Elements of Homology Theory (Graduate Studies in Mathematics 81)

Elements of Homology Theo ry v. V. Prasolov Graduate Studies in Mathematics Volume 81 _._,i&.� American Mathematical...

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Elements of Homology Theo ry v. V. Prasolov

Graduate Studies in Mathematics Volume 81

_._,i&.�

American Mathematical Society Providence, Rhode Island

David Cox (Chair) Walter Craig

N. V. Ivanov Steven G. Krantz

B. B. IIpaconoB 8JIEMEHThI TEOPI1I1 rOMOJIOrl1H MIIHMO, MocKBa, 2005 This work was originally published in Russian by MIIHMO under the title "8neMeHTLI TeOpH:" rOMOnOr"�"

©

2005. The present translation was created under license for the

American Mathematical Society and is published by p ermission. Translated from the Russian by Olga Sipacheva

2000

Mathematics Subject ClasSIfication.

Primary 55-01.

For additional information and updates on this book, visit www

.ams.org/bookpages/gsm-81

Library of Congress Cataloging-in-Publication Data Prasolov, V. V. (Viktor Vasil'evich)

[Elementy teorii gomologii. English]

Elements of homology theory / V. V. Prasolov. p. cm. - (Graduate studies in mathematics; v. 81) Includes bibliographical references and index. ISBN-13: 978-0-8218-3812-9 (alk. paper) ISBN-I0: 0-8218-3812-1 (alk. paper) 1. Homology theory. I. Title. QA612.3.P73

2007

514'.23---dc22

2006047074

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©

2007 by the American Mathematical Society. All rights reserved. The American Mathematical Society retains all rights except those granted to the United States Government. Printed in the United States of America.

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The paper used in this book is acid-free and falls within the guidelines established to ensure permanence and durability. Visit the AMS home page at http://vvv 1098 7 65 4 3 21

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BIDS

.orgl

12 11 10 09 08 07

Contents

Preface

vii

Notation Chapter 1.

ix

Simplicial Homology

1

§1.

Definition and Some Properties

1

§2.

Invariance of Homology

6

§3.

Relative Homology

12

§4.

Cohomology and Universal Coefficient Theorem

21

§5.

Calculations

35

§6.

The Euler Characteristic and the Lefschetz Theorem

51

Cohomology Rings

59

§1.

Multiplication in Cohomology

59

§2.

Homology and Cohomology of Manifolds

69

§3.

The Kiinneth Theorem

95

Chapter 2.

§1.

Homology and Homotopy

111 111

§2.

Characteristic Classes

131

§3.

Group Actions

173

§4.

Steenrod Squares

184

Chapter 3.

Chapter 4. §1.

Applications of Simplicial Homology

Singular Homology

Basic Definitions and Properties

195 195

-

v

Contents

vi

§2. §3.

The Poincare and Lefschetz Isomorphisms for Topological Manifolds

227

Characteristic Classes: Continuation

252

Chapter 5.

§1. §2. §3.

263

Sheaf Cohomology

263

De Rham Cohomology

275

The de Rham Theorem

289

Chapter 6.

§1. §2. §3. §4. §5.

Cech Cohomology and de Rham Cohomology

Miscellany

301

The Alexander Polynomial

301

The Arf Invariant

317

Embeddings and Immersions

325

Complex Manifolds

339

Lie Groups and H-Spaces

344

Hints and Solutions

365

Bibliography

403

Index

411

Preface

This book is a natural continuation of the author's earlier book Elements of Combinatorial and Differential Topology (American Mathematical Society, Providence, Rl, 2006), which we refer to as Part I here. (A corrected Russian version of Part I is available at http://www.mccme.ru/prasolov.) In Chapter 1, we define simplicial homology and cohomology and give many examples of their calculations and applications. At this point, the book diverges from standard modern courses in algebraic topology, which usually begin with introducing singular homology. Simplicial homology has a simpler and more natural definition. Moreover, it is simplicial homology that is usually involved in calculations. For this reason, we introduce singular homology near the end of the book and use it only when it is indeed necessary, mainly in studying topological manifolds. Homology and cohomology groups with arbitrary coefficients ale expressed in terms of integral homology by means of the functors Tor and Ext. The properties of these functors are very important for homology theory, so we discuss them in detail. We first prove the Poincare duality theorem for simplicial (co ) homology. This proof applies only to smooth (to be more precise, triangulable) manifolds. There is no triangularization theorem for topological manifolds, and the proof ofthe Poincare duality theorem for them uses, of necessity, singular (co) homology. This proof is given in Chapter 4; it is very cumbersome. Chapter 2 considers an important algebraic structure on cohomology, the cup product of Kolmogorov and Alexander. It is particularly useful in the case of manifolds. Multiplication in cohomology is related to many topological invariants of manifolds, such as the intersection form and signature.

-

vii

viii

Preface

One possible approach to constructing multiplication in cohomology is based on a theorem of Kiinneth, which expresses the (co)homology of X x Y in terms of those of X and Y and is of independent interest. Chapter 3 is devoted to various applications of (simplicial) homology and cohomology. Many of them are related to obstruction theory. One of such applications is the construction of the characteristic classes of vector bundles. Other approaches to constructing characteristic classes (namely, the universal bundle and axiomatic approaches) are also discussed. Then, we consider the (co ) homological properties of spaces with actions of groups; we construct transfers and Smith's exact sequences. We conclude the chapter with constructing Steenrod squares, which generalize multiplication in cohomology. In Chapter 4, we define singular (co)homology and describe some of its applications; in particular, we obtain certain properties of characteristic classes. (Technically, it is more convenient to prove them by using singular cohomology, although the assertions themselves can be stated for simplicial cohomology. ) Chapter 5 considers yet another approach to constructing cohomology theory, namely, Cech cohomology and de Rham cohomology, which are closely related to each other. For the de Rham cohomology, we prove the Poincare duality theorem. Then, we carryover the construction of de Rham, which was originally introduced for smooth manifolds, to arbitrary simplicial complexes. The final Chapter 6 is devoted to various applications of homology theory, largely to the topology of manifolds. We begin with a detailed account of the Alexander polynomials, which we construct by using the homology of cyclic coverings; the Arf invariant is also considered. Then, we prove the strong Whitney embedding theorem. We also give a formula for calculating the Chern classes of complete intersections and discuss some homological properties of Lie groups and H -spaces. The book contains many problems (with solutions) and exercises. The problems are based on the materials of topology seminars for second-year students held by the author at the Independent University of Moscow in 2003. The basic notation, as well as theorems and other assertions, of Part I are mostly used without explanations; in some cases, we give references to the corresponding places in Part I. This work was financially supported by the Russian Foundation for Basic Research (project nos. 05-01-01012a, 05-01-02805-NTsNIL_a, and 0501-02806-NTsNIL....a) .

Notation

Hk(Xj G), the k-dimensional homology group of X with coefficients in Gj Hk(Xj G), the k-dimensional cohomology group of X with coefficients in Gj

Hom(A, B), the group of homomorphisms A

--+

Bj

A ® B, the tensor product of the Abelian groups A and Bj

Tor(A, B), see p. 29j Ext (A, B), see p. 29j Coker a, the cokernel of the homomorphism a (see p. 15)j [Mn], the fundamental class of the manifold Mnj

X(X), the Euler characteristic of Xj AU), the Lefschetz number of the map Ij u(M4n), the signature of the manifold M4nj Ek, the k-dimensional trivial vector bundlej Wk({), the kth Stiefel Whitney class of the bundle {j Ck({), the kth Chern class of the bundle {j Pk({), the kth Pontryagin class of the bundle {j Sqi, the Steenrod square.

-

ix

•

Chapter 1

Simplicial Homology

1. Definition and Some Properties The homology groups of a topological space X can be defined in several ways; different definitions are equivalent only for sufficiently good l spaces. The simplest definition is that of simplicial homology. Unfortunately, it has the essential drawback of not being invariant; to be more precise, the proof of its invariance requires some effort. (By invariance here we mean that the homology groups of homeomorphic spaces are isomorphic.) However, the main ideas of homology theory are most transparent at the level of simplicial homology; for this reason, we begin with a detailed discussion of simplicial homology. 1.1. Definition of Homology Groups. Let K be a simplicial complex (see Part I, p. 99). We assume that all of its simplices are oriented, i.e., the vertices of each simplex are ordered (the orientations of different simplices are not assumed to be compatible). We denote a simplex with vertices ao, al, ... , an (in this order) by lao, al,"" an]. Two simplices lao, al, ... , an] and [au(O) , au(l),"" au(n)] are said to have the same orientation ifsignu = 1; if sign u = -1, then these simplices have opposite orientations.

We define the boundary of a simplex as 8[0,1, ... ,n]

= L(-I)i[O, ... ,i, ... ,n],

where [0, ... , i, ... , n] = [0, ... , i-I, i + 1, ... , n]. At this point, we begin to consider formal sums of simplices. To be more precise, we declare simplices with the same vertices and orientations to be equivalent and consider the 1 Most of the spaces that arise in topology from geometry are sufficiently good, as opposed to those arising from functional analysis.

-

1

1. Simplicial Homology

2

set of equivalence classes of simplices. Moreover, we assume that the sum of two simplices with the same vertices and opposite orientations is o. Thus, we can denote simplices with opposite orientations by ~ and -~. It is convenient to denote the simplex [a, bJ by an arrow from its starting vertex a to the end vertex b. In this notation, the boundary of the simplex [0,1, 2Jlooks as shown in Figure 1; this fully agrees with thE' intuitive idea of a boundary. 2

Figure 1. The boundary of the simplex [0,1,2]

The following assertion is very important for homology theory.

Theorem 1.1. For any

simplex~, OO~

= o.

Proof. Let i > j. The simplex [... ,], ... , i, ... J occurs in the expression for 00[0, . .. , nJ two times: in (-I)i o [... , i, ... J with the sign (_I)i+i and in (-I)io[ ... ,J, ... J with the opposite sign (_I)i+i-l. 0 We could consider not only linear combinations of simplices with integer coefficients but also finite 2 sums of the form E ai~~' where each ai is an element of some Abelian group G and each ~~ is a simplex of dimension k. The expression E ai~~ is called an n-chain (with coefficients in the group G). Chains can be added to each other; therefore, they form an Abelian group. The group of k-chains is denoted by Ck(K; G). For brevity, we often denote this group by Ck(K) or simply C k . We have defined the map 0 for simplices. Extending it by linearity, we obtain a map Ok: C k --+ Ck-l, which we call the boundary homomorphism. For the O-simplex ~o, we set 8o~o = o. A chain c E Ck is called a cycle if OkC = 0, i.e., c E Ker Ok. The group of k-dimensional cycles is denoted by Zk. A chain C E Ck is said to be a boundary if c = Ok+Ic' for some chain c' E CH1, i.e., C E Imok+I. The group of k-dimensional boundaries is denoted by Bk. It follows from 00 = a that Bk c Zk; therefore, we can consider the quotient group Hk(K) = Zk/ B k . its elements are equivalence classes of cycles: cycles are equivalent if their 2For infinite sums, we could not define the homomorphism is a face for infinitely many simplices.

a in the case wn

re one simplex

1. Definition and Some Properties

3

difference is a boundary; such cycles are said to be homologous. The group Hk(K) is called the k-dimensional simplicial homology group of the complex K. To indicate the group of coefficients, we use the notation Hk(K; G). Remark. For the first time, "homological numbers" were mentioned in the works of Riemann (1857) and Betti (1871). The correct definition was given by Poincare. He realized that, in defining chains, not only the geometric shapes of simplices but also their multiplicities should be taken into account. Initially, homology groups (with coefficients in Z) were described in terms of their ranks (Betti numbers) and torsion numbers. In the 1926 short paper [97], Emmy Noether made the important observation that the equivalence classes of cycles can be regarded as groups. It can be seen from the definition that if a simplicial complex K contains no k-simplices, then Hk(K) = a. Let us calculate homology groups for some simplicial complexes. Example 1. If K consists of n isolated points, then

Ho(K; G)

= ~

and

Hk(K; G)

=

a

for k ~ 1.

n

Proof. Clearly, in the situation under consideration, we have Co = G EEl ... EEl G, Ck = a for k ~ 1, Zo = Co, and Bo = a. 0 Example 2. Let Sl be the simplicial complex that is the boundary of the 2-simplex [0,1,2]. Then HO(Sl; G) = HI (Sl; G) = G.

Proof. Clearly, B1 = 0, whence HI = Zl. Consider an arbitrary I-chain c

= ao[l, 2] + ad2, 0] + a2[a, 1]

E Cl ,

where ao, aI, a2 E G. We have

oc = (al - a2)[a]

+ (a2 -

+ (ao - al)[2]. a[l, 2] + a[2, 0] + ala, 1], where

ao)[l]

Therefore, Zl consists of chains of the form a E G.

The equalities o(a[l, 2] + a[2, 0]) = a[a] - a[l] and o(a[2, 1] + a[l, 0]) = a[a]- a[2] show that any a-chain has the form a[a] (up to a boundary). On the other hand, if a[a] = oc = (al - a2)[a]

+ (a2 -

ao)[l]

+ (ao -

then a2 = ao and ao = a1; hence a = al - a2 = a. HO(Sl; G) = Col Eo is isomorphic to G.

at}[2], Thus, the group 0

The argument which we used to calculate HO(Sl) applies to any connected simplicial complex.


4

Theorem 1.2. If K is a connected simplicial complex, then Ho(K; G)

= G.

Proof. Arbitrary vertices [m] and [n] can be joined by I-simplices [m, ill, [iI, i2], ... , [ik' n]; therefore,

a[n] - arm]

= 8(a[m, ill + ali!, i2] + ... + a[ik' n]).

This means that, up to a boundary, any O-chain has the form arm], where [m] is a fixed vertex. It remains to verify that if the chain arm] is a boundary, then a = O. Suppose that

arm]

= 8(Laa[ia,ja]) = Laa[ja]- Laa[ia]. a

a

a

The sum of coefficients on the right-hand side vanishes. Therefore, a

=

O.

[]

Exercise. Suppose that S2 is represented as the boundary of the simplex [0,1,2,3]. Prove that Ker 82 consists of chains of the form a8[D, 1,2,3] and Ker8l = Im~. 1.2. Chain Complexes. A chain complex is defined as a family of Abelian groups C k and homomorphisms 8k: Ck --+ Ck-l satisfying the relations 8 k8k+1 = O. If Ck = a for k < 0, then the chain complex is said to be nonnegative. An Abelian group F is free if it contains a set of elements {fa} such that any f E F has a unique representation in the form f = naJal + ... +na/cfa/c , where n a• E Z and all fal" .. , falc are distinct. The set {fa} is called a basis of the group F. A free Abelian group can be equivalently defined as a (finite or infinite) direct sum of copies of Z. If all of the groups Ck are free, then the chain complex is said to be free. For any chain complex C .. , we can consider the homology groups Hk(C,,) = Ker8k/Im8k+1' A chain map of chain complexes C .. and C~ is a family of homomorphisms CPk: Ck --+ C~ satisfying the condition ~CPk = CPk-18k for each k. Each chain map CPk: Ck --+ C~ induces a family of homomorphisms cp .. : Hk(C,,) --+ Hk(C~) for which (cp'I/J) .. = CP .. 'I/J ... Every simplicial map f: K --+ L induces the map A: Ck(K) --+ Ck(L) defined by I ([ Jk ao,···,ak ]) = {[f(ao), . .. , f(ak)]

a

if f(ad -=I !(aj) for i -=I j, if !(ai) = !(aj) for some i -=I j.

If the dimension of the simplex with vertices !(ao), ... , !(ak) is equal to k

or less than k - 1, then, obviously,

8~A

= A-18k,

i.e., the map

IS

chain (in

1. Definition and Some Properties

5

the latter case, both sides of this equality vanish). Suppose that I(ao) = I(at) and the points I(at), . .. , I(ak) are different. In this case, we have !k([ao, ... , ak]) = 0, and

!k-I(8[ao, ... , ak])

=

[/(at), !(a2), ... , I(ak)]-[/(ao), l(a2), ... , I(ak)]

= 0,

because I(ao) = I(at). Thus, any simplicial map I: K -+ L induces a homomorphism I.: Hk(K)[O] -+ Hk(L)j the identity map induces the identity homomorphism, and (lg). = I.g.· If I, g: K -+ L are simplicial maps and there exists a family of homomorphisms D k : Ck(K) -+ CkH(L) satisfying the conditions

8kH D k + Dk-1 8 k = gk -

A,

then such a family D is called a chain homotopy between I and g. The notion of chain homotopy has a geometric origin. Namely, suppose that I and 9 are maps joined by a homotopy H, and these maps (including the homotopy) are simplicial. Then the homotopy H determines the chain homotopy that assigns the (k + I)-chain H(/:l.k) to each k-simplex /:l.k. This chain is the curvilinear prism swept out by the image of the simplex during the homotopy. The boundary of the prism consists of the bases I(/:l.k) and g(/:l.k) and the lateral surface H(8/:l. k ). Taking orientations into account, we obtain precisely the expression for the chain homotopy:

8H(/:l.k) - H(8/:l. k ) = g(/:l.k) - I(/:l.k). Theorem 1.3 (on chain homotopy). II simplicial maps I, g: K chain homotopic, then I. = g•.

-+

L are

Proof. Suppose that Zk E Ck(K) and 8 kz k = O. Then

gk(Zk) - Ik(Zk)

= 8kH DkZk + Dk- 18kZk = 8kH(Dkzk),

Le., the cycles gk(Zk) and A(Zk) are homologous.

o

1.3. Homology of Simplices and of Their Boundaries. Theorem 1.4. II k ~ 1, then Hk(/:l.n)

= o.

Proof. Note that 6,n = [b, aI, . .. , an] is a cone (over /:l.n-l = [al, ... , an]). To each simplex [ail' ... ,ai",] we assign the simplex [b, ail' ... ,ai",]. Extending this map by linearity, we get a homomorphism Ck_l(/:l.n-l) -+ Ck(6,n); we denote the image of Ck-l by [b, Ck-l]. It is easy to verify that 8[b, Ck] = Ck - [b, 8ck] for k ~ 1 and 8[b, eo] = eo - E(eo)b, where E(E npaip) = E np. Any chain Zk E Ck(6,n) has a unique representation in the form Zk = Ck + [b,dk-l], where Ck E Ck (6,n-l) and dk- 1 E Ck_l(/:l.n-l). Suppose that


6

8Zk = O. Then 8Ck+dk-l-[b,8dk-l] = 0 for k > 1 and act +cio-c:(do)b for k = 1. In both cases, 8Ck + dk-l = 0; therefore, alb, Ck]

= Ck -

[b,8ck]

=0

= Ck + [b, dk-l] = Zk,

i.e., any cycle Zk E Ck(.6. n ) with k ~ 1 is a boundary.

o

Corollary. Let a.6. n be the simplicial complex consisting of all simplices in .6. n except.6. n itself. Then H n _ 1 (8.6. n ) = G (ifn ~ 2) and Hk(8.6. n ) = 0 for 0 < k < n - 1. Proof. Up to the dimension n - 1, the homology groups of a.6. n coincide with those of .6.n . The complex 8.6. n contains no n-simplices; therefore, lman = 0, which means that H n _ 1 (8.6.n ) = Ker8n - 1 . But Hn(.6. n ) = 0; hence Ker 8 n - 1 = lm~, where ~ is the differential in the chain complex for .6.n. If.6. n = [0, 1, ... ,n], then 1m ~ consists of elements of the form

a(l:~o(-I)i[o, ... ,i, ... ,n]), where

a E G.

0

2. Invariance of Homology First, we prove a theorem about acyclic supports, which allows us to prove the chain homotopy of two chain maps in many cases. Then, we twice apply this theorem in different situations to prove the topological invariance of homology. Finally, using the same theorem, we prove the homotopy invariance of homology. We can define homology with coefficients in any Abelian group G. But the case most important for applications is the one in which G is the additive group of some commutative ring with identity (e.g., G = Z, Q, or Zp); on the other hand, some important facts about homology (and especially cohomology) groups can be proved only for such groups of coefficients. For this reason, in what follows, we usually assume the group of coefficients to be the additive group of some commutative ring with identity. 2.1. Acyclic Supports. A simplicial complex is said to be acyclic if its homology coincides with that of the singleton. For example, the cone over any simplicial complex is acyclic; this is proved in precisely the same way as the acyclicity of simplices (see Theorem 1.4 on p. 5; the proof of this theorem uses only the representation of a simplex as a cone). Suppose we have a chain Ck = l:ak.6.f E Ck(L). We refer to any sub complex L' c L containing all the simplices .6.~ as a support of Ck. We say that a chain map cP is augmentation-preserving3 if CPO (2: ai.6.?) =

l: bj.6.~, where l: ai = 2: bj .

3The origin of this term is explained on p. 17.

2. Invariance of Homology

7

Theorem 1.5 (on acyclic supports). Let 'Pk, 1/;k: Ck(K) -+ Ck(L) be augmentation-preserving chain maps. Suppose that to any simplex ~ C K there corresponds a complex L(~) c L so that the following conditions are satisfied:

(1) if~'

c

~,

then L(~')

(2) the complex

L(~)

c

L(~)j

is acyclicj

(3) the complex L(~k) is a support for both chains 'Pk(~k) and 1/;k(~k). Suppose also that the coefficient group G is the additive group of a ring with identity. Then the maps 'Pk and 1/;k are chain homotopicj in particular, 'P. = 1/;•. Proof. We construct a chain homotopy Dk by induction on k. First, suppose that k = O. Let ~o be a vertex of K. The complex L(~O) is a support for the chains 'Po(~O) and 1/;0(~0). Since the maps 'PO and 1/;0 are augmentation-preserving, it follows that ('Po -1/;o)(a~O) = Ebi~?' where E bi = O. In the acyclic complex L(~O), the chain E bi~?' where E bi = 0, is the boundary of some chainj therefore, ('Po - 1/;o)(a~O) = 81Do(a~0), where Do(a~O) is a I-chain for which L(~O) is a support. The equality Do(a~O) + Do(b~O) = Do«(a + b)~O) may be false. To make it true, we choose an identity element 1 in the ring G, fix a I-chain Do(1· ~O), and set Do(a~O) = aDo(I· ~O). In what follows, we denote chains of the form 1· ~ simply by ~ (note that this notation makes no sense if G is an arbitrary group of coefficients).

Suppose that the required homomorphisms Do, ... , Dk-l are already constructed and each L(~i) is a support for the chain Di(~i). The only requirement to the homomorphism Dk is that 8k+lDk(~) = Ck for any ksimplex ~ in K, where Ck = 'Pk(~) -1/;k(~) - Dk-18k(~)' All simplices 8k(~) are contained in ~j hence L(~) is a support for the chain 8k(~), and therefore L(~) is a support for the chain Dk-18k(~)' Thus, L(~) is a support for Ck, and 8k C k

= (8k¢k = (8k¢k -

8 k'Pk - 8kDk--18k)(~) 8 k 'Pk - (1/;k-18k - 'Pk-18k - Dk-28k-18k))(~)

= O.

We have assumed that k ~ 1. The acyclicity of the complex L(~) implies Hk(L(~)) = O. Hence there exists a chain Dk(~) for which L(~) is a support and 8k+l D k (6.) = Ck. 0 2.2. Topological Invariance of Homology. In this section, we prove that any continuous map f: IKI -+ ILl of simplicial complexes induces a homomorphism f.: Hk(K) -+ Hk(L). Moreover, (lg). = f.g., which


8

implies the isomorphism of the homology groups of homeomorphic simplicial complexes. To define f*, we use the simplicial approximation theorem (see Part I, p. 105). Theorem 1.6. Suppose that the coefficient group G is the additive group of a ring with identity. Let K' be the barycentric subdivision of a simplicial complex K. Then the homology groups Hk(K') and Hk(K) are isomorphic. Proof. First, consider the simplicial map j: K' - K defined as follows. If K = ~k = [0,1, ... , kJ, then any simplicial map K' - K can be defined by assigning the vertices (labels) 0,1, ... , k to the vertices of the complex K'. To each vertex of K' which is the barycenter of some simplex [io, ... , i p ] we assign one of the vertices io, ... , i p • An example of such an assignment is given in Figure 2. It is easy to show that precisely one k-simplex of K' has complete set of labels, and the orientation of this simplex coincides with that of K. Indeed, take the barycenter of K and consider the face whose vertices have labels different from those of the barycenter; take the barycenter of this face, and so on.

Figure 2. A set of labels

For general simplicial complexes, the map j: K' - K is defined similarly. Note that it is a simplicial approximation of the identity map. The map j induces a chain map jk: Ck(K') - Ck(K). Now, let us define a chain map ik: Ck(K) - Ck(K'); it is not induced by a simplicial map. We set io(v) = v and idvo, VI] = [b, VI]- [b, vol, where b is the barycenter of the simplex [vo, VI]. Formally, the definition of i 1 can be written as il(~I) = [b,ioo~I]. For k > 1, we set ik(~k) = [b,ik_IO~k], where b is the barycenter of the complex ~k. It remains to verify that ik is a chain map, i.e., Okik = ik-IOk. Clearly, okik(~k)

= 0k[b, ik_lOk~k] = ik_lOk(~k) - [b, Ok_Iik_IOk~k]. Therefore, if Ok-lik-I = ik-20k-l, then Ok-lik-lOk = ik-20k-lfh =

°

and Okik = ik-lOk· The map j takes exactly one k-simplex from the baryc{; tric subdivision of ~k to ~k and preserves its orientation; the remaining k simplices are

2. Invariance of Homology

9

mapped to simplices of dimension less than k. This implies that ikik is the identity map, and for a support of the chain ikik(fl.'), where fl.' is any simplex from K', we can take the barycentric subdivision of the simplex A from K that contains A'. The barycentric subdivision of A is the cone over 8Aj therefore, the barycentric subdivision of a simplex is an acyclic simplicial complex. Thus, the maps ikik and idck(K') have a common acyclic support. Clearly, the map ioio takes vertices to verticesj hence, it is augmentationpreserving. Therefore, ikik is chain homotopic to the identity. Thus, the induced homomorphisms i*i* and i*i* are the identity maps, and hence the groups Hk(K') and Hk(K) are isomorphic. 0 Remark 1. The map i*: Hk(K) - Hk(K') has a simple geometric meaning at the level of chains; namely, to every simplex it assigns the sum of simplices into which it decomposes under the barycentric subdivision. The inverse map i*: Hk(K') - Hk(K) is canonical4 only at the level of homology. Remark 2. It is easy to prove Theorem 1.6 without appealing to acyclic supports. Indeed, it suffices to verify that i*i* is the identity map. But all simplices of the barycentric subdivision of A must have the same coefficients in any cyclej therefore, the restriction of ikik to Zk is the identity.

Let I: /K/ - /L / be a continuous map of simplicial complexes. We define a homomorphism 1*: Hk(K) - Hk(L) as follows. Take the nth barycentric subdivision K(n) and consider a simplicial approximation 'P: K(n) - L of I. We set 1* = 'P*iin), where iin ): Hk(K) - Hk(K(n)) is the canonical isomorphism. We must verify that this map is well defined, i.e., if tP: K(m) L is another simplicial approximation of I, then 'P*iin ) = tP*iim ). First, suppose that m = n. Theorem 1.7. II 'P, tP: K I: /K/-/L/, then 'P* = tP*·

L are simplicial approximations 01 a map

Proof. We use only the following property of the simplicial maps 'P and tP· Let ao,· .. , ak be the vertices 01 some simplex in K. Then the points 4We say that a map is canonical if it is uniquely determined by construction or for some other reasons. For example, any group isomorphism Z:I -+ Z2 is canonical. Any ring isomorphism Z -+ Z is canonical also, but a group isomorphism Z -+ Z is canonical only if it is assumed to take 1 to 1 rather than to -l. In linear algebra, the most important example of a noncanonical isomorphism is the isomorphism between a linear space V and its dual space V·. This isomorphism depends on the choice of a basis in V. In topology, noncanonical isomorphisms often arise between bundles because the fibers of bundles are homeomorphic but the homeomorphisms are not canonical; the canonicity of the homeomorphisms between fibers means the triviality of the bundle. Therefore, the homotopy and homology groups of different fibers are isomorphic, but the isomorphisms are not always canonical because the products of the base and the fibers may be twisted.


10

cp(ao), . .. , cp(ak), .,p(ao), ... , .,p(ak) are the vertices of a simple:/J in L. To prove this assertion, take a point x E int[ao, ... , ak]. The point f(x) is interior for a unique simplex [bo, . .. ,bzl in L. According to the definition of a simplicial approximation, we have cp(x), .,p(x) E [bo, ... , bzl. Therefore, since the maps cp and .,p are simplicial, it follows that the simplices cp([ao, ... , ak]) and .,p([ao, .. . , ak]) belong to [bo, . .. ,bd. The maps cpo,.,po: Co(K) -+ Co(L) are augmentation-preserving. According to the acyclic support theorem, the maps CPk and .,pk are chain homotopic. D Now, let us prove the equality cp.i~n) = .,p.i~rn) for n =1= m. For definiteness, suppose that n > m. We can take K(rn) (instead of K) for the initial complex and consider its pth barycentric subdivision for p = n - m. Then the required equality takes the form cp.i~P) = .,p•. In the proof of Theorem 1.6, we constructed a map j: K(l) -+ K, which is a simplicial approximation of the identity map. Recall that a composition of simplicial approximations is a simplicial approximation of a composition (see Part I, the corollary of Theorem 3.17 on p. 104). Therefore, the identity map has a simplicial approximation jCP): K(p) -+ K; the maps j't) and i~P) are mutually inverse. Since .,pj(p) is a simplicial approximation of I, it follows that cp. = .,p.j(p), i.e., cp.i(p) = .,p•. Theorem 1.8. If I: then (gf). = g.I.·

IKI

-+

ILl

and g:

ILl

-+

IMI

are continuous maps,

Proof. First, we construct a simplicial approximation .,p: L' -+ M of g, and then, a simplicial approximation cp: K' -+ L' of I. Let jK: K' -+ K and jL: L' -+ L be simplicial approximations of the identity maps. These maps form the commutative diagram

IKI ~ ILl ~ IMI

r r IK'I ~ IL'I ~ IMI· jK

Since

.,pcP is

jL

a simplicial approximation of gl, we have

(gf). = (.,pcp).(j~)-l = .,p.CP.(j~)-l. The maps jLcp and .,p are simplicial approximations of f and g, resppctively; hence I. = jfcp.(j! m ~ 1; in particular, n ~ 2. If the spaces lRn and lRm are homeomorphic, then the spaces lRn \ {O} '" sn-l and lRm \ {O} '" sm-l are homeomorphic as well; hence they are homotopy equivalent. Thus, Z = Hn_l(sn-l; Z) = Hn_l(sm-l; Z) and n - 1 > 0; therefore, m = n. This contradiction completes the proof. 0

3. Relative Homology Let K be a simplicial complex and L c K a sub complex. Then Ck(L) c Ck(K), and we can consider the quotient group Ck(K, L) = Ck(K)/Ck(L); the elements of this group are called relative k-chains. The homomorphism 8k: Ck(K) -+ Ck-1(K) induces a homomorphism 8 k : Ck(K,L) -+ Ck-l(K, L) with the same property 88 = o. Therefore, we can again take the groups Zk(K,L) = Ker8k and Bk(K,L) = Im8k+l and consider the quotient group Hk(K, L) = Zk/ B k , which is called the k-dimensional relative homology group. Example 3. If each path-connected component of a complex K contains at least one point of a sub complex L, then Ho(K, L) = o. Proof. Let us join a vertex x E K with some vertex y E L by a polygonal line 0: formed by edges. Then 8 1 0: = [x] - [y] '" [x]; therefore, Bo ::J Zoo D Exercise. Prove that Ho(K, L; G)

=

--------

G EB··· EB G, where n is the number n

of path-connected components of K containing no points of L. Problem 1 (excision isomorphism). Suppose that K U M is a simplicial complex, K and M are its sub complexes, and L is a sub complex in K. Prove that Hk(K, L) ~ Hk(K U M, L U M) for all k. A map of simplicial pairs J: (K,L) -+ (K',L') induces the following homomorphism J.: H.(K, L) -+ H.(K', L') of relative homology groups. First, we take the map Ck(K)/Ck(L) -+ Ck(K')/ JCk(L), and then, using the inclusion JCk(L) c Ck(L'), apply the canonical projection Ck(K')/ JCk(L)

-+

(Ck(K')/ JCk(L»/(Ck(L')/ JCk(L» ~ Ck(K')/Ck(L').

As a result, we obtain a chain map J#: C.(K, L) a homomorphism between the homology groups.

-+

C.(K', L'). It induces

3.1. Exact Homology Sequences of Pairs. The inclusion i: L -+ K induces a homomorphism i.: Hk(L) -+ Hk(K). Moreover, any absolute cycle can be regarded as a relative cycle; thus, we have a homomorphism p.: Hk(K) -+ Hk(K, L). Let us construct a connecting homomorphism 8.: Hk(K, L) -+ Hk-l(L). Take a relative cycle Zk E Ck(K, L) and its absolute representative Zk; this means that Zk E Ck(K) and "'k = Zk+Ck(L).

13

3. Relative Homology

By assumption, (AZk = 0, i.e., OkZk E Ck-1(L). The map 0.: Hk(K, L) -+ H k-l (L) acting as Zk ........ OkZk is well defined because if Yk E Ck (L ), then

Ok(Zk

+ Yk) = OkZk + OkYk

rv

OkZk'

Theorem 1.12. The sequence of homomorphisms

... - - Hk(L) ~ Hk(K) ~ Hk(K,L) ~ Hk-l(L) - - ... is exact. Proof. (1) 1m i. C Kerp •. Any absolute cycle Zk E Ck(L) corresponds to the zero relative cycle. (2) Kerp. C Imi •. Suppose that an absolute cycle Zk E Ck(K) corresponds to a relative cycle homologous to zero. Then Zk = z~ + zZ, where z~ E Ck(L) and zZ = OZk+1 for Zk+1 E Ck+l(K). Therefore, the cycle Zk is homologous to the cycle z~ E Ck (L ). (3) Imp. C Kero•. Suppose that Zk E Zk(K) and Zk = Zk + Ck(L). Then Zk ........ OkZk = O. (4) Kero. C Imp•. Suppose that Zk = Zk + Ck(L) and OkZk = O. Then for a representative of the relative cycle Zk we can take the absolute cycle Zk. (5) Imo. C Keri •. If Zk-l homologous to zero in K.

= OkZk

E

Ck-l(L), then the cycle Zk-l is

(6) Keri. elmo•. If Zk-l E Ck-1(L) and Zk-l

Ck(K), then Zk-l

= O.(Zk + Ck(L)).

= OkZk,

where Zk E

0

Relative homology groups can be expressed in terms of absolute groups; namely, for k ~ 2, we have

where C L is the cone over L. The first isomorphism is obvious even at the level of relative chains. The second follows from the exact sequence

Hk(CL) ~ Hk(K U CL) ~ Hk(K U CL, CL) ~ Hk_1(CL) of a pair. Since the cone CL is contractible, it follows that Hk(CL) Hk_l(CL) = 0; therefore, the middle map is an isomorphism. Note that the complex K U CL is homotopy equivalent to K/ L because the complex CL is contractible and, hence, K U L (K U CL)/CL ~ K/ L. r-.J

Example 4. If n

~

1, then

Hk(D n , sn-l; G)

= {oG

for k for k

= n, 1= n.

14


Proof. The exact sequence

Hk(D n ) ---+ Hk(Dn,sn-l)

---+

Hk_1(sn- 1) ---+ Hk_l(D n )

shows that Hk(Dn,sn-l) ~ Hk_l(sn-l) for k ~ 2. For k = 0, the required assertion follows from Example 3. For k = 1, we have one of the exact sequences

and

o ---+ HI (Dl, SO)

---+

G EB G

---+

G

---+

O.

Here the map G - G is an isomorphism and G EB G - G has the form (a, b) 1--+ a + b. The former has zero kernelj the kernel of the latter consists of elements of the form (a, -a) and, therefore, is isomorphic to G. 0

Remark. The difficulties involved in the consideration of the case k = 1 can be avoided by turning to reduced homology, which is defined on p. 17. A generalization of exact homology sequences is the following algebraic construction. Suppose we have a short exact sequence of chain maps O ---+ C •'

i. ---+

C•

p. ---+

e".---+ 0

(for example, C~ = Ck(L), Ck = Ck(K), and C; = Ck(K)/Ck(L)). Then we can define a connecting homomorphism a.: Hk(C:) - Hk-l(C~), Namely, take zZ E C; for which a;zZ = O. We have zZ = PkCk for some Ck E Ck, and o = azzz = Pk(akCk). Therefore, akCk = ik-ldk_1 for some dk_ 1 E C~_l' We set a.z; = dk - I • It can be proved in precisely the same way as for exact sequences of pairs that this map is well defined at the homology level and the sequence

... ~ Hk(C~) ~ Hk(C.) ~ Hk(C:) ~ Hk-I(C~) ~ ... is exact.

Example 5. Let K be a simplicial complex, and let 0 - G' - G - Gil - 0 be a short exact sequence of Abelian groups. Then we have a short exact sequence of chain complexes

0---+ Ck(Kj G')

---+

Ck(Kj G)

---+

Ck(Kj Gil)

---+

O.

The connecting homomorphism {3.: Hk(Kj Gil) - Hk-1(Kj G) is called the Bockstein homomorphism. The most interesting examples of Bockstein homomorphisms can be obtained from the exact sequences

15


which determine homomorphisms Hk(Kj Zm) -+Hk-l(Kj Z) and Hk(Kj Zm)

Hk-I(KjZm). The cokernel of a homomorphism a: A A'/Ima. -+

A' is defined as Coker a

-+

=

Problem 2. Given a commutative diagram of Abelian groups with exact rows

0----+ A --+ B --+ C --+ 0

la

Ip

l~

o---+ A' ---+ B' ---+ C' ---+ 0, prove that there is an exact sequence

o -+ Ker a

-+

Ker {1

-+

Ker'Y

-+

Coker a

-+

Coker {1

-+

Coker'Y

-+

O.

For a triple of simplicial complexes LI C L2 C K, we can construct an exact homology sequence of the triple

as follows. The isomorphism

determines a short exact sequence

This short exact sequence of chain complexes induces the required exact sequence of homology groups. In dealing with exact sequences, the following assertion is often useful. Theorem 1.13 (Steenrod's five lemma). Suppose that

is a commutative diagram of Abelian groups with exact rows. If 'PI, 'P2, 'P4, and 'P5 are isomorphisms, then 'Pa is an isomorphism.

Proof. Consider the groups A~ = A2/Imal, A~ = Kera4, B~ = B2/Im{11, and B~ = Ker {14. It is sufficient to prove the required assertion for the

16


simpler diagram

,

02

03,

o ~ A2 ~ A3 ~ A4 ------+ 0

1~2 1~3, 1~4 I

, fJ2 fJ3 0~B2~B3~

B'4~ 0 ,

in which cP~ and cP~ are isomorphisms. Clearly, Ker CP3 C Ker(.B~CP3)

= Ker(cp~Q~) = Ker(Q~) = Im(Q~)j therefore, Kercp3 ~ Ker(cp3Q~) = Ker(.B~cp~) = 0, i.e., CP3 is a monomorphism. Moreover, Im(.B~CP3) = Im(cp~Q~), where Q~ is an epimorphism and cP~ is an isomorphism. Hence .B~CP3 is an epimorphism, and B3 = 1m CP3

+ Ker .B~.

Since

Ker.B~

it follows that B3

= Im.B~ = Im(.B~cp~) = Im(cp3Q~)

= 1m CP3,

C Imcp3,

D

i.e., CP3 is an epimorphism.

Remark. The five lemma remains valid if the diagram is commutative up to sign, i.e., CP2 Ql = ±.BICPl, etc. Indeed, the proof involves only the groups Ker and 1m, which do not change under the replacement of cP by -cpo Using the five lemma, we can easily prove the following theorem.

Theorem 1.14. Suppose that f: (K,L) -+ (K',L') is a map of pairs for which the induced maps K -+ K' and L -+ L' are homotopy equivalences. Then f.: Hk(K, L) -+ Hk(K', L') is an isomorphism. Proof. Consider the following commutative diagram with exact rows:

Hk(L)

~ Hk(K) ~ Hk(K, L) ~ Hk-l(L) ~ Hk-l(K)

'·l

' · 1 ~

isO

Hk(L')

.

iso

Hk(K')

'·1?

'·l

iSO

.

'·liSO

~ Hk(K', L') ~ Hk-l(L') ~ Hk-l(K').

The five lemma implies that the middle vertical arrow is an isomorphism.

D

Problem 3. Suppose that 02

03

0~A2~A3~A4~0

1~2 fJ2 1~3 fJ3 1~4 0~B2~B3~B4~0

is a commutative diagram of Abelian groups with exact rows. (a) Prove that if CP2 and CP4 are monomorphisms, then phism.

C(3

is a monomor-

17


(b) Prove that if r.p2 and r.p4 are epimorphisms, then r.p3 is an epimorphism. Problem 4. Given a commutative diagram

with exact rows, prove that r.p2 induces an isomorphism : Ker(r.p3Q2)/(KerQ2

+ Kerr.p2) --+ (Imr.p2 nlmlh)/Im(r.p2Qd.

Problem 5 ([83]). (a) Given two commutative diagrams satisfying the assumptions of Steenrod's five lemma in which the respective homomorphisms, except for the isomorphisms r.p~ = TIl and r.p~ = T/2, coincide, prove that for any x E A 3 . (b) Suppose that the diagrams in (a) are diagrams of rings and their homomorphisms. Prove that the isomorphisms TIl and Tl2 in these diagrams can be different if and only if there exists a nontrivial additive homomorphism 6: A3 --+ A3 such that dQ2 = 0, Q3d = 0, and d(xy) = (dX)Y + xdy + (dX)(dy) for all x, y E A 3 . (c) Give an example of two diagrams satisfying the assumptions of (a) for which the homomorphisms TIl and T/2 are different. 3.2. Reduced Homology. The statements and proofs of many theorems can be simplified by considering the reduced homology groups ih(K) instead of the homology groups Hk(K) themselvesj below, we give two equivalent definitions of reduced homology groups. Definition 1. The reduced homology group Hk(K) is the kernel of the homomorphism p*: Hk(K) --+ Hk(*), where p: K --+ * is the map from the complex K to the singleton. Definition 2. Let us replace the map 80: CoCK) --+ 0 by £: CoCK) --+ Z, where €(v) = 1 for each vertex v. If Cl E CI(K), then £alCI = OJ therefore, for the new chain complex, we can also define the reduced homology group Hk(K). The map £ is called an augmentation. This term is used because we augment the chain complex

18


to the chain complex ... ---+

where C_l(K)

Cl(K) ~ Co(K)

-=--. C-l(K) ---+ 0,

= Z.

Definitions 1 and 2 are equivalent because Hk(K) = Hk(K) for k ~ 1 and the map Po: Co (K) --+ Co (*) = Z coincides with c. For an arbitrary map i:

Ho(K)

* --+ X,

we have pi = id.; therefore,

= Imp. ffi Keri. = Ho(K)

ffi Z.

Thus, Ho(K) ~ Ho(K, *). For reduced homology, exact sequences of pairs are also defined. Example 4 shows that Hk(Dn, sn-l) ~ Hk(sn) for all k. We have already shown that Hk(K, L) ~ Hk(K U CL) for k ~ 2. Using reduced homology, we can also show that Hk(K, L) ~ Hk(KUCL) {or k = 0 and 1.

Exercise. Given a simplicial (possibly, disconnected) complex K embedded in R n , prove that Hi(Rn,K) ~ Hi l(K) for i ~ 2 and HI(Rn,K) ~ Ho(K). Moreover, a path between points of K represents the zero element of the group HI (Rn , K) if and only if these points belong to the same connected component of K. 3.3. The Mayer-Vietoris Sequence. The exact sequence of MayerViet oris relates the homology groups of the union of two simplicial complexes Ko and K l , of their intersection, and of these complexes themselves. In the simplest case, where Ko and Kl consist of finitely many points, we can use the well-known inclusion-exclusion formula N = No + Nl - NOl, where N is the number of points in Ko U K l , No and Nl are the numbers of points in Ko and K 1 , respectively, and NOl is the number of points in Ko n K 1 . In the language of exact sequences, this formula can be written as

0---+ Ho(Ko n KI)

---+

Ho(Ko) ffi Ho(Kd

---+

Ho(Ko U KI)

---+

O.

Another simple example of an exact sequence of this form arises in the homology of the wedge K = Ko V Kl of two simplicial complexes. In this case, we have the isomorphism Ck(K) ~ Ck(Ko) ffi Ck(KI) for k ~ 1 already at the level of chains. Moreover, the kernel of the boundary homomorphism on Cl(K) is the direct sum of the kernels of the boundary homomorphisms on Cl(Ko) and C1(Kt). Therefore, for any k ~ 1, we have the canonical isomorphism Hk(K) ~ Hk(Ko) ffi Hk(KI). These examples suggest the existence of an exact sequence

0 - Hk(Ko

n Kt)

---+

Hk(Ko) ffi Hk(Kd

---+

Hk(Ko U kl)

---+

o.

19


However, representing the circle S1 as the union of two arcs Ko and K1 with two-point intersection, we see that no such exact sequence exists for k = 1. In reality, the k-dimensional homology of a complex Ko U Kl depends not only on the k-dimensional homology ofthe complexes Ko n K1, K o, and K1 but also on their (k - 1)-dimensional homology; the correct exact sequence is as follows.

Theorem 1.15 (Mayer Vietoris [84, 142]). Suppose that K is a simplicial complex, Ko and Kl are subcomplexes of K such K = Ko U K 1, and L = Ko n K 1 . Then there is an exact sequence ... --+

Hk(L)

--+

Hk(Ko) ffi Hk(KJ)

--+

Hk(K)

--+

Hk-l(L)

--+ .. , .

Proof. The Mayer Vietoris sequence arises from the exact sequence

(1)

0 --+ C.(L)

(jOl-jI),

C.(Ko) ffi C.(KJ) ~ C.(K)

--+

0,

which we describe below. The complex C.(Ko) ffi C.(Kd consists of the groups Ck(Ko) ffi Ck(K1 ), and the boundary homomorphism in it is the direct sum of the boundary homomorphisms; i.e., a( co, c 1) = (ac O, 1 ). The maps jO/: L --+ KO/ and iO/: KO/ --+ K are the natural embeddings.

ac

Let us show that (1) is exact. The map (jo, -jl) is, obviously, a monomorphism. It follows from K = Ko U Kl that (io, i1) is an epimorphism. Indeed, take c = L ai~~ E Ck(K) and let cO be the sum of all terms for which ~f C Ko. Then Kl is a support of the chain c - co, and (io, i1)(CO, c - cO) = cO + (c - cO) = c. It remains to verify exactness in the middle term. The image of the map (jo, -JI) consists of chains of the form (c, -c), where c is a chain in L. The kernel of (io, iJ) consists of chains of the form (c, -c), where c is a chain in KonKl =L. Clearly, the homology of the chain complex C.(Ko) ffi C.(K1 ) is isomorphic to H.(Ko) ffi H.(K1 ). 0 Mayer Vietoris exact sequences exist also for reduced homology. Indeed, setting (jo, -jl)(n) = (n, -n) and (io, iJ)(m, n) = m + n, we obtain the commutative diagram

with exact rows. Using the Mayer Vietoris exact sequence for reduced homology, we can easily calculate the homology groups of a suspension (see Part I, p. 130).

20


Theorem 1.16 (the suspension isomorphism). Let K be a simplicial complex, and let EK be the suspension over K. Then, for any k ~ 1, there exists a canonical isomorphism Hk(EK) ~ Hk-l(K). Proof. We can represent EK in the form Ko U Kl, where Ko and Kl are cones over K and Ko n Kl = K. Let us write the Mayer Vietoris sequence for reduced homology:

Hk(Ko) ffi Hk(K l )

--+

Hk(EK)

--+

Hk-l(K)

--+

Hk-l(Ko) ffi Hk-l(Kl ).

The spaces Ko and Kl are contractible; therefore, the first and last terms in this sequence are trivial, and the middle map is an isomorphism. 0

Remark. We use reduced homology because for ordinary homology, the case k = 1 must be considered separately. Problem 6. Calculate the homology of the torus T2. Problem 7. Prove that Hk(SP x sq, SP V sq)

~

Hk(Sp+q) for all k

~

1.

Problem 8. Calculate the homology of the space SP x sq. Problem 9. (a) Calculate the homology of the complement of a knot in S3 (the answer does not depend on the knot). (b) Consider a link with n connected components in S3. Calculate the homology of the complement of this link (the answer depends only on n). Problem 10. Let M n be a smooth manifold. Prove that Hk(Mn\int Dn) ~ Hk(M n ) for 1 ~ k ~ n - 2. (Here D n is a ball contained in some chart of Mn.) Problem 11. Suppose that K is a simplicial complex and C = {L l , ... , Ln} is a family of its sub complexes such that they cover K and all their intersections are acyclic. Prove that the homology groups of the complex K are isomorphic to those of the nerve6 of the cover C. The Mayer Vietoris theorem has the following generalization (concerning the relative M ayer- Vietoris sequence).

Theorem 1.17. For any subcomplexes Lo C Ko and Ll C Kl of a simplicial complex K, there is an exact sequence --+

Hk(Ko n Kl, Lo n Lt) --+

--+

Hk(Ko, Lo) ffi Hk(Kl, L l )

Hk(Ko U Kl, Lo U Lt)

--+

Hk-l(Ko n Kl, Lo n Ll)

6The definition of the nerve of a cOVer is given in Part I on p. 108.

--+ .

21

4. Cohomology and Universal Coefficient Theorem

Proof. The relative Mayer Vietoris sequence arises from the short exact sequence (2)

o ---+ G.(Ko n KI) G.(L o n Ll)

(jo,-il)

- -__I

G.(Ko) G.(Lo)

$

G.(KI) G.(LI)

(io,h) --

G.(Ko U KI) G.(Lo U Ld

---+

0

defined as follows. For each quadruple of Abelian groups HI C H 2, G 1 C G2, where HI C Gl and H2 C G 2 , the canonical map Gd HI -+ G 2 / H2 of their quotients is defined. The maps io, il and io, il are such canonical maps. If c E G.(L o) n G.(Ld, then c E G.(Lo

n Ld. Therefore, (jo, -it) is a

monomorphism. We show that (io, id is an epimorphism. Take c= E ai6.~ EG.(KoUKd and let cO be the sum of all terms for which 6.~ C Ko. Then the chain c - cO is contained in C.(Kl)' Therefore, the pair (cO, c - cO) can be associated with an element of the middle group in (2). The image of this element under the map (io, it) coincides with c (mod G.(Lo U Ll))' It remains to verify the exactness in the middle term. The image of the map (jo, -il) consists of relative chains (c (mod G.(Lo)), -c (mod G.(Lt))), where c is a chain in KonKo. The kernel of (io, il) consists of relative chains (cO (mod G.(Lo)), c1 (mod G.(Ll))), where cO + c 1 E G.(L o U Ll)' Clearly, the image is contained in the kernel. Consider an element of the kernel. Let ifJ be the sum of all terms in the chain cO + c 1 that are contained in G. (L o), and let c1 = - (cO + & + c l ). Then cl E G. (L 1 ) j therefore, for c we can take cO + & = _ (c 1 + ( 1 ). 0

4. Cohomology and Universal Coefficient Theorem 4.1. Cohomology. Suppose that K is a simplicial complex, G an Abelian group, and Gk(K) = Ck(Kj Z). A homomorphism ck: Ck(K) -+ G is called a k-dimensional cochain with coefficients (or values) in G. The group of k-dimensional cochains is denoted by Gk(Kj G) = Hom(Gk(K), G). The text in small print below refers to infinite simplicial complexes. It can be skipped at the first reading. For any family of Abelian groups {Go}, the direct sum ffio Go and the Cartesian product ITo Go are defined. Both groups consist of indexed sets (go) under componentwise addition. The difference between them is that the group ffi", Go consists of the sets (go) in which go "10 only for finitely many indices a, whereas the group ITo Go consists of all such sets. For finite families of groups, Cartesian products coincide with direct sums. Suppose that the k-simplices in a complex K are indexed by a. According to the definition of groups of chains, we have Ck(Kj G) = ffio Go, where Go ~ G. But Ck(Kj G) = ITo Go for Go ~ G. Indeed, a cochain is a function on a set of

22


simplices with values in G. To each simplex ~~ any value can be assigned; infinitely many values may be nonzero. For the group of chains a direct sum rather than a Cartesian product is taken because boundary homomorphisms cannot be defined for Cartesian products. The reason for this is that the direct sums of free Abelian groups are free, while the Cartesian products are not. Suppose that G is the additive group of a ring with identity. Then to each simplex ~~ we can assign the k-cochain (~~). which takes the value 1 at ~~ and vanishes at all other k-simplices. Therefore, any k-cochain can be written as a formal sum L:go«~~)*, where go< E G. This sum can have infinitely many terms.

Take Ck E Ck(K) and ck E Ck(K; G). We denote the value of the homomorphism ck at the element Ck by (c k , Ck). Such a pairing allows us to associate with the boundary operator 0: Ck+1(K) --+ Ck(K) the dual operator c5: Ck (K; G) --+ Ck+ 1 (K; G) defined by

(c5c k ,Ck+1)

= (C k ,OCk+1)'

Note that the operator c5 increases dimension, whereas 0 decreases it. Exercise. Prove that the value of a cochain c5ck at the simplex [vo, . .. , Vk+1] equals k+I

L(ck, [vo, ... , Vi,"" Vk+1])' i=O

Remark. Sometimes, the operator c5: Ck(K; G) --+ CHI (K; G) is defined by

(c5c k ,Ck+1) = _(-l)k(ck,oCk+1)' This choice of sign is based on the following convention suggested by MacLane: when the n- and m-dimensional symbols are transposed, the righthand side must be multiplied by (-1 )mn; the maps 0 and c5 are then assigned the dimensions -1 and +1. We do not adopt this convention for two reasons. First, we use the duality between maps 0 and c5 very often, and the introduction of an additional sign would make it somewhat awkward. Second, to determine how the sign changes under transpositions of symbols, nontrivial considerations are needed every time. The equality 00

=0

implies c5c5

= O. Therefore, the cohomology groups

Hk(K; G) = Zk(K; G)j Bk(K; G)

and

HO(K; G)

= ZO(K; G)

are defined; here Zk and Bk+1 are the kernel ann. the image of the homomorphism c5: C k --+ C k+1, respectively. The elements of the groups Zk and Bk are called cocycles and co boundaries. Example 6. If K is a connected simplicial complex, then HO(K; G) = G.


23

Proof. Letel = [VO,VIJ. Then8q = [VIJ-[VOJ. Hence (deO,el) = (eO,8el) = (cO, [VI]) - (cO, [vo]). Thus, the equality deo = 0 means that the co chain cO takes the same value at any two vertices joined by an edge. For a connected simplicial complex K, this means that the cochain cO takes the same value at all vertices. Therefore, ZO(Kj G) = G. D Theorem 1.18. If G is the additive group of a field F, then Hk(Kj G) is the dual space of Hk(Kj G). Proof. The groups Ck (K j G) and C k (K j G) are linear spaces over the field F, and C k is the dual space of Ck. The map 8 is linear, and the map d is its dual. Therefore, we must prove that if A: U - V and B: V - W are linear maps for which BA = 0, then Ker A* / 1m B* is the dual space of Ker B/lmA. The linear functions on Ker B can be interpreted as linear functions on V considered up to functions of the form g(Bx)j the functions g(Bx) constitute the space 1m B*. The linear functions on Ker B / 1m A can be interpreted as the linear functions on Ker B satisfying the condition f(Ax) = 0 for all Xj the functions for which j(Ax) = 0 form the space Ker A*. D Corollary. If G is the additive group of a field F and one of the linear spaces Hk(K; G) and Hk(K; G) over F is finite-dimensional, then Hk(K; G) ~ Hk(K; G)i this isomorphism is not canonical. Any simplicial map cp: K - L induces a homomorphism cp*: Hk(Lj G) - Hk(K; G) acting in the opposite direction. Indeed, each cochain ek E Ck(L; G) corresponds to the cochain cpkek E Ck(K; G) defined by cp k ck(6.) = ek (cp(6.)). The reduced cohomology groups are defined by replacing the augmentatione: Co(K) -Zbythedualmap€: G-CO(K;G),where€(co) =ge(CO). The groups of relative cochains are defined as Ck(K, Lj G)

= Hom(Ck(K, L); G).

Eaeh group of relative cochains is a subgroup in the group of absolute eochains. It consists of the eochains vanishing on Ck(L). Recall that the relative chain group is a quotient of the absolute chain group. Under duality, quotient groups correspond to subgroups. In the relative case, the coboundary and boundary homomorphisms are dual also. The relative cohomology groups are defined in a natural way.

Exercise. Prove that if each connected component of IKI contains at least one connected component of ILl. then HO(K, L; G) = O.


24

The short exact sequence

(3) induces the dual sequence

(4)

0 - Ck(LjG)

L

Ck(KjG)

L

Ck(K,LjG) -

O.

This sequence is exact as well, but proving this requires some effort. Moreover, the proof of the exactness of the sequence (4) uses a special feature of the sequence (3), namely, its split property. An exact sequence

c

a -+ A :!..... B :!!... -+ a is said to be split if it satisfies any of the three equivalent conditions in Lemma 1.1. Lemma 1.1. Let a -+ A :!..... B :!!... following conditions are equivalent:

c -+ a be an exact sequence.

Then the

(a) this sequence has the forrn 1

(5)

a -----+ A .2...... A ffi C

~C

-----+

0,

where i is the natural embedding and p is the natural projectionj

(b) the homomorphism cp has a left inverse, i.e., there exists a homomorphism 4>: B -+ A for which 4>cp = idA j (c) the homomorphism,p has a right inverse, i.e., there exists a homomorphism \lI: C -+ B for which ,p\ll = ide. Proof. Clearly, conditions (b) and (c) hold for any sequence of the form (5)j to see this, it suffices to set 4>(a, e) = a and \lI(c) = (0, e). It remains to verify that if (b) or (c) holds, then the sequence has the form (5). Suppose that 4>cp = idA. Let us show that B = Imcp ffi Ker4>. Any element b E B can be represented as b = cp4>(b) + (b-cp4>(b)), where cp4>(b) E 1m cp and b - cp4>(b) E Ker 4>. Moreover, if b E 1m cp n Ker 4>, then b = cp(a) and a = 4>(b) = 4>cp(a) = aj hence b = O. Suppose that ,pw = ide. Let us show that B = Ker.,p ffi 1m \lI. Any element b E B can be represented as b = (b - \lI.,p{b)) + \lI,p{b), where b - w1j;(b) E Ker,p and W,p(b) Elm \lI. Moreover, if bE Ker,p n 1m W, then b = w(e) and a = ,pCb) = ,p\I!(e) = ej hence b = O. 0 7Formally, this means that there exists an isomorphism /: B diagram is commutative:

--+

AEBe for which the follOwing

25


Problem 12. Prove Lemma 1.1 using the five lemma. Problem 13. Let p: E --+ B be a fibration 8 with fiber F. (a) Prove that if there exists a section s: B --+ E (this means that po s = idB is the identity map on B), then 7rn (E) ~ 7rn (B) E9 7rn (F). (b) Prove that if there exists a retraction r: E --+ F, then 7rn (E) ~ 7rn (B) E9 7rn(F). (c) Prove that if the fiber F is contractible in the space E, then 7rn (B) ~ 7rn (E) E9 7rn-l(F). It is easy to verify that if the group C is free, then the exact sequence

o --+ A ~ B !t C

--+ 0 is split. Indeed, the map '1}1: C --+ B can be constructed as follows. Take a basis in C; for each basis element c, we set w(c) = b, where b is any element of 1/J-l(c). Since the group Ck(K, L) is free, it follows that the exact sequence (3) is split.

Theorem 1.19. (a) If a sequence A ~ B ~ C dual sequence Hom(A, G)

(b) If a sequence 0 the dual sequence

--+

0 is exact, then so is the

¢ ...!!.- Hom(B, G) . - Rom(C, G) . - o. --+

0.- Hom(A, G)

A ~ B

!t C

--+

0 is exact and split, then so is

¢ ...!!.- Hom(B, G) . - Hom(C, G) . - O.

Proof. (a) First, let us show that Kef'l,b = O. Suppose that C E Ker¢, i.e., 0= ifi(c) = Co1/J. This means that c(1/J(b)) = 0 for all b E B. By assumption, 1/J is an epimorphism; therefore, c = O. ~ow, let us show that 1m ¢ = Ker rj;. The equality 1/J 0 cp = 0 implies Ker rj;. Suppose that b E Ker rj;, i.e., 0 = rj;(b) = b 0 cpo _This means that b(lm cp) = 0; therefore, b induces a homomorphism b/ : B / 1m cp --+ G. The homomorphism 1/J induces an isomorphism 1/J/: B / 1m cp --+ C. Consider the homomorphism b' (1/J') -1 E Hom( C, G).

'l: 0 1/J = 0, whence 1m ifi c

Clearly,

8The definition of a fibration is given in Part I on p. 162.

26


because the diagram

B

Yl~

G +-- B / 1m r.p ~ C

is commutative. (b) Let cI>: B composition

-+

A be a homomorphism for which cI>r.p = idA. Then the

Hom(A,G)

~

-+

-

Hom(B,G) ~ Hom(A,G)

is the identity. Therefore, .:p is an epimorphism, i.e., the dual sequence is exact. Clearly, the homomorphism <J, splits this sequence. 0 Exercise. Prove that for the exact sequence 0 o - Hom(Z, Z) - Hom(Z, Z) is not exact.

-+

Z ~ Z, the dual sequence

In conclusion, we calculate the groups Hom(A, B) in some simple cases. (1) Hom(Z, G) ~ G because any homomorphism r.p: Z -+ G is completely determined by the element r.p(1) E G, and this element can be arbitrary. (2) Hom(Zn, Z) = 0 because if r.p: Zn nr.p(l) = 0 and, therefore, r.p(1) = O.

-+

Z is a homomorphism, then

(3) Hom(Zn, Zm) = Zd, where d = GCD(n, m). Indeed, any homomorphism r.p: Zn -+ Zm is completely determined by the element r.p(1), which must satisfy the relation nr.p(l) == 0 (mod m), i.e., r.p(1) == 0 (mod mid). Thus, m 2m (d-1)m} r.p(1) E { 0, d' d"'" d .

We also have Hom(Zn, G) ~ Ker(G ~ G)j the proof is similar to that of (3). Exercise. Write the Mayer Vietoris exact sequence for cohomology using Theorem 1.19. Problem 14. Let A be an Abelian group and mA that Hom(A, Z) ~ Hom(mA, Z).

= {ma I a E

A}. Prove

Problem 15. Prove that the group Hom(Q, Q/Z) is uncolUltable. (Here Q is the group of rational numbers under addition.) 4.2. Tensor Product and Homology with Arbitrary Coefficients. We have defined the cohomology groups Hk(Kj G) with arbitrary coefficients using chain groups with coefficients in Z. Similarly, the homology groups Hk(Kj G) can be defined using chain groups with codfi ients in Z. The


27

groups Hom(Ck(K), G) are replaced by Ck®G, where ® is the tensor product of Abelian groups, which is defined as follows. Suppose that A and B are Abelian groups, F(A, B) is the free Abelian group with basis A x B, and R(A, B) is the subgroup in F(A, B) generated by the elements of the form (a + a', b) - (a, b) - (a', b) and (a, b + b') - (a, b) - (a, b'). Then A ® B = F(A, B)I R(A, B). The coset containing an element Ca, b) is denoted by a®b. Less formally, A ® B is the group with generators a ® b and defining relations (a + a') ® b = a ® b + a' ® b and a ® (b + b') = a ® b + a ® b'. Remark. The group A ® B may contain an element al ® bl that cannot be represented in the form a ® b.

