space ofvectors (ð, ð£) â ðððð satisfying ð â ð£ = 0.
Exercise 4.2.iii. Let ðⶠðð â ðð be a ð¶â map and let ð = ð€ð. What is the tangent spacetoð at (ð, ð(ð))?
Exercise 4.2.iv. Let ðⶠððâ1 â ððâ1 be the antipodal map ð(ð¥) = âð¥. What is the derivativeof ð at ð â ððâ1?
Exercise 4.2.v. Letðð â ððð , ð = 1, 2, be an ðð-manifold and let ðð â ðð. Defineð to be theCartesian product
ð1 Ã ð2 â ðð1 Ã ðð2
and let ð = (ð1, ð2). Show that ððð â ðð1ð1 â ðð2ð2.
Exercise 4.2.vi. Let ð â ðð be an ð-manifold and for ð = 1, 2, let ðð ⶠðð â ð â© ðð be twoparametrizations. From these parametrizations one gets an overlap diagram
(4.2.16)ðâ11 (ð â© ð1 â© ð2) ðâ12 (ð â© ð1 â© ð2)
ð â© ð1 â© ð2 .
ðâ12 âð1
ð1 ð2
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§4.3 Vector fields & differential forms on manifolds 109
(1) Let ð â ðâ©ð and let ðð = ðâ1ð (ð). Derive from the overlap diagram (4.2.16) an overlapdiagram of linear maps
ðð1ðð ðð2ð
ð
ðððð .
ð(ðâ12 âð1)ð1
ð(ð1)ð1 ð(ð2)ð2
(2) Use overlap diagrams to give another proof that ððð is intrinsically defined.
4.3. Vector fields & differential forms on manifolds
A vector field on an open subset ð â ðð is a function ð which assigns to each ð â ð anelement ð(ð) of ððð, and a ð-form is a function ð which assigns to each ð â ð an elementðð of ð¬ð(ðâðð). These definitions have obvious generalizations to manifolds:
Definition 4.3.1. Let ð be a manifold. A vector field on ð is a function ð which assigns toeach ð â ð an element ð(ð) of ððð. A ð-form is a function ð which assigns to each ð â ðan element ðð of ð¬ð(ðâðð).
Weâll begin our study of vector fields and ð-forms on manifolds by showing that, liketheir counterparts on open subsets of ðð, they have nice pullback and pushforward proper-ties with respect to mappings. Letð and ð be manifolds and ðⶠð â ð a ð¶â mapping.
Definition 4.3.2. Given a vector field ð onð and a vector fieldð on ð, weâll say that ð andð are ð-related if for all ð â ð and ð = ð(ð) we have
(4.3.3) (ðð)ðð(ð) = ð(ð) .
In particular, if ð is a diffeomorphism, and weâre given a vector field ð on ð we candefine a vector field ð on ð by requiring that for every point ð â ð, the identity (4.3.3)holds at the point ð = ðâ1(ð). In this case weâll call ð the pushforward of by ð and denoteit byðâð. Similarly, given a vector fieldð onðwe can define a vector field ð onð by applyingthe same construction to the inverse diffeomorphism ðâ1 ⶠð â ð. We will call the vectorfield (ðâ1)âð the pullback of ð by ð (and also denote it by ðâð).
For differential forms the situation is even nicer. Just as in §2.6 we can define the pull-back operation on forms for any ð¶â map ðⶠð â ð. Specifically: Let ð be a ð-form on ð.For every ð â ð, and ð = ð(ð) the linear map
ððð ⶠððð â ðððinduces by (1.8.2)a pullback map
(ððð)â ⶠð¬ð(ðâð ð) â ð¬ð(ðâðð)and, as in §2.6, weâll define the pullback ðâð of ð toð by defining it at ð by the identity
(ðâð)(ð) = (ððð)âð(ð) .The following results about these operations are proved in exactly the same way as in
§2.6.
Proposition 4.3.4. Let ð, ð, and ð be manifolds and ðⶠð â ð and ðⶠð â ð ð¶â maps.Then if ð is a ð-form on ð
ðâ(ðâð) = (ð â ð)âð .
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110 Chapter 4: Manifolds & Forms on Manifolds
If ð is a vector field onð and ð and ð are diffeomorphisms, then(4.3.5) (ð â ð)âð£ = ðâ(ðâð£) .
Our first application of these identities will be to definewhat onemeans by a âð¶â vectorfieldâ and a âð¶â ð-formâ.
Definition4.3.6. Letðbe anð-manifold andð anopen subset ofð.The setð is a parametriz-able open set if there exists an open set ð0 in ðð and a diffeomorphism ð0 ⶠð0 â ð.
In other words,ð is parametrizable if there exists a parametrization havingð as its im-age. (Note thatð being a manifold means that every point is contained in a parametrizableopen set.)
Now let ð â ð be a parametrizable open set and ð0 ⶠð0 â ð a parametrization of ð.
Definition 4.3.7. A ð-form ð on ð is ð¶â or smooth if ðâ0ð is a ð¶â.
This definition appears to depend on the choice of the parametrization ð, but we claimit does not. To see this let ð1 ⶠð1 â ð be another parametrization of ð and let
ðⶠð0 â ð1be the composite map ðâ10 â ð0. Then ð0 = ð1 â ð and hence by Proposition 4.3.4
ðâ0ð = ðâðâ1ð ,so by (2.5.11) ðâ0ð is ð¶â if ðâ1ð is ð¶â. The same argument applied to ðâ1 shows that ðâ1ðis ð¶â if ðâ0ð is ð¶â.
The notion of âð¶ââ for vector fields is defined similarly:
Definition 4.3.8. A vector field ð on ð is smooth, or simply ð¶â, if ðâ0ð is ð¶â.
By Proposition 4.3.4 ðâ0ð = ðâðâ1ð, so, as above, this definition is independent of thechoice of parametrization.
We now globalize these definitions.
Definition 4.3.9. A ð-form ð on ð is ð¶â if, for every point ð â ð, ð is ð¶â on a neigh-borhood of ð. Similarly, a vector field ð on ð is ð¶â if, for every point ð â ð, ð is ð¶â on aneighborhood ofð.
We will also use the identities (4.3.5) and (4.3.18) to prove the following two results.
Proposition 4.3.10. Letð and ð be manifolds and ðⶠð â ð a ð¶â map. Then if ð is a ð¶âð-form on ð, the pullback ðâð is a ð¶â ð-form onð.
Proof. For ð â ð and ð = ð(ð) let ð0 ⶠð0 â ð and ð0 ⶠð0 â ð be parametrizations withð â ð and ð â ð. Shrinking ð if necessary we can assume that ð(ð) â ð. Let ðⶠð0 â ð0be the map ð = ðâ10 â ð â ð0. Then ð0 â ð = ð â ð0, so ðâðâ0 ð = ðâ0ðâð. Since ð is ð¶â, wesee that ðâ0 ð is ð¶â, so by equation (2.6.11) we have ðâðâ0 ð is ð¶â, and hence ðâ0ðâð is ð¶â.Thus by definition ðâð is ð¶â on ð. â¡
By exactly the same argument one proves.
Proposition 4.3.11. If ð is a ð¶â vector field on ð and ð is a diffeomorphism, then ðâð is að¶â vector field onð.
Notation 4.3.12.(1) Weâll denote the space of ð¶â ð-forms onð by ðºð(ð) .
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§4.3 Vector fields & differential forms on manifolds 111
(2) For ð â ðºð(ð) weâll define the support of ð bysupp(ð) â { ð â ð | ð(ð) â 0 }
and weâll denote by ðºðð (ð) â ðºð(ð) the space of compactly supported ð-forms.(3) For a vector field ð onð weâll define the support of ð to be the set
supp(ð) â { ð â ð | ð(ð) â 0 } .We will now review some of the results about vector fields and the differential forms
that we proved in Chapter 2 and show that they have analogues for manifolds.
Integral curvesLet ðŒ â ð be an open interval and ðŸâ¶ ðŒ â ð a ð¶â curve. For ð¡0 â ðŒ we will call
ᅵᅵ = (ð¡0, 1) â ðð¡0ð the unit vector in ðð¡0ð and if ð = ðŸ(ð¡0) we will call the vectorððŸð¡0(ᅵᅵ) â ððð
the tangent vector to ðŸ at ð. If ð is a vector field on ð we will say that ðŸ is an integral curveof ð if for all ð¡0 â ðŒ
ð(ðŸ(ð¡0)) = ððŸð¡0(ᅵᅵ) .Proposition 4.3.13. Let ð and ð be manifolds and ðⶠð â ð a ð¶â map. If ð and ð arevector fields on ð and ð which are ð-related, then integral curves of ð get mapped by ð ontointegral curves of ð.Proof. If the curve ðŸâ¶ ðŒ â ð is an integral curve of ð we have to show that ð â ðŸâ¶ ðŒ â ð isan integral curve of ð. If ðŸ(ð¡) = ð and ð = ð(ð) then by the chain rule
ð(ð) = ððð(ð(ð)) = ððð(ððŸð¡(ᅵᅵ))= ð(ð â ðŸ)ð¡(ᅵᅵ) . â¡
From this result it follows that the local existence, uniqueness and âsmooth dependenceon initial dataâ results about vector fields that we described in §2.1 are true for vector fieldson manifolds. More explicitly, let ð be a parametrizable open subset of ð and ðⶠð0 â ða parametrization. Sinceð0 is an open subset of ðð these results are true for the vector fieldð = ðâ0ð and hence since ð and ð are ð0-related they are true for ð. In particular:
Proposition 4.3.14 (local existence). For every ð â ð there exists an integral curve ðŸ(ð¡)defined for âð < ð¡ < ð, of ð with ðŸ(0) = ð.Proposition 4.3.15 (local uniqueness). For ð = 1, 2, let ðŸð ⶠðŒð â ð be integral curves of ðand let ðŒ = ðŒ1 â© ðŒ2. Suppose ðŸ2(ð¡) = ðŸ1(ð¡) for some ð¡ â ðŒ. Then there exists a unique integralcurve, ðŸâ¶ ðŒ1 ⪠ðŒ2 â ð with ðŸ = ðŸ1 on ðŒ1 and ðŸ = ðŸ2 on ðŒ2.Proposition 4.3.16 (smooth dependence on initial data). For every ð â ð, there exists aneighborhood ð of ð in ð, an interval (âð, ð) and a ð¶â map âⶠð à (âð, ð) â ð such thatfor every ð â ð the curve
ðŸð(ð¡) = â(ð, ð¡) , â ð < ð¡ < ð ,is an integral curve of ð with ðŸð(0) = ð.
As in Chapter 2 we will say that ð is complete if, for every ð â ð there exists an integralcurve ðŸ(ð¡) defined for ââ < ð¡ < â, with ðŸ(0) = ð. In Chapter 2 we showed that one simplecriterium for a vector field to be complete is that it be compactly supported. We will provethat the same is true for manifolds.
Theorem 4.3.17. Ifð is compact or, more generally, if ð is compactly supported, ð is complete.
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112 Chapter 4: Manifolds & Forms on Manifolds
Proof. Itâs not hard to prove this by the same argument that we used to prove this theoremfor vector fields on ðð, but weâll give a simpler proof that derives this directly from the ððresult. Supposeð is a submanifold of ðð. Then for ð â ð,
ððð â ðððð = { (ð, ð£) | ð£ â ðð } ,
so ð(ð) can be regarded as a pair (ð, ð(ð)), where ð(ð) is in ðð. Let
(4.3.18) ðð ⶠð â ðð
be the map ðð(ð) = ð£(ð). It is easy to check that ð is ð¶â if and only if ðð is ð¶â. (SeeExercise 4.3.xi.) Hence (see Appendix b) there exists a neighborhood, ð of ð and a mapðⶠâ ðð extending ðð. Thus the vector field ð on ð defined by ð(ð) = (ð, ð(ð)) extendsthe vector field ð toð. In other words if ð ⶠð ⪠ð is the inclusionmap ð andð are ð-related.Thus by Proposition 4.3.13 the integral curves of ð are just integral curves of ð that arecontained inð.
Suppose now that ð is compactly supported. Then there exists a function ð â ð¶â0 (ð)which is 1 on the support of ð, so, replacing ð by ðð, we can assume that ð is compactlysupported. Thus ð is complete. Let ðŸ(ð¡), defined for ââ < ð¡ < â, be an integral curve ofð. We will prove that if ðŸ(0) â ð, then this curve is an integral curve of ð. We first observe:
Lemma 4.3.19. The set of points ð¡ â ð for which ðŸ(ð¡) â ð is both open and closed.
Proof of Lemma 4.3.19. If ð â supp(ð) then ð(ð) = 0 so if ðŸ(ð¡) = ð, ðŸ(ð¡) is the constantcurve ðŸ = ð, and thereâs nothing to prove. Thus we are reduced to showing that the set
(4.3.20) { ð¡ â ð | ðŸ(ð¡) â supp(ð) }
is both open and closed. Since supp(ð) is compact this set is clearly closed. To show that itâsopen suppose ðŸ(ð¡0) â supp(ð). By local existence there exist an interval (âð + ð¡0, ð + ð¡0) andan integral curve ðŸ1(ð¡) of ð defined on this interval and taking the value ðŸ1(ð¡0) = ðŸ(ð¡0) at ð.However since ð andð are ð-related ðŸ1 is also an integral curve ofð and so it has to coincidewith ðŸ on the interval (âð + ð¡0, ð + ð¡0). In particular, for ð¡ on this interval ðŸ(ð¡) â supp(ð) sothe set (4.3.20) is open. â¡
To conclude the proof of Theorem 4.3.17 we note that since ð is connected it followsthat if ðŸ(ð¡0) â ð for some ð¡0 â ð then ðŸ(ð¡) â ð for all ð¡ â ð, and hence ðŸ is an integral curveof ð. Thus in particular every integral curve of ð exists for all time, so ð is complete. â¡
Sinceð is complete it generates a one-parameter group of diffeomorphismsðð¡ ⶠð ⥲ ð,defined for ââ < ð¡ < â, having the property that the curve
ðð¡(ð) = ðŸð(ð¡) , ââ < ð¡ < â
is the unique integral curve of ð with initial point ðŸð(0) = ð. But if ð â ð this curve is anintegral curve of ð, so the restriction
ðð¡ = ðð¡|ðis a one-parameter group of diffeomorphisms of ð with theproperty that for ð â ð thecurve
ðð¡(ð) = ðŸð(ð¡) , ââ < ð¡ < âis the unique integral curve of ð with initial point ðŸð(0) = ð.
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§4.3 Vector fields & differential forms on manifolds 113
The exterior differentiation operationLet ð be að¶â ð-form onð andð â ð a parametrizable open set. Given a parametriza-
tion ð0 ⶠð0 â ð we define the exterior derivative ðð of ð onð by the formula(4.3.21) ðð = (ðâ10 )âððâ0ð .(Notice that sinceð0 is an open subset of ðð and ðâ0ð a ð-form onð0, the âðâ on the right iswell-defined.) We claim that this definition does not depend on the choice of parametriza-tion. To see this let ð1 ⶠð1 â ð be another parametrization of ð and let ðⶠð0 â ð1 bethe diffeomorphism ðâ11 â ð0. Then ð0 = ð1 â ð and hence
ððâ0ð = ððâðâ1ð = ðâððâ1ð= ðâ0 (ðâ11 )âððâ1ð
so(ðâ10 )âððâ0ð = (ðâ11 )âððâ1ð
as claimed. We can, therefore, define the exterior derivative ðð globally by defining it to beequal to (4.3.21) on every parametrizable open set.
Itâs easy to see from the definition (4.3.21) that this exterior differentiation operationinherits from the exterior differentiation operation on open subsets of ðð the properties(2.4.3) and (2.4.4) and that for zero forms, i.e.,ð¶â functionsðⶠð â ð, ðð is the âintrinsicâðð defined in Section 2.1, i.e., for ð â ð, ððð is the derivative of ð
ððð ⶠððð â ð
viewed as an element of ð¬1(ðâðð). Letâs check that it also has the property (2.6.12).
Theorem 4.3.22. Letð and ð be manifolds and ðⶠð â ð a ð¶â map. Then for ð â ðºð(ð)we have
ðâ(ðð) = ð(ðâð) .
Proof. For every ð â ð weâll check that this equality holds in a neighborhood of ð. Letð = ð(ð) and let ð and ð be parametrizable neighborhoods of ð and ð. Shrinking ð ifnecessary we can assume ð(ð) â ð. Given parametrizations
ðⶠð0 â ðand
ðⶠð0 â ðwe get by composition a map
ðⶠð0 â ð0 , ð = ðâ1 â ð â ðwith the property ð â ð = ð â ð. Thus
ðâð(ðâð) = ððâðâð (by definition of ð)= ð(ð â ð)âð= ð(ð â ð)âð= ððâ(ðâð)= ðâððâð (by (2.6.12))= ðâðâðð (by definition of ð)= ðâðâðð .
Hence ððâð = ðâðð. â¡
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114 Chapter 4: Manifolds & Forms on Manifolds
The interior product and Lie derivative operationGiven a ð-form ð â ðºð(ð) and a ð¶â vector field ð we will define the interior product
(4.3.23) ððð â ðºðâ1(ð) ,as in §2.5, by setting
(ððð)ð = ðð(ð)ððand the Lie derivative ð¿ðð â ðºð(ð) by setting
(4.3.24) ð¿ðð â ðððð + ðððð .Itâs easily checked that these operations satisfy Properties 2.5.3 and Properties 2.5.10 (since,just as in §2.5, these identities are deduced from the definitions (4.3.23) and (4.3.3) by purelyformal manipulations). Moreover, if ð is complete and
ðð¡ ⶠð â ð , ââ < ð¡ < âis the one-parameter group of diffeomorphisms of ð generated by ð the Lie derivative op-eration can be defined by the alternative recipe
(4.3.25) ð¿ðð =ððð¡ðâð¡ ð|ð¡=0
as in (2.6.22). (Just as in §2.6 one proves this by showing that the operation (4.3.25) hassatisfies Properties 2.5.10 and hence that it agrees with the operation (4.3.24) provided thetwo operations agree on zero-forms.)
Exercises for §4.3Exercise 4.3.i. Letð â ð3 be the paraboloid defined by ð¥3 = ð¥21 +ð¥22 and letð be the vectorfield
ð â ð¥1ððð¥1+ ð¥2ððð¥2+ 2ð¥3ððð¥3
.
(1) Show that ð is tangent toð and hence defines by restriction a vector field ð onð.(2) What are the integral curves of ð?
Exercise 4.3.ii. Let ð2 be the unit 2-sphere, ð¥21 + ð¥22 + ð¥23 = 1, in ð3 and let ð be the vectorfield
ð â ð¥1ððð¥2â ð¥2ððð¥1
.
(1) Show that ð is tangent to ð2, and hence by restriction defines a vector field ð on ð2.(2) What are the integral curves of ð?
Exercise 4.3.iii. As in Exercise 4.3.ii let ð2 be the unit 2-sphere inð3 and letð be the vectorfield
ð â ððð¥3â ð¥3 (ð¥1
ððð¥1+ ð¥2ððð¥2+ ð¥3ððð¥3)
(1) Show that ð is tangent to ð2 and hence by restriction defines a vector field ð on ð2.(2) What do its integral curves look like?
Exercise 4.3.iv. Let ð1 be the unit circle, ð¥21 + ð¥22 = 1, in ð2 and let ð = ð1 à ð1 in ð4 withdefining equations
ð1 = ð¥21 + ð¥22 â 1 = 0ð2 = ð¥23 + ð¥24 â 1 = 0 .
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§4.3 Vector fields & differential forms on manifolds 115
(1) Show that the vector field
ð â ð¥1ððð¥2â ð¥2ððð¥1+ ð(ð¥4
ððð¥3â ð¥3ððð¥4) ,
ð â ð, is tangent toð and hence defines by restriction a vector field ð onð.(2) What are the integral curves of ð?(3) Show that ð¿ððð = 0.Exercise 4.3.v. For the vector field ð in Exercise 4.3.iv, describe the one-parameter groupof diffeomorphisms it generates.
Exercise 4.3.vi. Letð and ðbe as in Exercise 4.3.i and letðⶠð2 â ðbe themapð(ð¥1, ð¥2) =(ð¥1, ð¥2, ð¥21 + ð¥22). Show that if ð is the vector field,
ð â ð¥1ððð¥1+ ð¥2ððð¥2
,
then ðâð = ð.
Exercise 4.3.vii. Letð be a submanifold ofð in ðð and let ð andð be the vector fields onð and ð. Denoting by ð the inclusion map of ð into ð, show that ð and ð are ð-related ifand only if ð is tangent toð and its restriction toð is ð.
Exercise 4.3.viii. Letð be a submanifold of ðð and ð an open subset of ðð containingð,and let ð and ð be the vector fields on ð and ð. Denoting by ð the inclusion map of ð intoð, show that ð and ð are ð-related if and only if ð is tangent toð and its restriction toð isð.Exercise 4.3.ix. An elementary result in number theory asserts:
Theorem 4.3.26. A number ð â ð is irrational if and only if the set{ð + ðð |ð, ð â ð }
is a dense subset of ð.Let ð be the vector field in Exercise 4.3.iv. Using Theorem 4.3.26 prove that if ð is irra-
tional then for every integral curve ðŸ(ð¡), defined for ââ < ð¡ < â, of ð the set of points onthis curve is a dense subset ofð.Exercise 4.3.x. Let ð be an ð-dimensional submanifold of ðð. Prove that a vector field ðonð is ð¶â if and only if the map (4.3.18) is ð¶â.
Hint: Letð be a parametrizable open subset ofð and ðⶠð0 â ð a parametrization ofð. Composing ð with the inclusion map ð ⶠð â ðð one gets a map ð â ðⶠð â ðð. Showthat if
ðâð = âð£ðððð¥ð
thenðâðð = â
ððððð¥ðð£ð
where ð1,âŠ, ðð are the coordinates of the map ðð and ð1,âŠ, ðð the coordinates of ð â ð.Exercise 4.3.xi. Let ð be a vector field onð and let ðⶠð â ð be a ð¶â function. Show thatif the function
ð¿ðð = ððððis zero ð is constant along integral curves of ð.
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116 Chapter 4: Manifolds & Forms on Manifolds
Exercise 4.3.xii. Suppose that ðⶠð â ð is proper. Show that if ð¿ðð = 0, ð is complete.Hint: For ð â ð let ð = ð(ð). By assumption, ðâ1(ð) is compact. Let ð â ð¶â0 (ð) be a
âbumpâ function which is one on ðâ1(ð) and letð be the vector field ðð. By Theorem 4.3.17,ð is complete and since
ð¿ðð = ððððð = ððððð = 0ð is constant along integral curves of ð. Let ðŸ(ð¡), ââ < ð¡ < â, be the integral curve of ðwith initial point ðŸ(0) = ð. Show that ðŸ is an integral curve of ð.
4.4. Orientations
The last part of Chapter 4 will be devoted to the âintegral calculusâ of forms on mani-folds. In particularwewill provemanifold versions of two basic theorems of integral calculuson ðð, Stokesâ theorem and the divergence theorem, and also develop a manifold versionof degree theory. However, to extend the integral calculus to manifolds without getting in-volved in horrendously technical âorientationâ issues we will confine ourselves to a specialclass of manifolds: orientable manifolds. The goal of this section will be to explain what thisterm means.
Definition 4.4.1. Letð be an ð-manifold. An orientation ofð is a rule for assigning to eachð â ð an orientation of ððð.
Thus by Definition 1.9.3 one can think of an orientation as a âlabelingâ rule which, foreveryð â ð, labels one of the two components of the setð¬ð(ðâðð)â{0} byð¬ð(ðâðð)+, whichweâll henceforth call the âplusâ part of ð¬ð(ðâðð), and the other component by ð¬ð(ðâðð)â,which weâll henceforth call the âminusâ part of ð¬ð(ðâðð).
Definition 4.4.2. An orientation of a manifold ð is smooth, or simply ð¶â, if for everyð â ð, there exists a neighborhoodð of ð and a non-vanishing ð-form ð â ðºð(ð)with theproperty
ðð â ð¬ð(ðâð ð)+for every ð â ð.
Remark 4.4.3. If weâre given an orientation of ð we can define another orientation by as-signing to each ð â ð the opposite orientation to the orientation we already assigned, i.e.,by switching the labels on ð¬ð(ðâð )+ and ð¬ð(ðâð )â. We will call this the reversed orientationofð. We will leave for you to check as an exercise that ifð is connected and equipped witha smooth orientation, the only smooth orientations of ð are the given orientation and itsreversed orientation.
Hint: Given any smooth orientation ofð the set of points where it agrees with the givenorientation is open, and the set of points where it does not is also open. Therefore one ofthese two sets has to be empty.
Note that ifð â ðºð(ð) is a non-vanishing ð-form one gets fromð a smooth orientationofð by requiring that the âlabeling ruleâ above satisfy
ðð â ð¬ð(ðâðð)+for every ð â ð. If ð has this property we will call ð a volume form. Itâs clear from this defi-nition that ifð1 andð2 are volume forms onð thenð2 = ð2,1ð1 whereð2,1 is an everywherepositive ð¶â function.
Example 4.4.4. Let ð be an open subset ðð. We will usually assign to ð its standard orien-tation, by which we will mean the orientation defined by the ð-form ðð¥1 â§â¯ ⧠ðð¥ð.
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§4.4 Orientations 117
Example 4.4.5. Let ðⶠðð â ðð be a ð¶â map. If zero is a regular value of ð, the setð â ðâ1(0) is a submanifold ofðð of dimension ð â ðâð (byTheorem 4.2.13). Moreover,for ð â ð, ððð is the kernel of the surjective map
ððð ⶠðððð â ð0ðð
so we get from ððð a bijective linear map
(4.4.6) ðððð/ððð â ð0ðð .
As explained inExample 4.4.4,ðððð andð0ðð have âstandardâ orientations, hence if werequire that the map (4.4.6) be orientation preserving, this gives ðððð/ððð an orientationand, by Theorem 1.9.9, gives ððð an orientation. Itâs intuitively clear that since ððð variessmoothly with respect to ð this orientation does as well; however, this fact requires a proof,and weâll supply a sketch of such a proof in the exercises.
Example 4.4.7. A special case of Example 4.4.5 is the ð-sphereðð = { (ð¥1,âŠ, ð¥ð+1) â ðð+1 | ð¥21 +⯠+ ð¥2ð+1 = 1 } ,
which acquires an orientation from its defining map ðⶠðð+1 â ð given by
ð(ð¥) â ð¥21 +⯠+ ð¥2ð+1 â 1 .
Example 4.4.8. Let ð be an oriented submanifold of ðð. For every ð â ð, ððð sits insideðððð as a vector subspace, hence, via the identification ðððð â ðð, one can think of ðððas a vector subspace of ðð. In particular from the standard Euclidean inner product on ððone gets, by restricting this inner product to vectors in ððð, an inner product
ðµð ⶠððð à ððð â ð
on ððð. Let ðð be the volume element in ð¬ð(ðâðð) associated with ðµð (see Exercise 1.9.x)and let ð = ðð be the non-vanishing ð-form onð defined by the assignment
ð ⊠ðð .In the exercises at the end of this section weâll sketch a proof of the following.
Theorem 4.4.9. The form ðð is ð¶â and hence, in particular, is a volume form. (We will callthis form the Riemannian volume form.)
Example 4.4.10 (Möbius strip). The Möbius strip is a surface in ð3 which is not orientable.It is obtained from the rectangle
ð â [0, 1] à (â1, 1) = { (ð¥, ðŠ) â ð2 | 0 †ð¥ †1 and â 1 < ðŠ < 1 }by gluing the ends together in the wrong way, i.e., by gluing (1, ðŠ) to (0, âðŠ). It is easy to seethat theMöbius strip cannot be oriented by taking the standard orientation at ð = (1, 0) andmoving it along the line (ð¡, 0), 0 †ð¡ †1 to the point (0, 0) (which is also the point ð afterwe have glued the ends of the rectangle together).
Weâll next investigate the âcompatibilityâ question for diffeomorphisms between ori-ented manifolds. Let ð and ð be ð-manifolds and ðⶠð ⥲ ð a diffeomorphism. Supposeboth of these manifolds are equipped with orientations. We will say that ð is orientationpreserving if for all ð â ð and ð = ð(ð), the linear map
ððð ⶠððð â ððð
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118 Chapter 4: Manifolds & Forms on Manifolds
is orientation preserving. Itâs clear that if ð is a volume form on ð then ð is orientationpreserving if and only if ðâð is a volume form on ð, and from (1.9.5) and the chain ruleone easily deduces the following theorem.
Theorem 4.4.11. If ð is an oriented ð-manifold and ðⶠð â ð a diffeomorphism, then ifboth ð and ð are orientation preserving, so is ð â ð.
If ðⶠð ⥲ ð is a diffeomorphism then the set of points ð â ð at which the linear map
ððð ⶠððð â ððð , ð = ð(ð) ,
is orientation preserving is open, and the set of points at which is orientation reversing isopen as well. Hence if ð is connected, ððð has to be orientation preserving at all points ororientation reversing at all points. In the latter case weâll say that ð is orientation reversing.
If ð is a parametrizable open subset ofð and ðⶠð0 â ð a parametrization of ð weâllsay that this parametrization is an oriented parametrization if ð is orientation preservingwith respect to the standard orientation of ð0 and the given orientation on ð. Notice thatif this parametrization isnât oriented we can convert it into one that is by replacing everyconnected component ð0 of ð0 on which ð isnât orientation preserving by the open set
(4.4.12) ðâ¯0 = { (ð¥1,âŠ, ð¥ð) â ðð | (ð¥1,âŠ, ð¥ðâ1, âð¥ð) â ð0 }
and replacing ð by the map
(4.4.13) ð(ð¥1,âŠ, ð¥ð) = ð(ð¥1,âŠ, ð¥ðâ1, âð¥ð) .
If ðð ⶠðð â ð, ð = 0, 1, are oriented parametrizations of ð and ðⶠð0 â ð1 is thediffeomorphism ðâ11 â ð0, then by the theorem above ð is orientation preserving or in otherwords
(4.4.14) det(ððððð¥ð) > 0
at every point on ð0.Weâll conclude this section by discussing some orientation issues which will come up
when we discuss Stokesâ theorem and the divergence theorem in §4.6. First a definition.
Definition 4.4.15. An open subsetð· ofð is a smooth domain if(1) The boundary ðð· is an (ð â 1)-dimensional submanifold ofð, and(2) The boundary ofð· coincides with the boundary of the closure ofð·.
Examples 4.4.16.(1) The ð-ball, ð¥21 +⯠+ ð¥2ð < 1, whose boundary is the sphere, ð¥21 +⯠+ ð¥2ð = 1.(2) The ð-dimensional annulus,
1 < ð¥21 +⯠+ ð¥2ð < 2
whose boundary consists of the union of the two spheres
{ð¥21 +⯠+ ð¥2ð = 1ð¥21 +⯠+ ð¥2ð = 2 .
(3) Let ððâ1 be the unit sphere, ð¥21 +â¯+ ð¥22 = 1 and letð· = ðð â ððâ1. Then the boundaryofð· is ððâ1 butð· is not a smooth domain since the boundary of its closure is empty.
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§4.4 Orientations 119
(4) The simplest example of a smooth domain is the half-space
ðð = { (ð¥1,âŠ, ð¥ð) â ðð | ð¥1 < 0 }whose boundary
ððð = { (ð¥1,âŠ, ð¥ð) â ðð | ð¥1 = 0 }we can identify with ððâ1 via the map ððâ1 â ððð given by
(ð¥2,âŠ, ð¥ð) ⊠(0, ð¥2,âŠ, ð¥ð) .
We will show that every bounded domain looks locally like this example.
Theorem 4.4.17. Letð· be a smooth domain and ð a boundary point ofð·. Then there existsa neighborhood ð of ð in ð, an open set ð0 in ðð, and a diffeomorphism ðⶠð0 ⥲ ð suchthat ðmaps ð0 â© ðð onto ð â© ð·.
Proof. Let ð â ðð·. First we prove the following lemma.
Lemma4.4.18. For everyð â ð there exists an open setð inð containingð and a parametriza-tion
ðⶠð0 â ðof ð with the property
(4.4.19) ð(ð0 â© ððð) = ð â© ð .
Proof. Since ð is locally diffeomorphic at ð to an open subset of ðð so it suffices to provethis assertion for ð = ðð. However, if ð is an (ð â 1)-dimensional submanifold of ðð thenbyTheorem 4.2.15 there exists, for every ð â ð, a neighborhoodð of ð inðð and a functionð â ð¶â(ð) with the properties
(4.4.20) ð¥ â ð â© ð ⺠ð(ð¥) = 0and
(4.4.21) ððð â 0 .Without loss of generality we can assume by equation (4.4.21) that
ðððð¥1(ð) â 0 .
Hence if ðⶠð â ðð is the map
(4.4.22) ð(ð¥1,âŠ, ð¥ð) â (ð(ð¥), ð¥2,âŠ, ð¥ð)(ðð)ð is bijective, and hence ð is locally a diffeomorphism at ð. Shrinkingð we can assumethat ð is a diffeomorphism ofð onto an open setð0. By (4.4.20) and (4.4.22) ðmapsðâ©ðonto ð0 â© ððð hence if we take ð to be ðâ1, it will have the property (4.4.19). â¡
We will now prove Theorem 4.4.9. Without loss of generality we can assume that theopen set ð0 in Lemma 4.4.18 is an open ball with center at ð â ððð and that the diffeomor-phism ðmaps ð to ð. Thus for ðâ1(ð â© ð·) there are three possibilities:(i) ðâ1(ð â© ð·) = (ðð â ððð) â© ð0.(ii) ðâ1(ð â© ð·) = (ðð â ðð) â© ð0.(iii) ðâ1(ð â© ð·) = ðð â© ð0.However, (i) is excluded by the second hypothesis in Definition 4.4.15 and if (ii) occurs wecan rectify the situation by composing ð with the map (ð¥1,âŠ, ð¥ð) ⊠(âð¥1, ð¥2,âŠ, ð¥ð). â¡
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Definition4.4.23. Wewill call an open setðwith the properties above að·-adapted parame-trizable open set.
We will now show that ifð is oriented andð· â ð is a smooth domain then the bound-ary ð â ðð· ofð· acquires fromð a natural orientation. To see this we first observe:
Lemma 4.4.24. The diffeomorphism ðⶠð0 ⥲ ð in Theorem 4.4.17 can be chosen to beorientation preserving.
Proof. If it is not, then by replacing ð with the diffeomorphismðâ¯(ð¥1,âŠ, ð¥ð) â ð(ð¥1,âŠ, ð¥ðâ1, âð¥ð) ,
we get a ð·-adapted parametrization of ð which is orientation preserving. (See (4.4.12)â(4.4.13).) â¡
Letð0 = ð0â©ððâ1 be the boundary ofð0â©ðð. The restriction ofð toð0 is a diffeomor-phism of ð0 ontoðâ©ð, and we will orientðâ©ð by requiring that this map be an orientedparametrization. To show that this is an âintrinsicâ definition, i.e., does not depend on thechoice of ð. Weâll prove:
Theorem4.4.25. Ifðð ⶠðð â ð, ð = 0, 1, are oriented parametrizations ofðwith the propertyðð ⶠðð â© ðð â ð â© ð·
the restrictions of ðð to ðð â© ððâ1 induce compatible orientations on ð â© ð.
Proof. To prove this we have to prove that the map ðâ11 â ð0, restricted to ð â© ððð is anorientation preserving diffeomorphism ofð0 â©ððâ1 ontoð1 â©ððâ1. Thus we have to provethe following:
Proposition 4.4.26. Let ð0 and ð1 be open subsets of ðð and ðⶠð0 â ð1 an orientationpreserving diffeomorphism which maps ð0 â© ðð onto ð1 â© ðð. Then the restriction ð of ð tothe boundary ð0 â© ððâ1 of ð0 â© ðð is an orientation preserving diffeomorphism
ðⶠð0 â© ððâ1 ⥲ ð1 â© ððâ1 .Let ð(ð¥) = (ð1(ð¥),âŠ, ðð(ð¥)). By assumption ð1(ð¥1,âŠ, ð¥ð) is less than zero if ð¥1 is less
than zero and equal to zero if ð¥1 is equal to zero, hence
(4.4.27) ðð1ðð¥1(0, ð¥2,âŠ, ð¥ð) ⥠0
andðð1ðð¥ð(0, ð¥2,âŠ, ð¥ð) = 0
for ð > 1Moreover, since ð is the restriction of ð to the set ð¥1 = 0ððððð¥ð(0, ð¥2,âŠ, ð¥ð) =
ððððð¥ð(ð¥2,âŠ, ð¥1)
for ð, ð ⥠2. Thus on the set defined by ð¥1 = 0 we have
(4.4.28) det(ððððð¥ð) = ðð1ðð¥1
det(ððððð¥ð) .
