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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 ๐-tensors 111.5. The space ๐ฌ๐(๐โ) 171.6. The wedge product 201.7. The interior product 231.8. The pullback operation on ๐ฌ๐(๐โ) 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 ๐-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 Poincarรฉ lemma for compactly supported forms on rectangles 713.3. The Poincarรฉ lemma for compactly supported forms on open subsets of ๐๐ 763.4. 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: Sardโs theorem 92
Chapter 4. Manifolds & Forms on Manifolds 974.1. Manifolds 974.2. Tangent spaces 1044.3. Vector fields & differential forms on manifolds 109
iii
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iv Contents
4.4. Orientations 1164.5. Integration of forms on manifolds 1244.6. Stokesโ 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 MayerโVietoris Sequence 1585.3. Cohomology of Good Covers 1655.4. Poincarรฉ duality 1715.5. Thom classes & intersection theory 1765.6. The Lefschetz Theorem 1835.7. The Kรผnneth theorem 1915.8. ฤech 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 oneโs first encounterwith differential forms is the changeof variables formula in multivariable calculus, i.e. the formula
(1) โซ๐๐โ๐| det ๐ฝ๐|๐๐ฅ = โซ
๐๐๐๐ฆ
In this formula, ๐ and ๐ are bounded open subsets of ๐๐, ๐โถ ๐ โ ๐ is a bounded contin-uous function, ๐โถ ๐ โ ๐ is a bijective differentiable map, ๐โ๐โถ ๐ โ ๐ is the function๐ โ ๐, and det ๐ฝ๐(๐ฅ) is the determinant of the Jacobian matrix.
๐ฝ๐(๐ฅ) โ [๐๐๐๐๐ฅ๐(๐ฅ)] ,
As for the โ๐๐ฅโ and โ๐๐ฆโ, their presence in (1) can be accounted for by the fact that in single-variable calculus, with ๐ = (๐, ๐), ๐ = (๐, ๐), ๐โถ (๐, ๐) โ (๐, ๐), and ๐ฆ = ๐(๐ฅ) a ๐ถ1functionwith positive first derivative, the tautological equation ๐๐ฆ๐๐ฅ =
๐๐๐๐ฅ can be rewritten in the form
๐(๐โ๐ฆ) = ๐โ๐๐ฆ
and (1) can be written more suggestively as
(2) โซ๐๐โ(๐ ๐๐ฆ) = โซ
๐๐๐๐ฆ
One of the goals of this text on differential forms is to legitimize this interpretation of equa-tion (1) in ๐ dimensions and in fact, more generally, show that an analogue of this formulais true when ๐ and ๐ are ๐-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 ๐ be abijection and just require ๐ to be proper (i.e., that pre-images of compact subsets of๐ to becompact subsets of ๐) then the formula (1) can be replaced by
(3) โซ๐๐โ(๐ ๐๐ฆ) = deg(๐) โซ
๐๐๐๐ฆ
where deg(๐) is a topological invariant of ๐ that roughly speaking counts, with plus andminus signs, the number of pre-image points of a generically chosen point of ๐.1
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 ๐ an open subset of ๐2, we define ๐บ0(๐) 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 ๐ถโ functions on ๐. We define the vector space ๐บ1(๐) to be the space of formalsums(4) ๐1 ๐๐ฅ1 + ๐2 ๐๐ฅ2 ,where ๐1, ๐2 โ ๐ถโ(๐). We define the vector space ๐บ2(๐) to be the space of expressions ofthe form(5) ๐๐๐ฅ1 โง ๐๐ฅ2 ,where ๐ โ ๐ถโ(๐), and for ๐ > 2 define ๐บ๐(๐) to the zero vector space.
On these vector spaces one can define operators(6) ๐โถ ๐บ๐(๐) โ ๐บ๐+1(๐)by the recipes
(7) ๐๐ โ ๐๐๐๐ฅ1๐๐ฅ1 +๐๐๐๐ฅ2๐๐ฅ2
for ๐ = 0,
(8) ๐(๐1 ๐๐ฅ1 + ๐2 ๐๐ฅ2) = (๐๐2๐๐ฅ1โ ๐๐1๐๐ฅ2) ๐๐ฅ1 โง ๐๐ฅ
for ๐ = 1, and ๐ = 0 for ๐ > 1. It is easy to see that the operator
(9) ๐2 โถ ๐บ๐(๐) โ ๐บ๐+2(๐)is zero. Hence,
im(๐โถ ๐บ๐โ1(๐) โ ๐บ๐(๐)) โ ker(๐โถ ๐บ๐(๐) โ ๐บ๐+1(๐)) ,and this enables one to define the de Rham cohomology groups of ๐ as the quotient vectorspace
(10) ๐ป๐(๐) โ ker(๐โถ ๐บ๐(๐) โ ๐บ๐+1(๐))im(๐โถ ๐บ๐โ1(๐) โ ๐บ๐(๐))
.
It turns out that these cohomology groups are topological invariants of ๐ and are, infact, isomorphic to the cohomology groups of๐ 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 ๐๐ and, with a bit more effort, for arbitrary ๐ถโmanifolds (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 ๐๐ฅ
and ๐๐ฆ 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.Weโll 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 ๐-tensor and (the future key ingredient in our exposition ofthe theory of differential forms in Chapter 2) the concept of alternating ๐-tensor. Those ๐tensors come up in fact in two contexts: as alternating ๐-tensors, and as exterior forms, i.e.,in the first context as a subspace of the space of ๐-tensors and in the second as a quotientspace of the space of ๐-tensors. Both descriptions of ๐-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 ยง1.6), the interior product operation (see ยง1.7) and the pullback operation (see ยง1.8).We will alsomake use of these results in Section 1.9 to define the notion of an orientation foran ๐-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 ๐ฝ๐|.
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 theyโre expressions of this type. Weโllbegin this chapter, however, with the following more precise definition: Let ๐ be an opensubset of ๐๐. Then a ๐-form ๐ on ๐ is a โfunctionโ which to each ๐ โ ๐ assigns an elementof ๐ฌ๐(๐โ๐๐), ๐๐๐ being the tangent space to ๐ at ๐, ๐โ๐๐ its vector space dual, and ๐ฌ๐(๐โ๐ )the ๐th order exterior power of ๐โ๐๐. (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 ๐-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 weโve 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 Maxwellโs equation, as wellas to classical mechanisms and the HamiltonโJacobi 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,Laxโs 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 let๐ be a connected open set in ๐๐ and ๐ = ๐๐๐ฅ1 โงโฏโง ๐๐ฅ๐ a compactlysupported ๐-form on ๐. We will prove:
Theorem 11. The integral โซ๐ ๐ = โซ๐ ๐๐๐ฅ1โฏ๐๐ฅ๐ of ๐ over ๐ is zero if and only if ๐ = ๐๐where ๐ is a compactly supported (๐ โ 1)-form on ๐.
An easy corollary of this result is the following weak version of the degree theorem:Corollary 12. Let๐ and๐ be connected open subsets of๐๐ and๐โถ ๐ โ ๐ a propermapping.Then there exists a constant ๐ฟ(๐) with the property that for every compactly supported ๐-form๐ on ๐
โซ๐๐โ๐ = ๐ฟ(๐)โซ
๐๐ .
Thus to prove the degree theorem it suffices to show that this ๐ฟ(๐) is the deg(๐) 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 ๐ต๐ isthe closed unit ball in ๐๐ and ๐โถ ๐ต๐ โ ๐ต๐ is a continuous map, then ๐ has a fixed point,i.e. there is a point ๐ฅ0 โ ๐ต๐ that gets mapped onto itself by ๐. (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. If๐(๐ง) = ๐0 + ๐1๐ง +โฏ + ๐๐โ1๐ง๐โ1 + ๐ง๐
is a polynomial function of the complex variable ๐ง, then it has to have a complex root ๐ง0satisfying ๐(๐ง0) = 0. (Weโll 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 ๐๐. In this chapter weโll 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 ๐-dimensional manifolds as ๐-dimensional submanifolds of ๐๐,i.e., as ๐-dimensional generalizations of curves and surfaces in ๐3, whereas the tradition ingraduate level courses is to define them as abstract entities. Since this is intended to be a textbook for undergraduates, weโll 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 ๐๐๐to a manifold ๐ at a point ๐ and in sections Sections 4.3 and 4.4 show how to define dif-ferential forms on ๐ by defining a ๐-form to be, as in ยง2.4, a function ๐ which assigns toeach ๐ โ ๐ an element ๐๐ of ๐ฌ๐(๐โ๐๐). Then in Section 4.4 we will define what it meansfor a manifold to be oriented and in Section 4.5 show that if๐ is an oriented ๐-dimensionalmanifold and ๐ a compactly supported ๐-form, the integral of ๐ is well-defined. Moreover,weโll 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 ยง4.7) and a number of applications of this theorem(among them the JordanโBrouwer separation theorem, the GaussโBonnet theorem and theIndex theorem for vector fields (see ยงยง4.8 and 4.9).
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Organization ix
Chapter 5: Cohomology via FormsGiven an ๐-dimensional manifold let ๐บ๐(๐) be the space of differential forms on ๐ of
degree ๐, let ๐โถ ๐บ๐(๐) โ ๐บ๐+1(๐) be exterior differentiation and let ๐ป๐(๐) be the ๐th deRham cohomology group of๐: the quotient of the vector space
๐๐(๐) โ ker(๐โถ ๐บ๐(๐) โ ๐บ๐+1(๐))by the vector space
๐ต๐(๐) โ im(๐โถ ๐บ๐โ1(๐) โ ๐บ๐(๐)) .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 ๐-dimensionalmanifold the integration operator on forms that we discussed in Chapter 4 gives rise to apairing
๐ป๐(๐) ร ๐ป๐โ๐(๐) โ ๐ ,and in fact more generally if๐ is non-compact, a pairing
(13) ๐ป๐๐ (๐) ร ๐ป๐โ๐(๐) โ ๐ ,
where the groups ๐ป๐๐(๐) are the DeRham cohomology groups that one gets by replacingthe spaces of differential forms๐บ๐(๐) by the corresponding spaces๐บ๐๐(๐) of compactly sup-ported differential forms.
One problem one runs into in the definition of these cohomology groups is that sincethe spaces๐บ๐(๐) and๐บ๐๐ (๐) are infinite dimensional in general, thereโs no guarantee that thesame wonโt be the case for the spaces๐ป๐(๐) and๐ป๐๐ (๐), and weโll address this problem inSection 5.3.More explicitly, we will say that๐ has finite topology if it admits a finite coveringby open sets ๐1,โฆ๐๐ such that for every multi-index ๐ผ = (๐1,โฆ , ๐๐), where 1 โค ๐๐ โค ๐,the intersection(14) ๐๐ผ โ ๐๐1 โฉโฏ โฉ ๐๐๐is either empty or is diffeomorphic to a convex open subset of๐๐. We will show that if๐ hasfinite topology, then its cohomology groups are finite dimensional and, secondly, we willshow that many manifolds have finite topology. (For instance, if ๐ is compact then ๐ hasfinite topology; for details see ยงยง5.2 and 5.3.)
We mentioned above that if ๐ is not compact it has two types of cohomology groups:the groups๐ป๐(๐) and the groups๐ป๐โ๐๐ (๐). In Section 5.5 we will show that if๐ is orientedthen for ๐ โ ๐บ๐(๐) and ๐ โ ๐บ๐โ๐๐ (๐) the integration operation
(๐, ๐) โฆ โซ๐๐ โง ๐
gives rise to a pairing๐ป๐(๐) ร ๐ป๐โ๐๐ (๐) โ ๐ ,
and that if๐ is connected this pairing is non-degenerate, i.e., defines a bijective linear trans-formation
๐ป๐โ๐๐ (๐) โฅฒ ๐ป๐(๐)โ .This results in the Poincarรฉ 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 ๐ and๐ are closed oriented submanifolds of ๐ and ๐ is compact and of codimension equal to
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x Preface
the dimension of ๐, then the intersection number ๐ผ(๐, ๐) of ๐ and ๐ in ๐ is well-defined(no matter how badly ๐ and ๐ intersect). In Section 5.6 we will show that if๐ is a compactoriented manifold and ๐โถ ๐ โ ๐ a ๐ถโ map one can define the Lefschetz number of ๐ asthe intersection number ๐ผ(๐ฅ๐, graph(๐)) where ๐ฅ๐ is the diagonal in๐ ร ๐, and view thisas a โtopological countโ of the number of fixed points of the map ๐.
Finally in Section 5.8 we will show that if ๐ has finite topology (in the sense we de-scribed above), then one can define an alternative set of cohomology groups for๐, the ฤechcohomology groups of a cover U, and we will sketch a proof of the association that this ฤechcohomology 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 โimproper integralsโ and to extend differentiable maps ๐โถ ๐ โ ๐ on arbitrary subsets๐ โ ๐๐ to open neighborhoods of these sets.
In Appendix b we discuss another calculus issue: how to solve systems of equations
{{{{{
๐1(๐ฅ) = ๐ฆ1โฎ
๐๐(๐ฅ) = ๐ฆ๐for ๐ฅ in terms of ๐ฆ (here ๐1,โฆ, ๐๐ are differentiable functions on an open subset of ๐๐). 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 โfinite topologyโ results that we discussed in Chapter 5, namely the conceptof a โgood coverโ. In more detail let๐ be an ๐-dimensional manifold an let U = {๐๐ผ}๐ผโ๐ผ bea collection of open subsets of๐ with the property thatโ๐ผโ๐ผ๐๐ผ = ๐.
Then U is a good cover of๐ if for every finite subset
{๐ผ1,โฆ, ๐ผ๐} โ ๐ผ
the intersection ๐๐ผ1 โฉโฏ โฉ ๐๐ผ๐ is either empty or is diffeomorphic to a convex open subsetof ๐๐. 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.โ used to define a term; the term being defined appears to the left of the colonโ subsetโ denotes an map (of sets, vector spaces, etc.)โช inclusion mapโ surjective mapโฅฒ a map that is an isomorphismโ difference of sets: if ๐ด โ ๐, then๐ โ ๐ด is the complement of ๐ด in๐
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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 set๐ the elements of which we refer to as vectors.The set ๐ is equipped with two vector space operations:(1) Vector space addition: Given vectors ๐ฃ1, ๐ฃ2 โ ๐ one can add them to get a third vector,๐ฃ1 + ๐ฃ2.
(2) Scalar multiplication: Given a vector ๐ฃ โ ๐ and a real number ๐, one can multiply ๐ฃ by๐ to get a vector ๐๐ฃ.These operations satisfy a number of standard rules: associativity, commutativity, dis-
tributive laws, etc. which we assume youโre 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 ๐ is the unique vector 0 โ ๐ with the property that for every vector๐ฃ โ ๐, we have ๐ฃ + 0 = 0 + ๐ฃ = ๐ฃ and ๐๐ฃ = 0 if ๐ โ ๐ is nonzero.
(2) Linear independence: A (finite) set of vectors, ๐ฃ1,โฆ, ๐ฃ๐ โ ๐ is linearly independent ifthe map
(1.1.2) ๐๐ โ ๐ , (๐1,โฆ, ๐๐) โฆ ๐1๐ฃ1 +โฏ + ๐๐๐ฃ๐is injective.
(3) A (finite) set of vectors ๐ฃ1,โฆ, ๐ฃ๐ โ ๐ spans ๐ if the map (1.1.2) is surjective.(4) Vectors ๐ฃ1,โฆ, ๐ฃ๐ โ ๐ are a basis of ๐ if they span ๐ and are linearly independent; in
other words, if themap (1.1.2) is bijective.Thismeans that every vector ๐ฃ can be writtenuniquely as a sum
(1.1.3) ๐ฃ =๐โ๐=1๐๐๐ฃ๐ .
(5) If๐ is a vector space with a basis ๐ฃ1,โฆ, ๐ฃ๐, then๐ is said to be finite dimensional, and ๐is the dimension of๐. (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 weโll encounterwill be finite dimensional. We write dim(๐) for the dimension of ๐.
(6) A subset๐ โ ๐ is a subspace if it is vector space in its own right, i.e., for all ๐ฃ, ๐ฃ1, ๐ฃ2 โ ๐and ๐ โ ๐, both ๐๐ฃ and ๐ฃ1 + ๐ฃ2 are in ๐ (where the addition and scalar multiplicationis given as vectors of ๐).
(7) Let๐ and๐ be vector spaces. A map๐ดโถ ๐ โ ๐ is linear if for ๐ฃ, ๐ฃ1, ๐ฃ2 โ ๐ and ๐ โ ๐we have
(1.1.4) ๐ด(๐๐ฃ) = ๐๐ด๐ฃ1Such a course is a prerequisite for reading this text.
1
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2 Chapter 1: Multilinear Algebra
and๐ด(๐ฃ1 + ๐ฃ2) = ๐ด๐ฃ1 + ๐ด๐ฃ2 .
(8) Let ๐ดโถ ๐ โ ๐ be a linear map of vector spaces. The kernel of ๐ด is the subset of ๐defined by
ker(๐ด) โ { ๐ฃ โ ๐ |๐ด(๐ฃ) = 0 } ,i.e, is the set of vectors in ๐ which ๐ด sends to the zero vector in๐. By equation (1.1.4)and item 7, the subset ker(๐ด) is a subspace of ๐.
(9) By (1.1.4) and (7) the set-theoretic image im(๐ด)of ๐ด is a subspace of๐. We call im(๐ด)the image๐ด of๐ด.The following is an important rule for keeping track of the dimensionsof ker๐ด and im๐ด:
(1.1.5) dim(๐) = dim(ker(๐ด)) + dim(im(๐ด)) .(10) Linear mappings and matrices: Let ๐ฃ1,โฆ, ๐ฃ๐ be a basis of ๐ and๐ค1,โฆ,๐ค๐ a basis of๐.
Then by (1.1.3) ๐ด๐ฃ๐ can be written uniquely as a sum,
(1.1.6) ๐ด๐ฃ๐ =๐โ๐=1๐๐,๐๐ค๐ , ๐๐,๐ โ ๐ .
The ๐ ร ๐ matrix of real numbers, [๐๐,๐], is the matrix associated with ๐ด. Conversely,given such an๐ ร ๐matrix, there is a unique linear map ๐ด with the property (1.1.6).
(11) An inner product on a vector space is a map๐ตโถ ๐ ร ๐ โ ๐
with the following three properties:(a) Bilinearity: For vectors ๐ฃ, ๐ฃ1, ๐ฃ2, ๐ค โ ๐ and ๐ โ ๐ we have
๐ต(๐ฃ1 + ๐ฃ2, ๐ค) = ๐ต(๐ฃ1, ๐ค) + ๐ต(๐ฃ2, ๐ค)and
๐ต(๐๐ฃ, ๐ค) = ๐๐ต(๐ฃ, ๐ค) .(b) Symmetry: For vectors ๐ฃ, ๐ค โ ๐ we have ๐ต(๐ฃ, ๐ค) = ๐ต(๐ค, ๐ฃ).(c) Positivity: For every vector ๐ฃ โ ๐ we have ๐ต(๐ฃ, ๐ฃ) โฅ 0. Moreover, if ๐ฃ โ 0 then๐ต(๐ฃ, ๐ฃ) > 0.
Remark 1.1.7. Notice that by property (11b), property (11a) is equivalent to๐ต(๐ค, ๐๐ฃ) = ๐๐ต(๐ค, ๐ฃ)
and๐ต(๐ค, ๐ฃ1 + ๐ฃ2) = ๐ต(๐ค, ๐ฃ1) + ๐ต(๐ค, ๐ฃ2) .
Example 1.1.8. The map (1.1.2) is a linear map. The vectors ๐ฃ1,โฆ, ๐ฃ๐ span๐ if its image ofthis map is๐, and the ๐ฃ1,โฆ, ๐ฃ๐ are linearly independent if the kernel of this map is the zerovector in ๐๐.
The items on the list above are just a few of the topics in linear algebra that weโre assum-ing our readers are familiar with. We have highlighted them because theyโre easy to state.However, understanding them requires a heavy dollop of that indefinable quality โmathe-matical sophisticationโ, 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 ๐ ร ๐matrices, inverses of matrices, Cramerโs rule,recipes for solving systems of linear equations, etc. (See [10, ยงยง1.1 & 1.2] for a quick reviewof this material.)
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ยง1.2 Quotient spaces & dual spaces 3
Exercises for ยง1.1Exercise 1.1.i. Our basic example of a vector space in this course is ๐๐ equipped with thevector addition operation
(๐1,โฆ, ๐๐) + (๐1,โฆ, ๐๐) = (๐1 + ๐1,โฆ, ๐๐ + ๐๐)and the scalar multiplication operation
๐(๐1,โฆ, ๐๐) = (๐๐1,โฆ, ๐๐๐) .Check that these operations satisfy the axioms below.(1) Commutativity: ๐ฃ + ๐ค = ๐ค + ๐ฃ.(2) Associativity: ๐ข + (๐ฃ + ๐ค) = (๐ข + ๐ฃ) + ๐ค.(3) For the zero vector 0 โ (0,โฆ, 0) we have ๐ฃ + 0 = 0 + ๐ฃ.(4) ๐ฃ + (โ1)๐ฃ = 0.(5) 1๐ฃ = ๐ฃ.(6) Associative law for scalar multiplication: (๐๐)๐ฃ = ๐(๐๐ฃ).(7) Distributive law for scalar addition: (๐ + ๐)๐ฃ = ๐๐ฃ + ๐๐ฃ.(8) Distributive law for vector addition: ๐(๐ฃ + ๐ค) = ๐๐ฃ + ๐๐ค.Exercise 1.1.ii. Check that the standard basis vectors of๐๐: ๐1 = (1, 0,โฆ, 0), ๐2 = (0, 1, 0,โฆ, 0),etc. are a basis.
Exercise 1.1.iii. Check that the standard inner product on ๐๐
๐ต((๐1,โฆ, ๐๐), (๐1,โฆ, ๐๐)) =๐โ๐=1๐๐๐๐
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 weโll need for our treatment of multilinear al-gebra in ยงยง1.3 to 1.8.
The quotient spaces of a vector spaceDefinition 1.2.1. Let ๐ be a vector space and๐ a vector subspace of ๐. A๐-coset is a setof the form
๐ฃ +๐ โ { ๐ฃ + ๐ค |๐ค โ ๐} .It is easy to check that if ๐ฃ1 โ ๐ฃ2 โ ๐, the cosets ๐ฃ1 +๐ and ๐ฃ2 +๐, coincide while if
๐ฃ1โ๐ฃ2 โ ๐, the cosets ๐ฃ1+๐ and ๐ฃ2+๐ are disjoint.Thus the distinct๐-cosets decompose๐ into a disjoint collection of subsets of ๐.Notation 1.2.2. Let ๐ be a vector space and๐ โ ๐ a subspace. We write ๐/๐ for the setof distinct๐-cosets in ๐.Definition 1.2.3. Let ๐ be a vector space and๐ โ ๐ a subspace. Define a vector additionoperation on ๐/๐ by setting(1.2.4) (๐ฃ1 +๐) + (๐ฃ2 +๐) โ (๐ฃ1 + ๐ฃ2) + ๐and define a scalar multiplication operation on ๐/๐ by setting(1.2.5) ๐(๐ฃ +๐) โ (๐๐ฃ) + ๐ .These operations make ๐/๐ into a vector space, called quotient space of ๐ by๐.
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4 Chapter 1: Multilinear Algebra
It is easy to see that the operations on ๐/๐ from Definition 1.2.3 are well-defined. Forinstance, suppose ๐ฃ1 +๐ = ๐ฃโฒ1 +๐ and ๐ฃ2 +๐ = ๐ฃโฒ2 +๐. Then ๐ฃ1 โ ๐ฃโฒ1 and ๐ฃ2 โ ๐ฃโฒ2 are in๐, so
(๐ฃ1 + ๐ฃ2) โ (๐ฃโฒ1 + ๐ฃโฒ2) โ ๐ ,and hence (๐ฃ1 + ๐ฃ2) + ๐ = (๐ฃโฒ1 + ๐ฃโฒ2) + ๐.
Definition 1.2.6. Let ๐ be a vector space and๐ โ ๐ a subspace. Define a map quotientmap(1.2.7) ๐โถ ๐ โ ๐/๐by setting ๐(๐ฃ) โ ๐ฃ + ๐. It is clear from equations (1.2.4) and (1.2.5) that ๐ is linear, andthat it maps ๐ surjectively onto ๐/๐.
Observation 1.2.8. Note that the zero vector in the vector space ๐/๐ is the zero coset0 + ๐ = ๐. Hence ๐ฃ โ ker(๐) if and only if ๐ฃ + ๐ = ๐, i.e., ๐ฃ โ ๐. In other words,ker(๐) = ๐.
In the Definitions 1.2.3 and 1.2.6, ๐ and๐ do not have to be finite dimensional, but ifthey are then by equation (1.1.5) we have
dim(๐/๐) = dim(๐) โ dim(๐) .We leave the following easy proposition as an exercise.
Proposition 1.2.9. Let ๐ดโถ ๐ โ ๐ a linear map of vector spaces. If๐ โ ker(๐ด) there existsa unique linear map ๐ดโฏ โถ ๐/๐ โ ๐ with the property that ๐ด = ๐ดโฏ โ ๐, where ๐โถ ๐ โ ๐/๐is the quotient map.
The dual space of a vector spaceDefinition 1.2.10. Let๐ be a vector space.Write๐โ the set of all linear functions โโถ ๐ โ ๐.Define a vector space structure on ๐โ as follows: if โ1, โ2 โ ๐โ, then define the sum โ1 + โ2to be the map ๐ โ ๐ given by
(โ1 + โ2)(๐ฃ) โ โ1(๐ฃ) + โ2(๐ฃ) ,which is clearly is linear. If โ โ ๐โ is a linear function and ๐ โ ๐, define ๐โ by
(๐โ)(๐ฃ) โ ๐ โ โ(๐ฃ) ;then ๐โ is clearly linear.
The vector space ๐โ is called the dual space of ๐.
Suppose ๐ is ๐-dimensional, and let ๐1,โฆ, ๐๐ be a basis of ๐. Then every vector ๐ฃ โ ๐can be written uniquely as a sum
๐ฃ = ๐1๐1 +โฏ + ๐๐๐๐ , ๐๐ โ ๐ .Let(1.2.11) ๐โ๐ (๐ฃ) = ๐๐ .If ๐ฃ = ๐1๐1 +โฏ+๐๐๐๐ and ๐ฃโฒ = ๐โฒ1๐1 +โฏ+๐โฒ๐๐๐ then ๐ฃ+ ๐ฃโฒ = (๐1 + ๐โฒ1)๐1 +โฏ+ (๐๐ + ๐โฒ๐)๐๐, so
๐โ๐ (๐ฃ + ๐ฃโฒ) = ๐๐ + ๐โฒ๐ = ๐โ๐ (๐ฃ) + ๐โ๐ (๐ฃโฒ) .This shows that ๐โ๐ (๐ฃ) is a linear function of ๐ฃ and hence ๐โ๐ โ ๐โ.
Claim 1.2.12. If ๐ is an ๐-dimensional vector space with basis ๐1,โฆ, ๐๐, then ๐โ1 ,โฆ, ๐โ๐ is abasis of ๐โ.
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ยง1.2 Quotient spaces & dual spaces 5
Proof. First of all note that by (1.2.11)
(1.2.13) ๐โ๐ (๐๐) = {1, ๐ = ๐0, ๐ โ ๐ .
If โ โ ๐โ let ๐๐ = โ(๐๐) and let โโฒ = โ๐๐๐โ๐ . Then by (1.2.13)
โโฒ(๐๐) =๐โ๐=1๐๐๐โ๐ (๐๐) = ๐๐ = โ(๐๐) ,
i.e., โ and โโฒ take identical values on the basis vectors, ๐๐. Hence โ = โโฒ.Suppose next that โ๐๐=1 ๐๐๐โ๐ = 0. Then by (1.2.13), ๐๐ = (โ
๐๐=1 ๐๐๐โ๐ )(๐๐) = 0 for ๐ =
1,โฆ, ๐. Hence the vectors ๐โ1 ,โฆ, ๐โ๐ are linearly independent. โก
Let ๐ and๐ be vector spaces and ๐ดโถ ๐ โ ๐, a linear map. Given โ โ ๐โ the com-position, โ โ ๐ด, of ๐ด with the linear map, โโถ ๐ โ ๐, is linear, and hence is an element of๐โ. We will denote this element by ๐ดโโ, and we will denote by
๐ดโ โถ ๐โ โ ๐โ
the map, โ โฆ ๐ดโโ. It is clear from the definition that
๐ดโ(โ1 + โ2) = ๐ดโโ1 + ๐ดโโ2and that
๐ดโ(๐โ) = ๐๐ดโโ ,i.e., that ๐ดโ is linear.
Definition 1.2.14. Let๐ and๐ be vector spaces and๐ดโถ ๐ โ ๐, a linear map. We call themap ๐ดโ โถ ๐โ โ ๐โ defined above the transpose of the map ๐ด.
We conclude this section by giving a matrix description of ๐ดโ. Let ๐1,โฆ, ๐๐ be a basisof ๐ and ๐1,โฆ, ๐๐ a basis of๐; let ๐โ1 ,โฆ, ๐โ๐ and ๐โ1 ,โฆ, ๐โ๐ be the dual bases of ๐โ and๐โ. Suppose ๐ด is defined in terms of ๐1,โฆ, ๐๐ and ๐1,โฆ, ๐๐ by the๐ร๐matrix, [๐๐,๐], i.e.,suppose
๐ด๐๐ =๐โ๐=1๐๐,๐๐๐ .
Claim 1.2.15. The linear map๐ดโ is defined, in terms of๐โ1 ,โฆ, ๐โ๐ and ๐โ1 ,โฆ, ๐โ๐ by the trans-pose matrix (๐๐,๐).
Proof. Let
๐ดโ๐โ๐ =๐โ๐=1๐๐,๐๐โ๐ .
Then
๐ดโ๐โ๐ (๐๐) =๐โ๐=1๐๐,๐๐โ๐ (๐๐) = ๐๐,๐
by (1.2.13). On the other hand
๐ดโ๐โ๐ (๐๐) = ๐โ๐ (๐ด๐๐) = ๐โ๐ (๐โ๐=1๐๐,๐๐๐) =
๐โ๐=1๐๐,๐๐โ๐ (๐๐) = ๐๐,๐
so ๐๐,๐ = ๐๐,๐. โก
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6 Chapter 1: Multilinear Algebra
Exercises for ยง1.2Exercise 1.2.i. Let ๐ be an ๐-dimensional vector space and๐ a ๐-dimensional subspace.Show that there exists a basis, ๐1,โฆ, ๐๐ of ๐ with the property that ๐1,โฆ, ๐๐ is a basis of๐.
Hint: Induction on ๐ โ ๐. To start the induction suppose that ๐ โ ๐ = 1. Let ๐1,โฆ, ๐๐โ1be a basis of๐ and ๐๐ any vector in ๐ โ๐.
Exercise 1.2.ii. In Exercise 1.2.i show that the vectors ๐๐ โ ๐(๐๐+๐), ๐ = 1,โฆ, ๐ โ ๐ are abasis of ๐/๐, where ๐โถ ๐ โ ๐/๐ is the quotient map.
Exercise 1.2.iii. In Exercise 1.2.i let๐ be the linear span of the vectors ๐๐+๐ for ๐ = 1,โฆ, ๐โ๐.Show that the map
๐ โ ๐/๐ , ๐ข โฆ ๐(๐ข) ,is a vector space isomorphism, i.e., show that it maps ๐ bijectively onto ๐/๐.
Exercise 1.2.iv. Let ๐, ๐ and๐ be vector spaces and let ๐ดโถ ๐ โ ๐ and ๐ตโถ ๐ โ ๐ belinear mappings. Show that (๐ด๐ต)โ = ๐ตโ๐ดโ.
Exercise 1.2.v. Let ๐ = ๐2 and let๐ be the ๐ฅ1-axis, i.e., the one-dimensional subspace
{ (๐ฅ1, 0) | ๐ฅ1 โ ๐ }of ๐2.(1) Show that the๐-cosets are the lines, ๐ฅ2 = ๐, parallel to the ๐ฅ1-axis.(2) Show that the sum of the cosets, โ๐ฅ2 = ๐โ and โ๐ฅ2 = ๐โ is the coset โ๐ฅ2 = ๐ + ๐โ.(3) Show that the scalar multiple of the coset, โ๐ฅ2 = ๐โ by the number, ๐, is the coset, โ๐ฅ2 =๐๐โ.
Exercise 1.2.vi.(1) Let (๐โ)โ be the dual of the vector space, ๐โ. For every ๐ฃ โ ๐, let ev๐ฃ โถ ๐โ โ ๐ be the
evaluation function ev๐ฃ(โ) = โ(๐ฃ). Show that the ev๐ฃ is a linear function on ๐โ, i.e., anelement of (๐โ)โ, and show that the map
(1.2.16) ev = ev(โ) โถ ๐ โ (๐โ)โ , ๐ฃ โฆ ev๐ฃis a linear map of ๐ into (๐โ)โ.
(2) If ๐ is finite dimensional, show that the map (1.2.16) is bijective. Conclude that there isa natural identification of ๐ with (๐โ)โ, i.e., that ๐ and (๐โ)โ are two descriptions ofthe same object.
Hint: dim(๐โ)โ = dim๐โ = dim๐, so by equation (1.1.5) it suffices to show that(1.2.16) is injective.
Exercise 1.2.vii. Let๐ be a vector subspace of a finite dimensional vector space ๐ and let
๐โ = { โ โ ๐โ | โ(๐ค) = 0 for all ๐ค โ ๐} .๐โ is called the annihilator of๐ in ๐โ. Show that๐โ is a subspace of ๐โ of dimensiondim๐ โ dim๐.
Hint: By Exercise 1.2.i we can choose a basis, ๐1,โฆ, ๐๐ of ๐ such that ๐1,โฆ๐๐ is a basisof๐. Show that ๐โ๐+1,โฆ, ๐โ๐ is a basis of๐โ.
Exercise 1.2.viii. Let ๐ and ๐โฒ be vector spaces and ๐ดโถ ๐ โ ๐โฒ a linear map. Show that if๐ โ ker(๐ด), then there exists a linear map ๐ตโถ ๐/๐ โ ๐โฒ with the property that๐ด = ๐ต โ ๐(where ๐ is the quotient map (1.2.7)). In addition show that this linear map is injective ifand only if ker(๐ด) = ๐.
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ยง1.2 Quotient spaces & dual spaces 7
Exercise 1.2.ix. Let ๐ be a subspace of a finite dimensional vector space, ๐. From theinclusion map, ๐ โถ ๐โ โ ๐โ, one gets a transpose map,
๐โ โถ (๐โ)โ โ (๐โ)โ
and, by composing this with (1.2.16), a map
๐โ โ ev โถ ๐ โ (๐โ)โ .
Show that this map is onto and that its kernel is ๐. Conclude from Exercise 1.2.viii thatthere is a natural bijective linear map
๐โถ ๐/๐ โ (๐โ)โ
with the property ๐โ๐ = ๐โ โev. In other words,๐/๐ and (๐โ)โ are two descriptions of thesame object. (This shows that the โquotient spaceโ operation and the โdual spaceโ operationare closely related.)
Exercise 1.2.x. Let๐1 and๐2 be vector spaces and๐ดโถ ๐1 โ ๐2 a linear map. Verify that forthe transpose map ๐ดโ โถ ๐โ2 โ ๐โ1 we have:
ker(๐ดโ) = im(๐ด)โ
andim(๐ดโ) = ker(๐ด)โ .
Exercise 1.2.xi. Let ๐ be a vector space.(1) Let ๐ตโถ ๐ ร ๐ โ ๐ be an inner product on ๐. For ๐ฃ โ ๐ let
โ๐ฃ โถ ๐ โ ๐
be the function: โ๐ฃ(๐ค) = ๐ต(๐ฃ, ๐ค). Show that โ๐ฃ is linear and show that the map
(1.2.17) ๐ฟโถ ๐ โ ๐โ , ๐ฃ โฆ โ๐ฃis a linear mapping.
(2) If๐ is finite dimensional, prove that ๐ฟ bijective. Conclude that if๐ has an inner productone gets from it a natural identification of ๐ with ๐โ.
Hint: Since dim๐ = dim๐โ it suffices by equation (1.1.5) to show that ker(๐ฟ) = 0.Now note that if ๐ฃ โ 0 โ๐ฃ(๐ฃ) = ๐ต(๐ฃ, ๐ฃ) is a positive number.
Exercise 1.2.xii. Let ๐ be an ๐-dimensional vector space and ๐ตโถ ๐ ร ๐ โ ๐ an innerproduct on ๐. A basis, ๐1,โฆ, ๐๐ of ๐ is orthonormal if
(1.2.18) ๐ต(๐๐, ๐๐) = {1, ๐ = ๐0, ๐ โ ๐ .
(1) Show that an orthonormal basis exists.Hint: By induction let ๐1,โฆ, ๐๐ be vectors with the property (1.2.18) and let ๐ฃ be a
vector which is not a linear combination of these vectors. Show that the vector
๐ค = ๐ฃ โ๐โ๐=1๐ต(๐๐, ๐ฃ)๐๐
is nonzero and is orthogonal to the ๐๐โs. Now let ๐๐+1 = ๐๐ค, where ๐ = ๐ต(๐ค,๐ค)โ 12 .
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8 Chapter 1: Multilinear Algebra
(2) Let ๐1,โฆ๐๐ and ๐โฒ1,โฆ๐โฒ๐ be two orthonormal bases of ๐ and let
๐โฒ๐ =๐โ๐=1๐๐,๐๐๐ .
Show that
(1.2.19)๐โ๐=1๐๐,๐๐๐,๐ = {
1, ๐ = ๐0, ๐ โ ๐ .
(3) Let ๐ด be the matrix [๐๐,๐]. Show that equation (1.2.19) can be written more compactlyas the matrix identity
(1.2.20) ๐ด๐ดโบ = id๐where id๐ is the ๐ ร ๐ identity matrix.
(4) Let ๐1,โฆ, ๐๐ be an orthonormal basis of๐ and ๐โ1 ,โฆ, ๐โ๐ the dual basis of๐โ. Show thatthe mapping (1.2.17) is the mapping, ๐ฟ๐๐ = ๐โ๐ , ๐ = 1,โฆ๐.
1.3. Tensors
Definition 1.3.1. Let๐ be an ๐-dimensional vector space and let๐๐ be the set of all ๐-tuples(๐ฃ1,โฆ, ๐ฃ๐), where ๐ฃ1,โฆ, ๐ฃ๐ โ ๐, that is, the ๐-fold direct sum of ๐ with itself. A function
๐โถ ๐๐ โ ๐is said to be linear in its ๐th variable if, when we fix vectors, ๐ฃ1,โฆ, ๐ฃ๐โ1, ๐ฃ๐+1,โฆ, ๐ฃ๐, the map๐ โ ๐ defined by(1.3.2) ๐ฃ โฆ ๐(๐ฃ1,โฆ, ๐ฃ๐โ1, ๐ฃ, ๐ฃ๐+1,โฆ, ๐ฃ๐)is linear.
If ๐ is linear in its ๐th variable for ๐ = 1,โฆ, ๐ it is said to be ๐-linear, or alternativelyis said to be a ๐-tensor. We write L๐(๐) for the set of all ๐-tensors in ๐. We will agree that0-tensors are just the real numbers, that is L0(๐) โ ๐.
Let ๐1, ๐2 โถ ๐๐ โ ๐ be linear functions. It is clear from (1.3.2) that if ๐1 and ๐2 are๐-linear, so is ๐1 + ๐2. Similarly if ๐ is ๐-linear and ๐ is a real number, ๐๐ is ๐-linear. HenceL๐(๐) is a vector space. Note that for ๐ = 1, โ๐-linearโ just means โlinearโ, so L1(๐) = ๐โ.
Definition 1.3.3. Let ๐ and ๐ be positive integers. Amulti-index of ๐ of length ๐ is a ๐-tuple๐ผ = (๐1,โฆ, ๐๐) of integers with 1 โค ๐๐ โค ๐ for ๐ = 1,โฆ, ๐.
Example 1.3.4. Let ๐ be a positive integer.Themulti-indices of ๐ of length 2 are in bijectionthe square of pairs of integers integers
(๐, ๐) , 1 โค ๐, ๐ โค ๐ ,and there are exactly ๐2 of them.
We leave the following generalization as an easy exercise.
Lemma 1.3.5. Let ๐ and ๐ be positive integers. There are exactly ๐๐multi-indices of ๐ of length๐.
Now fix a basis ๐1,โฆ, ๐๐ of ๐. For ๐ โ L๐(๐) write(1.3.6) ๐๐ผ โ ๐(๐๐1 ,โฆ, ๐๐๐)for every multi-index ๐ผ of length ๐.
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ยง1.3 Tensors 9
Proposition 1.3.7. The real numbers ๐๐ผ determine ๐, i.e., if ๐ and ๐โฒ are ๐-tensors and ๐๐ผ =๐โฒ๐ผ for all ๐ผ, then ๐ = ๐โฒ.
Proof. By induction on ๐. For ๐ = 1 we proved this result in ยง1.1. Letโs prove that if thisassertion is true for ๐ โ 1, it is true for ๐. For each ๐๐ let ๐๐ be the (๐ โ 1)-tensor
(๐ฃ1,โฆ, ๐ฃ๐โ1) โฆ ๐(๐ฃ1,โฆ, ๐ฃ๐โ1, ๐๐) .Then for ๐ฃ = ๐1๐1 +โฏ๐๐๐๐
๐(๐ฃ1,โฆ, ๐ฃ๐โ1, ๐ฃ) =๐โ๐=1๐๐๐๐(๐ฃ1,โฆ, ๐ฃ๐โ1) ,
so the ๐๐โs determine ๐. Now apply the inductive hypothesis. โก
The tensor product operationDefinition 1.3.8. If ๐1 is a ๐-tensor and ๐2 is an โ-tensor, one can define a ๐ + โ-tensor,๐1 โ ๐2, by setting
(๐1 โ ๐2)(๐ฃ1,โฆ, ๐ฃ๐+โ) = ๐1(๐ฃ1,โฆ, ๐ฃ๐)๐2(๐ฃ๐+1,โฆ, ๐ฃ๐+โ) .This tensor is called the tensor product of ๐1 and ๐2.
We note that if ๐1 is a 0-tensor, i.e., scalar, then tensor product with ๐1 is just scalarmultiplication by ๐1, and similarly if ๐2 is a 0-tensor. That is ๐ โ ๐ = ๐ โ ๐ = ๐๐ for ๐ โ ๐and ๐ โ L๐(๐).
Properties 1.3.9. Suppose that we are given a ๐1-tensor ๐1, a ๐2-tensor ๐2, and a ๐3-tensor๐3 on a vector space ๐.(1) Associativity: One can define a (๐1 + ๐2 + ๐3)-tensor ๐1 โ ๐2 โ ๐3 by setting
(๐1 โ ๐2 โ ๐3)(๐ฃ1,โฆ, ๐ฃ๐+โ+๐) โ ๐1(๐ฃ1,โฆ, ๐ฃ๐)๐2(๐ฃ๐+1,โฆ, ๐ฃ๐+โ)๐3(๐ฃ๐+โ+1,โฆ, ๐ฃ๐+โ+๐) .Alternatively, one can define ๐1 โ ๐2 โ ๐3 by defining it to be the tensor product of(๐1 โ ๐2) โ ๐3 or the tensor product of ๐1 โ (๐2 โ ๐3). It is easy to see that both thesetensor products are identical with ๐1 โ ๐2 โ ๐3:
(๐1 โ ๐2) โ ๐3 = ๐1 โ ๐2 โ ๐3 = ๐1 โ (๐2 โ ๐3) .(2) Distributivity of scalar multiplication: We leave it as an easy exercise to check that if ๐ is
a real number then
๐(๐1 โ ๐2) = (๐๐1) โ ๐2 = ๐1 โ (๐๐2)(3) Left and right distributive laws: If ๐1 = ๐2, then
(๐1 + ๐2) โ ๐3 = ๐1 โ ๐3 + ๐2 โ ๐3and if ๐2 = ๐3, then
๐1 โ (๐2 + ๐3) = ๐1 โ ๐2 + ๐1 โ ๐3 .
A particularly interesting tensor product is the following. For ๐ = 1,โฆ, ๐ let โ๐ โ ๐โand let
(1.3.10) ๐ = โ1 โโฏ โ โ๐ .Thus, by definition,
(1.3.11) ๐(๐ฃ1,โฆ, ๐ฃ๐) = โ1(๐ฃ1)โฏโ๐(๐ฃ๐) .
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10 Chapter 1: Multilinear Algebra
A tensor of the form (1.3.11) is called a decomposable ๐-tensor. These tensors, as we willsee, play an important role in what follows. In particular, let ๐1,โฆ, ๐๐ be a basis of ๐ and๐โ1 ,โฆ, ๐โ๐ the dual basis of ๐โ. For every multi-index ๐ผ of length ๐ let
๐โ๐ผ = ๐โ๐1 โโฏ โ ๐โ๐๐ .
Then if ๐ฝ is another multi-index of length ๐,
(1.3.12) ๐โ๐ผ (๐๐1 ,โฆ, ๐๐๐) = {1, ๐ผ = ๐ฝ0, ๐ผ โ ๐ฝ
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 ๐ be a vector space with basis ๐1,โฆ, ๐๐ and 0 โค ๐ โค ๐ be an integer.The ๐-tensors ๐โ๐ผ of (1.3.12) are a basis of L๐(๐).
Proof. Given ๐ โ L๐(๐), let๐โฒ = โ
๐ผ๐๐ผ๐โ๐ผ
where the ๐๐ผโs are defined by (1.3.6). Then
(1.3.14) ๐โฒ(๐๐1 ,โฆ, ๐๐๐) = โ๐ผ๐๐ผ๐โ๐ผ (๐๐1 ,โฆ, ๐๐๐) = ๐๐ฝ
by (1.3.12); however, by Proposition 1.3.7 the ๐๐ฝโs determine ๐, so ๐โฒ = ๐. This proves thatthe ๐โ๐ผ โs are a spanning set of vectors for L๐(๐). To prove theyโre a basis, suppose
โ๐ผ๐ถ๐ผ๐โ๐ผ = 0
for constants, ๐ถ๐ผ โ ๐. Then by (1.3.14) with ๐โฒ = 0, ๐ถ๐ฝ = 0, so the ๐โ๐ผ โs are linearly indepen-dent. โก
As we noted in Lemma 1.3.5, there are exactly ๐๐ multi-indices of length ๐ and hence๐๐ basis vectors in the set {๐โ๐ผ }๐ผ, so we have proved
Corollary 1.3.15. Let ๐ be an ๐-dimensional vector space. Then dim(L๐(๐)) = ๐๐.
The pullback operationDefinition 1.3.16. Let ๐ and๐ be finite dimensional vector spaces and let ๐ดโถ ๐ โ ๐ bea linear mapping. If ๐ โ L๐(๐), we define
๐ดโ๐โถ ๐๐ โ ๐
to be the function
(1.3.17) (๐ดโ๐)(๐ฃ1,โฆ, ๐ฃ๐) โ ๐(๐ด๐ฃ1,โฆ,๐ด๐ฃ๐) .
It is clear from the linearity of ๐ด that this function is linear in its ๐th variable for all ๐, andhence is a ๐-tensor. We call ๐ดโ๐ the pullback of ๐ by the map ๐ด.
Proposition 1.3.18. The map
๐ดโ โถ L๐(๐) โ L๐(๐) , ๐ โฆ ๐ดโ๐ ,
is a linear mapping.
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ยง1.4 Alternating ๐-tensors 11
We leave this as an exercise. We also leave as an exercise the identity
(1.3.19) ๐ดโ(๐1 โ ๐2) = ๐ดโ(๐1) โ ๐ดโ(๐2)
for ๐1 โ L๐(๐) and ๐2 โ L๐(๐). Also, if ๐ is a vector space and ๐ตโถ ๐ โ ๐ a linearmapping, we leave for you to check that
(1.3.20) (๐ด๐ต)โ๐ = ๐ตโ(๐ดโ๐)
for all ๐ โ L๐(๐).
Exercises for ยง1.3Exercise 1.3.i. Verify that there are exactly ๐๐ multi-indices of length ๐.
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 ๐ดโถ ๐ โ ๐ be a linear map. Show that if โ๐, ๐ = 1,โฆ, ๐ are elements of๐โ
๐ดโ(โ1 โโฏ โ โ๐) = ๐ดโ(โ1) โ โฏ โ ๐ดโ(โ๐) .Conclude that ๐ดโ maps decomposable ๐-tensors to decomposable ๐-tensors.
Exercise 1.3.vi. Let ๐ be an ๐-dimensional vector space and โ๐, ๐ = 1, 2, elements of ๐โ.Show that โ1 โ โ2 = โ2 โ โ1 if and only if โ1 and โ2 are linearly dependent.
Hint: Show that if โ1 and โ2 are linearly independent there exist vectors, ๐ฃ๐, ๐ =, 1, 2 in๐ with property
โ๐(๐ฃ๐) = {1, ๐ = ๐0, ๐ โ ๐ .
Now compare (โ1 โ โ2)(๐ฃ1, ๐ฃ2) and (โ2 โ โ1)(๐ฃ1, ๐ฃ2). Conclude that if dim๐ โฅ 2 the tensorproduct operation is not commutative, i.e., it is usually not true that โ1 โ โ2 = โ2 โ โ1.
Exercise 1.3.vii. Let ๐ be a ๐-tensor and ๐ฃ a vector. Define ๐๐ฃ โถ ๐๐โ1 โ ๐ to be the map
(1.3.21) ๐๐ฃ(๐ฃ1,โฆ, ๐ฃ๐โ1) โ ๐(๐ฃ, ๐ฃ1,โฆ, ๐ฃ๐โ1) .
Show that ๐๐ฃ is a (๐ โ 1)-tensor.
Exercise 1.3.viii. Show that if ๐1 is an ๐-tensor and ๐2 is an ๐ -tensor, then if ๐ > 0,
(๐1 โ ๐2)๐ฃ = (๐1)๐ฃ โ ๐2 .
Exercise 1.3.ix. Let ๐ดโถ ๐ โ ๐ be a linear map mapping, and ๐ฃ โ ๐. Write ๐ค โ ๐ด๐ฃ. Showthat for ๐ โ L๐(๐), ๐ดโ(๐๐ค) = (๐ดโ๐)๐ฃ.
1.4. Alternating ๐-tensorsWewill discuss in this section a class of ๐-tensors which play an important role inmulti-
variable calculus. In this discussionwewill need some standard facts about the โpermutationgroupโ. 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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12 Chapter 1: Multilinear Algebra
PermutationsDefinition 1.4.1. Let ๐ด๐ be the ๐-element set ๐ด๐ โ {1, 2,โฆ, ๐}. A permutation of order๐ is a bijecton ๐โถ ๐ด๐ โฅฒ ๐ด๐. Given two permutations ๐1 and ๐2, their product ๐1๐2 is thecomposition of ๐1 โ ๐2, i.e., the map,
๐ โฆ ๐1(๐2(๐)) .For every permutation ๐, one denotes by ๐โ1 the inverse permutation given by the inversebijection of ๐, i.e., defined by
๐(๐) โ ๐ โบ ๐โ1(๐) = ๐ .Let ๐๐ be the set of all permutations of order ๐. One calls ๐๐ the permutation group of ๐ด๐ or,alternatively, the symmetric group on ๐ letters.
It is easy to check the following.
Lemma 1.4.2. The group ๐๐ has ๐! elements.
Definition 1.4.3. Let ๐ be a positive integer. For every 1 โค ๐ < ๐ โค ๐, let ๐๐,๐ be the permu-tation
๐๐,๐(โ) โ{{{{{
๐, โ = ๐๐, โ = ๐โ, โ โ ๐, ๐ .
The permuation ๐๐,๐ is called a transposition, and if ๐ = ๐+1, then ๐๐,๐ is called an elementarytransposition.
Theorem 1.4.4. Every permutation in ๐๐ can be written as a product of (a finite number of)transpositions.
Proof. We prove this by induction on ๐. The base case when ๐ = 2 is obvious.For the induction step, suppose that we know the claim for ๐๐โ1. Given ๐ โ ๐๐, we
have ๐(๐) = ๐ if and only if ๐๐,๐๐(๐) = ๐. Thus ๐๐,๐๐ is, in effect, a permutation of ๐ด๐โ1. Byinduction, ๐๐,๐๐ can be written as a product of transpositions, so
๐ = ๐๐,๐(๐๐,๐๐)can be written as a product of transpositions. โก
Theorem 1.4.5. Every transposition can be written as a product of elementary transpositions.
Proof. Let ๐ = ๐๐๐, ๐ < ๐. With ๐ fixed, argue by induction on ๐. Note that for ๐ > ๐ + 1๐๐๐ = ๐๐โ1,๐๐๐,๐โ1๐๐โ1,๐ .
Now apply the inductive hypothesis to ๐๐,๐โ1. โก
Corollary 1.4.6. Every permutation can be written as a product of elementary transpositions.
The sign of a permutationDefinition 1.4.7. Let ๐ฅ1,โฆ, ๐ฅ๐ be the coordinate functions on ๐๐. For ๐ โ ๐๐ we define
(1.4.8) (โ1)๐ โโ๐<๐
๐ฅ๐(๐) โ ๐ฅ๐(๐)๐ฅ๐ โ ๐ฅ๐
.
Notice that the numerator and denominator in (1.4.8) are identical up to sign. Indeed,if ๐ = ๐(๐) < ๐(๐) = ๐, the term, ๐ฅ๐ โ ๐ฅ๐ occurs once and just once in the numerator andonce and just once in the denominator; and if ๐ = ๐(๐) > ๐(๐) = ๐, the term, ๐ฅ๐ โ๐ฅ๐, occurs
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ยง1.4 Alternating ๐-tensors 13
once and just once in the numerator and its negative, ๐ฅ๐ โ ๐ฅ๐, once and just once in thenumerator. Thus
(โ1)๐ = ยฑ1 .In light of this, we call (โ1)๐ the sign of ๐.
Claim 1.4.9. For ๐, ๐ โ ๐๐ we have
(โ1)๐๐ = (โ1)๐(โ1)๐ .
That is, the sign defines a group homomorphism ๐๐ โ {ยฑ1}.
Proof. By definition,
(โ1)๐๐ = โ๐<๐
๐ฅ๐๐(๐) โ ๐ฅ๐๐(๐)๐ฅ๐ โ ๐ฅ๐
.
We write the right hand side as a product of
โ๐<๐
๐ฅ๐(๐) โ ๐ฅ๐(๐)๐ฅ๐ โ ๐ฅ๐
= (โ1)๐
and
(1.4.10) โ๐<๐
๐ฅ๐๐(๐) โ ๐ฅ๐๐(๐)๐ฅ๐(๐) โ ๐ฅ๐(๐)
For ๐ < ๐, let ๐ = ๐(๐) and ๐ = ๐(๐) when ๐(๐) < ๐(๐) and let ๐ = ๐(๐) and ๐ = ๐(๐) when๐(๐) < ๐(๐). Then
๐ฅ๐๐(๐) โ ๐ฅ๐๐(๐)๐ฅ๐(๐) โ ๐ฅ๐(๐)
=๐ฅ๐(๐) โ ๐ฅ๐(๐)๐ฅ๐ โ ๐ฅ๐
(i.e., if ๐(๐) < ๐(๐), the numerator and denominator on the right equal the numerator anddenominator on the left and, if ๐(๐) < ๐(๐) are negatives of the numerator and denominatoron the left). Thus equation (1.4.10) becomes
โ๐<๐
๐ฅ๐(๐) โ ๐ฅ๐(๐)๐ฅ๐ โ ๐ฅ๐
= (โ1)๐ . โก
Weโll leave for you to check that if ๐ is a transposition, (โ1)๐ = โ1 and to conclude fromthis:
Proposition 1.4.11. If ๐ is the product of an odd number of transpositions, (โ1)๐ = โ1 andif ๐ is the product of an even number of transpositions (โ1)๐ = +1.
AlternationDefinition 1.4.12. Let ๐ be an ๐-dimensional vector space and ๐ โ L๐(๐ฃ) a ๐-tensor. For๐ โ ๐๐, define ๐๐ โ L๐(๐) to be the ๐-tensor
(1.4.13) ๐๐(๐ฃ1,โฆ, ๐ฃ๐) โ ๐(๐ฃ๐โ1(1),โฆ, ๐ฃ๐โ1(๐)) .
Proposition 1.4.14.(1) If ๐ = โ1 โโฏ โ โ๐, โ๐ โ ๐โ, then ๐๐ = โ๐(1) โโฏ โ โ๐(๐).(2) The assignment ๐ โฆ ๐๐ is a linear map L๐(๐) โ L๐(๐).(3) If ๐, ๐ โ ๐๐, we have ๐๐๐ = (๐๐)๐.
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14 Chapter 1: Multilinear Algebra
Proof. To (1), we note that by equation (1.4.13)
(โ1 โโฏ โ โ๐)๐(๐ฃ1,โฆ, ๐ฃ๐) = โ1(๐ฃ๐โ1(1))โฏโ๐(๐ฃ๐โ1(๐)) .
Setting ๐โ1(๐) = ๐, the ๐th term in this product is โ๐(๐)(๐ฃ๐); so the product can be rewrittenas
โ๐(1)(๐ฃ1)โฆโ๐(๐)(๐ฃ๐)or
(โ๐(1) โโฏ โ โ๐(๐))(๐ฃ1,โฆ, ๐ฃ๐) .We leave the proof of (2) as an exercise.
Now we prove (3). Let ๐ = โ1 โโฏ โ โ๐. Then
๐๐ = โ๐(1) โโฏ โ โ๐(๐) = โโฒ1 โโฏ โ โโฒ๐where โโฒ๐ = โ๐(๐). Thus
(๐๐)๐ = โโฒ๐(1) โโฏ โ โโฒ๐(๐) .But if ๐(๐) = ๐, โโฒ๐(๐) = โ๐(๐(๐)). Hence
(๐๐)๐โ๐๐(1) โโฏ โ โ๐๐(๐) = ๐๐๐ . โก
Definition 1.4.15. Let ๐ be a vector space and ๐ โฅ 0 an integer. A ๐-tensor ๐ โ L๐(๐) isalternating if ๐๐ = (โ1)๐๐ for all ๐ โ ๐๐.
We denote by A๐(๐) the set of all alternating ๐-tensors in L๐(๐). By (2) this set is avector subspace of L๐(๐).
It is not easy to write down simple examples of alternating ๐-tensors. However, there isa method, called the alternation operation Alt for constructing such tensors.
Definition 1.4.16. Let ๐ be a vector space and ๐ a nonnegative integer. The alternationoperation on L๐(๐) is defined as follows: given ๐ โ L๐(๐) let
Alt(๐) โ โ๐โ๐๐(โ1)๐๐๐ .
The alternation operation enjoys the following properties.
Proposition 1.4.17. For ๐ โ L๐(๐) and ๐ โ ๐๐,(1) Alt(๐)๐ = (โ1)๐ Alt๐(2) If ๐ โ A๐(๐), then Alt๐ = ๐!๐.(3) Alt(๐๐) = Alt(๐)๐(4) The map
Alt โถ L๐(๐) โ L๐(๐) , ๐ โฆ Alt(๐)is linear.
Proof. To prove (1) we note that by Proposition 1.4.14:
Alt(๐)๐ = โ๐โ๐๐(โ1)๐๐๐๐ = (โ1)๐ โ
๐โ๐๐(โ1)๐๐๐๐๐ .
But as ๐ runs over ๐๐, ๐๐ runs over ๐๐, and hence the right hand side is (โ1)๐ Alt(๐).To prove (2), note that if ๐ โ A๐(๐)
Alt๐ = โ๐โ๐๐(โ1)๐๐๐ = โ
๐โ๐๐(โ1)๐(โ1)๐๐ = ๐!๐ .
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ยง1.4 Alternating ๐-tensors 15
To prove (3), we compute:
Alt(๐๐) = โ๐โ๐๐(โ1)๐๐๐๐ = (โ1)๐ โ
๐โ๐๐(โ1)๐๐๐๐๐
= (โ1)๐ Alt(๐) = Alt(๐)๐ .Finally, (4) is an easy corollary of (2) of Proposition 1.4.14. โก
Wewill use this alternation operation to construct a basis for A๐(๐). First, however, werequire some notation.
Definition 1.4.18. Let ๐ผ = (๐1,โฆ, ๐๐) be a multi-index of length ๐.(1) ๐ผ is repeating if ๐๐ = ๐๐ for some ๐ โ ๐ .(2) ๐ผ is strictly increasing if ๐1 < ๐2 < โฏ < ๐๐.(3) For ๐ โ ๐๐, write ๐ผ๐ โ (๐๐(1),โฆ, ๐๐(๐)).
Remark 1.4.19. If ๐ผ is non-repeating there is a unique ๐ โ ๐๐ so that ๐ผ๐ is strictly increasing.
Let ๐1,โฆ, ๐๐ be a basis of ๐ and let
๐โ๐ผ = ๐โ๐1 โโฏ โ ๐โ๐๐
and๐๐ผ = Alt(๐โ๐ผ ) .
Proposition 1.4.20.(1) ๐๐ผ๐ = (โ1)๐๐๐ผ.(2) If ๐ผ is repeating, ๐๐ผ = 0.(3) If ๐ผ and ๐ฝ are strictly increasing,
๐๐ผ(๐๐1 ,โฆ, ๐๐๐) = {1, ๐ผ = ๐ฝ0, ๐ผ โ ๐ฝ .
Proof. To prove (1) we note that (๐โ๐ผ )๐ = ๐โ๐ผ๐ ; soAlt(๐โ๐ผ๐) = Alt(๐โ๐ผ )๐ = (โ1)๐ Alt(๐โ๐ผ ) .
To prove (2), suppose ๐ผ = (๐1,โฆ, ๐๐) with ๐๐ = ๐๐ for ๐ โ ๐ . Then if ๐ = ๐๐๐,๐๐ , ๐โ๐ผ = ๐โ๐ผ๐ so
๐๐ผ = ๐๐ผ๐ = (โ1)๐๐๐ผ = โ๐๐ผ .To prove (3), note that by definition
๐๐ผ(๐๐1 ,โฆ, ๐๐๐) = โ๐(โ1)๐๐โ๐ผ๐(๐๐1 ,โฆ, ๐๐๐) .
But by (1.3.12)
(1.4.21) ๐โ๐ผ๐(๐๐1 ,โฆ, ๐๐๐) = {1, ๐ผ๐ = ๐ฝ0, ๐ผ๐ โ ๐ฝ .
Thus if ๐ผ and ๐ฝ are strictly increasing, ๐ผ๐ is strictly increasing if and only if ๐ผ๐ = ๐ผ, and(1.4.21) is nonzero if and only if ๐ผ = ๐ฝ. โก
Now let ๐ be in A๐(๐). By Theorem 1.3.13,
๐ = โ๐ฝ๐๐ฝ๐โ๐ฝ , ๐๐ฝ โ ๐ .
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16 Chapter 1: Multilinear Algebra
Since ๐!๐ = Alt(๐),๐ = 1๐!โ๐๐ฝ Alt(๐โ๐ฝ ) = โ๐๐ฝ๐๐ฝ .
We candiscard all repeating terms in this sum since they are zero; and for every non-repeatingterm, ๐ฝ, we can write ๐ฝ = ๐ผ๐, where ๐ผ is strictly increasing, and hence ๐๐ฝ = (โ1)๐๐๐ผ.Conclusion 1.4.22. We can write ๐ as a sum(1.4.23) ๐ = โ
๐ผ๐๐ผ๐๐ผ .
with ๐ผโs strictly increasing.
Claim 1.4.24. The ๐๐ผโs are unique.
Proof. For ๐ฝ strictly increasing
(1.4.25) ๐(๐๐1 ,โฆ, ๐๐๐) = โ๐ผ๐๐ผ๐๐ผ(๐๐1 ,โฆ, ๐๐๐) = ๐๐ฝ .
By (1.4.23) the ๐๐ผโs, ๐ผ strictly increasing, are a spanning set of vectors for A๐(๐), and by(1.4.25) they are linearly independent, so we have proved:
Proposition 1.4.26. The alternating tensors, ๐๐ผ, ๐ผ strictly increasing, are a basis for A๐(๐).
Thus dimA๐(๐) is equal to the number of strictly increasingmulti-indices ๐ผ of length ๐.We leave for you as an exercise to show that this number is equal to the binomal coefficient
(1.4.27) (๐๐) โ ๐!(๐ โ ๐)!๐!
if 1 โค ๐ โค ๐. โกHint: Show that every strictly increasingmulti-index of length ๐ determines a ๐ element
subset of {1,โฆ, ๐} and vice-versa.Note also that if ๐ > ๐ every multi-index
๐ผ = (๐1,โฆ, ๐๐)of length ๐ has to be repeating: ๐๐ = ๐๐ for some ๐ โ ๐ since the ๐๐โs lie on the interval 1 โค ๐ โค ๐.Thus by Proposition 1.4.17
๐๐ผ = 0for all multi-indices of length ๐ > 0 and
A๐(๐) = 0 .Exercises for ยง1.4
Exercise 1.4.i. Show that there are exactly ๐! permutations of order ๐.Hint: Induction on ๐: Let ๐ โ ๐๐, and let ๐(๐) = ๐, 1 โค ๐ โค ๐. Show that ๐๐,๐๐ leaves ๐
fixed and hence is, in effect, a permutation of ๐ด๐โ1.Exercise 1.4.ii. Prove that if ๐ โ ๐๐ is a transposition, (โ1)๐ = โ1 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 dimA๐(๐) is given by (1.4.27).
Exercise 1.4.v. Verify that for ๐ < ๐ โ 1๐๐,๐ = ๐๐โ1,๐๐๐,๐โ1, ๐๐โ1,๐ .
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ยง1.5 The space ๐ฌ๐(๐โ) 17
Exercise 1.4.vi. For ๐ = 3 show that every one of the six elements of ๐3 is either a transpo-sition or can be written as a product of two transpositions.
Exercise 1.4.vii. Let ๐ โ ๐๐ be the โcyclicโ permutation๐(๐) โ ๐ + 1 , ๐ = 1,โฆ, ๐ โ 1
and ๐(๐) โ 1. Show explicitly how to write ๐ as a product of transpositions and compute(โ1)๐.
Hint: Same hint as in Exercise 1.4.i.
Exercise 1.4.viii. In Exercise 1.3.vii show that if ๐ is in A๐(๐), ๐๐ฃ is in A๐โ1(๐). Show inaddition that for ๐ฃ, ๐ค โ ๐ and ๐ โ A๐(๐) we have (๐๐ฃ)๐ค = โ(๐๐ค)๐ฃ.
Exercise 1.4.ix. Let ๐ดโถ ๐ โ ๐ be a linear mapping. Show that if ๐ is in A๐(๐), ๐ดโ๐ is inA๐(๐).Exercise 1.4.x. In Exercise 1.4.ix show that if ๐ is in L๐(๐) then Alt(๐ดโ๐) = ๐ดโ(Alt(๐)),i.e., show that the Alt operation commutes with the pullback operation.
1.5. The space ๐ฌ๐(๐โ)
In ยง1.4 we showed that the image of the alternation operation, Alt โถ L๐(๐) โ L๐(๐) isA๐(๐). In this section we will compute the kernel of Alt.
Definition 1.5.1. A decomposable ๐-tensor โ1 โโฏโ โ๐, with โ1,โฆ, โ๐ โ ๐โ, is redundantif for some index ๐ we have โ๐ = โ๐+1.
Let I๐(๐) โ L๐(๐) be the linear span of the set of redundant ๐-tensors.Note that for ๐ = 1 the notion of redundant does not really make sense; a single vector
โ โ L1(๐โ) cannot be โredundantโ so we decreeI1(๐) โ 0 .
Proposition 1.5.2. If ๐ โ I๐(๐) then Alt(๐) = 0.Proof. Let ๐ = โ1 โโฏ โ โ๐ with โ๐ = โ๐+1. Then if ๐ = ๐๐,๐+1, ๐๐ = ๐ and (โ1)๐ = โ1. HenceAlt(๐) = Alt(๐๐) = Alt(๐)๐ = โAlt(๐); so Alt(๐) = 0. โกProposition 1.5.3. If ๐ โ I๐(๐) and ๐โฒ โ L๐ (๐) then ๐ โ ๐โฒ and ๐โฒ โ ๐ are in I๐+๐ (๐).Proof. We can assume that ๐ and ๐โฒ are decomposable, i.e., ๐ = โ1 โ โฏ โ โ๐ and ๐โฒ =โโฒ1 โโฏ โ โโฒ๐ and that ๐ is redundant: โ๐ = โ๐+1. Then
๐ โ ๐โฒ = โ1 โโฏโ๐โ1 โ โ๐ โ โ๐ โโฏโ๐ โ โโฒ1 โโฏ โ โโฒ๐ is redundant and hence in I๐+๐ . The argument for ๐โฒ โ ๐ is similar. โก
Proposition 1.5.4. If ๐ โ L๐(๐) and ๐ โ ๐๐, then(1.5.5) ๐๐ = (โ1)๐๐ + ๐where ๐ is in I๐(๐).Proof. We can assume๐ is decomposable, i.e.,๐ = โ1โโฏโโ๐. Letโs first look at the simplestpossible case: ๐ = 2 and ๐ = ๐1,2. Then
๐๐ โ (โ1)๐๐ = โ1 โ โ2 + โ2 โ โ1
= 12((โ1 + โ2) โ (โ1 + โ2) โ โ1 โ โ1 โ โ2 โ โ2) ,
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18 Chapter 1: Multilinear Algebra
and the terms on the right are redundant, and hence in I2(๐). Next let ๐ be arbitrary and๐ = ๐๐,๐+1. If ๐1 = โ1 โโฏ โ โ๐โ2 and ๐2 = โ๐+2 โโฏ โ โ๐. Then
๐ โ (โ1)๐๐ = ๐1 โ (โ๐ โ โ๐+1 + โ๐+1 โ โ๐) โ ๐2is in I๐(๐) by Proposition 1.5.3 and the computation above.The general case: By Theorem 1.4.5, ๐ can be written as a product of๐ elementary transpo-sitions, and weโll prove (1.5.5) by induction on๐.
We have just dealt with the case๐ = 1.The induction step: theโ๐โ 1โ case implies the โ๐โ case. Let ๐ = ๐๐ฝ where ๐ฝ is a product of๐ โ 1 elementary transpositions and ๐ is an elementary transposition. Then
๐๐ = (๐๐ฝ)๐ = (โ1)๐๐๐ฝ +โฏ= (โ1)๐(โ1)๐ฝ๐ +โฏ= (โ1)๐๐ +โฏ
where the โdotsโ are elements of I๐(๐), and the induction hypothesis was used in the secondline. โก
Corollary 1.5.6. If ๐ โ L๐(๐), then(1.5.7) Alt(๐) = ๐!๐ +๐ ,
where๐ is in I๐(๐).
Proof. By definition Alt(๐) = โ๐โ๐๐(โ1)๐๐๐, and by Proposition 1.5.4, ๐๐ = (โ1)๐๐ +๐๐,
with๐๐ โ I๐(๐). Thus
Alt(๐) = โ๐โ๐๐(โ1)๐(โ1)๐๐ + โ
๐โ๐๐(โ1)๐๐๐ = ๐!๐ +๐
where๐ = โ๐โ๐๐(โ1)๐๐๐. โก
Corollary 1.5.8. Let ๐ be a vector space and ๐ โฅ 1. Then
I๐(๐) = ker(Alt โถ L๐(๐) โ A๐(๐)) .
Proof. We have already proved that if ๐ โ I๐(๐), then Alt(๐) = 0. To prove the converseassertion we note that if Alt(๐) = 0, then by (1.5.7)
๐ = โ 1๐!๐ .
with๐ โ I๐(๐) . โก
Putting these results together we conclude:
Theorem 1.5.9. Every element ๐ โ L๐(๐) can be written uniquely as a sum ๐ = ๐1 + ๐2,where ๐1 โ A๐(๐) and ๐2 โ I๐(๐).
Proof. By (1.5.7), ๐ = ๐1 + ๐2 with
๐1 =1๐!
Alt(๐)
and๐2 = โ1๐!๐ .
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ยง1.5 The space ๐ฌ๐(๐โ) 19
To prove that this decomposition is unique, suppose ๐1 + ๐2 = 0, with ๐1 โ A๐(๐) and๐2 โ I๐(๐). Then
0 = Alt(๐1 + ๐2) = ๐!๐1so ๐1 = 0, and hence ๐2 = 0. โก
Definition 1.5.10. Let ๐ be a finite dimensional vector space and ๐ โฅ 0. Define
(1.5.11) ๐ฌ๐(๐โ) โ L๐(๐)/I๐(๐) ,i.e., let ๐ฌ๐(๐โ) be the quotient of the vector space L๐(๐) by the subspace I๐(๐). By (1.2.7)one has a linear map:
(1.5.12) ๐โถ L๐(๐) โ ๐ฌ๐(๐โ) , ๐ โฆ ๐ + I๐(๐)which is onto and has I๐(๐) as kernel.
Theorem 1.5.13. The map ๐โถ L๐(๐) โ ๐ฌ๐(๐โ)maps A๐(๐) bijectively onto ๐ฌ๐(๐โ).
Proof. ByTheorem 1.5.9 every I๐(๐) coset๐+I๐(๐) contains a unique element๐1 ofA๐(๐).Hence for every element of ๐ฌ๐(๐โ) there is a unique element of A๐(๐) which gets mappedonto it by ๐. โก
Remark 1.5.14. Since๐ฌ๐(๐โ) and A๐(๐) are isomorphic as vector spaces many treatmentsof multilinear algebra avoid mentioning ๐ฌ๐(๐โ), reasoning that A๐(๐) 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 betweenA๐(๐) and๐ฌ๐(๐โ), as we shall see in ยง1.6.
Exercises for ยง1.5Exercise 1.5.i. A ๐-tensor ๐ โ L๐(๐) is symmetric if ๐๐ = ๐ for all ๐ โ ๐๐. Show that theset ๐ฎ ๐(๐) of symmetric ๐ tensors is a vector subspace of L๐(๐).
Exercise 1.5.ii. Let ๐1,โฆ, ๐๐ be a basis of ๐. Show that every symmetric 2-tensor is of theform
โ1โค๐,๐โค๐๐๐,๐๐โ๐ โ ๐โ๐
where ๐๐,๐ = ๐๐,๐ and ๐โ1 ,โฆ, ๐โ๐ are the dual basis vectors of ๐โ.
Exercise 1.5.iii. Show that if ๐ is a symmetric ๐-tensor, then for ๐ โฅ 2, then ๐ is in I๐(๐).Hint: Let ๐ be a transposition and deduce from the identity, ๐๐ = ๐, that ๐ has to be in
the kernel of Alt.
Exercise 1.5.iv (a warning). In general ๐ฎ ๐(๐) โ I๐(๐). Show, however, that if ๐ = 2 thesetwo spaces are equal.
Exercise 1.5.v. Show that if โ โ ๐โ and ๐ โ L๐โ2(๐), then โ โ ๐ โ โ is in I๐(๐).
Exercise 1.5.vi. Show that if โ1 and โ2 are in๐โ and๐ โ L๐โ2(๐), then โ1โ๐โโ2+โ2โ๐โโ1is in I๐(๐).
Exercise 1.5.vii. Given a permutation ๐ โ ๐๐ and ๐ โ I๐(๐), show that ๐๐ โ I๐(๐).
Exercise 1.5.viii. Let W(๐) be a subspace of L๐(๐) having the following two properties.(1) For ๐ โ ๐ฎ 2(๐) and ๐ โ L๐โ2(๐), ๐ โ ๐ is in W(๐).
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20 Chapter 1: Multilinear Algebra
(2) For ๐ in W(๐) and ๐ โ ๐๐, ๐๐ is in W(๐).Show that W(๐) has to contain I๐(๐) and conclude that I๐(๐) is the smallest subspace
of L๐(๐) having properties (1) and (2).
Exercise 1.5.ix. Show that there is a bijective linear map
๐ผโถ ๐ฌ๐(๐โ) โฅฒ A๐(๐)with the property
(1.5.15) ๐ผ๐(๐) = 1๐!
Alt(๐)
for all๐ โ L๐(๐), and show that ๐ผ is the inverse of themap ofA๐(๐) onto๐ฌ๐(๐โ) describedin Theorem 1.5.13.
Hint: Exercise 1.2.viii.
Exercise 1.5.x. Let ๐ be an ๐-dimensional vector space. Compute the dimension of ๐ฎ ๐(๐).Hints:
(1) Introduce the following symmetrization operation on tensors ๐ โ L๐(๐):Sym(๐) = โ
๐โ๐๐๐๐ .
Prove that this operation has properties (2)โ(4) of Proposition 1.4.17 and, as a substitutefor (1), has the property: Sym(๐)๐ = Sym(๐).
(2) Let ๐๐ผ = Sym(๐โ๐ผ ), ๐โ๐ผ = ๐โ๐1 โโฏโ๐โ๐๐ . Prove that { ๐๐ผ | ๐ผ is non-decreasing } form a basis
of ๐๐(๐).(3) Conclude that dim(๐ฎ ๐(๐)) is equal to the number of non-decreasing multi-indices of
length ๐: 1 โค ๐1 โค ๐2 โค โฏ โค โ๐ โค ๐.(4) Compute this number by noticing that the assignment
(๐1,โฆ, ๐๐) โฆ (๐1 + 0, ๐2 + 1,โฆ, ๐๐ + ๐ โ 1)is a bijection between the set of these non-decreasing multi-indicesand the set of in-creasing multi-indices 1 โค ๐1 < โฏ < ๐๐ โค ๐ + ๐ โ 1.
1.6. The wedge product
The tensor algebra operations on the spaces L๐(๐)which we discussed in ยงยง1.2 and 1.3,i.e., the โtensor product operationโ and the โpullbackโ operation, give rise to similar oper-ations on the spaces, ๐ฌ๐(๐โ). We will discuss in this section the analogue of the tensorproduct operation.
Definition 1.6.1. Given ๐๐ โ ๐ฌ๐๐(๐โ), ๐ = 1, 2 we can, by equation (1.5.12), find a ๐๐ โL๐๐(๐) with ๐๐ = ๐(๐๐). Then ๐1 โ ๐2 โ L๐1+๐2(๐). The wedge product ๐1 โง ๐2 is defined by(1.6.2) ๐1 โง ๐2 โ ๐(๐1 โ ๐2) โ ๐ฌ๐1+๐2(๐โ) .Claim 1.6.3. This wedge product is well defined, i.e., does not depend on our choices of ๐1 and๐2.
Proof. Let ๐(๐1) = ๐(๐โฒ1 ) = ๐1. Then ๐โฒ1 = ๐1 +๐1 for some๐1 โ I๐1(๐), so๐โฒ1 โ ๐2 = ๐1 โ ๐2 +๐1 โ ๐2 .
But๐1 โ I๐1(๐) implies๐1 โ ๐2 โ I๐1+๐2(๐) and this implies:๐(๐โฒ1 โ ๐2) = ๐(๐1 โ ๐2) .
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ยง1.6 The wedge product 21
A symmetric argument shows that ๐1 โง ๐2 is well-defined, independent of the choice of๐2. โก
More generally let ๐๐ โ ๐ฌ๐๐(๐โ), for ๐ = 1, 2, 3, and let ๐๐ = ๐(๐๐), ๐๐ โ L๐๐(๐). Define
๐1 โง ๐2 โง ๐3 โ ๐ฌ๐1+๐2+๐3(๐โ)by setting
๐1 โง ๐2 โง ๐3 = ๐(๐1 โ ๐2 โ ๐3) .As above it is easy to see that this is well-defined independent of the choice of ๐1, ๐2 and ๐3.It is also easy to see that this triple wedge product is just the wedge product of ๐1 โง ๐2 with๐3 or, alternatively, the wedge product of ๐1 with ๐2 โง ๐3, i.e.,
๐1 โง ๐2 โง ๐3 = (๐1 โง ๐2) โง ๐3 = ๐1 โง (๐2 โง ๐3).We leave for you to check: for ๐ โ ๐
(1.6.4) ๐(๐1 โง ๐2) = (๐๐1) โง ๐2 = ๐1 โง (๐๐2)and verify the two distributive laws:
(1.6.5) (๐1 + ๐2) โง ๐3 = ๐1 โง ๐3 + ๐2 โง ๐3and
(1.6.6) ๐1 โง (๐2 + ๐3) = ๐1 โง ๐2 + ๐1 โง ๐3 .As we noted in ยง1.4, I๐(๐) = 0 for ๐ = 1, i.e., there are no nonzero โredundantโ ๐ tensorsin degree ๐ = 1. Thus
๐ฌ1(๐โ) = ๐โ = L1(๐).A particularly interesting example of a wedge product is the following. Let โ1,โฆ, โ๐ โ
๐โ = ๐ฌ1(๐โ). Then if ๐ = โ1 โโฏ โ โ๐(1.6.7) โ1 โงโฏ โง โ๐ = ๐(๐) โ ๐ฌ๐(๐โ) .We will call (1.6.7) a decomposable element of ๐ฌ๐(๐โ).
We will prove that these elements satisfy the following wedge product identity. For ๐ โ๐๐:(1.6.8) โ๐(1) โงโฏ โง โ๐(๐) = (โ1)๐โ1 โงโฏ โง โ๐ .
Proof. For every ๐ โ L๐(๐), ๐ = (โ1)๐๐ + ๐ for some๐ โ I๐(๐) by Proposition 1.5.4.Therefore since ๐(๐) = 0
๐(๐๐) = (โ1)๐๐(๐) .In particular, if ๐ = โ1 โโฏ โ โ๐, ๐๐ = โ๐(1) โโฏ โ โ๐(๐), so
๐(๐๐) = โ๐(1) โงโฏ โง โ๐(๐) = (โ1)๐๐(๐)= (โ1)๐โ1 โงโฏ โง โ๐ . โก
In particular, for โ1 and โ2 โ ๐โ
(1.6.9) โ1 โง โ2 = โโ2 โง โ1and for โ1, โ2 and โ3 โ ๐โ
โ1 โง โ2 โง โ3 = โโ2 โง โ1 โง โ3 = โ2 โง โ3 โง โ1 .More generally, it is easy to deduce from equation (1.6.8) the following result (which weโllleave as an exercise).
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22 Chapter 1: Multilinear Algebra
Theorem 1.6.10. If ๐1 โ ๐ฌ๐(๐โ) and ๐2 โ ๐ฌ๐ (๐โ) then
(1.6.11) ๐1 โง ๐2 = (โ1)๐๐ ๐2 โง ๐1 .
Hint: It suffices to prove this for decomposable elements i.e., for ๐1 = โ1 โงโฏ โง โ๐ and๐2 = โโฒ1 โงโฏ โง โโฒ๐ . Now make ๐๐ applications of (1.6.9).
Let ๐1,โฆ, ๐๐ be a basis of ๐ and let ๐โ1 ,โฆ, ๐โ๐ be the dual basis of ๐โ. For every multi-index ๐ผ of length ๐,
(1.6.12) ๐โ๐1 โงโฏ โง ๐โ๐๐ = ๐(๐
โ๐ผ ) = ๐(๐โ๐1 โโฏ โ ๐
โ๐๐) .
Theorem 1.6.13. The elements (1.6.12), with ๐ผ strictly increasing, are basis vectors of๐ฌ๐(๐โ).
Proof. The elements ๐๐ผ = Alt(๐โ๐ผ ), for ๐ผ strictly increasing, are basis vectors of A๐(๐) byProposition 1.4.26; so their images, ๐(๐๐ผ), are a basis of ๐ฌ๐(๐โ). But
๐(๐๐ผ) = ๐ (โ๐โ๐๐(โ1)๐(๐โ๐ผ )๐) = โ
๐โ๐๐(โ1)๐๐(๐โ๐ผ )๐
= โ๐โ๐๐(โ1)๐(โ1)๐๐(๐โ๐ผ ) = ๐!๐(๐โ๐ผ ) . โก
Exercises for ยง1.6Exercise 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 ๐ โ ๐ฌ๐(๐โ) let ๐๐ be the ๐-fold wedge product of ๐ with itself, i.e.,let ๐2 = ๐ โง ๐, ๐3 = ๐ โง ๐ โง ๐, etc.(1) Show that if ๐ is odd then for ๐ > 1, ๐๐ = 0.(2) Show that if ๐ is decomposable, then for ๐ > 1, ๐๐ = 0.
Exercise 1.6.iv. If ๐ and ๐ are in ๐ฌ๐(๐โ) prove:
(๐ + ๐)๐ =๐โโ=0(๐โ)๐โ โง ๐๐โโ .
Hint:As in freshman calculus, prove this binomial theoremby induction using the iden-tity: (๐โ) = (
๐โ1โโ1) + (
๐โ1โ ).
Exercise 1.6.v. Let ๐ be an element of๐ฌ2(๐โ). By definition the rank of ๐ is ๐ if ๐๐ โ 0 and๐๐+1 = 0. Show that if
๐ = ๐1 โง ๐1 +โฏ + ๐๐ โง ๐๐with ๐๐, ๐๐ โ ๐โ, then ๐ is of rank โค ๐.
Hint: Show that๐๐ = ๐!๐1 โง ๐1 โงโฏ โง ๐๐ โง ๐๐ .
Exercise 1.6.vi. Given ๐๐ โ ๐โ, ๐ = 1,โฆ, ๐ show that ๐1 โง โฏ โง ๐๐ โ 0 if and only if the ๐๐โsare linearly independent.
Hint: Induction on ๐.
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ยง1.7 The interior product 23
1.7. The interior product
Weโll describe in this section another basic product operation on the spaces๐ฌ๐(๐โ). Asabove weโll begin by defining this operator on the L๐(๐)โs.