+ ... + ak ® bk

It is easy to show that Z ® G ~ Gj the isomorphism is defined by n ® 9 1-+ ng. Indeed, n ® 9 = 1 ® 9 + ... + 1 ® 9 = 1 ® (ng). Therefore, any element of the group Z ® G ~ G has the form 1 ® gj there are no relations between these elements.

A little more involved argument proves that Zn ® G ~ GlnG. Indeed, any element of the group Zn ® G has the form 1 ® g, and 1 ® ng = O. In particular, Zn ® Zm ~ Zd, where d = GCD(n,m). Moreover, Zn ® Q = o. The very definition of tensor product readily implies the following two assertions. 1. !fa map ep: AxB -- C is bilinear, i.e., ep(a+a', b) = ep(a, b) +ep(a', b) and ep(a, b + b') = ep(a, b) + ep(a, b'), then there exists a homomorphism tj;: A ® B -- C for which tj;(a ® b) = ep(a, b). 2. For any two homomorphisms ep: A -- A' and 'If;: B -- B', there exists a homomorphism ep ® ¢: A ® B -- A' ® B' for which (ep ® ¢)(a ® b) ep(a) ® ¢(b). Indeed,

(ep ® ¢)[(a + a') ® b - a ® b - a' ® b]

= [ep(a) + ep(a')] ® ¢(b) -

ep(a) ® ¢(b) - ep(a') ® ¢(b)

= o.

For the tensor product, the following theorem is validj it is dual to Theorem 1.19. Theorem 1.20. (a) If a sequence A ~ B sequence

:!:... C

-- 0 is exact, then so is the

A®G~B®G~C®G--O, where 1

= idG.

(b) If a sequence 0 -- A ~ B the sequence

:!:...

C -- 0 is exact and split, then so is

0---+ A®G ~ B®G ~ C®G

---+

o.


28

Proof. (a) First, let us show that 1jJ ® 1 is an epimorphism and its kernel coincides with Ker1jJ ® G. The homomorphism 1jJ ® 1 induces a homomorphism

-----

1jJ ® 1: B ® G / (Ker 1jJ ® G)

-----

-+

C ® G.

It is sufficient to verify that 1jJ ® 1 is an isomorphism. Consider the map 'It: CxG -+ B®G/(Ker1jJ®G) defined by 'It(c,g) = b®g+Ker1jJ®G, where 1jJ(b) = c. This map is well defined because if 1jJ(b') = c, then b ® 9 - b' ® 9 = (b-b')®g, where b-b' E Ker1jJ. The map 'It is bilinear; therefore, it induces a homomorphism ~: C ® G are mutually inverse.

-+

B ® G / (Ker 1jJ ® G). The maps

¢®1 and ~

Thus, Ker(1jJ®l) = Ker1jJ®G = Imcp®G = Im(cp®l); the last equality holds because the group Im( cp ® 1) is generated by all elements of the form cp(a) ® g. (b) Let

~:

B

-+

(~ ® 1)

A be a homomorphism for which iPcp 0

(cp ® 1)

= iPcp ® 1 =

idA ® ide

= idA.

Then

= idA®e;

therefore, cp ® 1 is a monomorphism, and the map iP ® 1 splits the exact sequence of tensor products. 0 Exercise. Prove that the sequence 0 -+ Z ® Z2 exact sequence 0 -+ Z -+ Q is not exact.

-+

Q ® Z2 induced by the

4.3. The Groups Tor and Ext. The homology and cohomology groups with arbitrary coefficients can be expressed in terms of integral homology groups. The expressions include operations Tor and Ext, which assign Abelian groups Tor(A, B) and Ext(A, B) to pairs of Abelian groups A and B. In cmputations, it is sufficient to know Tor(A, B) and Ext(A, B) for A, B = Z, Zm, Q, and JR. The groups Tor(A, B) and Ext(A, B) are defined as follows. The Abelian group A can be specified by generators and defining relations. This means that there exists an exact sequence

(6)

i o -+ R -+ F

p

-+

A

-+

0,

where F and R are free Abelian groups (the group F is defined by generators and R, by relations; the group R is free because it is a subgroup of a free group). The exact sequence (6) is called a free resolution of the Abelian group A. The exact sequence (6) induces exact sequences

R®B ~ F®B ~ A®R

---+

0

and

Hom(R, B) ~ Hom(F, B)

J- Hom(A, B) ~ o.

29

4. Cohomology and Universal CoefIicient Theorem

We set Tor(A, B) = Ker(i ® 1) and Ext(A, B) = Cokeri' = Hom(R, B)/Imi' (note that both definitions use only the map i: R --+ F). Thus, the groups Tor(A, B) and Ext(A, B) complete the exact sequences under consideration to the exact sequences

o --. Tor(A, B) --. R

i®l

®B -

F ®B

p®l ----+

A ® B --. 0

and

o +-- Ext(A, B) +-- Hom(R, B) ~ Hom(F, B) ...L Hom(A, B) +-- O. First, we must check that the groups Tor(A, B) and Ext(A, B) thus defined do not depend on the choice of a free resolution. Lemma 1.2. (a) For an arbitrary homomorphism cP: A --+ A' of Abelian groups A and A', any free resolutions of A and A' can be completed to a commutative diagram i P O~R~F~A~O

l ~l

l~o

l~

i'

p' O~R'~F'~A'~O.

(b) If tPo and tPl are other homomorphisms completing the free resolutions to a commutative diagram, then there exists a homomorphism Do: F --+ R' for which i'Do = CPo - tPo and Doi = CPI - tPl . Proof. (a) Take bases {Ta.} and {ft3} in the free Abelian groups Rand F. The map p' is an epimorphism; therefore, F' contains elements f~ such that P'(J~) = cPp(Jt3). For the basis elements, we set cpo(Jt3) = f~; then, we extend CPo to a group homomorphism F --+ F'. We have p' CPo = cpp. By assumption, pi = 0; hence p'cpoi(ra) = cppi(Ta) = O. It follows from Ker p' = 1m i' that R' has an element T~ for which i'(T~) = CPOi(Ta). We set CPI(Ta ) = r~. Then i'CPI = cpoi. (b) Choose a basis {fa} in F. By assumption, p'(cpo-tPo)(fa) = cpp(Ja)cpp(Ja) = O. Therefore, R' has an element {T~} for which i'(T~) = (cpo Wo)(fa). We set Do(fa) = r~. Then i'Do = cpo - tPo, whence i'Doi = (cpo - tPo)i = i'(CPl -WI)' The map i is a monomorphism; therefore, Doi = CPI - tPl. 0 The awkward formulation of Lemma 1.2 has a very natural interpretation in the language of homological algebra. Namely, consider the sequence of homomorphisms

30


as a chain complex C. (here Co = F, Cl = R, and Ck = 0 for k > 1). Lemma 1.2 asserts that any homomorphism cp: A --+ A' induces a chain map C. --+ C~, which is unique up to chain homotopy. Thus, the induced homology map H.(C.) --+ H.(C~) is uniquely determined. Moreover, the corresponding homology group homomorphisms H .. (C. ® G) --+ H .. (C~ ® G) and H.(Hom(C~, G)) --+ H.(Hom(C., G)) are uniquely determined as well. It is these homomorphisms that we are interested in because HI (C. ® G) = Tor(A, G) and HI (Hom(C., G)) = Ext(A, G). Note also that Ho(C. ® G) = F®G/R®G ~ A®G and Ho(Hom(C.,G)) ~ Hom(A,G). Now, the relations (id). = id and (cp'I/J). = CP.'I/J. imply that the group Tor(A, B) is well defined. For Ext(A, B), the proof is similar; the only difference is that (cp'I/J). = 'I/J .. CP. for this group. Let us calculate the groups Tor(A, B) and Ext(A, B) in simple cases. Note that if the group A is free, then Tor(A, B) = 0 and Ext(A, B) = O. In particular, Tor(Z, B) = 0 and Ext(Z, B) = O. For the group A = Zn, consider the free resolution xn o ----+ Z ----+ Z ----+ Zn ----+ o.

First, we calculate Tor(Zn, B). Any element of the group Z ® B can be represented in the form 1 ® b. We must calculate the kernel of the map 1 ® b 1--+ n ® b = 1 ® nb. Clearly, it consists of the elements 1 ® b for which nb = O. Thus, the group Tor(Zn, B) is isomorphic to Ker(B ~ B). In particular, Tor(Zn, Z) = 0 and Tor(Zn, Zm) = Z(n,m)' Now, we calculate Ext(Zn, B). Clearly, Hom(Z, B) ~ B, and to multiplication by n in Z corresponds the homomorphism B --+ B that takes each b to nb. Therefore, Ext(Zn, B) = B/nB. In particular, Ext(Zn, Z) = Zn and Ext(Zn, Zm) = Z(n,m)' Problem 16. Let A = Zk EB T, where T is a finite Abelian group. Prove that Ext (A, Z) ~ T. Theorem 1.21. (a) If A is an Abelian group such that Ker(A ~ A) = 0 for any n E N, then Tor(A, B) = 0 for any group B. (b) If B is an Abelian group such that nB = B for any n E N, then Ext(A, B) = 0 for any group A. Proof. (a) Let A' be an arbitrary finitely generated subgroup in A. Then A' is a free Abelian group; therefore, for any free resolution .

o --+ R ~ F

p

--+

B

----+

0,


31

we have the commutative diagram

A' ® R

l'®i ----t

A' ® F

At~AL

(7)

in which the map l' ® i is a monomorphism. The vertical arrows in this diagram are monomorphisms too because Rand F are free groups. Let us prove that the map 1 ® i is a monomorphism. Take any element w = al ® n + ... + ak ® rk in A ® R. The group A' can be chosen so as to contain the elements al, ... , ak. Since the diagram (7) is commutative and, moreover, l' ® i and the right vertical arrow are monomorphisms, it follows that (1 ® i)(w) 1= o. (b) Any free resolution i 'P O~R~F~

A

~O

determines a homomorphism Hom(F, B) -+ Hom(R, B)j we have to prove that this is an epimorphism. Let 'P: R -+ B be a homomorphism. Take y E F \ R and extend 'P to the group generated by Rand y as follows: (1) if my ¢ R for all mEN, then we set cjS(y) = OJ (2) if my E R for some mEN, then we choose the least number n E N with this property and put cjS(y) = b, where b is the element of B such that nb = 'P(ny). An extension of'P over the entire group F can be constructed by induction (if the group F/ R is not finitely generated, then the induction is transfinite). 0 Corollary. If G = Q, JR, or C (we regard these fields as groups under addition), then Tor(G, B) = 0 and Ext(A, G) = 0 for any Abelian groups A andB. In calculating homology and cohomology groups with coefficients in G by using the universal coefficient theorem, the groups Tor(A, G) and Ext(A, G) arise. If G = Q, JR, or C, then we have Tor(A, G) = 0 because Tor(A, B) ~ Tor(B, A) for any Abelian groups A and B. Note that Ext does not have this propertyj for example, Ext(Z, Zn) = 0 and Ext(Zn, Z) = Zn. Problem 17. Prove that Tor(A, B) ~ Tor(B, A), and the isomorphism is canonical. An Abelian group T is said to be periodic if for any t E T, there exists a positive integer n for which nt = o.

32


=0

and T is a periodic group, then

Problem 19. Prove that if a sequence 0 the sequence

A :!.... B:!!.....C is exact, then so is

Problem 18. Prove that if Ext(T, Z)

T=O.

-

0----+ Hom(G,A) ~ Hom(G, B)

~

----+

Hom(G,C)

for any Abelian group G. An Abelian group G is said to be divisible if for any positive integer n, the map G - G defined by 9 1-----+ ng is an epimorphism.

Problem 20. Let 0 - A:!.... B:!!.....C - 0 be an exact sequence. (a) Prove that, for any free Abelian group F, the sequence

-

0----+ Hom(F, A) ~ Hom(F, B)

~

----+

Hom(F, C)

----+

0

is exact. (b) Prove that, for any divisible group G, the sequence

-

~

0-- Hom(A, G) ~ Hom(B,G) - - Hom(C,G) - - 0 is exact. A free resolution of an Abelian group A is often called a projective resolution because there is the dual notion of injective resolution; an injective resolution of an Abelian group A is an exact sequence 0 - A - G - H - 0 in which G and H are divisible Abelian groups. Any Abelian group A has an injective resolution. Indeed, any quotient of a divisible group is divisible. Therefore, it is sufficient to embed A in a divisible group G. Suppose that A = F / R, where F is a free group. The group F is a direct sum of a number of copies of Z; hence it is a subgroup of the group G defined as the direct sum of the same number of copies of Q. The group G has a subgroup corresponding to R. The quotient group G / R contains A as a subgroup. The group Ext(A, B) can be defined not only by using a projective resolution of A but also by using an injective resolution of B. Namely, let o - B - G - H - 0 be an injective resolution of the group B. According to Problem 19, the induced sequence 0 - Hom(A, B) _ Hom(A, G) Hom(A, H) is exact. Therefore, it can be completed to an exact sequence of the form

o ----+ Hom(A, B)

----+

Hom(A, G)

----+

Hom(A, H)

Problem 21. Prove that &t(A, B) ~ Ext(A, B).

----+

Ext(A, B)

----+

o.


33

Problem 22. Given an exact sequence of Abelian groups 0 - A _ B C - 0, prove that, for any Abelian group X, there are exact sequences

o ---+ Hom(X, A) ---+ Hom(X, B) ---+ Hom(X, C) ---+

Ext (X, A) -+ Ext(X, B)

---+

Ext(X, C)

---+

0

and

0-- Ext(A, X) - - Ext(B, X) - - Ext(C, X) - - Hom(A,X) - - Hom(B,X) - - Hom(C,X) - - O. 4.4. Universal Coefficient Theorem. Homology and cohomology with arbitrary coefficients can be expressed in terms of integral homology. Namely, the following theorem is valid.

Theorem 1.22 (universal coefficient formulas). For any Abelian group G, there are exact sequences

o ---+ Hk(K)

®G

---+

Hk(Kj G)

---+

Tor(Hk_l (K), G) -+ 0

and

0-- Hom(Hk(K), G) - - Hk(KjG) - - Ext(Hk-l(K),G) - - 0, where Hk(K)

= Hk(Kj Z).

These exact sequences splitj hence

and Hk(Kj G) ~ Hom(Hk(K), G) ffi Ext (Hk- 1 (K), G) (the isomorphisms are not canonical). Proof. Consider the exact sequence of chain complexes

It is assumed that Ck = Ck(Kj Z), etc. The group B k - 1 is freej therefore, tensoring with G, we obtain the exact sequence il8il Bl8il o ---+ Z k ® G --+ C k ® G ---+ B k- 1 ® G ---+ o.

34


Any short exact sequence of chain complexes induces an exact sequence of homology groups. Taking into account the fact that the homology groups of a chain complex with zero boundary homomorphisms coincide with the chain groups, we obtain the exact sequence -----+

Bk ® G

-----+

Zk ® G

-----+

Hk(C. ® G)

Bk

-----+

1

®G

-----+

Zk-l ® G

-----+ •

It follows directly from the definition of connecting homomorphisms in exact sequences of homology groups that the map Bk ® G --+ Zk ® G in this sequence is induced by the embedding j: Bk --+ Zk. Thus, we obtain the exact sequence 0--+ Zk®G/Bk®G

-----+

Hk(C.®G)

·1811

-----+

Ker(Bk-l®G ~ Zk-l®G)

Here Zk ® G/ Bk ® G = Hk ® G and Hk(C. ® G) = Hk(Kj G). ·1811

over, Ker(Bk-1 ® G ~ Zk-l ® G) sequence

o -----+ B k-l

-----+

= Tor(Hk _ 1 (K), G)

Z k-l

is a free resolution for the group H k -

1

-----+

H k-1

-----+

--+

O.

More-

because any exact

0

= H k - 1 (Kj Z).

Similarly, for cohomology, we have the exact sequence

0-- Hom(Hk(K) , G) - - Hk(KjG) - - Ext(Hk_1(K),G) - - O. It remains to show that these exact sequences split. In the exact sequence of homology groups, at the level of cycles, the homomorphism 'P: Hk(K) ® G --+ Hk(KjG) has the structure 'P((Eni~f) ® g) = Enig~f. It is required to construct a homomorphism ~: Hk(Kj G) --+ Hk(K) ® G for which ~'P = id. The exact sequence

o -----+ Z k

i -----+

Ck

8

-----+

B k-l

-----+

0

splits because the group Bk-1 is free. Suppose that I: Ck map, i.e., Ii = idz". Restricting the composition Ck ® G ~ Zk ® G

-----+

--+

Zk is a splitting

Hk ® G

to the cycles of the complex C. ® G, we obtain a map that induces a homomorphism ~: H k (K j G) --+ H k (K) ® G since I ® 1 takes boundaries to boundaries. Clearly, if Eni~f is a cycle in Ck, then ~'P((Eni~n ®g) = ~(E nig~n = (E ni~~) ® g. For cohomology, the fact that the exact sequpnce is split follows from Theorem 1.19, (b). 0

Corollary. IfG = Q, JR, ~ Hom(Hk(K), G).

ore,

then Hk(KjG) ~ Hk(K)®G and Hk(KjG)

35

5. Calculations

In calculating the groups Tor and Ext, the following fairly obvious isomorphisms are useful:

Ext(Al EB A 2, B)

~

Ext (AI, B) EB Ext(A2' B),

Tor(A l EB A 2 , B)

~

Tor(A 1 , B) EB Tor(A 2 , B).

Remark. For infinite families of Abelian groups, we have

Ext Tor

(~A.,B) = I] Ext (A., B),

Ext ( A,

II B) = II Ext(A, B

a ),

a

a

(~A.,B) = ~Tor(A.,B),

Problem 23. Prove that if H1(KjZ) then HI(Kj Z) ~ zr.

~

ZrEBT1, where Tl is a finite group,

Problem 24. Suppose that Hi(KjZ) = zn, EB11 and Hi(KjZ) = zm, EBr, where 11 and r are finite groups. Prove that mi = ni and r ~ 11-1' Problem 25. Derive the assertion of Problem 24 directly from the definition of homology and cohomology groups, without using the universal coefficient theorem. Problem 26. Let X and Y be finite simplicial complexes, and let f: X -- Y be a continuous map. Prove that if the map f.: HI (X) -- HI (Y) is zero, then so is f*: Hl(y) __ Hl(X).

5. Calculations 5.1. Fundamental Classes. Many homological properties of manifolds are consequences of the fact that for an orient able closed manifold Mn, the group Hn(Mnj Z) is nontrivial and generated by one homology classj a similar assertion is true for the Z2-homology of an arbitrary closed manifold. Pseudomanifolds 9 have the same homological properties.

Theorem 1.23. Let M n be a pseudomanifold, and let G be any Abelian group. (a) If Mn ha.~ nonempty boundary, then Hn(Mnj G) (b) If Mn is closed, then

Hn (Mn'G) ,

~

{G

Ker( G

= O.

if Mn is orientable, x2 --+

G)

if Mn is nonorientable.

9The definition of a pseudomanifold was given in Part I on p. 109. Recall that any smooth manifold can be triangulated, and a triangulated manifold is a pseudomanifold. Therefore, all theorems on pseudomanifolds presented below are also valid for smooth manifolds.

36


Proof. The group Hn(Mn; G) consists of chains en = E ai~f with ai E G for which 8en = o. Suppose that simplices ~f and ~j have a common face ~~-1. Since any pseudomanifold must be unramified, it follows that ~~-1 is a face for only these two simplices; therefore, it is contained in 8en with coefficient ±ai ± aj. The coefficient is ±(ai - aj) if the orientations of the simplices ~~ and ~j are compatible, and ±(ai + aj) otherwise. The coefficient ±ai ± aj can vanish only if ai = ±aj. Therefore, according to the strong connectedness condition, if 8en = 0, then en = E ±a~~, where the summation is over all n-simplices. If a simplex ~n-l belongs to the boundary of M n , then it is contained in 8en with coefficient ±a. Therefore, 8en = 0 implies a = 0; i.e., Hn(Mn; G) = o for any coefficient group G.

Changing (if necessary) the orientations of some simplices, we can assume that en = E a~f. If the pseudomanifold Mn is closed and 2a i- 0, then 8en = 0 if and only if the orientations of all simplices ~f are compatible; if 2a = 0, then 8en = O. Therefore, if M n is orientable, then Hn(Mn; G) ~ G, and if M n is nonorientable, then Hn(Mn; G) ~ Ker(G ~ G). 0 For any closed pseudo manifold M n , the homology class of the cycle E ~f in Hn(Mn; Z2) is called the fundamental class of Mn (more formally, this cycle should be written as E l·~f, where 1 E Z2). The term fundamental class is also used for the homology class of the cycle E ~f in Hn(Mn; Z) provided that the closed pseudomanifold Mn is oriented (the orientations of all simplices ~f are compatible with that of Mn). The fundamental class of a pseudomanifold M n is denoted by [Mn]; for the coefficient group Z2, the notation [Mnh is sometimes used.

Problem 27. Prove that if Mn is a closed orient able manifold, then the assertion of Problem 10 on p. 20 remains valid for k = n - 1. Is it essential that the manifold M n is closed? orientable? For pseudomanifolds with boundary, the same argument proves that if Mn is orient able, if Mn is nonorientable. Indeed, if a simplex ~n-l belongs to 8M n , then its contribution to the relative chain is zero. The definition of fundamental classes in relative homology is the same as in absolute homology. Let Mn and Nn be closed oriented pseudomanifolds. A map f: Mn -+ N n takes the fundamental class [Mn] to an element of the group HnCNn; Z), and any element of this group has the form k[Nn], where k E Z. The integer degf for which (degJ)[N n] = f*([Mnj) is called the degree of the map f. If

37

5. Calculations

the pseudomanifolds M n and N n are closed Cbut not necessarily orient able ), then the degree of f modulo 2 can be defined.

Remark. This definition is equivalent to that given in Part I on p. 111, where the degree of a map was defined as the number of preimages of points counted with signs.

If M n and Nn are pseudomanifolds with boundary, then we can also define the degree (or degree modulo 2) ofa map f: (Mn, aMn) -+ (Nn, aNn) by considering the image of the fundamental class [Mn] in the group Hn(N n , aNn;Z) (or in Hn(Nn,aNn;Z2))' Theorem 1.24. Let i.: HncaWn+l) -+ HnCwn+1) be the homomorphism induced by the natural embedding. If the pseudomanifold W n +1 is orientable and its boundary awn+l is connected, then i", = 0 for the coefficient groups Z and Z2, and if it is nonorientable, then i", = 0 for the coefficient group Z2. Proof. The group HnCalvn+l) is cyclic; therefore, it suffices to show that the fundamental cycle [awn+l] is the boundary of some cycle in W n+1 . Clearly, it is the boundary of the fundamental cycle [Wn+l]. If the pseudomanifold Wn+l is nonorientable, then its fundamental cyclt' is defined only for the coefficient group Z2. 0 Corollary. II I: awn+l -+ M n is the restriction of some continuous map F: Wn+l -+ Mn, the pseudomanilold wn+l is orientable, and its boundary awn+l is connected, then deg I = O. Proof. The homomorphism f",: Hncawn+l) as the composition f", = F",i",; hence f", = O.

-+

HnCMn) can be represented 0

Problem 28. Let f: sn -+ sn be a map, and letE/: ESn -+ Esn be the map whose restriction to sn x it} coincides with I for each t. Prove that deg I = deg Ef by using the suspension isomorphism. 5.2. Cellular Homology. If a simplicial complex K is endowed with the structure of a CW-complex Csee Part I, p. 118) X and each skeleton Xk is a subcomplex in K, then the integral homology of K can be calculated by using the chain complex C. in which every Ck is the free group whose generators are in one-to-one correspondence with the k-cells of the complex X. This substantially facilitates calculating the homology because the number of cells in X may be many times smaller than the number of simplices in K. For example, a minimal triangulation of the sphere sn contains 2n+l_l simplices (of all dimensions), while a minimal cell decomposition consists of only two cells (n-dimensional and zero-dimenSional).


38

Lemma. For i i= k, Hi(X k ,X k- 1) = 0, and H k (X k ,X k- 1) is the free Abelian group whose generators are in one-to-one correspondence with the k-cells of x. Proof. Any chain Ci E Ci (X k , X k-1) has a unique representation in the form of the (finite) sum of chains Cia:, where the {e~} are the k-cells of X. Moreover, 8Ci = 'E 8~a:. Any group Ci(e~, 8e~) has the form Ci(D k , Sk-l) for some triangulation of the disk Dk. Therefore, Hi (Xk, Xk-1) is the direct sum of the group~ H l (D k ,Sk-1). 0 The exact sequence

Hi+l(Xk+1,Xk) ____ Hi(Xk) ____ Hi(Xk+1) ____ Hi(Xk+1,Xk) for the pair (Xk+1,Xk) shows that Hi(Xk) ~ Hi(Xk+1) for i f. k,k Hence Hk_l(X k ) ~ H k_ 1 (X k+1) !:>! H k_ 1(Xk+2) ~ .. , ~ Hk-1(X).

+ 1.

Let Ck = Hk(Xk,Xk 1). Consider the map 8k : C k -+ C k- 1 defined as follows. We arrange the exact sequences for the pairs (Xk,X k- 1 ) and (X k - 1 , X k - 2 ) along horizontal and vertical lines:

The map 8k: Ck -+ Ck-l is the composition of the horizontal and vertical arrows. The group Hk(Xk) can be regarded as a subgroup in Ck. Moreover, we can identify Hk(Xk) with Ker 8k because if ep = hep' for a monomorphism h, then Kercp = Kerep'. Identifying Hk_l(X k - 1 ) with Ker8k_ 1 in a similar way, we obtain H k _ 1 (X k ) ~ Ker8k-I!Im8k. The map 8 k : C k

-+

Ck-l acts as follows. Suppose that to a generator

c~ E Ck corresponds the cell e~. Consider an absolute chain [e~] representing the relative fundamental class (e~, 8e~). At the level of H k-l ()( k-l ), we have 8[e~] = L: na:,6[e~-l]. Thus, 8~ = L: na:,6c~-l. The number na:,6 is equal to the degree of the map Sk-l -+ e~-l /8e~-1 = Sk-l induced by the

characteristic map of the cell e~. The equality 88

= 0 follows

from 88[e~1 = O.

39

5. Calculations

The homology with coefficients in Z2 is calculated similarlyj in this case, the degree modulo 2 should be considered. As an example, we calculate the homology of closed two-dimensional surfaces. We have already calculated the groups Ho and H 2 j let us calculate HI· Example 7. HI (nT2) = z2n and HI (nT2j Z2) = z~n (nT2 is a special case of the surface 8 2 #pT2 #q K #r p2 defined in Part I on p. 142). Proof. Consider the standard representation of nT2 by using a 4n-gon. For this representation, the complex for calculating the cellular homology has the form z ~ z2n ~ Z --. o.

Here 81 = 0 because there is one zero-dimensional cell, and (h = 0 because each I-cell occurs two times with opposite orientations in the boundary of the 2-ce11. For the coefficient group Z2, the calculation is similar. o For a sphere with n handles, the 2n cycles generating the one-dimensional homology group can be chosen as shown in Figure 3. To prove this, we must

Figure 3. Basis cycles

verify that these cycles are homologous to the sides of the 4n-gon from the sphere with handles is obtained. First, consider a handle from a disk is removed (see Figure 4). Clearly, to the cycles in Figure 3 spond those in Figure 5; the latter cycles are homologous to the sides polygon. Example 8. HI (mP2)

= zm-l ED Z2

and HI (mP2 j Z2)

which which correof the

= Zr.

Proof. The complex for calculating the cellular homology of mP2 has the form Z ~ zm ~ Z --. o. We again have 81 = 0, but this time, (h(!) = (2, ... ,2) because each I-cell occurs two times with the same orientation in the boundary of the 2-ce11. We

40


Figure 4. BlIBis cycles on a handle

Figure 5. BlIBis cycles on the polygon

must take the quotient group of zm modulo the subgroup generated by the element (2, ... ,2). Consider el - (1,0, ... ,0), ... , em-l = (0, ... ,0,1,0), and em = (1, ... ,1). We have

The condition (al, ... , am) E Imih is equivalent to al - am = 0, ... , am-lam = 0 and am == 0 (mod 2). Therefore, the quotient group zm / 1m ih is isomorphic to zm-l EB Z2; the group zm-l is generated by el. ... , em-I. and the group Z2 is generated by em. As before, for the coefficient group Z2, we have ih = O. 0 Cycles Q and (3 generating the integral homology group for the Klein bottle are shown in Figure 6. At the level of homology, we have 2(3 = 0 because the boundary of the Mobius band hatched in Figure 6a is homologous to 2(3. Problem 29. Calculate the homology groups of closed two-dimensional surfaces with coefficients in Zp for p f:. 2. Problem 30. Calculate the cohomology groups of closed two-dimensional surfaces with coefficients in Z.

41

5. Calculations

a

b

a

Figure 6. Basis cycles on the Klein bottle

Problem 31. Let H be a handlebody of genus g, that is, a three-dimensional body bounded by a sphere with 9 handles standardly embedded in JR3. Suppose that H is smoothly embedded in 8 3 and X = 8 3 \ H. Prove that Hl(X) ~ Z9; describe geometrically the generators of this group. Example 9. The following equality holds: Hk(cpn)

=

{Zo

if k = ~,2, 4, ... , 2n, otherwIse.

Proof. The space cpn has the structure of a CW-complex with cells Cpk \ Cpk-l (see Part I, pp. 118 119). We can construct a triangulation of the manifold cpn for which every skeleton X2k+l = X 2k = Cpk is a simplicial sub complex in the same way as the triangulation of a manifold with boundary was constructed in Part 1. The chain complex for calculating the cellular homology of the space cpn has the form Z

---+

0

---+

Z

---+

0

~

...

~

0

~

Z

~

0

---+

Z

---+

O.

The homology groups of this chain complex coincide with the chain groups.

o

n-l Example 10. (a) For k > 0, H2k(JRpn) = 0, and for 0 ~ k < -2-' H2k+l (lRpn) = Z2. (b) For 0 ~ k :::; n, Hk(JRpn j Z2) = Z2.

Proof. Suppose that IR.pn is endowed with the structure of the CW-complex with cells JRpk \ JR'pk - 1 (see Part I, pp. 118 119). Then Ck = Z for o ~ k ~ n. The central symmetry of the sphere 8 k - 1 preserves orientation for even k and reverses it for odd k. Therefore, the boundary homomorphism 8 k : Ck -+ Ck-l takes 1 to 1 + 1 = 2 for even k and to 1 - 1 = 0


42

for odd k. The chain complex for calculating the cellular homology of the complex lRpn for even n has the form Z

x2

---+

0

Z -- Z

x2

---+

Z

---+ ... ---+

x2

Z

---+

Z

0

---+

Z

---+

OJ

for odd n, it has the form Z ~Z ~Z

---+ ... ---+

Z ~Z ~Z

---+

O.

Therefore,

H.(lRpn) = { ;

if k = a or k = nand n is oddj is k is odd and a < k < nj otherwise.

For coefficients in Z2, all boundary homomorphisms are zero.

0

Problem 32. Calculate the integral homology groups of the n-torus Tn. Problem 33. Calculate the integral cohomology groups of the complex lRpn. Problem 34. Give an example of a noncontractible acyclic two-dimensional CW-complex with I-skeleton 8 1 V 8 1 . 5.3. The Intersection Number and the Poincare Duality Isomorphism. Suppose that a closed oriented manifold M n is decomposed into cells in two different ways so that there is a one-to-one correspondence between the open k-cells of one decomposition and the open (n - k)-cells of the other decomposition. We denote the corresponding open cells by Ui and ui, respectively. Suppose also that Ui n u; = 0 for i i= j and the cells Ui and ui transversally intersect each other in one point for all i. We assume that the cells Ui and ui are oriented so that if e1,· .. , ek and £1, ... , £n-k are positively oriented bases lO for Ui and ui, then e1, ... , ek, £1, ... ,£n-k is a positively oriented basis for Mn. Finally, suppose that the coefficient group is the additive group of an associative commutative ring R with identity. In this case, we can consider the intersection number for the chains E aiUi and E biUij it is defined as

The required pair of cell decompositions K and K* can be constructed, e.g., as follows. For K we take an arbitrary triangulation of the manifold Mn. Let K' be the barycentric subdivision of this triangulation. For each simplex 6. k in the triangulation of K, take all simplices from K' that intersect L\ k lOFor a simplex [vo •...• Vk). the basis Vl - Vo •...• Vk

-

Vo is oriented positively.

43

5. Calculations

precisely in its barycenter. Their union is a closed (n - k)-cell; we denote it by (.6. k) *. These cells form the decomposition K*. Using the intersection number, we can construct a bilinear map

cp: Hk(M n ; R) x Hn_k(M n ; R)

-4

R.

Namely, we take the cell decompositions K and K*, choose the cycles L: aiO'"i and L: bwi representing homology classes a E Hk(M n ; R) and /3 E Hn_k(M n ; R), and define cp(a, (3) to be the intersection number of these chains (cycles). We must verify that this intersection number does not depend on the cycles representing a and /3.

Lemma. If.6. i and.6.j are simplices of dimensions k and k-l, respectively, then Proof. First, note that the expressions on both sides of this equality are nonzero if and only if .6.j is a face of the simplex .6.i. Moreover, ((8.6. i , .6.;)) = ({±.6.j , .6.;)) = ±l and ((.6. i , 8(.6.;))) = ({.6.i' ±.6.;)) = ±l. It remains to determine the signs of .6.j in 8.6. i and of .6.; in 8(.6.;). Suppose that Vo is the barycenter of the simplex .6. i and vb is the barycenter of .6.j. Then.6. i and 8(.6.;) intersect in the point Vo, and 8.6. i and .6.; intersect in vb. Let V!, ... , Vk be the vertices of .6.j . We can assume that the orientations of the simplices [vo, vI, ... , VkJ and .6.i are compatible and, moreover, 8.6. i = .6.j + .. " i.e., .6.j = [Vb"" VkJ. Suppose that [vb, Vo, Vk+ I, .•. ,vn] is one of the simplices that form the cell .6.;, and its orientation is compatible with that of this cell. Then 8(.6.;) = [va, Vk+1,' .. , vn] + ... , where the dots denote the cells disjoint from .6.i. Augmenting a positively oriented basis of the simplex .6. i by a positively oriented basis of the cell 8(.6.;), we obtain a positively oriented basis of the simplex [vo, Vb ... , VnJ. We now find out what is the result of augmenting a positively oriented basis of .6.j = (VI"'" VkJ by a positively oriented basis of the simplex [vb, vo, Vk+1,"" VnJ. First, note that the simplex [vb, V2, .. ·, Vk] is contained in the simplex [VI, ... ,Vk], and the orientations of these simplices are compatible. Thus, we obtain a positively oriented basis of the simplex

[vb, V2, ... , Vk, Vk+l,"" vn] = (-l)k[vo, vb, V2, ... , vn]. It remains to note that [va, vb, V2," ., vn] is contained in [vo, Vb V2, . .. , Vn], and the orientations of these simplices are compatible.

0

It follows from this lemma that the map cp(a, (3) is well defined because if CI = 8d and 8C2 = 0, then {(Cl,C2)) = {(8d,c2)) = ±{(d,8c2)) = o.

Using the cell decompositions K and K*, we can obtain a correspondence between the chains L: aWi and 2: aiO'"i. However, this correspondence is not

44


interesting from the point of view of homology because the chains (8u)'" and 8(u"') belong to different spaces (if dimu = k, then dim(8u)'" = n - k + 1, whereas dim8(u"') = n - k - 1). It is more natural to associate each chain r . Ck = "L..J aiUi Wl·th a coch· aln Cn-k lor w h·lCh (n-k C 'Ui"') = ai, I.e., (Cn k,

L

btU;)

=L

atbi

=

((Ck'

L

bW;)}.

(In other words, the value of the cochain cn - k at the given chain is equal to the intersection number of cn k with ek; such a cochain exists and is unique.) We havE' (dcn k,d n HI}

= (cn - k ,8dn HI) = ((ck,8dn -k+l}) =

(-I)k((8ck,d n_k+l});

i.e., associated with each chain of the form 8Ck is the cochain ±dcn k. This means, in particular, that to cycles and boundaries correspond co cycles and coboundaries, respectively. Therefore, this isomorphism between the space of k-chains in K and the space of (n - k)-cochains in K'" induces an isomorphism between the groups Hk(M n ; R) and Hn-k(Mn; R). For the coefficient group Z2, we do not have to worry about orientations; therefore, for closed but not necessarily orient able manifolds, we can consider intersection numbers modulo 2 and obtain isomorphism Hk(M n ; Z2) ~ Hn-k(Mn; Z2). We have proved the following assertion. Theorem 1.25 (the Poincare duality isomorphism). (a) If M n is a closed orientable manifold and R is the additive group of a unital associative commutative ring, then Hk(M n ; R) ~ Hn-k(Mn; R). (b) If M n is a closed manifold, then H k (M n ;Z2) ~ Hn-k(M n ;Z2). Corollary. (a) If M n is a closed orientable manifold and F is the additive group of a field, then Hk(Mn; F) ~ Hn_k(M n ; F). (b) If Mn is a closed manifold, then H k (M n ;Z2) ~ H n _k(M n ;Z2). The subgroup formed by the finite-order elements in a group is called the torsion subgroup of this group. Problem 35 (Poincare duality). Suppose that Mn is a closed orientable manifold, Hk(M n ) ~ Zak EB Tk, and Hk(Mn) ~ Zbk EB Tk, where Tk and Tk are the torsion subgroups. Prove that ak = an-k, bk = bn - k , Tk ~ Tn-k-l, Tk ~ rn-k+l, and TI = o. The Poincare duality isomorphism and the universal coefficient theorem imply the following properties of the homology and coh mology groups of manifolds.

45

5. Calculations

Theorem 1.26. (a) If M n is a closed (connected) manifold and R is the additive group of a unital associative commutative ring, then

Hn(Mn. R) ,

~ {R R/2R

if Mn is orientable, if Mn is nonorientable.

(b) If Mn is a closed orientable manifold, then the group H n _l(Mn j '1.) is torsion-free, and if Mn is a closed connected (nonorientable) manifold, then the torsion subgroup of Hn_l(Mn j '1.) is isomorphic to '1.2. Proof. The group zn(Mn j R) is generated by co chains dual to n-simplices in Mn with the same orientation. Moreover, two cochains dual to n-simplices with the same orientation sharing an (n - 1)-face differ by a coboundary. Therefore, we have Hn(Mnj R) ~ R if the manifold is orientable and Hn(Mnj R) ~ R/2R otherwisej indeed, in the nonorientable case, every cochain cn dual to a simplex satisfies the relation 2cn = 0. For orientable manifolds, the Poincare duality isomorphism implies Tn-l ~ Tl = 0, and for nonorientable manifolds, Problem 24 implies Tn-l ~ Tn = ~. 0 Exercise. Let M4 be a closed orientable manifold. Prove that if the group Hl(M4) is torsion-free, then so are all other groups Hk(M4). Problem 36. Suppose that Mn is a closed orient able manifold and its suspension EMn is homeomorphic to a closed orientable manifold. Prove that Mn is a homology sphere, i.e., Hk(Mn) ~ Hk(8 n ) for all k. Remark. A homology sphere mayor may not be homeomorphic to the sphere. The best-known homology sphere not homeomorphic to the sphere is the Poincare 3-spherej its various descriptions can be found, e.g., in [108]. There are infinitely many pairwise nonhomeomorphic homology 3-spheres. Interestingly, the double suspension over any homology 3-sphere is homeomorphic to the sphere 8 5 (this was proved by Edwards). The bilinear map t.p: Hk(Mn) x Hn_k(Mn) - '1. constructed by using the intersection number has the following nondegeneracy property. Theorem 1.27. If Mn is a closed oriented manifold, then any homomorphism h: Hk(Mn) - Z can be represented as h(a) = t.p(a,f3h), where f3h E Hn_k(M n ). The element f3h is determined by the homomorphism h uniquely up to a finite-order element. Proof. The exact sequence


46

shows that the homomorphism h corresponds to an element of the group Hk(Mn; Z) determined up to an element of the group Ext(Hk_I(M n ), Z); this group is finite. Clearly, h(a) = cp(a, (3) for all a if and only if the co chain corresponding to a representative of the homology class (3 belongs to the cohomology class mapped to h. Note that if m{3 - 0, then mcp(a, (3)

-0.

= cp(a, m(3) = 0;

therefore, cp(a, (3) 0

If M n is a closed orient able manifold and the coefficient group F is the additive group of some field (or F = Z2 if M n nonorientable), then the bilinear map cp: Hk(M n ; F) x Hn_k(Mn; F) --+ F is nondegenerate in the usual sense.

Theorem 1.28. Any linear map h: Hk(Mn; F) --+ F can be represented as h(a) = cp(a, (3h), where (3h E Hn_k(M n ; F) is determined uniquely. Proof. The linear map h is an element of the space dual to Hk(M n ; F), that is, of Hk(Mn; F), which is isomorphic to Hn_k(M n ; F). The isomorphism takes h to (3h. 0 Corollary. In the spaces Hk(M n ; F) and Hn_k(M n ; F), there exist bases al, ... , am and {31, ... , {3m such that ((ai, {3j}} = dij. Proof. We choose an arbitrary basis aI, ... ,am and take the element corresponding to the linear map hj (E Xiai) = Xj for {3j. For these bases, we have ((ai, {3j)} = cp( ai, (3j) = hj (ai) = dij. 0

Similarly, if M n is a closed orient able manifold, then the free groups Hk(Mn)/Tk and Hn_k(Mn)/Tn_ k , where Tk and Tn - k are the torsion subgroups, have bases al,· .. , am and {31, ... , {3m for which ((ai, {3j}} = dij. Linking Number. Let CI and C2 be closed oriented curves in 8 3 . They determine one-dimensional cycles (with coefficients in Z), which we also denote by CI and C2. Let dl be a two-dimensional chain such that ad l = CI. The intersection number ((dl , C2)} is called the linking number of the curves CI and C2 and denoted by lk(cl, C2). It is easy to verify that the intersection number ((d l , C2)} does not depend on the choice of d l . Indeed, if d~ is another chain for which ad~ = ad l , then d~ - d l is a cycle and, hence, a boundary (because the two-dimensional homology group of the sphere 8 3 is trivial). Let e be a three-dimensional chain for which d~ - d 1 = ae. Then ((d~ - dl,C2)} = ((ae,c2}) = ((e,aC2}) = o. It is easy to show that this definition of linking number is equivalent to that given in PSknots, say. In [108], the linking number of closed oriented curves J and K was defined as follows. Consider the diagram of the oriented

47

5. Calculations

link formed by these curves. Look at the crosses where J passes under K. There are two types of such crosses (see Figure 7). For each cross under consideration, we set Ci = 1 or -1, depending on its type. The sum of all numbers Ci thus obtained is called the linking number of the closed oriented curves J and K.

IE

=1

IE

=-1

Figure 7. Two types of crosses

To show that the two definitions are equivalent, we arrange the curves and C2 so that they leave the plane of the diagram only in small neighborhoods of the crosses. We can also assume that the surface dl is located above the plane of the diagram, except in small neighborhoods of the crosses (see Figure 8). We are interested only in the crosses where the curve C2 passes Cl

Figure 8. The surface d 1

over Cl. These are precisely the places where C2 intersects the surface d1 . The signs do not cause any problem either. For example, in Figure 8, the basis eI, e2, el is oriented positively (we use the same notation as in the definition of the intersection number); this corresponds to a positive cross. There are many other definitions of the linking number of curves J and K. For example, consider the map f: J x K --+ 8 2 defined by

K(y) - J(x) f(x, y) = IIK(y) - J(x) II

(J and K are treated as circles and, simultaneously, as maps J, K: 8 1 IR3).

--+

48


Problem 37. Prove that deg f = ± Ik(J, K). Define orientations on the torus J x K and on the sphere S2 so that deg f = lk( J, K). Cohomology with Compact Support and Homology with Closed Support. The Poincare isomorphism theorem as stated above does not apply to noncompact manifolds. It is false even for the manifold lRn. Nevertheless, for noncompact manifolds, we can construct dual decompositions and try to establish a correspondence between chains Ck and co chains cn- k . Of course, such an effort cannot succeed, but we can see where it fails and we can try to remedy the situation. Recall that, by definition, any chain Ck is a finite sum of simplicesj therefore, when we assign a co chain cn- k to a chain Ck in the case of a noncompact manifold, we obtain only cochains that vanish on simplices outside some compact set, rather than all cochains. Such cochains are called cochains with compact supports. Clearly, applying the operator 5 to a cochflin with compact support, we obtain a cochain with compact support. Hence we can define cohomology groups H~(K) with compact supports for compactly supported cochains. If K and K* are dual decompositions of an orientable manifold, then we obtain an isomorphism Hk(Kj R) ~ H~-k(K*j R)j this isomorphism holds even at the level of (co ) chains. Now, let us try to assign a chain Ck to a cochain cn- k . When we assigned co chains to chains, we obtained something less than the set of all cochains. Now, on the contrary, we obtain something more than the set of all chains. Namely, instead of finite sums of simplices E ai~~' we obtain arbitrary sums of k-simplices, in which each simplex is contained with a certain coefficient but the number of simplices may be infinite. Such "chains" are called chains with closed supports because the union of all simplices with nonzero coefficients in such a chain is a closed set (possibly noncompact). Note that in defining chains with closed supports for any simplicial complexes, we must additionally require that each point belong to only finitely many simplices in the sum E ai~~. Otherwise, some simplex may be a face for infinitely many simplices in the sum Eai~~' and the chain a(Eai~~) is then not defined. But for triangulated manifolds, each simplex is a face for only finitely many simplices, and the two conditions are equivalent. Using chains with closed supports, we can construct homology groups Hkl(K) with closed supports. If K and K* are dual decompositions of an orient able manifold, we obtain an isomorphism Hkl(Kj R) ~ Hn-k(K*j R)j this isomorphism holds even at the level of (co)chains. Remark. Note that the groups H:(K) and H;I(K), in contrast to the ordinary cohomology and homology groups, are not homotopy invariants of the space K. For example, H~(lRnj Z) ~ Z and H~I(lRnj Z) ~ Z.

49

5. Calculations

5.4. Realization of Homology Classes of Surfaces. Any smooth curve I: 8 1 - M2 induces a homomorphism f*: H1 (81 ) - H1 (M2). We say that the curve I realizes the homology class 1.[81] E H1(M 2). It is fairly obvious that allowing self-intersecting curves, we can realize any homology class of a closed two-dimensional manifold M2. For this reason, we study realization of homology classes by only self-avoiding curves. We consider only orient able manifolds. We say that a homology class 0: over Z is primitive if 0: i= mj3, where mEN, m > 1, and f3 is a homology class over Z. Note that if mo: = 0, then 0: = (m + 1)0:; thus, an element of finite order cannot be primitive. Theorem 1.29. If a closed self-avoiding curve I realizes a homology class H 1 (M2), where M2 is a closed orientable manifold, then either 0: = 0 or 0: is a primitive homology class.

0: E

Proof (Samelson). If the curve I separates M2, then 0: = O. Indeed, let us represent one of the parts into which I divides M2 in the form of a twodimensional chain. The boundary of this chain is a representative of the homology class 0: or -0:. If the curve I does not divide M2, then there exists a closed self-avoiding curve g transversally intersecting I at precisely one point. To construct such a curve, it suffices to take a small interval transversally intersecting f and join its ends by a curve in M2 \ I (see Figure 9). Let ~ be the homology class realized by g. Then ((o:,~)) = ±1. On the other hand, if 0: = mf3, then ((o:,~)) = m((f3,~)). 0

Figure 9. The construction of the curve 9

Remark. The orient ability of the manifold was used to define the integer ((o:,~)). The generator of the group H1(JR'p 2 ) = Z2 is not primitive, but it is realized by a closed self-avoiding curve. Theorem 1.30 ([85]). Any primitive homology class of a closed orientable manifold M2 can be realized by a closed sell-avoiding curve.

[)U

1. ::ilmpllclal 11OmolOgy

Proof. To simplify notation, we denote curves and the homology classes they represent by the same symbols. Take curves 01. ... , Og, {31, ... , {3g, and 'YI, ... ,'Yg-I on the sphere with 9 handles as shown in Figure 10 (for 9 = 4). The homology classes 01, ... , 0g, {31, ... ,{3g generate the one-dimensional homology group. We denote the homology class alOI + .. ·+ago g+bg{31 + ... +bg{3g by (aI, bl , ... , ag, bg). This class is primitive if and only if GCD(al. bl , ... ,ag,bg) = 1.

Figure 10. Curves on a. sphere with handles

For each closed self-avoiding curve 'Y on a two-dimensional orient able surface M2, a homeomorphism M2 _ M2, called the Dehn twist along 'Y, is defined. Namely, consider a small neighborhood of'Y and choose one of the two parts into which it divides 'Y (in Figure 11, left, the chosen part is hatched). Let us make a cut along the curve 'Y and deform the chosen part of the neighborhood so that the points of the outer boundary remain fixed and the points of 'Y slide on 'Y. After a rotation through 360 0 , the points of the cut return to their initial positions; as a result, we obtain a homeomorphism. The Dchn twists are discussed in detail in [105, 108].

Figure 11. A Dehn twist

If a curve 0 transversally intersects 'Y in one point, then any Dehn twist along 'Y takes the homology class of 0 to 0 ± 'Y (thp. sign is determined by the direction of rotation in the Dehn twist). It is easy to verify that the homology class of 'Yk is ±(Ok - ok+d (see Figure 12). Thus the Dehn twists

51

6. The Euler Characteristic

Figure 12. Homologous cycles

act on homology classes as follows: f3k ~ o.k --+ o.k ± /Jk,

13k ~ 13k ± (o.k - o.k+!),

13k

Ok

--+

~

/Jk ± o.k i

13k+! ~ 13k+! ± (o.k - o.k+d·

Let (aI, bl, ... I ay, by) be an element of the group HI (M 2 ). By using Euclid's algorithm and employing twists along 0.1 and 131, we can transform this element into (dl' 0, a2, b2, ... , ay, by), where d l = ± GCD(al, bt}. The element thus obtained can be transformed into (d1 , 0, d 2 , 0, ... , d g , 0), where dk = ± GCD(ak, bk), by similar procedures. Using the operations (0, dk) ~ (d k , dk) ~ (dk,O) and (d k , 0, 0, dk+!) ~(dk' ±dk , 0, dk+! ± dk), we can apply Euclid's algorithm to the numbers dk and dk + 1 . As a result, we successively obtain the elements

(0,0, ± GCD(dl , d2 ), 0, d3, ... ), ... , (0,0, ... ,0, ± GCD(dI. ... , dy), 0). If the initial element (aI, b1 , ••• , ag, bg) is primitive, then

GCD(dI, ... , dy) = ±1; hence the resulting element is represented by one of the curves ±o.g • This means that there exists a homeomorphism h: M2 -+ M2 such that h.: HI (M2) ---. HI (M2) takes each element (al, ... , bg) to an element represented by a self-avoiding curve. The image of this curve under the homeomorphism h- l represents the element (al, ... , by). D

6. The Euler Characteristic and the Lefschetz Theorem 6.1. The Euler Characteristic. Let K be a finite simplicial complex of dimension n. Its Euler characteristic is defined as

X(K) where

ai

= ao -

al

+ a2 _

...

is the number of i-simplices in K.

+ (-l)na n ,

52


= dim Hk(Kj F),

Theorem 1.31. If bk field, then

where F is the additive group of a

Proof. The exactness of the sequence

.

0-+ Zk(Kj F) ~ Ck(Kj F) implies dim Zk

a

-+

Bk-l(Kj F)

-+

0

+ dim Bk-l = dim C k = ak. Therefore, X(K) = ~)-l)kdimZk + L(-l)kdimBk-l = L(-l)k(dimZk - dimB k )

= L(-l)kdimHk(KjF)

o

because Hk(Kj F) = Zk/ Bk.

Corollary. Homotopy equivalent simplicial complexes have equal Euler characteristics. Example 11. X(Dn)

= 1.

Example 12. X(S")

~ 1 + (-1)" ~ {~ o

for odd n, for even n.

Example 13. X(lRpn)

={

Example 14. X(cpn)

= n + 1.

1

for odd n, for even n.

Exercise. (a) Suppose that a finite simplicial complex K is covered by two sub complexes Kl and K2. Prove that

in two ways, by directly counting simplices in these complexes and by using the Mayer Vietoris sequence. (b) Suppose that a finite simplicial complex K is covered by sub complexes KI, K2, ... ,Km. Prove that

X(K)=LX(Kd-LX(KinKj)+ L x.(KinKjnKk)-···· i<jk, k~O

where cI>k is the number of k-simplices ~k C K' for which ~k C cp(~k) with signs taken into account.ll Proof. (a) It suffices to show that

trcpk = tr(CPkIBk ) + tr(CPk).

+ tr(CPk-lIBk_l)'

Let Ck = Zk ffi Ck. The linear operator CPk maps Zk to Zk; hence it induces an operator 'Pk: C k --+ Ck , and tr CPk = tr( CPklzk) + tr 'Pk. Let Zk = Bk ffi Zk. Then Zk ~ Hk(K; R), and the operator induced by CPk on Hk(Kj lR) coincides with (cp.)k· Therefore, tr(CPklzk) = tr(CPkIBk) +

tr(cp.)k. The map a: Ck --+ B k- 1 induces an isomorphism Ck --+ Bk-l' The equality aCPk = CPk-lO shows that the operator corresponding to 'Pk under this isomorphism is CPk-l. (b) The simplicial map cP: K' --+ K induces a chain map CPk: Ck(K'; JR) --+ Ck(K; JR). Consider also the chain map ik: Ck(Kj lR) --+ Ck(K'; lR) that takes each simplex ~ C K cut into simplices ~l,"" ~B to the sum ~1 + " .+ ~IJ (the orientations of these simplices are compatible with that of the simplex ~; a formal definition of the map ik for the barycentric subdivision was given on p. 8). The composition CPkik: Ck(K;R) --+ Ck(K;lR) takes each simplex ~ to the sum of simplices cp(~d + ... + CP(~8)' Therefore, cI>k = tr(cpkik)' According to (a), we have

~)-l)ktr(cpkik) = L(-l)ktr(cp.)k; k~O

k~O

this follows from the fact that, under the identification of Hk(K'; lR) with Hk(K; JR) by means of the isomorphism i., the map in homology induced by the chain map CPkik is identified with (cp.)k' 0 ~'

The Lefschetz theorem follows from part (b) of the lemma because if 0 for all simplices ~' C K', then cI>k = 0 for all k. 0

n cp(~') =

. 11 If the simplices ~k and 'P(~k) have the same orientation, then the plus sign is taken, and If their orientations are opposite, then they are taken with the minus sign.

58


Example 17. For a map f: sn - sn, we have A(f) = 1 hence if deg f -=1= (-1 )n+1, then the map f has a fixed point.

+ (-l)ndegf;

Corollary. On the sphere s2n, there is no continuous vector field without singular points. Proof. Suppose that v is a vector field on s2n, v(x) is the vector of this field at a point x E S2n, and f(x) is the intersection point of s2n with the ray from the center of the sphere to the endpoint of v(x) (see Figure 14). idS 2n, whence deg f = 1 -=1= -1 = (-1 )2n+1. Thus, the map f Clearly, f has a fixed point xo, and v(xo) = O. 0 f"V

Figure 14. The construction of the map

f

Remark. Recall that, in Part I (pp. 203 204), the nonexistence of a continuous vector field without singular points on the sphere s2n was proved by more elementary methods. Example 18. For the coefficient group JR, the space JRp2n is acyclic; therefore, any map f: JRp2n -- JRp2n has a fixed point. Problem 49 ([100]). Given a finite simplicial complex K, a continuous map f: K -- K, and a prime p, prove that A(fP) == A(f) (mod p) (here fP=~. P

•

Chapter 2

Cohomology Rings

1. Multiplication in Cohomology The set H*(K) = EB Hk(K) has not only a group structure but also a ring structure; it is this structure that is largely used in the applications of cohomology. The ring structure is determined by the K olmogorov Alexander multiplication, which is now known as the cup product. This operation was first introduced by Kolmogorov [72, 73J and Alexander [6J. Later, the definition was improved by Cech [21J and Whitney [151]. Before Kolmogorov and Alexander, there was a multiplication operation on the homology of manifolds (intersection of cycles); the Poincare duality isomorphism transforms it into Kolmogorov-Alexander multiplication. Certainly, for many applications of multiplication in cohomology, this longknown multiplication is sufficient, but there are also applications not related to manifolds. To define the cup product for a simplicial complex, we must order the vertices of this complex. Such a definition is not invariant at the level of cochains, but it is invariant at the cohomology level. The proof of its invariance uses an auxiliary object, a total chain complex. 1.1. Homology of Total Chain Complexes. In some cases, it is con-

venient to assume that the vertices of simplices are ordered. The most important instances of such situations are those of defining multiplication in cohomology and proving the isomorphism between singular and simplicial homology. Let us try to define the homology of a simplicial complex by considering, instead of oriented simplices, ordered sets (vo, ... , Vk), where Vo,·· ., Vk are pairwise distinct vertices of some simplex. A simple example shows that this attempt will be unsuccessfuLIndeed, for K = [vo, VI],

-

59

60

2. Cohomology Rings

Vo

Figure 1. A CW-complex homeomorphic to the circle

the group Ker81 is generated by the element (VO,VI) Im{h = o.

+ (VI,VO),

whereas

The reason for the failure is fairly clear: in fact, we try to calculate the homology of a CW -com plex homeomorphic to the circle (see Figure 1). Surprisingly, this situation is easy to remedy; it suffices to dispenfle with the seemingly natural condition that the vertices Vo, .. . , Vk are pairwise distinct. This is indeed an unexpected phenomenon, because now the calculation of the homology of the complex K = [vo, VI] involves an infinite chain complex with 2k+l generators of dimension k for each k = 0, 1,2, .... But in return, at least the one-dimensional homology group is as required. Indeed, the group Ker 8 1 is generated by the elements (vo, VI) + (VI, vo), (vo, vo), and (VI,Vl); allofthem are contained in Im{h: (vo,vo) =8(vo,vo,vo), (Vl,Vl) = 8(VI, VI. VI)' and (vo, VI) + (VI, Vo) = 8(Vo, VI, Vo) + 8( Vo, Vo, vo). The chain complex C.(K) generated by the ordered sets (vo, ... , Vk), where Vo, ... , Vk are vertices of the same simplex in K (some of them may coincide), is called a total, or ordered, chain complex. The boundary homomorphism for C.(K) is defined in precisely the same way as for C.(K).

Theorem 2.1. Let K be a simplicial complex with ordered vertices. Then the homology groups of the chain complexes C.(K) and C.(K) are canonically isomorphic. Proof. Consider the homomorphism cp: C.(K) -+ C.(K) taking (vo, ... ,Vk) to [vo, . .. , Vk] if the points Vo,· .. , Vk are pairwise distinct, and to 0 if some of these points coincide. This homomorphism is a chain map. The proof is similar to the argument showing that the simplicial map K -+ L induces a chain map C.(K) -+ C.(L); the case where there are precisely two coinciding points among Vo, .. . , Vk is again to be handled separately. We also consider the homomorphism cp is not the identity, but it is chain homotopic to the identity. To prove this, we use the same argument as in the proof of the acyclic support theorem (Theorem 1.5 on p. 7) with the only difference that instead of assigning the acyclic subcomplex L(~) to each simplex ~ c K, we associate with the ordered set of vertices v = (vo, ... , Vk) of each simplex ~"IJ C K a chain sub complex C.(~"IJ) c C.(K) so that the following conditions hold: (1) if ~'

c

~, then C.(~')

c

C.(~);

(2) the chain complex C.(~"IJ) is a support for the chains I!>cp(v) and v = id(v). If Vo is a vertex, then I!>cp(vo) = Vo = id(vo). Thus, it remains to verify that if z is a cycle in Ck(~"IJ) for k ~ 1, then z = 8d, where d E Ck+1(~"IJ). We associate with every chain c E 6k(~v) the chain (vo, c) E Ck+1(~"IJ). For k ~ 1, we have 8(vo, c) = c - (vo, 8c). If z is a cycle, then z = 8(vo, z) because 8z = o. 0 Any simplicial map /: K Ck(L) defined by

-+

L induces the homomorphism ik: Ck (K)

-+

It is easy to verify that ik is a chain map (it does not matter whether the

points /(Vi) and /(Vj) are different). It is also easy to check that the diagram

is commutative. Indeed, even at the level of chains, we have /kCP = CPik. The equality f.cp. = 'P.i. implies 'P./. = i.l!>. because I!>. = cp;l (but only at the level of homology, not at thE' level of chains). 1.2. Definition of Multiplication in Cohomology. Any total chain complex naturally determines a cochain complex. The cohomology groups of a total chain complex are canonically isomorphic to the simplicial cohomology groups because chain homotopic chain maps induce not only the same maps of homology groups but also the same maps of cohomology groupS. Let us prove this. First, note that any chain map 'Pk: C k --+ C k induces a co chain map cpk: Hom(Ck,G) --+ Hom(Ck , G). Suppose that 'Pk and 1/Jk are

62

2. Cohomology Rings

chain homotopic chain maps and let a

(acl, C2) = (c l , (aD

= cpk

+ Da)C2) =

- ¢k. If C2 is a cycle, then

(c l , aDC2) = (dcl, DC2).

Clearly, if cl is a cocycle, then the last quantity vanishes. Suppose that the coefficient group R is the additive group of a commutative associative ring with identity. Then the cup product cP '-" & E Cp+q(K; R) of cochains cP E CP(K; R) and cq E Cq(K; R) can be defined as

(cP '-" cq, (vo, ... , vp+q))

= (cP, (vo, ... , vp)) . (c q, (vp, vp+l, ... , vp+q));

here (cP, (vo, ... , vp)) and (c q, (vp, vp+l, ... , vp+q)) are multiplied as ring elements. Under the multiplication of co chains thus defined, the identity element is the cochain cO taking the value 1 E R at each vertex v. This is a cocycle. Indeed, if Cl = E advil, V,2], then = E aiVil - E aiVi2; therefore,

act

(dco,cl) = (CO,aCl) = Lai - Lai =

o.

Remark. The cup product can also be defined in a somewhat more general situation. Namely, the product of cP E CP(K; G) and cq E Cq(K; G') is the co chain cP '-" cq E Cp+q(K; G ® G') defined by (cP '-" cq, (vo, . .. , VP+q)) = (cP, (vo, . .. , vp)) ® (cq, (vp, Vp+l, . .. , vp+q)). If G = G' = R is the additive group of a unital commutative ring, then ring multiplication is a map R ® R - R, and we arrive at the preceding definition.

Clearly, the cup product is bilinear and associative. It turns out that it can be carried over to cohomology due to the following property.