Since ð is orientation preserving the left hand side of equation (4.4.28) is positive at allpoints (0, ð¥2,âŠ, ð¥ð) â ð0 â© ððâ1 hence by (4.4.27) the same is true for ðð1ðð¥1 and det ( ððððð¥ð ).Thus ð is orientation preserving. â¡
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§4.4 Orientations 121
Remark 4.4.29. For an alternative proof of this result see Exercise 3.2.viii andExercise 3.6.iv.
Wewill noworient the boundary ofð· by requiring that for everyð·-adapted parametriz-able open setð the orientation ofð coincides with the orientation ofðâ©ð that we describedabove.Wewill conclude this discussion of orientations by proving a global version of Propo-sition 4.4.26.
Proposition 4.4.30. For ð = 1, 2, let ðð be an oriented manifold, ð·ð â ðð a smooth domainandðð â ðð·ð its boundary. Then if ð is an orientation preserving diffeomorphism of (ð1, ð·1)onto (ð2, ð·2) the restriction ð of ð to ð1 is an orientation preserving diffeomorphism of ð1onto ð2.
Proof. Letðbe anopen subset ofð1 andðⶠð0 â ð an orientedð·1-compatible parametriza-tion ofð.Then ifð = ð(ð) themapððⶠð â ð is an orientedð·2-compatible parametriza-tion of ð and hence ðⶠð â© ð1 â ð â© ð2 is orientation preserving. â¡
Exercises for §4.4Exercise 4.4.i. Let ð be an oriented ð-dimensional vector space, ðµ an inner product on ðand ð1,âŠ, ðð â ð an oriented orthonormal basis. Given vectors ð£1,âŠ, ð£ð â ð, show that if
(4.4.31) ðð,ð = ðµ(ð£ð, ð£ð)and
ð£ð =ðâð=1ðð,ððð ,
the matrices ð = (ðð,ð) and ð = (ðð,ð) satisfy the identity:
ð = ðâºðand conclude that det(ð) = det(ð)2. (In particular conclude that det(ð) > 0.)
Exercise 4.4.ii. Let ð andð be oriented ð-dimensional vector spaces. Suppose that eachof these spaces is equipped with an inner product, and let ð1,âŠ, ðð â ð and ð1,âŠ, ðð â ðbe oriented orthonormal bases. Show that if ðŽâ¶ ð â ð is an orientation preserving linearmapping and ðŽðð = ð£ð then
ðŽâ volð = (det(ðð,ð))12 volð
where volð = ðâ1 â§â¯ ⧠ðâð , volð = ðâ1 â§â¯ ⧠ðâð and (ðð,ð) is the matrix (4.4.31).
Exercise 4.4.iii. Let ð be an oriented ð-dimensional submanifold of ðð, ð an open subsetof ð, ð0 an open subset of ðð, and ðⶠð0 â ð an oriented parametrization. Let ð1,âŠ, ððbe the coordinates of the map
ð0 â ð ⪠ðð .the second map being the inclusion map. Show that if ð is the Riemannian volume form onð then
ðâð = (det(ðð,ð))12 ðð¥1 â§â¯ ⧠ðð¥ð
where
(4.4.32) ðð,ð =ðâð=1
ððððð¥ðððððð¥ð
for 1 †ð, ð †ð .
Conclude that ð is a smooth ð-form and hence that it is a volume form.
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Hint: For ð â ð0 and ð = ð(ð) apply Exercise 4.4.ii with ð = ððð,ð = ðððð, ðŽ =(ðð)ð and ð£ð = (ðð)ð ( ððð¥ð )ð.
Exercise 4.4.iv. Given a ð¶â function ðⶠð â ð, its graph
ð â ð€ð = { (ð¥, ð(ð¥)) â ð à ð | ð¥ â ð }
is a submanifold of ð2 andðⶠð â ð , ð¥ ⊠(ð¥, ð(ð¥))
is a diffeomorphism. Orient ð by requiring that ð be orientation preserving and show thatif ð is the Riemannian volume form onð then
ðâð = (1 + (ðððð¥)2)12
ðð¥ .
Hint: Exercise 4.4.iii
Exercise 4.4.v. Given að¶â functionðⶠðð â ð the graph ð€ð ofð is a submanifold ofðð+1and
(4.4.33) ðⶠðð â ð , ð¥ ⊠(ð¥, ð(ð¥))is a diffeomorphism. Orient ð by requiring that ð is orientation preserving and show thatif ð is the Riemannian volume form onð then
ðâð = (1 +ðâð=1( ðððð¥ð)2)12
ðð¥1 â§â¯ ⧠ðð¥ð .
Hints:⢠Let ð£ = (ð1,âŠ, ðð) â ðð. Show that if ð¶â¶ ðð â ð is the linear mapping defined
by the matrix (ðððð) then ð¶ð£ = (âðð=1 ð2ð )ð£ and ð¶ð€ = 0 if ð€ â ð£ = 0 .
⢠Conclude that the eigenvalues of ð¶ are ð1 = â ð2ð and ð2 = ⯠= ðð = 0.⢠Show that the determinant of ðŒ + ð¶ is 1 + âðð=1 ð2ð .⢠Compute the determinant of the matrix (4.4.32) where ð is the mapping (4.4.33).
Exercise 4.4.vi. Let ð be an orientedð-dimensional vector space and âð â ðâ, ð = 1,âŠ, ð,ð linearly independent vectors in ðâ. Define
ð¿â¶ ð â ðð
to be the map ð£ ⊠(â1(ð£),âŠ, âð(ð£)).(1) Show that ð¿ is surjective and that the kernelð of ð¿ is of dimension ð = ð â ð.(2) Show that one gets from this mapping a bijective linear mapping
(4.4.34) ð/ð â ðð
and hence from the standard orientation on ðð an induced orientation onð/ð and onð.
Hint: Exercise 1.2.viii and Theorem 1.9.9.(3) Let ð be an element of ð¬ð(ðâ). Show that there exists a ð â ð¬ð(ðâ) with the property
(4.4.35) â1 â§â¯ ⧠âð ⧠ð = ð .Hint:Choose an oriented basis, ð1,âŠ, ðð ofð such thatð = ðâ1 â§â¯â§ðâð and âð = ðâð
for ð = 1,âŠ, ð, and let ð = ðâð+1 â§â¯ ⧠ðâð.
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§4.4 Orientations 123
(4) Show that if ð is an element of ð¬ð(ðâ) with the property
â1 â§â¯ ⧠âð ⧠ð = 0then there exist elements, ðð, of ð¬ðâ1(ðâ) such that
ð = ââð ⧠ðð .Hint: Same hint as in part (3).
(5) Show that if ð1 and ð2 are elements ofð¬ð(ðâ)with the property (4.4.35) and ð ⶠð â ðis the inclusion map then ðâð1 = ðâð2.
Hint: Let ð = ð1 â ð2. Conclude from part (4) that ðâð = 0.(6) Conclude that if ð is an element of ð¬ð(ðâ) satisfying (4.4.35) the element, ð = ðâð, ofð¬ð(ðâ) is intrinsically defined independent of the choice of ð.
(7) Show that ð lies in ð¬ð(ðâ)+.
Exercise 4.4.vii. Let ð be an open subset of ðð and ðⶠð â ðð a ð¶â map. If zero is aregular value of ð, the set,ð = ðâ1(0) is a manifold of dimension ð = ðâð. Show that thismanifold has a natural smooth orientation.
Some suggestions:⢠Let ð = (ð1,âŠ, ðð) and let
ðð1 â§â¯ ⧠ððð = âððŒðð¥ðŒsummed over multi-indices which are strictly increasing. Show that for every ð âð ððŒ(ð) â 0 for some multi-index, ðŒ = (ð1,âŠ, ðð), 1 †ð1 < ⯠< ðð †ð.
⢠Let ðœ = (ð1,âŠ, ðð), 1 †ð1 < ⯠< ðð †ð be the complementary multi-index to ðŒ,i.e., ðð â ðð for all ð and ð . Show that
ðð1 â§â¯ ⧠ððð ⧠ðð¥ðœ = ±ððŒðð¥1 â§â¯ ⧠ðð¥ðand conclude that the ð-form
ð = ± 1ððŒðð¥ðœ
is a ð¶â ð-form on a neighborhood of ð in ð and has the property:
ðð1 â§â¯ ⧠ððð ⧠ð = ðð¥1 â§â¯ ⧠ðð¥ð .
⢠Let ð ⶠð â ð be the inclusion map. Show that the assignment
ð ⊠(ðâð)ðdefines an intrinsic nowhere vanishing ð-form ð â ðºð(ð) onð.
⢠Show that the orientation of ð defined by ð coincides with the orientation thatwe described earlier in this section.
Exercise 4.4.viii. Let ðð be the ð-sphere and ð ⶠðð â ðð+1 the inclusion map. Show thatif ð â ðºð(ðð+1) is the ð-form ð = âð+1ð=1 (â1)ðâ1ð¥ððð¥1 ⧠⯠⧠ðð¥ð ⧠⯠⧠ðð¥ð+1, the ð-formðâð â ðºð(ðð) is the Riemannian volume form.
Exercise 4.4.ix. Let ðð+1 be the (ð + 1)-sphere and let
ðð+1+ = { (ð¥1,âŠ, ð¥ð+2) â ðð+1 | ð¥1 < 0 }be the lower hemisphere in ðð+1.(1) Prove that ðð+1+ is a smooth domain.(2) Show that the boundary of ðð+1+ is ðð.
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124 Chapter 4: Manifolds & Forms on Manifolds
(3) Show that the boundary orientation of ðð agrees with the orientation of ðð in Exer-cise 4.4.viii.
4.5. Integration of forms on manifolds
In this section we will show how to integrate differential forms over manifolds. In whatfollows ð will be an oriented ð-manifold andð an open subset of ð, and our goal will beto make sense of the integral
(4.5.1) â«ðð
whereð is a compactly supported ð-form.Weâll begin by showing how to define this integralwhen the support ofð is contained in a parametrizable open setð. Letð0 be an open subsetof ðð and ð0 ⶠð0 â ð a parametrization. As we noted in §4.4 we can assume without lossof generality that this parametrization is oriented. Making this assumption, weâll define
(4.5.2) â«ðð = â«ð0ðâ0ð
whereð0 = ðâ10 (ð â© ð). Notice that if ðâð = ððð¥1 â§â¯ ⧠ðð¥ð, then, by assumption, ð isin ð¶â0 (ð0). Hence since
â«ð0ðâ0ð = â«
ð0ððð¥1âŠðð¥ð
and since ð is a bounded continuous function and is compactly supported the Riemannintegral on the right is well-defined. (See Appendix b.) Moreover, if ð1 ⶠð1 â ð is anotheroriented parametrization of ð and ðⶠð0 â ð1 is the map ð = ðâ11 â ð0 then ð0 = ð1 â ð,so by Proposition 4.3.4
ðâ0ð = ðâðâ1ð .Moreover, by (4.3.18) ð is orientation preserving. Therefore since
ð1 = ð(ð0) = ðâ11 (ð â©ð)
Theorem 3.5.2 tells us that
(4.5.3) â«ð1ðâ1ð = â«
ð0ðâ0ð .
Thus the definition (4.5.2) is a legitimate definition. It does not depend on the parametriza-tion that we use to define the integral on the right. From the usual additivity propertiesof the Riemann integral one gets analogous properties for the integral (4.5.2). Namely forð1, ð2 â ðºðð (ð),
(4.5.4) â«ð(ð1 + ð2) = â«
ðð1 + â«
ðð2
and for ð â ðºðð (ð) and ð â ð
â«ððð = ðâ«
ðð .
We will next show how to define the integral (4.5.1) for any compactly supported ð-form. This we will do in more or less the same way that we defined improper Riemann inte-grals in Appendix b: by using partitions of unity. Weâll begin by deriving from the partitionof unity theorem in Appendix b a manifold version of this theorem.
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§4.5 Integration of forms on manifolds 125
Theorem 4.5.5. Let(4.5.6) U = {ððŒ}ðŒâðŒbe a covering of ð be open subsets. Then there exists a family of functions ðð â ð¶â0 (ð), forð ⥠1, with the properties(1) ðð ⥠0.(2) For every compact set ð¶ â ð there exists a positive integerð such that if ð > ð we have
supp(ðð) â© ð¶ = â .(3) ââð=1 ðð = 1.(4) For every ð ⥠1 there exists an index ðŒ â ðŒ such that supp(ðð) â ððŒ.
Remark 4.5.7. Conditions (1)â(3) say that the ððâs are a partition of unity and (4) says thatthis partition of unity is subordinate to the covering (4.5.5).
Proof. For each ð â ð and for some ððŒ containing a ð choose an open set ðð in ðð withð â ðð and with
(4.5.8) ðð â© ð â ððŒ .Let ð â âðâð ðð and let ðð â ð¶â0 (ð), for ð ⥠1, be a partition of unity subordinate to
the covering of ð by the ððâs. By (4.5.8) the restriction ðð of ðð to ð has compact supportand it is clear that the ððâs inherit from the ððâs the properties (1)â(4). â¡
Now let the covering (4.5.6) be any covering of ð by parametrizable open sets and letðð â ð¶â0 (ð), for ð ⥠1 be a partition of unity subordinate to this covering. Given ð â ðºðð (ð)we will define the integral of ð overð by the sum
(4.5.9)ââð=1â«ðððð .
Note that since each ðð is supported in some ððŒ the individual summands in this sum arewell-defined and since the support of ð is compact all but finitely many of these summandsare zero by part (2) of Theorem 4.5.5. Hence the sum itself is well-defined. Letâs show thatthis sum does not depend on the choice of U and the ððâs. Let Uâ² be another covering ofðby parametrizable open sets and (ðâ²ð )ðâ¥1 a partition of unity subordinate to Uâ². Then
(4.5.10)ââð=1â«ððâ²ðð =
ââð=1â«ð
ââð=1ðâ²ðððð =
ââð=1(ââð=1â«ððâ²ðððð)
by equation (4.5.4). Interchanging the orders of summation and resuming with respect tothe ðâs this sum becomes
ââð=1â«ð
ââð=1ðâ²ðððð
orââð=1â«ðððð .
Henceââð=1â«ððâ²ðð =
ââð=1â«ðððð ,
so the two sums are the same.From equations (4.5.4) and (4.5.9) one easily deduces:
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126 Chapter 4: Manifolds & Forms on Manifolds
Proposition 4.5.11. For ð1, ð2 â ðºðð (ð),
â«ðð1 + ð2 = â«
ðð1 + â«
ðð2
and for ð â ðºðð (ð) and ð â ðâ«ððð = ðâ«
ðð .
The definition of the integral (4.5.1) depends on the choice of an orientation of ð, butitâs easy to see how it depends on this choice. We pointed out in Section 4.4 that if ð isconnected, there is just one way to orient it smoothly other than by its given orientation,namely by reversing the orientation of ððð at each point ð and itâs clear from the definitions(4.5.2) and (4.5.9) that the effect of doing this is to change the sign of the integral, i.e., tochange â«ð ð to ââ«ð ð.
In the definition of the integral (4.5.1) we have allowedð to be an arbitrary open subsetofð but required ð to be compactly supported. This integral is also well-defined if we allowð to be an arbitrary element of ðºð(ð) but require the closure ofð in ð to be compact. Tosee this, note that under this assumption the sum (4.5.8) is still a finite sum, so the definitionof the integral still makes sense, and the double sum on the right side of (4.5.10) is still afinite sum so itâs still true that the definition of the integral does not depend on the choiceof partitions of unity. In particular if the closure ofð in ð is compact we will define thevolume ofð to be the integral,
vol(ð) = â«ððvol ,
where ðvol is the Riemannian volume form and ifð itself is compact weâll define its volumeto be the integral
vol(ð) = â«ððvol .
Weâll next prove a manifold version of the change of variables formula (3.5.3).
Theorem 4.5.12 (change of variables formula). Let ðâ² and ð be oriented ð-manifolds andðⶠðâ² â ð an orientation preserving diffeomorphism. If ð is an open subset of ð andðâ² â ðâ1(ð), then
(4.5.13) â«ðâ²ðâð = â«
ðð
for all ð â ðºðð (ð).
Proof. By (4.5.9) the integrand of the integral above is a finite sum of ð¶â forms, each ofwhich is supported on a parametrizable open subset, so we can assume that ð itself as thisproperty. Letð be a parametrizable open set containing the support ofð and let ð0 ⶠð ⥲ ðbe an oriented parameterization of ð. Since ð is a diffeomorphism its inverse exists and isa diffeomorphism ofð ontoð1. Let ðâ² â ðâ1(ð) and ðâ²0 â ðâ1 â ð0. Then ðâ²0 ⶠð ⥲ ðâ² isan oriented parameterization of ðâ². Moreover, ð â ðâ²0 = ð0 so ifð0 = ðâ10 (ð) we have
ð0 = (ðâ²0)â1(ðâ1(ð)) = (ðâ²0)â1(ðâ²)and by the chain rule we have
ðâ0ð = (ð â ðâ²0)âð = (ðâ²0)âðâðhence
â«ðð = â«ð0ðâ0ð = â«
ð0(ðâ²0)â(ðâð) = â«
ðâ²ðâð . â¡
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§4.5 Integration of forms on manifolds 127
As an exercise, show that if ðⶠðâ² â ð is orientation reversing
(4.5.14) â«ðâ²ðâð = ââ«
ðð .
Weâll conclude this discussion of âintegral calculus on manifoldsâ by proving a prelim-inary version of Stokesâ theorem.
Theorem 4.5.15. If ð is in ðºðâ1ð (ð) then
â«ððð = 0 .
Proof. Let (ðð)ðâ¥1 be a partition of unity with the property that each ðð is supported in aparametrizable open set ðð = ð. Replacing ð by ððð it suffices to prove the theorem forð â ðºðâ1ð (ð). Let ðⶠð0 â ð be an oriented parametrization of ð. Then
â«ððð = â«
ð0ðâðð = â«
ð0ð(ðâð) = 0
by Theorem 3.3.1. â¡
Exercises for §4.5Exercise 4.5.i. Letðⶠðð â ð be að¶â function.Orient the graphð â ð€ð ofð by requiringthat the diffeomorphism
ðⶠðð â ð , ð¥ ⊠(ð¥, ð(ð¥))be orientation preserving. Given a bounded open set ð in ðð compute the Riemannianvolume of the image
ðð = ð(ð)of ð inð as an integral over ð.
Hint: Exercise 4.4.v.
Exercise 4.5.ii. Evaluate this integral for the open subset ðð of the paraboloid defined byð¥3 = ð¥21 + ð¥22 , where ð is the disk ð¥21 + ð¥22 < 2.
Exercise 4.5.iii. In Exercise 4.5.i let ð ⶠð ⪠ðð+1 be the inclusion map ofð onto ðð+1.(1) If ð â ðºð(ðð+1) is the ð-form ð¥ð+1ðð¥1 ⧠⯠⧠ðð¥ð, what is the integral of ðâð over the
setðð? Express this integral as an integral over ð.(2) Same question for ð = ð¥2ð+1ðð¥1 â§â¯ ⧠ðð¥ð.(3) Same question for ð = ðð¥1 â§â¯ ⧠ðð¥ð.
Exercise 4.5.iv. Let ðⶠðð â (0, +â) be a positiveð¶â function,ð a bounded open subsetof ðð, andð the open set of ðð+1 defined by the inequalities
0 < ð¥ð+1 < ð(ð¥1,âŠ, ð¥ð)and the condition (ð¥1,âŠ, ð¥ð) â ð.(1) Express the integral of the (ð + 1)-form ð = ð¥ð+1ðð¥1 â§â¯â§ðð¥ð+1 overð as an integral
over ð.(2) Same question for ð = ð¥2ð+1ðð¥1 â§â¯ ⧠ðð¥ð+1.(3) Same question for ð = ðð¥1 â§â¯ ⧠ðð¥ð.
Exercise 4.5.v. Integrate the âRiemannian areaâ formð¥1ðð¥2 ⧠ðð¥3 + ð¥2ðð¥3 ⧠ðð¥1 + ð¥3ðð¥1 ⧠ðð¥2
over the unit 2-sphere ð2. (See Exercise 4.4.viii.)
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Hint: An easier problem: Using polar coordinates integrate ð = ð¥3ðð¥1 ⧠ðð¥2 over thehemisphere defined by ð¥3 = â1 â ð¥21 â ð¥22 , where ð¥21 + ð¥22 < 1.
Exercise 4.5.vi. Let ðŒ be the one-formâðð=1 ðŠððð¥ð in equation (2.8.2) and let ðŸ(ð¡), 0 †ð¡ †1,be a trajectory of the Hamiltonian vector field (2.8.2). What is the integral of ðŒ over ðŸ(ð¡)?
4.6. Stokesâ theorem & the divergence theorem
Letð be an oriented ð-manifold andð· â ð a smooth domain. We showed in §4.4 thatðð· acquires from ð· a natural orientation. Hence if ð ⶠðð· â ð is the inclusion map and ðis in ðºðâ1ð (ð), the integral
â«ðð·ðâð
is well-defined. We will prove:
Theorem 4.6.1 (Stokesâ theorem). For ð â ðºðâ1ð (ð) we have
(4.6.2) â«ðð·ðâð = â«
ð·ðð .
Proof. Let (ðð)ðâ¥1 be a partition of unity such that for each ð, the support of ðð is containedin a parametrizable open set ðð = ð of one of the following three types:(a) ð â int(ð·).(b) ð â ext(ð·).(c) There exists an open subset ð0 of ðð and an orientedð·-adapted parametrization
ðⶠð0 ⥲ ð .
Replacing ð by the finite sum ââð=1 ððð it suffices to prove (4.6.2) for each ððð separately. Inother words we can assume that the support of ð itself is contained in a parametrizable openset ð of type (a), (b), or (c). But if ð is of type (a)
â«ð·ðð = â«
ððð = â«
ððð
and ðâð = 0. Hence the left hand side of equation (4.6.2) is zero and, by Theorem 4.5.15,the right hand side is as well. If ð is of type (b) the situation is even simpler: ðâð is zeroand the restriction of ð toð· is zero, so both sides of equation (4.6.2) are automatically zero.Thus one is reduced to proving (4.6.2) whenð is an open subset of type (c). In this case therestriction of the map (4.6.2) to ð0 â© ððð is an orientation preserving diffeomorphism
(4.6.3) ðⶠð0 â© ððð â ð â© ð
and
(4.6.4) ðð â ð = ð â ðððâ1
where the maps ð = ðð andðððâ1 ⶠððâ1 ⪠ðð
are the inclusionmaps ofð intoð and ððð intoðð. (Here weâre identifying ððð withððâ1.)Thus
â«ð·ðð = â«
ðððâðð = â«
ððððâð
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§4.6 Stokesâ theorem & the divergence theorem 129
and by (4.6.4)
â«ððâðð = â«
ððâ1ðâðâðð
= â«ððâ1ðâððâ1ðâð
= â«ððððâððâ1ðâð .
Thus it suffices to prove Stokesâ theoremwith ð replaced by ðâð, or, in other words, to proveStokesâ theorem forðð; and this we will now do.
Theorem 4.6.5 (Stokesâ theorem forðð). Let
ð =ðâð=1(â1)ðâ1ðððð¥1 â§â¯ ⧠ðð¥ð â§â¯ ⧠ðð¥ð .
Then
ðð =ðâð=1
ððððð¥ððð¥1 â§â¯ ⧠ðð¥ð
and
â«ðððð =
ðâð=1â«ððððððð¥ððð¥1â¯ðð¥ð .
We will compute each of these summands as an iterated integral doing the integrationwith respect to ðð¥ð first. For ð > 1 the ðð¥ð integration ranges over the interval ââ < ð¥ð < âand hence since ðð is compactly supported
â«â
ââ
ððððð¥ððð¥ð = ðð(ð¥1,âŠ, ð¥ð,âŠ, ð¥ð)|
ð¥ð=+â
ð¥ð=ââ= 0 .
On the other hand the ðð¥1 integration ranges over the interval ââ < ð¥1 < 0 and
â«0
ââ
ðð1ðð¥1ðð¥1 = ð(0, ð¥2,âŠ, ð¥ð) .
Thus integrating with respect to the remaining variables we get
(4.6.6) â«ðððð = â«
ððâ1ð(0, ð¥2,âŠ, ð¥ð) ðð¥2 â§â¯ ⧠ðð¥ð .
On the other hand, since ðâððâ1ð¥1 = 0 and ðâððâ1 ð¥ð = ð¥ð for ð > 1,ðâððâ1ð = ð1(0, ð¥2,âŠ, ð¥ð) ðð¥2 â§â¯ ⧠ðð¥ð
so
(4.6.7) â«ððâ1ðâððâ1ð = â«
ððâ1ð(0, ð¥2,âŠ, ð¥ð) ðð¥2 â§â¯ ⧠ðð¥ð .
Hence the two sides, (4.6.6) and (4.6.7), of Stokesâ theorem are equal. â¡
One important variant of Stokesâ theorem is the divergence theorem: Let ð be inðºðð (ð)and let ð be a vector field onð. Then
ð¿ðð = ðððð + ðððð = ðððð ,hence, denoting by ðð the inclusion map of ð into ð we get from Stokesâ theorem, withð = ððð:
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130 Chapter 4: Manifolds & Forms on Manifolds
Theorem 4.6.8 (divergence theorem for manifolds).
â«ð·ð¿ðð = â«
ððâð(ððð) .
If ð· is an open domain in ðð this reduces to the usual divergence theorem of multi-variable calculus. Namely if ð = ðð¥1 â§â¯ ⧠ðð¥ð and ð = â
ðð=1 ð£ð
ððð¥ð
then by (2.5.14)
ð¿ððð¥1 â§â¯ ⧠ðð¥ð = div(ð) ðð¥1 â§â¯ ⧠ðð¥ðwhere
div(ð) =ðâð=1
ðð£ððð¥ð.
Thus if ð = ðð· and ðð is the inclusion ð ⪠ðð,
(4.6.9) â«ð·
div(ð)ðð¥ = â«ððâð(ðððð¥1 â§â¯ ⧠ðð¥ð) .
The right hand side of this identity can be interpreted as the âfluxâ of the vector field ðthrough the boundary ofð·. To see this let ðⶠðð â ð be a ð¶â defining function forð·, i.e.,a function with the properties
ð â ð· ⺠ð(ð) < 0and ððð â 0 if ð â ðð·. This second condition says that zero is a regular value of ð andhence that ð â ðð· is defined by the non-degenerate equation:
(4.6.10) ð â ð ⺠ð(ð) = 0 .Let ð be the vector field
ð â (ðâð=1( ðððð¥ð)2)â1 ðâð=1
ððððð¥ðððð¥ð
.
In view of (4.6.10), this vector field is well-defined on a neighborhood ð of ð and satisfies
ðððð = 1 .Now note that since ðð ⧠ðð¥1 â§â¯ ⧠ðð¥ð = 0
0 = ðð(ðð ⧠ðð¥1 â§â¯ ⧠ðð¥ð)= (ðððð)ðð¥1 â§â¯ ⧠ðð¥ð â ðð ⧠ðððð¥1 â§â¯ ⧠ðð¥ð= ðð¥1 â§â¯ ⧠ðð¥ð â ðð ⧠ðððð¥1 â§â¯ ⧠ðð¥ð ,
hence letting ð be the (ð â 1)-form ðððð¥1 â§â¯ ⧠ðð¥ð we get the identity
(4.6.11) ðð¥1 â§â¯ ⧠ðð¥ð = ðð ⧠ðand by applying the operation ðð to both sides of (4.6.11) the identity
(4.6.12) ðððð¥1 â§â¯ ⧠ðð¥ð = (ð¿ðð)ð â ðð ⧠ððð .Let ðð = ðâðð be the restriction of ð to ð. Since ðâð = 0, ðâððð = 0 and hence by (4.6.12)
ðâð(ðððð¥1 â§â¯ ⧠ðð¥ð) = ðâð(ð¿ðð)ðð ,
and the formula (4.6.9) now takes the form
(4.6.13) â«ð·
div(ð)ðð¥ = â«ðð¿ðððð
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§4.6 Stokesâ theorem & the divergence theorem 131
where the term on the right is by definition the flux of ð through ð. In calculus books thisis written in a slightly different form. Letting
ðð = (ðâð=1( ðððð¥ð)2)12
ðð
and letting
ᅵᅵ â (ðâð=1( ðððð¥ð)2)â 12( ðððð¥1,âŠ, ðððð¥ð)
and ð£ â (ð£1,âŠ, ð£ð) we haveð¿ððð = (ᅵᅵ â ð£)ðð
and hence
(4.6.14) â«ð·
div(ð)ðð¥ = â«ð(ᅵᅵ â ð£)ðð .
In three dimensions ðð is just the standard âinfinitesimal element of areaâ on the surface ðand ðð the unit outward normal to ð at ð, so this version of the divergence theorem is theversion one finds in most calculus books.
As an application of Stokesâ theorem, weâll give a very short alternative proof of theBrouwer fixed point theorem. As we explained in §3.6 the proof of this theorem basicallycomes down to proving the following theorem.
Theorem 4.6.15. Let ðµð be the closed unit ball in ðð and ððâ1 its boundary. Then the identitymap idððâ1 on ððâ1 cannot be extended to a ð¶â map ðⶠðµð â ððâ1.
Proof. Suppose that ð is such a map. Then for every (ð â 1)-form ð â ðºðâ1(ððâ1),
(4.6.16) â«ðµðð(ðâð) = â«
ððâ1(ðððâ1)âðâð .
Butð(ðâð) = ðâ(ðð) = 0 sinceð is an (ðâ1)-form and ððâ1 is an (ðâ1)-manifold, and sinceð is the identity map on ððâ1, (ðððâ1)
âðâð = (ð â ðððâ1)âð = ð. Thus for every ð â ðºðâ1(ððâ1),equation (4.6.16) says that the integral of ð over ððâ1 is zero. Since there are lots of (ð â 1)-forms for which this is not true, this shows that a map ð with the property above cannotexist. â¡
Exercises for §4.6Exercise 4.6.i. Let ðµð be the open unit ball in ðð and ððâ1 the unit (ð â 1)-sphere. Showthat vol(ððâ1) = ð vol(ðµð).
Hint: Apply Stokesâ theorem to the (ð â 1)-form
ð âðâð=1(â1)ðâ1ð¥ððð¥1 â§â¯ ⧠ðð¥ð â§â¯ ⧠ðð¥ð
and note by Exercise 4.4.ix that ð is the Riemannian volume form of ððâ1.
Exercise 4.6.ii. Letð· â ðð be a smooth domain. Show that there exists a neighborhood ðof ðð· in ðð and a ð¶â defining function ðⶠð â ð forð· with the properties:(1) ð â ð â© ð· if and only if ð(ð) < 0,(2) and ððð â 0 if ð â ð
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132 Chapter 4: Manifolds & Forms on Manifolds
Hint: Deduce from Theorem 4.4.9 that a local version of this result is true. Show thatyou can cover ð by a family U = {ððŒ}ðŒâðŒ of open subsets of ðð such that for each thereexists a function ððŒ ⶠððŒ â ð with the properties (1) and (2). Now let (ðð)ðâ¥1 be a partitionof unity subordinate to U and let ð = ââð=1 ððððŒð where supp(ðð) â ððŒð .
Exercise 4.6.iii. In Exercise 4.6.ii suppose ð is compact. Show that there exists a globaldefining function ðⶠðð â ð forð· with properties (1) and (2).
Hint: Let ð â ð¶â0 (ð) be a function such that 0 †ð(ð¥) †1 for all ð¥ â ð and ð isidentically 1 on a neighborhood of ð, and replace ð by the function
ð(ð¥) ={{{{{
ð(ð¥)ð(ð¥) + (1 â ð(ð¥)), ð¥ â ðð â ð·ð(ð¥), ð¥ â ðð·ð(ð¥) â (1 â ð(ð¥))ð(ð¥), ð¥ â int(ð·) .
Exercise 4.6.iv. Show that the form ð¿ðððð in equation (4.6.13) does not depend on whatchoice we make of a defining function ð forð·.
Hints:⢠Show that if ð is another defining function then, at ð â ð, ððð = ðððð, where ð
is a positive constant.⢠Show that if one replaces ððð by (ðð)ð the first term in the product (ð¿ðð)(ð)(ðð)ð
changes by a factor ð and the second term by a factor 1/ð.Exercise 4.6.v. Show that the form ðð is intrinsically defined in the sense that if ð is any(ð â 1)-form satisfying equation (4.6.11), then ðð = ðâðð.
Hint: Exercise 4.5.vi.
Exercise 4.6.vi. Show that the form ðð in equation (4.6.14) is the Riemannian volume formon ð.Exercise 4.6.vii. Show that the (ð â 1)-form
ð = (ð¥21 +⯠+ ð¥2ð)âððâð=1(â1)ðâ1ð¥ððð¥1 â§â¯ ⧠ðð¥ð â§â¯ðð¥ð
is closed and prove directly that Stokesâ theorem holds for the annulus ð < ð¥21 +â¯+ð¥2ð < ðby showing that the integral of ð over the sphere, ð¥21 + ⯠+ ð¥2ð = ð, is equal to the integralover the sphere, ð¥21 +⯠+ ð¥2ð = ð.Exercise 4.6.viii. Let ðⶠððâ1 â ð be an everywhere positive ð¶â function and let ð be abounded open subset of ððâ1. Verify directly that Stokesâ theorem is true ifð· is the domaindefined by
0 < ð¥ð < ð(ð¥1,âŠ, ð¥ðâ1) , (ð¥1,âŠ, ð¥ðâ1) â ðand ð an (ð â 1)-form of the form
ð(ð¥1,âŠ, ð¥ð)ðð¥1 â§â¯ ⧠ðð¥ðâ1where ð is in ð¶â0 (ðð).Exercise 4.6.ix. Letð be an oriented ð-manifold and ð a vector field onðwhich is complete.Verify that for ð â ðºðð (ð)
â«ðð¿ðð = 0
in the following ways:(1) Directly by using the divergence theorem.
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§4.7 Degree theory on manifolds 133
(2) Indirectly by showing that
â«ððâð¡ ð = â«
ðð
where ðð¡ ⶠð â ð, ââ < ð¡ < â, is the one-parameter group of diffeomorphisms ofðgenerated by ð.
Exercise 4.6.x. Letð be an orientedð-manifold andð· â ð a smooth domainwhose closureis compact. Show that if ð â ðð· is the boundary of ð· and ðⶠð â ð a diffeomorphism ðcannot be extended to a smooth map ðⶠð· â ð.
4.7. Degree theory on manifolds
In this section weâll show how to generalize to manifolds the results about the âdegreeâof a proper mapping that we discussed in Chapter 3. Weâll begin by proving the manifoldanalogue of Theorem 3.3.1.
Theorem 4.7.1. Let ð be an oriented connected ð-manifold and ð â ðºðð (ð) a compactlysupported ð-form. Then the following are equivalent(1) â«ð ð = 0.(2) ð = ðð for some ð â ðºðâ1ð (ð).
Proof. We have already verified the assertion (2)â(1) (see Theorem 4.5.15), so what is leftto prove is the converse assertion. The proof of this is more or less identical with the proofof the â(1)â(2)â part of Theorem 3.2.2:
Step 1. Letð be a connected parametrizable open subset ofð. If ð â ðºðð (ð) has property (1),then ð = ðð for some ð â ðºðâ1ð (ð).
Proof of Step 1. Let ðⶠð0 â ð be an oriented parametrization of ð. Then
â«ð0ðâð = â«
ðð = 0
and since ð0 is a connected open subset of ðð, ðâð = ðð for some ð â ðºðâ1ð (ð0) by Theo-rem 3.3.1. Let ð = (ðâ1)âð. Then ðð = (ðâ1)âðð = ð. â¡
Step 2. Fix a base point ð0 â ð and let ð be any point of ð. Then there exists a collectionof connected parametrizable open setsð1,âŠ,ðð with ð0 â ð1 and ð â ðð such that for1 †ð †ð â 1, the intersectionðð â©ðð+1 is non-empty.
Proof of Step 2. The set of points, ð â ð, for which this assertion is true is open and the setfor which it is not true is open. Moreover, this assertion is true for ð = ð0. â¡
Step 3. We deduce Theorem 4.7.1 from a slightly stronger result. Introduce an equivalencerelation on ðºðð (ð) by declaring that two ð-forms ð1, ð2 â ðºðð (ð) are equivalent if ð1 â ð2 âððºðâ1ð¥ (ð). Denote this equivalence relation by a ð1 ⌠ð2.