Definition 1.7.1. Let๐ be a vector space and ๐ a nonnegative integer. Given๐ โ L๐(๐) and๐ฃ โ ๐ let ๐๐ฃ๐ be the be the (๐ โ 1)-tensor which takes the value
(๐๐ฃ๐)(๐ฃ1,โฆ, ๐ฃ๐โ1) โ๐โ๐=1(โ1)๐โ1๐(๐ฃ1,โฆ, ๐ฃ๐โ1, ๐ฃ, ๐ฃ๐,โฆ, ๐ฃ๐โ1)
on the ๐ โ 1-tuple of vectors, ๐ฃ1,โฆ, ๐ฃ๐โ1, i.e., in the ๐th summand on the right, ๐ฃ gets in-serted between ๐ฃ๐โ1 and ๐ฃ๐. (In particular the first summand is ๐(๐ฃ, ๐ฃ1,โฆ, ๐ฃ๐โ1) and the lastsummand is (โ1)๐โ1๐(๐ฃ1,โฆ, ๐ฃ๐โ1, ๐ฃ).)
It is clear from the definition that if ๐ฃ = ๐ฃ1 + ๐ฃ2(1.7.2) ๐๐ฃ๐ = ๐๐ฃ1๐ + ๐๐ฃ2๐ ,and if ๐ = ๐1 + ๐2(1.7.3) ๐๐ฃ๐ = ๐๐ฃ๐1 + ๐๐ฃ๐2 ,and we will leave for you to verify by inspection the following two lemmas.
Lemma 1.7.4. If ๐ is the decomposable ๐-tensor โ1 โโฏ โ โ๐ then
(1.7.5) ๐๐ฃ๐ =๐โ๐=1(โ1)๐โ1โ๐(๐ฃ)โ1 โโฏ โ โ๐ โโฏ โ โ๐ ,
where the โhatโ over โ๐ means that โ๐ is deleted from the tensor product.
Lemma 1.7.6. If ๐1 โ L๐(๐) and ๐2 โ L๐(๐)(1.7.7) ๐๐ฃ(๐1 โ ๐2) = ๐๐ฃ๐1 โ ๐2 + (โ1)๐๐1 โ ๐๐ฃ๐2 .
We will next prove the important identity.
Lemma 1.7.8. Let ๐ be a vector space and ๐ โ L๐(๐). Then for all ๐ฃ โ ๐ we have
(1.7.9) ๐๐ฃ(๐๐ฃ๐) = 0 .
Proof. It suffices by linearity to prove this for decomposable tensors and since (1.7.9) istrivially true for ๐ โ L1(๐), we can by induction assume (1.7.9) is true for decomposabletensors of degree ๐ โ 1. Let โ1 โ โฏ โ โ๐ be a decomposable tensor of degree ๐. Setting๐ โ โ1 โโฏ โ โ๐โ1 and โ = โ๐ we have
๐๐ฃ(โ1 โโฏ โ โ๐) = ๐๐ฃ(๐ โ โ) = ๐๐ฃ๐ โ โ + (โ1)๐โ1โ(๐ฃ)๐by (1.7.7). Hence
๐๐ฃ(๐๐ฃ(๐ โ โ)) = ๐๐ฃ(๐๐ฃ๐) โ โ + (โ1)๐โ2โ(๐ฃ)๐๐ฃ๐ + (โ1)๐โ1โ(๐ฃ)๐๐ฃ๐ .But by induction the first summand on the right is zero and the two remaining summandscancel each other out. โก
From (1.7.9) we can deduce a slightly stronger result: For ๐ฃ1, ๐ฃ2 โ ๐(1.7.10) ๐๐ฃ1 ๐๐ฃ2 = โ๐๐ฃ2 ๐๐ฃ1 .
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24 Chapter 1: Multilinear Algebra
Proof. Let ๐ฃ = ๐ฃ1 + ๐ฃ2. Then ๐๐ฃ = ๐๐ฃ1 + ๐๐ฃ2 so
0 = ๐๐ฃ๐๐ฃ = (๐๐ฃ1 + ๐๐ฃ2)(๐๐ฃ1 + ๐๐ฃ2)= ๐๐ฃ1 ๐๐ฃ1 + ๐๐ฃ1 ๐๐ฃ2 + ๐๐ฃ2 ๐๐ฃ1 + ๐๐ฃ2 ๐๐ฃ2= ๐๐ฃ1 ๐๐ฃ2 + ๐๐ฃ2 ๐๐ฃ1
since the first and last summands are zero by (1.7.9). โก
Weโll now show how to define the operation ๐๐ฃ on ๐ฌ๐(๐โ). Weโll first prove:
Lemma 1.7.11. If ๐ โ L๐(๐) is redundant then so is ๐๐ฃ๐.
Proof. Let ๐ = ๐1 โ โ โ โ โ ๐2 where โ is in ๐โ, ๐1 is in L๐(๐) and ๐2 is in L๐(๐). Then byequation (1.7.7)
๐๐ฃ๐ = ๐๐ฃ๐1 โ โ โ โ โ ๐2 + (โ1)๐๐1 โ ๐๐ฃ(โ โ โ) โ ๐2 + (โ1)๐+2๐1 โ โ โ โ โ ๐๐ฃ๐2 .However, the first and the third terms on the right are redundant and
๐๐ฃ(โ โ โ) = โ(๐ฃ)โ โ โ(๐ฃ)โby equation (1.7.5). โก
Now let ๐ be the projection (1.5.12) of L๐(๐) onto ๐ฌ๐(๐โ) and for ๐ = ๐(๐) โ ๐ฌ๐(๐โ)define
(1.7.12) ๐๐ฃ๐ = ๐(๐๐ฃ๐) .To show that this definition is legitimate we note that if ๐ = ๐(๐1) = ๐(๐2), then ๐1 โ ๐2 โI๐(๐), so by Lemma 1.7.11 ๐๐ฃ๐1 โ ๐๐ฃ๐2 โ I๐โ1 and hence
๐(๐๐ฃ๐1) = ๐(๐๐ฃ๐2) .Therefore, (1.7.12) does not depend on the choice of ๐.
By definition, ๐๐ฃ is a linear map ๐ฌ๐(๐โ) โ ๐ฌ๐โ1(๐โ). We call this the interior productoperation. From the identities (1.7.2)โ(1.7.12) one gets, for ๐ฃ, ๐ฃ1, ๐ฃ2 โ ๐, ๐ โ ๐ฌ๐(๐โ), ๐1 โ๐ฌ๐(๐โ), and ๐2 โ ๐ฌ2(๐โ)
๐(๐ฃ1+๐ฃ2)๐ = ๐๐ฃ1๐ + ๐๐ฃ2๐(1.7.13)๐๐ฃ(๐1 โง ๐2) = ๐๐ฃ๐1 โง ๐2 + (โ1)๐๐1 โง ๐๐ฃ๐2(1.7.14)๐๐ฃ(๐๐ฃ๐) = 0(1.7.15)
and๐๐ฃ1 ๐๐ฃ2๐ = โ๐๐ฃ2 ๐๐ฃ1๐ .
Moreover if ๐ = โ1 โงโฏ โง โ๐ is a decomposable element of ๐ฌ๐(๐โ) one gets from (1.7.5)
๐๐ฃ๐ =๐โ๐=1(โ1)๐โ1โ๐(๐ฃ)โ1 โงโฏ โง โ๐ โงโฏ โง โ๐ .
In particular if ๐1,โฆ, ๐๐ is a basis of ๐, ๐โ1 ,โฆ, ๐โ๐ the dual basis of ๐โ and ๐๐ผ = ๐โ๐1 โงโฏโง ๐โ๐๐ ,
1 โค ๐1 < โฏ < ๐๐ โค ๐, then ๐๐๐๐๐ผ = 0 if ๐ โ ๐ผ and if ๐ = ๐๐
(1.7.16) ๐๐๐๐๐ผ = (โ1)๐โ1๐๐ผ๐
where ๐ผ๐ = (๐1,โฆ, ๐ค๐,โฆ, ๐๐) (i.e., ๐ผ๐ is obtained from the multi-index ๐ผ by deleting ๐๐).
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ยง1.8 The pullback operation on ๐ฌ๐(๐โ) 25
Exercises for ยง1.7Exercise 1.7.i. Prove Lemma 1.7.4.
Exercise 1.7.ii. Prove Lemma 1.7.6.
Exercise 1.7.iii. Show that if๐ โ A๐, ๐๐ฃ๐ = ๐๐๐ฃ where๐๐ฃ is the tensor (1.3.21). In particularconclude that ๐๐ฃ๐ โ A๐โ1(๐). (See Exercise 1.4.viii.)
Exercise 1.7.iv. Assume the dimension of ๐ is ๐ and let๐บ be a nonzero element of the onedimensional vector space ๐ฌ๐(๐โ). Show that the map(1.7.17) ๐โถ ๐ โ ๐ฌ๐โ1(๐โ) , ๐ฃ โฆ ๐๐ฃ๐บ ,is a bijective linear map. Hint: One can assume๐บ = ๐โ1 โงโฏโง ๐โ๐ where ๐1,โฆ, ๐๐ is a basis of๐. Now use equation (1.7.16) to compute this map on basis elements.
Exercise 1.7.v (cross-product). Let ๐ be a 3-dimensional vector space, ๐ต an inner producton ๐ and ๐บ a nonzero element of ๐ฌ3(๐โ). Define a map
โ ร โโถ ๐ ร ๐ โ ๐by setting
๐ฃ1 ร ๐ฃ2 โ ๐โ1(๐ฟ๐ฃ1 โง ๐ฟ๐ฃ2)where ๐ is the map (1.7.17) and ๐ฟโถ ๐ โ ๐โ the map (1.2.17). Show that this map is linearin ๐ฃ1, with ๐ฃ2 fixed and linear in ๐ฃ2 with ๐ฃ1 fixed, and show that ๐ฃ1 ร ๐ฃ2 = โ๐ฃ2 ร ๐ฃ1.Exercise 1.7.vi. For๐ = ๐3 let ๐1, ๐2 and ๐3 be the standard basis vectors and ๐ต the standardinner product. (See ยง1.1.) Show that if ๐บ = ๐โ1 โง ๐โ2 โง ๐โ3 the cross-product above is thestandard cross-product:
๐1 ร ๐2 = ๐3๐2 ร ๐3 = ๐1๐3 ร ๐1 = ๐2 .
Hint: If ๐ต is the standard inner product, then ๐ฟ๐๐ = ๐โ๐ .Remark 1.7.18. One can make this standard cross-product look even more standard byusing the calculus notation: ๐1 = ๐ค, ๐2 = ๐ฅ, and ๐3 = ๏ฟฝ๏ฟฝ.
1.8. The pullback operation on ๐ฌ๐(๐โ)Let ๐ and๐ be vector spaces and let ๐ด be a linear map of ๐ into๐. Given a ๐-tensor
๐ โ L๐(๐), recall that the pullback ๐ดโ๐ is the ๐-tensor(๐ดโ๐)(๐ฃ1,โฆ, ๐ฃ๐) = ๐(๐ด๐ฃ1,โฆ,๐ด๐ฃ๐)
in L๐(๐). (See ยง1.3 and equation (1.3.17).) In this section weโll show how to define a similarpullback operation on ๐ฌ๐(๐โ).
Lemma 1.8.1. If ๐ โ I๐(๐), then ๐ดโ๐ โ I๐(๐).Proof. It suffices to verify this when ๐ is a redundant ๐-tensor, i.e., a tensor of the form
๐ = โ1 โโฏ โ โ๐where โ๐ โ ๐โ and โ๐ = โ๐+1 for some index, ๐. But by equation (1.3.19),
๐ดโ๐ = ๐ดโโ1 โโฏ โ ๐ดโโ๐and the tensor on the right is redundant since ๐ดโโ๐ = ๐ดโโ๐+1. โก
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26 Chapter 1: Multilinear Algebra
Now let ๐ be an element of ๐ฌ๐(๐โ) and let ๐ = ๐(๐) where ๐ is in L๐(๐). We define
(1.8.2) ๐ดโ๐ = ๐(๐ดโ๐) .
Claim 1.8.3. The left hand side of equation (1.8.2) is well-defined.
Proof. If ๐ = ๐(๐) = ๐(๐โฒ), then ๐ = ๐โฒ + ๐ for some ๐ โ I๐(๐), and ๐ดโ๐ = ๐ดโ๐โฒ + ๐ดโ๐.But ๐ดโ๐ โ I๐(๐), so
๐(๐ดโ๐โฒ) = ๐(๐ดโ๐) . โก
Proposition 1.8.4. The map ๐ดโ โถ ๐ฌ๐(๐โ) โ ๐ฌ๐(๐โ) sending ๐ โฆ ๐ดโ๐ is linear. Moreover,(1) If ๐๐ โ ๐ฌ๐๐(๐โ), ๐ = 1, 2, then
๐ดโ(๐1 โง ๐2) = ๐ดโ(๐1) โง ๐ดโ(๐2) .
(2) If ๐ is a vector space and ๐ต โถ ๐ โ ๐ a linear map, then for ๐ โ ๐ฌ๐(๐โ),
๐ตโ๐ดโ๐ = (๐ด๐ต)โ๐ .
Weโll 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 weโll show how to use it to define the notionof determinant for a linear mapping.
Definition 1.8.5. Let ๐ be a ๐-dimensional vector space. Then dim๐ฌ๐(๐โ) = (๐๐) = 1;i.e., ๐ฌ๐(๐โ) is a one-dimensional vector space. Thus if ๐ดโถ ๐ โ ๐ is a linear mapping, theinduced pullback mapping:
๐ดโ โถ ๐ฌ๐(๐โ) โ ๐ฌ๐(๐โ) ,is just โmultiplication by a constantโ. We denote this constant by det(๐ด) and call it the de-terminant of ๐ด, Hence, by definition,
(1.8.6) ๐ดโ๐ = det(๐ด)๐
for all ๐ โ ๐ฌ๐(๐โ).
From equation (1.8.6) it is easy to derive a number of basic facts about determinants.
Proposition 1.8.7. If ๐ด and ๐ต are linear mappings of ๐ into ๐, then
det(๐ด๐ต) = det(๐ด) det(๐ต) .
Proof. By (2) and
(๐ด๐ต)โ๐ = det(๐ด๐ต)๐ = ๐ตโ(๐ดโ๐)= det(๐ต)๐ดโ๐ = det(๐ต) det(๐ด)๐ ,
so det(๐ด๐ต) = det(๐ด) det(๐ต). โก
Proposition 1.8.8. Write id๐ โถ ๐ โ ๐ for the identity map. Then det(id๐) = 1.
Weโll leave the proof as an exercise. As a hint, note that idโ๐ is the identity map on๐ฌ๐(๐โ).
Proposition 1.8.9. If ๐ดโถ ๐ โ ๐ is not surjective, then det(๐ด) = 0.
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ยง1.8 The pullback operation on ๐ฌ๐(๐โ) 27
Proof. Let๐ be the image of ๐ด. Then if ๐ด is not onto, the dimension of๐ is less than ๐,so ๐ฌ๐(๐โ) = 0. Now let ๐ด = ๐๐๐ต where ๐๐ is the inclusion map of๐ into ๐ and ๐ต is themapping, ๐ด, regarded as a mapping from ๐ to๐. Thus if ๐ is in ๐ฌ๐(๐โ), then by (2)
๐ดโ๐ = ๐ตโ๐โ๐๐and since ๐โ๐๐ is in ๐ฌ๐(๐โ) it is zero. โก
We will derive by wedge product arguments the familiar โmatrix formulaโ for the deter-minant. Let ๐ and๐ be ๐-dimensional vector spaces and let ๐1,โฆ, ๐๐ be a basis for ๐ and๐1,โฆ, ๐๐ a basis for๐. From these bases we get dual bases, ๐โ1 ,โฆ, ๐โ๐ and ๐โ1 ,โฆ, ๐โ๐ , for๐โand๐โ. Moreover, if ๐ด is a linear map of ๐ into๐ and [๐๐,๐] the ๐ ร ๐ matrix describing๐ด in terms of these bases, then the transpose map, ๐ดโ โถ ๐โ โ ๐โ, is described in terms ofthese dual bases by the ๐ ร ๐ transpose matrix, i.e., if
๐ด๐๐ =๐โ๐=1๐๐,๐๐๐ ,
then
๐ดโ๐โ๐ =๐โ๐=1๐๐,๐๐โ๐ .
(See ยง1.2.) Consider now ๐ดโ(๐โ1 โงโฏ โง ๐โ๐ ). By (1),
๐ดโ(๐โ1 โงโฏ โง ๐โ๐ ) = ๐ดโ๐โ1 โงโฏ โง ๐ดโ๐โ๐= โ1โค๐1,โฆ,๐๐โค๐
(๐1,๐1๐โ๐1) โง โฏ โง (๐๐,๐๐๐
โ๐๐) .
Thus,๐ดโ(๐โ1 โงโฏ โง ๐โ๐ ) = โ๐1,๐1โฆ๐๐,๐๐ ๐
โ๐1 โงโฏ โง ๐
โ๐๐ .
If the multi-index, ๐1,โฆ, ๐๐, is repeating, then ๐โ๐1 โงโฏโง๐โ๐๐ is zero, and if it is not repeating
then we can write๐๐ = ๐(๐) ๐ = 1,โฆ, ๐
for some permutation, ๐, and hence we can rewrite ๐ดโ(๐โ1 โงโฏ โง ๐โ๐ ) as
๐ดโ(๐โ1 โงโฏ โง ๐โ๐ ) = โ๐โ๐๐๐1,๐(1)โฏ๐๐,๐(๐) (๐โ1 โงโฏ โง ๐โ๐ )๐ .
But(๐โ1 โงโฏ โง ๐โ๐ )๐ = (โ1)๐๐โ1 โงโฏ โง ๐โ๐
so we get finally the formula
(1.8.10) ๐ดโ(๐โ1 โงโฏ โง ๐โ๐ ) = det([๐๐,๐])๐โ1 โงโฏ โง ๐โ๐where
(1.8.11) det([๐๐,๐]) = โ๐โ๐๐(โ1)๐๐1,๐(1)โฏ๐๐,๐(๐)
summed over ๐ โ ๐๐. The sum on the right is (as most of you know) the determinant of thematric [๐๐,๐].
Notice that if ๐ = ๐ and ๐๐ = ๐๐, ๐ = 1,โฆ, ๐, then ๐ = ๐โ1 โง โฏ โง ๐โ๐ = ๐โ1 โง โฏ โง ๐โ๐ ,hence by (1.8.6) and (1.8.10),
det(๐ด) = det[๐๐,๐] .
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28 Chapter 1: Multilinear Algebra
Exercises for ยง1.8Exercise 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 of๐ ร ๐matrices: 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 ๐ ร ๐matrices: If one interchanges two columns, then one changes the sign of thedeterminant.
Hint: Let ๐1,โฆ, ๐๐ be a basis of ๐ and let ๐ตโถ ๐ โ ๐ be the linear mapping: ๐ต๐๐ = ๐๐,๐ต๐๐ = ๐๐ and ๐ต๐โ = ๐โ, โ โ ๐, ๐. What is ๐ตโ(๐โ1 โงโฏ โง ๐โ๐ )?
Exercise 1.8.iv. Deduce from Propositions 1.8.3 and 1.8.4 another well-known fact aboutdeterminants of ๐ ร ๐matrix: If (๐๐,๐) is the inverse of [๐๐,๐], its determinant is the inverse ofthe determinant of [๐๐,๐].
Exercise 1.8.v. Extract from (1.8.11) a well-known formula for determinants of 2 ร 2 ma-trices:
det(๐11 ๐12๐21, ๐22) = ๐11๐22 โ ๐12๐21 .
Exercise 1.8.vi. Show that if ๐ด = [๐๐,๐] is an ๐ ร ๐ matrix and ๐ดโบ = [๐๐,๐] is its transposedet๐ด = det๐ดโบ.
Hint: You are required to show that the sums
โ๐โ๐๐(โ1)๐๐1,๐(1)โฆ๐๐,๐(๐)
andโ๐โ๐๐(โ1)๐๐๐(1),1โฆ๐๐(๐),๐
are the same. Show that the second sum is identical with
โ๐โ๐๐(โ1)๐โ1๐๐โ1(1),1โฆ๐๐โ1(๐),๐ .
Exercise 1.8.vii. Let ๐ด be an ๐ ร ๐matrix of the form
๐ด = (๐ต โ0 ๐ถ) ,
where ๐ต is a ๐ ร ๐ matrix and ๐ถ the โ ร โ matrix and the bottom โ ร ๐ block is zero. Showthat
det(๐ด) = det(๐ต) det(๐ถ) .Hint: Show that in equation (1.8.11) every nonzero term is of the form
(โ1)๐๐๐1,๐(1)โฆ๐๐,๐(๐)๐1,๐(1)โฆ๐โ,๐(โ)where ๐ โ ๐๐ and ๐ โ ๐โ.
Exercise 1.8.viii. Let ๐ and๐ be vector spaces and let ๐ดโถ ๐ โ ๐ be a linear map. Showthat if ๐ด๐ฃ = ๐ค then for ๐ โ ๐ฌ๐(๐โ),
๐ดโ๐๐ค๐ = ๐๐ฃ๐ดโ๐ .Hint:By equation (1.7.14) and Proposition 1.8.4 it suffices to prove this for๐ โ ๐ฌ1(๐โ),
i.e., for ๐ โ ๐โ.
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ยง1.9 Orientations 29
1.9. Orientations
Definition 1.9.1. We recall from freshman calculus that if โ โ ๐2 is a line through theorigin, then โ โ {0} has two connected components and an orientation of โ is a choice ofone of these components (as in the Figure 1.9.1 below).
โ
0
Figure 1.9.1. A line in ๐2
More generally, if ๐ฟ is a one-dimensional vector space then ๐ฟโ {0} consists of two com-ponents: namely if ๐ฃ is an element of ๐ฟ โ {0}, then these two components are
๐ฟ1 = { ๐๐ฃ | ๐ > 0 }and
๐ฟ2 = { ๐๐ฃ | ๐ < 0 } .
Definition 1.9.2. Let ๐ฟ be a one-dimensional vector space. An orientation of ๐ฟ is a choice ofone of a connected component of ๐ฟโ{0}. Usually the component chosen is denoted ๐ฟ+, andcalled the positive component of๐ฟโ{0} and the other component๐ฟโ the negative componentof ๐ฟ โ {0}.
Definition 1.9.3. Let (๐ฟ, ๐ฟ+) be an oriented one-dimensional vector space. A vector ๐ฃ โ ๐ฟis positively oriented if ๐ฃ โ ๐ฟ+.
Definition 1.9.4. Let ๐ be an ๐-dimensional vector space. An orientation of ๐ is an orien-tation of the one-dimensional vector space ๐ฌ๐(๐โ).
One important way of assigning an orientation to ๐ is to choose a basis, ๐1,โฆ, ๐๐ of ๐.Then, if ๐โ1 ,โฆ, ๐โ๐ is the dual basis, we can orient ๐ฌ๐(๐โ) by requiring that ๐โ1 โง โฏ โง ๐โ๐ bein the positive component of ๐ฌ๐(๐โ).
Definition 1.9.5. Let ๐ be an oriented ๐-dimensional vector space. We say that an orderedbasis (๐1,โฆ, ๐๐) of ๐ is positively oriented if ๐โ1 โง โฏ โง ๐โ๐ is in the positive component of๐ฌ๐(๐โ).
Suppose that ๐1,โฆ, ๐๐ and ๐1,โฆ, ๐๐ are bases of ๐ and that
(1.9.6) ๐๐ =๐โ๐=1๐๐,๐,๐๐ .
Then by (1.7.10)๐โ1 โงโฏ โง ๐โ๐ = det[๐๐,๐]๐โ1 โงโฏ โง ๐โ๐
so we conclude:
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30 Chapter 1: Multilinear Algebra
Proposition 1.9.7. If ๐1,โฆ, ๐๐ is positively oriented, then๐1,โฆ, ๐๐ is positively oriented if andonly if det[๐๐,๐] is positive.
Corollary 1.9.8. If ๐1,โฆ, ๐๐ is a positively oriented basis of ๐, then the basis
๐1,โฆ, ๐๐โ1, โ๐๐, ๐๐+1,โฆ, ๐๐is negatively oriented.
Now let ๐ be a vector space of dimension ๐ > 1 and๐ a subspace of dimension ๐ < ๐.We will use the result above to prove the following important theorem.
Theorem 1.9.9. Given orientations on๐ and๐/๐, one gets from these orientations a naturalorientation on๐.
Remark 1.9.10. What we mean by โnaturalโ will be explained in the course of the proof.
Proof. Let ๐ = ๐โ ๐ and let ๐ be the projection of๐ onto๐/๐ . By Exercises 1.2.i and 1.2.iiwe can choose a basis ๐1,โฆ, ๐๐ of ๐ such that ๐๐+1,โฆ, ๐๐ is a basis of๐ and ๐(๐1),โฆ, ๐(๐๐)a basis of๐/๐. Moreover, replacing ๐1 by โ๐1 if necessary we can assume by Corollary 1.9.8that ๐(๐1),โฆ, ๐(๐๐) is a positively oriented basis of ๐/๐ and replacing ๐๐ by โ๐๐ if neces-sary we can assume that ๐1,โฆ, ๐๐ is a positively oriented basis of ๐. Now assign to๐ theorientation associated with the basis ๐๐+1,โฆ, ๐๐.
Letโs show that this assignment is โnaturalโ(i.e., does not depend on our choice of basis๐1,โฆ, ๐๐). To see this let ๐1,โฆ, ๐๐ be another basis of ๐ with the properties above and let๐ด = [๐๐,๐] be thematrix (1.9.6) expressing the vectors ๐1,โฆ, ๐๐ as linear combinations of thevectors ๐1,โฆ๐๐. This matrix has to have the form
(1.9.11) ๐ด = (๐ต ๐ถ0 ๐ท)
where ๐ต is the ๐ร๐matrix expressing the basis vectors ๐(๐1),โฆ, ๐(๐๐) of๐/๐ as linear com-binations of ๐(๐1),โฆ, ๐(๐๐) and ๐ท the ๐ ร ๐matrix expressing the basis vectors ๐๐+1,โฆ, ๐๐of๐ as linear combinations of ๐๐+1,โฆ, ๐๐. Thus
det(๐ด) = det(๐ต) det(๐ท) .However, by Proposition 1.9.7, det๐ด and det๐ต are positive, so det๐ท is positive, and henceif ๐๐+1,โฆ, ๐๐ is a positively oriented basis of๐ so is ๐๐+1,โฆ, ๐๐. โก
As a special case of this theorem suppose dim๐ = ๐ โ 1. Then the choice of a vector๐ฃ โ ๐โ๐ gives one a basis vector ๐(๐ฃ) for the one-dimensional space๐/๐ and hence if๐is oriented, the choice of ๐ฃ gives one a natural orientation on๐.
Definition 1.9.12. Let๐ดโถ ๐1 โ ๐2 a bijective linear map of oriented ๐-dimensional vectorspaces. We say that ๐ด is orientation-preserving if, for ๐ โ ๐ฌ๐(๐โ2 )+, we have that ๐ดโ๐ is in๐ฌ๐(๐โ1 )+.
Example 1.9.13. If ๐1 = ๐2 then ๐ดโ๐ = det(๐ด)๐ so ๐ด is orientation preserving if and onlyif det(๐ด) > 0.
The following proposition weโll leave as an exercise.
Proposition 1.9.14. Let๐1,๐2, and๐3 be oriented ๐-dimensional vector spaces and๐ด๐ โถ ๐๐ โฅฒ๐๐+1 for ๐ = 1, 2 be bijective linear maps. Then if ๐ด1 and ๐ด2 are orientation preserving, so is๐ด2 โ ๐ด1.
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ยง1.9 Orientations 31
Exercises for ยง1.9Exercise 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 ๐ and๐ are oriented then these orientations induce a natural orientation on๐/๐.
Exercise 1.9.iii. Similarly show that if๐ and ๐/๐ are oriented these orientations inducea natural orientation on ๐.
Exercise 1.9.iv. Let ๐ be an ๐-dimensional vector space and๐ โ ๐ a ๐-dimensional sub-space. Let ๐ = ๐/๐ and let ๐ โถ ๐ โ ๐ and ๐โถ ๐ โ ๐ be the inclusion and projectionmaps. Suppose ๐ and ๐ are oriented. Let ๐ be in ๐ฌ๐โ๐(๐โ)+ and let ๐ be in ๐ฌ๐(๐โ)+. Showthat there exists a ๐ in๐ฌ๐(๐โ) such that ๐โ๐โง๐ = ๐. Moreover show that ๐โ๐ is intrinsicallydefined (i.e., does not depend on how we choose ๐) and sits in the positive part๐ฌ๐(๐โ)+ of๐ฌ๐(๐โ).
Exercise 1.9.v. Let ๐1,โฆ, ๐๐ be the standard basis vectors of๐๐.The standard orientation of๐๐ is, by definition, the orientation associated with this basis. Show that if๐ is the subspaceof ๐๐ defined by the equation ๐ฅ1 = 0, and ๐ฃ = ๐1 โ ๐ then the natural orientation of๐associated with ๐ฃ and the standard orientation of ๐๐ coincide with the orientation given bythe basis vectors ๐2,โฆ, ๐๐ of๐.
Exercise 1.9.vi. Let๐be anoriented๐-dimensional vector space and๐ an๐โ1-dimensionalsubspace. Show that if ๐ฃ and ๐ฃโฒ are in๐โ๐ then ๐ฃโฒ = ๐๐ฃ+๐ค, where๐ค is in๐ and๐ โ ๐โ{0}.Show that ๐ฃ and ๐ฃโฒ give rise to the same orientation of๐ if and only if ๐ 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 matrix๐ด expressing the vectors, ๐๐ as linear combinations of the vectors ๐๐ฝ, had to have the form(1.9.11). Why is this the case?
Exercise 1.9.ix.(1) Let๐ be a vector space,๐ a subspace of๐ and๐ดโถ ๐ โ ๐ a bijective linear map which
maps๐ onto๐. Show that one gets from ๐ด a bijective linear map๐ตโถ ๐/๐ โ ๐/๐
with the property๐๐ด = ๐ต๐ ,
where ๐โถ ๐ โ ๐/๐ is the quotient map.(2) Assume that ๐, ๐ and ๐/๐ are compatibly oriented. Show that if ๐ด is orientation-
preserving and its restriction to๐ is orientation preserving then ๐ต is orientation pre-serving.
Exercise 1.9.x. Let ๐ be a oriented ๐-dimensional vector space,๐ an (๐ โ 1)-dimensionalsubspace of ๐ and ๐ โถ ๐ โ ๐ the inclusion map. Given ๐ โ ๐ฌ๐(๐โ)+ and ๐ฃ โ ๐ โ๐ showthat for the orientation of๐ described in Exercise 1.9.v, ๐โ(๐๐ฃ๐) โ ๐ฌ๐โ1(๐โ)+.
Exercise 1.9.xi. Let ๐ be an ๐-dimensional vector space, ๐ตโถ ๐ ร ๐ โ ๐ an inner prod-uct and ๐1,โฆ, ๐๐ a basis of ๐ which is positively oriented and orthonormal. Show that thevolume element
vol โ ๐โ1 โงโฏ โง ๐โ๐ โ ๐ฌ๐(๐โ)
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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 ๐ be an oriented ๐-dimensional vector space and ๐ต an inner product on ๐. Fix an
oriented orthonormal basis, ๐1,โฆ, ๐๐, of ๐ and let ๐ดโถ ๐ โ ๐ be a linear map. Showthat if
๐ด๐๐ = ๐ฃ๐ =๐โ๐=1๐๐,๐๐๐
and ๐๐,๐ = ๐ต(๐ฃ๐, ๐ฃ๐), the matrices ๐๐ด = [๐๐,๐] and ๐๐ต = (๐๐,๐) are related by: ๐๐ต =๐โบ๐ด๐๐ด.
(2) Show that if vol is the volume form ๐โ1 โงโฏ โง ๐โ๐ , and ๐ด is orientation preserving
๐ดโ vol = det(๐๐ต)12 vol .
(3) ByTheorem 1.5.13 one has a bijective map๐ฌ๐(๐โ) โ A๐(๐) .
Show that the element, ๐บ, of ๐ด๐(๐) corresponding to the form vol has the property|๐บ(๐ฃ1,โฆ, ๐ฃ๐)|2 = det((๐๐,๐))
where ๐ฃ1,โฆ, ๐ฃ๐ are any ๐-tuple of vectors in ๐ and ๐๐,๐ = ๐ต(๐ฃ๐, ๐ฃ๐).
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CHAPTER 2
Differential Forms
The goal of this chapter is to generalize to ๐ 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 ๐ variables andintroduce their dual objects: one-forms. We will then take up in ยง2.3 the theory of ๐-formsfor ๐ > 1. We begin by fixing some notation.
Definition 2.1.1. Let ๐ โ ๐๐. The tangent space to ๐๐ at ๐ is the set of pairs๐๐๐๐ โ { (๐, ๐ฃ) | ๐ฃ โ ๐๐ } .
The identification(2.1.2) ๐๐๐๐ โฅฒ ๐๐ , (๐, ๐ฃ) โฆ ๐ฃmakes ๐๐๐๐ into a vector space. More explicitly, for ๐ฃ, ๐ฃ1, ๐ฃ2 โ ๐๐ and ๐ โ ๐ we define theaddition and scalar multiplication operations on ๐๐๐๐ by setting
(๐, ๐ฃ1) + (๐, ๐ฃ2) โ (๐, ๐ฃ1 + ๐ฃ2)and
๐(๐, ๐ฃ) โ (๐, ๐๐ฃ) .
Let ๐ be an open subset of ๐๐ and ๐โถ ๐ โ ๐๐ a ๐ถ1 map. We recall that the derivative๐ท๐(๐)โถ ๐๐ โ ๐๐
of ๐ at ๐ is the linear map associated with the๐ ร ๐matrix
[ ๐๐๐๐๐ฅ๐(๐)] .
It will be useful to have a โbase-pointedโ version of this definition aswell. Namely, if ๐ = ๐(๐)we will define
๐๐๐ โถ ๐๐๐๐ โ ๐๐๐๐
to be the map๐๐๐(๐, ๐ฃ) โ (๐,๐ท๐(๐)๐ฃ) .
It is clear from the way we have defined vector space structures on ๐๐๐๐ and ๐๐๐๐ that thismap is linear.
33
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34 Chapter 2: Differential Forms
Suppose that the image of ๐ is contained in an open set ๐, and suppose ๐โถ ๐ โ ๐๐ isa ๐ถ1 map. Then the โbase-pointedโ version of the chain rule asserts that
๐๐๐ โ ๐๐๐ = ๐(๐ โ ๐)๐ .(This is just an alternative way of writing๐ท๐(๐)๐ท๐(๐) = ๐ท(๐ โ ๐)(๐).)
In 3-dimensional vector calculus a vector field is a functionwhich attaches to each point๐ of ๐3 a base-pointed arrow (๐, ๐ฃ) โ ๐๐๐3. The ๐-dimensional version of this definition isessentially the same.
Definition 2.1.3. Let ๐ be an open subset of ๐๐. A vector field ๐ on ๐ is a function whichassigns to each point ๐ โ ๐ a vector ๐(๐) โ ๐๐๐๐.
Thus a vector field is a vector-valued function, but its value at ๐ is an element of a vectorspace ๐๐๐๐ that itself depends on ๐.
Examples 2.1.4.(1) Given a fixed vector ๐ฃ โ ๐๐, the function(2.1.5) ๐ โฆ (๐, ๐ฃ)
is a vector field on ๐๐. Vector fields of this type are called constant vector fields.(2) In particular let ๐1,โฆ, ๐๐ be the standard basis vectors of๐๐. If ๐ฃ = ๐๐ we will denote the
vector field (2.1.5) by ๐/๐๐ฅ๐. (The reason for this โderivation notationโ will be explainedbelow.)
(3) Given a vector field ๐ on๐ and a function, ๐โถ ๐ โ ๐ we denote by ๐๐ the vector fieldon ๐ defined by
๐ โฆ ๐(๐)๐(๐) .(4) Given vector fields ๐1 and ๐2 on ๐, we denote by ๐1 + ๐2 the vector field on ๐ defined
by๐ โฆ ๐1(๐) + ๐2(๐) .
(5) The vectors, (๐, ๐๐), ๐ = 1,โฆ, ๐, are a basis of ๐๐๐๐, so if ๐ is a vector field on ๐, ๐(๐)can be written uniquely as a linear combination of these vectors with real numbers๐1(๐),โฆ, ๐๐(๐) as coefficients. In other words, using the notation in example (2) above,๐ can be written uniquely as a sum
(2.1.6) ๐ =๐โ๐=1๐๐๐๐๐ฅ๐
where ๐๐ โถ ๐ โ ๐ is the function ๐ โฆ ๐๐(๐).
Definition 2.1.7. We say that ๐ is a ๐ถโ vector field if the ๐๐โs are in ๐ถโ(๐).
Definition 2.1.8. Abasic vector field operation is Lie differentiation. If๐ โ ๐ถ1(๐)we define๐ฟ๐๐ to be the function on ๐ whose value at ๐ is given by(2.1.9) ๐ฟ๐๐(๐) โ ๐ท๐(๐)๐ฃ ,where ๐(๐) = (๐, ๐ฃ).
Example 2.1.10. If ๐ is the vector field (2.1.6) then
๐ฟ๐๐ =๐โ๐=1๐๐๐๐๐๐ฅ๐
,
(motivating our โderivation notationโ for ๐).
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ยง2.1 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. Let๐ be an open subset of๐๐, ๐ a vector field on๐, and๐1, ๐2 โ ๐ถ1(๐). Then๐ฟ๐(๐1 โ ๐2) = ๐ฟ๐(๐1) โ ๐2 + ๐1 โ ๐ฟ๐(๐2) .
We now discuss a class of objects which are in some sense โdual objectsโ to vector fields.
Definition 2.1.12. Let ๐ โ ๐๐. The cotangent space to ๐๐ at ๐ is the dual vector space๐โ๐ ๐๐ โ (๐๐๐)โ .
An element of ๐โ๐ ๐๐ is called a cotangent vector to ๐๐ at ๐.
Definition 2.1.13. Let ๐ be an open subset of ๐๐. A differential one-form, or simply one-form on ๐ is a function ๐ which assigns to each point ๐ โ ๐ a vector ๐๐ in ๐โ๐ ๐๐.
Examples 2.1.14.(1) Let ๐โถ ๐ โ ๐ be a ๐ถ1 function. Then for ๐ โ ๐ and ๐ = ๐(๐) one has a linear map(2.1.15) ๐๐๐ โถ ๐๐๐๐ โ ๐๐๐
and by making the identification ๐๐๐ โฅฒ ๐ by sending (๐, ๐ฃ) โฆ ๐ฃ, the differential ๐๐๐can be regarded as a linear map from ๐๐๐๐ to ๐, i.e., as an element of ๐โ๐ ๐๐. Hence theassignment
๐ โฆ ๐๐๐defines a one-form on ๐ which we denote by ๐๐.
(2) Given a one-form๐ and a function๐โถ ๐ โ ๐ the pointwise product of๐with๐ definesone-form ๐๐ on ๐ by (๐๐)๐ โ ๐(๐)๐๐.
(3) Given two one-forms ๐1 and ๐2 their pointwise sum defines a one-form ๐1 + ๐2 by(๐1 + ๐2)๐ โ (๐1)๐ + (๐2)๐.
(4) The one-forms ๐๐ฅ1,โฆ, ๐๐ฅ๐ play a particularly important role. By equation (2.1.15)
(2.1.16) (๐๐ฅ๐) (๐๐๐ฅ๐)๐= ๐ฟ๐,๐
i.e., is equal to 1 if ๐ = ๐ and zero if ๐ โ ๐. Thus (๐๐ฅ1)๐,โฆ, (๐๐ฅ๐)๐ are the basis of๐โ๐ ๐๐ dual to the basis (๐/๐๐ฅ๐)๐. Therefore, if ๐ is any one-form on๐, ๐๐ can be writtenuniquely as a sum
๐๐ =๐โ๐=1๐๐(๐)(๐๐ฅ๐)๐ , ๐๐(๐) โ ๐ ,
and ๐ can be written uniquely as a sum
(2.1.17) ๐ =๐โ๐=1๐๐๐๐ฅ๐
where ๐๐ โถ ๐ โ ๐ is the function ๐ โฆ ๐๐(๐). We say that ๐ is a ๐ถโ one-form or smoothone-form if the functions ๐1,โฆ, ๐๐ are ๐ถโ.
We leave the following as an exercise.
Lemma 2.1.18. Let ๐ be an open subset of ๐๐. If ๐โถ ๐ โ ๐ is a ๐ถโ function, then
๐๐ =๐โ๐=1
๐๐๐๐ฅ๐๐๐ฅ๐ .
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36 Chapter 2: Differential Forms
Suppose that ๐ is a vector field and ๐ a one-form on ๐ โ ๐๐. Then for every ๐ โ ๐ thevectors ๐(๐) โ ๐๐๐๐ and ๐๐ โ (๐๐๐๐)โ can be paired to give a number ๐๐(๐)๐๐ โ ๐, andhence, as ๐ varies, an๐-valued function ๐๐๐which we will call the interior product of ๐with๐.Example 2.1.19. For instance if ๐ is the vector field (2.1.6) and ๐ the one-form (2.1.17),then
๐๐๐ =๐โ๐=1๐๐๐๐ .
Thus if ๐ and ๐ are ๐ถโ, so is the function ๐๐๐. Also notice that if ๐ โ ๐ถโ(๐), then as weobserved above
๐๐ =๐โ๐=1
๐๐๐๐ฅ๐๐๐๐ฅ๐
so if ๐ is the vector field (2.1.6) then we have
๐๐ ๐๐ =๐โ๐=1๐๐๐๐๐๐ฅ๐= ๐ฟ๐๐ .
Exercises for ยง2.1Exercise 2.1.i. Prove Lemma 2.1.18
Exercise 2.1.ii. Prove Lemma 2.1.11
Exercise 2.1.iii. Let ๐ be an open subset of ๐๐ and ๐1 and ๐2 vector fields on ๐. Show thatthere is a unique vector field ๐, on ๐ with the property
๐ฟ๐๐ = ๐ฟ๐1(๐ฟ๐2๐) โ ๐ฟ๐2(๐ฟ๐1๐)for all ๐ โ ๐ถโ(๐).Exercise 2.1.iv. The vector field ๐ in Exercise 2.1.iii is called the Lie bracket of the vectorfields ๐1 and ๐2 and is denoted by [๐1, ๐2]. Verify that the Lie bracket is skew-symmetric, i.e.,
[๐1, ๐2] = โ[๐2, ๐1] ,and satisfies the Jacobi identity
[๐1, [๐2, ๐3]] + [๐2, [๐3, ๐1]] + [๐3, [๐1, ๐2]] = 0 .(And thus defines the structure of a Lie algebra.)
Hint: Prove analogous identities for ๐ฟ๐1 , ๐ฟ๐2 and ๐ฟ๐3 .
Exercise 2.1.v. Let ๐1 = ๐/๐๐ฅ๐ and ๐2 = โ๐๐=1 ๐๐๐/๐๐ฅ๐. Show that
[๐1, ๐2] =๐โ๐=1
๐๐๐๐๐ฅ๐๐๐๐ฅ๐
.
Exercise 2.1.vi. Let ๐1 and ๐2 be vector fields and ๐ a ๐ถโ function. Show that[๐1, ๐๐2] = ๐ฟ๐1๐๐2 + ๐[๐1, ๐2] .
Exercise 2.1.vii. Let ๐ be an open subset of ๐๐ and let ๐พโถ [๐, ๐] โ ๐, ๐ก โฆ (๐พ1(๐ก),โฆ, ๐พ๐(๐ก))be a ๐ถ1 curve. Given a ๐ถโ one-form ๐ = โ๐๐=1 ๐๐๐๐ฅ๐ on๐, define the line integral of ๐ over๐พ to be the integral
โซ๐พ๐ โ๐โ๐=1โซ๐
๐๐๐(๐พ(๐ก))๐๐พ๐๐๐ก๐๐ก .
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ยง2.2 Integral Curves for Vector Fields 37
Show that if ๐ = ๐๐ for some ๐ โ ๐ถโ(๐)
โซ๐พ๐ = ๐(๐พ(๐)) โ ๐(๐พ(๐)) .
In particular conclude that if ๐พ is a closed curve, i.e., ๐พ(๐) = ๐พ(๐), this integral is zero.
Exercise 2.1.viii. Let ๐ be the ๐ถโ one-form on ๐2 โ {0} defined by
๐ = ๐ฅ1๐๐ฅ2 โ ๐ฅ2๐๐ฅ1๐ฅ21 + ๐ฅ22
,
and let ๐พโถ [0, 2๐] โ ๐2โ{0} be the closed curve ๐ก โฆ (cos ๐ก, sin ๐ก). Compute the line integralโซ๐พ ๐, and note that โซ๐พ ๐ โ 0. Conclude that ๐ is not of the form ๐๐ for ๐ โ ๐ถโ(๐2 โ {0}).
Exercise 2.1.ix. Let ๐ be the function
๐(๐ฅ1, ๐ฅ2) ={{{{{{{
arctan ๐ฅ2๐ฅ1 ๐ฅ1 > 0๐2 , ๐ฅ2 > 0 or ๐ฅ1 = 0arctan ๐ฅ2๐ฅ1 + ๐ ๐ฅ1 < 0 .
Recall that โ๐2 < arctan(๐ก) < ๐2 . Show that ๐ is ๐ถโ and that ๐๐ is the 1-form ๐ 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 weโll discuss some properties of vector fields which weโll need for themanifold segment of this text. Weโll begin by generalizing to ๐-variables the calculus notionof an โintegral curveโ of a vector field.
Definition 2.2.1. Let ๐ โ ๐๐ be open and ๐ a vector field on ๐. A ๐ถ1 curve ๐พโถ (๐, ๐) โ ๐is an integral curve of ๐ if for all ๐ก โ (๐, ๐) we have
๐(๐พ(๐ก)) = (๐พ(๐ก), ๐๐พ๐๐ก (๐ก)) .Remark 2.2.2. If ๐ is the vector field (2.1.6) and ๐โถ ๐ โ ๐๐ is the function (๐1,โฆ, ๐๐),the condition for ๐พ(๐ก) to be an integral curve of ๐ is that it satisfy the system of differentialequations
(2.2.3) ๐๐พ๐๐ก(๐ก) = ๐(๐พ(๐ก)) .
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 ๐ be an open subset of ๐๐ and ๐ a vectorfield on ๐. Given a point ๐0 โ ๐ and ๐ โ ๐, there exists an interval ๐ผ = (๐ โ ๐, ๐ + ๐), aneighborhood๐0 of ๐0 in๐, and for every ๐ โ ๐0 an integral curve ๐พ๐ โถ ๐ผ โ ๐ for ๐ such that๐พ๐(๐) = ๐.Theorem 2.2.5 (uniqueness of integral curves). Let๐ be an open subset of ๐๐ and ๐ a vectorfield on ๐. For ๐ = 1, 2, let ๐พ๐ โถ ๐ผ๐ โ ๐, ๐ = 1, 2 be integral curves for ๐. If ๐ โ ๐ผ1 โฉ ๐ผ2 and๐พ1(๐) = ๐พ2(๐) then ๐พ1|๐ผ1โฉ๐ผ2 = ๐พ2|๐ผ1โฉ๐ผ2 and the curve ๐พโถ ๐ผ1 โช ๐ผ2 โ ๐ defined by
๐พ(๐ก) โ {๐พ1(๐ก), ๐ก โ ๐ผ1๐พ2(๐ก), ๐ก โ ๐ผ2
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 ๐.
Theorem 2.2.6 (smooth dependence on initial data). Let ๐ โ ๐ โ ๐๐ be open subsets. Let๐ be a ๐ถโ-vector field on ๐, ๐ผ โ ๐ an open interval, ๐ โ ๐ผ, and โโถ ๐ ร ๐ผ โ ๐ a map withthe following properties.(1) โ(๐, ๐) = ๐.(2) For all ๐ โ ๐ the curve
๐พ๐ โถ ๐ผ โ ๐ , ๐พ๐(๐ก) โ โ(๐, ๐ก)is an integral curve of ๐.
Then the map โ is ๐ถโ.
One important feature of the system (2.2.3) is that it is an autonomous system of differ-ential equations: the function ๐(๐ฅ) is a function of ๐ฅ alone (it does not depend on ๐ก). Oneconsequence of this is the following.
Theorem 2.2.7. Let ๐ผ = (๐, ๐) and for ๐ โ ๐ let ๐ผ๐ = (๐ โ ๐, ๐ โ ๐). Then if ๐พโถ ๐ผ โ ๐ is anintegral curve, the reparametrized curve
(2.2.8) ๐พ๐ โถ ๐ผ๐ โ ๐ , ๐พ๐(๐ก) โ ๐พ(๐ก + ๐)is an integral curve.
We recall that ๐ถ1-function ๐โถ ๐ โ ๐ is an integral of the system (2.2.3) if for everyintegral curve ๐พ(๐ก), the function ๐ก โฆ ๐(๐พ(๐ก)) is constant. This is true if and only if for all ๐ก
0 = ๐๐๐ก๐(๐พ(๐ก)) = (๐ท๐)๐พ(๐ก) (
๐๐พ๐๐ก) = (๐ท๐)๐พ(๐ก)(๐ฃ)
where ๐(๐) = (๐, ๐ฃ). By equation (2.1.6) the term on the right is ๐ฟ๐๐(๐). Hence we conclude:
Theorem 2.2.9. Let ๐ be an open subset of ๐๐ and ๐ โ ๐ถ1(๐). Then ๐ is an integral of thesystem (2.2.3) if and only if ๐ฟ๐๐ = 0.
Definition 2.2.10. Let๐ be an open subset of ๐๐ and ๐ a vector field on๐. We say that ๐ iscomplete if for every ๐ โ ๐ there exists an integral curve ๐พโถ ๐ โ ๐ with ๐พ(0) = ๐, i.e., forevery ๐ there exists an integral curve that starts at ๐ and exists for all time.
To see what completeness involves, we recall that an integral curve
๐พโถ [0, ๐) โ ๐ ,
with ๐พ(0) = ๐ is called maximal if it cannot be extended to an interval [0, ๐โฒ), where ๐โฒ > ๐.(See for instance [2, ยง6.11].) For such curves it is known that either(1) ๐ = +โ,(2) |๐พ(๐ก)| โ +โ as ๐ก โ ๐,(3) or the limit set of
{ ๐พ(๐ก) | 0 โค ๐ก < ๐ }contains points on the boundary of ๐.Hence if we can exclude (2) and (3), we have shown that an integral curve with ๐พ(0) = ๐
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 ๐ถ1-function๐โถ ๐ โ ๐ with ๐ฟ๐๐ = 0.
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ยง2.2 Integral Curves for Vector Fields 39
Proof. The identity ๐ฟ๐๐ = 0 implies that ๐ is constant on ๐พ(๐ก), but if ๐(๐) = ๐ this impliesthat the curve ๐พ(๐ก) lies on the compact subset ๐โ1(๐) โ ๐, hence it cannot โrun off to infinityโas in scenario (2) or โrun off the boundaryโ as in scenario (3). โก
Applying a similar argument to the interval (โ๐, 0] we conclude:
Theorem 2.2.12. Suppose there exists a proper ๐ถ1-function ๐โถ ๐ โ ๐ with the property๐ฟ๐๐ = 0. Then the vector field ๐ is complete.
Example 2.2.13. Let ๐ = ๐2 and let ๐ฃ be the vector field
๐ฃ = ๐ฅ3 ๐๐๐ฆโ ๐ฆ ๐๐๐ฅ
.
Then ๐(๐ฅ, ๐ฆ) = 2๐ฆ2 + ๐ฅ4 is a proper function with the property above.
Another hypothesis on ๐ which excludes (2) and (3) is the following.
Definition 2.2.14. The support of ๐ is set
supp(๐) = { ๐ โ ๐ | ๐(๐) โ 0 } .
We say that ๐ is compactly supported if supp(๐) is compact.
We now show that compactly supported vector fields are complete.
Theorem 2.2.15. If a vector field ๐ is compactly supported, then ๐ is complete.
Proof. Notice first that if ๐(๐) = 0, the constant curve, ๐พ0(๐ก) = ๐, โโ < ๐ก < โ, satisfies theequation
๐๐๐ก๐พ0(๐ก) = 0 = ๐(๐) ,
so it is an integral curve of ๐. Hence if ๐พ(๐ก), โ๐ < ๐ก < ๐, is any integral curve of ๐ with theproperty ๐พ(๐ก0) = ๐ for some ๐ก0, it has to coincide with ๐พ0 on the interval (โ๐, ๐), and hencehas to be the constant curve ๐พ(๐ก) = ๐ on (โ๐, ๐).
Now suppose the support supp(๐) of ๐ is compact. Then either ๐พ(๐ก) is in supp(๐) for all๐ก or is in ๐ โ supp(๐) for some ๐ก0. But if this happens, and ๐ = ๐พ(๐ก0) then ๐(๐) = 0, so ๐พ(๐ก)has to coincide with the constant curve ๐พ0(๐ก) = ๐, for all ๐ก. In neither case can it go off toโor off to the boundary of ๐ as ๐ก โ ๐. โก
One useful application of this result is the following. Suppose ๐ is a vector field on๐ โ ๐๐, and one wants to see what its integral curves look like on some compact set ๐ด โ ๐.Let ๐ โ ๐ถโ0 (๐) be a bump function which is equal to one on a neighborhood of ๐ด. Thenthe vector field ๐ = ๐๐ is compactly supported and hence complete, but it is identical with๐ on ๐ด, so its integral curves on ๐ด coincide with the integral curves of ๐.
If ๐ is complete then for every ๐, one has an integral curve ๐พ๐ โถ ๐ โ ๐ with ๐พ๐(0) = ๐,so one can define a map
๐๐ก โถ ๐ โ ๐by setting๐๐ก(๐) โ ๐พ๐(๐ก). If ๐ is๐ถโ, thismapping is๐ถโ by the smooth dependence on initialdata theorem, and, by definition, ๐0 = id๐. We claim that the ๐๐กโs also have the property
(2.2.16) ๐๐ก โ ๐๐ = ๐๐ก+๐ .
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40 Chapter 2: Differential Forms
Indeed if ๐๐(๐) = ๐, then by the reparametrization theorem, ๐พ๐(๐ก) and ๐พ๐(๐ก + ๐) are bothintegral curves of ๐, and since ๐ = ๐พ๐(0) = ๐พ๐(๐) = ๐๐(๐), they have the same initial point,so
๐พ๐(๐ก) = ๐๐ก(๐) = (๐๐ก โ ๐๐)(๐)= ๐พ๐(๐ก + ๐) = ๐๐ก+๐(๐)
for all ๐ก. Since ๐0 is the identity it follows from (2.2.16) that ๐๐ก โ ๐โ๐ก is the identity, i.e.,
๐โ๐ก = ๐โ1๐ก ,
so ๐๐ก is a ๐ถโ diffeomorphism. Hence if ๐ is complete it generates a one-parameter group ๐๐ก,โโ < ๐ก < โ, of ๐ถโ-diffeomorphisms of ๐.
If ๐ is not complete there is an analogous result, but it is trickier to formulate precisely.Roughly speaking ๐ generates a one-parameter group of diffeomorphisms ๐๐ก but these dif-feomorphisms are not defined on all of ๐ nor for all values of ๐ก. Moreover, the identity(2.2.16) only holds on the open subset of ๐ where both sides are well-defined.
We devote the remainder of this section to discussing some โfunctorialโ properties ofvector fields and one-forms.
Definition 2.2.17. Let๐ โ ๐๐ and๐ โ ๐๐ be open, and let ๐โถ ๐ โ ๐ be a ๐ถโ map. If ๐is a ๐ถโ-vector field on ๐ and ๐ a ๐ถโ-vector field on๐ we say that ๐ and ๐ are ๐-relatedif, for all ๐ โ ๐ we have
๐๐๐(๐(๐)) = ๐(๐(๐)) .
Writing
๐ =๐โ๐=1๐ฃ๐๐๐๐ฅ๐
, ๐ฃ๐ โ ๐ถ๐(๐)
and
๐ =๐โ๐=1๐ค๐๐๐๐ฆ๐
, ๐ค๐ โ ๐ถ๐(๐)
this equation reduces, in coordinates, to the equation
๐ค๐(๐) =๐โ๐=1
๐๐๐๐๐ฅ๐(๐)๐ฃ๐(๐) .
In particular, if๐ = ๐ and๐ is a๐ถโ diffeomorphism, the formula (3.2) defines a๐ถโ-vectorfield on๐, i.e.,
๐ =๐โ๐=1๐ค๐๐๐๐ฆ๐
is the vector field defined by the equation
๐ค๐ =๐โ๐=1(๐๐๐๐๐ฅ๐๐ฃ๐) โ ๐โ1 .
Hence we have proved:
Theorem 2.2.18. If ๐โถ ๐ โฅฒ ๐ is a ๐ถโ diffeomorphism and ๐ a ๐ถโ-vector field on๐, thereexists a unique ๐ถโ vector field ๐ on๐ having the property that ๐ and ๐ are ๐-related.
Definition 2.2.19. In the setting of Theorem 2.2.18, we denote the vector field ๐ by ๐โ๐,and call ๐โ๐ the pushforward of ๐ by ๐ .
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ยง2.2 Integral Curves for Vector Fields 41
We leave the following assertions as easy exercises (but provide some hints).
Theorem 2.2.20. For ๐ = 1, 2, let ๐๐ be an open subset of ๐๐๐ , ๐๐ a vector field on ๐๐ and๐โถ ๐1 โ ๐2 a ๐ถโ-map. If ๐1 and ๐2 are ๐-related, every integral curve
๐พโถ ๐ผ โ ๐1of ๐1 gets mapped by ๐ onto an integral curve ๐ โ ๐พโถ ๐ผ โ ๐2 of ๐2.
Proof sketch. Theorem 2.2.20 follows from the chain rule: if ๐ = ๐พ(๐ก) and ๐ = ๐(๐)
๐๐๐ (๐๐๐ก๐พ(๐ก)) = ๐
๐๐ก๐(๐พ(๐ก)) . โก
Corollary 2.2.21. In the setting of Theorem 2.2.20, suppose ๐1 and ๐2 are complete. Let(๐๐,๐ก)๐กโ๐ โถ ๐๐ โฅฒ ๐๐ be the one-parameter group of diffeomorphisms generated by ๐๐. Then๐ โ ๐1,๐ก = ๐2,๐ก โ ๐.
Proof sketch. To deduce Corollary 2.2.21 from Theorem 2.2.20 note that for ๐ โ ๐, ๐1,๐ก(๐)is just the integral curve ๐พ๐(๐ก) of ๐1 with initial point ๐พ๐(0) = ๐.
The notion of ๐-relatedness can be very succinctly expressed in terms of the Lie differ-entiation operation. For ๐ โ ๐ถโ(๐2) let ๐โ๐ be the composition, ๐ โ ๐, viewed as a ๐ถโfunction on ๐1, i.e., for ๐ โ ๐1 let ๐โ๐(๐) = ๐(๐(๐)). Then
๐โ๐ฟ๐2๐ = ๐ฟ๐1๐โ๐ .
To see this, note that if ๐(๐) = ๐ then at the point ๐ the right hand side is(๐๐)๐ โ ๐๐๐(๐1(๐))
by the chain rule and by definition the left hand side is๐๐๐(๐2(๐)) .
Moreover, by definition๐2(๐) = ๐๐๐(๐1(๐))
so the two sides are the same. โก
Another easy consequence of the chain rule is:
Theorem 2.2.22. For ๐ = 1, 2, 3, let ๐๐ be an open subset of ๐๐๐ , ๐๐ a vector field on ๐๐, andfor ๐ = 1, 2 let ๐๐ โถ ๐๐ โ ๐๐+1 be a ๐ถโ-map. Suppose that ๐1 and ๐2 are ๐1-related and that ๐2and ๐3 are ๐2-related. Then ๐1 and ๐3 are (๐2 โ ๐1)-related.
In particular, if ๐1 and ๐2 are diffeomorphisms writing ๐ โ ๐1 we have(๐2)โ(๐1)โ๐ = (๐2 โ ๐1)โ๐ .
The results we described above have โdualโ analogues for one-forms.
Construction 2.2.23. Let ๐ โ ๐๐ and ๐ โ ๐๐ be open, and let ๐โถ ๐ โ ๐ be a ๐ถโ-map.Given a one-form ๐ on ๐ one can define a pullback one-form ๐โ๐ on ๐ by the followingmethod. For ๐ โ ๐, by definition ๐๐(๐) is a linear map
๐๐(๐) โถ ๐๐(๐)๐๐ โ ๐and by composing this map with the linear map
๐๐๐ โถ ๐๐๐๐ โ ๐๐(๐)๐๐
we get a linear map๐๐(๐) โ ๐๐๐ โถ ๐๐๐๐ โ ๐ ,
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42 Chapter 2: Differential Forms
i.e., an element ๐๐(๐) โ ๐๐๐ of ๐โ๐ ๐๐.The pullback one-form ๐โ๐ is defined by the assignment ๐ โฆ (๐๐(๐) โ ๐๐๐).
In particular, if ๐โถ ๐ โ ๐ is a ๐ถโ-function and ๐ = ๐๐ then๐๐ โ ๐๐๐ = ๐๐๐ โ ๐๐๐ = ๐(๐ โ ๐)๐
i.e.,๐โ๐ = ๐๐ โ ๐ .
In particular if ๐ is a one-form of the form ๐ = ๐๐, with ๐ โ ๐ถโ(๐), ๐โ๐ is ๐ถโ. From thisit is easy to deduce:
Theorem 2.2.24. If ๐ is any ๐ถโ one-form on๐, its pullback ๐โ๐ is ๐ถโ. (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 ๐ถโ maps,but the latter is only defined for diffeomorphisms.
Exercises for ยง2.2Exercise 2.2.i. Prove the reparameterization result Theorem 2.2.20.
Exercise 2.2.ii. Let๐ be an open subset of ๐๐,๐ an open subset of ๐๐ and ๐โถ ๐ โ ๐ a ๐ถ๐map.(1) Show that for ๐ โ ๐ถโ(๐) (2.2) can be rewritten
๐โ๐๐ = ๐๐โ๐ .(2) Let ๐ be the one-form
๐ =๐โ๐=1๐๐๐๐ฅ๐ , ๐๐ โ ๐ถโ(๐)
on ๐. Show that if ๐ = (๐1,โฆ, ๐๐) then
๐โ๐ =๐โ๐=1๐โ๐๐ ๐๐๐ .
(3) Show that if ๐ is ๐ถโ and ๐ is ๐ถโ, ๐โ๐ is ๐ถโ.
Exercise 2.2.iii. Let ๐ be a complete vector field on ๐ and ๐๐ก โถ ๐ โ ๐, the one parametergroup of diffeomorphisms generated by ๐. Show that if ๐ โ ๐ถ1(๐)
๐ฟ๐๐ =๐๐๐ก๐โ๐ก ๐ |๐ก=0
.
Exercise 2.2.iv.(1) Let ๐ = ๐2 and let ๐ be the vector field, ๐ฅ1๐/๐๐ฅ2 โ ๐ฅ2๐/๐๐ฅ1. Show that the curve
๐ก โฆ (๐ cos(๐ก + ๐), ๐ sin(๐ก + ๐)) ,for ๐ก โ ๐, is the unique integral curve of ๐ passing through the point, (๐ cos ๐, ๐ sin ๐),at ๐ก = 0.
(2) Let ๐ = ๐๐ and let ๐ be the constant vector field: โ๐๐=1 ๐๐๐/๐๐ฅ๐. Show that the curve
๐ก โฆ ๐ + ๐ก(๐1,โฆ, ๐๐) ,for ๐ก โ ๐, is the unique integral curve of ๐ passing through ๐ โ ๐๐ at ๐ก = 0.
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ยง2.2 Integral Curves for Vector Fields 43
(3) Let ๐ = ๐๐ and let ๐ be the vector field, โ๐๐=1 ๐ฅ๐๐/๐๐ฅ๐. Show that the curve
๐ก โฆ ๐๐ก(๐1,โฆ, ๐๐) ,for ๐ก โ ๐, is the unique integral curve of ๐ passing through ๐ at ๐ก = 0.
Exercise 2.2.v. Show that the following are one-parameter groups of diffeomorphisms:(1) ๐๐ก โถ ๐ โ ๐ , ๐๐ก(๐ฅ) = ๐ฅ + ๐ก(2) ๐๐ก โถ ๐ โ ๐ , ๐๐ก(๐ฅ) = ๐๐ก๐ฅ(3) ๐๐ก โถ ๐2 โ ๐2 , ๐๐ก(๐ฅ, ๐ฆ) = (๐ฅ cos(๐ก) โ ๐ฆ sin(๐ก), ๐ฅ sin(๐ก) + ๐ฆ cos(๐ก)).
Exercise 2.2.vi. Let ๐ดโถ ๐๐ โ ๐๐ be a linear mapping. Show that the series
exp(๐ก๐ด) โโโ๐=0
(๐ก๐ด)๐๐!= id๐ +๐ก๐ด +
๐ก22!๐ด2 + ๐ก
3
3!๐ด3 +โฏ
converges and defines a one-parameter group of diffeomorphisms of ๐๐.
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 fieldโ1โค๐,๐โค๐๐๐,๐๐ฅ๐๐๐๐ฅ๐
where (๐๐,๐) is the defining matrix of ๐ด.
Exercise 2.2.viii. Let ๐ be the vector field on ๐ given by ๐ฅ2 ๐๐๐ฅ . Show that the curve
๐ฅ(๐ก) = ๐๐ โ ๐๐ก
is an integral curve of ๐ with initial point ๐ฅ(0) = ๐. Conclude that for ๐ > 0 the curve
๐ฅ(๐ก) = ๐1 โ ๐๐ก
, 0 < ๐ก < 1๐
is a maximal integral curve. (In particular, conclude that ๐ is not complete.)
Exercise 2.2.ix. Let ๐ and ๐ be open subsets of ๐๐ and ๐โถ ๐ โฅฒ ๐ a diffeomorphism. If๐is a vector field on ๐, define the pullback of ๐ to ๐ to be the vector field
๐โ๐ โ (๐โ1โ ๐) .Show that if ๐ is a ๐ถโ function on ๐
๐โ๐ฟ๐๐ = ๐ฟ๐โ๐๐โ๐ .Hint: Equation (2.2.25).
Exercise 2.2.x. Let ๐ be an open subset of ๐๐ and ๐ and ๐ vector fields on ๐. Suppose ๐ isthe infinitesimal generator of a one-parameter group of diffeomorphisms
๐๐ก โถ ๐ โฅฒ ๐ , โโ < ๐ก < โ .Let ๐๐ก = ๐โ๐ก ๐. Show that for ๐ โ ๐ถโ(๐) we have
๐ฟ[๐,๐]๐ = ๐ฟ๏ฟฝ๏ฟฝ๐ ,where
๏ฟฝ๏ฟฝ = ๐๐๐ก๐โ๐ก ๐ |
๐ก=0.
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44 Chapter 2: Differential Forms
Hint: Differentiate the identity๐โ๐ก ๐ฟ๐๐ = ๐ฟ๐๐ก๐
โ๐ก ๐
with respect to ๐ก and show that at ๐ก = 0 the derivative of the left hand side is ๐ฟ๐๐ฟ๐๐ byExercise 2.2.iii, and the derivative of the right hand side is
๐ฟ๏ฟฝ๏ฟฝ + ๐ฟ๐(๐ฟ๐๐) .
Exercise 2.2.xi. Conclude from Exercise 2.2.x that
(2.2.25) [๐, ๐] = ๐๐๐ก๐โ๐ก ๐ |
๐ก=0.
2.3. Differential ๐-forms
One-forms are the bottom tier in a pyramid of objects whose ๐th tier is the space of ๐-forms. More explicitly, given ๐ โ ๐๐ we can, as in Section 1.5, form the ๐th exterior powers
๐ฌ๐(๐โ๐ ๐๐) , ๐ = 1, 2, 3,โฆ, ๐of the vector space ๐โ๐ ๐๐, and since
(2.3.1) ๐ฌ1(๐โ๐ ๐๐) = ๐โ๐ ๐๐
one can think of a one-form as a function which takes its value at ๐ in the space (2.3.1). Thisleads to an obvious generalization.
Definition 2.3.2. Let ๐ be an open subset of ๐๐. A ๐-form ๐ on ๐ is a function whichassigns to each point ๐ โ ๐ an element ๐๐ โ ๐ฌ๐(๐โ๐ ๐๐).
The wedge product operation gives us a way to construct lots of examples of such ob-jects.
Example 2.3.3. Let ๐1,โฆ, ๐๐ be one-forms. Then ๐1 โงโฏ โง ๐๐ is the ๐-form whose valueat ๐ is the wedge product(2.3.4) (๐1 โงโฏ โง ๐๐)๐ โ (๐1)๐ โงโฏ โง (๐๐)๐ .
Notice that since (๐๐)๐ is in ๐ฌ1(๐โ๐ ๐๐) the wedge product (2.3.4) makes sense and is anelement of ๐ฌ๐(๐โ๐ ๐๐).
Example 2.3.5. Let ๐1,โฆ, ๐๐ be a real-valued ๐ถโ functions on ๐. Letting ๐๐ = ๐๐๐ we getfrom (2.3.4) a ๐-form
๐๐1 โงโฏ โง ๐๐๐whose value at ๐ is the wedge product(2.3.6) (๐๐1)๐ โงโฏ โง (๐๐๐)๐ .
Since (๐๐ฅ1)๐,โฆ, (๐๐ฅ๐)๐ are a basis of ๐โ๐ ๐๐, the wedge products
(2.3.7) (๐๐ฅ๐1)๐ โงโฏ โง (๐๐ฅ1๐)๐ , 1 โค ๐1 < โฏ < ๐๐ โค ๐
are a basis of๐ฌ๐(๐โ๐ ). To keep our multi-index notation from getting out of hand, we denotethese basis vectors by (๐๐ฅ๐ผ)๐, where ๐ผ = (๐1,โฆ, ๐๐) and the multi-indices ๐ผ range over multi-indices of length ๐ which are strictly increasing. Since these wedge products are a basis of๐ฌ๐(๐โ๐ ๐๐), every element of ๐ฌ๐(๐โ๐ ๐๐) can be written uniquely as a sum
โ๐ผ๐๐ผ(๐๐ฅ๐ผ)๐ , ๐๐ผ โ ๐
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ยง2.3 Differential ๐-forms 45
and every ๐-form ๐ on ๐ can be written uniquely as a sum
(2.3.8) ๐ = โ๐ผ๐๐ผ๐๐ฅ๐ผ
where ๐๐ฅ๐ผ is the ๐-form ๐๐ฅ๐1 โงโฏ โง ๐๐ฅ๐๐ , and ๐๐ผ is a real-valued function ๐๐ผ โถ ๐ โ ๐.Definition 2.3.9. The ๐-form (2.3.8) is of class ๐ถ๐ if each function ๐๐ผ is in ๐ถ๐(๐).
Henceforth we assume, unless otherwise stated, that all the ๐-forms we consider are ofclass ๐ถโ. We denote the set of ๐-forms of class ๐ถโ on ๐ by ๐บ๐(๐).
We will conclude this section by discussing a few simple operations on ๐-forms. Let ๐be an open subset of ๐๐.(1) Given a function ๐ โ ๐ถโ(๐) and a ๐-form ๐ โ ๐บ๐(๐) we define ๐๐ โ ๐บ๐(๐) to be the๐-form defined by
๐ โฆ ๐(๐)๐๐ .(2) Given ๐1, ๐2 โ ๐บ๐(๐), we define ๐1 + ๐2 โ ๐บ๐(๐) to be the ๐-form
๐ โฆ (๐1)๐ + (๐2)๐ .
(Notice that this sum makes sense since each summand is in ๐ฌ๐(๐โ๐ ๐๐).)(3) Given๐1 โ ๐บ๐1(๐) and๐2 โ ๐บ๐2(๐)wedefine theirwedge product,๐1โง๐2 โ ๐บ๐1+๐2(๐)
to be the (๐1 + ๐2)-form๐ โฆ (๐1)๐ โง (๐2)๐ .
We recall that ๐ฌ0(๐โ๐ ๐๐) = ๐, so a zero-form is an ๐-valued function and a zero formof class ๐ถโ is a ๐ถโ function, i.e., ๐บ0(๐) = ๐ถโ(๐).The addition and multiplication by functions operations naturally give the sets ๐บ๐(๐)
the structures of๐-vector spacesโwe always regard๐บ๐(๐) as a vector space in thismanner.A fundamental operation on forms is the exterior differention operation which asso-
ciates to a function ๐ โ ๐ถโ(๐) the 1-form ๐๐. It is clear from the identity (2.2.3) that ๐๐is a 1-form of class ๐ถโ, so the ๐-operation can be viewed as a map
๐โถ ๐บ0(๐) โ ๐บ1(๐) .In the next section we show that an analogue of this map exists for every ๐บ๐(๐).
Exercises for ยง2.3Exercise 2.3.i. Let ๐ โ ๐บ2(๐4) be the 2-form ๐๐ฅ1 โง ๐๐ฅ2 + ๐๐ฅ3 โง ๐๐ฅ4. Compute ๐ โง ๐.Exercise 2.3.ii. Let ๐1, ๐2, ๐3 โ ๐บ1(๐3) be the 1-forms
๐1 = ๐ฅ2๐๐ฅ3 โ ๐ฅ3๐๐ฅ2 ,๐2 = ๐ฅ3๐๐ฅ1 โ ๐ฅ1๐๐ฅ3 ,๐3 = ๐ฅ1๐๐ฅ2 โ ๐ฅ2๐๐ฅ1 .
Compute the following.(1) ๐1 โง ๐2.(2) ๐2 โง ๐3.(3) ๐3 โง ๐1.(4) ๐1 โง ๐2 โง ๐3.Exercise 2.3.iii. Let ๐ be an open subset of ๐๐ and ๐1,โฆ, ๐๐ โ ๐ถโ(๐). Show that
๐๐1 โงโฏ โง ๐๐๐ = det[๐๐๐๐๐ฅ๐]๐๐ฅ1 โงโฏ โง ๐๐ฅ๐ .
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46 Chapter 2: Differential Forms
Exercise 2.3.iv. Let ๐ be an open subset of ๐๐. Show that every (๐ โ 1)-form ๐ โ ๐บ๐โ1(๐)can be written uniquely as a sum
๐โ๐=1๐๐๐๐ฅ1 โงโฏ โง ๐๐ฅ๐ โงโฏ โง ๐๐ฅ๐ ,
where ๐๐ โ ๐ถโ(๐) and ๐๐ฅ๐ indicates that ๐๐ฅ๐ is to be omitted from the wedge product๐๐ฅ1 โงโฏ โง ๐๐ฅ๐.
Exercise 2.3.v. Let ๐ = โ๐๐=1 ๐ฅ๐๐๐ฅ๐. Show that there exists an (๐โ1)-form,๐ โ ๐บ๐โ1(๐๐โ{0})with the property
๐ โง ๐ = ๐๐ฅ1 โงโฏ โง ๐๐ฅ๐ .
Exercise 2.3.vi. Let ๐ฝ be the multi-index (๐1,โฆ, ๐๐) and let ๐๐ฅ๐ฝ = ๐๐ฅ๐1 โง โฏ โง ๐๐ฅ๐๐ . Showthat ๐๐ฅ๐ฝ = 0 if ๐๐ = ๐๐ for some ๐ โ ๐ and show that if the numbers ๐1,โฆ, ๐๐ are all distinct,then
๐๐ฅ๐ฝ = (โ1)๐๐๐ฅ๐ผ ,where ๐ผ = (๐1,โฆ, ๐๐) is the strictly increasing rearrangement of (๐1,โฆ, ๐๐) and ๐ is the per-mutation
(๐1,โฆ, ๐๐) โฆ (๐1,โฆ, ๐๐) .
Exercise 2.3.vii. Let ๐ผ be a strictly increasingmulti-index of length ๐ and ๐ฝ a strictly increas-ing multi-index of length โ. What can one say about the wedge product ๐๐ฅ๐ผ โง ๐๐ฅ๐ฝ?
2.4. Exterior differentiation
Let๐ be an open subset of ๐๐. In this section we define an exterior differentiation oper-ation
(2.4.1) ๐โถ ๐บ๐(๐) โ ๐บ๐+1(๐) .
This operation is the fundamental operation in ๐-dimensional vector calculus.For ๐ = 0 we already defined the operation (2.4.1) in ยง2.1. Before defining it for ๐ > 0,
we list some properties that we require to this operation to satisfy.
Properties 2.4.2 (desired properties of exterior differentiation). Let๐ be an open subset of๐๐.(1) For ๐1 and ๐2 in ๐บ๐(๐), we have
๐(๐1 + ๐2) = ๐๐1 + ๐๐2 .
(2) For ๐1 โ ๐บ๐(๐) and ๐2 โ ๐บโ(๐) we have
(2.4.3) ๐(๐1 โง ๐2) = ๐๐1 โง ๐2 + (โ1)๐๐1 โง ๐๐2 .
(3) For all ๐ โ ๐บ๐(๐) we have
(2.4.4) ๐(๐๐) = 0
Let is point out a few consequences of these properties.
Lemma 2.4.5. Let ๐ be an open subset of ๐๐. For any functions ๐1,โฆ, ๐๐ โ ๐ถโ(๐) we have
๐(๐๐1 โงโฏ โง ๐๐๐) = 0 .
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ยง2.4 Exterior differentiation 47
Proof. We prove this by induction on ๐. For the base case note that by (3)(2.4.6) ๐(๐๐1) = 0for every function ๐1 โ ๐ถโ(๐).
For the induction step, suppose that we know the result for ๐ โ 1 functions and thatwe are given functions ๐1,โฆ, ๐๐ โ ๐ถโ(๐). Let ๐ = ๐๐2 โง โฏ โง ๐๐๐. Then by the inductionhypothesis ๐๐ = 0 Combining (2.4.3) with (2.4.6) we see that
๐(๐๐1 โง ๐๐2 โงโฏ โง ๐๐๐) = ๐(๐๐1 โง ๐)= ๐(๐๐1) โง ๐ + (โ1) ๐๐1 โง ๐๐= 0 . โก
Example 2.4.7. As a special case of Lemma 2.4.5, given a multi-index ๐ผ = (๐1,โฆ, ๐๐) with1 โค ๐๐ โค ๐ we have(2.4.8) ๐(๐๐ฅ๐ผ) = ๐(๐๐ฅ๐1 โงโฏ โง ๐๐ฅ๐๐) = 0 .
Recall that every ๐-form ๐ โ ๐บ๐(๐) can be written uniquely as a sum
๐ = โ๐ผ๐๐ผ๐๐ฅ๐ผ , ๐๐ผ โ ๐ถโ(๐)
where the multi-indices ๐ผ are strictly increasing. Thus by (2.4.3) and (2.4.8)
(2.4.9) ๐๐ = โ๐ผ๐๐๐ผ โง ๐๐ฅ๐ผ .
This shows that if there exists an operator ๐โถ ๐บโ(๐) โ ๐บโ+1(๐)with Properties 2.4.2, then๐ 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 ๐ be an open subset of ๐๐. There is a unique operator ๐โถ ๐บโ(๐) โ๐บโ+1(๐) satisfying Properties 2.4.2.Proof. The property (1) is obvious.
To verify (2) we first note that for ๐ผ strictly increasing equation (2.4.8) is a special caseof equation (2.4.9) (take ๐๐ผ = 1 and ๐๐ฝ = 0 for ๐ฝ โ ๐ผ). Moreover, if ๐ผ is not strictly increasingit is either repeating, in which case ๐๐ฅ๐ผ = 0, or non-repeating in which case there exists apermuation ๐ โ ๐๐ such that ๐ผ๐ is strictly increasing. Moreover,(2.4.11) ๐๐ฅ๐ผ = (โ1)๐๐๐ฅ๐ผ๐ .Hence (2.4.9) implies (2.4.8) for all multi-indices ๐ผ. The same argument shows that for anysum โ๐ผ ๐๐ผ๐๐ฅ๐ผ over multi-indices ๐ผ of length ๐, we have
(2.4.12) ๐(โ๐ผ ๐๐ผ๐๐ฅ๐ผ) = โ๐ผ๐๐๐ผ โง ๐๐ฅ๐ผ .
(As above we can ignore the repeatingmulti-indices ๐๐ฅ๐ผ = 0 if ๐ผ is repeating, and by (2.4.11)we can replace the non-repeating multi-indices by strictly increasing multi-indices.)
Suppose now that ๐1 โ ๐บ๐(๐) and ๐2 โ ๐บโ(๐). Write ๐1 โ โ๐ผ ๐๐ผ๐๐ฅ๐ผ and ๐2 โโ๐ฝ ๐๐ฝ๐๐ฅ๐ฝ, where ๐๐ผ, ๐๐ฝ โ ๐ถโ(๐). We see that the wedge product ๐1 โง ๐2 is given by
(2.4.13) ๐1 โง ๐2 = โ๐ผ,๐ฝ๐๐ผ๐๐ฝ๐๐ฅ๐ผ โง ๐๐ฅ๐ฝ ,
and by equation (2.4.12)
(2.4.14) ๐(๐1 โง ๐2) = โ๐ผ,๐ฝ๐(๐๐ผ๐๐ฝ) โง ๐๐ฅ๐ผ โง ๐๐ฅ๐ฝ .
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48 Chapter 2: Differential Forms
(Notice that if ๐ผ = (๐1,โฆ, ๐๐) and ๐ฝ = (๐๐,โฆ, ๐โ), we have ๐๐ฅ๐ผ โง ๐๐ฅ๐ฝ = ๐๐ฅ๐พ, where ๐พ โ(๐1,โฆ, ๐๐, ๐1,โฆ, ๐โ). Even if ๐ผ and ๐ฝ are strictly increasing, ๐พ 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)
๐(๐๐ผ๐๐ฝ) = ๐๐ฝ๐๐๐ผ + ๐๐ผ๐๐๐ฝ ,
and by the wedge product identities of ยง1.6,
๐๐๐ฝ โง ๐๐ฅ๐ผ = ๐๐๐ฝ โง ๐๐ฅ๐1 โงโฏ โง ๐๐ฅ๐๐ = (โ1)๐๐๐ฅ๐ผ โง ๐๐๐ฝ ,
so the sum (2.4.14) can be rewritten:
โ๐ผ,๐ฝ๐๐๐ผ โง ๐๐ฅ๐ผ โง ๐๐ฝ๐๐ฅ๐ฝ + (โ1)๐โ
๐ผ,๐ฝ๐๐ผ๐๐ฅ๐ผ โง ๐๐๐ฝ โง ๐๐ฅ๐ฝ ,
or
(โ๐ผ๐๐๐ผ โง ๐๐ฅ๐ผ) โง (โ
๐ฝ๐๐ฝ๐๐ฅ๐ฝ) + (โ1)๐(โ
๐ฝ๐๐๐ฝ โง ๐๐ฅ๐ฝ) โง (โ
๐ผ๐๐๐ผ โง ๐๐ฅ๐ผ) ,
or, finally,๐๐1 โง ๐2 + (โ1)๐๐1 โง ๐๐2 .
Thus the the operator ๐ defined by equation (2.4.9) has (2).Let is now check that ๐ satisfies (3). If ๐ = โ๐ผ ๐๐ผ๐๐ฅ๐ผ, where ๐๐ผ โ ๐ถโ(๐), then by
definition, ๐๐ = โ๐ผ ๐๐๐ผ โง ๐๐ฅ๐ผ and by (2.4.8) and (2.4.3)
๐(๐๐) = โ๐ผ๐(๐๐๐ผ) โง ๐๐ฅ๐ผ ,
so it suffices to check that ๐(๐๐๐ผ) = 0, i.e., it suffices to check (2.4.6) for zero forms ๐ โ๐ถโ(๐). However, by (2.2.3) we have
๐๐ =๐โ๐=1
๐๐๐๐ฅ๐๐๐ฅ๐
so by equation (2.4.9)
๐(๐๐) =๐โ๐=1๐( ๐๐๐๐ฅ๐)๐๐ฅ๐ =
๐โ๐=1(๐โ๐=1
๐2๐๐๐ฅ๐๐๐ฅ๐๐๐ฅ๐) โง ๐๐ฅ๐
= โ1โค๐,๐โค๐
๐2๐๐๐ฅ๐๐๐ฅ๐๐๐ฅ๐ โง ๐๐ฅ๐ .
Notice, however, that in this sum, ๐๐ฅ๐ โง ๐๐ฅ๐ = โ๐๐ฅ๐ โง ๐๐ฅ๐ and
๐2๐๐๐ฅ๐๐๐ฅ๐= ๐2๐๐๐ฅ๐๐๐ฅ๐
so the (๐, ๐) term cancels the (๐, ๐) term, and the total sum is zero. โก
Definition 2.4.15. Let ๐ be an open subset of ๐๐. A ๐-form ๐ โ ๐บ๐(๐) is closed if ๐๐ = 0and is exact if ๐ = ๐๐ for some ๐ โ ๐บ๐โ1(๐).
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ยง2.4 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 ๐ has the following property: For ๐ > 0 every closed ๐-form is exact.2
Some examples of spaces with this property are described in the exercises at the end ofยง2.6. We also sketch below a proof of the following result (and ask you to fill in the details).
Lemma 2.4.16 (Poincarรฉ lemma). If ๐ is a closed form on ๐ of degree ๐ > 0, then for everypoint ๐ โ ๐, there exists a neighborhood of ๐ on which ๐ is exact.
Proof. See Exercises 2.4.v and 2.4.vi. โกExercises for ยง2.4
Exercise 2.4.i. Compute the exterior derivatives of the following differential forms.(1) ๐ฅ1๐๐ฅ2 โง ๐๐ฅ3(2) ๐ฅ1๐๐ฅ2 โ ๐ฅ2๐๐ฅ1(3) ๐โ๐๐๐ where ๐ = โ๐๐=1 ๐ฅ2๐(4) โ๐๐=1 ๐ฅ๐๐๐ฅ๐(5) โ๐๐=1(โ1)๐๐ฅ๐๐๐ฅ1 โงโฏ โง ๐๐ฅ๐ โงโฏ โง ๐๐ฅ๐Exercise 2.4.ii. Solve the equation ๐๐ = ๐ for ๐ โ ๐บ1(๐3), where ๐ is the 2-form:(1) ๐๐ฅ2 โง ๐๐ฅ3(2) ๐ฅ2๐๐ฅ2 โง ๐๐ฅ3(3) (๐ฅ21 + ๐ฅ22)๐๐ฅ1 โง ๐๐ฅ2(4) cos(๐ฅ1)๐๐ฅ1 โง ๐๐ฅ3Exercise 2.4.iii. Let ๐ be an open subset of ๐๐.(1) Show that if ๐ โ ๐บ๐(๐) is exact and ๐ โ ๐บโ(๐) is closed then ๐ โง ๐ is exact.