Proof. The values of the cochains (dcP) '-" & and (-l)PcP '-" (d&) at (vo, . .. , Vp+q+l) are equal to

L

(-l)icP(vo, ... , Vi, ... , Vp+l)cq(Vp+1, ... , Vp+q+1)

O:O:;i:O:;p+1

and

(-l)P respectively. The last term in the first sum is canceled by the first term in the second sum, and the sum of the remaining terms in these sums equals the value of the cochain d(cP '-" cq) at (vo, . .. , vp+q+d. 0

1. Multiplication in Cohomology

63

The expression for d"(cP '-" cq) shows that if zP and zq are cocycles, then so is d"(zP '-" zq). Moreover, the cohomology class of the cup product of two co cycles depends only on the cohomology classes of these co cycles because

(zP

+ d"d'-l) '-" zq = zP '-" zq + d"(d'-l '-" zq)

and

Thus, cup product is defined in the entire cohomology. It follows directly from the definitions that if f: K --+ L is a simplicial map, then j*(cP '-" cq ) = (j*cP) '-" (j*c q ), where j* is the induced map of the cohomology groups of the total chain complex.

The isomorphism cp*: H*(K) --+ H*(C*(K)) allows us to transfer cup product to simplicial cohomology. Namely, for a, f3 E H*(K), we define a '-" f3 to be the element of H* (K) for which cp* (a '-" f3) = (cp* a) '-" (cp* f3). At the level of cocycles, the cup product in simplicial cohomology can be described as follows: the co cycle zP '-" zq belongs to the same cohomology class as the co cycle z determined by

(8)

(z, [vo, ... , VP+q])

= (zP, [vo, ... , vp]) . (zq, [vp, vp+l, ... , vp+q]).

It is assumed that the vertices of the simplicial complex K are ordered and the vertices of the oriented simplex are written in increasing order; otherwise, formula (8) does not define any cocycle z because the right-hand side changes unpredictably under permutations of the vertices Vo, ... , v p +q • The total chain complex is needed only for purely technical purposes. In simplicial cohomology, we can define the cup product by numbering the vertices and setting

(d' '-"

c q , [Vio, •.• , Vi p+q ])

= (d',

[Vio, ... , Vip]) • (c q , [Vip, Vip+l' ... , Vi p+q ]),

where io < il < ... < i p +q • Such a multiplication of co chains determines a well-defined multiplication of cohomology classes, which does not depend on the numbering. The total chain complex is only needed to prove this fact and the naturality of the cup product in the sense that if f: IKI --+ ILl is a continuous map, then f*(a '-" f3) = f*(a) '-" f*(f3). Multiplication in cohomology has the following anticommutativity (or skew-commutativity) property. Theorem 2.2. If a E HP(K) and f3 E Hq(K), then a '-" f3 = (-l)pqf3 '-" a. Proof. We take co cycles zP and zq representing a and f3, and order the vertices of K arbitrarily. We calculate zP '-" zq using this ordering, and

2. Cohomology Rings

64

zq '--' zP using the reverse ordering. We obtain (zP '--' zq, [va, ... , vp+q]) = (zP, [va, ... , vp]) . (zq, [vp, ... , vp+q]) and

(zq '--' zP, [vp+q, ... , va]) = (zq, [vp+q, ... , vp]) . (zP, [vp, ... , vo]). It is clear that [vr, ... , va] = (-It(r+1)/2[vo, ... , vr ] and (p + q)(p + q + 1) p(p + 1) - q(q + 1) = 2pq. 0 The definition of cup product in relative cohomology is precisely the same as in absolute cohomology. If a E HP(K, LI) and (3 E Hq(K, L 2), then a"""'" (3 E HP+q(K, L1 U L2)' Indeed, take cocycles zP and zq representing a and (3 and consider the simplex [va, .. . , vp+q] in L1 U L 2. This simplex, as well as any other simplex in L1 U L 2, is contained entirely either in L1 or in L 2. In the former case, [va, ... , vp] is contained in Ll, and in the latter, [vp, ... , vp+q] is contained in L 2. Therefore,

(zP '--' zq, [va, ... , vp+q])

= (zP, [va, ... , vp]) . (zq, [vp, ... , vp+q]) = O.

Thus, the cocycle zP '--' zq is contained in Cp+q(K, L1 U L2)' Problem 50. Given a,(3 E H*(K,L), prove that (aIK) '-" (3 = a........, (3, where alK is the image of a under the natural homomorphism H*(K, L) --+

H*(K). Problem 51. (a) Given K = U~=l Li, where the Li are contractible subcomplexes, prove that the product a1 '--' ... '-" an vanishes for any n elements ai E HP'(K), where Pi > O. (b) Prove that cohomology multiplication in EK is trivial, i.e., the cup product of any two cohomology classes of positive dimension vanishes. Problem 52. Given a manifold Mk embedded in sn, prove that the product of any two classes of positive dimension in the ring H*(sn, Mk) vanishes. Problem 53. Let a sub complex A

c X be a retract.

(a) Prove that H*(X) = Imi* ffi Kerr* and H*(X) = Keri* ffi Imr*, where i: A --+ X is the natural embedding and r: X --+ A is a retraction. (b) Prove that for cohomology, Ker i* is an ideal and 1m r* is a subring. Problem 54. Given multiplications in the rings H*(X) and H*(Y), describe the multiplication in the ring H*(X V Y). Problem 55. Given p, q 2 1, prove that SP V sq is not a retract of SP x sq.

=

(SP x {x }) U ({yo} x sq)


a

65

b

Figure 2. Basis co cycles on the torus

Problem 56. (a) Prove that for m C lRpn is not a retract.

< n, the standardly embedded space

~'pm

(b) Prove that for m is not a retract.

< n, the standardly embedded space cpm c cpn

1.3. Cohomology Rings of Two-Dimensional Surfaces. In this section, we calculate the cohomology rings of closed two-dimensional surfaces directly from the definition of cup multiplication. Note that the required result is easy to obtain by using the Poincare isomorphism and interpreting the multiplication of cocydes in the cohomology of a two-dimensional surface as the intersection of the dual cycles (see Section 2.2). For two-dimensional surfaces, we must calculate only the cup products of elements of HI, because the product of an element of HI and an element of H2 is contained in H3 = o. Before proceeding further, let us discuss the geometric meaning of cocydes and coboundaries. Each (oriented) k-simplex in a simplicial complex determines the dual k-cochain, which takes the value 1 at this simplex and vanishes at all of the other k-simplices. We consider only co chains that are sums of cochains dual to various simplices. We identify such cochains with sets of oriented simplices. The formula (dc I , C2) = (c I , 8C2) shows that a one-dimensional cochain c i of this form is a cocyde if and only if, for any 2-simplex, either none of the three I-simplices constituting its boundary is contained in c1 or c 1 contains precisely two of the three boundary simplices; moreover, in the latter case, if the coefficient group is Z, then the orientations of the two boundary cycles in c1 must be opposite to that of the boundary of the 2-simplex. An example of a cocyde on the torus is shown in Figure 2a; another cocyde, together with a cycle dual to it at the homology level, is shown in Figure 2b. The cycles dual to these two cocydes form a basis in the homology of the torus;

66

2. Cohomology Rings

MIV\ Figure 3. Basis co cycles on a handle

Figure 4. Calculation of

Cl! '-"

f3 for the torus

therefore, they form a basis in its cohomology (for both coefficient groups Z and Z2). A basis for the cohomology of a sphere with any number of handles is constructed similarly. In Figure 3, basis co cycles for a separate handle are presented; on the right, the orientations of the links of the polygonal line are shown. Any co chain dual to a 2-simplex generates the two-dimensional cohomology group of a sphere with handles. If simplices ~~, ... ,~; have the same orientation, then the cochain E ai(~n" is a generator if and only if E ai = ±1, because cochains dual to 2-simplices with the same orientation are cohomologous. For example, the difference of two co chains dual to 2simplices with the same orientation sharing a 1-simplex ~ 1 is the coboundary of the cochain dual to ~ 1 . The formula (0 ~ (3, [ijk]) = (0, [ij]) . ((3, [jk]) shows that in calculating o ~ (3, it suffices to consider only the 2-simplices adjacent to both 0 and

(3. For the chosen basis co cycles in the one-dimensional cohomology group of the torus, there are two such 2-simplices (they are hatched in Figure 4). We have

(0, [01]) . ((3, [12])

=1

and

(0, [02]) . ((3, [23])

=0

because ((3, [23]) = O. Thus, a ~ (3 generates the two-dime 11Sional cohomology group of the torus.


67

Figure 5. Cohomoiogou'l cocycles on the torus

For the coefficient group Z, the antisymmetry of the cup product implies a ........ a = 0 for any cohomology class of odd dimension. But for the coefficient group Z2, there are one-dimensional cohomology classes for which 'Y ........ 'Y 1= o. For this reason, below we give a proof of the equalities a ........ a = 0 and 13 . . . . 13 = 0 for the chosen basis co cycles a and 13 in the one-dimensional cohomology group of the torus that applies to both coefficient groups Z and Z2. In Figure 5, the co cycle a is cohomologous to the co cycle a ' . To prove this, we investigate the coboundary of a cochain that is dual to a point. The relations

(6cO, [VO, VI])

= (cO, a[VO, VI]) = (cO, VI) - (cO, vo)

show that the coboundary of a cochain dual to a point Wo is a sum of the co chains dual to I-simplices of the form [v, wo], i.e., intervals directed to wo0 Therefore, subtracting the coboundary of the sum of co chains dual to 0, 1, and 2 from a, we obtain the co cycle a ' . At the level of cohomology, we have a ........ a = a ........ a ' . However, none of the 2-simplices is adjacent simultaneously to a and a ' . Therefore, a ........ a ' = O. Applying a similar argument to separate handles and using the fact that any cochain dual to a 2-simplex is a generator of the two-dimensional cohomology group, we obtain the following theorem. Theorem 2.3. The one-dimensional cohomology group of a sphere with n handles (with coefficients in Z or Z2) has a basis aI, ... , an, 131, ... ,f3n such that ai ........ a] = 0 and f3i ........ f3j = 0 for all i and j, ai '-' f3j = 0 for i -I- j, and ai '-' f3i = A, where A is a genemtor of the two-dimensional cohomology group.

The cup product of co cycles pertaining to different handles vanishes already at the level of cochains for obvious reasons.

2. Cohomology Rings

68

Figure 6. A basis cycle on the Mobius band

~ -----_

....

Figure 7. The basis cocycle on the Mobius band

Now, consider the cohomology of nonorientable two-dimensional surfaces. Unlike the homology groups of a closed nonorientable surface with coefficients in Z, the two-dimensional cohomology group with coefficients in Z2 is nontrivial. Its generator is as follows. On the polygon from which the nonorientable two-dimensional surface is built by gluing, all simplices Ll~, ... ,Ll; can be endowed with compatible orientations. The cochains (Ll;)* and (LlJ)* are then cohomologous, as in the orient able case. But in the nonorientable case, there exist two simplices, say Ll~ and LlJ. which have opposite orientations on the surface itself. Hence, at the cohomology level, we have (Ll~)* = (Ll;)* and (Ll;)* = -(Ll;)*; therefore, 2(Ll~)* = O. Thus, a cochain E ai(Ll;)* generates the cohomology group if and only if the number E ai is odd; if E ai is even, then this cochain represents the zero cohomology class. Let us calculate the cohomology ring of the surface mP2 with coefficients in Z2. Each Mobius band attached to 8 2 corresponds to one direct summand Z2 in the one-dimensional homology and cohomology groups. For a Mobius band attached to the sphere, a basis cycle is represented by the diagonal of a rectangle (see Figure 6). In Figure 7, the basis cycle is shown by a dashed line and the dual co cycle a, by a polygonal line. The prod uet a '-' a can be calculated in the same way as in the case of a handle, i.e., by replacing a with a cocycle a' obtained from a by adding some coboundaries of cochains dual to points. However, in the nonorientable case, it is impossible to completely separate the cocycles, because A and B in Figure 8a represent the same point. Supplementing a with t e coboundaries of the co chains dual to the points AI, A 2 , •. • , A k , we obtain the co cycle a'

69

2. Homology and Cohomology of Manifolds

A

Az ... At.

a

b

Figure 8. The basis cocycle

0.'

on the Mobius band

shown in Figure 8b. The cocycles a and a' share their boundaries with three 2-simplices (they are hatched in Figure 8b). Let us number the vertices of these simplices. For the numbering shown in Figure 8b, we obtain

(a '-' a', [012]) = Ij (a '-' a', [123]) = 0, because (a', [23]) = OJ (a '-' a', [134])

= 0,

because (a', [34])

= o.

Therefore, the cocycle a '-' a' generates the two-dimensional cohomology group. We have proved the following theorem. Theorem 2.4. The one-dimensional Z2 -cohomology group of the sphere with m Mobius bands attached has generators a!, ... , am such that ai '-' aj = 0 for i =1= j and ai '-' ai = A, where A is the generator of the twodimensional cohomology group. Problem 57. Let Mt and M1 be closed orient able two-dimensional surfaces different from the sphere 8 2 • (a) Let I: Mt ~ M1 be a continuous map. Prove that the homomorphism f2: H2(M1) - H2(Mt) is completely determined by the homomorphism Ii: Hl(M1) ~ Hl(Mt). (b) Prove that there exists a homomorphism h: Hl(M1) ~ Hl(Mt) which is not induced by any continuous map I: Mt ~ Mr

2. Homology and Cohomology of Manifolds The most important role in studying the homology and cohomology of manifolds is played by the Poincare duality isomorphism, which can be expressed in terms of cap product. Cap product is defined for any simplicial complexes, but its applications are related mainly to manifolds. The identification of cohomology and homology by means of the Poincare isomorphism transforms the cap product of manifolds into the map Hk(Mn) x HI(M n ) ~ Hk+!_n(M n ) that sends each pair of cycles to their intersection. Historically,

2. Cohomology Rings

70

cap product arose as a generalization of intersections of cycles in manifolds to arbitrary complexes.

2.1. Cap Product. The cap product is a bilinear map

,-...: HP(K;R) x Hp+q(K;R)

--+

Hq(K;R),

where R is the additive group of an associative commutative ring with identity. We could define cap product for total chain complexes and then carry the definition over to simplicial homology and cohomology, but we shall not repeat this procedure; instead of checking that we have the well-defined notions (at the level of homology and cohomology), we order the vertices of our complex and consider simplicial chains and cochains. We assume that the simplicial complex K is path-connected. For cP E

CP(K; R), we set cP ,-... [va, ... ,vp+q] = (cP, [Vq, ... , vp+q]) [va, ... , vq]j it is assumed that the vertices Va, ... , vp+q are numbered in increasing order. For q = 0, we set

cP ,-... [va, ... , vp] = (cP, [va, ... , vp]) E R

~

Co(K; R).

Extending this map by linearity, we obtain a bilinear map

,-...: CP(K; R) x Cp+q(K; R)

--+

Cq(Kj R).

It induces a map of (co)homology groups due to the following property.

= (-1)q(&dP ,-... Cp+q) + dP ,-... aCp+q.

Lemma. a(dP,-... Cp+q)

Proof. It is sufficient to check the required equality for Cp+q = [va, . .. , vp+q]. By definition, we have

dP ,-... a[vo, ... ,vp+q]

L (_1)i(dP, [vq, ... , Vp+l]) [va, ... , Vi, ... , Vq] + L (_1)i (dP, [Vq-l, ... , Vi, ... ,vp+q]) [va, ... , Vq-l] O:O;i n, then any continuous map f: JRpm --+ lRpn induces the zero map Hk(JRpnj Z2) --+ Hk(JRpmj Z2) for each k ~ 1.

r:

(b) Prove that if m > n, then any continuous map f: lRpm induces the zero map f.: Hk(JRpmj Z2) --+ Hk(lRpn j Z2) for each k

JRpn ~ 1. --+

Problem 61. Prove that if m > n ~ 1, then there exists no continuous map --+ 8 n such that g( -x) = -g(x) for all x E 8 n (this fact is known as the Borsuk Ulam theorem).

g: 8 m

Problem 62. (a) Prove that lRpn and cpn cannot be represented as the union of n contractible subcomplexes. (b) Represent JRpn and cpn as the union of n + 1 contractible subcomplexes. Problem 63. Given a continuous map f: cpn 1 + >, + >,2 + ... + >,n, where>, E Z.

--+

cpn, prove that AU)

Problem 64. Prove that if n is even, then any continuous map f: cpn cpn has a fixed point. Problem 65. Prove that the degree of any map f: cpn form >,n, where>, E Z.

--+

= --+

cpn has the

Problem 66. (a) Prove that for even n, there exists no orientation-reversing diffeomorphism f: cpn --+ cpn. (b) Construct an orientation-reversing diffeomorphism cp2n+1 cp2n+l.

--+

Problem 67. Prove that the spaces CP2 and 8 2 V 8 4 are not homotopy equivalentj using this fact, show that the Hopf fibration 8 3 --+ 8 2 is not homotopic to a constant map. 2.3. Two Examples. In this section, we give two examples showing that cohomology rings are not determined by data that completely determine homology groups. First, CW-complexes with isomorphic cellular chain complexes may have nonisomorphic ring cohomologies. Thus, to define multiplication in cohomology, it is necessary to consider the simplicial structurej the


77

cell structure alone 1 is not sufficient for this purpose. Second, the cohomology ring with coefficients in Z does not determine cohomology rings with other coefficients; spaces with isomorphic integral cohomology rings may have nonisomorphic Z2-cohomology rings. We use the fact that for any coefficient ring and any i > 0, we have Hi(X V Y) = Hi(X) ffi Hi(y), and, moreover, (x + y) ........ (x' + y') = x ......... x' + y ......... y' for cohomology classes of positive dimension (see Problem 54). Example 19. The cellular chain complexes for the torus T2 and the space Sl V Sl V S2 are isomorphic, but the cohomology rings of these spaces are not. Proof. The cellular chain complexes for both spaces have the form Z

--+

Z ffi Z

--+

Z

--+

0;

all boundary homomorphisms are zero. For the space Sl V Sl V S2, the one-dimensional cohomology group is isomorphic to H1(Sl V Sl) because Hl(S2) = O. Therefore, the product of anyone-dimensional classes of co cycles vanishes. On the other hand, for the torus, the product of the generators of H1(T2) does not vanish. 0 Problem 6S. Prove that the spaces sm x sn and sm V sn V sn+m have equal cohomology groups but different cohomology rings. Example 20. The cohomology rings of the spaces Rp3 and Rp2 V S3 with coefficients in Z are isomorphic, whereas their cohomology rings with coefficients in Z2 are not. Proof. Both spaces have trivial one-dimensional integral cohomology group; therefore, multiplication in cohomology is trivial. The Z2-cohomology ring of space Rp3 has an element a for which a ........ a ........ a i- 0, whereas the Z2cohomology ring of p2 V S3 has no such element. Indeed, the only nonzero elements of this ring are a' E HI (lRp2; Z2), a' '-' a' E H 2(lRp2; Z2), and f3 E H3(S3; Z2). 0 2.4. The Lefschetz Isomorphism. The Lefschetz isomorphism generalizes the Poincare duality isomorphism to compact orient able manifolds with boundary. This is an isomorphism between the groups Hn-k(Mn) and HkCMn,8Mn) and between the groups Hn-k(Mn,8Mn) and Hk(Mn). Both isomorphisms are given by the same formula x ~ x r--.. [Mn], where 1 By the cell structure we mean the structure determined by the boundary homomorphism, i.e., that of the cellular chain complex. Certainly, if not only the boundary homomorphism but also the characteristic maps of cells are given, then the CW -complex is determined up to homeomorphism, and thereby its cohomology ring is determined uniquely.

2. Cohomology Rings

78

Figure 11. The manifold

1M

[MnJ E Hn(M n , aM n ) is the fundamental class. The point is that the cap multiplication ....... : Hn-k(K,Ll) x Hn(K,Ll U L2)

--+

Hk(K,L 2),

where K = Mn and Ll U L2 = aMn, can be defined in two different ways, by setting Ll = 0 and L2 = aM n or Ll = aM n and L2 = 0. To prove a theorem about the Lefschetz isomorphism, we need the following assertion, which is often useful in dealing with manifolds with boundary. Theorem 2.13 (on collar). If M n is a compact manifold with boundary aM n , then there exists a smooth embedding F: aM n x [0,1) --+ Mn such that F(x, 0) = x for x E aMn. Proof. Choose finitely many coordinate neighborhoods UI, ... , Urn in M n covering aM n . Let {~i} be a smooth partition of unity subordinate to the cover {Ui n aMn}. Each set Ui can be identified with a subset of ]Rn such that aMn is determined by the equation Xl = 0 and the vector el = (1,0, ... ,0) is directed inside Mn. Consider the vector fields W,(XI, ... , xn) = ~,(X2' ... ' Xn)ei on the Ui. On the set U = U~l Ui, the vector field E~l Wi is defined; it vanishes nowhere and is directed inside Mn on aMn. Let Ht(x) be the displacement of the point X during time t along the trajectory of this vector field. Since the manifold aM n is compact, we can choose E > 0 so that Ht(x) is defined for all x E aM n and all t E [0, EJ. The map F(x, t) = Ht(x) is the required smooth embedding aM n x [0, E) --+ Mn. 0

Let M be a compact manifold with boundary aM. Consider the closed manifold M obtained from two copies of M by identifyir g the respective points of their boundaries; we denote these copies by M and M'. The collar


79

theorem implies the existence of a submanifold N ;;::: 8M n x [-1,1] in £1 (8M corresponds to 8M x {O}) such that the closure of £1 \ N consists of two disjoint subsets Land L' homeomorphic to M (see Figure 11). The manifold £1 can be triangulated so that its submanifolds N, L, and L' are subcomplexes. Consider the relative Mayer Vietoris sequence

In this sequence, Hk(M, L n L') ~ Hk(M) because L n L' = 0. To transform the remaining homology groups, we use the isomorphism H.(A, B) ~ H.(AUG, BUG), where AUG is a simplicial complex, A and G are subcomplexes in An G, and B is a sub complex in A. This isomorphism is obvious even at the level of relative chains. Clearly, the pairs (£1, M') and (£1, L') are homotopy equivalent and (£I,M') = (MUM',8MUM'); therefore, Hk(M,L') ~ Hk(M, 8M). Moreover, (£1, L U L') = (N U (L U L'), 8N U (L U L')), whence Hk(M, L U L') ~ Hk(N,8N). Let us show that Hk(N, 8N) ~ Hk-l (8M) for k > 1. Attaching two copies of the cone over 8M to N, we obtain a simplicial complex homeomorphic to E(8M). Hence Hk(N, 8N) ~ Hk(E(8M), G U G'), where G';;::: G = G(8M). The exact sequence of the pair (E(8M), G U G') shows that Hk(N, 8N) ~ Hk(E(8M)) for k > 1. Moreover, Hk(E(8M)) ~ Hk-l(8M). In the new exact sequence

---+

Hk(M)

---+

Hk(M, 8M) ED Hk(M', 8M') ---+

the homomorphism Hk(M,8M) homomorphisms

--+

H k- 1 (8M)

---+

Hk-l(M)

---+,

Hk_l(8M) is the composition of the

therefore, it coincides with the homomorphism 8. sending each relative cycle to its absolute boundary. The remaining two homomorphisms are induced by inclusions. Consider the absolute Mayer-Victoris sequence

2. Cohomology Rings

~u

This exact sequence can be mapped to the one considered above as follows:

-+ Hn-k(M) --+ Hn-k(M) ffi Hn-k(M')

1~[M1E9[M'1

1~[Ml

-4

Hk(M)

-4

Hk(M, aM) ffi Hk(M', aM') ~

Hn-k(aM)

~

1~[BMl

-----+ Hk_1(aM)

Hn-k+l(M)

1~[Ml

~

• Hk-l(M) ~ .

In the resulting diagram, the left square is commutative because all operations in it (including the passage from [M] to [M] and [M'l) are induced by inclusions and restrictions. The middle square commutes as well because if zn-k is a cocycle, then

In the right square, the upper map Hn-k(8M) -+ Hn-k+1(M) acts as follows. Take a co cycle zn-k E C n- k (8M) and consider a co chain cn- k E Cn-k(M) whose restriction to 8M coincides with zn-k. Let rr-k+ 1 be the co chain in C n- k+1(M) which coincides with 6cn- k on M and vanishes outside M (it exists because 6cn- k vanishes on 8M). The map in question has the form [zn-k] ~ [rr- k+1]. The lower map i.: H k-l ( 8M) -+ H k-l (M) in this square is induced by an inclusion. Clearly, d n - k +1 '""' [M] = 6cn - k '""' [M] and

i.(zn-k,"", [8M]) = i.(i·(cn- k ) '""' [8M]) = cn- k '""' i.[8M] = cn- k ,-.. [8M]. Finally, the equality

shows that the chains (_1)k+16c n- k '""' [M] and cn- k ,-.. [8M] differ by a boundary, i.e., belong to the same homology class. Thus, the right square is commutative up to sign. We have constructed a diagram commutative up to sign. The maps

,-.. [U] and '""' [8M] in this diagram are isomorphisms by the Poincare ........ [MJ duality theorem. Therefore, by the five lemma, the map Hn-k(M) - - H k (M,8M) is an isomorphism as well.

81


Similarly, in the diagram

----+ Hn

k+l (aM)

----+ Hn k (M, aM) EB Hn-k (M', aM)

1~[M1EB[M'1

l~[aMl --~)

Hk(aM) - - - - - - + ) Hk(M) EB Hk(M')

----+ Hn-k(M)

~

l~[Ml -~)Hk(M)

the map Hn-k(M, 8M) the following theorem.

),

~[Mll Hk(M) is an isomorphism. We have proved

Theorem 2.14 (the Lefschetz isomorphism [78]). Let Mn be a compact

orientable manifold with boundary 8Mn. Then the maps Hn-k(M) k

Hk(M,8M) and Hn- (M,8M)

~[Ml

---+

~

. .

Hk(M) are zsomorphzsms.

The same isomorphisms hold in (co)homology with coefficients in R, where R is the additive group of a ring. In the case of a nonorientable manifold, such isomorphisms hold for (co)homology with coefficients in Z2.

Problem 69. Prove that if a manifold M n is contractible, then its boundary is a homology (n - I)-sphere, i.e., H k (8M n ) ~ Hk(sn-l) for all k. Problem 70. Given a compact orientable manifold M3 with nonempty boundary 8M3, prove that for its rational homology, the dimension of the image of the map 8: H 2 (M3, 8M 3) -+ Hl(8M 3) equals half the dimension of the space HI (8M3). 2.5. Alexander Duality. The Lefschetz isomorphism theorem has an important consequence, namely, the following duality theorem.

Theorem 2.15 (Alexander duality [4, 5]). If M S;; sn is a closed submanifold, then Hk(M) ~ H n _k_I(sn \ M) and iIk(M) ~ iI n- k- 1 (sn \ M) for O~k~n-1.

Proof. We begin by proving that Hk(sn, M) ~ Hn_k(sn \ M). Let us remove a point x ¢ M from sn, identify sn \ {x} with ]Rn, and apply the tubular neighborhood theorem to M c ]Rn. If M~ is an open tubular neighborhood of M, then the pair (sn, M) is homotopy equivalent to (sn, M~), and the space sn \ M is homotopy equivalent to sn \ Me. According to the Lefschetz isomorphism theorem, we have Hk(sn \ Me, 8Me) ~

2. Cohomology KIngs

Hn_k(sn \ Me). Moreover, Hk(sn \ Me, aMe) ~ Hk(sn, Me) ~ Hk(sn, M) and Hn_k(sn \ Me) ~ Hn_k(sn \ M). To obtain the first of the required isomorphisms, consider the exact sequence of the pair (sn, AI):

fIk(sn) _

fIk(M) ~ Hk+l(sn, M) _

fIk+l(sn).

For k f= n -1, we have fIk(sn) = fIk+l(sn) = 0, and 15* is an isomorphism. Hence fIk(M) ~ Hk+:(sn, M) ~ Hn_k_l(sn \ M)j here n - k -1 f= 0 and, therefore, Hn-k-l ~ Hn k-l. For k

=n-

1, the exact sequence takes the form

0 - fIn-I(M) ~ Hn(sn, M) ~ Hn(sn) ~ Z. Here the homomorphism i* is induced by the inclusion of pairs i: (sn, 0) 0, X2 = ... = xp = 0, Y = 0, I zl12 + :~ ~ 1 }

and

d

=

{(x,y,

z) I YI

~ 0,Y2 -

1

= Yq = 0, z = 0, IIxII 2 + ~ I}. n W2 = ae and W2 n W3 = ad. Therefore, ...

For relative chains, we have WI the triple Massey product (a P , {3q, -yr} corresponds to the one-dimensional relative cycle e n W3 + WI n d. It is easy to verify that

enW3

= {(x,y,z) I 0 ~ Xl

~ l,x2

= ... = Xp = O,y = O,z = O},

and WI

n d = {(X, y, z) I 0 ~

YI ~

1, Y2

= ... = Yq = 0, X = 0, z = O}.

Hence the relative cycle under consideration is a path joining different connected components: one of its endpoints belongs to Sj+r-l and the other to S~+q I. Such a cycle is a representative of a nonzero homology class. 2.7. Intersection Forms and the Signature of a Manifold. Let M 2n be a closed oriented manifold of even dimension, and let F be any field. On the space Hn(M2n; F), we define the bilinear form

f(a n , (3n)

=

(an '--' (3n, [M2n]) ,

where [M2n] is the fundamental class of the manifold M2n. For the opposite orientation, the fundamental class changes sign; therefore, the bilinear form f changes sign as well, Le., f = - f, where f is the form for the manifold with the opposite orientation. On dual space Hn(M2n; F), there is the dual form

!*(a n ,{3n) - ((an, {3n}}. Recall that the intersection number ((an, {3n}} depends on the orientation of the manifold, and for the opposite orientation, the intersection number is of the opposite sign. The bilinear forms f and 1* are called the intersection forms. The intersection form can be defined not only over the field F but also over the ring Z. For Z, Hn(M2n) must be replaced by the free Abelian group H n (M 2n )/Tn , where Tn is the torsion subgroup, because he intersection form vanishes at the elements of finite order.


89

We shall consider only intersection forms over lR and Z. The universal coefficient theorem implies

The elements of finite order do not affect the intersection number, and it is seen directly from the definition of the intersection number that if an, f3n E Cn (M 2n ) and ran, sf3n E Cn (M 2n j lR), then

Therefore, the intersection forms over lR and Z have the same matrix. This implies, in particular, that H n (M 2n j lR) has a basis in which the intersection form has integer coefficients. Since the cup product is anticommutative, it follows that the intersection form is symmetric for even nand antisymmetric for odd n. Let Q be the matrix of the intersection form over Z, and let m = dimHn (M 2n jlR). Theorem 2.10 can be reformulated as follows: there exist m x m matrices A and B over Z for which ATQB = 1m is the identity matrix. Therefore, det Q = ±1. The following theorem collects the properties of the intersection form proved above. Theorem 2.17. (a) The intersection form is determined by a nonsingular integer matrix.

(b) For a 4k-manifold, the matrix of the intersection form is symmetric, and for a (4k + 2) -manifold, it is antisymmetric. (c) For the same manifold with the opposite orientation, the intersection form has the opposite sign. Remark. The intersection form can also be defined for nonorientable and nonclosed manifolds, but it may be degenerate in this case. Moreover, the definitions of the intersection form in terms of multiplication in cohomology and in terms of intersections of cycles are not equivalent for such manifolds. For example, defining the intersection form for the cylinder I x 8 1 and for the Mobius band in terms of multiplication in cohomology, we obtain the zero quadratic form in both cases, while defining it in terms of intersections of cycles, we obtain the zero quadratic form for the cylinder and a nonzero form for the Mobius band. Indeed, on the cylinder, the cycles {ttl x 8 1 and {t2} x 8 1 with tl =I- t2 are disjoint, while on the Mobius band, the cycles a and a' shown in Figure 10 intersect at one point.

The properties of the intersection form specified in Theorem 2.17 have the following consequences.

90

2. Cohomology Rings

Theorem 2.18. Let M 4k +2 be a closed orientable manifold. dimension dim H2k+1 (M4k+2; JR) is even.

Then the

Proof. The matrix of the intersection form of the manifold M4k+2 is antisymmetric; therefore, it has even rank (see [104], p. 102). Since this matrix is nondegenerate, it follows that its rank is equal to dim H2k+l (l\14k+2; JR). 0 Corollary. If M4k+2 is a closed orientable manifold, then its Euler characteristic X(M 4k +2) is even. Proof. For a closed orient able manifold, we have X(M 2n )

== dimHn(M2n;JR)

(mod 2).

The proof of this relation is similar to that of Theorem 1.34 (for orient able manifolds, the argument applies not only to the coefficient group Z2 but also to JR). 0 The signature of a manifold M4k is defined to be equal to the signature of its intersection form. Recall that the signature of a symmetric bilinear form over JR is defined as follows. Using symmetry, we choose a basis for which f(E Xiei, E y~ei) = E AixiYi. The signature is equal to the difference between the numbers of positive and negative coefficients Ai. The signature of a form does not depend on the choice of a basis; therefore, the signature of the oriented manifold M4k, which we denote by 0'(M4k), is an invariant of this manifold. Moreover, 0'( _M4k) = _0'(M4k), where _M4k is the manifold M4k with the opposite orientation. We have proved the following theorem.

Theorem 2.19. If 0'(M4k) f. 0, then there is no orientation-reversing homeomorphism f: M4k _ M4k (i.e., such that /.[M4k] = _[M4k]). Theorem 2.19 implies that there exists no orientation-reversing homeomorphism cp2n _ cp2n because 0'(cp2n) = ±1. (Another proof of this assertion is given in the solution of Problem 66.) There is a distinguished orientation on the manifold cpm, because cpm \ cpm-l = C m and complex space does have a distinguished orientation (all complex transformations regarded as transformations over JR have positive determinants). It is usually assumed that cpm is endowed with this orientation. In this case, 0'(cp2n) = 1, because the orientation determined by a pair of transversally intersecting n-dimensional complex subspaces in C 2n is compatible with that of C 2n .

Problem 74 (Novikov Rokhlin). Given closed orient able manifolds Mt n and M~n, prove that a(Mtn # M~n) = a(Mtn) + a(M~n).

91


Using the Lefschetz isomorphism, we can prove the following property of the signature of a manifold that is the boundary of another manifold. Theorem 2.20 (Thorn [137]). If M4k is a closed orientable manifold which is the boundary of a compact orientable manifold W 4k +1, then 17(M4k) = O. Proof. Consider the cohomology and homology sequences of the pair (W4k+1 ,M4k) and the Poincare and Lefschetz isomorphisms from the first sequence to the second:

H2k(W4k+1)

1. .

i·

)

[W 4 11:+ 1 j

H2k(M4k) ~ H2k+1(W4k+1, M4k)

1. .

1. .

[M4kj

H2k+1(W4k+1,M4k) ~ H2k(M4k)

[W4k+lj

i.

)

H2k(W4k+1).

In the proof of the Lefschetz isomorphism theorem, we showed that this diagram is commutative up to sign. Exactness and commutativity imply Imi" ~ 1mB.. = Keri ... If 0,{3 E Imi", then (0 ........ (3,[M4k]) = O. Indeed, if 0 = i"(a) and (3 = i"(b), then

(0 ........ (3, [M4k]) = (i"(a ........ b), B.. [W 4k +1])

=

(5"i"(a ........ b), [W4k+1])

=0

because 5"i" = O. The next step of the proof is performed most conveniently in the absence of torsion, i.e., for (co)homology with coefficients in R. In this case, H2k(M4k) and H2k(W4k+l) are linear spaces, and they are dual to H2k(M4k) and H 2k (W 4k +1). Moreover, the maps i .. and i* are dual to each other. Lemma. If a map /*: W" W"/Ker/*.

-+

V" is dual to f: V

-+

W, then (1m f)"

~

Proof. The linear functions on 1m f coincide with the linear functions on W up to functions vanishing on 1m f. The equality (/*(11), v) = (11, f(v)) shows that Ker /* is a function on W vanishing on 1m f. D Using this lemma, we obtain (Imi")" ~ H 2k(M 4k ;JR)/Keri ... Since 1m i* ~ Keri .. , it follows that dim 1m i" = dim Ker i .. = dim H2k(M4k; JR). The intersection form is nondegenerate; therefore, the dimension of the subspace on which it vanishes is at most half the dimension of the entire space, and the equality holds only if the signature vanishes. D

!

2.8. The Bockstein Homomorphism and Poincare Isomorphism. The main theorem of this section is applied in the next section to classifying lens spaces up to homotopy equivalence.

92

2. Cohomology Rings

Suppose that K is a simplicial complex and 0 - G' - G - Gil - 0 is an exact sequence of Abelian groups. It induces an exact sequence of chain complexes

As we have mentioned earlier, the connecting homomorphism {3*: Hk(K; Gil) - Hk-l (K; G') is called the Bockstein homomorphism. For cochain complexes, we also have an exact sequence 0-- Hom(Ck,G') - - Hom(Ck, G) - - Hom(Ck, Gil) - - 0, where Ck = Ck(K). The connecting homomorphism {3*: Hk(K; Gil) Hk+1(K; G') of cohomology groups is also called the Bockstein homomorphism.

Theorem 2.21. For any closed orientable manifold Mn, the Bockstein homomorphism commutes with the Poincare isomorphism up to sign; i.e., if a k E Hk(Mn; Gil), then

where [Mnj E Hn(Mn; G') is a fundamental class. (It is assumed that G', G, and Gil are the additive groups of unital commutative rings.) Proof. At the level of (co )cycles, the homology and cohomology Bockstein homomorphisms can be described as follows. In the case of homology, for a cycle z~ with coefficients in Gil = G/G', a chain Zk with coefficients in G which represents this cycle is chosen, and to this chain 8Zk (which is a cycle with coefficients in G' c G) is assigned. In the case of cohomology, for a co cycle (Zk)" with values in Gil, a cochain zk with values in G which represents this cycle is chosen and sent to the co cycle &zk with values in

G'cG. At the level of (co) chains, the Poincare isomorphism acts as follows. Let K and K* be dual decompositions of Mn. The Poincare isomorphism associates each chain Ck E Ck(K) to the cochain cn - k E cn-k(K*), and associated with the chain 8q is the co chain ±&cn - k (to be more precise, the cochain (_l)k&c n - k ). Calculating the homology Bockstein homomorphism from the decomposition K and the cohomology Bockstein homomorphism from K*, we easily obtain the required result. 0

Problem 75. Given a P E HP(K; Gil) and bq E Hq(K; Gil), prove that


93

2.9. Lens Spaces. Suppose that p and q are coprime positive integers and p ~ 3. Consider the fixed-point-free action of the group Zp with generator u on the unit sphere 8 3 C ((:2 defined by

u(z, w)

= (/;' z, e 7

w).

The quotient of 8 3 by this action is a 3-manifoldj it is called a lens space and denoted by L(P, q). In [108], it was explained in detail that the lens space L(p, q) can be represented in the form of a CW-complex with one cell in each dimension 0, 1, 2, and 3 as follows. We take a ball n 3 , divide its equator into p equal parts, and identify the points of the lower hemisphere with the points of the upper hemisphere by rotating the lower hemisphere through an angle of 2rrq/p and reflecting it in the equator. Under this operation, the p points dividing the equator merge into one point; the p arcs into which the equator is partitioned are identified as well. As a result, we obtain a O-cell and a 1cellj for a 2-cell we take, e.g., the lower hemisphere. The boundary of the cell n 3 contains two copies of the 2-cell with opposite orientations. Therefore, the chain complex for calculating the cell homology of L(p, q) has the form

C3 where Ci

~

o

~

C2

Xp

~

Cl

0

~

Co

~

0,

Z, and the homology groups of L(p, q) with coefficients in Z are H3

= Z,

H2

= 0,

HI

= Zp,

Ho

= Z.

For L(p, q), the homology and cohomology with coefficients in Zp can easily be calculated without applying the universal coefficient theorem. Indeed, the chain complex with coefficients in Zp has a very simple form, namely, 000

C3 ~ C 2

~

Cl

~

Co

~

0,

where C i ~ Zp. Hence Hk(L(p, q)j Zp) ~ Hk(L(P, q); Zp) ~ Zp for k 2, 3.

= 0, 1,

The homology and cohomology groups are not sufficient to distinguish between the lens spaces L(p, ql) and L(p, q2). But we can distinguish between them in some cases by using the Bockstein homomorphism (3,. associated to the exact sequence

O ~Zp

Xp ~Zp2 ~Zp ~o.

Theorem 2.22. The group H2(L(p, q)j Zp) has an element a such that = ±1 E Zp.

q«a,(3,.(a)))

Proof. Let c be the 2-cell of the CW-complex L(p, q) described above, and let d be its 1-cell. For a we take the homology class represented by the cycle 1· c (modulo p), where 1 E Zp. The clement (3*(a) E HI (L(p, q)j Zp) is

2. Cohomology Rings

94

Figure 15. A 2-chain

constructed as follows. Consider the cycle 1· c as a chain with coefficient in Zp2, i.e., assume that 1 E Zp2 (1 can be replaced by any element of the form 1 + kp). By assumption, 8(1· c) is a cycle with coefficients in Zp C Zp2; the subgroup Zp of Zp2 consists of all elements of the form lp for l E Zp. Indeed, 8(1· c) = p. d. The element {3.(a) is represented by the cycle p. d treated as a cycle with coefficients in Zp (lp E Zp:l should be replaced by I E Zp). Thus, {3.(a) is represented by the cycle 1· d. The intersection number ((c, d)) cannot be calculated directly, because the cycles c and d are not transversal. For this reason, instead of the cycle d, we consider another cycle, which is homologous to qd and intersects c transversally. Suppose that Nand S are the north and the south poles, A is a point on the equator, and B is the point on the equator obtained by rotating A through the angle 27rq/p (see Figure 15). The boundary of the 2-chain hatched in Figure 15 consists of the interval NS and the arcs SA, AB, and BN. In L(p, q), the arcs SA and BN are identified, and they have opposite orientations; clearly, the points Nand S are identified as well. Therefore, the cycle N S is homologous to AB, which consists of q cycles d, and it transversally intersects c in one point; therefore, q((c, d))

=

((c,qd))

(all of the equalities are modulo p).

=

((c,NS))

= ±1

o

Corollary. The group Hl(L(P, q); Zp) has an element a such that qa '--" {3·(a) = ±Ap, where Ap E H 3 (L(P, q); Zp) is the element for which Ap " [L(P, q)]p = 1. Proof. The manifold L(p, q) is orient able because H 3 (L(P, q)) = Z. Therefore, the homology and cohomology Bockstein homomorr hisms commute with the Poincare isomorphism up to sign. 0

91

3. The K iinneth Theorem

Theorem 2.22 and its corollary readily imply the following assertion. Theorem 2.23. If lens spaces L(P, qt} and L(p, q2) are homotopy equivalent, then ql == ±a2q2 (mod pl. Proof. Choose elements 01 and 02 in Hl(L(p, qt}j Zp) and Hl(L(p, q2)j Zp), respectively, so that

qlOl '-" P"'(Ol)

= ±Ap,l

and

q202 '-" P"'(02)

= ±Ap,2.

Let f: L(P,qd - t L(p,q2) be a homotopy equivalence. Then j*(Ap ,2) ±Ap,l. Let us write 1*(02) in the form aOl, where a E Zp. Then

Qlol '-" P·(Ol) = ±Ap,l = ±/*(Ap,2) = ±a2q20 1 hence ql == ±a2 q2 (mod pl. (The equality easy to derive at the level of cochains.)

1*p.

=

'-"

P'" j*,

P"'(odj

which we use, is 0

Remark. In Part I (Chapter 5, Section 4.6), we proved the converse of Theorem 2.23: If ql == ±a2q2 (mod p), then the lens spaces L(p, ql) and L(p, Q2) are homotopy equivalent.

3. The Kiinneth Theorem The Kiinneth theorem allows us to express the homology of K x L in terms of "the homology of K and L. It consists of two parts. One, relatively simple, expresses the chain complex C",(K x L) in terms of the chain complexes C.(K) and C.(L)j the other, somewhat more complicated algebraic part, expresses the homology of the complex C",(K x L) in terms of the homology of C.(K) and C",(L). 3.1. The Chain Complex C.(K X L). Knowing chain complexes C",(K) and C.(L), we can construct the chain complex C.(K x L) for calculating the cellular homology of K x L. The cellular homology of K x L is more convenient to calculate than the simplicial homology because the product of any two open cells is an open cell, whereas the product of simplices may not be a simplex. We assume that K and L are finite CW-complexes because the topology of the product K x L of infinite CW-complexes may not coincide with that of K x L as a CW-complex. The open k-cells in K x L are the products uP x r q , where p+q = k and uP and r q are open cells in K and L. The main difficulty occurs when we try to express the boundary homomorphism in C.(K x L) in terms of the boundary homomorphisms in C.(K) and C.(L). It is clear that 8(DP x Dq) = (DP x 8Dq) U (8DP x Dq) and (DP x 8Dq) n (8DPxDQ) = Sp-l xSq-1j here 8 is the geometric boundary. The calculation

2. Cohomology Rings

96

o

Figure 16. The orientation of the boundary of a cell

of 8(uP X r q ) does not involve cells of dimension (p - 1) therefore,

+ (q -

1)

=

k - 2;

It remains to determine the signs.

First, we need an agreement on how the orientation of a cell or, more generally, of a manifold induces the orientation of its boundary. For a simplex [vo, ... , vnl, the orientation of the boundary is determined by 8[vo, ... , vnl = L?=o( _l)i[vO,"" Vi, ... , vnl. We assume that each simplex [Vo, ... , vnl is oriented so that the basis el = VI - Vo, ... , en = Vn - Vo has positive orientation. Moreover, the simplex [v!, ... , vnl, which is contained in the boundary of [vo, ... ,vnl, is oriented so that the basis e2 - el = V2 - VI, ... , en - el = Vn - VI has positive orientation (Le., the basis e2, .. . ,en has positive orientation). This convention on induced orientation is easy to transfer to any oriented manifold M n with boundary 8Mn. Namely, suppose that the orientation of a manifold Mn is determined by a basis el, . .. , en of the tangent space TzMn at some point x E 8M n , the vector el is directed inside the manifold, and the vectors e2, ... ,en belong to the tangent space to the boundary. Then the orientation of the boundary is determined by the basis e2,"" en· Suppose that the positive orientations of cells uP and r q are determined by bases el, .. " ep and Cl, ... , Cq. We endow the cell uP x r q with the orientation determined by the basis eI, ... , ep , CI, ... , Cq. The orientation of 8uP is determined by the basis e2 - el,"" ep - el (see Figure 16); the orientation of 8uP x r q is determined by the basis

(13) (this orientation coincides with that of 8(uP x r q »; and the orientation of 8rq is determined by the basis

uP x

(14)

3. The Kiinneth Theorem

9~

Let us move Cl - el in the basis (13) to the first position. The transitior matrix from the basis thus obtained to (14) is -1 0 0

-1 1 0

-1 0 1

-1 0 0

....................... 0

0

0

1

Therefore, the sign of the orientation of the basis (14) with respect to the basis (13) is (-l)P. As a result, we obtain

8(uP x r q) = 8uP x r q + (-l)PuP x 8r q. Now the problem of calculating the homology of K x L becomes purely algebraic.

3.2. The Algebraic Kiinneth Theorem. Let C~ and C: be nonnegative chain complexes with boundary homomorphisms 8' and 8". Their tensor product is the chain complex C., where Ck = ffip+q=k C~ ® C; and 8(c~ ®

C::) = a'c~ ® c~ + (-l)Pc~ ® a"c~.

The sign (-1)P in the second term ensures that 88

= O.

We denote the chain complex C. by C~ ® C:. A similar notation is used for any graded Abelian groups A. = ffi Ap and B. = ffi Bqj i.e., it is assumed that (A. ® B.)k = ffip+q=kAp ® B q. Similarly, we define (Tor(A., B.))k =ffip+q=k Tor(Ap, Bq).

Theorem 2.24. Let C~ and C: be free nonnegative chain complexes. Then the sequence

o ----+ (H.(C~) ® H.(C:))k

----+ Hk(C~ ® C:) ----+ (Tor(H.(C~), H.(C:)))k-l ----+ 0

is exact. This exact sequence splits, but not canonically. Proof. Consider the exact sequence

o ----+ Z p'

i' C'p ----+ 8' B'p-l ----+ O. ----+

This sequence splits because the group B~_l is free. Therefore, tensor multiplication by preserves exactness. The direct sum of two exact sequences is an exact sequence; hence the sequence (15)

C;

0----+

EB (z; ® C;) ~ ED

p+q-k is exact.

p+q=k

(C; ® C;)

~

E9 p+q=k-l

(B; ® C;) -------.0

98

2. Cohomology Rings

C:

Consider the chain complexes z~ ® and B~ ® C:; in these complexes, the boundary homomorphism has the form ±1 ® 8". Indeed, a(z~ ® d~) = ( -l)P z~ ® 8" d~, because a' z~ = O. The homomorphisms i' ® 1 and 8' ® 1 are chain maps. Thus, from the short exact sequence (15) we obtain the exact homology sequence (16)

... --+

Hk(B~ ® C:) ~ Hk(Z~ ® C:) --+

--+

Hk(C~ ® C:)

Hk-l(B~ ® C:) ~ Hk-l(Z~ ® C:)

--+ ....

Let us show that Hk(Z~ ® C:) ~ (Z~ ® H.(C:))k. The group Z~ is free; therefore, the exact sequences

"~ O --+ Z q~

a"

c"q---+ q B"q

1--+ 0

and

0--+ B:-l yield the exact sequences

--+

Z:-1

--+

H;_I(C:)

--+ 0

11818"

o --+ Z; ® Z; --+ Z; ® c; ~ Z; ® B:_ 1 --+ 0

(17) and (18)

0 --+

Z'P to. 'OJ B" q-l

--+

Z'P to. 'OJ z" q-l

--+

Z'P to. (G") 'OJ H" q-l.

--+ 0 .

The exact sequence (17) implies Ker(±l ® and

a::) = Ker(l ® a::) = Z; ® Z; a::) Z; ® B:-

a::)

Im(±l ® = Im(l ® = Hence the exact sequence (18) has the form 0--+

Im(±l ®

a;)

--+

Ker(±l ®

1·

a::-d --+ Z; ® H;_l(C:) --+ o.

It follows that Ker(±l ® a:-I)/Im(±l ® quently, Hk(Z~ ® C:) ~ (Z~ ® H.(C:))k.

a:;) ~ Z~ ® H:_1(C:) and, conse-

Similarly, Hk(B~ ® C:) ~ (B~ ® H.(C:))k. It follows directly from the definition of the connecting homomorphism Cik that, under the above identifications, this homomorphism takes the form j ® 1: (B~ ® H.(C:))k --+ (Z~ ® H.(C:))k,

where j: B~

--+ Z~

is the natural embedding. Therefore, the free resolution

o --+ B~ --+ Z~

--+ H(C~) --+ 0

determines the exact sequence Q --+ (Tor(H.(C~), H.(C:)))k --+ (B~ ® H.(C:))k

~ (Z~ ® H.(C:))k

--+

(H.(C~) ® H.(C:))k

--+ 0,


99

which implies that Kerak = (Tor(H.(C~), H.(C~»)k and Cokerak = (H.(C~)®H.(C~»k' Summarizing, we see that the exact homology sequence (16) gives rise to the exact sequence

o -- (H.(C~) ® H.(C:»k - - Hk(C~ ® C:) - - (Tor(H.(C~), H.(C:»))k-l - - O. It remains to verify that this exact sequence splits. Recall that the chain complex C~ is free (so far, we have used only the assumption that the chain complex C~ is free). Let I': C~ --+ Z~ and I": C~ --+ Z: be splitting maps. Then the restriction of I' ® I" to Z~ ® Z~ induces a homomorphism

H (C' •

G")

• ®.

--+

Z~ ® Z~ ~ H (G') H (G") Z" + Z'.\01. B" • • ® • •. .'01.

B'

/0\

/0\

This homomorphism is splitting.

D

3.3. The Homology of Products. Combining the algebraic Kiinneth theorem with the calculation of the chain complex C.(K x L), we obtain the following assertion. Theorem 2.25 (Kiinneth [74], [75]). If K and L are finite simplicial complexes, then the sequence

is exact. This exact sequence splits, but not canonically. Remark. Both Kiinneth's papers appeared before Emmy Noether noticed the group structure in homology. For this reason, Kiinneth considered the Betti and torsion numbers rather than groups. Example 21. The equality Hk(T"')

= Z(~)

holds.

Proof. We prove the required equality by induction on n. Note that Tn+! = T'" X SI. For n = 1, the assertion is obvious. We have Tor(H.(T n ), Z) = 0; therefore,

Hk(T"'+!) ~ (Hk(rn) ® HO(SI» ffi (Hk-l(rn) ® Hl(SI» ~

Hk(rn) ffi Hk-l (rn).

It remains to note that (ntl)

=

(~)

+ (k~l)'

D

The proof of the algebraic Kiinneth theorem remains valid for homology with coefficients in the additive group of the field F; moreover, the argument becomes simpler, because we deal with linear maps of vector spaces. For vector spaces V and W over F, the tensor product V ®F W is defined as the group obtained from V ® W by introducing the additional relation

100

2. Cohomology Rings

AV ® W = v ® AW for any A E F; the group V ®F W is a vector space over F. We repeat the proof of the algebraic Kiinneth theorem with the free chain complexes C~ and C: replaced by C~ ® F and C: ® F and the operation ®, by ®F. Up to the calculation of the kernel of Ok, no changes in the proof are needed. But now, the exact sequence obtained from

o -- B~®F - - Z~®F - - Hp(C~ ®F)

-- 0

has the form

o -- (B~ ® F) ®F tIq(C: ® F)

~ (Z~ ® F) ®F Hq(C: ® F) - - Hp(C~ ® F) ®F Hq(C: ® F) - - 0;

i.e., Ker Ok = 0 and the group Tor disappears. As a result, instead of a short exact sequence, we obtain a canonical isomorphism

E9

Hp(C~ ® F) ®F Hq(C: ® F)

-+

Hk(C~ ® C: ® F).

p+q k

Since C~ ® C: ® F ~ (C~ ® F) ®F (C: ® F), it follows that

EB

Hp(K; F) ®F Hq(L; F) ~ Hk(K x L; F).

p+q=k

Problem 76. Prove that the product of two closed manifolds is orient able if and only if both manifolds are orient able. Problem 77. Prove that the sphere sn is not a product of two manifolds of positive dimension. Problem 78. Given n > m > 1, prove that all homotopy groups of the spaces sn x lRpm and sm x lRpn are isomorphic, while the homology groups are different. Problem 79. Prove that all homotopy groups of the spaces S2 x lRPOO and lRP2 isomorphic, while the homology groups are different. Problem 80 ( [86]). Given finite connected simplicial complexes A and B, prove that Hr+1 (A

* B) =

E9 (H,(A) ® Hj(B)) E9 E9

Tor(Hi(A), Hj(B)).

i+j=r-l

3.4. The Kiinneth Theorem for Cohomology. The cohomology of a product can be calculated using the universal coefficient formulas and the Kiinneth theorem for homology, but it is also possible to directly obtain an exact sequence in cohomology similar to the Kiinneth exact sequence in homology. We consider only the coefficient groups Z and F he additive group of a field).

101


Suppose that C~ = C.(K;Z) and C: = C.(L;Z), where K and L are finite simplicial complexes. Consider the co chain complexes C'· = Hom(C~, Z) and C"· = Hom(C:, Z) and their tensor product C'· ® C"·. Any cochain complex C· can be regarded formally as the chain complex D. in which Dk = C- k (this change of numbering is needed for the boundary homomorphism to act in the required direction). The condition Dk = 0 for k $ c is equivalent to C' = 0 for l ~ -c; it holds for all cochain complexes under consideration, because K and L are finite simplicial complexes. We apply the algebraic Kiinneth theorem, formally treating cochain complexes as chain complexes. As a result, we obtain an exact sequence

0-- (H.(C'·) ® H.(C"·))k - - Hk(C'· ® C"·)

- - (Tor(H.(C'·), H.(C"·)))k+I - - O. Note that the number k - 1 is replaced by k + 1 because of the change of numbering: if (-p) + (-q) = (-k) - 1, then p + q = k + 1. Unfortunately, this exact sequence is not quite what we want. In this sequence, Hp(C'·) = HP(K; Z) and Hq(C"·) = Hq(L; Z), as required, but the cohomology of the product K x L is calculated using the cochain complex Hom(C~ ® C:, Z) rather than C'· ® C"· = Hom(C~, Z) ® Hom(C:, Z). Consider the homomorphism 0: Hom(C~,Z) ®Hom(C:,Z) - Hom(C~ ®C:,Z) defined by

(O(c'P ® e"q), e~ ® e:) = (c'P, e~)(e"q, e:); we assume that (c'P, c;.) = 0 if p f. rand (c',q, e:) = 0 if q f. s. Lemma. The homomorphism 0 is a cochain map, i.e., dO

= Od.

Proof. By definition, d(c'P ® c',q)

= d'c'P ® e"q + (-I)P c'P ® d"e"q.

Therefore,

(dO(e'P ® e"q), c~ ® c:)

= (O(c,P ® c"q), a(c;. ® c:)) = (e'P, 8' e~)(c',q, e:) + (-It (e'P, e~)(e"q, 8" e:)

and

+ (-I)P(e'P, ~)(d"e"q, e:) = (e'P, a' e~)(e"q, e:> + (-I)p(e'P, c~)(c',q, a" c:>.

(Od(e'P ® eM), e~ ® e:> = {d' e'P, e~> {eM, e:>

The second terms in these expressions may have different signs, but if p then (c'P, = o.

c;.)

f.

r, 0

2. Cohomology Rings

102

We show that (J is an isomorphism. First, note that if el,. , "en is a basis of the group then the maps Ei such that Ei (e J ) = dij form a basis in Hom(Zn, Z). Therefore, Hom(Zn, Z) ~ and the isomorphism is uniquely determined by the choice of a basis. Choosing bases in C; and we obtain isomorphisms Hom(C;, Z) ~ C; and Hom(C;, Z) ~ Bases in C; and C; determine bases in C; ® C;; hence

zn,

zn,

C;.

Hom(C;, Z) ® Hom(C:, Z) ~ C; ® C: ~ Hom(C; ® Clearly, the isomorphism thus defined coincides with

C;,

C:' Z).

(J,

As a result, we obtain the canonical isomorphism

Hk(C'* ® C"*) induced by

-+

Hk(K x L; Z)

(J.

If the coefficient group is the additive group of a field F rather than Z, then, repeating the argument for linear spaces, we obtain an isomorphism

EB

HP(K; F) ®F Hq(L; F)

-+

Hk(K x L; F).

p+q=k 3.5. Multiplication in Cohomology and the Kiinneth Theorem. The diagonal map d: IKI -+ IK x KI defined by d(x) = (x, x) induces homomorphisms H*(K)~H.(K x K) and H*(K x K) ~ H*(K) in homology and cohomology, respectively. For cohomology, we can consider the composition of d· and the canonical monomorphism H* (K) ® H* (K) -+ H* (K x K) from the Kiinneth theorem. As a result, we obtain a canonical homomorphism H*(K) ® H*(K) -+ H*(K). The image of an element a ® (3 under this homomorphism coincides with a '-"' (3, but the proof of this assertion is complicated because the map d is not simplicial. In this section, we give a proof based on acyclic models. This approach does not apply to homology because the homomorphism d* acts in the opposite direction:

H*(K) ® H*(K) ~ H.(K x K) ~ H*(K). For this reason, multiplication in homology, unlike in cohomology, can be defined only in some special cases. We proved the Kiinneth theorem not only for integer coefficients but also for coefficients in F, where F is the additive group of a field. The proof used the multiplicative structure of the field (e.g., in the definition of the homomorphism similar to (J for integer coefficients). Therefore, for cohomology with coefficients in F, as well as for integral cohomology, we have a canonically defined multiplication H*(K; F) ®F H*(K; F) -+ H*(K; F); it coincides with cup product also.


103

The Acyclic Model Theorem. The method of acyclic models, which was developed by Eilenberg and MacLane in [33], generalizes that of acyclic supports. The acyclic model theorem acquires the most natural formulation when stated in the language of categories and functors.

A category C consists of objects and morphisms hom (X, Y) defined for each pair of objects X and Y. For any two morphisms f E hom(X, Y) and 9 E homeY, Z), a morphism go f E hom(X, Z) must be defined; moreover, it is required that (1) (h

0

g) 0 f = h 0 (g 0 f) and

(2) hom (X, X) has an element id x such that f 0 idx = f and idx og = 9 for all f E hom (X, Y) and 9 E homeY, X). A covariant functor T from a category C to a category 'D takes each object X of the category C to an object T(X) of the category 'D and each morphism f E hom(X, Y) to a morphism T(f) E hom(T(X), T(Y)). It must satisfy the conditions T(idx) = idT(x) and T(f 0 g) = T(f) 0 T(g). A contravariant functor differs from a covariant functor in that T(f) E hom(T(Y) , T(X» and T(f 0 g) = T(g) 0 T(f). Let Tl and T2 be covariant functors from a category C to a category 'D. A natural transforrnationr: Tl ~ T2 assigns a morphism r(X): T1(X) ~ T2(X) to each object X of the category C in such a way that for any morphism f: X ~ Y (i.e., f E hom (X, V»~, the following diagram is commutative:

T 1 (X)

~ T1(Y)

IT(X)

T 2 (X)

IT(Y)

~ T2(Y).

A category with models is a category in which a class of objects M = {Ma} is distinguished; the distinguished objects are called models. A (covariant) functor T from a category with models to the category of free nonnegative chain complexes and chain maps is said to be free with respect to the models from M if for each k, there exists a subset Mk C M such that every group Ck(Ma ), where Ma E Mk, has an element e~ for which the elements T(f)(e~) E Ck(X) are pairwise distinct and form a basis of the group Ck(X). We assume that f ranges over all elements of the set hom(e~, X), and a ranges over all indices of the elements of Mk.