We will prove:
Theorem 4.7.2. For ð1 and ð2 â ðºðð (ð) the following are equivalent:(1) â«ð ð1 = â«ð ð2(2) ð1 ⌠ð2.
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134 Chapter 4: Manifolds & Forms on Manifolds
Applying this result to a form ð â ðºðð (ð) whose integral is zero, we conclude thatð ⌠0, which means that ð = ðð for some ð â ðºðâ1ð (ð). Hence Theorem 4.7.2 impliesTheorem 4.7.1. Conversely, if â«ð ð1 = â«ð ð2, then â«ð(ð1 â ð2) = 0, so ð1 â ð2 = ðð forsome ð â ðºðð (ð). Hence Theorem 4.7.2 implies Theorem 4.7.1.
Step 4. By a partition of unity argument it suffices to prove Theorem 4.7.2 for ð1 â ðºðð (ð1)and ð2 â ðºðð (ð2) where ð1 and ð2 are connected parametrizable open sets. Moreover, if theintegrals of ð1 and ð2 are zero then ðð = ððð for some ðð â ðºðð (ðð) by Step 1, so in this case,the theorem is true. Suppose on the other hand that
â«ðð1 = â«
ðð2 = ð â 0 .
Then dividing by ð, we can assume that the integrals of ð1 and ð2 are both equal to 1.
Step 5. Letð1,âŠ,ðð be, as in Step 2, a sequence of connected parametrizable open sets withthe property that the intersections,ð1â©ð1,ððâ©ð2 andððâ©ðð+1, for ð = 1,âŠ,ðâ1, are allnon-empty. Select ð-forms, ðŒ0 â ðºðð (ð1 â©ð1), ðŒð â ðºðð (ðð â© ð2) and ðŒð â ðºðð (ðð â©ðð+1),for ð = 1,âŠ,ð â 1, such that the integral of each ðŒð over ð is equal to 1 By Step 1 we see thatTheorem 4.7.1 is true for ð1, ð2 and theð1,âŠ,ðð, hence Theorem 4.7.2 is true for ð1, ð2and theð1,âŠ,ðð, so
ð1 ⌠ðŒ0 ⌠ðŒ1 ⌠⯠⌠ðŒð ⌠ð2and thus ð1 ⌠ð2. â¡
Just as in (3.4.2) we get as a corollary of the theorem above the following âdefinitionâtheoremâ of the degree of a differentiable mapping:
Theorem4.7.3. Letð andð be compact oriented ð-manifolds and letð be connected. Given aproper ð¶â mapping, ðⶠð â ð, there exists a topological invariant deg(ð) with the definingproperty:
(4.7.4) â«ððâð = deg(ð) â«
ðð .
Proof. As in the proof of Theorem 3.4.6 pick an ð-form ð0 â ðºðð (ð) whose integral over ðis one and define the degree of ð to be the integral overð of ðâð0, i.e., set
(4.7.5) deg(ð) â â«ððâð0 .
Now let ð be any ð-form in ðºðð (ð) and let
(4.7.6) ð â â«ðð .
Then the integral of ð â ðð0 over ð is zero so there exists an (ð â 1)-form ð â ðºðâ1ð (ð) forwhich ð â ðð0 = ðð. Hence ðâð = ððâð0 + ððâð, so
â«ððâð = ðâ«
ððâð0 = deg(ð) â«
ðð
by equations (4.7.5) and (4.7.6). â¡
Itâs clear from the formula (4.7.4) that the degree of ð is independent of the choice ofð0. (Just apply this formula to any ð â ðºðð (ð) having integral over ð equal to one.) Itâs alsoclear from (4.7.4) that âdegreeâ behaves well with respect to composition of mappings:
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§4.7 Degree theory on manifolds 135
Theorem 4.7.7. Let ð be an oriented, connected ð-manifold and ðⶠð â ð a proper ð¶âmap. Then
(4.7.8) deg(ð â ð) = deg(ð) deg(ð) .
Proof. Let ð â ðºðð (ð) be a compactly supported form with the property that â«ð ð = 1. Then
deg(ð â ð) = â«ð(ð â ð)âð = â«
ððâ â ðâð = deg(ð) â«
ððâð
= deg(ð) deg(ð) . â¡
We will next show how to compute the degree of ð by generalizing to manifolds theformula for deg(ð) that we derived in §3.6.
Definition 4.7.9. A point ð â ð is a critical point of ð if the map
(4.7.10) ððð ⶠððð â ðð(ð)ðis not bijective.
Weâll denote byð¶ð the set of all critical points of ð, and weâll call a point ð â ð a criticalvalue of ð if it is in the image ð(ð¶ð) of ð¶ð under ð and a regular value if itâs not. (Thus theset of regular values is the set ð â ð(ð¶ð).) If ð is a regular value, then as we observed in Sec-tion 3.6, the map (4.7.10) is bijective for every ð â ðâ1(ð) and hence by Theorem 4.2.13, ðmaps a neighborhoodðð of ð diffeomorphically onto a neighborhoodðð of ð. In particular,ðð â© ðâ1(ð) = ð. Since ð is proper the set ðâ1(ð) is compact, and since the sets ðð are acovering ofðâ1(ð), this covering must be a finite covering. In particular, the set ðâ1(ð) itselfhas to be a finite set. As in §2.7 we can shrink the ððâs so as to insure that they have thefollowing properties:(i) Each ðð is a parametrizable open set.(ii) ðð â© ððâ² is empty for ð â ðâ².(iii) ð(ðð) = ð(ððâ²) = ð for all ð and ðâ².(iv) ð is a parametrizable open set.(v) ðâ1(ð) = âðâðâ1(ð)ðð.
To exploit these properties let ð be an ð-form in ðºðð (ð) with integral equal to 1. Thenby (v):
deg(ð) = â«ððâð = â
ðâ«ðððâð .
But ðⶠðð â ð is a diffeomorphism, hence by (4.5.13) and (4.5.14)
â«ðððâð = â«
ðð
if ðⶠðð â ð is orientation preserving and
â«ðððâð = ââ«
ðð
if ðⶠðð â ð is orientation reversing. Thus we have proved:
Theorem 4.7.11. The degree of ð is equal to the sum
(4.7.12) deg(ð) = âðâðâ1(ð)ðð
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136 Chapter 4: Manifolds & Forms on Manifolds
where ðð = +1 if the map (4.7.10) is orientation preserving and ðð = â1 if the map (4.7.10)is orientation reversing.
We will next show that Sardâs Theorem is true for maps between manifolds and hencethat there exist lots of regular values. We first observe that if ð is a parametrizable opensubset ofð and ð a parametrizable open neighborhood of ð(ð) in ð, then Sardâs Theoremis true for the map ðⶠð â ð since, up to diffeomorphism, ð and ð are just open subsetsof ðð. Now let ð be any point in ð, let ðµ be a compact neighborhood of ð, and let ð be aparametrizable open set containing ðµ. Then if ðŽ = ðâ1(ðµ) it follows from Theorem 3.4.7thatðŽ can be covered by a finite collection of parametrizable open sets,ð1,âŠ,ðð such thatð(ðð) â ð. Hence since Sardâs Theorem is true for each of the maps ðⶠðð â ð and ðâ1(ðµ)is contained in the union of the ððâs we conclude that the set of regular values of ð intersectsthe interior of ðµ in an open dense set. Thus, since ð is an arbitrary point of ð, we have proved:
Theorem 4.7.13. If ð and ð are ð-manifolds and ðⶠð â ð is a proper ð¶â map the set ofregular values of ð is an open dense subset of ð.
Since there exist lots of regular values the formula (4.7.12) gives us an effective wayof computing the degree of ð. Weâll next justify our assertion that deg(ð) is a topologicalinvariant of ð. To do so, letâs generalize Definition 2.6.14 to manifolds.
Definition 4.7.14. Letð and ð be manifolds and ð0, ð1 ⶠð â ð be ð¶â maps. A ð¶â map
(4.7.15) ð¹â¶ ð à [0, 1] â ðis a homotopy betweenð0 andð1 if for all ð¥ â ðwe haveð¹(ð¥, 0) = ð0(ð¥) andð¹(ð¥, 1) = ð1(ð¥).Moreover, ifð0 andð1 are propermaps, the homotopy,ð¹, is a proper homotopy ifð¹ is properas a ð¶â map, i.e., for every compact set ð¶ of ð, ð¹â1(ð¶) is compact.
Letâs now prove the manifold analogue of Theorem 3.6.10.
Theorem 4.7.16. Let ð and ð be oriented ð-manifolds and assume that ð is connected. Ifð0, ð2 ⶠð â ð are proper maps and the map (4.7.8) is a proper homotopy, then deg(ð1) =deg(ð2).
Proof. Letð be an ð-form inðºðð (ð)whose integral overð is equal to 1, and letð¶ â supp(ð)be the support of ð. Then if ð¹ is a proper homotopy between ð0 and ð1, the set, ð¹â1(ð¶), iscompact and its projection onð(4.7.17) { ð¥ â ð | (ð¥, ð¡) â ð¹â1(ð¶) for some ð¡ â [0, 1] }is compact. Let ðð¡ ⶠð â ð be the map: ðð¡(ð¥) = ð¹(ð¥, ð¡). By our assumptions on ð¹, ðð¡ is aproper ð¶â map. Moreover, for all ð¡ the ð-form ðâð¡ ð is a ð¶â function of ð¡ and is supportedon the fixed compact set (4.7.17). Hence itâs clear from the definitions of the integral of aform and ðð¡ that the integral
â«ððâð¡ ð
is a ð¶â function of ð¡. On the other hand this integral is by definition the degree of ðð¡ andhence by Theorem 4.7.3, deg(ðð¡) is an integer, so it does not depend on ð¡. In particular,deg(ð0) = deg(ð1). â¡
Exercises for §4.7Exercise 4.7.i. Let ðⶠð â ð be the map ð¥ ⊠ð¥ð. Show that deg(ð) = 0 if ð is even and 1if ð is odd.
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§4.8 Applications of degree theory 137
Exercise 4.7.ii. Let ðⶠð â ð be the polynomial function,
ð(ð¥) = ð¥ð + ð1ð¥ðâ1 +⯠+ ððâ1ð¥ + ðð ,where the ððâs are in ð. Show that if ð is even, deg(ð) = 0 and if ð is odd, deg(ð) = 1.
Exercise 4.7.iii. Let ð1 be the unit circle in the complex plane and let ðⶠð1 â ð1 be themap ððð ⊠ðððð, whereð ias a positive integer. What is the degree of ð?
Exercise 4.7.iv. Let ððâ1 be the unit sphere in ðð and ðⶠððâ1 â ððâ1 the antipodal mapð¥ ⊠âð¥. What is the degree of ð?
Exercise 4.7.v. Let ðŽ â ð(ð) be an orthogonal ð à ðmatrix and let
ððŽ ⶠððâ1 â ððâ1
be the map ð¥ ⊠ðŽð¥. What is the degree of ððŽ?
Exercise 4.7.vi. A manifold ð is contractable if for some point ð0 â ð, the identity map ofð onto itself is homotopic to the constant map ðð0 ⶠð â ð given by ðð0(ðŠ) â ð0. Showthat if ð is an oriented contractable ð-manifold and ð an oriented connected ð-manifoldthen for every proper mapping ðⶠð â ð we have deg(ð) = 0. In particular show that ifð > 0 and ð is compact then ð cannot be contractable.
Hint: Let ð be the identity map of ð onto itself.
Exercise 4.7.vii. Let ð and ð be oriented connected ð-manifolds and ðⶠð â ð a properð¶â map. Show that if deg(ð) â 0, then ð is surjective.
Exercise 4.7.viii. Using Sardâs Theorem prove that if ð and ð are manifolds of dimensionð and â, with ð < â and ðⶠð â ð is a proper ð¶â map, then the complement of the imageofð in ð is open and dense.
Hint: Let ð â â â ð and apply Sardâs Theorem to the map
ðⶠð à ðð â ð , ð(ð¥, ð) â ð(ð¥) .
Exercise 4.7.ix. Prove that the 2-sphere ð2 and the torus ð1 Ã ð1 are not diffeomorphic.
4.8. Applications of degree theory
Thepurpose of this sectionwill be to describe a few typical applications of degree theoryto problems in analysis, geometry, and topology. The first of these applications will be yetanother variant of the Brouwer fixed point theorem.
Application 4.8.1. Let ð be an oriented (ð + 1)-dimensional manifold, ð· â ð a smoothdomain andð â ðð· the boundary ofð·. Assume that the closureð· = ðâªð· ofð· is compact(and in particular thatð is compact).
Theorem 4.8.2. Let ð be an oriented connected ð-manifold and ðⶠð â ð a ð¶â map. Sup-pose there exists a ð¶â map ð¹â¶ ð· â ð whose restriction to ð is ð. Then deg(ð) = 0.
Proof. Let ð be an element ofðºðð (ð). Then ðð = 0, so ðð¹âð = ð¹âðð = 0. On the other handif ð ⶠð â ð is the inclusion map,
â«ð·ðð¹âð = â«
ððâð¹âð = â«
ððâð = deg(ð) â«
ðð
by Stokesâ theorem since ð¹ â ð = ð. Hence deg(ð) has to be zero. â¡
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138 Chapter 4: Manifolds & Forms on Manifolds
Application 4.8.3 (a non-linear eigenvalue problem). This application is a non-linear gen-eralization of a standard theorem in linear algebra. Let ðŽâ¶ ðð â ðð be a linear map. If ð iseven, ðŽmay not have real eigenvalues. (For instance for the map
ðŽâ¶ ð2 â ð2 , (ð¥, ðŠ) ⊠(âðŠ, ð¥)the eigenvalues of ðŽ are ±ââ1.) However, if ð is odd it is a standard linear algebra fact thatthere exists a vector, ð£ â ðð â {0}, and a ð â ð such that ðŽð£ = ðð£. Moreover replacing ð£by ð£|ð£| one can assume that |ð£| = 1. This result turns out to be a special case of a much moregeneral result. Let ððâ1 be the unit (ð â 1)-sphere in ðð and let ðⶠððâ1 â ðð be a ð¶â map.
Theorem 4.8.4. There exists a vector ð£ â ððâ1 and a number ð â ð such that ð(ð£) = ðð£.
Proof. The proof will be by contradiction. If the theorem isnât true the vectors ð£ and ð(ð£),are linearly independent and hence the vector(4.8.5) ð(ð£) = ð(ð£) â (ð(ð£) â ð£)ð£is nonzero. Let
(4.8.6) â(ð£) = ð(ð£)|ð(ð£)|
.
By (4.8.5)â(4.8.6), |ð£| = |â(ð£)| = 1 and ð£ â â(ð£) = 0, i.e., ð£ and â(ð£) are both unit vectors andare perpendicular to each other. Let(4.8.7) ðŸð¡ ⶠððâ1 â ððâ1 , 0 †ð¡ †1be the map(4.8.8) ðŸð¡(ð£) = cos(ðð¡)ð£ + sin(ðð¡)â(ð£) .For ð¡ = 0 this map is the identity map and for ð¡ = 1, it is the antipodal map ð(ð£) = ð£, hence(4.8.7) asserts that the identity map and the antipodal map are homotopic and thereforethat the degree of the antipodal map is one. On the other hand the antipodal map is therestriction to ððâ1 of the map (ð¥1,âŠ, ð¥ð) ⊠(âð¥1,âŠ, âð¥ð) and the volume form ð on ððâ1is the restriction to ððâ1 of the (ð â 1)-form
(4.8.9)ðâð=1(â1)ðâ1ð¥ððð¥ð â§â¯ ⧠ðð¥ð â§â¯ ⧠ðð¥ð .
If we replace ð¥ð by âð¥ð in equation (4.8.9) the sign of this form changes by (â1)ð henceðâð = (â1)ðð. Thus if ð is odd, ð is an orientation reversing diffeomorphism of ððâ1 ontoððâ1, so its degree is â1, and this contradicts what we just deduced from the existence of thehomotopy (4.8.8). â¡
From this argument we can deduce another interesting fact about the sphere ððâ1 whenð â 1 is even. For ð£ â ððâ1 the tangent space to ððâ1 at ð£ is just the space,
ðð£ððâ1 = { (ð£, ð€) | ð€ â ðð and ð£ â ð€ = 0 } ,so a vector field on ððâ1 can be viewed as a function ðⶠððâ1 â ðð with the property
ð(ð£) â ð£ = 0for all ð£ â ððâ1. If this function is nonzero at all points, then, letting â be the function (4.8.6),and arguing as above, weâre led to a contradiction. Hence we conclude:
Theorem 4.8.10. If ð â 1 is even and ð is a vector field on the sphere ððâ1, then there exists apoint ð â ððâ1 at which ð(ð) = 0.
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§4.8 Applications of degree theory 139
Note that if ð â 1 is odd this statement is not true. Thevector field
ð¥1ððð¥2â ð¥2ððð¥1+⯠+ ð¥2ðâ1
ððð¥2ðâ ð¥2ð
ððð¥2ðâ1
is a counterexample. It is nowhere vanishing and at ð â ððâ1 is tangent to ððâ1.Application 4.8.11 (The JordanâBrouwer separation theorem). Let ð be a compact ori-ented (ðâ1)-dimensional submanifold ofðð. In this subsection of §4.8 weâll outline a proofof the following theorem (leaving the details as a string of exercises).
Theorem 4.8.12. Ifð is connected, the complement of ðð âð ofð has exactly two connectedcomponents.
This theorem is known as the JordanâBrouwer separation theorem (and in two dimen-sions as the Jordan curve theorem). For simple, easy to visualize, submanifolds ofðð like the(ð â 1)-sphere this result is obvious, and for this reason itâs easy to be misled into thinkingof it as being a trivial (and ot very interesting) result. However, for submanifolds of ðð likethe curve in ð2 depicted in the Figure 4.8.2 itâs much less obvious. (In ten seconds or less: isthe point ð in this figure inside this curve or outside?)
To determine whether a point ð â ððâð is insideð or outsideð, one needs a topolog-ical invariant to detect the difference; such an invariant is provided by the winding number.
Definition 4.8.13. For ð â ðð â ð letðŸð ⶠð â ððâ1
be the mapðŸð(ð¥) =
ð¥ â ð|ð¥ â ð|
.
The winding numberð(ð, ð) ofð about ð is the degreeð(ð, ð) â deg(ðŸð) .
We show below thatð(ð, ð) = 0 ifð is outside, and ifð is insideð, thenð(ð, ð) = ±1(depending on the orientation of ð). Hence the winding number tells us which of the twocomponents of ðð â ð the point ð is contained in.
Exercises for §4.8Exercise 4.8.i. Letð be a connected component of ðð âð. Show that if ð0 and ð1 are inð,ð(ð, ð0) = ð(ð, ð1).
Hints:⢠First suppose that the line segment,
ðð¡ = (1 â ð¡)ð0 + ð¡ð1 , 0 †ð¡ †1lies in ð. Conclude from the homotopy invariance of degree that
ð(ð, ð0) = ð(ð, ðð¡) = ð(ð, ð1) .⢠Show that there exists a sequence of points ð1,âŠ, ðð â ð, with ð1 = ð0 andðð = ð1, such that the line segment joining ðð to ðð+1 is in ð.
Exercise 4.8.ii. Let ð be a connected (ð â 1)-dimensional submanifold of ðð. Show thatðð â ð has at most two connected components.
Hints:⢠Show that if ð is in ð there exists a small ð-ball ðµð(ð) centered at ð such thatðµð(ð) â ð has two components. (See Theorem 4.2.15.
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140 Chapter 4: Manifolds & Forms on Manifolds
⢠Show that if ð is in ðð â ð, there exists a sequence ð1,âŠ, ðð â ðð â ð such thatð1 = ð, ðð â ðµð(ð) and the line segments joining ðð to ðð+1 are in ðð â ð.
Exercise 4.8.iii. For ð£ â ððâ1, show that ð¥ â ð is in ðŸâ1ð (ð£) if and only if ð¥ lies on the ray
(4.8.14) ð + ð¡ð£ , 0 < ð¡ < â .
Exercise 4.8.iv. Let ð¥ â ð be a point on this ray. Show that
(4.8.15) (ððŸð)ð¥ ⶠððð â ðð£ððâ1
is bijective if and only if ð£ â ððð, i.e., if and only if the ray (4.8.14) is not tangent toð at ð¥.Hint: ðŸð ⶠð â ððâ1 is the composition of the maps
ðð ⶠð â ðð â {0} , ð¥ ⊠ð¥ â ð ,
andðⶠðð â {0} â ððâ1 , ðŠ ⊠ðŠ
|ðŠ|.
Show that if ð(ðŠ) = ð£, then the kernel of (ðð)ð is the one-dimensional subspace of ððspanned by ð£. Conclude that if ðŠ = ð¥ â ð and ð£ = ðŠ/|ðŠ| the composite map
(ððŸð)ð¥ = (ðð)ðŠ â (ððð)ð¥is bijective if and only if ð£ â ðð¥ð.
Exercise 4.8.v. From Exercises 4.8.iii and 4.8.iv deduce that ð£ is a regular value of ðŸð ifand only if the ray (4.8.14) intersects ð in a finite number of points and at each point ofintersection is not tangent toð at that point.
Exercise 4.8.vi. In Exercise 4.8.v show that the map (4.8.15) is orientation preserving ifthe orientations of ðð¥ð and ð£ are compatible with the standard orientation of ðððð. (SeeExercise 1.6.v.)
Exercise 4.8.vii. Conclude that deg(ðŸð) counts (with orientations) the number of pointswhere the ray (4.8.14) intersectsð.
Exercise 4.8.viii. Let ð1 â ðð â ð be a point on the ray (4.8.14). Show that if ð£ â ððâ1 is aregular value of ðŸð, it is a regular value of ðŸð1 and show that the number
deg(ðŸð) â deg(ðŸð1) = ð(ð, ð) âð(ð, ð1)
counts (with orientations) the number of points on the ray lying between ð and ð1.Hint: Exercises 4.8.v and 4.8.vii.
Exercise 4.8.ix. Let ð¥ â ð be a point on the ray (4.8.14). Suppose ð¥ = ð + ð¡ð£. Show that ifð is a small positive number and
ð± = ð + (ð¡ ± ð)ð£
thenð(ð, ð+) = ð(ð, ðâ) ± 1 ,
and from Exercise 4.8.i conclude that ð+ and ðâ lie in different components of ðð â ð. Inparticular conclude that ðð â ð has exactly two components.
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§4.8 Applications of degree theory 141
Exercise 4.8.x. Finally show that if ð is very large the difference
ðŸð(ð¥) âð|ð|, ð¥ â ð ,
is very small, i.e., ðŸð is not surjective and hence the degree of ðŸð is zero. Conclude that forð â ððâð, ð is in the unbounded component ofððâð ifð(ð, ð) = 0 and in the boundedcomponent ifð(ð, ð) = ±1 (the â±â depending on the orientation ofð).
Remark 4.8.16. The proof of JordanâBrouwer sketched above gives us an effective way ofdecidingwhether the pointð in Figure 4.8.2, is insideð or outsideð. Draw a non-tangentialray from ð. If it intersectsð in an even number of points, ð is outsideð and if it intersectsð is an odd number of points ð inside.
ᅵ
ᅵ
ᅵ
inside
outside
Figure 4.8.1. A Jordan curve with ð inside of the curve and ð outside
Application 4.8.17 (TheGaussâBonnet theorem). Letð â ðð be a compact, connected, ori-ented (ðâ1)-dimensional submanifold. By the JordanâBrouwer theoremð is the boundaryof a bounded smooth domain, so for each ð¥ â ð there exists a unique outward pointing unitnormal vector ðð¥. The Gauss map
ðŸâ¶ ð â ððâ1
is the map ð¥ ⊠ðð¥. Let ð be the Riemannian volume form of ððâ1, or, in other words, therestriction to ððâ1 of the form,
ðâð=1(â1)ðâ1ð¥ððð¥1 â§â¯ ⧠ðð¥ð â§â¯ ⧠ðð¥ð ,
and let ðð be the Riemannian volume form ofð. Then for each ð â ð(4.8.18) (ðŸâð)ð = ðŸ(ð)(ðð)ðwhere ðŸ(ð) is the scalar curvature of ð at ð. This number measures the extent to whichâð is curvedâ at ð. For instance, if ðð is the circle, |ð¥| = ð in ð2, the Gauss map is the mapð â ð/ð, so for all ð we have ðŸð(ð) = 1/ð, reflecting the fact that for ð < ð, ðð is morecurved thanðð.
The scalar curvature can also be negative. For instance for surfaces ð in ð3, ðŸ(ð) ispositive at ð ifð is convex at ð and negative ifð is convexâconcave at ð. (See Figures 4.8.2
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142 Chapter 4: Manifolds & Forms on Manifolds
and 4.8.3 below. The surface in Figure 4.8.2 is convex at ð, and the surface in Figure 4.8.3 isconvexâconcave.)
ᅵ(ᅵ)
ᅵ ᅵ
Figure 4.8.2. Positive scalar curvature at ð
ᅵ(ᅵ)ᅵ
ᅵ
Figure 4.8.3. Negative scalar curvature at ð
Let vol(ððâ1) be the Riemannian volume of the (ð â 1)-sphere, i.e., let
vol(ððâ1) = 2ðð/2
ð€(ð/2)(where ð€ is the Gamma function). Then by (4.8.18) the quotient
(4.8.19)â«ðŸðð
vol(ððâ1)is the degree of the Gauss map, and hence is a topological invariant of the surface ofð. For
ᅵ
surface ᅵ of genus ᅵ
ᅵ1 ᅵ2 ᅵ0ᅵᅵ
the sphere ᅵ2
⊠ᅵ0 ᅵ
Figure 4.8.4. The Gauss map from a surface of genus ð to the sphere
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§4.9 The index of a vector field 143
ð = 3 the GaussâBonnet theorem asserts that this topological invariant is just 1 â ð whereð is the genus of ð or, in other words, the ânumber of holesâ. Figure 4.8.4 gives a pictorialproof of this result. (Notice that at the points ð1,âŠ, ðð the surfaceð is convexâconcave sothe scalar curvature at these points is negative, i.e., the Gauss map is orientation reversing.On the other hand, at the point ð0 the surface is convex, so the Gauss map at this point isorientation preserving.)
4.9. The index of a vector field
Let ð· be a bounded smooth domain in ðð and ð = âðð=1 ð£ðð/ðð¥ð a ð¶â vector field de-fined on the closureð· ofð·.Wewill define below a topological invariant which, for âgenericâðâs, counts (with appropriate ±-signs) the number of zeros of ð. To avoid complications weâllassume ð has no zeros on the boundary ofð·.
Definition 4.9.1. Let ðð· be the boundary ofð· and let
ðð ⶠðð· â ððâ1
be the mapping
ð ⊠ð(ð)|ð(ð)|
,
where ð(ð) = (ð£1(ð),âŠ, ð£ð(ð)). The index of ð with respect toð· is by definition the degreeof ðð.
Weâll denote this index by ind(ð, ð·) and as a first step in investigating its propertiesweâll prove:
Theorem 4.9.2. Letð·1 be a smooth domain in ðð whose closure is contained inð·. Then
ind(ð, ð·) = ind(ð, ð·1)
provided that ð has no zeros in the setð· â ð·1.
Proof. Letð = ð· â ð·1. Then the map ðð ⶠðð· â ððâ1 extends to a ð¶â map
ð¹â¶ ð â ððâ1 , ð ⊠ð(ð)|ð(ð)|
.
Moreover,ðð = ð ⪠ðâ1
whereð is the boundary ofð·with its natural boundary orientation andðâ1 is the boundaryof ð·1 with its boundary orientation reversed. Let ð be an element of ðºðâ1(ððâ1) whoseintegral over ððâ1 is 1. Then if ð = ðð and ð1 = (ð1)ð are the restrictions of ð¹ to ð and ð1we get from Stokesâ theorem (and the fact that ðð = 0) the identity
0 = â«ðð¹âðð = â«
ððð¹âð
= â«ððâð â â«
ð1ðâ1 ð = deg(ð) â deg(ð1) .
Henceind(ð, ð·) = deg(ð) = deg(ð1) = ind(ð, ð·) . â¡
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144 Chapter 4: Manifolds & Forms on Manifolds
Suppose now that ð has a finite number of isolated zeros ð1,âŠ, ðð in ð·. Let ðµð(ðð) bethe open ball of radius ð with center at ðð. By making ð small enough we can assume thateach of these balls is contained in ð· and that theyâre mutually disjoint. We will define thelocal index ind(ð, ðð) of ð at ðð to be the index of ð with respect to ðµð(ðð). By Theorem 4.9.2these local indices are unchanged if we replace ð by a smaller ð, and by applying this theoremto the domain
ð·1 =ðâð=1ðµðð(ð)
we get, as a corollary of the theorem, the formula
(4.9.3) ind(ð, ð·) =ðâð=1
ind(ð, ðð)
which computes the global index of ð with respect toð· in terms of these local indices.Letâs now see how to compute these local indices. Letâs first suppose that the pointð = ðð
is at the origin of ðð. Then near ð = 0
ð = ðlin + ðâ²
whereð = ðlin = â
1â€ð,ðâ€ððð,ðð¥ðððð¥ð
the ðð,ðâs being constants and
ðð =ðâð=1ðððððð¥ð
where the ðððâs vanish to second order near zero, i.e., satisfy
(4.9.4) |ðð,ð(ð¥)| †ð¶|ð¥|2
for some constant ð¶ > 0.
Definition 4.9.5. Weâll say that the point ð = 0 is a non-degenerate zero of ð is the matrixðŽ = (ðð,ð) is non-singular.
This means in particular that
(4.9.6)ðâð=1|ðâð=1ðð,ðð¥ð| ⥠ð¶1|ð¥|
form some constant ð¶1 > 0. Thus by (4.9.4) and (4.9.6) the vector field
ðð¡ = ðlin + ð¡ðâ² , 0 †ð¡ †1
has no zeros in the ball ðµð(0), other than the point ð = 0 itself, provided that ð is smallenough. Therefore if we letðð be the boundary of ðµð(0) we get a homotopy
ð¹â¶ ðð à [0, 1] â ððâ1 , (ð¥, ð¡) ⊠ððð¡(ð¥)
between the maps ððlin ⶠðð â ððâ1 and ðð ⶠðð â ððâ1, thus by Theorem 4.7.16 we see that
deg(ðð) = deg(ððlin). Therefore,
(4.9.7) ind(ð, ð) = ind(ðlin, ð) .
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§4.9 The index of a vector field 145
Wehave assumed in the above discussion thatð = 0, but by introducing atð a translatedcoordinate system for which ð is the origin, these comments are applicable to any zero ð ofð. More explicitly if ð1,âŠ, ðð are the coordinates of ð then as above
ð = ðlin + ðâ²
where
ðlin â â1â€ð,ðâ€ððð,ð(ð¥ð â ðð)
ððð¥ð
and ðâ² vanishes to second order at ð, and if ð is a non-degenerate zero of ð, i.e., if the matrixðŽ = (ðð,ð) is non-singular, then exactly the same argument as before shows that
ind(ð, ð) = ind(ðlin, ð) .
We will next prove:
Theorem 4.9.8. If ð is a non-degenerate zero of ð, the local index ind(ð, ð) is +1 or â1 de-pending on whether the determinant of the matrix, ðŽ = (ðð,ð) is positive or negative.
Proof. As above we can, without loss of generality, assume that ð = 0. By (4.9.7) we canassume ð = ðlin. Letð· be the domain defined by
(4.9.9)ðâð=1(ðâð=1ðð,ðð¥ð)
2
< 1 ,
and let ð â ðð· be its boundary. Since the only zero of ðlin inside this domain is ð = 0 weget from (4.9.3) and (4.9.4) that
ind(ð, ð) = ind(ðlin, ð) = ind(ðlin, ð·) .
Moreover, the mapððlin ⶠð â ð
ðâ1
is, in view of (4.9.9), just the linear map ð£ ⊠ðŽð£, restricted to ð. In particular, since ðŽ isa diffeomorphism, this mapping is a diffeomorphism as well, so the degree of this map is+1 or â1 depending on whether this map is orientation preserving or not. To decide whichof these alternatives is true let ð = âðð=1(â1)ðð¥ððð¥1 â§â¯ ⧠ðð¥ð â§â¯ðð¥ð be the Riemannianvolume form on ððâ1 then
â«ððâðlinð = deg(ððlin) vol(ð
ðâ1) .
Since ð is the restriction of ðŽ toð. By Stokesâ theorem this is equal to
â«ð·ðŽâðð = ðâ«
ð·ðŽâðð¥1 â§â¯ ⧠ðð¥ð
= ð det(ðŽ) â«ð·ðð¥1 â§â¯ ⧠ðð¥ð
= ð det(ðŽ) vol(ð·)
which gives us the formula
deg(ððlin , ð·) =ð detðŽ vol(ð·)
vol(ððâ1). â¡
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146 Chapter 4: Manifolds & Forms on Manifolds
Weâll briefly describe the various types of non-degenerate zeros that can occur for vectorfields in two dimensions. To simplify this description a bit weâll assume that the matrix
ðŽ = (ð11 ð12ð21 ð22)
is diagonalizable and that its eigenvalues are not purely imaginary. Then the following fivescenarios can occur:(1) Theeigenvalues ofðŽ are real and positive. In this case the integral curves of ðlin in a neigh-
borhood of ð look like the curves in Figure 4.9.1 and hence, in a small neighborhood
Figure 4.9.1. Real and positive eigenvalues
of ð, the integral sums of ð itself look approximately like the curve in Figure 4.9.1.(2) The eigenvalues ofðŽ are real and negative. In this case the integral curves of ðlin look like
the curves in Figure 4.9.1, but the arrows are pointing into ð, rather than out of ð, i.e.,
Figure 4.9.2. Real and negative eigenvalues
(3) The eigenvalues of ðŽ are real, but one is positive and one is negative. In this case theintegral curves of ðlin look like the curves in Figure 4.9.3.
(4) The eigenvalues of ðŽ are complex and of the form ð ± ââð, with ð positive. In this casethe integral curves of ðlin are spirals going out of ð as in Figure 4.9.4.
(5) The eigenvalues of ðŽ are complex and of the form ð ± ââð, with ð negative. In this casethe integral curves are as in Figure 4.9.4 but are spiraling into ð rather than out of ð.
Definition 4.9.10. Zeros of ð of types (1) and (4) are called sources; those of types (2) and(5) are called sinks and those of type (3) are called saddles.
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§4.9 The index of a vector field 147
Figure 4.9.3. Real eigenvalues, one positive and one negative
Figure 4.9.4. Complex eigenvalues of the form ð ± ââð, with ð positive.
Thus in particular the two-dimensional version of equation (4.9.3) tells us:
Theorem 4.9.11. If the vector field ð on ð· has only non-degenerate zeros then ind(ð, ð·) isequal to the number of sources plus the number of sinks minus the number of saddles.
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CHAPTER 5
Cohomology via forms
5.1. The de Rham cohomology groups of a manifold
In the last four chapters weâve frequently encountered the question: When is a closedð-formon an open subset ofðð (or,more generally on a submanifold ofðð) exact? To inves-tigate this question more systematically than weâve done heretofore, letð be an ð-manifoldand let
ðð(ð) â {ð â ðºð(ð) | ðð = 0 }and
ðµð(ð) â ð(ðºðâ1(ð)) â ðºð(ð)be the vector spaces of closed and exact ð-forms. Since ðµð(ð) is a vector subspace ofðð(ð),we can form the quotient space
ð»ð(ð) â ðð(ð)/ðµð(ð) ,and the dimension of this space is a measure of the extent to which closed forms fail to beexact. We will call this space the ðth de Rham cohomology group of the manifold ð. Sincethe vector spaces ðð(ð) and ðµð(ð) are both infinite dimensional in general, there is noguarantee that this quotient space is finite dimensional, however, weâll show later in thischapter that it is in a lot of interesting cases.
The spacesð»ð(ð) also have compactly supported counterparts. Namely let
ððð (ð) â {ð â ðºðð (ð) | ðð = 0 }and
ðµðð (ð) â ð(ðºðâ1ð (ð)) â ðºðð (ð) .Then as above ðµðð (ð) is a vector subspace of ððð (ð) and the vector space quotient
ð»ðð (ð) â ððð (ð)/ðµðð (ð)is the ðth compactly supported de Rham cohomology group ofð.
Given a closed ð-form ð â ðð(ð) we will denote by [ð] the image of ð in the quotientspace ð»ð(ð) = ðð(ð)/ðµð(ð) and call [ð] the cohomology class of ð. We will also use thesame notation for compactly supported cohomology. If ð is in ððð (ð) weâll denote by [ð]the cohomology class of ð in the quotient spaceð»ðð (ð) = ððð (ð)/ðµðð (ð).