Hint: Equation (2.4.3).(2) In particular, ๐๐ฅ1 is exact, so if ๐ โ ๐บโ(๐) is closed ๐๐ฅ1 โง ๐ = ๐๐. What is ๐?Exercise 2.4.iv. Let ๐ be the rectangle (๐1, ๐1) ร โฏ ร (๐๐, ๐๐). Show that if ๐ is in ๐บ๐(๐),then ๐ is exact.
Hint: Let ๐ = ๐๐๐ฅ1 โงโฏ โง ๐๐ฅ๐ with ๐ โ ๐ถโ(๐) and let ๐ be the function
๐(๐ฅ1,โฆ, ๐ฅ๐) = โซ๐ฅ1
๐1๐(๐ก, ๐ฅ2,โฆ, ๐ฅ๐)๐๐ก .
Show that ๐ = ๐(๐๐๐ฅ2 โงโฏ โง ๐๐ฅ๐).Exercise 2.4.v. Let ๐ be an open subset of ๐๐โ1, ๐ด โ ๐ an open interval and (๐ฅ, ๐ก) productcoordinates on ๐ ร ๐ด. We say that a form ๐ โ ๐บโ(๐ ร ๐ด) is reduced if ๐ can be written as asum(2.4.17) ๐ = โ
๐ผ๐๐ผ(๐ฅ, ๐ก)๐๐ฅ๐ผ ,
(i.e., with no terms involving ๐๐ก).(1) Show that every form, ๐ โ ๐บ๐(๐ ร ๐ด) can be written uniquely as a sum:(2.4.18) ๐ = ๐๐ก โง ๐ผ + ๐ฝ
where ๐ผ and ๐ฝ are reduced.
2For ๐ = 0, ๐๐ = 0 does not imply that ๐ is exact. In fact โexactnessโ does not make much sense for zeroforms since there are not any โ(โ1)-formsโ. However, if ๐ โ ๐ถโ(๐) and ๐๐ = 0 then ๐ is constant on connectedcomponents of ๐ (see Exercise 2.2.iii).
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50 Chapter 2: Differential Forms
(2) Let ๐ be the reduced form (2.4.17) and let๐๐๐๐ก= โ ๐๐๐ก๐๐ผ(๐ฅ, ๐ก)๐๐ฅ๐ผ
and
๐๐๐ = โ๐ผ(๐โ๐=1
๐๐๐ฅ๐๐๐ผ(๐ฅ, ๐ก)๐๐ฅ๐) โง ๐๐ฅ๐ผ .
Show that๐๐ = ๐๐ก โง ๐๐
๐๐ก+ ๐๐๐ .
(3) Let ๐ be the form (2.4.18). Show that
๐๐ = โ๐๐ก โง ๐๐๐ผ + ๐๐ก โง๐๐ฝ๐๐ก+ ๐๐๐ฝ
and conclude that ๐ is closed if and only if
(2.4.19) {๐๐ฝ๐๐ก = ๐๐๐ผ๐๐๐ฝ = 0 .
(4) Let ๐ผ be a reduced (๐ โ 1)-form. Show that there exists a reduced (๐ โ 1)-form ๐ suchthat
(2.4.20) ๐๐๐๐ก= ๐ผ .
Hint: Let ๐ผ = โ๐ผ ๐๐ผ(๐ฅ, ๐ก)๐๐ฅ๐ผ and ๐ = โ๐๐ผ(๐ฅ, ๐ก)๐๐ฅ๐ผ. The equation (2.4.20) reducesto the system of equations
(2.4.21) ๐๐๐ก๐๐ผ(๐ฅ, ๐ก) = ๐๐ผ(๐ฅ, ๐ก) .
Let ๐ be a point on the interval, ๐ด, and using calculus show that equation (2.4.21) has aunique solution, ๐๐ผ(๐ฅ, ๐ก), with ๐๐ผ(๐ฅ, ๐) = 0.
(5) Show that if ๐ is the form (2.4.18) and ๐ a solution of (2.4.20) then the form
(2.4.22) ๐ โ ๐๐is reduced.
(6) Let๐พ = โโ๐ผ(๐ฅ, ๐ก)๐๐ฅ๐ผ
be a reduced ๐-form. Deduce from (2.4.19) that if ๐พ is closed then ๐๐พ๐๐ก = 0 and ๐๐๐พ = 0.Conclude that โ๐ผ(๐ฅ, ๐ก) = โ๐ผ(๐ฅ) and that
๐พ = โ๐ผโ๐ผ(๐ฅ)๐๐ฅ๐ผ
is effectively a closed ๐-form on๐. Prove that if every closed ๐-form on๐ is exact, thenevery closed ๐-form on ๐ ร ๐ด is exact.
Hint: Let ๐ be a closed ๐-form on ๐ ร ๐ด and let ๐พ be the form (2.4.22).
Exercise 2.4.vi. Let ๐ โ ๐๐ be an open rectangle. Show that every closed form on ๐ ofdegree ๐ > 0 is exact.
Hint: Let ๐ = (๐1, ๐1) ร โฏ ร (๐๐, ๐๐). Prove this assertion by induction, at the ๐th stageof the induction letting ๐ = (๐1, ๐1) ร โฏ ร (๐๐โ1, ๐๐โ1) and ๐ด = (๐๐, ๐๐).
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ยง2.5 The interior product operation 51
2.5. The interior product operation
In ยง2.1 we explained how to pair a one-form ๐ and a vector field ๐ to get a function ๐๐๐.This pairing operation generalizes.
Definition 2.5.1. Let ๐ be an open subset of ๐๐, ๐ a vector field on ๐, and ๐ โ ๐บ๐(๐)Theinterior product of ๐ with ๐ is (๐ โ 1)-form ๐๐๐ on ๐ defined by declaring the value of ๐๐๐value at ๐ โ ๐ to be the interior product(2.5.2) ๐๐(๐)๐๐ .
Note that ๐(๐) is in๐๐๐๐ and๐๐ in๐ฌ๐(๐โ๐ ๐๐), so by definition of interior product (see ยง1.7),the expression (2.5.2) is an element of ๐ฌ๐โ1(๐โ๐ ๐๐).
From the properties of interior product on vector spaces which we discussed in ยง1.7,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. Let๐ be an open subset of๐๐, ๐ and๐ vector fields on๐, ๐1 and ๐2 both๐-forms on ๐, and ๐ a ๐-form and ๐ an โ-form on ๐.(1) Linearity in the form: we have
๐๐(๐1 + ๐2) = ๐๐๐1 + ๐๐๐2 .(2) Linearity in the vector field: we have
๐๐+๐๐ = ๐๐๐ + ๐๐๐ .(3) Derivation property: we have
๐๐(๐ โง ๐) = ๐๐๐ โง ๐ + (โ1)๐๐ โง ๐๐๐(4) The interior product satisfies the identity
๐๐(๐๐๐) = โ๐๐(๐๐๐) .(5) As a special case of (4), the interior product satisfies the identity
๐๐(๐๐๐) = 0 .(6) Moreover, if ๐ is decomposable, i.e., is a wedge product of one-forms
๐ = ๐1 โงโฏ โง ๐๐ ,then
๐๐๐ =๐โ๐=1(โ1)๐โ1๐๐(๐๐)๐1 โงโฏ โง ๏ฟฝ๏ฟฝ๐ โงโฏ โง ๐๐ .
We also leave for you to prove the following two assertions, both of which are specialcases of (6).
Example 2.5.4. If ๐ = ๐/๐๐ฅ๐ and ๐ = ๐๐ฅ๐ผ = ๐๐ฅ๐1 โงโฏ โง ๐๐ฅ๐๐ then
(2.5.5) ๐๐๐ =๐โ๐=1(โ1)๐โ1๐ฟ๐,๐๐๐๐ฅ๐ผ๐
where
๐ฟ๐,๐๐ โ {1, ๐ = ๐๐0, ๐ โ ๐๐ .
and ๐ผ๐ = (๐1,โฆ, ๐ค๐,โฆ, ๐๐).
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52 Chapter 2: Differential Forms
Example 2.5.6. If ๐ = โ๐๐=1 ๐๐ ๐/๐๐ฅ๐ and ๐ = ๐๐ฅ1 โงโฏ โง ๐๐ฅ๐ then
(2.5.7) ๐๐๐ =๐โ๐=1(โ1)๐โ1๐๐ ๐๐ฅ1 โงโฏ๐๐ฅ๐โฏโง ๐๐ฅ๐ .
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., for๐ถโ functions, we defined this operation by the formula (2.1.16). For ๐-formswe define it by a slightly more complicated formula.Definition 2.5.8. Let ๐ be an open subset of ๐๐, ๐ a vector field on ๐, and ๐ โ ๐บ๐(๐). TheLie derivative of ๐ with respect to ๐ is the ๐-form(2.5.9) ๐ฟ๐๐ โ ๐๐(๐๐) + ๐(๐๐๐) .
Notice that for zero-forms the second summand is zero, so (2.5.9) and (2.1.16) agree.Properties 2.5.10. Let ๐ be an open subset of ๐๐, ๐ a vector field on ๐, ๐ โ ๐บ๐(๐), and๐ โ ๐บโ(๐). The Lie derivative enjoys the following properties:(1) Commutativity with exterior differentiation: we have(2.5.11) ๐(๐ฟ๐๐) = ๐ฟ๐(๐๐) .(2) Interaction with wedge products: we have(2.5.12) ๐ฟ๐(๐ โง ๐) = ๐ฟ๐๐ โง ๐ + ๐ โง ๐ฟ๐๐
From Properties 2.5.10 it is fairly easy to get an explicit formula for ๐ฟ๐๐. Namely let ๐be the ๐-form
๐ = โ๐ผ๐๐ผ๐๐ฅ๐ผ , ๐๐ผ โ ๐ถโ(๐)
and ๐ the vector field
๐ =๐โ๐=1๐๐๐๐๐ฅ๐
, ๐๐ โ ๐ถโ(๐) .
By equation (2.5.12)๐ฟ๐(๐๐ผ๐๐ฅ๐ผ) = (๐ฟ๐๐๐ผ)๐๐ฅ๐ผ + ๐๐ผ(๐ฟ๐๐๐ฅ๐ผ)
and
๐ฟ๐๐๐ฅ๐ผ =๐โ๐=1๐๐ฅ๐1 โงโฏ โง ๐ฟ๐๐๐ฅ๐๐ โงโฏ โง ๐๐ฅ๐๐ ,
and by equation (2.5.11)๐ฟ๐๐๐ฅ๐๐ = ๐๐ฟ๐๐ฅ๐๐
so to compute ๐ฟ๐๐ one is reduced to computing ๐ฟ๐๐ฅ๐๐ and ๐ฟ๐๐๐ผ.However by equation (2.5.12)
๐ฟ๐๐ฅ๐๐ = ๐๐๐and
๐ฟ๐๐๐ผ =๐โ๐=1๐๐๐๐๐ผ๐๐ฅ๐
.
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. Let๐ โ ๐๐ be open, let ๐1,โฆ, ๐๐ โ ๐ถโ(๐), and set ๐ โ โ๐๐=1 ๐๐ ๐/๐๐ฅ๐. Then
(2.5.14) ๐ฟ๐(๐๐ฅ1 โงโฏ โง ๐๐ฅ๐) =๐โ๐=1(๐๐๐๐๐ฅ๐)๐๐ฅ1 โงโฏ โง ๐๐ฅ๐ .
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ยง2.5 The interior product operation 53
Exercises for ยง2.5Exercise 2.5.i. Verify the Properties 2.5.3 (1)โ(6).
Exercise 2.5.ii. Show that if ๐ is the ๐-form ๐๐ฅ๐ผ and ๐ the vector field ๐/๐๐ฅ๐, then ๐๐๐ isgiven by equation (2.5.5).
Exercise 2.5.iii. Show that if๐ is the ๐-form๐๐ฅ1โงโฏโง๐๐ฅ๐ and ๐ the vector fieldโ๐๐=1 ๐๐๐/๐๐ฅ๐,then ๐๐๐ is given by equation (2.5.7).
Exercise 2.5.iv. Let ๐ be an open subset of ๐๐ and ๐ a ๐ถโ vector field on ๐. Show that for๐ โ ๐บ๐(๐)
๐๐ฟ๐๐ = ๐ฟ๐๐๐and
๐๐(๐ฟ๐๐) = ๐ฟ๐(๐๐๐) .Hint: Deduce the first of these identities from the identity ๐(๐๐) = 0 and the second
from the identity ๐๐(๐๐๐) = 0.
Exercise 2.5.v. Given ๐๐ โ ๐บ๐๐(๐), for ๐ = 1, 2 show that๐ฟ๐(๐1 โง ๐2) = ๐ฟ๐๐1 โง ๐2 + ๐1 โง ๐ฟ๐๐2 .
Hint: Plug ๐ = ๐1 โง ๐2 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 ๐1 and ๐2 be vector fields on๐ and let๐ be their Lie bracket. Show thatfor ๐ โ ๐บ๐(๐)
๐ฟ๐๐ = ๐ฟ๐1(๐ฟ๐2๐) โ ๐ฟ๐2(๐ฟ๐1๐) .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 ๐ฟ๐ satisfies the product identity (2.5.12).
Exercise 2.5.vii. Prove the divergence formula (2.5.14).
Exercise 2.5.viii.(1) Let ๐ = ๐บ๐(๐๐) be the form
๐ = โ๐ผ๐๐ผ(๐ฅ1,โฆ, ๐ฅ๐)๐๐ฅ๐ผ
and ๐ the vector field ๐/๐๐ฅ๐. Show that
๐ฟ๐๐ = โ๐๐๐ฅ๐๐๐ผ(๐ฅ1,โฆ, ๐ฅ๐)๐๐ฅ๐ผ .
(2) Suppose ๐๐๐ = ๐ฟ๐๐ = 0. Show that ๐ only depends on ๐ฅ1,โฆ, ๐ฅ๐โ1 and ๐๐ฅ1,โฆ, ๐๐ฅ๐โ1,i.e., is effectively a ๐-form on ๐๐โ1.
(3) Suppose ๐๐๐ = ๐๐ = 0. Show that ๐ is effectively a closed ๐-form on ๐๐โ1.(4) Use these results to give another proof of the Poincarรฉ lemma for๐๐. Prove by induction
on ๐ that every closed form on ๐๐ is exact.Hints:
โข Let ๐ be the form in part (1) and let
๐๐ผ(๐ฅ1,โฆ, ๐ฅ๐) = โซ๐ฅ๐
0๐๐ผ(๐ฅ1,โฆ, ๐ฅ๐โ1, ๐ก)๐๐ก .
Show that if ๐ = โ๐ผ ๐๐ผ๐๐ฅ๐ผ, then ๐ฟ๐๐ = ๐.
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54 Chapter 2: Differential Forms
โข Conclude that(2.5.15) ๐ โ ๐๐๐๐ = ๐๐ ๐๐ .
โข Suppose ๐๐ = 0. Conclude from equation (2.5.15) and (5) of Properties 2.5.3 thatthe form ๐ฝ = ๐๐๐๐ satisfies ๐๐ฝ = ๐(๐)๐ฝ = 0.
โข By part (3), ๐ฝ is effectively a closed form on ๐๐โ1, and by induction, ๐ฝ = ๐๐ผ. Thusby equation (2.5.15)
๐ = ๐๐๐๐ + ๐๐ผ .
2.6. The pullback operation on forms
Let๐ be an open subset of๐๐,๐ an open subset of๐๐ and๐โถ ๐ โ ๐ a๐ถโmap. Thenfor ๐ โ ๐ and the derivative of ๐ at ๐
๐๐๐ โถ ๐๐๐๐ โ ๐๐(๐)๐๐
is a linear map, so (as explained in ยง1.7) we get from a pullback map
(2.6.1) ๐๐โ๐ โ (๐๐๐)โ โถ ๐ฌ๐(๐โ๐(๐)๐๐) โ ๐ฌ๐(๐โ๐ ๐๐) .In particular, let ๐ be a ๐-form on ๐. Then at ๐(๐) โ ๐, ๐ takes the value
๐๐(๐) โ ๐ฌ๐(๐โ๐ ๐๐) ,so we can apply to it the operation (2.6.1), and this gives us an element:
(2.6.2) ๐๐โ๐ ๐๐(๐) โ ๐ฌ๐(๐โ๐ ๐๐) .In fact we can do this for every point ๐ โ ๐, and the assignment
(2.6.3) ๐ โฆ (๐๐๐)โ๐๐(๐)defines a ๐-form on ๐, which we denote by ๐โ๐. We call ๐โ๐ the pullback of ๐ along themap ๐. A few of its basic properties are described below.
Properties 2.6.4. Let ๐ be an open subset of ๐๐, ๐ an open subset of ๐๐ and ๐โถ ๐ โ ๐ a๐ถโ map.(1) Let ๐ be a zero-form, i.e., a function ๐ โ ๐ถโ(๐). Since
๐ฌ0(๐โ๐ ) = ๐ฌ0(๐โ๐(๐)) = ๐the map (2.6.1) is just the identity map of ๐ onto ๐ when ๐ = 0. Hence for zero-forms
(2.6.5) (๐โ๐)(๐) = ๐(๐(๐)) ,that is, ๐โ๐ is just the composite function, ๐ โ ๐ โ ๐ถโ(๐).
(2) Let ๐ โ ๐บ0(๐) and let ๐ โ ๐บ1(๐) be the 1-form ๐ = ๐๐. By the chain rule (2.6.2)unwinds to:
(2.6.6) (๐๐๐)โ๐๐๐ = (๐๐)๐ โ ๐๐๐ = ๐(๐ โ ๐)๐and hence by (2.6.5) we have
(2.6.7) ๐โ ๐๐ = ๐๐โ๐ .(3) If ๐1, ๐2 โ ๐บ๐(๐) from equation (2.6.2) we see that
(๐๐๐)โ(๐1 + ๐2)๐ = (๐๐๐)โ(๐1)๐ + (๐๐๐)โ(๐2)๐ ,and hence by eq:2.5.3 we have
๐โ(๐1 + ๐2) = ๐โ๐1 + ๐โ๐2 .
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ยง2.6 The pullback operation on forms 55
(4) We observed in ยง1.7 that the operation (2.6.1) commutes with wedge-product, hence if๐1 โ ๐บ๐(๐) and ๐2 โ ๐บโ(๐)
๐๐โ๐ (๐1)๐ โง (๐2)๐ = ๐๐โ๐ (๐1)๐ โง ๐๐โ๐ (๐2)๐ .In other words
(2.6.8) ๐โ๐1 โง ๐2 = ๐โ๐1 โง ๐โ๐2 .(5) Let๐ be an open subset of ๐๐ and ๐โถ ๐ โ ๐ a ๐ถโ map. Given a point ๐ โ ๐, let๐ = ๐(๐) and ๐ค = ๐(๐). Then the composition of the map
(๐๐๐)โ โถ ๐ฌ๐(๐โ๐ ) โ ๐ฌ๐(๐โ๐ )and the map
(๐๐๐)โ โถ ๐ฌ๐(๐โ๐ค) โ ๐ฌ๐(๐โ๐ )is the map
(๐๐๐ โ ๐๐๐)โ โถ ๐ฌ๐(๐โ๐ค) โ ๐ฌ๐(๐โ๐ )by formula (1.7.5). However, by the chain rule
(๐๐๐) โ (๐๐)๐ = ๐(๐ โ ๐)๐so this composition is the map
๐(๐ โ ๐)โ๐ โถ ๐ฌ๐(๐โ๐ค) โ ๐ฌ๐(๐โ๐ ) .
Thus if ๐ โ ๐บ๐(๐)๐โ(๐โ๐) = (๐ โ ๐)โ๐ .
Let us see what the pullback operation looks like in coordinates. Using multi-indexnotation we can express every ๐-form ๐ โ ๐บ๐(๐) as a sum over multi-indices of length ๐(2.6.9) ๐ = โ
๐ผ๐๐ผ๐๐ฅ๐ผ ,
the coefficient ๐๐ผ of ๐๐ฅ๐ผ being in ๐ถโ(๐). Hence by (2.6.5)
๐โ๐ = โ๐โ๐๐ผ๐โ(๐๐ฅ๐ผ)where ๐โ๐๐ผ is the function of ๐ โ ๐.
What about ๐โ๐๐ฅ๐ผ? If ๐ผ is the multi-index, (๐1,โฆ, ๐๐), then by definition๐๐ฅ๐ผ = ๐๐ฅ๐1 โงโฏ โง ๐๐ฅ๐๐
so๐โ(๐๐ฅ๐ผ) = ๐โ(๐๐ฅ๐1) โง โฏ โง ๐
โ(๐๐ฅ๐๐)by (2.6.8), and by (2.6.7), we have
๐โ๐๐ฅ๐ = ๐๐โ๐ฅ๐ = ๐๐๐where ๐๐ is the ๐th coordinate function of the map ๐. Thus, setting
๐๐๐ผ โ ๐๐๐1 โงโฏ โง ๐๐๐๐ ,we see that for each multi-index ๐ผ we have(2.6.10) ๐โ(๐๐ฅ๐ผ) = ๐๐๐ผ ,and for the pullback of the form (2.6.9)
(2.6.11) ๐โ๐ = โ๐ผ๐โ๐๐ผ ๐๐๐ผ .
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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) ๐(๐โ๐) = ๐โ(๐๐) .To prove this we recall that by (2.3.6) we have ๐(๐๐๐ผ) = 0, hence by (2.3.1) and (2.6.10) wehave
๐(๐โ๐) = โ๐ผ๐(๐โ๐๐ผ) โง ๐๐๐ผ = โ
๐ผ๐โ(๐๐๐ผ) โง ๐โ(๐๐ฅ๐ผ)
= ๐โโ๐ผ๐๐๐ผ โง ๐๐ฅ๐ผ = ๐โ(๐๐) .
A special case of formula (2.6.10) will be needed in Chapter 4: Let ๐ and ๐ be opensubsets of ๐๐ and let ๐ = ๐๐ฅ1 โงโฏ โง ๐๐ฅ๐. Then by (2.6.10) we have
๐โ๐๐ = (๐๐1)๐ โงโฏ โง (๐๐๐)๐for all ๐ โ ๐. However,
(๐๐๐)๐ = โ๐๐๐๐๐ฅ๐(๐)(๐๐ฅ๐)๐
and hence by formula (1.7.10)
๐โ๐๐ = det[๐๐๐๐๐ฅ๐(๐)] (๐๐ฅ1 โงโฏ โง ๐๐ฅ๐)๐ .
In other words
(2.6.13) ๐โ(๐๐ฅ1 โงโฏ โง ๐๐ฅ๐) = det[๐๐๐๐๐ฅ๐]๐๐ฅ1 โงโฏ โง ๐๐ฅ๐ .
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 ๐ be an open subset of ๐๐, ๐ an open subset of ๐๐, ๐ด โ ๐ an openinterval containing 0 and 1 and ๐0, ๐1 โถ ๐ โ ๐ two ๐ถโ maps. A ๐ถโ map ๐นโถ ๐ร๐ด โ ๐ isa ๐ถโ homotopy between ๐0 and ๐1 if ๐น(๐ฅ, 0) = ๐0(๐ฅ) and ๐น(๐ฅ, 1) = ๐1(๐ฅ).
If there exists a homotopy between ๐0 and ๐1, we say that ๐0 and ๐1 are homotopic andwrite ๐0 โ ๐1.
Intuitively, ๐0 and ๐1 are homotopic if there exists a family of ๐ถโ maps ๐๐ก โถ ๐ โ ๐,where ๐๐ก(๐ฅ) = ๐น(๐ฅ, ๐ก), which โsmoothly deform ๐0 into ๐1โ. In Exercise 2.6.iv and equa-tion (2.6.6) you will be asked to verify that for ๐0 and ๐1 to be homotopic they have tosatisfy the following criteria.
Theorem 2.6.15. Let๐ be an open subset of ๐๐,๐ an open subset of ๐๐, and ๐0, ๐1 โถ ๐ โ ๐two ๐ถโ maps. If ๐0 and ๐1 are homotopic then for every closed form ๐ โ ๐บ๐(๐) the form๐โ1 ๐ โ ๐โ0 ๐ is exact.
This theorem is closely related to the Poincarรฉ lemma, and, in fact, one gets from ita slightly stronger version of the Poincarรฉ lemma than that described in Exercise 2.3.ivand equation (2.3.7).
Definition 2.6.16. An open subset ๐ of ๐๐ is contractible if, for some point ๐0 โ ๐, theidentity map id๐ โถ ๐ โ ๐ is homotopic to the constant map
๐0 โถ ๐ โ ๐ , ๐0(๐) = ๐0at ๐0.
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ยง2.6 The pullback operation on forms 57
From Theorem 2.6.15 it is easy to see that the Poincarรฉ lemma holds for contractableopen subsets of๐๐. If๐ is contractable every closed ๐-form on๐ of degree ๐ > 0 is exact. Tosee this, note that if ๐ โ ๐บ๐(๐), then for the identity map idโ๐ ๐ = ๐ and if ๐ is the constantmap at a point of ๐, then ๐โ๐ = 0.
Exercises for ยง2.6Exercise 2.6.i. Let ๐โถ ๐3 โ ๐3 be the map
๐(๐ฅ1, ๐ฅ2, ๐ฅ3) = (๐ฅ1๐ฅ2, ๐ฅ2๐ฅ23 , ๐ฅ33) .Compute the pullback, ๐โ๐ for the following forms.(1) ๐ = ๐ฅ2๐๐ฅ3(2) ๐ = ๐ฅ1๐๐ฅ1 โง ๐๐ฅ3(3) ๐ = ๐ฅ1๐๐ฅ1 โง ๐๐ฅ2 โง ๐๐ฅ3Exercise 2.6.ii. Let ๐โถ ๐2 โ ๐3 be the map
๐(๐ฅ1, ๐ฅ2) = (๐ฅ21 , ๐ฅ22 , ๐ฅ1๐ฅ2) .Complete the pullback, ๐โ๐, for the following forms.(1) ๐ = ๐ฅ2๐๐ฅ2 + ๐ฅ3๐๐ฅ3(2) ๐ = ๐ฅ1๐๐ฅ2 โง ๐๐ฅ3(3) ๐ = ๐๐ฅ1 โง ๐๐ฅ2 โง ๐๐ฅ3Exercise 2.6.iii. Let ๐ be an open subset of ๐๐, ๐ an open subset of ๐๐, ๐โถ ๐ โ ๐ a ๐ถโmap and ๐พโถ [๐, ๐] โ ๐ a ๐ถโ curve. Show that for ๐ โ ๐บ1(๐)
โซ๐พ๐โ๐ = โซ
๐พ1๐
where ๐พ1 โถ [๐, ๐] โ ๐ is the curve, ๐พ1(๐ก) = ๐(๐พ(๐ก)). (See Exercise 2.2.viii.)
Exercise 2.6.iv. Let ๐ be an open subset of ๐๐, ๐ด โ ๐ an open interval containing thepoints, 0 and 1, and (๐ฅ, ๐ก) product coordinates on ๐ ร ๐ด. Recall from Exercise 2.3.v that aform ๐ โ ๐บโ(๐ ร ๐ด) is reduced if it can be written as a sum
๐ = โ๐ผ๐๐ผ(๐ฅ, ๐ก)๐๐ฅ๐ผ
(i.e., none of the summands involve ๐๐ก). For a reduced form ๐, let ๐๐ โ ๐บโ(๐) be the form
(2.6.17) ๐๐ โ (โ๐ผโซ1
0๐๐ผ(๐ฅ, ๐ก)๐๐ก)๐๐ฅ๐ผ
and let ๐1, ๐2 โ ๐บโ(๐), be the forms
๐0 = โ๐ผ๐๐ผ(๐ฅ, 0)๐๐ฅ๐ผ
and๐1 = โ๐๐ผ(๐ฅ, 1)๐๐ฅ๐ผ .
Now recall from Exercise 2.4.v that every form ๐ โ ๐บ๐(๐ ร ๐ด) can be written uniquely as asum(2.6.18) ๐ = ๐๐ก โง ๐ผ + ๐ฝwhere ๐ผ and ๐ฝ 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) ๐ฝ1 โ ๐ฝ0 = ๐๐๐ผ .
Hint: Equation (2.4.19).Let ๐0 and ๐1 be themaps of๐ into๐ร๐ด defined by ๐0(๐ฅ) = (๐ฅ, 0)and ๐1(๐ฅ) = (๐ฅ, 1). Show that (2.6.20) can be rewritten
(2.6.21) ๐โ1๐ โ ๐โ0๐ = ๐๐๐ผ .
Exercise 2.6.v. Let๐ be an open subset of ๐๐ and ๐0, ๐1 โถ ๐ โ ๐ be ๐ถโ maps. Suppose ๐0and ๐1 are homotopic. Show that for every closed form, ๐ โ ๐บ๐(๐), ๐โ1 ๐ โ ๐โ0 ๐ is exact.
Hint: Let ๐นโถ ๐ร๐ด โ ๐ be a homotopy between ๐0 and ๐1 and let ๐ = ๐นโ๐. Show that๐ is closed and that ๐โ0 ๐ = ๐โ0๐ and ๐โ1 ๐ = ๐โ1๐. Conclude from equation (2.6.21)that
๐โ1 ๐ โ ๐โ0 ๐ = ๐๐๐ผ
where ๐ = ๐๐ก โง ๐ผ + ๐ฝ and ๐ผ and ๐ฝ are reduced.
Exercise 2.6.vi. Show that if ๐ โ ๐๐ is a contractable open set, then the Poincarรฉ lemmaholds: every closed form of degree ๐ > 0 is exact.
Exercise 2.6.vii. An open subset, ๐, of ๐๐ is said to be star-shaped if there exists a point๐0 โ ๐, with the property that for every point ๐ โ ๐, the line segment,
{ ๐ก๐ + (1 โ ๐ก)๐0 | 0 โค ๐ก โค 1 } ,
joining ๐ to ๐0 is contained in ๐. Show that if ๐ is star-shaped it is contractable.
Exercise 2.6.viii. Show that the following open sets are star-shaped.(1) The open unit ball
{ ๐ฅ โ ๐๐ | |๐ฅ| < 1 } .(2) The open rectangle, ๐ผ1 รโฏ ร ๐ผ๐, where each ๐ผ๐ is an open subinterval of ๐.(3) ๐๐ itself.(4) Product sets
๐1 ร ๐2 โ ๐๐ โ ๐๐1 ร ๐๐2
where ๐๐ is a star-shaped open set in ๐๐๐ .
Exercise 2.6.ix. Let ๐ be an open subset of ๐๐, ๐๐ก โถ ๐ โ ๐, ๐ก โ ๐, a one-parameter groupof diffeomorphisms and ๐ฃ its infinitesimal generator. Given ๐ โ ๐บ๐(๐) show that at ๐ก = 0
(2.6.22) ๐๐๐ก๐โ๐ก ๐ = ๐ฟ๐๐ .
Here is a sketch of a proof:(1) Let ๐พ(๐ก) be the curve, ๐พ(๐ก) = ๐๐ก(๐), and let ๐ be a zero-form, i.e., an element of ๐ถโ(๐).
Show that๐โ๐ก ๐(๐) = ๐(๐พ(๐ก))
and by differentiating this identity at ๐ก = 0 conclude that (2.6.22) holds for zero-forms.(2) Show that if (2.6.22) holds for ๐ it holds for ๐๐.
Hint: Differentiate the identity
๐โ๐ก ๐๐ = ๐๐โ๐ก ๐
at ๐ก = 0.
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ยง2.7 Divergence, curl, and gradient 59
(3) Show that if (2.6.22) holds for ๐1 and ๐2 it holds for ๐1 โง ๐2.Hint: Differentiate the identity
๐โ๐ก (๐1 โง ๐2) = ๐โ๐ก ๐1 โง ๐โ๐ก ๐2at ๐ก = 0.
(4) Deduce (2.6.22) from (1)โ(3).Hint: Every ๐-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 ๐ก
(2.6.23) ๐๐๐ก๐โ๐ก ๐ = ๐โ๐ก ๐ฟ๐๐ = ๐ฟ๐๐โ๐ก ๐ .
Hint: By the definition of a one-parameter group, ๐๐ +๐ก = ๐๐ โ ๐๐ก = ๐๐ โ ๐๐ , hence:๐โ๐ +๐ก๐ = ๐โ๐ก (๐โ๐ ๐) = ๐โ๐ (๐โ๐ก ๐) .
Prove the first assertion by differentiating the first of these identities with respect to ๐ andthen setting ๐ = 0, and prove the second assertion by doing the same for the second of theseidentities.
In particular conclude that๐โ๐ก ๐ฟ๐๐ = ๐ฟ๐๐โ๐ก ๐ .
Exercise 2.6.xi.(1) By massaging the result above show that
๐๐๐ก๐โ๐ก ๐ = ๐๐๐ก๐ + ๐๐ก๐๐
where๐๐ก๐ = ๐โ๐ก ๐๐๐ .
Hint: Equation (2.5.9).(2) Let
๐๐ = โซ1
0๐โ๐ก ๐๐๐๐๐ก .
Prove the homotopy identity
๐โ1 ๐ โ ๐โ0 ๐ = ๐๐๐ + ๐๐๐ .
Exercise 2.6.xii. Let๐ be an open subset of ๐๐, ๐ an open subset of ๐๐, ๐ a vector field on๐, ๐ a vector field on ๐ and ๐โถ ๐ โ ๐ a ๐ถโ map. Show that if ๐ and ๐ are ๐-related
๐๐๐โ๐ = ๐โ๐๐๐ .
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) ๐โถ ๐บ๐(๐3) โ ๐บ๐+1(๐3)in degrees ๐ = 0, 1, 2. However, only the gradient and divergence generalize to ๐ dimensions.(They are essentially the ๐-operations in degrees zero and ๐ โ 1.) And, unfortunately, thereis no simple description in terms of vector fields for the other ๐-operations. This is one ofthe main reasons why an adequate theory of vector calculus in ๐-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 Maxwellโs 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 Maxwellโs 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 ๐3 which preserve div and curlare translations and otations, whereas the operations (2.7.1) admit all diffeomorphisms of๐3 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 ๐ต(๐ฃ, ๐ค) = โ๐ฃ๐๐ค๐ on ๐๐. From thisinner product one gets by Exercise 1.2.ix a bijective linear map
๐ฟโถ ๐๐ โฅฒ (๐๐)โ
with the defining property: ๐ฟ(๐ฃ) = โ if and only if โ(๐ค) = ๐ต(๐ฃ, ๐ค). Via the identification(2.1.2) ๐ต and ๐ฟ can be transferred to ๐๐๐๐, giving one an inner product ๐ต๐ on ๐๐๐๐ and abijective linear map
๐ฟ๐ โถ ๐๐๐๐ โฅฒ ๐โ๐ ๐๐ .Hence if we are given a vector field ๐ on ๐ we can convert it into a 1-form ๐โฏ by setting(2.7.2) ๐โฏ(๐) โ ๐ฟ๐๐(๐)and this sets up a bijective correspondence between vector fields and 1-forms.
For instance๐ = ๐๐๐ฅ๐โบ ๐โฏ = ๐๐ฅ๐ ,
(see Exercise 2.7.iii) and, more generally,
(2.7.3) ๐ =๐โ๐=1๐๐๐๐๐ฅ๐โบ ๐โฏ =
๐โ๐=1๐๐๐๐ฅ๐ .
Example 2.7.4. In particular if๐ is a๐ถโ function on๐ โ ๐๐ the gradient of๐ is the vectorfield
grad(๐) โ๐โ๐=1
๐๐๐๐ฅ๐
,
and this gets converted by (2.7.5) into the 1-form ๐๐. Thus the gradient operation in vectorcalculus is basically just the exeterior derivative operation ๐โถ ๐บ0(๐) โ ๐บ1(๐).
The second way of converting vector fields into forms is via the interior product oper-ation. Namely let๐บ be the ๐-form ๐๐ฅ1 โงโฏโง ๐๐ฅ๐. Given an open subset ๐ of ๐๐ and a ๐ถโvector field
(2.7.5) ๐ =๐โ๐=1๐๐๐๐๐ฅ๐
on ๐ the interior product of ๐ฃ with ๐บ is the (๐ โ 1)-form
(2.7.6) ๐๐๐บ =๐โ๐=1(โ1)๐โ1๐๐๐๐ฅ1 โงโฏ โง ๐๐ฅ๐โฏโง ๐๐ฅ๐ .
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ยง2.7 Divergence, curl, and gradient 61
Moreover, every (๐ โ 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 (๐ โ 1)-forms. Under this corre-spondence the ๐-operation gets converted into an operation on vector fields
๐ โฆ ๐๐๐๐บ .Moreover, by (2.5.9)
๐๐๐๐บ = ๐ฟ๐๐บand by (2.5.14)
๐ฟ๐๐บ = div(๐)๐บwhere
(2.7.7) div(๐) =๐โ๐=1
๐๐๐๐๐ฅ๐
.
In other words, this correspondence between (๐ โ 1)-forms and vector fields converts the๐-operation into the divergence operation (2.7.7) on vector fields.
Notice that divergence and gradient are well-defined as vector calculus operations in ๐dimensions, 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 ๐ on an open subset ๐ โ ๐3 can bewritten uniquely as an interior product,(2.7.8) ๐ = ๐๐๐๐ฅ1 โง ๐๐ฅ2 โง ๐๐ฅ3 ,for some vector field ๐, and the left-hand side of this formula determines ๐ uniquely.
Definition 2.7.9. Let๐ be an open subset of ๐3 and ๐ a vector field on๐. From ๐ we get by(2.7.3) a 1-form ๐โฏ, and hence by (2.7.8) a vector field ๐ satisfying
๐๐โฏ = ๐๐๐๐ฅ1 โง ๐๐ฅ2 โง ๐๐ฅ3 .The curl of ๐ is the vector field
curl(๐) โ ๐ .
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 ๐ is the vector field
๐ = ๐1๐๐๐ฅ1+ ๐2๐๐๐ฅ2+ ๐3๐๐๐ฅ3
thencurl(๐) = ๐1
๐๐๐ฅ1+ ๐2๐๐๐ฅ2+ ๐3๐๐๐ฅ3
where
๐1 =๐๐2๐๐ฅ3โ ๐๐3๐๐ฅ2
๐2 =๐๐3๐๐ฅ1โ ๐๐1๐๐ฅ3
(2.7.10)
๐3 =๐๐1๐๐ฅ2โ ๐๐2๐๐ฅ1
.
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 ๐ต๐ to the tan-gent space ๐๐๐๐ for each ๐ โ ๐; to define divergence we had to equip ๐ with the 3-form๐บ and to define curl, the most complicated of these three operations, we needed the innerproducts ๐ต๐ and the form ๐บ. 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.
Maxwellโs equations and differential formsWe conclude this section by showing how Maxwellโs equations, which are usually for-
mulated in terms of divergence and curl, can be reset into โformโ language. The discussionbelow is an abbreviated version of [5, ยง1.20].
Maxwellโs equations assert:div(๐๐ธ) = ๐(2.7.11)
curl(๐๐ธ) = โ๐๐๐ก๐๐(2.7.12)
div(๐๐) = 0(2.7.13)
๐2 curl(๐๐) = ๐ +๐๐๐ก๐๐ธ(2.7.14)
where ๐๐ธ and ๐๐ are the electric andmagnetic fields, ๐ is the scalar charge density,๐ is thecurrent density and ๐ is the velocity of light. (To simplify (2.7.15) slightly we assume thatour units of spaceโtime are chosen so that ๐ = 1.) As above let๐บ = ๐๐ฅ1 โง ๐๐ฅ2 โง ๐๐ฅ3 and let
๐๐ธ = ๐(๐๐ธ)๐บand
๐๐ = ๐(๐๐)๐บ .We can then rewrite equations (2.7.11) and (2.7.13) in the form(2.7.11โฒ) ๐๐๐ธ = ๐๐บand(2.7.13โฒ) ๐๐๐ = 0 .
What about (2.7.12) and (2.7.14)?We leave the following โformโ versions of these equa-tions as an exercise:
(2.7.12โฒ) ๐๐โฏ๐ธ = โ๐๐๐ก๐๐
and
(2.7.14โฒ) ๐๐โฏ๐ = ๐๐๐บ +๐๐๐ก๐๐ธ
where the 1-forms, ๐โฏ๐ธ and ๐โฏ๐, are obtained from ๐๐ธ and ๐๐ by the operation (2.7.2).
These equations can be written more compactly as differential form identities in 3 + 1dimensions. Let ๐๐ and ๐๐ธ be the 2-forms
๐๐ = ๐๐ โ ๐โฏ๐ธ โง ๐๐ก
and(2.7.15) ๐๐ธ = ๐๐ธ โ ๐
โฏ๐ โง ๐๐ก
and let ๐ฌ be the 3-form๐ฌ โ ๐๐บ + ๐๐๐บ โง ๐๐ก .
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ยง2.8 Symplectic geometry & classical mechanics 63
We will leave for you to show that the four equations (2.7.11)โ(2.7.14) are equivalentto two elegant and compact (3 + 1)-dimensional identities
๐๐๐ = 0and
๐๐๐ธ = ๐ฌ .
Exercises for ยง2.7Exercise 2.7.i. Verify that the curl operation is given in coordinates by equation (2.7.10).
Exercise 2.7.ii. Verify thatMaxwellโs 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)-dimensionsMaxwellโs equations take the form of equa-tions (2.7.10) and (2.7.11).
Exercise 2.7.iv. Let ๐ be an open subset of ๐3 and ๐ a vector field on ๐. Show that if ๐ isthe gradient of a function, its curl has to be zero.
Exercise 2.7.v. If ๐ is simply connected prove the converse: If the curl of ๐ vanishes, ๐ isthe gradient of a function.
Exercise 2.7.vi. Let ๐ = curl(๐). Show that the divergence of ๐ is zero.
Exercise 2.7.vii. Is the converse to Exercise 2.7.vi true? Suppose the divergence of๐ is zero.Is ๐ = curl(๐) for some vector field ๐?
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 ๐ฅ1,โฆ, ๐ฅ2๐ be the standard coordinate functions on ๐2๐ and for ๐ =1,โฆ, ๐ let ๐ฆ๐ โ ๐ฅ๐+๐. The two-form
๐ โ๐โ๐=1๐๐ฅ๐ โง ๐๐ฆ๐ =
๐โ๐=1๐๐ฅ๐ โง ๐๐ฅ๐+๐
is called the Darboux form on ๐2๐.From the identity
(2.8.2) ๐ = โ๐ (โ๐๐=1 ๐ฆ๐๐๐ฅ๐) .it follows that ๐ is exact. Moreover computing the ๐-fold wedge product of ๐ with itself weget
๐๐ = (๐โ๐๐=1๐๐ฅ๐1 โง ๐๐ฆ๐1) โงโฏ โง(
๐โ๐๐=1๐๐ฅ๐๐ โง ๐๐ฆ๐๐)
= โ๐1,โฆ,๐๐๐๐ฅ๐1 โง ๐๐ฆ๐1 โงโฏ โง ๐๐ฅ๐๐ โง ๐๐ฆ๐๐ .
We can simplify this sum by noting that if the multi-index ๐ผ = (๐1,โฆ, ๐๐) is repeating, thewedge product(2.8.3) ๐๐ฅ๐1 โง ๐๐ฆ๐1 โงโฏ โง ๐๐ฅ๐๐ โง ๐๐ฅ๐๐
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64 Chapter 2: Differential Forms
involves two repeating ๐๐ฅ๐๐ and hence is zero, and if ๐ผ is non-repeating we can permute thefactors and rewrite (2.8.3) in the form
๐๐ฅ1 โง ๐๐ฆ1 โงโฏ โง ๐๐ฅ๐ โง ๐๐ฆ๐ .(See Exercise 1.6.v.)
Hence since these are exactly ๐! non-repeating multi-indices
๐๐ = ๐!๐๐ฅ1 โง ๐๐ฆ1 โงโฏ โง ๐๐ฅ๐ โง ๐๐ฆ๐ ,i.e.,
(2.8.4) 1๐!๐๐ = ๐บ
where
(2.8.5) ๐บ = ๐๐ฅ1 โง ๐๐ฆ1 โงโฏ โง ๐๐ฅ๐ โง ๐๐ฆ๐is the symplectic volume form on ๐2๐.
Definition 2.8.6. Let ๐ and ๐ be open subsets of ๐2๐. A diffeomorphism ๐โถ ๐ โฅฒ ๐ is asymplectic diffeomorphism or symplectomorphism if ๐โ๐ = ๐, where ๐ denotes the Dar-boux form on ๐2๐.
Definition 2.8.7. Let ๐ be an open subset of ๐2๐, let(2.8.8) ๐๐ก โถ ๐ โฅฒ ๐ , โโ < ๐ก < โbe a one-parameter group of diffeomorphisms of ๐, and ๐ be the vector field generating(2.8.8) We say that ๐ is a symplectic vector field if the diffeomorphisms (2.8.8) are symplec-tomorphisms, i.e., for all ๐ก we have ๐โ๐ก ๐ = ๐.
Let us see what such vector fields have to look like. Note that by (2.6.23)
(2.8.9) ๐๐๐ก๐โ๐ก ๐ = ๐โ๐ก ๐ฟ๐๐ ,
hence if ๐โ๐ก ๐ = ๐ for all ๐ก, the left hand side of (2.8.9) is zero, so
๐โ๐ก ๐ฟ๐๐ = 0 .In particular, for ๐ก = 0, ๐๐ก is the identity map so ๐โ๐ก ๐ฟ๐๐ = ๐ฟ๐๐ = 0. Conversely, if ๐ฟ๐๐ = 0,then ๐โ๐ก ๐ฟ๐๐ = 0 so by (2.8.9) ๐โ๐ก ๐ does not depend on ๐ก. However, since ๐โ๐ก ๐ = ๐ for ๐ก = 0we conclude that ๐โ๐ก ๐ = ๐ for all ๐ก. To summarize, we have proved:
Theorem 2.8.10. Let ๐๐ก โถ ๐ โ ๐ be a one-parameter group of diffeomorphisms and ๐ theinfinitesmal generator of this group. Then ๐ is symplectic of and only if ๐ฟ๐๐ = 0.
There is an equivalent formulation of Theorem 2.8.10 in terms of the interior product๐๐๐. By (2.5.9)
๐ฟ๐๐ = ๐๐๐๐ + ๐๐ ๐๐ .But by (2.8.2) we have ๐๐ = 0, so
๐ฟ๐๐ = ๐๐๐๐ .Thus we have shown:
Theorem 2.8.11. A vector field ๐ on an open subset of ๐2๐ is symplectic if and only if the form๐๐๐ is closed.
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ยง2.8 Symplectic geometry & classical mechanics 65
Let๐ be an open subset of ๐2๐ and ๐ a vector field on๐. If ๐๐๐ is not only closed but isexact we say that ๐ is a Hamiltonian vector field. In other words, ๐ is Hamiltonian if(2.8.12) ๐๐๐ = ๐๐ปfor some ๐ถโ function๐ป โ ๐ถโ(๐).
Let is see what this condition looks like in coordinates. Let
(2.8.13) ๐ =๐โ๐=1(๐๐๐๐๐ฅ๐+ ๐๐๐๐๐ฆ๐) .
Then๐๐๐ = โ
1โค๐,๐โค๐๐๐ ๐๐/๐๐ฅ๐(๐๐ฅ๐ โง ๐๐ฆ๐) + โ
1โค๐,๐โค๐๐๐ ๐๐/๐๐ฆ๐(๐๐ฅ๐ โง ๐๐ฆ๐) .
But
๐๐/๐๐ฅ๐๐๐ฅ๐ = {1, ๐ = ๐0, ๐ โ ๐
and๐๐/๐๐ฅ๐๐๐ฆ๐ = 0
so the first summand above is โ๐๐=1 ๐๐ ๐๐ฆ๐, and a similar argument shows that the secondsummand is โโ๐๐=1 ๐๐๐๐ฅ๐.
Hence if ๐ is the vector field (2.8.13), then
(2.8.14) ๐๐๐ =๐โ๐=1(๐๐ ๐๐ฆ๐ โ ๐๐๐๐ฅ๐) .
Thus since
๐๐ป =๐โ๐=1(๐๐ป๐๐ฅ๐๐๐ฅ๐ +๐๐ป๐๐ฆ๐๐๐ฆ๐)
we get from (2.8.12)โ(2.8.14)
๐๐ =๐๐ป๐๐ฆ๐
and ๐๐ = โ๐๐ป๐๐ฅ๐
so ๐ has the form:
(2.8.15) ๐ =๐โ๐=1(๐๐ป๐๐ฆ๐๐๐๐ฅ๐โ ๐๐ป๐๐ฅ๐๐๐๐ฆ๐) .
In particular if ๐พ(๐ก) = (๐ฅ(๐ก), ๐ฆ(๐ก)) is an integral curve of ๐ it has to satisfy the system ofdifferential equations
(2.8.16){{{{{{{
๐๐ฅ๐๐๐ก= ๐๐ป๐๐ฆ๐(๐ฅ(๐ก), ๐ฆ(๐ก))
๐๐ฆ๐๐๐ก= โ๐๐ป๐๐ฅ๐(๐ฅ(๐ก), ๐ฆ(๐ก)) .
The formulas (2.8.13) and (2.8.14) exhibit an important property of the Darboux form ๐.Every one-form on ๐ can be written uniquely as a sum
๐โ๐=1(๐๐ ๐๐ฆ๐ โ ๐๐๐๐ฅ๐)
with ๐๐ and ๐๐ in ๐ถโ(๐) and hence (2.8.13) and (2.8.14) imply:
Theorem 2.8.17. The map ๐ โฆ ๐๐๐ defines a bijective between vector field and one-forms.
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66 Chapter 2: Differential Forms
In particular, for every ๐ถโ function๐ป, we get by correspondence a unique vector field๐ = ๐๐ป with the property (2.8.12). We next note that by (1.7.9)
๐ฟ๐๐ป = ๐๐ ๐๐ป = ๐๐(๐๐๐) = 0 .Thus(2.8.18) ๐ฟ๐๐ป = 0i.e.,๐ป is an integral ofmotion of the vector field ๐. In particular, if the function๐ปโถ ๐ โ ๐ isproper, then by Theorem 2.2.12 the vector field, ๐, 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, then๐โ๐ก ๐๐ = ๐โ๐ก ๐ โงโฏ โง ๐โ๐ก ๐ = ๐๐, so by (2.8.4)(2.8.20) ๐โ๐ก ๐บ = ๐บ ,where ๐บ 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 of๐ interacting point-masses.The configuration space of such a system is
๐๐ = ๐3 รโฏ ร ๐3 ,where there are ๐ copies of ๐3 in the product, with position coordinates ๐ฅ1,โฆ, ๐ฅ๐ andthe phase space is ๐2๐ with position coordinates ๐ฅ1,โฆ, ๐ฅ๐ and momentum coordinates,๐ฆ1,โฆ, ๐ฆ๐. The kinetic energy of this system is a quadratic function of the momentum co-ordinates
12
๐โ๐=1
1๐๐๐ฆ2๐ ,
and for simplicity we assume that the potential energy is a function ๐(๐ฅ1,โฆ, ๐ฅ๐) of theposition coordinates alone, i.e., it does not depend on themomenta and is time-independentas well. Let
(2.8.22) ๐ป = 12
๐โ๐=1
1๐๐๐ฆ2๐ + ๐(๐ฅ1,โฆ, ๐ฅ๐)
be the total energy of the system.
We now show that Newtonโs 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 ๐๐ป.Proof. For the function (2.8.22) the equations (2.8.16) become
{{{{{{{{{
๐๐ฅ๐๐๐ก= 1๐๐๐ฆ๐
๐๐ฆ๐๐๐ก= โ ๐๐๐๐ฅ๐
.
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) ๐๐๐2๐ฅ๐๐๐ก2= โ๐๐๐๐ฅ๐
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ยง2.8 Symplectic geometry & classical mechanics 67
and interpreting the term on the right as the force exerted on the ๐th point-mass and theterm on the left as mass times acceleration this equation becomes Newtonโs second law. โก
In classical mechanics the equations (2.8.16) are known as the HamiltonโJacobi 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 toMaxwellโs equations can be found in [5].
We conclude this chapter bymentioning a few implications of the Hamiltonian descrip-tion (2.8.16) of Newtonโs 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 ๐, hence the energy of the system (2.8.16) does notchange in time.(2) Noetherโs principle: Let ๐พ๐ก โถ ๐2๐ โ ๐2๐ be a one-parameter group of diffeomorphisms of
phase space and ๐ its infinitesimal generator. Then (๐พ๐ก)๐กโ๐ is a symmetry of the systemabove if each ๐พ๐ก preserves the function (2.8.22) and the vector field ๐ is Hamiltonian.The Hamiltonian condition means that
๐๐๐ = ๐๐บfor some ๐ถโ function ๐บ, and what Noetherโs principle asserts is that this function is anintegral of motion of the system (2.8.16), i.e., satisfies ๐ฟ๐๐บ = 0. In other words statedmore succinctly: symmetries of the system (2.8.16) give rise to integrals of motion.
(3) Poincarรฉ recurrence: An important theorem of Poincarรฉ asserts that if the function
๐ปโถ ๐2๐ โ ๐defined 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 ๐ก0, 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 ยง2.8Exercise 2.8.i. Let ๐๐ป be the vector field (2.8.15). Prove that div(๐๐ป) = 0.
Exercise 2.8.ii. Let ๐ be an open subset of ๐๐, ๐๐ก โถ ๐ โ ๐ a one-parameter group ofdiffeomorphisms of ๐ and ๐ the infinitesmal generator of this group. Show that if ๐ผ is a ๐-form on ๐ then ๐โ๐ก ๐ผ = ๐ผ for all ๐ก if and only if ๐ฟ๐๐ผ = 0 (i.e., generalize to arbitrary ๐-formsthe result we proved above for the Darboux form).
Exercise 2.8.iii (the harmonic oscillator). Let๐ป be the functionโ๐๐=1๐๐(๐ฅ2๐ +๐ฆ2๐ )where the๐๐โs are positive constants.(1) Compute the integral curves of ๐๐ป.(2) Poincarรฉ recurrence: Show that if (๐ฅ(๐ก), ๐ฆ(๐ก)) is an integral curvewith initial point (๐ฅ0, ๐ฆ0) โ(๐ฅ(0), ๐ฆ(0)) and ๐ an arbitrarily small neighborhood of (๐ฅ0, ๐ฆ0), then for every ๐ > 0there exists a ๐ก > ๐ such that (๐ฅ(๐ก), ๐ฆ(๐ก)) โ ๐.
Exercise 2.8.iv. Let ๐ be an open subset of ๐2๐ and let๐ป1, ๐ป2 โ ๐ถโ(๐). Show that
[๐๐ป1 , ๐๐ป2] = ๐๐ปwhere
(2.8.25) ๐ป =๐โ๐=1
๐๐ป1๐๐ฅ๐๐๐ป2๐๐ฆ๐โ ๐๐ป2๐๐ฅ๐๐๐ป1๐๐ฆ๐
.
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68 Chapter 2: Differential Forms
Exercise 2.8.v. The expression (2.8.25) is known as the Poisson bracket of๐ป1 and๐ป2, de-noted by {๐ป1, ๐ป2}. Show that it is anti-symmetric
{๐ป1, ๐ป2} = โ{๐ป2, ๐ป1}and satisfies the Jacobi identity
0 = {๐ป1, {๐ป2, ๐ป3}} + {๐ป2, {๐ป3, ๐ป1}} + {๐ป3, {๐ป1, ๐ป2}} .Exercise 2.8.vi. Show that
{๐ป1, ๐ป2} = ๐ฟ๐๐ป1๐ป2 = โ๐ฟ๐๐ป2๐ป1 .
Exercise 2.8.vii. Prove that the following three properties are equivalent.(1) {๐ป1, ๐ป2} = 0.(2) ๐ป1 is an integral of motion of ๐2.(3) ๐ป2 is an integral of motion of ๐1.Exercise 2.8.viii. Verify Noetherโs principle.
Exercise 2.8.ix (conservation of linearmomentum). Suppose the potential๐ in equation (2.8.22)is invariant under the one-parameter group of translations
๐๐ก(๐ฅ1,โฆ, ๐ฅ๐) = (๐ฅ1 + ๐ก,โฆ, ๐ฅ๐ + ๐ก) .(1) Show that the function (2.8.22) is invariant under the group of diffeomorphisms
๐พ๐ก(๐ฅ, ๐ฆ) = (๐๐ก๐ฅ, ๐ฆ) .(2) Show that the infinitesmal generator of this group is the Hamiltonian vector field ๐๐บ
where ๐บ = โ๐๐=1 ๐ฆ๐.(3) Conclude from Noetherโs principle that this function is an integral of the vector field๐๐ป, i.e., that โtotal linear momentโ is conserved.
(4) Show that โtotal linear momentumโ is conserved if ๐ is the Coulomb potential
โ๐โ ๐
๐๐|๐ฅ๐ โ ๐ฅ๐|
.
Exercise 2.8.x. Let ๐ ๐๐ก โถ ๐2๐ โ ๐2๐ be the rotation which fixes the variables, (๐ฅ๐, ๐ฆ๐), ๐ โ ๐and rotates (๐ฅ๐, ๐ฆ๐) by the angle, ๐ก:
๐ ๐๐ก(๐ฅ๐, ๐ฆ๐) = (cos ๐ก ๐ฅ๐ + sin ๐ก ๐ฆ๐ ,โ sin ๐ก ๐ฅ๐ + cos ๐ก ๐ฆ๐) .(1) Show that ๐ ๐๐ก, โโ < ๐ก < โ, is a one-parameter group of symplectomorphisms.(2) Show that its generator is the Hamiltonian vector field, ๐๐ป๐ , where๐ป๐ = (๐ฅ2๐ + ๐ฆ2๐ )/2.(3) Let๐ป be the harmonic oscillator Hamiltonian from Exercise 2.8.iii. Show that the ๐ ๐๐ก โs
preserve๐ป.(4) What does Noetherโs principle tell one about the classical mechanical system with en-
ergy function๐ป?Exercise 2.8.xi. Show that if๐ is an open subset of๐2๐ and ๐ is a symplectic vector field on๐then for every point ๐0 โ ๐ there exists a neighborhood๐0 of ๐0 onwhich ๐ is Hamiltonian.
Exercise 2.8.xii. Deduce from Exercises 2.8.iv and 2.8.xi that if ๐1 and ๐2 are symplecticvector fields on an open subset ๐ of ๐2๐ their Lie bracket [๐1, ๐2] is a Hamiltonian vectorfield.
Exercise 2.8.xiii. Let ๐ผ = โ๐๐=1 ๐ฆ๐๐๐ฅ๐.(1) Show that ๐ = โ๐๐ผ.
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ยง2.8 Symplectic geometry & classical mechanics 69
(2) Show that if ๐ผ1 is any one-form on ๐2๐ with the property ๐ = โ๐๐ผ1, then๐ผ = ๐ผ1 + ๐๐น
for some ๐ถโ function ๐น.(3) Show that ๐ผ = ๐๐๐ where ๐ is the vector field
๐ โ โ๐โ๐=1๐ฆ๐๐๐๐ฆ๐
.
Exercise 2.8.xiv. Let ๐ be an open subset of ๐2๐ and ๐ a vector field on ๐. Show that ๐ hasthe property, ๐ฟ๐๐ผ = 0, if and only if
๐๐๐ = ๐๐๐๐ผ .In particular conclude that if ๐ฟ๐๐ผ = 0 then ๐ฃ is Hamiltonian.
Hint: Equation (2.8.2).
Exercise 2.8.xv. Let๐ป be the function
(2.8.26) ๐ป(๐ฅ, ๐ฆ) =๐โ๐=1๐๐(๐ฅ)๐ฆ๐ ,
where the ๐๐โs are ๐ถโ functions on ๐๐. Show that(2.8.27) ๐ฟ๐๐ป๐ผ = 0 .
Exercise 2.8.xvi. Conversely show that if ๐ป is any ๐ถโ function on ๐2๐ satisfying equa-tion (2.8.27) it has to be a function of the form (2.8.26).Hints:(1) Let ๐ be a vector field on ๐2๐ satisfying ๐ฟ๐๐ผ = 0. By equation (2.8.16) we have ๐ = ๐๐ป,
where๐ป = ๐๐๐ผ.(2) Show that๐ป has to satisfy the equation
๐โ๐=1๐ฆ๐๐๐ป๐๐ฆ๐= ๐ป .
(3) Conclude that if๐ป๐ = ๐๐ป๐๐ฆ๐ then๐ป๐ has to satisfy the equation๐โ๐=1๐ฆ๐๐๐๐ฆ๐๐ป๐ = 0 .
(4) Conclude that๐ป๐ has to be constant along the rays (๐ฅ, ๐ก๐ฆ), for 0 โค ๐ก < โ.(5) Conclude finally that๐ป๐ has to be a function of ๐ฅ alone, i.e., does not depend on ๐ฆ.Exercise 2.8.xvii. Show that if ๐๐๐ is a vector field
๐โ๐=1๐๐(๐ฅ)๐๐๐ฅ๐
on configuration space there is a unique lift of ๐๐๐ to phase space
๐ =๐โ๐=1๐๐(๐ฅ)๐๐๐ฅ๐+ ๐๐(๐ฅ, ๐ฆ)
๐๐๐ฆ๐
satisfying ๐ฟ๐๐ผ = 0.
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CHAPTER 3
Integration of Forms
3.1. Introduction
The change of variables formula asserts that if ๐ and ๐ are open subsets of ๐๐ and๐โถ ๐ โ ๐ a๐ถ1 diffeomorphism then, for every continuous function๐โถ ๐ โ ๐ the integral
โซ๐๐(๐ฆ)๐๐ฆ
exists if and only if the integral
โซ๐(๐ โ ๐)(๐ฅ) |det๐ท๐(๐ฅ)|๐๐ฅ
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 ยง 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 ยง3.5 canbe couched entirely in the language of elementary multivariable calculus.
The virtue of Laxโs 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 ยง3.6of these notes.
Remark 3.1.1. The proof we are about to describe is somewhat simpler and more transpar-ent if we assume that ๐ is a ๐ถโ diffeomorphism. Weโll henceforth make this assumption.
3.2. The Poincarรฉ lemma for compactly supported forms on rectangles
Definition 3.2.1. Let ๐ be a ๐-form on ๐๐. We define the support of ๐ bysupp(๐) โ { ๐ฅ โ ๐๐ | ๐๐ฅ โ 0 } ,
and we say that ๐ is compactly supported if supp(๐) is compact.We will denote by ๐บ๐๐ (๐๐) the set of all ๐ถโ ๐-forms which are compactly supported,
and if๐ is an open subset of ๐๐, we will denote by๐บ๐๐ (๐) the set of all compactly supported๐-forms whose support is contained in ๐.
Let ๐ = ๐๐๐ฅ1 โงโฏโง ๐๐ฅ๐ be a compactly supported ๐-form with ๐ โ ๐ถโ0 (๐๐). We willdefine the integral of ๐ over ๐๐:
โซ๐๐๐
to be the usual integral of ๐ over ๐๐
โซ๐๐๐๐๐ฅ .
71
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72 Chapter 3: Integration of Forms
(Since ๐ is ๐ถโ and compactly supported this integral is well-defined.)Now let ๐ be the rectangle
[๐1, ๐1] ร โฏ ร [๐๐, ๐๐] .The Poincarรฉ lemma for rectangles asserts:
Theorem3.2.2 (Poincarรฉ lemma for rectangles). Let๐ be a compactly supported ๐-formwithsupp(๐) โ int(๐). Then the following assertions are equivalent.(1) โซ๐ = 0.(2) There exists a compactly supported (๐ โ 1)-form ๐ with supp(๐) โ int(๐) satisfying ๐๐ =๐.
We will first prove that (2)โ(1).
Proof that (2)โ(1). Let
๐ =๐โ๐=1๐๐ ๐๐ฅ1 โงโฏ โง ๐๐ฅ๐ โงโฏ โง ๐๐ฅ๐ ,
(the โhatโ over the ๐๐ฅ๐ meaning that ๐๐ฅ๐ has to be omitted from the wedge product). Then
๐๐ =๐โ๐=1(โ1)๐โ1 ๐๐๐๐๐ฅ๐๐๐ฅ1 โงโฏ โง ๐๐ฅ๐ ,
and to show that the integral of ๐๐ is zero it suffices to show that each of the integrals
(3.2.2)๐ โซ๐๐๐๐๐๐๐ฅ๐๐๐ฅ
is zero. By Fubini we can compute (3.2.2)๐ by first integrating with respect to the variable,๐ฅ๐, and then with respect to the remaining variables. But
โซ ๐๐๐๐๐ฅ๐๐๐ฅ๐ = ๐(๐ฅ)|
๐ฅ๐=๐๐
๐ฅ๐=๐๐= 0
since ๐๐ is supported on ๐. โก
We will prove that (1)โ(2) by proving a somewhat stronger result. Let ๐ be an opensubset of ๐๐. Weโll say that ๐ has property ๐ if every form ๐ โ ๐บ๐๐ (๐) such that โซ๐ ๐ = 0we have ๐ โ ๐๐บ๐โ1๐ (๐). We will prove:
Theorem 3.2.3. Let ๐ be an open subset of ๐๐โ1 and ๐ด โ ๐ an open interval. Then if ๐ hasproperty ๐, ๐ ร ๐ด does as well.
Remark 3.2.4. It is very easy to see that the open interval ๐ด itself has property ๐. (SeeExercise 3.2.i below.) Hence it follows by induction from Theorem 3.2.3 that
int๐ = ๐ด1 รโฏ ร ๐ด๐ , ๐ด๐ = (๐๐, ๐๐)has property ๐, and this proves โ(1)โ(2)โ.
Proof of Theorem 3.2.3. Let (๐ฅ, ๐ก) = (๐ฅ1,โฆ, ๐ฅ๐โ1, ๐ก) be product coordinates on๐ร๐ด. Given๐ โ ๐บ๐๐ (๐ร๐ด)we can express๐ as a wedge product, ๐๐กโง๐ผwith ๐ผ = ๐(๐ฅ, ๐ก) ๐๐ฅ1โงโฏโง๐๐ฅ๐โ1and ๐ โ ๐ถโ0 (๐ ร ๐ด). Let ๐ โ ๐บ๐โ1๐ (๐) be the form
(3.2.5) ๐ = (โซ๐ด๐(๐ฅ, ๐ก)๐๐ก) ๐๐ฅ1 โงโฏ โง ๐๐ฅ๐โ1 .
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ยง3.2 The Poincarรฉ lemma for compactly supported forms on rectangles 73
Thenโซ๐๐โ1๐ = โซ๐๐๐(๐ฅ, ๐ก)๐๐ฅ๐๐ก = โซ
๐๐๐
so if the integral of ๐ is zero, the integral of ๐ is zero. Hence since ๐ has property ๐, ๐ = ๐๐for some ๐ โ ๐บ๐โ2๐ (๐). Let ๐ โ ๐ถโ(๐) be a bump function which is supported on ๐ด andwhose integral over ๐ด is 1. Setting
๐ = โ๐(๐ก)๐๐ก โง ๐we have
๐๐ = ๐(๐ก)๐๐ก โง ๐๐ = ๐(๐ก)๐๐ก โง ๐ ,and hence
๐ โ ๐๐ = ๐๐ก โง (๐ผ โ ๐(๐ก)๐)= ๐๐ก โง ๐ข(๐ฅ, ๐ก)๐๐ฅ1 โงโฏ โง ๐๐ฅ๐โ1
where๐ข(๐ฅ, ๐ก) = ๐(๐ฅ, ๐ก) โ ๐(๐ก) โซ
๐ด๐(๐ฅ, ๐ก)๐๐ก
by (3.2.5). Thus
(3.2.6) โซ๐ข(๐ฅ, ๐ก) ๐๐ก = 0 .
Let ๐ and ๐ be the end points of ๐ด and let
(3.2.7) ๐ฃ(๐ฅ, ๐ก) = โซ๐ก
๐๐ข(๐ฅ, ๐ ) ๐๐ .
By equation (3.2.6) ๐ฃ(๐, ๐ฅ) = ๐ฃ(๐, ๐ฅ) = 0, so ๐ฃ is in ๐ถโ0 (๐ ร ๐ด) and by (3.2.7), ๐๐ฃ/๐๐ก = ๐ข.Hence if we let ๐พ be the form, ๐ฃ(๐ฅ, ๐ก) ๐๐ฅ1 โงโฏ โง ๐๐ฅ๐โ1, we have:
๐๐พ = ๐๐ก โง ๐ข(๐ฅ, ๐ก) ๐๐ฅ1 โงโฏ โง ๐๐ฅ๐โ1 = ๐ โ ๐๐ and
๐ = ๐(๐พ + ๐ ) .Since ๐พ and ๐ are both in๐บ๐โ1๐ (๐ ร๐ด) this proves that ๐ is in ๐๐บ๐โ1๐ (๐ ร๐ด) and hence
that ๐ ร ๐ด has property ๐. โก
Exercises for ยง3.2Exercise 3.2.i. Let ๐โถ ๐ โ ๐ be a compactly supported function of class ๐ถ๐ with supporton the interval (๐, ๐). Show that the following are equivalent.
(1) โซ๐๐ ๐(๐ฅ)๐๐ฅ = 0.(2) There exists a function ๐โถ ๐ โ ๐ of class ๐ถ๐+1 with support on (๐, ๐) with ๐๐๐๐ฅ = ๐.
Hint: Show that the function
๐(๐ฅ) = โซ๐ฅ
๐๐(๐ )๐๐
is compactly supported.
Exercise 3.2.ii. Let ๐ = ๐(๐ฅ, ๐ฆ) be a compactly supported function on ๐๐ ร ๐โ with theproperty that the partial derivatives
๐๐๐๐ฅ๐(๐ฅ, ๐ฆ) , for ๐ = 1,โฆ, ๐ ,
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74 Chapter 3: Integration of Forms
exist and are continuous as functions of ๐ฅ and ๐ฆ. Prove the following โdifferentiation underthe integral signโ theorem (which we implicitly used in our proof of Theorem 3.2.3).
Theorem 3.2.8. The function ๐(๐ฅ) โ โซ๐โ ๐(๐ฅ, ๐ฆ)๐๐ฆ is of class ๐ถ1 and๐๐๐๐ฅ๐(๐ฅ) = โซ ๐๐
๐๐ฅ๐(๐ฅ, ๐ฆ)๐๐ฆ .
Hints: For ๐ฆ fixed and โ โ ๐๐,๐(๐ฅ + โ, ๐ฆ) โ ๐(๐ฅ, ๐ฆ) = ๐ท๐ฅ๐(๐, ๐ฆ)โ
for some point ๐ on the line segment joining ๐ฅ to ๐ฅ+โ. Using the fact that๐ท๐ฅ๐ is continuousas a function of ๐ฅ and ๐ฆ and compactly supported, conclude the following.
Lemma 3.2.9. Given ๐ > 0 there exists a ๐ฟ > 0 such that for |โ| โค ๐ฟ|๐(๐ฅ + โ, ๐ฆ) โ ๐(๐ฅ, ๐ฆ) โ ๐ท๐ฅ๐(๐ฅ, ๐ฆ)โ| โค ๐|โ| .
Now let ๐ โ ๐โ be a rectangle with supp(๐) โ ๐๐ ร ๐ and show that
|๐(๐ฅ + โ) โ ๐(๐ฅ) โ (โซ๐ท๐ฅ๐(๐ฅ, ๐ฆ) ๐๐ฆ) โ| โค ๐ vol(๐)|โ| .
Conclude that ๐ is differentiable at ๐ฅ and that its derivative is
โซ๐ท๐ฅ๐(๐ฅ, ๐ฆ)๐๐ฆ .
Exercise 3.2.iii. Let ๐โถ ๐๐ร๐โ โ ๐ be a compactly supported continuous function. Provethe following theorem.
Theorem 3.2.10. If all the partial derivatives of ๐(๐ฅ, ๐ฆ) with respect to ๐ฅ of order โค ๐ existand are continuous as functions of ๐ฅ and ๐ฆ the function
๐(๐ฅ) = โซ๐(๐ฅ, ๐ฆ)๐๐ฆ
is of class ๐ถ๐.Exercise 3.2.iv. Let๐ be an open subset of ๐๐โ1,๐ด โ ๐ an open interval and (๐ฅ, ๐ก) productcoordinates on ๐ ร ๐ด. Recall from Exercise 2.3.v that every form ๐ โ ๐บ๐(๐ ร ๐ด) can bewritten uniquely as a sum ๐ = ๐๐ก โง ๐ผ + ๐ฝ where ๐ผ and ๐ฝ are reduced, i.e., do not contain afactor of ๐๐ก.(1) Show that if ๐ is compactly supported on ๐ ร ๐ด then so are ๐ผ and ๐ฝ.(2) Let ๐ผ = โ๐ผ ๐๐ผ(๐ฅ, ๐ก)๐๐ฅ๐ผ. Show that the form
(3.2.11) ๐ = โ๐ผ(โซ๐ด๐๐ผ(๐ฅ, ๐ก) ๐๐ก) ๐๐ฅ๐ผ
is in ๐บ๐โ1๐ (๐).(3) Show that if ๐๐ = 0, then ๐๐ = 0.
Hint: By equation (3.2.11),
๐๐ = โ๐ผ,๐(โซ๐ด
๐๐๐ผ๐๐ฅ๐(๐ฅ, ๐ก) ๐๐ก) ๐๐ฅ๐ โง ๐๐ฅ๐ผ = โซ
๐ด(๐๐๐ผ) ๐๐ก
and by equation (2.4.19) we have ๐๐๐ผ =๐๐ฝ๐๐ก .
Exercise 3.2.v. In Exercise 3.2.iv, show that if ๐ is in ๐๐บ๐โ2๐ (๐) then ๐ is in ๐๐บ๐โ1๐ (๐ ร ๐ด)as follows.
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ยง3.2 The Poincarรฉ lemma for compactly supported forms on rectangles 75
(1) Let ๐ = ๐๐, with ๐ โ ๐บ๐โ2๐ (๐) and let๐ โ ๐ถโ(๐) be a bump functionwhich is supportedon ๐ด and whose integral over ๐ด is one. Setting ๐ = โ๐(๐ก)๐๐ก โง ๐ show that
๐ โ ๐๐ = ๐๐ก โง (๐ผ โ ๐(๐ก)๐) + ๐ฝ= ๐๐ก โง (โ๐ผ ๐ข๐ผ(๐ฅ, ๐ก)๐๐ฅ๐ผ) + ๐ฝ ,
where
๐ข๐ผ(๐ฅ, ๐ก) = ๐๐ผ(๐ฅ, ๐ก) โ ๐(๐ก) โซ๐ด๐๐ผ(๐ฅ, ๐ก)๐๐ก .
(2) Let ๐ and ๐ be the end points of ๐ด and let
๐ฃ๐ผ(๐ฅ, ๐ก) = โซ๐ก
๐๐ข๐ผ(๐ฅ, ๐ก) ๐๐ก .
Show that the form โ๐ผ ๐ฃ๐ผ(๐ฅ, ๐ก) ๐๐ฅ๐ผ is in ๐บ๐โ1๐ (๐ ร ๐ด) and that
๐๐พ = ๐ โ ๐๐ โ ๐ฝ + ๐๐๐พ .
(3) Conclude that the form ๐ โ ๐(๐ + ๐พ) is reduced.(4) Prove that if ๐ โ ๐บ๐๐ (๐ ร ๐ด) is reduced and ๐๐ = 0 then ๐ = 0.
Hint: Let ๐ = โ๐ผ ๐๐ผ(๐ฅ, ๐ก) ๐๐ฅ๐ผ. Show that ๐๐ = 0 โ ๐๐๐ก๐๐ผ(๐ฅ, ๐ก) = 0 and exploit thefact that for fixed ๐ฅ, ๐๐ผ(๐ฅ, ๐ก) is compactly supported in ๐ก.
Exercise 3.2.vi. Let๐ be an open subset of๐๐.We say that๐ has property๐๐, for 1 โค ๐ < ๐,if every closed ๐-form ๐ โ ๐บ๐๐ (๐) is in ๐๐บ๐โ1๐ (๐). Prove that if the open set ๐ โ ๐๐โ1 inExercise 3.2.iv has property ๐๐โ1 then ๐ ร ๐ด has property ๐๐.
Exercise 3.2.vii. Show that if ๐ is the rectangle [๐1, ๐1] ร โฏ ร [๐๐, ๐๐] and ๐ = int๐ then๐ has property ๐๐.
Exercise 3.2.viii. Let๐๐ be the half-space
(3.2.12) ๐๐ โ { (๐ฅ1,โฆ, ๐ฅ๐) โ ๐๐ | ๐ฅ1 โค 0 }
and let ๐ โ ๐บ๐๐ (๐๐) be the ๐-form ๐ โ ๐๐๐ฅ1 โงโฏ โง ๐๐ฅ๐ with ๐ โ ๐ถโ0 (๐๐). Define
(3.2.13) โซ๐๐๐ โ โซ
๐๐๐(๐ฅ1,โฆ, ๐ฅ๐)๐๐ฅ1โฏ๐๐ฅ๐
where the right hand side is the usual Riemann integral of ๐ over ๐๐. (This integral makessense since ๐ is compactly supported.) Show that if ๐ = ๐๐ for some ๐ โ ๐บ๐โ1๐ (๐๐) then
(3.2.14) โซ๐๐๐ = โซ๐๐โ1๐โ๐
where ๐ โถ ๐๐โ1 โ ๐๐ is the inclusion map
(๐ฅ2,โฆ, ๐ฅ๐) โฆ (0, ๐ฅ2,โฆ, ๐ฅ๐) .
Hint: Let ๐ = โ๐ ๐๐ ๐๐ฅ1 โง โฏ๐๐ฅ๐โฏ โง ๐๐ฅ๐. Mimicking the โ(2)โ(1)โ part of the proofof Theorem 3.2.2 show that the integral (3.2.13) is the integral over ๐๐โ1 of the function
โซ0
โโ
๐๐1๐๐ฅ1(๐ฅ1, ๐ฅ2,โฆ, ๐ฅ๐)๐๐ฅ1 .
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76 Chapter 3: Integration of Forms
3.3. The Poincarรฉ lemma for compactly supported forms on open subsets of ๐๐
In this section we will generalize Theorem 3.2.2 to arbitrary connected open subsets of๐๐.Theorem3.3.1 (Poincarรฉ lemma for compactly supported forms). Let๐ be a connected opensubset of ๐๐ and let ๐ be a compactly supported ๐-form with supp(๐) โ ๐. The the followingassertions are equivalent:(1) โซ๐๐ ๐ = 0.(2) There exists a compactly supported (๐ โ 1)-form ๐ with supp ๐ โ ๐ and ๐ = ๐๐.Proof that (2)โ(1). The support of ๐ is contained in a large rectangle, so the integral of ๐๐is zero by Theorem 3.2.2. โกProof that (1)โ(2). Let ๐1 and ๐2 be compactly supported ๐-forms with support in ๐. Wewill write
๐1 โผ ๐2as shorthand notation for the statement: There exists a compactly supported (๐ โ 1)-form,๐, with support in ๐ and with ๐1 โ ๐2 = ๐๐. We will prove that (1)โ(2) by proving anequivalent statement: Fix a rectangle, ๐0 โ ๐ and an ๐-form, ๐0, with supp๐0 โ ๐0 andintegral equal to one.
Theorem 3.3.2. If ๐ is a compactly supported ๐-form with supp(๐) โ ๐ and ๐ = โซ๐ then๐ โผ ๐๐0.
Thus in particular if ๐ = 0, Theorem 3.3.2 says that ๐ โผ 0 proving that (1)โ(2). โกTo prove Theorem 3.3.2 let ๐๐ โ ๐, ๐ = 1, 2, 3,โฆ, be a collection of rectangles with
๐ = โโ๐=1 int(๐๐) and let ๐๐ be a partition of unity with supp(๐๐) โ int(๐๐). Replacing ๐by the finite sum โ๐๐=1 ๐๐๐ for ๐ large, it suffices to prove Theorem 3.3.2 for each of thesummands ๐๐๐. In other words we can assume that supp(๐) is contained in one of the openrectangles int(๐๐). Denote this rectangle by ๐. We claim that one can join ๐0 to ๐ by asequence of rectangles as in the figure below.
๏ฟฝ
๏ฟฝ0
Figure 3.3.1. A sequence of rectangles joining the rectangles ๐0 and ๐
Lemma3.3.3. There exists a sequence of rectangles๐ 0,โฆ, ๐ ๐+1 such that๐ 0 = ๐0,๐ ๐+1 = ๐and int(๐ ๐) โฉ int(๐ ๐+1) is nonempty.Proof. Denote by ๐ด the set of points, ๐ฅ โ ๐, for which there exists a sequence of rectangles,๐ ๐, ๐ = 0,โฆ,๐ + 1 with ๐ 0 = ๐0, with ๐ฅ โ int๐ ๐+1 and with int๐ ๐ โฉ int๐ ๐+1 nonempty.It is clear that this set is open and that its complement is open; so, by the connectivity of ๐,๐ = ๐ด. โก
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ยง3.4 The degree of a differentiable mapping 77
To prove Theorem 3.3.2 with supp๐ โ ๐, select, for each ๐, a compactly supported๐-form ๐๐ with supp(๐๐) โ int(๐ ๐) โฉ int(๐ ๐+1) and with โซ ๐๐ = 1. The difference, ๐๐ โ ๐๐+1is supported in int๐ ๐+1, and its integral is zero. So by Theorem 3.2.2, ๐๐ โผ ๐๐+1. Similarly,๐0 โผ ๐0 and, setting ๐ โ โซ๐, we have ๐ โผ ๐๐๐. Thus
๐๐0 โผ ๐๐0 โผ โฏ โผ ๐๐๐ = ๐proving the theorem.
3.4. The degree of a differentiable mapping
Definition 3.4.1. Let๐ and๐ be open subsets of๐๐ and๐๐. A continuous map ๐โถ ๐ โ ๐,is proper if for every compact subset ๐พ โ ๐, the preimage ๐โ1(๐พ) is compact in ๐.
Proper mappings have a number of nice properties which will be investigated in theexercises below. One obvious property is that if ๐ is a ๐ถโ mapping and ๐ is a compactlysupported ๐-form with support on ๐, ๐โ๐ is a compactly supported ๐-form with supporton ๐. Our goal in this section is to show that if ๐ and ๐ are connected open subsets of ๐๐and๐โถ ๐ โ ๐ is a proper๐ถโmapping then there exists a topological invariant of๐, whichwe will call its degree (and denote by deg(๐)), such that the โchange of variablesโ formula:
(3.4.2) โซ๐๐โ๐ = deg(๐) โซ
๐๐
holds for all ๐ โ ๐บ๐๐ (๐).Before we prove this assertion letโs see what this formula says in coordinates. If
๐ = ๐(๐ฆ)๐๐ฆ1 โงโฏ โง ๐๐ฆ๐then at ๐ฅ โ ๐
๐โ๐ = (๐ โ ๐)(๐ฅ) det(๐ท๐(๐ฅ))๐๐ฅ1 โงโฏ โง ๐๐ฅ๐ .Hence, in coordinates, equation (3.4.2) takes the form
(3.4.3) โซ๐๐(๐ฆ)๐๐ฆ = deg(๐) โซ
๐๐ โ ๐(๐ฅ) det(๐ท๐(๐ฅ))๐๐ฅ .
Proof of equation (3.4.2). Let ๐0 be a compactly-supported ๐-form with supp(๐0) โ ๐ andwith โซ๐0 = 1. If we set deg๐ โ โซ๐ ๐
โ๐0 then (3.4.2) clearly holds for ๐0. We will provethat (3.4.2) holds for every compactly supported ๐-form ๐ with supp(๐) โ ๐. Let ๐ โ โซ๐ ๐.Then by Theorem 3.3.1 ๐ โ ๐๐0 = ๐๐, where ๐ is a compactly supported (๐ โ 1)-form withsupp ๐ โ ๐. Hence
๐โ๐ โ ๐๐โ๐0 = ๐โ๐๐ = ๐๐โ๐ ,and by part (1) of Theorem 3.3.1
โซ๐๐โ๐ = ๐โซ๐โ๐0 = deg(๐) โซ
๐๐ . โก
We will show in ยง3.6 that the degree of ๐ is always an integer and explain why it is aโtopologicalโ invariant of ๐.
Proposition 3.4.4. For the moment, however, weโll content ourselves with pointing out a sim-ple but useful property of this invariant. Let ๐, ๐ and๐ be connected open subsets of ๐๐ and๐โถ ๐ โ ๐ and ๐โถ ๐ โ ๐ proper ๐ถโ maps. Then
(3.4.5) deg(๐ โ ๐) = deg(๐) deg(๐) .
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78 Chapter 3: Integration of Forms
Proof. Let ๐ be a compactly supported ๐-form with support on๐. Then
(๐ โ ๐)โ๐ = ๐โ๐โ๐ ;so
โซ๐(๐ โ ๐)โ๐ = โซ
๐๐โ(๐โ๐) = deg(๐) โซ
๐๐โ๐
= deg(๐) deg(๐) โซ๐๐ . โก
From this multiplicative property it is easy to deduce the following result (which wewill need in the next section).
Theorem 3.4.6. Let ๐ด be a non-singular ๐ ร ๐matrix and ๐๐ด โถ ๐๐ โ ๐๐ the linear mappingassociated with ๐ด. Then deg(๐๐ด) = +1 if det๐ด is positive and โ1 if det๐ด is negative.
A proof of this result is outlined in Exercises 3.4.v to 3.4.ix below.
Exercises for ยง3.4Exercise 3.4.i. Let๐ be an open subset of ๐๐ and (๐๐)๐โฅ1 be a partition of unity on๐. Showthat the mapping ๐โถ ๐ โ ๐ defined by
๐ โโโ๐=1๐๐๐
is a proper ๐ถโ mapping.