A functor T is said to be acyclic with respect to M if Hk(T(Ma» = 0 for k > o. If a functor is free with respect to the models from M, then it is also free with respect to the models from any M' ::> M, and if a functor is acyclic

2. Cohomology Rings

104

with respect to the models from M, then it is also acyclic with respect to the models from any M' eM. Example 22. Consider the category of simplicial complexes with ordered vertices and simplicial maps preserving the order of vertices. The functor that takes each simplicial complex to its chain complex is free and acyclic with respect to the models from M = {[O, 1, ... , k]}, where k = 0,1, .... Theorem 2.26 (on acyclic models). 1fT is a free functor andT' i.s a functor acyclic with respect to the models from M, then any natural transformation cp: Ho(T) - Ho(T') is induced by some natural chain map T: T - T'. Moreover, any two natural chain maps T, T: T - T' inducing cp are joined by a natural chain homotopy. Proof. The group Ho is a quotient of the zero-dimensional chains; hence there exist epimorphisms p: To(Ma,) - Ho(T(Ma)) and p': T6(MD ) Ho(T'(Ma)). For each element e~ E To(Ma), there exists a To(e~) E To(Ma) such that p'To(e~) = cpp(e~). The elements T(f)(e~), where f E hom(Ma, X), form a basis of the group To(X). The formula To(T(f)(e~)) = T'(f)(To(e~)) uniquely determines a homomorphism TO: To(X) - To(X). Since the transformation cp is natural and T(f) and T'(f) are chain maps, the homomorphism TO induces the initial map cp: Ho(To(X)) - Ho(To(X)) in the zero-dimensional homology group. We construct homomorphisms Tk: Tk(X) - Tk(X) for which aTk = Tk la by induction on k. For each element e~ E Tk(Ma ), consider Tk-l(ae~) E Tk-l (Ma). There exists a chain ~ E Hk(Ma) with a~ = 7l:-1 (ae~). Indeed, for k = 1 it exists because TO induces a map of zero-dimensional homology groups, and for k > 1, because aTk-l(ae~) = Tk-2(aae~) = 0 and, by assumption, Hk-1(T'(Ma )) = O. We set 7l:(e~) = ~ and define a homomorphism 7l:: Tk(X) - Tk(X) by using the construction described above. For chain maps T, T: T - T' that induce the same natural transformation cp: Ho(T) - Ho(T'), the chain homotopy D k : Tk(X) - Tk+1 (X) is constructed in a similar way. For the equalities aDo = TO - TO and aDk = 7l: - Tk - Dk-1a to hold for k > 1, it is sufficient for the chain 7l:(e~) - Tk(e~) - Dk-lae~ (the chain To(e~) - To(e~) for k = 0) to be the boundary for any element e~. This condition holds for k = 0 because TO and TO induce the same map in homology; for k > 1, this follows from the equality aTk - aTk - aDk-la

= Tk-la -

and the acyclicity of the functor T'.

Tk-1a - aDk-la

=

DA-2aa

=0

o


105

The Alexander-Whitney Diagonal Approximation. Let d: IK x KI be a diagonal map, and let

d.: H.(K)

-+

H.(K x K)

~

IKI

-+

H.(C.(K) ® C.(K))

be the induced map of homology groups. At the level of chains, do takes each vertex v to the chain v ® v. We refer to any natural chain map T: C.(K) -+ C.(K) ® C.(K) that takes v to v ® vasa diagonal approximation. The functor C.(K) is free with respect to the models from M = {~k} (see Example 22), and the functor C.(K) ® C.(K) is acyclic with respect to these models. Therefore, according to the acyclic model theorem, any diagonal approximation induces a map d. in homology. In calculations, it is convenient to use the Alexander Whitney diagonal approximation, which is defined by n

T([VO, v}, ... , v n ])

= ~)Vo, ... , Vi]

® [Vi, ... , Vn ].

i=O

We must verify only that this is a chain map, i.e., Ta = aT. Clearly,

T(a[a, 1, ... , n])

= T(~) -1)i[a, ... , i, . .. , n]) =

L(-I)i[a, ... ,i, ... ,i] ® [j, ... ,n] i<j

+ L(-I)i[a, ... ,i] ® [j, ... ,i, ... ,n] i<j

and

a(T[a, 1, ... , n])

= a L[a, . .. ,i] ® [j, ... , n] = L(-I)i[a] ® [0, ... ,i, ... ,n] i

L

+

(-I)i[a, ... ,i, ... ,i]®[j, ... ,n]

i:$.j,jofO

L (-I)i[a,···,i]®[j, ... ,i, ... ,n] + L( -1)i[a, ... , i, ... , n] ® in]. +

i?j,jofn

i

2. Cohomology Rings

106

The second expression differs from the first by the additional sum

+ [0] ® [l, ... ,n] - [0] ® [1, ... , n] + [0,1] ® [2, ... ,n]

- [0,1] ® [2, ... , n] + [0,1,2] ® [3, ... , n]

+ (_l)n 1[0,1, ... ,n - 2] ® [n -l,n] + (_l)n-l[O, 1, ... ,n -1] ® [n] + (-l)n[O, 1, ... , n - 1] ® [n], but this sum vanishes. The map C*(K)

-+

C*(K x K) defined as the composition

L 0, ... ,i] ® [i, ... ,n]1-+ L[O, ... ,i] x [i, ... ,n]

[0,1, ... ,n]1-+

induces the map d* in homology and d* in cohomology. It easily follows that the composition

H*(K) ® H*(K)

---+

H*(K x K)

.!£... H*(K)

is a cup product. Indeed, at the level of cochains, the co chain cfJ+q corresponding to cP ® c q has the property

(c'P+ q, [va, Vb ... , Vp+q]) = (O(c'P ® cq), L[vo, ... , Vi] ® [Vi, ... , Vp+q])

= L (c'P, [Va, ... , Vp]) (c q, [vp, . .. ,vp+q]). The last equality holds because the remaining terms vanish. Thus, cp +q r!' '--' cq , as required.

=

3.6. Cohomology Cross Product. Suppose that K and L are finite simplicial complexes, C*(K) and C*(L) are simplicial co chain complexes, and C*(K x L) is the cellular cochain complex. Two cochains r!' E CP(K) and & E Cq(L) determine the cochain r!' x cq E Cp+q(K x L) by the rule

(cfJ x cq, t!!"P' x tl" q') = ~P'P'~qq' (c'P, tl,,'P')(cq, tl" q'). The equality 8(tl"p x tl"q) = 8tl"p x tl"q + (-l)Ptl"p x 8tl"q implies ~(r!' x cq) = ~cfJx&+( -l)Pr!'x~&. We obtain a group homomorphism HP(K) xHq(L) -+ H'P+q(K x L), which is called the cohomology cross product, or external cup product.

It is easy to show that cohomology cross product has the following properties: (1) (0 x (3) x -y

=0

x ({3 x -y);

K is the natural projection, then P'K(oP) = oP x 1L ; (3) if a map T: X x Y -+ Y x X is given by T(x, y) = (y, x), then T*({3q x oP) = (-l) pq oP x {3q. (2) if PK: K x L

-+


107

The cohomology cross product is closely related to the cup product. Namely, o:P ......., {3q = d*(a P x {3q), where d*: Hp+q(K x K) - Hp+q(K) is induced by the diagonal map d. Indeed, (d*(o:P x {3q), ~p+q) = (o:P X {3q, d(~p+q)). The map d is not cellular; therefore, we must replace d(~p+q) by the Alexander Whitney diagonal approximation. As a result, we obtain

(d*(aP x {3q), [va, . .. , vp +q]) = =

(

~q

aP x {3q, t;[va, . .. , Vi]

X [Vi, ••• ,Vp +q ]

)

(aP x {3q, [va, ... ,VpJ x [vp , ••• , Vp +q ]}

= (aP '-" {3Q, [va, ... , VP+q]) , as required.

In calculating the cohomology rings of products of simplicial complexes, the following assertion is often useful.

Theorem 2.27. In the ring H*(K xL), (afl x {3fl) '-" (~2

X {3~2)

= (-1 )q1P2 (afl ......., ~2)

X

({3'r '-"

{3~2).

Proof. We have (afl

X

{3il ) ......., (~2

X {3~2)

= di< x £ ( (afl

= di<x£(af l

X X

{3il )

{3fl

X

X (~2 X {3~2))

~2

X

{3~2)

and (afl '-" ~2) Clearly, dKx£

=

X

({3fl '-" {3~2)

(idK xT x id£)

(afl x {3fl) '-" (~2

0

= di< (afl

x ~2)

X

dL({3i l

X {3~2).

(dK x d£). Therefore,

X {3~2)

= (dK X d£)*(idK X T X id£)*(afl X {3fl X ~2 = (_1)qlP2(dK X d£)*(afl X ~ X {3fl x {3?) = (-1)QIP2(dK x d£)*di«afl

x~) X

d£({3i l

Corollary. If 0: E H*(K) and {3 E H*(L), then Pi.([snJ) is the homotopy class of the map ar, ... , a k be the simplices in sn for which cp(ai) i= Xo. If k ~ 2, then the required assertion follows directly from the definitions. For k > 2, we use induction on k.

Sn!!....K. Let

Thus, we must prove that cI>.([snJ) - O. For any simplex a n +1 C Ln+l, the element cI>.(aa n+1 ) corresponds to the homotopy class of the map aan+l ~ K. This map is null-homotopic because it can be extended over a n+l . Therefore, cI>. (a a n+ 1 ) = O. Let Cn+l - E ai af+ 1 . Then cI>.([snJ) = cI>.(aCn+d = E alcI>.(aa~+l) - 0, as required. 0 Remark. The proof of the Hurewicz theorem given above follows largely the paper [lllJ by Rokhlin. In the original proof of Hurewicz, more general spaces than simplicial complexes were considered. Example 23. The dunce hat is the CW-complex K obtained by identifying the edges of the triangle as shown in Figure 2. The dunce hat is contractible.

Figure 2. The dunce hat

Proof. The fundamental group 71"1 (K) has one generator a, and this generator satisfies the relation a 2 a- 1 = 1, i.e., a = 1. Thus, the complex K is simply connected. The chain complex for calculating the cell homology of the complex K has the form· .. -+ 0 -+ 0 -+ C 2 .!!..... C 1 -+ 0, where C 2 ~ Z, Cl ~ Z, and a is an isomorphism. Therefore, Hi(K) = 0 for i ~ 1. It follows from the Hurewicz theorem that 7I"i(K) = 0 for all i ~ 1. Therefore, according to the Whitehead theorem (see Part I, p. 179), K is contractible. 0 Problem 86. (a) Prove that the suspension over any acyclic CW-complex is contractible. (b) Give an example of a noncontractible space whose suspension is contractible.

3. Applications of Simplicial Homology

116

1.2. Obstruction Theory. Suppose that X is a path-connected topological space, K is a simplicial complex, L is a subcomplex of K, and Kn is the n-skeleton of K. Let f: kn --+ X be a map defined on kn = K n U L. We want to determine whether this map can be extended to kn+1. For n = 0, the construction of an extension of a map to a path-connected space is obvious; thus, we assume that n ~ l. We will need to add elements of groups 1I"n(X, x) with different base points x E X; for this reason, we assume that the space X is n-simple. 2 For n = 1, this means that the group 11"1 (X, x) is commutative. We associate with a cochain en+1(J) E C n+1(K;1I"n(X)) the map f: kn --+ X as follows. The consideration of chains and co chains for K assumes that all simplices in K are oriented. An orientation of a simplex ~ n+1 C K induces an orientation of the boundary a~n+1 ~ sn; hence the restriction of f to a~n+1 determines an element of the group 1I"n(X). We assume that the value of the co chain en +1 (J) at the simplex ~ n+1 is equal to this element of 1I"n(X). It follows directly from the definition that the map f can be extended to a simplex ~n+1 C K if and only if en+1(J)(~n+1) = O. In particular, if ~n+1 C L, then en+1(J)(~n+1) = 0 because f is defined on L. Therefore, en+1(J) E C n+1(K, L; 1I"n(X» C Cn+l(K; 1I"n(X». In what follows, we treat the co chain en + 1 (J) as a relative cochain. It is called an obstruction to extending f over kn+1.

Problem 87. Prove that SP x sq = (SP V sq) Uj Dp+q, where the map f: Sp+q-l --+ SP V sq is not null-homotopic. Theorem 3.3. We have the equality 8en+1(J) cn+1(J) E zn+1(K,L;1I"n(X)) is a relative cocycle.

=

0, which means that

Proof. We must show that cn+1(J)(a~n+2) = 0 for any (n + 2)-simplex in K. First, note that the n-skeleton of the simplicial complex a~n+2 ~ sn+1 is (n -I)-connected. This skeleton is homotopy equivalent to the punctured sphere sn+ 1 (from each (n+ I)-face of the simplex ~n+2, one interior point is removed), but puncturing an (n + I)-manifold Mn+1 does not affect the homotopy groups 11"0,11"1. ••• ,1I"n-l. Indeed, for k ~ n, any map Sk --+ M n+1 is homotopic to a map not involving the removed point Yo, and for k ~ n-I, any homotopy between two maps Sk --+ Mn+ 1\ {YO} in Mn+ 1 call be replaced by a homotopy that does not involve Yo· Let a~n+2 = E~~+1. By definition, the element en+1(J)(~~+1) of

~n(X)

corresponds to the map a~~+1 Consider the simplicial complex B n that is the n-skeleton of a~n+2. Take the element Qi of 1I"n(Bn)

Lx.

2The definition of n-simple spaces was given in Part I, p. 168.

117

1. Homology and Homotopy

sn

that corresponds to an orientation-preserving homeomorphism a~~+1. Clearly, the homomorphism I.: Trn(Bn) --+ Trn(X) takes cn+1(J)(~~+1). This means that the composition the cycle a~~+1

----+

a class in Hn(Bn)

h- 1 ----+

0i

--+

to

f

Trn(Bn) ~ Trn(X),

where h is the Hurewicz homomorphism, yields the element cn+1(J)(~~+1). Indeed, since the space B n is (n - 1)-connected, it follows that h is an isomorphism for n ~ 2. If n = 1, then l.h- 1 is well defined since, by assumption, the group Trl (X) is Abelian and, therefore, I. maps the commutator subgroup of Trl(B2) to O. Thus, the element h-1(z) is determined up to the addition of [Trl(B2), Trl(B2)], and the element l.h-1(z) is determined uniquely. The element cn+1(J)(a~n+2) is equal to the image of the homology class of the cycle E a~~+1 = aa~n+2 = 0 under the homomorphism l.h-1j therefore, cn+1(J)(a~n+2) = o. 0 We denote the cohomology class of the co cycle cn+1 (J) by C n+1 (J). Theorem 3.4 (Eilenberg [31]). The equality C n+1(J) = 0 holds il and only il the map I: kn --+ X can be extended over kn+1 after having been changed on kn \ kn-l. Proof. Let I, g: kn --+ X be two maps coinciding on kn-l. We define the difference cochain ~(J, g) E Cn(K, Lj Trn(X») as follows. We take two copies of the simplex ~n C kn and glue them together along the boundary a~n c kn-l. Then we identify the first copy with the upper hemisphere and the second copy with the lower hemispherej the orientation of the sphere is assumed to be compatible with that of the second simplex. We apply I to the first copy of ~ nand g to the second. The maps I and 9 coincide on a~nj thus, we obtain a continuous map sn --+ X. The corresponding element of the group Trn(X) is the value ofthe cochain ~(J,g) on ~n. The equality ~(J, g)(~n) = 0 holds if and only if the map a(~n x I) --+ X, which coincides with I on ~n X {1}, with 9 on ~n X {O}, and with 118.6.n = gl8.6.n on a~n x it} for all t, can be extended to ~n X I, i.e., the maps ~n L X and ~n !!..... X can be connected by a homotopy which is the identity on a~n. In particular, we have d:n(J, g)(~n) = 0 if ~n C Lj thus, ~(J,g)(~n) is indeed a relative co chain. Moreover, ~(J,g)(~n) = 0 if and only if the maps f and 9 can be connected by a homotopy which is the identity on kn-l. Lemma 1. For any map f: kn --+ X and cochain ~ E Cn(K, Lj Trn(X», there exists a map g: kn --+ X such that it coincides with f on k n - 1 and d!"(J, g) = an.


118

Proof. For each simplex D.. n c k n , there exists a map 9 such that the map sn -+ X which coincides with / on the upper hemisphere and with 9 on the lower hemisphere represents a given element of 7l"n(X)j such a map for the simplex D.. n c L represents the zero element of 7l"n(X), so that 9 = /. The desired map 9 can always be constructed because 7l"n(X) is a group. 0 Lemma 2. M:'(f,g)

= cn+l(g) -

cn+l(f).

Proof. Any element of the form (cn+l(g) - cn+1(f))(D..n+l) is equal to the difference of the homotopy classes of the maps 8D.. n+l ~ X and 8D.. n+l L X. Let 8D.. n+l = ED..i. Then the element dn (f,g)(8D.. n+l) equals the sum of n + 2 homotopy classes of maps sn -+ X each of which coincides with D..i L X on the upper hemisphere and with D..~ ~ X on the lower hemisphere (i = 1, ... ,n + 2). For instance, if n = 1, then one element is represented by the SJJIU of the homotopy classes of the two curves shown in Figure 3a and the other,

b

a

Figure 3. Equal homotopy classes

by the sum of the homotopy classes of the three curves shown in Figure 3b. These elements are equal because, by assumption, the sum of the homotopy classes of two curves does not depend on the base point. For an arbitrary n, the homotopy classes coincide as well. 0 Now we can proceed to the proof of the theorem. Suppose Cn+l(f) = 0, i.e., cn+l(f) = ~d!' for some cochain an. According to Lemma 1, there exists a map g: kn -+ X such that it coincides with / on k n- 1 and d!'(f, g) = _dn . By Lemma 2, we have cn+l(g)

= ~cr(f,g) + cn+l(f) =

-~cr

+ ~cr = OJ

thus, the map 9 can be extended to kn+l. Now suppose that there exists a map g: kn -+ X such that it coincides with / on k n - 1 and can be extended to kn+ 1. Then c.,. 1 (g) 0 and cn+1(f) = cn+1(f) - cn+l(g) = -~dn(f,g), i.e., Cn+l(f) = O. 0


119

Remark. In what follows, we use the equality dn(J, g)+~(g, h)+dn(h, f) = O. It is proved by the same argument as Lemma 2. 1.3. The Hopf-Whitney Theorem. In Part I, p. 231, we proved the Hopf theorem about a one-to-one correspondence between the set of homotopy classes of maps M n -+ sn and the group Z (to each map I: M n -+ sn corresponds the integer deg I E Z). This theorem can be substantially generalized by methods of obstruction theory. We use the notation [X, Y] for the set of classes of homotopy equivalent maps X -+ Y. For spaces with base points, we consider only maps that take base points to base points and study homotopies in the class of such maps. The following theorem was proved by Hopf [59], but Hopf considered homology rather than cohomology. For this reason, in the case where the (n - I)-dimensional homology group of the complex K has torsion, the statement was fairly intricate. Whitney [150] noticed that, in terms of cohomology, not only the statement is simpler but also the correspondence acquires a more natural description. Theorem 3.5 (Hopf Whitney). For any n-dimensional simplicial complex K (with a base vertex) and any (n - 1) -connected space X (with a base point xo), there is a one-to-one correspondence

Proof. Consider the set of maps of pairs (K, K n - 1 ) -+ (X, xo). We declare two such maps to be equivalent if they are joined by a homotopy fixed on Kn-2. We denote the set of equivalence classes by [K, Xn. Let us show that the natural map (): [K, Xn -+ [K, X] is one-to-one.

I:

X, we construct a homotopy between the maps K O -+ xo and K O L X and extend it to a homotopy between Kn-l -+ xo and K n - 1 L X using the (n-l)-connectedness of X. According to the Borsuk lemma (see Part I, p. 164), this homotopy can be extended to a homotopy of the map I: K -+ X. As a result, we obtain a map g: K -+ X which is homotopic to I and tak('s K n - 1 to xo. Therefore, the map () is surjective. Now consider two maps I, g: (K, Kn-l) -+ (X, xo) joined by a homotopy Ho: K x 1 -+ X. We must prove that they can be joined by a homotopy fixed on K n - 2. By assumption, we have Ho(K n - 1 x 81) = xo; this means that Ho(K n - 2 x 81) = Xo. Let us extend the map Kn-2 x 81 x 1 -+ Xo EX to a map K n - 2 x 1 x 1 - X which coincides with Ho on K n - 2 x 1 x {O} and is constant on Kn-2 x 1 x {I}. This can be done by induction on the dimension of the skeleton because the dimension of the CW-complex K n - 2 x 1 x I is at most n and the space X is (n-l )-connected. The homotopy For an arbitrary map

K

-+

120


constructed on Kn-2 x I can be extended to a homotopy H t of the map Ho. For !'(Y) = Hl(Y, 0) and g'(y) = H 1 (y, 1), we have I"'!' '" g' '" g, and all the homotopies involved are fixed on Kn-2. To each map I: (K, K n 1) --+ (X, xo) we assign the cohomology class k(J) E Hn(K;7rn(X)) corresponding to the co cycle d"'(J,*), where * is the map of the complex K to Xo (the difference cochain d"'(J, *) is a co cycle because dimK = n). Let us show that the map

k: [K, X~

--+

Hn(K; 7rn (X))

thus defined is one-to-one. Let I, g: (K, K n- 1 ) --+ (X, xo) be maps for which k(J) = k(g). Then the co cycle dn(J,g) = cF(J,*) - dn(g,*) is a coboundary, i.e., I and 9 are joined by a homotopy fixed on Kn-2. Therefore, the map k is injective. Consider any co cycle cF E zn(K; 7rn (X)). It can be realized as the difference cochain dn(J, *) for some map I: (K, K n- 1 ) --4 (X, xo). Moreover, kf:.f} is the cohomology class of the co cycle d"' = cF(J, *). Therefore, the map k is surjective. 0 For X = sn, the correspondence [K, sn] +-----+ Hn(K; Z) can be defined explicitly: corresponding to each map I: K --+ sn is the element E Hn(K; Z), where a is the generator of the group Hn(sn; Z) (determined by orientation). Indeed, we can assume that I takes K n - 1 to the base point *. By definition, the value of the difference cochain cF(J, *) at each simplex ~n C K n = K is equal to the degree of the map ~n/a~n L sn. The co cycle representing the cohomology class takes the same value at this simplex. Thus, the Hopf Whitney theorem has the following corollary.

rOo

rOo

Theorem 3.6. II K is an n-dimensional simplicial complex, then any element 01 the group Hn(K; Z) can be represented in the lorm where a is the generator 01 Hn(sn; Z) and I: K --+ sn is a map determined uniquely up to homotopy.

rOo,

1.4. Algebraically Trivial Maps. A map I: X --+ Y is said to be algeHi(y) --+ Hi(X) braically trivial if the maps I.: H I (X) --+ H,(Y) and are trivial for all i ~ l.

r:

An example of an algebraically trivial map is an arbitrary map I: sn --+ sm, where n i- m. In particular, the Hopf fibration p: S3 ---+ S2 is algebraically trivial. We use this observation to obtain a homotopy classification of algebraically trivial maps I: X --+ S2, where X is a three-dimensional simplicial complex; to be more precise, we establish a one-to-one correspondence between the homotopy classes of algebraically trivial maps X --+ S2 and the elements of H 3 (X). Namely, according to the Hopf-Whitn y theorem, We have [X, S3] ~ H3(X, 7r3(S3)) = H 3(X). Therefore, each c•. h mology class


121

a E H3(X) corresponds to a homotopy class of maps F: X -+ 8 3 . We associate a class a E H3(X) to the map 1 = pF, where p: 8 3 -+ 8 2 is the Hopf fibration. The map 1 is algebraically trivial because so is p. A classification of algebraically trivial maps I: X -+ 8 3 for a threedimensional simplicial complex X is obtained from a more general classification for an arbitrary simplicial complex X.

Theorem 3.7 (Pontryagin). For any simplicial complex X, [F]I--+ (PF] is a one-to-one correspondence between the homotopy classes 01 maps X -+ 8 3 and the algebraically trivial maps X -+ 8 2 • Proof (see [63]). As usual, the proof that the correspondence is one-to-one consists of two steps: (1) any map belongs to the image; (2) if the images of two' maps are homotopic, then so are the initial maps. Step 1. For any algebraically trivial map F: X -+ 8 3 for which pF = I.

I:

X

-+

8 2 , there exists a map

The map p: 8 3 -+ 8 2 is a locally trivial fibration; hence there exists an open cover {Uo.} over which the fibration is trivial. Consider a refinement of the triangulation of X such that the image of each simplex under 1 is contained in some U0.. We set Ik = Ilxk. Let us construct maps Fk: Xk -+ 8 3 such that pFk = Ik for k = 2, 3, .... We start with F2. The map h is algebraically trivial because it is a composition of the form X 2 -+ X L 8 2 and 1 is algebraically trivial. It follows that h is null-homotopic. Indeed, the homotopy class of the map h: X 2 -+ 8 2 corresponds to the element I;a E H2(X2), where Q is the generator of the group H2(8 2), and this element is zero because the map h is algebraically trivial. Consider a homotopy H between h and a constant map. Obviously, the constant map admits a lifting; therefore, we can construct a lifting of the homotopy H. As a result, we obtain a map F2 : X 2 -+ 8 3 for which pF2 = h· Suppose that the map Fk-l is constructed for some k > 2. We construct Fk as follows. Let (I = t:J..k be a k-simplex in X. On its boundary 8t:J.. k ~ 8 k - 1 , the map F k - 1 is defined; hence we can consider the map

8t:J.. k ~ p-l(Uu ) ~ Uu x 8 1 ~ 8 1 . Here Uu is the set from the cover that contains the image of t:J.. k , hu is the homeomorphism from the definition of a locally trivial fibration, and 7ru is the projection onto the second factor. We have constructed a map 8 k - 1 -+ 8 1 , where k - 1 ~ 2. Any such map is null-homotopic; therefore, it


122

can be extended to a map gO": /:l.k

Fk(X) =

{

--+

8 1 . Now we can define Fk as follows:

D (x) I'k-l h;l(fk(X),90"(x))

l'f X

E Xk 1

,

if x E /:l.k C Xk.

Finally, the map F: X --+ 8 3 is defined by the condition that its restriction to Xk coincides with Fk for any k ~ 2. Step 2. If maps F, G: X

--+

8 3 are such that pF ~ pG, then F ~ G.

If pF ~ pG, then, by the covering homotopy theorem, we can lift the homotopy between these maps, obtaining a map F': X --+ 8 3 such that it is homotopic to F and pF' = pG. So, we can assume from the very beginning that pF =pG.

Consider the sphere 8 3 as the group of unit quaternions. Then the Hopf fibration is the map taking each unit quaternion to the corresponding coset modulo the subgroup 8 1 of complex numbers with modulus 1. 'Let H: X --+ 8 3 be the map defined by

H{x) = F(x) . [G(x)]-I. Since pF = pG, the quaternions F(x) and G{x) belong to the same coset modulo 8 1 . Therefore, the map H takes X to a proper subset 8 1 c 8 3 , which means that H is null-homotopic, i.e., there exists a homotopy H t : X --+ 8 3 , where Ho = Hand H l (X) = 1 E 8 3 . Let cI>t(x) = Ht(x) . G(x). Then cI>o required homotopy.

= F

and cI>1

= G,

and cI>t is the 0

1.5. The Eilenberg MacLane Spaces. Suppose that n is a positive integer and 7r is a group; for n ~ 2, we assume that 7r is Abelian. A CW -complex X is called a K(7r, n) space, or an Eilenberg MacLane space, if

1I'k(X) = {11'

o

for k for k

= n, i- n.

These spaces were introduced in [32, 34]. Example 24. The circle 8 1 is a K(Z, 1) space. Example 25. Any closed 2-manifold K(7r, 1) space for 11' = 11'1 (M 2 ).

M2

different from 8 2 and

R'p 2

is a

Example 26. The space lRPao is K(Z2' 1). Example 27. The space Cpoo is K(Z, 2). Example 28. Let Xn be the set of all points of C n with pairwise different coordinates. Then Xn is a K(Pn , 1) space for some group Pn (This group P n is called the colored braid group on n strands.)


123

Proof. Consider the map Xn --+ Xn 1 forgetting the last coordinate. This map is a locally trivial fibration. Ifs fiber F is ([; with n - 1 points deleted. Thus, it is homotopy equivalent to the wedge of n - 1 circles. In particular, 7rk(F) = 0 for k ~ 2. The exact sequence 7rk(F) --+ 7rk(Xn) --+ 7rk(Xn-t} shows that if k ~ 2 and 7rk(Xn-1) = 0, then 1fk(Xn) = O. If remains to note that the space Xl = ([; is contractible. 0 Example 29. Let Y n be the quotient of Xn by the action of the group Sn (of permutations of coordinates). Then Y n is a K(Bn, l) space for some group Bn. (This group Bn is called the braid group on n strands.) Proof. The map Xn 7rk(Yn) for k ~ 2.

-+

Xn/ Sn = Y n is a covering. Therefore, 7rk(Xn)

~

0

Remark. It can be proved that the group Bn is defined by generators bl, ... , bn-l and the relations

bibi+lb, = b'+lbib'+l

for

1~i ~ n- 1

and for Ii - il ~ 2 (the latter relations are called the far commutativity relations). bibj = bjb,

We regard Soo as the subset of ([;00 consisting of the points (Zl' Z2, ... ) for which E Iz,1 2 = 1. On Soo, we introduce the equivalence relation defined by (ZI. Z2,"') '" (EZI, EZ2, . .. ), where E = exp(27ri/m). The space L~ = Soo / '" is called the infinite-dimensional lens space.

Example 30. The space L~ is K(Zm, 1). Proof. The covering Soo --+ Lc;{ is universal because the space Soo is contractible (see Part I, Problem 35). This covering is a regular covering with automorphism group Zm; hence 7r1(Lc;{) = Zm. Moreover, 7ri(Lc;{) = 7ri(SOO) = 0 for i ~ 2. 0 Problem 88. Prove that

H.(L::;)

~ {~

for odd n, for even n

> O.

Problem 89. Given a 3-manifold M3 with infinite fundamental group for which 7r2(M3) = 0, prove that M3 is a K(7r, 1) space. Theorem 3.8. For any nand 7r, there exists a K(1f, n) simplicial complex. Proof. Consider a free resolution 0 -+ R -+ F -+ 7r -+ 0, where Rand F are free groups (for n = 1) or free Abelian groups (for n ~ 2). In other words, let 1f be a group (Abelian for n ~ 2) with generators {bj I j E J}

124


and relations {ai liE I}; here {ai liE J} and {b j I j E J} are bases of the groups Rand F. Consider the wedge Kn of triangulated n-spheres Sj, where j E J. The group 7I"n(Kn) is canonically isomorphic to F. For each basis relation ai ERe F = 7I"n(Kn), take a continuous map fi: Sf -+ K n representing the homotopy class ai E 7I"n(Kn) and consider its simplicial approximation 'Pi. The triangulation of Sf can be naturally extended to a triangulation of D;+l ::J Sf (to each n-simplex we add one vertex at the center of the ball). Let K n +1 be the simplicial complex obtained from K n by attaching the balls D;+l via the maps 'Pi. If n = 1, then, according to Theorem 6.1 in Part I, we have 7I"1(K 2) ~ 71". If n > 1, then the Hurewicz theorem can be applied. Indeed, it follows from the cellular approximation theorem that the space Kn+l is (n-l)-connected. Clearly, Hn(Kn+l) ~ 71". Therefore, by the Hurewicz theorem, 7I"n(Kn+l) ~ 71". Now, we choose a system of generators in 71"n+l (Kn+l ), consider the simplicial map 1/;: sn+l -+ Kn+l corresponding to each generator I and attach Dn+2 to Kn+l via the map 1/;. As a result, we obtain a simplicial complex Kn+2 whose homotopy groups up to dimension n + 1 are as required. This is implied by the following lemma. Lemma. Suppose that a CW -complex X' is obtained from a CW -complex X by attaching an (n + 2) -cell via a map 1/;: sn+l -+ xn+l. Then the inclusion i: X ....... X' induces an isomorphism i.: 7I"k(X) -+ 7I"k(X') for k ~ n; for k - n + 1, it induces an epimorphism i.: 7I"k(X) -+ 7I"k(X') whose kernel contains the subgroup generated by the homotopy class of 1/;. Proof. According to the cellular approximation theorem, for k ~ n+ 1, any map from the sphere Sk into a CW -complex Y is homotopic to a map into the (n + I)-skeleton yn+l, and for k ~ n, any homotopy in Y between two maps Sk -+ yn+l can be replaced by a homotopy in yn+l. This implies that i. is an isomorphisms for Ir ~ n and an epimorphism for k = n + 1. The composition sn+l J!.... X ~ X' is null-homotopic; the contraction is performed along the cell Dn+2. D Then, we kill the group 7I"n+2(Kn+2) in the same way as the (n dimensional homotopy group of Kn+l, and so on. Theorem 3.9. Any two K(7I", n) spaces (with the same topy equivalent.

71"

+ 1)D

and n) are homo-

Proof. It is sufficient to verify that any K(7I", n) space X is homotopy equivalent to the simplicial complex K constructed in the proof of Theorem 3.8. Let us construct a continuous map K ~ X i.nducing isomorphisms 7I"k(K) ~ 7I"k(X) for all k. First, we define this m8P on K n = VSf.


125

Each sphere Sj corresponds to a generator bj of the group 7rn (X). Choose a map Sj -+ X representing the element bj E 1Tn(X). As a result, we obtain a map Kn -+ X. We extend it over Kn+l as follows. For each (n + 1)-cell D~+l, the characteristic map S; -+ K n corresponds to a basis relation ai ERe F = 7rn (Kn). The map K n -+ X induces an epimorphism 7rn(Kn) -+ 7rn (X) with kernel R. Therefore, the composition S:' -+ K n -+ X is homotopic to a constant map and can therefore be extended over the entire cell D~+l. We have obtained a map Kn+l -+ X. Consider the composition Kn -+ Kn+l -+ X and the induced group homomorphisms 7rn(Kn) -+ 7rn(Kn+l) -+ 7rn(X). We know that both homomorphisms 7rn(Kn) -+ 7rn(Kn+l) and 1Tn(Kn) -+ 7rn(X) are epimorphisms with kernel R. It follows that 7rn(Kn+l) -+ 7rn (X) is an isomorphism. There are no obstructions to extending the map Kn+l -+ X over cells of dimension higher than n + 1 because 1Tn(X) = 0 for k ~ n + 1. The extension K -+ X induces group isomorphisms 7rk(K) -+ 7rk(X) for all k; for k = n, this is so because the inclusion Kn+l ~ K induces an isomorphism 7rn(Kn+l) -+ 7rn (K), and for k i- n, because 7rk(K) = 1Tk(X) = O. By assumption, both spaces K and X are CW-complexes; thus, we can apply the Whitehead theorem (Theorem 4.21 in Part I). 0 Problem 90. Let X be a finite-dimensional K(7r,l) simplicial complex. Prove that the group 7r contains no elements of finite order. 1.6. Cohomology and Maps to K(7r, n) Spaces. Let K be an (n-1)connected simplicial complex, where n ~ 2. Then the group Hom(Hn(K), 7rn (K» contains the homomorphism h- 1 inverse to the Hurewicz homomorphism h: 1Tn(K) -+ Hn(K). Moreover, Hn-l(K) = 0; hence the exact sequence

o --+ Ext(Hn-l (K), G) --+ Hn(K; G)

--+

Hom(Hn(K), G)

--+

0

implies the isomorphism Hn(K; G) ~ Hom(Hn(K), G). The cohomology class FK E Hn(K; 1Tn(K» corresponding to h- 1 is called the fundamental cohomology class of the space K. The most important case is where K is a K(1T, n) simplicial complex. In this case, the fundamental cohomology class is denoted by F-n; E Hn(K; 7r) ~ Hom(Hn(K), 1T); note that Hn(K) ~ 7r and F-n; corresponds to the identity map 7r -+ 7r. For the K(7r, n) simplicial complex K constructed in the proof of Theorem 3.8, the fundamental cohomology class F-n; is represented by the co cycle en defined as follows. Recall that Kn = V sn where each sphere S!" correI , I sponds to a generator ai E 7r. To the fundamental class of Sf the co cycle en assigns the value ai because the Hurewicz homomorphism assigns to this fundamental class the element ai E 7rn (K).

126


Theorem 3.10. Let L be a K(1I",n) simplicial complex.

simplicial complex K, the map [K, L] !*(F1r ) E Hn(Kj1l") is one-to-one.

---+

Then, for any n H (K, 11") that takes f: K ---+ L to

Proof. According to the Hopf Whitney theorem, there is a one-to-one correspondence

We must only verify that this correspondence has the required form and is determined by the n-skeleton Kn. Consider the diagram

(21)

In this diagram both horizontal arrows are induced by the natural embedding i: K n ---+ K, and kO- 1 is the map from the Hopf-Whitney theorem. The map i*: Hn(Kj 11") ---+ Hn(Knj 11") is injective because the co chains of dimension at most n in K are identified with the co chains in K n and the equality zn = 8w n- 1 carries over from K n to K. The map i#: [K, L] ---+ [Kn, L] is injective as well. Indeed, if maps f, g: K ---+ L have homotopic restrictions to K''', then they are themselves homotopic because all obstructions to extending the homotopies lie in cocycles with values in 1I"m(L) with m > n, whereas, by assumption, 11"m (L) = 0 for m > n. To prove that the diagram (21) defines a one-to-one map n). If such an extension exists, then cn+ 1 (f) = OJ here en+l E C n+1(Kj 11") is the obstruction to extending the map f. The equality cn+1(f) = 0 is equivalent to 8dn (f, *) = 0 because o~(f, *) = cn+l(*) - cn+1(f) and cn+l(*) = O. By definition, k(f) is the cohomology class of the co cycle dn(f, *) in Kn. The cochain dn(f, *) is al 0 a co cycle in K; therefore, k(f) E 1m i#, as required.


127

The map cp is natural in the sense that for any h: K - K', the diagram h#

[K',L]

1~

Hn(K'j7r)

~

h* ~

[K,L]

1~

H n (Kj7r)

is commutative. We can assume that L is the K (7r, n) simplicial complex that was constructed in the proof of Theorem 3.8. Then the skeleton Ln-l consists of one point, and it follows directly from the definitions that cp(idL) = F1r • We set K' = Land h = f. Then cpU) = cpU#(idL)) = !*(cp(id L )) = !*(F1r ), i.e., the map cp does have the required form. 0

Corollary. There is a one-to-one correspondence

[K(7r, n), K(7r', n)] ~ Hom(7r, 7r'), under which the homotopy class of any map f: K (7r, n) - K (7r', n) corresponds to the homomorphism f.: 7r - 7r' induced by this map. Proof. According to Theorem 3.10, we have

[K(7r, n), K(7r', n)]

~

Hn(K(7r, n)j 7r').

The universal coefficient theorem implies the isomorphism Hn(K(7r, n)j 7r') ~ Hom(Hn(K(7r, n)), 7r'). Finally, from the Hurewicz theorem, it follows that Hn(K(7r, n)) ~ 7r. 0 Theorem 3.10 allows us to describe all cohomology operations, which are defined as follows. Take Abelian groups G and H and positive integers m and n. A cohomology operation of type (m, n, G, H) is a family of maps (not necessarily homomorphisms) 8 K: H m (K j G) - Hn( K; H) that are defined for each simplicial complex K so that the diagram

Hm(Lj G) ~ Hn(Lj H)

11*

11*

Hm(Kj G) ~ Hn(Kj H) is commutative for any continuous map f: K - L. Example 31. A homomorphism of groups G _ H induces a homomorphism Hffl(Kj G) - Hm(K; H), which is a dimension-preserving cohomology operation. Example 32. If R is the additive group of a unital ring, then for any k, the map a t---+ a k is a cohomology operation of type (m, mk, R, R).

128


Example 33. The Bockstein homomorphism f3*: Hk(Kj G")

-+

H k+1(Kj G f )

corresponding to an exact sequence

o-+ G

f

-+

G

-+

G"

-+

0

is a cohomology operation that increases the dimension by 1. Theorem 3.11. There exists a one-to-one correspondence between the cohomology operations of type (m, n, G, H) and the elements of the group Hn(K(G, m)j H). This correspondence is defined by 8 I-t 8(Fa), where Fa E Hm(K(G, m)j G) is the fundamental cohomology class. Proof. An element a E Hm(Kj G) can be represented in the form a f*(Fa), where f: K -+ K(Gjm) is a map. Therefore, 8(a) = e(f*(Fa)) = f*(8(Fa)), i.e., the cohomology operation e is completely determiIlAld by the element 8(Fa). It remains to show that any element () E H n (K (G, m) j H) can be represented in the form () = 8(Fa) for some cohomology operation e. The element () corresponds to a map cp: K(G, m) -+ K(H, n) for which () = cp*(FH)' Taking the composition of every g: K -+ K(G, m) with cp, we obtain a map [K,K(G,m)] -+ [K, K(H, n)]. This map corresponds to the cohomology operation 8: Hm(KjG) -+ Hn(KjH) that takes g*(Fa) to g*cp*(FH)' In particular, 8(Fa) = cp*(FH) = (). 0

Corollary 1. Cohomology operations cannot decrease dimension. Proof. If n < m, then H n (K (G, m) j H) = 0 because the space K (G, m) is (m - I)-connected. 0 Corollary 2. All dimension-preserving cohomology operations are induced by homomorphisms G -+ H. Proof. In the proof of the corollary of Theorem 3.10, it was mentioned that Hn(K(G,n)jH)

~

Hom(Hn(K(G,n)),H)

~

Hom(G,H).

0

1. 7. The Moore Spaces. The definition of the Moore spaces is similar to that of the Eilenberg MacLane spaces, with the only difference that homotopy groups are replaced by homology groups. Namely, for an Abelian group G and a positive integer n, a Moore space M(G,n) is a simplicial complex X for which if k = n, if k =I n.

For n > 1, it is also assumed that

71'1 (X)

= o.


129

Let us show that, for any Abelian group G and any positive integer n, there exists a Moore space M(G, n). For M(Z, n) we take the sphere sn, and for M(Zm, n) we can take the sphere sn to which the (n+l)-cell D n+1 is attached by a map 8Dn+1 ---+ sn of degree m. In this case, the chain complex for calculating the cellular homology has the form Z ~ Z ---+ 0 ---+ •••• If the group G is finitely generated, then we can take the wedge of several spaces M(Z, n) and M(Zm, n) for M(G, n). For an arbitrary Abelian group G, the space M(G, n) can be constructed as follows. There exists an epimorphism F ---+ G, where F is a free Abelian group. Let H be its kernel. The group H, being a subgroup of the free Abelian group F, is itself a free Abelian group. Let {fa,} be a basis of the group F, and let {hp} be a basis of H. Then hp = ~a no.pfo.. We set xn = VaS;:. To xn we attach cells {D3+1} via the maps Xp: 8D3+1 ---+ xn defined as follows. First, 8D3+1 is mapped to the wedge of kp copies of the n-sphere, where kp is the number of nonzero coefficients nap, by contracting kp - 1 (n - I)-spheres in 8D3+l. Then each of the spheres in the wedge is mapped to the corresponding sphere S;: by a map of degree naP' The resulting CW-complex can easily be made into a simplicial complex Xj this is the Moore space M(G, n). Indeed, the chain complex for calculating the i cellular homology of the complex X has the form H ---+ F ---+ 0 ---+ ••• , where i is the inclusion of the subgroup H into the group F. If GI, G2, ... are any Abelian groups, then there exists a simplicial complex X for which Hi(X) = Gij for X we can take the wedge ViM(Gi,i). Surprisingly, for cohomology, a similar assertion is false. An example was constructed by Kan and Whitehead [67]. Theorem 3.12 (Kan Whitehead). For no positive integer n, there exists a simplicial complex X for which Hn-1(Xj Z) = 0 and Hn(Xj Z) = Q, where Q is the additive group of rational numbers. Proof. Suppose there exists a simplicial complex X for which H n - 1 (X j Z) Then the universal coefficient formulas yield

o and Hn(x j Z) = Q. (22)

0

= Hn-l(X)

Q

= Hn(x)

~ Hom(Hn - 1 (X), Z) EB Ext(Hn_2(X), Z)

and

(23)

~ Hom(Hn(X), Z) EB Ext(Hn-1(X), Z).

Relations (22) imply, in particular,that Hom(Hn_ 1(X), Z) = O.

=

130


Lemma. For any Abelian group A such that Hom(A, Z) = 0, the following assertions are valid. (a) The group Ext(A, Z) is divisible if and only if A is torsion-free; (b) The group Ext(A, Z) is torsion-free if and only if A is divisible. Proof. First, we introduce the notation

rnA

= {b E A I b = rna, a E A},

rnA = {a

E

A I rna = O},

Am = A/rnA. Note that if B is an Abelian group such that rnB = 0, then Hom(B, Z) = Indeed, rp(b) = n implies 0 = rp(rnb) = rnn. Both groups rnA and Am have this property; hence Hom(mA, Z) = 0 and Hom(Am, Z) - o. Taking into account this observation and applying Problem 22 to the exact sequem;-es. c xm o ---+ m A ---+ A ---+ rnA ---+ 0, 0---+ rnA ---+ A ---+ Am ---+ 0,

o.

we obtain the exact sequences

o ---+ Ext (rnA, Z)

---+

Ext(A, Z)

---+

Ext(mA, Z)

---+

0

and

o ---+ Hom(A, Z)

---+

Hom(rnA, Z)

---+

Ext(Am, Z)

---+

Ext(A, Z)

---+

Ext (rnA, Z)

---+

o.

By assumption, Hom(A, Z) = o. Moreover, according to Problem 14, we have Hom(rnA, Z) ~ Hom(A, Z). Thus, the second exact sequence becomes

o ---+ Ext(Am, Z)

---+

Ext(A, Z)

---+

Ext (rnA, Z)

---+

o.

Using these exact sequences, we construct the exact sequence

o ---+ Ext(Am, Z)

---+

Ext(A, Z) ~ Ext(A, Z)

---+

Ext(mA, Z)

---+

0,

in which rp is the composition of the maps Ext (A, Z) -+ Ext(rnA, Z) -+ Ext(A, Z) induced by the inclusion rnA c A and by the map A~rnA. Hence, rp is the multiplication by rn, which implies the formulas mExt(A, Z) = Ext(Am, Z) and Ext(A, Z)m = Ext(mA, Z). For a periodic group T, the equality Ext(T, Z) = 0 holds if and only if T = 0 (see Problem 18). Hence mA = 0 if and only if Ext(mA, Z) = Ext(A, Z)m = 0, and Am = 0 if and only if Ext(A m , Z) = mExt(A, Z) = o. 'l'o prove the lemma, it remains to note that, first, an Abplian group B has no torsion if and only if mB = a for all rn > 1 and, second, an Abelian group B is divisible if and only if Bm = a for all m > 1. 0

2. Characteristic Classes

131

We continue the proof of Theorem 3.12. Consider equality (23). Note that the group Q cannot be represented as a direct sum of two nonzero groups. Indeed, suppose that Q = A ffi B and a and b are nonzero elements of A and B. Then pa i= qb for any p, q E Z such that pa i= 0 and qb i= O. But no rational numbers a and b have this property. Therefore, either Hom(Hn(X),Z) = Q or Ext(Hn_l(X),Z) = Q. The equality Hom(Hn(X), Z) = Q cannot hold because, for no Abelian group A, the group Hom(A, Z) is divisible. Indeed, suppose that 'P(a) = k i= O. Choose a positive integer m so that kim is not an integer. If m'I/J = 'P for some 'I/J E Hom(A, Z), then m'I/J(a) = 'P(a) = k, i.e., 'I/J(a) = kim; hence m'I/J never equals 'P. Thus, Ext(Hn l(X), Z) - Q is a torsion-free divisible group. According to the lemma, the group Hn l(X) is torsion-free and divisible. Such a group necessarily contains Q as a subgroup, i.e., there exists a monomorphism 0 - Q - H n - 1 (X). This implies the existence of an epimorphism Ext(Hn- 1 (X), Z) - Ext(Q, Z) - O. To obtain a contradiction, it remains to prove that the group Ext(Q, Z) is uncountable, and an epimorphism Q = Ext(Hn l(X),Z) - Ext(Q,Z) - 0 cannot exist. The group Ext(A, Z) can be calculated by using an injective resolution 0 - Z - Q - Q/Z - O. As a result, for A = Q, we obtain an exact sequence

o ----+ Hom(Q, Q)

----+

Hom(Q, Q/Z)

----+

Ext(Q, Z)

----+

O.

The group Hom(Q, Q) ~ Q is countable, whereas Hom(Q, Q/Z) is uncountable (see Problem 15). Therefore, the group Ext(Q, Z) is uncountable. 0

2. Characteristic Classes 2.1. Vector Bundles. A locally trivial fiber bundle p: E - B is called an n-dimensional vector bundle if its fiber F is the linear space ]Rn, each set p-l(b) (where b E B) is endowed with the structure of an n-dimensional linear space, and the homeomorphisms h: UxF _ p-l(U) have the property that for each point b E U, the map x ~ h(b,x) E p-l(b) is a linear space isomorphism between F =]Rn and p-l(b). A vector bundle is said to be smooth if E and B are smooth manifolds, p is a smooth map, and the homeomorphisms hare diffeomorphisms. Example 34. Let T M n be the tangent bundle3 of a closed manifold Mn. Then the natural projection p: T M n _ M n is a smooth vector bundle. We denote this bundle by TMn. A map s: B - E is called a section of the bundle if ps idB; the sections of the tangent bundle are vector fields on the manifold. For any 3The definition of the tangent bundle is given in Part I on p. 202.

132


vector bundle, the zero section is defined, which takes each point b E B to the zero vector of the space p I (b). Vector bundles PI: EI --+ Band P2: E2 --+ B are isomorphic if there exists a homeomorphism cp: EI --+ E2 which isomorphically maps pII(b) onto p2"I(b) for all b E B. Isomorphic vector bundles are also called equivalent. Note that, for isomorphic vector bundles, the images of the zero sections have homeomorphic complements in the total spaces of the bundles.

Theorem 3.13. Suppose that PI: EI --+ Band P2: E2 --+ B are vector bundles and f: El --+ E2 is a continuous map which induces an isomorphism of linear spaces PI I(b) and p2"I(b) for each b E B. Then f is a homeomorphism, i.e., the bundles PI and P2 are isomorphic. Proof. For each point b E B, we choose neighborhoods Ui (i = 1,21 and homeomorphisms hi: Ui x]Rn --+ p;I(U,). It is sufficient to verify that the map h2"1 fh I : (UI n U2) x ]Rn --+ (UI n U2) x ]Rn is a homeomorphism. By assumption, this map takes (b, v) to (b, A(b)v), where A(b) is a nondegenerate matrix whose elements continuously depend on b. The inverse map has the form (b,w) 1-+ (b,A-I(b)w). The elements of the matrix A-I(b) continuously depend on b as well; therefore, the inverse map is continuous. D The natural projection Bx]Rn --+ B is an n-dimensional vector bundle. It is called a trivial, or a product vector bundle. Any vector bundle isomorphic to a trivial bundle is also said to be trivial.

Problem 91. Prove that an n-dimensional vector bundle p: E --+ B is trivial if and only if there exists a continuous map 7r: E --+ ]Rn whose restriction to each fiber p I(b), bE B, is a linear space isomorphism. We say that a closed manifold Mn is parallelizable if its tangent bundle p: T M n --+ Mn is trivial. Theorem 3.14. A manifold M n is parallelizable if and only if there exist n continuous vector fields VI (x), ... , Vn (x) on M n that are linearly independent at each point x E Mn.

Proof. Let cp: T M n --+ M n x ]Rn be a homeomorphism that isomorphic ally maps TxMn to {x} x IRn. We choose a basis eI, ... , en in ]Rn and put Vi(X) = cp-I(x, ei). Now suppose that VI (x), . .. , vn(x) are linearly independent vector fields. Consider the map M n x IRn --+ TMn defined by (x,a) 1-+ alvl(x) + ... + anvn(x). According to Theorem 3.13, this map is a homeomorphism. D

133


Example 35. The spheres SI, S3, and S7 are parallelizable manifolds. The construction of linearly independent vector fields on these spheres is discussed in detail in [104, Section 41]. Problem 92. Prove that the tangent bundle TS3 has infinitely many pairwise nonhomotopic 4 trivializations. Problem 93. Prove that there exist infinitely many pairwise nonhomotopic 5 vector fields without singular points on S3. Example 36. The torus

rn is a parallelizable manifold.

Now we discuss some constructions related to vector bundles. Pullbacks. Let p: E - B be a vector bundle, and let f: B' - B be a continuous map. Then we can construct a pullback, or an induced bundle p' = f*p: E' - B', where

E'

= {(e, b')

E E

x B' I pee)

= feb')}

and p' (e, b') = b'. For the pullback, the diagram

E'~E

lp,

lp

B'~B, where lee, b') = e, is commutative. Let us show that the pullback is indeed a vector bundle. The linear space structure on the fiber (p')-I(b) is determined by

Al(el. b')

+ A2(e2, b') = (AIel + A2e2, b');

here peed = p(e2) = feb'), i.e., el and e2 belong to the same linear space p-l(J(b'». To a neighborhood U C B and a homeomorphism h: U x]Rn _ p-I(U) we assign the neighborhood U' = f-I(U) C B' and the homeomorphism h': U' x ]Rn - (p')-I(U') defined by (b', v) ...... (e, b'), where e = h(J(b'), v). The condition pee) = feb') does hold. Exercise. Prove that a vector bundle is trivial if and only if it is a pullback of a bundle over a point. 4Two trivializations of a bundle are homotopic if they can be joined by a continuous family of trivializations. 5Two vector fields without singular points are said to be homotopic if they caD be joined by a continuous family of vector fields without singular points.


134

Direct Products of Bundles. The direct product of vector bundles PI: EI - BI and P2: E2 - B2 is defined as PI x P2: EI x E2 the fiber (PI XP2)-I(b l ,b2) structure.

BI

X

B2j

= Pll(bl ) xP2"1(b2) has the natural vector space

Example 37. Let MI and M2 be closed manifolds, and let M = MI X M2. Then the tangent bundle TM is isomorphic to the direct product TMl x TM2' The Whitney Sum. The Whitney sum of two bundles can be defined as the bundle whose fibers are direct sums of the fibers of these bundles. The formal definition is as follows. Suppose that PI: EI - Band P2: E2 - B are two vector bundles over the same base. Their Whitney, or direct, sum is the bundle d*(Pl x P2), where d: B - B x B is the diagonal map defined by d(b) = (b, b). It is often convenient to denote the bundle p: E -:t B by one letter ~j the Whitney sum of bundles 6 and 6 is denoted by 6 ffi 6· This notation is used because the fiber of the bundle 6 ffi 6 over a point b is canonically isomorphic to Pll(b) ffi p2"l(b). Example 3S. The Whitney sum of the tangent bundle over the sphere sn and the one-dimensional trivial bundle is the (n + I)-dimensional trivial bundle. Proof. The trivial (n+l)-dimensional vector bundle." over sn can be represented as follows. Consider the standard embedding sn _ lRn +1 and assume that each point x E sn is supplied with the linear space lRn+1 obtained by translating the origin to x. Under this interpretation, the total space E(.,,) of the bundle." contains the total space E( Tsn) of the tangent bundle. Consider the one-dimensional vector bundle v for which the total space E(v) C E(.,,) consists of all pairs (x, v), where x E sn and v-.lTxsn (Le., v is the normal vector to sn at x). It is easy to verify that v is trivial. Indeed, treating x as a unit vector, we can write v in the form v = ,xx, where ,x E lRj then the map (x, v) 1-+ (x,,x) is an isomorphism between v and the trivial bundle. Consider the map E( T ffi v) - E(.,,) defined by (x, Vb V2) 1-+ (x, VI + V2). According to Theorem 3.13, this is an isomorphism of bundlE'S. 0 Problem 94. Prove that snl X sn2 X Euclidean space of dimension nl + ... +

'"

nk

X snk can be embedded in the + 1.

Remark. In [3], an assertion similar to Problem 94 for embeddings of products of 2-manifolds in the Euclidean space was proved. Example 39. For any closed manifolds MI and M 2 , TMIXM'J ~ where the 7l'i: Ml x M2 - Mi are natural projection .

7r2TM2'

7riTM1 ffi

135


Proof. The fiber of the bundle 7riTMl ffi 7r2TM2 over the point (x, y) E MI X M2 is canonically isomorphic to TxMI ffi TyM2 ~ Tex,y)(MI x M2). 0 Problem 95*. Prove that the manifold snl X ••. X snlc, where k ~ 2, is parallelizable if and only if at least one of the numbers nl, ... ,nk is odd. The Bundle Hom(el, e2). Let 6 and 6 be two vector bundles over B. They determine the bundle Hom(6, 6) whose fiber over each point bE B consists of homomorphisms (linear maps) from the fiber of 6 over b to that of 6. To construct this bundle, we must endow the total space with a topology. Let U be an open set over which the bundles 6 and 6 are trivial, and let hi: U x R n , - p;I(U) be the coordinate homeomorphisms. When a point b E U is fixed, the map hI identifies R n , with the fiber of ~i over b. Thus, we have a one-to-one map h: U x Hom(Rnl,Rn2) - p-I(U). It is required that such maps be homeomorphisms for all U and that all the sets p-I(U) be open. These conditions uniquely determine a topology on E. We must verify only that it is consistent, i.e., that the map

(Un U') x Hom(Rnl,Rn2)

h

loh'. (un U') x Hom(R nl,R n2)

is continuous for any two intersecting coordinate neighborhoods U and U'. The continuity of this map follows from the continuity of the maps

(U n U') x R n ,

h

•

I.

loh'

(U n U') x Rni.

A similar construction can be applied to obtain the bundle 6 ® 6. Applying it to copies of the same bundle, we obtain the bundle Ake (the exterior power of

o.

2.2. Cohomology with Local Coefficients. The constructions we used in obstruction theory (Section 1.2) show that it is sometimes natural to consider cohomology the coefficients of which are not in a fixed group but rather each point has its own coefficient group; the groups corresponding to different points are isomorphic, but the isomorphisms are not canonical. Using this kind of cohomology with local coefficients, we can develop obstruction theory that applies to a larger class of spaces. We use it to construct linearly independent sections of vector bundles.

Remark. In fact, the construction of characteristic classes of vector bundles involves only trivial systems of local coefficients, which can be handled without using cohomology with local coefficients. But the reason for this is that some characteristic classes of vector bundles are obtained by reducing nontrivial systems of local coefficients modulo 2. To elucidate the relationship between obstruction theory and the characteristic classes of vector bundles, we present the theory of cohomology with local coefficients.

136


Suppose that associated with each point x of a topological space X is an Abelian group G x (all the groups G x are isomorphic), and with each path 'Y from x to y, an isomorphism 'Y.: G x --+ G y so that the same isomorphisms correspond to homotopic paths, and to any composition of paths corresponds the composition of the respective isomorphisms. Then we say that X is endowed with a local system of Abelian groups. A local system of Abelian groups on X determines an action of the group 7I'"l(X,X) on G x because each loop 'Y based at x is associated with an isomorphism 'Y: G x --+ G x . If the space X is path-connected, then the local system of Abelian groups is uniquely determined (up to a natural equivalence) by the group G x associated with one point x and by the action of 7I'"l(X,X) on this group G x • Indeed, for each point y E X, we can fix a path Wy from x to y and assign the same group G = G x to all points of X. To a path 'Yyz from Y to z we assign the isomorphism G x --+ G x induced by the loop Wy'YyzW;l. Example 40. The groups 7I'"n(X, x), where n ~ 2, form a local system of Abelian groups. A path from x to y induces an isomorphism 7I'"n(X, x) --+ 7I'"n(X, y) in a standard way. Example 41. On a manifold Mn, consider a local system of Abelian groups G x ~ Z on which every element of the fundamental group acts as either the map a ....... -a or the identity map, depending on whether or not the orientation changes when going along the representative path. This local system of Abelian groups is called an orientation system of groups and denoted by ZOr Mn. If the space X is locally path-connected and locally simply connected, then there exists a canonical isomorphism between groups from a local system of Abelian groups in sufficiently small neighborhoods. Thus, for a simplicial complex, such isomorphisms exist on each simplex. It follows that a local system of Abelian groups on a simplicial complex can be specified by choosing a group G v for each vertex v and defining an isomorphism 'Yuv: G v --+ G u for any vertices u and v belonging to the same simplex so that 'Yuv'Yvw = 'Yuw whenever u, v, and ware vertices of the same simplex. Now we can define the homology and cohomology with coefficients in a local system C!5 of Abelian groups on a simplicial complex K. Let 6. be a simplex in K. The groups G x are canonically isomorphic; therefore, we can use the notation Gl:!... The group Ck(K; C!5) of k-chains consists of finite formal sums L ai6.f, where Ui E G l:!..". The boundary homomorphism 8k: Ck --+ C k - 1 is given by •


137

This definition makes sense because if fl.1 C ofl. c fl., then the group Gll' is canonically isomorphic to Gll. The proof of the equality 00 = 0 is completely similar to that for usual simplicial chains. The homology group Hk(K; 0) is defined to be the quotient group Kerok/ImOk+1' The group C k (K; 0) of k-cochains consists of maps taking each simplex fl.k to an element of the group Gllle. The coboundary homomorphism 6 is given by (6c k ) [vo, ... , Vk+1]

= L( _l)ick ([vO, ... , Vi,""

Vk+lJ).

The definition of the cohomology group Hk(K; 0) is standard. If a simplicial complex is endowed with the structure of a CW-complex (the cells are unions of simplices), then its cellular homology and cohomology can be defined in the standard manner. It must be taken into account that if x: nk -+ X is a characteristic map of a cell, then the groups G z are canonically isomorphic for all x E n k , but different points x, y E nk may be mapped to the same point. In this case, the two copies of the group G associated with the points x and y are identified by the action on G of the element of the fundamental group represented by the image of the interval xy.

Example 42. Let G z ~ Z be a local system of Abelian groups on Rpn with n ~ 2 on which the generator of the fundamental group acts as a 1-+ -a. We denote this local system of Abelian groups by ZT (for twisted). We have

Z Hk(lRpn;ZT)= { Z2

o

if k = nand n is even, ifk n. Now, suppose that n than 1. Then

+1

= 2 k m, where m is an odd number larger

(1 + O')n+l = (1 + O' 2k )m = 1 + mO' 2k + m(m - 1) O' 2.2k + ... , 2

where

0 2 '"

i= 0 and m ;! 0

Corollary 1. If JRpn is parallelizable, then n Corollary 2. Ifn+1 O' 2k

i=

0

(mod 2).

= 2k -

1 for some k.

= 2k m,

where m is an odd number, then W2",(JRpn) = OJ hence there exist no 2k linearly independent vector fields on lRpn.

Problem 105. Prove that if JRpn is immersed in JRn+1, then n has the form 2 r - 1 or 2 r - 2.

Theorem 3.25. Suppose that 11.: JRn x JRn

-+ JRn is a bilinear multiplication without divisors of zero. Suppose also that there exists a left identity, i. e., an element e such that 11.( e, x) = x for all x E lRn (the multiplication is not assumed to be associative). Then n = 2k for some k.

Proof. Consider a basis el = e, e2,.'" en in JRn. At each point x E sn-l C JRn, the vectors x = p(e,x),p(e2,x), ... ,I1.(e n ,x) are linearly independent because otherwise I1.(Y, x) = 0 for some nonzero vectors y = >'1 el + ... + >'nen and x. The projections of l1.(e2, x), . .. , peen, x) to the tangent space to sn-l at x are linearly independent. These projections form linearly independent vector fields on the sphere. Since l1.(ei, -x) = -p(ei, x), it follows that these vector fields can be transferred to JR,pn-l. Thus, the manifold JRpn-l is parallelizable. It remains to apply Corollary 1 of Theorem 3.24. 0

Theorem 3.26. If n lR2n - 2 .