Some cohomology groups of manifolds weâve already computed in the previous chap-ters (although we didnât explicitly describe these computations as âcomputing cohomol-ogyâ). Weâll make a list below of some of the properties that we have already discoveredabout the de Rham cohomology of a manifoldð:(1) If ð is connected, ð»0(ð) = ð. To see this, note that a closed zero form is a functionð â ð¶â(ð) having the property ðð = 0, and ifð is connected the only such functionsare constants.
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(2) If ð is connected and non-compactð»0ð (ð) = 0. To see this, note that if ð is in ð¶â0 (ð)andð is non-compact, ð has to be zero at some point, and hence if ðð = 0, then ð hasto be identically zero.
(3) Ifð is ð-dimensional,ðºð(ð) = ðºðð (ð) = 0
for ð < 0 or ð > ð, henceð»ð(ð) = ð»ðð (ð) = 0
for ð < 0 or ð > ð.(4) Ifð is an oriented, connected ð-manifold, the integration operation is a linear map
(5.1.1) â«ðⶠðºðð (ð) â ð
and, by Theorem 4.8.2, the kernel of this map is ðµðð (ð). Moreover, in degree ð,ððð (ð) =ðºðð (ð) and hence by the definition ofð»ðð (ð) as the quotientððð (ð)/ðµðð (ð), we get from(5.1.1) a bijective map
(5.1.2) ðŒð ⶠð»ðð (ð) ⥲ ð .
In other wordsð»ðð (ð) = ð.(5) Let ð be a star-shaped open subset of ðð. In Exercises 2.6.iv to 2.6.vii we sketched a
proof of the assertion: For ð > 0 every closed form ð â ðð(ð) is exact. Translating thisassertion into cohomology language, we showed thatð»ð(ð) = 0 for ð > 0.
(6) Letð â ðð be an open rectangle. In Exercises 3.2.iv to 3.2.vii we sketched a proof of theassertion: If ð â ðºðð (ð) is closed and ð is less than ð, then ð = ðð for some (ðâ1)-formð â ðºðâ1ð (ð). Hence we showedð»ðð (ð) = 0 for ð < ð.
(7) Poincaré lemma for manifolds: Let ð be an ð-manifold and ð â ðð(ð), ð > 0 a closedð-form. Then for every point ð â ð there exists a neighborhood ð of ð and a (ð â 1)-form ð â ðºðâ1(ð) such that ð = ðð on ð. To see this note that for open subsets of ððwe proved this result in Section 2.4 and sinceð is locally diffeomorphic at ð to an opensubset of ðð this result is true for manifolds as well.
(8) Letðbe the unit sphere ðð inðð+1. Since ðð is compact, connected andorientedð»0(ðð) =ð»ð(ðð) = ð.
We will show that for ð â , 0, ðð»ð(ðð) = 0 .
To see this let ð â ðºð(ðð) be a closed ð-form and let ð = (0,âŠ, 0, 1) â ðð be the ânorthpoleâ of ðð. By the Poincaré lemma there exists a neighborhoodð of ð in ðð and a ðâ1-form ð â ðºðâ1(ð) with ð = ðð on ð. Let ð â ð¶â0 (ð) be a âbump functionâ which isequal to one on a neighborhood ð0 of ð in ð. Then
(5.1.3) ð1 = ð â ðððis a closed ð-form with compact support in ðð â {ð}. However stereographic projectiongives one a diffeomorphism
ðⶠðð â ðð â {ð}(see Exercise 5.1.i), and hence ðâð1 is a closed compactly supported ð-form onðð withsupport in a large rectangle. Thus by (5.1.3) ðâð = ðð, for some ð â ðºðâ1ð (ðð), and by(5.1.3)
(5.1.4) ð = ð(ðð + (ðâ1)âð)
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§5.1 The de Rham cohomology groups of a manifold 151
with (ðâ1)âð â ðºðâ1ð (ððâ{ð}) â ðºð(ðð), so weâve proved that for 0 < ð < ð every closedð-form on ðð is exact.We will next discuss some âpullbackâ operations in de Rham theory. Let ð and ð be
manifolds and ðⶠð â ð a ð¶â map. For ð â ðºð(ð), ððâð = ðâðð, so if ð is closed, ðâðis as well. Moreover, if ð = ðð, ðâð = ððâð, so if ð is exact, ðâð is as well. Thus we havelinear maps
ðâ ⶠðð(ð) â ðð(ð)and
(5.1.5) ðâ ⶠðµð(ð) â ðµð(ð)and composing ðâ ⶠðð(ð) â ðð(ð) with the projection
ðⶠðð(ð) â ðð(ð)/ðµð(ð)we get a linear map
(5.1.6) ðð(ð) â ð»ð(ð) .In view of (5.1.5), ðµð(ð) is in the kernel of this map, so by Proposition 1.2.9 one gets aninduced linear map
ð⯠ⶠð»ð(ð) â ð»ð(ð) ,such that ð⯠â ð is the map (5.1.6). In other words, if ð is a closed ð-form on ð, then ð⯠hasthe defining property
(5.1.7) ðâ¯[ð] = [ðâð] .This âpullbackâ operation on cohomology satisfies the following chain rule: Let ð be a
manifold and ðⶠð â ð a ð¶â map. Then if ð is a closed ð-form on ð(ð â ð)âð = ðâðâð
by the chain rule for pullbacks of forms, and hence by (5.1.7)
(5.1.8) (ð â ð)â¯[ð] = ðâ¯(ðâ¯[ð]) .The discussion above carries over verbatim to the setting of compactly supported de
Rham cohomology: If ðⶠð â ð is a proper ð¶â map ð induces a pullback map on coho-mology
ð⯠ⶠð»ðð (ð) â ð»ðð (ð)and if ðⶠð â ð and ðⶠð â ð are proper ð¶â maps then the chain rule (5.1.8) holds forcompactly supported de Rham cohomology as well as for ordinary de Rham cohomology.Notice also that if ðⶠð â ð is a diffeomorphism, we can take ð to be ð itself and ð to beðâ1, and in this case the chain rule tells us that the inducedmapsmapsð⯠ⶠð»ð(ð) â ð»ð(ð)and ð⯠ⶠð»ðð (ð) â ð»ðð (ð) are bijections, i.e.,ð»ð(ð) andð»ð(ð) andð»ðð (ð) andð»ðð (ð) areisomorphic as vector spaces.
We will next establish an important fact about the pullback operation ðâ¯; weâll showthat itâs a homotopy invariant of ð. Recall that two ð¶â maps
(5.1.9) ð0, ð1 ⶠð â ðare homotopic if there exists a ð¶â map
ð¹â¶ ð à [0, 1] â ðwith the property ð¹(ð, 0) = ð0(ð) and ð¹(ð, 1) = ð1(ð) for all ð â ð. We will prove:
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152 Chapter 5: Cohomology via forms
Theorem 5.1.10. If the maps (5.1.9) are homotopic then, for the maps they induce on coho-mology
(5.1.11) ðâ¯0 = ðâ¯1 .
Our proof of this will consist of proving this for an important special class of homo-topies, and then by âpullbackâ tricks deducing this result for homotopies in general. Let ðbe a complete vector field onð and let
ðð¡ ⶠð â ð , ââ < ð¡ < âbe the one-parameter group of diffeomorphisms that ð generates. Then
ð¹â¶ ð à [0, 1] â ð , ð¹(ð, ð¡) â ðð¡(ð) ,is a homotopy between ð0 and ð1, and weâll show that for this homotopic pair (5.1.11) istrue.
Proof. Recall that for ð â ðºð(ð)ððð¡ðâð¡ ð |ð¡=0= ð¿ð = ðððð + ðððð
and more generally for all ð¡ we haveððð¡ðâð¡ ð =
ððð ðâð +ð¡ð |
ð =0= ððð (ðð â ðð¡)âð |
ð =0
= ððð ðâð¡ ðâð ð |
ð =0= ðâð¡ððð ðâð ð |ð =0
= ðâð¡ ð¿ðð= ðâð¡ ðððð + ððâð¡ ððð .
Thus if we setðð¡ð â ðâð¡ ððð
we get from this computation:ððð¡ðâð = ððð¡ + ðð¡ðð
and integrating over 0 †ð¡ †1:(5.1.12) ðâ1 ð â ðâ0 ð = ððð + ððð
where ðⶠðºð(ð) â ðºðâ1(ð) is the operator
ðð = â«1
0ðð¡ððð¡ .
The identity (5.1.11) is an easy consequence of this âchain homotopyâ identity. If ð is inðð(ð), then ðð = 0, so
ðâ1 ð â ðâ0 ð = ðððand
ðâ¯1 [ð] â ðâ¯0 [ð] = [ðâ1 ð â ðâ0 ð] = 0 . â¡
Weâll now describe how to extract from this result a proof of Theorem 5.1.10 for anypair of homotopic maps. Weâll begin with the following useful observation.
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Proposition 5.1.13. If ð0, ð1 ⶠð â ð are homotopic ð¶â mappings, then there exists a ð¶âmap
ð¹â¶ ð à ð â ðsuch that the restriction of ð¹ toð à [0, 1] is a homotopy between ð0 and ð1.
Proof. Let ð â ð¶â0 (ð), ð ⥠0, be a bump function which is supported on the interval [ 14 ,34 ]
and is positive at ð¡ = 12 . Then
ð(ð¡) =â«ð¡ââ ð(ð )ðð â«âââ ð(ð )ðð
is a function which is zero on the interval (ââ, 14 ], is 1 on the interval [ 34 ,â), and, for all ð¡,lies between 0 and 1. Now let
ðºâ¶ ð à [0, 1] â ðbe a homotopy between ð0 and ð1 and let ð¹â¶ ð à ð â ð be the map(5.1.14) ð¹(ð¥, ð¡) = ðº(ð¥, ð(ð¡)) .This is a ð¶â map and since
ð¹(ð¥, 1) = ðº(ð¥, ð(1)) = ðº(ð¥, 1) = ð1(ð¥) ,and
ð¹(ð¥, 0) = ðº(ð¥, ð(0)) = ðº(ð¥, 0) = ð0(ð¥) ,it gives one a homotopy between ð0 and ð1. â¡
Weâre now in position to deduce Theorem 5.1.10 from the version of this result that weproved above.
Proof of Theorem 5.1.10. LetðŸð¡ ⶠð à ð â ð à ð , ââ < ð¡ < â
be the one-parameter group of diffeomorphismsðŸð¡(ð¥, ð) = (ð¥, ð + ð¡)
and let ð = ð/ðð¡ be the vector field generating this group. For a ð-form ð â ðºð(ð à ð), wehave by (5.1.12) the identity
ðŸâ1 ð â ðŸâ0 ð = ðð€ð + ð€ððwhere
(5.1.15) ð€ð = â«1
0ðŸâð¡ (ðð/ðð¡ð)ðð¡ .
Now let ð¹, as in Proposition 5.1.13, be a ð¶â mapð¹â¶ ð à ð â ð
whose restriction toð à [0, 1] is a homotopy between ð0 and ð1. Then for ð â ðºð(ð)ðŸâ1ð¹âð â ðŸâ0ð¹âð = ðð€ð¹âð + ð€ð¹âðð
by the identity (5.1.14). Now let ð ⶠð â ð à ð be the inclusion ð ⊠(ð, 0), and note that(ð¹ â ðŸ1 â ð)(ð) = ð¹(ð, 1) = ð1(ð)
and(ð¹ â ðŸ0 â ð)(ð) = ð¹(ð, 0) = ð0(ð)
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154 Chapter 5: Cohomology via forms
i.e.,ð¹ â ðŸ1 â ð = ð1
andð¹ â ðŸ0 â ð = ð0 .
Thusðâ(ðŸâ1ð¹âð â ðŸâ0ð¹âð) = ðâ1 ð â ðâ0 ð
and on the other hand by (5.1.15)
ðâ(ðŸâ1ð¹âð â ðŸâ0ð¹âð) = ððâð€ð¹âð + ðâð€ð¹âðð .
Lettingðⶠðºð(ð) â ðºðâ1(ð)
be the âchain homotopyâ operator
(5.1.16) ðð â ðâð€ð¹âð
we can write the identity above more succinctly in the form
(5.1.17) ðâ1 ð â ðâ0 ð = ððð + ððð
and from this deduce, exactly as we did earlier, the identity (5.1.11).This proof can easily be adapted to the compactly supported setting. Namely the oper-
ator (5.1.16) is defined by the integral
(5.1.18) ðð = â«1
0ðâðŸâð¡ (ðð/ðð¡ð¹âð)ðð¡ .
Hence if ð is supported on a setðŽ in ð, the integrand of (5.1.17) at ð¡ is supported on the set
(5.1.19) { ð â ð | ð¹(ð, ð¡) â ðŽ } ,
and hence ðð is supported on the set
ð(ð¹â1(ðŽ) â© ð à [0, 1])
where ðⶠð à [0, 1] â ð is the projection map ð(ð, ð¡) = ð. â¡
Suppose now that ð0 and ð1 are proper mappings and
ðºâ¶ ð à [0, 1] â ð
a proper homotopy between ð0 and ð1, i.e., a homotopy between ð0 and ð1 which is properas a ð¶â map. Then if ð¹ is the map (5.1.14) its restriction to ð à [0, 1] is also a proper map,so this restriction is also a proper homotopy between ð0 and ð1. Hence if ð is in ðºðð (ð) andðŽ is its support, the set (5.1.19) is compact, so ðð is in ðºðâ1ð (ð). Therefore all summandsin the âchain homotopyâ formula (5.1.17) are compactly supported. Thus weâve proved:
Theorem 5.1.20. If ð0, ð1 ⶠð â ð are proper ð¶â maps which are homotopic via a properhomotopy, the induced maps on cohomology
ðâ¯0 , ðâ¯1 ⶠð»ðð (ð) â ð»ðð (ð)
are the same.
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§5.1 The de Rham cohomology groups of a manifold 155
Weâll conclude this section by noting that the cohomology groupsð»ð(ð) are equippedwith a natural product operation. Namely, suppose ðð â ðºðð(ð), ð = 1, 2, is a closed formand that ðð = [ðð] is the cohomology class represented by ðð. We can then define a productcohomology class ð1 â ð2 inð»ð1+ð2(ð) by the recipe(5.1.21) ð1 â ð2 â [ð1 ⧠ð2] .To show that this is a legitimate definition we first note that since ð2 is closed
ð(ð1 ⧠ð2) = ðð1 ⧠ð2 + (â1)ð1ð1 ⧠ðð2 = 0 ,so ð1 ⧠ð2 is closed and hence does represent a cohomology class. Moreover if we replaceð1 by another representative ð1 + ðð1 = ðâ² of the cohomology class ð1, then
ðâ²1 ⧠ð2 = ð1 ⧠ð2 + ðð1 ⧠ð2 .But since ð2 is closed,
ðð1 ⧠ð2 = ð(ð1 ⧠ð2) + (â1)ð1ð1 ⧠ðð2= ð(ð1 ⧠ð2)
soðâ²1 ⧠ð2 = ð1 ⧠ð2 + ð(ð1 ⧠ð2)
and [ðâ²1 ⧠ð2] = [ð1 ⧠ð2]. Similarly (5.1.21) is unchanged if we replace ð2 by ð2 + ðð2,so the definition of (5.1.21) depends neither on the choice of ð1 nor ð2 and hence is anintrinsic definition as claimed.
There is a variant of this product operation for compactly supported cohomology classes,and weâll leave for you to check that itâs also well defined. Suppose ð1 is inð»
ð1ð (ð) and ð2 is inð»ð2(ð) (i.e., ð1 is a compactly supported class and ð2 is an ordinary cohomology class). Letð1 be a representative of ð1 inðº
ð1ð (ð) and ð2 a representative of ð2 inðºð2(ð). Then ð1 â§ð2is a closed form in ðºð1+ð2ð (ð) and hence defines a cohomology class(5.1.22) ð1 â ð2 â [ð1 ⧠ð2]
in ð»ð1+ð2ð (ð). Weâll leave for you to check that this is intrinsically defined. Weâll also leavefor you to check that (5.1.22) is intrinsically defined if the roles of ð1 and ð2 are reversed, i.e.,if ð1 is inð»ð1(ð) and ð2 inð»
ð2ð (ð) and that the products (5.1.21) and (5.1.22) both satisfy
ð1 â ð2 = (â1)ð1ð2ð2 â ð1 .Finally we note that ifð is anothermanifold andðⶠð â ð að¶âmap then forð1 â ðºð1(ð)and ð2 â ðºð2(ð)
ðâ(ð1 ⧠ð2) = ðâð1 ⧠ðâð2by (2.6.8) and hence if ð1 and ð2 are closed and ðð = [ðð], then(5.1.23) ðâ¯(ð1 â ð2) = ðâ¯ð1 â ðâ¯ð2 .
Exercises for §5.1Exercise 5.1.i (stereographic projection). Let ð â ðð be the point ð = (0,âŠ, 0, 1). Showthat for every point ð¥ = (ð¥1,âŠ, ð¥ð+1) of ðð â {ð} the ray
ð¡ð¥ + (1 â ð¡)ð , ð¡ > 0intersects the plane ð¥ð+1 = 0 in the point
ðŸ(ð¥) = 11 â ð¥ð+1
(ð¥1,âŠ, ð¥ð)
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156 Chapter 5: Cohomology via forms
and that the map ðŸâ¶ ðð â {ð} â ðð is a diffeomorphism.
Exercise 5.1.ii. Show that the operator
ðð¡ ⶠðºð(ð) â ðºðâ1(ð)in the integrand of equation (5.1.18), i.e., the operator
ðð¡ð = ðâðŸâð¡ (ðð/ðð¡)ð¹âðhas the following description. Let ð be a point of ð and let ð = ðð¡(ð). The curve ð ⊠ðð (ð)passes through ð at time ð = ð¡. Let ð£(ð) â ððð be the tangent vector to this curve at ð¡. Showthat
(5.1.24) (ðð¡ð)(ð) = (ððâð¡ )ððð£(ð)ðð .
Exercise 5.1.iii. Letð be a star-shaped open subset of ðð, i.e., a subset of ðð with the prop-erty that for every ð â ð the ray { ð¡ð | 0 †ð¡ †1 } is in ð.(1) Let ð be the vector field
ð =ðâð=1ð¥ðððð¥ð
and ðŸð¡ ⶠð â ð the map ð ⊠ð¡ð. Show that for every ð-form ð â ðºð(ð)ð = ððð + ððð
where
ðð = â«1
0ðŸâð¡ ððððð¡ð¡
.
(2) Show that if ð = âðŒ ððŒ(ð¥)ðð¥ðŒ then
(5.1.25) ðð = âðŒ,ð(â« ð¡ðâ1(â1)ðâ1ð¥ððððŒ(ð¡ð¥)ðð¡) ðð¥ðŒð
where ðð¥ðŒð â ðð¥ð1 â§â¯ ⧠ðð¥ðð â§â¯ðð¥ððExercise 5.1.iv. Let ð and ð be oriented connected ð-manifolds, and ðⶠð â ð a propermap. Show that the linear map ð¿ defined by the commutative square
ð»ðð (ð) ð»ðð (ð)
ð ð
ðâ¯
ðŒð â ðŒðâ
ð¿
is multiplication by deg(ð).
Exercise 5.1.v. A homotopy equivalence betweenð and ð is a pair of maps ðⶠð â ð andðⶠð â ð such that ð â ð â idð and ð â ð â idð. Show that if ð and ð are homotopyequivalent their cohomology groups are the same âup to isomorphismâ, i.e., ð and ð induceinverse isomorphisms ð⯠ⶠð»ð(ð) â ð»ð(ð) and ð⯠ⶠð»ð(ð) â ð»ð(ð).
Exercise 5.1.vi. Show that ðð â {0} and ððâ1 are homotopy equivalent.
Exercise 5.1.vii. What are the cohomology groups of the ð-sphere with two points deleted?Hint: The ð-sphere with one point deleted is homeomorphic to ðð.
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Exercise 5.1.viii. Let ð and ð be manifolds and ð0, ð1, ð2 ⶠð â ð three ð¶â maps. Showthat if ð0 and ð1 are homotopic and ð1 and ð2 are homotopic then ð0 and ð2 are homotopic.
Hint: The homotopy (5.1.7) has the property thatð¹(ð, ð¡) = ðð¡(ð) = ð0(ð)
for 0 †ð¡ †14 andð¹(ð, ð¡) = ðð¡(ð) = ð1(ð)
for 34 †ð¡ < 1. Show that two homotopies with these properties: a homotopy between ð0and ð1 and a homotopy between ð1 and ð2, are easy to âglue togetherâ to get a homotopybetween ð0 and ð2.Exercise 5.1.ix.(1) Letð be an ð-manifold. Given points ð0, ð1, ð2 â ð, show that if ð0 can be joined to ð1
by að¶â curve ðŸ0 ⶠ[0, 1] â ð, andð1 can be joined toð2 by að¶â curve ðŸ1 ⶠ[0, 1] â ð,then ð0 can be joined to ð2 by a ð¶â curve, ðŸâ¶ [0, 1] â ð.
Hint: Að¶â curve, ðŸâ¶ [0, 1] â ð, joining ð0 to ð2 can be thought of as a homotopybetween the maps
ðŸð0 ⶠâ â ð , â ⊠ð0and
ðŸð1 ⶠâ â ð , â ⊠ð1where â is the zero-dimensional manifold consisting of a single point.
(2) Show that if a manifold ð is connected it is arc-wise connected: any two points can byjoined by a ð¶â curve.
Exercise 5.1.x. Letð be a connected ð-manifold and ð â ðº1(ð) a closed one-form.(1) Show that if ðŸâ¶ [0, 1] â ð is a ð¶â curve there exists a partition:
0 = ð0 < ð1 < ⯠< ðð = 1of the interval [0, 1] and open sets ð1,âŠ,ðð inð such that ðŸ ([ððâ1, ðð]) â ðð and suchthat ð|ðð is exact.
(2) In part (1) show that there exist functions ðð â ð¶â(ðð) such that ð|ðð = ððð andðð (ðŸ(ðð)) = ðð+1 (ðŸ(ðð)).
(3) Show that if ð0 and ð1 are the end points of ðŸ
ðð(ð1) â ð1(ð0) = â«1
0ðŸâð .
(4) Let(5.1.26) ðŸð ⶠ[0, 1] â ð , 0 †ð †1
be a homotopic family of curves with ðŸð (0) = ð0 and ðŸð (1) = ð1. Prove that the integral
â«1
0ðŸâð ð
is independent of ð 0.Hint: Let ð 0 be a point on the interval, [0, 1]. For ðŸ = ðŸð 0 choose ððâs and ððâs as in
parts (a)â(b) and show that for ð close to ð 0, ðŸð [ððâ1, ðð] â ðð.(5) A connected manifold ð is simply connected if for any two curves ðŸ0, ðŸ1 ⶠ[0, 1] â ð
with the same end-points ð0 and ð1, there exists a homotopy (5.1.22) with ðŸð (0) = ð0and ðŸð (1) = ð1, i.e., ðŸ0 can be smoothly deformed into ðŸ1 by a family of curves all havingthe same end-points. Prove the following theorem.
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158 Chapter 5: Cohomology via forms
Theorem 5.1.27. Ifð is simply-connected, thenð»1(ð) = 0.
Exercise 5.1.xi. Show that the product operation (5.1.21) is associative and satisfies left andright distributive laws.
Exercise 5.1.xii. Letð be a compact oriented 2ð-dimensional manifold. Show that themap
ðµâ¶ ð»ð(ð) à ð»ð(ð) â ðdefined by
ðµ(ð1, ð2) = ðŒð(ð1 â ð2)is a bilinear form onð»ð(ð) and that ðµ is symmetric if ð is even and alternating if ð is odd.
5.2. The MayerâVietoris Sequence
In this section weâll develop some techniques for computing cohomology groups ofmanifolds. (These techniques are known collectively as âdiagram chasingâ and themasteringof these techniques is more akin to becoming proficient in checkers or chess or the Sundayacrostics in the New York Times than in the areas of mathematics to which theyâre applied.)Let ð¶0, ð¶1, ð¶2,⊠be vector spaces and ðⶠð¶ð â ð¶ð+1 a linear map. The sequence of vectorspaces and maps
(5.2.1) ð¶0 ð¶1 ð¶2 ⯠.ð ð ð
is called a cochain complex, or simply a complex, if ð2 = 0, i.e., if for ð â ð¶ð, ð(ðð) = 0. Forinstance ifð is a manifold the de Rham complex
(5.2.2) ðº0(ð) ðº1(ð) ðº2(ð) ⯠.ð ð ð
is an example of a complex, and the complex of compactly supported de Rham forms
(5.2.3) ðº0ð (ð) ðº1ð (ð) ðº2ð (ð) ⯠.ð ð ð
is another example. One defines the cohomology groups of the complex (5.2.1) in exactlythe same way that we defined the cohomology groups of the complexes (5.2.2) and (5.2.3)in §5.1. Let
ðð â ker(ðⶠð¶ð â ð¶ð+1) = { ð â ð¶ð | ðð = 0 }and
ðµð â ðð¶ðâ1
i.e., let ð be in ðµð if and only if ð = ðð for some ð â ð¶ðâ1. Then ðð = ð2ð = 0, so ðµð is avector subspace of ðð, and we define ð»ð(ð¶)â the ðth cohomology group of the complex(5.2.1) â to be the quotient space
(5.2.4) ð»ð(ð¶) â ðð/ðµð .Given ð â ðð we will, as in §5.1, denote its image inð»ð(ð¶) by [ð] and weâll call ð a represen-tative of the cohomology class [ð].
We will next assemble a small dictionary of âdiagram chasingâ terms.
Definition 5.2.5. Let ð0, ð1, ð2,⊠be vector spaces and ðŒð ⶠðð â ðð+1 linear maps. Thesequence
ð0 ð1 ð2 ⯠.ðŒ0 ðŒ1 ðŒ2
is an exact sequence if, for each ð, the kernel of ðŒð+1 is equal to the image of ðŒð.
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§5.2 The MayerâVietoris Sequence 159
For example the sequence (5.2.1) is exact if ðð = ðµð for all ð, or, in other words, ifð»ð(ð¶) = 0 for all ð. A simple example of an exact sequence that weâll encounter a lot belowis a sequence of the form
0 ð1 ð2 ð3 0 ,ðŒ1 ðŒ2
i.e., a five term exact sequence whose first and last terms are the vector space, ð0 = ð4 = 0,and hence ðŒ0 = ðŒ3 = 0. This sequence is exact if and only if the following conditions hold:
(1) ðŒ1 is injective,(2) the kernel of ðŒ2 equals the image of ðŒ1, and(3) ðŒ2 is surjective.
We will call an exact sequence of this form a short exact sequence. (Weâll also encountera lot below an even shorter example of an exact sequence, namely a sequence of the form
0 ð1 ð2 0 .ðŒ1
This is an exact sequence if and only if ðŒ1 is bijective.)Another basic notion in the theory of diagram chasing is the notion of a commutative
diagram. The square diagram of vector spaces and linear maps
ðŽ ðµ
ð¶ ð·
ð
ð
ð
ð
commutes if ð â ð = ð â ð, and a more complicated diagram of vector spaces and linear mapslike the diagram
ðŽ1 ðŽ2 ðŽ3
ðµ1 ðµ2 ðµ3
ð¶1 ð¶2 ð¶3
commutes if every sub-square in the diagram commutes.We now have enough âdiagram chasingâ vocabulary to formulate the MayerâVietoris
theorem. For ð = 1, 2, 3 let
(5.2.6) ð¶0ð ð¶1ð ð¶2ð ⯠.ð ð ð
be a complex and, for fixed ð, let
(5.2.7) 0 ð¶ð1 ð¶ð2 ð¶ð3 0ð ð
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160 Chapter 5: Cohomology via forms
be a short exact sequence. Assume that the diagram below commutes:
â® â® â®
0 ð¶ðâ11 ð¶ðâ12 ð¶ðâ13 0
0 ð¶ð1 ð¶ð2 ð¶ð3 0
0 ð¶ð+11 ð¶ð+12 ð¶ð+13 0
â® â® â® .
ð
ð
ð
ð
ð
ð
ð
ð
ð
ð
ð ð
That is, assume that in the left hand squares, ðð = ðð, and in the right hand squares, ðð = ðð.The MayerâVietoris theorem addresses the following question: If one has information
about the cohomology groups of two of the three complexes (5.2.6), what information aboutthe cohomology groups of the third can be extracted from this diagram? Let us first observethat the maps ð and ð give rise to mappings between these cohomology groups. Namely, forð = 1, 2, 3 let ððð be the kernel of the map ðⶠð¶ðð â ð¶ð+1ð , and ðµðð the image of the mapðⶠð¶ðâ1ð â ð¶ðð . Since ðð = ðð, ðmaps ðµð1 into ðµð2 and ðð1 into ðð2 , therefore by (5.2.4) it givesrise to a linear mapping
ð⯠ⶠð»ð(ð¶1) â ð»ð(ð¶2) .Similarly since ðð = ðð, ðmaps ðµð2 into ðµð3 and ðð2 into ðð3 , and so by (5.2.4) gives rise to alinear mapping
ð⯠ⶠð»ð(ð¶2) â ð»ð(ð¶3) .Moreover, since ð â ð = 0 the image of ð⯠is contained in the kernel of ðâ¯. Weâll leave as anexercise the following sharpened version of this observation:
Proposition 5.2.8. The kernel of ð⯠equals the image of ðâ¯, i.e., the three term sequence
(5.2.9) ð»ð(ð¶1) ð»ð(ð¶2) ð»ð(ð¶3)ð⯠ðâ¯
is exact.
Since (5.2.7) is a short exact sequence one is tempted to conjecture that (5.2.9) is alsoa short exact sequence (which, if it were true, would tell us that the cohomology groupsof any two of the complexes (5.2.6) completely determine the cohomology groups of thethird). Unfortunately, this is not the case. To see how this conjecture can be violated Let ustry to show that the mapping ð⯠is surjective. Let ðð3 be an element of ðð3 representing thecohomology class [ðð3] in ð»3(ð¶3). Since (5.2.7) is exact there exists a ðð2 in ð¶ð2 which getsmapped by ð onto ðð3 , and if ðð3 were in ðð2 this would imply
ðâ¯[ðð2] = [ððð2] = [ðð3] ,
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§5.2 The MayerâVietoris Sequence 161
i.e., the cohomology class [ðð3] would be in the image of ðâ¯. However, since thereâs no reasonfor ðð2 to be in ðð2 , thereâs also no reason for [ðð3] to be in the image of ðâ¯. What we can say,however, is that ðððð2 = ðððð2 = ððð3 = 0 since ðð3 is in ðð3 . Therefore by the exactness of(5.2.7) in degree ð + 1 there exists a unique element ðð+11 in ð¶ð+11 with property(5.2.10) ððð2 = ððð+11 .Moreover, since
0 = ð(ððð2) = ðð(ðð+11 ) = ð ððð+11and ð is injective, ððð+11 = 0, i.e.,(5.2.11) ðð+11 â ðð+11 .Thus via (5.2.10) and (5.2.11) weâve converted an element ðð3 of ðð3 into an element ðð+11 ofðð+11 and hence set up a correspondence(5.2.12) ðð3 â ðð3 â ðð+11 â ðð+11 .Unfortunately this correspondence isnât, strictly speaking, a map of ðð3 into ðð+11 ; the ðð1 in(5.2.12) isnât determined by ðð3 alone but also by the choice we made of ðð2 . Suppose, how-ever, that we make another choice of a ðð2 with the property ð(ðð2) = ðð3 . Then the differencebetween our two choices is in the kernel of ð and hence, by the exactness of (5.2.6) at levelð, is in the image of ð. In other words, our two choices are related by
(ðð2)new = (ðð2)old + ð(ðð1)for some ðð1 in ð¶ð1 , and hence by (5.2.10)
(ðð+11 )new = (ðð+11 )old + ððð1 .Therefore, even though the correspondence (5.2.12) isnât strictly speaking amap it does giverise to a well-defined map
ðð3 â ð»ð+1(ð¶1) , ðð3 ⊠[ðð+13 ] .Moreover, if ðð3 is in ðµð3 , i.e., ðð3 = ðððâ13 for some ððâ13 â ð¶ðâ13 , then by the exactness of (5.2.6)at level ðâ1, ððâ13 = ð(ððâ12 ) for some ððâ12 â ð¶ðâ12 and hence ðð3 = ð(ðððâ22 ). In other words wecan take the ðð2 above to be ðððâ12 in which case the ðð+11 in equation (5.2.10) is just zero.Thusthe map (5.2.12) maps ðµð3 to zero and hence by Proposition 1.2.9 gives rise to a well-definedmap
ð¿â¶ ð»ð(ð¶3) â ð»ð+1(ð¶1)mapping [ðð3] â [ðð+11 ]. We will leave it as an exercise to show that this mapping measuresthe failure of the arrow ð⯠in the exact sequence (5.2.9) to be surjective (and hence the failureof this sequence to be a short exact sequence at its right end).
Proposition 5.2.13. The image of the map ð⯠ⶠð»ð(ð¶2) â ð»ð(ð¶3) is equal to the kernel ofthe map ð¿â¶ ð»ð(ð¶3) â ð»ð+1(ð¶1).
Hint: Suppose that in the correspondence (5.2.12), ðð+11 is in ðµð+11 . Then ðð+11 = ððð1 forsome ðð1 in ð¶ð1 . Show that ð(ðð2 â ð(ðð1)) = ðð3 and ð(ðð2 â ð(ðð1)) = 0, i.e., ðð2 â ð(ðð1) is in ðð2 andhence ðâ¯[ðð2 â ð(ðð1)] = [ðð3].
Let us next explore the failure of themap ð⯠ⶠð»ð+1(ð¶1) â ð»ð+1(ð¶2), to be injective. Letðð+11 be in ðð+11 and suppose that its cohomology class, [ðð+11 ], gets mapped by ð⯠into zero.This translates into the statement(5.2.14) ð(ðð+11 ) = ððð2
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162 Chapter 5: Cohomology via forms
for some ðð2 â ð¶ð2 . Moreover since ððð2 = ð(ðð+11 ), ð(ððð2) = 0. But if
(5.2.15) ðð3 â ð(ðð2)then ððð3 = ðð(ðð2) = ð(ððð2) = ð(ð(ðð+11 )) = 0, so ðð3 is in ðð3 , and by (5.2.14), (5.2.15) and thedefinition of ð¿
[ðð+11 ] = ð¿[ðð3] .In other words the kernel of themap ð⯠ⶠð»ð+1(ð¶1) â ð»ð+1(ð¶2) is contained in the image ofthe map ð¿â¶ ð»ð(ð¶3) â ð»ð+1(ð¶1). We will leave it as an exercise to show that this argumentcan be reversed to prove the converse assertion and hence to prove
Proposition 5.2.16. The image of the map ð¿â¶ ð»ð(ð¶1) â ð»ð+1(ð¶1) is equal to the kernel ofthe map ð⯠ⶠð»ð+1(ð¶1) â ð»ð+1(ð¶2).
Putting together the Propositions 5.2.8, 5.2.13 and 5.2.16 we obtain the main result ofthis section: the MayerâVietoris theorem. The sequence of cohomology groups and linearmaps
(5.2.17) ⯠ð»ð(ð¶1) ð»ð(ð¶2) ð»ð(ð¶3) ð»ð+1(ð¶1) â¯ð¿ ð⯠ð⯠ð¿ ðâ¯
is exact.
Remark 5.2.18. To see an illuminating real world example of an application of these ideas,we strongly recommend that our readers stare at Exercise 5.2.v with gives amethod for com-puting the de Rham cohomology of the ð-sphere ðð in terms of the de Rham cohomologiesof ðð â {(0,âŠ, 0, 1)} and ðð â {(0,âŠ, 0, â1)} via an exact sequence of the form (5.2.17).
Remark 5.2.19. In view of the ââ¯ââs this sequence can be a very long sequence and is com-monly referred to as the long exact sequence in cohomology associated to the short exactsequence of complexes (5.2.7).