Exercise 3.4.ii. Let ๐ and ๐ be open subsets of ๐๐ and ๐๐ and let ๐โถ ๐ โ ๐ be a propercontinuous mapping. Prove:
Theorem 3.4.7. If ๐ต is a compact subset of ๐ and ๐ด = ๐โ1(๐ต) then for every open subset ๐0with ๐ด โ ๐0 โ ๐, there exists an open subset ๐0 with ๐ต โ ๐0 โ ๐ and ๐โ1(๐0) โ ๐0.
Hint: Let ๐ถ be a compact subset of ๐ with ๐ต โ int๐ถ. Then the set๐ = ๐โ1(๐ถ) โ ๐0 iscompact; so its image ๐(๐) is compact. Show that ๐(๐) and ๐ต are disjoint and let
๐0 = int๐ถ โ ๐(๐) .
Exercise 3.4.iii. Show that if ๐โถ ๐ โ ๐ is a proper continuous mapping and๐ is a closedsubset of ๐, then ๐(๐) is closed.
Hint: Let๐0 = ๐โ๐. Show that if ๐ is in๐โ๐(๐), then ๐โ1(๐) is contained in๐0 andconclude from Exercise 3.4.ii that there exists a neighborhood ๐0 of ๐ such that ๐โ1(๐0) iscontained in ๐0. Conclude that ๐0 and ๐(๐) are disjoint.
Exercise 3.4.iv. Let ๐โถ ๐๐ โ ๐๐ be the translation ๐(๐ฅ) = ๐ฅ + ๐. Show that deg(๐) = 1.Hint: Let ๐โถ ๐ โ ๐ be a compactly supported ๐ถโ function. For ๐ โ ๐, the identity
(3.4.8) โซ๐๐(๐ก) ๐๐ก = โซ
๐๐(๐ก โ ๐) ๐๐ก
is easy to prove by elementary calculus, and this identity proves the assertion above in di-mension one. Now let
(3.4.9) ๐(๐ฅ) = ๐(๐ฅ1)โฏ๐(๐ฅ๐)and compute the right and left sides of equation (3.4.3) by Fubiniโs theorem.
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ยง3.4 The degree of a differentiable mapping 79
Exercise 3.4.v. Let ๐ be a permutation of the numbers, 1,โฆ, ๐ and let ๐๐ โถ ๐๐ โ ๐๐ be thediffeomorphism, ๐๐(๐ฅ1,โฆ, ๐ฅ๐) = (๐ฅ๐(1),โฆ, ๐ฅ๐(๐)). Prove that deg๐๐ = (โ1)๐.
Hint: Let ๐ be the function (3.4.9). Show that if ๐ = ๐(๐ฅ) ๐๐ฅ1 โงโฏโง ๐๐ฅ๐, then we have๐โ๐ = (โ1)๐๐.Exercise 3.4.vi. Let ๐โถ ๐๐ โ ๐๐ be the mapping
๐(๐ฅ1,โฆ, ๐ฅ๐) = (๐ฅ1 + ๐๐ฅ2, ๐ฅ2,โฆ, ๐ฅ๐).Prove that deg(๐) = 1.
Hint: Let ๐ = ๐(๐ฅ1,โฆ, ๐ฅ๐) ๐๐ฅ1 โง โฏ โง ๐๐ฅ๐ where ๐โถ ๐๐ โ ๐ is compactly supportedand of class ๐ถโ. Show that
โซ๐โ๐ = โซ๐(๐ฅ1 + ๐๐ฅ2, ๐ฅ2,โฆ, ๐ฅ๐) ๐๐ฅ1โฏ๐๐ฅ๐and evaluate the integral on the right by Fubiniโs theorem; i.e., by first integrating with re-spect to the ๐ฅ1 variable and then with respect to the remaining variables. Note that by equa-tion (3.4.8)
โซ๐(๐ฅ1 + ๐๐ฅ2, ๐ฅ2,โฆ, ๐ฅ๐) ๐๐ฅ1 = โซ๐(๐ฅ1, ๐ฅ2,โฆ, ๐ฅ๐) ๐๐ฅ1 .
Exercise 3.4.vii. Let ๐โถ ๐๐ โ ๐๐ be the mapping๐(๐ฅ1,โฆ, ๐ฅ๐) = (๐๐ฅ1, ๐ฅ2,โฆ, ๐ฅ๐)
with ๐ โ 0. Show that deg๐ = +1 if ๐ is positive and โ1 if ๐ is negative.Hint: In dimension 1 this is easy to prove by elementary calculus techniques. Prove it
in ๐-dimensions by the same trick as in the previous exercise.
Exercise 3.4.viii.(1) Let ๐1,โฆ, ๐๐ be the standard basis vectors of ๐๐ and ๐ด, ๐ต and ๐ถ the linear mappings
defined by
๐ด๐๐ = {๐1, ๐ = 1โ๐๐=1 ๐๐,๐๐๐, ๐ โ 1 ,
๐ต๐๐ = {โ๐๐=1 ๐๐๐๐, ๐ = 1๐๐, ๐ โ 1 ,
(3.4.10)
๐ถ๐๐ = {๐1, ๐ = 1๐๐ + ๐๐๐1, ๐ โ 1 .
Show that
๐ต๐ด๐ถ๐๐ = {โ๐๐=1 ๐๐๐๐, ๐ = 1โ๐๐=1(๐๐,๐ + ๐๐๐๐)๐๐ + ๐๐๐1๐1, ๐ โ 1 .
for ๐ > 1.(2) Let ๐ฟโถ ๐๐ โ ๐๐ be the linear mapping
(3.4.11) ๐ฟ๐๐ =๐โ๐=1โ๐,๐๐๐ , ๐ = 1,โฆ, ๐ .
Show that if โ1,1 โ 0 one can write ๐ฟ as a product, ๐ฟ = ๐ต๐ด๐ถ, where๐ด, ๐ต and๐ถ are linearmappings of the form (3.4.10).
Hint: First solve the equationsโ๐,1 = ๐๐
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80 Chapter 3: Integration of Forms
for ๐ = 1,โฆ, ๐. Next solve the equations
โ1,๐ = ๐1๐๐for ๐ > 1. Finally, solve the equations
โ๐,๐ = ๐๐,๐ + ๐๐๐๐for ๐, ๐ > 1.
(3) Suppose ๐ฟ is invertible. Conclude that ๐ด, ๐ต and ๐ถ are invertible and verify that Theo-rem 3.4.6 holds for ๐ต and ๐ถ using the previous exercises in this section.
(4) Show by an inductive argument that Theorem 3.4.6 holds for ๐ด and conclude from(3.4.5) that it holds for ๐ฟ.
Exercise 3.4.ix. To show that Theorem 3.4.6 holds for an arbitrary linear mapping ๐ฟ of theform (3.4.11) weโll need to eliminate the assumption: โ1,1 โ 0. Show that for some ๐, โ๐,1 isnonzero, and show how to eliminate this assumption by considering ๐๐1,๐ โ ๐ฟ where ๐1,๐ isthe transposition 1 โ ๐.
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 ๐ฟโบ๐ฟ = id๐.
Hints:โข Show that ๐ฟโ(๐ฅ21 +โฏ + ๐ฅ2๐) = ๐ฅ21 +โฏ + ๐ฅ2๐.โข Show that ๐ฟโ(๐๐ฅ1 โง โฏ โง ๐๐ฅ๐) is equal to ๐๐ฅ1 โง โฏ โง ๐๐ฅ๐ or โ๐๐ฅ1 โง โฏ โง ๐๐ฅ๐
depending on whether ๐ฟ is orientation-preserving or orinetation reversing. (SeeExercise 1.2.x.)
โข Let ๐ be as in Exercise 3.4.iv and let ๐ be the form
๐ = ๐(๐ฅ21 +โฏ + ๐ฅ2๐) ๐๐ฅ1 โงโฏ โง ๐๐ฅ๐ .Show that ๐ฟโ๐ = ๐ if ๐ฟ is orientation-preserving and ๐ฟโ๐ = โ๐ if ๐ฟ is orientationreversing.
(2) Prove Theorem 3.4.6 for linear mappings which are self-adjoint (satisfy ๐ฟโบ = ๐ฟ).Hint: A self-adjoint linear mapping is diagonizable: there exists an intervertible lin-
ear mapping,๐โถ ๐๐ โ ๐๐ such that
(3.4.12) ๐โ1๐ฟ๐๐๐ = ๐๐๐๐ , ๐ = 1,โฆ, ๐ .(3) Prove that every invertible linearmapping, ๐ฟ, can bewritten as a product, ๐ฟ = ๐ต๐ถwhere๐ต is orthogonal and ๐ถ is self-adjoint.
Hints:โข Show that the mapping, ๐ด = ๐ฟโบ๐ฟ, is self-adjoint and that it is eigenvalues (the ๐๐โs
in equation (3.4.12)) are positive.โข Show that there exists an invertible self-adjoint linear mapping, ๐ถ, such that ๐ด =๐ถ2 and ๐ด๐ถ = ๐ถ๐ด.
โข Show that the mapping ๐ต = ๐ฟ๐ถโ1 is orthogonal.
3.5. The change of variables formula
Let ๐ and ๐ be connected open subsets of ๐๐. If ๐๐ โฅฒ ๐ is a diffeomorphism, thedeterminant of๐ท๐(๐ฅ) at ๐ฅ โ ๐ is nonzero, and hence, since๐ท๐(๐ฅ) is a continuous functionof ๐ฅ, its sign is the same at every point. We will say that ๐ is orientation-preserving if thissign is positive and orientation reversing if it is negative. We will prove below:
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ยง3.5 The change of variables formula 81
Theorem 3.5.1. The degree of ๐ is +1 if ๐ is orientation-preserving and โ1 if ๐ is orientationreversing.
We will then use this result to prove the following change of variables formula for dif-feomorphisms.
Theorem 3.5.2. Let ๐โถ ๐ โ ๐ be a compactly supported continuous function. Then
(3.5.3) โซ๐(๐ โ ๐)(๐ฅ)|det(๐ท๐(๐ฅ))| = โซ
๐๐(๐ฆ) ๐๐ฆ .
Proof of Theorem 3.5.1. Given a point ๐1 โ ๐, let ๐2 = โ๐(๐1) and for ๐ = 1, 2 let ๐๐ โถ ๐๐ โ๐๐ be the translation, ๐๐(๐ฅ) = ๐ฅ + ๐๐. By equation (3.4.2) and Exercise 3.4.iv the compositediffeomorphism(3.5.4) ๐2 โ ๐ โ ๐1has the samedegree as๐, so it suffices to prove the theorem for thismapping.Notice howeverthat this mapping maps the origin onto the origin. Hence, replacing ๐ by this mapping, wecan, without loss of generality, assume that 0 is in the domain of ๐ and that ๐(0) = 0.
Next notice that if ๐ดโถ ๐๐ โฅฒ ๐๐ is a bijective linear mapping the theorem is true for ๐ด(by Exercise 3.4.ix), and hence if we can prove the theorem for๐ดโ1 โ ๐, equation (3.4.2) willtell us that the theorem is true for ๐. In particular, letting ๐ด = ๐ท๐(0), we have
๐ท(๐ดโ1 โ ๐)(0) = ๐ดโ1๐ท๐(0) = id๐where id๐ is the identity mapping. Therefore, replacing ๐ by ๐ดโ1 โ ๐, we can assume thatthe mapping ๐ (for which we are attempting to prove Theorem 3.5.1) has the properties:๐(0) = 0 and ๐ท๐(0) = id๐. Let ๐(๐ฅ) = ๐(๐ฅ) โ ๐ฅ. Then these properties imply that ๐(0) = 0and๐ท๐(0) = 0. โก
Lemma 3.5.5. There exists a ๐ฟ > 0 such that |๐(๐ฅ)| โค 12 |๐ฅ| for |๐ฅ| โค ๐ฟ.
Proof. Let ๐(๐ฅ) = (๐1(๐ฅ),โฆ, ๐๐(๐ฅ)). Then๐๐๐๐๐ฅ๐(0) = 0 ;
so there exists a ๐ฟ > 0 such that
| ๐๐๐๐๐ฅ๐(๐ฅ)| โค 12
for |๐ฅ| โค ๐ฟ. However, by the mean value theorem,
๐๐(๐ฅ) =๐โ๐=1
๐๐๐๐๐ฅ๐(๐)๐ฅ๐
for ๐ = ๐ก0๐ฅ, 0 < ๐ก0 < 1. Thus, for |๐ฅ| < ๐ฟ,
|๐๐(๐ฅ)| โค12
sup |๐ฅ๐| =12|๐ฅ| ,
so|๐(๐ฅ)| = sup |๐๐(๐ฅ)| โค
12|๐ฅ| . โก
Let ๐ be a compactly supported ๐ถโ function with 0 โค ๐ โค 1 and with ๐(๐ฅ) = 0 for|๐ฅ| โฅ ๐ฟ and ๐(๐ฅ) = 1 for |๐ฅ| โค ๐ฟ2 and let ๐ โถ ๐๐ โ ๐๐ be the mapping
๐(๐ฅ) โ ๐ฅ + ๐(๐ฅ)๐(๐ฅ) .
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82 Chapter 3: Integration of Forms
It is clear that(3.5.6) ๐(๐ฅ) = ๐ฅ for |๐ฅ| โฅ ๐ฟand, since ๐(๐ฅ) = ๐ฅ + ๐(๐ฅ),
(3.5.7) ๐(๐ฅ) = ๐(๐ฅ) for |๐ฅ| โค ๐ฟ2
.
In addition, for all ๐ฅ โ ๐๐:
(3.5.8) | ๐(๐ฅ)| โฅ 12|๐ฅ| .
Indeed, by (3.5.6), | ๐(๐ฅ)| โฅ |๐ฅ| for |๐ฅ| โฅ ๐ฟ, and for |๐ฅ| โค ๐ฟ| ๐(๐ฅ)| โฅ |๐ฅ| โ ๐(๐ฅ)|๐(๐ฅ)|
โฅ |๐ฅ| โ |๐(๐ฅ)| โฅ |๐ฅ| โ 12|๐ฅ|
= 12|๐ฅ|
by Lemma 3.5.5.Now let ๐๐ be the cube ๐๐ โ {๐ฅ โ ๐๐ | |๐ฅ| โค ๐ }, and let ๐๐๐ โ ๐๐ โ ๐๐. From (3.5.8)
we easily deduce that
(3.5.9) ๐โ1(๐๐) โ ๐2๐for all ๐, and hence that ๐ is proper. Also notice that for ๐ฅ โ ๐๐ฟ,
| ๐(๐ฅ)| โค |๐ฅ| + |๐(๐ฅ)| โค 32|๐ฅ|
by Lemma 3.5.5 and hence
(3.5.10) ๐โ1(๐๐32 ๐ฟ) โ ๐๐๐ฟ .
We will now prove Theorem 3.5.1.
Proof of Theorem 3.5.1. Since ๐ is a diffeomorphism mapping 0 to 0, it maps a neighbor-hood ๐0 of 0 in ๐ diffeomorphically onto a neighborhood ๐0 of 0 in ๐, and, by shrinking๐0 if necessary, we can assume that ๐0 is contained in ๐๐ฟ/2 and ๐0 contained in ๐๐ฟ/4. Let ๐be an ๐-form with support in ๐0 whose integral over ๐๐ is equal to one. Then ๐โ๐ is sup-ported in๐0 and hence in๐๐ฟ/2. Also by (3.5.9) ๐โ๐ is supported in๐๐ฟ/2.Thus both of theseforms are zero outside ๐๐ฟ/2. However, on ๐๐ฟ/2, ๐ = ๐ by (3.5.7), so these forms are equaleverywhere, and hence
deg(๐) = โซ๐โ๐ = โซ ๐โ๐ = deg( ๐) .
Next let ๐ be a compactly supported ๐-form with support in ๐๐3๐ฟ/2 and with integralequal to one. Then ๐โ๐ is supported in๐๐๐ฟ by (3.5.10), and hence since ๐(๐ฅ) = ๐ฅ on๐๐๐ฟ, wehave ๐โ๐ = ๐. Thus
deg( ๐) = โซ ๐โ๐ = โซ๐ = 1 .
Putting these two identities together we conclude that deg(๐) = 1. โก
If the function, ๐, in equation (3.5.4) is a ๐ถโ function, the identity (3.5.3) is an im-mediate consequence of the result above and the identity (3.4.3). If ๐ is not ๐ถโ, but is justcontinuous, we will deduce equation (3.5.4) from the following result.
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ยง3.5 The change of variables formula 83
Theorem 3.5.11. Let ๐ be an open subset of ๐๐. If ๐โถ ๐๐ โ ๐ is a continuous function ofcompact support with supp๐ โ ๐; then for every ๐ > 0 there exists a ๐ถโ function of compactsupport, ๐โถ ๐๐ โ ๐ with supp๐ โ ๐ and
sup |๐(๐ฅ) โ ๐(๐ฅ)| < ๐ .Proof. Let ๐ด be the support of ๐ and let ๐ be the distance in the sup norm from ๐ด to thecomplement of ๐. Since ๐ is continuous and compactly supported it is uniformly continu-ous; so for every ๐ > 0 there exists a ๐ฟ > 0 with ๐ฟ < ๐2 such that |๐(๐ฅ) โ ๐(๐ฆ)| < ๐ when|๐ฅ โ ๐ฆ| โค ๐ฟ. Now let ๐ be the cube: |๐ฅ| < ๐ฟ and let ๐โถ ๐๐ โ ๐ be a non-negative ๐ถโfunction with supp ๐ โ ๐ and
(3.5.12) โซ๐(๐ฆ)๐๐ฆ = 1 .
Set๐(๐ฅ) = โซ๐(๐ฆ โ ๐ฅ)๐(๐ฆ)๐๐ฆ .
By Theorem 3.2.10 ๐ is a ๐ถโ function. Moreover, if ๐ด๐ฟ is the set of points in ๐๐ whosedistance in the sup norm from ๐ด is โค ๐ฟ then for ๐ฅ โ ๐ด๐ฟ and ๐ฆ โ ๐ด , |๐ฅ โ ๐ฆ| > ๐ฟ and hence๐(๐ฆ โ ๐ฅ) = 0. Thus for ๐ฅ โ ๐ด๐ฟ
โซ๐(๐ฆ โ ๐ฅ)๐(๐ฆ)๐๐ฆ = โซ๐ด๐(๐ฆ โ ๐ฅ)๐(๐ฆ)๐๐ฆ = 0 ,
so ๐ is supported on the compact set ๐ด๐ฟ. Moreover, since ๐ฟ < ๐2 , supp๐ is contained in ๐.Finally note that by (3.5.12) and Exercise 3.4.iv
(3.5.13) โซ๐(๐ฆ โ ๐ฅ) ๐๐ฆ = โซ๐(๐ฆ) ๐๐ฆ = 1
and hence๐(๐ฅ) = โซ๐(๐ฅ)๐(๐ฆ โ ๐ฅ)๐๐ฆ
so๐(๐ฅ) โ ๐(๐ฅ) = โซ(๐(๐ฅ) โ ๐(๐ฆ))๐(๐ฆ โ ๐ฅ)๐๐ฆ
and|๐(๐ฅ) โ ๐(๐ฅ)| โค โซ |๐(๐ฅ) โ ๐(๐ฆ)|๐(๐ฆ โ ๐ฅ)๐๐ฆ .
But ๐(๐ฆ โ ๐ฅ) = 0 for |๐ฅ โ ๐ฆ| โฅ ๐ฟ; and |๐(๐ฅ) โ ๐(๐ฆ)| < ๐ for |๐ฅ โ ๐ฆ| โค ๐ฟ, so the integrand onthe right is less than
๐ โซ ๐(๐ฆ โ ๐ฅ) ๐๐ฆ ,
and hence by equation (3.5.13) we have|๐(๐ฅ) โ ๐(๐ฅ)| โค ๐ . โก
To prove the identity (3.5.3), let ๐พโถ ๐๐ โ ๐ be a ๐ถโ cut-off function which is one ona neighborhood ๐1 of the support of ๐ is non-negative, and is compactly supported withsupp ๐พ โ ๐, and let
๐ = โซ ๐พ(๐ฆ) ๐๐ฆ .
By Theorem 3.5.11 there exists, for every ๐ > 0, a ๐ถโ function ๐, with support on ๐1 satis-fying
(3.5.14) |๐ โ ๐| โค ๐2๐
.
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84 Chapter 3: Integration of Forms
Thus
|โซ๐(๐ โ ๐)(๐ฆ)๐๐ฆ| โค โซ
๐|๐ โ ๐|(๐ฆ)๐๐ฆ
โค โซ๐๐พ|๐ โ ๐|(๐ฅ๐ฆ)๐๐ฆ
โค ๐2๐โซ ๐พ(๐ฆ)๐๐ฆ โค ๐
2so
(3.5.15) |โซ๐๐(๐ฆ) ๐๐ฆ โ โซ
๐๐(๐ฆ) ๐๐ฆ| โค ๐
2.
Similarly, the expression
|โซ๐(๐ โ ๐) โ ๐(๐ฅ)|det๐ท๐(๐ฅ)| ๐๐ฅ|
is less than or equal to the integral
โซ๐(๐พ โ ๐)(๐ฅ)|(๐ โ ๐) โ ๐(๐ฅ)| |det๐ท๐(๐ฅ)| ๐๐ฅ
and by (3.5.14), |(๐ โ ๐) โ ๐(๐ฅ)| โค ๐2๐ , so this integral is less than or equal to๐2๐โซ(๐พ โ ๐)(๐ฅ)|det๐ท๐(๐ฅ)| ๐๐ฅ
and hence by (3.5.3) is less than or equal to ๐2 . Thus
(3.5.16) |โซ๐(๐ โ ๐)(๐ฅ) | det๐ท๐(๐ฅ)|๐๐ฅ โ โซ
๐๐ โ ๐(๐ฅ)| det๐ท๐(๐ฅ)| ๐๐ฅ| โค ๐
2.
Combining (3.5.15), (3.5.16) and the identity
โซ๐๐(๐ฆ) ๐๐ฆ = โซ๐ โ ๐(๐ฅ)|det๐ท๐(๐ฅ)| ๐๐ฅ
we get, for all ๐ > 0,
|โซ๐๐(๐ฆ) ๐๐ฆ โ โซ
๐(๐ โ ๐)(๐ฅ)|det๐ท๐(๐ฅ)| ๐๐ฅ| โค ๐
and henceโซ๐(๐ฆ) ๐๐ฆ = โซ(๐ โ ๐)(๐ฅ)|det๐ท๐(๐ฅ)| ๐๐ฅ .
Exercises for ยง3.5Exercise 3.5.i. Let โโถ ๐ โ ๐ be a non-negative continuous function. Show that if theimproper integral
โซ๐โ(๐ฆ)๐๐ฆ
is well-defined, then the improper integral
โซ๐(โ โ ๐)(๐ฅ)|det๐ท๐(๐ฅ)|๐๐ฅ
is well-defined and these two integrals are equal.Hint: If (๐๐)๐โฅ1 is a partition of unity on ๐ then ๐๐ = ๐๐ โ ๐ is a partition of unity on ๐
andโซ๐๐โ๐๐ฆ = โซ๐๐(โ โ ๐(๐ฅ))|det๐ท๐(๐ฅ)|๐๐ฅ .
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ยง3.6 Techniques for computing the degree of a mapping 85
Now sum both sides of this identity over ๐.Exercise 3.5.ii. Show that the result above is true without the assumption that โ is non-negative.
Hint: โ = โ+ โ โโ, where โ+ = max(โ, 0) and โโ = max(โโ, 0).Exercise 3.5.iii. Show that in equation (3.4.3) one can allow the function ๐ to be a continu-ous compactly supported function rather than a ๐ถโ compactly supported function.
Exercise 3.5.iv. Let๐๐ be the half-space (3.2.12) and๐ and๐ open subsets of ๐๐. Suppose๐โถ ๐ โ ๐ is an orientation-preserving diffeomorphism mapping ๐ โฉ ๐๐ onto ๐ โฉ ๐๐.Show that for ๐ โ ๐บ๐๐ (๐)
(3.5.17) โซ๐โฉ๐๐๐โ๐ = โซ
๐โฉ๐๐๐ .
Hint: Interpret the left and right hand sides of this formula as improper integrals over๐ โฉ int(๐๐) and ๐ โฉ int(๐๐).Exercise 3.5.v. The boundary of๐๐ is the set
๐๐๐ โ { (0, ๐ฅ2,โฆ, ๐ฅ๐) | (๐ฅ2,โฆ, ๐ฅ๐) โ ๐๐โ1 }so the map
๐ โถ ๐๐โ1 โ ๐๐ , (๐ฅ2,โฆ, ๐ฅ๐) โฆ (0, ๐ฅ2,โฆ, ๐ฅ๐)in Exercise 3.2.viii maps ๐๐โ1 bijectively onto ๐๐๐.(1) Show that the map ๐โถ ๐ โ ๐ in Exercise 3.5.iv maps ๐ โฉ ๐๐๐ onto ๐ โฉ ๐๐๐.(2) Let๐โฒ = ๐โ1(๐) and๐โฒ = ๐โ1(๐). Conclude from (1) that the restriction of๐ to๐โฉ๐๐๐
gives one a diffeomorphism๐โถ ๐โฒ โ ๐โฒ
satisfying:๐ โ ๐ = ๐ โ ๐ .
(3) Let ๐ be in ๐บ๐โ1๐ (๐). Conclude from equations (3.2.14) and (3.5.17):
โซ๐โฒ๐โ๐โ๐ = โซ
๐โฒ๐โ๐
and in particular show that the diffeomorphism ๐โถ ๐โฒ โ ๐โฒ is orientation-preserving.
3.6. Techniques for computing the degree of a mapping
Let ๐ and ๐ be open subsets of ๐๐ and ๐โถ ๐ โ ๐ a proper ๐ถโ mapping. In thissectionwewill show how to compute the degree of๐ and, in particular, show that it is alwaysan integer. From this fact we will be able to conclude that the degree of ๐ is a topologicalinvariant of ๐: if we deform ๐ smoothly, its degree doesnโt change.
Definition 3.6.1. A point ๐ฅ โ ๐ is a critical point of ๐ if the derivative๐ท๐(๐ฅ)โถ ๐๐ โ ๐๐
fails to be bijective, i.e., if det(๐ท๐(๐ฅ)) = 0.We will denote the set of critical points of ๐ by ๐ถ๐. It is clear from the definition that
this set is a closed subset of๐ and hence, by Exercise 3.4.iii,๐(๐ถ๐) is a closed subset of๐.Wewill call this image the set of critical values of ๐ and the complement of this image the set ofregular values of ๐. Notice that ๐ โ ๐(๐) is contained in ๐(๐) โ ๐(๐ถ๐), so if a point ๐ โ ๐is not in the image of ๐, it is a regular value of ๐ โby defaultโ, i.e., it contains no points of
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86 Chapter 3: Integration of Forms
๐ in the pre-image and hence, a fortiori, contains no critical points in its pre-image. Noticealso that ๐ถ๐ can be quite large. For instance, if ๐ โ ๐ and ๐โถ ๐ โ ๐ is the constant mapwhich maps all of ๐ onto ๐, then ๐ถ๐ = ๐. However, in this example, ๐(๐ถ๐) = {๐}, so the setof regular values of ๐ is ๐ โ {๐}, and hence (in this example) is an open dense subset of ๐.We will show that this is true in general.Theorem 3.6.2 (Sard). If ๐ and ๐ are open subsets of ๐๐ and ๐โถ ๐ โ ๐ a proper ๐ถโ map,the set of regular values of ๐ is an open dense subset of ๐.
We will defer the proof of this to Section 3.7 and in this section explore some of itsimplications. Picking a regular value ๐ of ๐ we will prove:Theorem 3.6.3. The set ๐โ1(๐) is a finite set. Moreover, if ๐โ1(๐) = {๐1,โฆ, ๐๐} there existconnected open neighborhoods๐๐ of ๐๐ in ๐ and an open neighborhood๐ of ๐ in๐ such that:(1) for ๐ โ ๐ the sets ๐๐ and ๐๐ are disjoint;(2) ๐โ1(๐) = ๐1 โชโฏ โช ๐๐,(3) ๐maps ๐๐ diffeomorphically onto๐.Proof. If ๐ โ ๐โ1(๐) then, since ๐ is a regular value, ๐ โ ๐ถ๐; so
๐ท๐(๐)โถ ๐๐ โ ๐๐
is bijective. Hence by the inverse function theorem, ๐maps a neighborhood,๐๐ of ๐ diffeo-morphically onto a neighborhood of ๐. The open sets
{ ๐๐ | ๐ โ ๐โ1(๐) }are a covering of ๐โ1(๐); and, since ๐ is proper, ๐โ1(๐) is compact. Thus we can extract afinite subcovering
{๐๐1 ,โฆ,๐๐๐}and since ๐๐ is the only point in ๐๐๐ which maps onto ๐, we have that ๐โ1(๐) = {๐1,โฆ, ๐๐}.
Without loss of generality we can assume that the๐๐๐ โs are disjoint from each other; for,if not, we can replace them by smaller neighborhoods of the ๐๐โs which have this property.By Theorem 3.4.7 there exists a connected open neighborhood๐ of ๐ in ๐ for which
๐โ1(๐) โ ๐๐1 โชโฏ โช ๐๐๐ .
To conclude the proof let ๐๐ โ ๐โ1(๐) โฉ ๐๐๐ . โก
The main result of this section is a recipe for computing the degree of ๐ by countingthe number of ๐๐โs above, keeping track of orientation.Theorem 3.6.4. For each ๐๐ โ ๐โ1(๐) let ๐๐๐ = +1 if ๐โถ ๐๐ โ ๐ is orientation-preservingand โ1 if ๐โถ ๐๐ โ๐ is orientation reversing. Then
(3.6.5) deg(๐) =๐โ๐=1๐๐๐ .
Proof. Let ๐ be a compactly supported ๐-form on๐ whose integral is one. Then
deg(๐) = โซ๐๐โ๐ =
๐โ๐=1โซ๐๐๐โ๐ .
Since ๐โถ ๐๐ โ๐ is a diffeomorphism
โซ๐๐๐โ๐ = ยฑโซ
๐๐ = {1, ๐ is orientation-preservingโ1, ๐ is not orientation-preserving .
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ยง3.6 Techniques for computing the degree of a mapping 87
Thus deg(๐) is equal to the sum (3.6.5). โกAs we pointed out above, a point ๐ โ ๐ can qualify as a regular value of ๐ โby defaultโ,
i.e., by not being in the image of ๐. In this case the recipe (3.6.5) for computing the degreegives โby defaultโ the answer zero. Letโs corroborate this directly.Theorem 3.6.6. If ๐โถ ๐ โ ๐ is not surjective, then deg(๐) = 0.Proof. By Exercise 3.4.iii, ๐ โ ๐(๐) is open; so if it is nonempty, there exists a compactlysupported ๐-form ๐ with support in ๐ โ ๐(๐) and with integral equal to one. Since ๐ = 0on the image of ๐, ๐โ๐ = 0; so
0 = โซ๐๐โ๐ = deg(๐) โซ
๐๐ = deg(๐) . โก
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(๐) โ 0, then ๐maps ๐ surjectively onto ๐.
In other words if deg(๐) โ 0 the equation๐(๐ฅ) = ๐ฆ
has a solution, ๐ฅ โ ๐ for every ๐ฆ โ ๐.We will now show that the degree of ๐ is a topological invariant of ๐: if we deform ๐ by
a โhomotopyโ we do not change its degree. To make this assertion precise, letโs recall whatwe mean by a homotopy between a pair of ๐ถโ maps. Let ๐ be an open subset of ๐๐, ๐an open subset of ๐๐, ๐ด an open subinterval of ๐ containing 0 and 1, and ๐1, ๐2 โถ ๐ โ ๐a pair of ๐ถโ maps. Then a ๐ถโ map ๐นโถ ๐ ร ๐ด โ ๐ is a homotopy between ๐0 and ๐1 if๐น(๐ฅ, 0) = ๐0(๐ฅ) and ๐น(๐ฅ, 1) = ๐1(๐ฅ). (See Definition 2.6.14.) Suppose now that ๐0 and ๐1 areproper.Definition 3.6.9. A homotopy ๐น between ๐0 and ๐1 is a proper homotopy if the map
๐นโฏ โถ ๐ ร ๐ด โ ๐ ร ๐ดdefined by (๐ฅ, ๐ก) โฆ (๐น(๐ฅ, ๐ก), ๐ก) is proper.
Note that if ๐น is a proper homotopy between ๐0 and ๐1, then for every ๐ก between 0 and1, the map
๐๐ก โถ ๐ โ ๐ , ๐๐ก(๐ฅ) โ ๐น(๐ฅ, ๐ก)is proper.
Now let ๐ and ๐ be open subsets of ๐๐.Theorem 3.6.10. If ๐0 and ๐1 are properly homotopic, then deg(๐0) = deg(๐1).Proof. Let
๐ = ๐(๐ฆ)๐๐ฆ1 โงโฏ โง ๐๐ฆ๐be a compactly supported ๐-form on ๐ whose integral over ๐ is 1. The the degree of ๐๐ก isequal to
(3.6.11) โซ๐๐(๐น1(๐ฅ, ๐ก),โฆ, ๐น๐(๐ฅ, ๐ก)) det๐ท๐ฅ๐น(๐ฅ, ๐ก)๐๐ฅ .
The integrand in (3.6.11) is continuous and for 0 โค ๐ก โค 1 is supported on a compact subsetof ๐ ร [0, 1], hence (3.6.11) is continuous as a function of ๐ก. However, as weโve just proved,deg(๐๐ก) is integer valued so this function is a constant. โก
(For an alternative proof of this result see Exercise 3.6.ix below.)
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88 Chapter 3: Integration of Forms
ApplicationsWeโll conclude this account of degree theory by describing a couple applications.
Application 3.6.12 (The Brouwer fixed point theorem). Let ๐ต๐ be the closed unit ball in๐๐:๐ต๐ โ { ๐ฅ โ ๐๐ | โ๐ฅโ โค 1 } .
Theorem 3.6.13. If ๐โถ ๐ต๐ โ ๐ต๐ is a continuous mapping then ๐ has a fixed point, i.e., mapssome point, ๐ฅ0 โ ๐ต๐ onto itself.
The idea of the proof will be to assume that there isnโt a fixed point and show that thisleads to a contradiction. Suppose that for every point ๐ฅ โ ๐ต๐ we have ๐(๐ฅ) โ ๐ฅ. Considerthe ray through ๐(๐ฅ) in the direction of ๐ฅ:
๐(๐ฅ) + ๐ (๐ฅ โ ๐(๐ฅ)) , ๐ โ [0,โ) .This ray intersects the boundary ๐๐โ1 โ ๐๐ต๐ in a unique point ๐พ(๐ฅ) (see Figure 3.6.1 below);and one of the exercises at the end of this section will be to show that the mapping ๐พโถ ๐ต๐ โ๐๐โ1 given by ๐ฅ โฆ ๐พ(๐ฅ), is a continuous mapping. Also it is clear from Figure 3.6.1 that๐พ(๐ฅ) = ๐ฅ if ๐ฅ โ ๐๐โ1, so we can extend ๐พ to a continuous mapping of ๐๐ into ๐๐ by letting ๐พbe the identity for โ๐ฅโ โฅ 1. Note that this extended mapping has the property
โ๐พ(๐ฅ)โ โฅ 1for all ๐ฅ โ ๐๐ and(3.6.14) ๐พ(๐ฅ) = ๐ฅfor all โ๐ฅโ โฅ 1. To get a contradiction weโll show that ๐พ can be approximated by a ๐ถโ mapwhich has similar properties. For this wewill need the following corollary ofTheorem 3.5.11.
Lemma 3.6.15. Let ๐ be an open subset of ๐๐, ๐ถ a compact subset of ๐ and ๐โถ ๐ โ ๐ acontinuous function which is ๐ถโ on the complement of ๐ถ. Then for every ๐ > 0, there exists a๐ถโ function ๐โถ ๐ โ ๐, such that ๐ โ ๐ has compact support and |๐ โ ๐| < ๐.Proof. Let ๐ be a bump function which is in ๐ถโ0 (๐) and is equal to 1 on a neighborhoodof ๐ถ. By Theorem 3.5.11 there exists a function ๐0 โ ๐ถโ0 (๐) such that |๐๐ โ ๐0| < ๐. Tocomplete the proof, let ๐(1 โ ๐)๐ + ๐0, and note that
๐ โ ๐ = (1 โ ๐)๐ + ๐๐ โ (1 โ ๐)๐ โ ๐0= ๐๐ โ ๐0 . โก
By applying Lemma 3.6.15 to each of the coordinates of the map ๐พ, one obtains a ๐ถโmap ๐โถ ๐๐ โ ๐๐ such that
โ๐ โ ๐พโ < ๐ < 1and such that ๐ = ๐พ on the complement of a compact set. However, by (3.6.14), this meansthat ๐ is equal to the identity on the complement of a compact set and hence (see Exer-cise 3.6.ix) that ๐ is proper and deg(๐) = 1. On the other hand by (3.6.19) and (3.6.14) wehave โ๐(๐ฅ)โ > 1 โ ๐ for all ๐ฅ โ ๐๐, so 0 โ im(๐) and hence by Theorem 3.6.4 we havedeg(๐) = 0, which provides the desired contradiction.
Application 3.6.16 (The fundamental theorem of algebra). Let๐(๐ง) = ๐ง๐ + ๐๐โ1๐ง๐โ1 +โฏ + ๐1๐ง + ๐0
be a polynomial of degree ๐ with complex coefficients. If we identify the complex plane๐ โ { ๐ง = ๐ฅ + ๐๐ฆ | ๐ฅ, ๐ฆ โ ๐ }
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ยง3.6 Techniques for computing the degree of a mapping 89
x
๏ฟฝ(๏ฟฝ)
๏ฟฝ(๏ฟฝ)
Figure 3.6.1. Brouwer fixed point theorem
with ๐2 via the map ๐2 โ ๐ given by (๐ฅ, ๐ฆ) โฆ ๐ง = ๐ฅ + ๐๐ฆ, we can think of ๐ as defining amapping
๐โถ ๐2 โ ๐2 , ๐ง โฆ ๐(๐ง) .We will prove:
Theorem 3.6.17. The mapping ๐โถ ๐2 โ ๐2 is proper and deg(๐) = ๐.
Proof. For ๐ก โ ๐ let
๐๐ก(๐ง) โ (1 โ ๐ก)๐ง๐ + ๐ก๐(๐ง) = ๐ง๐ + ๐ก๐โ1โ๐=0๐๐๐ง๐ .
We will show that the mapping
๐โถ ๐2 ร ๐ โ ๐2 , (๐ง, ๐ก) โฆ ๐๐ก(๐ง)is a proper homotopy. Let
๐ถ = sup0โค๐โค๐โ1|๐๐| .
Then for |๐ง| โฅ 1 we have
|๐0 +โฏ + ๐๐โ1๐ง๐โ1| โค |๐0| + |๐1||๐ง| + โฏ + |๐๐โ1| |๐ง|๐โ1
โค ๐ถ๐|๐ง|๐โ1 ,and hence, for |๐ก| โค ๐ and |๐ง| โฅ 2๐๐ถ๐,
|๐๐ก(๐ง)| โฅ |๐ง|๐ โ ๐๐ถ๐|๐ง|๐โ1
โฅ ๐๐ถ๐|๐ง|๐โ1 .If ๐ด โ ๐ is compact, then for some ๐ > 0, ๐ด is contained in the disk defined by |๐ค| โค ๐ ,and hence the set
{ ๐ง โ ๐ | (๐ก, ๐๐ก(๐ง)) โ [โ๐, ๐] ร ๐ด }is contained in the compact set
{ ๐ง โ ๐ | ๐๐ถ|๐ง|๐โ1 โค ๐ } .This shows that ๐ is a proper homotopy.
Thus for each ๐ก โ ๐, the map ๐๐ก โถ ๐ โ ๐ is proper and
deg(๐๐ก) = deg(๐1) = deg(๐) = deg(๐0)However, ๐0 โถ ๐ โ ๐ is just the mapping ๐ง โฆ ๐ง๐ and an elementary computation (seeExercises 3.6.v and 3.6.vi below) shows that the degree of this mapping is ๐. โก
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90 Chapter 3: Integration of Forms
In particular for ๐ > 0 the degree of ๐ is nonzero; so byTheorem 3.6.4 we conclude that๐โถ ๐ โ ๐ is surjective and hence has zero in its image.
Theorem 3.6.18 (Fundamental theorem of algebra). Every positive-degree polynomial
๐(๐ง) = ๐ง๐ + ๐๐โ1๐ง๐โ1 +โฏ + ๐0with complex coefficients has a complex root: ๐(๐ง0) = 0 for some ๐ง0 โ ๐.
Exercises for ยง3.6Exercise 3.6.i. Let๐ be a subset of ๐๐ and let ๐(๐ฅ), ๐(๐ฅ) and ๐(๐ฅ) be real-valued functionson๐ of class ๐ถ๐. Suppose that for every ๐ฅ โ ๐ the quadratic polynomial
๐(๐ฅ)๐ 2 + ๐(๐ฅ)๐ + ๐(๐ฅ)
has two distinct real roots, ๐ +(๐ฅ) and ๐ โ(๐ฅ), with ๐ +(๐ฅ) > ๐ โ(๐ฅ). Prove that ๐ + and ๐ โ arefunctions of class ๐ถ๐.
Hint: What are the roots of the quadratic polynomial: ๐๐ 2 + ๐๐ + ๐?
Exercise 3.6.ii. Show that the function ๐พ(๐ฅ) defined in Figure 3.6.1 is a continuous surjec-tion ๐ต๐ โ ๐2๐โ1.
Hint: ๐พ(๐ฅ) lies on the ray,
๐(๐ฅ) + ๐ (๐ฅ โ ๐(๐ฅ)) , ๐ โ [0,โ)
and satisfies โ๐พ(๐ฅ)โ = 1. Thus
๐พ(๐ฅ) = ๐(๐ฅ) + ๐ 0(๐ฅ โ ๐(๐ฅ)) ,
where ๐ 0 is a non-negative root of the quadratic polynomial
โ๐(๐ฅ) + ๐ (๐ฅ โ ๐(๐ฅ))โ2 โ 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 isnโt true if one replaces theclosed unit ball by the open unit ball.
Hint: Let ๐ be the open unit ball (i.e., the interior of ๐ต๐). Show that the map
โโถ ๐ โ ๐๐ , โ(๐ฅ) โ ๐ฅ1 โ โ๐ฅโ2
is a diffeomorphism of ๐ onto ๐๐, and show that there are lots of mappings of ๐๐ onto ๐๐which do not have fixed points.
Exercise 3.6.iv. Show that the fixed point in the Brouwer theorem doesnโt have to be aninterior point of ๐ต๐, i.e., show that it can lie on the boundary.
Exercise 3.6.v. If we identify ๐ with ๐2 via the mapping (๐ฅ, ๐ฆ) โฆ ๐ฅ + ๐๐ฆ, we can think of a๐-linear mapping of ๐ into itself, i.e., a mapping of the form
๐ง โฆ ๐๐ง
for a fixed ๐ โ ๐, as an ๐-linear mapping of ๐2 into itself. Show that the determinant of thismapping is |๐|2.
Exercise 3.6.vi.
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ยง3.6 Techniques for computing the degree of a mapping 91
(1) Let ๐โถ ๐ โ ๐ be the mapping ๐(๐ง) โ ๐ง๐. Show that๐ท๐(๐ง) is the linear map
๐ท๐(๐ง) = ๐๐ง๐โ1
given by multiplication by ๐๐ง๐โ1.Hint: Argue from first principles. Show that for โ โ ๐ = ๐2
(๐ง + โ)๐ โ ๐ง๐ โ ๐๐ง๐โ1โ|โ|
tends to zero as |โ| โ 0.(2) Conclude from Exercise 3.6.v that
det(๐ท๐(๐ง)) = ๐2|๐ง|2๐โ2 .(3) Show that at every point ๐ง โ ๐ โ {0}, ๐ is orientation preserving.(4) Show that every point, ๐ค โ ๐ โ {0} is a regular value of ๐ and that
๐โ1(๐ค) = {๐ง1,โฆ, ๐ง๐}with ๐๐ง๐ = +1.
(5) Conclude that the degree of ๐ is ๐.
Exercise 3.6.vii. Prove that the map ๐ from Exercise 3.6.vi has degree ๐ by deducing thisdirectly from the definition of degree.
Hints:โข Show that in polar coordinates, ๐ is the map (๐, ๐) โฆ (๐๐, ๐๐).โข Let ๐ be the 2-form ๐ โ ๐(๐ฅ2 +๐ฆ2)๐๐ฅ โง ๐๐ฆ, where ๐(๐ก) is a compactly supported๐ถโ function of ๐ก. Show that in polar coordinates ๐ = ๐(๐2)๐๐๐โง๐๐, and computethe degree of ๐ by computing the integrals of ๐ and ๐โ๐ in polar coordinates andcomparing them.
Exercise 3.6.viii. Let ๐ be an open subset of ๐๐, ๐ an open subset of ๐๐, ๐ด an open subin-terval of ๐ containing 0 and 1, ๐0, ๐1 โถ ๐ โ ๐ a pair of ๐ถโ mappings, and ๐นโถ ๐ ร ๐ด โ ๐a homotopy between ๐0 and ๐1.(1) In Exercise 2.4.iv you proved that if ๐ โ ๐บ๐(๐) and ๐๐ = 0, then(3.6.19) ๐โ0 ๐ โ ๐โ1 ๐ = ๐๐
where ๐ is the (๐ โ 1)-form ๐๐ผ in equation (2.6.17). Show (by careful inspection of thedefinition of ๐๐ผ) that if ๐น is a proper homotopy and ๐ โ ๐บ๐๐ (๐) then ๐ โ ๐บ๐โ1๐ (๐).
(2) Suppose in particular that ๐ and ๐ are open subsets of ๐๐ and ๐ is in ๐บ๐๐ (๐). Deducefrom equation (3.6.19) that
โซ๐โ0 ๐ = โซ๐โ1 ๐
and deduce directly from the definition of degree that degree is a proper homotopyinvariant.
Exercise 3.6.ix. Let ๐ be an open connected subset of ๐๐ and ๐โถ ๐ โ ๐ a proper ๐ถโmap. Prove that if ๐ is equal to the identity on the complement of a compact set ๐ถ, then ๐is proper and deg(๐) = 1.
Hints:โข Show that for every subset ๐ด โ ๐, we have ๐โ1(๐ด) โ ๐ด โช ๐ถ, and conclude from
this that ๐ is proper.โข Use the recipe (1.6.2) to compute deg(๐) with ๐ โ ๐ โ ๐(๐ถ).
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92 Chapter 3: Integration of Forms
Exercise 3.6.x. Let (๐๐,๐) be an ๐ร๐matrix and๐ดโถ ๐๐ โ ๐๐ the linear mapping associatedwith this matrix. Frobeniusโ Theorem asserts: If the ๐๐,๐ are non-negative then ๐ด has a non-negative eigenvalue. In other words there exists a ๐ฃ โ ๐๐ and a ๐ โ ๐, ๐ โฅ 0, such that๐ด๐ฃ = ๐๐ฃ. Deduce this linear algebra result from the Brouwer fixed point theorem.
Hints:โข We can assume that ๐ด is bijective, otherwise 0 is an eigenvalue. Let ๐๐โ1 be the(๐ โ 1)-sphere, defined by |๐ฅ| = 1, and ๐โถ ๐๐โ1 โ ๐๐โ1 the map,
๐(๐ฅ) = ๐ด๐ฅโ๐ด๐ฅโ
.
Show that ๐maps the set
๐ โ { (๐ฅ1,โฆ, ๐ฅ๐) โ ๐๐โ1 | ๐ฅ1,โฆ, ๐ฅ๐ โฅ 0 }into itself.
โข It is easy to prove that ๐ is homeomorphic to the unit ball ๐ต๐โ1, i.e., that thereexists a continuous map ๐โถ ๐ โ ๐ต๐โ1 which is invertible and has a continuousinverse. Without bothering to prove this fact deduce from it Frobeniusโ theorem.
3.7. Appendix: Sardโs theorem
The version of Sardโs theorem stated in ยง3.5 is a corollary of the following more generalresult.
Theorem 3.7.1. Let๐ be an open subset of ๐๐ and ๐โถ ๐ โ ๐๐ a ๐ถโ map. Then ๐๐ โ๐(๐ถ๐)is dense in ๐๐.
Before undertaking to prove this wewill make a few general comments about this result.
Remark 3.7.2. If (๐๐)๐โฅ1 are open dense subsets of ๐๐, the intersectionโ๐โฅ1๐๐ is densein ๐๐. (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 (๐ด๐)๐โฅ1 is a covering of๐ by compact sets,๐๐ โ ๐๐ โ๐(๐ถ๐ โฉ๐ด๐) is open,so if we can prove that it is dense then by Remark 3.7.2 we will have proved Sardโs theorem.Hence since we can always cover ๐ by a countable collection of closed cubes, it suffices toprove: for every closed cube ๐ด โ ๐, the subspace ๐๐ โ ๐(๐ถ๐ โฉ ๐ด) is dense in ๐๐.
Remark 3.7.4. Let ๐โถ ๐ โ ๐ be a diffeomorphism and let โ = ๐ โ ๐. Then
(3.7.5) ๐(๐ถ๐) = โ(๐ถโ)so Sardโs theorem for โ implies Sardโs theorem for ๐.
We will first prove Sardโs theorem for the set of super-critical points of ๐, the set:
๐ถโฏ๐ โ {๐ โ ๐ |๐ท๐(๐) = 0 } .
Proposition 3.7.6. Let ๐ด โ ๐ be a closed cube. Then the open set ๐๐ โ ๐(๐ด โฉ ๐ถโฏ๐) is a densesubset of ๐๐.
Weโll deduce this from the lemma below.
Lemma 3.7.7. Given ๐ > 0 one can cover ๐(๐ด โฉ ๐ถโฏ๐) by a finite number of cubes of totalvolume less than ๐.
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ยง3.7 Appendix: Sardโs theorem 93
Proof. Let the length of each of the sides of๐ด be โ. Given ๐ฟ > 0 one can subdivide๐ด into๐๐cubes, each of volume (โ/๐)๐, such that if ๐ฅ and ๐ฆ are points of any one of these subcubes
(3.7.8) | ๐๐๐๐๐ฅ๐(๐ฅ) โ ๐๐๐๐๐ฅ๐(๐ฆ)| < ๐ฟ .
Let ๐ด1,โฆ,๐ด๐ be the cubes in this collection which intersect ๐ถโฏ๐. Then for ๐ง0 โ ๐ด๐ โฉ ๐ถโฏ๐,
๐๐๐๐๐ฅ๐(๐ง0) = 0, so for ๐ง โ ๐ด๐ we have
(3.7.9) | ๐๐๐๐๐ฅ๐(๐ง)| < ๐ฟ
by equation (3.7.8). If ๐ฅ and ๐ฆ are points of ๐ด๐ then by the mean value theorem there existsa point ๐ง on the line segment joining ๐ฅ to ๐ฆ such that
๐๐(๐ฅ) โ ๐๐(๐ฆ) =๐โ๐=1
๐๐๐๐๐ฅ๐(๐ง)(๐ฅ๐ โ ๐ฆ๐)
and hence by (3.7.9)
|๐๐(๐ฅ) โ ๐๐(๐ฆ)| โค ๐ฟ๐โ๐=1|๐ฅ๐ โ ๐ฆ๐| โค ๐๐ฟ
โ๐
.
Thus ๐(๐ถ๐ โฉ๐ด๐) is contained in a cube ๐ต๐ of volume (๐ ๐ฟโ๐ )๐, and ๐(๐ถ๐ โฉ๐ด) is contained in
a union of cubes ๐ต๐ of total volume less than
๐๐๐๐ ๐ฟ๐โ๐๐๐= ๐๐๐ฟ๐โ๐
so if we choose ๐ฟ such that ๐๐๐ฟ๐โ๐ < ๐, weโre done. โก
Proof of Proposition 3.7.6. To prove Proposition 3.7.6 we have to show that for every point๐ โ ๐๐ and neighborhood,๐ of ๐, the set๐โ ๐(๐ถโฏ๐ โฉ ๐ด) is nonempty. Suppose
(3.7.10) ๐ โ ๐(๐ถโฏ๐ โฉ ๐ด) .Without loss of generality we can assume๐ is a cube of volume ๐, but the lemma tells usthat ๐(๐ถโฏ๐ โฉ๐ด) can be covered by a finite number of cubes whose total volume is less than ๐,and hence by (3.7.10)๐ can be covered by a finite number of cubes of total volume less than๐, so its volume is less than ๐. This contradiction proves that the inclusion (3.7.10) cannothold. โก
Now we prove Theorem 3.7.1.
Proof of Theorem 3.7.1. Let ๐๐,๐ be the subset of ๐ where ๐๐๐๐๐ฅ๐ โ 0. Then
๐ = ๐ถโฏ๐ โช โ1โค๐,๐โค๐๐๐,๐ ,
so to prove the theorem it suffices to show that ๐๐ โ ๐(๐๐,๐ โฉ ๐ถ๐) is dense in ๐๐, i.e., itsuffices to prove the theorem with ๐ replaced by ๐๐,๐. Let ๐๐ โถ ๐๐ โฅฒ ๐๐ be the involutionwhich interchanges ๐ฅ1 and ๐ฅ๐ and leaves the remaining ๐ฅ๐โs fixed. Letting ๐new = ๐๐๐old๐๐and ๐new = ๐๐๐old, we have, for ๐ = ๐new and ๐ = ๐new
(3.7.11) ๐๐1๐๐ฅ1(๐) โ 0
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94 Chapter 3: Integration of Forms
for all ๐ โ ๐ so weโre reduced to proving Theorem 3.7.1 for maps ๐โถ ๐ โ ๐๐ having theproperty (3.7.11). Let ๐โถ ๐ โ ๐๐ be defined by
(3.7.12) ๐(๐ฅ1,โฆ, ๐ฅ๐) = (๐1(๐ฅ), ๐ฅ2,โฆ, ๐ฅ๐) .Then
๐โ๐ฅ1 = ๐โ๐ฅ1 = ๐1(๐ฅ1,โฆ, ๐ฅ๐)and
det(๐ท๐) = ๐๐1๐๐ฅ1โ 0 .
Thus, by the inverse function theorem, ๐ is locally a diffeomorphism at every point ๐ โ ๐.This means that if ๐ด is a compact subset of ๐ we can cover ๐ด by a finite number of opensubsets ๐๐ โ ๐ such that ๐ maps ๐๐ diffeomorphically onto an open subset ๐๐ in ๐๐. Toconclude the proof of the theorem weโll show that ๐๐ โ ๐(๐ถ๐ โฉ ๐๐ โฉ ๐ด) is a dense subset of๐๐. Let โโถ ๐๐ โ ๐๐ be themapโ = ๐โ๐โ1. To prove this assertion it suffices byRemark 3.7.4to prove that the set ๐๐ โ โ(๐ถโ) is dense in ๐๐. This we will do by induction on ๐. First notethat for ๐ = 1, we have ๐ถ๐ = ๐ถ
โฏ๐, so weโve already proved Theorem 3.7.1 in dimension one.
Now note that by (3.7.12) we have โโ๐ฅ1 = ๐ฅ1, i.e., โ is a mapping of the form
โ(๐ฅ1,โฆ, ๐ฅ๐) = (๐ฅ1, โ2(๐ฅ),โฆ, โ๐(๐ฅ)) .Thus if we let
๐๐ โ { (๐ฅ2,โฆ, ๐ฅ๐) โ ๐๐โ1 | (๐, ๐ฅ2,โฆ, ๐ฅ๐) โ ๐๐ }and let โ๐ โถ ๐๐ โ ๐๐โ1 be the map
โ๐(๐ฅ2,โฆ, ๐ฅ๐) = (โ2(๐, ๐ฅ2,โฆ, ๐ฅ๐),โฆ, โ๐(๐, ๐ฅ2,โฆ, ๐ฅ๐)) .Then
det(๐ทโ๐)(๐ฅ2,โฆ, ๐ฅ๐) = det(๐ทโ)(๐, ๐ฅ2,โฆ, ๐ฅ๐) ,and hence
(3.7.13) (๐, ๐ฅ) โ ๐๐ โฉ ๐ถโ โบ ๐ฅ โ ๐ถโ๐ .Now let ๐0 = (๐, ๐ฅ0) be a point in ๐๐. We have to show that every neighborhood ๐ of ๐0contains a point ๐ โ ๐๐โโ(๐ถโ). Let๐๐ โ ๐๐โ1 be the set of points ๐ฅ for which (๐, ๐ฅ) โ ๐. Byinduction ๐๐ contains a point ๐ฅ โ ๐๐โ1 โ โ๐(๐ถโ๐) and hence ๐ = (๐, ๐ฅ) is in ๐ by definitionand in ๐๐ โ โ(๐ถโ) by (3.7.13). โก
Exercises for ยง3.7Exercise 3.7.i. What are the set of critical points and the image of the set of critical pointsfor the following maps ๐ โ ๐?(1) The map ๐1(๐ฅ) = (๐ฅ2 โ 1)2.(2) The map ๐2(๐ฅ) = sin(๐ฅ) + ๐ฅ.(3) The map
๐3(๐ฅ) = {0, ๐ฅ โค 0๐โ 1๐ฅ , ๐ฅ > 0 .
Exercise 3.7.ii (Sardโs theorem for affine maps). Let ๐โถ ๐๐ โ ๐๐ be an affine map, i.e., amap of the form
๐(๐ฅ) = ๐ด(๐ฅ) + ๐ฅ0where ๐ดโถ ๐๐ โ ๐๐ is a linear map and ๐ฅ0 โ ๐๐. Prove Sardโs theorem for ๐.
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ยง3.7 Appendix: Sardโs theorem 95
Exercise 3.7.iii. Let๐โถ ๐ โ ๐be a๐ถโ functionwhich is supported in the interval (โ1/2, 1/2)and has amaximum at the origin. Let (๐๐)๐โฅ1 be an enumeration of the rational numbers, andlet ๐โถ ๐ โ ๐ be the map
๐(๐ฅ) =โโ๐=1๐๐๐(๐ฅ โ ๐) .
Show that ๐ is a ๐ถโ map and show that the image of ๐ถ๐ is dense in ๐.The moral of this example: Sardโs theorem says that the complement of ๐ถ๐ is dense in ๐,
but ๐ถ๐ 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 ๐ โ ๐๐ and every neighborhood ๐ of ๐,
๐โฉโ๐โฅ1๐๐ is nonempty. Construct, by induction, a family of closed balls (๐ต๐)๐โฅ1 such thatโข ๐ต๐ โ ๐,โข ๐ต๐+1 โ ๐ต๐,โข ๐ต๐ โ โ๐โค๐๐๐,โข The radius of ๐ต๐ is less than 1๐ ,
and show thatโ๐โฅ1 ๐ต๐ โ โ .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: Stokesโ theorem and the divergence theorem. In this section weโll 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 ๐3, i.e., a manifold isbasically just an ๐-dimensional surface in some high dimensional Euclidean space. Tomakethis definition precise let๐ be a subset of ๐๐, ๐ a subset of ๐๐ and ๐โถ ๐ โ ๐ a continuousmap. We recall:
Definition 4.1.1. The map ๐ is a ๐ถโ map if for every ๐ โ ๐, there exists a neighborhood๐๐ of ๐ in ๐๐ and a ๐ถโ map ๐๐ โถ ๐๐ โ ๐๐ which coincides with ๐ on ๐๐ โฉ ๐.
We also recall:
Theorem 4.1.2. If ๐โถ ๐ โ ๐ is a ๐ถโ map, there exists a neighborhood๐, of๐ in ๐๐ and a๐ถโ map ๐โถ ๐ โ ๐๐ such that ๐ coincides with ๐ on๐.
(A proof of this can be found in Appendix a.)We will say that ๐ is a diffeomorphism if ๐ is a bijection and ๐ and ๐โ1 are both ๐ถโ
maps. In particular if ๐ is an open subset of ๐๐,๐ is an example of an object which we willcall a manifold. More generally:
Definition 4.1.3. Let๐ and ๐ be nonnegative integers with ๐ โค ๐. A subset ๐ โ ๐๐ is an๐-manifold if for every ๐ โ ๐ there exists a neighborhood ๐ of ๐ in ๐๐, an open subset ๐in ๐๐, and a diffeomorphism ๐โถ ๐ โฅฒ ๐ โฉ ๐.
Thus๐ is an ๐-manifold if, locally near every point ๐,๐ โlooks likeโ an open subset of๐๐.
Examples 4.1.4.(1) Graphs of functions: Let ๐ be an open subset of ๐๐ and ๐โถ ๐ โ ๐ a ๐ถโ function. Its
graph๐ค๐ = { (๐ฅ, ๐(๐ฅ)) โ ๐๐+1 | ๐ฅ โ ๐ }
is an ๐-manifold in ๐๐+1. In fact the map
๐โถ ๐ โ ๐๐+1 , ๐ฅ โฆ (๐ฅ, ๐(๐ฅ))is a diffeomorphism of ๐ onto ๐ค๐. (Itโs clear that ๐ is a ๐ถโ map, and it is a diffeomor-phism since its inverse is the map ๐โถ ๐ค๐ โ ๐ given by ๐(๐ฅ, ๐ก) โ ๐ฅ, which is also clearly๐ถโ.)
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(2) Graphs of mappings: More generally if ๐โถ ๐ โ ๐๐ is a ๐ถโ map, its graph ๐ค๐ is an๐-manifold in ๐๐+๐.
(3) Vector spaces: Let๐ be an ๐- dimensional vector subspace of ๐๐, and (๐1,โฆ, ๐๐) a basisof ๐. Then the linear map
(4.1.5) ๐โถ ๐๐ โ ๐ , (๐ฅ1,โฆ, ๐ฅ๐) โฆ๐โ๐=1๐ฅ๐๐๐
is a diffeomorphism of ๐๐ onto ๐. Hence every ๐-dimensional vector subspace of ๐๐is automatically an ๐-dimensional submanifold of ๐๐. Note, by the way, that if๐ is any๐-dimensional vector space, not necessarily a subspace of ๐๐, the map (4.1.5) givesus an identification of ๐ with ๐๐. This means that we can speak of subsets of ๐ asbeing ๐-dimensional submanifolds if, via this identification, they get mapped onto ๐-dimensional submanifolds of ๐๐. (This is a trivial, but useful, observation since a lot ofinteresting manifolds occur โin natureโ as subsets of some abstract vector space ratherthan explicitly as subsets of some ๐๐. An example is the manifold, ๐(๐), of orthogonal๐ ร ๐ matrices. This manifold occurs in nature as a submanifold of the vector space of๐ by ๐matrices.)
(4) Affine subspaces of ๐๐: These are manifolds of the form ๐ + ๐, where ๐ is a vector sub-space of ๐๐ and ๐ is a specified point in ๐๐. In other words, they are diffeomorphiccopies of vector subspaces with respect to the diffeomorphism
๐๐ โถ ๐๐ โฅฒ ๐๐ , ๐ฅ โฆ ๐ + ๐ฅ .
If๐ is an arbitrary submanifold of ๐๐ its tangent space a point ๐ โ ๐, is an example ofa manifold of this type. (Weโll have more to say about tangent spaces in Section 4.2.)
(5) Product manifolds: For ๐ = 1, 2, let ๐๐ be an ๐๐-dimensional submanifold of ๐๐๐ . Thenthe Cartesian product of๐1 and๐2
๐1 ร ๐2 = { (๐ฅ1, ๐ฅ2) | ๐ฅ๐ โ ๐๐ }is an (๐1 + ๐2)-dimensional submanifold of ๐๐1+๐2 = ๐๐1 ร ๐๐2 .
We will leave for you to verify this fact as an exercise.Hint: For ๐๐ โ ๐๐, ๐ = 1, 2, there exists a neighborhood,๐๐, of ๐๐ in ๐๐๐ , an open set,
๐๐ in ๐๐๐ , and a diffeomorphism ๐โถ ๐๐ โ ๐๐ โฉ ๐๐. Let ๐ = ๐1 ร ๐2, ๐ = ๐1 ร ๐2 and๐ = ๐1 ร ๐2, and let ๐โถ ๐ โ ๐ โฉ ๐ be the product diffeomorphism, (๐(๐1), ๐2(๐2)).
(6) The unit ๐-sphere: This is the set of unit vectors in ๐๐+1:๐๐ = { ๐ฅ โ ๐๐+1 | ๐ฅ21 +โฏ + ๐ฅ2๐+1 = 1 } .
To show that ๐๐ is an ๐-manifold, let๐ be the open subset of ๐๐+1 on which ๐ฅ๐+1 ispositive. If๐ is the open unit ball in๐๐ and๐โถ ๐ โ ๐ is the function,๐(๐ฅ) = (1โ(๐ฅ21 +โฏ + ๐ฅ2๐))1/2, then ๐๐ โฉ ๐ is just the graph ๐ค๐ of ๐. Hence we have a diffeomorphism
๐โถ ๐ โ ๐๐ โฉ ๐ .More generally, if ๐ = (๐ฅ1,โฆ, ๐ฅ๐+1) is any point on the unit sphere, then ๐ฅ๐ is nonzerofor some ๐. If ๐ฅ๐ is positive, then letting ๐ be the transposition, ๐ โ ๐+1 and๐๐ โถ ๐๐+1 โ๐๐+1, the map
๐๐(๐ฅ1,โฆ, ๐ฅ๐) = (๐ฅ๐(1),โฆ, ๐ฅ๐(๐))one gets a diffeomorphism๐๐โ๐ of๐ onto a neighborhood of๐ in ๐๐ and if๐ฅ๐ is negativeone gets such a diffeomorphism by replacing ๐๐ by โ๐๐. In either case we have shownthat for every point ๐ in ๐๐, there is a neighborhood of ๐ in ๐๐ which is diffeomorphicto ๐.
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(7) The 2-torus: In calculus books this is usually described as the surface of rotation in ๐3obtained by taking the unit circle centered at the point, (2, 0), in the (๐ฅ1, ๐ฅ3) plane androtating it about the ๐ฅ3-axis. However, a slightly nicer description of it is as the productmanifold ๐1 ร ๐1 in ๐4. (As an exercise, reconcile these two descriptions.)
Weโll now turn to an alternative way of looking at manifolds: as solutions of systems ofequations. Let ๐ be an open subset of ๐๐ and ๐โถ ๐ โ ๐๐ a ๐ถโ map.
Definition 4.1.6. A point ๐ โ ๐๐ is a regular value of ๐ if for every point ๐ โ ๐โ1(๐), themap ๐ is a submersion at ๐.
Note that for ๐ to be a submersion at ๐, the differential ๐ท๐(๐)โถ ๐๐ โ ๐๐ has to besurjective, and hence ๐ has to be less than or equal to ๐. Therefore this notion of regularvalue is interesting only if๐ โฅ ๐.
Theorem4.1.7. Let ๐ โ ๐โ๐. If ๐ is a regular value of๐, the set๐ โ ๐โ1(๐) is an ๐-manifold.
Proof. Replacing ๐ by ๐โ๐ โ ๐ we can assume without loss of generality that ๐ = 0. Let ๐ โ๐โ1(0). Since ๐ is a submersion at ๐, the canonical submersion theorem (see Theorem b.16)tells us that there exists a neighborhood ๐ of 0 in ๐๐, a neighborhood ๐0, of ๐ in ๐ and adiffeomorphism ๐ โ ๐ โฅฒ ๐0 such that
(4.1.8) ๐ โ ๐ = ๐
where ๐ is the projection map
๐๐ = ๐๐ ร ๐๐ โ ๐๐ , (๐ฅ, ๐ฆ) โฆ ๐ฅ .
Hence๐โ1(0) = {0}ร๐๐ = ๐๐ and by equation (4.1.5)), ๐maps๐โฉ๐โ1(0) diffeomorphicallyonto ๐0 โฉ ๐โ1(0). However, ๐ โฉ ๐โ1(0) is a neighborhood of 0 in ๐๐ and ๐0 โฉ ๐โ1(0) is aneighborhood of๐ in๐, and, as remarked, these two neighborhoods are diffeomorphic. โก
Some examples:
Examples 4.1.9.(1) The ๐-sphere: Let
๐โถ ๐๐+1 โ ๐be the map,
(๐ฅ1,โฆ, ๐ฅ๐+1) โฆ ๐ฅ21 +โฏ + ๐ฅ2๐+1 โ 1 .Then
๐ท๐(๐ฅ) = 2(๐ฅ1,โฆ, ๐ฅ๐+1)so, if ๐ฅ โ 0, then ๐ is a submersion at ๐ฅ. In particular, ๐ is a submersion at all points ๐ฅon the ๐-sphere
๐๐ = ๐โ1(0)so the ๐-sphere is an ๐-dimensional submanifold of ๐๐+1.
(2) Graphs: Let ๐โถ ๐๐ โ ๐๐ be a ๐ถโ map and, as in (2), let
๐ค๐ โ { (๐ฅ, ๐ฆ) โ ๐๐ ร ๐๐ | ๐ฆ = ๐(๐ฅ) } .
We claim that ๐ค๐ is an ๐-dimensional submanifold of ๐๐+๐ = ๐๐ ร ๐๐.
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Proof. Let๐โถ ๐๐ ร ๐๐ โ ๐๐
be the map ๐(๐ฅ, ๐ฆ) โ ๐ฆ โ ๐(๐ฅ). Then
๐ท๐(๐ฅ, ๐ฆ) = (โ๐ท๐(๐ฅ) id๐)
where id๐ is the identity map of ๐๐ onto itself. This map is always of rank ๐. Hence๐ค๐ = ๐โ1(0) is an ๐-dimensional submanifold of ๐๐+๐. โก
(3) Let M ๐ be the set of all ๐ ร ๐ matrices and let ๐ฎ๐ be the set of all symmetric ๐ ร ๐matrices, i.e., the set
๐ฎ๐ = {๐ด โM ๐ | ๐ด = ๐ดโบ } .The map
๐ด = (๐๐,๐) โฆ (๐1,1, ๐1,2,โฆ, ๐1,๐, ๐2,1,โฆ, ๐2,๐,โฆ)gives us an identification
M ๐ โฅฒ ๐๐2
and the map
๐ด = (๐๐,๐) โฆ (๐1,1,โฆ, ๐1,๐, ๐2,2,โฆ, ๐2,๐, ๐3,3,โฆ, ๐3,๐,โฆ)gives us an identification
๐ฎ๐ โฅฒ ๐๐(๐+1)2 .
(Note that if๐ด is a symmetric matrix, then ๐๐,๐ = ๐๐,๐, so this map avoids redundancies.)Let
๐(๐) = {๐ด โM ๐ | ๐ดโบ๐ด = id๐ } .This is the set of orthogonal ๐ ร ๐matrices, and we will leave for you as an exercise toshow that itโs an ๐(๐ โ 1)/2-manifold.
Hint: Let ๐โถ M ๐ โ ๐ฎ๐ be the map ๐(๐ด) = ๐ดโบ๐ด โ id๐. Then
๐(๐) = ๐โ1(0) .
These examples show that lots of interestingmanifolds arise as zero sets of submersions๐โถ ๐ โ ๐๐. This is, in fact, not just an accident. We will show that locally every manifoldarises this way. More explicitly let ๐ โ ๐๐ be an ๐-manifold, ๐ a point of ๐, ๐ a neighbor-hood of 0 in ๐๐, ๐ a neighborhood of ๐ in ๐๐ and ๐โถ (๐, 0) โฅฒ (๐ โฉ ๐, ๐) a diffeomor-phism. We will for the moment think of ๐ as a ๐ถโ map ๐โถ ๐ โ ๐๐ whose image happensto lie in๐.
Lemma 4.1.10. The linear map๐ท๐(0)โถ ๐๐ โ ๐๐ is injective.
Proof. Since๐โ1 โถ ๐ โฉ ๐ โฅฒ ๐ is a diffeomorphism, shrinking๐ if necessary, we can assumethat there exists a ๐ถโ map ๐โถ ๐ โ ๐ which coincides with ๐โ1on ๐ โฉ ๐. Since ๐maps ๐onto ๐ โฉ ๐, ๐ โ ๐ = ๐โ1 โ ๐ is the identity map on ๐. Therefore,
๐ท(๐ โ ๐)(0) = (๐ท๐)(๐)๐ท๐(0) = id๐by the chain rule, and hence if๐ท๐(0)๐ฃ = 0, it follows from this identity that ๐ฃ = 0. โก
Lemma 4.1.10 says that ๐ is an immersion at 0, so by the canonical immersion theorem(see Theorem b.18) there exists a neighborhood ๐0 of 0 in ๐, a neighborhood ๐๐ of ๐ in ๐,and a diffeomorphism
(4.1.11) ๐โถ (๐๐, ๐) โฅฒ (๐0 ร ๐๐โ๐, 0)
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such that ๐ โ ๐ = ๐, where ๐ is, as in Appendix b, the canonical immersion
(4.1.12) ๐ โถ ๐0 โ ๐0 ร ๐๐โ๐ , ๐ฅ โฆ (๐ฅ, 0) .By (4.1.11) ๐ maps ๐(๐0) diffeomorphically onto ๐(๐0). However, by (4.1.8) and (4.1.11)๐(๐0) is defined by the equations, ๐ฅ๐ = 0, ๐ = ๐ + 1,โฆ,๐. Hence if ๐ = (๐1,โฆ, ๐๐) the set,๐(๐0) = ๐๐ โฉ ๐ is defined by the equations
(4.1.13) ๐๐ = 0 , ๐ = ๐ + 1,โฆ,๐ .
Let โ = ๐ โ ๐, let๐โถ ๐๐ = ๐๐ ร ๐โ โ ๐โ
be the canonical submersion,
๐(๐ฅ1,โฆ, ๐ฅ๐) = (๐ฅ๐+1,โฆ, ๐ฅ๐)and let ๐ = ๐ โ ๐. Since ๐ is a diffeomorphism, ๐ is a submersion and (4.1.12) can beinterpreted as saying that
(4.1.14) ๐๐ โฉ ๐ = ๐โ1(0) .Thus to summarize we have proved:
Theorem 4.1.15. Let๐ be an ๐-dimensional submanifold of ๐๐ and let โ โ ๐โ ๐. Then forevery ๐ โ ๐ there exists a neighborhood ๐๐ of ๐ in ๐๐ and a submersion
๐โถ (๐๐, ๐) โ (๐โ, 0)such that๐ โฉ ๐๐ 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, ๐. More specifically if ๐ = (๐1,โฆ, ๐๐) we can think of ๐โ1(๐) as being the set ofsolutions of the system of equations
๐๐(๐ฅ) = ๐๐ , ๐ = 1,โฆ, ๐and the condition that ๐ be a regular value of ๐ can be interpreted as saying that for everysolution ๐ of this system of equations the vectors
(๐๐๐)๐ =๐โ๐=1
๐๐๐๐๐ฅ๐(0)๐๐ฅ๐
in ๐โ๐ ๐๐ are linearly independent, i.e., the system (4.1.14) is an โindependent system ofdefining equationsโ for๐.
Exercises for ยง4.1Exercise 4.1.i. Show that the set of solutions of the system of equations
{๐ฅ21 +โฏ + ๐ฅ2๐ = 1๐ฅ1 +โฏ + ๐ฅ๐ = 0
is an (๐ โ 2)-dimensional submanifold of ๐๐.
Exercise 4.1.ii. Let ๐๐โ1 โ ๐๐ be the (๐ โ 1)-sphere and let
๐๐ = { ๐ฅ โ ๐๐โ1 | ๐ฅ1 +โฏ + ๐ฅ๐ = ๐ } .For what values of ๐ is๐๐ an (๐ โ 2)-dimensional submanifold of ๐๐โ1?
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Exercise 4.1.iii. Show that if๐๐ is an ๐๐-dimensional submanifold of ๐๐๐ , for ๐ = 1, 2, then๐1 ร ๐2 โ ๐๐1 ร ๐๐2
is an (๐1 + ๐2)-dimensional submanifold of ๐๐1 ร ๐๐2 .Exercise 4.1.iv. Show that the set
๐ = { (๐ฅ, ๐ฃ) โ ๐๐โ1 ร ๐๐ | ๐ฅ โ ๐ฃ = 0 }is a (2๐โ2)-dimensional submanifold of๐๐ร๐๐. (Here ๐ฅโ ๐ฃ โ โ๐๐=1 ๐ฅ๐๐ฃ๐ is the dot product.)
Exercise 4.1.v. Let ๐โถ ๐๐ โ ๐๐ be a ๐ถโ map and let ๐ = ๐ค๐ be the graph of ๐. Provedirectly that๐ is an ๐-manifold by proving that the map
๐พ๐ โถ ๐๐ โ ๐ , ๐ฅ โฆ (๐ฅ, ๐(๐ฅ))is a diffeomorphism.
Exercise 4.1.vi. Prove that the orthogonal group ๐(๐) is an ๐(๐ โ 1)/2-manifold.Hints:
โข Let ๐โถ M ๐ โ ๐ฎ๐ be the map๐(๐ด) = ๐ดโบ๐ด โ id๐ .
Show that ๐(๐) = ๐โ1(0).โข Show that
๐(๐ด + ๐๐ต) = ๐ดโบ๐ด + ๐(๐ดโบ๐ต + ๐ตโบ๐ด) + ๐2๐ตโบ๐ต โ id๐ .โข Conclude that the derivative of ๐ at ๐ด is the map given by
(4.1.16) ๐ต โฆ ๐ดโบ๐ต + ๐ตโบ๐ด .โข Let ๐ด be in ๐(๐). Show that if ๐ถ is in ๐ฎ๐ and ๐ต = ๐ด๐ถ/2 then the map (4.1.16)
maps ๐ต onto ๐ถ.โข Conclude that the derivative of ๐ is surjective at ๐ด.โข Conclude that 0 is a regular value of the mapping ๐.
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 ๐1,โฆ, ๐๐ be the standard basis of ๐๐ and let๐ = span(๐๐+1,โฆ, ๐๐). Prove that if ๐
is a ๐-dimensional subspace of ๐๐ and(4.1.17) ๐ โฉ๐ = 0 ,
then one can find a unique basis of ๐ of the form
(4.1.18) ๐ฃ๐ = ๐๐ +โโ๐=1๐๐,๐๐๐+๐ , ๐ = 1,โฆ, ๐ ,
where โ = ๐ โ ๐.(2) Let ๐บ๐ be the set of ๐-dimensional subspaces of ๐๐ having the property (4.1.17) and
let M ๐,โ be the vector space of ๐ ร โ matrices. Show that one gets from the identities(4.1.18) a bijective map:
(4.1.19) ๐พโถ M ๐,โ โ ๐บ๐ .Exercise 4.1.viii. Let ๐๐ be the vector space of linear mappings of ๐๐ into itself which areself-adjoint, i.e., have the property ๐ด = ๐ดโบ.
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(1) Given a ๐-dimensional subspace,๐ of๐๐ let๐๐ โถ ๐๐ โ ๐๐ be the orthogonal projectionof ๐๐ onto ๐. Show that ๐๐ is in ๐๐ and is of rank ๐, and show that (๐๐)2 = ๐๐.
(2) Conversely suppose๐ด is an element of ๐๐ which is of rank ๐ and has the property,๐ด2 =๐ด. Show that if ๐ is the image of ๐ด in ๐๐, then ๐ด = ๐๐.
Definition 4.1.20. We call an๐ด โ ๐๐ of the form๐ด = ๐๐ above a rank ๐ projection operator.
Exercise 4.1.ix. Composing the map
๐โถ ๐บ๐ โ ๐๐ , ๐ โฆ ๐๐with the map (4.1.19) we get a map
๐โถ M ๐,โ โ ๐๐ , ๐ = ๐ โ ๐พ .Prove that ๐ is ๐ถโ.
Hints:โข By GramโSchmidt one can convert (4.1.18) into an orthonormal basis
๐1,๐ต,โฆ, ๐๐,๐ตof ๐. Show that the ๐๐,๐ตโs are ๐ถโ functions of the matrix ๐ต = (๐๐,๐).
โข Show that ๐๐ is the linear mapping
๐ฃ โฆ๐โ๐=1(๐ฃ โ ๐๐,๐ต)๐๐,๐ต .
Exercise 4.1.x. Let ๐0 = span(๐1,โฆ, ๐๐) and let ๐บ๐ = ๐(๐บ๐). Show that ๐maps a neighbor-hood of ๐0 in M ๐,โ diffeomorphically onto a neighborhood of ๐๐0 in ๐บ๐.
Hints: ๐๐ is in ๐บ๐ if and only if ๐ satisfies (4.1.17). For 1 โค ๐ โค ๐ let
๐ค๐ = ๐๐(๐๐) =๐โ๐=1๐๐,๐๐๐ +
โโ๐=1๐๐,๐๐๐+๐ .
โข Show that if the matrix ๐ด = (๐๐,๐) is invertible, ๐๐ is in ๐บ๐.โข Let ๐ โ ๐บ๐ be the set of all ๐๐โs for which ๐ด is invertible. Show that ๐โ1 โถ ๐ โ
M ๐,โ is the map๐โ1(๐๐) = ๐ต = ๐ดโ1๐ถ
where ๐ถ = [๐๐,๐].
Exercise 4.1.xi. Let๐บ(๐, ๐) โ ๐๐ be the set of rank ๐ projection operators. Prove that๐บ(๐, ๐)is a ๐โ-dimensional submanifold of the Euclidean space ๐๐ โ ๐
๐(๐+1)2 .
Hints:โข Show that if ๐ is any ๐-dimensional subspace of ๐๐ there exists a linear mapping,๐ด โ ๐(๐)mapping ๐0 to ๐.
โข Show that ๐๐ = ๐ด๐๐0๐ดโ1.
โข Let ๐พ๐ด โถ ๐๐ โ ๐๐ be the linear mapping,
๐พ๐ด(๐ต) = ๐ด๐ต๐ดโ1 .Show that
๐พ๐ด โ ๐โถ M ๐,โ โ ๐๐maps a neighborhood of ๐0 in M ๐,โ diffeomorphically onto a neighborhood of๐๐ in ๐บ(๐, ๐).
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Remark 4.1.21. Let Gr(๐, ๐) be the set of all ๐-dimensional subspaces of ๐๐. The identifi-cation of Gr(๐, ๐) with ๐บ(๐, ๐) given by ๐ โฆ ๐๐ allows us to restate the result above in theform.
Theorem4.1.22. TheGrassmannianGr(๐, ๐) of ๐-dimensional subspaces of๐๐ is a ๐โ-dimens-ional submanifold of ๐๐ โ ๐
๐(๐+1)2 .
Exercise 4.1.xii. Show that Gr(๐, ๐) is a compact submanifold of ๐๐.Hint: Show that itโs closed and bounded.
4.2. Tangent spaces
We recall that a subset ๐ of ๐๐ is an ๐-dimensional manifold if for every ๐ โ ๐ thereexists an open set๐ โ ๐๐, a neighborhood๐ of๐ in๐๐, and a๐ถโ-diffeomorphism๐โถ ๐ โฅฒ๐ โฉ ๐.
Definition 4.2.1. We will call ๐ a parametrization of๐ at ๐.
Our goal in this section is to define the notion of the tangent space ๐๐๐ to ๐ at ๐ anddescribe some of its properties. Before giving our official definitionweโll discuss some simpleexamples.
Example 4.2.2. Let ๐โถ ๐ โ ๐ be a ๐ถโ function and let๐ = ๐ค๐ be the graph of ๐.
๏ฟฝ๏ฟฝ
๏ฟฝ
๏ฟฝ
โ
๏ฟฝ0
Figure 4.2.1. The tangent line โ to the graph of ๐ at ๐0
Then in this figure above the tangent line โ to ๐ at ๐0 = (๐ฅ0, ๐ฆ0) is defined by theequation
๐ฆ โ ๐ฆ0 = ๐(๐ฅ โ ๐ฅ0)where ๐ = ๐โฒ(๐ฅ0) In other words if ๐ is a point on โ then ๐ = ๐0 + ๐๐ฃ0 where ๐ฃ0 = (1, ๐)and ๐ โ ๐. We would, however, like the tangent space to ๐ at ๐0 to be a subspace of thetangent space to ๐2 at ๐0, i.e., to be the subspace of the space: ๐๐0๐
2 = {๐0} ร ๐2, and thisweโll achieve by defining
๐๐0๐ โ { (๐0, ๐๐ฃ0) | ๐ โ ๐ } .
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ยง4.2 Tangent spaces 105
Example 4.2.3. Let ๐2 be the unit 2-sphere in ๐3. The tangent plane to ๐2 at ๐0 is usuallydefined to be the plane
{ ๐0 + ๐ฃ | ๐ฃ โ ๐3 and ๐ฃ โ ๐0 } .However, this tangent plane is easily converted into a subspace of๐๐๐3 via themap๐0+๐ฃ โฆ(๐0, ๐ฃ) and the image of this map
{ (๐0, ๐ฃ) | ๐ฃ โ ๐3 and ๐ฃ โ ๐0 } .will be our definition of ๐๐0๐
2.
Letโs now turn to the general definition. As above let ๐ be an ๐-dimensional submani-fold of ๐๐, ๐ a point of๐, ๐ a neighborhood of ๐ in ๐๐, ๐ an open set in ๐๐ and
๐โถ (๐, ๐) โฅฒ (๐ โฉ ๐, ๐)a parameterization of๐. We can think of ๐ as a ๐ถโ map
๐โถ (๐, ๐) โ (๐, ๐)whose image happens to lie in๐ โฉ ๐ and we proved in ยง4.1 that its derivative at ๐
(๐๐)๐ โถ ๐๐๐๐ โ ๐๐๐๐
is injective.
Definition 4.2.4. The tangent space ๐๐๐ to๐ at ๐ is the image of the linear map
(๐๐)๐ โถ ๐๐๐๐ โ ๐๐๐๐ .
In other words, ๐ค โ ๐๐๐๐ is in ๐๐๐ if and only if ๐ค = ๐๐๐(๐ฃ) for some ๐ฃ โ ๐๐๐๐. Moresuccinctly,
(4.2.5) ๐๐๐ โ (๐๐๐)(๐๐๐๐) .
(Since ๐๐๐ is injective, ๐๐๐ is an ๐-dimensional vector subspace of ๐๐๐๐.)