=

2k, then the manifold JRpn cannot be immersed in

Proof. If n = 2k, then w(~'pn) = (1 + 0')(1 + 0')2" = (1 + a)(l + O' 2k ) = . 1 + a + a 2k + a 2"'+1 -- 1 + a + an. By assumption, we have an +1 = OJ therefore, (1 + a + a n)(l + a + a 2 + '" + an-I) = 1. Thus, w(JRpn) =

l+a+a 2 + ... +0' n-l .

ii. ApplIcatIOns ot :iimpllclal .HOmolOgy

Suppose that ~pn is immersed in ~n+m. Then the dimension of the normal bundle cannot be lower than the maximal dimension of a nonzero dual Stiefel Whitney class. In the case under consideration, we have m ~ n - 1, i.e., ~pn cannot be immersed in ~2n-2. 0 Theorem 3.27. If a number n has the binary representation 2kl + ... + 2k s, then the manifold ~p2kl x ... X ~p2ks cannot be immersed in R 2n - s - 1. Proof. To avoid cumbersome notation, we assume that s = 2. We set 2kl = 1fp and 1fq be the projections of the product RPP X RPq onto the first and second factors. Then w(~PP x ~pq) = W(1f;(11RPP) E9 1f;(11RPQ» = 1f;(w(~PP» '-' 1f~(w(~pq» = 1f;(1 + ap)P+! '--' 1f~(1 + ap)q+1. In the proof of Theorem 3.26, we showed that ((1 + ap)P+I )-1 = 1 + a p + ... +a:; 1. For coefficients in Z2, we have (a x (3) ........ (r x 8) = (a '--' r) x ({3 '--' 8); thus, w(~PP x ~pq) = (1 + {3p + ... + (3C- 1) x (1 + {3q + ... + (3~-1), where {3p = 1f;ap and (3q = 1f~aq. The leading dimension of a nonzero 1, it is constructed by applying the same construction to each plane generated by vectors of the form ea and en+cJ. For each t, the map .,pn,t = (1 - t)in,o.,po + tin,l.,pl determines a map In,t: B -+ G(2n, k). Indeed, since .,po and .,pI are monomorphisms from p l(b) to lRn , the map .,pn,t is also a monomorphism because the vectors in,o.,po(v) and in,l.,pl(V) belong to orthogonal subspaces. Thus, the maps In,o = in,o/o = inlo and In,1 = in,til ~ inh are homotopic; therefore, so are the maps inlo and inh. 0 Problem 108. Given a k-dimensional vector bundle {k over a compact CW-complex B of dimension less than k and the trivIal one-dimensional vector bundle £1 over B, prove that the bundle {k is trivial if and only if so is the bundle {k E9 £1.

A manifold Mn is said to be stably parallelizable if there exists a trivial k-dimensional bundle £k over Mn for which the bundle TMn E9 £k is trivial. Problem 109. Prove that a manifold M n is stably parallelizable if and only if one of the following conditions holds:

(a) the bundle TMn E9 £1 is trivial; (b) the normal bundle for an embedding of Mn into 2n + 1, is trivial.

]RN,

where N ~

2.7. Stable Cohomology of Grassmann Manifolds. The one-to-one correspondence between the k-dimensional vector bundles over B and the

157


homotopy classes of maps B -+ G(oc, k) implies that the cohomology groups of the manifolds G(n, k) and their behavior under the embeddings G(n, k) -+ G(n, k + a) are of great importance in the study of vector bundles. Indeed, every cohomology class stable with respect to such embeddings allows us to associate to any bundle { over B a cohomology class of the space B, namely, the characteristic class of the bundle {. In this section, we calculate the homology and cohomology groups with coefficients in Z2 of the (real) Grassmann manifolds G(n, k). We shall use the coordinate neighborhoods U/ and open Schubert cells e(u) introduced in Part I on p. 196. Any open Schubert cell e(u) is contained entirely in the chart U/, where I = Uj moreover, it is determined by a system of equations of the form Xi = 0 in this chart. Its closure e(u) intersects only those charts U/ for which i1 ~ U1, • .. ,ik ~ Uk. In any such chart, the set e(u) n U/ is determined by equations of the same form. Therefore, e(a) is a submanifold in G(n, k). It is called the Schubert manifold. The open Schubert cells form a cell decomposition of the Grassmann manifold, and their closures are submanifolds. Hence there exists a triangulation of G(n, k) that induces a triangulation on each Schubert manifold. Given such a triangulation, we can calculate cellular homology. First, we must determine the structure of the characteristic maps of Schubert cells. The dimension of the cell e(u) equals (Ul -1) + (U2 - 2) + ... + (Uk - k). We are interested in the coefficients of the cells of dimension lower by 1 in 8[e(u)]. If a cell e(u' ) is contained in e(u), then u~ ~ Ul,'" ,u~ ~ Uk. Therefore, for such a cell, the equality ui + ... + u~ = Ul + ... + Uk - 1 can hold only if u~ = Ui - 1 and uj = Uj for j i= ij in this case, Ui-1 = Ui - 1. In an appropriate coordinate system, the subset e(u' ) of e(u) is determined by the equation Vi,a, = 0 (see Figure 5). Depending on the orientations of the

V"a,

= 0

Figure 5. The boundary of a Schubert cell

two parts of e(u) adjacent to e(u' ), the coefficient of the cell e(u /) in 8[e(u)] is equal to ±2 or 0, which means that 8[e(u)] = 0 for Z2-homology. Thus, Hr(G(n, k); Z2) ~ Z~(r,n,k), where N(r, n, k) is the number of r-dimensional Schubert cells in the manifold Gen, k),


158

Each Schubert symbol U can be associated with the sequence of numbers U1 - 1, U2 - 2, ... , Uk - k. Removing all zeros from this sequence, we obtain a sequence iI, i2,' .. , is, where s $ k and 1 $ i1 $ i2 $ . " $ is $ n - k (the inequality ia $ ia+1 follows from uf3 < uf3+1)' An unordered set of positive integers iI, i2," ., is is called a partition of an integer r if i1 + .. , + is = r. Every unordered set of numbers uniquely determines the set of the same numbers arranged in increasing order. Therefore, N(r, n, k) is equal to the number of partitions of r into at most k positive integers not exceeding n - k. In particular, if both numbers k and n - k are not smaller than r, then N(r, n, k) = p(r) equals the number of all partitions of r. If n ~ r + k, then the embedding G(n, k) - G(n + 1, k) induces an isomorphism Hr(G(n, k); Z2) - Hr(G(n + 1, k); Z2) of cohomology groups and an isomorphism Hr(G(n + 1, k); Z2) - Hr(G(n, k); Z2) of dual cohomology spaces; here we have in mind the duality of linear spaces rather th-an Poincare duality. For n ~ r + k, we denote the group Hr(G{n, k); Z2) by Hr(G(oo, k); Z2).

We determine the multiplicative structure of the ring H*(G(oo, k); Z2)' To do this, consider the embedding

f: !Rpn

1 X ...

x

1~'pn-1" _

G(nk, k)

'"k that identifies every set of vectors VI, ... , Vk E lRn with the subspace of JRn x ·· ·xlRn spanned by (VI, 0, ... ,0), (0, V2, 0, ... ,0), ... , (0,0, ... , Vk). This map induces a ring homomorphism

H*(G(nk, k); Z2) _ H*(lRpn-1 x ... x lRp n- 1; Z2). The ring H*(JRpn-1 x " . x JRp n- 1; Z2) is isomorphic to the quotient of the polynomial ring Z2[Ol, ... , Ok] by the relations of = 0, where i = 1, ... , k. For r $ n, the embedding lRpn - lRpn+1 induces a ring isomorphism Hr(~'pn+1; Z2) _ Hr(lRpn; Z2). Hence we can introduce the ring H*(lRPoo; Z2) ~ Z2[O]; for every fixed r, we simply treat lRp oo as lRp n with sufficiently large n. Similarly, we introduce the ring H*(lRpoo x ... x lRpoo ; Z2) ~ Z2[Ol, ... , Ok] and define the homomorphism

r: H*(G(oo, k); Z2) -

H*(lRPoo x ... x lRp oo ; Z2) ~ Z2[Ot, ... , Ok].

Theorem 3.32. The homomorphism f* is an isomorphism onto a subring of symmetric polynomials. Moreover, f*(Wi(r~,)) = Ui(Ol, .. " Ok), where Ui is the ith elementary symmetric function. Proof. First, we show that the dimension of the space Hr(G(oo, k); Z2) coincides with that of the space of degree r symmetric polYnomials in k


159

variables over the field Z2. To this end, we must establish a one-to-one correspondence between the partitions of r into at most k positive numbers and the polynomials of the form U~k, where r1 + 2r2 + ... + krk = r (r1 ~ 0). Note that for such ri, the sum of the numbers rk ~ rk + rk-1 ~ ... ~ rk + ... + r1 equals r. Conversely, given r = 81 + ... + 8k, where 81 ~ 82 ~ .•. ~ Sk ~ 0, we set rk = Sk, rk-1 = Sk-1 - sk, ... , r1 = 81 - 82·

urI ...

It is easy to verify that !*,':x, ~ ,~ x ... x '~i to be more precise, !*'~k ~ '~-1 X ... X '~-1· Indeed, the fiber of the bundle !*'~k over a point (Xl, ... , Xk), where Xi E R.n, consists of vectors ).lX1 + ... + ).kXk. Using the naturality property, we obtain !*(w(r':x,» = w(!*(r~J) = w(,~ x ... x ,~) = (1 + ad ... (1 + ak), where ai is the generator of the cohomology ring of the ith copy oflRpoo . Thus, !*(w(r':x,» = 0",(a1, ... , ak). It follows that !* is an epimorphism from the ring H*(G(oo, k); Z2) onto the ring of symmetric polynomials in a!, ... , ak over the field Z2. Therefore, !* is an isomorphism between Hr(G(oo, k); Z2) and the space of symmetric polynomials of degree r. 0

Corollary (an axiomatic approach to the Stiefel Whitney classes). Suppose that to any vector bundle ~ over any simplicial complex B we associate cohomology classes Wi(~) E Hi(Bi Z2) so that wo(~) = 1, Wi(~) = 0 for i > dim~, W1('~) = a, the natumlity condition holds, and the Whitney formula (see (26) on p. 147) is valid. Then the Wi(~) = Wi(~) are the Stiefel Whitney classes. Proof. The proof of Theorem 3.32 uses only the properties of Stiefel Whitney classes mentioned in the statement. Therefore, the same argument as in that proof implies !*(Wi(r':x,» = !*(Wi(r':x,». Since is an isomorphism, it follows that Wi(r':x,) = Wi(r~), and since the bundle -y':x, is universal, it follows that Wi(~) = Wi(~) for any k-dimensional bundle~. 0

r

Using Theorem 3.32, we can obtain an expression for the characteristic classes of the bundle ~m ®."n (where m and n are the dimensions of the bundles) in terms of those of the bundles ~m and ."n.

Theorem 3.33. Let U1, ... , Um and O"~, ••• ,O"~ be elementary symmetric functions of variables t1, ... ,tm and t~, ... ,t~ over the field Z2, and let Pm,n be the (unique) polynomial in m + n variables for which m

Pm,n(UI, ... , Um, O"~,

••• ,

u~)

n

= II II (1 + ti + tj). i=l j=l

Then w(~m ® ."n) = Pm,n(W!, ... , Wm , wi, ... , w~), where Wi = Wi(~m) and wj = Wj(."n).


160

Proof. First, note that WI (~1 ® .,,1) = WI (e) + WI (.,,1) for one-dimensional bundles. Indeed, for a one-dimensional bundle, the class WI is completely determined by the restriction of this bundle to the I-skeleton; therefore, it is sufficient to check the required equality for one-dimensional circle bundles. For such bundles, ®.,,1 is orientable if and only if the bundles and .,,1 are both orientable or both nonorientable (if the orientations of the fibers of and .,,1 are multiplied by Cl and C2 when the circle is traversed, then the orientation of the fiber of ®.,,1 is mUltiplied by clc2).

e

e

e

e

This equality can also be derived from the fact that, for any I-cell, the value of any co cycle representing, say, the class WI (~ 1 ) is congrul'nt to the number of nondegenerate zeros of a section of over this cell modulo 2. Choose sections of the bundles and .,,1 over the I-skeleton so that they have no common zeros and all of their zeros are nondegenerate. The tensor product of these sections is a section of the bundle ~1 ®.,,1 whose zeros are those of the sections of ~1 and .,,1.

e

e

Step 1. The required assertion is true for direct sums of one-dimensional bundles, i.e., in the case of ~m = 6 EB ... EB ~m and ."n = EB ... EB where are one-dimensional. all bundles ~i and

"'n,

"'1

"'i

We set ti = Wl(~i) and tj TI(1 + tj), i.e., Wi = Ui and wj

= WI (."d· Then W = = uj. Moreover,

TI(I

+ ti)

anti Wi -

(6 ffi .. , ffi ~m) ® ("'1 EB ... ffi "'n) = E9(~i ® "'i)' i,i

Therefore, n

w(~m ® ."n) =

II W(~i ® .",) = II (1 + ti + tj). i,i

i,i

By assumption, n

II (1 + ti + tj) = Pm,n(Ul,""

Um, u~, ... , U~),

i,i

i.e., w(~m ® ."n) = Pm,n(W}"'" Wm, w~, ... , w~). Step 2. It is sufficient to prove the required assertion for dirl'Ct sums of one-dimensional bundles. Let PI and P2 be the projections of G(oo, m) x G(oo, n) to the first and second factors, respectively. We set ,i = pi,m and,'!J.' = P2,m. The bundle lYi®''!J.' over G(oo, m) xG(oo, n) is universal for bundles of the form ~m®."n. Indeed, if ~m = fi,m and ."n = /2,n, then ~m ®."n = (II x h)*(,i ® ,r); here (II x h)(b) = (lI(b), h(b)).


161

Thus,

w({m®rt)

E

(/1 x htH*(G(oo,m) x G(OO,n)jZ2)j

to shorten the notation, we do not indicate the coefficient groups Z2. If O! E H*(G(oo,m)) and /3 E H*(G(oo,n)), then O! X /3 =piO!""'" P'2/3. According to the K tinneth theorem, the map O! ® /3 f-+ O! x /3 induces a cohomology isomorphism

H*(G(oo, m)) ® H*(G(oo, n))

-+

H*(G(oo, m) x G(oo, n)).

Therefore, the ring H*(G(oo, m) x G(oo, n)) is multiplicatively generated by the free generators Wi hI") = Wi and WI (2) = wj. Hence the total Stiefel Whitney class of the bundle 1'1" ® 1'2 has a unique representation in the form Pm,n(Wl,"" W m , w~, ... , w~). 0

Proof. We have n~1 +t~)

+ ....

n; 1(1 + ti + tj) = 1 + n(tl + ... + t m ) + m(t~ + ... 0

Theorem 3.33 can be applied to calculate the Stiefel Whitney classes of Grassmann manifolds. Let I'~ be the bundle over G(n, k) whose fiber over rrk E G(n,k) is the orthogonal complement of the subspace rrk. Then the bundle I'~ E91'~ is trivial, and therefore wh~) = w(I'~).

Theorem 3.34.

TG(n,k)

~ Homb~, I'~).

Proof. A tangent vector at a point nk E G(n, k) is determined by a curve = rrk. For small t, the subspace rrk(t) is uniquely determined by the linear map CPt: rrk - (rrk)l... Indeed, choose a basis el, ... ,ek in rrk. Let e~, ... , e~ be vectors in n k(t) which are mapped to el, ... ,ek under the orthogonal projection onto rrk. Suppose that e1, ... , eZ are the orthogonal projections of the vectors e~, ... ,e~ on (rrk)l... The formula cpt(ei) = e~' defines the linear maps CPt: rrk - (rrk)l..j for t = 0, we have CPO = O. Each map CPt uniquely determines the subspace rrk(t); it is spanned by the vectors el + cpt(el), ... , ek + cpt(ek)' Thus, any curve in G(n, k) passing through the point IIk can be viewed as a curve in Hom(rrk , (IIk)l..) passing through the origin for small t. The tangent vectors to such curves are in one-to-one correspondence with the elements of the space Hom(rrk , (IIk)l..). Thus, we have proved the isomorphism TG(n,k) ~ Homb~, I'~). 0

rrk(t) such that rrk(O)

It is known from linear algebra (see, e.g., [104, p. 120]) that Hom(I'~, I'~) ~ b~)* ® I'~' where * denotes the passage to the dual space. Endowing the


It>;l

bundle 'Y~ with a Riemannian metric, we get the isomorphism ('Y~). ~ 'Y~. Thus, TG(n,k) ~ 'Y~ ® 'Y~; therefore,

where WI

k

-

= w('Yn ® 'Y~) = Pk,n-k(Wl, ... , Wk, WI,··., Wn-k), = Wi('Y~) and Wj = wJ(-y~) = Wj(-y~).

w(G(n, k))

Example 47. The Grassmann manifold G(n, k) is orient able if and only if n is even. Proof. Applying Example 46, we obtain wI(G(n, k)) = (n - k)WI('Y~) + kWl('Y~) = nWI(-y~) because WI = WI. Moreover, WI(-y~) -=I- o. 0 2.8. The Chern Characteristic Classes. By analogy with real vector bundles, we can define complex vector bundles, whose fibers are vector spaces over C. Applying the constructions of obstruction theory to complex vector bundles, we obtain characteristic classes. For complex buool~, the situation is simpler in many respects. One of the reasons for this is that the realification of a complex space has a canonical orientation, and therefore the realification of any complex vector bundle is orientable. Below, we explain this in more detail. Let V be a vector space over C with basis el, ... ,en. It can be assigned the vector space VIll over lR with basis el, ... , en, iel, ... ,ien . If the transition matrix between the bases e and c has the form A + iB, where A and Bare real matrices, then the transition matrix from e, ie to c, ic has the form (~ .f). The determinant of this matrix is equal to 1det(A + iB) 12 > 0 because

/~

-B/ = /A+iB A B

-B+iA/ = /A+iB 0 / A B A-iB·

(These equalities of determinants are obtained as follows: first, we add the (n + k)th row multiplied by i to the kth row for each 1 ~ k ~ n, and then, subtract the (n + k)th column multiplied by i from the kth row for each 1 ~ k ~ n.) Therefore, the space VIll has a canonical orientation. Let w be a complex bundle with realification WIll; this bundle is oriented. We construct the universal bundle in the complex case in analogy with the real case. The base of this bundle is the complex Grassmann manifold Gc(oo,k), where k = dimcw = ~dimIllWIll. We can endow every fiber of a complex vector bundle w over a compact base with a Hermitian metric and construct the bundle Wk whose fiber is the complex Stiefel manifold Vc(n, k); the points of this manifold are k-element sets of vectors in cn orthonormal with respect to Hermitian inner product. To construct the characteristic class of the bundle Wk, we must ~alculate the first nontrivial homotopy group 7l"i(Vc(n, k)).


16~

k)) = O. The first nontrivia homotopy group 1T"2(n k)+l (Vc( n, k)) is isomorphic to Z.

Theorem 3.35. If i

~

2( n - k), then

1T"t (Vc( n,

Proof. The proof of this theorem follows the same scheme as that of Theorem 3.16 (see p. 140). In the complex case, the argument becomes simpler because we do not have to calculate the homomorphism 8*. For k = 1, the manifold Vc(n,1) is homeomorphic to s2n 1; clearly, 1T"i(s2n 1) = 0 for i ~ 2(n - 1). The induction step uses the locally trivial bundle Vc(n,k + 1) -+ Vc(n,k) with fiber s2(n k) 1; as a result, we obtain 1T"t(Vc(n, k + 1)) = 0 for i ~ 2(n - k - 1). Using the locally trivial bundle Vc(n + 1, k + 1) -+ Vc(n + 1, 1) = S2 n+l with fiber Vc(n, k), we can show that 1T"t(Vc(n, k)) '" 1T"t(Vc(n + 1, k + 1)) for i ~ 2n - 1. Therefore, for i ~ 2(n - k) + 1, we have 1T"t(Vc(n - k + 1, 1)) ~ 1T"i(Vc(n - k + 2, 2)) ::: .... Thus, we see that 1T"2(n-k)+1 (Vc(n, k)) ~ 1T"2(n-k)+l(Vc(n - k + 1,1)) = 1T"2 n k)+l(S2(n k)+l) = Z. 0 The canonical orientation of C n determines a canonical isomorphism 1T"2(n-k)+1(Vc(n, k)) -+ Z. Thus, instead of a cohomology with local coefficients, we obtain the ordinary integral cohomology. The Chern characteristic class Ct+l (w) is defined as the obstruction to extending n - i linearly independent sections of the bundle w over the (2i + 2)-dimensional skeleton; this obstruction belongs to the group H2i+2(B; {1T"2i+I(Vc(n,n - i))}) = H2i+2(B; Z).

The definition directly implies the following properties of the Chern classes. (i) If dimcw = n, then Cn(w) = e(WJR); i.e., the highest Chern class coincides with the Euler class of the realification of the bundle w. (ii) The homomorphism Hi(B; Z) -+ Hi(B; Z2) maps the total Chern class c(w) to the total Stiefel Whitney class w(WJR). In particular, W2j+l (WJR) = 0 for all j. In the same way as for the Stiefel Whitney classes, the following a..')sertions are proved. (i) The Chern classes satisfy the naturality condition. (ii) The Chern classes of stably equivalent bundles coincide. Let ,~ be the bundle over cpn whose total space consists of all pairs (x E Cpn,v = AX), where).. E C. Theorem 3.36. The Chern class

Cl

(-y~) is a generator a

of the group

H2(cpn;z).

ern·

Proof. First, we calculate Cl To this end, we describe the structure of the bundle ,f in more detail. Let us represent Cpl = {(Zl : Z2)} as


104

the union of the sets Ul and U2 determined by the inequalities IZII ~ IZ21 and IZII ~ IZ21. On Ul and U2, we introduce the coordinates WI = Zl/Z2 and W2 = Z2/Zl. The fiber of the bundle 'Yi over a point (Zl : Z2) consists of all vectors (AZI : AZ2), where A E C. The restrictions of 'Yi to Ul and U2 are trivial; the trivializations have the forms (AWl, A) and (J.t, J.tW2). We identify the fibers over WI and W2 = w i l by setting (AWl, A) = (J.t, J.tW2), Le., J.t = AWl. The section determined by A = 1 over the circle IWll = 1 is identified with that determined by J.t = WI over W2 = WI 1. We obtain a map 8 1 --+ 8 1 of degree -1. Therefore, extending this section inside U2, we obtain a section with one nondegenerate zero. It follows that cl(Ti) = e((Ti)lR) is a generator of the group H2(cpn; Z) ~ Z. Let i: Cpl --+ cpn be the natural embedding. It is seen directly from the definition that i*(T~) ~ 'Yf; therefore, i*CI(T~) = q(Ti). Clearly, i induces an isomorphism i*: H2(cpn; Z) --+ H2(Cpl; Z). Hence cl(T~) is a generator of the group H2(cpn; Z). f] Remark. Since the degree of the map 8 1 --+ 8 1 under consideration equals -1, it follows that cl(T~) is the generator of H2(cpn; Z) that takes the value -1 for Cpl with canonical orientation (rather than the generator taking the value +1). Theorem 3.37. Ifw and cp are complex vector bundles over simplicial complexes X and Y, then Ck(W x cp) = Li+i=k q(w) x ci(CP). Proof. The proof of this theorem closely resembles that of the generalized Whitney formula (see (26) on p. 147), but there are substantial differences. Again, consider the embeddings ix and jy. The bundle wlx' has dim w - [i/2] linearly independent sections; therefore, it is stably equivalent to a bundle Wo of dimension [i/2]. Applying Theorem 3.21 instead of Theorem 3.20, we see that the element Ck(W x cp) - LO+.8=k co(w) x c.8(cp) belongs to the kernel of the homomorphism (i x x jy) * for all i and j such that [i/2] + [j/2] = k. We prove that this element vanishes. Note that if i + j = 2k + 1, then [i /2] + [j /2] = k. Indeed, i = 2i' and j = 2/ + 1 (or vice versa); therefore, [i/2] + [j/2] = i'+/ = (i+j-l)/2 = k. We show that the direct sum of the maps (ix x jy)* over all i and j such that i + j = 2k + 1 is a monomorphism. For the coefficient group Z, the Kiinneth theorem implies that the group H2k(X x Y) is the direct sum ofthe groups

EB o+.8=2k

HO(X) ® H.8(y)

and

EB o+.8=2k+l

Tor(HO(X), Ii (Y».


165

Similarly, the group H2k(Xi x yi) is the direct sum of the groups

EB

Ho(Xi) ® Hf3(yi) ~ (Hi(Xi) ® Hi-l(yi)) E9 (Hi-I(Xi) ® Hi(yi))

o+f3=2k and

EB

Tor(HO(Xi), Hf3(yi))

= Tor (Hi (Xi) , Hi (yi)).

o+f3=2k+1

Both maps from HO(X) ®Hf3(y) to HO(XO) ®Hf3(yf3+l) and to HO(xo+l) ® Hf3(yf3) are monomorphisms. For the Tor summand, the corresponding maps are monomorphisms also. 0 The properties of Chern classes just proved allow us to prove the following assertion using the scheme of the proof of Theorem 3.32.

Theorem 3.38. The homomorphism

r: H*(Gdoo, k)j Z)

-+

H*(CpOO x '" x CPOOj Z)

~

Z[Ol,"" Ok]

is an isomorphism onto the subring of symmetric polynomials. Moreover, Ui is the ith elementary symmetric function.

f*(Ci(;~)) = Ui(Ol, ... , on), where

The only essential difference between the complex and real cases is that for the complex Grassmann manifolds, all Schubert cells have even dimension, and therefore the boundary homomorphism for integer coefficients is zero for obvious reasons. The Chern characteristic classes, as well as the Stiefel Whitney classes, can be defined axiomaticallYj the corresponding reformulation of the corollary to Theorem 3.32 is fairly obvious. The expression for the Chern class c(~m ® "In) in terms of the classes c(~m) and c(."n) is precisely the same as for the Stiefel Whitney classes (Theorem 3.33). The only essential difference in the proof is that the following assertion is needed in the case of Chern classes.

Lemma. If w~ and wJ are one-dimensional complex bundles over the same base, then Cl(W~ ® wJ) = Cl(W~) + Cl(WJ). Proof. For a one-dimensional bundle, the class CI is the obstruction to extending a nonzero section to the 2-skeleton. The value of the cocycle representing this class for a 2-cell is equal to the sum of the indices of the singular points of the section (it is assumed that all singular points are nondegenerate). Choose generic sections (that is, sections with pairwise different zeros) of the bundles w~ and wJ. The tensor product of these sections vanishes at all points at which one of the sections vanishes. We


166

show that the zeros of the section and those of the tensor product of sections have the same sign over each singular point. It is sufficient to consider the case where one of the sections is constant and the other is determined by a linear map. Suppose that, in some basis, the section of the bundle is given by (~~) ......... (:~~ +:~~: its value at any vector xel + ye2 is determined by linearity. The section of the bundle w~ is constant; suppose that it takes all vectors to ~ + J.Li. Then the section of the bundle ® w~ is determined by

);

wI

wI

(all ( e1) e2 ......... (a21

+ a12i) ® (~ + J.Li)) = (~all + a22 i ) ® (~ + J.Li) ~a21 -

+ (~a12 + J.Lall)i) + (~a22 + J.L a2t)i . equal to (~2 + J.L2)(alla22 J.La12 J.L a22

The determinant of the matrix of this map is a12a2t). Therefore, the singular point of the section of index as that of the section of ® w~.

wI

wI

has the same 0

Similarly to real bundles, it can be proved that the complex k-dimensional vector bundles over any compact Hausdorff space B are in one-to-one correspondence with the homotopy classes of the maps [B, Gc(oo, k)], where Gc(oo, k) is the infinite complex Grassmann manifold (it is defined byanalogy with the real case). For one-dimensional bundles, these two classification theorems (real and complex) can be substantially improved by using the fact that G(oo, 1) = Rpoo and Gc(oo, 1) = Cpoo are K(7I', n) spaces; namely, Rpoo = K(Z2' 1) and Cpoo = K(Z,2). Therefore, according to Theorem 3.10 (see p. 125), if X is a finite simplicial complex, then the elements of the sets [X, JRpOOj and [X, CPOOj are in one-to-one correspondence with those of the groups Hl(X; Z2) and H2(X; Z). On the other hand, the elements of the sets [X, JRpOOj and [X, CPOOj are in one-to-one correspondence with the one-dimensional real and complex bundlf's over X considered up to isomorphism. This correspondence can be described explicitly by using characteristic classes. Namely, suppose that a bundle ~ is determined by a map f: X -+ JRpoo or f: X -+ Cpoo. Theorem 3.10 assigns to the map f the element J*(F1r ) E Hn(x; 71'); here n = 1 and 71' = Z2 (in the real case) or n = 2 and 71' = Z (in the complex case). Moreover, the cohomology class F1r is a generator of the group HI (JRpoo; Z2) or H2(CpOO; Z), which is seen from its description. Thus, the one-to-one correspondence between the one-dimensional bundles and the elements of HI (JRpoo; Z2) or H2(CpOO; Z) is defined by ~ ......... WI (~) or ~ ......... CI (~), respectively, and the addition of cohomology classes corresponds to tensor multiplication of the bundles. As a result, we obtain the following theorem. Theorem 3.39. Let X be a finite-dimensional simplicial cnmplex.


167

(a) One-dimensional real bundles ~ and." over X are isomorphic if and only if Wl(~) = Wl("'); moreover, for each element n E H 1 (XjIl2), there exists a one-dimensional real bundle ~ over X such that wJ(~) = n.

(b) One-dimensional complex bundles ~ and." over X are isomorphic if and only ifcl(~) - CI(.,,)j moreover, for each element n E H2(Xjll), there exists a one-dimensional complex bundle ~ over X such that Cl (~) = n. Note that a one-dimensional complex bundle is the same object as an oriented real two-dimensional bundle with a fixed Riemannian metric. Indeed, to specify a complex structure on the plane R2, we must define multiplication by i, i.e., rotation through 90 0 • It is also clear that the Euler class of an oriented two-dimensional real bundle coincides with the first Chern class of the corresponding one-dimensional complex bundle. Recall that the Euler class of a bundle changes sign under the change of orientation. Theorem 3.39 (b) implies the following assertion. Theorem 3.40. Oriented two-dimensional real bundles ~ and." over a finite simplicial complex X are isomorphic if and only if e(~) = e(.,,)j moreover, for any element n E H2(Xj Il), there exists an oriented two-dimensional real bundle ~ such that e(~) = n. 2.9. Splitting Maps. Let ~ be a vector bundle over a base B. A continuous map f: Bl - B is said to be splitting for ~ if f*(~) is a direct sum of one-dimensional bundles and f*: H*(B) - H*(BJ) is a monomorphism. Using splitting maps makes it possible to reduce the proofs of many assertions about characteristic classes to the case of direct sums of onedimensional bundles. To prove the existence of splitting maps, we need the following construction involving vector bundles. To each vector bundle ~ over a base B we can assign its projectivization which is a locally trivial bundle over B. The fiber of the bundle p~ over a point b E B is obtained from the fiber V of ~ by identifying (in V \ {a}) all points of a straight line passing through zero. Thus, the fiber of the bundle p~ coincides with Rpn-l or cpn-l, where n = dim~. p~,

Problem 110. Let E(P/~) - G(n, k) be the projectivization of the canonical bundle I~ over G(n, k), and let E(P/~-l) _ G(n, k - 1) be the projectivization of the orthogonal complement of the canonical bundle 1~-1 over G(n, k - 1). Prove that E(P/~) ~ E(P'Y~-l). Suppose that a bundle Pf. is determined by a map q: E(P~) - B. Consider the induced vector bundle q*(f.) over E(P~). The points of the space E(P~) are the lines 1 passing through zero. Therefore, q* (f.) has a one-dimensional sub bundle A~ for which the total space consists of all pairs (v, I), where vEL. If the base B is compact, then the bundle q*(f.)

168


admits a Riemannian (Hermitian in the complex case) metric; therefore, q*({) ~ >'e ffi ue, where ue is the orthogonal complement of >'e. The bundle >'e, as well as any other one-dimensional bundle, is obtained from the canonical bundle -yl over lRPCO (or CPOO); i.e., there exists a map f: E({) ----. lRpco for which j"'(-yl) ~ >'e. Let a be the generator of the group HI (lRPCOj Z2) (in the real case) or HI(Cpco j Z) (in the complex case). Consider the element ae = j"'(a) E H*(E(pe))j it depends only on the bundle because if two maps E(P{) ----. lRp co induce isomorphIc bundles when applied to -yl, then these maps are homotopic.

e

e

Theorem 3.41 (Leray Hirsh). Let be a real or complex vector bundle of dimension n over a compact base B(e).

(a) In the real case, the linear space (I, ae, ... ,ae'-l) generated by the elements 1, ae, a~, . .. , ae'-l has dimension n, and

The space H*(B(e)j Z2) can be identified with the image of the homomorphism q*: H*(B(e)j Z2) ----. H*(E(pe)j Z2) so that each element {J ® a~ is identified with (J '-' a~.

(b) In the complex case, the free A belian group (I, ae, ... , ae'-l) generated by the elements 1, ae, a~, ... ,ae'-l in H*(E(pe); Z) has rank n, and H*(E(pe)j Z) ~ H*(B(e)j Z) ® (1, ae,"" ae'-l). This decomposition is also compatible with the homomorphism q* . Proof. For each point b E B, consider the map ib: Rpn-l ----. E(pe) that is the inclusion of the fiber in the total space of the bundle (in the complex case, lRp n I is replaced by cpn-l). It follows directly from the definitions that ib (>'d is the canonical bundle -Y~-l over Rpn-l or cpn-l. The bundle ib(f*(-yl)) coincides with -Y~-lj therefore, the map fib is homotopic to the natural embedding lRpn-l ----. lRPCO or Cpn-l ----. Cpco. This implies the independence of the elements ib (1), ... , ib (ae'-l ).

The required isomorphisms are proved as follows. Let K be a subcomplex in B over which the bundle pe is trivial (for example, a simplex), and let EK = q-l (K). Then there is a homeomorphism K x F ----. EK compatible with the projection q. An isomorphism H*(EK) ~ H*(K)®H*(F) compatible with q exists by the Kiinneth theorem. In the real case, H* (K) ® H* (F) is understood as the tensor product of vector spaces over Z2. I the complex n-l) case, the proof uses the freeness of the group H*(F) = (1, at,· ,ae .


169

Formally, an isomorphism between H*(EK) and H*(K) ® H*(F) compatible with the projection can be written as follows. Consider the homomorphism cpr; defined by E f3k t--+ E q*(f3k) '-" where f3k E Hk(K) and 1 is such that k + 1 = m (in the real case) or k + 2l = m (in the complex case). In the real case, the inequalities 0 ~ m - k ~ n - 1 must hold; in the complex case, the inequalities 0 ~ m - k ~ 2n - 2 must hold and the number m - k must be even. Consider the group ll m(K), which coincides with EB~=l-n+m Hk(K) in the real case and with EBk=2-2n+m Hk(K) in the

ak'

complex case. The map m.

cpr;: llm(K) -+ Hm(EK)

m

k even

is an isomorphism for all

The group llm(L) and the homomorphism cpT can be defined for any sub complex L c B. We must prove that cp'J] is an isomorphism for any m. To do this, we use the Mayer Vietoris exact sequence and the five lemma. The Mayer Vietoris exact sequence holds for the groups 1l* because any direct sum of exact sequences is an exact sequence. The commutative diagram

1

1

cpo;+l

Hm+1(EKnL)

1

CPKffiCP't

~

CPKUL

Hm(EK) E9 Hm(EL)

~

1

1

cpO;

CPKnL

~

Hm(EKnL)

Hm(EKUL)

~

IffiCPT- 1

Hm-l(EK) E9 Hm-1(EL)

shows that if CPK, CPL, and CPKnL are isomorphisms, then so is CPKUL. This implies the required assertion. 0 The existence of splitting maps follows easily from Theorem 3.41. Theorem 3.42. For any vector bundle { over a compact simplicial complex B, there exists a splitting map f: B 1 -+ B. Proof. We prove the theorem by induction on n = dim{. For n = 1, the map idE is splitting. Suppose that for any vector bundle of dimension less than n, a splitting map exists. According to Theorem 3.41, the map q*: H*(B) -+ H*(E(P{)) is a monomorphism. Moreover, q*({) = A{ E9 u{, where A{ is a one-dimensional bundle. By the induction hypothesis, for u{, a splitting map g: Bl -+ E(P{) exists. We show that the map f = qg is splitting as well. Indeed, f* = g*q* is a monomorphism because g* and q* are

170


monomorphisms. Moreover, r(~) = g.(>'~) ffi g.((]"~), and, by assumption, is a direct sum of one-dimensional bundles. D

g.((]"~)

To illustrate applications of splitting maps, we give a new proof of an assertion which we have proved earlier by a different method. Namely, we show that the existence of splitting maps implies the uniqueness of Stiefel Whitney and Chern classes (under the axiomatic approach; see p. 159). Indeed, for a one-dimensional bundle, the Stiefel Whitney or Chern class is uniquely determined by the naturality condition, the condition on WI ('Y 1), and the universality of the bundle 'Yl. Let f: BI --+ B be a splitting map for the bundle~. Then r(~) = 6ffi·· ·ffi~n, where the bundles 6, ... , ~n are onedimensional. Therefore, the class r(w(~)) - I1(1 + WI(~I)) is determined uniquely. Since the map is a monomorphism, the class w(~) is unique.

r

We have already discussed two approaches to the Stiefel Whitney and Chern characteristic classes, namely, treating these classes as obstruetions to extending sections and the axiomatic approach. Theorem .a.41 suggests yet another construction of the characteristic classes. According to this theorem, the element can be uniquely written as E~=l (_I)i+lxi(~)a~-i, where each Xi(~) belongs to Hi(B; Z2) (in the real case) or to H2i(B; Z) (in the complex case). The Stiefel Whitney and Chern characteristic classes can be constructed using the following theorem.

ar

Theorem 3.43. Ifxo(~) = 1 andxi(~) = Ofori > dim~, thenxi(~) in the real case and Xi(~) = £;(~) in the complex case.

= Wi(~)

It is most convenient for us to prove Theorem 3.43 using singular cohomology, so we postpone the proof until p. 252.

Any vector space V over C can be associated with a vector space V over C. Each vector v E V corresponds to v E V, and multiplication by complex numbers in V is defined by >.v = >'v; i.e., multiplication by >. is replaced with multiplication by X. Applying this operation to every fiber of a complex vector bundle w, we obtain a bundle w, which is said to be conjugate to w. Example 48. The tangent bundle

Tl

of Cpl is not isomorphic to

TI.

Proof. Let f: ~ --+ V be an isomorphism over C. If f(v) = w, then f(>.v) = >.w = Xw. This means that if we identify corresponding elements of the spaces V and V, then f(>.v) = Xf(v). If dime V = 1, then f reverses the orientation of VR. Thus, geometrically, f is the symmetry with respect to some straight line; this line uniquely determines the isomorphism f·

8 2.

The realification of the bundle Tl is the tangent bundle of the sphere Therefore, if the bundles TI and Tl were isomorphic, ~h n we would

171


obtain a continuous line-element field on 8 2 , which does not exist. Indeed, a closed manifold admits a continuous line-element field if and only if it admits a continuous vector field without singular points (see Problem 84 in Part I). 0 The conjugate bundle W is closely related to the dual bundle w· Homc(w, C). If a space V over C is endowed with a Hermitian inner product for which (~v, J.Lw) = Xji(v, w), then the formula v ~ ~(x) = (x, v) defines an isomorphism V --+ Homc(V, C). Therefore, in the presence of a Hermitian metric on the bundle w, the conjugate bundle W is canonically isomorphic to the dual bundle Homc(w, C). Theorem 3.44. The bundle w· ® w has a nonzero section. In particular, if dime w = 1, then the bundle is trivial. Proof. There is a canonical isomorphism V· ® V --+ Hom(V, V) (see, for example, [104, p. 120]). The space Hom(V, V) has a distinguished element, the identity map. Hence the bundle w" ® w has a nonzero section. 0 Theorem 3.45. Ck(W) Proof. If dimw

=

=

(-I)kck(w).

1, then Cl(W)

=

e(wlR.) and Cl(W)

= e(wJR). The = -CI(W).

bundles

WJR and WJR have opposite orientations; therefore, CI(W)

Suppose that W is a complex vector bundle of dimension nand splitting map for w. Then r(w) = WI ffi ... ffi Wn ; hence

r

f is a

(c(w)) = c(r (w)) = C(Wl) ... c(wn )

= (1- Cl(Wl))··· (1 = ~)-I)kck(r(W))

CI(Wn )) = r(L(-I)kck(w)).

This implies the required assertion because Corollary. Ifw ~ w, then 2C2k+1(W) = Proof. If W ~ w, we have C2k+1(W) -C2k+1(W).

r

is a monomorphism.

0

o. =

C2k+1(W)

o

The Chern classes of a complex manifold are defined as the Chern classes of its tangent bundles. The calculation of the Chern classes of complex projective spaces is similar to that of the Stiefel Whitney classes of real projective spaces (see the corollary to Theorem 3.23). Example 49. We have c(cpn) = (1 generator of the group H2(cpn; Z).

+ a)n+1,

where a = -Cl(-y~) is a

J. fippllCatlOns

1.1.t;

ot :)lmpllClal nomolOgy

Proof. The fiber of the bundle 'Y~ over a point l E cpn is the line l in C n+!. Let wn be the bundle over cpn whose fiber over the point l is the orthogonal complement l1. of the line l in C n +! with respect to Hermitian product. We claim that rep" ~ Homd'Y~,wn). Indeed, each tangent vector to cpn is determined by a pair (x, v), where x E l \ {O} and v is a vector orthogonal to x; the pairs (Ax, AV) with A E C \ {O} determine the same tangent vector. Therefore, each tangent vector is uniquely determined by the linear map l - l1. taking x to v.

The bundle Homd'Y~, 'Y~) is trivial because it has a zero section, which corresponds to the identity map. Let £k be a trivial k-dimensional bundle over CP. Then rep" ffi £1 ~ Homd'Y~,wn ffi 'Y~) ~ Homd'Y~,en+l); therefore, the bundle rep" ffi £1 is isomorphic to the direct sum of n + 1 copies of the bundle Homd'Y~,£l) ~ 'YA. According to Theorem 3.45, we have Cl ('YA) = -Cl ('Y~). Therefore, Theorems 3.37 and 3.36 imply c(cpn) = (1- Cl('Y~))n+! = (1 + a)n+!. -- 0 To each vector space V over lR we can assign its complexification V ®Ill C. As a space over JR, it is canonically isomorphic to the space consisting of the vectors v+iw, where v,w E V. Multiplication by i is defined as i(v+iw) = iv-w. If ~ is a real bundle, then we can complexify each of its fibers. We denote the bundle thus obtained by ~ ® C. Lemma. The bundle

~

® C is isomorphic to the conjugate bundle { ® C.

Proof. Consider the map f given by f(v + iw) = v - iw. It is easy to verify that f(i(v + iw)) = -if(v + iw). Therefore, the map f determines a fiberwise isomorphism of the bundles { ® C and ~ ® C. 0 Corollary. 2C2k+!(~ ® C)

= O.

The Pontryagin characteristic classes of an arbitrary real bundle ~ are defined by Pk(~) = (_1)kc2k(~ ® C) E H4k(B; Z). The elements of order 2 (namely, C2k+ 1 ({ ® C)) are ignored. The Pontryagin classes have the naturality property; the Pontryagin classes of stably equivalent bundles coincide. However, the equality p({ ffi",) = p(~)p(",) does not hold because we have discarded the elements of order 2. We can assert only that the relation given in the following theorem is valid. Theorem 3.46. The equality 2(p(~ ffi",) - p(~)p(",))

= 0 holds.

Proof. The bundle (~ffi",) ® C is isomorphic (~® C) ffi (", ® C). Therefore,

Ck((~ ffi",) ® C)

=

L i+j=k

Ci(~ ® C)Cj('" ® C).

3. Group Actions

173

Discarding the odd-dimensional Chern classes yields 2C2k(({ $ "') ® C)

=2

L

C2i({ ® C)C2j('" ® C).

i+j=k

The required equality is obtained by multiplying both sides by (_I)k = (_I)i( -1)i. 0 Theorem 3.47. Suppose that w is a complex n-dimensional vector bundle, Pk = Pk(WJR), and Ck = Ck(W). Then n ( n ) n L( -1)kpk = L( _1)k q L Ck· kO

kO

k-O

Proof. By definition, (_I)kpk = C2k(WJR ®C). Therefore, is suffices to verify that WJR ® C ~ W $ w. Let W be a complex space and V = WK. The space V ® C is canonically isomorphic to V $ V under the operation J(x,y) = (-y,x), which corresponds to multiplication by i. In the space W, multiplication by i is given. Thus, the two maps I±(x) = (x, =fix) from V to V $ V are defined. It is easy to verify that I±(ix) = ±J(f±(x)), i.e., 1+ is complexlinear and I_is antilinear. Moreover, V $ V = 1m 1+ $ 1m I_because (x, y) = 1+ ( U) + 1- ( X-;iU). The space 1m 1+ is canonically isomorphic to W, and 1m I-is canonically isomorphic to w. Having constructed such canonical decompositions for all fibers of the bundle WJR ® c, we obtain the required isomorphism. 0

xt

Corollary. We have p(cpn) = (1 group H2(cpnj Z).

+ ( 2)n+1,

where a is a generator

01 the

Proof. Let r be the tangent bundle of cpn treated as a complex manifold, and 7lR its realification, i.e., the tangent bundle of cpn as a real manifold. We already know that c(r) = (1 + a)n+1 = E Ck. Therefore, E( -1)kck = (1- a)n+1, whence E(-I)k pk (7lR) = ((1- 0)(1 + a))n+1 = (1- ( 2)n+1. It follows that EPk(7lR) = (1 + ( 2)n+1. 0

3. Group Actions 3.1. Simplicial Actions. A topological space X with an action of a group G is called a G-space. A G-space map f: X - Y is said to be equivariant if I(g(x)) = g(f(x)) for all 9 E G and all x E X. Let G be a finite group acting on the space IKI, where K is a simplicial complex. This action is said to be simplicial if the map g: IKI _ IKI is simplicial for any 9 E G. A simplicial complex K with a fixed simplicial action of a finite group G is called a simplicial G-complex.

lf4

J. Applications of Simplicial Homology

Any simplicial map of simplicial complexes induces a simplicial map of the barycentric subdivisions of these complexes. Therefore, the barycentric subdivision of a simplicial G-complex is a simplicial G-complex as well. For a simplicial G-complex K, the simplicial complex K/G is defined. Its vertices are the orbits of the action of G on the vertex set of Kj i.e., each vertex of K/G has the form v* = G(v), where v is a vertex of K. Vertices va' .. . , v~ are the vertices of a simplex in K/G if and only if the orbits va, ... , v~ contain vertices vo, ... , Vn that are the vertices of a simplex in K. In this case, we say that [vo, ... ,vnl is a simplex over the simplex [va, ... ,v~l. Note that other vertices in the orbits va' ... ' v~ do not necessarily determine a simplex. Assigning to each vertex V the orbit v*, we obtain a simplicial map K --+ K/G and, therefore, a continuous map IKI--+ IK/GI. The group G acts on the topological space IKI: an element 9 E G takes each point E ),iVi to the point E ),igvi (we use the assumption that 9 takes simplices to ~impliteg.). Thus, the orbit space IKI/G is defined. Assigning the point L),iG(Vi) = E),IVi to each orbit G(E),iV,), we obtain a map IKI/G --+ IK/GI. This map is one-to-one if and only if for every two simplices [vo, ... , vnl and [govO, ... , gnvnl in K, there exists an element 9 E G such that gVi = giVi for i = 0,1, ... ,no Many theorems about simplicial G-complexes can be proved only under additional conditions on the G-complexes. But, as rule, these conditions can be fulfilled by refining the triangulation (e.g., by passing to the first or second barycentric subdivision). They include the properties (R) and (A) specified below. A simplicial G-complex K is said to be regular if any subgroup H c G has the following property (R):

Suppose that ho, ... , h n E Hand [vo, ... , vnl and [hovo, ... , hnvnl are two simplices in K. Then there exists an element h E H such that hVi = hiVi for i = 0,1, ... , n. We say that a simplicial G-complex has property (A) if it satisfies any of the two conditions specified in the following lemma. Lemma. For a simplicial G-complex K, the following conditions are equivalent. (i) For each element 9 E G and every simplex 6. from K, all points of the set 6. n g6. are fixed under the action of g. (ii) If v and gv are vertices of the same simplex, then v = gv. Proof. Suppose that (i) holds. Let v and gv be vertices of the 2, then r n (mod 2). (b) If (K, L) is a Zp-homology n-disk, then (KG, L G) is a Zp-homology r-disk, where 1 S r S n. If p > 2, then r = n (mod 2). Proof (see [36]). (a) Applying Theorem 3.55 to L = 0, we obtain ~i>odimHt(KG) < ~i>odimHi(K) = 2 and X(K G) == X(K) (modp). But X(K) is equal to 0 or 2. Therefore, the equality ~i 0 dim H,(K G) = 1 cannot hold. Thus, ~i>O dim Hz (KG) = 0 or 2. In the former case, we have KG = 0, and in the latter case, KG is a Zp-homology sphere of some nonnegative dimension r.

The inequality r S n follows from

L dimHi(KG) S L dimH,(K) i~n+l i>n+l If p

(_l)n

= O.

> 2, then the congruence X(KG) == X(K) (mod p) (i.e., (-It + 1 ==

+1

(mod p)) implies r

== n

(mod 2).

(b) For homology disks, the proof is even simpler: it suffices to apply the relations

LdimHz(KG,L G) S LdimHi(K,L) = 1, 1>0

L '~n+l

gx

i~O

dimHt(KG,L G) S L

dimHi(K,L)

= O.

o

i~n+l

The action of a group G on a topological space X is said to be free if for any point x E X and any nonidentity clement 9 E G.

f:. x

Problem 111 (Smith). Prove that finite groups cannot act freely on ~n.


HS4

Exercise. Prove that if G = Zp (for prime p) and K is an acyclic regular G-complex, then KG is an acyclic complex. (In particular, KG i- 0.) 4. Steenrod Squares In this section, we construct the Steenrod squares and prove some of their properties. We use them in Section 3.2 to derive the Thorn and Wu formulas and prove the Stiefel theorem on the parallelizability of orient able 3-manifolds. 4.1. Construction of the Steenrod Squares. The Steenrod squares are additional algebraic operations on cohomology rings, which generalize cup product. Recall that cup product can be constructed using any diagonal approximation AJ: C.(K) - C.(K) ® C.(K) that takes every vertex v to v ® v. Such a diagonal approximation is not unique. Two diagonal approximations Do and TDo for a I-simplex K are shown in Figure 6. The diagonal approximation T Do is symmetric to Do with respect to the diagonal. Formally, symmetry with respect to the diagonal can be defined as the chain transformation T of C.(K) ® C.(K) given by T(u P ® r q) = (-l)pqr q ® uP; the sign is chosen so as to make T the chain map. The diagonal approximation T Do is the composition of the diagonal approximation Do and the chain transformation T. TDo

r---------------

I

D

Figure 6. Two diagonal approximations

The diagonal map d: KI- IKxKI is not cellular; therefore, TDo i- Do. and we can consider the difference Do - T Do. It turns out that the maps Do and T Do are chain homotopic, i.e., Do - T Do = BDI + DIB for some homomorphism D I : C.(K) - C.(K) ® C.(K) increasing the dimension of the chain by 1. Moreover, extending this construction, we obtain a sequence of homomorphisms D k : C.(K) - (C.(K) ® C.(K))iH' The Steenrod squares are constructed using the homomorphisms Dk. Let us study these homomorphisms in more detail. Consider the chain complex W corresponding to the CW-complex soo having two cells of each dimension from 0 to 00 (the k-skelpton of the complex SOO is the sphere Sk). The chain transformation T Cdl1 be treated as

4. Steenrod Squares

185

an action of the generator T of the group Z2. The generator T of Z2 acts on as the symmetry about the center of the sphere. Therefore, the chain complex W has two generators, Wk and TWk, in each dimension k ~ O. The generators can be chosen so that OWl = Wo - Two, OW2 = Wl + TWI, and in general OWk = Wk-l + (-1)kTwk_l (T preserves the orientation of cells for odd k and reverses it for even k). Let K be an arbitrary simplicial complex. We define the action of the group Z2 on the chain complex W ® G.(K) by T(Wk ® a) = (TWk ® a); its action on G.(K) ® G.(K) has already been defined. Let G.(K x K) be a chain complex for calculating cellular homology. It can be identified with G.(K) ® G.(K). The chain subcomplex G.(a x a) is defined for any cell (simplex) a in K; we denote by the same symbol the corresponding sub complex in C. (K) ® G. (K).

soo

We set .c(Wk ® a) = G.(a x a) c G.(K) ® G.(K). The sUbcomplex .c(wk®a) is acyclic, and T.c(wk®a) = .c(wk®a). This allows us to construct a chain map dimK (by assumption). Therefore, Hk(L) = o. Now, consider singular homology. We have Hk(IL/) = O. By construction, X c ILl; hence z is a singular cycle in ILl, and therefore z = aw~where w is a singular chain in ILl cU. 0

Remark. The analog of Theorem 4.8 for closed sets does not hold. In [10], an example of a closed subset in ]R3 with nontrivial homology groups of arbitrarily high dimensions was constructed. Using the properties of nerves of covers described in Part I, we can easily derive from Theorem 4.8 the following well-known theorem of ReIly.

Theorem 4.9 (ReIly [48]). If UI. ... , Un+! are open convex sets in lRn and any n of them have a common point, then all these sets have a common point. Proof. We set X = U I U ... U Un+! and consider the cover U = {Ui } (i = 1, ... , n + 1) of X. Let N be the nerve of the cover U. According to Theorem 3.21 from Part I, the space X is homotopy equivalent to INI. Suppose that UI n ... nUn+! = 0. Then N consists of the n-faces of the simplex An+l. Therefore, INI row sn; in particular, Hn(X) ~ Hn(/N/) !:!:!. Hn(sn) i- O. This contradicts Theorem 4.8. 0 Helly's theorem (and even a more general assertion, which was also proved by ReIly) can be derived from Theorem 4.8 without using the properties of nerves of covers. For this purpose, the following lemma is needed.

Lemma 4.1. Let Xl. ... , Xm be acyclic open subsets of a topological space X and anyr of them (1::::; r::::; m-1) have acyclic (in particular, nonempty) intersection.

(a) If Xl n··· nXm = 0, then the group Hq(XI U··· UXm) is nontrivial 1!recisely when q = m - 2. (b) If Xln·· ·nXm i- 0, then Hq(XIU .. ·UXm) ~ Hq_m+I(Xl .. ·nxm ) for all q.

1. Basic Definitions and Properties

209

Proof. We prove the lemma by induction on m. First, consider the case m = 2. If Xl n X 2 = 0, then Xl U X 2 is a union of two disjoint acyclic spaces. Therefore, HO(XI U X 2) f. 0 and Hq(XI U X2) = 0 for q Xl n X2 f. 0, then the reduced Mayer Vietoris sequence

f. o.

If

Hq(Xd El3 Hq(X2) --- Hq(XI U X 2) --- Hq-l(XI n X 2) --- Hq I(Xl) ffi Hq I(X2) implies Hq(XI U X2) ~ Hq-I(X I n X 2) because Hq leX,) = 0 for all q. Consider an arbitrary m. By the induction hypothesis, the space UI = Xl U ... U Xm I is acyclic because Xl n ... n Xm I f. 0. The reduced Mayer Vietoris sequence implies Hq(UI UXm) rv Hq I(UI nXm). Moreover, UI n Xm = (Xl n Xm) U ... U (Xm I n Xm). Let us apply the induction hypothesis to the sets Xl n X m , ... , Xm In X m . Clearly, the intersection of r such sets is the intersection of r + 1 of the sets Xl. .. . , X m . Suppose that Xl n··· n Xm = 0 (or, equivalently, the intersection of the sets Xl n X m , ... , Xm-l n Xm is empty). By the induction hypothesis, the group Hq-l(UI nXm) is nontrivial if and only if q-I = (m-I) -2, i.e., q = m-2. If xln·· ·nxm f. 0, then Hq I(UlnXm ) ~ HCq-l)-Cm-I)+l(Xln .. ·nXm). But (q - 1) - (m - 1) + 1 = q - m + 1. 0 Now, we can prove the following theorem; the classical Helly's theorem is its special case.

Theorem 4.10 (ReIly [49]). Suppose that Xl! ... , Xm is a finite family of open subsets of R n and the intersection of any r of these sets is nonempty for r ~ n + 1 and acyclic for r ~ n. Then the space Xl n ... n Xm is acyclic (in particular, it is nonempty).

Proof (see [29]). Suppose that the required assertion does not hold for sets X I, ... , Xm and the number m is minimal. The minimality of m implies that the intersection of any r ~ m - 1 of the sets Xl, ... , Xm is acyclic. Suppose that Xl n··· n Xm = 0. This can happen only if m > n + 1. Applying assertion (a) of Lemma 4.1, we obtain H m - 2(X I U ... U Xm) i- o. On the other hand, m - 2 ~ n; therefore, according to Theorem 4.8, we have H m- 2(XI U··· U Xm) = o. We have arrived at a contradiction. Now, suppose that Xl n ... n Xm i- 0. By assumption, the space Xl n··· n Xm is not acyclic. This can happen only if m > n. There exists a number p ~ 0 for which Hp(XI U ... U Xm) i- O. Let us write p in the form p = q - m + 1. According to assertion (b) of Lemma 4.1, we have Hq(XI n ... n Xm) i- o. Note that q = p + m - 1 ~ n. This contradicts Theorem 4.8. 0

4. Singular Homology

210

Cellular Homology. For any CW-complex X, we can define cellular homology in analogy with the case of simplicial complexes with the only difference that simplicial homology should be replaced by singular homology. Using Problem 116, we obtain Hi(Xk,X k- 1) ~ Hi(Xk/X k- 1) ~ Hi(V Q S~) for i ?: 1. Therefore, Hi(X k , X k- 1) = 0 for i i- k; for i = k, this is a free 3 Abelian group whose generators are in one-to-one correspondence with the k-cells of the complex X. This implies, in particular, that Hk+l(Xk) = o. Indeed, the exact sequence of a pair •• , -----+

HHl(X k , X k- 1)

-----+

Hi(X k- 1) -----+

Ha(Xk)

shows that H,(X k ) ~ Hi(X k- 1 ) for i ?: k Hk+1 (X k- 1) ~ ... ~ Hk+1 (XQ) = o.

-----+

+ 1.

Hi(X k , X k- 1 )

-----+ .••

Therefore, Hk+1(Xk) ~

As in the simplicial case, we set Ck = Hi(X k , X k- 1 ) and' define a homomorphism 8 k : C k --+ Ck-l. Then, we prove that H k- 1 ~ Ker8k_I/Im8k and Hk-l(X) ~ Hk_1(X k ). The proof of the latter isomorphism uses the fact that any singular chain is contained in some finite-dimensional skeleton XN. When cellular homology is used, the following assertion becomes obvious. Let X be a CW -complex containing mk cells of dimension k. Then the rank of the group Hk(X) is at most mk. We need it to prove the Morse inequali ties. 1.5. The Morse Inequalities. Suppose that M n is a closed manifold and

f is a Morse function on M n with pairwise distinct critical values. For each k = 0,1, ... , n, we define rk to be the rank of Hk(Mn) and mk to be the number of index k critical points of f. These numbers are related by certain equalities and inequalities. First, note that E (-I)krk = X(Mn) =

E (-I)kmk

(see Part I, p. 244). Moreover, it is easy to see that rk ~ mk. Indeed, in the complex for calculating the cellular homology of M n , the chain group Ck coincides with zm" (see Part I, Theorem 5.44). Clearly, zm" has a quotient of rank at most mk. A more detailed study of this chain complex leads to the following sharper inequality. Theorem 4.11. Let tk be the minimum number of generators in the torsion subgroup Tk of Hk(Mn). Then

3This group contains only finite sums of cells because the image of 1LIl) simplex under a continuous map is compact and any compact subset of a CW-complex intersects only finitely many open cells.


211

Proof. The groups Tk and Tk-l are quotients of free Abelian subgroups Fk and Fk-l of ranks tk and tk-l in Ck and Ck-l. The group Fk-l is the epimorphic image under the map 8k of a rank-tk_l free Abelian group F~_l C Ck, and the group Fk is the image of a similar group F~ C CHI. Thus, the group Ck contains the subgroups Fk and F~ 1; it has also a free subgroup Sk of rank rk, which is contained in Ker 8 k and has trivial intersection with 1m 8k+l; this subgroup corresponds to a free summand of Hk(Mn). The group Fk is contained in Im8k+l; therefore, the sum Fk + Sk is direct. Moreover, we have 8 k (Fk ffi Sk) = 0 and 8kF~ 1 = Fk-l, where F~_l and Fk 1 are free Abelian groups of the same rank. Hence the sum FLI + Fk ffi Sk is direct as well. Thus, the group Ck, whose rank equals mk, contains a free Abelian subgroup of rank tk 1 + tk + rkj therefore, tk-l + tk + rk ~ mk. 0 The numbers rk and mk are also related by the following series of inequalities (Morse inequalities).

= 0,1, ... , n,

Theorem 4.12. (a) For every k rk - rk-l

+ rk-2 -

... ± ro

~

mk - mk-l

+ mk-2 -

.. , ± mo

(Jor k = n, this inequality becomes an equality). (b) Suppose that M(t) = Emktk and R(t) = Erktk. Then the polynomial M(t) - R(t) is divisible by 1 + t and the coefficients of the polynomial M(tt;~(t) are nonnegative.

Proof. It is easy to show that (a) follows from (b). Indeed, suppose that - rk)t k = E (1 + t)dktk, where dk ~ O. Then mo - ro = do ~ 0, ml - rl - (mo - ro) = d 1 ~ 0, m2 - r2 - (ml - rl) + (mo - ro) = d2 ~ 0, etc. Clearly, d m = O. Thus, it suffices to prove (b). Let al < a2 < ... < am be the critical values of the function f. Choose numbers b1, ... , bm+l so that -00 < b1 < al < b2 < ... < am < bm+l < 00 and consider the manifold Mp = {x E M n I f(x) ~ bp}, where p = 1,2, ... , m + 1. Consider also the polynomials Rp(t) = E rkHk(Mp)tk and Rpq(t) = ErkHk(Mp,Mq)tk for p > q. Let us write the exact sequence for the pair (Mp, Mq):

E (mk

Hk+l(Mp,Mq)

8 k +1,

Hk(M n )

~

Hk(Mp)

~ Hk(Mp,Mq) ~

Hk-l(Mq).

Clearly, and rk 1m 11'

= rkHk(Mp) -

rklmi

= rkHk(Mp) -

(rkHk(Mq) - rklmak).

212


Therefore,

rkHk(Mp, Mq) - (rkHk(Mp) - rkHk(Mq)) = rkIm8k and hence

+ rk 1m 8k+1,

Rpq(t) - (Rp(t) - Rq(t)) = (1 + t)dpq(t),

where dpq is a polynomial with nonnegative coefficients. Obviously, m

L

(Rp+1(t) - Rp(t))

= Rm+1(t) = R(t);

p-l

thus, it remains to prove that E;=l Rp+1,p(t) = M(f). To this end, it suffices to verify that ~l p(t) = t~, where I-L is the index of the critical point between the levels f(x) = bp+1 and f(x) = bp. We have Mp+1 ::::: MpU(S~ x Dn-~); therefore, Hk(Mp+1' Mp) ~ Hk(MpU(S~ x D n ~~k{p) ~ iIk(S~ X Dn-~) ~ iIk(S~), 0

1.6. Multiplications. The Eilenberg-Zilber Theorem. The Eilenberg Zilber theorem claims that Hk(X x Y) ~ Hk(C.(X) ®C.(Y)) for any topological spaces X and Y. Its proof, which is based on the acyclic model theorem, is presented below. Let T be the category of topological spaces and continuous maps, and let

TxT be the category whose objects are ordered pairs of topological spaces (X, Y) and morphisms are ordered pairs of continuous maps f: X -+ X', g: Y -+ Y'. Let us define two functors, T and T', from the category TxT to the category of nonnegative chain complexes. (i) The functor T takes each pair (X, Y) to the singular chain complex

C.(X x Y); (ii) the functor T' takes each pair (X, Y) to the tensor product C.(X) ®

C.(Y). Recall that the tensor product of chain complexes is defined by

(C.(X) ® C.(Y))k =

EB

Cp(X) ® Cq(Y), p+q-k 8(ep ® cq) = 8ep ® cq + (-1)Pep ® 8cq.