Before we discuss the applications of this result, we will introduce some vector spacenotation. Given vector spaces ð1 and ð2, weâll denote by ð1 â ð2 the vector space sum of ð1and ð2, i.e., the set of all pairs
(ð¢1, ð¢2) , ð¢ð â ððwith the addition operation
(ð¢1, ð¢2) + (ð£1 + ð£2) â (ð¢1 + ð£1, ð¢2 + ð£2)and the scalar multiplication operation
ð(ð¢1, ð¢2) â (ðð¢1, ðð¢2) .Now letð be a manifold and letð1 andð2 be open subsets ofð. Then one has a linear map
ð ⶠðºð(ð1 ⪠ð2) â ðºð(ð1) â ðºð(ð2)defined by(5.2.20) ð ⊠(ð|ð1 , ð|ð2)where ð|ðð is the restriction of ð to ðð. Similarly one has a linear map
ðⶠðºð(ð1) â ðºð(ð2) â ðºð(ð1 â© ð2)defined by
(ð1, ð2) ⊠ð1|ð1â©ð2 â ð2|ð1â©ð2 .We claim:
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§5.2 The MayerâVietoris Sequence 163
Theorem 5.2.21. The sequence
0 ðºð(ð1 â© ð2) ðºð(ð1) â ðºð(ð2) ðºð(ð1 â© ð2) 0 .ð ð
is a short exact sequence.
Proof. If the right hand side of (5.2.20) is zero, ð itself has to be zero so the map (5.2.20) isinjective. Moreover, if the right hand side of (5.2) is zero,ð1 andð2 are equal on the overlap,ð1 â© ð2, so we can glue them together to get a ð¶â ð-form on ð1 ⪠ð2 by setting ð = ð1 onð1 and ð = ð2 onð2. Thus by (5.2.20) ð(ð) = (ð1, ð2), and this shows that the kernel of ð isequal to the image of ð. Hence to complete the proof we only have to show that ð is surjective,i.e., that every form ð onðºð(ð1â©ð2) can be written as a difference, ð1|ð1â©ð2âð2|ð1â©ð2,where ð1 is inðºð(ð1) and ð2 in inðºð(ð2). To prove this weâll need the following variant ofthe partition of unity theorem.
Theorem 5.2.22. There exist functions, ððŒ â ð¶â(ð1 ⪠ð2), ðŒ = 1, 2, such that support ððŒ iscontained in ððŒ and ð1 + ð2 = 1.
Before proving this Let us use it to complete our proof of Theorem 5.2.21. Given ð âðºð(ð1 â© ð2) let
(5.2.23) ð1 â {ð2ð, on ð1 â© ð20, on ð1 â ð1 â© ð2
and let
(5.2.24) ð2 â {âð1ð, on ð1 â© ð20, on ð2 â ð1 â© ð2 .
Since ð2 is supported on ð2 the form defined by (5.2.23) is ð¶â on ð1 and since ð1 is sup-ported on ð1 the form defined by (5.2.24) is ð¶â on ð2 and since ð1 + ð2 = 1, ð1 â ð2 =(ð1 + ð2)ð = ð on ð1 â© ð2. â¡
Proof of Theorem 5.2.22. Let ðð â ð¶â0 (ð1âªð2), ð = 1, 2, 3,⊠be a partition of unity subordi-nate to the cover {ð1, ð2} ofð1 âªð2 and let ð1 be the sum of the ððâs with support onð1 andð2 the sum of the remaining ððâs. Itâs easy to check (using part (b) of Theorem 4.6.1) that ððŒis supported in ððŒ and (using part (a) of Theorem 4.6.1) that ð1 + ð2 = 1. â¡
Now let
(5.2.25) 0 ð¶01 ð¶11 ð¶21 ⯠.ð ð ð
be the de Rham complex of ð1 ⪠ð2, let
(5.2.26) 0 ð¶03 ð¶13 ð¶23 ⯠.ð ð ð
be the de Rham complex of ð1 â© ð2, and let
(5.2.27) 0 ð¶02 ð¶12 ð¶22 ⯠.ð ð ð
be the vector space direct sum of the de Rham complexes of ð1 and ð2, i.e., the complexwhose ðth term is
ð¶ð2 = ðºð(ð1) â ðºð(ð2)
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164 Chapter 5: Cohomology via forms
with ðⶠð¶ð2 â ð¶ð+12 defined to be the map ð(ð1, ð2) = (ðð1, ðð2). Since ð¶ð1 = ðºð(ð1 ⪠ð2)and ð¶ð3 = ðºð(ð1 â© ð2) we have, by Theorem 5.2.21, a short exact sequence
0 ð¶ð1 ð¶ð2 ð¶ð3 0 ,ð ð
and itâs easy to see that ð and ð commute with the ðâs:ðð = ðð and ðð = ðð .
Hence weâre exactly in the situation to which MayerâVietoris applies. Since the cohomologygroups of the complexes (5.2.25) and (5.2.26) are the de Rham cohomology groupsð»ð(ð1âªð2) and ð»ð(ð1 â© ð2), and the cohomology groups of the complex (5.2.27) are the vectorspace direct sums,ð»ð(ð1) â ð»ð(ð2), we obtain from the abstract MayerâVietoris theorem,the following de Rham theoretic version of MayerâVietoris.
Theorem 5.2.28. Letting ð = ð1 ⪠ð2 and ð = ð1 â© ð2 one has a long exact sequence in deRham cohomology:
⯠ð»ð(ð) ð»ðð (ð1) â ð»ðð (ð2) ð»ðð (ð) ð»ð+1ð (ð) ⯠.ð¿ ð⯠ð⯠ð¿ ðâ¯
This result also has an analogue for compactly supported de Rham cohomology. Let(5.2.29) ð ⶠðºðð (ð1 â© ð2) â ð»ðð (ð1) â ðºðð (ð2)be the map
ð(ð) â (ð1, ð2)where
(5.2.30) ðð â {ð, on ð1 â© ð20, on ðð â ð1 â© ð2 .
(Since ð is compactly supported on ð1 â© ð2 the form defined by equation (5.2.29) is a ð¶âform and is compactly supported on ðð.) Similarly, let
ðⶠðºðð (ð1) â ðºðð (ð2) â ðºðð (ð1 ⪠ð2)be the map
ð(ð1, ð2) â ᅵᅵ1 â ᅵᅵ2where:
ᅵᅵð â {ðð, on ðð0, on (ð1 ⪠ð2) â ðð .
As above itâs easy to see that ð is injective and that the kernel of ð is equal to the image of ð.Thus if we can prove that ð is surjective weâll have proved
Theorem 5.2.31. The sequence
0 ðºðð (ð1 â© ð2) ðºðð (ð1) â ðºðð (ð2) ðºðð (ð1 â© ð2) 0 .ð ð
is a short exact sequence.
Proof. To prove the surjectivity of ð we mimic the proof above. Given ð â ðºðð (ð1 ⪠ð2), letð1 â ð1ð|ð1
andð2 â âð2ð|ð2 .
Then by (5.2.30) we have thatð = ð(ð1, ð2). â¡
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§5.3 Cohomology of Good Covers 165
Thus, applying MayerâVietoris to the compactly supported versions of the complexes(5.2.6), we obtain the following theorem.
Theorem 5.2.32. Letting ð = ð1 ⪠ð2 and ð = ð1 â© ð2 there exists a long exact sequence incompactly supported de Rham cohomology
⯠ð»ðð (ð) ð»ðð (ð1) â ð»ðð (ð2) ð»ðð (ð) ð»ð+1ð (ð) ⯠.ð¿ ð⯠ð⯠ð¿ ðâ¯
Exercises for §5.2Exercise 5.2.i. Prove Proposition 5.2.8.
Exercise 5.2.ii. Prove Proposition 5.2.13.
Exercise 5.2.iii. Prove Proposition 5.2.16.
Exercise 5.2.iv. Show that if ð1, ð2 and ð1 â© ð2 are non-empty and connected the firstsegment of the MayerâVietoris sequence is a short exact sequence
0 ð»0(ð1 ⪠ð2) ð»0(ð1) â ð»0(ð2) ð»0(ð1 â© ð2) 0 .ð⯠ðâ¯
Exercise 5.2.v. Letð = ðð and letð1 andð2 be the open subsets of ðð obtained by removingfrom ðð the points, ð1 = (0,âŠ, 0, 1) and ð2 = (0,âŠ, 0, â1).(1) Using stereographic projection show that ð1 and ð2 are diffeomorphic to ðð.(2) Show thatð1âªð2 = ðð andð1â©ð2 is homotopy equivalent to ððâ1. (See Exercise 5.1.v.)
Hint: ð1 â© ð2 is diffeomorphic to ðð â {0}.(3) Deduce from the MayerâVietoris sequence thatð»ð+1(ðð) = ð»ð(ððâ1) for ð ⥠1.(4) Using part (3) give an inductive proof of a result that we proved by other means in
Section 5.1:ð»ð(ðð) = 0 for 1 †ð < ð.
Exercise 5.2.vi. Using the MayerâVietoris sequence of Exercise 5.2.v with cohomology re-placed by compactly supported cohomology show that
ð»ðð (ðð â {0}) â {ð, ð = 1, ð0, otherwise .
Exercise 5.2.vii. Let ð a positive integer and let
0 ð1 ð2 ⯠ððâ1 ðð 0ð1 ððâ1
be an exact sequence of finite dimensional vector spaces. Prove that âðð=1 dim(ðð) = 0.
5.3. Cohomology of Good Covers
In this section we will show that for compact manifolds (and for lots of other manifoldsbesides) the de Rham cohomology groups which we defined in §5.1 are finite dimensionalvector spaces and thus, in principle, âcomputableâ objects. A key ingredient in our proof ofthis fact is the notion of a good cover of a manifold.
Definition 5.3.1. Letð be an ð-manifold, and let U = {ððŒ}ðŒâðŒ be an open cover ofð. ThenU is a good cover if for every finite set of indices ðŒ1,âŠ, ðŒð â ðŒ the intersectionððŒ1 â©â¯â©ððŒðis either empty or is diffeomorphic to ðð.
One of our first goals in this section will be to show that good covers exist. We willsketch below a proof of the following.
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166 Chapter 5: Cohomology via forms
Theorem 5.3.2. Every manifold admits a good cover.
The proof involves an elementary result about open convex subsets of ðð.
Proposition 5.3.3. A bounded open convex subset ð â ðð is diffeomorphic to ðð.
A proof of this will be sketched in Exercises 5.3.i to 5.3.iv. One immediate consequenceof Proposition 5.3.3 is an important special case of Theorem 5.3.2.
Theorem 5.3.4. Every open subset ð of ðð admits a good cover.
Proof. For each ð â ð let ðð be an open convex neighborhood of ð in ð (for instance anð-ball centered at ð). Since the intersection of any two convex sets is again convex the cover,{ðð}ðâð is a good cover by Proposition 5.3.3. â¡
For manifolds the proof of Theorem 5.3.2 is somewhat trickier. The proof requires amanifold analogue of the notion of convexity and there are several serviceable candidates.The one we will use is the following. Let ð â ðð be an ð-manifold and for ð â ð let ðððbe the tangent space toð at ð. Recalling that ððð sits inside ðððð and that
ðððð = { (ð, ð£) | ð£ â ðð }
we get a mapððð ⪠ðððð â ðð , (ð, ð¥) ⊠ð + ð¥ ,
and this map maps ððð bijectively onto an ð-dimensional âaffineâ subspace ð¿ð ofðð whichis tangent to ð at ð. Let ðð ⶠð â ð¿ð be, as in the figure below, the orthogonal projectionofð onto ð¿ð.
ᅵ
ᅵᅵᅵᅵ(ᅵ)ᅵ
Figure 5.3.1. The orthogonal projection ofð onto ð¿ð
Definition 5.3.5. An open subset ð ofð is convex if for every ð â ð the map ðð ⶠð â ð¿ðmaps ð diffeomorphically onto a convex open subset of ð¿ð.
Itâs clear from this definition of convexity that the intersection of two open convex sub-sets of ð is an open convex subset of ð and that every open convex subset of ð is diffeo-morphic to ðð. Hence to prove Theorem 5.3.2 it suffices to prove that every point ð â ð iscontained in an open convex subset ðð of ð. Here is a sketch of how to prove this. In thefigure above let ðµð(ð) be the ball of radius ð about ð in ð¿ð centered at ð. Since ð¿ð and ððare tangent at ð the derivative of ðð at ð is just the identity map, so for ð small ðð maps aneighborhood, ððð of ð inð diffeomorphically onto ðµð(ð). We claim:
Proposition 5.3.6. For ð small, ððð is a convex subset ofð.
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§5.3 Cohomology of Good Covers 167
Intuitively this assertion is pretty obvious: if ð is in ððð and ð is small the map
ðµðð ððð ð¿ððâ1ð ðð
is to order ð2 equal to the identity map, so itâs intuitively clear that its image is a slightlywarped, but still convex, copy of ðµð(ð). We wonât, however, bother to write out the detailsthat are required to make this proof rigorous.
A good cover is a particularly good âgood coverâ if it is a finite cover. Weâll codify thisproperty in the definition below.
Definition 5.3.7. An ð-manifold ð is said to have finite topology if ð admits a finite cov-ering by open sets ð1,âŠ,ðð with the property that for every multi-index, ðŒ = (ð1,âŠ, ðð),1 †ð1 †ð2⯠< ð🠆ð, the set(5.3.8) ððŒ â ðð1 â©â¯ â© ðððis either empty or is diffeomorphic to ðð.
If ð is a compact manifold and U = {ððŒ}ðŒâðŒ is a good cover of ð then by the HeineâBorel theorem we can extract from U a finite subcover ð1,âŠ,ðð, where
ðð â ððŒð , for ðŒ1,âŠ, ðŒð â ðŒ ,hence we conclude:
Theorem 5.3.9. Every compact manifold has finite topology.
More generally, for any manifold ð, let ð¶ be a compact subset of ð. Then by HeineâBorel we can extract from the cover U a finite subcollection ð1,âŠ,ðð, where
ðð â ððŒð , for ðŒ1,âŠ, ðŒð â ðŒ ,that covers ð¶, hence letting ð â ð1 âªâ¯ ⪠ðð, weâve proved:
Theorem5.3.10. Ifð is an ð-manifold andð¶ a compact subset ofð, then there exists an openneighborhood ð of ð¶ inð with finite topology.
We can in fact even strengthen this further. Let ð0 be any open neighborhood of ð¶ inð. Then in the theorem above we can replaceð by ð0 to conclude:
Theorem5.3.11. Letð be amanifold,ð¶ a compact subset ofð, andð0 an open neighborhoodof ð¶ in ð. Then there exists an open neighborhood ð of ð¶ in ð, ð contained in ð0, so that ðhas finite topology.
We will justify the term âfinite topologyâ by devoting the rest of this section to proving.
Theorem 5.3.12. Let ð be an ð-manifold. If ð has finite topology the de Rham cohomologygroups ð»ð(ð), for ð = 0,âŠ, ð, and the compactly supported de Rham cohomology groupsð»ðð (ð), for ð = 0,âŠ, ð are finite dimensional vector spaces.
The basic ingredients in the proof of this will be the MayerâVietoris techniques that wedeveloped in §5.2 and the following elementary result about vector spaces.
Lemma 5.3.13. Let ð1, ð2, and ð3 be vector spaces and
(5.3.14) ð1 ð2 ð3ðŒ ðœ
an exact sequence of linear maps. Then if ð1 and ð3 are finite dimensional, so is ð2.
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168 Chapter 5: Cohomology via forms
Proof. Since ð3 is finite dimensional, the im(ðœ) is of finite dimension ð. Hence there existvectors ð£1,âŠ, ð£ð â ð2 such that
im(ðœ) = span(ðœ(ð£1),âŠ, ðœ(ð£ð)) .
Now let ð£ â ð2.Then ðœ(ð£) is a linear combination ðœ(ð£) = âðð=1 ðððœ(ð£ð), where ð1,âŠ, ðð âð. So
(5.3.15) ð£â² â ð£ âðâð=1ððð£ð
is in the kernel of ðœ and hence, by the exactness of (5.3.14), in the image of ðŒ. Butð1 is finitedimensional, so im(ðŒ) is finite dimensional. Letting ð£ð+1,âŠ, ð£ð be a basis of im(ðŒ) we canby (5.3.15)) write ð£ as a sum ð£ = âðð=1 ððð£ð. In other words ð£1,âŠ, ð£ð is a basis of ð2. â¡
Proof of Theorem 5.3.4. Our proof will be by induction on the number of open sets in a goodcover ofð. More specifically, let
U = {ð1,âŠ,ðð}
be a good cover of ð. Ifð = 1, ð = ð1 and hence ð is diffeomorphic to ðð, soð»ð(ð) = 0for ð > 0, and ð»0(ð) â ð, so the theorem is certainly true in this case. Let us now proveitâs true for arbitraryð by induction. Let ð â ð2 âªâ¯ âªðð. Then ð is a submanifold ofð,and, thinking ofð as a manifold in its own right, {ð2,âŠ,ðð} is a good cover ofð involvingonlyðâ1 sets. Hence the cohomology groups ofð are finite dimensional by the inductionhypothesis. The manifold ð â© ð1 has a good cover given by {ð â© ð2,âŠ,ð â© ðð}, so by theinduction hypothesis the cohomology groups ofðâ©ð1 are finite-dimensional. To prove thatthe theorem is true for ð we note that ð = ð1 ⪠ð, and and the MayerâVietoris sequencegives an exact sequence
ð»ðâ1(ð1 â© ð) ð»ð(ð) ð»ð(ð1) â ð»ð(ð) .ð¿ ðâ¯
Since the right-hand and left-hand terms are finite dimensional it follows fromLemma5.3.13that the middle term is also finite dimensional. â¡
The proof works practically verbatim for compactly supported cohomology. Forð = 1
ð»ðð (ð) = ð»ðð (ð1) = ð»ðð (ðð)
so all the cohomology groups ofð»ð(ð) are finite in this case, and the induction âðâ 1âââðâ follows from the exact sequence
ð»ðð (ð1) â ð»ðð (ð) ð»ðð (ð) ð»ð+1ð (ð1 â© ð) .ð⯠ð¿
Remark 5.3.16. A careful analysis of the proof above shows that the dimensions of thevector spacesð»ð(ð) are determined by the intersection properties of the open sets ðð, i.e.,by the list of multi-indices ðŒ for which th intersections (5.3.8) are non-empty.
This collection of multi-indices is called the nerve of the cover U, and this remarksuggests that there should be a cohomology theory which has as input the nerve of U andas output cohomology groups which are isomorphic to the de Rham cohomology groups.Such a theory does exist, and we further address it in §5.8. (A nice account of it can also befound in [13, Ch. 5]).
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§5.3 Cohomology of Good Covers 169
Exercises for §5.3Exercise 5.3.i. Let ð be a bounded open subset of ðð. A continuous function
ðⶠð â [0,â)is called an exhaustion function if it is proper as a map of ð into [0,â); i.e., if, for everyð > 0, ðâ1([0, ð]) is compact. For ð¥ â ð let
ð(ð¥) = inf{ |ð¥ â ðŠ| | ðŠ â ðð â ð } ,i.e., let ð(ð¥) be the âdistanceâ from ð¥ to the boundary ofð. Show that ð(ð¥) > 0 and that ð(ð¥)is continuous as a function of ð¥. Conclude that ð0 = 1/ð is an exhaustion function.
Exercise 5.3.ii. Show that there exists að¶â exhaustion function, ð0 ⶠð â [0,â), with theproperty ð0 ⥠ð20 where ð0 is the exhaustion function in Exercise 5.3.i.
Hints: For ð = 2, 3,⊠let
ð¶ð = { ð¥ â ð | 1ð †ð(ð¥) â€1ðâ1 }
andðð = { ð¥ â ð | 1ð+1 < ð(ð¥) <
1ðâ2 } .
Let ðð â ð¶â0 (ðð), ðð ⥠0, be a âbumpâ function which is identically one on ð¶ð and let ð0 =ââð=2 ð2ðð + 1.
Exercise 5.3.iii. Letð be a bounded open convex subset of ðð containing the origin. Showthat there exists an exhaustion function
ðⶠð â ð , ð(0) = 1 ,having the property that ð is a monotonically increasing function of ð¡ along the ray, ð¡ð¥,0 †ð¡ †1, for all points ð¥ â ð.
Hints:⢠Let ð(ð¥), 0 †ð(ð¥) †1, be a ð¶â function which is one outside a small neighbor-
hood of the origin in ð and is zero in a still smaller neighborhood of the origin.Modify the function, ð0, in the previous exercise by setting ð(ð¥) = ð(ð¥)ð0(ð¥) andlet
ð(ð¥) = â«1
0ð(ð ð¥)ðð ð + 1 .
Show that for 0 †ð¡ †1ðððð¡(ð¡ð¥) = ð(ð¡ð¥)/ð¡
and conclude from equation (5.3.15) thatð is monotonically increasing along theray, ð¡ð¥, 0 †ð¡ †1.
⢠Show that for 0 < ð < 1,ð(ð¥) ⥠ðð(ðŠ)
where ðŠ is a point on the ray, ð¡ð¥, 0 †ð¡ †1 a distance less than ð|ð¥| fromð.⢠Show that there exist constants, ð¶0 and ð¶1, ð¶1 > 0 such that
ð(ð¥) = ð¶1ð(ð¥)+ ð¶0 .
(Take ð to be equal to 12ð(ð¥)/|ð¥|.)
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170 Chapter 5: Cohomology via forms
Exercise 5.3.iv. Show that every bounded, open convex subset, ð, of ðð is diffeomorphicto ðð.
Hints:⢠Let ð(ð¥) be the exhaustion function constructed in Exercise 5.3.iii and let
ðⶠð â ðð
be the map: ð(ð¥) = ð(ð¥)ð¥. Show that this map is a bijective map of ð onto ðð.⢠Show that for ð¥ â ð and ð£ â ðð
(ðð)ð¥ð£ = ð(ð¥)ð£ + ððð¥(ð£)ð¥and conclude that ððð¥ is bijective at ð¥, i.e., that ð is locally a diffeomorphism of aneighborhood of ð¥ in ð onto a neighborhood of ð(ð¥) in ðð.
⢠Putting these together show that ð is a diffeomorphism of ð onto ðð.Exercise 5.3.v. Let ð â ð be the union of the open intervals, ð < ð¥ < ð + 1, ð an integer.Show that ð does not have finite topology.
Exercise 5.3.vi. Let ð â ð2 be the open set obtained by deleting from ð2 the points, ðð =(0, ð), for every integer ð. Show that ð does not have finite topology.
Hint: Let ðŸð be a circle of radius 12 centered about the point ðð. Using Exercises 2.1.viiiand 2.1.ix show that there exists a closed smooth one-form, ðð on ð with the property thatâ«ðŸð ðð = 1 and â«ðŸð ðð = 0 forð â ð.
Exercise 5.3.vii. Letð be an ð-manifold and U = {ð1, ð2} a good cover ofð. What are thecohomology groups ofð if the nerve of this cover is:(1) {1}, {2}.(2) {1}, {2}, {1, 2}.Exercise 5.3.viii. Let ð be an ð-manifold and U = {ð1, ð2, ð3} a good cover of ð. Whatare the cohomology groups ofð if the nerve of this cover is:(1) {1}, {2}, {3}.(2) {1}, {2}, {3}, {1, 2}.(3) {1}, {2}, {3}, {1, 2}, {1, 3}(4) {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}.(5) {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}.Exercise 5.3.ix. Let ð1 be the unit circle inð3 parametrized by arc length: (ð¥, ðŠ) = (cos ð, sin ð).Let ð1 be the set: 0 < ð < 2ð3 , ð2 the set: ð2 < ð <
3ð2 , and ð3 the set: â 2ð3 < ð <
ð3 .
(1) Show that the ððâs are a good cover of ð1.(2) Using the previous exercise compute the cohomology groups of ð1.
Exercise 5.3.x. Let ð2 be the unit 2-sphere in ð3. Show that the setsðð = { (ð¥1, ð¥2, ð¥3) â ð2 | ð¥ð > 0 }
ð = 1, 2, 3 andðð = { (ð¥1, ð¥2, ð¥3) â ð2 | ð¥ðâ3 < 0 } ,
ð = 4, 5, 6, are a good cover of ð2. What is the nerve of this cover?
Exercise 5.3.xi. Letð andð be manifolds. Show that if they both have finite topology, theirproductð Ã ð does as well.
Exercise 5.3.xii.
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§5.4 Poincaré duality 171
(1) Letð be amanifold and letð1,âŠ,ðð, be a good cover ofð. Show thatð1Ãð,âŠ,ððÃðis a good cover ofð à ð and that the nerves of these two covers are the same.
(2) By Remark 5.3.16,ð»ð(ð à ð) â ð»ð(ð) .
Verify this directly using homotopy techniques.(3) More generally, show that for all â ⥠0
ð»ð(ð à ðâ) â ð»ð(ð)(a) by concluding that this has to be the case in view of the Remark 5.3.16,(b) and by proving this directly using homotopy techniques.
5.4. Poincaré duality
In this chapter weâve been studying two kinds of cohomology groups: the ordinary deRham cohomology groupsð»ð and the compactly supported de Rham cohomology groupsð»ðð . It turns out that these groups are closely related. In fact if ð is a connected orientedð-manifold and has finite topology, then ð»ðâðð (ð) is the vector space dual of ð»ð(ð). Wegive a proof of this later in this section, however, before we do we need to review some basiclinear algebra.
Given finite dimensional vector spaces ð andð, a bilinear pairing between ð andðis a map(5.4.1) ðµâ¶ ð Ãð â ðwhich is linear in each of its factors. In other words, for fixed ð€ â ð, the map
âð€ ⶠð â ð , ð£ ⊠ðµ(ð£, ð€)is linear, and for ð£ â ð, the map
âð£ ⶠð â ð , ð€ ⊠ðµ(ð£, ð€)is linear. Therefore, from the pairing (5.4.1) one gets a map(5.4.2) ð¿ðµ ⶠð â ðâ , ð€ ⊠âð€and since âð€1 + âð€2(ð£) = ðµ(ð£, ð€1 + ð€2) = âð€1+ð€2(ð£), this map is linear. Weâll say that (5.4.1)is a non-singular pairing if (5.4.2) is bijective. Notice, by the way, that the roles of ð andð can be reversed in this definition. Letting ðµâ¯(ð€, ð£) â ðµ(ð£, ð€) we get an analogous linearmap
ð¿ðµâ¯ ⶠð â ðâ
and in fact(5.4.3) (ð¿ðµâ¯(ð£))(ð€) = (ð¿ðµ(ð€))(ð£) = ðµ(ð£, ð€) .Thus if
ðⶠð â (ðâ)â
is the canonical identification of ð with (ðâ)â given by the recipeð(ð£)(â) = â(ð£)
for ð£ â ð and â â ðâ, we can rewrite equation (5.4.3) more suggestively in the form(5.4.4) ð¿ðµâ¯ = ð¿âðµði.e., ð¿ðµ and ð¿ðµâ¯ are just the transposes of each other. In particular ð¿ðµ is bijective if and onlyif ð¿ðµâ¯ is bijective.
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172 Chapter 5: Cohomology via forms
Let us now apply these remarks to de Rham theory. Let ð be a connected, orientedð-manifold. If ð has finite topology the vector spaces, ð»ðâðð (ð) and ð»ð(ð) are both finitedimensional. We will show that there is a natural bilinear pairing between these spaces, andhence by the discussion above, a natural linear mapping ofð»ð(ð) into the vector space dualofð»ðâ1ð (ð). To see this let ð1 be a cohomology class inð»ðâðð (ð) and ð2 a cohomology classinð»ð(ð). Then by (5.1.22) their product ð1 â ð2 is an element ofð»ðð (ð), and so by (5.1.2) wecan define a pairing between ð1 and ð2 by setting(5.4.5) ðµ(ð1, ð2) â ðŒð(ð1 â ð2) .Notice that if ð1 â ðºðâðð (ð) and ð2 â ðºð(ð) are closed forms representing the cohomologyclasses ð1 and ð2, respectively, then by (5.1.22) this pairing is given by the integral
ðµ(ð1, ð2) â â«ðð1 ⧠ð2 .
Weâll next show that this bilinear pairing is non-singular in one important special case:Proposition 5.4.6. Ifð is diffeomorphic to ðð the pairing defined by (5.4.5) is non-singular.
Proof. To verify this there is very little to check. The vector spaces ð»ð(ðð) and ð»ðâðð (ðð)are zero except for ð = 0, so all we have to check is that the pairing
ð»ðð (ð) à ð»0(ð) â ðis non-singular. To see this recall that every compactly supported ð-form is closed and thatthe only closed zero-forms are the constant functions, so at the level of forms, the pairing(5.4.5) is just the pairing
ðºð(ð) à ð â ð , (ð, ð) ⊠ðâ«ðð ,
and this is zero if and only if ð is zero or ð is in ððºðâ1ð (ð). Thus at the level of cohomologythis pairing is non-singular. â¡
We will now show how to prove this result in general.Theorem 5.4.7 (Poincaré duality). Letð be an oriented, connected ð-manifold having finitetopology. Then the pairing (5.4.5) is non-singular.
The proof of this will be very similar in spirit to the proof that we gave in the last sec-tion to show that ifð has finite topology its de Rham cohomology groups are finite dimen-sional. Like that proof, it involves MayerâVietoris plus some elementary diagram-chasing.The âdiagram-chasingâ part of the proof consists of the following two lemmas.Lemma 5.4.8. Let ð1, ð2 and ð3 be finite dimensional vector spaces, and let
ð1 ð2 ð3ðŒ ðœ
be an exact sequence of linear mappings. Then the sequence of transpose maps
ðâ3 ðâ2 ðâ1ðœâ ðŒâ
is exact.Proof. Given a vector subspace,ð2, of ð2, let
ðâ2 = { â â ðâ2 | â(ð€) = 0 for all ð€ â ð} .Weâll leave for you to check that ifð2 is the kernel of ðœ, thenðâ2 is the image of ðœâ and thatifð2 is the image of ðŒ,ðâ2 is the kernel of ðŒâ. Hence if ðœ = im(ðŒ), im(ðœâ) = ker(ðŒâ). â¡
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§5.4 Poincaré duality 173
Lemma 5.4.9 (5-lemma). Consider a commutative diagram of vector spaces
(5.4.10)
ðµ1 ðµ2 ðµ3 ðµ4 ðµ5
ðŽ1 ðŽ2 ðŽ3 ðŽ4 ðŽ5 .
ðœ1
ðŸ1
ðœ2
ðŸ2
ðœ3
ðŸ3
ðœ4
ðŸ4 ðŸ5
ðŒ1 ðŒ2 ðŒ3 ðŒ4
satisfying the following conditions:(1) All the vector spaces are finite dimensional.(2) The two rows are exact.(3) The linear maps, ðŸ1, ðŸ2, ðŸ4, and ðŸ5 are bijections.Then the map ðŸ3 is a bijection.
Proof. Weâll show that ðŸ3 is surjective. Given ð3 â ðŽ3 there exists a ð4 â ðµ4 such thatðŸ4(ð4) = ðŒ3(ð3) since ðŸ4 is bijective.Moreover, ðŸ5(ðœ4(ð4)) = ðŒ4(ðŒ3(ð3)) = 0, by the exactnessof the top row. Therefore, since ðŸ5 is bijective, ðœ4(ð4) = 0, so by the exactness of the bottomrow ð4 = ðœ3(ð3) for some ð3 â ðµ3, and hence
ðŒ3(ðŸ3(ð3)) = ðŸ4(ðœ3(ð3)) = ðŸ4(ð4) = ðŒ3(ð3) .Thus ðŒ3(ð3 â ðŸ3(ð3)) = 0, so by the exactness of the top row
ð3 â ðŸ3(ð3) = ðŒ2(ð2)for some ð2 â ðŽ2. Hence by the bijectivity of ðŸ2 there exists a ð2 â ðµ2 with ð2 = ðŸ2(ð2), andhence
ð3 â ðŸ3(ð3) = ðŒ2(ð2) = ðŒ2(ðŸ2(ð2)) = ðŸ3(ðœ2(ð2)) .Thus finally
ð3 = ðŸ3(ð3 + ðœ2(ð2)) .Since ð3 was any element of ðŽ3 this proves the surjectivity of ðŸ3.
One can prove the injectivity of ðŸ3 by a similar diagram-chasing argument, but one canalso prove this with less duplication of effort by taking the transposes of all the arrows in(5.4.10) and noting that the same argument as above proves the surjectivity of ðŸâ3 ⶠðŽâ3 âðµâ3 . â¡
To prove Theorem 5.4.7 we apply these lemmas to the diagram below. In this diagramð1 and ð2 are open subsets of ð,ð = ð1 ⪠ð2, we write ð1,2 â ð1 â© ð2, and the verticalarrows are the mappings defined by the pairing (5.4.5). We will leave for you to check thatthis is a commutative diagram âup to signâ. (To make it commutative one has to replacesome of the vertical arrows, ðŸ, by their negatives: âðŸ.) This is easy to check except for thecommutative square on the extreme left. To check that this square commutes, some seriousdiagram-chasing is required.
⯠ð»ðâ1ð (ð) ð»ðð (ð1,2) ð»ðð (ð1) â ð»ðð (ð2) ð»ðð (ð) â¯
⯠ð»ðâ(ðâ1)(ð)â ð»ðâð(ð1,2)â ð»ðâð(ð1)â â ð»ðâð(ð2)â ð»ðâð(ð)â â¯
By MayerâVietoris the bottom row of the diagram is exact and by MayerâVietoris andLemma 5.4.8 the top row of the diagram is exact. Hence we can apply the âfive lemmaâ tothe diagram to conclude:
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174 Chapter 5: Cohomology via forms
Lemma 5.4.11. If the mapsð»ð(ð) â ð»ðâðð (ð)â
defined by the pairing (5.4.5) are bijective forð1, ð2 andð1,2 = ð1â©ð2, they are also bijectiveforð = ð1 ⪠ð2.
Thus to prove Theorem 5.4.7 we can argue by induction as in §5.3. Let ð1, ð2,âŠ,ððbe a good cover of ð. If ð = 1, then ð = ð1 and, hence, since ð1 is diffeomorphic toðð, the map (5.4.12) is bijective by Proposition 5.4.6. Now Let us assume the theorem istrue for manifolds involving good covers by ð open sets where ð is less than ð. Let ðâ² =ð1 âªâ¯ ⪠ððâ1 and ðâ³ = ðð. Since
ðâ² â© ðâ³ = ð1 â© ðð âªâ¯ ⪠ððâ1 â© ððit can be covered by a good cover by ð open sets, ð < ð, and hence the hypotheses of thelemma are true for ðâ², ðâ³ and ðâ² â© ðâ³. Thus the lemma says that (5.4.12) is bijective forthe unionð of ðâ² and ðâ³. â¡
Exercises for §5.4Exercise 5.4.i (proper pushforward for compactly supported de Rham cohomology). Letð be an ð-manifold, ð an ð-manifold and ðⶠð â ð a ð¶â map. Suppose that both ofthese manifolds are oriented and connected and have finite topology. Show that there existsa unique linear map
(5.4.12) ð! ⶠð»ðâðð (ð) â ð»ðâðð (ð)with the property(5.4.13) ðµð(ð!ð1, ð2) = ðµð(ð1, ðâ¯ð2)for all ð1 â ð»ðâðð (ð) and ð2 â ð»ð(ð). (In this formula ðµð is the bilinear pairing (5.4.5) onð and ðµð is the bilinear pairing (5.4.5) on ð.)Exercise 5.4.ii. Suppose that the map ð in Exercise 5.4.i is proper. Show that there exists aunique linear map
ð! ⶠð»ðâð(ð) â ð»ðâð(ð)with the property
ðµð(ð1, ð!ð2) = (â1)ð(ðâð)ðµð(ðâ¯ð1, ð2)for all ð1 â ð»ðð (ð) and ð2 â ð»ðâð(ð), and show that, ifð and ð are compact, this mappingis the same as the mapping ð! in Exercise 5.4.i.
Exercise 5.4.iii. Let ð be an open subset of ðð and let ðⶠð à ð â ð be the projection,ð(ð¥, ð¡) = ð¥. Show that there is a unique linear mapping
ðâ ⶠðºð+1ð (ð à ð) â ðºðð (ð)with the property
(5.4.14) â«ððâð ⧠ð = â«
ðÃðð ⧠ðâð
for all ð â ðºð+1ð (ð à ð) and ð â ðºðâð(ð).Hint: Let ð¥1,âŠ, ð¥ð and ð¡ be the standard coordinate functions on ðð à ð. By Exer-
cise 2.3.v every (ð + 1)-form ð â ðºð+1ð (ð Ãð) can be written uniquely in âreduced formâ asa sum
ð = âðŒððŒðð¡ ⧠ðð¥ðŒ +â
ðœððœðð¥ðœ
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§5.4 Poincaré duality 175
over multi-indices, ðŒ and ðœ, which are strictly increasing. Let
ðâð = âðŒ(â«ðððŒ(ð¥, ð¡)ðð¡) ðð¥ðŒ .