One problem with this definition is that it appears to depend on the choice of ๐. Toget around this problem, weโll give an alternative definition of ๐๐๐. In ยง4.1 we showed thatthere exists a neighborhood ๐ of ๐ in ๐๐ (which we can without loss of generality take tobe the same as ๐ above) and a ๐ถโ map
(4.2.6) ๐โถ (๐, ๐) โ (๐๐, 0) , where ๐ = ๐ โ ๐ ,such that ๐ โฉ ๐ = ๐โ1(0) and such that ๐ is a submersion at all points of ๐ โฉ ๐, and inparticular at ๐. Thus
๐๐๐ โถ ๐๐๐๐ โ ๐0๐๐
is surjective, and hence the kernel of ๐๐๐ has dimension ๐. Our alternative definition of๐๐๐is
(4.2.7) ๐๐๐ โ ker(๐๐๐) .
Lemma 4.2.8. The spaces (4.2.5) and (4.2.7) are both ๐-dimensional subspaces of ๐๐๐๐, andwe claim that these spaces are the same.
(Notice that the definition (4.2.7) of ๐๐๐ does not depend on ๐, so if we can show thatthese spaces are the same, the definitions (4.2.5) and (4.2.7)) will depend neither on ๐ noron ๐.)
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106 Chapter 4: Manifolds & Forms on Manifolds
Proof. Since ๐(๐) is contained in๐ โฉ ๐ and๐ โฉ ๐ is contained in ๐โ1(0), ๐ โ ๐ = 0, so bythe chain rule(4.2.9) ๐๐๐ โ ๐๐๐ = ๐(๐ โ ๐)๐ = 0 .Hence if ๐ฃ โ ๐๐๐๐ and ๐ค = ๐๐๐(๐ฃ), ๐๐๐(๐ค) = 0. This shows that the space (4.2.5) iscontained in the space (4.2.7). However, these two spaces are ๐-dimensional so they co-incide. โก
From the proof above one can extract a slightly stronger result.
Theorem 4.2.10. Let๐ be an open subset of ๐โ and โโถ (๐, ๐) โ (๐๐, ๐) a ๐ถโ map. Sup-pose โ(๐) is contained in๐. Then the image of the map
๐โ๐ โถ ๐๐๐โ โ ๐๐๐๐
is contained in ๐๐๐.Proof. Let ๐ be the map (4.2.6). We can assume without loss of generality that โ(๐) is con-tained in๐, and so, by assumption, โ(๐) โ ๐โฉ๐. Therefore, as above, ๐โโ = 0, and hence๐โ๐(๐๐๐โ) is contained in the kernel of ๐๐๐. โก
This result will enable us to define the derivative of a mapping between manifolds. Ex-plicitly: Let ๐ be a submanifold of ๐๐, ๐ a submanifold of ๐๐, and ๐โถ (๐, ๐) โ (๐, ๐ฆ0)a ๐ถโ map. By Definition 4.1.1 there exists a neighborhood ๐ of ๐ in ๐๐ and a ๐ถโ map๐โถ ๐ โ ๐๐ extending to ๐. We will define(4.2.11) (๐๐)๐ โถ ๐๐๐ โ ๐๐ฆ0๐to be the restriction of the map
(๐๐)๐ โถ ๐๐๐๐ โ ๐๐ฆ0๐๐
to ๐๐๐. There are two obvious problems with this definition:(1) Is the space (๐๐)๐(๐๐๐) contained in ๐๐ฆ0๐?(2) Does the definition depend on ๐?
To show that the answer to (1) is yes and the answer to (2) is no, let๐โถ (๐, ๐ฅ0) โฅฒ (๐ โฉ ๐, ๐)
be a parametrization of ๐, and let โ = ๐ โ ๐. Since ๐(๐) โ ๐, โ(๐) โ ๐ and hence byTheorem 4.2.10
๐โ๐ฅ0(๐๐ฅ0๐๐) โ ๐๐ฆ0๐ .
But by the chain rule๐โ๐ฅ0 = ๐๐๐ โ ๐๐๐ฅ0 ,
so by (4.2.5)(๐๐๐)(๐๐๐) โ ๐๐๐
and(๐๐๐)(๐๐๐) = (๐โ)๐ฅ0(๐๐ฅ0๐
๐)Thus the answer to (1) is yes, and since โ = ๐ โ ๐ = ๐ โ ๐, 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 ๐ be a submanifold of๐โ and ๐โถ (๐, ๐ฆ0) โ (๐, ๐ง0) a ๐ถโ map. Then ๐๐๐ฆ0 โ ๐๐๐ = ๐(๐ โ ๐)๐.
Wewill next provemanifold versions of the inverse function theorem and the canonicalimmersion and submersion theorems.
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ยง4.2 Tangent spaces 107
Theorem 4.2.13 (Inverse function theorem for manifolds). Let๐ and๐ be ๐-manifolds and๐โถ ๐ โ ๐ a ๐ถโ map. Suppose that at ๐ โ ๐ the map
๐๐๐ โถ ๐๐๐ โ ๐๐๐ , ๐ โ ๐(๐) ,
is bijective. Then ๐maps a neighborhood ๐ of ๐ in ๐ diffeomorphically onto a neighborhood๐ of ๐ in ๐.
Proof. Let๐ and๐ be open neighborhoods of๐ in๐ and ๐ in๐ and let๐0 โถ (๐0, 0) โ (๐, ๐)and ๐0 โถ (๐0, 0) โ (๐, ๐) be parametrizations of these neighborhoods. Shrinking ๐0 and ๐we can assume that ๐(๐) โ ๐. Let
๐โถ (๐0, ๐0) โ (๐0, ๐0)
be the map ๐โ10 โ ๐ โ ๐0. Then ๐0 โ ๐ = ๐ โ ๐, so by the chain rule
(๐๐0)๐0 โ (๐๐)๐0 = (๐๐)๐ โ (๐๐0)๐0 .
Since (๐๐0)๐0 and (๐๐0)๐0 are bijective itโs clear from this identity that if ๐๐๐ is bijectivethe same is true for (๐๐)๐0 . Hence by the inverse function theorem for open subsets of ๐๐,๐ maps a neighborhood of ๐0 in ๐0 diffeomorphically onto a neighborhood of ๐0 in ๐0.Shrinking๐0 and๐0 we assume that these neighborhoods are๐0 and๐0 and hence that ๐ isa diffeomorphism. Thus since ๐โถ ๐ โ ๐ is the map ๐0 โ ๐ โ ๐โ10 , it is a diffeomorphism aswell. โก
Theorem 4.2.14 (The canonical submersion theorem for manifolds). Let ๐ and ๐ be man-ifolds of dimension ๐ and๐, where๐ < ๐, and let ๐โถ ๐ โ ๐ be a ๐ถโ map. Suppose that at๐ โ ๐ the map
๐๐๐ โถ ๐๐๐ โ ๐๐๐ , ๐ โ ๐(๐) ,is surjective. Then there exists an open neighborhood ๐ of ๐ in ๐, and open neighborhood, ๐of ๐(๐) in ๐ and parametrizations ๐0 โถ (๐0, 0) โ (๐, ๐) and ๐0 โถ (๐0, 0) โ (๐, ๐) such thatin the square
๐0 ๐0
๐ ๐
๐0
๐โ10 ๐๐0
๐0
๐
the map ๐โ10 โ ๐ โ ๐0 is the canonical submersion ๐.
Proof. Let ๐ and ๐ be open neighborhoods of ๐ and ๐ and ๐0 โถ (๐0, 0) โ (๐, ๐) and๐0 โถ (๐0, 0) โ (๐, ๐) be parametrizations of these neighborhoods. Composing ๐0 and ๐0with the translations we can assume that ๐0 is the origin in ๐๐ and ๐0 the origin in ๐๐,and shrinking ๐ we can assume ๐(๐) โ ๐. As above let ๐โถ (๐0, 0) โ (๐0, 0) be the map๐โ10 โ ๐ โ ๐0. By the chain rule
(๐๐0)0 โ (๐๐)0 = ๐๐๐ โ (๐๐0)0 ,
therefore, since (๐๐0)0 and (๐๐0)0 are bijective it follows that (๐๐)0 is surjective. Hence, byTheorem 4.1.2, we can find an open neighborhood ๐ of the origin in ๐๐ and a diffeomor-phism ๐1 โถ (๐1, 0) โฅฒ (๐0, 0) such that ๐ โ ๐1 is the canonical submersion. Now replace ๐0by ๐1 and ๐0 by ๐0 โ ๐1. โก
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108 Chapter 4: Manifolds & Forms on Manifolds
Theorem 4.2.15 (The canonical immersion theorem for manifolds). Let ๐ and ๐ be mani-folds of dimension ๐ and๐, where ๐ < ๐, and ๐โถ ๐ โ ๐ a ๐ถโ map. Suppose that at ๐ โ ๐the map
๐๐๐ โถ ๐๐๐ โ ๐๐๐ , ๐ = ๐(๐)
is injective. Then there exists an open neighborhood ๐ of ๐ in ๐, an open neighborhood ๐ of๐(๐) in ๐ and parametrizations ๐0 โถ (๐0, 0) โ (๐, ๐) and ๐0 โถ (๐0, 0) โ (๐, ๐) such that inthe square
๐0 ๐0
๐ ๐
๐0
๐โ10 ๐๐0
๐0
๐
the map ๐โ10 โ ๐ โ ๐0 is the canonical immersion ๐.
Proof. The proof is identical with the proof of Theorem 4.2.14 except for the last step. In thelast step one converts ๐ into the canonical immersion via a map ๐1 โถ (๐1, 0) โ (๐0, 0) withthe property ๐ โ ๐1 = ๐ and then replaces ๐0 by ๐0 โ ๐1. โก
Exercises for ยง4.2Exercise 4.2.i. What is the tangent space to the quadric
๐ = { (๐ฅ1,โฆ, ๐ฅ๐) โ ๐๐ | ๐ฅ๐ = ๐ฅ21 +โฏ + ๐ฅ2๐โ1 }
at the point (1, 0,โฆ, 0, 1)?
Exercise 4.2.ii. Show that the tangent space to the (๐ โ 1)-sphere ๐๐โ1 at ๐ is the space ofvectors (๐, ๐ฃ) โ ๐๐๐๐ satisfying ๐ โ ๐ฃ = 0.
Exercise 4.2.iii. Let ๐โถ ๐๐ โ ๐๐ be a ๐ถโ map and let ๐ = ๐ค๐. What is the tangent spaceto๐ at (๐, ๐(๐))?
Exercise 4.2.iv. Let ๐โถ ๐๐โ1 โ ๐๐โ1 be the antipodal map ๐(๐ฅ) = โ๐ฅ. What is the derivativeof ๐ at ๐ โ ๐๐โ1?
Exercise 4.2.v. Let๐๐ โ ๐๐๐ , ๐ = 1, 2, be an ๐๐-manifold and let ๐๐ โ ๐๐. Define๐ to be theCartesian product
๐1 ร ๐2 โ ๐๐1 ร ๐๐2
and let ๐ = (๐1, ๐2). Show that ๐๐๐ โ ๐๐1๐1 โ ๐๐2๐2.
Exercise 4.2.vi. Let ๐ โ ๐๐ be an ๐-manifold and for ๐ = 1, 2, let ๐๐ โถ ๐๐ โ ๐ โฉ ๐๐ be twoparametrizations. From these parametrizations one gets an overlap diagram
(4.2.16)๐โ11 (๐ โฉ ๐1 โฉ ๐2) ๐โ12 (๐ โฉ ๐1 โฉ ๐2)
๐ โฉ ๐1 โฉ ๐2 .
๐โ12 โ๐1
๐1 ๐2
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ยง4.3 Vector fields & differential forms on manifolds 109
(1) Let ๐ โ ๐โฉ๐ and let ๐๐ = ๐โ1๐ (๐). Derive from the overlap diagram (4.2.16) an overlapdiagram of linear maps
๐๐1๐๐ ๐๐2๐
๐
๐๐๐๐ .
๐(๐โ12 โ๐1)๐1
๐(๐1)๐1 ๐(๐2)๐2
(2) Use overlap diagrams to give another proof that ๐๐๐ is intrinsically defined.
4.3. Vector fields & differential forms on manifolds
A vector field on an open subset ๐ โ ๐๐ is a function ๐ which assigns to each ๐ โ ๐ anelement ๐(๐) of ๐๐๐, and a ๐-form is a function ๐ which assigns to each ๐ โ ๐ an element๐๐ of ๐ฌ๐(๐โ๐๐). These definitions have obvious generalizations to manifolds:
Definition 4.3.1. Let ๐ be a manifold. A vector field on ๐ is a function ๐ which assigns toeach ๐ โ ๐ an element ๐(๐) of ๐๐๐. A ๐-form is a function ๐ which assigns to each ๐ โ ๐an element ๐๐ of ๐ฌ๐(๐โ๐๐).
Weโll begin our study of vector fields and ๐-forms on manifolds by showing that, liketheir counterparts on open subsets of ๐๐, they have nice pullback and pushforward proper-ties with respect to mappings. Let๐ and ๐ be manifolds and ๐โถ ๐ โ ๐ a ๐ถโ mapping.
Definition 4.3.2. Given a vector field ๐ on๐ and a vector field๐ on ๐, weโll say that ๐ and๐ are ๐-related if for all ๐ โ ๐ and ๐ = ๐(๐) we have
(4.3.3) (๐๐)๐๐(๐) = ๐(๐) .
In particular, if ๐ is a diffeomorphism, and weโre given a vector field ๐ on ๐ we candefine a vector field ๐ on ๐ by requiring that for every point ๐ โ ๐, the identity (4.3.3)holds at the point ๐ = ๐โ1(๐). In this case weโll call ๐ the pushforward of by ๐ and denoteit by๐โ๐. Similarly, given a vector field๐ on๐we can define a vector field ๐ on๐ by applyingthe same construction to the inverse diffeomorphism ๐โ1 โถ ๐ โ ๐. We will call the vectorfield (๐โ1)โ๐ the pullback of ๐ by ๐ (and also denote it by ๐โ๐).
For differential forms the situation is even nicer. Just as in ยง2.6 we can define the pull-back operation on forms for any ๐ถโ map ๐โถ ๐ โ ๐. Specifically: Let ๐ be a ๐-form on ๐.For every ๐ โ ๐, and ๐ = ๐(๐) the linear map
๐๐๐ โถ ๐๐๐ โ ๐๐๐induces by (1.8.2)a pullback map
(๐๐๐)โ โถ ๐ฌ๐(๐โ๐ ๐) โ ๐ฌ๐(๐โ๐๐)and, as in ยง2.6, weโll define the pullback ๐โ๐ of ๐ to๐ by defining it at ๐ by the identity
(๐โ๐)(๐) = (๐๐๐)โ๐(๐) .The following results about these operations are proved in exactly the same way as in
ยง2.6.
Proposition 4.3.4. Let ๐, ๐, and ๐ be manifolds and ๐โถ ๐ โ ๐ and ๐โถ ๐ โ ๐ ๐ถโ maps.Then if ๐ is a ๐-form on ๐
๐โ(๐โ๐) = (๐ โ ๐)โ๐ .
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110 Chapter 4: Manifolds & Forms on Manifolds
If ๐ is a vector field on๐ and ๐ and ๐ are diffeomorphisms, then(4.3.5) (๐ โ ๐)โ๐ฃ = ๐โ(๐โ๐ฃ) .
Our first application of these identities will be to definewhat onemeans by a โ๐ถโ vectorfieldโ and a โ๐ถโ ๐-formโ.
Definition4.3.6. Let๐be an๐-manifold and๐ anopen subset of๐.The set๐ is a parametriz-able open set if there exists an open set ๐0 in ๐๐ and a diffeomorphism ๐0 โถ ๐0 โ ๐.
In other words,๐ is parametrizable if there exists a parametrization having๐ as its im-age. (Note that๐ being a manifold means that every point is contained in a parametrizableopen set.)
Now let ๐ โ ๐ be a parametrizable open set and ๐0 โถ ๐0 โ ๐ a parametrization of ๐.
Definition 4.3.7. A ๐-form ๐ on ๐ is ๐ถโ or smooth if ๐โ0๐ is a ๐ถโ.
This definition appears to depend on the choice of the parametrization ๐, but we claimit does not. To see this let ๐1 โถ ๐1 โ ๐ be another parametrization of ๐ and let
๐โถ ๐0 โ ๐1be the composite map ๐โ10 โ ๐0. Then ๐0 = ๐1 โ ๐ and hence by Proposition 4.3.4
๐โ0๐ = ๐โ๐โ1๐ ,so by (2.5.11) ๐โ0๐ is ๐ถโ if ๐โ1๐ is ๐ถโ. The same argument applied to ๐โ1 shows that ๐โ1๐is ๐ถโ if ๐โ0๐ is ๐ถโ.
The notion of โ๐ถโโ for vector fields is defined similarly:
Definition 4.3.8. A vector field ๐ on ๐ is smooth, or simply ๐ถโ, if ๐โ0๐ is ๐ถโ.
By Proposition 4.3.4 ๐โ0๐ = ๐โ๐โ1๐, so, as above, this definition is independent of thechoice of parametrization.
We now globalize these definitions.
Definition 4.3.9. A ๐-form ๐ on ๐ is ๐ถโ if, for every point ๐ โ ๐, ๐ is ๐ถโ on a neigh-borhood of ๐. Similarly, a vector field ๐ on ๐ is ๐ถโ if, for every point ๐ โ ๐, ๐ is ๐ถโ on aneighborhood of๐.
We will also use the identities (4.3.5) and (4.3.18) to prove the following two results.
Proposition 4.3.10. Let๐ and ๐ be manifolds and ๐โถ ๐ โ ๐ a ๐ถโ map. Then if ๐ is a ๐ถโ๐-form on ๐, the pullback ๐โ๐ is a ๐ถโ ๐-form on๐.
Proof. For ๐ โ ๐ and ๐ = ๐(๐) let ๐0 โถ ๐0 โ ๐ and ๐0 โถ ๐0 โ ๐ be parametrizations with๐ โ ๐ and ๐ โ ๐. Shrinking ๐ if necessary we can assume that ๐(๐) โ ๐. Let ๐โถ ๐0 โ ๐0be the map ๐ = ๐โ10 โ ๐ โ ๐0. Then ๐0 โ ๐ = ๐ โ ๐0, so ๐โ๐โ0 ๐ = ๐โ0๐โ๐. Since ๐ is ๐ถโ, wesee that ๐โ0 ๐ is ๐ถโ, so by equation (2.6.11) we have ๐โ๐โ0 ๐ is ๐ถโ, and hence ๐โ0๐โ๐ is ๐ถโ.Thus by definition ๐โ๐ is ๐ถโ on ๐. โก
By exactly the same argument one proves.
Proposition 4.3.11. If ๐ is a ๐ถโ vector field on ๐ and ๐ is a diffeomorphism, then ๐โ๐ is a๐ถโ vector field on๐.
Notation 4.3.12.(1) Weโll denote the space of ๐ถโ ๐-forms on๐ by ๐บ๐(๐) .
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ยง4.3 Vector fields & differential forms on manifolds 111
(2) For ๐ โ ๐บ๐(๐) weโll define the support of ๐ bysupp(๐) โ { ๐ โ ๐ | ๐(๐) โ 0 }
and weโll denote by ๐บ๐๐ (๐) โ ๐บ๐(๐) the space of compactly supported ๐-forms.(3) For a vector field ๐ on๐ weโll define the support of ๐ to be the set
supp(๐) โ { ๐ โ ๐ | ๐(๐) โ 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 ๐ผ โ ๐ be an open interval and ๐พโถ ๐ผ โ ๐ a ๐ถโ curve. For ๐ก0 โ ๐ผ we will call
๏ฟฝ๏ฟฝ = (๐ก0, 1) โ ๐๐ก0๐ the unit vector in ๐๐ก0๐ and if ๐ = ๐พ(๐ก0) we will call the vector๐๐พ๐ก0(๏ฟฝ๏ฟฝ) โ ๐๐๐
the tangent vector to ๐พ at ๐. If ๐ is a vector field on ๐ we will say that ๐พ is an integral curveof ๐ if for all ๐ก0 โ ๐ผ
๐(๐พ(๐ก0)) = ๐๐พ๐ก0(๏ฟฝ๏ฟฝ) .Proposition 4.3.13. Let ๐ and ๐ be manifolds and ๐โถ ๐ โ ๐ a ๐ถโ map. If ๐ and ๐ arevector fields on ๐ and ๐ which are ๐-related, then integral curves of ๐ get mapped by ๐ ontointegral curves of ๐.Proof. If the curve ๐พโถ ๐ผ โ ๐ is an integral curve of ๐ we have to show that ๐ โ ๐พโถ ๐ผ โ ๐ isan integral curve of ๐. If ๐พ(๐ก) = ๐ and ๐ = ๐(๐) then by the chain rule
๐(๐) = ๐๐๐(๐(๐)) = ๐๐๐(๐๐พ๐ก(๏ฟฝ๏ฟฝ))= ๐(๐ โ ๐พ)๐ก(๏ฟฝ๏ฟฝ) . โก
From this result it follows that the local existence, uniqueness and โsmooth dependenceon initial dataโ results about vector fields that we described in ยง2.1 are true for vector fieldson manifolds. More explicitly, let ๐ be a parametrizable open subset of ๐ and ๐โถ ๐0 โ ๐a parametrization. Since๐0 is an open subset of ๐๐ these results are true for the vector field๐ = ๐โ0๐ and hence since ๐ and ๐ are ๐0-related they are true for ๐. In particular:
Proposition 4.3.14 (local existence). For every ๐ โ ๐ there exists an integral curve ๐พ(๐ก)defined for โ๐ < ๐ก < ๐, of ๐ with ๐พ(0) = ๐.Proposition 4.3.15 (local uniqueness). For ๐ = 1, 2, let ๐พ๐ โถ ๐ผ๐ โ ๐ be integral curves of ๐and let ๐ผ = ๐ผ1 โฉ ๐ผ2. Suppose ๐พ2(๐ก) = ๐พ1(๐ก) for some ๐ก โ ๐ผ. Then there exists a unique integralcurve, ๐พโถ ๐ผ1 โช ๐ผ2 โ ๐ with ๐พ = ๐พ1 on ๐ผ1 and ๐พ = ๐พ2 on ๐ผ2.Proposition 4.3.16 (smooth dependence on initial data). For every ๐ โ ๐, there exists aneighborhood ๐ of ๐ in ๐, an interval (โ๐, ๐) and a ๐ถโ map โโถ ๐ ร (โ๐, ๐) โ ๐ such thatfor every ๐ โ ๐ the curve
๐พ๐(๐ก) = โ(๐, ๐ก) , โ ๐ < ๐ก < ๐ ,is an integral curve of ๐ with ๐พ๐(0) = ๐.
As in Chapter 2 we will say that ๐ is complete if, for every ๐ โ ๐ there exists an integralcurve ๐พ(๐ก) defined for โโ < ๐ก < โ, with ๐พ(0) = ๐. 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. If๐ is compact or, more generally, if ๐ is compactly supported, ๐ is complete.
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112 Chapter 4: Manifolds & Forms on Manifolds
Proof. Itโs not hard to prove this by the same argument that we used to prove this theoremfor vector fields on ๐๐, but weโll give a simpler proof that derives this directly from the ๐๐result. Suppose๐ is a submanifold of ๐๐. Then for ๐ โ ๐,
๐๐๐ โ ๐๐๐๐ = { (๐, ๐ฃ) | ๐ฃ โ ๐๐ } ,
so ๐(๐) can be regarded as a pair (๐, ๐(๐)), where ๐(๐) is in ๐๐. Let
(4.3.18) ๐๐ โถ ๐ โ ๐๐
be the map ๐๐(๐) = ๐ฃ(๐). It is easy to check that ๐ is ๐ถโ if and only if ๐๐ is ๐ถโ. (SeeExercise 4.3.xi.) Hence (see Appendix b) there exists a neighborhood, ๐ of ๐ and a map๐โถ โ ๐๐ extending ๐๐. Thus the vector field ๐ on ๐ defined by ๐(๐) = (๐, ๐(๐)) extendsthe vector field ๐ to๐. In other words if ๐ โถ ๐ โช ๐ is the inclusionmap ๐ and๐ are ๐-related.Thus by Proposition 4.3.13 the integral curves of ๐ are just integral curves of ๐ that arecontained in๐.
Suppose now that ๐ is compactly supported. Then there exists a function ๐ โ ๐ถโ0 (๐)which is 1 on the support of ๐, so, replacing ๐ by ๐๐, we can assume that ๐ is compactlysupported. Thus ๐ is complete. Let ๐พ(๐ก), defined for โโ < ๐ก < โ, be an integral curve of๐. We will prove that if ๐พ(0) โ ๐, then this curve is an integral curve of ๐. We first observe:
Lemma 4.3.19. The set of points ๐ก โ ๐ for which ๐พ(๐ก) โ ๐ is both open and closed.
Proof of Lemma 4.3.19. If ๐ โ supp(๐) then ๐(๐) = 0 so if ๐พ(๐ก) = ๐, ๐พ(๐ก) is the constantcurve ๐พ = ๐, and thereโs nothing to prove. Thus we are reduced to showing that the set
(4.3.20) { ๐ก โ ๐ | ๐พ(๐ก) โ supp(๐) }
is both open and closed. Since supp(๐) is compact this set is clearly closed. To show that itโsopen suppose ๐พ(๐ก0) โ supp(๐). By local existence there exist an interval (โ๐ + ๐ก0, ๐ + ๐ก0) andan integral curve ๐พ1(๐ก) of ๐ defined on this interval and taking the value ๐พ1(๐ก0) = ๐พ(๐ก0) at ๐.However since ๐ and๐ are ๐-related ๐พ1 is also an integral curve of๐ and so it has to coincidewith ๐พ on the interval (โ๐ + ๐ก0, ๐ + ๐ก0). In particular, for ๐ก on this interval ๐พ(๐ก) โ supp(๐) sothe set (4.3.20) is open. โก
To conclude the proof of Theorem 4.3.17 we note that since ๐ is connected it followsthat if ๐พ(๐ก0) โ ๐ for some ๐ก0 โ ๐ then ๐พ(๐ก) โ ๐ for all ๐ก โ ๐, and hence ๐พ is an integral curveof ๐. Thus in particular every integral curve of ๐ exists for all time, so ๐ is complete. โก
Since๐ is complete it generates a one-parameter group of diffeomorphisms๐๐ก โถ ๐ โฅฒ ๐,defined for โโ < ๐ก < โ, having the property that the curve
๐๐ก(๐) = ๐พ๐(๐ก) , โโ < ๐ก < โ
is the unique integral curve of ๐ with initial point ๐พ๐(0) = ๐. But if ๐ โ ๐ this curve is anintegral curve of ๐, so the restriction
๐๐ก = ๐๐ก|๐is a one-parameter group of diffeomorphisms of ๐ with theproperty that for ๐ โ ๐ thecurve
๐๐ก(๐) = ๐พ๐(๐ก) , โโ < ๐ก < โis the unique integral curve of ๐ with initial point ๐พ๐(0) = ๐.
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ยง4.3 Vector fields & differential forms on manifolds 113
The exterior differentiation operationLet ๐ be a๐ถโ ๐-form on๐ and๐ โ ๐ a parametrizable open set. Given a parametriza-
tion ๐0 โถ ๐0 โ ๐ we define the exterior derivative ๐๐ of ๐ on๐ by the formula(4.3.21) ๐๐ = (๐โ10 )โ๐๐โ0๐ .(Notice that since๐0 is an open subset of ๐๐ and ๐โ0๐ a ๐-form on๐0, the โ๐โ on the right iswell-defined.) We claim that this definition does not depend on the choice of parametriza-tion. To see this let ๐1 โถ ๐1 โ ๐ be another parametrization of ๐ and let ๐โถ ๐0 โ ๐1 bethe diffeomorphism ๐โ11 โ ๐0. Then ๐0 = ๐1 โ ๐ and hence
๐๐โ0๐ = ๐๐โ๐โ1๐ = ๐โ๐๐โ1๐= ๐โ0 (๐โ11 )โ๐๐โ1๐
so(๐โ10 )โ๐๐โ0๐ = (๐โ11 )โ๐๐โ1๐
as claimed. We can, therefore, define the exterior derivative ๐๐ globally by defining it to beequal to (4.3.21) on every parametrizable open set.
Itโs easy to see from the definition (4.3.21) that this exterior differentiation operationinherits from the exterior differentiation operation on open subsets of ๐๐ the properties(2.4.3) and (2.4.4) and that for zero forms, i.e.,๐ถโ functions๐โถ ๐ โ ๐, ๐๐ is the โintrinsicโ๐๐ defined in Section 2.1, i.e., for ๐ โ ๐, ๐๐๐ is the derivative of ๐
๐๐๐ โถ ๐๐๐ โ ๐
viewed as an element of ๐ฌ1(๐โ๐๐). Letโs check that it also has the property (2.6.12).
Theorem 4.3.22. Let๐ and ๐ be manifolds and ๐โถ ๐ โ ๐ a ๐ถโ map. Then for ๐ โ ๐บ๐(๐)we have
๐โ(๐๐) = ๐(๐โ๐) .
Proof. For every ๐ โ ๐ weโll check that this equality holds in a neighborhood of ๐. Let๐ = ๐(๐) and let ๐ and ๐ be parametrizable neighborhoods of ๐ and ๐. Shrinking ๐ ifnecessary we can assume ๐(๐) โ ๐. Given parametrizations
๐โถ ๐0 โ ๐and
๐โถ ๐0 โ ๐we get by composition a map
๐โถ ๐0 โ ๐0 , ๐ = ๐โ1 โ ๐ โ ๐with the property ๐ โ ๐ = ๐ โ ๐. Thus
๐โ๐(๐โ๐) = ๐๐โ๐โ๐ (by definition of ๐)= ๐(๐ โ ๐)โ๐= ๐(๐ โ ๐)โ๐= ๐๐โ(๐โ๐)= ๐โ๐๐โ๐ (by (2.6.12))= ๐โ๐โ๐๐ (by definition of ๐)= ๐โ๐โ๐๐ .
Hence ๐๐โ๐ = ๐โ๐๐. โก
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114 Chapter 4: Manifolds & Forms on Manifolds
The interior product and Lie derivative operationGiven a ๐-form ๐ โ ๐บ๐(๐) and a ๐ถโ vector field ๐ we will define the interior product
(4.3.23) ๐๐๐ โ ๐บ๐โ1(๐) ,as in ยง2.5, by setting
(๐๐๐)๐ = ๐๐(๐)๐๐and the Lie derivative ๐ฟ๐๐ โ ๐บ๐(๐) by setting
(4.3.24) ๐ฟ๐๐ โ ๐๐๐๐ + ๐๐๐๐ .Itโs easily checked that these operations satisfy Properties 2.5.3 and Properties 2.5.10 (since,just as in ยง2.5, these identities are deduced from the definitions (4.3.23) and (4.3.3) by purelyformal manipulations). Moreover, if ๐ is complete and
๐๐ก โถ ๐ โ ๐ , โโ < ๐ก < โis the one-parameter group of diffeomorphisms of ๐ generated by ๐ the Lie derivative op-eration can be defined by the alternative recipe
(4.3.25) ๐ฟ๐๐ =๐๐๐ก๐โ๐ก ๐|๐ก=0
as in (2.6.22). (Just as in ยง2.6 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 ยง4.3Exercise 4.3.i. Let๐ โ ๐3 be the paraboloid defined by ๐ฅ3 = ๐ฅ21 +๐ฅ22 and let๐ be the vectorfield
๐ โ ๐ฅ1๐๐๐ฅ1+ ๐ฅ2๐๐๐ฅ2+ 2๐ฅ3๐๐๐ฅ3
.
(1) Show that ๐ is tangent to๐ and hence defines by restriction a vector field ๐ on๐.(2) What are the integral curves of ๐?
Exercise 4.3.ii. Let ๐2 be the unit 2-sphere, ๐ฅ21 + ๐ฅ22 + ๐ฅ23 = 1, in ๐3 and let ๐ be the vectorfield
๐ โ ๐ฅ1๐๐๐ฅ2โ ๐ฅ2๐๐๐ฅ1
.
(1) Show that ๐ is tangent to ๐2, and hence by restriction defines a vector field ๐ on ๐2.(2) What are the integral curves of ๐?
Exercise 4.3.iii. As in Exercise 4.3.ii let ๐2 be the unit 2-sphere in๐3 and let๐ be the vectorfield
๐ โ ๐๐๐ฅ3โ ๐ฅ3 (๐ฅ1
๐๐๐ฅ1+ ๐ฅ2๐๐๐ฅ2+ ๐ฅ3๐๐๐ฅ3)
(1) Show that ๐ is tangent to ๐2 and hence by restriction defines a vector field ๐ on ๐2.(2) What do its integral curves look like?
Exercise 4.3.iv. Let ๐1 be the unit circle, ๐ฅ21 + ๐ฅ22 = 1, in ๐2 and let ๐ = ๐1 ร ๐1 in ๐4 withdefining equations
๐1 = ๐ฅ21 + ๐ฅ22 โ 1 = 0๐2 = ๐ฅ23 + ๐ฅ24 โ 1 = 0 .
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ยง4.3 Vector fields & differential forms on manifolds 115
(1) Show that the vector field
๐ โ ๐ฅ1๐๐๐ฅ2โ ๐ฅ2๐๐๐ฅ1+ ๐(๐ฅ4
๐๐๐ฅ3โ ๐ฅ3๐๐๐ฅ4) ,
๐ โ ๐, is tangent to๐ and hence defines by restriction a vector field ๐ on๐.(2) What are the integral curves of ๐?(3) Show that ๐ฟ๐๐๐ = 0.Exercise 4.3.v. For the vector field ๐ in Exercise 4.3.iv, describe the one-parameter groupof diffeomorphisms it generates.
Exercise 4.3.vi. Let๐ and ๐be as in Exercise 4.3.i and let๐โถ ๐2 โ ๐be themap๐(๐ฅ1, ๐ฅ2) =(๐ฅ1, ๐ฅ2, ๐ฅ21 + ๐ฅ22). Show that if ๐ is the vector field,
๐ โ ๐ฅ1๐๐๐ฅ1+ ๐ฅ2๐๐๐ฅ2
,
then ๐โ๐ = ๐.
Exercise 4.3.vii. Let๐ be a submanifold of๐ in ๐๐ and let ๐ and๐ be the vector fields on๐ and ๐. Denoting by ๐ the inclusion map of ๐ into ๐, show that ๐ and ๐ are ๐-related ifand only if ๐ is tangent to๐ and its restriction to๐ is ๐.
Exercise 4.3.viii. Let๐ be a submanifold of ๐๐ and ๐ an open subset of ๐๐ containing๐,and let ๐ and ๐ be the vector fields on ๐ and ๐. Denoting by ๐ the inclusion map of ๐ into๐, show that ๐ and ๐ are ๐-related if and only if ๐ is tangent to๐ and its restriction to๐ is๐.Exercise 4.3.ix. An elementary result in number theory asserts:
Theorem 4.3.26. A number ๐ โ ๐ is irrational if and only if the set{๐ + ๐๐ |๐, ๐ โ ๐ }
is a dense subset of ๐.Let ๐ be the vector field in Exercise 4.3.iv. Using Theorem 4.3.26 prove that if ๐ is irra-
tional then for every integral curve ๐พ(๐ก), defined for โโ < ๐ก < โ, of ๐ the set of points onthis curve is a dense subset of๐.Exercise 4.3.x. Let ๐ be an ๐-dimensional submanifold of ๐๐. Prove that a vector field ๐on๐ is ๐ถโ if and only if the map (4.3.18) is ๐ถโ.
Hint: Let๐ be a parametrizable open subset of๐ and ๐โถ ๐0 โ ๐ a parametrization of๐. Composing ๐ with the inclusion map ๐ โถ ๐ โ ๐๐ one gets a map ๐ โ ๐โถ ๐ โ ๐๐. Showthat if
๐โ๐ = โ๐ฃ๐๐๐๐ฅ๐
then๐โ๐๐ = โ
๐๐๐๐๐ฅ๐๐ฃ๐
where ๐1,โฆ, ๐๐ are the coordinates of the map ๐๐ and ๐1,โฆ, ๐๐ the coordinates of ๐ โ ๐.Exercise 4.3.xi. Let ๐ be a vector field on๐ and let ๐โถ ๐ โ ๐ be a ๐ถโ function. Show thatif the function
๐ฟ๐๐ = ๐๐๐๐is zero ๐ is constant along integral curves of ๐.
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Exercise 4.3.xii. Suppose that ๐โถ ๐ โ ๐ is proper. Show that if ๐ฟ๐๐ = 0, ๐ is complete.Hint: For ๐ โ ๐ let ๐ = ๐(๐). By assumption, ๐โ1(๐) is compact. Let ๐ โ ๐ถโ0 (๐) be a
โbumpโ function which is one on ๐โ1(๐) and let๐ be the vector field ๐๐. By Theorem 4.3.17,๐ is complete and since
๐ฟ๐๐ = ๐๐๐๐๐ = ๐๐๐๐๐ = 0๐ is constant along integral curves of ๐. Let ๐พ(๐ก), โโ < ๐ก < โ, be the integral curve of ๐with initial point ๐พ(0) = ๐. Show that ๐พ is an integral curve of ๐.
4.4. Orientations
The last part of Chapter 4 will be devoted to the โintegral calculusโ of forms on mani-folds. In particularwewill provemanifold versions of two basic theorems of integral calculuson ๐๐, Stokesโ 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 โorientationโ 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. Let๐ be an ๐-manifold. An orientation of๐ is a rule for assigning to each๐ โ ๐ an orientation of ๐๐๐.
Thus by Definition 1.9.3 one can think of an orientation as a โlabelingโ rule which, forevery๐ โ ๐, labels one of the two components of the set๐ฌ๐(๐โ๐๐)โ{0} by๐ฌ๐(๐โ๐๐)+, whichweโll henceforth call the โplusโ part of ๐ฌ๐(๐โ๐๐), and the other component by ๐ฌ๐(๐โ๐๐)โ,which weโll henceforth call the โminusโ part of ๐ฌ๐(๐โ๐๐).
Definition 4.4.2. An orientation of a manifold ๐ is smooth, or simply ๐ถโ, if for every๐ โ ๐, there exists a neighborhood๐ of ๐ and a non-vanishing ๐-form ๐ โ ๐บ๐(๐)with theproperty
๐๐ โ ๐ฌ๐(๐โ๐ ๐)+for every ๐ โ ๐.
Remark 4.4.3. If weโre given an orientation of ๐ we can define another orientation by as-signing to each ๐ โ ๐ the opposite orientation to the orientation we already assigned, i.e.,by switching the labels on ๐ฌ๐(๐โ๐ )+ and ๐ฌ๐(๐โ๐ )โ. We will call this the reversed orientationof๐. We will leave for you to check as an exercise that if๐ is connected and equipped witha smooth orientation, the only smooth orientations of ๐ are the given orientation and itsreversed orientation.
Hint: Given any smooth orientation of๐ 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 if๐ โ ๐บ๐(๐) is a non-vanishing ๐-form one gets from๐ a smooth orientationof๐ by requiring that the โlabeling ruleโ above satisfy
๐๐ โ ๐ฌ๐(๐โ๐๐)+for every ๐ โ ๐. If ๐ has this property we will call ๐ a volume form. Itโs clear from this defi-nition that if๐1 and๐2 are volume forms on๐ then๐2 = ๐2,1๐1 where๐2,1 is an everywherepositive ๐ถโ function.
Example 4.4.4. Let ๐ be an open subset ๐๐. We will usually assign to ๐ its standard orien-tation, by which we will mean the orientation defined by the ๐-form ๐๐ฅ1 โงโฏ โง ๐๐ฅ๐.
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ยง4.4 Orientations 117
Example 4.4.5. Let ๐โถ ๐๐ โ ๐๐ be a ๐ถโ map. If zero is a regular value of ๐, the set๐ โ ๐โ1(0) is a submanifold of๐๐ of dimension ๐ โ ๐โ๐ (byTheorem 4.2.13). Moreover,for ๐ โ ๐, ๐๐๐ is the kernel of the surjective map
๐๐๐ โถ ๐๐๐๐ โ ๐0๐๐
so we get from ๐๐๐ a bijective linear map
(4.4.6) ๐๐๐๐/๐๐๐ โ ๐0๐๐ .
As explained inExample 4.4.4,๐๐๐๐ and๐0๐๐ have โstandardโ orientations, hence if werequire that the map (4.4.6) be orientation preserving, this gives ๐๐๐๐/๐๐๐ an orientationand, by Theorem 1.9.9, gives ๐๐๐ an orientation. Itโs intuitively clear that since ๐๐๐ variessmoothly with respect to ๐ this orientation does as well; however, this fact requires a proof,and weโll supply a sketch of such a proof in the exercises.
Example 4.4.7. A special case of Example 4.4.5 is the ๐-sphere๐๐ = { (๐ฅ1,โฆ, ๐ฅ๐+1) โ ๐๐+1 | ๐ฅ21 +โฏ + ๐ฅ2๐+1 = 1 } ,
which acquires an orientation from its defining map ๐โถ ๐๐+1 โ ๐ given by
๐(๐ฅ) โ ๐ฅ21 +โฏ + ๐ฅ2๐+1 โ 1 .
Example 4.4.8. Let ๐ be an oriented submanifold of ๐๐. For every ๐ โ ๐, ๐๐๐ sits inside๐๐๐๐ as a vector subspace, hence, via the identification ๐๐๐๐ โ ๐๐, one can think of ๐๐๐as a vector subspace of ๐๐. In particular from the standard Euclidean inner product on ๐๐one gets, by restricting this inner product to vectors in ๐๐๐, an inner product
๐ต๐ โถ ๐๐๐ ร ๐๐๐ โ ๐
on ๐๐๐. Let ๐๐ be the volume element in ๐ฌ๐(๐โ๐๐) associated with ๐ต๐ (see Exercise 1.9.x)and let ๐ = ๐๐ be the non-vanishing ๐-form on๐ defined by the assignment
๐ โฆ ๐๐ .In the exercises at the end of this section weโll sketch a proof of the following.
Theorem 4.4.9. The form ๐๐ is ๐ถโ and hence, in particular, is a volume form. (We will callthis form the Riemannian volume form.)
Example 4.4.10 (Mรถbius strip). The Mรถbius strip is a surface in ๐3 which is not orientable.It is obtained from the rectangle
๐ โ [0, 1] ร (โ1, 1) = { (๐ฅ, ๐ฆ) โ ๐2 | 0 โค ๐ฅ โค 1 and โ 1 < ๐ฆ < 1 }by gluing the ends together in the wrong way, i.e., by gluing (1, ๐ฆ) to (0, โ๐ฆ). It is easy to seethat theMรถbius strip cannot be oriented by taking the standard orientation at ๐ = (1, 0) andmoving it along the line (๐ก, 0), 0 โค ๐ก โค 1 to the point (0, 0) (which is also the point ๐ afterwe have glued the ends of the rectangle together).
Weโll next investigate the โcompatibilityโ question for diffeomorphisms between ori-ented manifolds. Let ๐ and ๐ be ๐-manifolds and ๐โถ ๐ โฅฒ ๐ a diffeomorphism. Supposeboth of these manifolds are equipped with orientations. We will say that ๐ is orientationpreserving if for all ๐ โ ๐ and ๐ = ๐(๐), the linear map
๐๐๐ โถ ๐๐๐ โ ๐๐๐
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118 Chapter 4: Manifolds & Forms on Manifolds
is orientation preserving. Itโs clear that if ๐ is a volume form on ๐ then ๐ is orientationpreserving if and only if ๐โ๐ is a volume form on ๐, and from (1.9.5) and the chain ruleone easily deduces the following theorem.
Theorem 4.4.11. If ๐ is an oriented ๐-manifold and ๐โถ ๐ โ ๐ a diffeomorphism, then ifboth ๐ and ๐ are orientation preserving, so is ๐ โ ๐.
If ๐โถ ๐ โฅฒ ๐ is a diffeomorphism then the set of points ๐ โ ๐ at which the linear map
๐๐๐ โถ ๐๐๐ โ ๐๐๐ , ๐ = ๐(๐) ,
is orientation preserving is open, and the set of points at which is orientation reversing isopen as well. Hence if ๐ is connected, ๐๐๐ has to be orientation preserving at all points ororientation reversing at all points. In the latter case weโll say that ๐ is orientation reversing.
If ๐ is a parametrizable open subset of๐ and ๐โถ ๐0 โ ๐ a parametrization of ๐ weโllsay that this parametrization is an oriented parametrization if ๐ is orientation preservingwith respect to the standard orientation of ๐0 and the given orientation on ๐. Notice thatif this parametrization isnโt oriented we can convert it into one that is by replacing everyconnected component ๐0 of ๐0 on which ๐ isnโt orientation preserving by the open set
(4.4.12) ๐โฏ0 = { (๐ฅ1,โฆ, ๐ฅ๐) โ ๐๐ | (๐ฅ1,โฆ, ๐ฅ๐โ1, โ๐ฅ๐) โ ๐0 }
and replacing ๐ by the map
(4.4.13) ๐(๐ฅ1,โฆ, ๐ฅ๐) = ๐(๐ฅ1,โฆ, ๐ฅ๐โ1, โ๐ฅ๐) .
If ๐๐ โถ ๐๐ โ ๐, ๐ = 0, 1, are oriented parametrizations of ๐ and ๐โถ ๐0 โ ๐1 is thediffeomorphism ๐โ11 โ ๐0, then by the theorem above ๐ is orientation preserving or in otherwords
(4.4.14) det(๐๐๐๐๐ฅ๐) > 0
at every point on ๐0.Weโll conclude this section by discussing some orientation issues which will come up
when we discuss Stokesโ theorem and the divergence theorem in ยง4.6. First a definition.
Definition 4.4.15. An open subset๐ท of๐ is a smooth domain if(1) The boundary ๐๐ท is an (๐ โ 1)-dimensional submanifold of๐, and(2) The boundary of๐ท coincides with the boundary of the closure of๐ท.
Examples 4.4.16.(1) The ๐-ball, ๐ฅ21 +โฏ + ๐ฅ2๐ < 1, whose boundary is the sphere, ๐ฅ21 +โฏ + ๐ฅ2๐ = 1.(2) The ๐-dimensional annulus,
1 < ๐ฅ21 +โฏ + ๐ฅ2๐ < 2
whose boundary consists of the union of the two spheres
{๐ฅ21 +โฏ + ๐ฅ2๐ = 1๐ฅ21 +โฏ + ๐ฅ2๐ = 2 .
(3) Let ๐๐โ1 be the unit sphere, ๐ฅ21 +โฏ+ ๐ฅ22 = 1 and let๐ท = ๐๐ โ ๐๐โ1. Then the boundaryof๐ท is ๐๐โ1 but๐ท is not a smooth domain since the boundary of its closure is empty.
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ยง4.4 Orientations 119
(4) The simplest example of a smooth domain is the half-space
๐๐ = { (๐ฅ1,โฆ, ๐ฅ๐) โ ๐๐ | ๐ฅ1 < 0 }whose boundary
๐๐๐ = { (๐ฅ1,โฆ, ๐ฅ๐) โ ๐๐ | ๐ฅ1 = 0 }we can identify with ๐๐โ1 via the map ๐๐โ1 โ ๐๐๐ given by
(๐ฅ2,โฆ, ๐ฅ๐) โฆ (0, ๐ฅ2,โฆ, ๐ฅ๐) .
We will show that every bounded domain looks locally like this example.
Theorem 4.4.17. Let๐ท be a smooth domain and ๐ a boundary point of๐ท. Then there existsa neighborhood ๐ of ๐ in ๐, an open set ๐0 in ๐๐, and a diffeomorphism ๐โถ ๐0 โฅฒ ๐ suchthat ๐maps ๐0 โฉ ๐๐ onto ๐ โฉ ๐ท.
Proof. Let ๐ โ ๐๐ท. First we prove the following lemma.
Lemma4.4.18. For every๐ โ ๐ there exists an open set๐ in๐ containing๐ and a parametriza-tion
๐โถ ๐0 โ ๐of ๐ with the property
(4.4.19) ๐(๐0 โฉ ๐๐๐) = ๐ โฉ ๐ .
Proof. Since ๐ is locally diffeomorphic at ๐ to an open subset of ๐๐ so it suffices to provethis assertion for ๐ = ๐๐. However, if ๐ is an (๐ โ 1)-dimensional submanifold of ๐๐ thenbyTheorem 4.2.15 there exists, for every ๐ โ ๐, a neighborhood๐ of ๐ in๐๐ and a function๐ โ ๐ถโ(๐) with the properties
(4.4.20) ๐ฅ โ ๐ โฉ ๐ โบ ๐(๐ฅ) = 0and
(4.4.21) ๐๐๐ โ 0 .Without loss of generality we can assume by equation (4.4.21) that
๐๐๐๐ฅ1(๐) โ 0 .
Hence if ๐โถ ๐ โ ๐๐ is the map
(4.4.22) ๐(๐ฅ1,โฆ, ๐ฅ๐) โ (๐(๐ฅ), ๐ฅ2,โฆ, ๐ฅ๐)(๐๐)๐ is bijective, and hence ๐ is locally a diffeomorphism at ๐. Shrinking๐ we can assumethat ๐ is a diffeomorphism of๐ onto an open set๐0. By (4.4.20) and (4.4.22) ๐maps๐โฉ๐onto ๐0 โฉ ๐๐๐ hence if we take ๐ to be ๐โ1, it will have the property (4.4.19). โก
We will now prove Theorem 4.4.9. Without loss of generality we can assume that theopen set ๐0 in Lemma 4.4.18 is an open ball with center at ๐ โ ๐๐๐ and that the diffeomor-phism ๐maps ๐ to ๐. Thus for ๐โ1(๐ โฉ ๐ท) there are three possibilities:(i) ๐โ1(๐ โฉ ๐ท) = (๐๐ โ ๐๐๐) โฉ ๐0.(ii) ๐โ1(๐ โฉ ๐ท) = (๐๐ โ ๐๐) โฉ ๐0.(iii) ๐โ1(๐ โฉ ๐ท) = ๐๐ โฉ ๐0.However, (i) is excluded by the second hypothesis in Definition 4.4.15 and if (ii) occurs wecan rectify the situation by composing ๐ with the map (๐ฅ1,โฆ, ๐ฅ๐) โฆ (โ๐ฅ1, ๐ฅ2,โฆ, ๐ฅ๐). โก
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120 Chapter 4: Manifolds & Forms on Manifolds
Definition4.4.23. Wewill call an open set๐with the properties above a๐ท-adapted parame-trizable open set.
We will now show that if๐ is oriented and๐ท โ ๐ is a smooth domain then the bound-ary ๐ โ ๐๐ท of๐ท acquires from๐ a natural orientation. To see this we first observe:
Lemma 4.4.24. The diffeomorphism ๐โถ ๐0 โฅฒ ๐ in Theorem 4.4.17 can be chosen to beorientation preserving.
Proof. If it is not, then by replacing ๐ with the diffeomorphism๐โฏ(๐ฅ1,โฆ, ๐ฅ๐) โ ๐(๐ฅ1,โฆ, ๐ฅ๐โ1, โ๐ฅ๐) ,
we get a ๐ท-adapted parametrization of ๐ which is orientation preserving. (See (4.4.12)โ(4.4.13).) โก
Let๐0 = ๐0โฉ๐๐โ1 be the boundary of๐0โฉ๐๐. The restriction of๐ to๐0 is a diffeomor-phism of ๐0 onto๐โฉ๐, and we will orient๐โฉ๐ by requiring that this map be an orientedparametrization. To show that this is an โintrinsicโ definition, i.e., does not depend on thechoice of ๐. Weโll prove:
Theorem4.4.25. If๐๐ โถ ๐๐ โ ๐, ๐ = 0, 1, are oriented parametrizations of๐with the property๐๐ โถ ๐๐ โฉ ๐๐ โ ๐ โฉ ๐ท
the restrictions of ๐๐ to ๐๐ โฉ ๐๐โ1 induce compatible orientations on ๐ โฉ ๐.
Proof. To prove this we have to prove that the map ๐โ11 โ ๐0, restricted to ๐ โฉ ๐๐๐ is anorientation preserving diffeomorphism of๐0 โฉ๐๐โ1 onto๐1 โฉ๐๐โ1. Thus we have to provethe following:
Proposition 4.4.26. Let ๐0 and ๐1 be open subsets of ๐๐ and ๐โถ ๐0 โ ๐1 an orientationpreserving diffeomorphism which maps ๐0 โฉ ๐๐ onto ๐1 โฉ ๐๐. Then the restriction ๐ of ๐ tothe boundary ๐0 โฉ ๐๐โ1 of ๐0 โฉ ๐๐ is an orientation preserving diffeomorphism
๐โถ ๐0 โฉ ๐๐โ1 โฅฒ ๐1 โฉ ๐๐โ1 .Let ๐(๐ฅ) = (๐1(๐ฅ),โฆ, ๐๐(๐ฅ)). By assumption ๐1(๐ฅ1,โฆ, ๐ฅ๐) is less than zero if ๐ฅ1 is less
than zero and equal to zero if ๐ฅ1 is equal to zero, hence
(4.4.27) ๐๐1๐๐ฅ1(0, ๐ฅ2,โฆ, ๐ฅ๐) โฅ 0
and๐๐1๐๐ฅ๐(0, ๐ฅ2,โฆ, ๐ฅ๐) = 0
for ๐ > 1Moreover, since ๐ is the restriction of ๐ to the set ๐ฅ1 = 0๐๐๐๐๐ฅ๐(0, ๐ฅ2,โฆ, ๐ฅ๐) =
๐๐๐๐๐ฅ๐(๐ฅ2,โฆ, ๐ฅ1)
for ๐, ๐ โฅ 2. Thus on the set defined by ๐ฅ1 = 0 we have
(4.4.28) det(๐๐๐๐๐ฅ๐) = ๐๐1๐๐ฅ1
det(๐๐๐๐๐ฅ๐) .
Since ๐ is orientation preserving the left hand side of equation (4.4.28) is positive at allpoints (0, ๐ฅ2,โฆ, ๐ฅ๐) โ ๐0 โฉ ๐๐โ1 hence by (4.4.27) the same is true for ๐๐1๐๐ฅ1 and det ( ๐๐๐๐๐ฅ๐ ).Thus ๐ is orientation preserving. โก
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ยง4.4 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 of๐ท by requiring that for every๐ท-adapted parametriz-able open set๐ the orientation of๐ coincides with the orientation of๐โฉ๐ 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 ๐ = 1, 2, let ๐๐ be an oriented manifold, ๐ท๐ โ ๐๐ a smooth domainand๐๐ โ ๐๐ท๐ its boundary. Then if ๐ is an orientation preserving diffeomorphism of (๐1, ๐ท1)onto (๐2, ๐ท2) the restriction ๐ of ๐ to ๐1 is an orientation preserving diffeomorphism of ๐1onto ๐2.
Proof. Let๐be anopen subset of๐1 and๐โถ ๐0 โ ๐ an oriented๐ท1-compatible parametriza-tion of๐.Then if๐ = ๐(๐) themap๐๐โถ ๐ โ ๐ is an oriented๐ท2-compatible parametriza-tion of ๐ and hence ๐โถ ๐ โฉ ๐1 โ ๐ โฉ ๐2 is orientation preserving. โก
Exercises for ยง4.4Exercise 4.4.i. Let ๐ be an oriented ๐-dimensional vector space, ๐ต an inner product on ๐and ๐1,โฆ, ๐๐ โ ๐ an oriented orthonormal basis. Given vectors ๐ฃ1,โฆ, ๐ฃ๐ โ ๐, show that if
(4.4.31) ๐๐,๐ = ๐ต(๐ฃ๐, ๐ฃ๐)and
๐ฃ๐ =๐โ๐=1๐๐,๐๐๐ ,
the matrices ๐ = (๐๐,๐) and ๐ = (๐๐,๐) satisfy the identity:
๐ = ๐โบ๐and conclude that det(๐) = det(๐)2. (In particular conclude that det(๐) > 0.)
Exercise 4.4.ii. Let ๐ and๐ be oriented ๐-dimensional vector spaces. Suppose that eachof these spaces is equipped with an inner product, and let ๐1,โฆ, ๐๐ โ ๐ and ๐1,โฆ, ๐๐ โ ๐be oriented orthonormal bases. Show that if ๐ดโถ ๐ โ ๐ is an orientation preserving linearmapping and ๐ด๐๐ = ๐ฃ๐ then
๐ดโ vol๐ = (det(๐๐,๐))12 vol๐
where vol๐ = ๐โ1 โงโฏ โง ๐โ๐ , vol๐ = ๐โ1 โงโฏ โง ๐โ๐ and (๐๐,๐) is the matrix (4.4.31).
Exercise 4.4.iii. Let ๐ be an oriented ๐-dimensional submanifold of ๐๐, ๐ an open subsetof ๐, ๐0 an open subset of ๐๐, and ๐โถ ๐0 โ ๐ an oriented parametrization. Let ๐1,โฆ, ๐๐be the coordinates of the map
๐0 โ ๐ โช ๐๐ .the second map being the inclusion map. Show that if ๐ is the Riemannian volume form on๐ then
๐โ๐ = (det(๐๐,๐))12 ๐๐ฅ1 โงโฏ โง ๐๐ฅ๐
where
(4.4.32) ๐๐,๐ =๐โ๐=1
๐๐๐๐๐ฅ๐๐๐๐๐๐ฅ๐
for 1 โค ๐, ๐ โค ๐ .
Conclude that ๐ is a smooth ๐-form and hence that it is a volume form.
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122 Chapter 4: Manifolds & Forms on Manifolds
Hint: For ๐ โ ๐0 and ๐ = ๐(๐) apply Exercise 4.4.ii with ๐ = ๐๐๐,๐ = ๐๐๐๐, ๐ด =(๐๐)๐ and ๐ฃ๐ = (๐๐)๐ ( ๐๐๐ฅ๐ )๐.
Exercise 4.4.iv. Given a ๐ถโ function ๐โถ ๐ โ ๐, its graph
๐ โ ๐ค๐ = { (๐ฅ, ๐(๐ฅ)) โ ๐ ร ๐ | ๐ฅ โ ๐ }
is a submanifold of ๐2 and๐โถ ๐ โ ๐ , ๐ฅ โฆ (๐ฅ, ๐(๐ฅ))
is a diffeomorphism. Orient ๐ by requiring that ๐ be orientation preserving and show thatif ๐ is the Riemannian volume form on๐ then
๐โ๐ = (1 + (๐๐๐๐ฅ)2)12
๐๐ฅ .
Hint: Exercise 4.4.iii
Exercise 4.4.v. Given a๐ถโ function๐โถ ๐๐ โ ๐ the graph ๐ค๐ of๐ is a submanifold of๐๐+1and
(4.4.33) ๐โถ ๐๐ โ ๐ , ๐ฅ โฆ (๐ฅ, ๐(๐ฅ))is a diffeomorphism. Orient ๐ by requiring that ๐ is orientation preserving and show thatif ๐ is the Riemannian volume form on๐ then
๐โ๐ = (1 +๐โ๐=1( ๐๐๐๐ฅ๐)2)12
๐๐ฅ1 โงโฏ โง ๐๐ฅ๐ .
Hints:โข Let ๐ฃ = (๐1,โฆ, ๐๐) โ ๐๐. Show that if ๐ถโถ ๐๐ โ ๐ is the linear mapping defined
by the matrix (๐๐๐๐) then ๐ถ๐ฃ = (โ๐๐=1 ๐2๐ )๐ฃ and ๐ถ๐ค = 0 if ๐ค โ ๐ฃ = 0 .
โข Conclude that the eigenvalues of ๐ถ are ๐1 = โ ๐2๐ and ๐2 = โฏ = ๐๐ = 0.โข Show that the determinant of ๐ผ + ๐ถ is 1 + โ๐๐=1 ๐2๐ .โข Compute the determinant of the matrix (4.4.32) where ๐ is the mapping (4.4.33).
Exercise 4.4.vi. Let ๐ be an oriented๐-dimensional vector space and โ๐ โ ๐โ, ๐ = 1,โฆ, ๐,๐ linearly independent vectors in ๐โ. Define
๐ฟโถ ๐ โ ๐๐
to be the map ๐ฃ โฆ (โ1(๐ฃ),โฆ, โ๐(๐ฃ)).(1) Show that ๐ฟ is surjective and that the kernel๐ of ๐ฟ is of dimension ๐ = ๐ โ ๐.(2) Show that one gets from this mapping a bijective linear mapping
(4.4.34) ๐/๐ โ ๐๐
and hence from the standard orientation on ๐๐ an induced orientation on๐/๐ and on๐.
Hint: Exercise 1.2.viii and Theorem 1.9.9.(3) Let ๐ be an element of ๐ฌ๐(๐โ). Show that there exists a ๐ โ ๐ฌ๐(๐โ) with the property
(4.4.35) โ1 โงโฏ โง โ๐ โง ๐ = ๐ .Hint:Choose an oriented basis, ๐1,โฆ, ๐๐ of๐ such that๐ = ๐โ1 โงโฏโง๐โ๐ and โ๐ = ๐โ๐
for ๐ = 1,โฆ, ๐, and let ๐ = ๐โ๐+1 โงโฏ โง ๐โ๐.
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ยง4.4 Orientations 123
(4) Show that if ๐ is an element of ๐ฌ๐(๐โ) with the property
โ1 โงโฏ โง โ๐ โง ๐ = 0then there exist elements, ๐๐, of ๐ฌ๐โ1(๐โ) such that
๐ = โโ๐ โง ๐๐ .Hint: Same hint as in part (3).
(5) Show that if ๐1 and ๐2 are elements of๐ฌ๐(๐โ)with the property (4.4.35) and ๐ โถ ๐ โ ๐is the inclusion map then ๐โ๐1 = ๐โ๐2.
Hint: Let ๐ = ๐1 โ ๐2. Conclude from part (4) that ๐โ๐ = 0.(6) Conclude that if ๐ is an element of ๐ฌ๐(๐โ) satisfying (4.4.35) the element, ๐ = ๐โ๐, of๐ฌ๐(๐โ) is intrinsically defined independent of the choice of ๐.
(7) Show that ๐ lies in ๐ฌ๐(๐โ)+.
Exercise 4.4.vii. Let ๐ be an open subset of ๐๐ and ๐โถ ๐ โ ๐๐ a ๐ถโ map. If zero is aregular value of ๐, the set,๐ = ๐โ1(0) is a manifold of dimension ๐ = ๐โ๐. Show that thismanifold has a natural smooth orientation.
Some suggestions:โข Let ๐ = (๐1,โฆ, ๐๐) and let
๐๐1 โงโฏ โง ๐๐๐ = โ๐๐ผ๐๐ฅ๐ผsummed over multi-indices which are strictly increasing. Show that for every ๐ โ๐ ๐๐ผ(๐) โ 0 for some multi-index, ๐ผ = (๐1,โฆ, ๐๐), 1 โค ๐1 < โฏ < ๐๐ โค ๐.
โข Let ๐ฝ = (๐1,โฆ, ๐๐), 1 โค ๐1 < โฏ < ๐๐ โค ๐ be the complementary multi-index to ๐ผ,i.e., ๐๐ โ ๐๐ for all ๐ and ๐ . Show that
๐๐1 โงโฏ โง ๐๐๐ โง ๐๐ฅ๐ฝ = ยฑ๐๐ผ๐๐ฅ1 โงโฏ โง ๐๐ฅ๐and conclude that the ๐-form
๐ = ยฑ 1๐๐ผ๐๐ฅ๐ฝ
is a ๐ถโ ๐-form on a neighborhood of ๐ in ๐ and has the property:
๐๐1 โงโฏ โง ๐๐๐ โง ๐ = ๐๐ฅ1 โงโฏ โง ๐๐ฅ๐ .
โข Let ๐ โถ ๐ โ ๐ be the inclusion map. Show that the assignment
๐ โฆ (๐โ๐)๐defines an intrinsic nowhere vanishing ๐-form ๐ โ ๐บ๐(๐) on๐.
โข Show that the orientation of ๐ defined by ๐ coincides with the orientation thatwe described earlier in this section.
Exercise 4.4.viii. Let ๐๐ be the ๐-sphere and ๐ โถ ๐๐ โ ๐๐+1 the inclusion map. Show thatif ๐ โ ๐บ๐(๐๐+1) is the ๐-form ๐ = โ๐+1๐=1 (โ1)๐โ1๐ฅ๐๐๐ฅ1 โง โฏ โง ๐๐ฅ๐ โง โฏ โง ๐๐ฅ๐+1, the ๐-form๐โ๐ โ ๐บ๐(๐๐) is the Riemannian volume form.
Exercise 4.4.ix. Let ๐๐+1 be the (๐ + 1)-sphere and let
๐๐+1+ = { (๐ฅ1,โฆ, ๐ฅ๐+2) โ ๐๐+1 | ๐ฅ1 < 0 }be the lower hemisphere in ๐๐+1.(1) Prove that ๐๐+1+ is a smooth domain.(2) Show that the boundary of ๐๐+1+ is ๐๐.
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124 Chapter 4: Manifolds & Forms on Manifolds
(3) Show that the boundary orientation of ๐๐ agrees with the orientation of ๐๐ 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 ๐ will be an oriented ๐-manifold and๐ an open subset of ๐, and our goal will beto make sense of the integral
(4.5.1) โซ๐๐
where๐ is a compactly supported ๐-form.Weโll begin by showing how to define this integralwhen the support of๐ is contained in a parametrizable open set๐. Let๐0 be an open subsetof ๐๐ and ๐0 โถ ๐0 โ ๐ a parametrization. As we noted in ยง4.4 we can assume without lossof generality that this parametrization is oriented. Making this assumption, weโll define
(4.5.2) โซ๐๐ = โซ๐0๐โ0๐
where๐0 = ๐โ10 (๐ โฉ ๐). Notice that if ๐โ๐ = ๐๐๐ฅ1 โงโฏ โง ๐๐ฅ๐, then, by assumption, ๐ isin ๐ถโ0 (๐0). Hence since
โซ๐0๐โ0๐ = โซ
๐0๐๐๐ฅ1โฆ๐๐ฅ๐
and since ๐ is a bounded continuous function and is compactly supported the Riemannintegral on the right is well-defined. (See Appendix b.) Moreover, if ๐1 โถ ๐1 โ ๐ is anotheroriented parametrization of ๐ and ๐โถ ๐0 โ ๐1 is the map ๐ = ๐โ11 โ ๐0 then ๐0 = ๐1 โ ๐,so by Proposition 4.3.4
๐โ0๐ = ๐โ๐โ1๐ .Moreover, by (4.3.18) ๐ is orientation preserving. Therefore since
๐1 = ๐(๐0) = ๐โ11 (๐ โฉ๐)
Theorem 3.5.2 tells us that
(4.5.3) โซ๐1๐โ1๐ = โซ
๐0๐โ0๐ .
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 for๐1, ๐2 โ ๐บ๐๐ (๐),
(4.5.4) โซ๐(๐1 + ๐2) = โซ
๐๐1 + โซ
๐๐2
and for ๐ โ ๐บ๐๐ (๐) and ๐ โ ๐
โซ๐๐๐ = ๐โซ
๐๐ .
We will next show how to define the integral (4.5.1) for any compactly supported ๐-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. Weโll begin by deriving from the partitionof unity theorem in Appendix b a manifold version of this theorem.
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ยง4.5 Integration of forms on manifolds 125
Theorem 4.5.5. Let(4.5.6) U = {๐๐ผ}๐ผโ๐ผbe a covering of ๐ be open subsets. Then there exists a family of functions ๐๐ โ ๐ถโ0 (๐), for๐ โฅ 1, with the properties(1) ๐๐ โฅ 0.(2) For every compact set ๐ถ โ ๐ there exists a positive integer๐ such that if ๐ > ๐ we have
supp(๐๐) โฉ ๐ถ = โ .(3) โโ๐=1 ๐๐ = 1.(4) For every ๐ โฅ 1 there exists an index ๐ผ โ ๐ผ such that supp(๐๐) โ ๐๐ผ.
Remark 4.5.7. Conditions (1)โ(3) say that the ๐๐โs are a partition of unity and (4) says thatthis partition of unity is subordinate to the covering (4.5.5).
Proof. For each ๐ โ ๐ and for some ๐๐ผ containing a ๐ choose an open set ๐๐ in ๐๐ with๐ โ ๐๐ and with
(4.5.8) ๐๐ โฉ ๐ โ ๐๐ผ .Let ๐ โ โ๐โ๐ ๐๐ and let ๐๐ โ ๐ถโ0 (๐), for ๐ โฅ 1, be a partition of unity subordinate to
the covering of ๐ by the ๐๐โs. By (4.5.8) the restriction ๐๐ of ๐๐ to ๐ has compact supportand it is clear that the ๐๐โs inherit from the ๐๐โs the properties (1)โ(4). โก
Now let the covering (4.5.6) be any covering of ๐ by parametrizable open sets and let๐๐ โ ๐ถโ0 (๐), for ๐ โฅ 1 be a partition of unity subordinate to this covering. Given ๐ โ ๐บ๐๐ (๐)we will define the integral of ๐ over๐ by the sum
(4.5.9)โโ๐=1โซ๐๐๐๐ .
Note that since each ๐๐ is supported in some ๐๐ผ the individual summands in this sum arewell-defined and since the support of ๐ 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. Letโs show thatthis sum does not depend on the choice of U and the ๐๐โs. Let Uโฒ be another covering of๐by parametrizable open sets and (๐โฒ๐ )๐โฅ1 a partition of unity subordinate to Uโฒ. Then
(4.5.10)โโ๐=1โซ๐๐โฒ๐๐ =
โโ๐=1โซ๐
โโ๐=1๐โฒ๐๐๐๐ =
โโ๐=1(โโ๐=1โซ๐๐โฒ๐๐๐๐)
by equation (4.5.4). Interchanging the orders of summation and resuming with respect tothe ๐โs this sum becomes
โโ๐=1โซ๐
โโ๐=1๐โฒ๐๐๐๐
orโโ๐=1โซ๐๐๐๐ .
Henceโโ๐=1โซ๐๐โฒ๐๐ =
โโ๐=1โซ๐๐๐๐ ,
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 ๐1, ๐2 โ ๐บ๐๐ (๐),
โซ๐๐1 + ๐2 = โซ
๐๐1 + โซ
๐๐2
and for ๐ โ ๐บ๐๐ (๐) and ๐ โ ๐โซ๐๐๐ = ๐โซ
๐๐ .
The definition of the integral (4.5.1) depends on the choice of an orientation of ๐, butitโs easy to see how it depends on this choice. We pointed out in Section 4.4 that if ๐ isconnected, there is just one way to orient it smoothly other than by its given orientation,namely by reversing the orientation of ๐๐๐ at each point ๐ and itโs 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 โซ๐ ๐ to โโซ๐ ๐.
In the definition of the integral (4.5.1) we have allowed๐ to be an arbitrary open subsetof๐ but required ๐ to be compactly supported. This integral is also well-defined if we allow๐ to be an arbitrary element of ๐บ๐(๐) but require the closure of๐ in ๐ 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 itโs still true that the definition of the integral does not depend on the choiceof partitions of unity. In particular if the closure of๐ in ๐ is compact we will define thevolume of๐ to be the integral,
vol(๐) = โซ๐๐vol ,
where ๐vol is the Riemannian volume form and if๐ itself is compact weโll define its volumeto be the integral
vol(๐) = โซ๐๐vol .
Weโll next prove a manifold version of the change of variables formula (3.5.3).
Theorem 4.5.12 (change of variables formula). Let ๐โฒ and ๐ be oriented ๐-manifolds and๐โถ ๐โฒ โ ๐ an orientation preserving diffeomorphism. If ๐ is an open subset of ๐ and๐โฒ โ ๐โ1(๐), then
(4.5.13) โซ๐โฒ๐โ๐ = โซ
๐๐
for all ๐ โ ๐บ๐๐ (๐).
Proof. By (4.5.9) the integrand of the integral above is a finite sum of ๐ถโ forms, each ofwhich is supported on a parametrizable open subset, so we can assume that ๐ itself as thisproperty. Let๐ be a parametrizable open set containing the support of๐ and let ๐0 โถ ๐ โฅฒ ๐be an oriented parameterization of ๐. Since ๐ is a diffeomorphism its inverse exists and isa diffeomorphism of๐ onto๐1. Let ๐โฒ โ ๐โ1(๐) and ๐โฒ0 โ ๐โ1 โ ๐0. Then ๐โฒ0 โถ ๐ โฅฒ ๐โฒ isan oriented parameterization of ๐โฒ. Moreover, ๐ โ ๐โฒ0 = ๐0 so if๐0 = ๐โ10 (๐) we have
๐0 = (๐โฒ0)โ1(๐โ1(๐)) = (๐โฒ0)โ1(๐โฒ)and by the chain rule we have
๐โ0๐ = (๐ โ ๐โฒ0)โ๐ = (๐โฒ0)โ๐โ๐hence
โซ๐๐ = โซ๐0๐โ0๐ = โซ
๐0(๐โฒ0)โ(๐โ๐) = โซ
๐โฒ๐โ๐ . โก
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ยง4.5 Integration of forms on manifolds 127
As an exercise, show that if ๐โถ ๐โฒ โ ๐ is orientation reversing
(4.5.14) โซ๐โฒ๐โ๐ = โโซ
๐๐ .
Weโll conclude this discussion of โintegral calculus on manifoldsโ by proving a prelim-inary version of Stokesโ theorem.
Theorem 4.5.15. If ๐ is in ๐บ๐โ1๐ (๐) then
โซ๐๐๐ = 0 .
Proof. Let (๐๐)๐โฅ1 be a partition of unity with the property that each ๐๐ is supported in aparametrizable open set ๐๐ = ๐. Replacing ๐ by ๐๐๐ it suffices to prove the theorem for๐ โ ๐บ๐โ1๐ (๐). Let ๐โถ ๐0 โ ๐ be an oriented parametrization of ๐. Then
โซ๐๐๐ = โซ
๐0๐โ๐๐ = โซ
๐0๐(๐โ๐) = 0
by Theorem 3.3.1. โก
Exercises for ยง4.5Exercise 4.5.i. Let๐โถ ๐๐ โ ๐ be a๐ถโ function.Orient the graph๐ โ ๐ค๐ of๐ by requiringthat the diffeomorphism
๐โถ ๐๐ โ ๐ , ๐ฅ โฆ (๐ฅ, ๐(๐ฅ))be orientation preserving. Given a bounded open set ๐ in ๐๐ compute the Riemannianvolume of the image
๐๐ = ๐(๐)of ๐ in๐ as an integral over ๐.
Hint: Exercise 4.4.v.
Exercise 4.5.ii. Evaluate this integral for the open subset ๐๐ of the paraboloid defined by๐ฅ3 = ๐ฅ21 + ๐ฅ22 , where ๐ is the disk ๐ฅ21 + ๐ฅ22 < 2.
Exercise 4.5.iii. In Exercise 4.5.i let ๐ โถ ๐ โช ๐๐+1 be the inclusion map of๐ onto ๐๐+1.(1) If ๐ โ ๐บ๐(๐๐+1) is the ๐-form ๐ฅ๐+1๐๐ฅ1 โง โฏ โง ๐๐ฅ๐, what is the integral of ๐โ๐ over the
set๐๐? Express this integral as an integral over ๐.(2) Same question for ๐ = ๐ฅ2๐+1๐๐ฅ1 โงโฏ โง ๐๐ฅ๐.(3) Same question for ๐ = ๐๐ฅ1 โงโฏ โง ๐๐ฅ๐.
Exercise 4.5.iv. Let ๐โถ ๐๐ โ (0, +โ) be a positive๐ถโ function,๐ a bounded open subsetof ๐๐, and๐ the open set of ๐๐+1 defined by the inequalities
0 < ๐ฅ๐+1 < ๐(๐ฅ1,โฆ, ๐ฅ๐)and the condition (๐ฅ1,โฆ, ๐ฅ๐) โ ๐.(1) Express the integral of the (๐ + 1)-form ๐ = ๐ฅ๐+1๐๐ฅ1 โงโฏโง๐๐ฅ๐+1 over๐ as an integral
over ๐.(2) Same question for ๐ = ๐ฅ2๐+1๐๐ฅ1 โงโฏ โง ๐๐ฅ๐+1.(3) Same question for ๐ = ๐๐ฅ1 โงโฏ โง ๐๐ฅ๐.
Exercise 4.5.v. Integrate the โRiemannian areaโ form๐ฅ1๐๐ฅ2 โง ๐๐ฅ3 + ๐ฅ2๐๐ฅ3 โง ๐๐ฅ1 + ๐ฅ3๐๐ฅ1 โง ๐๐ฅ2
over the unit 2-sphere ๐2. (See Exercise 4.4.viii.)
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128 Chapter 4: Manifolds & Forms on Manifolds
Hint: An easier problem: Using polar coordinates integrate ๐ = ๐ฅ3๐๐ฅ1 โง ๐๐ฅ2 over thehemisphere defined by ๐ฅ3 = โ1 โ ๐ฅ21 โ ๐ฅ22 , where ๐ฅ21 + ๐ฅ22 < 1.
Exercise 4.5.vi. Let ๐ผ be the one-formโ๐๐=1 ๐ฆ๐๐๐ฅ๐ in equation (2.8.2) and let ๐พ(๐ก), 0 โค ๐ก โค 1,be a trajectory of the Hamiltonian vector field (2.8.2). What is the integral of ๐ผ over ๐พ(๐ก)?
4.6. Stokesโ theorem & the divergence theorem
Let๐ be an oriented ๐-manifold and๐ท โ ๐ a smooth domain. We showed in ยง4.4 that๐๐ท acquires from ๐ท a natural orientation. Hence if ๐ โถ ๐๐ท โ ๐ is the inclusion map and ๐is in ๐บ๐โ1๐ (๐), the integral
โซ๐๐ท๐โ๐
is well-defined. We will prove:
Theorem 4.6.1 (Stokesโ theorem). For ๐ โ ๐บ๐โ1๐ (๐) we have
(4.6.2) โซ๐๐ท๐โ๐ = โซ
๐ท๐๐ .
Proof. Let (๐๐)๐โฅ1 be a partition of unity such that for each ๐, the support of ๐๐ is containedin a parametrizable open set ๐๐ = ๐ of one of the following three types:(a) ๐ โ int(๐ท).(b) ๐ โ ext(๐ท).(c) There exists an open subset ๐0 of ๐๐ and an oriented๐ท-adapted parametrization
๐โถ ๐0 โฅฒ ๐ .
Replacing ๐ by the finite sum โโ๐=1 ๐๐๐ it suffices to prove (4.6.2) for each ๐๐๐ separately. Inother words we can assume that the support of ๐ itself is contained in a parametrizable openset ๐ of type (a), (b), or (c). But if ๐ is of type (a)
โซ๐ท๐๐ = โซ
๐๐๐ = โซ
๐๐๐
and ๐โ๐ = 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 ๐ is of type (b) the situation is even simpler: ๐โ๐ is zeroand the restriction of ๐ to๐ท is zero, so both sides of equation (4.6.2) are automatically zero.Thus one is reduced to proving (4.6.2) when๐ is an open subset of type (c). In this case therestriction of the map (4.6.2) to ๐0 โฉ ๐๐๐ is an orientation preserving diffeomorphism
(4.6.3) ๐โถ ๐0 โฉ ๐๐๐ โ ๐ โฉ ๐
and
(4.6.4) ๐๐ โ ๐ = ๐ โ ๐๐๐โ1
where the maps ๐ = ๐๐ and๐๐๐โ1 โถ ๐๐โ1 โช ๐๐
are the inclusionmaps of๐ into๐ and ๐๐๐ into๐๐. (Here weโre identifying ๐๐๐ with๐๐โ1.)Thus
โซ๐ท๐๐ = โซ
๐๐๐โ๐๐ = โซ
๐๐๐๐โ๐
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ยง4.6 Stokesโ theorem & the divergence theorem 129
and by (4.6.4)
โซ๐๐โ๐๐ = โซ
๐๐โ1๐โ๐โ๐๐
= โซ๐๐โ1๐โ๐๐โ1๐โ๐
= โซ๐๐๐๐โ๐๐โ1๐โ๐ .
Thus it suffices to prove Stokesโ theoremwith ๐ replaced by ๐โ๐, or, in other words, to proveStokesโ theorem for๐๐; and this we will now do.
Theorem 4.6.5 (Stokesโ theorem for๐๐). Let
๐ =๐โ๐=1(โ1)๐โ1๐๐๐๐ฅ1 โงโฏ โง ๐๐ฅ๐ โงโฏ โง ๐๐ฅ๐ .
Then
๐๐ =๐โ๐=1
๐๐๐๐๐ฅ๐๐๐ฅ1 โงโฏ โง ๐๐ฅ๐
and
โซ๐๐๐๐ =
๐โ๐=1โซ๐๐๐๐๐๐๐ฅ๐๐๐ฅ1โฏ๐๐ฅ๐ .
We will compute each of these summands as an iterated integral doing the integrationwith respect to ๐๐ฅ๐ first. For ๐ > 1 the ๐๐ฅ๐ integration ranges over the interval โโ < ๐ฅ๐ < โand hence since ๐๐ is compactly supported
โซโ
โโ
๐๐๐๐๐ฅ๐๐๐ฅ๐ = ๐๐(๐ฅ1,โฆ, ๐ฅ๐,โฆ, ๐ฅ๐)|
๐ฅ๐=+โ
๐ฅ๐=โโ= 0 .
On the other hand the ๐๐ฅ1 integration ranges over the interval โโ < ๐ฅ1 < 0 and
โซ0
โโ
๐๐1๐๐ฅ1๐๐ฅ1 = ๐(0, ๐ฅ2,โฆ, ๐ฅ๐) .
Thus integrating with respect to the remaining variables we get
(4.6.6) โซ๐๐๐๐ = โซ
๐๐โ1๐(0, ๐ฅ2,โฆ, ๐ฅ๐) ๐๐ฅ2 โงโฏ โง ๐๐ฅ๐ .
On the other hand, since ๐โ๐๐โ1๐ฅ1 = 0 and ๐โ๐๐โ1 ๐ฅ๐ = ๐ฅ๐ for ๐ > 1,๐โ๐๐โ1๐ = ๐1(0, ๐ฅ2,โฆ, ๐ฅ๐) ๐๐ฅ2 โงโฏ โง ๐๐ฅ๐
so
(4.6.7) โซ๐๐โ1๐โ๐๐โ1๐ = โซ
๐๐โ1๐(0, ๐ฅ2,โฆ, ๐ฅ๐) ๐๐ฅ2 โงโฏ โง ๐๐ฅ๐ .
Hence the two sides, (4.6.6) and (4.6.7), of Stokesโ theorem are equal. โก
One important variant of Stokesโ theorem is the divergence theorem: Let ๐ be in๐บ๐๐ (๐)and let ๐ be a vector field on๐. Then
๐ฟ๐๐ = ๐๐๐๐ + ๐๐๐๐ = ๐๐๐๐ ,hence, denoting by ๐๐ the inclusion map of ๐ into ๐ we get from Stokesโ theorem, with๐ = ๐๐๐:
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130 Chapter 4: Manifolds & Forms on Manifolds
Theorem 4.6.8 (divergence theorem for manifolds).
โซ๐ท๐ฟ๐๐ = โซ
๐๐โ๐(๐๐๐) .
If ๐ท is an open domain in ๐๐ this reduces to the usual divergence theorem of multi-variable calculus. Namely if ๐ = ๐๐ฅ1 โงโฏ โง ๐๐ฅ๐ and ๐ = โ
๐๐=1 ๐ฃ๐
๐๐๐ฅ๐
then by (2.5.14)
๐ฟ๐๐๐ฅ1 โงโฏ โง ๐๐ฅ๐ = div(๐) ๐๐ฅ1 โงโฏ โง ๐๐ฅ๐where
div(๐) =๐โ๐=1
๐๐ฃ๐๐๐ฅ๐.
Thus if ๐ = ๐๐ท and ๐๐ is the inclusion ๐ โช ๐๐,
(4.6.9) โซ๐ท
div(๐)๐๐ฅ = โซ๐๐โ๐(๐๐๐๐ฅ1 โงโฏ โง ๐๐ฅ๐) .
The right hand side of this identity can be interpreted as the โfluxโ of the vector field ๐through the boundary of๐ท. To see this let ๐โถ ๐๐ โ ๐ be a ๐ถโ defining function for๐ท, i.e.,a function with the properties
๐ โ ๐ท โบ ๐(๐) < 0and ๐๐๐ โ 0 if ๐ โ ๐๐ท. This second condition says that zero is a regular value of ๐ andhence that ๐ โ ๐๐ท is defined by the non-degenerate equation:
(4.6.10) ๐ โ ๐ โบ ๐(๐) = 0 .Let ๐ be the vector field
๐ โ (๐โ๐=1( ๐๐๐๐ฅ๐)2)โ1 ๐โ๐=1
๐๐๐๐๐ฅ๐๐๐๐ฅ๐
.
In view of (4.6.10), this vector field is well-defined on a neighborhood ๐ of ๐ and satisfies
๐๐๐๐ = 1 .Now note that since ๐๐ โง ๐๐ฅ1 โงโฏ โง ๐๐ฅ๐ = 0
0 = ๐๐(๐๐ โง ๐๐ฅ1 โงโฏ โง ๐๐ฅ๐)= (๐๐๐๐)๐๐ฅ1 โงโฏ โง ๐๐ฅ๐ โ ๐๐ โง ๐๐๐๐ฅ1 โงโฏ โง ๐๐ฅ๐= ๐๐ฅ1 โงโฏ โง ๐๐ฅ๐ โ ๐๐ โง ๐๐๐๐ฅ1 โงโฏ โง ๐๐ฅ๐ ,
hence letting ๐ be the (๐ โ 1)-form ๐๐๐๐ฅ1 โงโฏ โง ๐๐ฅ๐ we get the identity
(4.6.11) ๐๐ฅ1 โงโฏ โง ๐๐ฅ๐ = ๐๐ โง ๐and by applying the operation ๐๐ to both sides of (4.6.11) the identity
(4.6.12) ๐๐๐๐ฅ1 โงโฏ โง ๐๐ฅ๐ = (๐ฟ๐๐)๐ โ ๐๐ โง ๐๐๐ .Let ๐๐ = ๐โ๐๐ be the restriction of ๐ to ๐. Since ๐โ๐ = 0, ๐โ๐๐๐ = 0 and hence by (4.6.12)
๐โ๐(๐๐๐๐ฅ1 โงโฏ โง ๐๐ฅ๐) = ๐โ๐(๐ฟ๐๐)๐๐ ,
and the formula (4.6.9) now takes the form
(4.6.13) โซ๐ท
div(๐)๐๐ฅ = โซ๐๐ฟ๐๐๐๐
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ยง4.6 Stokesโ theorem & the divergence theorem 131
where the term on the right is by definition the flux of ๐ through ๐. In calculus books thisis written in a slightly different form. Letting
๐๐ = (๐โ๐=1( ๐๐๐๐ฅ๐)2)12
๐๐
and letting
๏ฟฝ๏ฟฝ โ (๐โ๐=1( ๐๐๐๐ฅ๐)2)โ 12( ๐๐๐๐ฅ1,โฆ, ๐๐๐๐ฅ๐)
and ๐ฃ โ (๐ฃ1,โฆ, ๐ฃ๐) we have๐ฟ๐๐๐ = (๏ฟฝ๏ฟฝ โ ๐ฃ)๐๐
and hence
(4.6.14) โซ๐ท
div(๐)๐๐ฅ = โซ๐(๏ฟฝ๏ฟฝ โ ๐ฃ)๐๐ .