The functors T and T' are acyclic4 with respect to the models from M = {(~p, ~q)}, where p and q range over all nonnegative integers. Indeed, the acyclicity of T is obvious because the space ~p x ~ q is contractible. To prove the acyclicity of T', is suffices to show that if chain complexes C~ and C: are acyclic, then so is the chain complex C~ ® Thb follows easily from the algebraic Kiinneth theorem.

C:.

4The definition of a functor acyclic with respect to models was given on p. 103.


213

The functor T' is free 5 with respect to the models from M. Indeed, the group Cp(X) ® Cq(Y) is freely generated by the singular chains of the form Cp ® cq , which are in one-to-one correspondence with the pairs of maps I: /:l.P --+ X, g: /:l. q --+ Y. Therefore, for the element ep,q E C* (~P) ® C* (/:l. q) we can take the singular chain /:l.P x /:l. q • The functor T is free with respect to the models from {(/:l.k, /:l.k)} eM. Indeed, the group Ck(X x Y) is freely generated by singular chains that are in one-to-one correspondence with the maps F: /:l.k --+ X X Y. Any such map is determined by a pair of maps I: /:l. k --+ X, g: /:l. k --+ Y. Let dk: /:l.k --+ /:l.k X ~k be the diagonal map. Then F can be represented as the composition /:l.k :!!:..... /:l.k X /:l.k ~ X X Y. Therefore, for the element ek E C*(/:l.k x /:l.k) we can take the singular chain corresponding to the map dk. The algebraic Kiinneth theorem implies the existence of a canonical isomorphism Ho(C*(X) ® C*(Y)) --+ Ho(X) ® Ho(Y). Clearly, the pathconnected components of X x Yare the products of those of X and Y. Hence there exists a natural isomorphism cp: Ho(X)®Ho(Y) --+ Ho(C*(X)® C*(Y)). Applying the acyclic model theorem to the maps cp and cp-l, we obtain chain maps T: C*(XxY) --+ C*(X)®C*(Y) and T: C*(X)®C*(Y) --+ C.. (X x Y), which induce the maps cp and cp-I, respectively, in the 0dimensional homology groups. The chain map TOT induces the identity map of the group Ho(X x Y). Hence, according to the acyclic model theorem, the map TOT is chain homotopic to the identity. Similarly, TOT is chain homotopic to the identity as well. Thus, the map T*: H*(X x Y) --+ H*(C.. (X) ® C*(Y)) is an isomorphism. We have proved the following theorem. Theorem 4.13 (Eilenberg Zilber [35]). For any topological spaces X and Y, the groups Hk(X x Y) and Hk(C*(X) ® C*(Y)) are isomorphic.

Combining the Eilenberg Zilber theorem with the algebraic Kiinneth theorem, we obtain the Kiinneth theorem lor singular homology. Theorem 4.14. For any topological spaces X and Y,

Hk(X x Y) ~ (H*(X) ® H*(Y))k E9 (Tor(H*(X), H .. (Y)))k-l. The Alexander-Whitney Diagonal Approximation. Suppose X is a topological space, d: X --+ XxX is a diagonal map, d*: C*(X) --+ C*(XxX) is the induced chain map, and T: C*(X x X) --+ C*(X) ®C*(X) is the functorial chain map inducing an isomorphism of homology groups. Consider 5The definition of a functor free with respect to models was given on p. 103.

214


the composition

C.(X) ~ C.(X x X) ~ C.(X) ® C.(X). The chain map rd.: C.(X) -+ C.(X) ® C.(X) is functorial and takes every zero-dimensional simplex v to v ® v. We refer to any functorial chain map Do: C.(X) -+ C.(X)®C.(X) that takes v to v®v as a diagonal approximation. The acyclic model theorem implies that any diagonal approximation is chain homotopic to rd•. Indeed, the functor C.(X) is free with respect to the models from M = {6. k }, and the functor C.(X) ® C.(X) is acyclic with respect to these models. For singular cohomology, the cup product is defined as follows. If cP E CP(X; R) and c q E cq(X; R), where R is a commutative ring with identity, then the cochain cP '-" & assigns (cP ® c q, DocP+q) to the chain cP+ q E Cp+q(X; R); here

(cP ® c q , Ci ® Cj) = dpidqj(cP, Ci)(Cq , Cj}. At the level of cochains, the cup product depends on the choice of the diagonal approximation Do, but at the level of cohomology, it is invariant. For calculations, it is convenient to have a diagonal approximation specified by some simple formula. Best suited is the Alexander Whitney diagonal approximation n

[0,1, ... ,n]

t-+

~ [0, ... ,i] ® [i, ... ,n]; i=O

here [0, 1, ... ,n] is a singular simplex f: [0, 1, ... , n] -+ X, and [0, ... , i] and [i, .. . , n] are the restrictions of f to the corresponding faces. The proof that this is a chain map is the same as in the simplicial case (see p. 105). At the level of cochains, the cup product determined by the AlexanderWhitney diagonal approximation has the form

(cP '-" cq , [0, 1, ... ,p + q]) = (cP, [0, ... ,p])(cq , [p, ... ,p + q]). Now let us define the cap product. Any cochain cP E CP(X; R) can be regarded as a homomorphism c?: C. (X) -+ R (for i i- p, this homomorphism vanishes at the i-dimensional chains). Consider the composition

C.(X) ~ C.(X) ® C.(X) ~ C.(X) ® R. We take the tensor product of the groups involved in this composition and R and apply the multiplication J.L: R ® R -+ R in the ring R to the result:

C.(X) ® R ~ C.(X) ® C.(X) ® R id®c1'®id l

C.(X) ® R ® R id®~.. C.(X) ® R.


215

We denote the image of a chain Cp+q E Cp+q(X; R) = Cp+q(X) ® R by cP ,.-... Cp+q' At the level of (co)homology, the cap product does not depend on the choice of the diagonal approximation Do. For the Alexander Whitney diagonal approximation, we obtain the usual cap product:

cfJ,.-... [O,1, ... ,p+q]

= (cfJ,[q, ... ,p+q])[O, ... ,q].

The Relative Case. In the relative case, the functorial chain equivalence between C.(XxY) and C.(X)®C.(Y) is replaced by that between C.(XxY, (X x B) U (A x Y)) and C.(X, A) ® C.(Y, B) (the spaces {X x B, A x Y} must satisfy the excision axiom). This chain equivalence is constructed as follows. If the spaces {X x B, A x Y} satisfy the excision axiom, then the natural embedding

C.(X x B)

+ C.(A x

Y) ~ C .. «X x B) U (A x Y))

induces an isomorphism of homology groups, and therefore so does the natural map

C.(X x Y) C.(X x Y) C .. (X x B) +C.(A x Y) ~ C.«X x B) U (A x y))' The functorial chain equivalence C.(X)®C.(Y) ~ C .. (XxY) maps C.(X)® C .. (B) to C .. (X x B) and C.(A) ® C .. (Y) to C.. (A x Y). We have obtained c.(XxY) d . . I b t a ch am eqmva ence e ween C.(XxB)+C.(AxY) an

___ C.~(X---.:.....)®-=--C.~(_Y!..-)_ _ '" C.(X) C .. (Y) C.(X) ® C.(B) + C.(A) ® C.(Y) = C .. (A) ffi -C..-(B-), The composition of these two chain equivalences is the required chain equivalence. This chain equivalence implies the Kiinneth theorem for relative homology:

Hk(X x Y, (X x B) U (A x Y)) ~

(H. (X, A) ® H.(Y, B))k ffi (Tor (H. (X, A), H.(Y, B)))k-l'

Moreover, it allows us to define the homology cross product

Hp(X,A) ® Hq(Y,B) ~ Hp+q(X x Y, (X x B) U (A x Y)) and the cohomology cross product

HP(X, A) ® Hq(y, B) ~ Hp+q(X x Y, (X x B) U (A x Y)). The cup product construction in the relative case, i.e., for HP(X, AI) and Hq(X, A2)' involves the cohomology of the complex of cochains vanishing on C .. (A I ) + C.(A2), and the cap product construction involves the homology


216

of the complex c.(A~il~:(A2)· Therefore, if sets {AI. A2} in X satisfy the excision axiom, then the cup product

HP(X, AI) ® Hq(X, A 2) --+ Hp+q(X, Al U A 2) and the cap product

HP(X, Ad ® Hp+q(X, Al U A 2) --+ Hq(X, A 2) are defined. Cohomology of Products: A Special Case. The proof of the Kiinneth theorem for cohomology is more involved than for homology (moreover, the cohomology theorem is valid only under certain finiteness conditions). We shall not prove the Kiinneth theorem in the general form; instead~we shall prove a special case of thIS theorem, namely, the expression fqt the cohomology of the pair (X x lR n , X x (lRn \ {O} )) in terms of the cohomology of X and of the pair (lRn , lR n \ {O} ), which plays an important role in what follows (in the proof of the Thorn isomorphism). First, we calculate the cohomology

H*(lRn,lR n \ {O}). The cohomology sequence for the pair (lRn ,lRn \ {O}) implies that Hk(lRn,lR n \ {O}) ~ iIk l(lRn \ {O}) ~ iIk-l(sn-I). Therefore, the group Hn(lR n , lR n \ {O}) is isomorphic to the additive group of the coefficient ring, and all of the other groups Hk(lRn,lR n \ {O}),k i- n are trivial. To simplify notation, we set lR8 = lRn \ {O}, lR+ = {x E lR I x > O}, and lR_ - {x E lR I x < O}. The excision isomorphism HO (lR+) ~ HO (lRo, lR_) holds. Let us write the exact sequence for the triple (lR, lRo, lR_):

HO(lR,lR_) __ HO(lRo,lR ) ~ HI(lR,lRO) __ H1(lR,lR_). We claim that HO (lR, lR ) = HI (lR, lR ) = 0, and therefore d is an isomorphism. Since lR is connected, H°(lR,lR_) = 0, and since the sequence

Hk (lR) __ Hk (lR,lR_) __ Hk+ I (lR_) for the pair (lR,lR ) is exact, Hk(lR, lR ) = 0 for k > o. Let e be the image of the element 1 E HO(lR+) corresponding to the identity element of the coefficient ring under the composition of isomorphisms

HO(lR+) ~ HO(lRo,lR_) ~ H1(lR, lRo). The group HI (lR,lRo) is isomorphic to the additive group of the coefficient ring, and e is the identity element of this ring. The cohomology cross product of the cohomology clasbes of the pairs (X, A) and (Y, B) belongs to the cohomology of the pair (X x Y, (X x B) u (A x Y)). For brevity, we denote this pair by (X, A) x (Y, B).

217


It is easy to verify that

(IRn,lRQ) x (IRm,IR

o)= (IRn+m,IR(j+m).

Indeed, take u E IRn and v E IRm. Then IRn x IRQ consists of all pairs Cu, v) in which v :/: 0, and IR~ x IR m consists of all pairs (u, v) in which u :/: O. Clearly, the union of these sets consists of all nonzero vectors in IRn+m. Consider the element

en

= ex···

x e,

~

n

which belongs to

Hn(IRn,IR~).

Theorem 4.15. If A

defined by a

t-+

c

X is an open set, then the map

a x en is an isomorphism.

Proof. First, suppose that n = 1 and A and consider the diagram

= 0.

Take an element a E Hk(X)

where the horizontal arrows in the left square are the excision isomorphisms and 0' is the homomorphism from the cohomology sequence of the triple (X x IR, X x RQ, X x IR_). The left square of the diagram is commutative, and the right one is commutative up to sign because o'(axx) = (-l)k axox. The homomorphism 0' is an isomorphism because

for i > 0 (both spaces X x IR and X x IR_ are deformation retracts of the same space X x {-I}). Thus, the element a x e E Hk+l(X X IR, X x IRo) is the image of a E Hk(X) under the composition of isomorphisms.


218

Now, suppose that n = 1 but A i- 0. Let Z E ZI(IR, IRa) be a co cycle representing the cohomology class e. Consider the commutative diagram 0----+) Ck(X, A) ----~) Ck(X)

1

1

xz

o ~ Ck+l ((X, A)

xz

x (IR, IRa)) ~ Ck+1 (X x IR, X x IRa) ---~)

Ck(A) ---~) 0

1

xz

~ Ck+I(A

x IR, A x IRa)

~

0

with exact rows. The vertical arrows are cochain maps bec~use "(.c.c: z) = (de) x z. Thus, we obtain the following commutative diagram for cohomology groups: ----+)

Hk(X, A) ----~) Hk(X)

1

1

xe

~ Hk+l ((X, A)

x (IR, IRa))

xe

~ Hk+1 (X

x JR, X x IRa) --~)

Hk(A) --~

1

xe

~

Hk+1(A x IR,A x IRa) ~.

We have already proved that the vertical maps are isomorphisms for absolute cohomology. By the five lemma, they remain isomorphisms for relative cohomology. The required assertion for an arbitrary n follows from if s validity for n - 1 because the associativity of the cross product implies that a x en =

(axen-I)xe.

0

1.7. The Hopflnvariant. The interest in the homotopy groups of spheres was caused by Hopf's example of a map S3 --+ S2 not homotopic to a constant [58]. To prove that this map is nontrivial, Hopf defined a homotopy invariant for maps f: S3 --+ S2, which is now known as the Hopf invariant. In [60]' Hopf defined a similar invariant for maps f: s2n-1 --+ sn with any n. In the original definition, Hopf used the linking number. Steenrod [126] suggested a definition in terms of multiplication in cohomology. Let n 2': 2. Consider the oriented n-sphere sn and a continuous map f: s2n-1 --+ sn. We regard the oriented sphere s2n-l as the boundary of


219

the oriented disk D2n. Thus, the space XI = sn U f D 2n is defined. The homotopy type of this space depends only on the homotopy class of the map f (see Part I, p. 248). Throughout this section, we consider only integral cohomology. Let us show that Hk(Xf ) = Z for k = 0, n, and 2n; all of the other cohomology groups are trivial. Consider the cohomology sequence of the pair (XI, sn): ... --+

Hk(X" sn)

--+

Hk(Xf)

--+

Hk(sn)

--+ ....

The excision theorem implies Hk(Xf' sn) ~ Hk(s2n) for k > O. Therefore, for k i- 0, n, 2n, the group Hk(Xf) is surrounded by trivial groups; for k = n and 2n, we obtain isomorphisms Hn(Xf) Hn(sn) and H2n(s2n) 2n H (Xf)· Let 0 and (3 be the generators of the groups Hn(Xf) and H 2n(xf ) corresponding to the orientations of the sphere sn and the disk D2n. Then o '--' 0 = H(f){3, where H(f) is an integer. This integer is called the Hopf invariant of the map f. It follows directly from the definition that the change of the orientation of sn does not affect the Hopf invariant, whereas the change of the orientation of D2n changes its sign.

.=.

Exercise. Given maps g: sn _ sn and h: s2n _ s2n, prove that H(gf) (degg)2 H(f) and H(fh) = (deg h)H(f).

.=.

=

If n is odd, then 0 '--' 0 = - 0 '--' 0, and this element belongs to Z. Hence 0 '--' 0 = 0, and therefore H(f) = 0 for any map f. For the Hopf fibration f: S3 - S2, we have X f = CP2. In this case, a '--' a = (3 and H(f) = 1.

Exercise. Using the quaternions and the Cayley numbers, construct maps f: S7 - S4 and f: S15 _ S8 with Hopf invariant equal to 1. Remark. Adams [2] proved that the maps f: s2n-l _ sn with Hopf invariant equal to 1 exist only for n = 2,4,8.

The Hopf invariant can be defined in many different ways. Below, we give yet another co homological definition of the Hopf invariant (it essentially coincides with that given above but is formulated in a different language). Let [a] and [b] be the generators of the groups Hn(sn) and H 2n - 1(S2n-l) corresponding to the orientations of the spheres sn and s2n-l, and let a and b be co cycles representing them. The cohomology class of the cocycle a '--' a belongs to H2n(sn) = 0; hence there exists a cochain c E C 2n - 1(sn) for which 5c = a '--' a. The cohomology class of the co cycle f#(a) belongs to Hn(s2n-l) = 0; hence there exists a co chain e E cn(s2n-l) for which


220

eSe = f#(a). It is easy to verify that the cochain f#(c) - e '-"' f#(a) is a cocycle. Indeed, eS(f#(c) - e '-"' f#(a»

= f#(a '-"' a) - eS(e '-" eSe) = f#(a) '-"' f#(a) - Je '-"' eSe = O.

Therefore, the cohomology class of f#(c) - e '-"' f#(a) is defined. This class belongs to the group H 2n-l(s2n-l); hence it is proportional to the class [bJ. The coefficient of proportionality is the Hopf invariant H(f). Let us prove the equivalence of the two definitions of the Hopf invariant given above. Consider the auxiliary space Sf = snUf(s2n-l x I) (the sphere sn attached to the cylinder s2n-l X I along one of its bases) and the pair ( Sf, s2n-l ), which contains the other base of the cylinder. The excision theorem implies Hk(Sf, s2n-l) ~ Hk(Xf' *) ~ Hk(Xf) for 1£ ~ 1. Consider the diagram Hn(sn) ~ I Hn(Sf) ~ Hn(sj, s2n-l) ~ H 2n (Sf, s2n-l) +-- H 2n - 1 (s2n-l). ~

Here sq(x) = x '-"' x, and the remaining maps are natural isomorphisms (the first is induced by the projection p: Sf --+ sn, which is a homotopy equivalence, and the two others are taken from the cohomology sequence of the pair). It is fairly easy to extract the first definition of the Hopf invariant from this diagram: the class [aJ is identified with a class 0 by means of the first two isomorphisms; applying the map sq to Q (sq(o) = 0 '-"' 0), we obtain H(f)(3, where (3 is the image of the class [bJ under the natural isomorphism. Let us show how to obtain the second definition from this diagram considered at the level of cocycles. At the level of cocycles, the isomorphism p. takes the co cycle a to af = p# (a). Moreover, f# (a) is the restriction of the co cycle af to the set of singular simplices s2n-I, i.e., f#(a) = i(af), where i: c·(s2n-l) --+ C·(Sf) is the natural embedding (the restriction map to the subset S2n-l). Take a cochain c E c·(s2n) such that eSc = a '-"' a and consider cf = p#(c) and f#(c) = i(cf). By assumption, we have i(af) = f#(a) = eSe for some e E c·(s2n-l). Let e' be an extension of e to Sf, i.e., a co chain such that e = i(e'). The difference af - eSe' is a co cycle in Cn(Sf)' and it vanishes on c n (s2n-l). It can be considered as a cocycle in C n (Sf,S2n-l); in Hn(Sf' S2n-l), this co cycle represents the class corresponding to raJ. Let us apply the map sq to this cocycle. For a representative of the resulting class we can take the co cycle (af - eSe') '-" af because this prouuct vanishes on cn(S2n-l) and eSe' is a coboundary.


221

Clearly, f#(c) - e '--' f#(u) = i(cf) - ie' '--' i(uf) = i(cf - e' -- af). Therefore, the class of the co cycle f#(c) - e '--' f#(a) is identified with the class of the co cycle 5(cl-e' '-' a,) = a, '-' af-(5e') '--' a, = (a,-5e') '-' a" as required. 1.8. Simplicial Volume (the Gromov Norm). Let M n be a closed orientable manifold. Each singular chain E ad, E Ck(Mnj R) is assigned its norm /IE adi" = E Iai I· This norm carries over to the homology groups Hk(Mn jR)j namely, for a homology class ( E Hk(Mn j R), 11(11

= inf{llzlll z

is a cycle in the class (}.

The simplicial volume, or Gromov norm, of a manifold Mn is defined as II [Mn]ll, where [Mn] E Hn(MnjR) is the fundamental class of the manifold Mn. We denote the simplicial volume of Mn by IIMnllj its properties are discussed in detail in [45]. Example 55. IIS 1 11

= o.

Proof. Let fn: [0,1] -+ SI be the map defined by fn(t) = e2'1!"lnt. Then fn is a cycle representing the homology class n[SI]. Therefore, the class [SI] is represented by the cycle !fn, whose norm equals ~. D The construction from Example 55 applies to any closed orient able manifold M n admitting maps Mn -+ M n of arbitrarily large degrees. Such maps are easy to construct, e.g., for the sphere sn and the torus Tn (it suffices to construct a map of degree larger than 1). Therefore, IIsnll = 0 and IITnll =0. Example 56. If M; is a sphere with 9 handles, where 9 ~ 2, then IIM;II = 2Ix(M;)1 = 2(2g - 2).

M;

Proof. The manifold can be obtained by gluing together sides of a 4g-gon. This 4g-gon can be cut into 4g - 2 triangles. Therefore, (32)

M;

Applying (32) to a manifold N2 which covers nlIM;1I ~ IIN 2 11 ~ 2Ix(N 2 )1 + 2 = 2Ix(N 2 )1 + 2 ting n tend to 00, we see that IIM;II ~ 2Ix(M;)I.

=

n-to-one, we obtain 2Inx(M;)1 + 2. Let-

Now let us prove the inequality IIM;II ~ 2Ix(M;)I. Consider a singular 2-simplex f: .£l2 -+ M; and the universal covering p: H2 -+ M;, where H2 is the hyperbolic plane; we assume that M; is a Riemannian manifold endowed with a metric compatible with that on the hyperbolic plane. The


222

map

f

admits a lifting

j:

~ 2 --t H2 for which the diagram

H2

Ylp

~2~M; is commutative. Let us replace the singular simplex

j

by a singular simplex

l' whose image is the triangle in the hyperbolic plane with the same vertices.

If a cycle L adi represents the fundamental class [M;J, then so does the cycle L oW(];). Moreover, 27rlx(M;)I = vol(M;) :::; L la~1 vol(p(]I(~2))), where vol denotes area in the hyperbolic plane. The inequality arises because some overlapping parts of simplices may cancel each other; if there are no such cancellations, we obtain an equality. Finally, since the ar~ Qi any triangle in the hyperbolic plane does not exceed 7r, we have yol(p(]I(~2))) :::; 7r. Therefore, 2Ix(M;)1 :::; L lail :::; IIM;II. 0

Example 56 admits the following generalization. Let M n be a closed orientable hyperbolic manifold, that is, a Riemannian manifold such that the universal covering over it is the hyperbolic space Hn. Then II Mn II = vol(Mn)/vol(~n), where ~n is a simplex of maximum volume in Hn. This was proved by Thurston [139]; see also [95]. 1.9. Cohomology with Noncommutative Coefficients and the van Kampen Theorem. Zero- and one-dimensional cohomology groups with coefficients in a non-Abelian group G can be defined as follows [98]. A co chain en E Cn(X; G) (of any dimension) is defined as a function taking each singular simplex --t X to an element of G (a relative eochain en E Cn(X, Y; G) must vanish at the simplices ~n --t Y). As previously, we use additive notation for the group operation, although it is no longer commutative. We call a one-dimensional cochain zl E Cn(X, Y; G) a cocycle if, for any singular simplex f: ~ 2 --t X,

.an

zl(JC:5) - zl(Jc:n

+ zl(Jc:~) = 0,

where c:j: ~k-l --t ~k is the map defined on p. 195. A eoehain zO E CO(X, Y; G) is called a cocycle if z°(JC:6) = z°(Jd) for any singular simplex f: ~l --t X. Two one-dimensional co cycles zl and zl are sl1id to be cohomologous if there exists a O-cochain eO such that Zl(J)

= -e°(Jd) + zl(J) + c°(JC:6)

for all singular 1-simplices f: ~l --t X. (Note that we cannut take zl(f) to the left-hand side because the group operation is noncommutative.)


223

It is easy to show that the relation of being cohomologous is an equivalence. The set of equivalence classes is Hl(X, Y; G). It is not a group, but it contains a distinguished (zero) element, namely, the equivalence class of the trivial cocycle, which vanishes at each singular I-simplex. We put HO(X, Yj G) = ZO(X, Y; G); this set is a group.

Let Xl and X2 be subsets of X = X I UX2. Define HQ(C!X 1 ,X2 } (X, Y); G) for q = 0 and 1 by considering only singular simplices that are contained entirely in Xl or X 2. To each cochain there corresponds its restriction, that is, a cochain defined only for some of the simplices. Let us show that if X = int Xl U int X 2, then this correspondence induces a one-to-one map (33) that takes the zero element to the zero element (q

= 0,1).

For q = 0, this assertion is obvious. Indeed, on the left-hand side, the co cycles take equal values at the endpoints of any curve, and on the righthand side, they take equal values at the endpoints of any curve contained entirely in Xl or X2. But any curve can be divided into parts contained entirely in int Xl or int X 2. To prove the required assertion for q = 1, consider the map

defined as follows. Suppose that zl is a cocycle defined only for simplices in Xl and X 2, and I: ~l ---t X is an arbitrary singular simplex. To define the value of k(zl) at I, we divide the interval ~l into M = 2m equal intervals h, ... ,1M (the intervals are numbered in the natural order). If m is sufficiently large, then any interval 101. is contained in either Xl or X 2. In this case, we set k(zl )(J) = zl(JIIJ + ... + zl(JIIM). The map k(zd is well defined because zl is a cocycle. Indeed, if an interval I contained in Xl or X2 is partitioned into subintervals If and I", then zl(Jll') _zl (JII) +zl(JII") =

o.

Let us show that k(zl) is a co cycle. Consider any singular simplex ---t X. Choose m so that the image of the restriction of I to any simplex from the mth barycentric subdivision is contained entirely in X I or X2. Then, for every simplex from the mth barycentric subdivision, the co cyclicity condition holds. Let us show that this implies cocyclicity for the simplices from the (m - l)th barycentric subdivision. Summing the co cyclicity conditions for the simplices 1,2, ... ,6 (see Figure 3), we obtain x+al +a2+ bl +b2+CI +C2- X = 0 (here ai is the value of the co cycle at the simplex ai, etc.). Any element conjugate to the identity element in a group is itself the identity. Therefore, al + a2 + bl + b2 + Cl + C2 = o. Moreo"er, by

I:

~2


Figure 3. Summation of the cocyclicity conditions

construction, the value of the co cycle k(zl) at an interval I partitioned into two equal intervals I' and I" is equal to the sum of its values at I' and I". Clearly, k takes cohomologous co cycles to cohomologous cocycle§. "fhus, we have obtained a map

k*: HI (C!Xl,X2} (X, Y)j G)

-+

HI(X, Yj G).

It follows directly from the definition that k*i* = id and i* k* = id. The statement that the map (33) is one-to-one is an analog of the excision axiom for noncommutative cohomology. Let us construct an analog of the Mayer Vietoris sequence. Suppose that Xl and X2 are sets, X = Xl U X2, and Xo E Xl n X2. We assume that the natural map

i*: Hq(X, Yj G)

-+

is one-to-one. This is so if, e.g., X

Hq(C!Xl,X 2}(X, Y)j G)

= int Xl U int X 2 •

As in the commutative case, any continuous map f: (X, Y) -+ (X', Y') induces a map Hq(X', Y'j G) -+ Hq(X, Yj G) for q = 0,1, which takes the distinguished element to the distinguished element. This yields the maps

r:

I·

H I (C!Xl,X2}(X,X2)jG) ~ HI(XI, XI nX2 jG)

121 HI (C!Xl,X2} (X, XO)j G). The map li is one-to-one because there is no essential difference between cochains taking nonzero values at simplices in Xl and X2 and vanishing at all simplices in X2 and those taking nonzero values at simplices in Xl and vanishing at all simplices in Xl n X 2 • For any triple of spaces Z eYe X, we define a map d' HO(y, Zj G) -+ HI (X, Yj G) as follows. Given a cohomology class 0:0 E H (Y, Zj G), we define a cochain cO E CO(X, Yj G) by setting cO(x) = o:O(x) for x E Yand

1. Basie Definitions and Properties

221

= 0 for x f/. Y. Then, we define a co cycle zl E Zl(X, Y; G) by settin! zl(f) = -e°(fct} + e°(fcfi) for any singular simplex f: 6. 1 -+ X. eO(x)

Consider the diagram

(34)

where the maps i~ and j~ are induced by inclusions and 6. is the composition

By the kernel of a map to a set with a distinguished element we mean the preimage of the distinguished element.

Theorem 4.16. (a) Kerii n Keri; = 1m 6.. (b) For a E HI (Xl. xo; G) and b E H I (X2, Xoi G), ji(a) = j2(b) if and only if there exists e E HI(XI UX2, xo; G) for which ii(e) = a and i;(e) = b. Moreover, if HO(XI n X 2 , xo; G) = 0, then such an element e is determined uniquely. Proof. Consider the map 6.' = 1;(ii)- 1 6 (it is obtained from 6. by removing the last isomorphism). It is sufficient to prove the required assertion for the diagram

(35)

(to simplify notation, we do not indicate the coefficient group G). Both sets 1m 6.' and Ker(ii)* n Ker(i~)* consist of equivalence classes of cocycles vanishing at the I-simplices contained in Xl n X 2 • The equality ji(a) = ji(b) is equivalent to the existence of a c' E HI (C!Xl,X2} (X, xo) for which (ii)*(c') = a and (i~)*(c') = b; this is verified directly. The equality HO(XI n X2, xo) = 0 means that the space Xl n X2 is path-connected (provided that G::f:. 0). The uniqueness of c' follows from the path-connectedness ~~n~.

0


226

Theorem 4.16 allows us to obtain the most general version of van Kampen's theorem about the fundamental group of a union of two sets. For this purpose, we need a dual definition of an amalgam of groups. Namely, a group G is the amalga.m of two groups G I and G 2 over a group Go with respect to homomorphisms 'PI: Go --+ G 1 and 'P2: Go --+ G2 if the commutative diagram

,p0----+ Hom(G, G') ~ Hom(GI, G')

,pi

1

_ 1¥?t

Hom(G 2 , G') ~ Hom(Go, G) is exact for any group G'. This definition of an amalgam is equivalent to that given in Part I.

Theorem 4.17 (Dlum [98]). Suppose that spaces UI, U2, and Ul n U2 are path-connected and Xo E UI n U2. The group 7l"l(Ul U U2, xo) is the amalgam of the groups 7l"1(Ul, xo) and 7l"1(U2, xo) over 7l"l(Ul U U2, xo) with respect to the homomorphisms induced by the inclusions UlnU2 c Ul and UlnU2 c U2 if and only if the natural map i*: Hl(UI U U2, Xoj G') --+ HI (C!Ul,U2} (U1 U U2, XO)j G') is one-to-one for any group G'.

Proof. First, suppose that the map i* is one-to-one for any group G'. Then we have the commutative diagram (34) (with G replaced by G'). Moreover, the path-connectedness of Ul n U2 implies HO(UI U U2, Xoj G') = O. Suppose that X is a path-connected space and Xo EX. Take an element wE 7l"1(X, xo) and consider a sequence of I-simplices 6L ... , 6l which form a loop belonging to the homotopy class of w. For zl E ZI(X, Xo; G'), we set zl(w) = zl(6l) + ... + zl(6l). It is easy to show that zl(w) depends only on the homotopy class of the loopj moreover, it depends only on the cohomology class of the cycle zl, i.e., we have a map H1(X,xo;G') --+ Hom(7l"l(X, xo), G'). This map is one-to-one. Applying Theorem 4.16 and replacing HI by Hom, we see that 7l"1(Ul U U2 , xo) is the required amalgam. Now, suppose that 7l"1(Ul U U2 , xo) is such an amalgam. Replacing Hom by HI, we obtain the exact commutative diagram

2. The Poincare and Lefschetz Isomorphisms

227

for any group C ' . Comparing it with the diagram (35) in which Xi is replaced by Ui, we see that the map i" is one-to-one. D

2. The Poincare and Lefschetz Isomorphisms for Topological Manifolds The Poincare isomorphism does hold for topological manifolds, but this cannot be proved by the same method as for smooth manifolds because not all topological manifolds are triangulable. The proofs of the Poincare and Lefschetz isomorphisms given below follow largely [113J and [141J. 2.1. Fundamental Classes. Before constructing fundamental classes for topological manifolds, we prove theorems about the vanishing of the kdimensional homology groups of a topological manifold M n for k > nand of the n-dimensional homology groups of a noncompact topological manifold M n without boundary. The latter is needed to construct fundamental classes. Theorem 4.18. If Mn is a topological manifold, then Hk(M n ) k > n.

o for

Proof. Take a singular cycle Zk E Zk(Mn). The union of the images of singular simplices in Zk is a compact set; therefore, it is contained in the union of finitely many open sets UI, ... , Urn each of which is homeomorphic to an open subset of JRn. We put Vj = U1 U U2 U ... U Uj . Let us show that Hk(Vj) = O. For j = 1, this follows from Theorem 4.8. For j > 1, consider the Mayer Vietoris exact sequence for Vj-l U Uj = Vj: ••• -----t

Hk(Vj-I) ffi Hk(Uj )

-----t

Hk(Vj)

-----t

Hk-I(Vj-1 n Uj )

-----t • • . .

The sets Uj and Vj-l n Uj are homeomorphic to open subsets of JRn; hence Hk(Uj) = 0 and Hk-I(Vj-1 n Uj ) = O. Moreover, Hk(Vj-I) = 0 by the induction hypothesis. Therefore, Hk(Vj) = O. This means that zk is the boundary of a singular chain in Vj C Mn. D For connected noncompact manifolds, Theorem 4.18 can be strengthened; namely, we can guarantee that Hk(M n ) = 0 for k ~ n. To prove this, we need the following auxiliary lemma. Lemma 4.2. Suppose that U C JRn (n ~ 2) is an open set, a E Hn(JRn, U), and for each point x E JRn \ U, we have (ix) .. (a) = 0, where ix: (lRn , U) _ (JRn, lRn \ {x}) is the natural embedding. Then a = O. Proof. Consider the exact sequence of the pair (lRn , U):

.,. __ Hn(JR n )

-----t

Hn(JR n , U) ~ Hn-I(U)

-----t

Hn(JR n ) -----t

., . .


228

Here B. is an isomorphismj therefore, it suffices to prove that B.a = O. Let z E Zn leU) be a cycle representing the homology class B.a. The support of the chain z is compactj hence we can choose an open set V so that V c U, V is compact, and the support of z is contained in V. For this V, Hn-I(V) contains an element {3 such that i.{3 = B.a, where i: V - U is the natural embedding. Using the compactness of V, we choose an open cube Q containing V in JRn and put K = Q \ (Q n V). We have K n V = 0; therefore, for each point x E K, we can choose a closed cube P so that x E P and P n V = 0. The compact set K is covered by finitely many such cubes PI, ... , Pm. By assumption, we have (ix).a

= 0,

so the commutative diagram

shows that the image of (3 under the inclusion homomorphism Hn-I(V) Hn-l (JRn \ {x}) is the zero element. Now, let us show that the image of {3 under the inclusion homomorphism Hn-l (V) - Hn-l (Q \ U:l p,) is zero as well. Consider the sets Qk = Q \ U7-1 Pi (k = 0,1, ... , m). We prove the required assertion by induction on k. For k = 0, it is obvious since Qo = Q and Hn-I(Q) = O. To make the induction step, consider the Mayer Vietoris exact sequence for Q and JRn \ PHI (it is defined because both sets are open): Hn(Q U (JR n \ PHd)

---t

Hn-I(QHd ~ Hn l(Q) EB Hn_I(JR n \ PHd.

Since the set Q U (JRn \ PHd is open, its n-dimensional homology group is trivial. Therefore, j. is a monomorphism. It remains to note that the images of (3 under the homomorphisms induced by the inclusions V C Qk and V C JRn \ Pk+1 are zero. For the former homomorphism, this is so by the induction hypothesis, and for the latter because Hn_l(JRn \ Pk+1) 9:! H n_l(JRn \ {x}). Since the set Qm is contained in U, we can represent the homomorphism Hn-l(V) - Hn-I(U) as the composition of homomorphisms Hn-I(V) Hn-I(Qm) - Hn-I(U). The image of {3 under the first homomorphism is the zero element. Therefore, i.{3 = 0, i.e., B*a = O. 0 The exact sequence of the pair (JRn, JRn \ {O}) implies H k (JRn JRn \ {O}) ~ HdJR n \ {O}) ~ Hk_1(sn-l) for all k > 1. Hence Hn(JRn,JR n \ {al) ~ Z and Hn(JRn, JRn \ {O}j Z2) ~ Z2 for n ~ 2.


229

If U is an open chart homeomorphic to ]Rn and containing a point x E n M , then Hn(M n , M n \ {x}) ~ Hn(U, U\ {x}) ~ Z (or Z2, if the coefficient group is Z2). To prove this, it suffices to apply the excision theorem to the sets M n \ U c M n \ {x} C Mn. (The inclusion Mn \ U c M n \ {x} follows from Mn \ U = Mn \ U and x E U.) The groups Hn(Mn,M n \ {x}) play an important role in constructing fundamental classes.

Theorem 4.19. If Mn is a connected noncompact manifold without boundary, then Hn(Mn) = o. Proof. We show that the homomorphismp. : Hn(Mn)--+Hn(Mn, Mn\{x}) is zero for any x E Mn. Consider a homology class [zn] represented by a cycle Zn E Zn(Mn) with compact support C. First, suppose that x E M n \ C. Then Zn is the image of some cycle zn under the homomorphism i#: Zn(M n \ {x}) --+ Zn(Mn). Therefore, p .. [zn] = p.i.[zn] = 0 because p.i. = O. Now suppose that x E C. By assumption, the manifold Mn is not compact. In particular, Mn =I- C. Therefore, we can choose a point y E Mn \ C. Consider a homeomorphism h: M n --+ Mn which is isotopic to the identity and takes x to y (such a homeomorphism exists for any connected topological manifold without boundary; the proof is similar to that for smooth manifolds (see the lemma on the homogeneity of manifolds in Part I, p. 223). The homeomorphism h induces a homeomorphism M n \ {x} --+ M n \ {y}; hence we have the commutative diagram Hn(Mn)

h.=idl Hn(Mn) The equality (Px).

= 0 follows

~ Hn(Mn,Mn \

1

{x})

h.

~ Hn(Mn,Mn \ {y}). from (Py).

= o.

As in the proof of Theorem 4.18, we consider each singular cycle Zn E Zn(Mn) separately, which allows us to replace Mn by a finite union of open coordinate neighborhoods Ul, ... , Urn homeomorphic to ]Rn. We set Vk = U7=1 Ui· Let us show that the cycle Zn is the boundary of some cycle in Vrn , i.e., Hn(Vrn) = O. We prove this by induction on k. For k = 1, the required assertion is obvious because U1 is contractible. To make the induction step, consider the Mayer Vietoris sequence for the open sets Vk and Uk+!:

Hn(Vk) E9 Hn(Uk+d - - Hn(Vk+d - - Hn-1(Vk n Uk+d - - Hn-1(Vk) E9 Hn-l(Uk+1).


230

By the induction hypothesis, we have Hn(Vk) = O. Moreover, the contractibility of Uk+l implies Hn(Uk+d = 0 and H n - 1(Uk+d = O. Therefore, Hn(Vk+d = 0 if and only if the homomorphism i.: Hn-l(Vk n Uk+d Hn l(Vk) is a monomorphism. Suppose that {3 E H n- 1(Vk

n Uk+d

and i.{3

= o.

Consider the diagram

The second row and the second column in this diagram are segments of exact sequences of pairs. The equality i.{3 = 0 implies {3 = {3" for some (3" E Hn(Vk, Vk n Uk+!). Moreover, {3 = a~{3' for some {3' E Hn(Uk+l' Vk n Uk+d because a~ is an epimorphism. Consider the element (3 = i~{3' - i~{3". Let us write a part of the exact sequence of a pair:

8':

Hn(Vk+d

~ Hn(Vk+l' Vk n Uk+!) ~ Hn-1(Vk n Uk+!).

Clearly, a.i~ = a~ and a.i~ = a~. Hence a.~ = a.i~{3' - a.i~{3" = EY.{3' a~{3" = {3 - (3 = 0, and therefore a.{3 = p.a. for some a. E Hn(Vk+!). As at the beginning of the proof, the noncom pact ness of M n implies that the homomorphism (Px).: Hn(Vk+d - Hn(Mn,M n \ {x}) is zero for each point x E Mn. In particular, p.a. = O. Let x E Uk+! \ (Vk n Uk+l). Then Vk n Uk+! C Mn \ {x}j hence the homomorphism (Px). can be represented as the composition of homomorphisms

Hn(Vk+d ~ Hn(Vk+l, Vk n Uk+!) ~ Hn(Mn,M n \ {x}). Therefore, 0 = (Px).a. = l.p.a. = l.~ = l.i~{3' - l.i~{3". Since x does not belong to Vk, we can replace l.i~ by the composition of homomorphisms

Thus, l.i~ = O. This implies l.i~{3' = o. Under the identification of Hn(M n , M n \ {x}) with H n (Uk+l, Uk+! \ {x}), the homomorphism l.i~ transforms into

(i x ).: Hn(Uk+l' Vk n Uk+l) - Hn(Uk+l, Uk+! \ {'I;}). Applying Lemma 4.2, we obtain {3' = OJ hence {3 =

o.

o

231


Corollary 1. If M n is a connected topological mamfold, then P.: Hn(Mn) --+ H n (Mn , Mn \ {x}) is a monomorphism. Proof. Suppose 0: E Hn(Mn) and P.O: O. Then the group Hn(Mn \ {x}) has an element {3 for which i.{3 - 0:. But the manifold M n \ {x} is not compactj therefore, Hn(Mn \ {x}) O. 0 Corollary 2. If Mn is a connected topological manifold, then either Hn(Mn) - a or Hn(Mn) '" Z and, for any x E Mn, the homomorphismp.: Hn(Mn) --+ Hn(Mn, M n \ {x}) is a monomorphism. Proof. By virtue of Corollary 1, the homomorphism P.: Hn(Mn) --+ Hn(Mn, M n \ {x}) '" Z is a monomorphism. Therefore, either Hn(Mn) = a or Hn(Mn) '" Z and the image of P. is the group mZ C Z. It remains to show that m ± 1. Suppose that m > 1. The proof of Theorem 4.19 (and Corollary 1) is valid not only for the coefficient group Z but also for an arbitrary Abelian coefficient group G. Let G - Zm. Consider the commutative diagram

Hn(Mn) ®Zm ~ Hn(Mn,Mn \ {x}) ®Zm

0, then for the map ri+l on ~~+l we take a map cp for which a :::; 8( cp) < 2a. We set N = U~ 0 Ni and define r: N --+ M n to coincide with ri on each N i . It is seen from the construction that this map is continuous at all points of the set N \ Mn. It remains to show that for each point x E Mn, the map r is continuous at x and, moreover, N contains an open neighborhood of x. Lemma 4.5. For any c > 0, there exists a 8 > 0 such that if x E ~k \ M n and d(x, Mn) < 8, then the diameter of any k-simplex from K containing x is less than c. Proof. Take a positive integer m such that the diameter of any simplex from the mth barycentric subdivision of ~k is less than c. Let Km be the sub complex in K consisting of all simplices from the mth barycentric subdivision of ~k that do not intersect Mn. If Km = 0, then for 8 we take any positive number, e.g., c. If Km =I- 0, then we set 8 = d(IKml, Mn). This number is positive because the sets IKml and M n are compact and disjoint. Suppose that d(x, Mn) < 8. Then x tJ. IKml; therefore, the k-simplex in the mth barycentric subdivision of ~k that contains x must intersect Mn. By construction, such a simplex undergoes the barycentric subdivision. 7Since Mn C y E Mn for which

]Rk

is compact it follows that for any x E]Rk there exists at least one point = d(x, un). Such poi~ts y are said to be nearest to x in Mn.

IIx - yll


241

Thus, the k-simplex from K containing x is obtained not earlier than at the (m + l)th step. The diameter of any such simplex is less than c. D For each point x E Mn c ]Rk and every c > 0, we construct a system of open neighborhoods Va ::J Vi ::J ... ::J Vk+1 in ]Rk so that Vk+1 c Nand r(Vk+d C D! e' where D! e = {y E ]Rk I IIx - yll < c}. For Va we take the open subset ~f D! e containing x and such that Vo n M n is an admissible subset homeomor~hic to ]Rn. We assume that Va c ~ k • For i ~ 1, the neighborhood Vi is constructed from V; 1 as follows. First, we choose a number c(i) > 0 so that D!,5e(,) eVil. Then, using Lemma 4.5, we choose ~(i) < c(i) such that if y E ~k \ Mn and IIx yll < ~(i), then the diameter of any k-simplex from K containing y is less than c(i). Finally, we choose an open set Vi C ~k in ]Rk such that it is contained in D!,6(,) (and contains x) and Vi n M n is an admissible set homeomorphic to ]Rn. Suppose that y E Vk+1 \ Mn and ~k(y) is the k-simplex from K containing y. By construction, all vertices of ~k(y) belong to D!,26(k+1) C Vk. If v is a vertex of ~k(y), then Ilv - r(v)11 ~ Ilv - xii; therefore, the images of all vertices of ~k(y) under the map r belong to D!,46(k+1) C Vk. Since the set Vk n Mn is homeomorphic to ]Rn, we can extend the map r, which is defined on the vertex set of the simplex ~k(y), to the edges ~~. Hence each ~~ is contained in N. There is an extension of r to ~~ for which the diameter of the image of ~~ is less than 2~(k + 1); therefore, the diameter of r(~~) is less than 4~(k + 1) < ~(k). Since the endvertices of the edge ~~ belong to Vk, it follows that r(~~) C Vk-l n Mn. Applying the same argument to the 2-skeleton, 3-skeleton, and so on, we obtain ~k(y) eN and r(~k(y» eva = D!,e. In particular, r(y) E D;,e. This means that the map r is continuous at the point y. Moreover, Vk+1 C N. D Lemma 4.6. For any closed topological manifold M n , there exists an open neighborhood W of d(Mn) in Mn x M n for which the maps pIlw and P21w are homotopic.

Proof. According to Theorem 4.23, we can assume that Mn is embedded in ]Rk for some k, and, moreover, Mn is a retract of its neighborhood U in ]Rk. Since M n is compact, we can choose c > 0 such that if the distance between points x, y E Mn in]Rk be less than c, then the line segment joining x and y is contained in U. The maps PI and P2 coincide on the set d(Mn). Therefore, using the compactness of Mn, we can choose a neighborhood W of d(Mn) in M n x M n so that for each w E W, the distance between PI ( w) and P2 (w) be less than c. The segment with endpoints PI (w) and P2 (w) is contained in U; therefore,


242

the maps PI Iwand P21 ware homotopic in U. We construct a homotopy in Mn from this homotopy in U by using a retraction r: U ---+ Mn. D Consider the map T: M n x M n ---+ M n x Mn defined by T(x, y) This map induces a map of pairs T: Mnx ---+ Mnx.

= (y, x).

Lemma 4.7. If a E H"'(Mnx), then T"'a - ( 1)na . Proof. Let V be an admissible open subset of M n homeomorphic to JRn . Consider the commutative diagram

Here i'" is the homomorphism induced by the inclusion of pairs (V x V, V x V \ d(V)) c Mnx. It is easy to see that Hn(v x V, V x V \ d(V)) '" z. Indeed, the exact sequence of a pair implies that Hn (V x V, V x V \ d(V)) '" H n I(V x V\d(V)). Moreover, V x V\d(V) ~ JRn x (JRn \ {O}) '" JRn \ {OJ '" sn I. Therefore, Hn(v x V, V x V \ d(V)) ~ H n 1(sn-l) ~ Z. This group is generated by (3 - i"'(U), where i'" is the homomorphism induced by the inclusion V c Mn. Before proving the equality T"'{3 = (-l)n{3, we show that the pair (V x V, V x V \ d(V)) is homotopy equivalent to (JRn x JRn, JRn x (JRn \ {O} )) '" (JRn,JRn \ {OJ). The homotopy equivalence is constructed as follows. We identify V with JRn and, in each layer {Yo} x JRn, move the point (Yo, Yo), which belongs to d(V), to the point (YO,O) in such a way that the layer remains invariant. Moreover, the homotopy ("shift") can be constructed so that (Yo, 0) moves to (Yo, -Yo). To prove the equality T'" (3 = (-1) n {3, it suffices to track the action of T on one layer, say, Yo - o. The map T takes (0, xo) to (.co, 0). The homotopy shifts it to (xo, -xo). Projecting this point onto the initial layer, we obtain (0, xo). Therefore, the homomorphism T"': Hn(JRn, JRn \ {O}) ---+ Hn(JR n , JRn \ {O}) is induced by the map xo ......... -xo. The degree of this map equals (-l)n. Let W be the open neighborhood of d(Mn) mentioned in Lemma 4.6. Replacing W by TW n W if necessary, we can assume that the neighborhood W is invariant with respect to T. We set U = M n x M n \ W. Then U c M n x Mn \ d(Mn); therefore, we can apply the excision theorem and obtain an isomorphism H"'(W, W \ d(Mn)) ~ H"'(Mnx). This allows us to represent the Thorn isomorphism cp'" as the composition

H*(Mn)

(P1Iw)",

H*(W) ~ H*(W, W \ d(Mn)) ~ H*(Mnx),


243

where U' is the restriction of the Thorn class to W. Let a E H"'(Mnx). Using the equality PIT = P2, we obtain T"'((pIlw)"'(a) '--' U') - (P2 w)*(a)

But (PIlw) '" (p2Iw); hence T"'<J»"'(a) (-l)na because <J»'" is an isomorphism.

l)n<J»"'(a), and therefore T"'a

o

We are almost ready to prove the Poincare isomorphism theorem. It only remains to verify that the homology groups of a closed topological manifold are finitely generated. Theorem 4.24. If Mn is a closed topological manifold, then the group H",(M n ) is finitely generated. Proof. We assume the manifold Mn to be embedded in Rk. Consider an open set U ::J M n in Rk for which M n is a retract. Let l:!..k be a simplex containing U in Rk. The distance d between the compact sets M n and l:!..k\U is positive; hence there exists an m such that the diameter of any simplex in the mth barycentric subdivision of l:!..k is less than d. Consider the simplicial complex K consisting of the simplices from the mth barycentric subdivision of l:!..k that intersect Mn. Clearly, Mn c IKI c U. Let r: IK I ---+ Mn be the restriction of a retraction U ---+ Mn. The composition M n ~ IKI ..!:... M n is the identity map; therefore, r",: H",(IKI) ---+ H",(M n ) is an epimorphism. Since the group H .. (IKI) is finitely generated, it follows that so is the group H",(Mn). 0 Theorem 4.25 (the Poincare isomorphism). Let Mn be a closed orientable topological manifold with an orientation s, and let [Mn] E Hn(Mn) be its fundamental class corresponding to s. Then the map D: Hk(Mn) ---+ Hn_k(Mn) defined by D(a k ) = a k ----- [Mn] is an isomorphism for any k. Proof. Let a' E Cn(M n ) be a cycle representing the fundamental class [Mn]. Consider the map D#: ck(Mn) ---+ Cn_k(M n ) defined by D#a' = a' ----- a', The equality aa' = 0 implies aD# = (_1)n-kD#6 because a(D#a') = a(a' ----- a') = (-1)n- k 6a' ----- a'+a' ----- aa' = (-1)n- k 6a' ----- 0/ = (-l)n- k D#(c5a'). Hence the map D# takes co cycles to cycles and, therefore, induces a homomorphism of (co)homology groups, which coincides with D. First, we shall prove the Poincare isomorphism theorem for the case of coefficients in Zp, and then reduce the proof for the integral (co)homology to this case. So, instead of (co)homology with coefficients in Z, we shall consider (co)homology with coefficients in an arbitrary commutative ring R with identity.


244

For homology classes a,13 E H*(M n ; R), let a x 13 denote the image of a®13 in H*(M n x Mn; R) under the monomorphism from the Kiinneth theorem; for cohomology classes a, b E H*(Mn; R), a x b denotes the element pi(a) p;(b) E H*(Mn x Mn; R). As above, U E Hn(Mnx) is the Thorn class. The notation U - i*(U) is used for the element of the group Hn(Mn x Mn) that corresponds to U under the homomorphism i*: Hn(Mnx) Hn(Mn X Mn). Under a ring homomorphism Z _ R taking 1 to the identity element of R, the element fl transforms into an element of the ring Hn(Mn x Mn; R), which we denote by the same symbol fl. Finally, for an arbitrary point x E M n , [x] denotes the generator of Ho(Mn) determined by x. The element [x] x [Mn] is represented in Hn(Mn x Mn) by the cycle (lx)*([M n]); therefore, the homomorphism i*: Hn(Mn X Mn) _ ~¥nx) takes [x] x [Mn] to (lx)*(s(x)). Thus, -...J

(fl, [x] x [Mn]) - (i*(U), [x] x [Mn])

= (U, (lx)*(s(x))) = 1.

Since the classes [x] x [Mn] and [Mn] x [x] are obtained from each other by means of T*, it follows that [Mn] x [x] = (-l)n[x] x [Mn]; hence (36) Let us prove one more formula: If aP E HP(M n ; R) and bq E Hq(M n ; R), then (37) it is assumed that we have

(-l)nu

-...J

UE

Hn(Mn x Mn; R). Indeed, according to Lemma 4.7,

(a P x bq) = T*(U

-...J

= T*(U)

(a P x bq)) T*(a P x bq) = (-l)n+pqu '-' (b q x aP ).

-...J

This equality for U implies equality (37) for fl. The main purpose of our calculations is to express

(U, D(a k ) x 13k) = (fl, (a k r--- [Mn]) x 13k) in terms of (ak, 13k). To do this, we need, in addition to (36) and (37), expressions for (a k x 1) r--- ([Mn] x 13k) and (1 x a k ) r--- ([Mn] x 13k)' By definition, we have a k x 1 = pi(a k ) '-' P2(1) = pi(a k ) '-' 1 and 1 x ak = 1 '-' p;(a k ). Therefore,

(a k x 1) and

r---

([Mn] x 13k)

= pi(ak)

r---

([Mn] x fJk)


245

Let [mo, ... , mnl and [b o, ... , bkl be simplices contained in representatives of the homology classes [Mnl and 13k. Then [mo, ... , mnl x [b o, ... , bkl is the sum of simplices [vo, ... , vn+k], where Vi = (mM(i) , bB(i») and either (i) Vi+! = (mM(i)+!' bB(i») or (ii) v~+l = (mM(I), bB(i)+l). By definition,

pt(a k ) ,....., [vo, . .. , vn+kl = (pHa k ), [v n , ... ,vn+kl)[vo, ... , vnl = (a k , pt[vn, ... ,Vn+kl) [va, ... ,vnl = (a k , [mM(n) , ... , mM(n+kll) [vo, ... , vnl

and

p;(a k ),....., [vo, ... ,vn+kl- (ak,[bB(n), ... ,bB(n+kll)[vo, ... ,vnl. Nonzero expressions are obtained only if the simplices [mM(n) , ... ,mM(n+kll and [bB(n) , ... , bB(n+k)l are nondegenerate. In case (ii), the simplex [vo, ... , Vn+kl is uniquely determined by this condition; therefore, p;(ak) ,....., ([Mnl x 13k) = (a k , 13k} [Mnl x [bolo We can take [xl instead of [bol because all points of the connected manifold Mn determine the same zero-dimensional homology class. In case (i), only the simplex [v n , ... , vn+kl is determined uniquely. The only conditions on the simplex [vo, ... , vn+kl are Vo = (mo, bo) and Vn = (mn-k, bk ). The sum of all such simplices is equal to [mo, ... , mn-kl x [bo, .. . , bkl. Therefore, PiCak) ,....., ([Mnl x 13k) = (ak ,-... ([Mn]) x 13k. We obtain

(U, (a k ,....., [Mn]) x 13k) = (U, (a k x 1) ,....., ([Mnl x 13k)

= (U '-' (a k x 1), [Mnl x 13k) (~) (U '-' (1 x ak ), [Mnl x 13k} = (U, (1 x ak ) ,....., ([Mnl x 13k)} = (U, (a k , 13k) [Mnl x [xl) (~) (_1)n(a k , 13k), i.e., (38)

(U, D(a k ) x 13k}

=

(-1)n(a k ,13k).

Relation (38) leads us to conclude that if a k -I- 0, then D(a k ) -I- o. Hence, the maps Hk(Mn; R) ~ Hn_k(Mn; R) and Hn-k(Mn; R)~Hk(Mn; R) are monomorphisms. If R is a field, then Hk(Mn; R) is the linear space over R dual to Hk(M n ; R). These spaces are finite-dimensional by Theorem 4.24; hence they are isomorphic to the same linear space Vk • Thus, we have two monomorphisms, Vk -+ Vn- k and Vn-k -+ Vk, of finite-dimensional linear spaces; clearly, they must be isomorphisms. Thus, we have proved the Poincare isomorphism theorem for (co ) homology with coefficients in Zp, where p is a prime.


246

Let us prove it for the integral (co )homology. To this end, we construct a chain complex with chain groups C n k - cHI(Mn) ffi Cn_k(M n ), i.e., Ck cn HI(Mn) ffi Ck(Mn), and define the boundary homomorphism 0: Ck - t Ck I by o(an k+l,f3k) = ((-1)k6a n k+l,of3k + D#(a n HI)). It is easy to see that 00

o.

Indeed,

oo(an k+l,f3k) - (-66a n HI, O(of3k

+ D#(a n

k+l))

+(

l)k D#(6a n HI)) - 0,

because oD#(an HI) _ ( 1)HID#6(an k+l). We have a short exact sequence 0---+ Ck(M n ) ---+ C n k+l(Mn) ffi Ck(M n )

---+

C n HI(Mnr't:~

o.

Such short exact sequences of groups form an exact sequence of chain complexes. The map C* - t C*(Mn) is chain only up to sign, but this is sufficient for constructing an exact homology sequence for chain complexes. Let us show that the connecting homomorphism Hn-k+I(Mn) - t Hk I(M n ) in this exact sequence coincides with D. Suppose that a E cn k+l(Mn) and 6a - O. Then (a, 0) E Ck belongs to the preimage of a. Let o(a, 0) = (0, D#(a)). The preimage ofthis element in Ck(M n ) is equal to D#(a). The map a 1-+ D#(a) is the required connecting homomorphism (at the level of chains). Applying the same construction to the coefficient group Zp, we obtain the exact sequence ... ---+

Hn-k(Mn; Zp)

J!...... Hk(M n ; Zp) ---+

Hk(C*; Zp)

---+

H n k+l(Mn; Zp)

J!...... ... ,

in which the homomorphisms D are isomorphisms. Therefore, H*(C*; Zp) O. Since all groups H*(C*) are finitely generated, the universal coefficient theorem shows that H*(C*) = o. Thus, D is an isomorphism for integer coefficients too. 0 Any closed connected topological manifold is orientable with respect to the coefficient group Z2; therefore, a similar (but somewhat simpler) argument shows that the map D: Hk(Mn; Z2) - t Hn_k(M n ; Z2) defined by D(a k ) = a k ,....., [Mnh is an isomorphism. 2.4. The Lefschetz Isomorphism. For topological manifolds, the theorem on the Lefschetz isomorphism is derived from the Poincare isomorphism theorem in the same way as for smooth manifolds. However, the reduction requires some effort, for it uses the collar theorem, whose proof for topological manifolds is more complicated than in the smooth case. Moreover, we

247

2. The Poincare and LeEschetz Isomorphisms

have to construct the fundamental class of an oriented topological manifold with boundary and prove its properties. The collar theorem for topological manifolds was proved by Brown [20]; we follow the exposition of [24]. Theorem 4.26 (on a collar). Any compact topological manifold M n with boundary B 8M n has an open subset U whzch contains B and is homeomorphic to B x [0,1). Moreover, the homeomorphism h: U - B x [0,1) can be chosen so that h(x) (x, 0) for all x E B. Proof. For each point x E B, the boundary B contains an open subset U:r; for which there exists an embedding h:r;: U:r; x [0, 1] - M n such that h:r;(Y, 0) - Y for all y E U:r;. Since B is compact, the cover {U:r;} has a finite subcover {U:r;l' ... ' U:r;m}. For short, we denote its elements by UI , ... , Um and the corresponding embeddings by hI, ... , h m . Being Hausdorff and compact, the topological space B is normal. Therefore, the cover {Ud has an open refinement {\Ii} such that Vi C U,. The main idea of the proof of the collar theorem is to consider the topological space M+ obtained by attaching B x [-1, 0] to Mn via the map cp: B x {a} - M n that takes each point (x, 0) to x E B c Mn. Then, the required homeomorphism from M n on M+ is constructed by induction on m involving the "local collars" hi: U i X [0,1] _ Mn. Consider the embeddings Hi: U i

H.(x t) "

=

X

{h,(X' t) (x, t)

[-1, 1] - M+ defined by if t E [0,1], ift E [-1, 0].

These embeddings are well defined because hi(x, 0) = (x, 0) on U i x {a}. We use them to construct functions Ii: B - [-1, 0] and embeddings gi: M n M+ (i - 0, 1, ... ,m) with the following properties:

x

E

(a) f,(x) B);

=

-1 for x E U~=I Vj (in particular, Im(x)

=

-1 for all

= (x, li(x» for x E B; (c) g,(M n ) = Mn U {(x, t) I t 2: I,(x)}. (b) g,(x)

If we constructed such maps, we would obtain the required collar because gm must be a homeomorphism from M n to M+.

We construct them by induction on i. We set go = idMn and 10 == o. Suppose that Ii 1 and g, 1 are already constructed. Let us construct Ii and gi. Consider the set Hi l(g, I(Mn» C U i x[-I,I]. Sincegi_I(Mn) contains Mn, we have H,-1(gi_1(Mn» ::J U i x [0,1]; this is shown schematically in Figure 5. Let us construct an embedding 'Pi: Hi-1 (gi_l(M n » - U. x [-1, 1] by blowing up the fibers over x E B until the fiber over each point x E::: Vi


248

u,

coincides with the entire interval [-1,1]; formally, this can be written as 'P~(H~ l(y,_l(Vi ))) = Ui x [ 1,0]. Moreover, we want the maPt~.. to be the identity on (U, \ U,) x [-1,1] and on Ui x {O}. To construct such an embedding, we apply the Urysohn lemma (see Part I, Theorem 3.6). The closed subsets Ui \ Ui and V, of the normal space B are disjoint. Hence there exists a function Ai: U, --+ [0,1] taking the value on U i \Ui and 1 on Vi. Let Lz be the affine map from the interval [fi-l (x), 1] onto [( 1 - Ai (x)) fi-l (x) + Ai(X)( -1),1]. For x E V" we obtain the interval [-1,1]' and for x E Ui \ Ui, we obtain [f,-l(x), 1]; thus, in the latter case, Lz is the identity map. For (x, t) E Hi-l (Yi-l (Mn)), we set 'P,(x, t) = (x, Lx(t)); the map 'Pi thus defined is continuous. Next, we define the map 4>,: Yi(Mn) --+ M+ by

°

4>,(x) = {H''P,H,-I(X) x

if x E y, I(Mn ) n H,(Ui x [-1,1]), for all other x E Yi(Mn).