Exercise 5.4.iv. Show that the mapping ðâ in Exercise 5.4.iii satisfies ðâðð = ððâð.Exercise 5.4.v. Show that if ð is a closed compactly supported (ð + 1)-form on ð Ã ð then
[ðâð] = ð![ð]where ð! is the mapping (5.4.13) and ðâ the mapping (5.4.14).
Exercise 5.4.vi.(1) Let ð be an open subset of ðð and let ðⶠð à ðâ â ð be the projection, ð(ð¥, ð¡) = ð¥.
Show that there is a unique linear mapping
ðâ ⶠðºð+âð (ð à ðâ) â ðºðð (ð)with the property
â«ððâð ⧠ð = â«
ðÃðâð ⧠ðâð
for all ð â ðºð+âð (ð à ðâ) and ð â ðºðâð(ð).Hint: Exercise 5.4.iii plus induction on â.
(2) Show that for ð â ðºð+âð (ð à ðâ)ððâð = ðâðð .
(3) Show that if ð is a closed, compactly supported (ð + â)-form onð Ã ðâ
[ðâð] = ð![ð]where ð! ⶠð»ð+âð (ð à ðâ) â ð»ðð (ð) is the map (5.4.13).
Exercise 5.4.vii. Let ð be an ð-manifold and ð anð-manifold. Assume ð and ð are com-pact, oriented and connected, and orient ð Ã ð by giving it its natural product orientation.Let
ðⶠð à ð â ðbe the projection map ð(ð¥, ðŠ) = ðŠ. Given
ð â ðºð(ð à ð)and ð â ð, let
(5.4.15) ðâð(ð) â â«ððâðð ,
where ðð ⶠð â ð à ð is the inclusion map ðð(ð¥) = (ð¥, ð).(1) Show that the function ðâð defined by equation (5.4.15) is ð¶â, i.e., is in ðº0(ð).(2) Show that if ð is closed this function is constant.(3) Show that if ð is closed
[ðâð] = ð![ð]where ð! ⶠð»ð(ð à ð) â ð»0(ð) is the map (5.4.13).
Exercise 5.4.viii.(1) Letð be an ð-manifoldwhich is compact, connected and oriented. CombiningPoincaré
duality with Exercise 5.3.xii show that
ð»ð+âð (ð à ðâ) â ð»ðð (ð) .
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(2) Show, moreover, that if ðⶠð à ðâ â ð is the projection ð(ð¥, ð) = ð¥, thenð! ⶠð»ð+âð (ð à ðâ) â ð»ðð (ð)
is a bijection.
Exercise 5.4.ix. Let ð and ð be as in Exercise 5.4.i. Show that the pushforward operation(5.4.13) satisfies the projection formula
ð!(ð1 â ðâ¯ð2) = ð!(ð1) â ð2 ,for ð1 â ð»ðð (ð) and ð2 â ð»â(ð).
5.5. Thom classes & intersection theory
Let ð be a connected, oriented ð-manifold. If ð has finite topology its cohomologygroups are finite-dimensional, and since the bilinear pairing ðµ defined by (5.4.5) is non-singular we get from this pairing bijective linear maps
(5.5.1) ð¿ðµ ⶠð»ðâðð (ð) â ð»ð(ð)â
and(5.5.2) ð¿âðµ ⶠð»ðâð(ð) â ð»ðð (ð)â .In particular, if âⶠð»ð(ð) â ð is a linear function (i.e., an element of ð»ð(ð)â), then by(5.5.1) we can convert â into a cohomology class
(5.5.3) ð¿â1ðµ (â) â ð»ðâðð (ð) ,and similarly if âð ⶠð»ðð (ð) â ð is a linear function, we can convert it by (5.5.2) into acohomology class
(ð¿âðµ)â1(â) â ð»ðâð(ð) .One way that linear functions like this arise in practice is by integrating forms over
submanifolds of ð. Namely let ð be a closed oriented ð-dimensional submanifold of ð.Since ð is oriented, we have by (5.1.2) an integration operation in cohomology
ðŒð ⶠð»ðð (ð) â ð ,and since ð is closed the inclusion map ðð of ð intoð is proper, so we get from it a pullbackoperation on cohomology
ðâ¯ð ⶠð»ðð (ð) â ð»ðð (ð)and by composing these two maps, we get a linear map âð = ðŒð â ð
â¯ð ⶠð»ðð (ð) â ð. The
cohomology classðð â ð¿â1ðµ (âð) â ð»ðð (ð)
associated with âð is called the Thom class of the manifold ð and has the defining property
(5.5.4) ðµ(ðð, ð) = ðŒð(ðâ¯ðð)
for ð â ð»ðð (ð). Let us see what this defining property looks like at the level of forms. Letðð â ðºðâð(ð) be a closed ð-form representing ðð. Then by (5.4.5), the formula (5.5.4) forð = [ð], becomes the integral formula
(5.5.5) â«ððð ⧠ð = â«
ððâðð .
In other words, for every closed form ð â ðºðâðð (ð), the integral of ð over ð is equal tothe integral overð of ðð ⧠ð. A closed form ðð with this âreproducingâ property is called aThom form for ð. Note that if we add to ðð an exact (ð â ð)-form ð â ððºðâðâ1(ð), we get
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§5.5 Thom classes & intersection theory 177
another representative ðð +ð of the cohomology class ðð, and hence another form with thisreproducing property. Also, since equation (5.5.5) is a direct translation into form languageof equation (5.5.4), any closed (ð â ð)-form ðð with the reproducing property (5.5.5) is arepresentative of the cohomology class ðð.
These remarks make sense as well for compactly supported cohomology. Suppose ð iscompact. Then from the inclusion map we get a pullback map
ðâ¯ð ⶠð»ð(ð) â ð»ð(ð)and since ð is compact, the integration operation ðŒð is a mapð»ð(ð) â ð, so the composi-tion of these two operations is a map,
âð ⶠð»ð(ð) â ðwhich by (5.5.3) gets converted into a cohomology class
ðð = ð¿â1ðµ (âð) â ð»ðâðð (ð) .Moreover, if ðð â ðºðâðð (ð) is a closed form, it represents this cohomology class if and onlyif it has the reproducing property
(5.5.6) â«ððð ⧠ð = â«
ððâðð
for closed formsð â ðºðâð(ð). (Thereâs a subtle difference, however, between formula (5.5.5)and formula (5.5.6). In (5.5.5) ð has to be closed and compactly supported and in (5.5.6) itjust has to be closed.)
As above we have a lot of latitude in our choice of ðð: we can add to it any element ofððºðâðâ1ð (ð). One consequence of this is the following.
Theorem 5.5.7. Given a neighborhood ð of ð in ð, there exists a closed form ðð â ðºðâðð (ð)with the reproducing property
(5.5.8) â«ððð ⧠ð = â«
ððâðð
for all closed forms ð â ðºð(ð).
Hence, in particular, ðð has the reproducing property (5.5.6) for all closed forms ð âðºðâð(ð). This result shows that the Thom form ðð can be chosen to have support in anarbitrarily small neighborhood of ð.
Proof of Theorem 5.5.7. By Theorem 5.3.9 we can assume that ð has finite topology andhence, in our definition of ðð, we can replace the manifold ð by the open submanifold ð.This gives us a Thom form ðð with support in ð and with the reproducing property (5.5.8)for closed forms ð â ðºðâð(ð). â¡
Let us see what Thom forms actually look like in concrete examples. Suppose ð is de-fined globally by a system of â independent equations, i.e., suppose there exists an openneighborhood ð of ð in ð, a ð¶â map ðⶠð â ðâ, and a bounded open convex neighbor-hood ð of the origin in ðð satisfying the following properties.
Properties 5.5.9.(1) The origin is a regular value of ð.(2) ðâ1(ð) is closed inð.(3) ð = ðâ1(0).
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Then by (1) and (3) ð is a closed submanifold of ð and by (2) itâs a closed submanifoldofð. Moreover, it has a natural orientation: For every ð â ð the map
ððð ⶠððð â ð0ðâ
is surjective, and its kernel is ððð, so from the standard orientation of ð0ðâ one gets anorientation of the quotient space
ððð/ððð ,and hence since ððð is oriented, one gets by Theorem 1.9.9 an orientation on ððð. (SeeExample 4.4.5.) Now let ð be an element of ðºâð(ð). Then ðâð is supported in ðâ1(ð) andhence by property (2) we can extend it to ð by setting it equal to zero outside ð. We willprove:
Theorem 5.5.10. If â«ð ð = 1, then ðâð is a Thom form for ð.
To prove this weâll first prove that if ðâð has property (5.5.5) for some choice of ð it hasthis property for every choice of ð.
Lemma 5.5.11. Let ð1 and ð2 be forms inðºâð(ð) such that â«ð ð1 = â«ð ð2 = 1. Then for everyclosed ð-form ð â ðºðð (ð) we have
â«ððâð1 ⧠ð = â«
ððâð2 ⧠ð .
Proof. By Theorem 3.2.2, ð1 â ð2 = ððœ for some ðœ â ðºââ1ð (ð), hence, since ðð = 0(ðâð1 â ðâð2) ⧠ð = ððâðœ ⧠ð = ð(ðâðœ ⧠ð) .
Therefore, by Stokes theorem, the integral overð of the expression on the left is zero. â¡Now suppose ð = ð(ð¥1,âŠ, ð¥â)ðð¥1 â§â¯ ⧠ðð¥â, for ð in ð¶â0 (ð). For ð¡ †1 let
ðð¡ = ð¡âð (ð¥1ð¡,âŠ, ð¥âð¡) ðð¥1 â§â¯ðð¥â .
This form is supported in the convex set ð¡ð, so by Lemma 5.5.11
â«ððâðð¡ ⧠ð = â«
ððâð ⧠ð
for all closed forms ð â ðºðð (ð). Hence to prove that ðâð has the property (5.5.5), it sufficesto prove that
(5.5.12) limð¡â0â«ðâðð¡ ⧠ð = â«
ððâðð .
We prove this by proving a stronger result.
Lemma 5.5.13. The assertion (5.5.12) is true for every ð-form ð â ðºðð (ð).Proof. The canonical sumbersion theorem (see Theorem b.17) says that for every ð â ðthere exists a neighborhoodðð of ð in ð, a neighborhood,ð of 0 in ðð, and an orientationpreserving diffeomorphism ðⶠ(ð, 0) â (ðð, ð) such that(5.5.14) ð â ð = ðwhere ðⶠðð â ðâ is the canonical submersion, ð(ð¥1,âŠ, ð¥ð) = (ð¥1,âŠ, ð¥â). Let U be thecover of ð by the open sets, ð â ð and the ððâs. Choosing a partition of unity subordinateto this cover it suffices to verify (5.5.12) for ð in ðºðð (ð â ð) and ð in ðºðð (ðð). Let us firstsuppose ð is in ðºðð (ð â ð). Then ð(supp ð) is a compact subset of ðâ â {0} and hence for ð¡
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§5.5 Thom classes & intersection theory 179
small ð(supp ð) is disjoint from ð¡ð, and both sides of (5.5.12) are zero. Next suppose that ðis in ðºðð (ðð). Then ðâð is a compactly supported ð-form onð so we can write it as a sum
ðâð = âðŒâðŒ(ð¥)ðð¥ðŒ , âðŒ â ð¶â0 (ð)
the ðŒâs being strictly increasing multi-indices of length ð. Let ðŒ0 = (â + 1, â2 + 2,âŠ, ð). Then
(5.5.15) ðâðð¡ ⧠ðâð = ð¡âð (ð¥1ð¡,âŠ, ð¥âð¡) âðŒ0(ð¥1,âŠ, ð¥ð)ðð¥ð â§â¯ðð¥ð
and by (5.5.14)ðâ(ðâðð¡ ⧠ð) = ðâðð¡ ⧠ðâð
and hence since ð is orientation preserving
â«ððâðð¡ ⧠ð = â«
ðððâðð¡ ⧠ð = ð¡â â«
ððð (ð¥1ð¡,âŠ, ð¥âð¡) âðŒ0(ð¥1,âŠ, ð¥ð)ðð¥
= â«ððð(ð¥1,âŠ, ð¥â)âðŒ0(ð¡ð¥1,âŠ, ð¡ð¥â, ð¥â+1,âŠ, ð¥ð)ðð¥
and the limit of this expression as ð¡ tends to zero is
â«ð(ð¥1,âŠ, ð¥â)âðŒ0(0,âŠ, 0, ð¥â+1,âŠ, ð¥ð)ðð¥1â¯ðð¥ðor
(5.5.16) â«âðŒ(0,âŠ, 0, ð¥â+1,âŠ, ð¥ð) ðð¥â+1â¯ðð¥ð .
This, however, is just the integral ofðâð over the setðâ1(0)â©ð. By (5.5.12),ðmapsðâ1(0)â©ð diffeomorphically onto ð â© ðð and by our recipe for orienting ð this diffeomorphism isan orientation-preserving diffeomorphism, so the integral (5.5.16) is equal to â«ð ð. â¡
Weâll now describe some applications of Thom forms to topological intersection theory.Let ð and ð be closed, oriented submanifolds of ð of dimensions ð and â where ð + â = ð,and Let us assume one of them (say ð) is compact. We will show below how to define anintersection number ðŒ(ð, ð), which on the one hand will be a topological invariant of ðand ð and on the other hand will actually count, with appropriate ±-signs, the number ofpoints of intersection of ð and ð when they intersect non-tangentially. (Thus this notion issimilar to the notion of the degree deg(ð) for a ð¶â mapping ð. On the one hand deg(ð)is a topological invariant of ð. Itâs unchanged if we deform ð by a homotopy. On the otherhand if ð is a regular value ofð, then deg(ð) counts with appropriate ±-signs the number ofpoints in the set ðâ1(ð).)
Weâll first give the topological definition of this intersection number. This is by the for-mula(5.5.17) ðŒ(ð, ð) = ðµ(ðð, ðð)where ðð â ð»â(ð) and ðð â ð»ðð (ð) and ðµ is the bilinear pairing (5.4.5). If ðð â ðºâ(ð)and ðð â ðºðð (ð) are Thom forms representing ðð and ðð, (5.5.17) can also be defined as theintegral
ðŒ(ð, ð) = â«ððð ⧠ðð
or by (5.5.8), as the integral over ð,
(5.5.18) ðŒ(ð, ð) = â«ððâððð
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or, since ðð ⧠ðð = (â1)ðâðð ⧠ðð, as the integral over ð
ðŒ(ð, ð) = (â1)ðâ â«ððâððð .
In particularðŒ(ð, ð) = (â1)ðâðŒ(ð, ð).
As a test case for our declaring ðŒ(ð, ð) to be the intersection number ofð andðwe will firstprove:
Proposition 5.5.19. If ð and ð do not intersect, then ðŒ(ð, ð) = 0.
Proof. Ifð andð donât intersect then, sinceð is closed,ð = ðâð is an open neighborhoodof ð in ð, therefore since ð is compact there exists by Theorem 5.5.7 a Thom form ðð âðºâð(ð). Thus ðâððð = 0, and so by (5.5.18) we have ðŒ(ð, ð) = 0. â¡
Weâll next indicate howone computes ðŒ(ð, ð)whenð andð intersect ânon-tangentiallyâ,or, to use terminology more in current usage, when their intersection is transversal. Recallthat at a point of intersection, ð â ð â© ð, ððð and ððð are vector subspaces of ððð.
Definition 5.5.20. ð and ð intersect transversally if for every ð â ð â© ð we haveððð â© ððð = 0 .
Since ð = ð + â = dimððð + dimððð = dimððð, this condition is equivalent to
(5.5.21) ððð = ððð â ððð ,
i.e., every vector, ð¢ â ððð, can be written uniquely as a sum, ð¢ = ð£ + ð€, with ð£ â ððð andð€ â ððð. Since ð, ð and ð are oriented, their tangent spaces at ð are oriented, and weâllsay that these spaces are compatibly oriented if the orientations of the two sides of (5.5.21)agree. (In other words if ð£1,âŠ, ð£ð is an oriented basis of ððð and ð€1,âŠ,ð€â is an orientedbasis ofððð, the ð vectors, ð£1,âŠ, ð£ð,ð€1,âŠ,ð€â, are an oriented basis ofððð.)Wewill definethe local intersection number ðŒð(ð, ð) of ð and ð at ð to be equal to +1 if ð, ð, and ð arecompatibly oriented at ð and to be equal to â1 if theyâre not. With this notation weâll prove:
Theorem 5.5.22. If ð and ð intersect transversally then ð â© ð is a finite set and
ðŒ(ð, ð) = âðâðâ©ððŒð(ð, ð) .
To prove this we first need to show that transverse intersections look nice locally.
Theorem 5.5.23. If ð and ð intersect transversally, then for every ð â ð â© ð, there existsan open neighborhood ðð of ð in ð, an open neighborhood ðð of the origin in ðð, and anorientation-preserving diffeomorphism ðð ⶠðð ⥲ ðð which maps ðð â© ð diffeomorphicallyonto the subset of ðð defined by the equations: ð¥1 = ⯠= ð¥â = 0, and maps ð â© ð onto thesubset of ðð defined by the equations: ð¥â+1 = ⯠= ð¥ð = 0.
Proof. Since this result is a local result, we can assume that ð = ðð and hence by Theo-rem 4.2.15 that there exists a neighborhood ðð of ð in ðð and submersions ðⶠ(ðð, ð) â(ðâ, 0) and ðⶠ(ðð, ð) â (ðð, 0) with the properties
(5.5.24) ðð â© ð = ðâ1(0)and(5.5.25) ðð â© ð = ðâ1(0) .
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§5.5 Thom classes & intersection theory 181
Moreover, by (4.3.5) ððð = (ððð)â1(0) and ððð = (ððð)â1(0). Hence by (5.5.21), the equa-tions(5.5.26) ððð(ð£) = ððð(ð£) = 0for ð£ â ððð imply that ð£ = 0. Now let ðð ⶠðð â ðð be the map
(ð, ð) ⶠðð â ðâ à ðð = ðð .Then by (5.5.26), ððð is bijective, therefore, shrinking ðð if necessary, we can assume thatðð maps ðð diffeomorphically onto a neighborhood ðð of the origin in ðð, and hence by(5.5.24) and (5.5.25)), ðð mapsðð â©ð onto the set: ð¥1 = ⯠= ð¥â = 0 and mapsðð â©ð ontothe set: ð¥â+1 = ⯠= ð¥ð = 0. Finally, if ð isnât orientation preserving, we can make it so bycomposing it with the involution, (ð¥1,âŠ, ð¥ð) ⊠(ð¥1, ð¥2,âŠ, ð¥ðâ1, âð¥ð). â¡
From this result we deduce:
Theorem 5.5.27. If ð and ð intersect transversally, their intersection is a finite set.
Proof. By Theorem 5.5.23 the only point of intersection in ðð is ð itself. Moreover, since ðis closed and ð is compact, ð â© ð is compact. Therefore, since the ððâs cover ð â© ð we canextract a finite subcover by the HeineâBorel theorem. However, since no two ððâs cover thesame point of ð â© ð, this cover must already be a finite subcover. â¡
We will now prove Theorem 5.5.22.
Proof of Theorem 5.5.22. Since ð is closed, the map ðð ⶠð ⪠ð is proper, so by Theo-rem 3.4.7 there exists a neighborhoodð ofð inð such thatðâ©ð is contained in the unionof the open sets ðð above. Moreover by Theorem 5.5.7 we can choose ðð to be supportedin ð and by Theorem 5.3.2 we can assume that ð has finite topology, so weâre reduced toproving the theorem withð replaced by ð and ð replaced by ð â© ð. Let
ð â ð â© âðâððð ,
let ðⶠð â ðâ be the map whose restriction to ðð â© ð is ð â ðð where ð is, as in equa-tion (5.5.14), the canonical submersion of ðð onto ðâ, and finally let ð be a bounded con-vex neighborhood of ðâ, whose closure is contained in the intersection of the open sets,ð â ðð(ðð â© ð). Then ðâ1(ð) is a closed subset of ð, so if we replace ð by ð and ð byðâ©ð, the data (ð, ð, ð) satisfy Properties 5.5.9. Thus to prove Theorem 5.5.22 it suffices byTheorem 5.5.10 to prove this theorem with
ðð = ðð(ð)ðâðon ðð â© ð where ðð(ð) = +1 or â1 depending on whether the orientation of ð â© ðð inTheorem 5.5.10 coincides with the given orientation of ð or not. Thus
ðŒ(ð, ð) = (â1)ðâðŒ(ð, ð)
= (â1)ðâ âðâððð(ð) â«
ððâððâð
= (â1)ðâ âðâððð(ð) â«
ððâððâððâð
= âðâð(â1)ðâðð(ð) â«
ðâ©ðð(ð â ðð â ðð)âð .
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182 Chapter 5: Cohomology via forms
But ð â ðð â ðð maps an open neighborhood of ð in ðð â© ð diffeomorphically onto ð, andð is compactly supported in ð, so since â«ð ð = 1,
â«ðâ©ðð(ð â ðð â ðð)âð = ðð(ð) â«
ðð = ðð(ð) ,
where ðð(ð) = +1 or â1 depending on whether ð â ðð â ðð is orientation preserving or not.Thus finally
ðŒ(ð, ð) = âðâð(â1)ðâðð(ð)ðð(ð) .
We will leave as an exercise the task of unraveling these orientations and showing that
(â1)ðâðð(ð)ðð(ð) = ðŒð(ð, ð)
and hence that ðŒ(ð, ð) = âðâð ðŒð(ð, ð). â¡
Exercises for §5.5Exercise 5.5.i. Letðbe a connected orientedð-manifold,ð a connected oriented â-dimensionalmanifold, ðⶠð â ð a ð¶â map, and ð a closed submanifold ofð of dimension ð â ð â â.Suppose ð is a âlevel setâ of the map ð, i.e., suppose that ð is a regular value of ð and thatð = ðâ1(ð). Show that if ð is in ðºâð(ð) and its integral over ð is 1, then one can orient ð sothat ðð = ðâð is a Thom form for ð.
Hint: Theorem 5.5.10.
Exercise 5.5.ii. In Exercise 5.5.i show that if ð â ð is a compact oriented â-dimensionalsubmanifold ofð then
ðŒ(ð, ð) = (â1)ðâ deg(ð â ðð) .
Exercise 5.5.iii. Let ð1 be another regular value of the map ðⶠð â ð and let ð1 = ðâ1(ð).Show that
ðŒ(ð, ð) = ðŒ(ð1, ð) .
Exercise 5.5.iv.(1) Show that if ð is a regular value of the map ð â ðð ⶠð â ð, then ð and ð intersect
transversally.(2) Show that this is an âif and only if â proposition: If ð and ð intersect transversally, thenð is a regular value of the map ð â ðð.
Exercise 5.5.v. Suppose ð is a regular value of the map ð â ðð. Show that ð is in ðâ©ð if andonly if ð is in the preimage (ð â ðð)â1(ð) of ð and that
ðŒð(ð, ð) = (â1)ðâððwhere ðð is the orientation number of the map ð â ðð, at ð, i.e., ðð = 1 if ð â ðð is orientation-preserving at ð and ðð = â1 if ð â ðð is orientation-reversving at ð.
Exercise 5.5.vi. Suppose the map ðⶠð â ð is proper. Show that there exists a neighbor-hood, ð, of ð inð having the property that all points of ð are regular values of ð.
Hint: Since ð is a regular value of ð there exists, for every ð â ðâ1(ð) a neighborhoodðð of ð on which ð is a submersion. Conclude, by Theorem 3.4.7, that there exists a neigh-borhood ð of ð with ðâ1(ð) â âðâðâ1(ð)ðð.
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§5.6 The Lefschetz Theorem 183
Exercise 5.5.vii. Show that in every neighborhoodð1 of ð inð there exists a point ð1 whosepreimage
ð1 â ðâ1(ð1)intersects ð transversally. Conclude that one can âdeform ð an arbitrarily small amount sothat it intersects ð transversallyâ.
Hint: Exercise 5.5.iv plus Sardâs theorem.
Exercise 5.5.viii (Intersection theory for mappings). Let ð be an oriented, connected ð-manifold, ð a compact, oriented â-dimensional submanifold, ð an oriented manifold ofdimension ð â ð â â and ðⶠð â ð a proper ð¶â map. Define the intersection number ofð with ð to be the integral
ðŒ(ð, ð) â â«ððâðð .
(1) Show that ðŒ(ð, ð) is a homotopy invariant of ð, i.e., show that if ð0, ð1 ⶠð â ð areproper ð¶â maps and are properly homotopic, then
ðŒ(ð0, ð) = ðŒ(ð1, ð) .(2) Show that if ð is a closed submanifold of ð of dimension ð = ð â â and ðð ⶠð â ð is
the inclusion map, then ðŒ(ðð, ð) = ðŒ(ð, ð).
Exercise 5.5.ix.(1) Let ð be an oriented connected ð-manifold and let ð be a compact zero-dimensional
submanifold consisting of a single point ð§0 â ð. Show that if ð is in ðºðð (ð) then ð is aThom form for ð if and only if its integral is 1.
(2) Let ð be an oriented ð-manifold and ðⶠð â ð a ð¶â map. Show that for ð = {ð§0} asin part (1) we have ðŒ(ð, ð) = deg(ð).
5.6. The LefschetzTheorem
In this section weâll apply the intersection techniques that we developed in §5.5 to aconcrete problem in dynamical systems: counting the number of fixed points of a differen-tiable mapping. The Brouwer fixed point theorem, which we discussed in §3.6, told us thata ð¶â map of the unit ball into itself has to have at least one fixed point. The Lefschetz the-orem is a similar result for manifolds. It will tell us that a ð¶â map of a compact manifoldinto itself has to have a fixed point if a certain topological invariant of the map, its globalLefschetz number, is nonzero.
Before stating this result, we will first show how to translate the problem of countingfixed points of a mapping into an intersection number problem. Let ð be an oriented com-pact ð-manifold and ðⶠð â ð a ð¶â map. Define the graph of ð in ð à ð to be theset
ð€ð â { (ð¥, ð(ð¥)) | ð¥ â ð } â ð à ð .Itâs easy to see that this is an ð-dimensional submanifold ofð Ãð and that this manifold isdiffeomorphic toð itself. In fact, in one direction, there is a ð¶â map
(5.6.1) ðŸð ⶠð â ð€ð , ð¥ ⊠(ð¥, ð(ð¥)) ,and, in the other direction, a ð¶â map
ðⶠð€ð â ð , (ð¥, ð(ð¥)) ⊠ð¥ ,and itâs obvious that these maps are inverses of each other and hence diffeomorphisms. Wewill orient ð€ð by requiring that ðŸð and ð be orientation-preserving diffeomorphisms.
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184 Chapter 5: Cohomology via forms
An example of a graph is the graph of the identity map of ð onto itself. This is thediagonal inð Ã ð
ð¥ â { (ð¥, ð¥) | ð¥ â ð } â ð à ðand its intersection with ð€ð is the set
ð€ð â© ð¥ = { (ð¥, ð¥) | ð(ð¥) = ð¥ } ,which is just the set of fixed points ofð. Hence a natural way to count the fixed points ofð isas the intersection number of ð€ð and ð¥ inðÃð. To do so we need these three manifolds tobe oriented, but, as we noted above, ð€ð and ð¥ acquire orientations from the identifications(5.6.1) and, as forðÃð, weâll give it its natural orientation as a product of orientedmanifolds.(See §4.5.)
Definition 5.6.2. The global Lefschetz number of ð is the intersection number
ð¿(ð) â ðŒ(ð€ð, ð¥) .
In this section weâll give two recipes for computing this number: one by topologicalmethods and the other by making transversality assumptions and computing this numberas a sum of local intersection numbers viaTheorem 5.5.22.We first showwhat one gets fromthe transversality approach.
Definition 5.6.3. The map ðⶠð â ð is a Lefschetz map, or simply Lefschetz, if ð€ð and ð¥intersect transversally.
Let us see what being Lefschetz entails. Supposeð is a fixed point ofð.Then at the pointð = (ð, ð) of ð€ð(5.6.4) ðð(ð€ð) = (ððŸð)ðððð = { (ð£, ððð(ð£)) | ð£ â ððð}and, in particular, for the identity map,
ðð(ð¥) = { (ð£, ð£) | ð£ â ððð} .Therefore, if ð¥ and ð€ð are to intersect transversally, the intersection of ðð(ð€ð) â©ðð(ð¥) insideðð(ð à ð) has to be the zero space. In other words if
(5.6.5) (ð£, ððð(ð£)) = (ð£, ð£)then ð£ = 0. But the identity (5.6.5) says that ð£ is a fixed point of ððð, so transversality at ðamounts to the assertion
ððð(ð£) = ð£ ⺠ð£ = 0 ,or in other words the assertion that the map
(5.6.6) (idððð âððð) ⶠððð â ððð
is bijective.
Proposition5.6.7. The local intersection number ðŒð(ð€ð, ð¥) is 1 if (5.6.6) is orientation-preservingand â1 otherwise.
In other words ðŒð(ð€ð, ð¥) is the sign of det(idððð âððð).
Proof. To prove this let ð1,âŠ, ðð be an oriented basis of ððð and let
(5.6.8) ððð(ðð) =ðâð=1ðð,ððð .
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§5.6 The Lefschetz Theorem 185
Now set ð£ð â (ðð, 0) â ðð(ð à ð) and ð€ð â (0, ðð) â ðð(ð à ð). Then by the definition ofthe product orientation on ð à ð, we have that (ð£1,âŠ, ð£ð, ð€1,âŠ,ð€ð) is an oriented basisof ðð(ð à ð), and by (5.6.4)
(ð£1 + âðð=1 ðð,ðð€ð,âŠ, ð£ð + â
ðð=1 ðð,ðð€ð)
is an oriented basis of ððð€ð and ð£1 + ð€1,âŠ, ð£ð + ð€ð is an oriented basis of ððð¥. ThusðŒð(ð€ð, ð¥) = ±1, depending on whether or not the basis
(ð£1 + âðð=1 ðð,ðð€ð,âŠ, ð£ð + â
ðð=1 ðð,ðð€ð, ð£1 + ð€1,âŠ, ð£ð + ð€ð)
of ðð(ð à ð) is compatibly oriented with the basis (5.6.8). Thus ðŒð(ð€ð, ð¥) = ±1, the signdepending on whether the determinant of the 2ð à 2ðmatrix relating these two bases:
ð â (idð ðŽidð idð)
is positive or negative, whereðŽ = (ðð,ð). However, by the block matrix determinant formula,we see that to see that
det(ð) = det(idð âðŽ idâ1ð idð) det(idð) = det(idð âðŽ) ,hence by (5.6.8) we see that det(ð) = det(idð âððð). (If you are not familiar with this, it iseasy to see that by elementary row operations ð can be converted into the matrix
(idð ðŽ0 idð âðŽ
)
apply Exercise 1.8.vii.) â¡
Let us summarize what we have shown so far.
Theorem 5.6.9. A ðⶠð â ð, is a Lefschetz map if and only if for every fixed point ð of ð,the map
idððð âððð ⶠððð â ðððis bijective. Moreover for Lefschetz maps ð we have
ð¿(ð) = âð=ð(ð)ð¿ð(ð)
where ð¿ð(ð) = +1 if idððð âððð is orientation-preserving and ð¿ð(ð) = â1 if idððð âððð isorientation-reversing.
Weâll next describe how to compute ð¿(ð) as a topological invariant of ð. Let ðð€ð be theinclusion map of ð€ð into ð à ð and let ðð¥ â ð»ð(ð à ð) be the Thom class of ð¥. Then by(5.5.18)
ð¿(ð) = ðŒð€ð(ðâð€ððð¥)
and hence since themapping ðŸð ⶠð â ðÃð defined by (5.6.1) is an orientation-preservingdiffeomorphism ofð ⥲ ð€ð we have
(5.6.10) ð¿(ð) = ðŒð(ðŸâððð¥) .To evaluate the expression on the right weâll need to know some facts about the cohomologygroups of product manifolds.Themain result on this topic is the KÃŒnneth theorem, and weâlltake up the formulation and proof of this theorem in §5.7. First, however, weâll describe aresult which follows from the KÃŒnneth theorem and which will enable us to complete ourcomputation of ð¿(ð).
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186 Chapter 5: Cohomology via forms
Let ð1 and ð2 be the projection ofðÃð onto its first and second factors, i.e., for ð = 1, 2let
ðð ⶠð à ð â ðbe the projection ðð(ð¥1, ð¥2) â ð¥ð. Then by equation (5.6.1) we have
(5.6.11) ð1 â ðŸð = idðand
(5.6.12) ð2 â ðŸð = ð .
Lemma 5.6.13. If ð1 and ð2 are in ðºð(ð) then
(5.6.14) â«ðÃððâ1ð1 ⧠ðâ2ð2 = (â«
ðð1)(â«
ðð2) .
Proof. By a partition of unity argument we can assume that ðð has compact support in aparametrizable open set, ðð. Let ðð be an open subset of ðð and ðð ⶠðð â ðð an orientation-preserving diffeomorphism. Then
ðâð ð = ðððð¥1 â§â¯ ⧠ðð¥ðwith ðð â ð¶â0 (ðð), so the right hand side of (5.6.14) is the product of integrals over ðð:
(5.6.15) (â«ððð1(ð¥)ðð¥)(â«
ððð2(ð¥)ðð¥) .
Moreover, sinceð Ã ð is oriented by its product orientation, the map
ðⶠð1 à ð2 â ð1 à ð2is given by (ð¥, ðŠ) ⊠(ð1(ð¥) ,ð2(ðŠ)) is an orientation-preserving diffeomorphism and sinceðð â ð = ðð
ðâ(ðâ1ð1 ⧠ðâ2ð2) = ðâ1ð1 ⧠ðâ2ð2= ð1(ð¥)ð2(ðŠ)ðð¥1 â§â¯ ⧠ðð¥ð ⧠ððŠ1 â§â¯ ⧠ððŠð
and hence the left hand side of (5.6.14) is the integral over ð2ð of the function, ð1(ð¥)ð2(ðŠ),and therefore, by integration by parts, is equal to the product (5.6.15). â¡
As a corollary of this lemma we get a product formula for cohomology classes.
Lemma 5.6.16. If ð1 and ð2 are inð»ð(ð) thenðŒðÃð(ðâ1 ð1 â ðâ2 ð2) = ðŒð(ð1)ðŒð(ð2) .
Now let ðð â dimð»ð(ð) and note that since ð is compact, Poincaré duality tells usthat ðð = ðâ when â = ð â ð. In fact it tells us even more. Let ðð1,âŠ, ðððð be a basis ofð»ð(ð).Then, since the pairing (5.4.5) is non-singular, there exists for â = ð â ð a âdualâ basis
ðâð , ð = 1,âŠ, ðâofð»â(ð) satisfying(5.6.17) ðŒð(ððð â ðâð ) = ð¿ð,ð .
Lemma 5.6.18. The cohomology classes
ðâ¯1ðâð â ðâ¯2ððð , ð + â = ð
for ð = 0,âŠ, ð and 1 †ð, ð †ðð, are a basis forð»ð(ð à ð).
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§5.6 The Lefschetz Theorem 187
This is the corollary of the KÃŒnneth theorem that we alluded to above (and whose proofweâll give in §5.7). Using these results we prove the following.
Theorem 5.6.19. The Thom class ðð¥ â ð»ð(ð à ð) is given explicitly by the formula
(5.6.20) ðð¥ = âð+â=ð(â1)â
ððâð=1ðâ¯1ððð â ð
â¯2ðâð .
Proof. Wehave to check that for every cohomology class ð â ð»ð(ðÃð), the classðð¥ definedby (5.6.20) has the reproducing property
(5.6.21) ðŒðÃð(ðð¥ â ð) = ðŒð¥(ðâ¯ð¥ð)
where ðð¥ is the inclusion map of ð¥ intoð à ð. However the map
ðŸð¥ ⶠð â ð à ð , ð¥ ⊠(ð¥, ð¥)is an orientation-preserving diffeomorphism ofð onto ð¥, so it suffices to show that
(5.6.22) ðŒðÃð(ðð¥ â ð) = ðŒð(ðŸâ¯ð¥ð)
and by Lemma 5.6.18 it suffices to verify (5.6.22) for ðâs of the form
ð = ðâ¯1ðâð â ðâ¯2ððð .
The product of this class with a typical summand of (5.6.20), for instance, the summand
(5.6.23) (â1)ââ²ðâ¯1ððâ²ð â ðâ¯2ðââ²ð , ðâ² + ââ² = ð ,
is equal, up to sign to,ðâ¯1ðð
â²ð â ðâð â ð
â¯2ððð â ðâ
â²ð .