In three dimensions ๐๐ is just the standard โinfinitesimal element of areaโ on the surface ๐and ๐๐ the unit outward normal to ๐ at ๐, so this version of the divergence theorem is theversion one finds in most calculus books.
As an application of Stokesโ theorem, weโll give a very short alternative proof of theBrouwer fixed point theorem. As we explained in ยง3.6 the proof of this theorem basicallycomes down to proving the following theorem.
Theorem 4.6.15. Let ๐ต๐ be the closed unit ball in ๐๐ and ๐๐โ1 its boundary. Then the identitymap id๐๐โ1 on ๐๐โ1 cannot be extended to a ๐ถโ map ๐โถ ๐ต๐ โ ๐๐โ1.
Proof. Suppose that ๐ is such a map. Then for every (๐ โ 1)-form ๐ โ ๐บ๐โ1(๐๐โ1),
(4.6.16) โซ๐ต๐๐(๐โ๐) = โซ
๐๐โ1(๐๐๐โ1)โ๐โ๐ .
But๐(๐โ๐) = ๐โ(๐๐) = 0 since๐ is an (๐โ1)-form and ๐๐โ1 is an (๐โ1)-manifold, and since๐ is the identity map on ๐๐โ1, (๐๐๐โ1)
โ๐โ๐ = (๐ โ ๐๐๐โ1)โ๐ = ๐. Thus for every ๐ โ ๐บ๐โ1(๐๐โ1),equation (4.6.16) says that the integral of ๐ over ๐๐โ1 is zero. Since there are lots of (๐ โ 1)-forms for which this is not true, this shows that a map ๐ with the property above cannotexist. โก
Exercises for ยง4.6Exercise 4.6.i. Let ๐ต๐ be the open unit ball in ๐๐ and ๐๐โ1 the unit (๐ โ 1)-sphere. Showthat vol(๐๐โ1) = ๐ vol(๐ต๐).
Hint: Apply Stokesโ theorem to the (๐ โ 1)-form
๐ โ๐โ๐=1(โ1)๐โ1๐ฅ๐๐๐ฅ1 โงโฏ โง ๐๐ฅ๐ โงโฏ โง ๐๐ฅ๐
and note by Exercise 4.4.ix that ๐ is the Riemannian volume form of ๐๐โ1.
Exercise 4.6.ii. Let๐ท โ ๐๐ be a smooth domain. Show that there exists a neighborhood ๐of ๐๐ท in ๐๐ and a ๐ถโ defining function ๐โถ ๐ โ ๐ for๐ท with the properties:(1) ๐ โ ๐ โฉ ๐ท if and only if ๐(๐) < 0,(2) and ๐๐๐ โ 0 if ๐ โ ๐
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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 ๐ by a family U = {๐๐ผ}๐ผโ๐ผ of open subsets of ๐๐ such that for each thereexists a function ๐๐ผ โถ ๐๐ผ โ ๐ with the properties (1) and (2). Now let (๐๐)๐โฅ1 be a partitionof unity subordinate to U and let ๐ = โโ๐=1 ๐๐๐๐ผ๐ where supp(๐๐) โ ๐๐ผ๐ .
Exercise 4.6.iii. In Exercise 4.6.ii suppose ๐ is compact. Show that there exists a globaldefining function ๐โถ ๐๐ โ ๐ for๐ท with properties (1) and (2).
Hint: Let ๐ โ ๐ถโ0 (๐) be a function such that 0 โค ๐(๐ฅ) โค 1 for all ๐ฅ โ ๐ and ๐ isidentically 1 on a neighborhood of ๐, and replace ๐ by the function
๐(๐ฅ) ={{{{{
๐(๐ฅ)๐(๐ฅ) + (1 โ ๐(๐ฅ)), ๐ฅ โ ๐๐ โ ๐ท๐(๐ฅ), ๐ฅ โ ๐๐ท๐(๐ฅ) โ (1 โ ๐(๐ฅ))๐(๐ฅ), ๐ฅ โ int(๐ท) .
Exercise 4.6.iv. Show that the form ๐ฟ๐๐๐๐ in equation (4.6.13) does not depend on whatchoice we make of a defining function ๐ for๐ท.
Hints:โข Show that if ๐ is another defining function then, at ๐ โ ๐, ๐๐๐ = ๐๐๐๐, where ๐
is a positive constant.โข Show that if one replaces ๐๐๐ by (๐๐)๐ the first term in the product (๐ฟ๐๐)(๐)(๐๐)๐
changes by a factor ๐ and the second term by a factor 1/๐.Exercise 4.6.v. Show that the form ๐๐ is intrinsically defined in the sense that if ๐ is any(๐ โ 1)-form satisfying equation (4.6.11), then ๐๐ = ๐โ๐๐.
Hint: Exercise 4.5.vi.
Exercise 4.6.vi. Show that the form ๐๐ in equation (4.6.14) is the Riemannian volume formon ๐.Exercise 4.6.vii. Show that the (๐ โ 1)-form
๐ = (๐ฅ21 +โฏ + ๐ฅ2๐)โ๐๐โ๐=1(โ1)๐โ1๐ฅ๐๐๐ฅ1 โงโฏ โง ๐๐ฅ๐ โงโฏ๐๐ฅ๐
is closed and prove directly that Stokesโ theorem holds for the annulus ๐ < ๐ฅ21 +โฏ+๐ฅ2๐ < ๐by showing that the integral of ๐ over the sphere, ๐ฅ21 + โฏ + ๐ฅ2๐ = ๐, is equal to the integralover the sphere, ๐ฅ21 +โฏ + ๐ฅ2๐ = ๐.Exercise 4.6.viii. Let ๐โถ ๐๐โ1 โ ๐ be an everywhere positive ๐ถโ function and let ๐ be abounded open subset of ๐๐โ1. Verify directly that Stokesโ theorem is true if๐ท is the domaindefined by
0 < ๐ฅ๐ < ๐(๐ฅ1,โฆ, ๐ฅ๐โ1) , (๐ฅ1,โฆ, ๐ฅ๐โ1) โ ๐and ๐ an (๐ โ 1)-form of the form
๐(๐ฅ1,โฆ, ๐ฅ๐)๐๐ฅ1 โงโฏ โง ๐๐ฅ๐โ1where ๐ is in ๐ถโ0 (๐๐).Exercise 4.6.ix. Let๐ be an oriented ๐-manifold and ๐ a vector field on๐which is complete.Verify that for ๐ โ ๐บ๐๐ (๐)
โซ๐๐ฟ๐๐ = 0
in the following ways:(1) Directly by using the divergence theorem.
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ยง4.7 Degree theory on manifolds 133
(2) Indirectly by showing that
โซ๐๐โ๐ก ๐ = โซ
๐๐
where ๐๐ก โถ ๐ โ ๐, โโ < ๐ก < โ, is the one-parameter group of diffeomorphisms of๐generated by ๐.
Exercise 4.6.x. Let๐ be an oriented๐-manifold and๐ท โ ๐ a smooth domainwhose closureis compact. Show that if ๐ โ ๐๐ท is the boundary of ๐ท and ๐โถ ๐ โ ๐ a diffeomorphism ๐cannot be extended to a smooth map ๐โถ ๐ท โ ๐.
4.7. Degree theory on manifolds
In this section weโll show how to generalize to manifolds the results about the โdegreeโof a proper mapping that we discussed in Chapter 3. Weโll begin by proving the manifoldanalogue of Theorem 3.3.1.
Theorem 4.7.1. Let ๐ be an oriented connected ๐-manifold and ๐ โ ๐บ๐๐ (๐) a compactlysupported ๐-form. Then the following are equivalent(1) โซ๐ ๐ = 0.(2) ๐ = ๐๐ for some ๐ โ ๐บ๐โ1๐ (๐).
Proof. We have already verified the assertion (2)โ(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 โ(1)โ(2)โ part of Theorem 3.2.2:
Step 1. Let๐ be a connected parametrizable open subset of๐. If ๐ โ ๐บ๐๐ (๐) has property (1),then ๐ = ๐๐ for some ๐ โ ๐บ๐โ1๐ (๐).
Proof of Step 1. Let ๐โถ ๐0 โ ๐ be an oriented parametrization of ๐. Then
โซ๐0๐โ๐ = โซ
๐๐ = 0
and since ๐0 is a connected open subset of ๐๐, ๐โ๐ = ๐๐ for some ๐ โ ๐บ๐โ1๐ (๐0) by Theo-rem 3.3.1. Let ๐ = (๐โ1)โ๐. Then ๐๐ = (๐โ1)โ๐๐ = ๐. โก
Step 2. Fix a base point ๐0 โ ๐ and let ๐ be any point of ๐. Then there exists a collectionof connected parametrizable open sets๐1,โฆ,๐๐ with ๐0 โ ๐1 and ๐ โ ๐๐ such that for1 โค ๐ โค ๐ โ 1, the intersection๐๐ โฉ๐๐+1 is non-empty.
Proof of Step 2. The set of points, ๐ โ ๐, for which this assertion is true is open and the setfor which it is not true is open. Moreover, this assertion is true for ๐ = ๐0. โก
Step 3. We deduce Theorem 4.7.1 from a slightly stronger result. Introduce an equivalencerelation on ๐บ๐๐ (๐) by declaring that two ๐-forms ๐1, ๐2 โ ๐บ๐๐ (๐) are equivalent if ๐1 โ ๐2 โ๐๐บ๐โ1๐ฅ (๐). Denote this equivalence relation by a ๐1 โผ ๐2.
We will prove:
Theorem 4.7.2. For ๐1 and ๐2 โ ๐บ๐๐ (๐) the following are equivalent:(1) โซ๐ ๐1 = โซ๐ ๐2(2) ๐1 โผ ๐2.
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134 Chapter 4: Manifolds & Forms on Manifolds
Applying this result to a form ๐ โ ๐บ๐๐ (๐) whose integral is zero, we conclude that๐ โผ 0, which means that ๐ = ๐๐ for some ๐ โ ๐บ๐โ1๐ (๐). Hence Theorem 4.7.2 impliesTheorem 4.7.1. Conversely, if โซ๐ ๐1 = โซ๐ ๐2, then โซ๐(๐1 โ ๐2) = 0, so ๐1 โ ๐2 = ๐๐ forsome ๐ โ ๐บ๐๐ (๐). 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 ๐1 โ ๐บ๐๐ (๐1)and ๐2 โ ๐บ๐๐ (๐2) where ๐1 and ๐2 are connected parametrizable open sets. Moreover, if theintegrals of ๐1 and ๐2 are zero then ๐๐ = ๐๐๐ for some ๐๐ โ ๐บ๐๐ (๐๐) by Step 1, so in this case,the theorem is true. Suppose on the other hand that
โซ๐๐1 = โซ
๐๐2 = ๐ โ 0 .
Then dividing by ๐, we can assume that the integrals of ๐1 and ๐2 are both equal to 1.
Step 5. Let๐1,โฆ,๐๐ be, as in Step 2, a sequence of connected parametrizable open sets withthe property that the intersections,๐1โฉ๐1,๐๐โฉ๐2 and๐๐โฉ๐๐+1, for ๐ = 1,โฆ,๐โ1, are allnon-empty. Select ๐-forms, ๐ผ0 โ ๐บ๐๐ (๐1 โฉ๐1), ๐ผ๐ โ ๐บ๐๐ (๐๐ โฉ ๐2) and ๐ผ๐ โ ๐บ๐๐ (๐๐ โฉ๐๐+1),for ๐ = 1,โฆ,๐ โ 1, such that the integral of each ๐ผ๐ over ๐ is equal to 1 By Step 1 we see thatTheorem 4.7.1 is true for ๐1, ๐2 and the๐1,โฆ,๐๐, hence Theorem 4.7.2 is true for ๐1, ๐2and the๐1,โฆ,๐๐, so
๐1 โผ ๐ผ0 โผ ๐ผ1 โผ โฏ โผ ๐ผ๐ โผ ๐2and thus ๐1 โผ ๐2. โก
Just as in (3.4.2) we get as a corollary of the theorem above the following โdefinitionโtheoremโ of the degree of a differentiable mapping:
Theorem4.7.3. Let๐ and๐ be compact oriented ๐-manifolds and let๐ be connected. Given aproper ๐ถโ mapping, ๐โถ ๐ โ ๐, there exists a topological invariant deg(๐) with the definingproperty:
(4.7.4) โซ๐๐โ๐ = deg(๐) โซ
๐๐ .
Proof. As in the proof of Theorem 3.4.6 pick an ๐-form ๐0 โ ๐บ๐๐ (๐) whose integral over ๐is one and define the degree of ๐ to be the integral over๐ of ๐โ๐0, i.e., set
(4.7.5) deg(๐) โ โซ๐๐โ๐0 .
Now let ๐ be any ๐-form in ๐บ๐๐ (๐) and let
(4.7.6) ๐ โ โซ๐๐ .
Then the integral of ๐ โ ๐๐0 over ๐ is zero so there exists an (๐ โ 1)-form ๐ โ ๐บ๐โ1๐ (๐) forwhich ๐ โ ๐๐0 = ๐๐. Hence ๐โ๐ = ๐๐โ๐0 + ๐๐โ๐, so
โซ๐๐โ๐ = ๐โซ
๐๐โ๐0 = deg(๐) โซ
๐๐
by equations (4.7.5) and (4.7.6). โก
Itโs clear from the formula (4.7.4) that the degree of ๐ is independent of the choice of๐0. (Just apply this formula to any ๐ โ ๐บ๐๐ (๐) having integral over ๐ equal to one.) Itโs alsoclear from (4.7.4) that โdegreeโ behaves well with respect to composition of mappings:
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ยง4.7 Degree theory on manifolds 135
Theorem 4.7.7. Let ๐ be an oriented, connected ๐-manifold and ๐โถ ๐ โ ๐ a proper ๐ถโmap. Then
(4.7.8) deg(๐ โ ๐) = deg(๐) deg(๐) .
Proof. Let ๐ โ ๐บ๐๐ (๐) be a compactly supported form with the property that โซ๐ ๐ = 1. Then
deg(๐ โ ๐) = โซ๐(๐ โ ๐)โ๐ = โซ
๐๐โ โ ๐โ๐ = deg(๐) โซ
๐๐โ๐
= deg(๐) deg(๐) . โก
We will next show how to compute the degree of ๐ by generalizing to manifolds theformula for deg(๐) that we derived in ยง3.6.
Definition 4.7.9. A point ๐ โ ๐ is a critical point of ๐ if the map
(4.7.10) ๐๐๐ โถ ๐๐๐ โ ๐๐(๐)๐is not bijective.
Weโll denote by๐ถ๐ the set of all critical points of ๐, and weโll call a point ๐ โ ๐ a criticalvalue of ๐ if it is in the image ๐(๐ถ๐) of ๐ถ๐ under ๐ and a regular value if itโs not. (Thus theset of regular values is the set ๐ โ ๐(๐ถ๐).) If ๐ is a regular value, then as we observed in Sec-tion 3.6, the map (4.7.10) is bijective for every ๐ โ ๐โ1(๐) and hence by Theorem 4.2.13, ๐maps a neighborhood๐๐ of ๐ diffeomorphically onto a neighborhood๐๐ of ๐. In particular,๐๐ โฉ ๐โ1(๐) = ๐. Since ๐ is proper the set ๐โ1(๐) is compact, and since the sets ๐๐ are acovering of๐โ1(๐), this covering must be a finite covering. In particular, the set ๐โ1(๐) itselfhas to be a finite set. As in ยง2.7 we can shrink the ๐๐โs so as to insure that they have thefollowing properties:(i) Each ๐๐ is a parametrizable open set.(ii) ๐๐ โฉ ๐๐โฒ is empty for ๐ โ ๐โฒ.(iii) ๐(๐๐) = ๐(๐๐โฒ) = ๐ for all ๐ and ๐โฒ.(iv) ๐ is a parametrizable open set.(v) ๐โ1(๐) = โ๐โ๐โ1(๐)๐๐.
To exploit these properties let ๐ be an ๐-form in ๐บ๐๐ (๐) with integral equal to 1. Thenby (v):
deg(๐) = โซ๐๐โ๐ = โ
๐โซ๐๐๐โ๐ .
But ๐โถ ๐๐ โ ๐ is a diffeomorphism, hence by (4.5.13) and (4.5.14)
โซ๐๐๐โ๐ = โซ
๐๐
if ๐โถ ๐๐ โ ๐ is orientation preserving and
โซ๐๐๐โ๐ = โโซ
๐๐
if ๐โถ ๐๐ โ ๐ is orientation reversing. Thus we have proved:
Theorem 4.7.11. The degree of ๐ is equal to the sum
(4.7.12) deg(๐) = โ๐โ๐โ1(๐)๐๐
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136 Chapter 4: Manifolds & Forms on Manifolds
where ๐๐ = +1 if the map (4.7.10) is orientation preserving and ๐๐ = โ1 if the map (4.7.10)is orientation reversing.
We will next show that Sardโs Theorem is true for maps between manifolds and hencethat there exist lots of regular values. We first observe that if ๐ is a parametrizable opensubset of๐ and ๐ a parametrizable open neighborhood of ๐(๐) in ๐, then Sardโs Theoremis true for the map ๐โถ ๐ โ ๐ since, up to diffeomorphism, ๐ and ๐ are just open subsetsof ๐๐. Now let ๐ be any point in ๐, let ๐ต be a compact neighborhood of ๐, and let ๐ be aparametrizable open set containing ๐ต. Then if ๐ด = ๐โ1(๐ต) it follows from Theorem 3.4.7that๐ด can be covered by a finite collection of parametrizable open sets,๐1,โฆ,๐๐ such that๐(๐๐) โ ๐. Hence since Sardโs Theorem is true for each of the maps ๐โถ ๐๐ โ ๐ and ๐โ1(๐ต)is contained in the union of the ๐๐โs we conclude that the set of regular values of ๐ intersectsthe interior of ๐ต in an open dense set. Thus, since ๐ is an arbitrary point of ๐, we have proved:
Theorem 4.7.13. If ๐ and ๐ are ๐-manifolds and ๐โถ ๐ โ ๐ is a proper ๐ถโ map the set ofregular values of ๐ is an open dense subset of ๐.
Since there exist lots of regular values the formula (4.7.12) gives us an effective wayof computing the degree of ๐. Weโll next justify our assertion that deg(๐) is a topologicalinvariant of ๐. To do so, letโs generalize Definition 2.6.14 to manifolds.
Definition 4.7.14. Let๐ and ๐ be manifolds and ๐0, ๐1 โถ ๐ โ ๐ be ๐ถโ maps. A ๐ถโ map
(4.7.15) ๐นโถ ๐ ร [0, 1] โ ๐is a homotopy between๐0 and๐1 if for all ๐ฅ โ ๐we have๐น(๐ฅ, 0) = ๐0(๐ฅ) and๐น(๐ฅ, 1) = ๐1(๐ฅ).Moreover, if๐0 and๐1 are propermaps, the homotopy,๐น, is a proper homotopy if๐น is properas a ๐ถโ map, i.e., for every compact set ๐ถ of ๐, ๐นโ1(๐ถ) is compact.
Letโs now prove the manifold analogue of Theorem 3.6.10.
Theorem 4.7.16. Let ๐ and ๐ be oriented ๐-manifolds and assume that ๐ is connected. If๐0, ๐2 โถ ๐ โ ๐ are proper maps and the map (4.7.8) is a proper homotopy, then deg(๐1) =deg(๐2).
Proof. Let๐ be an ๐-form in๐บ๐๐ (๐)whose integral over๐ is equal to 1, and let๐ถ โ supp(๐)be the support of ๐. Then if ๐น is a proper homotopy between ๐0 and ๐1, the set, ๐นโ1(๐ถ), iscompact and its projection on๐(4.7.17) { ๐ฅ โ ๐ | (๐ฅ, ๐ก) โ ๐นโ1(๐ถ) for some ๐ก โ [0, 1] }is compact. Let ๐๐ก โถ ๐ โ ๐ be the map: ๐๐ก(๐ฅ) = ๐น(๐ฅ, ๐ก). By our assumptions on ๐น, ๐๐ก is aproper ๐ถโ map. Moreover, for all ๐ก the ๐-form ๐โ๐ก ๐ is a ๐ถโ function of ๐ก and is supportedon the fixed compact set (4.7.17). Hence itโs clear from the definitions of the integral of aform and ๐๐ก that the integral
โซ๐๐โ๐ก ๐
is a ๐ถโ function of ๐ก. On the other hand this integral is by definition the degree of ๐๐ก andhence by Theorem 4.7.3, deg(๐๐ก) is an integer, so it does not depend on ๐ก. In particular,deg(๐0) = deg(๐1). โก
Exercises for ยง4.7Exercise 4.7.i. Let ๐โถ ๐ โ ๐ be the map ๐ฅ โฆ ๐ฅ๐. Show that deg(๐) = 0 if ๐ is even and 1if ๐ is odd.
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ยง4.8 Applications of degree theory 137
Exercise 4.7.ii. Let ๐โถ ๐ โ ๐ be the polynomial function,
๐(๐ฅ) = ๐ฅ๐ + ๐1๐ฅ๐โ1 +โฏ + ๐๐โ1๐ฅ + ๐๐ ,where the ๐๐โs are in ๐. Show that if ๐ is even, deg(๐) = 0 and if ๐ is odd, deg(๐) = 1.
Exercise 4.7.iii. Let ๐1 be the unit circle in the complex plane and let ๐โถ ๐1 โ ๐1 be themap ๐๐๐ โฆ ๐๐๐๐, where๐ ias a positive integer. What is the degree of ๐?
Exercise 4.7.iv. Let ๐๐โ1 be the unit sphere in ๐๐ and ๐โถ ๐๐โ1 โ ๐๐โ1 the antipodal map๐ฅ โฆ โ๐ฅ. What is the degree of ๐?
Exercise 4.7.v. Let ๐ด โ ๐(๐) be an orthogonal ๐ ร ๐matrix and let
๐๐ด โถ ๐๐โ1 โ ๐๐โ1
be the map ๐ฅ โฆ ๐ด๐ฅ. What is the degree of ๐๐ด?
Exercise 4.7.vi. A manifold ๐ is contractable if for some point ๐0 โ ๐, the identity map of๐ onto itself is homotopic to the constant map ๐๐0 โถ ๐ โ ๐ given by ๐๐0(๐ฆ) โ ๐0. Showthat if ๐ is an oriented contractable ๐-manifold and ๐ an oriented connected ๐-manifoldthen for every proper mapping ๐โถ ๐ โ ๐ we have deg(๐) = 0. In particular show that if๐ > 0 and ๐ is compact then ๐ cannot be contractable.
Hint: Let ๐ be the identity map of ๐ onto itself.
Exercise 4.7.vii. Let ๐ and ๐ be oriented connected ๐-manifolds and ๐โถ ๐ โ ๐ a proper๐ถโ map. Show that if deg(๐) โ 0, then ๐ is surjective.
Exercise 4.7.viii. Using Sardโs Theorem prove that if ๐ and ๐ are manifolds of dimension๐ and โ, with ๐ < โ and ๐โถ ๐ โ ๐ is a proper ๐ถโ map, then the complement of the imageof๐ in ๐ is open and dense.
Hint: Let ๐ โ โ โ ๐ and apply Sardโs Theorem to the map
๐โถ ๐ ร ๐๐ โ ๐ , ๐(๐ฅ, ๐) โ ๐(๐ฅ) .
Exercise 4.7.ix. Prove that the 2-sphere ๐2 and the torus ๐1 ร ๐1 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 ๐ be an oriented (๐ + 1)-dimensional manifold, ๐ท โ ๐ a smoothdomain and๐ โ ๐๐ท the boundary of๐ท. Assume that the closure๐ท = ๐โช๐ท of๐ท is compact(and in particular that๐ is compact).
Theorem 4.8.2. Let ๐ be an oriented connected ๐-manifold and ๐โถ ๐ โ ๐ a ๐ถโ map. Sup-pose there exists a ๐ถโ map ๐นโถ ๐ท โ ๐ whose restriction to ๐ is ๐. Then deg(๐) = 0.
Proof. Let ๐ be an element of๐บ๐๐ (๐). Then ๐๐ = 0, so ๐๐นโ๐ = ๐นโ๐๐ = 0. On the other handif ๐ โถ ๐ โ ๐ is the inclusion map,
โซ๐ท๐๐นโ๐ = โซ
๐๐โ๐นโ๐ = โซ
๐๐โ๐ = deg(๐) โซ
๐๐
by Stokesโ theorem since ๐น โ ๐ = ๐. Hence deg(๐) has to be zero. โก
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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 ๐ดโถ ๐๐ โ ๐๐ be a linear map. If ๐ iseven, ๐ดmay not have real eigenvalues. (For instance for the map
๐ดโถ ๐2 โ ๐2 , (๐ฅ, ๐ฆ) โฆ (โ๐ฆ, ๐ฅ)the eigenvalues of ๐ด are ยฑโโ1.) However, if ๐ is odd it is a standard linear algebra fact thatthere exists a vector, ๐ฃ โ ๐๐ โ {0}, and a ๐ โ ๐ such that ๐ด๐ฃ = ๐๐ฃ. Moreover replacing ๐ฃby ๐ฃ|๐ฃ| one can assume that |๐ฃ| = 1. This result turns out to be a special case of a much moregeneral result. Let ๐๐โ1 be the unit (๐ โ 1)-sphere in ๐๐ and let ๐โถ ๐๐โ1 โ ๐๐ be a ๐ถโ map.
Theorem 4.8.4. There exists a vector ๐ฃ โ ๐๐โ1 and a number ๐ โ ๐ such that ๐(๐ฃ) = ๐๐ฃ.
Proof. The proof will be by contradiction. If the theorem isnโt true the vectors ๐ฃ and ๐(๐ฃ),are linearly independent and hence the vector(4.8.5) ๐(๐ฃ) = ๐(๐ฃ) โ (๐(๐ฃ) โ ๐ฃ)๐ฃis nonzero. Let
(4.8.6) โ(๐ฃ) = ๐(๐ฃ)|๐(๐ฃ)|
.
By (4.8.5)โ(4.8.6), |๐ฃ| = |โ(๐ฃ)| = 1 and ๐ฃ โ โ(๐ฃ) = 0, i.e., ๐ฃ and โ(๐ฃ) are both unit vectors andare perpendicular to each other. Let(4.8.7) ๐พ๐ก โถ ๐๐โ1 โ ๐๐โ1 , 0 โค ๐ก โค 1be the map(4.8.8) ๐พ๐ก(๐ฃ) = cos(๐๐ก)๐ฃ + sin(๐๐ก)โ(๐ฃ) .For ๐ก = 0 this map is the identity map and for ๐ก = 1, it is the antipodal map ๐(๐ฃ) = ๐ฃ, 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 ๐๐โ1 of the map (๐ฅ1,โฆ, ๐ฅ๐) โฆ (โ๐ฅ1,โฆ, โ๐ฅ๐) and the volume form ๐ on ๐๐โ1is the restriction to ๐๐โ1 of the (๐ โ 1)-form
(4.8.9)๐โ๐=1(โ1)๐โ1๐ฅ๐๐๐ฅ๐ โงโฏ โง ๐๐ฅ๐ โงโฏ โง ๐๐ฅ๐ .
If we replace ๐ฅ๐ by โ๐ฅ๐ in equation (4.8.9) the sign of this form changes by (โ1)๐ hence๐โ๐ = (โ1)๐๐. Thus if ๐ is odd, ๐ is an orientation reversing diffeomorphism of ๐๐โ1 onto๐๐โ1, so its degree is โ1, and this contradicts what we just deduced from the existence of thehomotopy (4.8.8). โก
From this argument we can deduce another interesting fact about the sphere ๐๐โ1 when๐ โ 1 is even. For ๐ฃ โ ๐๐โ1 the tangent space to ๐๐โ1 at ๐ฃ is just the space,
๐๐ฃ๐๐โ1 = { (๐ฃ, ๐ค) | ๐ค โ ๐๐ and ๐ฃ โ ๐ค = 0 } ,so a vector field on ๐๐โ1 can be viewed as a function ๐โถ ๐๐โ1 โ ๐๐ with the property
๐(๐ฃ) โ ๐ฃ = 0for all ๐ฃ โ ๐๐โ1. If this function is nonzero at all points, then, letting โ be the function (4.8.6),and arguing as above, weโre led to a contradiction. Hence we conclude:
Theorem 4.8.10. If ๐ โ 1 is even and ๐ is a vector field on the sphere ๐๐โ1, then there exists apoint ๐ โ ๐๐โ1 at which ๐(๐) = 0.
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ยง4.8 Applications of degree theory 139
Note that if ๐ โ 1 is odd this statement is not true. Thevector field
๐ฅ1๐๐๐ฅ2โ ๐ฅ2๐๐๐ฅ1+โฏ + ๐ฅ2๐โ1
๐๐๐ฅ2๐โ ๐ฅ2๐
๐๐๐ฅ2๐โ1
is a counterexample. It is nowhere vanishing and at ๐ โ ๐๐โ1 is tangent to ๐๐โ1.Application 4.8.11 (The JordanโBrouwer separation theorem). Let ๐ be a compact ori-ented (๐โ1)-dimensional submanifold of๐๐. In this subsection of ยง4.8 weโll outline a proofof the following theorem (leaving the details as a string of exercises).
Theorem 4.8.12. If๐ is connected, the complement of ๐๐ โ๐ of๐ has exactly two connectedcomponents.
This theorem is known as the JordanโBrouwer separation theorem (and in two dimen-sions as the Jordan curve theorem). For simple, easy to visualize, submanifolds of๐๐ like the(๐ โ 1)-sphere this result is obvious, and for this reason itโs easy to be misled into thinkingof it as being a trivial (and ot very interesting) result. However, for submanifolds of ๐๐ likethe curve in ๐2 depicted in the Figure 4.8.2 itโs much less obvious. (In ten seconds or less: isthe point ๐ in this figure inside this curve or outside?)
To determine whether a point ๐ โ ๐๐โ๐ is inside๐ or outside๐, one needs a topolog-ical invariant to detect the difference; such an invariant is provided by the winding number.
Definition 4.8.13. For ๐ โ ๐๐ โ ๐ let๐พ๐ โถ ๐ โ ๐๐โ1
be the map๐พ๐(๐ฅ) =
๐ฅ โ ๐|๐ฅ โ ๐|
.
The winding number๐(๐, ๐) of๐ about ๐ is the degree๐(๐, ๐) โ deg(๐พ๐) .
We show below that๐(๐, ๐) = 0 if๐ is outside, and if๐ is inside๐, then๐(๐, ๐) = ยฑ1(depending on the orientation of ๐). Hence the winding number tells us which of the twocomponents of ๐๐ โ ๐ the point ๐ is contained in.
Exercises for ยง4.8Exercise 4.8.i. Let๐ be a connected component of ๐๐ โ๐. Show that if ๐0 and ๐1 are in๐,๐(๐, ๐0) = ๐(๐, ๐1).
Hints:โข First suppose that the line segment,
๐๐ก = (1 โ ๐ก)๐0 + ๐ก๐1 , 0 โค ๐ก โค 1lies in ๐. Conclude from the homotopy invariance of degree that
๐(๐, ๐0) = ๐(๐, ๐๐ก) = ๐(๐, ๐1) .โข Show that there exists a sequence of points ๐1,โฆ, ๐๐ โ ๐, with ๐1 = ๐0 and๐๐ = ๐1, such that the line segment joining ๐๐ to ๐๐+1 is in ๐.
Exercise 4.8.ii. Let ๐ be a connected (๐ โ 1)-dimensional submanifold of ๐๐. Show that๐๐ โ ๐ has at most two connected components.
Hints:โข Show that if ๐ is in ๐ there exists a small ๐-ball ๐ต๐(๐) centered at ๐ such that๐ต๐(๐) โ ๐ has two components. (See Theorem 4.2.15.
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140 Chapter 4: Manifolds & Forms on Manifolds
โข Show that if ๐ is in ๐๐ โ ๐, there exists a sequence ๐1,โฆ, ๐๐ โ ๐๐ โ ๐ such that๐1 = ๐, ๐๐ โ ๐ต๐(๐) and the line segments joining ๐๐ to ๐๐+1 are in ๐๐ โ ๐.
Exercise 4.8.iii. For ๐ฃ โ ๐๐โ1, show that ๐ฅ โ ๐ is in ๐พโ1๐ (๐ฃ) if and only if ๐ฅ lies on the ray
(4.8.14) ๐ + ๐ก๐ฃ , 0 < ๐ก < โ .
Exercise 4.8.iv. Let ๐ฅ โ ๐ be a point on this ray. Show that
(4.8.15) (๐๐พ๐)๐ฅ โถ ๐๐๐ โ ๐๐ฃ๐๐โ1
is bijective if and only if ๐ฃ โ ๐๐๐, i.e., if and only if the ray (4.8.14) is not tangent to๐ at ๐ฅ.Hint: ๐พ๐ โถ ๐ โ ๐๐โ1 is the composition of the maps
๐๐ โถ ๐ โ ๐๐ โ {0} , ๐ฅ โฆ ๐ฅ โ ๐ ,
and๐โถ ๐๐ โ {0} โ ๐๐โ1 , ๐ฆ โฆ ๐ฆ
|๐ฆ|.
Show that if ๐(๐ฆ) = ๐ฃ, then the kernel of (๐๐)๐ is the one-dimensional subspace of ๐๐spanned by ๐ฃ. Conclude that if ๐ฆ = ๐ฅ โ ๐ and ๐ฃ = ๐ฆ/|๐ฆ| the composite map
(๐๐พ๐)๐ฅ = (๐๐)๐ฆ โ (๐๐๐)๐ฅis bijective if and only if ๐ฃ โ ๐๐ฅ๐.
Exercise 4.8.v. From Exercises 4.8.iii and 4.8.iv deduce that ๐ฃ is a regular value of ๐พ๐ ifand only if the ray (4.8.14) intersects ๐ in a finite number of points and at each point ofintersection is not tangent to๐ 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 ๐๐ฅ๐ and ๐ฃ are compatible with the standard orientation of ๐๐๐๐. (SeeExercise 1.6.v.)
Exercise 4.8.vii. Conclude that deg(๐พ๐) counts (with orientations) the number of pointswhere the ray (4.8.14) intersects๐.
Exercise 4.8.viii. Let ๐1 โ ๐๐ โ ๐ be a point on the ray (4.8.14). Show that if ๐ฃ โ ๐๐โ1 is aregular value of ๐พ๐, it is a regular value of ๐พ๐1 and show that the number
deg(๐พ๐) โ deg(๐พ๐1) = ๐(๐, ๐) โ๐(๐, ๐1)
counts (with orientations) the number of points on the ray lying between ๐ and ๐1.Hint: Exercises 4.8.v and 4.8.vii.
Exercise 4.8.ix. Let ๐ฅ โ ๐ be a point on the ray (4.8.14). Suppose ๐ฅ = ๐ + ๐ก๐ฃ. Show that if๐ is a small positive number and
๐ยฑ = ๐ + (๐ก ยฑ ๐)๐ฃ
then๐(๐, ๐+) = ๐(๐, ๐โ) ยฑ 1 ,
and from Exercise 4.8.i conclude that ๐+ and ๐โ lie in different components of ๐๐ โ ๐. Inparticular conclude that ๐๐ โ ๐ has exactly two components.
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ยง4.8 Applications of degree theory 141
Exercise 4.8.x. Finally show that if ๐ is very large the difference
๐พ๐(๐ฅ) โ๐|๐|, ๐ฅ โ ๐ ,
is very small, i.e., ๐พ๐ is not surjective and hence the degree of ๐พ๐ is zero. Conclude that for๐ โ ๐๐โ๐, ๐ is in the unbounded component of๐๐โ๐ if๐(๐, ๐) = 0 and in the boundedcomponent if๐(๐, ๐) = ยฑ1 (the โยฑโ depending on the orientation of๐).
Remark 4.8.16. The proof of JordanโBrouwer sketched above gives us an effective way ofdecidingwhether the point๐ in Figure 4.8.2, is inside๐ or outside๐. Draw a non-tangentialray from ๐. If it intersects๐ in an even number of points, ๐ is outside๐ and if it intersects๐ is an odd number of points ๐ inside.
๏ฟฝ
๏ฟฝ
๏ฟฝ
inside
outside
Figure 4.8.1. A Jordan curve with ๐ inside of the curve and ๐ outside
Application 4.8.17 (TheGaussโBonnet theorem). Let๐ โ ๐๐ be a compact, connected, ori-ented (๐โ1)-dimensional submanifold. By the JordanโBrouwer theorem๐ is the boundaryof a bounded smooth domain, so for each ๐ฅ โ ๐ there exists a unique outward pointing unitnormal vector ๐๐ฅ. The Gauss map
๐พโถ ๐ โ ๐๐โ1
is the map ๐ฅ โฆ ๐๐ฅ. Let ๐ be the Riemannian volume form of ๐๐โ1, or, in other words, therestriction to ๐๐โ1 of the form,
๐โ๐=1(โ1)๐โ1๐ฅ๐๐๐ฅ1 โงโฏ โง ๐๐ฅ๐ โงโฏ โง ๐๐ฅ๐ ,
and let ๐๐ be the Riemannian volume form of๐. Then for each ๐ โ ๐(4.8.18) (๐พโ๐)๐ = ๐พ(๐)(๐๐)๐where ๐พ(๐) is the scalar curvature of ๐ at ๐. This number measures the extent to whichโ๐ is curvedโ at ๐. For instance, if ๐๐ is the circle, |๐ฅ| = ๐ in ๐2, the Gauss map is the map๐ โ ๐/๐, so for all ๐ we have ๐พ๐(๐) = 1/๐, reflecting the fact that for ๐ < ๐, ๐๐ is morecurved than๐๐.
The scalar curvature can also be negative. For instance for surfaces ๐ in ๐3, ๐พ(๐) ispositive at ๐ if๐ is convex at ๐ and negative if๐ is convexโconcave at ๐. (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 ๐, and the surface in Figure 4.8.3 isconvexโconcave.)
๏ฟฝ(๏ฟฝ)
๏ฟฝ ๏ฟฝ
Figure 4.8.2. Positive scalar curvature at ๐
๏ฟฝ(๏ฟฝ)๏ฟฝ
๏ฟฝ
Figure 4.8.3. Negative scalar curvature at ๐
Let vol(๐๐โ1) be the Riemannian volume of the (๐ โ 1)-sphere, i.e., let
vol(๐๐โ1) = 2๐๐/2
๐ค(๐/2)(where ๐ค is the Gamma function). Then by (4.8.18) the quotient
(4.8.19)โซ๐พ๐๐
vol(๐๐โ1)is the degree of the Gauss map, and hence is a topological invariant of the surface of๐. For
๏ฟฝ
surface ๏ฟฝ of genus ๏ฟฝ
๏ฟฝ1 ๏ฟฝ2 ๏ฟฝ0๏ฟฝ๏ฟฝ
the sphere ๏ฟฝ2
โฆ ๏ฟฝ0 ๏ฟฝ
Figure 4.8.4. The Gauss map from a surface of genus ๐ to the sphere
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ยง4.9 The index of a vector field 143
๐ = 3 the GaussโBonnet theorem asserts that this topological invariant is just 1 โ ๐ where๐ is the genus of ๐ or, in other words, the โnumber of holesโ. Figure 4.8.4 gives a pictorialproof of this result. (Notice that at the points ๐1,โฆ, ๐๐ the surface๐ is convexโconcave sothe scalar curvature at these points is negative, i.e., the Gauss map is orientation reversing.On the other hand, at the point ๐0 the surface is convex, so the Gauss map at this point isorientation preserving.)
4.9. The index of a vector field
Let ๐ท be a bounded smooth domain in ๐๐ and ๐ = โ๐๐=1 ๐ฃ๐๐/๐๐ฅ๐ a ๐ถโ vector field de-fined on the closure๐ท of๐ท.Wewill define below a topological invariant which, for โgenericโ๐โs, counts (with appropriate ยฑ-signs) the number of zeros of ๐. To avoid complications weโllassume ๐ has no zeros on the boundary of๐ท.
Definition 4.9.1. Let ๐๐ท be the boundary of๐ท and let
๐๐ โถ ๐๐ท โ ๐๐โ1
be the mapping
๐ โฆ ๐(๐)|๐(๐)|
,
where ๐(๐) = (๐ฃ1(๐),โฆ, ๐ฃ๐(๐)). The index of ๐ with respect to๐ท is by definition the degreeof ๐๐.
Weโll denote this index by ind(๐, ๐ท) and as a first step in investigating its propertiesweโll prove:
Theorem 4.9.2. Let๐ท1 be a smooth domain in ๐๐ whose closure is contained in๐ท. Then
ind(๐, ๐ท) = ind(๐, ๐ท1)
provided that ๐ has no zeros in the set๐ท โ ๐ท1.
Proof. Let๐ = ๐ท โ ๐ท1. Then the map ๐๐ โถ ๐๐ท โ ๐๐โ1 extends to a ๐ถโ map
๐นโถ ๐ โ ๐๐โ1 , ๐ โฆ ๐(๐)|๐(๐)|
.
Moreover,๐๐ = ๐ โช ๐โ1
where๐ is the boundary of๐ทwith its natural boundary orientation and๐โ1 is the boundaryof ๐ท1 with its boundary orientation reversed. Let ๐ be an element of ๐บ๐โ1(๐๐โ1) whoseintegral over ๐๐โ1 is 1. Then if ๐ = ๐๐ and ๐1 = (๐1)๐ are the restrictions of ๐น to ๐ and ๐1we get from Stokesโ theorem (and the fact that ๐๐ = 0) the identity
0 = โซ๐๐นโ๐๐ = โซ
๐๐๐นโ๐
= โซ๐๐โ๐ โ โซ
๐1๐โ1 ๐ = deg(๐) โ deg(๐1) .
Henceind(๐, ๐ท) = deg(๐) = deg(๐1) = ind(๐, ๐ท) . โก
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144 Chapter 4: Manifolds & Forms on Manifolds
Suppose now that ๐ has a finite number of isolated zeros ๐1,โฆ, ๐๐ in ๐ท. Let ๐ต๐(๐๐) bethe open ball of radius ๐ with center at ๐๐. By making ๐ small enough we can assume thateach of these balls is contained in ๐ท and that theyโre mutually disjoint. We will define thelocal index ind(๐, ๐๐) of ๐ at ๐๐ to be the index of ๐ with respect to ๐ต๐(๐๐). By Theorem 4.9.2these local indices are unchanged if we replace ๐ by a smaller ๐, and by applying this theoremto the domain
๐ท1 =๐โ๐=1๐ต๐๐(๐)
we get, as a corollary of the theorem, the formula
(4.9.3) ind(๐, ๐ท) =๐โ๐=1
ind(๐, ๐๐)
which computes the global index of ๐ with respect to๐ท in terms of these local indices.Letโs now see how to compute these local indices. Letโs first suppose that the point๐ = ๐๐
is at the origin of ๐๐. Then near ๐ = 0
๐ = ๐lin + ๐โฒ
where๐ = ๐lin = โ
1โค๐,๐โค๐๐๐,๐๐ฅ๐๐๐๐ฅ๐
the ๐๐,๐โs being constants and
๐๐ =๐โ๐=1๐๐๐๐๐๐ฅ๐
where the ๐๐๐โs vanish to second order near zero, i.e., satisfy
(4.9.4) |๐๐,๐(๐ฅ)| โค ๐ถ|๐ฅ|2
for some constant ๐ถ > 0.
Definition 4.9.5. Weโll say that the point ๐ = 0 is a non-degenerate zero of ๐ is the matrix๐ด = (๐๐,๐) is non-singular.
This means in particular that
(4.9.6)๐โ๐=1|๐โ๐=1๐๐,๐๐ฅ๐| โฅ ๐ถ1|๐ฅ|
form some constant ๐ถ1 > 0. Thus by (4.9.4) and (4.9.6) the vector field
๐๐ก = ๐lin + ๐ก๐โฒ , 0 โค ๐ก โค 1
has no zeros in the ball ๐ต๐(0), other than the point ๐ = 0 itself, provided that ๐ is smallenough. Therefore if we let๐๐ be the boundary of ๐ต๐(0) we get a homotopy
๐นโถ ๐๐ ร [0, 1] โ ๐๐โ1 , (๐ฅ, ๐ก) โฆ ๐๐๐ก(๐ฅ)
between the maps ๐๐lin โถ ๐๐ โ ๐๐โ1 and ๐๐ โถ ๐๐ โ ๐๐โ1, thus by Theorem 4.7.16 we see that
deg(๐๐) = deg(๐๐lin). Therefore,
(4.9.7) ind(๐, ๐) = ind(๐lin, ๐) .
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ยง4.9 The index of a vector field 145
Wehave assumed in the above discussion that๐ = 0, but by introducing at๐ a translatedcoordinate system for which ๐ is the origin, these comments are applicable to any zero ๐ of๐. More explicitly if ๐1,โฆ, ๐๐ are the coordinates of ๐ then as above
๐ = ๐lin + ๐โฒ
where
๐lin โ โ1โค๐,๐โค๐๐๐,๐(๐ฅ๐ โ ๐๐)
๐๐๐ฅ๐
and ๐โฒ vanishes to second order at ๐, and if ๐ is a non-degenerate zero of ๐, i.e., if the matrix๐ด = (๐๐,๐) is non-singular, then exactly the same argument as before shows that
ind(๐, ๐) = ind(๐lin, ๐) .
We will next prove:
Theorem 4.9.8. If ๐ is a non-degenerate zero of ๐, the local index ind(๐, ๐) is +1 or โ1 de-pending on whether the determinant of the matrix, ๐ด = (๐๐,๐) is positive or negative.
Proof. As above we can, without loss of generality, assume that ๐ = 0. By (4.9.7) we canassume ๐ = ๐lin. Let๐ท be the domain defined by
(4.9.9)๐โ๐=1(๐โ๐=1๐๐,๐๐ฅ๐)
2
< 1 ,
and let ๐ โ ๐๐ท be its boundary. Since the only zero of ๐lin inside this domain is ๐ = 0 weget from (4.9.3) and (4.9.4) that
ind(๐, ๐) = ind(๐lin, ๐) = ind(๐lin, ๐ท) .
Moreover, the map๐๐lin โถ ๐ โ ๐
๐โ1
is, in view of (4.9.9), just the linear map ๐ฃ โฆ ๐ด๐ฃ, restricted to ๐. In particular, since ๐ด isa diffeomorphism, this mapping is a diffeomorphism as well, so the degree of this map is+1 or โ1 depending on whether this map is orientation preserving or not. To decide whichof these alternatives is true let ๐ = โ๐๐=1(โ1)๐๐ฅ๐๐๐ฅ1 โงโฏ โง ๐๐ฅ๐ โงโฏ๐๐ฅ๐ be the Riemannianvolume form on ๐๐โ1 then
โซ๐๐โ๐lin๐ = deg(๐๐lin) vol(๐
๐โ1) .
Since ๐ is the restriction of ๐ด to๐. By Stokesโ theorem this is equal to
โซ๐ท๐ดโ๐๐ = ๐โซ
๐ท๐ดโ๐๐ฅ1 โงโฏ โง ๐๐ฅ๐
= ๐ det(๐ด) โซ๐ท๐๐ฅ1 โงโฏ โง ๐๐ฅ๐
= ๐ det(๐ด) vol(๐ท)
which gives us the formula
deg(๐๐lin , ๐ท) =๐ det๐ด vol(๐ท)
vol(๐๐โ1). โก
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146 Chapter 4: Manifolds & Forms on Manifolds
Weโll briefly describe the various types of non-degenerate zeros that can occur for vectorfields in two dimensions. To simplify this description a bit weโll assume that the matrix
๐ด = (๐11 ๐12๐21 ๐22)
is diagonalizable and that its eigenvalues are not purely imaginary. Then the following fivescenarios can occur:(1) Theeigenvalues of๐ด are real and positive. In this case the integral curves of ๐lin in a neigh-
borhood of ๐ look like the curves in Figure 4.9.1 and hence, in a small neighborhood
Figure 4.9.1. Real and positive eigenvalues
of ๐, the integral sums of ๐ itself look approximately like the curve in Figure 4.9.1.(2) The eigenvalues of๐ด are real and negative. In this case the integral curves of ๐lin look like
the curves in Figure 4.9.1, but the arrows are pointing into ๐, rather than out of ๐, i.e.,
Figure 4.9.2. Real and negative eigenvalues
(3) The eigenvalues of ๐ด are real, but one is positive and one is negative. In this case theintegral curves of ๐lin look like the curves in Figure 4.9.3.
(4) The eigenvalues of ๐ด are complex and of the form ๐ ยฑ โโ๐, with ๐ positive. In this casethe integral curves of ๐lin are spirals going out of ๐ as in Figure 4.9.4.
(5) The eigenvalues of ๐ด are complex and of the form ๐ ยฑ โโ๐, with ๐ negative. In this casethe integral curves are as in Figure 4.9.4 but are spiraling into ๐ rather than out of ๐.
Definition 4.9.10. Zeros of ๐ 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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ยง4.9 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 ๐ ยฑ โโ๐, with ๐ positive.
Thus in particular the two-dimensional version of equation (4.9.3) tells us:
Theorem 4.9.11. If the vector field ๐ on ๐ท has only non-degenerate zeros then ind(๐, ๐ท) 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 weโve frequently encountered the question: When is a closed๐-formon an open subset of๐๐ (or,more generally on a submanifold of๐๐) exact? To inves-tigate this question more systematically than weโve done heretofore, let๐ be an ๐-manifoldand let
๐๐(๐) โ {๐ โ ๐บ๐(๐) | ๐๐ = 0 }and
๐ต๐(๐) โ ๐(๐บ๐โ1(๐)) โ ๐บ๐(๐)be the vector spaces of closed and exact ๐-forms. Since ๐ต๐(๐) is a vector subspace of๐๐(๐),we can form the quotient space
๐ป๐(๐) โ ๐๐(๐)/๐ต๐(๐) ,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 ๐th de Rham cohomology group of the manifold ๐. Sincethe vector spaces ๐๐(๐) and ๐ต๐(๐) are both infinite dimensional in general, there is noguarantee that this quotient space is finite dimensional, however, weโll show later in thischapter that it is in a lot of interesting cases.
The spaces๐ป๐(๐) also have compactly supported counterparts. Namely let
๐๐๐ (๐) โ {๐ โ ๐บ๐๐ (๐) | ๐๐ = 0 }and
๐ต๐๐ (๐) โ ๐(๐บ๐โ1๐ (๐)) โ ๐บ๐๐ (๐) .Then as above ๐ต๐๐ (๐) is a vector subspace of ๐๐๐ (๐) and the vector space quotient
๐ป๐๐ (๐) โ ๐๐๐ (๐)/๐ต๐๐ (๐)is the ๐th compactly supported de Rham cohomology group of๐.
Given a closed ๐-form ๐ โ ๐๐(๐) we will denote by [๐] the image of ๐ in the quotientspace ๐ป๐(๐) = ๐๐(๐)/๐ต๐(๐) and call [๐] the cohomology class of ๐. We will also use thesame notation for compactly supported cohomology. If ๐ is in ๐๐๐ (๐) weโll denote by [๐]the cohomology class of ๐ in the quotient space๐ป๐๐ (๐) = ๐๐๐ (๐)/๐ต๐๐ (๐).
Some cohomology groups of manifolds weโve already computed in the previous chap-ters (although we didnโt explicitly describe these computations as โcomputing cohomol-ogyโ). Weโll make a list below of some of the properties that we have already discoveredabout the de Rham cohomology of a manifold๐:(1) If ๐ is connected, ๐ป0(๐) = ๐. To see this, note that a closed zero form is a function๐ โ ๐ถโ(๐) having the property ๐๐ = 0, and if๐ is connected the only such functionsare constants.
149
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150 Chapter 5: Cohomology via forms
(2) If ๐ is connected and non-compact๐ป0๐ (๐) = 0. To see this, note that if ๐ is in ๐ถโ0 (๐)and๐ is non-compact, ๐ has to be zero at some point, and hence if ๐๐ = 0, then ๐ hasto be identically zero.
(3) If๐ is ๐-dimensional,๐บ๐(๐) = ๐บ๐๐ (๐) = 0
for ๐ < 0 or ๐ > ๐, hence๐ป๐(๐) = ๐ป๐๐ (๐) = 0
for ๐ < 0 or ๐ > ๐.(4) If๐ is an oriented, connected ๐-manifold, the integration operation is a linear map
(5.1.1) โซ๐โถ ๐บ๐๐ (๐) โ ๐
and, by Theorem 4.8.2, the kernel of this map is ๐ต๐๐ (๐). Moreover, in degree ๐,๐๐๐ (๐) =๐บ๐๐ (๐) and hence by the definition of๐ป๐๐ (๐) as the quotient๐๐๐ (๐)/๐ต๐๐ (๐), we get from(5.1.1) a bijective map
(5.1.2) ๐ผ๐ โถ ๐ป๐๐ (๐) โฅฒ ๐ .
In other words๐ป๐๐ (๐) = ๐.(5) Let ๐ be a star-shaped open subset of ๐๐. In Exercises 2.6.iv to 2.6.vii we sketched a
proof of the assertion: For ๐ > 0 every closed form ๐ โ ๐๐(๐) is exact. Translating thisassertion into cohomology language, we showed that๐ป๐(๐) = 0 for ๐ > 0.
(6) Let๐ โ ๐๐ be an open rectangle. In Exercises 3.2.iv to 3.2.vii we sketched a proof of theassertion: If ๐ โ ๐บ๐๐ (๐) is closed and ๐ is less than ๐, then ๐ = ๐๐ for some (๐โ1)-form๐ โ ๐บ๐โ1๐ (๐). Hence we showed๐ป๐๐ (๐) = 0 for ๐ < ๐.
(7) Poincarรฉ lemma for manifolds: Let ๐ be an ๐-manifold and ๐ โ ๐๐(๐), ๐ > 0 a closed๐-form. Then for every point ๐ โ ๐ there exists a neighborhood ๐ of ๐ and a (๐ โ 1)-form ๐ โ ๐บ๐โ1(๐) such that ๐ = ๐๐ on ๐. To see this note that for open subsets of ๐๐we proved this result in Section 2.4 and since๐ is locally diffeomorphic at ๐ to an opensubset of ๐๐ this result is true for manifolds as well.
(8) Let๐be the unit sphere ๐๐ in๐๐+1. Since ๐๐ is compact, connected andoriented๐ป0(๐๐) =๐ป๐(๐๐) = ๐.
We will show that for ๐ โ , 0, ๐๐ป๐(๐๐) = 0 .
To see this let ๐ โ ๐บ๐(๐๐) be a closed ๐-form and let ๐ = (0,โฆ, 0, 1) โ ๐๐ be the โnorthpoleโ of ๐๐. By the Poincarรฉ lemma there exists a neighborhood๐ of ๐ in ๐๐ and a ๐โ1-form ๐ โ ๐บ๐โ1(๐) with ๐ = ๐๐ on ๐. Let ๐ โ ๐ถโ0 (๐) be a โbump functionโ which isequal to one on a neighborhood ๐0 of ๐ in ๐. Then
(5.1.3) ๐1 = ๐ โ ๐๐๐is a closed ๐-form with compact support in ๐๐ โ {๐}. However stereographic projectiongives one a diffeomorphism
๐โถ ๐๐ โ ๐๐ โ {๐}(see Exercise 5.1.i), and hence ๐โ๐1 is a closed compactly supported ๐-form on๐๐ withsupport in a large rectangle. Thus by (5.1.3) ๐โ๐ = ๐๐, for some ๐ โ ๐บ๐โ1๐ (๐๐), and by(5.1.3)
(5.1.4) ๐ = ๐(๐๐ + (๐โ1)โ๐)
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ยง5.1 The de Rham cohomology groups of a manifold 151
with (๐โ1)โ๐ โ ๐บ๐โ1๐ (๐๐โ{๐}) โ ๐บ๐(๐๐), so weโve proved that for 0 < ๐ < ๐ every closed๐-form on ๐๐ is exact.We will next discuss some โpullbackโ operations in de Rham theory. Let ๐ and ๐ be
manifolds and ๐โถ ๐ โ ๐ a ๐ถโ map. For ๐ โ ๐บ๐(๐), ๐๐โ๐ = ๐โ๐๐, so if ๐ is closed, ๐โ๐is as well. Moreover, if ๐ = ๐๐, ๐โ๐ = ๐๐โ๐, so if ๐ is exact, ๐โ๐ is as well. Thus we havelinear maps
๐โ โถ ๐๐(๐) โ ๐๐(๐)and
(5.1.5) ๐โ โถ ๐ต๐(๐) โ ๐ต๐(๐)and composing ๐โ โถ ๐๐(๐) โ ๐๐(๐) with the projection
๐โถ ๐๐(๐) โ ๐๐(๐)/๐ต๐(๐)we get a linear map
(5.1.6) ๐๐(๐) โ ๐ป๐(๐) .In view of (5.1.5), ๐ต๐(๐) is in the kernel of this map, so by Proposition 1.2.9 one gets aninduced linear map
๐โฏ โถ ๐ป๐(๐) โ ๐ป๐(๐) ,such that ๐โฏ โ ๐ is the map (5.1.6). In other words, if ๐ is a closed ๐-form on ๐, then ๐โฏ hasthe defining property
(5.1.7) ๐โฏ[๐] = [๐โ๐] .This โpullbackโ operation on cohomology satisfies the following chain rule: Let ๐ be a
manifold and ๐โถ ๐ โ ๐ a ๐ถโ map. Then if ๐ is a closed ๐-form on ๐(๐ โ ๐)โ๐ = ๐โ๐โ๐
by the chain rule for pullbacks of forms, and hence by (5.1.7)
(5.1.8) (๐ โ ๐)โฏ[๐] = ๐โฏ(๐โฏ[๐]) .The discussion above carries over verbatim to the setting of compactly supported de
Rham cohomology: If ๐โถ ๐ โ ๐ is a proper ๐ถโ map ๐ induces a pullback map on coho-mology
๐โฏ โถ ๐ป๐๐ (๐) โ ๐ป๐๐ (๐)and if ๐โถ ๐ โ ๐ and ๐โถ ๐ โ ๐ are proper ๐ถโ 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 ๐โถ ๐ โ ๐ is a diffeomorphism, we can take ๐ to be ๐ itself and ๐ to be๐โ1, and in this case the chain rule tells us that the inducedmapsmaps๐โฏ โถ ๐ป๐(๐) โ ๐ป๐(๐)and ๐โฏ โถ ๐ป๐๐ (๐) โ ๐ป๐๐ (๐) are bijections, i.e.,๐ป๐(๐) and๐ป๐(๐) and๐ป๐๐ (๐) and๐ป๐๐ (๐) areisomorphic as vector spaces.
We will next establish an important fact about the pullback operation ๐โฏ; weโll showthat itโs a homotopy invariant of ๐. Recall that two ๐ถโ maps
(5.1.9) ๐0, ๐1 โถ ๐ โ ๐are homotopic if there exists a ๐ถโ map
๐นโถ ๐ ร [0, 1] โ ๐with the property ๐น(๐, 0) = ๐0(๐) and ๐น(๐, 1) = ๐1(๐) for all ๐ โ ๐. 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) ๐โฏ0 = ๐โฏ1 .
Our proof of this will consist of proving this for an important special class of homo-topies, and then by โpullbackโ tricks deducing this result for homotopies in general. Let ๐be a complete vector field on๐ and let
๐๐ก โถ ๐ โ ๐ , โโ < ๐ก < โbe the one-parameter group of diffeomorphisms that ๐ generates. Then
๐นโถ ๐ ร [0, 1] โ ๐ , ๐น(๐, ๐ก) โ ๐๐ก(๐) ,is a homotopy between ๐0 and ๐1, and weโll show that for this homotopic pair (5.1.11) istrue.
Proof. Recall that for ๐ โ ๐บ๐(๐)๐๐๐ก๐โ๐ก ๐ |๐ก=0= ๐ฟ๐ = ๐๐๐๐ + ๐๐๐๐
and more generally for all ๐ก we have๐๐๐ก๐โ๐ก ๐ =
๐๐๐ ๐โ๐ +๐ก๐ |
๐ =0= ๐๐๐ (๐๐ โ ๐๐ก)โ๐ |
๐ =0
= ๐๐๐ ๐โ๐ก ๐โ๐ ๐ |
๐ =0= ๐โ๐ก๐๐๐ ๐โ๐ ๐ |๐ =0
= ๐โ๐ก ๐ฟ๐๐= ๐โ๐ก ๐๐๐๐ + ๐๐โ๐ก ๐๐๐ .
Thus if we set๐๐ก๐ โ ๐โ๐ก ๐๐๐
we get from this computation:๐๐๐ก๐โ๐ = ๐๐๐ก + ๐๐ก๐๐
and integrating over 0 โค ๐ก โค 1:(5.1.12) ๐โ1 ๐ โ ๐โ0 ๐ = ๐๐๐ + ๐๐๐
where ๐โถ ๐บ๐(๐) โ ๐บ๐โ1(๐) is the operator
๐๐ = โซ1
0๐๐ก๐๐๐ก .
The identity (5.1.11) is an easy consequence of this โchain homotopyโ identity. If ๐ is in๐๐(๐), then ๐๐ = 0, so
๐โ1 ๐ โ ๐โ0 ๐ = ๐๐๐and
๐โฏ1 [๐] โ ๐โฏ0 [๐] = [๐โ1 ๐ โ ๐โ0 ๐] = 0 . โก
Weโll now describe how to extract from this result a proof of Theorem 5.1.10 for anypair of homotopic maps. Weโll begin with the following useful observation.
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ยง5.1 The de Rham cohomology groups of a manifold 153
Proposition 5.1.13. If ๐0, ๐1 โถ ๐ โ ๐ are homotopic ๐ถโ mappings, then there exists a ๐ถโmap
๐นโถ ๐ ร ๐ โ ๐such that the restriction of ๐น to๐ ร [0, 1] is a homotopy between ๐0 and ๐1.
Proof. Let ๐ โ ๐ถโ0 (๐), ๐ โฅ 0, be a bump function which is supported on the interval [ 14 ,34 ]
and is positive at ๐ก = 12 . Then
๐(๐ก) =โซ๐กโโ ๐(๐ )๐๐ โซโโโ ๐(๐ )๐๐
is a function which is zero on the interval (โโ, 14 ], is 1 on the interval [ 34 ,โ), and, for all ๐ก,lies between 0 and 1. Now let
๐บโถ ๐ ร [0, 1] โ ๐be a homotopy between ๐0 and ๐1 and let ๐นโถ ๐ ร ๐ โ ๐ be the map(5.1.14) ๐น(๐ฅ, ๐ก) = ๐บ(๐ฅ, ๐(๐ก)) .This is a ๐ถโ map and since
๐น(๐ฅ, 1) = ๐บ(๐ฅ, ๐(1)) = ๐บ(๐ฅ, 1) = ๐1(๐ฅ) ,and
๐น(๐ฅ, 0) = ๐บ(๐ฅ, ๐(0)) = ๐บ(๐ฅ, 0) = ๐0(๐ฅ) ,it gives one a homotopy between ๐0 and ๐1. โก
Weโre now in position to deduce Theorem 5.1.10 from the version of this result that weproved above.
Proof of Theorem 5.1.10. Let๐พ๐ก โถ ๐ ร ๐ โ ๐ ร ๐ , โโ < ๐ก < โ
be the one-parameter group of diffeomorphisms๐พ๐ก(๐ฅ, ๐) = (๐ฅ, ๐ + ๐ก)
and let ๐ = ๐/๐๐ก be the vector field generating this group. For a ๐-form ๐ โ ๐บ๐(๐ ร ๐), wehave by (5.1.12) the identity
๐พโ1 ๐ โ ๐พโ0 ๐ = ๐๐ค๐ + ๐ค๐๐where
(5.1.15) ๐ค๐ = โซ1
0๐พโ๐ก (๐๐/๐๐ก๐)๐๐ก .
Now let ๐น, as in Proposition 5.1.13, be a ๐ถโ map๐นโถ ๐ ร ๐ โ ๐
whose restriction to๐ ร [0, 1] is a homotopy between ๐0 and ๐1. Then for ๐ โ ๐บ๐(๐)๐พโ1๐นโ๐ โ ๐พโ0๐นโ๐ = ๐๐ค๐นโ๐ + ๐ค๐นโ๐๐
by the identity (5.1.14). Now let ๐ โถ ๐ โ ๐ ร ๐ be the inclusion ๐ โฆ (๐, 0), and note that(๐น โ ๐พ1 โ ๐)(๐) = ๐น(๐, 1) = ๐1(๐)
and(๐น โ ๐พ0 โ ๐)(๐) = ๐น(๐, 0) = ๐0(๐)
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154 Chapter 5: Cohomology via forms
i.e.,๐น โ ๐พ1 โ ๐ = ๐1
and๐น โ ๐พ0 โ ๐ = ๐0 .
Thus๐โ(๐พโ1๐นโ๐ โ ๐พโ0๐นโ๐) = ๐โ1 ๐ โ ๐โ0 ๐
and on the other hand by (5.1.15)
๐โ(๐พโ1๐นโ๐ โ ๐พโ0๐นโ๐) = ๐๐โ๐ค๐นโ๐ + ๐โ๐ค๐นโ๐๐ .
Letting๐โถ ๐บ๐(๐) โ ๐บ๐โ1(๐)
be the โchain homotopyโ operator
(5.1.16) ๐๐ โ ๐โ๐ค๐นโ๐
we can write the identity above more succinctly in the form
(5.1.17) ๐โ1 ๐ โ ๐โ0 ๐ = ๐๐๐ + ๐๐๐
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) ๐๐ = โซ1
0๐โ๐พโ๐ก (๐๐/๐๐ก๐นโ๐)๐๐ก .
Hence if ๐ is supported on a set๐ด in ๐, the integrand of (5.1.17) at ๐ก is supported on the set
(5.1.19) { ๐ โ ๐ | ๐น(๐, ๐ก) โ ๐ด } ,
and hence ๐๐ is supported on the set
๐(๐นโ1(๐ด) โฉ ๐ ร [0, 1])
where ๐โถ ๐ ร [0, 1] โ ๐ is the projection map ๐(๐, ๐ก) = ๐. โก
Suppose now that ๐0 and ๐1 are proper mappings and
๐บโถ ๐ ร [0, 1] โ ๐
a proper homotopy between ๐0 and ๐1, i.e., a homotopy between ๐0 and ๐1 which is properas a ๐ถโ map. Then if ๐น is the map (5.1.14) its restriction to ๐ ร [0, 1] is also a proper map,so this restriction is also a proper homotopy between ๐0 and ๐1. Hence if ๐ is in ๐บ๐๐ (๐) and๐ด is its support, the set (5.1.19) is compact, so ๐๐ is in ๐บ๐โ1๐ (๐). Therefore all summandsin the โchain homotopyโ formula (5.1.17) are compactly supported. Thus weโve proved:
Theorem 5.1.20. If ๐0, ๐1 โถ ๐ โ ๐ are proper ๐ถโ maps which are homotopic via a properhomotopy, the induced maps on cohomology
๐โฏ0 , ๐โฏ1 โถ ๐ป๐๐ (๐) โ ๐ป๐๐ (๐)
are the same.
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ยง5.1 The de Rham cohomology groups of a manifold 155
Weโll conclude this section by noting that the cohomology groups๐ป๐(๐) are equippedwith a natural product operation. Namely, suppose ๐๐ โ ๐บ๐๐(๐), ๐ = 1, 2, is a closed formand that ๐๐ = [๐๐] is the cohomology class represented by ๐๐. We can then define a productcohomology class ๐1 โ ๐2 in๐ป๐1+๐2(๐) by the recipe(5.1.21) ๐1 โ ๐2 โ [๐1 โง ๐2] .To show that this is a legitimate definition we first note that since ๐2 is closed
๐(๐1 โง ๐2) = ๐๐1 โง ๐2 + (โ1)๐1๐1 โง ๐๐2 = 0 ,so ๐1 โง ๐2 is closed and hence does represent a cohomology class. Moreover if we replace๐1 by another representative ๐1 + ๐๐1 = ๐โฒ of the cohomology class ๐1, then
๐โฒ1 โง ๐2 = ๐1 โง ๐2 + ๐๐1 โง ๐2 .But since ๐2 is closed,
๐๐1 โง ๐2 = ๐(๐1 โง ๐2) + (โ1)๐1๐1 โง ๐๐2= ๐(๐1 โง ๐2)
so๐โฒ1 โง ๐2 = ๐1 โง ๐2 + ๐(๐1 โง ๐2)
and [๐โฒ1 โง ๐2] = [๐1 โง ๐2]. Similarly (5.1.21) is unchanged if we replace ๐2 by ๐2 + ๐๐2,so the definition of (5.1.21) depends neither on the choice of ๐1 nor ๐2 and hence is anintrinsic definition as claimed.
There is a variant of this product operation for compactly supported cohomology classes,and weโll leave for you to check that itโs also well defined. Suppose ๐1 is in๐ป
๐1๐ (๐) and ๐2 is in๐ป๐2(๐) (i.e., ๐1 is a compactly supported class and ๐2 is an ordinary cohomology class). Let๐1 be a representative of ๐1 in๐บ
๐1๐ (๐) and ๐2 a representative of ๐2 in๐บ๐2(๐). Then ๐1 โง๐2is a closed form in ๐บ๐1+๐2๐ (๐) and hence defines a cohomology class(5.1.22) ๐1 โ ๐2 โ [๐1 โง ๐2]
in ๐ป๐1+๐2๐ (๐). Weโll leave for you to check that this is intrinsically defined. Weโll also leavefor you to check that (5.1.22) is intrinsically defined if the roles of ๐1 and ๐2 are reversed, i.e.,if ๐1 is in๐ป๐1(๐) and ๐2 in๐ป
๐2๐ (๐) and that the products (5.1.21) and (5.1.22) both satisfy
๐1 โ ๐2 = (โ1)๐1๐2๐2 โ ๐1 .Finally we note that if๐ is anothermanifold and๐โถ ๐ โ ๐ a๐ถโmap then for๐1 โ ๐บ๐1(๐)and ๐2 โ ๐บ๐2(๐)
๐โ(๐1 โง ๐2) = ๐โ๐1 โง ๐โ๐2by (2.6.8) and hence if ๐1 and ๐2 are closed and ๐๐ = [๐๐], then(5.1.23) ๐โฏ(๐1 โ ๐2) = ๐โฏ๐1 โ ๐โฏ๐2 .
Exercises for ยง5.1Exercise 5.1.i (stereographic projection). Let ๐ โ ๐๐ be the point ๐ = (0,โฆ, 0, 1). Showthat for every point ๐ฅ = (๐ฅ1,โฆ, ๐ฅ๐+1) of ๐๐ โ {๐} the ray
๐ก๐ฅ + (1 โ ๐ก)๐ , ๐ก > 0intersects the plane ๐ฅ๐+1 = 0 in the point
๐พ(๐ฅ) = 11 โ ๐ฅ๐+1
(๐ฅ1,โฆ, ๐ฅ๐)
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156 Chapter 5: Cohomology via forms
and that the map ๐พโถ ๐๐ โ {๐} โ ๐๐ is a diffeomorphism.
Exercise 5.1.ii. Show that the operator
๐๐ก โถ ๐บ๐(๐) โ ๐บ๐โ1(๐)in the integrand of equation (5.1.18), i.e., the operator
๐๐ก๐ = ๐โ๐พโ๐ก (๐๐/๐๐ก)๐นโ๐has the following description. Let ๐ be a point of ๐ and let ๐ = ๐๐ก(๐). The curve ๐ โฆ ๐๐ (๐)passes through ๐ at time ๐ = ๐ก. Let ๐ฃ(๐) โ ๐๐๐ be the tangent vector to this curve at ๐ก. Showthat
(5.1.24) (๐๐ก๐)(๐) = (๐๐โ๐ก )๐๐๐ฃ(๐)๐๐ .
Exercise 5.1.iii. Let๐ be a star-shaped open subset of ๐๐, i.e., a subset of ๐๐ with the prop-erty that for every ๐ โ ๐ the ray { ๐ก๐ | 0 โค ๐ก โค 1 } is in ๐.(1) Let ๐ be the vector field
๐ =๐โ๐=1๐ฅ๐๐๐๐ฅ๐
and ๐พ๐ก โถ ๐ โ ๐ the map ๐ โฆ ๐ก๐. Show that for every ๐-form ๐ โ ๐บ๐(๐)๐ = ๐๐๐ + ๐๐๐
where
๐๐ = โซ1
0๐พโ๐ก ๐๐๐๐๐ก๐ก
.
(2) Show that if ๐ = โ๐ผ ๐๐ผ(๐ฅ)๐๐ฅ๐ผ then
(5.1.25) ๐๐ = โ๐ผ,๐(โซ ๐ก๐โ1(โ1)๐โ1๐ฅ๐๐๐๐ผ(๐ก๐ฅ)๐๐ก) ๐๐ฅ๐ผ๐
where ๐๐ฅ๐ผ๐ โ ๐๐ฅ๐1 โงโฏ โง ๐๐ฅ๐๐ โงโฏ๐๐ฅ๐๐Exercise 5.1.iv. Let ๐ and ๐ be oriented connected ๐-manifolds, and ๐โถ ๐ โ ๐ a propermap. Show that the linear map ๐ฟ defined by the commutative square
๐ป๐๐ (๐) ๐ป๐๐ (๐)
๐ ๐
๐โฏ
๐ผ๐ โ ๐ผ๐โ
๐ฟ
is multiplication by deg(๐).
Exercise 5.1.v. A homotopy equivalence between๐ and ๐ is a pair of maps ๐โถ ๐ โ ๐ and๐โถ ๐ โ ๐ such that ๐ โ ๐ โ id๐ and ๐ โ ๐ โ id๐. Show that if ๐ and ๐ are homotopyequivalent their cohomology groups are the same โup to isomorphismโ, i.e., ๐ and ๐ induceinverse isomorphisms ๐โฏ โถ ๐ป๐(๐) โ ๐ป๐(๐) and ๐โฏ โถ ๐ป๐(๐) โ ๐ป๐(๐).
Exercise 5.1.vi. Show that ๐๐ โ {0} and ๐๐โ1 are homotopy equivalent.
Exercise 5.1.vii. What are the cohomology groups of the ๐-sphere with two points deleted?Hint: The ๐-sphere with one point deleted is homeomorphic to ๐๐.
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ยง5.1 The de Rham cohomology groups of a manifold 157
Exercise 5.1.viii. Let ๐ and ๐ be manifolds and ๐0, ๐1, ๐2 โถ ๐ โ ๐ three ๐ถโ maps. Showthat if ๐0 and ๐1 are homotopic and ๐1 and ๐2 are homotopic then ๐0 and ๐2 are homotopic.
Hint: The homotopy (5.1.7) has the property that๐น(๐, ๐ก) = ๐๐ก(๐) = ๐0(๐)
for 0 โค ๐ก โค 14 and๐น(๐, ๐ก) = ๐๐ก(๐) = ๐1(๐)
for 34 โค ๐ก < 1. Show that two homotopies with these properties: a homotopy between ๐0and ๐1 and a homotopy between ๐1 and ๐2, are easy to โglue togetherโ to get a homotopybetween ๐0 and ๐2.Exercise 5.1.ix.(1) Let๐ be an ๐-manifold. Given points ๐0, ๐1, ๐2 โ ๐, show that if ๐0 can be joined to ๐1
by a๐ถโ curve ๐พ0 โถ [0, 1] โ ๐, and๐1 can be joined to๐2 by a๐ถโ curve ๐พ1 โถ [0, 1] โ ๐,then ๐0 can be joined to ๐2 by a ๐ถโ curve, ๐พโถ [0, 1] โ ๐.
Hint: A๐ถโ curve, ๐พโถ [0, 1] โ ๐, joining ๐0 to ๐2 can be thought of as a homotopybetween the maps
๐พ๐0 โถ โ โ ๐ , โ โฆ ๐0and
๐พ๐1 โถ โ โ ๐ , โ โฆ ๐1where โ is the zero-dimensional manifold consisting of a single point.
(2) Show that if a manifold ๐ is connected it is arc-wise connected: any two points can byjoined by a ๐ถโ curve.
Exercise 5.1.x. Let๐ be a connected ๐-manifold and ๐ โ ๐บ1(๐) a closed one-form.(1) Show that if ๐พโถ [0, 1] โ ๐ is a ๐ถโ curve there exists a partition:
0 = ๐0 < ๐1 < โฏ < ๐๐ = 1of the interval [0, 1] and open sets ๐1,โฆ,๐๐ in๐ such that ๐พ ([๐๐โ1, ๐๐]) โ ๐๐ and suchthat ๐|๐๐ is exact.
(2) In part (1) show that there exist functions ๐๐ โ ๐ถโ(๐๐) such that ๐|๐๐ = ๐๐๐ and๐๐ (๐พ(๐๐)) = ๐๐+1 (๐พ(๐๐)).
(3) Show that if ๐0 and ๐1 are the end points of ๐พ
๐๐(๐1) โ ๐1(๐0) = โซ1
0๐พโ๐ .
(4) Let(5.1.26) ๐พ๐ โถ [0, 1] โ ๐ , 0 โค ๐ โค 1
be a homotopic family of curves with ๐พ๐ (0) = ๐0 and ๐พ๐ (1) = ๐1. Prove that the integral
โซ1
0๐พโ๐ ๐
is independent of ๐ 0.Hint: Let ๐ 0 be a point on the interval, [0, 1]. For ๐พ = ๐พ๐ 0 choose ๐๐โs and ๐๐โs as in
parts (a)โ(b) and show that for ๐ close to ๐ 0, ๐พ๐ [๐๐โ1, ๐๐] โ ๐๐.(5) A connected manifold ๐ is simply connected if for any two curves ๐พ0, ๐พ1 โถ [0, 1] โ ๐
with the same end-points ๐0 and ๐1, there exists a homotopy (5.1.22) with ๐พ๐ (0) = ๐0and ๐พ๐ (1) = ๐1, i.e., ๐พ0 can be smoothly deformed into ๐พ1 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. If๐ is simply-connected, then๐ป1(๐) = 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. Let๐ be a compact oriented 2๐-dimensional manifold. Show that themap
๐ตโถ ๐ป๐(๐) ร ๐ป๐(๐) โ ๐defined by
๐ต(๐1, ๐2) = ๐ผ๐(๐1 โ ๐2)is a bilinear form on๐ป๐(๐) and that ๐ต is symmetric if ๐ is even and alternating if ๐ is odd.
5.2. The MayerโVietoris Sequence
In this section weโll develop some techniques for computing cohomology groups ofmanifolds. (These techniques are known collectively as โdiagram chasingโ 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 theyโre applied.)Let ๐ถ0, ๐ถ1, ๐ถ2,โฆ be vector spaces and ๐โถ ๐ถ๐ โ ๐ถ๐+1 a linear map. The sequence of vectorspaces and maps
(5.2.1) ๐ถ0 ๐ถ1 ๐ถ2 โฏ .๐ ๐ ๐
is called a cochain complex, or simply a complex, if ๐2 = 0, i.e., if for ๐ โ ๐ถ๐, ๐(๐๐) = 0. Forinstance if๐ is a manifold the de Rham complex
(5.2.2) ๐บ0(๐) ๐บ1(๐) ๐บ2(๐) โฏ .๐ ๐ ๐
is an example of a complex, and the complex of compactly supported de Rham forms
(5.2.3) ๐บ0๐ (๐) ๐บ1๐ (๐) ๐บ2๐ (๐) โฏ .๐ ๐ ๐
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 ยง5.1. Let
๐๐ โ ker(๐โถ ๐ถ๐ โ ๐ถ๐+1) = { ๐ โ ๐ถ๐ | ๐๐ = 0 }and
๐ต๐ โ ๐๐ถ๐โ1
i.e., let ๐ be in ๐ต๐ if and only if ๐ = ๐๐ for some ๐ โ ๐ถ๐โ1. Then ๐๐ = ๐2๐ = 0, so ๐ต๐ is avector subspace of ๐๐, and we define ๐ป๐(๐ถ)โ the ๐th cohomology group of the complex(5.2.1) โ to be the quotient space
(5.2.4) ๐ป๐(๐ถ) โ ๐๐/๐ต๐ .Given ๐ โ ๐๐ we will, as in ยง5.1, denote its image in๐ป๐(๐ถ) by [๐] and weโll call ๐ a represen-tative of the cohomology class [๐].
We will next assemble a small dictionary of โdiagram chasingโ terms.
Definition 5.2.5. Let ๐0, ๐1, ๐2,โฆ be vector spaces and ๐ผ๐ โถ ๐๐ โ ๐๐+1 linear maps. Thesequence
๐0 ๐1 ๐2 โฏ .๐ผ0 ๐ผ1 ๐ผ2
is an exact sequence if, for each ๐, the kernel of ๐ผ๐+1 is equal to the image of ๐ผ๐.
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ยง5.2 The MayerโVietoris Sequence 159
For example the sequence (5.2.1) is exact if ๐๐ = ๐ต๐ for all ๐, or, in other words, if๐ป๐(๐ถ) = 0 for all ๐. A simple example of an exact sequence that weโll encounter a lot belowis a sequence of the form
0 ๐1 ๐2 ๐3 0 ,๐ผ1 ๐ผ2
i.e., a five term exact sequence whose first and last terms are the vector space, ๐0 = ๐4 = 0,and hence ๐ผ0 = ๐ผ3 = 0. This sequence is exact if and only if the following conditions hold:
(1) ๐ผ1 is injective,(2) the kernel of ๐ผ2 equals the image of ๐ผ1, and(3) ๐ผ2 is surjective.
We will call an exact sequence of this form a short exact sequence. (Weโll also encountera lot below an even shorter example of an exact sequence, namely a sequence of the form
0 ๐1 ๐2 0 .๐ผ1
This is an exact sequence if and only if ๐ผ1 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
๐ด ๐ต
๐ถ ๐ท
๐
๐
๐
๐
commutes if ๐ โ ๐ = ๐ โ ๐, and a more complicated diagram of vector spaces and linear mapslike the diagram
๐ด1 ๐ด2 ๐ด3
๐ต1 ๐ต2 ๐ต3
๐ถ1 ๐ถ2 ๐ถ3
commutes if every sub-square in the diagram commutes.We now have enough โdiagram chasingโ vocabulary to formulate the MayerโVietoris
theorem. For ๐ = 1, 2, 3 let
(5.2.6) ๐ถ0๐ ๐ถ1๐ ๐ถ2๐ โฏ .๐ ๐ ๐
be a complex and, for fixed ๐, let
(5.2.7) 0 ๐ถ๐1 ๐ถ๐2 ๐ถ๐3 0๐ ๐
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160 Chapter 5: Cohomology via forms
be a short exact sequence. Assume that the diagram below commutes:
โฎ โฎ โฎ
0 ๐ถ๐โ11 ๐ถ๐โ12 ๐ถ๐โ13 0
0 ๐ถ๐1 ๐ถ๐2 ๐ถ๐3 0
0 ๐ถ๐+11 ๐ถ๐+12 ๐ถ๐+13 0
โฎ โฎ โฎ .