Finally, we set Y~ = 4>,Yi-l. The map 4>i (and hence y,) is well defined because 'Pi is the identity map on (U, \ U,) x [-1,1] and Ui x {o}. The map 4>i (and hence y,) is an embedding because each map Lx is one-to-one and, therefore, 'Pi is an embedding. Moreover, we have y, l(M n ) n H,(Ui x [-1,1]) - H,(Ui

X

[0,1]) U {(x, t) I t

~

Ii-l(x), x E Ui }

because (c) holds for Yi-l. The function flex) is determined by (b). Conditions (a) and (c) hold by construction. 0 Let Mn be a compact topological manifold with boundary alvIn. Consider the closed topological manifold Mn obtained from two copies of M n by identifying the respective points of their boundaries. For all x E M n \ aMn, the groups Tx are defined (see p. 231). They determine a covering T --+ M n \ aM n . The manifold M n is said to be orientable if this covering has a section s such that sex) generates the group Tx for every x E M n \ aMn. The section s is then called an orientation. It is easy to verify that a manifold Mn is orient able if and only if Mn


249

is orient able. Moreover, if Mn is orientable, then so is aM n (i.e., each connected component of aM n is orientable). Let ax be a generator of the group Tx = Hn(M n , Mn\{x}). A homology class a E Hn(Mn, aM n ) is called fundamental if the map p.: Hn(M n , aMn ) --+ Hn (Mn, M n \ {x}) = Tx takes a to ax for all x E M n \ aMn.

Theorem 4.27. Let Mn be a compact connected orientable topological manifold with boundary aMn, and let s be an orientation of Mn. Then there exists a unique fundamental class a E Hn(Mn, aMn) such that ax = sex) for all x E Mn \ aMn. Also, the connecting homomorphism a.: Hn(Mn, aM n ) --+ Hn_l(aM n ) takes a to a fundamental class of the closed manifold aM n , i.e., a.(a) determines a fundamental class on each connected component of the manifold aMn. Proof. The orientation s of lof n determines an orientation s of jjn. Since the manifold jjn is closed and orient able, there exists a unique fundamental class a E Hn(jjn) such that (px).(a) = sex) for all x E Mn. We shall construct the class a from o. For this purpose, we need an isomorphism

We might have obtained such an isomorphism by excising (Mn)' \ aMn, where (Mn)' is the second copy of Mn. But we cannot do this because the condition

does not hold. Instead, we construct the required isomorphism using the collar theorem; namely, we cut off (Mn)' \ aM n from the union of (Mn)' and a collar of M n rather than from (Mn)'. To do this, some obvious homotopy equivalences are needed. We define the class a as the image of a under the composition of homomorphisms

Obviously, ax = sex). Let x E aM n . Take a closed ball nn-l centered at x in some chart of the manifold aM n and consider the subset E

= {(x, t) I x

E nn-l, t E [0,1/2]}

of the open set U = aM n x [0,1) mentioned in the statement of the collar theorem. We put ME = Mn \ int E. We have the following commutative


250

diagram: I.

H n (M n ,8M n )

la.

, I.

Hn 1(8Mn)

1

) Hn l(M'E) (

{x})~Hn l(M'E,M'E

k

H n (E,8E)

~ lao

la.

1

p ).

Hn 1(8Mn,8Mn

) Hn(Mn,M'E) (

k~

Hn 1(8E)

~l {x})~Hn 1(8E,8E {x}).

The horizontal isomorphisms in this diagram arise from the excision theorem (in some cases, excision should be performed with care, by using obvious homotopy equivalences). ~

Take y E intE. We represent the homomorphibm (py).: Hn(Mn,Mn \ {y}) as the composition

H n (M n ,8Mn ) ~ Hn(Mn,M'E) _

Hn~,

8Mn)

Hn(Mn,M n \ {y}).

The element (py).(a) s(y) generates the group Hn(M n , M n \ {y}) ~ Zj hence i.(a) generates the group Hn(Mn, M'E) ~ Z. Therefore, the group Hn(E, 8E) ~ Z has a generator a' for which k.(a') - i.(a). The commutativity of the diagram implies k~8.(a') = i~8.(a). Therefore, the images of 8. (a) and 8.(a') in the group Hn l(M'E,M'E \ {x}) ~ Z coincide. Moreover, the image of 8.(a') is a generatorj hence so is the image of 8.(0:). Thus, the element (px).8.(a) generates the group Hn 1(8Mn ,8Mn \ {x}) ~ Z. This means that 8.(a) is a fundamental class of the manifold 8 M n . Now it is easy to prove the remaining part of the theorem, which concerns the uniqueness of the fundamental class a. Suppose that al and a2 are two fundamental classes in Hn(M n , 8Mn) corresponding to the same orientation. Then 8.(aI) and 8.(a2) are two fundamental classes in H n _ 1 (8M n ) corresponding to the same orientation. Therefore, al - a2 E Ker 8 •. The collar theorem implies that the spaces M n and M n \8Mn are homotopy equivalent to the same space M n \ U, where U = 8M n x [0,1). Hence Hn(Mn) ~ Hn(Mn 8Mn) - o. Therefore, the left term in the exact sequence Hn(Mn) _ Hn(M n , 8Mn) ~ Hn 1(8Mn ) is trivial, and Ker 8.

= o.

o

The same argument as the one used for smooth manifolds (in the triangulable case) proves the following theorem.

Theorem 4.28 (the Lefschetz isomorphism). Let M n be a compact orientable topological manifold with boundary 8M n , and [Mn] E Hn(M n , 8Mn)

2. The Poincare and Le£,>chetz Isomorphisms

251

the fundamental class of Mn. Then the homomorphisms D: Hk(Mn,aMn) Hn k(M n ) and D: Hk(Mn) --+ Hn k(Mn, aM n ) that take each class to its cap product with [Mnj are isomorphisms.

--+

In the nonorientable case, the Lefschetz isomorphism theorem is valid for homology and cohomology with coefficients in Z2. Problem 118 ([113]). Let Mn be a compact orient able topological manifold with boundary aMn. Suppose that aM n is partitioned into disjoint sets VI and V2, each consisting of several connected components of the manifold aMn. Prove that the homomorphism Hk(Mn, VI) --+ Hn k(Mn, V2) taking each class to its cap product with [Mnj is an isomorphism. 2.5. A Generalization of Helly's Theorem. Results obtained in Section 2.1 (namely, Theorems 4.18 and 4.19) make it possible to carryover the generalized Helly theorem (Theorem 4.10 on p. 209) to an arbitrary manifold. Theorem 4.29 (Debrunner [29]). Suppose that Mn is a topological nmanifold, and XI. ... , Xm are open subsets of Mn such that the intersection of any r < n + 1 of these sets is nonempty and has trivial reduced homology in all dimensions up to n - r. (a) If M n is not a homology n-sphere, then the space Xl n ... n Xm is acyclic (and, in particular, nonempty). (b) If M n is a homology n-sphere and no n + 2 of the sets Xl, ... , Xm cover Mn, then the space Xl n··· n Xm is acyclic. Proof. Suppose that the assertion of the theorem does not hold for sets X I, ... , X m , and let m be the minimum number of such sets. The minimality of m implies that the intersection of any r ~ m - 1 of the sets Xl, ... , Xm is acyclic. Therefore, Lemma 4.1 applies (see p. 208). First, suppose that Xl n ... n Xm = 0. By assumption, this is possible only if m ~ n + 2. Let Y = Xl U··· U X m . According to Lemma 4.1(a), the group Hq(Y) - 0 is nontrivial precisely when q = m - 2. On the other hand, by Theorem 4.18, we have Hq(Y) = 0 for q > nj therefore, m-2 ~ n. Comparing this with the inequality m ~ n + 2, we obtain n = m - 2. Thus, Hn(Y) i= OJ this is possible only when Y = Mn. (Indeed, if Y i= Mn, then Y is a noncom pact n-manifold without boundary, and Theorem 4.10 implies Hn(Y) = 0.) Consequently, the group Hq(Mn) is nontrivial precisely when q = n, Le., k[n is a homology sphere. This contradicts the assumption that if M n is a homology sphere, then Y i= Mn. Now suppose that the intersection Xl n··· n Xm is nonempty (but not acyclic). Let p ~ 0 be the minimum number for which Hp(X1n.· ·nXm ) i= o.

252


We set q = p+m -1. According to Lemma 4.1, (b), Hq(XI U··· UXm) i- O. Applying Theorem 4.10 to the n-manifold Xl U ... U X m , we obtain q ~ n. Moreover, q - m + 1 = p ~ 0; hence m ~ q + 1 < n + 1. By assumption, the space Xl U··· U Xm has trivial reduced homology in all dimensions up to n - m. Therefore, p > n - m, i.e., q - m + 1 > n - m. Comparing this inequality with q + 1 ~ n + 1, we obtain q = n. The minimality of p (and q) implies that XIU" ,UXm is ahomologyn-sphere. Thus, M n - XIU·· ·UXm , which contradicts the assumption. D Problem 119. GIVe an example of n + 2 open subsets of sn such that the intersection of any r ~ n + 1 of them is a (nonempty) contractible set and the union of these sets coincides with sn (in particular, is not acyclic).

3. Characteristic Classes: Continuation In proving many properties of characteristic classes, it is convenient to use singular homology and cohomology. Most of these properties are related to the Thorn isomorphism, but we start by proving Theorem 3.43 formulated on p. 170. In the proof, we use the notation introduced in and before the formulation of this theorem. Proof of Theorem 3.43. It is sufficient to verify that Xi({) satisfies all of the conditions that must be satisfied by the characteristic classes. The relations xo({) = 1 and Xi({) = 0 for i > dim{ are included in the definition. Let us show that Xi(g*({)) = g*(Xi({)) for an arbitrary map g: BI --+ B. The fibers of the bundles g*(~) and { over the points bl and g(b l ) are canonically identified; therefore, the fibers of the bundles Pg*(~) and P{ over these points are canonically identified as well. Thus, a bundle map g: E(Pg*~) --+ E(P~) is defined. The diagram

is commutative; therefore, the bundle g(>,~) is isomorphic to the bundle Ag.~, and g(ad = ag.~ in the ring H*(E(Pg*~)). By definition, we have n _ ",n-l ( 1)i+1 (*c) n-i ag.~ - L...i=l Xi 9 .. ag.~ an d

253

3. Characteristic Classes: Continuation

the last equality holds because g*(.8a~) = g*(,8)g*(a~) for any ,8 E H*(B). The uniqueness of the decompositions implies Xi (g* €) = g* (Xi (€)), as required. We proceed to the last (and most difficult to prove) condition, namely, that the Whitney formula Xk(€ EEl TJ) = I:~+i k x~(€)xi(TJ) must hold. (The Whitney formula can be written differently, but here it is more convenient to prove it for bundles over the same base.) The linear space Vi EEl \12 contains the subspaces VI EEl {O} and {O} EEl V2, which are isomorphic to Vi and \12. Therefore, the topological space E(P(€EElTJ)) contains the canonically defined subspaces E(P€) and E(PTJ). These subspaces are deformation retracts of U = E(P(€ EEl TJ)) \ E(P€) and V E(P(€ EEl TJ)) \ E(PTJ)' respectively. In the cohomology ring of E(P(€ EEl TJ)), take the elements m

01

=

L ( l)ixi(€)ym'

n

and

i=O

02

=

L ( l)i xi (TJ)yn

i,

3 0

where y = Xl(>'(E!)Tj), m = dim~, and n = dimTJ. To simplify notation, we set E = E(P(~ EEl TJ)). Consider the cohomology sequence of the pair (E, V): ... - +

Hm(E, V)

L

Hm(E) ~ Hm(v)

- + ., ..

The map i* implements the restriction of the cohomology classes to V. Since E(P€) is a deformation retract of V, we can consider restriction to E(P€) rather than to V. Let us show that the restriction of the class 01 to E(P~) is I::"o (-l)ixi(€)ar- i = O. Indeed, suppose that A( = j*('l)· The restriction of the class y = Xl (A(E!)Tj) to E(P~) is Xl(A() = X1U*'Yl) = j*Xl('Y1) = j*(o) = a( (we use some of the properties already proved, namely, naturality and the relation Xl ("tl) = 0). Since the sequence of a pair is exact, there exists an element o~ E Hm(E, V) for which p*o~ = 01. Similarly, there exists an a~ E Hm(E, U) for which p*o~ = a2. The element o~ '-" o~ belongs to H*(E, U U V) = 0; hence o~ '--' a~ = O. Therefore, 01 '--' a2 = (p*aD '--' (P*o~) = O. As a result, we obtain

(t (-l)iXi(~)ym-i) (t ,-0

(-l)i Xi (1J)yn- i ) =

o.

3=0

For the Stiefel Whitney classes, we consider cohomology with coefficients in Z2, in which multiplication is commutative (rather than anticommutative). For the Chern classes, Xi and y have even dimension; thus, for these classes, multiplication is commutative also. Therefore, we can rewrite the obtained equality in the form


254

On the other hand, there exist unique elements Xk(' E9 TJ) for which m+n

L

(-l)k xk (e E917)ym+n k -

o.

k 0

Thus, Xk(e E917)

= L'+i

D

k x,(e) ........ x J (TJ), as required.

3.1. The Thorn Isomorphism for Bundles. Every smooth manifold has a tangent bundle, whereas topological manifolds have no tangent bundles. For this reason, Milnor [91] introduced the notion of a microbundle, which generalizes that of a vector bundle. Each topological manifold has a tangent microbundle; tangent micro bundles mimic tangent bundles of smooth manifolds in many respects. For example, the proof of the Poincare isomorphism for topological manifolds given above essentially uses microbund~

A micro bundle of dimension n is a map p: E properties:

(i) there exists a map i: B

--+

B with the following

E for which pi - idB;

(ii) the map p is locally trivial in the sense that for each point b E B, there exist open neighborhoods U 3 b and V 3 i(b) and a homeomorphism hb: U x IRn --+ Vnp leU) such that phb(u,v) = u for all (u,v) E U x IRn and hb(U, v) = i(u) for all u E U. Example 57. A vector bundle p: E section i(b) - (b,O) can be taken.

--+

B is a microbundle. For i the zero

Example 58. Let M be a topological manifold without boundary. Then the projection p: M x M --+ M onto the first factor is a microbundle. For i the diagonal map d: M --+ M x M can be taken. The proof that the map from Example 58 is indeed a micro bundle is essentially contained in the proof of Lemma 4.3 on p. 235. This microbundle is called the tangent micro bundle of the topological manifold M. The tangent microbundle of a topological manifold M with boundary Let us attach the direct product aM x [0,1) to the boundary of M. The collar theorem implies that the result is a topological manifold M'. This manifold has no boundary. The tangent microbundle of M is the restriction to M x M of the tangent microbundle of M'. Microbundles p: E --+ B are usually considered together with given maps i:B--+E. Two micro bundles p: E --+ Band p': E' --+ B (over the same base B) with fixed maps i: B --+ E and i': B --+ E' are said to be equivalent if the sets i(B) and i'(B) have open neighborhoods U and U' in E and E'

aM can be defined as follows.

25E

3. Characteristic Cla.. n = dim{, then SqiUe = o. Let us verify naturality. Suppose that f: B' ---. B is a map, { is a bundle over B, f*({) is its pullback over B', and g: E ---. E' is the fiberwise map of bundles induced by f. Then g*(U{) = UrCe), and we have the following commutative diagram for the Thorn isomorphisms: Hi(B) ~ Hi+n(E, Eo)

1r,

19·

Hl(B') ~ Hi+n(E', Eb).

9This means that the orientations of all fibers S .. -1 are compatible.


258

Thus, ep'f*

g*ep, i.e., f*ep

r(Wt(~))

1

= (ep')-lg*. Therefore,

rep l(SqtU~) - (ep') 19*(Sqtu~) - (ep') lSqi(g*(Ue)) - (ep') lSqtUr(e) = Wt(f*(~)).

Let us show that the Whitney formula holds. This time, it is more convenient to prove it in the form w(~ x "I) - w(~) x w(rJ). Consider the cross product of the Thorn classes Ue E Hm(E, Eo) and UTI E Hn(E', Eo). The class Ue x UTI belongs to the group Hn+m(E x E', (E x Eo) U (Eo x E')). The set (E x Eo) U (Eo x E') consists precisely of the nonzero vectors in E x E'. Therefore, the class Ue x UTI belongs to the same group as the Thorn class U~XT/' Let us show that these two cohomology classes coincide. To this end, it suffices to verify that the restriction of Ue x UTI to (IR n +m, lRo+m) , where IR n +m is the fiber of the bundle ~ x "I, is a nonzero cohomology chtsfiJ'or the fiber of the bundle ~ x "lover each point. The class undeuonsideration equals the cross product of the restrictions of the classes Ue and UTI to (IRm,IRW) and (IRn,IR~); according to Theorem 4.15 on p. 217, this product is nonzero. To prove the Whitney formula, we must also verify that the Thorn isomorphisms of the bundles ~, "I, and ~ x "I are related by ep~ (a) x epT/ (/3) = ep~XT/(a X /3), i.e.,

(Pea '-' U~) x (P~/3 '-' U~) = (PeXT/ (a x (3)) '-' U~XT/' In the case of coefficients in Z2, signs do not matter; thus,

(Pea '-' U~) x (p~/3 '-' Ue ) - (Pea x p~/3) '-' (U~ x Ue). The equality Ue x UTI - U~XT/ has already been proved. Let PB and PB' be the projections of B x B' onto the first and second factors, and let PE and PE' be similar maps for E x E'. Then PeXT/(a x (3) = PeXT/(PBa '-' PB,/3) = (PBP~xT/)*a '-' (PB'P~xT/)* /3 and Pea xp;/3 - (P'EPea) '-' (PE'P~/3) (pePE)*a '-' (PT/PE' )*/3. But PBPexT/ = PePE (both maps coincide with the natural projection of Ex E' onto B), and PB'P~xT/ = P~PE" Thus, in the case that w(~) = ep l(SqU~), we have ep~XT/(w(rJ) x w(~)) epeXT/((epZl SqUe ) x (ep;jl SqUT/)) - SqUexSqUT/ = Sq(UexUT/) = SqU~XT/ epeXT/(w(rJ x ~)), which means that w(rJ) x w(~) = w(rJ x ~). It remains to perform calculations for the bundle ,I over Rpl ~ 8 1 • For this bundle, the pair (E, Eo) consists of the Mobius band }.;[ and the Mobius band from which the central circle is removed. It has a deform ailion retract of the form (M, oeM), where oeM is the curve at distance e from the boundary of the Mobius band (parallel to the boundary). The excision isomorphism implies H*(M,oe M ) ~ H*(lRp2,D 2), and th~ cohomology sequence of the pair (lRp2, D2) implies Hi (lRp2 , D2) ~ Ht (lRP2)

3. Characteristic Classes: COIltiIluatioIl

25£

'Yt

for i > 1. Thus, the Thorn class U of the bundle is identified with the unique nonzero element a E HI (lRp2). We have Sq1 a = a '--" a i- 0; therefore, SqIU i- 0, and cp I(SqIU) i- O. 0 The Thorn formula is valid for all vector bundles, but its application is hindered by the necessity of calculating the Thorn class. For the Stiefel Whitney classes of tangent bundles of manifolds, there exists a more explicit formula, which does not involve the Thorn class. Let SqT: H*(X) -+ H.(X) be the operation dual to Sq E,>o Sqi, i.e., such that (a, SqT,8) Sq a,,8. For any closed manifold Mn, the cohomology class v D ISqT[Mn] E H*(Mn) is defined, where [Mn] is the fundamental class (with coefficients in Z2) and D: Hk(Mn) -+ Hn k(M n ) is the Poincare isomorphism. The class v is called the Wu class of the manifold

Mn. Theorem 4.33 (Wu [155]). The Wu class of any closed manifold Mn has

the following properties: (i) (a '--" v, [Mn]) Sqa, [Mn]) for any element a E H·(Mn); (ii) w(Mn) Sq v.

= SqT[Mn]. Therefore, (Sq a, [Mn]) - (a, SqT[Mn]) = (a, Dv) = (a, v [Mn]) = (a '--" v, [Mn]).

Proof. By definition, Dv

r--,

To prove the equality w(Mn) = Sq v, it suffices to verify that (w(Mn),,B) = (Sq v,,B) for any element ,8 E H. (Mn). According to the Thorn formula, we have SqU = cp(w(Mn)). Instead of the Thorn class of the tangent bundle, we take the Thorn class of the tangent microbundle, which belongs to Hn(Mn x M n , M n x Mn \ d(Mn)). The Thorn isomorphism has the form cp(a) = pia '--" U, where PI: Mn x M n -+ M n is the projection onto the first factor. Hence pia = a x 1, where 1 E HO(Mn) is the identity element. Therefore, SqU = pi(w(Mn)) '--" U = (w(Mn) x 1) '--" U. Let us use the class U E Hn(Mn x Mn) defined on p. 244 and its properties (36) and (37) (see p. 244). Similarly to U, this class satisfies the relation Sqii = (w(Mn) x 1) '--" U. Clearly,

(Sq v,,B)

= (v, SqT {J) (~) (ii, SqT,8 x Dv) = (U, SqT {J x SqT[Mn]) = (U,SqT(fJ x [Mn])) = (Sqii,,8 x [MnJ) = ((w(Mn) x 1) '-' ii, {J x [Mn]) (~) ((1 x w(Mn)) '--" U, {J x [MnJ) = (U, (1 x w(Mn)) (,8 x [Mn])) = (ii, {J x (w(Mn) [Mn])) r--,

= (U, {J

x Dw(Mn) (~) (w(Mn), {J).

r--,

0

260


It follows directly from the definition of the Wu class that if v = Vo + where Vi E H'(Mn), then Vi = 0 for i > [n/2]. Indeed, let SqT[Mn] = mo + mi + ... + m n , where mi E H,(Mn). If 0: E Hk(Mn), then (o:,mk) = (Sqo:, [Mn]) = (sqn-ko:, [Mn]). Moreover, Sqn ko: = 0 if n - k > k, i.e., 2k < n. Therefore, mk = 0 for 2k < n. This means that Vn k = 0 for 2k < n, i.e., v, = 0 for 2i > n. VI

+ ... + V n ,

Problem 121. Let V = Vo + VI + ... + V[n 2] be the Wu class of a closed manifold M n , and let W = 1 + WI + ... + Wn be its Stiefel Whitney class. (a) Prove that if WI = ... = Vk+I = Wk+I· (b) Prove that if WI - ... Wk+I - Wk+2 = ... - Wn - O.

Wk -

Wk

0, then

= 0 and

VI =

n -

... -

Vk =

2k or 2k

0 and

+ 1,

then

Problem 122 ([88]). Given a closed (2k I)-connected manifold M 4 k, prove that the diagonal elements of its intersection form are even if and only if W2k(M4k) = O. Problem 121 can be used to prove the following theorem. Theorem 4.34 (Stiefel [132]). Any closed orientable 3-manifold is parallelizable. Proof. To prove the parallelizability of a manifold M 3 , we must construct three linearly independent sections of its tangent bundle. Since the manifold is orientable, it suffices to construct two independent sections. In constructing k independent sections of an n-dimensional real bundle, the first obstruction is involved in the extension over the (n - k + I)-skeleton. If n - k is odd and k ~ 2, then this obstruction coincides with the class Wn k+1' In the case under consideration, this is the class W2. If we construct two independent sections on the 2-skeleton, then we shall a('complish the task because the obstruction to extending them over the 3-skeleton lies in the cohomology with coefficients in the group 11"2 (V(3, 2)), which is trivial because V(3,2) ~ ~'p3. Thus, it is sufficient prove that w2(M 3 ) = O. For an orient able manifold, we have WI - O. According to Problem 121, (b), the equality WI = 0 implies W2 = 0 and W3 = O. 0 Remark. Stiefel proved the parallelizability of a closed orientable 3-manifold under the assumption that w2(M3) = 0; he had suggested no convincing proof of this equality. Stiefel considered the general problem of constructing n - 1 linearly independent vector fields on an orient able n-manifold, that i... the parallelizability problem; he constructed the theory of characteristic classes [132] in order to solve this problem. On Hopf's advice, Stiefel started with the

3. Characteristic Classes: Continuation

261

simplest nontrivial case n = 3. But all of the numerous examples of closed orientable 3-manifolds which he considered turned out to be parallelizable. Hopf made a report about Stiefel's results at the First International Topological Conference in Moscow (1935). Whitney, who attended his talk, said that many of these results were contained in his paper [149] published just before the conference; Stiefel and Hopf were unaware of it at that time. 3.3. Obstructions to Embeddings. Suppose that Mn is a closed manifold immersed in ]Rn+k, and liMn is the normal bundle for this immersion. According to Whitney's theorem (Theorem 3.22 on p. 149), we have Wi(IIMn) = w,(Mn). Therefore, w,(Mn) = 0 for i > k because dim liMn = k. If the manifold Mn is embedded in ]Rn+k, the additional condition wk(M n ) = o arises; let us prove it. The tubular neighborhood theorem implies that if a closed manifold Mn is embedded in ]Rn+k, then, for sufficiently small € > 0, the set Me consisting of the points at distance at most € from Mn is homeomorphic to the total space E(IIMn) = E of the bundle. Thus, the homeomorphism of pairs (E, Eo) ~ (Me, Me \ Mn) arises. Using the excision isomorphism, we obtain H*(E, Eo) ~ H*(Me, Me \ Mn) ~ H*(]Rn+k, ]Rn+k \ Mn). Consider the composition

(39)

Hk(]Rn+k, ]Rn+k \ Mn)

--+

Hk(]Rn+k)

--+

Hk(Jvl n )

of restriction homomorphisms. Obviously, this composition is the zero homomorphism because Hk(]Rn+k) = O. Therefore, it suffices to prove the following lemma.

Lemma. The composition (39) takes the class corresponding to the Thom class U E Hk(E,Eo;Z2) to Wk(IIMn). Proof. The zero section s: M Hic(M n ). The composition

Hk(E, Eo)

--+

E induces an isomorphism s*: Hk(E)--+

--+

Hk(E) ~ Hk(Mn)

takes the Thom class U to Wk(IIMn). Indeed, the Thom isomorphism cp: Hk(Mn) --+ H2k(E, Eo) takes the class S*(UIE) to P*S*(UIE) '-" U = (UIE) '-" U = U '-" U = SqkU; therefore, S*(UIE) = cp-lSqkU = Wk(IIMn). Replacing the pair (E, Eo) by the homeomorphic pair (Me, Me \ M n ), we see that the composition

Hk(Me, Me \ Mn)

--+

Hk(Me)

--+

Hk(Mn)

of restriction homomorphisms takes the class corresponding to the Thorn class U to Wk(IIMn).


262

The composition (39) fits into the commutative diagram

Hk(JRn+k,JRn+k \ Mn) -------+ Hk(JRn+k)

1~

1

Therefore, the composition (39) takes the class in Hk(JRn+k, JRn+k \ Mn) corresponding to the Thorn class U to Wk(VMn). 0 Problem 123. Prove that if n - 2m , then JRpn cannot be embedded in JR2n 1

Chapter 5

....

Cech Cohomology and de Rham Cohomology

1. Sheaf Cohomology 1.1. Sheaves and Presheaves. Suppose that X is a topological space, associated with each open set U C X is an Abelian group F(U) so that F(0) = 0, and associated with each inclusion of open sets V C U is a homomorphism T~: F(U) -+ F(V), which is called the restriction homomorphism. Suppose also that the correspondences F and T satisfy the conditions (i) T~ = idF(u), and

(ii) W eVe U implies T{{. = TWT~. Then we say that F is a presheaf of Abelian groups on X. Let F and Q be presheaves on the same space X. Suppose that for any open set U C X, a homomorphism hu: F(U) -+ Q(U) is defined so that the diagram F(U)

~Q(U)

lr~

lr~

F(V)~Q(V)

is commutative whenever V cU. Then we say that a homomorphism of presheaves, h: F -+ Q, is given. Presheaves of rings or R-modules (where R is a fixed ring) and homomorphisms of such presheaves are defined in a similar way.

-

263

264

5. tech and de Rham Cohomology

Example 59. For any Abelian group G, we can consider the presheaf that assigns the group G to every nonempty open set U C X and the trivial group to the empty set. This presheaf is called constant. Example 60. The set F(U) of continuous functions on U is a presheafj the group operation is the pointwise addition of functions.

We denote the element r~(J), where I E F(U), by Ilv. A sheaf is a presheaf satisfying certain additional conditions. These conditions are as follows. Let U = {U} be a family of open subsets of X. A family {/u E F(U)}, where U E U, is said to be consistent if lulunu' = lu' Iu U' for any U, U' E U Each family U of open sets determines the open set V - UUEU U. A presheaf F is a sheal if for any family U of open sets, the following conditions hold:

(i) if I E F(V) is such that Ilu = 0 for all U E U, then I = OJ (ii) for any consistent family {Iu}, there exists an I E F(V) such that I Iu = lu for all U E U. Example 61. Let X be the union of two disjoint open sets Ul and U2 • Then no (nonzero) constant presheaf on X is a sheaf.

Proof. Let U = {Ul, U2 }. The family consisting of 0 E F(U1 ) and nonzero E F(U2) is consistent, but there exists no f' E F(X) for which 1'luI = 0 and f'lu2 = I i= O. 0

I

Example 62. Suppose that X = lR and F is the presheaf of bounded functions, i.e., F(U) consists of all functions bounded on U. This presheaf is not a sheaf.

Proof. Let U = {UkhEN, where Uk = (-k,k). Then the family {A}, where A(x) = x, is consistent, but there exists no bounded function I on lR = U~l Uk coinciding with A on each set Uk. Indeed, such a function must be I(x) = x, which is not bounded. 0 With any presheaf we can associate a sheaf by passing to the direct limit. Below, we recall the definition of direct limit. A set J is said to be directed if a relation> between some of its elements is defined and the following conditions hold: (i) a > a for any a E Jj (ii) if a > f3 and f3 > I, then a > Ij (iii) for any a, f3 E J, there exists a 8 E J for which a > 8 and f3 > 8. Example 63. Let J be a family of open sets that contain a fixed point x E X, and let V < U in J if and only if V c U. Then J is a directed set.

265

1. Sheaf Cohomology

Let J be a directed set. A set of Abelian groups {Go}, where a E J, is said to be directed if for any pair a > (3, a homomorphism lap: Go. --+ Gp is defined and for any a < (3 < 'Y, lao. = id and I p.., a lap = Io.-y. Example 64. Suppose that F is a presheaf on a space X, J is the directed set from Example 63, G u = F(U), and luv r~. Then {Gu}, where U 3 x, is a directed set of Abelian groups. The direct limit lim -o.E J Go. of a directed set of Abelian groups is defined as follows. Consider the disjoint union of all groups Go. and take its quotient by the following equivalence relation: go '" gp if 10.6(go.) = Ip6(gp) for some 5 (here go. E Go. and gp E Gp). The group operation is defined by {go.} + {gp} = {/0.6(go.) + Ip6(gp)}, where a > 5 and (3 > 5. For the directed set of Abelian groups from Example 64, the direct limit consists of germs at the point x, which are defined as follows. We declare elements I E F(U) and 9 E F(V), where U and V are open neighborhoods of x, to be equivalent if Ilw = glw for some open set W C Un V. The equivalence class containing I E F(U) is called the germ of I at x and denoted by Ix. The addition of germs is defined in a natural way. We set Fx

=

F(U) (thus, Fx is the group of germs at the point x) and F = UXEX Fx. The set F is endowed with a topology with respect to which the natural projection p: F --+ X taking Fx to x is a local homeomorphism. Let U c X be an open set, and let I E F(U). Consider the set of all germs Ix for x E U. Such sets form a base for a topology on F. Clearly, if F is endowed with this topology, then p is a local homeomorphism. A section of a presheaf F over an open set U C X is defined as a continuous map s: U --+ F for which pas = id u . The set of sections over U is denoted by r(u, F); this set is a group under the operation defined by (SI + S2)(X) = SI(X) + S2(X) for each point x E U. For a family of groups r(u, F) with different U, the natural restriction homomorphisms are defined; they satisfy the axioms for a presheaf. Moreover, setting F(U) = r(u, F), we obtain a presheaf F. This presheaf is a sheaf; it is called the sheaf generated by, or associated with, F. limu -

3x

We can associate with each element I E F(U) a section S E r(u, F) by setting s(x) = Ix. Thereby, we obtain a homomorphism of presheaves T: F --+ F. Theorem 5.1. II F is a sheaf, then T is an isomorphism 01 sheaves, i.e., the map T(j: F(U) --+ F(U) is an isomorphism lor any open set U eX. Proof. First, we verify that TrJ is a monomorphism. Suppose that I, 9 E F(U) and Ix = gx for all x E U. It follows from the definition of a germ

266

5. Cech and de Rham Cohomology

that each point x E U has a neighborhood V :3 x such that flv (J - g)lv = o. Therefore, the first sheaf axiom implies f = g.

= glv,

Le.,

Now let us show that T[j is an epimorphism. Take s E r(U, i"). Each point x E U has a neighborhood V :3 x such that there exists an f E F(V) for which fx = sex). This means that the sections sand rv(J) coincide at x. Two sections coinciding at a point x must coincide in somE' neighborhood W of this point. Therefore, slw = 7W(Jlw). To each point x E U we have E F(Wz ) assigned an open set Wx :3 x (contained in U) and an element so that f: = s(y) for all y E W z . The family of elements {r E F(Wx )} is consistent, Le., rlw",nw.. = rlw",nw... Indeed, we have I: = s(y) = g for all y E Wx n W z . Therefore, 7W",nw.. (rlw",nw.. - rlw", w..) = 0; recall that T is already proved to be a monomorphism. Using the second sheaf axiom, 0 we obtain an element f E F(U) such that fx = sex) for all x E U ~

r

It is often convenient to calculate direct limits over some subsets Jo of a directed set J rather than over the entire set J. A subset Jo c J is cofinal in a directed set J if for any a E J, Jo contains an element tS such that a> tS.

Theorem 5.2. Suppose that {G a }, where a E J, is a directed set of Abelian groups and Jo c J is its cofinal subset. Then J o is a directed set and lim JGa ~ lim Ga. ---+aE ---+aE .. o 7

Proof. For any a, (3 E J o, we can choose tS' E J so that a > tS' and (3 > tS'. For this tS', we can choose tS E Jo so that tS' > tS. Therefore, Jo is a directed set.

The identity maps G a -+ G a , where a E J o, commute with the maps fa6; therefore, they determine a homomorphism lim J G a -+ lim J, Ga· ---+aE ---+aE 0 This homomorphism is an epimorphism because for any element ga E G a with a E J, we can choose tS E Jo so that a> tS. For these a and tS, the map fa6: G a -+ G6 is defined, and ga '" g6 = fa6(g6). Now, let us check that this homomorphism is a monomorphism. Suppose that ga '" O~ for some (3 E J, Le., there exists a tS E J such that fa6(ga) = f~6(0~) = 06. Choose 'Y E Jo so that tS > 'Y. We have fa..,(ga) = 16.., 0 fa6(ga) = f6..,(0..,) = 0.." Le., ga '" 0.., for some 'Y E J o . 0 For more information about sheaves, see [12, 66]. 1.2. Cech Cohomology. Suppose that U = {Ua } is all open cover of a space X and F is a presheaf on X. We assume that all sets Ua are pairwise distinct. We write UaO ... ak to denote uaon·· ·nuak . A cochain ck E Ck(U; F) assigns to each ordered set Uao , ... , Uak an element d'(Uao ,···, Uak ) E

1. Sheaf Cohomology

267

F(Uoo ... o,J. The coboundary homomorphism is defined by k+1 k

_'"'

k ..... (dC )(Uoo,,,,,UOA:+l) - L.,,(-l) iC (Uoo,,,,,Uol,,,,,UOII:+Jluao ... all:+l· i=O

A simple standard calculation shows that dd = O. Thus, we can define the cohomology group iik (Uj F), which is called the Cech cohomology group uf the cover U with coefficients in the presheaf F. Example 65. If F is the constant presheaf corresponding to an Abelian group G, then iik(Uj F) ~ Hk(N(U)j G), where N(U) is the nerve of the cover U. Passing to the direct limit, we obtain the Cech cohomology of the space X, which does not depend on a particular cover U of this space. To this end, we construct a directed family of cohomology groups as follows. Suppose U < V, i.e., U = {Uo I a E A} is a refinement of a cover V = {V/j I J3 E B}. By the definition of a refinement, there exists a map ~: A -+ B such that Uo C V~(o) for all £lEA. To each cochain rJc E Ck(Vj F) we assign a cochain ~#rJc E Ck(Uj F) such that

(~#ck)(Uoo,"" It is easy to verify that ).,# is a

= ck(V~(oo)"'" V~(OA:»' cochain map, i.e., ~#d = d~#.

U Ot )

The cochain map ).,# induces a homomorphism~·: bk(Vj F) -+ bk(Uj F) of cohomology groups. The homomorphism ~. does not depend on the choice of~. Before proving this assertion in the general form, we explain why it is true for a constant sheaf F. Example 66. For a constant sheaf F, the homomorphism ~.: bk(VjF)

-+

bk(UjF)

is induced by the simplicial map X: N(U) -+ N(V) taking a to ~(a). If 1-': A -+ B is another map for which Uo C V"Co), then the maps Xand {.t are homotopic. Proof. If a point x belongs to a simplex [ao, ... , an], then the points X(x) and {.t(x) belong to the simplex with vertices ~(ao), ... , ~(an), I-'(ao), ... , I-'(an)j this simplex exists because

n(V~(OI) n

V"COI» :::)

n

UOi

f.

0.

Joining such points X(x) and ji(x) by segments, we obtain the required homotopy. 0 Let us prove a general assertion.

268


Theorem 5.3. If >',J.L: A --+ B are maps for which Ua C V~(a) and Ua C VJL(a), then there exists a cochain homotopy between>. # and J.L#. Proof. Consider the map D: Ck(V;F)

--+

Ck-I(U;F) defined by

k-l k

(Dc )(Uao ,···, Ua/c

J-

_""' _

ik

LJ ( 1) c (V~(ao),···, V~(a.), VJL(a.).···, VJL(a/c d)· ,=0

Let us show first that 6D + D5 = J.L# - >.#. In the expression (5(Dck) D(5ck))(Uo, . .. , Uk), the restrictions of the cochains .

k

+

-

(-1)'( -l)J c (V~(O),··· ,~(j) , VJLU ),· .. , UII · · · , VJL(k») and . k (-1)3( -1) 1+1 C"-(V~(O), .. ·, V~u), VJL(j), ... , VJL(i), ... , VJ.l(k)~

to the set V~(O) ... ~(J)JLu) ... JL(I) ... IL(k) cancel each other (we assume that i > j; the case of i < j is treated similarly). The only remaining terms are ck(V~(O)' VIL(O), ... , VIL(k») and (_l)k( -l)k+1ck(V~(o), ... , V~(k)' VIL(k»). 0 We have constructed a directed set of Abelian groups; so we can consider the direct limits !!!!l bk(U; F) = bk(X; F), which are called the tech cohomology groups of the space X with coefficients in the presheaf F. We denote the Cech cohomology groups of X with coefficients in the constant presheaf corresponding to an Abelian group G by bk(X; G). Theorem 5.4. If K is a finite simplicial complex, then Hk(K;G).

bk(IKI; G)

~

Proof. Let Uo be the cover of the space IKI by the open sets st Vi, where each Vi is a vertex of K. The nerve N(Uo) is identified with K; therefore, bk(Uo; G) ~ Hk(N(Uo); G) ~ Hk(K; G). Suppose that K' is the barycentric subdivision of the complex K, hI: K' K is a simplicial approximation ofthe identity map (hI takes the barycenter of each face to one of the vertices of this face), and UI is the cover of IKI by the stars of vertices of the simplicial complex K'. Identifying N(UI) with K', we can regard hI as a map N(UI) --+ N(Uo). This map induces an isomorphism of cohomology groups. --+

Similarly, for the cover Urn of IKI by the stars of the vertices of the mth barycentric subdivision of K, we construct a map N(Urn) --+ N(Urn-I), which induces an isomorphism of the cohomology groups. We have constructed the directed set Uo > UI > ... and the directed set of Abelian groups bk(Urn ; G), in which all homomorphIsms fa{3 are isomorphisms. We have !!!!l b k (Urn; G) ~ Hk (K; G).

269

1. Sheaf Cohomology

To apply Theorem 5.2, we must show that the covers Uo > UI > ... form a cofinal subset in the set of all covers of IKI. Take an arbitrary open cover U of IKI; let d be its Lebesgue number (we assume that the simplicial complex K is embedded in Euclidean space). Choose m so that the diameter of any simplex in the mth barycentric subdivision of K is less than d. We haveUm U~ > form a cofinal subset in the family of covers of Y by open sets in X. Let U be a cover of Y by open sets in X. First, we show that U covers some of the sets IYnl. Suppose that each IYnl has a point Yn not covered by U. IThis means that there exists a homeomorphism X --> IKI, where K is a Simplicial complex. to Problem 30 from Part I, it suffices to take the barycentric subdivision.

2 According

1. Sheaf Cohomology

271

Since X is compact, the sequence {Yn} contains a convergent subsequence {Yn,.,}. Its limit point Y belongs to IYnl = Y. This point is covered by some open set U from the cover U. We have Yn" E U for sufficiently large k, which contradicts the assumption.

n:'-l

The same argument as the one used to prove the Lebesgue theorem about open covers shows that there exists a positive number 0 such that any B of diameter less than 0 intersecting IYnl is contained in an open set from U. Suppose that 11m < 0/2 and m :S n. Let us show that the cover U.:n is a refinement of U. The diameter of the star of any vertex of Ym is less than 21m < Moreover, this star intersects IYnl because IYml c IYnl. Therefore, it is contained in an open set from U. 0

o.

Using the Cech cohomology, we can state and prove the following, more general, version of the Alexander duality theorem. Theorem 5.7 (Alexander Pontryagin duality). If A ~ gn is a closed set, then jJk(A) ~ Hn_k_1(sn \ A) for 0 :S k :S n - 1. (Here jJ. is the reduced tech cohomology and iI. is the reduced singular homology.)

Proof. Let K be a triangulation of the sphere gn so fine that it contains nsimplices disjoint from A, and let K(m) be the mth barycentric subdivision of K. Consider the subcomplex Mm of K(m) consisting of the simplices intersecting A. The simplicial complex Mm is not necessarily a manifold; it may have singular points (or simplices). But all of its singular points are outside A; therefore, we can turn Mm into a manifold containing A by removing small neighborhoods of singular points (see Figure 1).

Figure 1. The correction of the complex Mm

In this way, we construct a sequence of triangulated manifolds Ml ::) M2 ::) ... such that Mi = A. For each M i , the Alexander duality iIk(Mi) ~ iIn- k- 1 (sn \ M,) holds; its proof is similar to that in the smooth (triangulable) case. This isomorphism commutes with the homomorphism ind uced by the inclusion Mi+l C M i ; hence

n:l

~iik(Mi) ~ ~Hn_k_l(sn \ Md.


272

According to Theorem 5.6, the direct limit on the left-hand side is the Cech cohomology group j{k(A). Let us show that the group on the right-hand side is the singular homology group H n_k_l(sn \ A). The homomorphisms Hn-k 1 (sn \ M,) the inclusions determines a homomorphism

~Hn_k_l(sn \ M,)

--+

--+

Hn k

1 (sn

\ A) induced by

Hn-k-l(sn \ A).

Let us show that this is an isomorphism. The support of any singular chain in en k_l(sn \ A) i.s compact. The increasing open sets sn \ Mi cover it; therefore, it is contained in one of these sets. This implies surjectivity. Injectivity follows from the compactness of the support of any singular (n - k)-chain whose boundary is the difference of two given singular (n - k - I)-chains. 0 1.3. Bundles with Structure Groups. Suppose that G is a topological group, B is a topological space, and the group G acts effectively on a topological space F. A locally trivial bundle p: E --+ B with fiber F is called a bundle with structure group G if the homeomorphisms hi: p-l (Ui) --+ Ui X F have the property that, for any pair of indices i, j, there exists a map 9,]: U, n Uj --+ G such that hzht(u, J) = (U,9ij(U)J) for all u E Ui n Uj. Since the action of G on F is effective, each map 9ij is uniquely determined by the homeomorphisms h, and hj . Such a map is called a transition function. The transition functions determine how the "pillars" Ui x F and Uj x F are attached to each other (see Figure 2).

Figure 2. Transition functions

1. Sheaf Cohomology

273

This definition of a bundle with a structure group has the essential drawback of being dependent on the cover {Ui}' To overcome it, consider all homeomorphisms hu: p-l(U) --+ U x F that can be added to the homeomorphisms hi, namely, those with the property that, for any Ui, there exists a map gU,i: Un Ui --+ G such that hUh-;l(u,f) = (U,gU,i(U)f) for all u E Un Ui. Such homeomorphisms are called admissible charts. We assume that two covers {Ui} and {Uj} with homeomorphisms {hI} and {hj} specify the same bundle with structure group G if they determine the same set of admissible charts. Now we introduce the notion of an isomorphism between two bundles with structure group G over the same base B. Let p: E --+ Band p': E' --+ B be bundles with structure group G. A homeomorphism cp: E --+ E' is called an isomorphism of bundles with structure group G if, for each point bE B, the following conditions hold: (i) each fiber p-l(b) is mapped to (p') l(b);

(ii) there exists a neighborhood U 3 b, a map gu: U --+ G, an admissible chart hu: p I(U) --+ U x F for the bundle E, and an admissible chart hf,: (P')-l(U) --+ U x F for the bundle E' such that hf,cphc/(u, f) = (u, gu(u)f) for all u E U. For example, an n-dimensional vector bundle is the same thing as an JRn_ bundle with structure group GL(n, JR). An isomorphism of vector bundles is the same as an isomorphism of bundles with this structure group. If a vector bundle admits a Riemannian metric (e.g., if its base is compact), then we can assume that the structure group is O(n). Such a bundle is orient able if and only if for the structure group the group SO(n) can be taken. Example 67. Suppose that the manifold cpn is covered by charts Ui, i = 0, ... ,n, which are specified by the equations Zi = 0 in homogeneous coordinates (zQ : '" : zn). Then the canonical bundle 'Y~ over cpn is determined by the transition functions 9ij = zd Zj.

Proof. The fiber of the canonical bundle over a point (zQ : .. , : zn) is the complex line (AZQ, ... , AZn). The homeomorphism hi: p-l (UI ) --+ Ui x C can be defined by hi (AZQ, ... , AZn )

=

((zQ : ... : zn), Ai),

where Ai = AZi, i.e., (AZQ, ... , AZn ) = Ai (!O., ... , !n.). By definition, we have z, Z'I Ai = gijAj, i.e., 91j = At/Aj = Zi/Zj. 0 Example 68. Suppose that an n-dimensional vector bundle is given by transition functions 9ij (Le., at each point, 9ij is a matrix of order n). Then the dual bundle is given by the transition functions (9&)-1.

274


vt,

Proof. Each map 91j: VI --+ V2 induces a dual map V2* --+ which is determined by the matrix 9~. We are interested in the inverse map Vt --+ V2*. 0

Clearly, if ~ is a one-dimensional bundle with transition functions 9ij, then the dual bundle C is determined by the transition functions 1/gjj , and if a one-dimensional bundle 7J is determined by transition functions hij, then the bundle ~ ® 7J is determined by the transition functions 9j,hij . In particular, the bundle ~ ®~. is trivial. Example 69. Consider the map

p: cpn+1 \ (0 : ... : 0 : 1)

--+

cpn

given by p(zo : ... : Zn+1) = (zo : ... : zn). This map is the vector ~undle dual to the canonical bundle 'Y~. Proof. The homeomorphism hj: p-l(U,) --+ Uj xC can be defined by hj(zo : ... : Zn+1) = «zo : ... : zn), ~,), where ~i = Zn+1/Zj. Thus,9ij = ~j/~j = Zj/Zj. 0 1.4. Noncommutative Cech Cohomology. We can define 8 sheaf of non-Abelian groups on a space X in precisely the same way as a sheaf of Abelian groups. In the non-Abelian case, the cohomology groups iIk(X; F) for k > 1 cannot be defined, but there is a cohomology set iII (X; F) with a distinguished element. Moreover, if F is a sheaf of Abelian groups, then this set coincides with the Cech cohomology group, and the distinguished element is the zero element of the cohomology group. Recall that if F is a sheaf of Abelian groups on a topological space X and U = {UDo} is an open cover of X, then to every ordered set Uj, Uj any cochain c 1 E Cl(U; F) assigns an element Cl(Ui' Uj ) E F(UjnUj) = r(UinUj ; F). A cochain c l is a co cycle iffor any ordered triple Uj , Uj , Uk, we have c l (Uj, Uj) + cl(Uj , Uk) = cl(Uj , Uk) on Uj n Uj n Uk. A co chain c l is a coboundary if there exists a cochain cO E CO(U; F) such that c1(Uj, Uj) = cO(Uj ) - cO(Ui ) on Uj n Uj. For sheaves of non-Abelian groups, the corresponding definitions are as follows. A (one-dimensional) cocycle I assigns an element lij E F(UjnU,) = f(UjnUj;F) to each ordered pair Ui, Uj so that lij/jk = lik on U,nUjnUk. Two co cycles I and I' are equivalent (differ by a coboundary) if for each set Ui, there exists an element 9i E qUj; F) such that IIj = 9il/i,9j on Ui n Uj. The cohomology set iIl(U; F) is the set of classes of equivalent cocycles. The distinguished element of this set is the class of the co cycle 1 that takes all ordered sets Ui, Uj to the identity element of the group

qUi n Uj; F).

2. De Rham Cohomology

275

As in the commutative case, the cohomology set iII(Xj:F) is constructed as a direct limit. Suppose that a cover U is a refinement of a cover V and Ui C V~(i). Then every co cycle 1 E Zl (Vj:F) can be associated with a co cycle )..# 1 E ZI(Uj:F) for which ()..# J)(UinUj) = I(V~(i)nV~(j». The map )..# induces a map)..*: iII(Vj:F) ~ iII(Uj :F). Let us show that if Ui c V/J(i), then J1.* = )..*. The equivalence of co cycles )..#1 and J.L# 1 is established by . means of the O-cochain gi = I~(i),/J('). Indeed, by the definition of a cocycle, we have 1/J(i),~(,)/~(,),~(j)h(j),/J(j) = I/J(I),/J(j), i.e., g;l()..# f)iJgj = (J1.# f),j. Let Q be the sheaf on B for which qu, Q) is the group of all continuous maps from U to Gj in other words, Q is the sheaf of germs of continuous functions taking values in G. Given an open cover U = {U,} of B and a co cycle 9 = {gij} E ZI(Uj Q), we can construct a bundle Eg over B with structure group G and fiber F as follows. We take the disjoint union of the sets Ui x F and for each u E Ui n Uj ' identify the points (u, f) E Ui X F and (U,gij(U)f) E Uj x F. We can do this due to the equality in the definition of a cocycle. For the space Eg thus obtained, the projection Pg: Eg ~ B is induced by the natural projection Ui x F ~ Ui. Given two co cycles 9 E ZI(Uj Q) and g' E ZI(Uj Q), the bundles Pg: Eg Band Pg': E g, ~ B are isomorphic (as bundles with structure group G) if and only if the co cycles 9 and g' correspond to the same element of the cohomology set iI1(Bj Q). ~

Clearly, any bundle over B with structure group G and fiber F can be represented as Eg for some co cycle g. We have proved the following theorem.

Theorem 5.B. Suppose that a topological group G acts effectively and continuously on a space F. Then there is a natural one-to-one correspondence between the classes 01 isomorphic bundles over B with structure group G and fiber F and the elements 01 the cohomology set iI 1 (Bj Q). Moreover, the trivial bundle B x F corresponds to the distinguished element.

2. De Rham Cohomology Recall that a differential k-form W on a manifold Mn is a polylinear skewsymmetric function of k vector fields 6, ... , ~k on Mn. We denote the linear space of k-forms on a manifold Mn br nk(Mn). For two differential forms WI E np(Mn) and W2 E nq(Mn), their exterior product WI 1\ W2 E np +q (Mn) is definedj it has the property WI 1\ W2 = (-1 )1JqW2 1\ WI. Exterior multiplication makes the linear space n*(Mn) = ffik>o nk(Mn) into an algebra. Any smooth map I: Mm ~ Nn induces a linear map nk(N n ) ~ nk(Mm) by the rule (j*w)(~1. ... , ~k) = w(j.6, ... , I*~k)' where I. = dl

r:

3In Part I on p. 205, the notation II. k MR was used.


276

is the differential of I. In local coordinates, this map looks as follows. Let Xl, ... ,X n and Yl, ... ,Ym be local coordinates on Nn and Mm. Then

a 111 ... a II/c 1 * (~d ~ ail ... i/c Xi1"···" dXi/c ) -- ~ ~ ai1 ... i/c""i)-:a dYjl " ... " dYj/c· Y]l Y]/c The map n*(N n ) --+ n*(Mm) is a homomorphism of algebras. Let us construct a polylinear map d: nk(Mn) --+ nk+l(Mn) such that (i) dod = 0 and (ii) d is polylinear with respect to smooth functions, i.e., dw(cp~o, 6, ... , ~p) = cp dw(~o, .. . , ~p) for any cp E coo(Mn). For this purpose, we need the notion of the commutator of vector fields. Let ~ and TJ be vector fields on M n (we consider them as differential operators), and let cp E coo(Mn). We set [~, TJl(cp) - ~(TJ(CP)) - TJ(~(cp)). Using the relation ox,ax; ~x = ~x82 8xJ 8x, ,we obtain the following expreS6iqp. in local coordinates:

r:

Thus, in a fixed local coordinate system, [e, TJl is a vector field. The operator [~, TJl, which acts on smooth functions, does not depend on the choice of a local coordinate system; so we obtain a vector field [~, TJ]' which is called the commutator of the vector fields ~ and TJ. It follows directly from the definition that [~, TJl = -[TJ, ~l. Moreover, if = cp[e, TJl - TJ(cp)~; this follows from the relation

cP E coo(M n ), then [cp~, TJl

a(cp~l) = cp a~i aXj

aXj

+ ~I acp . aXj

Now we can define the map d. Let wE nk(Mn). We set k

dw(~o, . .. , ~k) =

L

(-l)i~,(w(~o, ... , t" ... , ~k))

i-a

+

L

(-l)i+jW([~i,~j], ... ,ti, ... ,tj' ... '~k).

O'pw is the intersection of the compact set supp wand the closed set supp >'p; therefore, it is compact (any closed subset of a compact space is compact). Hence >'pw E nc(Up). Clearly, i~(>'IW, >'2W) = w. 0 2.1. The Stokes Theorem. Homotopy Invariance. Integration of Forms. Let us choose coordinates in ]Rn and write a form W E nn(JRn) as W - ~(XI' ... ' xn) dXI /\ ... /\ dx n . We shall use the abbreviated notation W = ~(x) dx. Passing to another coordinate system y(x), we obtain dYI /\ ... /\ dYn = det(~) dXI /\ ... /\ dx n , or, in the abbreviated notation, dy = det(~) dx. In the new coordinates, we have W = J ¢(y) dy. Let us determine the relation between the functions ¢(y) and ~(x). Clearly, ~(x) dx = W = ¢(y) dy = ¢(y) det(~) dx; therefore, ~(x) = J

¢(y(x)) det(~).

The formula for a change of variables in an integral implies

Ian ¢(y) dy = !an ¢(y(x))ldet(!::) Idx = ± Ian ~(x) dx, where the sign ± coincides with that of the determinant det (~ ). Thus, we can define the integral of a form as follows. Suppose that ]Rn is oriented. Let W E n~(]Rn) be a compactly supported form. We set W = Jan ~(x) dx, where x is a coordinate system whose orientation is compatible with that of ]Rn.

Jan

We define the integral of a form W E n~ (Mn), where M n is an oriented manifold, as follows. Take an arbitrary locally finite atlas {Ua } with orientation-preserving charts fa: Ua - ]Rn. Let {>'a} be a smooth partition of unity subordinate to it. We set

{ lMn

W=

L Jan { >'aU;;I)*W = L 1 >'aW a

a

Ua

The support of the form >'o<w is compact because it is closed in the compact set supp >'0 1, we represent the union under consideration in the form U' U Uk, where Ul = U1 U··· U Uk-I. The Poincare isomorphism theorem is valid for U' and Uk by the induction hypothesis. It remains to show that it is valid for U' n Uk = (UI n Uk) U··· U (Uk-l n Uk). According to (ii), the theorem is true for each set Ui n Uk; therefore, it is also true for U' n Uk. 0 Lemma 5.2. Let U be a base for the topology of Mn with property (i). Then, for any (not necessarily finite!) disjoint union of elements ofU, the Poincare isomorphism theorem is valid. Proof. Suppose that {Ua.}aEA C U and Ua n U{j = 0 for any a, fj E A. The natural embeddings nk (U aEA Ua ) -. fIaEA nk(Ua ) and ffiaEA n~-k(Ua) n~-k(UaEA Ua ) are isomorphisms. The second map acts on the direct sum rather than product because a compactly supported form can differ from zero only on finitely many sets Ua . Dualizing the second map, we obtain an isomorphism n~-k (UaEA Ua )* -. fIaEA n~-k(Ua)* (this time, to the direct product). The situation is the same as for chains and cochains: each chain


288

contains only finitely many simplices, but a co chain may assign nonzero values to arbitrarily many simplices. Passing from forms to cohomology, we obtain isomorphisms of cohomology groups, which form the commutative diagram

"1

llQEADQ

k(

UUa)

$

~

aEA

II H~-k(Ua). aEA

D

Hence D is an isomorphism.

Let U c ]Rn be an arbitrary open set. It can be represented as U = U~ 1 Vi, where Vi C Uj for ~ we take open balls of rational radii centered at points with rational coordinates. We set Kl = VI. The boundary of Kl is a compact subset of Uj hence it can be covered by finitely many sets ViI' ... , Vim· We put K2 = V 2UVii U· . ·UVtm . In a similar way, we construct K 3 , and so on. By construction, we have U:l Ki = U and Ki C int K i +1. The inclusion Ki C int K'+1 implies the existence of a compact set Li for which Ki C int Li C Li C int KH 1 (see Figure 3). L.

Lj

I (

,, \. \

/

U.----JY K.

K.

2 ,,, , ,,, , ,,, , ,,, ,

I

Figure 3. The sets K. and L.

We cover Ki \ K i - 1 with the base sets from U that are contained entirply n int(Li \Li - I ). Since Ki \ K i - I is compact, it follows that this cover has a inite subcover. Let Ui be the union of all sets from this subcover. According o Lemma 5.1, the Poincare isomorphism theorem is valid for Ui. Moreover, ly construction, we have U:I Ui = U and Ui n Uj = 0 for z- j I > 1. Ne set WI = U:o U2i +1 and W2 = U:I U2i. By Lemma 5.2, the Poincare somorphism theorem is valid for WI and W2. We have U = WI U W2· Thus,

3. The de Rham Theorem

289

it remains to prove the theorem for WI nW2. The set WI nW2 is the disjoint union of the sets Ui n Ui+I, i = 1,2, .... Each of these sets is a finite union of elements of U. By Lemma 5.1, the Poincare isomorphism theorem is valid for Ui n Ui+l, and by Lemma 5.2, it is valid for WI n W2. The proof of the Poincare isomorphism theorem for an arbitrary manifold Mn follows approximately the same scheme as that for an open set U C JR n . The only essential difference is related to the choice of a base U for the topology of Mn. For U we take the family of all open sets homeomorphic to JRn (they cover Mn) and their finite intersections. This base has property (ii) by definitionj (i) holds because any set from U can be considered as an open subset of lRn , and for such sets, the Poincare isomorphism is already proved. Instead of balls with rational radii and centers, we take a suitable countable base for the topology of Mn. Otherwise the proof is the same. 0

3. The de Rham Theorem The de Rham theorem [109] asserts that, for any closed manifold M n , the de Rham cohomology HDR(M n ) is isomorphic to the simplicial (or singular) cohomology H*(Mnj JR). They are isomorphic as algebras, that is, '--"products of co chains correspond to I\-products of forms. There exist many different proofs of the de Rham theorem, but none of them is simple. We give a proof which goes back to Whitney [154] (see also [121]). Other proofs of the de Rham theorem can be found in [28, 51, 114, 146, 147]. We also prove an analog of the de Rham theorem for simplicial complexes. 3.1. Proof of the de Rham Theorem. A triangulation f: IKI -+ M n is said to be smooth if for any n-simplex ,6 n in K, there exists an open set U ::J ,6n in JRn and an extension of the map fl~n to U such that this extension is a (smooth) embedding of the manifold U into Mn. We assume that Mn is endowed with a fixed smooth triangulation (its existence was proved in Part I). Let ck(Mn) = Ck(Kj JR) be the group of simplicial k-cochains. Consider the map p: nk(Mn) -+ ck(Mn) defined by (p(w k ),,6k) = J~kwk. According to the Stokes theorem, we have pck.J = (d"p)Wj therefore, p induces a homomorphism p*: Hf)R(M n ) -+ Hk(MnjR) of linear spaces. Step 1. The map p* is an epimorphism. We shall show that this map is an epimorphism even at the level of cochainsj i.e., for each co chain d'= E ck(Mn), there is a form wk E nk(Mn) such that p(w k ) = d'=. To construct such a form for a given cochain, we need a special partition of unity subordinate to the cover {st Vi}, where VI, ... , Vr are the vertices of the complex K. Namely, let Xi be the barycentric coordinate corresponding to the vertex Vi (it is assumed that Xi = 0 for any

290


point outside the simplices for which Vi is a vertex). Suppose that Pi is the set of all points for which Xi 2 n~l and G i is the set of all points for which Xi :::; n~2 (see Figure 4). Let).i be a nonnegative smooth function which is positive on Pi and vanishes on G i . The sets Pi cover Mn. Indeed, the sum of the n + 1 barycentric coordinates of any point X E M n equals

Figure 4. The sets F. and G i

1; therefore, one of these coordinates is at least n~l' Thus, the function ).(x) = ).l(X) + ... + ).n(x) is positive. Hence the functions {J.Li} , where J.Li(X) = ).i(X)/).(X) , form a smooth partition of unity subordinate to the cover {Mn \ Gi}. Consider the simplex .6.! = [Vio"'" Vi",] and the cochain c~ dual to .6.!, i.e., such that (c!, .6.~) = 8sr . To this co chain ~ we assign the form k

cp to···,,,, o

0

= k! ~ (-l)j r'J Ho dHo ~ r'o

/\ ... /\;;;; . r,] /\ ... /\ dHo rtk

j=O

As a result, we obtain a linear map cp: ck(Mn) -+ nk(Mn) (any linear map is determined by its values at the elements of a basis). Below, we prove several properties of this map, which imply that p* is an epimorphism. Property 1. dcp(c!)

=

cp(8~).