Notice, however, that if ð â ðâ² this product is zero: For ð < ðâ², ðâ² + â is greater than ð + âand hence greater than ð. Therefore
ððâ²ð â ðâð â ð»ðâ²+â(ð)
is zero since ð is of dimension ð, and for ð > ðâ², ââ² is greater than â and ððð â ðââ²ð is zero for
the same reason. Thus in taking the product of ðð¥ with ð we can ignore all terms in the sumexcept for the terms, ðâ² = ð and ââ² = â. For these terms, the product of (5.6.23) with ð is
(â1)ðâðâ¯1ððð â ðâð â ðâ¯2ðð3 â ðâð .
(Exercise: Check this. Hint: (â1)â(â1)â2 = 1.) Thus
ðð¥ â ð = (â1)ðâððâð=1ðâ¯1ððð â ðâð â ð
â¯2ððð â ðâð
and hence by Lemma 5.6.13 and (5.6.17)
ðŒðÃð(ðð¥ â ð) = (â1)ðâððâð=1ðŒð(ððð â ðâð)ðŒð(ððð â ðâð )
= (â1)ðâððâð=1ð¿ð,ðð¿ð,ð
= (â1)ðâð¿ð,ð .
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188 Chapter 5: Cohomology via forms
On the other hand for ð = ðâ¯1ðâð â ðâ¯2ððð ðŸâ¯ð¥ð = ðŸ
â¯ð¥ðâ¯1ðâð â ðŸ
â¯ð¥ðâ¯2ððð
= (ð1 â ðŸð¥)â¯ðâð(ð2 â ðŸð¥)â¯ððð = ðâð â ððð
sinceð1 â ðð¥ = ð2 â ðŸð¥ = idð .
SoðŒð(ðŸâ¯ð¥ð) = ðŒð(ðâð â ððð ) = (â1)ðâð¿ð,ð
by (5.6.17). Thus the two sides of (5.6.21) are equal. â¡
Weâre now in position to compute ð¿(ð) , i.e., to compute the expression ðŒð(ðŸâððð¥) onthe right hand side of (5.6.10). Since ðâ1,âŠ, ðâðâ is a basis ofð»â(ð) the linear mapping
(5.6.24) ð⯠ⶠð»â(ð) â ð»â(ð)can be described in terms of this basis by a matrix [ðâð,ð] with the defining property
ðâ¯ðâð =ðââð=1ðâð,ððâð .
Thus by equations (5.6.11), (5.6.12) and (5.6.20) we have
ðŸâ¯ððð¥ = ðŸâ¯ð âð+â=ð(â1)â
ððâð=1ðâ¯1ððð â ð
â¯2ðâð
= âð+â=ð(â1)â
ððâð=1(ð1 â ðŸð)â¯ððð â (ð2 â ðð)â¯ðâð
= âð+â=ð(â1)â
ððâð=1ððð â ðâ¯ðâð
= âð+â=ð(â1)â
ððâð=1ðâð,ðððð â ðâð .
Thus by (5.6.17)
ðŒð(ðŸâ¯ððð¥) = â
ð+â=ð(â1)â â
1â€ð,ðâ€ðâðâð,ððŒð(ððð â ðâð )
= âð+â=ð(â1)â â
1â€ð,ðâ€ðâðâð,ðð¿ð,ð
=ðââ=0(â1)â
ðââð=1ðâð,ð .
But âðâð=1 ðâð,ð is just the trace of the linear mapping (5.6.24) (see Exercise 5.6.xii), so we endup with the following purely topological prescription of ð¿(ð).
Theorem 5.6.25. The Lefschetz number ð¿(ð) is the alternating sum
ð¿(ð) =ðââ=0(â1)â tr(ð⯠| ð»â(ð))
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§5.6 The Lefschetz Theorem 189
where tr(ð⯠| ð»â(ð)) is the trace of the map ð⯠ⶠð»â(ð) â ð»â(ð).Exercises for §5.6
Exercise 5.6.i. Show that if ð0 ⶠð â ð and ð1 ⶠð â ð are homotopic ð¶â mappingsð¿(ð0) = ð¿(ð1).Exercise 5.6.ii.(1) The Euler characteristic ð(ð) ofð is defined to be the intersection number of the diag-
onal with itself inð à ð, i.e., the âself-intersectionâ numberð(ð) â ðŒ(ð¥, ð¥) = ðŒðÃð(ðð¥, ðð¥) .
Show that if a ð¶â map ðⶠð â ð is homotopic to the identity, then ð¿(ð) = ð(ð).(2) Show that
ð(ð) =ðââ=0(â1)â dimð»â(ð) .
(3) Show that ifð is an odd-dimensional (compact) manifold, then ð(ð) = 0.Exercise 5.6.iii.(1) Let ðð be the unit ð-sphere in ðð+1. Show that if ðⶠðð â ðð is a ð¶â map, then
ð¿(ð) = 1 + (â1)ð deg(ð) .(2) Conclude that if deg(ð) â (â1)ð+1, then ð has to have a fixed point.Exercise 5.6.iv. Let ð be a ð¶â mapping of the closed unit ball ðµð+1 into itself and letðⶠðð â ðð be the restriction of ð to he boundary of ðµð+1. Show that if deg(ð) â (â1)ð+1then the fixed point of ð predicted by Brouwerâs theorem can be taken to be a point on theboundary of ðµð+1.Exercise 5.6.v.(1) Show that if ðⶠðð â ðð is the antipodal map ð(ð¥) â âð¥, then deg(ð) = (â1)ð+1.(2) Conclude that the result in Exercise 5.6.iv is sharp. Show that the map
ðⶠðµð+1 â ðµð+1 , ð(ð¥) = âð¥ ,has only one fixed point, namely the origin, and in particular has no fixed points on theboundary.
Exercise 5.6.vi. Let ð be a vector field on a compact manifold ð. Since ð is compact, ðgenerates a one-parameter group of diffeomorphisms(5.6.26) ðð¡ ⶠð â ð , ââ < ð¡ < â .(1) Let ðŽð¡ be the set of fixed points of ðð¡. Show that this set contains the set of zeros of ð,
i.e., the points ð â ð where ð(ð) = 0.(2) Suppose that for some ð¡0, ðð¡0 is Lefschetz. Show that for all ð¡, ðð¡ maps ðŽð¡0 into itself.(3) Show that for |ð¡| < ð, ð small, the points of ðŽð¡0 are fixed points of ðð¡.(4) Conclude that ðŽð¡0 is equal to the set of zeros of ð£.(5) In particular, conclude that for all ð¡ the points of ðŽð¡0 are fixed points of ðð¡.Exercise 5.6.vii.(1) Let ð be a finite dimensional vector space and
ð¹(ð¡) ⶠð â ð , ââ < ð¡ < âa one-parameter group of linear maps of ð onto itself. Let ðŽ = ðð¹ðð¡ (0) and show thatð¹(ð¡) = exp ð¡ðŽ. (See Exercise 2.2.viii.)
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190 Chapter 5: Cohomology via forms
(2) Show that if idð âð¹(ð¡0) ⶠð â ð is bijective for some ð¡0, then ðŽâ¶ ð â ð is bijective.Hint: Show that if ðŽð£ = 0 for some ð£ â ð â {0}, ð¹(ð¡)ð£ = ð£.
Exercise 5.6.viii. Let ð be a vector field on a compact manifold ð and let (5.6.26) be theone-parameter group of diffeomorphisms generated by ð. If ð(ð) = 0 then by part (1) ofExercise 5.6.vi, ð is a fixed point of ðð¡ for all ð¡.(1) Show that
(ððð¡) ⶠððð â ðððis a one-parameter group of linear mappings of ððð onto itself.
(2) Conclude from Exercise 5.6.vii that there exists a linear map(5.6.27) ð¿ð(ð)ⶠððð â ððð
with the propertyexp ð¡ð¿ð(ð) = (ððð¡)ð .
Exercise 5.6.ix. Suppose ðð¡0 is a Lefschetz map for some ð¡0. Let ð = ð¡0/ð where ð is apositive integer. Show that ðð is a Lefschetz map.
Hints:⢠Show that
ðð¡0 = ðð â ⯠â ðð = ððð
(i.e., ðð composed with itselfð times).⢠Show that if ð is a fixed point of ðð, it is a fixed pointof ðð¡0 .⢠Conclude from Exercise 5.6.vi that the fixed points of ðð are the zeros of ð£.⢠Show that if ð is a fixed point of ðð,
(ððð¡0)ð = (ððð)ðð .
⢠Conclude that if (ððð)ðð£ = ð£ for some ð£ â ððð â {0}, then (ððð¡0)ðð£ = ð£.Exercise 5.6.x. Show that for all ð¡, ð¿(ðð¡) = ð(ð).
Hint: Exercise 5.6.ii.
Exercise 5.6.xi (Hopf Theorem). A vector field ð on a compact manifold ð is a Lefschetzvector field if for some ð¡0 â ð the map ðð¡0 is a Lefschetz map.(1) Show that if ð is a Lefschetz vector field then it has a finite number of zeros and for each
zero ð the linear map (5.6.27) is bijective.(2) For a zero ð of ð let ðð(ð) = +1 if the map (5.6.27) is orientation-preserving and letðð(ð) = â1 if itâs orientation-reversing. Show that
ð(ð) = âð(ð)=0ðð(ð) .
Hint: Apply the Lefschetz theorem to ðð, ð = ð¡0/ð,ð large.
Exercise 5.6.xii (review of the trace). For ðŽ = (ðð,ð) an ð à ðmatrix define
tr(ðŽ) âðâð=1ðð,ð .
(1) Show that if ðŽ and ðµ are ð à ðmatricestr(ðŽðµ) = tr(ðµðŽ) .
(2) Show that if ðµ is an invertible ð à ðmatrixtr(ðµðŽðµâ1) = tr(ðŽ) .
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§5.7 The KÌnneth theorem 191
(3) Let ð be and ð-dimensional vector space and ð¿â¶ ð â ð a linear map. Fix a basisð£1,âŠ, ð£ð of ð and define the trace of ð¿ to be the trace of ðŽ where ðŽ is the definingmatrix for ð¿ in this basis, i.e.,
ð¿ð£ð =ðâð=1ðð,ðð£ð .
Show that this is an intrinsic definition not depending on the basis ð£1,âŠ, ð£ð.
5.7. The KÃŒnneth theorem
Let ð be an ð-manifold and ð an ð-dimensional manifold, both of these manifoldshaving finite topology. Let
ðⶠð à ð â ðbe the projection map ð(ð¥, ðŠ) = ð¥ and
ðⶠð à ð â ð
the projectionmap (ð¥, ðŠ) â ðŠ. Sinceð andð have finite topology their cohomology groupsare finite dimensional vector spaces. For 0 †ð †ð let
ððð , 1 †ð †dimð»ð(ð) ,
be a basis ofð»ð(ð) and for 0 †â †ð let
ðâð , 1 †ð †dimð»â(ð)
be a basis of ð»â(ð). Then for ð + â = ð the product, ðâ¯ððð â ðâ¯ðâð , is in ð»ð(ð à ð). TheKÃŒnneth theorem asserts
Theorem 5.7.1. The product manifold ð à ð has finite topology and hence the cohomologygroupsð»ð(ð à ð) are finite dimensional. Moreover, the products over ð + â = ð
ðâ¯ððð â ðâ¯ðâð , 0 †ð †dimð»ð(ð) , 0 †ð †dimð»â(ð) ,(5.7.2)
are a basis for the vector spaceð»ð(ð à ð).
The fact thatðÃð has finite topology is easy to verify. Ifð1,âŠ,ðð, is a good cover ofð and ð1,âŠ,ðð, is a good cover of ð the products of these open setsðð Ãðð, for 1 †ð †ðand 1 †ð †ð is a good cover of ð à ð: For every multi-index ðŒ, ððŒ is either empty ordiffeomorphic to ðð, and for every multi-index ðœ,ððœ is either empty or diffeomorphic to ðð,hence for any productmulti-index (ðŒ, ðœ), the productððŒÃððœ is either empty or diffeomorphicto ðð Ãðð. The tricky part of the proof is verifying that the products from (5.7.2) are a basisof ð»ð(ð à ð), and to do this it will be helpful to state the theorem above in a form thatavoids our choosing specified bases forð»ð(ð) andð»â(ð). To do so weâll need to generalizeslightly the notion of a bilinear pairing between two vector space.
Definition 5.7.3. Letð1,ð2 andð be finite dimensional vector spaces. Amapðµâ¶ ð1Ãð2 âð of sets is a bilinear map if it is linear in each of its factors, i.e., for ð£2 â ð2 the map
ð£1 â ð1 ⊠ðµ(ð£1, ð£2)
is a linear map of ð1 intoð and for ð£1 â ð1 so is the map
ð£2 â ð2 ⊠ðµ(ð£1, ð£2) .
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192 Chapter 5: Cohomology via forms
Itâs clear that if ðµ1 and ðµ2 are bilinear maps of ð1 Ã ð2 intoð and ð1 and ð2 are realnumbers the function
ð1ðµ1 + ð2ðµ2 ⶠð1 à ð2 âðis also a bilinear map of ð1 à ð2 intoð, so the set of all bilinear maps of ð1 à ð2 intoðforms a vector space. In particular the set of all bilinear maps of ð1 à ð2 into ð is a vectorspace, and since this vector space will play an essential role in our intrinsic formulation ofthe KÃŒnneth theorem, weâll give it a name. Weâll call it the tensor product ofðâ1 andðâ2 anddenote it by ðâ1 â ðâ2 . To explain where this terminology comes from we note that if â1 andâ2 are vectors in ðâ1 and ðâ2 then one can define a bilinear map
â1 â â2 ⶠð1 à ð2 â ðby setting (â1 â â2)(ð£1, ð£2) â â1(ð£1)â2(ð£2). In other words one has a tensor product map:
(5.7.4) ðâ1 Ã ðâ2 â ðâ1 â ðâ2mapping (â1, â2) to â1 â â2. We leave for you to check that this is a bilinear map of ðâ1 Ã ðâ2into ðâ1 â ðâ2 and to check as well
Proposition 5.7.5. If â11 ,âŠ, â1ð is a basis ofðâ1 and â21 ,âŠ, â2ð is a basis ofðâ2 then â1ð â â2ð , for1 †ð †ð and 1 †ð †ð, is a basis of ðâ1 â ðâ2 .
Hint: If ð1 and ð2 are the same vector space you can find a proof of this in §1.3 and theproof is basically the same if theyâre different vector spaces.
Corollary 5.7.6. Let ð1 and ð2 be finite-dimensional vector spaces. Then
dim(ðâ1 â ðâ2 ) = dim(ðâ1 ) dim(ðâ2 ) = dim(ð1) dim(ð2) .
Weâll now perform some slightly devious maneuvers with âdualityâ operations. Firstnote that for any finite dimensional vector space ð, the evaluation pairing
ð à ðâ â ð , (ð£, â) ⊠â(ð£)is a non-singular bilinear pairing, so, as we explained in §5.4 it gives rise to a bijective linearmapping
(5.7.7) ð â (ðâ)â .Next note that if
(5.7.8) ð¿â¶ ð1 à ð2 âðis a bilinear mapping and âⶠð â ð a linear mapping (i.e., an element ofðâ), then thecomposition of â and ð¿ is a bilinear mapping
â â ð¿â¶ ð1 à ð2 â ðand hence by definition an element of ðâ1 â ðâ2 . Thus from the bilinear mapping (5.7.8) weget a linear mapping
(5.7.9) ð¿â¯ ⶠðâ â ðâ1 â ðâ2 .
Weâll now define a notion of tensor product for the vector spaces ð1 andð2 themselves.
Definition 5.7.10. The vector space ð1 â ð2 is the vector space dual of ðâ1 â ðâ2 , i.e., is thespace
(5.7.11) ð1 â ð2 â (ðâ1 â ðâ2 )â .
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§5.7 The KÌnneth theorem 193
One implication of (5.7.11) is that there is a natural bilinear map
(5.7.12) ð1 Ã ð2 â ð1 â ð2 .
(In (5.7.4) replace ðð by ðâð and note that by (5.7.7) (ðâð )â = ðð.) Another is the following:
Proposition 5.7.13. Let ð¿ be a bilinear map of ð1 Ã ð2 intoð. Then there exists a uniquelinear map
ð¿# ⶠð1 â ð2 âðwith the property
(5.7.14) ð¿#(ð£1 â ð£2) = ð¿(ð£1, ð£2)
where ð£1 â ð£2 is the image of (ð£1, ð£2) with respect to (5.7.12).
Proof. Let ð¿# be the transpose of the map ð¿â¯ in (5.7.9) and note that by (5.7.7) (ðâ)â =ð. â¡
Notice that by Proposition 5.7.13 the property (5.7.14) is the defining property of ð¿#;it uniquely determines this map. (This is in fact the whole point of the tensor product con-struction. Its purpose is to convert bilinear maps, which are not linear, into linear maps.)
After this brief digression (into an area of mathematics which some mathematiciansunkindly refer to as âabstract nonsenseâ), let us come back to our motive for this digression:an intrinsic formulation of the KÃŒnneth theorem. As above let ð and ð be manifolds ofdimension ð and ð, respectively, both having finite topology. For ð+â = ð one has a bilinearmap
ð»ð(ð) à ð»â(ð) â ð»ð(ð à ð)mapping (ð1, ð2) to ðâð1 â ðâð2, and hence by Proposition 5.7.13 a linear map
(5.7.15) ð»ð(ð) â ð»â(ð) â ð»ð(ð à ð) .
Letð»ð1 (ð à ð) â âš
ð+â=ðð»ð(ð) â ð»â(ð) .
The maps (5.7.15) can be combined into a single linear map
(5.7.16) ð»ð1 (ð à ð) â ð»ð(ð à ð)
and our intrinsic version of the KÃŒnneth theorem asserts the following.
Theorem 5.7.17. The map (5.7.16) is bijective.
Here is a sketch of how to prove this. (Filling in the details will be left as a series ofexercises.) Let ð be an open subset ofð which has finite topology and let
Hð1 (ð) â âšð+â=ðð»ð(ð) â ð»â(ð)
andHð2 (ð) â ð»ð(ð à ð) .
As weâve just seen thereâs a KÃŒnneth map
ð ⶠHð1 (ð) â Hð2 (ð) .
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Exercises for §5.7Exercise 5.7.i. Let ð1 and ð2 be open subsets of ð, both having finite topology, and letð = ð1 ⪠ð2. Show that there is a long exact sequence:
⯠Hð1 (ð) Hð1 (ð1) âHð1 (ð2) Hð1 (ð1 â© ð2) Hð+11 (ð) ⯠,ð¿ ð¿
Hint: Take the usual MayerâVietoris sequence
⯠ð»ð(ð) ð»ð(ð1) â ð»ð(ð2) ð»ð(ð1 â© ð2) ð»ð+1(ð) ⯠,ð¿ ð¿
tensor each term in this sequence withð»â(ð), and sum over ð + â = ð.
Exercise 5.7.ii. Show that for H2 there is a similar sequence.Hint: Apply MayerâVietoris to the open subsets ð1 Ã ð and ð2 Ã ð ofð.
Exercise 5.7.iii. Show that the diagram below commutes:
⯠Hð1 (ð) Hð1 (ð1) âHð1 (ð2) Hð1 (ð1 â© ð2) Hð+11 (ð) â¯
⯠Hð2 (ð) Hð2 (ð1) âHð2 (ð2) Hð2 (ð1 â© ð2) Hð+12 (ð) â¯
ð¿
ð ð
ð¿
ð ð
ð¿ ð¿
(This looks hard but is actually very easy: just write down the definition of each arrow in thelanguage of forms.)
Exercise 5.7.iv. Conclude from Exercise 5.7.iii that if the KÃŒnneth map is bijective for ð1,ð2 and ð1 â© ð2 it is bijective for ð.
Exercise 5.7.v. Prove the KÃŒnneth theorem by induction on the number of open sets in agood cover ofð. To get the induction started, note that
ð»ð(ðð à ð) â ð»ð(ð) .(See Exercise 5.3.xi.)
5.8. Äech Cohomology
We pointed out at the end of §5.3 that if a manifoldð admits a finite good cover
U = {ð1,âŠ,ðð} ,then in principle its cohomology can be read off from the intersection properties of theððâs.In this section weâll explain in more detail how to do this. More explicitly, we will show howto construct from the intersection properties of the ððâs a sequence of cohomology groupsð»ð(U; ð), and we will prove that these Äech cohomology groups are isomorphic to the deRham cohomology groups of ð. (However we will leave for the reader a key ingredient inthe proof: a diagram chasing argument that is similar to the diagram chasing arguments thatwere used to prove the MayerâVietoris theorem in §5.2 and the five lemma in §5.4.)
We will denote byðð(U) the set of all multi-indices
ðŒ = (ð0,âŠ, ðð) , 1 †ð0,âŠ, ðð †ðwith the property that the intersection
ððŒ â ðð0 â©â¯ â© ððð
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§5.8 Äech Cohomology 195
is nonempty. (For example, for 1 †ð †ð, themulti-index ðŒ = (ð0,âŠ, ðð)with ð0 = ⯠= ðð = ðis inðð(U) since ððŒ = ðð.) The disjoint unionð(U) â âðâ¥0ðð(U) is called the nerve ofthe cover U andðð(U) is the ð-skeleton of the nerveð(U). Note that if we delete ðð froma multi-index ðŒ = (ð0,âŠ, ðð) inðð(U), i.e., replace ðŒ by
ðŒð â (ð0,âŠ, ððâ1, ðð+1,âŠ, ðð) ,then ðŒð â ððâ1(U).
We now associate a cochain complex to ð(U) by defining ð¶ð(U; ð) to be the finitedimensional vector space consisting of all (set-theoretic) maps ðⶠðð(U) â ð:
ð¶ð(U; ð) â {ðⶠðð(U) â ð} ,with addition and scalar multiplication of functions. Define a coboundary operator
ð¿â¶ ð¶ðâ1(U; ð) â ð¶ð(U; ð)by setting
(5.8.1) ð¿(ð)(ðŒ) âðâð=0(â1)ðð(ðŒð) .
We claim that ð¿ actually is a coboundary operator, i.e., ð¿ â ð¿ = 0. To see this, we note thatfor ð â ð¶ðâ1(U; ð) and ðŒ â ðð(U), by (5.8.1) we have
ð¿(ð¿(ð))(ðŒ) =ð+1âð=0(â1)ðð¿(ð)(ðŒð)
=ð+1âð=0(â1)ð(â
ð <ð(â1)ð ð(ðŒð,ð ) + â
ð >ð(â1)ð â1ð(ðŒð,ð )) .
Thus the term ð(ðŒð,ð ) occurs twice in the sum, but occurs oncewith opposite signs, and henceð¿(ð¿(ð))(ðŒ) = 0.
The cochain complex
(5.8.2) 0 ð¶0(U; ð) ð¶1(U; ð) ⯠ð¶ð(U; ð) â¯ð¿ ð¿ ð¿ ð¿
is called theÄech cochain complex of the coverU.TheÄech cohomology groups of the coverU are the cohomology groups of the Äech cochain complex:
ð»ð(U; ð) â ker(ð¿â¶ ð¶ð(U; ð) â ð¶ð+1(U; ð))im(ð¿â¶ ð¶ðâ1(U; ð) â ð¶ð(U; ð))
.
The rest of the section is devoted to proving the following theorem:
Theorem 5.8.3. Let ð be a manifold and U a finite good cover of ð. Then for all ð ⥠0 wehave isomorphisms
ð»ð(U; ð) â ð»ð(ð) .
Remarks 5.8.4.(1) The definition ofð»ð(U; ð) only involves the nerveð(U) of this cover soTheorem 5.8.3
is in effect a proof of the claim we made above: the cohomology ofð is in principle deter-mined by the intersection properties of the ððâs.
(2) Theorem 5.8.3 gives us another proof of an assertion we proved earlier: if ð admits afinite good cover the cohomology groups ofð are finite-dimensional.
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Theproof ofTheorem 5.8.3 will involve an interesting de Rham theoretic generalizationof the Äech cochain complex (5.8.2). Namely, for ð and â nonnegative integers we define aÄech cochain of degree ð with values inðºâ to be a map ð which assigns each ðŒ â ðð(U) anâ-form ð(ðŒ) â ðºâ(ððŒ).
The set of these cochains forms an infinite dimensional vector space ð¶ð(U; ðºâ). Wewill show how to define a coboundary operator
ð¿â¶ ð¶ðâ1(U; ðºâ) â ð¶ð(U; ðºâ)similar to the Äech coboundary operator (5.8.1). To define this operator, let ðŒ â ðð(U) andfor 0 †ð †ð let ðŸð ⶠðºâ(ððŒð) â ðº
â(ððŒ) be the restriction map defined byð ⊠ð|ððŒ .
(Note that the restiction ð|ððŒ is well-defined since ððŒ â ððŒð .) Thus given a Äech cochainð â ð¶ðâ1(U; ðºâ) we can, mimicking (5.8.1), define a Äech cochain ð¿(ð) â ð¶ð(U; ðºâ) bysetting
(5.8.5) ð¿(ð)(ðŒ) âðâð=0ðŸð(ð(ðŒð)) =
ðâð=0ð(ðŒð)|ððŒ .
In other words, except for the ðŸðâs the definition of this cochain is formally identical withthe definition (5.8.1). We leave it as an exercise that the operators
ð¿â¶ ð¶ðâ1(U; ðºâ) â ð¶ð(U; ðºâ)satisfy ð¿ â ð¿ = 0.1
Thus, to summarize, for every nonnegative integer âwehave constructed aÄech cochaincomplex with values in ðºâ
(5.8.6) 0 ð¶0(U; ðºâ) ð¶1(U; ðºâ) ⯠ð¶ð(U; ðºâ) ⯠.ð¿ ð¿ ð¿ ð¿
Our next task in this sectionwill be to compute the cohomology of the complex (5.8.6).Weâllbegin with the 0th cohomology group of (5.8.6), i.e., the kernel of the coboundary operator
ð¿â¶ ð¶0(U; ðºâ) â ð¶1(U; ðºâ) .An element ð â ð¶0(U; ðºâ) is, by definition, a map which assigns to each ð â {1,âŠ, ð} anelement ð(ð) â ðð of ðºâ(ðð).
Now let ðŒ = (ð0, ð1) be an element ofð1(U). Then, by definition, ðð0 â© ðð1 is nonemptyand
ð¿(ð)(ðŒ) = ðŸð0ðð1 â ðŸð1ðð0 .Thus ð¿(ð) = 0 if and only if
ðð0 |ðð0â©ðð1 = ðð1 |ðð0â©ðð1for all ðŒ â ð1(U). This is true if and only if each â-form ðð â ðºâ(ðð) is the restriction to ððof a globally defined â-form ð on ð1 âªâ¯ ⪠ðð = ð. Thus
ker(ð¿â¶ ð¶0(U; ðºâ) â ð¶1(U; ðºâ)) = ðºâ(ð) .Inserting ðºâ(ð) into the sequence (5.8.6) we get a new sequence
(5.8.7) 0 ðºâ(ð) ð¶0(U; ðºâ) ⯠ð¶ð(U; ðºâ) ⯠,ð¿ ð¿ ð¿
and we prove:
1Hint: Except for keeping track of the ðŸðâs the proof is identical to the proof of the analogous result for thecoboundary operator (5.8.1).
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§5.8 Äech Cohomology 197
Theorem5.8.8. Letð be amanifold andU afinite good cover ofð.Then for each nonnegativeinteger â, the sequence (5.8.7) is exact.
Weâve just proved that the sequence (5.8.7) is exact in the first spot. To prove that (5.8.7)is exact in its ðth spot, we will construct a chain homotopy operator
ðⶠð¶ð(U; ðºâ) â ð¶ð+1(U; ðºâ)with the property that(5.8.9) ð¿ð(ð) + ðð¿(ð) = ðfor all ð â ð¶ð(U; ðºâ). To define this operator weâll need a slight generalization of Theo-rem 5.2.22.
Theorem 5.8.10. Let ð be a manifold with a finite good cover {ð1,âŠ,ðð}. Then there existfunctions ð1,âŠ, ðð â ð¶â(ð) such that supp(ðð) â ðð for ð = 1,âŠ,ð and âðð=1 ðð = 1.Proof of Theorem 5.8.10. Let (ðð)ðâ¥1, where ðð â ð¶â0 (ð), be a partition of unity subordinateto the cover U and let ð1 be the sum of the ððâs with support in ð1:
ð1 â âðâ¥1
supp(ðð)âð1
ðð .
Deleting those ðð such that supp(ðð) â ð1 from the sequence (ðð)ðâ¥1, we get a new sequence(ðâ²ð )ðâ¥1. Let ð2 be the sum of the ðâ²ð âs with support in ð1:
ð2 â âðâ¥1
supp(ðâ²ð )âð2
ðâ²ð .
Now we delete those ðâ²ð such that supp(ðâ²ð ) â ð2 from the sequence (ðâ²ð )ðâ¥1 and constructð3 by the same method we used to construct ð1 and ð2, and so on. â¡
Now we define the operator ð and prove Theorem 5.8.8.
Proof of Theorem 5.8.8. To define the operator (5.8.9) let ð â ð¶ð(U; ðºâ). Then we defineðð â ð¶ðâ1(U; ðºâ) by defining its value at ðŒ â ððâ1(ðŒ) to be
ðð(ðŒ) âðâð=0ððð(ð, ð0,âŠ, ððâ1) ,
where ðŒ = (ð0,âŠ, ððâ1) and the ðth summand on the right is defined by be equal to 0whenððâ©ððŒ is empty and defined to be the product of the function ðð and the â-form ð(ð, ð0,âŠ, ððâ1)whenððâ©ððŒ is nonempty. (Note that in this case, the multi-index (ð, ð0,âŠ, ððâ1) is inðð(U)and so by definition ð(ð, ð0,âŠ, ððâ1) is an element of ðºâ(ðð â© ððŒ). However, since ðð is sup-ported on ðð, we can extend the â-form ððð(ð, ð0,âŠ, ððâ1) ti ððŒ by setting it equal to zerooutside of ðð â© ððŒ.)
To prove (5.8.9) we note that for ð â ð¶ð(U; ðºâ) we have
ðð¿ð(ð0,âŠ, ðð) =ðâð=1ððð¿ð(ð, ð0,âŠ, ðð) ,
so by (5.8.5) the right hand side is equal to
(5.8.11)ðâð=1ðððŸðð(ð0,âŠ, ðð) +
ðâð=1
ðâð =0(â1)ð +1ðððŸðð ð(ð, ð0,âŠ, ð€ð ,âŠ, ðð) ,
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where (ð, ð0,âŠ, ð€ð ,âŠ, ðð) is the multi-index (ð, ð0,âŠ, ðð) with ðð deleted. Similarly,
ð¿ðð(ð0,âŠ, ðð) =ðâð =0(â1)ð ðŸðð ðð(ð0,âŠ, ð€ð ,âŠ, ðð)
=ðâð=1
ðâð =0(â1)ð ðððŸðð ð(ð, ð0,âŠ, ð€ð ,âŠ, ðð) .
However, this is the negative of the second summand of (5.8.11); so, by adding these twosummands we get
(ðð¿ð + ð¿ðð)(ð0,âŠ, ðð) =ðâð=1ðððŸðð(ð0,âŠ, ðð)
and since âðð=1 ðð = 1, this identity reduced to
(ðð¿ð + ð¿ðð)(ðŒ) = ð(ðŒ) ,or, since ðŒ â ðð(U) is any element, ðð¿ð + ð¿ðð = ð.
Finally, note the Theorem 5.8.8 is an immediate consequence of the existence of thischain homotopy operator. Namely, if ð â ð¶ð(U; ðºâ) is in the kernel of ð¿, then
ð = ðð¿(ð) + ð¿ð(ð) = ð¿ð(ð) ,so ð is in the image of ð¿â¶ ð¶ðâ1(U; ðºâ) â ð¶ð(U; ðºâ). â¡
To prove Theorem 5.8.3, we note that in addition to the exact sequence
0 ðºâ(ð) ð¶0(U; ðºâ) ⯠ð¶ð(U; ðºâ) ⯠,ð¿ ð¿ ð¿
we have another exact sequence(5.8.12)0 ð¶ð(U; ð) ð¶ð(U; ðº0) ð¶ð(U; ðº1) ⯠ð¶ð(U; ðºâ) ⯠,ð ð ð ð ð
where the ðâs are defined as follows: given ð â ð¶ð(U; ðºâ) and ðŒ â ðð(U), we have thatð(ðŒ) â ðºâ(ððŒ) and ð(ð(ðŒ)) â ðºâ+1(ððŒ), and we define ð(ð) â ð¶ð(U; ðºâ+1) to be the mapgiven by
ðŒ ⊠ð(ð(ðŒ)) .It is clear from this definition that ð(ð(ð)) = 0 so the sequence
(5.8.13) ð¶ð(U; ðº0) ð¶ð(U; ðº1) ⯠ð¶ð(U; ðºâ) â¯ð ð ð ð
is a complex and the exactness of this sequence follows from the fact that, since U is a goodcover, the sequence
ðº0(ððŒ) ðº1(ððŒ) ⯠ðºâ(ððŒ) â¯ð ð ð ð
is exact. Moreover, if ð â ð¶ð(U; ðº0) and ðð = 0, then for every ðŒ â ðð(U), we haveðð(ðŒ) = 0, i.e., the 0-form ð(ðŒ) in ðº0(ððŒ) is constant. Hence the map given by ðŒ ⊠ð(ðŒ) isjust a Äech cochain of the type we defined earlier, i.e., a map ðð(U) â ð. Therefore, thekernel of this map ðⶠð¶ð(U; ðº0) â ð¶ð(U; ðº1) is ð¶0(U; ð); and, adjoining this term to thesequence (5.8.13) we get the exact sequence (5.8.12).
Lastly we note that the two operations weâve defined above, the ð operationðⶠð¶ð(U; ðºâ) â ð¶ð(U; ðºâ+1)
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§5.8 Äech Cohomology 199
and the ð¿ operation ð¿â¶ ð¶ð(U; ðºâ) â ð¶ð+1(U; ðºâ), commute. i.e., for ð â ð¶ð(U; ðºâ) wehave
(5.8.14) ð¿ðð = ðð¿ð .
Proof of equation (5.8.14). For ðŒ â ðð+1(U)
ð¿ð(ðŒ) =ðâð=0(â1)ððŸðð(ðŒð) ,
so if ð(ðŒð) â ððŒð â ðºâ(ððŒð), then
ð¿ð(ðŒ) =ðâð=0(â1)ðððŒð |ððŒ
and
ðð¿(ðŒ) =ðâð=0ðððŒð |ððŒ = ð¿ðð(ðŒ) . â¡
Putting these results together we get the following commutative diagram:
â® â® â® â®
0 ðº2(ð) ð¶0(U; ðº2) ð¶1(U; ðº2) ð¶2(U; ðº2) â¯
0 ðº1(ð) ð¶0(U; ðº1) ð¶1(U; ðº1) ð¶2(U; ðº1) â¯
0 ðº0(ð) ð¶0(U; ðº0) ð¶1(U; ðº0) ð¶2(U; ðº0) â¯
0 ð¶0(U; ð) ð¶1(U; ð) ð¶2(U; ð) â¯
0 0 0 .
ð
ð¿ ð¿
ð
ð¿
ð
ð¿
ð ð
ð
ð¿
ð
ð¿
ð ð
ð¿ ð¿
In this diagram all columns are exact except for the extreme left and column, which is theusual de Rham complex, and all rows are exact except for the bottom row which is the usualÄech cochain complex.
Exercises for §5.8Exercise 5.8.i. Deduce from this diagram that if a manifold ð admits a finite good coverU, thenð»ð(U; ð) â ð»ð(ð).
Hint: Let ð â ðºð+1(ð) be a form such that ðð = 0. Show that one can, by a sequence ofâchess movesâ (of which the first few stages are illustrated in the (5.8.15) below), convert ð
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200 Chapter 5: Cohomology via forms
into an element ð of ð¶ð+1(U; ð) satisfying ð¿( ð) = 0
(5.8.15)
ð ðð+1,0
ðð,0 ðð,1
ððâ1,1 ððâ1,2
ððâ2,2 ððâ2,3
ððâ3,3 â¯
ð
ð¿
ð
ð¿
ð
ð¿
ð
ð¿
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APPENDIX a
Bump Functions & Partitions of Unity
Wewill discuss in this section a number of âglobal to localâ techniques inmulti-variablecalculus: techniques which enable one to reduce global problems on manifolds and largeopen subsets of ðð to local problems on small open subsets of these sets. Our starting pointwill be the function ð on ð defined by
(a.1) ð(ð¥) â {ðâ 1ð¥ , ð¥ > 00, 𥠆0 ,
which is positive for ð¥ positive, zero for ð¥ negative and everywhere ð¶â. (We will sketch aproof of this last assertion at the end of this appendix). From ð one can construct a numberof other interesting ð¶â functions.