๐
๐
๐
๐
๐
๐
๐
๐
๐
๐
๐ ๐
That is, assume that in the left hand squares, ๐๐ = ๐๐, and in the right hand squares, ๐๐ = ๐๐.The MayerโVietoris 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 ๐ and ๐ give rise to mappings between these cohomology groups. Namely, for๐ = 1, 2, 3 let ๐๐๐ be the kernel of the map ๐โถ ๐ถ๐๐ โ ๐ถ๐+1๐ , and ๐ต๐๐ the image of the map๐โถ ๐ถ๐โ1๐ โ ๐ถ๐๐ . Since ๐๐ = ๐๐, ๐maps ๐ต๐1 into ๐ต๐2 and ๐๐1 into ๐๐2 , therefore by (5.2.4) it givesrise to a linear mapping
๐โฏ โถ ๐ป๐(๐ถ1) โ ๐ป๐(๐ถ2) .Similarly since ๐๐ = ๐๐, ๐maps ๐ต๐2 into ๐ต๐3 and ๐๐2 into ๐๐3 , and so by (5.2.4) gives rise to alinear mapping
๐โฏ โถ ๐ป๐(๐ถ2) โ ๐ป๐(๐ถ3) .Moreover, since ๐ โ ๐ = 0 the image of ๐โฏ is contained in the kernel of ๐โฏ. Weโll leave as anexercise the following sharpened version of this observation:
Proposition 5.2.8. The kernel of ๐โฏ equals the image of ๐โฏ, i.e., the three term sequence
(5.2.9) ๐ป๐(๐ถ1) ๐ป๐(๐ถ2) ๐ป๐(๐ถ3)๐โฏ ๐โฏ
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 ๐โฏ is surjective. Let ๐๐3 be an element of ๐๐3 representing thecohomology class [๐๐3] in ๐ป3(๐ถ3). Since (5.2.7) is exact there exists a ๐๐2 in ๐ถ๐2 which getsmapped by ๐ onto ๐๐3 , and if ๐๐3 were in ๐๐2 this would imply
๐โฏ[๐๐2] = [๐๐๐2] = [๐๐3] ,
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ยง5.2 The MayerโVietoris Sequence 161
i.e., the cohomology class [๐๐3] would be in the image of ๐โฏ. However, since thereโs no reasonfor ๐๐2 to be in ๐๐2 , thereโs also no reason for [๐๐3] to be in the image of ๐โฏ. What we can say,however, is that ๐๐๐๐2 = ๐๐๐๐2 = ๐๐๐3 = 0 since ๐๐3 is in ๐๐3 . Therefore by the exactness of(5.2.7) in degree ๐ + 1 there exists a unique element ๐๐+11 in ๐ถ๐+11 with property(5.2.10) ๐๐๐2 = ๐๐๐+11 .Moreover, since
0 = ๐(๐๐๐2) = ๐๐(๐๐+11 ) = ๐ ๐๐๐+11and ๐ is injective, ๐๐๐+11 = 0, i.e.,(5.2.11) ๐๐+11 โ ๐๐+11 .Thus via (5.2.10) and (5.2.11) weโve converted an element ๐๐3 of ๐๐3 into an element ๐๐+11 of๐๐+11 and hence set up a correspondence(5.2.12) ๐๐3 โ ๐๐3 โ ๐๐+11 โ ๐๐+11 .Unfortunately this correspondence isnโt, strictly speaking, a map of ๐๐3 into ๐๐+11 ; the ๐๐1 in(5.2.12) isnโt determined by ๐๐3 alone but also by the choice we made of ๐๐2 . Suppose, how-ever, that we make another choice of a ๐๐2 with the property ๐(๐๐2) = ๐๐3 . Then the differencebetween our two choices is in the kernel of ๐ and hence, by the exactness of (5.2.6) at level๐, is in the image of ๐. In other words, our two choices are related by
(๐๐2)new = (๐๐2)old + ๐(๐๐1)for some ๐๐1 in ๐ถ๐1 , and hence by (5.2.10)
(๐๐+11 )new = (๐๐+11 )old + ๐๐๐1 .Therefore, even though the correspondence (5.2.12) isnโt strictly speaking amap it does giverise to a well-defined map
๐๐3 โ ๐ป๐+1(๐ถ1) , ๐๐3 โฆ [๐๐+13 ] .Moreover, if ๐๐3 is in ๐ต๐3 , i.e., ๐๐3 = ๐๐๐โ13 for some ๐๐โ13 โ ๐ถ๐โ13 , then by the exactness of (5.2.6)at level ๐โ1, ๐๐โ13 = ๐(๐๐โ12 ) for some ๐๐โ12 โ ๐ถ๐โ12 and hence ๐๐3 = ๐(๐๐๐โ22 ). In other words wecan take the ๐๐2 above to be ๐๐๐โ12 in which case the ๐๐+11 in equation (5.2.10) is just zero.Thusthe map (5.2.12) maps ๐ต๐3 to zero and hence by Proposition 1.2.9 gives rise to a well-definedmap
๐ฟโถ ๐ป๐(๐ถ3) โ ๐ป๐+1(๐ถ1)mapping [๐๐3] โ [๐๐+11 ]. We will leave it as an exercise to show that this mapping measuresthe failure of the arrow ๐โฏ 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 ๐โฏ โถ ๐ป๐(๐ถ2) โ ๐ป๐(๐ถ3) is equal to the kernel ofthe map ๐ฟโถ ๐ป๐(๐ถ3) โ ๐ป๐+1(๐ถ1).
Hint: Suppose that in the correspondence (5.2.12), ๐๐+11 is in ๐ต๐+11 . Then ๐๐+11 = ๐๐๐1 forsome ๐๐1 in ๐ถ๐1 . Show that ๐(๐๐2 โ ๐(๐๐1)) = ๐๐3 and ๐(๐๐2 โ ๐(๐๐1)) = 0, i.e., ๐๐2 โ ๐(๐๐1) is in ๐๐2 andhence ๐โฏ[๐๐2 โ ๐(๐๐1)] = [๐๐3].
Let us next explore the failure of themap ๐โฏ โถ ๐ป๐+1(๐ถ1) โ ๐ป๐+1(๐ถ2), to be injective. Let๐๐+11 be in ๐๐+11 and suppose that its cohomology class, [๐๐+11 ], gets mapped by ๐โฏ into zero.This translates into the statement(5.2.14) ๐(๐๐+11 ) = ๐๐๐2
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162 Chapter 5: Cohomology via forms
for some ๐๐2 โ ๐ถ๐2 . Moreover since ๐๐๐2 = ๐(๐๐+11 ), ๐(๐๐๐2) = 0. But if
(5.2.15) ๐๐3 โ ๐(๐๐2)then ๐๐๐3 = ๐๐(๐๐2) = ๐(๐๐๐2) = ๐(๐(๐๐+11 )) = 0, so ๐๐3 is in ๐๐3 , and by (5.2.14), (5.2.15) and thedefinition of ๐ฟ
[๐๐+11 ] = ๐ฟ[๐๐3] .In other words the kernel of themap ๐โฏ โถ ๐ป๐+1(๐ถ1) โ ๐ป๐+1(๐ถ2) is contained in the image ofthe map ๐ฟโถ ๐ป๐(๐ถ3) โ ๐ป๐+1(๐ถ1). 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 ๐ฟโถ ๐ป๐(๐ถ1) โ ๐ป๐+1(๐ถ1) is equal to the kernel ofthe map ๐โฏ โถ ๐ป๐+1(๐ถ1) โ ๐ป๐+1(๐ถ2).
Putting together the Propositions 5.2.8, 5.2.13 and 5.2.16 we obtain the main result ofthis section: the MayerโVietoris theorem. The sequence of cohomology groups and linearmaps
(5.2.17) โฏ ๐ป๐(๐ถ1) ๐ป๐(๐ถ2) ๐ป๐(๐ถ3) ๐ป๐+1(๐ถ1) โฏ๐ฟ ๐โฏ ๐โฏ ๐ฟ ๐โฏ
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 ๐-sphere ๐๐ in terms of the de Rham cohomologiesof ๐๐ โ {(0,โฆ, 0, 1)} and ๐๐ โ {(0,โฆ, 0, โ1)} via an exact sequence of the form (5.2.17).
Remark 5.2.19. In view of the โโฏโโs 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 ๐1 and ๐2, weโll denote by ๐1 โ ๐2 the vector space sum of ๐1and ๐2, i.e., the set of all pairs
(๐ข1, ๐ข2) , ๐ข๐ โ ๐๐with the addition operation
(๐ข1, ๐ข2) + (๐ฃ1 + ๐ฃ2) โ (๐ข1 + ๐ฃ1, ๐ข2 + ๐ฃ2)and the scalar multiplication operation
๐(๐ข1, ๐ข2) โ (๐๐ข1, ๐๐ข2) .Now let๐ be a manifold and let๐1 and๐2 be open subsets of๐. Then one has a linear map
๐ โถ ๐บ๐(๐1 โช ๐2) โ ๐บ๐(๐1) โ ๐บ๐(๐2)defined by(5.2.20) ๐ โฆ (๐|๐1 , ๐|๐2)where ๐|๐๐ is the restriction of ๐ to ๐๐. Similarly one has a linear map
๐โถ ๐บ๐(๐1) โ ๐บ๐(๐2) โ ๐บ๐(๐1 โฉ ๐2)defined by
(๐1, ๐2) โฆ ๐1|๐1โฉ๐2 โ ๐2|๐1โฉ๐2 .We claim:
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ยง5.2 The MayerโVietoris Sequence 163
Theorem 5.2.21. The sequence
0 ๐บ๐(๐1 โฉ ๐2) ๐บ๐(๐1) โ ๐บ๐(๐2) ๐บ๐(๐1 โฉ ๐2) 0 .๐ ๐
is a short exact sequence.
Proof. If the right hand side of (5.2.20) is zero, ๐ itself has to be zero so the map (5.2.20) isinjective. Moreover, if the right hand side of (5.2) is zero,๐1 and๐2 are equal on the overlap,๐1 โฉ ๐2, so we can glue them together to get a ๐ถโ ๐-form on ๐1 โช ๐2 by setting ๐ = ๐1 on๐1 and ๐ = ๐2 on๐2. Thus by (5.2.20) ๐(๐) = (๐1, ๐2), and this shows that the kernel of ๐ isequal to the image of ๐. Hence to complete the proof we only have to show that ๐ is surjective,i.e., that every form ๐ on๐บ๐(๐1โฉ๐2) can be written as a difference, ๐1|๐1โฉ๐2โ๐2|๐1โฉ๐2,where ๐1 is in๐บ๐(๐1) and ๐2 in in๐บ๐(๐2). To prove this weโll need the following variant ofthe partition of unity theorem.
Theorem 5.2.22. There exist functions, ๐๐ผ โ ๐ถโ(๐1 โช ๐2), ๐ผ = 1, 2, such that support ๐๐ผ iscontained in ๐๐ผ and ๐1 + ๐2 = 1.
Before proving this Let us use it to complete our proof of Theorem 5.2.21. Given ๐ โ๐บ๐(๐1 โฉ ๐2) let
(5.2.23) ๐1 โ {๐2๐, on ๐1 โฉ ๐20, on ๐1 โ ๐1 โฉ ๐2
and let
(5.2.24) ๐2 โ {โ๐1๐, on ๐1 โฉ ๐20, on ๐2 โ ๐1 โฉ ๐2 .
Since ๐2 is supported on ๐2 the form defined by (5.2.23) is ๐ถโ on ๐1 and since ๐1 is sup-ported on ๐1 the form defined by (5.2.24) is ๐ถโ on ๐2 and since ๐1 + ๐2 = 1, ๐1 โ ๐2 =(๐1 + ๐2)๐ = ๐ on ๐1 โฉ ๐2. โก
Proof of Theorem 5.2.22. Let ๐๐ โ ๐ถโ0 (๐1โช๐2), ๐ = 1, 2, 3,โฆ be a partition of unity subordi-nate to the cover {๐1, ๐2} of๐1 โช๐2 and let ๐1 be the sum of the ๐๐โs with support on๐1 and๐2 the sum of the remaining ๐๐โs. Itโs easy to check (using part (b) of Theorem 4.6.1) that ๐๐ผis supported in ๐๐ผ and (using part (a) of Theorem 4.6.1) that ๐1 + ๐2 = 1. โก
Now let
(5.2.25) 0 ๐ถ01 ๐ถ11 ๐ถ21 โฏ .๐ ๐ ๐
be the de Rham complex of ๐1 โช ๐2, let
(5.2.26) 0 ๐ถ03 ๐ถ13 ๐ถ23 โฏ .๐ ๐ ๐
be the de Rham complex of ๐1 โฉ ๐2, and let
(5.2.27) 0 ๐ถ02 ๐ถ12 ๐ถ22 โฏ .๐ ๐ ๐
be the vector space direct sum of the de Rham complexes of ๐1 and ๐2, i.e., the complexwhose ๐th term is
๐ถ๐2 = ๐บ๐(๐1) โ ๐บ๐(๐2)
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164 Chapter 5: Cohomology via forms
with ๐โถ ๐ถ๐2 โ ๐ถ๐+12 defined to be the map ๐(๐1, ๐2) = (๐๐1, ๐๐2). Since ๐ถ๐1 = ๐บ๐(๐1 โช ๐2)and ๐ถ๐3 = ๐บ๐(๐1 โฉ ๐2) we have, by Theorem 5.2.21, a short exact sequence
0 ๐ถ๐1 ๐ถ๐2 ๐ถ๐3 0 ,๐ ๐
and itโs easy to see that ๐ and ๐ commute with the ๐โs:๐๐ = ๐๐ and ๐๐ = ๐๐ .
Hence weโre exactly in the situation to which MayerโVietoris applies. Since the cohomologygroups of the complexes (5.2.25) and (5.2.26) are the de Rham cohomology groups๐ป๐(๐1โช๐2) and ๐ป๐(๐1 โฉ ๐2), and the cohomology groups of the complex (5.2.27) are the vectorspace direct sums,๐ป๐(๐1) โ ๐ป๐(๐2), we obtain from the abstract MayerโVietoris theorem,the following de Rham theoretic version of MayerโVietoris.
Theorem 5.2.28. Letting ๐ = ๐1 โช ๐2 and ๐ = ๐1 โฉ ๐2 one has a long exact sequence in deRham cohomology:
โฏ ๐ป๐(๐) ๐ป๐๐ (๐1) โ ๐ป๐๐ (๐2) ๐ป๐๐ (๐) ๐ป๐+1๐ (๐) โฏ .๐ฟ ๐โฏ ๐โฏ ๐ฟ ๐โฏ
This result also has an analogue for compactly supported de Rham cohomology. Let(5.2.29) ๐ โถ ๐บ๐๐ (๐1 โฉ ๐2) โ ๐ป๐๐ (๐1) โ ๐บ๐๐ (๐2)be the map
๐(๐) โ (๐1, ๐2)where
(5.2.30) ๐๐ โ {๐, on ๐1 โฉ ๐20, on ๐๐ โ ๐1 โฉ ๐2 .
(Since ๐ is compactly supported on ๐1 โฉ ๐2 the form defined by equation (5.2.29) is a ๐ถโform and is compactly supported on ๐๐.) Similarly, let
๐โถ ๐บ๐๐ (๐1) โ ๐บ๐๐ (๐2) โ ๐บ๐๐ (๐1 โช ๐2)be the map
๐(๐1, ๐2) โ ๏ฟฝ๏ฟฝ1 โ ๏ฟฝ๏ฟฝ2where:
๏ฟฝ๏ฟฝ๐ โ {๐๐, on ๐๐0, on (๐1 โช ๐2) โ ๐๐ .
As above itโs easy to see that ๐ is injective and that the kernel of ๐ is equal to the image of ๐.Thus if we can prove that ๐ is surjective weโll have proved
Theorem 5.2.31. The sequence
0 ๐บ๐๐ (๐1 โฉ ๐2) ๐บ๐๐ (๐1) โ ๐บ๐๐ (๐2) ๐บ๐๐ (๐1 โฉ ๐2) 0 .๐ ๐
is a short exact sequence.
Proof. To prove the surjectivity of ๐ we mimic the proof above. Given ๐ โ ๐บ๐๐ (๐1 โช ๐2), let๐1 โ ๐1๐|๐1
and๐2 โ โ๐2๐|๐2 .
Then by (5.2.30) we have that๐ = ๐(๐1, ๐2). โก
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ยง5.3 Cohomology of Good Covers 165
Thus, applying MayerโVietoris to the compactly supported versions of the complexes(5.2.6), we obtain the following theorem.
Theorem 5.2.32. Letting ๐ = ๐1 โช ๐2 and ๐ = ๐1 โฉ ๐2 there exists a long exact sequence incompactly supported de Rham cohomology
โฏ ๐ป๐๐ (๐) ๐ป๐๐ (๐1) โ ๐ป๐๐ (๐2) ๐ป๐๐ (๐) ๐ป๐+1๐ (๐) โฏ .๐ฟ ๐โฏ ๐โฏ ๐ฟ ๐โฏ
Exercises for ยง5.2Exercise 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 ๐1, ๐2 and ๐1 โฉ ๐2 are non-empty and connected the firstsegment of the MayerโVietoris sequence is a short exact sequence
0 ๐ป0(๐1 โช ๐2) ๐ป0(๐1) โ ๐ป0(๐2) ๐ป0(๐1 โฉ ๐2) 0 .๐โฏ ๐โฏ
Exercise 5.2.v. Let๐ = ๐๐ and let๐1 and๐2 be the open subsets of ๐๐ obtained by removingfrom ๐๐ the points, ๐1 = (0,โฆ, 0, 1) and ๐2 = (0,โฆ, 0, โ1).(1) Using stereographic projection show that ๐1 and ๐2 are diffeomorphic to ๐๐.(2) Show that๐1โช๐2 = ๐๐ and๐1โฉ๐2 is homotopy equivalent to ๐๐โ1. (See Exercise 5.1.v.)
Hint: ๐1 โฉ ๐2 is diffeomorphic to ๐๐ โ {0}.(3) Deduce from the MayerโVietoris sequence that๐ป๐+1(๐๐) = ๐ป๐(๐๐โ1) for ๐ โฅ 1.(4) Using part (3) give an inductive proof of a result that we proved by other means in
Section 5.1:๐ป๐(๐๐) = 0 for 1 โค ๐ < ๐.
Exercise 5.2.vi. Using the MayerโVietoris sequence of Exercise 5.2.v with cohomology re-placed by compactly supported cohomology show that
๐ป๐๐ (๐๐ โ {0}) โ {๐, ๐ = 1, ๐0, otherwise .
Exercise 5.2.vii. Let ๐ a positive integer and let
0 ๐1 ๐2 โฏ ๐๐โ1 ๐๐ 0๐1 ๐๐โ1
be an exact sequence of finite dimensional vector spaces. Prove that โ๐๐=1 dim(๐๐) = 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 ยง5.1 are finite dimensionalvector spaces and thus, in principle, โcomputableโ objects. A key ingredient in our proof ofthis fact is the notion of a good cover of a manifold.
Definition 5.3.1. Let๐ be an ๐-manifold, and let U = {๐๐ผ}๐ผโ๐ผ be an open cover of๐. ThenU is a good cover if for every finite set of indices ๐ผ1,โฆ, ๐ผ๐ โ ๐ผ the intersection๐๐ผ1 โฉโฏโฉ๐๐ผ๐is either empty or is diffeomorphic to ๐๐.
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 ๐๐.
Proposition 5.3.3. A bounded open convex subset ๐ โ ๐๐ is diffeomorphic to ๐๐.
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 ๐ of ๐๐ admits a good cover.
Proof. For each ๐ โ ๐ let ๐๐ be an open convex neighborhood of ๐ in ๐ (for instance an๐-ball centered at ๐). Since the intersection of any two convex sets is again convex the cover,{๐๐}๐โ๐ is a good cover by Proposition 5.3.3. โก
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 ๐ โ ๐๐ be an ๐-manifold and for ๐ โ ๐ let ๐๐๐be the tangent space to๐ at ๐. Recalling that ๐๐๐ sits inside ๐๐๐๐ and that
๐๐๐๐ = { (๐, ๐ฃ) | ๐ฃ โ ๐๐ }
we get a map๐๐๐ โช ๐๐๐๐ โ ๐๐ , (๐, ๐ฅ) โฆ ๐ + ๐ฅ ,
and this map maps ๐๐๐ bijectively onto an ๐-dimensional โaffineโ subspace ๐ฟ๐ of๐๐ whichis tangent to ๐ at ๐. Let ๐๐ โถ ๐ โ ๐ฟ๐ be, as in the figure below, the orthogonal projectionof๐ onto ๐ฟ๐.
๏ฟฝ
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ(๏ฟฝ)๏ฟฝ
Figure 5.3.1. The orthogonal projection of๐ onto ๐ฟ๐
Definition 5.3.5. An open subset ๐ of๐ is convex if for every ๐ โ ๐ the map ๐๐ โถ ๐ โ ๐ฟ๐maps ๐ diffeomorphically onto a convex open subset of ๐ฟ๐.
Itโs clear from this definition of convexity that the intersection of two open convex sub-sets of ๐ is an open convex subset of ๐ and that every open convex subset of ๐ is diffeo-morphic to ๐๐. Hence to prove Theorem 5.3.2 it suffices to prove that every point ๐ โ ๐ iscontained in an open convex subset ๐๐ of ๐. Here is a sketch of how to prove this. In thefigure above let ๐ต๐(๐) be the ball of radius ๐ about ๐ in ๐ฟ๐ centered at ๐. Since ๐ฟ๐ and ๐๐are tangent at ๐ the derivative of ๐๐ at ๐ is just the identity map, so for ๐ small ๐๐ maps aneighborhood, ๐๐๐ of ๐ in๐ diffeomorphically onto ๐ต๐(๐). We claim:
Proposition 5.3.6. For ๐ small, ๐๐๐ is a convex subset of๐.
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ยง5.3 Cohomology of Good Covers 167
Intuitively this assertion is pretty obvious: if ๐ is in ๐๐๐ and ๐ is small the map
๐ต๐๐ ๐๐๐ ๐ฟ๐๐โ1๐ ๐๐
is to order ๐2 equal to the identity map, so itโs intuitively clear that its image is a slightlywarped, but still convex, copy of ๐ต๐(๐). We wonโt, however, bother to write out the detailsthat are required to make this proof rigorous.
A good cover is a particularly good โgood coverโ if it is a finite cover. Weโll codify thisproperty in the definition below.
Definition 5.3.7. An ๐-manifold ๐ is said to have finite topology if ๐ admits a finite cov-ering by open sets ๐1,โฆ,๐๐ with the property that for every multi-index, ๐ผ = (๐1,โฆ, ๐๐),1 โค ๐1 โค ๐2โฏ < ๐๐พ โค ๐, the set(5.3.8) ๐๐ผ โ ๐๐1 โฉโฏ โฉ ๐๐๐is either empty or is diffeomorphic to ๐๐.
If ๐ is a compact manifold and U = {๐๐ผ}๐ผโ๐ผ is a good cover of ๐ then by the HeineโBorel theorem we can extract from U a finite subcover ๐1,โฆ,๐๐, where
๐๐ โ ๐๐ผ๐ , for ๐ผ1,โฆ, ๐ผ๐ โ ๐ผ ,hence we conclude:
Theorem 5.3.9. Every compact manifold has finite topology.
More generally, for any manifold ๐, let ๐ถ be a compact subset of ๐. Then by HeineโBorel we can extract from the cover U a finite subcollection ๐1,โฆ,๐๐, where
๐๐ โ ๐๐ผ๐ , for ๐ผ1,โฆ, ๐ผ๐ โ ๐ผ ,that covers ๐ถ, hence letting ๐ โ ๐1 โชโฏ โช ๐๐, weโve proved:
Theorem5.3.10. If๐ is an ๐-manifold and๐ถ a compact subset of๐, then there exists an openneighborhood ๐ of ๐ถ in๐ with finite topology.
We can in fact even strengthen this further. Let ๐0 be any open neighborhood of ๐ถ in๐. Then in the theorem above we can replace๐ by ๐0 to conclude:
Theorem5.3.11. Let๐ be amanifold,๐ถ a compact subset of๐, and๐0 an open neighborhoodof ๐ถ in ๐. Then there exists an open neighborhood ๐ of ๐ถ in ๐, ๐ contained in ๐0, so that ๐has finite topology.
We will justify the term โfinite topologyโ by devoting the rest of this section to proving.
Theorem 5.3.12. Let ๐ be an ๐-manifold. If ๐ has finite topology the de Rham cohomologygroups ๐ป๐(๐), for ๐ = 0,โฆ, ๐, and the compactly supported de Rham cohomology groups๐ป๐๐ (๐), for ๐ = 0,โฆ, ๐ are finite dimensional vector spaces.
The basic ingredients in the proof of this will be the MayerโVietoris techniques that wedeveloped in ยง5.2 and the following elementary result about vector spaces.
Lemma 5.3.13. Let ๐1, ๐2, and ๐3 be vector spaces and
(5.3.14) ๐1 ๐2 ๐3๐ผ ๐ฝ
an exact sequence of linear maps. Then if ๐1 and ๐3 are finite dimensional, so is ๐2.
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168 Chapter 5: Cohomology via forms
Proof. Since ๐3 is finite dimensional, the im(๐ฝ) is of finite dimension ๐. Hence there existvectors ๐ฃ1,โฆ, ๐ฃ๐ โ ๐2 such that
im(๐ฝ) = span(๐ฝ(๐ฃ1),โฆ, ๐ฝ(๐ฃ๐)) .
Now let ๐ฃ โ ๐2.Then ๐ฝ(๐ฃ) is a linear combination ๐ฝ(๐ฃ) = โ๐๐=1 ๐๐๐ฝ(๐ฃ๐), where ๐1,โฆ, ๐๐ โ๐. So
(5.3.15) ๐ฃโฒ โ ๐ฃ โ๐โ๐=1๐๐๐ฃ๐
is in the kernel of ๐ฝ and hence, by the exactness of (5.3.14), in the image of ๐ผ. But๐1 is finitedimensional, so im(๐ผ) is finite dimensional. Letting ๐ฃ๐+1,โฆ, ๐ฃ๐ be a basis of im(๐ผ) we canby (5.3.15)) write ๐ฃ as a sum ๐ฃ = โ๐๐=1 ๐๐๐ฃ๐. In other words ๐ฃ1,โฆ, ๐ฃ๐ is a basis of ๐2. โก
Proof of Theorem 5.3.4. Our proof will be by induction on the number of open sets in a goodcover of๐. More specifically, let
U = {๐1,โฆ,๐๐}
be a good cover of ๐. If๐ = 1, ๐ = ๐1 and hence ๐ is diffeomorphic to ๐๐, so๐ป๐(๐) = 0for ๐ > 0, and ๐ป0(๐) โ ๐, so the theorem is certainly true in this case. Let us now proveitโs true for arbitrary๐ by induction. Let ๐ โ ๐2 โชโฏ โช๐๐. Then ๐ is a submanifold of๐,and, thinking of๐ as a manifold in its own right, {๐2,โฆ,๐๐} is a good cover of๐ involvingonly๐โ1 sets. Hence the cohomology groups of๐ are finite dimensional by the inductionhypothesis. The manifold ๐ โฉ ๐1 has a good cover given by {๐ โฉ ๐2,โฆ,๐ โฉ ๐๐}, so by theinduction hypothesis the cohomology groups of๐โฉ๐1 are finite-dimensional. To prove thatthe theorem is true for ๐ we note that ๐ = ๐1 โช ๐, and and the MayerโVietoris sequencegives an exact sequence
๐ป๐โ1(๐1 โฉ ๐) ๐ป๐(๐) ๐ป๐(๐1) โ ๐ป๐(๐) .๐ฟ ๐โฏ
Since the right-hand and left-hand terms are finite dimensional it follows fromLemma5.3.13that the middle term is also finite dimensional. โก
The proof works practically verbatim for compactly supported cohomology. For๐ = 1
๐ป๐๐ (๐) = ๐ป๐๐ (๐1) = ๐ป๐๐ (๐๐)
so all the cohomology groups of๐ป๐(๐) are finite in this case, and the induction โ๐โ 1โโโ๐โ follows from the exact sequence
๐ป๐๐ (๐1) โ ๐ป๐๐ (๐) ๐ป๐๐ (๐) ๐ป๐+1๐ (๐1 โฉ ๐) .๐โฏ ๐ฟ
Remark 5.3.16. A careful analysis of the proof above shows that the dimensions of thevector spaces๐ป๐(๐) are determined by the intersection properties of the open sets ๐๐, i.e.,by the list of multi-indices ๐ผ 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 ยง5.8. (A nice account of it can also befound in [13, Ch. 5]).
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ยง5.3 Cohomology of Good Covers 169
Exercises for ยง5.3Exercise 5.3.i. Let ๐ be a bounded open subset of ๐๐. A continuous function
๐โถ ๐ โ [0,โ)is called an exhaustion function if it is proper as a map of ๐ into [0,โ); i.e., if, for every๐ > 0, ๐โ1([0, ๐]) is compact. For ๐ฅ โ ๐ let
๐(๐ฅ) = inf{ |๐ฅ โ ๐ฆ| | ๐ฆ โ ๐๐ โ ๐ } ,i.e., let ๐(๐ฅ) be the โdistanceโ from ๐ฅ to the boundary of๐. Show that ๐(๐ฅ) > 0 and that ๐(๐ฅ)is continuous as a function of ๐ฅ. Conclude that ๐0 = 1/๐ is an exhaustion function.
Exercise 5.3.ii. Show that there exists a๐ถโ exhaustion function, ๐0 โถ ๐ โ [0,โ), with theproperty ๐0 โฅ ๐20 where ๐0 is the exhaustion function in Exercise 5.3.i.
Hints: For ๐ = 2, 3,โฆ let
๐ถ๐ = { ๐ฅ โ ๐ | 1๐ โค ๐(๐ฅ) โค1๐โ1 }
and๐๐ = { ๐ฅ โ ๐ | 1๐+1 < ๐(๐ฅ) <
1๐โ2 } .
Let ๐๐ โ ๐ถโ0 (๐๐), ๐๐ โฅ 0, be a โbumpโ function which is identically one on ๐ถ๐ and let ๐0 =โโ๐=2 ๐2๐๐ + 1.
Exercise 5.3.iii. Let๐ be a bounded open convex subset of ๐๐ containing the origin. Showthat there exists an exhaustion function
๐โถ ๐ โ ๐ , ๐(0) = 1 ,having the property that ๐ is a monotonically increasing function of ๐ก along the ray, ๐ก๐ฅ,0 โค ๐ก โค 1, for all points ๐ฅ โ ๐.
Hints:โข Let ๐(๐ฅ), 0 โค ๐(๐ฅ) โค 1, be a ๐ถโ function which is one outside a small neighbor-
hood of the origin in ๐ and is zero in a still smaller neighborhood of the origin.Modify the function, ๐0, in the previous exercise by setting ๐(๐ฅ) = ๐(๐ฅ)๐0(๐ฅ) andlet
๐(๐ฅ) = โซ1
0๐(๐ ๐ฅ)๐๐ ๐ + 1 .
Show that for 0 โค ๐ก โค 1๐๐๐๐ก(๐ก๐ฅ) = ๐(๐ก๐ฅ)/๐ก
and conclude from equation (5.3.15) that๐ is monotonically increasing along theray, ๐ก๐ฅ, 0 โค ๐ก โค 1.
โข Show that for 0 < ๐ < 1,๐(๐ฅ) โฅ ๐๐(๐ฆ)
where ๐ฆ is a point on the ray, ๐ก๐ฅ, 0 โค ๐ก โค 1 a distance less than ๐|๐ฅ| from๐.โข Show that there exist constants, ๐ถ0 and ๐ถ1, ๐ถ1 > 0 such that
๐(๐ฅ) = ๐ถ1๐(๐ฅ)+ ๐ถ0 .
(Take ๐ to be equal to 12๐(๐ฅ)/|๐ฅ|.)
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170 Chapter 5: Cohomology via forms
Exercise 5.3.iv. Show that every bounded, open convex subset, ๐, of ๐๐ is diffeomorphicto ๐๐.
Hints:โข Let ๐(๐ฅ) be the exhaustion function constructed in Exercise 5.3.iii and let
๐โถ ๐ โ ๐๐
be the map: ๐(๐ฅ) = ๐(๐ฅ)๐ฅ. Show that this map is a bijective map of ๐ onto ๐๐.โข Show that for ๐ฅ โ ๐ and ๐ฃ โ ๐๐
(๐๐)๐ฅ๐ฃ = ๐(๐ฅ)๐ฃ + ๐๐๐ฅ(๐ฃ)๐ฅand conclude that ๐๐๐ฅ is bijective at ๐ฅ, i.e., that ๐ is locally a diffeomorphism of aneighborhood of ๐ฅ in ๐ onto a neighborhood of ๐(๐ฅ) in ๐๐.
โข Putting these together show that ๐ is a diffeomorphism of ๐ onto ๐๐.Exercise 5.3.v. Let ๐ โ ๐ be the union of the open intervals, ๐ < ๐ฅ < ๐ + 1, ๐ an integer.Show that ๐ does not have finite topology.
Exercise 5.3.vi. Let ๐ โ ๐2 be the open set obtained by deleting from ๐2 the points, ๐๐ =(0, ๐), for every integer ๐. Show that ๐ does not have finite topology.
Hint: Let ๐พ๐ be a circle of radius 12 centered about the point ๐๐. Using Exercises 2.1.viiiand 2.1.ix show that there exists a closed smooth one-form, ๐๐ on ๐ with the property thatโซ๐พ๐ ๐๐ = 1 and โซ๐พ๐ ๐๐ = 0 for๐ โ ๐.
Exercise 5.3.vii. Let๐ be an ๐-manifold and U = {๐1, ๐2} a good cover of๐. What are thecohomology groups of๐ if the nerve of this cover is:(1) {1}, {2}.(2) {1}, {2}, {1, 2}.Exercise 5.3.viii. Let ๐ be an ๐-manifold and U = {๐1, ๐2, ๐3} a good cover of ๐. Whatare the cohomology groups of๐ 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 ๐1 be the unit circle in๐3 parametrized by arc length: (๐ฅ, ๐ฆ) = (cos ๐, sin ๐).Let ๐1 be the set: 0 < ๐ < 2๐3 , ๐2 the set: ๐2 < ๐ <
3๐2 , and ๐3 the set: โ 2๐3 < ๐ <
๐3 .
(1) Show that the ๐๐โs are a good cover of ๐1.(2) Using the previous exercise compute the cohomology groups of ๐1.
Exercise 5.3.x. Let ๐2 be the unit 2-sphere in ๐3. Show that the sets๐๐ = { (๐ฅ1, ๐ฅ2, ๐ฅ3) โ ๐2 | ๐ฅ๐ > 0 }
๐ = 1, 2, 3 and๐๐ = { (๐ฅ1, ๐ฅ2, ๐ฅ3) โ ๐2 | ๐ฅ๐โ3 < 0 } ,
๐ = 4, 5, 6, are a good cover of ๐2. What is the nerve of this cover?
Exercise 5.3.xi. Let๐ and๐ be manifolds. Show that if they both have finite topology, theirproduct๐ ร ๐ does as well.
Exercise 5.3.xii.
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ยง5.4 Poincarรฉ duality 171
(1) Let๐ be amanifold and let๐1,โฆ,๐๐, be a good cover of๐. Show that๐1ร๐,โฆ,๐๐ร๐is a good cover of๐ ร ๐ and that the nerves of these two covers are the same.
(2) By Remark 5.3.16,๐ป๐(๐ ร ๐) โ ๐ป๐(๐) .
Verify this directly using homotopy techniques.(3) More generally, show that for all โ โฅ 0
๐ป๐(๐ ร ๐โ) โ ๐ป๐(๐)(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. Poincarรฉ duality
In this chapter weโve been studying two kinds of cohomology groups: the ordinary deRham cohomology groups๐ป๐ and the compactly supported de Rham cohomology groups๐ป๐๐ . It turns out that these groups are closely related. In fact if ๐ is a connected oriented๐-manifold and has finite topology, then ๐ป๐โ๐๐ (๐) is the vector space dual of ๐ป๐(๐). 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 ๐ and๐, a bilinear pairing between ๐ and๐is a map(5.4.1) ๐ตโถ ๐ ร๐ โ ๐which is linear in each of its factors. In other words, for fixed ๐ค โ ๐, the map
โ๐ค โถ ๐ โ ๐ , ๐ฃ โฆ ๐ต(๐ฃ, ๐ค)is linear, and for ๐ฃ โ ๐, the map
โ๐ฃ โถ ๐ โ ๐ , ๐ค โฆ ๐ต(๐ฃ, ๐ค)is linear. Therefore, from the pairing (5.4.1) one gets a map(5.4.2) ๐ฟ๐ต โถ ๐ โ ๐โ , ๐ค โฆ โ๐คand since โ๐ค1 + โ๐ค2(๐ฃ) = ๐ต(๐ฃ, ๐ค1 + ๐ค2) = โ๐ค1+๐ค2(๐ฃ), this map is linear. Weโll say that (5.4.1)is a non-singular pairing if (5.4.2) is bijective. Notice, by the way, that the roles of ๐ and๐ can be reversed in this definition. Letting ๐ตโฏ(๐ค, ๐ฃ) โ ๐ต(๐ฃ, ๐ค) we get an analogous linearmap
๐ฟ๐ตโฏ โถ ๐ โ ๐โ
and in fact(5.4.3) (๐ฟ๐ตโฏ(๐ฃ))(๐ค) = (๐ฟ๐ต(๐ค))(๐ฃ) = ๐ต(๐ฃ, ๐ค) .Thus if
๐โถ ๐ โ (๐โ)โ
is the canonical identification of ๐ with (๐โ)โ given by the recipe๐(๐ฃ)(โ) = โ(๐ฃ)
for ๐ฃ โ ๐ and โ โ ๐โ, we can rewrite equation (5.4.3) more suggestively in the form(5.4.4) ๐ฟ๐ตโฏ = ๐ฟโ๐ต๐i.e., ๐ฟ๐ต and ๐ฟ๐ตโฏ are just the transposes of each other. In particular ๐ฟ๐ต is bijective if and onlyif ๐ฟ๐ตโฏ is bijective.
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172 Chapter 5: Cohomology via forms
Let us now apply these remarks to de Rham theory. Let ๐ be a connected, oriented๐-manifold. If ๐ has finite topology the vector spaces, ๐ป๐โ๐๐ (๐) and ๐ป๐(๐) 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 of๐ป๐(๐) into the vector space dualof๐ป๐โ1๐ (๐). To see this let ๐1 be a cohomology class in๐ป๐โ๐๐ (๐) and ๐2 a cohomology classin๐ป๐(๐). Then by (5.1.22) their product ๐1 โ ๐2 is an element of๐ป๐๐ (๐), and so by (5.1.2) wecan define a pairing between ๐1 and ๐2 by setting(5.4.5) ๐ต(๐1, ๐2) โ ๐ผ๐(๐1 โ ๐2) .Notice that if ๐1 โ ๐บ๐โ๐๐ (๐) and ๐2 โ ๐บ๐(๐) are closed forms representing the cohomologyclasses ๐1 and ๐2, respectively, then by (5.1.22) this pairing is given by the integral
๐ต(๐1, ๐2) โ โซ๐๐1 โง ๐2 .
Weโll next show that this bilinear pairing is non-singular in one important special case:Proposition 5.4.6. If๐ is diffeomorphic to ๐๐ the pairing defined by (5.4.5) is non-singular.
Proof. To verify this there is very little to check. The vector spaces ๐ป๐(๐๐) and ๐ป๐โ๐๐ (๐๐)are zero except for ๐ = 0, so all we have to check is that the pairing
๐ป๐๐ (๐) ร ๐ป0(๐) โ ๐is non-singular. To see this recall that every compactly supported ๐-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
๐บ๐(๐) ร ๐ โ ๐ , (๐, ๐) โฆ ๐โซ๐๐ ,
and this is zero if and only if ๐ is zero or ๐ is in ๐๐บ๐โ1๐ (๐). Thus at the level of cohomologythis pairing is non-singular. โก
We will now show how to prove this result in general.Theorem 5.4.7 (Poincarรฉ duality). Let๐ be an oriented, connected ๐-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 if๐ has finite topology its de Rham cohomology groups are finite dimen-sional. Like that proof, it involves MayerโVietoris plus some elementary diagram-chasing.The โdiagram-chasingโ part of the proof consists of the following two lemmas.Lemma 5.4.8. Let ๐1, ๐2 and ๐3 be finite dimensional vector spaces, and let
๐1 ๐2 ๐3๐ผ ๐ฝ
be an exact sequence of linear mappings. Then the sequence of transpose maps
๐โ3 ๐โ2 ๐โ1๐ฝโ ๐ผโ
is exact.Proof. Given a vector subspace,๐2, of ๐2, let
๐โ2 = { โ โ ๐โ2 | โ(๐ค) = 0 for all ๐ค โ ๐} .Weโll leave for you to check that if๐2 is the kernel of ๐ฝ, then๐โ2 is the image of ๐ฝโ and thatif๐2 is the image of ๐ผ,๐โ2 is the kernel of ๐ผโ. Hence if ๐ฝ = im(๐ผ), im(๐ฝโ) = ker(๐ผโ). โก
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ยง5.4 Poincarรฉ duality 173
Lemma 5.4.9 (5-lemma). Consider a commutative diagram of vector spaces
(5.4.10)
๐ต1 ๐ต2 ๐ต3 ๐ต4 ๐ต5
๐ด1 ๐ด2 ๐ด3 ๐ด4 ๐ด5 .
๐ฝ1
๐พ1
๐ฝ2
๐พ2
๐ฝ3
๐พ3
๐ฝ4
๐พ4 ๐พ5
๐ผ1 ๐ผ2 ๐ผ3 ๐ผ4
satisfying the following conditions:(1) All the vector spaces are finite dimensional.(2) The two rows are exact.(3) The linear maps, ๐พ1, ๐พ2, ๐พ4, and ๐พ5 are bijections.Then the map ๐พ3 is a bijection.
Proof. Weโll show that ๐พ3 is surjective. Given ๐3 โ ๐ด3 there exists a ๐4 โ ๐ต4 such that๐พ4(๐4) = ๐ผ3(๐3) since ๐พ4 is bijective.Moreover, ๐พ5(๐ฝ4(๐4)) = ๐ผ4(๐ผ3(๐3)) = 0, by the exactnessof the top row. Therefore, since ๐พ5 is bijective, ๐ฝ4(๐4) = 0, so by the exactness of the bottomrow ๐4 = ๐ฝ3(๐3) for some ๐3 โ ๐ต3, and hence
๐ผ3(๐พ3(๐3)) = ๐พ4(๐ฝ3(๐3)) = ๐พ4(๐4) = ๐ผ3(๐3) .Thus ๐ผ3(๐3 โ ๐พ3(๐3)) = 0, so by the exactness of the top row
๐3 โ ๐พ3(๐3) = ๐ผ2(๐2)for some ๐2 โ ๐ด2. Hence by the bijectivity of ๐พ2 there exists a ๐2 โ ๐ต2 with ๐2 = ๐พ2(๐2), andhence
๐3 โ ๐พ3(๐3) = ๐ผ2(๐2) = ๐ผ2(๐พ2(๐2)) = ๐พ3(๐ฝ2(๐2)) .Thus finally
๐3 = ๐พ3(๐3 + ๐ฝ2(๐2)) .Since ๐3 was any element of ๐ด3 this proves the surjectivity of ๐พ3.
One can prove the injectivity of ๐พ3 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 ๐พโ3 โถ ๐ดโ3 โ๐ตโ3 . โก
To prove Theorem 5.4.7 we apply these lemmas to the diagram below. In this diagram๐1 and ๐2 are open subsets of ๐,๐ = ๐1 โช ๐2, we write ๐1,2 โ ๐1 โฉ ๐2, 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 โup to signโ. (To make it commutative one has to replacesome of the vertical arrows, ๐พ, by their negatives: โ๐พ.) This is easy to check except for thecommutative square on the extreme left. To check that this square commutes, some seriousdiagram-chasing is required.
โฏ ๐ป๐โ1๐ (๐) ๐ป๐๐ (๐1,2) ๐ป๐๐ (๐1) โ ๐ป๐๐ (๐2) ๐ป๐๐ (๐) โฏ
โฏ ๐ป๐โ(๐โ1)(๐)โ ๐ป๐โ๐(๐1,2)โ ๐ป๐โ๐(๐1)โ โ ๐ป๐โ๐(๐2)โ ๐ป๐โ๐(๐)โ โฏ
By MayerโVietoris the bottom row of the diagram is exact and by MayerโVietoris andLemma 5.4.8 the top row of the diagram is exact. Hence we can apply the โfive lemmaโ tothe diagram to conclude:
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174 Chapter 5: Cohomology via forms
Lemma 5.4.11. If the maps๐ป๐(๐) โ ๐ป๐โ๐๐ (๐)โ
defined by the pairing (5.4.5) are bijective for๐1, ๐2 and๐1,2 = ๐1โฉ๐2, they are also bijectivefor๐ = ๐1 โช ๐2.
Thus to prove Theorem 5.4.7 we can argue by induction as in ยง5.3. Let ๐1, ๐2,โฆ,๐๐be a good cover of ๐. If ๐ = 1, then ๐ = ๐1 and, hence, since ๐1 is diffeomorphic to๐๐, 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 ๐ open sets where ๐ is less than ๐. Let ๐โฒ =๐1 โชโฏ โช ๐๐โ1 and ๐โณ = ๐๐. Since
๐โฒ โฉ ๐โณ = ๐1 โฉ ๐๐ โชโฏ โช ๐๐โ1 โฉ ๐๐it can be covered by a good cover by ๐ open sets, ๐ < ๐, and hence the hypotheses of thelemma are true for ๐โฒ, ๐โณ and ๐โฒ โฉ ๐โณ. Thus the lemma says that (5.4.12) is bijective forthe union๐ of ๐โฒ and ๐โณ. โก
Exercises for ยง5.4Exercise 5.4.i (proper pushforward for compactly supported de Rham cohomology). Let๐ be an ๐-manifold, ๐ an ๐-manifold and ๐โถ ๐ โ ๐ a ๐ถโ map. Suppose that both ofthese manifolds are oriented and connected and have finite topology. Show that there existsa unique linear map
(5.4.12) ๐! โถ ๐ป๐โ๐๐ (๐) โ ๐ป๐โ๐๐ (๐)with the property(5.4.13) ๐ต๐(๐!๐1, ๐2) = ๐ต๐(๐1, ๐โฏ๐2)for all ๐1 โ ๐ป๐โ๐๐ (๐) and ๐2 โ ๐ป๐(๐). (In this formula ๐ต๐ is the bilinear pairing (5.4.5) on๐ and ๐ต๐ is the bilinear pairing (5.4.5) on ๐.)Exercise 5.4.ii. Suppose that the map ๐ in Exercise 5.4.i is proper. Show that there exists aunique linear map
๐! โถ ๐ป๐โ๐(๐) โ ๐ป๐โ๐(๐)with the property
๐ต๐(๐1, ๐!๐2) = (โ1)๐(๐โ๐)๐ต๐(๐โฏ๐1, ๐2)for all ๐1 โ ๐ป๐๐ (๐) and ๐2 โ ๐ป๐โ๐(๐), and show that, if๐ and ๐ are compact, this mappingis the same as the mapping ๐! in Exercise 5.4.i.
Exercise 5.4.iii. Let ๐ be an open subset of ๐๐ and let ๐โถ ๐ ร ๐ โ ๐ be the projection,๐(๐ฅ, ๐ก) = ๐ฅ. Show that there is a unique linear mapping
๐โ โถ ๐บ๐+1๐ (๐ ร ๐) โ ๐บ๐๐ (๐)with the property
(5.4.14) โซ๐๐โ๐ โง ๐ = โซ
๐ร๐๐ โง ๐โ๐
for all ๐ โ ๐บ๐+1๐ (๐ ร ๐) and ๐ โ ๐บ๐โ๐(๐).Hint: Let ๐ฅ1,โฆ, ๐ฅ๐ and ๐ก be the standard coordinate functions on ๐๐ ร ๐. By Exer-
cise 2.3.v every (๐ + 1)-form ๐ โ ๐บ๐+1๐ (๐ ร๐) can be written uniquely in โreduced formโ asa sum
๐ = โ๐ผ๐๐ผ๐๐ก โง ๐๐ฅ๐ผ +โ
๐ฝ๐๐ฝ๐๐ฅ๐ฝ
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ยง5.4 Poincarรฉ duality 175
over multi-indices, ๐ผ and ๐ฝ, which are strictly increasing. Let
๐โ๐ = โ๐ผ(โซ๐๐๐ผ(๐ฅ, ๐ก)๐๐ก) ๐๐ฅ๐ผ .
Exercise 5.4.iv. Show that the mapping ๐โ in Exercise 5.4.iii satisfies ๐โ๐๐ = ๐๐โ๐.Exercise 5.4.v. Show that if ๐ is a closed compactly supported (๐ + 1)-form on ๐ ร ๐ then
[๐โ๐] = ๐![๐]where ๐! is the mapping (5.4.13) and ๐โ the mapping (5.4.14).
Exercise 5.4.vi.(1) Let ๐ be an open subset of ๐๐ and let ๐โถ ๐ ร ๐โ โ ๐ be the projection, ๐(๐ฅ, ๐ก) = ๐ฅ.
Show that there is a unique linear mapping
๐โ โถ ๐บ๐+โ๐ (๐ ร ๐โ) โ ๐บ๐๐ (๐)with the property
โซ๐๐โ๐ โง ๐ = โซ
๐ร๐โ๐ โง ๐โ๐
for all ๐ โ ๐บ๐+โ๐ (๐ ร ๐โ) and ๐ โ ๐บ๐โ๐(๐).Hint: Exercise 5.4.iii plus induction on โ.
(2) Show that for ๐ โ ๐บ๐+โ๐ (๐ ร ๐โ)๐๐โ๐ = ๐โ๐๐ .
(3) Show that if ๐ is a closed, compactly supported (๐ + โ)-form on๐ ร ๐โ
[๐โ๐] = ๐![๐]where ๐! โถ ๐ป๐+โ๐ (๐ ร ๐โ) โ ๐ป๐๐ (๐) is the map (5.4.13).
Exercise 5.4.vii. Let ๐ be an ๐-manifold and ๐ an๐-manifold. Assume ๐ and ๐ are com-pact, oriented and connected, and orient ๐ ร ๐ by giving it its natural product orientation.Let
๐โถ ๐ ร ๐ โ ๐be the projection map ๐(๐ฅ, ๐ฆ) = ๐ฆ. Given
๐ โ ๐บ๐(๐ ร ๐)and ๐ โ ๐, let
(5.4.15) ๐โ๐(๐) โ โซ๐๐โ๐๐ ,
where ๐๐ โถ ๐ โ ๐ ร ๐ is the inclusion map ๐๐(๐ฅ) = (๐ฅ, ๐).(1) Show that the function ๐โ๐ defined by equation (5.4.15) is ๐ถโ, i.e., is in ๐บ0(๐).(2) Show that if ๐ is closed this function is constant.(3) Show that if ๐ is closed
[๐โ๐] = ๐![๐]where ๐! โถ ๐ป๐(๐ ร ๐) โ ๐ป0(๐) is the map (5.4.13).
Exercise 5.4.viii.(1) Let๐ be an ๐-manifoldwhich is compact, connected and oriented. CombiningPoincarรฉ
duality with Exercise 5.3.xii show that
๐ป๐+โ๐ (๐ ร ๐โ) โ ๐ป๐๐ (๐) .
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176 Chapter 5: Cohomology via forms
(2) Show, moreover, that if ๐โถ ๐ ร ๐โ โ ๐ is the projection ๐(๐ฅ, ๐) = ๐ฅ, then๐! โถ ๐ป๐+โ๐ (๐ ร ๐โ) โ ๐ป๐๐ (๐)
is a bijection.
Exercise 5.4.ix. Let ๐ and ๐ be as in Exercise 5.4.i. Show that the pushforward operation(5.4.13) satisfies the projection formula
๐!(๐1 โ ๐โฏ๐2) = ๐!(๐1) โ ๐2 ,for ๐1 โ ๐ป๐๐ (๐) and ๐2 โ ๐ปโ(๐).
5.5. Thom classes & intersection theory
Let ๐ be a connected, oriented ๐-manifold. If ๐ has finite topology its cohomologygroups are finite-dimensional, and since the bilinear pairing ๐ต defined by (5.4.5) is non-singular we get from this pairing bijective linear maps
(5.5.1) ๐ฟ๐ต โถ ๐ป๐โ๐๐ (๐) โ ๐ป๐(๐)โ
and(5.5.2) ๐ฟโ๐ต โถ ๐ป๐โ๐(๐) โ ๐ป๐๐ (๐)โ .In particular, if โโถ ๐ป๐(๐) โ ๐ is a linear function (i.e., an element of ๐ป๐(๐)โ), then by(5.5.1) we can convert โ into a cohomology class
(5.5.3) ๐ฟโ1๐ต (โ) โ ๐ป๐โ๐๐ (๐) ,and similarly if โ๐ โถ ๐ป๐๐ (๐) โ ๐ is a linear function, we can convert it by (5.5.2) into acohomology class
(๐ฟโ๐ต)โ1(โ) โ ๐ป๐โ๐(๐) .One way that linear functions like this arise in practice is by integrating forms over
submanifolds of ๐. Namely let ๐ be a closed oriented ๐-dimensional submanifold of ๐.Since ๐ is oriented, we have by (5.1.2) an integration operation in cohomology
๐ผ๐ โถ ๐ป๐๐ (๐) โ ๐ ,and since ๐ is closed the inclusion map ๐๐ of ๐ into๐ is proper, so we get from it a pullbackoperation on cohomology
๐โฏ๐ โถ ๐ป๐๐ (๐) โ ๐ป๐๐ (๐)and by composing these two maps, we get a linear map โ๐ = ๐ผ๐ โ ๐
โฏ๐ โถ ๐ป๐๐ (๐) โ ๐. The
cohomology class๐๐ โ ๐ฟโ1๐ต (โ๐) โ ๐ป๐๐ (๐)
associated with โ๐ is called the Thom class of the manifold ๐ and has the defining property
(5.5.4) ๐ต(๐๐, ๐) = ๐ผ๐(๐โฏ๐๐)
for ๐ โ ๐ป๐๐ (๐). Let us see what this defining property looks like at the level of forms. Let๐๐ โ ๐บ๐โ๐(๐) be a closed ๐-form representing ๐๐. Then by (5.4.5), the formula (5.5.4) for๐ = [๐], becomes the integral formula
(5.5.5) โซ๐๐๐ โง ๐ = โซ
๐๐โ๐๐ .
In other words, for every closed form ๐ โ ๐บ๐โ๐๐ (๐), the integral of ๐ over ๐ is equal tothe integral over๐ of ๐๐ โง ๐. A closed form ๐๐ with this โreproducingโ property is called aThom form for ๐. Note that if we add to ๐๐ an exact (๐ โ ๐)-form ๐ โ ๐๐บ๐โ๐โ1(๐), we get
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ยง5.5 Thom classes & intersection theory 177
another representative ๐๐ +๐ of the cohomology class ๐๐, 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 (๐ โ ๐)-form ๐๐ with the reproducing property (5.5.5) is arepresentative of the cohomology class ๐๐.
These remarks make sense as well for compactly supported cohomology. Suppose ๐ iscompact. Then from the inclusion map we get a pullback map
๐โฏ๐ โถ ๐ป๐(๐) โ ๐ป๐(๐)and since ๐ is compact, the integration operation ๐ผ๐ is a map๐ป๐(๐) โ ๐, so the composi-tion of these two operations is a map,
โ๐ โถ ๐ป๐(๐) โ ๐which by (5.5.3) gets converted into a cohomology class
๐๐ = ๐ฟโ1๐ต (โ๐) โ ๐ป๐โ๐๐ (๐) .Moreover, if ๐๐ โ ๐บ๐โ๐๐ (๐) is a closed form, it represents this cohomology class if and onlyif it has the reproducing property
(5.5.6) โซ๐๐๐ โง ๐ = โซ
๐๐โ๐๐
for closed forms๐ โ ๐บ๐โ๐(๐). (Thereโs a subtle difference, however, between formula (5.5.5)and formula (5.5.6). In (5.5.5) ๐ 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 ๐๐: we can add to it any element of๐๐บ๐โ๐โ1๐ (๐). One consequence of this is the following.
Theorem 5.5.7. Given a neighborhood ๐ of ๐ in ๐, there exists a closed form ๐๐ โ ๐บ๐โ๐๐ (๐)with the reproducing property
(5.5.8) โซ๐๐๐ โง ๐ = โซ
๐๐โ๐๐
for all closed forms ๐ โ ๐บ๐(๐).
Hence, in particular, ๐๐ has the reproducing property (5.5.6) for all closed forms ๐ โ๐บ๐โ๐(๐). This result shows that the Thom form ๐๐ can be chosen to have support in anarbitrarily small neighborhood of ๐.
Proof of Theorem 5.5.7. By Theorem 5.3.9 we can assume that ๐ has finite topology andhence, in our definition of ๐๐, we can replace the manifold ๐ by the open submanifold ๐.This gives us a Thom form ๐๐ with support in ๐ and with the reproducing property (5.5.8)for closed forms ๐ โ ๐บ๐โ๐(๐). โก
Let us see what Thom forms actually look like in concrete examples. Suppose ๐ is de-fined globally by a system of โ independent equations, i.e., suppose there exists an openneighborhood ๐ of ๐ in ๐, a ๐ถโ map ๐โถ ๐ โ ๐โ, and a bounded open convex neighbor-hood ๐ of the origin in ๐๐ satisfying the following properties.
Properties 5.5.9.(1) The origin is a regular value of ๐.(2) ๐โ1(๐) is closed in๐.(3) ๐ = ๐โ1(0).
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178 Chapter 5: Cohomology via forms
Then by (1) and (3) ๐ is a closed submanifold of ๐ and by (2) itโs a closed submanifoldof๐. Moreover, it has a natural orientation: For every ๐ โ ๐ the map
๐๐๐ โถ ๐๐๐ โ ๐0๐โ
is surjective, and its kernel is ๐๐๐, so from the standard orientation of ๐0๐โ one gets anorientation of the quotient space
๐๐๐/๐๐๐ ,and hence since ๐๐๐ is oriented, one gets by Theorem 1.9.9 an orientation on ๐๐๐. (SeeExample 4.4.5.) Now let ๐ be an element of ๐บโ๐(๐). Then ๐โ๐ is supported in ๐โ1(๐) andhence by property (2) we can extend it to ๐ by setting it equal to zero outside ๐. We willprove:
Theorem 5.5.10. If โซ๐ ๐ = 1, then ๐โ๐ is a Thom form for ๐.
To prove this weโll first prove that if ๐โ๐ has property (5.5.5) for some choice of ๐ it hasthis property for every choice of ๐.
Lemma 5.5.11. Let ๐1 and ๐2 be forms in๐บโ๐(๐) such that โซ๐ ๐1 = โซ๐ ๐2 = 1. Then for everyclosed ๐-form ๐ โ ๐บ๐๐ (๐) we have
โซ๐๐โ๐1 โง ๐ = โซ
๐๐โ๐2 โง ๐ .
Proof. By Theorem 3.2.2, ๐1 โ ๐2 = ๐๐ฝ for some ๐ฝ โ ๐บโโ1๐ (๐), hence, since ๐๐ = 0(๐โ๐1 โ ๐โ๐2) โง ๐ = ๐๐โ๐ฝ โง ๐ = ๐(๐โ๐ฝ โง ๐) .
Therefore, by Stokes theorem, the integral over๐ of the expression on the left is zero. โกNow suppose ๐ = ๐(๐ฅ1,โฆ, ๐ฅโ)๐๐ฅ1 โงโฏ โง ๐๐ฅโ, for ๐ in ๐ถโ0 (๐). For ๐ก โค 1 let
๐๐ก = ๐กโ๐ (๐ฅ1๐ก,โฆ, ๐ฅโ๐ก) ๐๐ฅ1 โงโฏ๐๐ฅโ .
This form is supported in the convex set ๐ก๐, so by Lemma 5.5.11
โซ๐๐โ๐๐ก โง ๐ = โซ
๐๐โ๐ โง ๐
for all closed forms ๐ โ ๐บ๐๐ (๐). Hence to prove that ๐โ๐ has the property (5.5.5), it sufficesto prove that
(5.5.12) lim๐กโ0โซ๐โ๐๐ก โง ๐ = โซ
๐๐โ๐๐ .
We prove this by proving a stronger result.
Lemma 5.5.13. The assertion (5.5.12) is true for every ๐-form ๐ โ ๐บ๐๐ (๐).Proof. The canonical sumbersion theorem (see Theorem b.17) says that for every ๐ โ ๐there exists a neighborhood๐๐ of ๐ in ๐, a neighborhood,๐ of 0 in ๐๐, and an orientationpreserving diffeomorphism ๐โถ (๐, 0) โ (๐๐, ๐) such that(5.5.14) ๐ โ ๐ = ๐where ๐โถ ๐๐ โ ๐โ is the canonical submersion, ๐(๐ฅ1,โฆ, ๐ฅ๐) = (๐ฅ1,โฆ, ๐ฅโ). Let U be thecover of ๐ by the open sets, ๐ โ ๐ and the ๐๐โs. Choosing a partition of unity subordinateto this cover it suffices to verify (5.5.12) for ๐ in ๐บ๐๐ (๐ โ ๐) and ๐ in ๐บ๐๐ (๐๐). Let us firstsuppose ๐ is in ๐บ๐๐ (๐ โ ๐). Then ๐(supp ๐) is a compact subset of ๐โ โ {0} and hence for ๐ก
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ยง5.5 Thom classes & intersection theory 179
small ๐(supp ๐) is disjoint from ๐ก๐, and both sides of (5.5.12) are zero. Next suppose that ๐is in ๐บ๐๐ (๐๐). Then ๐โ๐ is a compactly supported ๐-form on๐ so we can write it as a sum
๐โ๐ = โ๐ผโ๐ผ(๐ฅ)๐๐ฅ๐ผ , โ๐ผ โ ๐ถโ0 (๐)
the ๐ผโs being strictly increasing multi-indices of length ๐. Let ๐ผ0 = (โ + 1, โ2 + 2,โฆ, ๐). Then
(5.5.15) ๐โ๐๐ก โง ๐โ๐ = ๐กโ๐ (๐ฅ1๐ก,โฆ, ๐ฅโ๐ก) โ๐ผ0(๐ฅ1,โฆ, ๐ฅ๐)๐๐ฅ๐ โงโฏ๐๐ฅ๐
and by (5.5.14)๐โ(๐โ๐๐ก โง ๐) = ๐โ๐๐ก โง ๐โ๐
and hence since ๐ is orientation preserving
โซ๐๐โ๐๐ก โง ๐ = โซ
๐๐๐โ๐๐ก โง ๐ = ๐กโ โซ
๐๐๐ (๐ฅ1๐ก,โฆ, ๐ฅโ๐ก) โ๐ผ0(๐ฅ1,โฆ, ๐ฅ๐)๐๐ฅ
= โซ๐๐๐(๐ฅ1,โฆ, ๐ฅโ)โ๐ผ0(๐ก๐ฅ1,โฆ, ๐ก๐ฅโ, ๐ฅโ+1,โฆ, ๐ฅ๐)๐๐ฅ
and the limit of this expression as ๐ก tends to zero is
โซ๐(๐ฅ1,โฆ, ๐ฅโ)โ๐ผ0(0,โฆ, 0, ๐ฅโ+1,โฆ, ๐ฅ๐)๐๐ฅ1โฏ๐๐ฅ๐or
(5.5.16) โซโ๐ผ(0,โฆ, 0, ๐ฅโ+1,โฆ, ๐ฅ๐) ๐๐ฅโ+1โฏ๐๐ฅ๐ .
This, however, is just the integral of๐โ๐ over the set๐โ1(0)โฉ๐. By (5.5.12),๐maps๐โ1(0)โฉ๐ diffeomorphically onto ๐ โฉ ๐๐ and by our recipe for orienting ๐ this diffeomorphism isan orientation-preserving diffeomorphism, so the integral (5.5.16) is equal to โซ๐ ๐. โก
Weโll now describe some applications of Thom forms to topological intersection theory.Let ๐ and ๐ be closed, oriented submanifolds of ๐ of dimensions ๐ and โ where ๐ + โ = ๐,and Let us assume one of them (say ๐) is compact. We will show below how to define anintersection number ๐ผ(๐, ๐), which on the one hand will be a topological invariant of ๐and ๐ and on the other hand will actually count, with appropriate ยฑ-signs, the number ofpoints of intersection of ๐ and ๐ when they intersect non-tangentially. (Thus this notion issimilar to the notion of the degree deg(๐) for a ๐ถโ mapping ๐. On the one hand deg(๐)is a topological invariant of ๐. Itโs unchanged if we deform ๐ by a homotopy. On the otherhand if ๐ is a regular value of๐, then deg(๐) counts with appropriate ยฑ-signs the number ofpoints in the set ๐โ1(๐).)
Weโll first give the topological definition of this intersection number. This is by the for-mula(5.5.17) ๐ผ(๐, ๐) = ๐ต(๐๐, ๐๐)where ๐๐ โ ๐ปโ(๐) and ๐๐ โ ๐ป๐๐ (๐) and ๐ต is the bilinear pairing (5.4.5). If ๐๐ โ ๐บโ(๐)and ๐๐ โ ๐บ๐๐ (๐) are Thom forms representing ๐๐ and ๐๐, (5.5.17) can also be defined as theintegral
๐ผ(๐, ๐) = โซ๐๐๐ โง ๐๐
or by (5.5.8), as the integral over ๐,
(5.5.18) ๐ผ(๐, ๐) = โซ๐๐โ๐๐๐
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180 Chapter 5: Cohomology via forms
or, since ๐๐ โง ๐๐ = (โ1)๐โ๐๐ โง ๐๐, as the integral over ๐
๐ผ(๐, ๐) = (โ1)๐โ โซ๐๐โ๐๐๐ .
In particular๐ผ(๐, ๐) = (โ1)๐โ๐ผ(๐, ๐).
As a test case for our declaring ๐ผ(๐, ๐) to be the intersection number of๐ and๐we will firstprove:
Proposition 5.5.19. If ๐ and ๐ do not intersect, then ๐ผ(๐, ๐) = 0.
Proof. If๐ and๐ donโt intersect then, since๐ is closed,๐ = ๐โ๐ is an open neighborhoodof ๐ in ๐, therefore since ๐ is compact there exists by Theorem 5.5.7 a Thom form ๐๐ โ๐บโ๐(๐). Thus ๐โ๐๐๐ = 0, and so by (5.5.18) we have ๐ผ(๐, ๐) = 0. โก
Weโll next indicate howone computes ๐ผ(๐, ๐)when๐ and๐ intersect โnon-tangentiallyโ,or, to use terminology more in current usage, when their intersection is transversal. Recallthat at a point of intersection, ๐ โ ๐ โฉ ๐, ๐๐๐ and ๐๐๐ are vector subspaces of ๐๐๐.
Definition 5.5.20. ๐ and ๐ intersect transversally if for every ๐ โ ๐ โฉ ๐ we have๐๐๐ โฉ ๐๐๐ = 0 .
Since ๐ = ๐ + โ = dim๐๐๐ + dim๐๐๐ = dim๐๐๐, this condition is equivalent to
(5.5.21) ๐๐๐ = ๐๐๐ โ ๐๐๐ ,
i.e., every vector, ๐ข โ ๐๐๐, can be written uniquely as a sum, ๐ข = ๐ฃ + ๐ค, with ๐ฃ โ ๐๐๐ and๐ค โ ๐๐๐. Since ๐, ๐ and ๐ are oriented, their tangent spaces at ๐ are oriented, and weโllsay that these spaces are compatibly oriented if the orientations of the two sides of (5.5.21)agree. (In other words if ๐ฃ1,โฆ, ๐ฃ๐ is an oriented basis of ๐๐๐ and ๐ค1,โฆ,๐คโ is an orientedbasis of๐๐๐, the ๐ vectors, ๐ฃ1,โฆ, ๐ฃ๐,๐ค1,โฆ,๐คโ, are an oriented basis of๐๐๐.)Wewill definethe local intersection number ๐ผ๐(๐, ๐) of ๐ and ๐ at ๐ to be equal to +1 if ๐, ๐, and ๐ arecompatibly oriented at ๐ and to be equal to โ1 if theyโre not. With this notation weโll prove:
Theorem 5.5.22. If ๐ and ๐ intersect transversally then ๐ โฉ ๐ is a finite set and
๐ผ(๐, ๐) = โ๐โ๐โฉ๐๐ผ๐(๐, ๐) .
To prove this we first need to show that transverse intersections look nice locally.
Theorem 5.5.23. If ๐ and ๐ intersect transversally, then for every ๐ โ ๐ โฉ ๐, there existsan open neighborhood ๐๐ of ๐ in ๐, an open neighborhood ๐๐ of the origin in ๐๐, and anorientation-preserving diffeomorphism ๐๐ โถ ๐๐ โฅฒ ๐๐ which maps ๐๐ โฉ ๐ diffeomorphicallyonto the subset of ๐๐ defined by the equations: ๐ฅ1 = โฏ = ๐ฅโ = 0, and maps ๐ โฉ ๐ onto thesubset of ๐๐ defined by the equations: ๐ฅโ+1 = โฏ = ๐ฅ๐ = 0.
Proof. Since this result is a local result, we can assume that ๐ = ๐๐ and hence by Theo-rem 4.2.15 that there exists a neighborhood ๐๐ of ๐ in ๐๐ and submersions ๐โถ (๐๐, ๐) โ(๐โ, 0) and ๐โถ (๐๐, ๐) โ (๐๐, 0) with the properties
(5.5.24) ๐๐ โฉ ๐ = ๐โ1(0)and(5.5.25) ๐๐ โฉ ๐ = ๐โ1(0) .
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ยง5.5 Thom classes & intersection theory 181
Moreover, by (4.3.5) ๐๐๐ = (๐๐๐)โ1(0) and ๐๐๐ = (๐๐๐)โ1(0). Hence by (5.5.21), the equa-tions(5.5.26) ๐๐๐(๐ฃ) = ๐๐๐(๐ฃ) = 0for ๐ฃ โ ๐๐๐ imply that ๐ฃ = 0. Now let ๐๐ โถ ๐๐ โ ๐๐ be the map
(๐, ๐) โถ ๐๐ โ ๐โ ร ๐๐ = ๐๐ .Then by (5.5.26), ๐๐๐ is bijective, therefore, shrinking ๐๐ if necessary, we can assume that๐๐ maps ๐๐ diffeomorphically onto a neighborhood ๐๐ of the origin in ๐๐, and hence by(5.5.24) and (5.5.25)), ๐๐ maps๐๐ โฉ๐ onto the set: ๐ฅ1 = โฏ = ๐ฅโ = 0 and maps๐๐ โฉ๐ ontothe set: ๐ฅโ+1 = โฏ = ๐ฅ๐ = 0. Finally, if ๐ isnโt orientation preserving, we can make it so bycomposing it with the involution, (๐ฅ1,โฆ, ๐ฅ๐) โฆ (๐ฅ1, ๐ฅ2,โฆ, ๐ฅ๐โ1, โ๐ฅ๐). โก
From this result we deduce:
Theorem 5.5.27. If ๐ and ๐ intersect transversally, their intersection is a finite set.
Proof. By Theorem 5.5.23 the only point of intersection in ๐๐ is ๐ itself. Moreover, since ๐is closed and ๐ is compact, ๐ โฉ ๐ is compact. Therefore, since the ๐๐โs cover ๐ โฉ ๐ we canextract a finite subcover by the HeineโBorel theorem. However, since no two ๐๐โs cover thesame point of ๐ โฉ ๐, this cover must already be a finite subcover. โก
We will now prove Theorem 5.5.22.
Proof of Theorem 5.5.22. Since ๐ is closed, the map ๐๐ โถ ๐ โช ๐ is proper, so by Theo-rem 3.4.7 there exists a neighborhood๐ of๐ in๐ such that๐โฉ๐ is contained in the unionof the open sets ๐๐ above. Moreover by Theorem 5.5.7 we can choose ๐๐ to be supportedin ๐ and by Theorem 5.3.2 we can assume that ๐ has finite topology, so weโre reduced toproving the theorem with๐ replaced by ๐ and ๐ replaced by ๐ โฉ ๐. Let
๐ โ ๐ โฉ โ๐โ๐๐๐ ,
let ๐โถ ๐ โ ๐โ be the map whose restriction to ๐๐ โฉ ๐ is ๐ โ ๐๐ where ๐ is, as in equa-tion (5.5.14), the canonical submersion of ๐๐ onto ๐โ, and finally let ๐ be a bounded con-vex neighborhood of ๐โ, whose closure is contained in the intersection of the open sets,๐ โ ๐๐(๐๐ โฉ ๐). Then ๐โ1(๐) is a closed subset of ๐, so if we replace ๐ by ๐ and ๐ by๐โฉ๐, the data (๐, ๐, ๐) satisfy Properties 5.5.9. Thus to prove Theorem 5.5.22 it suffices byTheorem 5.5.10 to prove this theorem with
๐๐ = ๐๐(๐)๐โ๐on ๐๐ โฉ ๐ where ๐๐(๐) = +1 or โ1 depending on whether the orientation of ๐ โฉ ๐๐ inTheorem 5.5.10 coincides with the given orientation of ๐ or not. Thus
๐ผ(๐, ๐) = (โ1)๐โ๐ผ(๐, ๐)
= (โ1)๐โ โ๐โ๐๐๐(๐) โซ
๐๐โ๐๐โ๐
= (โ1)๐โ โ๐โ๐๐๐(๐) โซ
๐๐โ๐๐โ๐๐โ๐
= โ๐โ๐(โ1)๐โ๐๐(๐) โซ
๐โฉ๐๐(๐ โ ๐๐ โ ๐๐)โ๐ .
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But ๐ โ ๐๐ โ ๐๐ maps an open neighborhood of ๐ in ๐๐ โฉ ๐ diffeomorphically onto ๐, and๐ is compactly supported in ๐, so since โซ๐ ๐ = 1,
โซ๐โฉ๐๐(๐ โ ๐๐ โ ๐๐)โ๐ = ๐๐(๐) โซ
๐๐ = ๐๐(๐) ,
where ๐๐(๐) = +1 or โ1 depending on whether ๐ โ ๐๐ โ ๐๐ is orientation preserving or not.Thus finally
๐ผ(๐, ๐) = โ๐โ๐(โ1)๐โ๐๐(๐)๐๐(๐) .
We will leave as an exercise the task of unraveling these orientations and showing that
(โ1)๐โ๐๐(๐)๐๐(๐) = ๐ผ๐(๐, ๐)
and hence that ๐ผ(๐, ๐) = โ๐โ๐ ๐ผ๐(๐, ๐). โก
Exercises for ยง5.5Exercise 5.5.i. Let๐be a connected oriented๐-manifold,๐ a connected oriented โ-dimensionalmanifold, ๐โถ ๐ โ ๐ a ๐ถโ map, and ๐ a closed submanifold of๐ of dimension ๐ โ ๐ โ โ.Suppose ๐ is a โlevel setโ of the map ๐, i.e., suppose that ๐ is a regular value of ๐ and that๐ = ๐โ1(๐). Show that if ๐ is in ๐บโ๐(๐) and its integral over ๐ is 1, then one can orient ๐ sothat ๐๐ = ๐โ๐ is a Thom form for ๐.
Hint: Theorem 5.5.10.
Exercise 5.5.ii. In Exercise 5.5.i show that if ๐ โ ๐ is a compact oriented โ-dimensionalsubmanifold of๐ then
๐ผ(๐, ๐) = (โ1)๐โ deg(๐ โ ๐๐) .
Exercise 5.5.iii. Let ๐1 be another regular value of the map ๐โถ ๐ โ ๐ and let ๐1 = ๐โ1(๐).Show that
๐ผ(๐, ๐) = ๐ผ(๐1, ๐) .
Exercise 5.5.iv.(1) Show that if ๐ is a regular value of the map ๐ โ ๐๐ โถ ๐ โ ๐, then ๐ and ๐ intersect
transversally.(2) Show that this is an โif and only if โ proposition: If ๐ and ๐ intersect transversally, then๐ is a regular value of the map ๐ โ ๐๐.
Exercise 5.5.v. Suppose ๐ is a regular value of the map ๐ โ ๐๐. Show that ๐ is in ๐โฉ๐ if andonly if ๐ is in the preimage (๐ โ ๐๐)โ1(๐) of ๐ and that
๐ผ๐(๐, ๐) = (โ1)๐โ๐๐where ๐๐ is the orientation number of the map ๐ โ ๐๐, at ๐, i.e., ๐๐ = 1 if ๐ โ ๐๐ is orientation-preserving at ๐ and ๐๐ = โ1 if ๐ โ ๐๐ is orientation-reversving at ๐.
Exercise 5.5.vi. Suppose the map ๐โถ ๐ โ ๐ is proper. Show that there exists a neighbor-hood, ๐, of ๐ in๐ having the property that all points of ๐ are regular values of ๐.
Hint: Since ๐ is a regular value of ๐ there exists, for every ๐ โ ๐โ1(๐) a neighborhood๐๐ of ๐ on which ๐ is a submersion. Conclude, by Theorem 3.4.7, that there exists a neigh-borhood ๐ of ๐ with ๐โ1(๐) โ โ๐โ๐โ1(๐)๐๐.
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ยง5.6 The Lefschetz Theorem 183
Exercise 5.5.vii. Show that in every neighborhood๐1 of ๐ in๐ there exists a point ๐1 whosepreimage
๐1 โ ๐โ1(๐1)intersects ๐ transversally. Conclude that one can โdeform ๐ an arbitrarily small amount sothat it intersects ๐ transversallyโ.
Hint: Exercise 5.5.iv plus Sardโs theorem.
Exercise 5.5.viii (Intersection theory for mappings). Let ๐ be an oriented, connected ๐-manifold, ๐ a compact, oriented โ-dimensional submanifold, ๐ an oriented manifold ofdimension ๐ โ ๐ โ โ and ๐โถ ๐ โ ๐ a proper ๐ถโ map. Define the intersection number of๐ with ๐ to be the integral
๐ผ(๐, ๐) โ โซ๐๐โ๐๐ .
(1) Show that ๐ผ(๐, ๐) is a homotopy invariant of ๐, i.e., show that if ๐0, ๐1 โถ ๐ โ ๐ areproper ๐ถโ maps and are properly homotopic, then
๐ผ(๐0, ๐) = ๐ผ(๐1, ๐) .(2) Show that if ๐ is a closed submanifold of ๐ of dimension ๐ = ๐ โ โ and ๐๐ โถ ๐ โ ๐ is
the inclusion map, then ๐ผ(๐๐, ๐) = ๐ผ(๐, ๐).
Exercise 5.5.ix.(1) Let ๐ be an oriented connected ๐-manifold and let ๐ be a compact zero-dimensional
submanifold consisting of a single point ๐ง0 โ ๐. Show that if ๐ is in ๐บ๐๐ (๐) then ๐ is aThom form for ๐ if and only if its integral is 1.
(2) Let ๐ be an oriented ๐-manifold and ๐โถ ๐ โ ๐ a ๐ถโ map. Show that for ๐ = {๐ง0} asin part (1) we have ๐ผ(๐, ๐) = deg(๐).
5.6. The LefschetzTheorem
In this section weโll apply the intersection techniques that we developed in ยง5.5 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 ยง3.6, told us thata ๐ถโ 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 ๐ถโ 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 ๐ be an oriented com-pact ๐-manifold and ๐โถ ๐ โ ๐ a ๐ถโ map. Define the graph of ๐ in ๐ ร ๐ to be theset
๐ค๐ โ { (๐ฅ, ๐(๐ฅ)) | ๐ฅ โ ๐ } โ ๐ ร ๐ .Itโs easy to see that this is an ๐-dimensional submanifold of๐ ร๐ and that this manifold isdiffeomorphic to๐ itself. In fact, in one direction, there is a ๐ถโ map
(5.6.1) ๐พ๐ โถ ๐ โ ๐ค๐ , ๐ฅ โฆ (๐ฅ, ๐(๐ฅ)) ,and, in the other direction, a ๐ถโ map
๐โถ ๐ค๐ โ ๐ , (๐ฅ, ๐(๐ฅ)) โฆ ๐ฅ ,and itโs obvious that these maps are inverses of each other and hence diffeomorphisms. Wewill orient ๐ค๐ by requiring that ๐พ๐ and ๐ be orientation-preserving diffeomorphisms.
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An example of a graph is the graph of the identity map of ๐ onto itself. This is thediagonal in๐ ร ๐
๐ฅ โ { (๐ฅ, ๐ฅ) | ๐ฅ โ ๐ } โ ๐ ร ๐and its intersection with ๐ค๐ is the set
๐ค๐ โฉ ๐ฅ = { (๐ฅ, ๐ฅ) | ๐(๐ฅ) = ๐ฅ } ,which is just the set of fixed points of๐. Hence a natural way to count the fixed points of๐ isas the intersection number of ๐ค๐ and ๐ฅ in๐ร๐. To do so we need these three manifolds tobe oriented, but, as we noted above, ๐ค๐ and ๐ฅ acquire orientations from the identifications(5.6.1) and, as for๐ร๐, weโll give it its natural orientation as a product of orientedmanifolds.(See ยง4.5.)
Definition 5.6.2. The global Lefschetz number of ๐ is the intersection number
๐ฟ(๐) โ ๐ผ(๐ค๐, ๐ฅ) .
In this section weโll 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 ๐โถ ๐ โ ๐ is a Lefschetz map, or simply Lefschetz, if ๐ค๐ and ๐ฅintersect transversally.
Let us see what being Lefschetz entails. Suppose๐ is a fixed point of๐.Then at the point๐ = (๐, ๐) of ๐ค๐(5.6.4) ๐๐(๐ค๐) = (๐๐พ๐)๐๐๐๐ = { (๐ฃ, ๐๐๐(๐ฃ)) | ๐ฃ โ ๐๐๐}and, in particular, for the identity map,
๐๐(๐ฅ) = { (๐ฃ, ๐ฃ) | ๐ฃ โ ๐๐๐} .Therefore, if ๐ฅ and ๐ค๐ are to intersect transversally, the intersection of ๐๐(๐ค๐) โฉ๐๐(๐ฅ) inside๐๐(๐ ร ๐) has to be the zero space. In other words if
(5.6.5) (๐ฃ, ๐๐๐(๐ฃ)) = (๐ฃ, ๐ฃ)then ๐ฃ = 0. But the identity (5.6.5) says that ๐ฃ is a fixed point of ๐๐๐, so transversality at ๐amounts to the assertion
๐๐๐(๐ฃ) = ๐ฃ โบ ๐ฃ = 0 ,or in other words the assertion that the map
(5.6.6) (id๐๐๐ โ๐๐๐) โถ ๐๐๐ โ ๐๐๐
is bijective.
Proposition5.6.7. The local intersection number ๐ผ๐(๐ค๐, ๐ฅ) is 1 if (5.6.6) is orientation-preservingand โ1 otherwise.
In other words ๐ผ๐(๐ค๐, ๐ฅ) is the sign of det(id๐๐๐ โ๐๐๐).
Proof. To prove this let ๐1,โฆ, ๐๐ be an oriented basis of ๐๐๐ and let
(5.6.8) ๐๐๐(๐๐) =๐โ๐=1๐๐,๐๐๐ .
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ยง5.6 The Lefschetz Theorem 185
Now set ๐ฃ๐ โ (๐๐, 0) โ ๐๐(๐ ร ๐) and ๐ค๐ โ (0, ๐๐) โ ๐๐(๐ ร ๐). Then by the definition ofthe product orientation on ๐ ร ๐, we have that (๐ฃ1,โฆ, ๐ฃ๐, ๐ค1,โฆ,๐ค๐) is an oriented basisof ๐๐(๐ ร ๐), and by (5.6.4)
(๐ฃ1 + โ๐๐=1 ๐๐,๐๐ค๐,โฆ, ๐ฃ๐ + โ
๐๐=1 ๐๐,๐๐ค๐)
is an oriented basis of ๐๐๐ค๐ and ๐ฃ1 + ๐ค1,โฆ, ๐ฃ๐ + ๐ค๐ is an oriented basis of ๐๐๐ฅ. Thus๐ผ๐(๐ค๐, ๐ฅ) = ยฑ1, depending on whether or not the basis
(๐ฃ1 + โ๐๐=1 ๐๐,๐๐ค๐,โฆ, ๐ฃ๐ + โ
๐๐=1 ๐๐,๐๐ค๐, ๐ฃ1 + ๐ค1,โฆ, ๐ฃ๐ + ๐ค๐)
of ๐๐(๐ ร ๐) is compatibly oriented with the basis (5.6.8). Thus ๐ผ๐(๐ค๐, ๐ฅ) = ยฑ1, the signdepending on whether the determinant of the 2๐ ร 2๐matrix relating these two bases:
๐ โ (id๐ ๐ดid๐ id๐)
is positive or negative, where๐ด = (๐๐,๐). However, by the block matrix determinant formula,we see that to see that
det(๐) = det(id๐ โ๐ด idโ1๐ id๐) det(id๐) = det(id๐ โ๐ด) ,hence by (5.6.8) we see that det(๐) = det(id๐ โ๐๐๐). (If you are not familiar with this, it iseasy to see that by elementary row operations ๐ can be converted into the matrix
(id๐ ๐ด0 id๐ โ๐ด
)
apply Exercise 1.8.vii.) โก
Let us summarize what we have shown so far.
Theorem 5.6.9. A ๐โถ ๐ โ ๐, is a Lefschetz map if and only if for every fixed point ๐ of ๐,the map
id๐๐๐ โ๐๐๐ โถ ๐๐๐ โ ๐๐๐is bijective. Moreover for Lefschetz maps ๐ we have
๐ฟ(๐) = โ๐=๐(๐)๐ฟ๐(๐)
where ๐ฟ๐(๐) = +1 if id๐๐๐ โ๐๐๐ is orientation-preserving and ๐ฟ๐(๐) = โ1 if id๐๐๐ โ๐๐๐ isorientation-reversing.
Weโll next describe how to compute ๐ฟ(๐) as a topological invariant of ๐. Let ๐๐ค๐ be theinclusion map of ๐ค๐ into ๐ ร ๐ and let ๐๐ฅ โ ๐ป๐(๐ ร ๐) be the Thom class of ๐ฅ. Then by(5.5.18)
๐ฟ(๐) = ๐ผ๐ค๐(๐โ๐ค๐๐๐ฅ)
and hence since themapping ๐พ๐ โถ ๐ โ ๐ร๐ defined by (5.6.1) is an orientation-preservingdiffeomorphism of๐ โฅฒ ๐ค๐ we have
(5.6.10) ๐ฟ(๐) = ๐ผ๐(๐พโ๐๐๐ฅ) .To evaluate the expression on the right weโll need to know some facts about the cohomologygroups of product manifolds.Themain result on this topic is the Kรผnneth theorem, and weโlltake up the formulation and proof of this theorem in ยง5.7. First, however, weโll describe aresult which follows from the Kรผnneth theorem and which will enable us to complete ourcomputation of ๐ฟ(๐).
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186 Chapter 5: Cohomology via forms
Let ๐1 and ๐2 be the projection of๐ร๐ onto its first and second factors, i.e., for ๐ = 1, 2let
๐๐ โถ ๐ ร ๐ โ ๐be the projection ๐๐(๐ฅ1, ๐ฅ2) โ ๐ฅ๐. Then by equation (5.6.1) we have
(5.6.11) ๐1 โ ๐พ๐ = id๐and
(5.6.12) ๐2 โ ๐พ๐ = ๐ .
Lemma 5.6.13. If ๐1 and ๐2 are in ๐บ๐(๐) then
(5.6.14) โซ๐ร๐๐โ1๐1 โง ๐โ2๐2 = (โซ
๐๐1)(โซ
๐๐2) .
Proof. By a partition of unity argument we can assume that ๐๐ has compact support in aparametrizable open set, ๐๐. Let ๐๐ be an open subset of ๐๐ and ๐๐ โถ ๐๐ โ ๐๐ an orientation-preserving diffeomorphism. Then
๐โ๐ ๐ = ๐๐๐๐ฅ1 โงโฏ โง ๐๐ฅ๐with ๐๐ โ ๐ถโ0 (๐๐), so the right hand side of (5.6.14) is the product of integrals over ๐๐:
(5.6.15) (โซ๐๐๐1(๐ฅ)๐๐ฅ)(โซ
๐๐๐2(๐ฅ)๐๐ฅ) .
Moreover, since๐ ร ๐ is oriented by its product orientation, the map
๐โถ ๐1 ร ๐2 โ ๐1 ร ๐2is given by (๐ฅ, ๐ฆ) โฆ (๐1(๐ฅ) ,๐2(๐ฆ)) is an orientation-preserving diffeomorphism and since๐๐ โ ๐ = ๐๐
๐โ(๐โ1๐1 โง ๐โ2๐2) = ๐โ1๐1 โง ๐โ2๐2= ๐1(๐ฅ)๐2(๐ฆ)๐๐ฅ1 โงโฏ โง ๐๐ฅ๐ โง ๐๐ฆ1 โงโฏ โง ๐๐ฆ๐
and hence the left hand side of (5.6.14) is the integral over ๐2๐ of the function, ๐1(๐ฅ)๐2(๐ฆ),and therefore, by integration by parts, is equal to the product (5.6.15). โก
As a corollary of this lemma we get a product formula for cohomology classes.
Lemma 5.6.16. If ๐1 and ๐2 are in๐ป๐(๐) then๐ผ๐ร๐(๐โ1 ๐1 โ ๐โ2 ๐2) = ๐ผ๐(๐1)๐ผ๐(๐2) .
Now let ๐๐ โ dim๐ป๐(๐) and note that since ๐ is compact, Poincarรฉ duality tells usthat ๐๐ = ๐โ when โ = ๐ โ ๐. In fact it tells us even more. Let ๐๐1,โฆ, ๐๐๐๐ be a basis of๐ป๐(๐).Then, since the pairing (5.4.5) is non-singular, there exists for โ = ๐ โ ๐ a โdualโ basis
๐โ๐ , ๐ = 1,โฆ, ๐โof๐ปโ(๐) satisfying(5.6.17) ๐ผ๐(๐๐๐ โ ๐โ๐ ) = ๐ฟ๐,๐ .
Lemma 5.6.18. The cohomology classes
๐โฏ1๐โ๐ โ ๐โฏ2๐๐๐ , ๐ + โ = ๐
for ๐ = 0,โฆ, ๐ and 1 โค ๐, ๐ โค ๐๐, are a basis for๐ป๐(๐ ร ๐).
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ยง5.6 The Lefschetz Theorem 187
This is the corollary of the Kรผnneth theorem that we alluded to above (and whose proofweโll give in ยง5.7). Using these results we prove the following.
Theorem 5.6.19. The Thom class ๐๐ฅ โ ๐ป๐(๐ ร ๐) is given explicitly by the formula
(5.6.20) ๐๐ฅ = โ๐+โ=๐(โ1)โ
๐๐โ๐=1๐โฏ1๐๐๐ โ ๐
โฏ2๐โ๐ .
Proof. Wehave to check that for every cohomology class ๐ โ ๐ป๐(๐ร๐), the class๐๐ฅ definedby (5.6.20) has the reproducing property
(5.6.21) ๐ผ๐ร๐(๐๐ฅ โ ๐) = ๐ผ๐ฅ(๐โฏ๐ฅ๐)
where ๐๐ฅ is the inclusion map of ๐ฅ into๐ ร ๐. However the map
๐พ๐ฅ โถ ๐ โ ๐ ร ๐ , ๐ฅ โฆ (๐ฅ, ๐ฅ)is an orientation-preserving diffeomorphism of๐ onto ๐ฅ, so it suffices to show that
(5.6.22) ๐ผ๐ร๐(๐๐ฅ โ ๐) = ๐ผ๐(๐พโฏ๐ฅ๐)
and by Lemma 5.6.18 it suffices to verify (5.6.22) for ๐โs of the form
๐ = ๐โฏ1๐โ๐ โ ๐โฏ2๐๐๐ .
The product of this class with a typical summand of (5.6.20), for instance, the summand
(5.6.23) (โ1)โโฒ๐โฏ1๐๐โฒ๐ โ ๐โฏ2๐โโฒ๐ , ๐โฒ + โโฒ = ๐ ,
is equal, up to sign to,๐โฏ1๐๐
โฒ๐ โ ๐โ๐ โ ๐
โฏ2๐๐๐ โ ๐โ
โฒ๐ .
Notice, however, that if ๐ โ ๐โฒ this product is zero: For ๐ < ๐โฒ, ๐โฒ + โ is greater than ๐ + โand hence greater than ๐. Therefore
๐๐โฒ๐ โ ๐โ๐ โ ๐ป๐โฒ+โ(๐)
is zero since ๐ is of dimension ๐, and for ๐ > ๐โฒ, โโฒ is greater than โ and ๐๐๐ โ ๐โโฒ๐ is zero for
the same reason. Thus in taking the product of ๐๐ฅ with ๐ we can ignore all terms in the sumexcept for the terms, ๐โฒ = ๐ and โโฒ = โ. For these terms, the product of (5.6.23) with ๐ is
(โ1)๐โ๐โฏ1๐๐๐ โ ๐โ๐ โ ๐โฏ2๐๐3 โ ๐โ๐ .
(Exercise: Check this. Hint: (โ1)โ(โ1)โ2 = 1.) Thus
๐๐ฅ โ ๐ = (โ1)๐โ๐๐โ๐=1๐โฏ1๐๐๐ โ ๐โ๐ โ ๐
โฏ2๐๐๐ โ ๐โ๐
and hence by Lemma 5.6.13 and (5.6.17)
๐ผ๐ร๐(๐๐ฅ โ ๐) = (โ1)๐โ๐๐โ๐=1๐ผ๐(๐๐๐ โ ๐โ๐)๐ผ๐(๐๐๐ โ ๐โ๐ )
= (โ1)๐โ๐๐โ๐=1๐ฟ๐,๐๐ฟ๐,๐
= (โ1)๐โ๐ฟ๐,๐ .
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188 Chapter 5: Cohomology via forms
On the other hand for ๐ = ๐โฏ1๐โ๐ โ ๐โฏ2๐๐๐ ๐พโฏ๐ฅ๐ = ๐พ
โฏ๐ฅ๐โฏ1๐โ๐ โ ๐พ
โฏ๐ฅ๐โฏ2๐๐๐
= (๐1 โ ๐พ๐ฅ)โฏ๐โ๐(๐2 โ ๐พ๐ฅ)โฏ๐๐๐ = ๐โ๐ โ ๐๐๐
since๐1 โ ๐๐ฅ = ๐2 โ ๐พ๐ฅ = id๐ .
So๐ผ๐(๐พโฏ๐ฅ๐) = ๐ผ๐(๐โ๐ โ ๐๐๐ ) = (โ1)๐โ๐ฟ๐,๐
by (5.6.17). Thus the two sides of (5.6.21) are equal. โก
Weโre now in position to compute ๐ฟ(๐) , i.e., to compute the expression ๐ผ๐(๐พโ๐๐๐ฅ) onthe right hand side of (5.6.10). Since ๐โ1,โฆ, ๐โ๐โ is a basis of๐ปโ(๐) the linear mapping
(5.6.24) ๐โฏ โถ ๐ปโ(๐) โ ๐ปโ(๐)can be described in terms of this basis by a matrix [๐โ๐,๐] with the defining property
๐โฏ๐โ๐ =๐โโ๐=1๐โ๐,๐๐โ๐ .
Thus by equations (5.6.11), (5.6.12) and (5.6.20) we have
๐พโฏ๐๐๐ฅ = ๐พโฏ๐ โ๐+โ=๐(โ1)โ
๐๐โ๐=1๐โฏ1๐๐๐ โ ๐
โฏ2๐โ๐
= โ๐+โ=๐(โ1)โ
๐๐โ๐=1(๐1 โ ๐พ๐)โฏ๐๐๐ โ (๐2 โ ๐๐)โฏ๐โ๐
= โ๐+โ=๐(โ1)โ
๐๐โ๐=1๐๐๐ โ ๐โฏ๐โ๐
= โ๐+โ=๐(โ1)โ
๐๐โ๐=1๐โ๐,๐๐๐๐ โ ๐โ๐ .
Thus by (5.6.17)
๐ผ๐(๐พโฏ๐๐๐ฅ) = โ
๐+โ=๐(โ1)โ โ
1โค๐,๐โค๐โ๐โ๐,๐๐ผ๐(๐๐๐ โ ๐โ๐ )
= โ๐+โ=๐(โ1)โ โ
1โค๐,๐โค๐โ๐โ๐,๐๐ฟ๐,๐
=๐โโ=0(โ1)โ
๐โโ๐=1๐โ๐,๐ .
But โ๐โ๐=1 ๐โ๐,๐ 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 ๐ฟ(๐).
Theorem 5.6.25. The Lefschetz number ๐ฟ(๐) is the alternating sum
๐ฟ(๐) =๐โโ=0(โ1)โ tr(๐โฏ | ๐ปโ(๐))
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ยง5.6 The Lefschetz Theorem 189
where tr(๐โฏ | ๐ปโ(๐)) is the trace of the map ๐โฏ โถ ๐ปโ(๐) โ ๐ปโ(๐).Exercises for ยง5.6
Exercise 5.6.i. Show that if ๐0 โถ ๐ โ ๐ and ๐1 โถ ๐ โ ๐ are homotopic ๐ถโ mappings๐ฟ(๐0) = ๐ฟ(๐1).Exercise 5.6.ii.(1) The Euler characteristic ๐(๐) of๐ is defined to be the intersection number of the diag-
onal with itself in๐ ร ๐, i.e., the โself-intersectionโ number๐(๐) โ ๐ผ(๐ฅ, ๐ฅ) = ๐ผ๐ร๐(๐๐ฅ, ๐๐ฅ) .
Show that if a ๐ถโ map ๐โถ ๐ โ ๐ is homotopic to the identity, then ๐ฟ(๐) = ๐(๐).(2) Show that
๐(๐) =๐โโ=0(โ1)โ dim๐ปโ(๐) .
(3) Show that if๐ is an odd-dimensional (compact) manifold, then ๐(๐) = 0.Exercise 5.6.iii.(1) Let ๐๐ be the unit ๐-sphere in ๐๐+1. Show that if ๐โถ ๐๐ โ ๐๐ is a ๐ถโ map, then
๐ฟ(๐) = 1 + (โ1)๐ deg(๐) .(2) Conclude that if deg(๐) โ (โ1)๐+1, then ๐ has to have a fixed point.Exercise 5.6.iv. Let ๐ be a ๐ถโ mapping of the closed unit ball ๐ต๐+1 into itself and let๐โถ ๐๐ โ ๐๐ be the restriction of ๐ to he boundary of ๐ต๐+1. Show that if deg(๐) โ (โ1)๐+1then the fixed point of ๐ predicted by Brouwerโs theorem can be taken to be a point on theboundary of ๐ต๐+1.Exercise 5.6.v.(1) Show that if ๐โถ ๐๐ โ ๐๐ is the antipodal map ๐(๐ฅ) โ โ๐ฅ, then deg(๐) = (โ1)๐+1.(2) Conclude that the result in Exercise 5.6.iv is sharp. Show that the map
๐โถ ๐ต๐+1 โ ๐ต๐+1 , ๐(๐ฅ) = โ๐ฅ ,has only one fixed point, namely the origin, and in particular has no fixed points on theboundary.
Exercise 5.6.vi. Let ๐ be a vector field on a compact manifold ๐. Since ๐ is compact, ๐generates a one-parameter group of diffeomorphisms(5.6.26) ๐๐ก โถ ๐ โ ๐ , โโ < ๐ก < โ .(1) Let ๐ด๐ก be the set of fixed points of ๐๐ก. Show that this set contains the set of zeros of ๐,
i.e., the points ๐ โ ๐ where ๐(๐) = 0.(2) Suppose that for some ๐ก0, ๐๐ก0 is Lefschetz. Show that for all ๐ก, ๐๐ก maps ๐ด๐ก0 into itself.(3) Show that for |๐ก| < ๐, ๐ small, the points of ๐ด๐ก0 are fixed points of ๐๐ก.(4) Conclude that ๐ด๐ก0 is equal to the set of zeros of ๐ฃ.(5) In particular, conclude that for all ๐ก the points of ๐ด๐ก0 are fixed points of ๐๐ก.Exercise 5.6.vii.(1) Let ๐ be a finite dimensional vector space and
๐น(๐ก) โถ ๐ โ ๐ , โโ < ๐ก < โa one-parameter group of linear maps of ๐ onto itself. Let ๐ด = ๐๐น๐๐ก (0) and show that๐น(๐ก) = exp ๐ก๐ด. (See Exercise 2.2.viii.)
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190 Chapter 5: Cohomology via forms
(2) Show that if id๐ โ๐น(๐ก0) โถ ๐ โ ๐ is bijective for some ๐ก0, then ๐ดโถ ๐ โ ๐ is bijective.Hint: Show that if ๐ด๐ฃ = 0 for some ๐ฃ โ ๐ โ {0}, ๐น(๐ก)๐ฃ = ๐ฃ.
Exercise 5.6.viii. Let ๐ be a vector field on a compact manifold ๐ and let (5.6.26) be theone-parameter group of diffeomorphisms generated by ๐. If ๐(๐) = 0 then by part (1) ofExercise 5.6.vi, ๐ is a fixed point of ๐๐ก for all ๐ก.(1) Show that
(๐๐๐ก) โถ ๐๐๐ โ ๐๐๐is a one-parameter group of linear mappings of ๐๐๐ onto itself.
(2) Conclude from Exercise 5.6.vii that there exists a linear map(5.6.27) ๐ฟ๐(๐)โถ ๐๐๐ โ ๐๐๐
with the propertyexp ๐ก๐ฟ๐(๐) = (๐๐๐ก)๐ .
Exercise 5.6.ix. Suppose ๐๐ก0 is a Lefschetz map for some ๐ก0. Let ๐ = ๐ก0/๐ where ๐ is apositive integer. Show that ๐๐ is a Lefschetz map.
Hints:โข Show that
๐๐ก0 = ๐๐ โ โฏ โ ๐๐ = ๐๐๐
(i.e., ๐๐ composed with itself๐ times).โข Show that if ๐ is a fixed point of ๐๐, it is a fixed pointof ๐๐ก0 .โข Conclude from Exercise 5.6.vi that the fixed points of ๐๐ are the zeros of ๐ฃ.โข Show that if ๐ is a fixed point of ๐๐,
(๐๐๐ก0)๐ = (๐๐๐)๐๐ .
โข Conclude that if (๐๐๐)๐๐ฃ = ๐ฃ for some ๐ฃ โ ๐๐๐ โ {0}, then (๐๐๐ก0)๐๐ฃ = ๐ฃ.Exercise 5.6.x. Show that for all ๐ก, ๐ฟ(๐๐ก) = ๐(๐).
Hint: Exercise 5.6.ii.
Exercise 5.6.xi (Hopf Theorem). A vector field ๐ on a compact manifold ๐ is a Lefschetzvector field if for some ๐ก0 โ ๐ the map ๐๐ก0 is a Lefschetz map.(1) Show that if ๐ is a Lefschetz vector field then it has a finite number of zeros and for each
zero ๐ the linear map (5.6.27) is bijective.(2) For a zero ๐ of ๐ let ๐๐(๐) = +1 if the map (5.6.27) is orientation-preserving and let๐๐(๐) = โ1 if itโs orientation-reversing. Show that
๐(๐) = โ๐(๐)=0๐๐(๐) .
Hint: Apply the Lefschetz theorem to ๐๐, ๐ = ๐ก0/๐,๐ large.
Exercise 5.6.xii (review of the trace). For ๐ด = (๐๐,๐) an ๐ ร ๐matrix define
tr(๐ด) โ๐โ๐=1๐๐,๐ .
(1) Show that if ๐ด and ๐ต are ๐ ร ๐matricestr(๐ด๐ต) = tr(๐ต๐ด) .
(2) Show that if ๐ต is an invertible ๐ ร ๐matrixtr(๐ต๐ด๐ตโ1) = tr(๐ด) .
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ยง5.7 The Kรผnneth theorem 191
(3) Let ๐ be and ๐-dimensional vector space and ๐ฟโถ ๐ โ ๐ a linear map. Fix a basis๐ฃ1,โฆ, ๐ฃ๐ of ๐ and define the trace of ๐ฟ to be the trace of ๐ด where ๐ด is the definingmatrix for ๐ฟ in this basis, i.e.,
๐ฟ๐ฃ๐ =๐โ๐=1๐๐,๐๐ฃ๐ .
Show that this is an intrinsic definition not depending on the basis ๐ฃ1,โฆ, ๐ฃ๐.
5.7. The Kรผnneth theorem
Let ๐ be an ๐-manifold and ๐ an ๐-dimensional manifold, both of these manifoldshaving finite topology. Let
๐โถ ๐ ร ๐ โ ๐be the projection map ๐(๐ฅ, ๐ฆ) = ๐ฅ and
๐โถ ๐ ร ๐ โ ๐
the projectionmap (๐ฅ, ๐ฆ) โ ๐ฆ. Since๐ and๐ have finite topology their cohomology groupsare finite dimensional vector spaces. For 0 โค ๐ โค ๐ let
๐๐๐ , 1 โค ๐ โค dim๐ป๐(๐) ,
be a basis of๐ป๐(๐) and for 0 โค โ โค ๐ let
๐โ๐ , 1 โค ๐ โค dim๐ปโ(๐)
be a basis of ๐ปโ(๐). Then for ๐ + โ = ๐ the product, ๐โฏ๐๐๐ โ ๐โฏ๐โ๐ , is in ๐ป๐(๐ ร ๐). TheKรผnneth theorem asserts
Theorem 5.7.1. The product manifold ๐ ร ๐ has finite topology and hence the cohomologygroups๐ป๐(๐ ร ๐) are finite dimensional. Moreover, the products over ๐ + โ = ๐
๐โฏ๐๐๐ โ ๐โฏ๐โ๐ , 0 โค ๐ โค dim๐ป๐(๐) , 0 โค ๐ โค dim๐ปโ(๐) ,(5.7.2)
are a basis for the vector space๐ป๐(๐ ร ๐).
The fact that๐ร๐ has finite topology is easy to verify. If๐1,โฆ,๐๐, is a good cover of๐ and ๐1,โฆ,๐๐, is a good cover of ๐ the products of these open sets๐๐ ร๐๐, for 1 โค ๐ โค ๐and 1 โค ๐ โค ๐ is a good cover of ๐ ร ๐: For every multi-index ๐ผ, ๐๐ผ is either empty ordiffeomorphic to ๐๐, and for every multi-index ๐ฝ,๐๐ฝ is either empty or diffeomorphic to ๐๐,hence for any productmulti-index (๐ผ, ๐ฝ), the product๐๐ผร๐๐ฝ is either empty or diffeomorphicto ๐๐ ร๐๐. The tricky part of the proof is verifying that the products from (5.7.2) are a basisof ๐ป๐(๐ ร ๐), and to do this it will be helpful to state the theorem above in a form thatavoids our choosing specified bases for๐ป๐(๐) and๐ปโ(๐). To do so weโll need to generalizeslightly the notion of a bilinear pairing between two vector space.
Definition 5.7.3. Let๐1,๐2 and๐ be finite dimensional vector spaces. Amap๐ตโถ ๐1ร๐2 โ๐ of sets is a bilinear map if it is linear in each of its factors, i.e., for ๐ฃ2 โ ๐2 the map
๐ฃ1 โ ๐1 โฆ ๐ต(๐ฃ1, ๐ฃ2)
is a linear map of ๐1 into๐ and for ๐ฃ1 โ ๐1 so is the map
๐ฃ2 โ ๐2 โฆ ๐ต(๐ฃ1, ๐ฃ2) .
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192 Chapter 5: Cohomology via forms
Itโs clear that if ๐ต1 and ๐ต2 are bilinear maps of ๐1 ร ๐2 into๐ and ๐1 and ๐2 are realnumbers the function
๐1๐ต1 + ๐2๐ต2 โถ ๐1 ร ๐2 โ๐is also a bilinear map of ๐1 ร ๐2 into๐, so the set of all bilinear maps of ๐1 ร ๐2 into๐forms a vector space. In particular the set of all bilinear maps of ๐1 ร ๐2 into ๐ is a vectorspace, and since this vector space will play an essential role in our intrinsic formulation ofthe Kรผnneth theorem, weโll give it a name. Weโll call it the tensor product of๐โ1 and๐โ2 anddenote it by ๐โ1 โ ๐โ2 . To explain where this terminology comes from we note that if โ1 andโ2 are vectors in ๐โ1 and ๐โ2 then one can define a bilinear map
โ1 โ โ2 โถ ๐1 ร ๐2 โ ๐by setting (โ1 โ โ2)(๐ฃ1, ๐ฃ2) โ โ1(๐ฃ1)โ2(๐ฃ2). In other words one has a tensor product map:
(5.7.4) ๐โ1 ร ๐โ2 โ ๐โ1 โ ๐โ2mapping (โ1, โ2) to โ1 โ โ2. We leave for you to check that this is a bilinear map of ๐โ1 ร ๐โ2into ๐โ1 โ ๐โ2 and to check as well
Proposition 5.7.5. If โ11 ,โฆ, โ1๐ is a basis of๐โ1 and โ21 ,โฆ, โ2๐ is a basis of๐โ2 then โ1๐ โ โ2๐ , for1 โค ๐ โค ๐ and 1 โค ๐ โค ๐, is a basis of ๐โ1 โ ๐โ2 .
Hint: If ๐1 and ๐2 are the same vector space you can find a proof of this in ยง1.3 and theproof is basically the same if theyโre different vector spaces.
Corollary 5.7.6. Let ๐1 and ๐2 be finite-dimensional vector spaces. Then
dim(๐โ1 โ ๐โ2 ) = dim(๐โ1 ) dim(๐โ2 ) = dim(๐1) dim(๐2) .
Weโll now perform some slightly devious maneuvers with โdualityโ operations. Firstnote that for any finite dimensional vector space ๐, the evaluation pairing
๐ ร ๐โ โ ๐ , (๐ฃ, โ) โฆ โ(๐ฃ)is a non-singular bilinear pairing, so, as we explained in ยง5.4 it gives rise to a bijective linearmapping
(5.7.7) ๐ โ (๐โ)โ .Next note that if
(5.7.8) ๐ฟโถ ๐1 ร ๐2 โ๐is a bilinear mapping and โโถ ๐ โ ๐ a linear mapping (i.e., an element of๐โ), then thecomposition of โ and ๐ฟ is a bilinear mapping
โ โ ๐ฟโถ ๐1 ร ๐2 โ ๐and hence by definition an element of ๐โ1 โ ๐โ2 . Thus from the bilinear mapping (5.7.8) weget a linear mapping
(5.7.9) ๐ฟโฏ โถ ๐โ โ ๐โ1 โ ๐โ2 .
Weโll now define a notion of tensor product for the vector spaces ๐1 and๐2 themselves.
Definition 5.7.10. The vector space ๐1 โ ๐2 is the vector space dual of ๐โ1 โ ๐โ2 , i.e., is thespace
(5.7.11) ๐1 โ ๐2 โ (๐โ1 โ ๐โ2 )โ .
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ยง5.7 The Kรผnneth theorem 193
One implication of (5.7.11) is that there is a natural bilinear map
(5.7.12) ๐1 ร ๐2 โ ๐1 โ ๐2 .
(In (5.7.4) replace ๐๐ by ๐โ๐ and note that by (5.7.7) (๐โ๐ )โ = ๐๐.) Another is the following:
Proposition 5.7.13. Let ๐ฟ be a bilinear map of ๐1 ร ๐2 into๐. Then there exists a uniquelinear map
๐ฟ# โถ ๐1 โ ๐2 โ๐with the property
(5.7.14) ๐ฟ#(๐ฃ1 โ ๐ฃ2) = ๐ฟ(๐ฃ1, ๐ฃ2)
where ๐ฃ1 โ ๐ฃ2 is the image of (๐ฃ1, ๐ฃ2) with respect to (5.7.12).
Proof. Let ๐ฟ# be the transpose of the map ๐ฟโฏ in (5.7.9) and note that by (5.7.7) (๐โ)โ =๐. โก
Notice that by Proposition 5.7.13 the property (5.7.14) is the defining property of ๐ฟ#;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 โabstract nonsenseโ), let us come back to our motive for this digression:an intrinsic formulation of the Kรผnneth theorem. As above let ๐ and ๐ be manifolds ofdimension ๐ and ๐, respectively, both having finite topology. For ๐+โ = ๐ one has a bilinearmap
๐ป๐(๐) ร ๐ปโ(๐) โ ๐ป๐(๐ ร ๐)mapping (๐1, ๐2) to ๐โ๐1 โ ๐โ๐2, and hence by Proposition 5.7.13 a linear map
(5.7.15) ๐ป๐(๐) โ ๐ปโ(๐) โ ๐ป๐(๐ ร ๐) .
Let๐ป๐1 (๐ ร ๐) โ โจ
๐+โ=๐๐ป๐(๐) โ ๐ปโ(๐) .
The maps (5.7.15) can be combined into a single linear map
(5.7.16) ๐ป๐1 (๐ ร ๐) โ ๐ป๐(๐ ร ๐)
and our intrinsic version of the Kรผnneth 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 ๐ be an open subset of๐ which has finite topology and let
H๐1 (๐) โ โจ๐+โ=๐๐ป๐(๐) โ ๐ปโ(๐)
andH๐2 (๐) โ ๐ป๐(๐ ร ๐) .
As weโve just seen thereโs a Kรผnneth map
๐ โถ H๐1 (๐) โ H๐2 (๐) .
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194 Chapter 5: Cohomology via forms
Exercises for ยง5.7Exercise 5.7.i. Let ๐1 and ๐2 be open subsets of ๐, both having finite topology, and let๐ = ๐1 โช ๐2. Show that there is a long exact sequence:
โฏ H๐1 (๐) H๐1 (๐1) โH๐1 (๐2) H๐1 (๐1 โฉ ๐2) H๐+11 (๐) โฏ ,๐ฟ ๐ฟ
Hint: Take the usual MayerโVietoris sequence
โฏ ๐ป๐(๐) ๐ป๐(๐1) โ ๐ป๐(๐2) ๐ป๐(๐1 โฉ ๐2) ๐ป๐+1(๐) โฏ ,๐ฟ ๐ฟ
tensor each term in this sequence with๐ปโ(๐), and sum over ๐ + โ = ๐.
Exercise 5.7.ii. Show that for H2 there is a similar sequence.Hint: Apply MayerโVietoris to the open subsets ๐1 ร ๐ and ๐2 ร ๐ of๐.
Exercise 5.7.iii. Show that the diagram below commutes:
โฏ H๐1 (๐) H๐1 (๐1) โH๐1 (๐2) H๐1 (๐1 โฉ ๐2) H๐+11 (๐) โฏ
โฏ H๐2 (๐) H๐2 (๐1) โH๐2 (๐2) H๐2 (๐1 โฉ ๐2) H๐+12 (๐) โฏ
๐ฟ
๐ ๐
๐ฟ
๐ ๐
๐ฟ ๐ฟ
(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 Kรผnneth map is bijective for ๐1,๐2 and ๐1 โฉ ๐2 it is bijective for ๐.
Exercise 5.7.v. Prove the Kรผnneth theorem by induction on the number of open sets in agood cover of๐. To get the induction started, note that
๐ป๐(๐๐ ร ๐) โ ๐ป๐(๐) .(See Exercise 5.3.xi.)
5.8. ฤech Cohomology
We pointed out at the end of ยง5.3 that if a manifold๐ admits a finite good cover
U = {๐1,โฆ,๐๐} ,then in principle its cohomology can be read off from the intersection properties of the๐๐โs.In this section weโll explain in more detail how to do this. More explicitly, we will show howto construct from the intersection properties of the ๐๐โs a sequence of cohomology groups๐ป๐(U; ๐), and we will prove that these ฤech cohomology groups are isomorphic to the deRham cohomology groups of ๐. (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 MayerโVietoris theorem in ยง5.2 and the five lemma in ยง5.4.)
We will denote by๐๐(U) the set of all multi-indices
๐ผ = (๐0,โฆ, ๐๐) , 1 โค ๐0,โฆ, ๐๐ โค ๐with the property that the intersection
๐๐ผ โ ๐๐0 โฉโฏ โฉ ๐๐๐
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ยง5.8 ฤech Cohomology 195
is nonempty. (For example, for 1 โค ๐ โค ๐, themulti-index ๐ผ = (๐0,โฆ, ๐๐)with ๐0 = โฏ = ๐๐ = ๐is in๐๐(U) since ๐๐ผ = ๐๐.) The disjoint union๐(U) โ โ๐โฅ0๐๐(U) is called the nerve ofthe cover U and๐๐(U) is the ๐-skeleton of the nerve๐(U). Note that if we delete ๐๐ froma multi-index ๐ผ = (๐0,โฆ, ๐๐) in๐๐(U), i.e., replace ๐ผ by
๐ผ๐ โ (๐0,โฆ, ๐๐โ1, ๐๐+1,โฆ, ๐๐) ,then ๐ผ๐ โ ๐๐โ1(U).
We now associate a cochain complex to ๐(U) by defining ๐ถ๐(U; ๐) to be the finitedimensional vector space consisting of all (set-theoretic) maps ๐โถ ๐๐(U) โ ๐:
๐ถ๐(U; ๐) โ {๐โถ ๐๐(U) โ ๐} ,with addition and scalar multiplication of functions. Define a coboundary operator
๐ฟโถ ๐ถ๐โ1(U; ๐) โ ๐ถ๐(U; ๐)by setting
(5.8.1) ๐ฟ(๐)(๐ผ) โ๐โ๐=0(โ1)๐๐(๐ผ๐) .
We claim that ๐ฟ actually is a coboundary operator, i.e., ๐ฟ โ ๐ฟ = 0. To see this, we note thatfor ๐ โ ๐ถ๐โ1(U; ๐) and ๐ผ โ ๐๐(U), by (5.8.1) we have
๐ฟ(๐ฟ(๐))(๐ผ) =๐+1โ๐=0(โ1)๐๐ฟ(๐)(๐ผ๐)
=๐+1โ๐=0(โ1)๐(โ
๐ <๐(โ1)๐ ๐(๐ผ๐,๐ ) + โ
๐ >๐(โ1)๐ โ1๐(๐ผ๐,๐ )) .
Thus the term ๐(๐ผ๐,๐ ) occurs twice in the sum, but occurs oncewith opposite signs, and hence๐ฟ(๐ฟ(๐))(๐ผ) = 0.
The cochain complex
(5.8.2) 0 ๐ถ0(U; ๐) ๐ถ1(U; ๐) โฏ ๐ถ๐(U; ๐) โฏ๐ฟ ๐ฟ ๐ฟ ๐ฟ
is called theฤech cochain complex of the coverU.Theฤech cohomology groups of the coverU are the cohomology groups of the ฤech cochain complex:
๐ป๐(U; ๐) โ ker(๐ฟโถ ๐ถ๐(U; ๐) โ ๐ถ๐+1(U; ๐))im(๐ฟโถ ๐ถ๐โ1(U; ๐) โ ๐ถ๐(U; ๐))
.
The rest of the section is devoted to proving the following theorem:
Theorem 5.8.3. Let ๐ be a manifold and U a finite good cover of ๐. Then for all ๐ โฅ 0 wehave isomorphisms
๐ป๐(U; ๐) โ ๐ป๐(๐) .
Remarks 5.8.4.(1) The definition of๐ป๐(U; ๐) only involves the nerve๐(U) of this cover soTheorem 5.8.3
is in effect a proof of the claim we made above: the cohomology of๐ is in principle deter-mined by the intersection properties of the ๐๐โs.
(2) Theorem 5.8.3 gives us another proof of an assertion we proved earlier: if ๐ admits afinite good cover the cohomology groups of๐ are finite-dimensional.
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196 Chapter 5: Cohomology via forms
Theproof ofTheorem 5.8.3 will involve an interesting de Rham theoretic generalizationof the ฤech cochain complex (5.8.2). Namely, for ๐ and โ nonnegative integers we define aฤech cochain of degree ๐ with values in๐บโ to be a map ๐ which assigns each ๐ผ โ ๐๐(U) anโ-form ๐(๐ผ) โ ๐บโ(๐๐ผ).
The set of these cochains forms an infinite dimensional vector space ๐ถ๐(U; ๐บโ). Wewill show how to define a coboundary operator
๐ฟโถ ๐ถ๐โ1(U; ๐บโ) โ ๐ถ๐(U; ๐บโ)similar to the ฤech coboundary operator (5.8.1). To define this operator, let ๐ผ โ ๐๐(U) andfor 0 โค ๐ โค ๐ let ๐พ๐ โถ ๐บโ(๐๐ผ๐) โ ๐บ
โ(๐๐ผ) be the restriction map defined by๐ โฆ ๐|๐๐ผ .
(Note that the restiction ๐|๐๐ผ is well-defined since ๐๐ผ โ ๐๐ผ๐ .) Thus given a ฤech cochain๐ โ ๐ถ๐โ1(U; ๐บโ) we can, mimicking (5.8.1), define a ฤech cochain ๐ฟ(๐) โ ๐ถ๐(U; ๐บโ) bysetting
(5.8.5) ๐ฟ(๐)(๐ผ) โ๐โ๐=0๐พ๐(๐(๐ผ๐)) =
๐โ๐=0๐(๐ผ๐)|๐๐ผ .
In other words, except for the ๐พ๐โs the definition of this cochain is formally identical withthe definition (5.8.1). We leave it as an exercise that the operators
๐ฟโถ ๐ถ๐โ1(U; ๐บโ) โ ๐ถ๐(U; ๐บโ)satisfy ๐ฟ โ ๐ฟ = 0.1
Thus, to summarize, for every nonnegative integer โwehave constructed aฤech cochaincomplex with values in ๐บโ
(5.8.6) 0 ๐ถ0(U; ๐บโ) ๐ถ1(U; ๐บโ) โฏ ๐ถ๐(U; ๐บโ) โฏ .๐ฟ ๐ฟ ๐ฟ ๐ฟ
Our next task in this sectionwill be to compute the cohomology of the complex (5.8.6).Weโllbegin with the 0th cohomology group of (5.8.6), i.e., the kernel of the coboundary operator
๐ฟโถ ๐ถ0(U; ๐บโ) โ ๐ถ1(U; ๐บโ) .An element ๐ โ ๐ถ0(U; ๐บโ) is, by definition, a map which assigns to each ๐ โ {1,โฆ, ๐} anelement ๐(๐) โ ๐๐ of ๐บโ(๐๐).
Now let ๐ผ = (๐0, ๐1) be an element of๐1(U). Then, by definition, ๐๐0 โฉ ๐๐1 is nonemptyand
๐ฟ(๐)(๐ผ) = ๐พ๐0๐๐1 โ ๐พ๐1๐๐0 .Thus ๐ฟ(๐) = 0 if and only if
๐๐0 |๐๐0โฉ๐๐1 = ๐๐1 |๐๐0โฉ๐๐1for all ๐ผ โ ๐1(U). This is true if and only if each โ-form ๐๐ โ ๐บโ(๐๐) is the restriction to ๐๐of a globally defined โ-form ๐ on ๐1 โชโฏ โช ๐๐ = ๐. Thus
ker(๐ฟโถ ๐ถ0(U; ๐บโ) โ ๐ถ1(U; ๐บโ)) = ๐บโ(๐) .Inserting ๐บโ(๐) into the sequence (5.8.6) we get a new sequence
(5.8.7) 0 ๐บโ(๐) ๐ถ0(U; ๐บโ) โฏ ๐ถ๐(U; ๐บโ) โฏ ,๐ฟ ๐ฟ ๐ฟ
and we prove:
1Hint: Except for keeping track of the ๐พ๐โs the proof is identical to the proof of the analogous result for thecoboundary operator (5.8.1).
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ยง5.8 ฤech Cohomology 197
Theorem5.8.8. Let๐ be amanifold andU afinite good cover of๐.Then for each nonnegativeinteger โ, the sequence (5.8.7) is exact.
Weโve just proved that the sequence (5.8.7) is exact in the first spot. To prove that (5.8.7)is exact in its ๐th spot, we will construct a chain homotopy operator
๐โถ ๐ถ๐(U; ๐บโ) โ ๐ถ๐+1(U; ๐บโ)with the property that(5.8.9) ๐ฟ๐(๐) + ๐๐ฟ(๐) = ๐for all ๐ โ ๐ถ๐(U; ๐บโ). To define this operator weโll need a slight generalization of Theo-rem 5.2.22.
Theorem 5.8.10. Let ๐ be a manifold with a finite good cover {๐1,โฆ,๐๐}. Then there existfunctions ๐1,โฆ, ๐๐ โ ๐ถโ(๐) such that supp(๐๐) โ ๐๐ for ๐ = 1,โฆ,๐ and โ๐๐=1 ๐๐ = 1.Proof of Theorem 5.8.10. Let (๐๐)๐โฅ1, where ๐๐ โ ๐ถโ0 (๐), be a partition of unity subordinateto the cover U and let ๐1 be the sum of the ๐๐โs with support in ๐1:
๐1 โ โ๐โฅ1
supp(๐๐)โ๐1
๐๐ .
Deleting those ๐๐ such that supp(๐๐) โ ๐1 from the sequence (๐๐)๐โฅ1, we get a new sequence(๐โฒ๐ )๐โฅ1. Let ๐2 be the sum of the ๐โฒ๐ โs with support in ๐1:
๐2 โ โ๐โฅ1
supp(๐โฒ๐ )โ๐2
๐โฒ๐ .
Now we delete those ๐โฒ๐ such that supp(๐โฒ๐ ) โ ๐2 from the sequence (๐โฒ๐ )๐โฅ1 and construct๐3 by the same method we used to construct ๐1 and ๐2, and so on. โก
Now we define the operator ๐ and prove Theorem 5.8.8.
Proof of Theorem 5.8.8. To define the operator (5.8.9) let ๐ โ ๐ถ๐(U; ๐บโ). Then we define๐๐ โ ๐ถ๐โ1(U; ๐บโ) by defining its value at ๐ผ โ ๐๐โ1(๐ผ) to be
๐๐(๐ผ) โ๐โ๐=0๐๐๐(๐, ๐0,โฆ, ๐๐โ1) ,
where ๐ผ = (๐0,โฆ, ๐๐โ1) and the ๐th summand on the right is defined by be equal to 0when๐๐โฉ๐๐ผ is empty and defined to be the product of the function ๐๐ and the โ-form ๐(๐, ๐0,โฆ, ๐๐โ1)when๐๐โฉ๐๐ผ is nonempty. (Note that in this case, the multi-index (๐, ๐0,โฆ, ๐๐โ1) is in๐๐(U)and so by definition ๐(๐, ๐0,โฆ, ๐๐โ1) is an element of ๐บโ(๐๐ โฉ ๐๐ผ). However, since ๐๐ is sup-ported on ๐๐, we can extend the โ-form ๐๐๐(๐, ๐0,โฆ, ๐๐โ1) ti ๐๐ผ by setting it equal to zerooutside of ๐๐ โฉ ๐๐ผ.)
To prove (5.8.9) we note that for ๐ โ ๐ถ๐(U; ๐บโ) we have
๐๐ฟ๐(๐0,โฆ, ๐๐) =๐โ๐=1๐๐๐ฟ๐(๐, ๐0,โฆ, ๐๐) ,
so by (5.8.5) the right hand side is equal to
(5.8.11)๐โ๐=1๐๐๐พ๐๐(๐0,โฆ, ๐๐) +
๐โ๐=1
๐โ๐ =0(โ1)๐ +1๐๐๐พ๐๐ ๐(๐, ๐0,โฆ, ๐ค๐ ,โฆ, ๐๐) ,
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198 Chapter 5: Cohomology via forms
where (๐, ๐0,โฆ, ๐ค๐ ,โฆ, ๐๐) is the multi-index (๐, ๐0,โฆ, ๐๐) with ๐๐ deleted. Similarly,
๐ฟ๐๐(๐0,โฆ, ๐๐) =๐โ๐ =0(โ1)๐ ๐พ๐๐ ๐๐(๐0,โฆ, ๐ค๐ ,โฆ, ๐๐)
=๐โ๐=1
๐โ๐ =0(โ1)๐ ๐๐๐พ๐๐ ๐(๐, ๐0,โฆ, ๐ค๐ ,โฆ, ๐๐) .
However, this is the negative of the second summand of (5.8.11); so, by adding these twosummands we get
(๐๐ฟ๐ + ๐ฟ๐๐)(๐0,โฆ, ๐๐) =๐โ๐=1๐๐๐พ๐๐(๐0,โฆ, ๐๐)
and since โ๐๐=1 ๐๐ = 1, this identity reduced to
(๐๐ฟ๐ + ๐ฟ๐๐)(๐ผ) = ๐(๐ผ) ,or, since ๐ผ โ ๐๐(U) is any element, ๐๐ฟ๐ + ๐ฟ๐๐ = ๐.
Finally, note the Theorem 5.8.8 is an immediate consequence of the existence of thischain homotopy operator. Namely, if ๐ โ ๐ถ๐(U; ๐บโ) is in the kernel of ๐ฟ, then
๐ = ๐๐ฟ(๐) + ๐ฟ๐(๐) = ๐ฟ๐(๐) ,so ๐ is in the image of ๐ฟโถ ๐ถ๐โ1(U; ๐บโ) โ ๐ถ๐(U; ๐บโ). โก
To prove Theorem 5.8.3, we note that in addition to the exact sequence
0 ๐บโ(๐) ๐ถ0(U; ๐บโ) โฏ ๐ถ๐(U; ๐บโ) โฏ ,๐ฟ ๐ฟ ๐ฟ
we have another exact sequence(5.8.12)0 ๐ถ๐(U; ๐) ๐ถ๐(U; ๐บ0) ๐ถ๐(U; ๐บ1) โฏ ๐ถ๐(U; ๐บโ) โฏ ,๐ ๐ ๐ ๐ ๐
where the ๐โs are defined as follows: given ๐ โ ๐ถ๐(U; ๐บโ) and ๐ผ โ ๐๐(U), we have that๐(๐ผ) โ ๐บโ(๐๐ผ) and ๐(๐(๐ผ)) โ ๐บโ+1(๐๐ผ), and we define ๐(๐) โ ๐ถ๐(U; ๐บโ+1) to be the mapgiven by
๐ผ โฆ ๐(๐(๐ผ)) .It is clear from this definition that ๐(๐(๐)) = 0 so the sequence
(5.8.13) ๐ถ๐(U; ๐บ0) ๐ถ๐(U; ๐บ1) โฏ ๐ถ๐(U; ๐บโ) โฏ๐ ๐ ๐ ๐
is a complex and the exactness of this sequence follows from the fact that, since U is a goodcover, the sequence
๐บ0(๐๐ผ) ๐บ1(๐๐ผ) โฏ ๐บโ(๐๐ผ) โฏ๐ ๐ ๐ ๐
is exact. Moreover, if ๐ โ ๐ถ๐(U; ๐บ0) and ๐๐ = 0, then for every ๐ผ โ ๐๐(U), we have๐๐(๐ผ) = 0, i.e., the 0-form ๐(๐ผ) in ๐บ0(๐๐ผ) is constant. Hence the map given by ๐ผ โฆ ๐(๐ผ) isjust a ฤech cochain of the type we defined earlier, i.e., a map ๐๐(U) โ ๐. Therefore, thekernel of this map ๐โถ ๐ถ๐(U; ๐บ0) โ ๐ถ๐(U; ๐บ1) is ๐ถ0(U; ๐); and, adjoining this term to thesequence (5.8.13) we get the exact sequence (5.8.12).
Lastly we note that the two operations weโve defined above, the ๐ operation๐โถ ๐ถ๐(U; ๐บโ) โ ๐ถ๐(U; ๐บโ+1)
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ยง5.8 ฤech Cohomology 199
and the ๐ฟ operation ๐ฟโถ ๐ถ๐(U; ๐บโ) โ ๐ถ๐+1(U; ๐บโ), commute. i.e., for ๐ โ ๐ถ๐(U; ๐บโ) wehave
(5.8.14) ๐ฟ๐๐ = ๐๐ฟ๐ .
Proof of equation (5.8.14). For ๐ผ โ ๐๐+1(U)
๐ฟ๐(๐ผ) =๐โ๐=0(โ1)๐๐พ๐๐(๐ผ๐) ,
so if ๐(๐ผ๐) โ ๐๐ผ๐ โ ๐บโ(๐๐ผ๐), then
๐ฟ๐(๐ผ) =๐โ๐=0(โ1)๐๐๐ผ๐ |๐๐ผ
and
๐๐ฟ(๐ผ) =๐โ๐=0๐๐๐ผ๐ |๐๐ผ = ๐ฟ๐๐(๐ผ) . โก
Putting these results together we get the following commutative diagram:
โฎ โฎ โฎ โฎ
0 ๐บ2(๐) ๐ถ0(U; ๐บ2) ๐ถ1(U; ๐บ2) ๐ถ2(U; ๐บ2) โฏ
0 ๐บ1(๐) ๐ถ0(U; ๐บ1) ๐ถ1(U; ๐บ1) ๐ถ2(U; ๐บ1) โฏ
0 ๐บ0(๐) ๐ถ0(U; ๐บ0) ๐ถ1(U; ๐บ0) ๐ถ2(U; ๐บ0) โฏ
0 ๐ถ0(U; ๐) ๐ถ1(U; ๐) ๐ถ2(U; ๐) โฏ
0 0 0 .
๐
๐ฟ ๐ฟ
๐
๐ฟ
๐
๐ฟ
๐ ๐
๐
๐ฟ
๐
๐ฟ
๐ ๐
๐ฟ ๐ฟ
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 usualฤech cochain complex.
Exercises for ยง5.8Exercise 5.8.i. Deduce from this diagram that if a manifold ๐ admits a finite good coverU, then๐ป๐(U; ๐) โ ๐ป๐(๐).
Hint: Let ๐ โ ๐บ๐+1(๐) be a form such that ๐๐ = 0. Show that one can, by a sequence ofโchess movesโ (of which the first few stages are illustrated in the (5.8.15) below), convert ๐
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200 Chapter 5: Cohomology via forms
into an element ๐ of ๐ถ๐+1(U; ๐) satisfying ๐ฟ( ๐) = 0
(5.8.15)
๐ ๐๐+1,0
๐๐,0 ๐๐,1
๐๐โ1,1 ๐๐โ1,2
๐๐โ2,2 ๐๐โ2,3
๐๐โ3,3 โฏ
๐
๐ฟ
๐
๐ฟ
๐
๐ฟ
๐
๐ฟ
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APPENDIX a
Bump Functions & Partitions of Unity
Wewill discuss in this section a number of โglobal to localโ techniques inmulti-variablecalculus: techniques which enable one to reduce global problems on manifolds and largeopen subsets of ๐๐ to local problems on small open subsets of these sets. Our starting pointwill be the function ๐ on ๐ defined by
(a.1) ๐(๐ฅ) โ {๐โ 1๐ฅ , ๐ฅ > 00, ๐ฅ โค 0 ,
which is positive for ๐ฅ positive, zero for ๐ฅ negative and everywhere ๐ถโ. (We will sketch aproof of this last assertion at the end of this appendix). From ๐ one can construct a numberof other interesting ๐ถโ functions.
Example a.2. For ๐ > 0 the function ๐(๐ฅ) + ๐(๐ โ ๐ฅ) is positive for all ๐ฅ so the quotient๐๐(๐ฅ) of ๐(๐ฅ) by this function is a well-defined ๐ถโ function with the properties:
{{{{{
๐๐(๐ฅ) = 0, for ๐ฅ โค 00 โค ๐๐(๐ฅ) โค 1๐๐(๐ฅ) = 1, for ๐ฅ โฅ ๐ .
Example a.3. Let ๐ผ be the open interval (๐, ๐), and ๐๐ผ the product of ๐(๐ฅ โ ๐) and ๐(๐ โ ๐ฅ).Then ๐๐ผ(๐ฅ) is positive for ๐ฅ โ ๐ผ and zero on the complement of ๐ผ.
Example a.4. More generally let ๐ผ1,โฆ, ๐ผ๐ be open intervals and let ๐ = ๐ผ1 รโฏ ร ๐ผ๐ be theopen rectangle in ๐๐ having these intervals as sides. Then the function
๐๐(๐ฅ) โ ๐๐ผ1(๐ฅ1)โฏ๐๐ผ๐(๐ฅ๐)is a ๐ถโ function on ๐๐ thatโs positive on ๐ and zero on the complement of ๐.
Using these functions we will prove:
Lemma a.5. Let ๐ถ be a compact subset of ๐๐ and ๐ an open set containing ๐ถ. Then thereexists function ๐ โ ๐ถโ0 (๐) such that
{๐(๐ฅ) โฅ 0, for all ๐ฅ โ ๐๐(๐ฅ) > 0, for ๐ฅ โ ๐ถ .
Proof. For each ๐ โ ๐ถ let๐๐ be an open rectangle with ๐ โ ๐๐ and๐๐ โ ๐. The๐๐โs cover๐ถ๐ so, by HeineโBorel there exists a finite subcover๐๐1 ,โฆ,๐๐๐ . Now let ๐ โ โ๐๐=1 ๐๐๐๐ โก
This result can be slightly strengthened. Namely we claim:
Theorem a.6. There exists a function ๐ โ ๐ถโ0 (๐) such that 0 โค ๐ โค 1 and ๐ = 1 on ๐ถ.201
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202 Appendix a: Bump Functions & Partitions of Unity
Proof. Let ๐ be as in Lemma a.5 and let ๐ > 0 be the greatest lower bound of the restrictionof ๐ to ๐ถ. Then if ๐๐ is the function in Example a.2 the function ๐๐ โ ๐ has the propertiesindicated above. โก
Remark a.7. The function๐ in this theorem is an example of a bump function. If one wantsto study the behavior of a vector field ๐ or a ๐-form ๐ on the set ๐ถ, then by multiplying ๐(or ๐) by ๐ one can, without loss or generality, assume that ๐ (or ๐) is compactly supportedon a small neighborhood of ๐ถ.
Bump functions are one of the standard tools in calculus for converting global problemsto local problems. Another such tool is partitions of unity.
Let ๐ be an open subset of ๐๐ and U = {๐๐ผ}๐ผโ๐ผ a covering of ๐ by open subsets (in-dexed by the elements of the index set ๐ผ). Then the partition of unity theorem asserts:
Theorem a.8. There exists a sequence of functions ๐1, ๐2,โฆ โ ๐ถโ0 (๐) such that(1) ๐๐ โฅ 0(2) For every ๐ there is an ๐ผ โ ๐ผ with ๐๐ โ ๐ถโ0 (๐๐ผ)(3) For every ๐ โ ๐ there exists a neighborhood ๐๐ of ๐ in ๐ and an ๐๐ > 0 such that๐๐|๐๐ = 0 for ๐ข > ๐๐.
(4) โ๐๐ = 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 ๐ by open rectangles (๐๐)๐โฅ1 such that(1) ๐๐ โ ๐(2) For each ๐ there is an ๐ผ โ ๐ผ with ๐๐ โ ๐๐ผ(3) For every ๐ โ ๐ there exists a neighborhood ๐๐ of ๐ in ๐ and๐๐ > 0 such that ๐๐ โฉ ๐๐
is empty for ๐ > ๐๐We first note that Theorem a.10 implies Theorem a.8. To see this note that the func-
tions ๐๐๐ in Example a.4 above have all the properties indicated in Theorem a.8 except forproperty (4). Moreover since the ๐๐โs are a covering of ๐ the sum โโ๐=1 ๐๐๐ is everywherepositive. Thus we get a sequence of ๐๐โs satisfying (1)โ(4) by taking ๐๐ to be the quotient of๐๐๐ by this sum.
Proof of Theorem a.10. Let ๐(๐ฅ, ๐๐) be the distance of a point ๐ฅ โ ๐ to the complement๐๐ โ ๐๐ โ ๐ of ๐ in ๐๐, and let ๐ด๐ be the compact subset of ๐ consisting of points, ๐ฅ โ ๐,satisfying ๐(๐ฅ, ๐๐) โฅ 1/๐ and |๐ฅ| โค ๐. By HeineโBorel we can find, for each ๐, a collection ofopen rectangles,๐๐,๐, ๐ = 1,โฆ,๐๐, such that๐๐,๐ is contained in int(๐ด๐+1โ๐ด๐โ2) and in some๐๐ผ and such that the๐๐, ๐โs are a covering of๐ด๐โint(๐ด๐โ1).Thus the๐๐,๐โs have the propertieslisted in Theorem a.10, and by relabelling, i.e., setting๐๐ = ๐1,๐ for 1 โค ๐ โค ๐๐,๐๐1+๐ โ ๐2,๐,for 1 โค ๐ โค ๐2, etc., we get a sequence, ๐1, ๐2,โฆ with the desired properties. โก
ApplicationsWe will next describe a couple of applications of Theorem a.10.
Application a.11 (Improper integrals). Let ๐โถ ๐ โ ๐ be a continuous function. We willsay that ๐ is integrable over ๐ if the infinite sum
(a.12)โโ๐=1โซ๐๐๐(๐ฅ) |๐(๐ฅ)| ๐๐ฅ
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203
converges and if so we will define the improper integral of ๐ over ๐ to be the sum
(a.13)โโ๐=1โซ๐๐๐(๐ฅ)๐(๐ฅ)๐๐ฅ
(Notice that each of the summands in this series is the integral of a compactly supportedcontinuous function over ๐๐ so the individual summands are well-defined. Also itโs clearthat if ๐ itself is a compactly supported function on ๐๐, (a.12) is bounded by the integral of|๐| over ๐๐, so for every ๐ โ ๐ถ0 (๐๐) the improper integral of ๐ over ๐ is well-defined.)
Application a.14 (An extension theorem for ๐ถโ maps). Let ๐ be a subset of ๐๐ and ๐ โถ๐ โ ๐๐ a continuous map. We will say that ๐ is ๐ถโ if, for every ๐ โ ๐, there exists anopen neighborhood, ๐๐, of ๐ in ๐๐ and a ๐ถโ map, ๐๐ โถ ๐๐ โ ๐๐ such that ๐๐ = ๐ on๐๐ โฉ ๐.
Theorem a.15 (Extension Theorem). If ๐โถ ๐ โ ๐๐ is ๐ถโ there exists an open neighbor-hood ๐ of๐ in ๐๐ and a ๐ถโ map ๐โถ ๐ โ ๐๐, such that ๐ = ๐ on๐.Proof. Let ๐ = โ๐โ๐๐๐ and let (๐๐)๐โฅ1 be a partition of unity subordinate to the coveringU = {๐๐}๐โ๐ of๐. Then, for each ๐, there exists a ๐ such that the support of ๐๐ is containedin ๐๐. Let
๐๐(๐ฅ) = {๐๐๐๐(๐ฅ), ๐ฅ โ ๐๐0, ๐ฅ โ ๐๐๐ .
Then ๐ = โโ๐=1 ๐๐ is well defined by item (3) of Theorem a.8 and the restriction of ๐ to๐ isgiven by โโ๐=1 ๐๐๐, which is equal to ๐. โก
Exercises for Appendix aExercise a.i. Show that the function (a.1) is ๐ถโ.
Hints:(1) From the Taylor series expansion
๐๐ฅ =โโ๐=0
๐ฅ๐๐!
conclude that for ๐ฅ > 0๐๐ฅ โฅ ๐ฅ
๐+๐
(๐ + ๐)!.
(2) Replacing ๐ฅ by 1/๐ฅ conclude that for ๐ฅ > 0,
๐1/๐ฅ โฅ 1(๐ + ๐)!
1๐ฅ๐+๐
.
(3) From this inequality conclude that for ๐ฅ > 0,
๐โ 1๐ฅ โค (๐ + ๐)!๐ฅ๐+๐ .(4) Let ๐๐(๐ฅ) be the function
๐๐(๐ฅ) โ {๐โ 1๐ฅ ๐ฅโ๐, ๐ฅ > 00, ๐ฅ โค 0 .
Conclude from (3) that for ๐ฅ > 0,๐๐(๐ฅ) โค (๐ + ๐)!๐ฅ๐
for all ๐
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204 Appendix a: Bump Functions & Partitions of Unity
(5) Conclude that ๐๐ is ๐ถ1 differentiable.(6) Show that
๐๐๐ฅ๐๐ = ๐๐+2 โ ๐๐๐+1
(7) Deduce by induction that the ๐๐โs are ๐ถ๐ differentiable for all ๐ and ๐.Exercise a.ii. Show that the improper integral (a.13) is well-defined independent of thechoice of partition of unity.
Hint: Let (๐โฒ๐ )๐โฅ1 be another partition of unity. Show that (a.13) is equal toโโ๐=1
โโ๐=1โซ๐๐๐(๐ฅ)๐โฒ๐ (๐ฅ)๐(๐ฅ)๐๐ฅ .
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APPENDIX b
The Implicit FunctionTheorem
Let ๐ be an open neighborhood of the origin in ๐๐ and let ๐1,โฆ, ๐๐ be ๐ถโ functionson ๐ with the property ๐1(0) = โฏ = ๐๐(0) = 0. Our goal in this appendix is to prove thefollowing implicit function theorem.
Theorem b.1. Suppose the matrix
(b.2) [ ๐๐๐๐๐ฅ๐ (0)]
is non-singular.Then there exists a neighborhood๐0 of 0 in๐๐ and a diffeomorphism๐โถ (๐0, 0) โช(๐, 0) of ๐0 onto an open neighborhood of 0 in ๐ such that
๐โ๐๐ = ๐ฅ๐ , ๐ = 1,โฆ, ๐and
๐โ๐ฅ๐ = ๐ฅ๐ , ๐ = ๐ + 1,โฆ, ๐ .
Remarks b.3.(1) Let ๐(๐ฅ) = (๐1(๐ฅ),โฆ, ๐๐(๐ฅ)). Then the second set of equations just say that
๐๐(๐ฅ1,โฆ, ๐ฅ๐) = ๐ฅ๐for ๐ = ๐ + 1,โฆ, ๐, so the first set of equations can be written more concretely in theform
(b.4) ๐๐(๐1(๐ฅ1,โฆ, ๐ฅ๐),โฆ, ๐๐(๐ฅ1,โฆ, ๐ฅ๐), ๐ฅ๐+1,โฆ, ๐ฅ๐) = ๐ฅ๐for ๐ = 1,โฆ, ๐.
(2) Letting ๐ฆ๐ = ๐๐(๐ฅ1,โฆ, ๐ฅ๐) the equations (b.4) become(b.5) ๐๐(๐ฆ1,โฆ, ๐ฆ๐, ๐ฅ๐+1,โฆ, ๐ฅ๐) = ๐ฅ๐ .
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 ๐ฆ๐โs in terms of the ๐ฅ๐โs.
(3) By a linear change of coordinates:
๐ฅ๐ โฆ๐โ๐=1๐๐,๐๐ฅ๐ , ๐ = 1,โฆ, ๐
we can arrange without loss of generality for the matrix (b.2) to be the identity matrix,i.e.,
(b.6) ๐๐๐๐๐ฅ๐(0) = ๐ฟ๐,๐ , 1 โค ๐, ๐ โค ๐ .
Our proof of this theorem will be by induction on ๐.First, letโs suppose that ๐ = 1 and, for the moment, also suppose that ๐ = 1. Then the
theorem above is just the inverse function theoremof freshman calculus. Namely if๐(0) = 0and ๐๐/๐๐ฅ(0) = 1, there exists an interval (๐, ๐) about the origin on which ๐๐/๐๐ฅ is greater
205
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206 Appendix b: The Implicit Function Theorem
than 1/2, so ๐ is strictly increasing on this interval and its graph looks like the curve in thefigure below with ๐ = ๐(๐) โค โ 12๐ and ๐ = ๐(๐) โฅ
12๐.
๏ฟฝ
๏ฟฝ
๏ฟฝ
๏ฟฝ
๏ฟฝ
๏ฟฝ
Figure b.1. The implicit function theorem
The graph of the inverse function ๐โถ [๐, ๐] โ [๐, ๐] is obtained from this graph by justrotating it through ninety degrees, i.e., making the ๐ฆ-axis the horizontal axis and the ๐ฅ-axisthe vertical axis. (From the picture itโs clear that ๐ฆ = ๐(๐ฅ) โบ ๐ฅ = ๐(๐ฆ) โบ ๐(๐(๐ฆ)) =๐ฆ.)
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 (๐ฅ, ๐ฆ) is equal to ๐ the slope ofthe rotated curve at (๐ฆ, ๐ฅ) is 1/๐, so from this proof one concludes that if ๐ฆ = ๐(๐ฅ)
(b.7) ๐๐๐๐ฆ(๐ฆ) = (๐๐๐๐ฅ(๐ฅ))โ1= (๐๐๐๐ฅ(๐(๐ฆ)))
โ1.
Since ๐ is a continuous function, its graph is a continuous curve and, therefore, since thegraph of ๐ is the same curve rotated by ninety degrees, ๐ is also a continuous functions.Hence by (b.7), ๐ is also a ๐ถ1 function and hence by (b.7), ๐ is a ๐ถ2 function, etc. In otherwords ๐ is in ๐ถโ([๐, ๐]).
Letโs now prove that the implicit function theorem with ๐ = 1 and ๐ arbitrary. Thisamounts to showing that if the function ๐ in the discussion above depends on the param-eters, ๐ฅ2,โฆ๐ฅ๐ in a ๐ถโ fashion, then so does its inverse ๐. More explicitly letโs suppose๐(๐ฅ1,โฆ, ๐ฅ๐) is a ๐ถโ function on a subset of ๐๐ of the form [๐, ๐] ร๐ where๐ is a compact,convex neighborhood of the origin in ๐๐ and satisfies ๐๐/๐๐ฅ1 โฅ 12 on this set. Then by theargument above there exists a function, ๐ = ๐(๐ฆ, ๐ฅ2,โฆ, ๐ฅ๐) defined on the set [๐/2, ๐/2]ร๐with the property
(b.8) ๐(๐ฅ1, ๐ฅ2,โฆ, ๐ฅ๐) = ๐ฆ โบ ๐(๐ฆ, ๐ฅ2,โฆ, ๐ฅ๐) = ๐ฅ1 .Moreover by (b.7) ๐ is a ๐ถโ function of ๐ฆ and
(b.9) ๐๐๐๐ฆ(๐ฆ, ๐ฅ2,โฆ, ๐ฅ๐) =
๐๐๐๐ฅ1(๐ฅ1, ๐ฅ2,โฆ, ๐ฅ๐)โ1
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207
at ๐ฅ1 = ๐(๐ฆ). In particular, since ๐๐๐๐ฅ1 โฅ12
0 < ๐๐๐๐ฆ< 2
and hence
(b.10) |๐(๐ฆโฒ, ๐ฅ2,โฆ, ๐ฅ๐) โ ๐(๐ฆ, ๐ฅ2,โฆ, ๐ฅ๐)| < 2|๐ฆโฒ โ ๐ฆ|for points ๐ฆ and ๐ฆโฒ in the interval [๐/2, ๐/2].
The ๐ = 1 case of Theorem b.1 is almost implied by (b.8) and (b.9) except that we muststill show that ๐ is a ๐ถโ function, not just of ๐ฆ, but of all the variables, ๐ฆ, ๐ฅ2,โฆ, ๐ฅ๐, and thisweโll 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 ๐ variables). Let ๐ be a convex open set in ๐๐ and๐โถ ๐ โ ๐ a ๐ถโ function. Then for ๐, ๐ โ ๐ there exists a point ๐ on the line interval joining๐ to ๐ such that
๐(๐) โ ๐(๐) =๐โ๐=1
๐๐๐๐ฅ๐(๐)(๐๐ โ ๐๐) .
Proof. Apply the one-dimensional mean value theorem to the function
โ(๐ก) โ ๐((1 โ ๐ก)๐ + ๐ก๐) . โก
Letโs now show that the function ๐ in (b.8) is a ๐ถโ function of the variables, ๐ฆ and ๐ฅ2.To simplify our notation weโll suppress the dependence of ๐ on ๐ฅ3,โฆ, ๐ฅ๐ and write ๐ as๐(๐ฅ1, ๐ฅ2) and ๐ as ๐(๐ฆ, ๐ฅ2). For โ โ (โ๐, ๐), ๐ small, we have
๐ฆ = ๐(๐(๐ฆ, ๐ฅ2 + โ), ๐ฅ2 + โ) = ๐(๐(๐ฆ, ๐ฅ2), ๐ฅ2) ,and, hence, setting ๐ฅโฒ1 = ๐(๐ฆ, ๐ฅ2 + โ), ๐ฅ1 = ๐(๐ฆ, ๐ฅ2) and ๐ฅโฒ2 = ๐ฅ2 + โ, we get from the meanvalue theorem
0 = ๐(๐ฅโฒ1, ๐ฅโฒ2) โ ๐(๐ฅ1, ๐ฅ2)
= ๐๐๐๐ฅ1(๐)(๐ฅโฒ1 โ ๐ฅ1) +
๐๐๐๐ฅ2(๐)(๐ฅโฒ2 โ ๐ฅ2)
and therefore
(b.12) ๐(๐ฆ, ๐ฅ2 + โ) โ ๐(๐ฆ, ๐ฅ2) = (โ๐๐๐๐ฅ1(๐))โ1 ๐๐๐๐ฅ2(๐) โ
for some ๐ on the line segment joining (๐ฅ1, ๐ฅ2) to (๐ฅโฒ1, ๐ฅโฒ2). Letting โ tend to zero we can drawfrom this identity a number of conclusions:(1) Since ๐ is a ๐ถโ function on the compact set [๐, ๐] ร ๐ its derivatives are bounded on
this set, so the right hand side of (b.12) tends to zero as โ tends to zero, i.e., for fixed ๐ฆ,๐ is continuous as a function of ๐ฅ2.
(2) Nowdivide the right hand side of (b.12) by โ and let โ tend to zero. Since๐ is continuousin ๐ฅ2, the quotient of (b.12) by โ tends to its value at (๐ฅ1, ๐ฅ2). Hence for fixed ๐ฆ ๐ isdifferentiable as a function of ๐ฅ2 and
(b.13) ๐๐๐๐ฅ2(๐ฆ, ๐ฅ2) = โ(
๐๐๐๐ฅ1)โ1(๐ฅ1, ๐ฅ2)
๐๐๐๐ฅ2(๐ฅ1, ๐ฅ2)
where ๐ฅ1 = ๐(๐ฆ, ๐ฅ2).
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208 Appendix b: The Implicit Function Theorem
(3) Moreover, by the inequality (b.10) and the triangle inequality
|๐(๐ฆโฒ, ๐ฅโฒ2) โ ๐(๐ฆ, ๐ฅ2)| โค |๐(๐ฆโฒ, ๐ฅโฒ2) โ ๐(๐ฆ, ๐ฅโฒ2)| + |๐(๐ฆ, ๐ฅโฒ2) โ ๐(๐ฆ, ๐ฅ2)|โค 2|๐ฆโฒ โ ๐ฆ| + |๐(๐ฆ, ๐ฅโฒ2) โ ๐(๐ฆ, ๐ฅ2)| ,
hence ๐ is continuous as a function of ๐ฆ and ๐ฅ2.(4) Hence by (b.9) and (b.13), ๐ is a ๐ถ1 function of ๐ฆ and ๐ฅ2, and henceโฆ.
In other words ๐ is a ๐ถโ function of ๐ฆ and ๐ฅ2. This argument works, more or lessverbatim, for more than two ๐ฅ๐โs and proves that ๐ is a ๐ถโ function of ๐ฆ, ๐ฅ2,โฆ, ๐ฅ๐. Thuswith ๐ = ๐1 and ๐ = ๐1Theorem b.1 is proved in the special case ๐ = 1.
Proof of Theorem b.1 for ๐ > 1. Weโll now prove Theorem b.1 for arbitrary ๐ by inductionon ๐. By induction we can assume that there exists a neighborhood ๐โฒ0 , of the origin in ๐๐and a ๐ถโ diffeomorphism ๐โถ (๐โฒ0 , 0) โช (๐, 0) such that
(b.14) ๐โ๐๐ = ๐ฅ๐for 2 โค ๐ โค ๐ and(b.15) ๐โ๐ฅ๐ = ๐ฅ๐for ๐ = 1 and ๐ + 1 โค ๐ โค ๐. Moreover, by (b.6)
( ๐๐๐ฅ1๐โ๐1) (0) =
๐โ๐=1
๐๐๐๐๐ฅ1(0) ๐๐๐ฅ๐๐๐(0) =
๐๐1๐๐ฅ1= 1
since๐1 = ๐โ๐ฅ1 = ๐ฅ1.Thereforewe can applyTheoremb.1, with ๐ = 1, to the function,๐โ๐1,to conclude that there exists a neighborhood ๐0 of the origin in ๐๐ and a diffeomorphism๐โถ (๐0, 0) โ (๐โฒ0 , 0) such that ๐โ๐โ๐1 = ๐ฅ1 and ๐โ๐โ๐ฅ๐ = ๐ฅ๐ for 1 โค ๐ โค ๐. Thus by (b.14)๐โ๐โ๐๐ = ๐โ๐ฅ๐ = ๐ฅ๐ for 2 โค ๐ โค ๐, and by (b.15) ๐โ๐โ๐ฅ๐ = ๐โ๐ฅ๐ = ๐ฅ๐ for ๐ + 1 โค ๐ โค ๐.Hence if we let ๐ = ๐ โ ๐ we see that:
๐โ๐๐ = (๐ โ ๐)โ๐๐ = ๐โ๐โ๐๐ = ๐ฅ๐for ๐ โค ๐ โค ๐ and
๐โ๐ฅ๐ = (๐ โ ๐)โ๐ฅ๐ = ๐โ๐โ๐ฅ๐ = ๐ฅ๐for ๐ + 1 โค ๐ โค ๐. โก
Weโll derive a number of subsidiary results from Theorem b.1. The first of these is the๐-dimensional version of the inverse function theorem:
Theoremb.16. Let๐ and๐ be open subsets of๐๐ and๐โถ (๐, ๐) โ (๐, ๐) a๐ถโmap. Supposethat the derivative of ๐ at ๐
๐ท๐(๐)โถ ๐๐ โ ๐๐
is bijective. Then ๐ maps a neighborhood of ๐ in ๐ diffeomorphically onto a neighborhood of๐ in ๐.
Proof. By pre-composing and post-composing ๐ by translations we can assume that ๐ =๐ = 0. Let ๐ = (๐1,โฆ, ๐๐). Then the condition that ๐ท๐(0) be bijective is the condition thatthe matrix (b.2) be non-singular. Hence, by Theorem b.1, there exists a neighborhood๐0 of0 in ๐, a neighborhood ๐0 of 0 in ๐, and a diffeomorphism ๐โถ ๐0 โ ๐0 such that
๐โ๐๐ = ๐ฅ๐for ๐ = 1,โฆ, ๐. However, these equations simply say that ๐ is the inverse of ๐, and hencethat ๐ is a diffeomorphism. โก
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209
A second result which weโll extract from Theorem b.1 is the canonical submersion the-orem.
Theorem b.17. Let ๐ be an open subset of ๐๐ and ๐โถ (๐, ๐) โ (๐๐, 0) a ๐ถโ map. Suppose๐ is a submersion at ๐, i.e., suppose its derivative
๐ท๐(๐)โถ ๐๐ โ ๐๐
is onto. Then there exists a neighborhood ๐ of ๐ in ๐, a neighborhood ๐0 of the origin in ๐๐,and a diffeomorphism ๐โถ (๐0, 0) โ (๐, ๐) such that the map ๐ โ ๐โถ (๐0, 0) โ (๐๐, 0) is therestriction to ๐0 of the canonical submersion:
๐โถ ๐๐ โ ๐๐ , ๐(๐ฅ1,โฆ, ๐ฅ๐) โ (๐ฅ1,โฆ, ๐ฅ๐) .Proof. Let ๐ = (๐1,โฆ, ๐๐). Composing ๐ with a translation we can assume ๐ = 0 and by apermutation of the variables ๐ฅ1,โฆ, ๐ฅ๐ we can assume that the matrix (b.2) is non-singular.By Theorem b.1 we conclude that there exists a diffeomorphism ๐โถ (๐0, 0) โช (๐, ๐) withthe properties ๐โ๐๐ = ๐ฅ๐ for ๐ = 1,โฆ, ๐, and hence,
๐ โ ๐(๐ฅ1,โฆ, ๐ฅ๐) = (๐ฅ1,โฆ, ๐ฅ๐) . โก
As a third application of Theorem b.1 weโll prove a theorem which is similar in spirit toTheorem b.17, the canonical immersion theorem.
Theorem b.18. Let ๐ be an open neighborhood of the origin in ๐๐ and ๐โถ (๐, 0) โ (๐๐, ๐)a ๐ถโ map. Suppose that the derivative of ๐ at 0
๐ท๐(0)โถ ๐๐ โ ๐๐
is injective. Then there exists a neighborhood ๐0 of 0 in ๐, a neighborhood ๐ of ๐ in ๐๐, anda diffeomorphism.
๐โถ ๐ โ ๐0 ร ๐๐โ๐
such that the map ๐ โ ๐โถ ๐0 โ ๐0 ร ๐๐โ๐ is the restriction to ๐0 of the canonical immersion๐ โถ ๐๐ โ ๐๐ ร ๐๐โ๐ , ๐(๐ฅ1,โฆ, ๐ฅ๐) = (๐ฅ1,โฆ, ๐ฅ๐, 0,โฆ, 0) .
Proof. Let ๐ = (๐1,โฆ, ๐๐). By a permutation of the ๐๐โs we can arrange that the matrix
[ ๐๐๐๐๐ฅ๐ (0)] , 1 โค ๐, ๐ โค ๐
is non-singular and by composing ๐ with a translation we can arrange that ๐ = 0. Hence byTheorem b.16 the map
๐โถ (๐, 0) โ (๐๐, 0) ๐ฅ โฆ (๐1(๐ฅ),โฆ, ๐๐(๐ฅ))maps a neighborhood ๐0 of 0 diffeomorphically onto a neighborhood ๐0 of 0 in ๐๐. Let๐โถ (๐0, 0) โ (๐0, 0) be its inverse and let
๐พโถ ๐0 ร ๐๐โ๐ โ ๐0 ร ๐๐โ๐
be the map๐พ(๐ฅ1,โฆ, ๐ฅ๐) โ (๐(๐ฅ1,โฆ, ๐ฅ๐), ๐ฅ๐+1,โฆ, ๐ฅ๐) .
Then(๐พ โ ๐)(๐ฅ1,โฆ, ๐ฅ๐) = ๐พ(๐(๐ฅ1,โฆ, ๐ฅ๐), ๐๐+1(๐ฅ),โฆ, ๐๐(๐ฅ))(b.19)
= (๐ฅ1,โฆ, ๐ฅ๐, ๐๐+1(๐ฅ),โฆ, ๐๐(๐ฅ)) .Now note that the map
โโถ ๐0 ร ๐๐โ๐ โ ๐0 ร ๐๐โ๐
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210 Appendix b: The Implicit Function Theorem
defined byโ(๐ฅ1,โฆ, ๐ฅ๐) โ (๐ฅ1,โฆ, ๐ฅ๐, ๐ฅ๐+1 โ ๐๐+1(๐ฅ),โฆ, ๐ฅ๐ โ ๐๐(๐ฅ))
is a diffeomorphism. (Why? What is its inverse?) and, by (b.19),(โ โ ๐พ โ ๐)(๐ฅ1,โฆ, ๐ฅ๐) = (๐ฅ1,โฆ, ๐ฅ๐, 0,โฆ, 0) ,
i.e., (โ โ ๐พ) โ ๐ = ๐. Thus we can take the ๐ in Theorem b.18 to be ๐0 ร ๐๐โ๐ and the ๐ to beโ โ ๐พ. โกRemark b.20. The canonical submersion and immersion theorems can be succinctly sum-marized as saying that every submersion โlooks locally like the canonical submersionโ andevery immersion โlooks locally like the canonical immersionโ.
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APPENDIX c
Good Covers & ConvexityTheorems
Let ๐ be an ๐-dimensional submanifold of ๐๐. Our goal in this appendix is to provethat๐ admits a good cover. To do so weโll need some basic facts about convexity.
Definition c.1. Let ๐ be an open subset of ๐๐ and ๐ a ๐ถโ function on ๐. We say that ๐ isconvex if the matrix(c.2) [ ๐
2๐๐๐ฅ๐๐๐ฅ๐(๐)]
is positive-definite for all ๐ โ ๐.Suppose now that๐ itself is convex and ๐ is a convex function on๐which has as image
the half open interval [0, ๐) and is a proper map ๐ โ [0, ๐). (In other words, for 0 โค ๐ < ๐,the set { ๐ฅ โ ๐ | ๐(๐ฅ) โค ๐ } is compact).
Theorem c.3. For 0 < ๐ < ๐, the set๐๐ โ { ๐ฅ โ ๐ | ๐(๐ฅ) < ๐ }
is convex.
Proof. For ๐, ๐ โ ๐๐๐ with ๐ โ ๐, let๐(๐ก) โ ๐((1 โ ๐ก)๐ + ๐ก๐) ,
for 0 โค ๐ก โค 1. We claim that
(c.4) ๐2๐๐๐ก2(๐ก) > 0
and
(c.5) ๐๐๐๐ก(0) < 0 < ๐๐
๐๐ก(1) .
To prove (c.4) we note that๐2๐๐๐ก2(๐ก) = โ1โค๐,๐โค๐
๐2๐๐๐ฅ๐๐๐ฅ๐((1 โ ๐ก)๐ + ๐ก๐)(๐๐ โ ๐๐)(๐๐ โ ๐๐)
and that the right-hand side is positive because of the positive-definiteness of (c.2).To prove (c.5) we note that ๐๐๐๐ก is strictly increasing by (c.4). Hence ๐๐๐๐ก (0) > 0 would
imply that ๐๐๐๐ก (๐ก) > 0 for all ๐ก โ [0, 1] and hence would imply that ๐ is strictly increasing.But
๐(0) = ๐(1) = ๐(๐) = ๐(๐) = ๐ ,so this would be a contradiction. We conclude that ๐๐๐๐ก (0) < 0. A similar argument showsthat ๐๐๐๐ก (1) > 0.
Now we use equations (c.4) and (c.5) to show that ๐๐ is convex. Since ๐๐๐๐ก is strictlyincreasing it follows from (c.5) that ๐๐๐๐ก (๐ก0) = 0 for some ๐ก0 โ (0, 1) and that ๐๐๐๐ก (๐ก) < 0 for
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212 Appendix c: Good Covers & Convexity Theorems
๐ก โ [0, ๐ก0) and๐๐๐๐ก (๐ก) > 0 for ๐ก โ (๐ก0, 1]. Therefore, since ๐(0) = ๐(1) = ๐, we have ๐(๐ก) < ๐
for all ๐ก โ (0, 1); so ๐(๐ฅ) < ๐ on the line segment defined by (1โ ๐ก)๐+ ๐ก๐ for ๐ก โ (0, 1). Hence๐๐ is convex. โก
Coming back to the manifold ๐ โ ๐๐, let ๐ โ ๐ and let ๐ฟ๐ โ ๐๐๐. We denote by ๐๐the orthogonal projection
๐๐ โถ ๐ โ ๐ฟ๐ + ๐ ,i.e., for ๐ โ ๐ and ๐ฅ = ๐๐(๐), we have that ๐ โ ๐ฅ is orthogonal to ๐ฟ๐ (itโs easy to see that ๐๐is defined by this condition). We will prove that this projection has the following convexityproperty.
Theorem c.6. There exists a positive constant ๐(๐) such that if ๐ โ ๐ satisfies |๐ โ ๐|2 < ๐(๐)and ๐ < ๐(๐), then the set
(c.7) ๐(๐, ๐) โ { ๐โฒ โ ๐ | |๐โฒ โ ๐| < ๐ }is mapped diffeomorphically by ๐๐ onto a convex open set in ๐ฟ๐ + ๐.
Proof. We can, without loss of generality, assume that ๐ = 0 and that
๐ฟ = ๐๐ = ๐๐ ร {0}
in ๐๐ ร ๐๐ = ๐๐, where ๐ โ ๐ โ ๐. Thus ๐ can be described, locally over a convex neigh-borhood ๐ of the origin in ๐๐ as the graph of a ๐ถโ function ๐โถ (๐, 0) โ (๐๐, 0) and since๐๐ ร {0} is tangent to๐ at 0, we have
(c.8) ๐๐๐๐ฅ๐= 0 for ๐ = 1,โฆ, ๐ .
Given ๐ = (๐ฅ๐, ๐(๐ฅ๐)) โ ๐, let ๐๐ โถ ๐ โ ๐ by the function
(c.9) ๐๐(๐ฅ) โ |๐ฅ โ ๐ฅ๐|2 + |๐(๐ฅ) โ ๐(๐ฅ๐)|2 .
Then by (c.7) ๐๐(๐(๐, ๐)) is just the set
{ ๐ฅ โ ๐ | ๐๐(๐ฅ) < ๐ } ,
so to prove the theorem it suffices to prove that if ๐(๐) is sufficiently small and ๐ < ๐(๐), thisset is convex. However, at ๐ = 0, by equations (c.8) and (c.9) we have
๐2๐๐๐๐ฅ๐๐๐ฅ๐= ๐ฟ๐,๐ ,
and hence by continuity ๐๐ is convex on the set { ๐ฅ โ ๐ | |๐ฅ|2 < ๐ฟ } provided that ๐ฟ and|๐| are sufficiently small. Hence if ๐(๐) is sufficiently small, then ๐๐(๐(๐, ๐)) is convex for๐ < ๐(๐). โก
We will now use Theorem c.6 to prove that๐ admits a good cover: for every ๐ โ ๐, let
๐(๐) โ โ๐(๐)3
and let๐๐ โ { ๐ โ ๐ | |๐ โ ๐| < ๐(๐) } .
Theorem c.10. The ๐๐โs are a good cover of๐.
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213
Proof. Suppose that the intersection(c.11) ๐๐1 โฉโฏ โฉ ๐๐๐is nonempty and that
๐(๐1) โฅ ๐(๐2) โฅ โฏ โฅ ๐(๐๐) .Then if ๐ is a point in the intersection๐๐1 โฉโฏโฉ๐๐๐ , then for ๐ = 1,โฆ, ๐ we have |๐ โ ๐๐| <๐(๐1), hence |๐๐โ๐1| < 2๐(๐1).Moreover, if ๐ โ ๐๐๐ and |๐โ๐๐| < ๐(๐1), then |๐โ๐1| < 3๐(๐1)and so |๐ โ๐1|2 < ๐(๐1). Therefore the set๐๐๐ is contained in๐(๐๐, ๐(๐1)) and consequentlyby Theorem c.6 is mapped diffeomorphically onto a convex open subset of ๐ฟ๐1 + ๐1 by theprojection ๐๐1 . Consequently the intersection (c.11) is mapped diffeomorphically onto aconvex open subset of ๐ฟ๐1 + ๐1 by ๐๐1 . โก
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Bibliography
1. Vladimir Igorevich Arnolโd, 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-
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497โ501.9. , Change of variables in multiple integrals II, The American Mathematical Monthly 108 (2001), no. 2,
115โ119.10. 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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vol. 94, Springer-Verlag New York, 1983.
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Index of Notation
A๐(๐), 14Alt, 14
๐ต๐(๐), 149๐ต๐๐ (๐), 149(๐๐), 16
๐ถ๐(U; ๐บโ), 196๐ถ๐(U; ๐), 195๐(๐), 189
deg, 77๐ท๐, 33๐๐, 33dim, 1๐๐ฅ๐, 35๐/๐๐ฅ๐, 34
๐ป๐(๐), 149๐ป๐(U; ๐), 195๐ป๐๐ (๐), 149โ, 56
I๐(๐), 17im, 2๐๐ฃ, 23๐๐, 36
ker, 2
๐ฟ(๐), 184L๐(๐), 8
๐ฌ๐(๐โ), 19๐ฟ๐, 34
๐(U), 195๐๐(U), 194
๐บ๐(๐), 110๐บ๐๐ (๐), 111
๐โ๐, 40
๐๐, 12๐ด๐, 12(โ1)๐, 12supp, 39, 71
๐๐๐๐, 33๐๐(๐), 104, 105โ, 9๐๐, 176๐๐, 13๐๐, 176
๐(๐, ๐), 139โง, 20
๐๐(๐), 149๐๐๐ (๐), 149
217
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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, 166convexโconcave, 141coset, 3cotangent space
to ๐๐, 35critial point, 135critical
point, 85value, 85
critical value, 135current density, 62ฤech
cochain complex, 195cohomology groups, 195
Darboux form, 63de Rham cohomology, 149
compactly supported, 149
decomposable element, 21degree
of a map, 77derivative
of a ๐ถโ map between manifolds, 106determinant, 26diagonal, 184diffeomorphism, 97
orientation preserving, 117orientation reversing, 118
differential formclosed, 48compactly supported, 71, 111exact, 48on an open subset of ๐๐, 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
๐-relatedvector fields, 40vector fields on manifolds, 109
finite topology, 167Frobeniusโ Theorem, 92
genus, 143
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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
๐-formon an open subset of ๐๐, 44๐-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, 97MayerโVietoris Theorem, 162multi-index, 8
repeating, 15strictly increasing, 15
nerve, 168nerve of a cover, 195Noetherโs principle, 67
one-parameter groupassociated to a complete vector field,
40open set
parametrizable, 110star-shaped, 58
orientationcompatible, 180of a line in ๐2, 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 set๐ท-adapted, 120
parametrization, 104oriented, 118
permutation, 12transposition, 12
elementary, 12phase space, 66Poincarรฉ lemma
for compactly supported forms, 76for manifolds, 150for rectangles, 72
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Poincarรฉ 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, 146Sardโs theorem, 86scalar charge density, 62scalar curvature, 141sign
of a permutation, 13simply connected, 157sink, 146smooth domain, 118source, 146submersion, 209sumbersion
canonical, 209support
of a differential formon ๐๐, 71on a manifold, 111
of a vector field, 39symmetric group, 12symplectomorphism, 64
tangent spaceof a manifold, 98, 104, 105to ๐๐, 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