Example a.2. For ð > 0 the function ð(ð¥) + ð(ð â ð¥) is positive for all ð¥ so the quotientðð(ð¥) of ð(ð¥) by this function is a well-defined ð¶â function with the properties:
{{{{{
ðð(ð¥) = 0, for 𥠆00 †ðð(ð¥) †1ðð(ð¥) = 1, for ð¥ ⥠ð .
Example a.3. Let ðŒ be the open interval (ð, ð), and ððŒ the product of ð(ð¥ â ð) and ð(ð â ð¥).Then ððŒ(ð¥) is positive for ð¥ â ðŒ and zero on the complement of ðŒ.
Example a.4. More generally let ðŒ1,âŠ, ðŒð be open intervals and let ð = ðŒ1 Ã⯠à ðŒð be theopen rectangle in ðð having these intervals as sides. Then the function
ðð(ð¥) â ððŒ1(ð¥1)â¯ððŒð(ð¥ð)is a ð¶â function on ðð thatâs positive on ð and zero on the complement of ð.
Using these functions we will prove:
Lemma a.5. Let ð¶ be a compact subset of ðð and ð an open set containing ð¶. Then thereexists function ð â ð¶â0 (ð) such that
{ð(ð¥) ⥠0, for all ð¥ â ðð(ð¥) > 0, for ð¥ â ð¶ .
Proof. For each ð â ð¶ letðð be an open rectangle with ð â ðð andðð â ð. Theððâs coverð¶ð so, by HeineâBorel there exists a finite subcoverðð1 ,âŠ,ððð . Now let ð â âðð=1 ðððð â¡
This result can be slightly strengthened. Namely we claim:
Theorem a.6. There exists a function ð â ð¶â0 (ð) such that 0 †ð †1 and ð = 1 on ð¶.201
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202 Appendix a: Bump Functions & Partitions of Unity
Proof. Let ð be as in Lemma a.5 and let ð > 0 be the greatest lower bound of the restrictionof ð to ð¶. Then if ðð is the function in Example a.2 the function ðð â ð has the propertiesindicated above. â¡
Remark a.7. The functionð in this theorem is an example of a bump function. If one wantsto study the behavior of a vector field ð or a ð-form ð on the set ð¶, then by multiplying ð(or ð) by ð one can, without loss or generality, assume that ð (or ð) is compactly supportedon a small neighborhood of ð¶.
Bump functions are one of the standard tools in calculus for converting global problemsto local problems. Another such tool is partitions of unity.
Let ð be an open subset of ðð and U = {ððŒ}ðŒâðŒ a covering of ð by open subsets (in-dexed by the elements of the index set ðŒ). Then the partition of unity theorem asserts:
Theorem a.8. There exists a sequence of functions ð1, ð2,⊠â ð¶â0 (ð) such that(1) ðð ⥠0(2) For every ð there is an ðŒ â ðŒ with ðð â ð¶â0 (ððŒ)(3) For every ð â ð there exists a neighborhood ðð of ð in ð and an ðð > 0 such thatðð|ðð = 0 for ð¢ > ðð.
(4) âðð = 1
Remark a.9. Because of item (3) the sum in item (4) is well defined. We will derive thisresult from a somewhat simpler set theoretical result.
Theorem a.10. There exists a countable covering of ð by open rectangles (ðð)ðâ¥1 such that(1) ðð â ð(2) For each ð there is an ðŒ â ðŒ with ðð â ððŒ(3) For every ð â ð there exists a neighborhood ðð of ð in ð andðð > 0 such that ðð â© ðð
is empty for ð > ððWe first note that Theorem a.10 implies Theorem a.8. To see this note that the func-
tions ððð in Example a.4 above have all the properties indicated in Theorem a.8 except forproperty (4). Moreover since the ððâs are a covering of ð the sum ââð=1 ððð is everywherepositive. Thus we get a sequence of ððâs satisfying (1)â(4) by taking ðð to be the quotient ofððð by this sum.
Proof of Theorem a.10. Let ð(ð¥, ðð) be the distance of a point ð¥ â ð to the complementðð â ðð â ð of ð in ðð, and let ðŽð be the compact subset of ð consisting of points, ð¥ â ð,satisfying ð(ð¥, ðð) ⥠1/ð and |ð¥| †ð. By HeineâBorel we can find, for each ð, a collection ofopen rectangles,ðð,ð, ð = 1,âŠ,ðð, such thatðð,ð is contained in int(ðŽð+1âðŽðâ2) and in someððŒ and such that theðð, ðâs are a covering ofðŽðâint(ðŽðâ1).Thus theðð,ðâs have the propertieslisted in Theorem a.10, and by relabelling, i.e., settingðð = ð1,ð for 1 †ð †ðð,ðð1+ð â ð2,ð,for 1 †ð †ð2, etc., we get a sequence, ð1, ð2,⊠with the desired properties. â¡
ApplicationsWe will next describe a couple of applications of Theorem a.10.
Application a.11 (Improper integrals). Let ðⶠð â ð be a continuous function. We willsay that ð is integrable over ð if the infinite sum
(a.12)ââð=1â«ððð(ð¥) |ð(ð¥)| ðð¥
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203
converges and if so we will define the improper integral of ð over ð to be the sum
(a.13)ââð=1â«ððð(ð¥)ð(ð¥)ðð¥
(Notice that each of the summands in this series is the integral of a compactly supportedcontinuous function over ðð so the individual summands are well-defined. Also itâs clearthat if ð itself is a compactly supported function on ðð, (a.12) is bounded by the integral of|ð| over ðð, so for every ð â ð¶0 (ðð) the improper integral of ð over ð is well-defined.)
Application a.14 (An extension theorem for ð¶â maps). Let ð be a subset of ðð and ð â¶ð â ðð a continuous map. We will say that ð is ð¶â if, for every ð â ð, there exists anopen neighborhood, ðð, of ð in ðð and a ð¶â map, ðð ⶠðð â ðð such that ðð = ð onðð â© ð.
Theorem a.15 (Extension Theorem). If ðⶠð â ðð is ð¶â there exists an open neighbor-hood ð ofð in ðð and a ð¶â map ðⶠð â ðð, such that ð = ð onð.Proof. Let ð = âðâððð and let (ðð)ðâ¥1 be a partition of unity subordinate to the coveringU = {ðð}ðâð ofð. Then, for each ð, there exists a ð such that the support of ðð is containedin ðð. Let
ðð(ð¥) = {ðððð(ð¥), ð¥ â ðð0, ð¥ â ððð .
Then ð = ââð=1 ðð is well defined by item (3) of Theorem a.8 and the restriction of ð toð isgiven by ââð=1 ððð, which is equal to ð. â¡
Exercises for Appendix aExercise a.i. Show that the function (a.1) is ð¶â.
Hints:(1) From the Taylor series expansion
ðð¥ =ââð=0
ð¥ðð!
conclude that for ð¥ > 0ðð¥ ⥠ð¥
ð+ð
(ð + ð)!.
(2) Replacing ð¥ by 1/ð¥ conclude that for ð¥ > 0,
ð1/ð¥ ⥠1(ð + ð)!
1ð¥ð+ð
.
(3) From this inequality conclude that for ð¥ > 0,
ðâ 1𥠆(ð + ð)!ð¥ð+ð .(4) Let ðð(ð¥) be the function
ðð(ð¥) â {ðâ 1ð¥ ð¥âð, ð¥ > 00, 𥠆0 .
Conclude from (3) that for ð¥ > 0,ðð(ð¥) †(ð + ð)!ð¥ð
for all ð
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204 Appendix a: Bump Functions & Partitions of Unity
(5) Conclude that ðð is ð¶1 differentiable.(6) Show that
ððð¥ðð = ðð+2 â ððð+1
(7) Deduce by induction that the ððâs are ð¶ð differentiable for all ð and ð.Exercise a.ii. Show that the improper integral (a.13) is well-defined independent of thechoice of partition of unity.
Hint: Let (ðâ²ð )ðâ¥1 be another partition of unity. Show that (a.13) is equal toââð=1
ââð=1â«ððð(ð¥)ðâ²ð (ð¥)ð(ð¥)ðð¥ .
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APPENDIX b
The Implicit FunctionTheorem
Let ð be an open neighborhood of the origin in ðð and let ð1,âŠ, ðð be ð¶â functionson ð with the property ð1(0) = ⯠= ðð(0) = 0. Our goal in this appendix is to prove thefollowing implicit function theorem.
Theorem b.1. Suppose the matrix
(b.2) [ ððððð¥ð (0)]
is non-singular.Then there exists a neighborhoodð0 of 0 inðð and a diffeomorphismðⶠ(ð0, 0) âª(ð, 0) of ð0 onto an open neighborhood of 0 in ð such that
ðâðð = ð¥ð , ð = 1,âŠ, ðand
ðâð¥ð = ð¥ð , ð = ð + 1,âŠ, ð .
Remarks b.3.(1) Let ð(ð¥) = (ð1(ð¥),âŠ, ðð(ð¥)). Then the second set of equations just say that
ðð(ð¥1,âŠ, ð¥ð) = ð¥ðfor ð = ð + 1,âŠ, ð, so the first set of equations can be written more concretely in theform
(b.4) ðð(ð1(ð¥1,âŠ, ð¥ð),âŠ, ðð(ð¥1,âŠ, ð¥ð), ð¥ð+1,âŠ, ð¥ð) = ð¥ðfor ð = 1,âŠ, ð.
(2) Letting ðŠð = ðð(ð¥1,âŠ, ð¥ð) the equations (b.4) become(b.5) ðð(ðŠ1,âŠ, ðŠð, ð¥ð+1,âŠ, ð¥ð) = ð¥ð .
Hence what the implicit function theorem is saying is that, modulo the assumption (b.2)the system of equations (b.5) can be solved for the ðŠðâs in terms of the ð¥ðâs.
(3) By a linear change of coordinates:
ð¥ð âŠðâð=1ðð,ðð¥ð , ð = 1,âŠ, ð
we can arrange without loss of generality for the matrix (b.2) to be the identity matrix,i.e.,
(b.6) ððððð¥ð(0) = ð¿ð,ð , 1 †ð, ð †ð .
Our proof of this theorem will be by induction on ð.First, letâs suppose that ð = 1 and, for the moment, also suppose that ð = 1. Then the
theorem above is just the inverse function theoremof freshman calculus. Namely ifð(0) = 0and ðð/ðð¥(0) = 1, there exists an interval (ð, ð) about the origin on which ðð/ðð¥ is greater
205
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206 Appendix b: The Implicit Function Theorem
than 1/2, so ð is strictly increasing on this interval and its graph looks like the curve in thefigure below with ð = ð(ð) †â 12ð and ð = ð(ð) â¥
12ð.
ᅵ
ᅵ
ᅵ
ᅵ
ᅵ
ᅵ
Figure b.1. The implicit function theorem
The graph of the inverse function ðⶠ[ð, ð] â [ð, ð] is obtained from this graph by justrotating it through ninety degrees, i.e., making the ðŠ-axis the horizontal axis and the ð¥-axisthe vertical axis. (From the picture itâs clear that ðŠ = ð(ð¥) ⺠ð¥ = ð(ðŠ) ⺠ð(ð(ðŠ)) =ðŠ.)
Most elementary text books regard this intuitive argument as being an adequate proofof the inverse function theorem; however, a slightly beefed-up version of this proof (which iscompletely rigorous) can be found in [12, Ch. 12].Moreover, as Spivak points out in [12, Ch.12], if the slope of the curve in the figure above at the point (ð¥, ðŠ) is equal to ð the slope ofthe rotated curve at (ðŠ, ð¥) is 1/ð, so from this proof one concludes that if ðŠ = ð(ð¥)
(b.7) ððððŠ(ðŠ) = (ðððð¥(ð¥))â1= (ðððð¥(ð(ðŠ)))
â1.
Since ð is a continuous function, its graph is a continuous curve and, therefore, since thegraph of ð is the same curve rotated by ninety degrees, ð is also a continuous functions.Hence by (b.7), ð is also a ð¶1 function and hence by (b.7), ð is a ð¶2 function, etc. In otherwords ð is in ð¶â([ð, ð]).
Letâs now prove that the implicit function theorem with ð = 1 and ð arbitrary. Thisamounts to showing that if the function ð in the discussion above depends on the param-eters, ð¥2,âŠð¥ð in a ð¶â fashion, then so does its inverse ð. More explicitly letâs supposeð(ð¥1,âŠ, ð¥ð) is a ð¶â function on a subset of ðð of the form [ð, ð] Ãð whereð is a compact,convex neighborhood of the origin in ðð and satisfies ðð/ðð¥1 ⥠12 on this set. Then by theargument above there exists a function, ð = ð(ðŠ, ð¥2,âŠ, ð¥ð) defined on the set [ð/2, ð/2]Ãðwith the property
(b.8) ð(ð¥1, ð¥2,âŠ, ð¥ð) = ðŠ ⺠ð(ðŠ, ð¥2,âŠ, ð¥ð) = ð¥1 .Moreover by (b.7) ð is a ð¶â function of ðŠ and
(b.9) ððððŠ(ðŠ, ð¥2,âŠ, ð¥ð) =
ðððð¥1(ð¥1, ð¥2,âŠ, ð¥ð)â1
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at ð¥1 = ð(ðŠ). In particular, since ðððð¥1 â¥12
0 < ððððŠ< 2
and hence
(b.10) |ð(ðŠâ², ð¥2,âŠ, ð¥ð) â ð(ðŠ, ð¥2,âŠ, ð¥ð)| < 2|ðŠâ² â ðŠ|for points ðŠ and ðŠâ² in the interval [ð/2, ð/2].
The ð = 1 case of Theorem b.1 is almost implied by (b.8) and (b.9) except that we muststill show that ð is a ð¶â function, not just of ðŠ, but of all the variables, ðŠ, ð¥2,âŠ, ð¥ð, and thisweâll do by quoting another theorem from freshman calculus (this time a theorem from thesecond semester of freshman calculus).
Theorem b.11 (Mean Value Theorem in ð variables). Let ð be a convex open set in ðð andðⶠð â ð a ð¶â function. Then for ð, ð â ð there exists a point ð on the line interval joiningð to ð such that
ð(ð) â ð(ð) =ðâð=1
ðððð¥ð(ð)(ðð â ðð) .
Proof. Apply the one-dimensional mean value theorem to the function
â(ð¡) â ð((1 â ð¡)ð + ð¡ð) . â¡
Letâs now show that the function ð in (b.8) is a ð¶â function of the variables, ðŠ and ð¥2.To simplify our notation weâll suppress the dependence of ð on ð¥3,âŠ, ð¥ð and write ð asð(ð¥1, ð¥2) and ð as ð(ðŠ, ð¥2). For â â (âð, ð), ð small, we have
ðŠ = ð(ð(ðŠ, ð¥2 + â), ð¥2 + â) = ð(ð(ðŠ, ð¥2), ð¥2) ,and, hence, setting ð¥â²1 = ð(ðŠ, ð¥2 + â), ð¥1 = ð(ðŠ, ð¥2) and ð¥â²2 = ð¥2 + â, we get from the meanvalue theorem
0 = ð(ð¥â²1, ð¥â²2) â ð(ð¥1, ð¥2)
= ðððð¥1(ð)(ð¥â²1 â ð¥1) +
ðððð¥2(ð)(ð¥â²2 â ð¥2)
and therefore
(b.12) ð(ðŠ, ð¥2 + â) â ð(ðŠ, ð¥2) = (âðððð¥1(ð))â1 ðððð¥2(ð) â
for some ð on the line segment joining (ð¥1, ð¥2) to (ð¥â²1, ð¥â²2). Letting â tend to zero we can drawfrom this identity a number of conclusions:(1) Since ð is a ð¶â function on the compact set [ð, ð] à ð its derivatives are bounded on
this set, so the right hand side of (b.12) tends to zero as â tends to zero, i.e., for fixed ðŠ,ð is continuous as a function of ð¥2.
(2) Nowdivide the right hand side of (b.12) by â and let â tend to zero. Sinceð is continuousin ð¥2, the quotient of (b.12) by â tends to its value at (ð¥1, ð¥2). Hence for fixed ðŠ ð isdifferentiable as a function of ð¥2 and
(b.13) ðððð¥2(ðŠ, ð¥2) = â(
ðððð¥1)â1(ð¥1, ð¥2)
ðððð¥2(ð¥1, ð¥2)
where ð¥1 = ð(ðŠ, ð¥2).
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208 Appendix b: The Implicit Function Theorem
(3) Moreover, by the inequality (b.10) and the triangle inequality
|ð(ðŠâ², ð¥â²2) â ð(ðŠ, ð¥2)| †|ð(ðŠâ², ð¥â²2) â ð(ðŠ, ð¥â²2)| + |ð(ðŠ, ð¥â²2) â ð(ðŠ, ð¥2)|†2|ðŠâ² â ðŠ| + |ð(ðŠ, ð¥â²2) â ð(ðŠ, ð¥2)| ,
hence ð is continuous as a function of ðŠ and ð¥2.(4) Hence by (b.9) and (b.13), ð is a ð¶1 function of ðŠ and ð¥2, and henceâŠ.
In other words ð is a ð¶â function of ðŠ and ð¥2. This argument works, more or lessverbatim, for more than two ð¥ðâs and proves that ð is a ð¶â function of ðŠ, ð¥2,âŠ, ð¥ð. Thuswith ð = ð1 and ð = ð1Theorem b.1 is proved in the special case ð = 1.
Proof of Theorem b.1 for ð > 1. Weâll now prove Theorem b.1 for arbitrary ð by inductionon ð. By induction we can assume that there exists a neighborhood ðâ²0 , of the origin in ððand a ð¶â diffeomorphism ðⶠ(ðâ²0 , 0) ⪠(ð, 0) such that
(b.14) ðâðð = ð¥ðfor 2 †ð †ð and(b.15) ðâð¥ð = ð¥ðfor ð = 1 and ð + 1 †ð †ð. Moreover, by (b.6)
( ððð¥1ðâð1) (0) =
ðâð=1
ððððð¥1(0) ððð¥ððð(0) =
ðð1ðð¥1= 1
sinceð1 = ðâð¥1 = ð¥1.Thereforewe can applyTheoremb.1, with ð = 1, to the function,ðâð1,to conclude that there exists a neighborhood ð0 of the origin in ðð and a diffeomorphismðⶠ(ð0, 0) â (ðâ²0 , 0) such that ðâðâð1 = ð¥1 and ðâðâð¥ð = ð¥ð for 1 †ð †ð. Thus by (b.14)ðâðâðð = ðâð¥ð = ð¥ð for 2 †ð †ð, and by (b.15) ðâðâð¥ð = ðâð¥ð = ð¥ð for ð + 1 †ð †ð.Hence if we let ð = ð â ð we see that:
ðâðð = (ð â ð)âðð = ðâðâðð = ð¥ðfor ð †ð †ð and
ðâð¥ð = (ð â ð)âð¥ð = ðâðâð¥ð = ð¥ðfor ð + 1 †ð †ð. â¡
Weâll derive a number of subsidiary results from Theorem b.1. The first of these is theð-dimensional version of the inverse function theorem:
Theoremb.16. Letð andð be open subsets ofðð andðⶠ(ð, ð) â (ð, ð) að¶âmap. Supposethat the derivative of ð at ð
ð·ð(ð)ⶠðð â ðð
is bijective. Then ð maps a neighborhood of ð in ð diffeomorphically onto a neighborhood ofð in ð.
Proof. By pre-composing and post-composing ð by translations we can assume that ð =ð = 0. Let ð = (ð1,âŠ, ðð). Then the condition that ð·ð(0) be bijective is the condition thatthe matrix (b.2) be non-singular. Hence, by Theorem b.1, there exists a neighborhoodð0 of0 in ð, a neighborhood ð0 of 0 in ð, and a diffeomorphism ðⶠð0 â ð0 such that
ðâðð = ð¥ðfor ð = 1,âŠ, ð. However, these equations simply say that ð is the inverse of ð, and hencethat ð is a diffeomorphism. â¡
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209
A second result which weâll extract from Theorem b.1 is the canonical submersion the-orem.
Theorem b.17. Let ð be an open subset of ðð and ðⶠ(ð, ð) â (ðð, 0) a ð¶â map. Supposeð is a submersion at ð, i.e., suppose its derivative
ð·ð(ð)ⶠðð â ðð
is onto. Then there exists a neighborhood ð of ð in ð, a neighborhood ð0 of the origin in ðð,and a diffeomorphism ðⶠ(ð0, 0) â (ð, ð) such that the map ð â ðⶠ(ð0, 0) â (ðð, 0) is therestriction to ð0 of the canonical submersion:
ðⶠðð â ðð , ð(ð¥1,âŠ, ð¥ð) â (ð¥1,âŠ, ð¥ð) .Proof. Let ð = (ð1,âŠ, ðð). Composing ð with a translation we can assume ð = 0 and by apermutation of the variables ð¥1,âŠ, ð¥ð we can assume that the matrix (b.2) is non-singular.By Theorem b.1 we conclude that there exists a diffeomorphism ðⶠ(ð0, 0) ⪠(ð, ð) withthe properties ðâðð = ð¥ð for ð = 1,âŠ, ð, and hence,
ð â ð(ð¥1,âŠ, ð¥ð) = (ð¥1,âŠ, ð¥ð) . â¡
As a third application of Theorem b.1 weâll prove a theorem which is similar in spirit toTheorem b.17, the canonical immersion theorem.
Theorem b.18. Let ð be an open neighborhood of the origin in ðð and ðⶠ(ð, 0) â (ðð, ð)a ð¶â map. Suppose that the derivative of ð at 0
ð·ð(0)ⶠðð â ðð
is injective. Then there exists a neighborhood ð0 of 0 in ð, a neighborhood ð of ð in ðð, anda diffeomorphism.
ðⶠð â ð0 à ððâð
such that the map ð â ðⶠð0 â ð0 à ððâð is the restriction to ð0 of the canonical immersionð ⶠðð â ðð à ððâð , ð(ð¥1,âŠ, ð¥ð) = (ð¥1,âŠ, ð¥ð, 0,âŠ, 0) .
Proof. Let ð = (ð1,âŠ, ðð). By a permutation of the ððâs we can arrange that the matrix
[ ððððð¥ð (0)] , 1 †ð, ð †ð
is non-singular and by composing ð with a translation we can arrange that ð = 0. Hence byTheorem b.16 the map
ðⶠ(ð, 0) â (ðð, 0) ð¥ ⊠(ð1(ð¥),âŠ, ðð(ð¥))maps a neighborhood ð0 of 0 diffeomorphically onto a neighborhood ð0 of 0 in ðð. Letðⶠ(ð0, 0) â (ð0, 0) be its inverse and let
ðŸâ¶ ð0 à ððâð â ð0 à ððâð
be the mapðŸ(ð¥1,âŠ, ð¥ð) â (ð(ð¥1,âŠ, ð¥ð), ð¥ð+1,âŠ, ð¥ð) .
Then(ðŸ â ð)(ð¥1,âŠ, ð¥ð) = ðŸ(ð(ð¥1,âŠ, ð¥ð), ðð+1(ð¥),âŠ, ðð(ð¥))(b.19)
= (ð¥1,âŠ, ð¥ð, ðð+1(ð¥),âŠ, ðð(ð¥)) .Now note that the map
âⶠð0 à ððâð â ð0 à ððâð
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210 Appendix b: The Implicit Function Theorem
defined byâ(ð¥1,âŠ, ð¥ð) â (ð¥1,âŠ, ð¥ð, ð¥ð+1 â ðð+1(ð¥),âŠ, ð¥ð â ðð(ð¥))
is a diffeomorphism. (Why? What is its inverse?) and, by (b.19),(â â ðŸ â ð)(ð¥1,âŠ, ð¥ð) = (ð¥1,âŠ, ð¥ð, 0,âŠ, 0) ,
i.e., (â â ðŸ) â ð = ð. Thus we can take the ð in Theorem b.18 to be ð0 à ððâð and the ð to beâ â ðŸ. â¡Remark b.20. The canonical submersion and immersion theorems can be succinctly sum-marized as saying that every submersion âlooks locally like the canonical submersionâ andevery immersion âlooks locally like the canonical immersionâ.
Draft: March 28, 2018
APPENDIX c
Good Covers & ConvexityTheorems
Let ð be an ð-dimensional submanifold of ðð. Our goal in this appendix is to provethatð admits a good cover. To do so weâll need some basic facts about convexity.
Definition c.1. Let ð be an open subset of ðð and ð a ð¶â function on ð. We say that ð isconvex if the matrix(c.2) [ ð
2ððð¥ððð¥ð(ð)]
is positive-definite for all ð â ð.Suppose now thatð itself is convex and ð is a convex function onðwhich has as image
the half open interval [0, ð) and is a proper map ð â [0, ð). (In other words, for 0 †ð < ð,the set { ð¥ â ð | ð(ð¥) †ð } is compact).
Theorem c.3. For 0 < ð < ð, the setðð â { ð¥ â ð | ð(ð¥) < ð }
is convex.
Proof. For ð, ð â ððð with ð â ð, letð(ð¡) â ð((1 â ð¡)ð + ð¡ð) ,
for 0 †ð¡ †1. We claim that
(c.4) ð2ððð¡2(ð¡) > 0
and
(c.5) ðððð¡(0) < 0 < ðð
ðð¡(1) .
To prove (c.4) we note thatð2ððð¡2(ð¡) = â1â€ð,ðâ€ð
ð2ððð¥ððð¥ð((1 â ð¡)ð + ð¡ð)(ðð â ðð)(ðð â ðð)
and that the right-hand side is positive because of the positive-definiteness of (c.2).To prove (c.5) we note that ðððð¡ is strictly increasing by (c.4). Hence ðððð¡ (0) > 0 would
imply that ðððð¡ (ð¡) > 0 for all ð¡ â [0, 1] and hence would imply that ð is strictly increasing.But
ð(0) = ð(1) = ð(ð) = ð(ð) = ð ,so this would be a contradiction. We conclude that ðððð¡ (0) < 0. A similar argument showsthat ðððð¡ (1) > 0.
Now we use equations (c.4) and (c.5) to show that ðð is convex. Since ðððð¡ is strictlyincreasing it follows from (c.5) that ðððð¡ (ð¡0) = 0 for some ð¡0 â (0, 1) and that ðððð¡ (ð¡) < 0 for
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212 Appendix c: Good Covers & Convexity Theorems
ð¡ â [0, ð¡0) andðððð¡ (ð¡) > 0 for ð¡ â (ð¡0, 1]. Therefore, since ð(0) = ð(1) = ð, we have ð(ð¡) < ð
for all ð¡ â (0, 1); so ð(ð¥) < ð on the line segment defined by (1â ð¡)ð+ ð¡ð for ð¡ â (0, 1). Henceðð is convex. â¡
Coming back to the manifold ð â ðð, let ð â ð and let ð¿ð â ððð. We denote by ððthe orthogonal projection
ðð ⶠð â ð¿ð + ð ,i.e., for ð â ð and ð¥ = ðð(ð), we have that ð â ð¥ is orthogonal to ð¿ð (itâs easy to see that ððis defined by this condition). We will prove that this projection has the following convexityproperty.
Theorem c.6. There exists a positive constant ð(ð) such that if ð â ð satisfies |ð â ð|2 < ð(ð)and ð < ð(ð), then the set
(c.7) ð(ð, ð) â { ðâ² â ð | |ðâ² â ð| < ð }is mapped diffeomorphically by ðð onto a convex open set in ð¿ð + ð.
Proof. We can, without loss of generality, assume that ð = 0 and that
ð¿ = ðð = ðð Ã {0}
in ðð à ðð = ðð, where ð â ð â ð. Thus ð can be described, locally over a convex neigh-borhood ð of the origin in ðð as the graph of a ð¶â function ðⶠ(ð, 0) â (ðð, 0) and sinceðð à {0} is tangent toð at 0, we have
(c.8) ðððð¥ð= 0 for ð = 1,âŠ, ð .
Given ð = (ð¥ð, ð(ð¥ð)) â ð, let ðð ⶠð â ð by the function
(c.9) ðð(ð¥) â |ð¥ â ð¥ð|2 + |ð(ð¥) â ð(ð¥ð)|2 .
Then by (c.7) ðð(ð(ð, ð)) is just the set
{ ð¥ â ð | ðð(ð¥) < ð } ,
so to prove the theorem it suffices to prove that if ð(ð) is sufficiently small and ð < ð(ð), thisset is convex. However, at ð = 0, by equations (c.8) and (c.9) we have
ð2ðððð¥ððð¥ð= ð¿ð,ð ,
and hence by continuity ðð is convex on the set { ð¥ â ð | |ð¥|2 < ð¿ } provided that ð¿ and|ð| are sufficiently small. Hence if ð(ð) is sufficiently small, then ðð(ð(ð, ð)) is convex forð < ð(ð). â¡
We will now use Theorem c.6 to prove thatð admits a good cover: for every ð â ð, let
ð(ð) â âð(ð)3
and letðð â { ð â ð | |ð â ð| < ð(ð) } .
Theorem c.10. The ððâs are a good cover ofð.
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213
Proof. Suppose that the intersection(c.11) ðð1 â©â¯ â© ðððis nonempty and that
ð(ð1) ⥠ð(ð2) ⥠⯠⥠ð(ðð) .Then if ð is a point in the intersectionðð1 â©â¯â©ððð , then for ð = 1,âŠ, ð we have |ð â ðð| <ð(ð1), hence |ððâð1| < 2ð(ð1).Moreover, if ð â ððð and |ðâðð| < ð(ð1), then |ðâð1| < 3ð(ð1)and so |ð âð1|2 < ð(ð1). Therefore the setððð is contained inð(ðð, ð(ð1)) and consequentlyby Theorem c.6 is mapped diffeomorphically onto a convex open subset of ð¿ð1 + ð1 by theprojection ðð1 . Consequently the intersection (c.11) is mapped diffeomorphically onto aconvex open subset of ð¿ð1 + ð1 by ðð1 . â¡
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Bibliography
1. Vladimir Igorevich Arnolâd, Mathematical methods of classical mechanics, Vol. 60, Springer Science & BusinessMedia, 2013.
2. Garrett Birkhoff and Gian-Carlo Rota, Ordinary differential equations, 4th ed., John Wiley & Sons, 1989.3. Raoul Bott and Loring W. Tu, Differential forms in algebraic topology, Vol. 82, Springer Science & Business
Media, 2013.4. Victor Guillemin and Alan Pollack, Differential topology, Vol. 370, American Mathematical Society, 2010.5. Victor Guillemin and Shlomo Sternberg, Symplectic techniques in physics, Cambridge University Press, 1990.6. Nikolai V. Ivanov, A differential forms perspective on the Lax proof of the change of variables formula, The Amer-
ican Mathematical Monthly 112 (2005), no. 9, 799â806.7. John L. Kelley, General topology, Graduate Texts in Mathematics, vol. 27, Springer, 1975.8. Peter D. Lax, Change of variables in multiple integrals, The American Mathematical Monthly 106 (1999), no. 6,
497â501.9. , Change of variables in multiple integrals II, The American Mathematical Monthly 108 (2001), no. 2,
115â119.10. James R. Munkres, Analysis on manifolds, Westview Press, 1997.11. , Topology, 2nd ed., Prentice Hall, 2000 (English).12. Michael Spivak, Calculus on manifolds: a modern approach to classical theorems of advanced calculus, Mathe-
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Index of Notation
Að(ð), 14Alt, 14
ðµð(ð), 149ðµðð (ð), 149(ðð), 16
ð¶ð(U; ðºâ), 196ð¶ð(U; ð), 195ð(ð), 189
deg, 77ð·ð, 33ðð, 33dim, 1ðð¥ð, 35ð/ðð¥ð, 34
ð»ð(ð), 149ð»ð(U; ð), 195ð»ðð (ð), 149â, 56
Ið(ð), 17im, 2ðð£, 23ðð, 36
ker, 2
ð¿(ð), 184Lð(ð), 8
ð¬ð(ðâ), 19ð¿ð, 34
ð(U), 195ðð(U), 194
ðºð(ð), 110ðºðð (ð), 111
ðâð, 40
ðð, 12ðŽð, 12(â1)ð, 12supp, 39, 71
ðððð, 33ðð(ð), 104, 105â, 9ðð, 176ðð, 13ðð, 176
ð(ð, ð), 139â§, 20
ðð(ð), 149ððð (ð), 149
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Glossary of Terminology
alternation operation, 14annihilator of a subspace, 6
basis, 1positively oriented, 29
bilinear, 191bilinear pairing, 171
non-singular, 171
canonical immersion theorem, 108, 209canonical submersion theorem, 107, 209chain rule
for manifolds, 106change of variables formula
for manifolds, 126cochain complex, 158cohomology class, 149configuration space, 66contractable, 56
manifold, 137convex, 141
open set, 166convexâconcave, 141coset, 3cotangent space
to ðð, 35critial point, 135critical
point, 85value, 85
critical value, 135current density, 62Äech
cochain complex, 195cohomology groups, 195
Darboux form, 63de Rham cohomology, 149
compactly supported, 149
decomposable element, 21degree
of a map, 77derivative
of a ð¶â map between manifolds, 106determinant, 26diagonal, 184diffeomorphism, 97
orientation preserving, 117orientation reversing, 118
differential formclosed, 48compactly supported, 71, 111exact, 48on an open subset of ðð, 44one-form, 35
dimensionof a vector space, 1
divergence theoremfor manifolds, 130
dual vector space, 4
electric field, 62energy
conservation of, 67kinetic, 66total, 66
Euler characteristic, 189exact sequence, 158
short, 159exterior derivative, 46
ð-relatedvector fields, 40vector fields on manifolds, 109
finite topology, 167Frobeniusâ Theorem, 92
genus, 143
219
Draft: March 28, 2018
220 Glossary of Terminology
global Lefschetz number, 184good cover, 165gradient, 60
homotopic, 56homotopy, 56
equivalence, 156invariant, 151proper, 87
imageof a linear map, 2
immersioncanonical, 209
implicit function theorem, 205index
local, 144of a vector field, 143
inner product, 2integral
of a system, 38integral curve, 37, 111
maximal, 38interior product
for vector spaces, 24of differential forms, 51of forms, 36
intersectiontransverse, 180
intersection numberlocal, 180of a map with a submanifold, 183
inverse function theoremfor manifolds, 107
Jacobi identityfor vector fields, 36
ð-formon an open subset of ðð, 44ð-skeleton, 195kernel
of a linear map, 2
Lefschetz map, 184Lie algebra, 36Lie bracket
of vector fields, 36Lie derivative, 34
of a differential form, 52
line integral, 36linear independence, 1long exact sequence in cohomology, 162
magnetic field, 62manifold, 97MayerâVietoris Theorem, 162multi-index, 8
repeating, 15strictly increasing, 15
nerve, 168nerve of a cover, 195Noetherâs principle, 67
one-parameter groupassociated to a complete vector field,
40open set
parametrizable, 110star-shaped, 58
orientationcompatible, 180of a line in ð2, 29of a manifold, 116of a one-dimensional vector space, 29of a vector space (in general), 29reversed, 116smooth, 116standard, 116
orientation preservingdiffeomorphism, 117
orientation reversingdiffeomorphism, 118
orientation-preservingmap of vector spaces, 30
orthonormal vectors, 7
parametrizable open setð·-adapted, 120
parametrization, 104oriented, 118
permutation, 12transposition, 12
elementary, 12phase space, 66Poincaré lemma
for compactly supported forms, 76for manifolds, 150for rectangles, 72
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221
Poincaré recurrence, 67Poisson bracket, 68projection formula, 176proper
map, 77pullback
of a one-form, 41of forms, 54of vector fields, 43, 109
pushforwardof a vector field, 40of vector fields, 109
quotientvector space, 3
regular value, 85, 99, 135
saddle, 146Sardâs theorem, 86scalar charge density, 62scalar curvature, 141sign
of a permutation, 13simply connected, 157sink, 146smooth domain, 118source, 146submersion, 209sumbersion
canonical, 209support
of a differential formon ðð, 71on a manifold, 111
of a vector field, 39symmetric group, 12symplectomorphism, 64
tangent spaceof a manifold, 98, 104, 105to ðð, 33
tensor, 8alternating, 14decomposable, 10redundant, 17symmetric, 19
tensor product, 192of linear maps, 9
Thomclass, 176form, 176
vector, 1positively oriented, 29
vector field, 34compactly supported, 39complete, 38, 111Hamiltonian, 65Lefschetz, 190on a manifold, 109smooth, 110symplectic, 64
vector space, 1volume element
of a vector space, 31volume form
on a manifold, 116Riemannian, 117symplectic, 64
wedge productof forms, 45
winding number, 139
zeronon-degenerate, 144