Clearly, dcp(~) = (k + 1)! dJ.Lio /\ ... /\ dJ.Li",. Let us calculate cp(8~). The co chain 8~ is the sum of the cochains dual to simplices of the form [vp, Vio"'" Vi",]. Therefore, the cochain (k~1)!cp(8c!) is equal to

L' (J.LP dJ.Lio /\ ... /\ dJ.Li", -

t(

-l)j J.Lij dJ.Lp /\ dJ.Lio /\ .•. /\ d;i.i] 1\ ., 1\ dJ.Lt"') ,

3=0

291


where E' denotes summation over all (k+1)-simplices [vp, Viol ••• I Vi,.]. Note that if p ¢ {io, ... ,ik} but no simplex is spanned by vp,Vio, ••. ,Vi/c, then J.LpdJ.Lio /\ •.. /\ dJ.Li/c = o. Indeed, if x ¢ stvp, then J.Lp(x) = O. If x E stvp, then xp =1= O. In the latter case, XiJ = 0 for some j (otherwise, xp =1= 0, Xio =1= 0, ... ,Xi/c =1= 0, and x belongs to the interior of the simplex [vp, Vio' ••. , Vi/c]). Consider the open set U consisting of all points Y E M" for which Yi J < "~2. The point x belongs to U, and J.LiJ vanishes on U because U C Gi J. Therefore, dJ.LiJ = O. Thus,

L' J.Lp dJ.Lio /\ ... /\ dJ.Lilo = L

J.Lp dJ.L,o /\ •.. /\ dJ.Li/c.

p!t{ io, ... ,i/c}

Combining this equality with obtain

E J.Li(X) =

1 (which implies

E dJ.Li

= 0), we

k

L' L (-1)i J.LiJ dJ.Lp /\ dJ.Lio /\ ... /\ dj;,iJ /\ .•. /\ dJ.Li/c i=O k

=L

(-1)i L' J.LiJ dJ.Lp /\ dJ.Lio /\ •.• /\ dj;,iJ /\ ... /\ dJ.Li/c

i=O k

=L

(-1)i L

i=O

J.Li; dJ.Lp /\ dJ.Lio /\ .•. /\ dj;,i J /\ ... /\ dJ.Li/c

p!t{io, ... ,i/c}

k

=

L (-1)i L i=O

J.LiJ dJ.Lp /\ dJ.Lio /\ ... /\ dj;,iJ

/\ ••• /\

dJ.Li/c

p~i;

k

= ~ (-1)i J.LiJ ( ~ dJ.Lp) dJ.Lio k

=L

/\ ... /\ dj;,i J /\ •.. /\ dJ.Li/c

P~'J

3=0

(-1)i J.LiJ ( -dJ.LiJ) dJ.Lio /\ .•. /\ dj;,iJ /\ •.. /\ dJ.Li/c

;=0

k

=-

L

J.LiJ dJ.Lio /\ •.. /\ dJ.Li/c.

;=0

Thus, !t'(b"c!)

=

(k

+ 1)! LJ.LpdJ.Lio /\ ... /\ dJ.Li/c p

= (k + 1)! dJ.Lio /\ ... /\ dJ.Li/c Property 2. If (cO, Vi) to 1.

= 1 for each vertex Vi,

= d!t'(c!).

then !t'(cO) is identically equal


292

Clearly, cO = L:i c?, where c? is the cochain dual to the simplex [ViJ. Therefore, cp(cO) = L:i J.L, - 1. Property 3. If c~ is the cochain dual to a simplex 6.~ - [Vio' ..• , Vik], then the form cp(~) vanishes identically in a neighborhood of the set M n \ st 6.~. By definition, k

cp( c~)

= k! L

( l)j J.LtJ dJ.Lio 1\ ... 1\ dlttJ 1\ ... 1\ dJ.Ltk. ]°

For each i, the function J.Li and the form dJ.Li vanish identically on G t ; therefore, the form cp(~) vanishes on the set G iO U ... U G tk . This set contains a neighborhood of Mn \ st 6.!. Property 4. p 0 cp

= id.

We prove this property by induction on k. First, consider the case k = O. Let c; be the cochain dual to the simplex [ViJ. Then cp(c?) = J.Li and (pcp(c?) , [v]]) - Jv J] J.Li = J.Lt(v]). If i =1= j, then Vj stvi; outside st Vi, the function J.Li vanishes identically. Moreover, J.Li( Vi) = L:j J.Li( Vj) = 1. Therefore, the cochain pcp (c?) is dual to [Vt], i.e., pcp (c?) = c? Now, suppose that k ~ 1 and Property 4 holds for all cochains of dimension k - 1. Let c~ be the cochain dual to a simplex 6.~ - [Vio'···' VikJ. According to Property 3, the form cp(c~) vanishes outside st 6.~; therefore, it vanishes identically on any k-simplex different from 6.~. It remains to show that (pcp(c!) , 6.!) = 1. Let c~/ be the co chain dual to the simplex

rt

~~,jl = [Vio' ..• ' Vi J , .•. , VtkJ. Then 8c~,OI = ~+cf+·· ·+c~, where c~, . .. , ~ are the cochains dual to k-simplices containing ~~ 0 1 as a face (and different from ~~). But we just proved that (pcp(cf) , ~!) ~ 0 for t =1= s. Therefore, (pcp(8c~ 1), ~~) _ (pcp(~), ~~). On the other hand, Property 1 and the Stokes formula imply

°

(the last equality follows from Property 4 for cochains of dimension k - 1). This completes the proof of the surjectivity of p.. Indeed, Property 1 implies that a induces the homomorphism cp*: Hk(Mn; JR) - t HtR(Mn) , and Property 4 implies p* 0 cp* = id. Therefore, p* is an epiluo phism. Step 2. The map p* is a monomorphism.


Suppose that m assertions are valid.

~

293

1 and .6.m is a simplex in IRn. Then the following

(A k ), k ~ O. Let w k be a closed form defined in a neighborhood of m 8.6. ; if m = k + 1, then we additionally assume that fall'" w k = O. Then there exists a closed form a k defined in a neighborhood of .6. m and such that a k = w k in a neighborhood of 8.6. m. (Bk), k ~ 1. Let wk be a closed form defined in a neighborhood of the simplex .6. m, and let a k - 1 be a form defined in a neighborhood of its boundary 8.6. m. Suppose that w k = da k - 1 in a neighborhood of 8.6. m. If m = k, then we additionally assume that faLlIe a k- l = fLlIe wk. Then there exists a form pk-l = a k- 1 defined in a neighborhood of 8.6. m and such that f3 k- l = a k- 1 in a neighborhood of 8.6. m and df3 k- l = wk in a neighborhood of .6.m . The additional assumptions are needed because the Stokes theorem must hold for the required form a, and this imposes certain constraints on the ini tial form w. We start by proving (Ao); then, we prove that (Ak t) ~ (Bk) and (Bk) ~ (Ak)' (Ao) The closedness of the O-form (function) wo means that this form is constant on each connected component. If m ~ 2, then the set 8.6. m is connected; we can assume that a neighborhood of 8.6. m is connected as well. In this case, the form wo is equal to a constant c, and we set aO = c. If m = 1, then 8.6. m is disconnected; but then m = k + 1, and the condition faLll wO = 0 must hold. Let.6. 1 = [vo, VIJ. Then this condition has the form wO(vt} = wO(vo); therefore, the form wO takes the same value c on both connected components, and we again set aO = c. (Ak-l) ~ (Bk) Let wk be a closed form defined in a neighborhood of .6. m . According to the Poincare lemma, we have w k = df3f- 1 for some form f3f- l (we assume that the neighborhood of the simplex .6.m is diffeomorphic to the open n-disk). In a neighborhood of 8{)"m, consider the form -yk-l = a k- l f3f- l . We have d-yk-l = wk - wk = O. Moreover, if m = k, then faLlIe -yk-l = faLlIe a k- 1 - faLlIe f3~-1 = fLlIe dw k - fLlIe df3~-l = O. Applying (Ak-l) to the form -yk-l, we see that there exists a closed form -y:-l defined in a neighborhood of {)"m such that -y~-l = -yk-l = a k - 1 -f3f- l in a neighborhood of 8.6.m . The form f3 k- l = f3~-1 + -y~-l has the required properties. (Bk) ~ (Ak) Let wk be a closed form defined in a neighborhood of 8{)"m, where {)"m = [vo, ... , vmJ. Consider the set (8{)"m) \ [VI, .. . , vmJ, which is homeomorphic to the open (m - I)-disk. We can assume that its neighborhood is homeomorphic to the open n-disk. Applying the Poincare lemma,

294


we obtain a form fjk-I defined in this neighborhood and satisfying the equality dfjk-I = wk. In particular, dfjk-I = w k in a neighborhood of 86.m - l , where 6. m - 1 = [VI, ..• , v m ]. If m > 1, then we can apply assertion (Bk) to the forms w k and fjk-I and the simplex 6. m 1. We must only verify that ILlm. 1 w k = IaLl'" 1 fjk-I for m -1 = k. Let c = 86. m - 6.m - 1 . Then 8c = _86. m - 1 and any simplex from the chain c is contained in the neighborhood in which the form fjk-I is defined. Therefore, ILl'" 1 w k - IaLl'" 1 fjk-I = ILlm 1 W k + Iacfjk-l = ILl'" 1 W k + Ic wk = IaLl'" W k = O. Applying (Bk-l), we obtain a form fj~ I defined in a neighborhood of the simplex 6. m - 1 and such that fj~-I = fjk-l in a neighborhood of 86. m - 1. The forms fj~-l and fjk-l coincide on the intersection of their domains; hence we can sew them together so as to obtain a form et~-l defined in a neighborhood of 86.m - 1 and satisfying the equality det~-l = Wk. Let us extend it to a form etk - 1 defined in a neighborhood of 6. m. Consider a function >. that takes the value 1 in a small neighborhood of 86. m and vanishes outside a slightly larger neighborhood of 86.m. The form et k - I = >'et~-l has the required properties. It remains to consider the case m = 1. In this case, the form w k is defined in neighborhoods of the vertices Va and VI; we can assume them to be disjoint (and homeomorphic to the open n-disk). According to the Poincare lemma, there exists a form et~-l defined in these neighborhoods and satisfying the equality det~-l = wk. A form et k - I defined in a neighborhood of the segment [vo, VI] is constructed from et~-I in precisely the same way as above. The injectivity of p. is implied by the following lemma, which by using (Bk).

IS

proved

Lemma. Suppose that wk is a closed form on Mn and p(w k ) = 6ck- 1 for some cochain ck- 1 E C k - I (Mn; lR) . Then there exists a form et k - 1 on Mn for which det k - 1 = w k and p(et k - 1 ) = ck - 1 .

Proof. Recall that we have fixed a triangulation f: IKI --+ Mn. We construct the et k - 1 by induction on the dimension ofthe skeleton of the complex K. To be more precise, we shall construct a sequence of forms et~-l , ... ,et~ I such that (i) each form et~-l is defined in a neighborhood of the m-skeleton of K, and det~-l = w k in this neighborhood;

(ii) the form et~11 coincides with et~-l in a neighborhood of the skeleton;

111.-

(iii) p(et~-_\) = ck - I • The last equality is understood as follows. The map p is defined only for forms defined on the entire complex K, but in reality, only the restrictions of k-forms to the k-skeleton is used. Therefore, we can assume that p is


295

defined for the forms a~ 1 with m ~ k - 1; moreover, p(a~ \) and for a k - 1 we can take the form a~ 1.

= p(a~ 1),

The form a~ 1 is constructed as follows. Take disjoint neighborhoods of the vertices VI, ... , Vr of K. In each of these neighborhoods, we apply the Poincare lemma and construct a form with differential wk. If k > 1, then these forms determine the required form a~ 1. For k 1, we must ensure that p(ag) = co, i.e., ag(v~) - (eO,v~). This is easy to achieve by adding a constant Ci to the function ag in a neighborhood of each point VI. We can construct the form a~ 1 from a~ 11 for each simplex ~ m separately. Indeed, the common domain of the forms thus constructed is contained in a neighborhood of the (m I)-skeleton, and these forms coincide with a~-\ on this domain. The construction of a~ 1 is based on applying (Bk) to the forms Wk and a~ \. To apply (B k ), we must check that if k = m, then Ja~ka~ \ - J~kwk. But J~kwk = p(w k ) - (6e k 1,~k) (e k 1, a~k) - (p(aZ a~k) Ja~k a~ll. If m = k - 1, then it is also required that J~k 1 aZ ~ = (e k 1,~k-l). Let us try to achieve this by adding a closed form {3k 1 to the form already constructed. The form to be added must be defined in a neighborhood of the (k - I)-skeleton and vanish in a neighborhood of the (k - 2)-skeleton, and the integral J~k 1 (3k 1 must take a given value for each simplex ~k 1; in other words, p({3k 1) must coincide with a given co chain dk 1. Let us show that the form (3k-l = ep(dk- l ), where ep: ck(Mn) --+ nk(Mn) is the map defined on p. 290, has the required properties. Recall that by Property 3, the form ep(e!) vanishes identically in a neighborhood of Mn \ st~!. In particular, it vanishes in a neighborhood of the (k -I)-skeleton. Therefore, the form {3k 1 vanishes in a neighborhood of the (k - 2)-skeleton, and d{3k-1 = ep(6d k 1) vanishes in a neighborhood of the (k - 1)-skeleton. 0

D,

Finally, let us show that I\-products of forms correspond to '--'-products of cohomology classes under the de Rham isomorphism. Suppose that forms and w~ correspond to cohomology classes a P and {3q. Consider the form (piwi) 1\ (P2w~) on the manifold M n x Mn; here PI,P2: M n x M n --+ M n are the natural projections onto the first and second factor. The Fubini theorem implies that this form corresponds to the cohomology class a P ® {3q because

wi

{

~rx~

(Piwi) 1\ (P2W~) = 6pr6qs ( [

~r

wi) ({~~ w~).

Let d: M n --+ M n x M n be the diagonal map. Then d* (a P ® (3q) and d*((piwf) 1\ (P2w~» = wi 1\ w~.

= aP

'--'

(3q

3.2. The Simplicial de Rham Theorem. Differential forms can be considered not only on smooth manifolds but also on simplicial complexes. The

296


idea of constructing such forms goes back to Whitney [154] and Thom [138]. But the most important role in the development of this theory was played by Sullivan's work [135], in which piecewise polynomial differential forms were applied to solving some problems of homotopy topology. Moreover, Sullivan developed the theory not only over the field R but also over the field Q (and, in general, over an arbitrary field of characteristic zero). Let XQ, .•• , Xn be the barycentric coordinates on a simplex assumed that E x, - 1. We refer to any expression of the form

~n;

it is

where Pil, ... ,i,. is a polynomial with rational coefficients, as a polynomial k-form on ~n. Polynomials with real coefficients can also be considered. Again, it is assumed that E Xi = 1 and, accordingly, E dXi = O. In addition to polynomial differential forms, we consider smooth differential k-forms on the simplex ~ n. They are expressed as

where I'l, ... ,i,. is a smooth function on an open subset of lRn containing ~n. A polynomial differential form on a simplicial complex K is obtained by sewing together polynomial differential forms defined on the simplices of K. These forms must be compatible in the sense that if two simplices ~l and ~2 have a common face ~12 and forms WI and W2 are defined on these simplices, then the restrictions of WI and W2 to ~12 must coincide. The restriction of a form to a face XI = 0 is defined by setting Xi = 0 and dXi - o. Smooth differential forms on a simplicial complex K are defined similarly. Note that smooth forms on a triangulation of a manifold are not the same as smooth forms on this manifold. For example, smooth functions (O-forms) on a triangulation of the circle are piecewise smooth (they may be nondifferentiable at the vertices of the triangulation). We denote the linear spaces of polynomial and smooth k-forms on a simplicial complex K by Ak(K) and A~oc (K), respectively. On the linear spaces Ak(K) and A~oc (K), the operator d acts and exterior product is defined. Importantly, for polynomial forms over Q, the form d(Pil, ... ,i,. dXil 1\ . .. 1\ dXi,.) can be expressed in terms of the partial derivatives of the polynomial Pi1, ... ,i,., which are also polynomials with rational coefficients. Polynomial forms have properties similar to those of dlfferential forms on manifolds.

297


Lemma 1. Suppose that K is a finite simplicial complex, C K is a cone over K, k ~ 1, and w k E Ak(CK) is a closed form. Then w k = do k - 1 for some form ok-l E Ak-1(CK).

Proof. Let us represent each point of the cone CK as >.x + (1- >.)a, where x E K, a is the vertex of the cone, and 0 :$ >. :$ 1. Consider the map p.:CKxI--+CK, p.(>.x

+ (1 -

>.)a) = >'(1 - t)x +

(>. + t(I -

>.))a,

and the form p.·(w k ) on CK x I. Let ~ be a simplex in K. The restriction of p.·(w k ) to CK x I can be represented as Ep>o(tPop(~) + tP{3p(~) /\ dt), where op(~) and (3p(~) are forms on ~ not containing t and dt. These forms determine forms op and (3p on the entire complex K. For t = 0, p. is the identity map; thus, for t = 0 and dt = 0, the form p.·(w k ) coincides with wk. This means that 00 = wk. For t = 1, the map p. is constant; thus, for t = 1 and dt = 0, the form p.·(w k ) is zero (any constant map induces the zero map of k-forms for k ~ 1; the condition k ~ 1 is essential). This means that Ep~o op = O. Now it is time to use the closed ness of the form w k , which implies d(p.·(w k )) = p.·(dw k ) = 0, i.e., d(Ep~o tPo p +tP{3p/\dt) = O. Here d(tPop) = tPdo p + (_I)degopptp-lop /\ dt and d(t P{3p /\ dt) = t P d{3p /\ dt. Thus, dop = 0 and (_I)degoppop + d{3p-l = 0 for all p ~ O. The above equalities imply

~I)

d(L(-I)degQP p~O

P

=L(_I)degop d!PI =-Lop=oo=w k . 0 p~O

p~l

P

Lemma 2 (on extension). For any form w E Ak(a~n), there exists a form

n E Ak(~n)

such that its restriction to a~n coincides with w.

Proof. Consider the projection of the simplex ~n = [vo, . .. , v n ] minus the vertex Vo, from the point Vo onto the opposite face [Vb ... , v n ]. In barycentric coordinates, this map is written as (xo, ... ,xn )

~ (~, ... ,~). 1- Xo

1- Xo

We denote it by 7To· Let Wo be the restriction of the form w to the face [Vb ... , v n ]. Consider the form 7Towo on ~n \ {xo}. If Wo

= L ~1 ..• ilt(YI"'"

Yn) dYil 1\ ... /\ Yilt,

where YI. ... , Yn are the barycentric coordinates on [VI, ... , V n ], then

7Towo = L

Pil ... ilt

(-1 Xl , ... , -1 Xn ) d(I Xii ) - xo - Xo - Xo

1\ ... /\ d(I Xilt

- Xo

)\

298


But d ( 1 X~o) -

(1

X,i d:'o1:' dxo. Hence there exists a positive integer N

such

that the form (1- xo)N 7rowo = wo is polynomial. The form wo is well defined on the entire simplex ~n; indeed, Xo = 1 at the vertex Vo, so wo vanishes at

Vo· The face [VI, ... , vnl is determined by the equation Xo = 0; therefore, the form Wo coincides with won this face. Thus, the form w - wo 8l!..n vanishes on the face [VI, ... ,vnl. By construction, if the form w vanishes on a face XI - 0, then the form w - Wol8l!..n also vanishes on this face. In other words, given a form w on a~n that vanishes on several faces, we can construct a form WI on ~n that vanishes on the same faces and, in addition, coincides with won some other face. Now we can construct the required form by induction because the form w = a can be extended to ~ n in an obvious way. D

Remark. A similar assertion is valid for smooth functions on ~ n. Its proof coincides with the proof of the extension lemma with the only difference that the function (1- xo)N should be replaced by a smooth function taking the value 1 in a neighborhood of the face [Vb ... , vnl and vanishing in a neighborhood of the vertex vo. Lemma 3. Let w k be a closed polynomial form on a simplex ~ n vanishing on a~n; for k = n, we assume in addition that Il!..n w k = a. Then w k = da k - I for some form a k - I E Ak-I(~n) vanishing on a~n (w k = a for k = 0). Proof. For n = 0, only the zero form exists. Suppose that n = 1 and ~ 1 = [0,11. If k = 0, then there exists a polynomial P(x) such that P'(x) = a for all X and p(a) - P(I) - a. Therefore, P(x) = a for all x. If k = 1, then the I-form P(x) dx is defined, and, by assumption, IOI P(x) dx = a. Let Q(x) = P(t) dt = a. Then d(Q(x)) = P(x) dx and Q(a) = Q(I) = a. Let us apply induction on n. The implication (An-I) =? (An) (where (An) is the assertion of the lemma for ~n) follows from (An-I) =? (Bn) =? (An), where (Bn) is the following assertion: Suppose that k ~ 1 and w k is a closed polynomial form on a~n; for k = n - 1, we assume in addition that I8l!..n w n- I = a. Then w k = da k- I for some polynomial form a k - I on a~n. First, we prove the implication (An-I) =? (Bn). Let w k be a closed polynomial form on a~ n. Consider the face [VI, ... ,vn1 of the simplex ~ n = [vo, ... , vn1· We set K = a[VI,"" vn1 and denote the union of all faces of ~n except [VI, ... , vn1 by CK. The Poincare lemma (Lemma 1) implies that wklcK = d/3k-I for some form /3k-I E Ak(CK). Applymg the extension lemma, we extend the form /3k-I to a form {Jk-I on a~n. The form w k {Jk-I is closed, and it may differ from zero only on the face [VI,' .. 'Vn1 j on

Ie:


K

= 8[vl, ... , V n ], 0=

299

this form vanishes. If k - n - 1, then

faoan w k = faoan(w k -

pk-l)

= {

(w k - pk-l).

j[Vl •... ,Vn]

Thus, we can apply (An-I) to the restriction of the form w k - pk-l to the simplex [VI, ... ,vnl and obtain (w k - pk l)![Vl, ...• Vn ] = 8,l-1 for some form ,k-l on [VI, ... , vnl vanishing on 8[Vl, ... , vnl. The form ,k-l can be extended to a form .:yk-l on 8fj.n vanishing on CK. Moreover, w k = d(pk-l +.:yk 1). Now, let us prove the implication (Bn) ~ (An). Suppose that w k is a closed polynomial form on ~n vanishing on 8fj.n; for k = n-I, we assume in addition that foa n w k = O. Obviously, if k = 0, then w k = O. Suppose that k ~ 1. In this case, we can apply the Poincare lemma and construct a form fJ k - l E Ak-l(fj.n) for which d{jk 1 _ wk. The form {jk-l may be nonzero on 8fj. n. We make it vanish as follows. Suppose that k = 1 and n ~ 2. Then the function {jk-l is constant on 8fj.n. Therefore, we can assume that {jk-l vanishes on 8fj.n. Now, suppose that k ~ 2. If k = n, then {

joan

{jk-l =

{

jan

d{jk-l =

{

jan

dJ...Jk =

o.

Thus, we can apply the assertion (Bn) and obtain {jk-ll oan = d.:yk-2, where .:yk-2 is a polylinear form on 8fj. n. Take an extension ,k-2 of .:yk-2 over fj.n and let o:k-l = {jk-l - d,k-2. We have do: k - l = d{jk-l = w k and o:k-ll oan = (jk-ll oan - d.:yk-2 = o. 0 We are ready to formulate and prove the simplicial de Rham theorem. Let p: A"'(K) _ C"'(K; Q) be the map defined by (p(w), fj.n) = fan w n . According to the Stokes theorem, p dJ...J = 5pw; hence p induces the map p'" of cohomology algebras. Theorem 5.13. The map p'" is an algebra isomorphism. Proof. We start by proving that p'" is an isomorphism of vector spaces; then, we prove that p'" preserves multiplication.

First, let us show that p is an epimorphism. Consider the form wi = dXl 1\ ... 1\ dXn on the simplex fj.i. This form is closed because there are no nonzero (n + I)-forms on an n-simplex. Let us verify that wi loa:, = o. This is obvious for the faces given by the equations Xl = 0, ... ,Xn = 0; for the face given by Xo = 0, this follows from the relation dXl + ... + dX n = -dxo = O. Thus , we can take a form on the n-skeleton which coincides with w!' on fj.!'I I and vanishes on the other n-simplices. Applying the extension lemma, we extend this form to the entire simplicial complex K. As a result, we obtain


300

a form n~ E An(K) such that (p(ni), ~j) = c6iJ , where c =1= 0 is a rational number. It follows that p is an epimorphism. We have obtained the short exact sequence

o ----t Ker p ----t A*(K) ...!!..... C*(K; Q)

----t

O.

Here Ker p consists of all forms wn E A*(K) such that L~n wn = 0 for any simplex ~ n in K. To prove that p* is an isomorphism, it suffices to verify that H*(Ker p) = 0; in other words, we must show that if wn E A*(K), dw n = 0, and J~n w n = 0 for all simplices ~n in K, then w n = da n - 1 for some form an 1 E Ker p (this means that J~n 1 an 1 - 0 for any simplex ~n-l in K (if n > 0) or wn = 0 (if n = 0)). According to Lemma 3, there exists a form (Jr- 1 E An-l(~i) for which wnb~ = d{J~ 1 and {J~-118~~ = O. Sewing all such forms {J~-1 together, we obtain a form ~n-l on the n-skeleton of K which vanishes on the (n - 1)-skeleton. Using the extension lemma, we extend the form ~n 1 to a form {In-l on the entire K. The closed form wn - d{Jn-l vanishes on the n-skeleton, i.e., on the boundary of any (n + 1)-simplex. Again applying Lemma 3, we construct a form 'Yj-l E An+l) lor r d'Yjn-l and 'Yjn-11 8~n+l = 0 . A n-l( i..JJ.j w h'ICh (n W - dan-1)1 tJ ~n+l = ]

]

Then, we sew together the forms 'Yj-l so as to obtain a form .:yn-l on the (n+1)-skeleton, which we extend to a form 'Yn - 1 on the entire K. The closed form w n - d({Jn-l + 'Yn - 1) vanishes on the (n + 1)-skeleton. Thus, we can repeat the construction. At the end, we obtain a n - 1 = {In-l + 'Y n - 1 + .... This sum is well defined because the form obtained at the kth step vanishes on the (n + k - 2)-skeleton. The form a n - 1 has the required properties because the forms {In-l, 'Yn - 1, . " vanish on the (n - 1)-skeleton. Now let us prove that p* preserves multiplication. Take a, (J E Z*(K; Q). Recall that the class [a] '-' [(J] E H*(K; Q) is the image of the class [a ® (J] under the map induced by the diagonal map d: IKI -+ IK x KI. Take wi,wr E A*(K). Let L be the triangulation of IK x KI whose vertices are the products of vertices of K. We define the form wi x w2 E An+m(L) as follows. Any simplex ~ in L is contained in the product ~l x ~2 of two simplices from K. Consider the canonical projections Pi: ~1 x ~2 -+ ~i (i = 1,2). Sewing together the restrictions of the forms (Piwi) 1\ (P2wr) (which are defined on ~l x ~2) to ~ for all simplices ~, we obtain a form on L, which we denote by wi x wr. The map p: A *(L) -+ C* (L; Q) takes this form wi x wr to a cochain p(wi x wr) for which (p(wi x wr),Sf x ~~) = J~ix~~(Piwi) 1\ (P2 wr) = 6np6mq(J~n wi) (J~m wr); the second equality follows from the Fubini theo1 2 rem. Thus, p(wi x wr) = p(wi) ® p(wr). Moreover, it follows directly from the definition that d*(wi x wr) = wi 1\ wr. 0

Chapter 6

Miscellany

1. The Alexander Polynomial A natural way to construct an invariant of a link L C 8 3 is to consider the space 8 3 \ L. The homology groups of this space form an invariant that is too rough because these groups are the same for all links with the same number of components. There is a finer invariant related to homology, namely, the Alexander polynomial. It can be defined in several ways. One of them is to paste up the link L with a surface F (i.e., take a self-avoiding surface F with boundary L) and construct the so-called Seifert form by using this surface. This approach involves certain difficulties, which arise in the proof of the independence of the invariant thus constructed of the choice of F. An approach avoiding these difficulties is as follows. We construct an infinite cyclic covering Xoo over 8 3 \ L. There is a natural automorphism t: Xoo ---+ X oo , which depends only on L. This automorphism defines the structure of a module over the ring Z[C 1 , t] on Hl(Xoo). It turns out that this module structure is closely related to the matrix determining the Seifert form (see Theorem 6.2). Such is the program. Let us implement it. 1.1. The Seifert Form. In this section, for a surface F embedded in the sphere 8 3 , we construct a bilinear form on Hl(F). We use this form in what follows to define the Alexander polynomial.

Let F be a sphere with g handles from which n ~ 1 open disks are removed (we assume that F is compact). Then X(F) = 2 - 2g - n. Clearly, the surface F is homotopy equivalent to a wedge of circles. The Euler characteristic of the wedge of k circles is equal to k - 1; therefore, the surface F is homotopy equivalent to the wedge of 2g + n - 1 circles. Thus,

-

301

6. Miscellany

302

HI (F) '" z2g+n 1. For generators of the group HI (X) we can take the homology classes of the cycles shown in two equivalent ways in Figures I and 2. Figure I shows a sphere with 9 handles with n disks removed, which is standardly embedded in 3-space. The parallels and meridians on the handles represent the generators of the I-dimensional homology groups; the remaining generating cycles are homologous to the boundaries of n - I removed disks. This surface is homeomorphic to a disk with 2g pairs of bands attached to the upper part and n - I bands attached to the lower part, as shown in Figure 2. Indeed, both surfaces are orientable, their Euler characteristics coincide, and their boundaries have equally many components. It is easy to explicitly construct a homeomorphism between the surfaces and see that the system of generating cycles on one surface is mapped to a system of generating cycles on the other surface. It is seen from Figure 2 that each of the cycles shown in this figure corresponds to a circle from a wedge of circles homotopy equivalent to the surface under consideration.

Figure 1. A homology basis

Figure 2. A homology basis (an equivalent presentation)

1. The Alexander Polynomial

303

Suppose that the surface F is embedded in S3. Then, according to the Alexander duality theorem, we have HI (Fj Z) ~ HI (S3 \ F). The space F is homotopy equivalent to the wedge of 2g + n - 1 circlesj therefore, HI(FjZ) ~ z2g+n I ~ HI(F). The isomorphism HI(S3 \ F) ~ H1(F) can also be proved directly. Let us do this in order to obtain an explicit description of this isomorphism. The most important point in the proof is that the bilinear form f3: H 1(S3 \ F) x H1(F) - Z (which assigns to any two closed oriented curves in S3 \ F and in F the linking number of these curves) is nondegenerate.

If E > 0 is sufficiently small, then the closed E-neighborhood V of the surface F in S3 is a 3-disk to which 2g + n - 1 handles D2 x I are attached. Moreover, the inclusion F c V is a homotopy equivalence. The boundary 8V is a sphere with 2g + n - 1 handles. Corresponding to each handle are two generators, c, and I:, in the group HI (8V); c, is the boundary of a small disk transversally intersecting I, in one point and disjoint from!; for j =1= i, and I: is the element of HI (8V) mapped to Ii under the homomorphism of homology groups induced by the inclusion 8V C V (this homomorphism takes Ci to 0). We choose the orientation of Ci so that lk(ci' I,) = 1. Then lk(ci' I,) = dij. Let V' be the closure of the complement S3 \ V. Then the inclusion V' C S3 \ F is a homotopy equivalence because so is 8V c V \ F. In the Mayer Vietoris sequence H2(S3)

--+

HI (8V)

--+

HI(V) EB HI (V')

--+

H 1(S3),

the first and last terms are zero; therefore, HI (8V) ~ HI (V) EB HI (V'). In particular, HI (V) and HI (V') are free Abelian groups. We know the ranks of the groups H 1(8V) and H1(V); thus, HI(V') ~ z2g+n-l. This group is generated by the images of the elements Ci under the homomorphism induced by the inclusion 8V C V'. Thus, the groups H1(V) ~ H1(F) and HI(V') ~ H1(S3 \ F) have bases h, ... , !2g+n-l and CI, .•. , e2g+n-l such that lk(ci' I,) = dij. The bilinear form f3: H 1 (S3 \ F) x HI(F) - Z determines the bilinear form a: HI(F) x H1(F) - Z defined as follows. Since the surface F is orientable, we can choose a positively oriented basis VI, V2 at each of its points. Let n be the unit normal vector to F for which the basis VI, V2, n is positively oriented. Note that it is not always possible to choose the vectors VI and V2 so that they depend continuously on the corresponding point of F, but the vector n depends only on the orientation determined by VI and V2, rather than on these vectors themselves. Therefore, the vector n continuously depends on the point. Choose a sufficiently small number E > o. For each curve 'Y on the surface F, consider the curve 'Y+ obtained by translating each point of'Y in the direction of the chosen normal by a distance of Ej the curve obtained by applying a similar translation in the opposite

6. Miscellany

304

direction is denoted by 'Y-. We set a(x, y) = (3(x-, y). The bilinear form a is called the Seifert form. For the basis of HI (F) we again take !I, ... , !2g+n-I. In this basis, the Seifert form is determined by the Seifert matrix A = (aij), where aij = a(J"fJ) = Ik(J,-,fJ ) -lk(Ji,f/). Let el, ... ,e2g+n 1 be the basis of the group H 1 (S3 \ F) dual to !I, ... , !2g+n-l with respect to the form {3. Then I, - L J aiJej and I J+ = Li aije,. 1.2. Infinite Cyclic Coverings. Consider a compact orient able surface F with boundary OF = L embedded in 8 3 . Let us construct a space Xoo that is an infinite covering of 8 3 \ L. To this end, we cut the sphere S3 along F. To be more precise, we take the space 8 3 \ F, add L to it, and for each a E F \ L, add points a+ and a-, the respective limits of points converging to a from each of the two sides of F. As a result, we obtain a 3-manifold M3 with boundary F+ U F-; corresponding to each point a E F \ L are two points, a+ E F+ and a- E F- (since the surface F is orient able, we can choose compatible signs + and - for all points). We take countably many copies M,3 (i E Z) of this manifold and identify Fi~l with F i- (by and a; that correspond to the same point a). identifying the points This construction is schematically shown in Figure 3. The space Xoo thus obtained covers 8 3 \L. The maps Ml- Mi~t-1 taking each point of Ml to the corresponding point of the manifold Ml-tl ~ Ml induce an automorphism t: Xoo - Xoo of this covering. The automorphism group of the covering is generated by t and is isomorphic to Z; it acts transitively on the covering.

a;_l

a

b

c

d

Figure 3. The space Xoo

We assume that the surface F is oriented and the points a+ and acorrespond to the positive and negative directions of the normal. We have constructed the covering Xoo _ 8 3 \ L by using the surface F. It turns out that this construction does not depend on the choice of F; it depends only on the oriented link L (the boundary of F). Indeed, any covering over 8 3 \ L is uniquely determined by the image of the fundamental group of the covering space in 71"1 (8 3 \L). Let us show that this image consists of those loops 'Y for which the linking number with L is zero (we assume that if L has connected components L1, ... ,Ln , then lk(-y, L) = L lk(-y, Li)).


305

Clearly, the image consists of the loops I whose liftings to Xoo are closed. In other words, going from any point of Mg along the lifting of I, we must return to Mg. The transitions between M,3 and Mi~H correspond to the intersection points of I with Fj the sign is the same as in the definition of the intersection number. We arrive at the condition that the intersection number of the curve I with the surface F equals o. According to one of the definitions of linking number (see [108]), this is equivalent to the vanishing of the linking number of I with L = 8F. The automorphism of the covering t: Xoo -+ Xoo does not depend on the surface F either. It is determined by those loops in S3 \ L for which the linking number with L equals 1 (the starting point of the lifting of I is mapped to the ending point of this lifting). For each oriented link L C S3, there exists a connected oriented surface F embedded in S3 so that its boundary is L (see, e.g., [105]). This surface is called a Seifert surface of the link L. Using the surface F, we can construct a covering Xoo -+ S3\L and an automorphism t: Xoo -+ Xoo ofthis covering which depends only on L. Thus, with the link L we can associate the group HI (Xoo) on which the automorphism t. is defined. This means that HI (Xoo) is a module over the ring Z[C I , tJ, whose elements are polynomials in t I and t with integer coefficients. To apply this algebraic structure, we need a matrix describing the specification of a module by generators and relations (a presentation matrix). Below, we give a definition of such a matrix and prove some of its properties. Let M be a module over a commutative ring R with identityj it is assumed that 1m = m for all m E M. The module M is said to be free if there exist elements ml, ... ,mn E M such that any element m E M has a unique representation in the form m = rlml + .. ·+rnmn , where ri E Rj the elements ml, ... ,mn form a basis of the module M. A module M is finitely generated if there exists an exact sequence

(40)

F

Q

---+

E

'P

---+

M

---+

0,

where E and F are free modules (over the same ring R). Take bases em and h, ... , fn in E and F. Let A = (aij) be the matrix of the map a in these bases (Le., aUi) = E.i=l ajiej). The matrix A is called a presentation matrix for the module M. Each of the m rows of A corresponds to a generator of M, and each of the n columns corresponds to a relation between generators. For a given module M, the exact sequence (40) is not uniquej moreover, we can take different bases in E and F. eI, . . . ,

Theorem 6.1. Let A and A' be two presentation matrices for the same module M. Then they can be obtained from each other by the following transformations (and their inverses):

6. Miscellany

306

(1) a permutation 01 rows or columns; (2) the replacement 01 a matrix B by the matrix ( ~ (3) the addition 01 the zero column; (4) the addition to a row (column) 01 another row (column) multiplied by a number.

g );

Proof. The exact sequences corresponding to the matrices A and A' give the commutative diagram

F~E~M----+O

l'

19 ~' }d

a'

F' -----+ E' -----+ M -----+ 0 . Indeed, the homomorphisms 9 and I can be specified on the bases and then extended by linearity. The element g(ej) E E' is defined as follows. The map cp' is an epimorphism; hence there exists an element ej E E' for which cp'(e~) = cp(eJ ). We set g(eJ ) = ej; then cp'g(ej) = cp'(ej) = cp(ej). The element I(fi) E F' is defined as follows. The equality cp' ga(f,) = cpo:(fi) = 0 implies the existence of an element I: E F' for which a'(fD = ga(fi). We set I(fl) = II; then a'I(f,) = a'(fD = ga(f,). Let X and Y be the matrices of the maps I and 9 in the bases in which the matrices A and A' are written. The equality ga = a'i means that XA = A'y. The maps f' and g' with matrices X' and Y' for which X' A' = AY' are constructed in the same way as I and g. The equalities cp' 9 = cp and cpg' = cp' imply cpg' 9 = cp' 9 = cpo Therefore, Im(g'9 - idE) C Ker cp = 1m a. Using this property and the freeness of the module E, we can construct a homomorphism h: E --+ F such that ah = g'g - idE. For matrices, this equality takes the form AZ = X'X - I, where Z is the matrix of h. We shall write P '" Q if the matrix Q can be obtained from the matrix P by applying transformations (1) (4) and their inverses. Clearly,

A ~ (A 0) ~ (A X')

o

I

I

0

(3):...,(4)

Taking into account the relation X' A' =

X' X' A') ~ (A (A o I A' 0

Next, using the equality

(~

X' I

X' I

(A X' X' A'\ I

0

o

A'

X') ~ (A0

X X

I

J.

AY', we obtain X' o XX') 0) (3):...,(4)

(A 0

XX' = AZ + I, we obtain A'

A'

X' I

0 A'

I

A'

X

.


307

Therefore,

o

as required.

Let A be a presentation matrix of a module M over a ring R. If A has m rows and n columns, then the ideal of R generated by all minors of order m - r + 1 of the matrix A is called the rth elementary ideal of the module M and denoted by er. Using standard properties of determinants and applying Theorem 6.1, we see that the ideal er does not depend on the choice of A. We assume that er = R for r > m and er = 0 for r < o. It is easy to show that r 1 C er. Note that if m - n, then el is the ideal in Z[t 1, tl generated by the polynomial det A.

e

Problem 124. Let G be a finite Abelian group regarded as a Z-module. Prove that for this module, the ideal el is generated by the number IGI equal to the order of G. 1.3. Fundamental Theorem. The fundamental theorem, which relates the Seifert matrix A to the structure of the module HI (Xoo) over the ring Z[t 1, tl, is as follows. Theorem 6.2. The matrix tA - AT is a presentation matrix of the module H 1 (Xoo ) over the ring Z[C 1 , tl. Proof. We use the notation of the preceding section. Let us represent the space Xoo as the union of its subspaces M' = UMii+1 and M" = UMii . The intersection of these subspaces is a countable set of copies of the Seifert surface. In the Mayer Vietoris sequence ... --+

Hi(M' n M")

--+

Hi(M') EB H,(M")

--+

Hi(Xoo )

--+ ... ,

all groups are modules over the ring Z[t- 1 , tl. The action of t interchanges the groups Hi(M') and Hi(M"); hence it is not defined on any of these groups, but it is defined on Hi(M') EB Hi(M"). The maps in the Mayer Vietoris sequence are module homomorphisms. We show that the segment

Hl(M' n M")

J..:.... HI (M')

EB HI (M") ~ HI (Xoo)

of the Mayer Vietoris sequence determines a presentation of the module HI (Xoo) by generators and relations. First, we verify that i. is an epimorphism. This is so if and only if Ker 8. = HI (Xoo) , i.e., 1m 8. = o. The latter equality is equivalent to the map j.: Ho(M' n M") --+ Ho(M') EB Ho(M")

6. Miscellany

308

being a monomorphism. The groups Ha(M' n M") and Ha(M') ffi Ha(M") are direct sums of the groups Ha(M: n M:+1) and Ha(M:), and each of these groups is isomorphic to Z because the Seifert surface F is connected. Therefore, the modules Ha(M' n M") and Ha(M') ffi Ha(M") are isomorphic to Z[t 1, tj; the element t k corresponds to 1 E Ha(M: n M:+1) and 1 E Ha(M:). Under this identification, the map i" takes t k to t k + t k +1. In particular, i,,(I) - 1 + t i- OJ therefore, i" is a monomorphism. Choose a basis {I,} in HI(F) = HI(M: n M:+1)' and let {ei} be the basis in H I (S3 \ F) - HI(M:) dual to {I,} with respect to the form (3. The modules H 1 (M' n M") and HI(M') ffi HI (M") are free. Their bases consist of the elements 1 ® J.;, and 1 ® ei. Corresponding to the elements I, E H 1 (M: n M:+1) and e, E HI (M:) are t k ® I, and t k ® e,. It is easy to show that

i,,(1 ® I,) = t ® li+ - 1 ® li-· Indeed, consider I, E H 1 (MJ n MP). The restriction of I, to MJ C M" is li- (the curve is translated in the direction opposite to that of the normal), and the restriction of II to MP c M' is It (the curve is translated in the direction of the normal). By definition, i" = (l, -iff); thus, l(1 ® Ii) E HI(MP) = t ® HI (MJ) and i"(1 ® II) E HI (MJ). Using the expressions for It and li- in terms of the Seifert matrix A = (aij), we obtain

i,,(1 ® I,) =

L aji t ® ej - L aij ® ej. j

j

Thus, the matrix of the map i" in the chosen bases is tA - AT. But the matrix of i" coincides with the presentation matrix of HI (Xoo). 0 The elementary ideal £r of the module Hl(Xoo ) over the ring Z[rl, tj is called the rth Alexander ideal of the oriented link L. A generator of the minimal principal ideal containing the rth Alexander ideal is said to be an rth Alexander polynomial of the oriented link L. The first Alexander polynomial is called simply the Alexander polynomial; it is denoted by ~L(t). A generator of the principal ideal is determined up to multiplication by a unity (an invertible element) of the ring. In the ring Z[rl, tj, the unities are ±tn , where n E Z. Therefore, the rth Alexander polynomial is determined up to multiplication by ±tn. According to Theorem 6.2, the module HI(Xoo ) has a square presentation matrix, namely, tA - AT. Therefore, the first elementarv ideal of this module is principal, and hence ~dt) ~ det(tA - AT); here and in what follows, the symbol ~ denotes equality up to multiplication by ±tn.

309


1.4. Properties of the Alexander Polynomial. First, note that if K is a trivial knot, then ~K(t) ~ 1. Indeed, in this case, we have Xoo = D2 X JR, whence HI (Xoo) = o. Calculating the Alexander polynomial reduces to calculating the Seifert matrix A = (aij), where aij = lk(/i, it). Let us calculate, for example, the Seifert matrix for the trefoil. By cutting and gluing, we can easily show that the surface in Figure 4 is a torus with a hole (see [105] for more details). Hence, to calculate the Seifert matrix for the trefoil, we can choose generators of the homology groups as in Figure 4. It is easy to show that the curves and are arranged as shown in Figure 5. Therefore, A = (~ -t) and tA - AT = (tIl t=t1 ). Thus, the Alexander polynomial for the trefoil is equal to t 2 - t + 1.

rt

ii

Figure 4. The trefoil

Figure 5. The curves and It

I.

Note that, in calculating the Seifert matrix, it is very convenient to use the "plane" Seifert surface. This surface for the trefoil is shown in Figure 6. Such a Seifert surface exists for any link. Indeed, look again at Figure 2. Let

Figure 6. The plane Seifert surface

us deform the Seifert surface so that the handles attached to the disk become long and narrow. Were it not for the possible twisting of the handles (see Figure 7a) it would be obvious how to arrange the Seifert surface. Since the surface is oriented, it follows that the number of twistings of each handle

6. Miscellany

310

---X'--~> N(T). Since T is maximal, we have N(g)e = T. Therefore, N(g)e contains all maximal tori containing g. Now suppose that n E N(g)e. Consider a maximal torus S of N(g)e containing n. Any element of S commutes with g; this allows us to use Lemma 1, which implies the existence of a maximal torus T containing S and g. This maximal torus contains nand g. D Theorem 6.30. An element 9 EGis regular of and only if the dimension of its normalizer N(g) is equal to that of a maximal torus. For any singular element, the dimension of its normalizer is larger than that of a maximal torus.

Proof. First, suppose that 9 is regular, i.e., belongs to precisely one maximal torus T. Then, according to Lemma 2, we have N(g)e = T, whence dimN(g) = dimN(g)e = dimT. Now, suppose that 9 belongs to two different maximal tori, Tl and T 2 • According to Lemma 2, the normalizer N(g) contains both Tl and T2· Thprefore, the tangent space N (g) at e contains the sum of the tangent spaces to Tl and T2. These spaces are diffcrent, and hence the dimcnslon of their sum is strictly larger than that of a maximal torus. 0

5. Lie Groups and H -Spaces

355

Roots. Suppose that G is a compact connected Lie group, 9 = TeG is its Lie algebra, and (u, v) 1 is an inner product in g. Consider the inner product defined by (u, v) = IgEc(Adg u, Adg vh. This new inner product is invariant with respect to Adg; that is, the map Adg : 9 -+ 9 is an orthogonal transformation for any g E G. In what follows, we assume that 9 is endowed with this inner product. Theorem 6.31. Let T c G be a torus. Then the space 9 decomposes into a direct sum as 9 = VoffiE~l Yr, where the spaces Vo, VI,··., Vm are invariant with respect to the action of T by means of Ad and this action is trivial on Vo, each of the spaces Vr with 1 ~ r ~ m is two-dimensional, and, in some orthogonal basis, the action of each element t E T on Yr is determined by the matrix

where Or: T

-+

Hi/Z is a nontrivial homomorphism.

Proof. Take a topological generator to of the torus T. The transformation Ad to is orthogonal; therefore, in some orthonormal basis, its matrix is the direct sum of the identity matrix and several matrices of the form (

21TClr sin 21TCl r

COS

- sin 21TClr) cos 21TCl r .

Clearly, Adtko = (Adto)k, and the map Ad t depends continuously on t. Hence the matrix of Ad t in the same orthonormal basis has the required form. 0 The set of numbers {±Cll, ... , ±Clm} that appear in the proof of Theorem 6.31 is determined uniquely; therefore, so is the set of homomorphisms {±Ol, ... , ±Om} in the statement of this theorem. For a maximal torus T, these homomorphisms Or: T -+ lR/Z are called roots. Together with the homomorphisms Or. the linear maps Or: t -+ lR for which the diagram

is commutative are often considered. The map Or is the lifting of Or 0 exp for which Or(O) = O. Such a lifting exists and is unique because the topological space t is simply connected. The linear maps Or: t -+ Hi are called roots too. Theorem 6.32. A torus T is maximal if and only if Vo

= t.

Proof. Clearly, t c Vo for any torus. Suppose that T is a maximal torus and Vo f= t, Le., there exists an X E Vo such that X ¢ t. On the group

356

6. Miscellany

H consisting of the elements etX , where t E JR, the torus T acts triviallyj therefore, the subgroup generated by T and H is connected and Abelian, and it contains T as a proper subset. This contradicts the maximality of T.

Now, suppose that Vo = i, T and T' are tori, and T is contained in T'. Then T acts trivially on t', and therefore t' c Yo. Hence t c t' c Vo = t, which means that dim T = dim T'. Thus, T = T', and the torus T is maximal. 0 Corollary. The dimension of any compact connected Lie group G is equal to 1+ 2m, where 1 is the rank of G and 2m is the number of roots. For each root Or: T --+ JR/Z, we denote its kernel by Ur . Clearly, Ur is a subgroup in T of dimension 1 - 1, where 1 = dim T (= the rank of the Lie group G). Theorem 6.33. If an element t E T belongs to precisely J.L groups Ur , then dim N(t) = 1 + 2J.L. Proof. Let V C 9 be the subspace on which the given element t acts trivially. By Theorem 6.31, we have V = Vo EB LUr 3t If,.j therefore, dim V = 1 + 2J.L. Thus, it is sufficient to prove that V = TeN(t). The elements of N(t) commute with tj hence tnC 1 = n for any n E N(t), which means that t acts trivially on TeN(t). Therefore, TeN(t) C V. Now, suppose that X E V. Then tXC l = X, and hence te>'xC l = e>'x for any ~ E lR. Therefore, e>'x E N(t) for all ~ E lRj thus, X E TeN(t). 0 Corollary. An element t E T is regular if and only if it does not belong to any of the subgroups Ur . Theorem 6.34. The singular elements of a Lie group G form a subset of codimension at least 3 in G in the sense that this set is the image of one or several smooth compact manifolds of dimension at most dim G - 3 under a smooth map. Proof. First, let us show that the group Ur has a topological generator. The connected component (Ur)e of the identity in Ur is a torus of co dimension 1 in the torus T. The quotient of a torus by a torus is a compact connected commutative Lie group, that is, a torus. Therefore, T / (Ur)e is a torus of dimension 1, i.e., the circle S1. The group Ur/(Ur)e is a finite subgroup in T/(Ur)e ~ Slj hence Ur/(Ur)e ~ Zk for some positive integer k. Let t be a topological generator of the torus (Ur )e, and let 9 E Ur be an element such that the group Zk is generated by the coset g(Ur)e. fis in the proof of Lemma 1 on p. 354, we choose an element 8 E (Ur)e for whirh tg- k = 8 k and consider h = g8. The same argument shows that h is a topological generator of Ur .


357

Let h be a topological generator of Ur . According to Theorem 6.33, we have dimN(h) ~ l + 2. If n E N(h), then n commutes with h m for any integer mj therefore, n commutes with all elements of Ur . Consider the map f: G x Ur - G defined by f(g, u) = gug- 1 . If n E N(h) and u E Ur, then nun- 1 = Uj hence f(gn, u) = f(g, u). Thus, in fact, we have a map fr: G/N(h) x Ur - G. The image of fr consists of all elements of G conjugate to the elements of Ur . The dimension of the manifold G/N(h) x Ur is equal to dimG - dimN(h)

+ dimUr

~

dimG - (I

+ 2) + (I-I) = dimG -

3.

An element of the group G is regular if and only if it does not belong to the image of any of the maps h, ... , fm. Indeed, the normalizers of conjugate elements always have the same dimension, and therefore an element is regular if and only if none of the elements of the maximal torus T conjugate to it belongs to any of the subgroups Ub ... , Um . 0 Proof of the Equality 7r2 (G) = o. Let Greg be the set of regular elements of a compact connected Lie group G, and let Gsing = G \ Greg. By Theorem 6.34, the set Gsing has co dimension at least 3 in G. Therefore, any smooth map 8 2 - G admits an arbitrarily close approximation whose image is contained in Greg. This means that the equality 7l"2(G) = 0 for a compact connected Lie group G is implied by the following theorem (as applied to X = 8 2).

Theorem 6.35. Any continuous map from a simply connected space X to a compact connected Lie group G whose image is contained in Greg is nullhomotopic. Proof. Let T be a maximal torus in G with Lie algebra t. Consider the map cp: G x t - G defined by cp(g, X) = ge X g-l. If t E T and X E t, then te X C I = eXj we obtain a map cp: (G/T) x t _ G, where G/T is the space of left cosets gT. Let treg be the set of those X E t for which eX E Greg. Clearly, X E

treg if and only if ge X g-1 E Greg for any 9 E G. Thus, we have a map CPreg: (G/T) x treg - Greg, and this map is surjective because so is cp. Let us show that cpreg is a local homeomorphism. It suffices to verify that the Jacobian cpreg vanishes nowhere.

Lemma. A point (g,X) is regular for the map cp: (G/T) x t only if X E trego

G if and

6. Miscellany

358

Proof. Since eX and dX commute, we have e X + dX over, (g + dg)-l = 9 - 9 l(dg)g-l. Therefore, (g

+ dg)e X +dX (g + dg)-1 _

= eX + eX dX.

More-

ge X g-1

= (dg)e x 9 1 _ ge X g-l(dg)g-1 + g(e X dX)g-l = g(g-1 dg _ eX (g 1 dg)e- X + eX (dX)e X)e X g-l. Here dg is an element of the tangent space at gj hence dg is an element of the Lie algebra. Thus, g-1 dg = dz.

= 9 dz,

where dz

Below, we use the notation of Theorem 6.31. The map id - Adex vanishes on Va = t, and it is nonsingular on v;. (r = 1, ... , m) if and only if Or(e X ) i- 1, i.e., eX ¢ Ur . Thus, the restriction of id - Adex to E:.n=1 Vr is nonsingular if and only if the element eX is regular, i.e., X E trego Clearly, the restriction of Adex to t is an automorphism, so it is always nonsingular. D Consider the commutative diagram

7 X

(G/T) x

ireg

l~reg

~ (G/T) x t C

I Greg

U

1~ I G.

Here ii. is a lifting of u. Similarly to the case of coverings, it is constructed using liftings of paths starting at a base point Xo EX. The lifting does not depend on the choice of paths from Xo to x, since X is simply connected. Using natural embeddings, we obtain the commutative diagram

(G/T) x t

X

/

1~

U

IG.

The space t is contractible, and therefore the map ii. is homotopic to a map X - (G/T) x {O}. Since cp(g, 0) = e is the identity element of G, it follows that cp takes (G/T) x {O} to the identity element e. D 5.5. H-Spaces and Hopf Algebras. A topological space X with a base point Xo E X is called an H-space if it is equipped with a continuous map /L: X x X - X such that the maps x ~ /L(x,xo) and x ~ /L(xo,x) are homotopic to the identity map (it is assumed that /Ltxo. xo) = xo). The map /L is called a multiplication. The Hopf theorem (Theorem 6.41) shows that the cohomology algebra of an H -space over a field of characteristic zero has a very special form.


359

One of the corollaries of this theorem is that H2(G; JR) = 0 for any simply connected Lie group G. Natural examples of H-spaces are Lie groups and topological groups. But there exist other examples. Example 80. The spaces Cpoo and Rp oo are H-spaces. Proof. Let us identify Coo \ {OJ with the space of polynomials by assigning the polynomial ao + alz + ... + anz n to each point (ao, al, ... , an, 0, ... ). Multiplication of polynomials induces a multiplication in Coo \ {OJ for which (1,0,0, ... ) is an identity element. Identifying proportional polynomials, we obtain a multiplication in the space Cpoo with the identity element (1 : 0 : 0 : ... ). For JRpoo , the construction is similar. 0 Example 81. The sphere 8 7 is an H-space. Proof. Let us represent 8 7 as the set of Cayley numbers with unit norm. Multiplication of Cayley numbers induces a (nonassociative) multiplication on S7. 0 Theorem 6.36. If X is an H -space with base point xo, then the group xo) is Abelian.

71'1 (X,

Proof. A homotopy between loops afJ and fJa can be constructed as follows. Consider the map of the inner rhombus in Figure 33 defined by (s, t) 1-+ Q

fJ

Figure 33. The construction of the homotopy

J.L(a(s), fJ(t)). Maps of the remaining four triangles are constructed by using homotopies between the maps s 1-+ a(s) and S 1-+ J.L(a(s), xo) and between the maps t 1-+ fJ(t) and t 1-+ J.L(xo, fJ(t)). 0 Theorem 6.36 has the following generalization. Theorem 6.37. Any H -space X is simple. Proof. Clearly, 71'n(X X Y, (xo, Yo)) ~ 71'n(X, xo) x 71'n(Y, Yo), and the action of the group 71'1 (X x Y, (xo, Yo)) corresponds to a product of actions; namely, (a,fJ)(u,v) = (au.fJ v ).

6. Miscellany

360

The multiplication J.t: X x X - X induces the commutative diagram

1I"1(X x X)

X

1I"n(X

X

X)

~

1I"n(X

l~·x~.

X

X)

1~·

1I"1(X) x 1I"n(X)

) 1I"n(X),

where the horizontal arrows represent the action of the group

11"1

on 1I"n.

Let el and en be the identity elements of 11"1 (X) and 1I"n(X). Then, for any a E 1I"1(X) and U E 1I"n(X), we have (a, el)(e n , u) = (ae n , elu) = (en, u). Therefore, J.t*(a, edJ.t*(e n , u) = J.t*(en, u), and hence au = u. D Let F be the additive group of some field, and let X be a connected H-space for which all groups Hi(X; F) ~ Hi(X; F) are finite-dimensional spaces over F. Consider the cohomology algebra A = ffi,>o Ai, where Ai = Hi(X; F). The diagonal map d: X - X x X determines a multiplication A ® A - A in the cohomology algebra. The multiplication J.t: X x X - X induces a map ~ - J.t*: A - A ® A. This map is called a coproduct. The coproduct ~ has the following properties. 1. The map ~: A - A ® A is a homomorphism of algebras, i.e., ~(1) 1 ® 1, and if ~(a) - Eali ® a2i and ~(b) = Eb 1j ® b2j , then ~(a ~ b)

= L(_1)dimb1J dima 2 '(ali

=

~ b1j ) ® (a2i ~ b2j).

I,j

This follows from the Kiinneth theorem for cohomology with coefficients in a field and from the multiplication law in the algebra H*(X x X). 2. Let PI: A®A - A be the map defined by a P®l 1-+ a P and aP®a q 1-+ 0 for q > 0, and let P2: A ® A - A be a similar map taking 1 ® a P to a P. Then both compositions in the diagram

A--~~)A®A

1

lpl

~

P2

A®A

)A

are the identity. In other words, we have ~(aP) = aP®l+l~aP+ E ar®a B for r, s ~ 1. This is so because both compositions in the diagram

X(

~

~r

XxX

r~(·,XO)

X x X (~(xo,·) X are homotopic to the identity map.


361

3. We have ~(AP) C EBi+i=p Ai®Ai. This follows from the isomorphism HP(X x X) ~ EBi+i-pHi(X) ® Hi (X) for cohomology with coefficients in a field. A graded5 associative algebra A = EBi>O Ai for which a coproduct ~ with properties 1 3 is defined is called a Hopf algebra. If AO coincides with the ground field F, then the Hopf algebra is said to be connected (this condition corresponds to the path-connectedness of the space X). Theorem 6.38. Suppose that char F = 0 and A is a connected Hopf algebra generated multiplicatively by an element x E Ak, where k ~ 1 and k is even. Then A is the polynomial algebra F[x] . Proof. Clearly, ~(x) = 1 ® x + x ® 1 because there are no elements of positive dimension strictly less than k. Since ~ is an algebra homomorphism, ~(xn) = E~-o xr ® xn-r. (Note that for odd k, the corresponding equality is quite different. For example, ~(x2) = 1®x 2+x 2®1 because the two terms x ® x have opposite signs in this case.) Suppose that xn = 0 for n ~ 2 and n is the minimum number with this property. Then 0 = E~~i xr ® x n - r , which is impossible. D Remark. If k is odd, then x 2 = o. Therefore, A = A(x) is an exterior algebraj it consists of all elements of the form a + bx, where a, bE F. Example 82. The sphere s2n is not an H-space. Proof. Suppose that the sphere s2n is an H-space. Then A = H*(S2nj JR) is a connected Hopf algebra. The algebra A is generated multiplicatively by an element x E H2n(s2nj JR). Therefore, by Theorem 6.38, we have xk i- 0 for all k. On the other hand, x 2 = o. D

For a field F with characteristic Pi- 0, Theorem 6.38 is false because the equality xn = 0 may hold for n = pm. To prove a version of Theorem 6.38 for fields of nonzero characteristic, we need the following lemma. Lemma. Suppose that n = E nipi and r Then (~) == TI, (~:) (mod p).

= E ripi,

where 0 ~ ni, ri ~ p-l.

Proof. Suppose that 1 ~ k ~ p - 1. Then (~) is divisible by p. Therefore, (1 +x)P == 1 +xP (mod p). Applying induction on k, we obtain the relation 5We assume that ab = (-l)pqba for a E AP and b E Aq.

6. Miscellany

362

(1

+ X)pk

1 + Xpk (mod p). Hence

(1

+ x)n

+ x)no+nlP+··+nmpm (1 + x)n0(1 + x)n 1P ... (1 + x)nmpm (1 + x)n0(1 + xP)n 1 ••• (1 + xpTn)n", (1

-

IT f= (~t)xkP' t

(mod p).

0k 0

Thus, the coefficient of xr = xro+rlP+,,+rmpm in the expression for (1 + x)n is congruent to (~~) (~:) ... (~::) modulo p. On the other hand, it equals ~. 0 This lemma implies, in particular, that (P;) = 0 (mod p) for 0 < r < pm. Therefore, L~ xr I2l xn r - 0 for n = pm. Moreover, if L~-t xr I2l xn r _ 0 and xr i= 0 for 1 :=:; r :=:; n - 1, then n - pm for some m. Indeed, suppose that n = no + nIP + ... + nsps, where ns i= 0 and either nq i= 0 for q < s or ns i= 1. In the former case, we set r = nqpq and obtain (~) == (~:) (mod p) and (~:) - 1. In the latter case, we set r = pS and obtain r < n,

t

(~)

= (~.)

(mod p), and ns) = n s , where 2 :=:; ns :=:; p - 1. These congruences imply the following version of Theorem 6.38.

Theorem 6.39. Suppose that char F - p, where p is an odd number, and A is a connected Hopf algebra generated multiplicatively by x E Ak, where k ~ 1. Then A = A(x) if k is odd, and either A = F[x] or A = F[xJl(x pm ) if k is even. This theorem has an analog for p = 2; its formulation and proof are similar to those of the above theorem, except that the equality x 2 = 0 may not hold for odd k and A(x) = F[xJl(x 2). Thus, for a field of characteristic 2, the analog of Theorem 6.38 is as follows. Theorem 6.40. Suppose that char F = 2 and A is a connected Hopf algebra generated multiplicatively by an element x E Ak, where k ~ 1. Then either A = F[x] or A = F[xJl(x 2Tn ). Theorem 6.41 (Hopf [61]). Suppose that A is a connected Hopf algebra over a field F of characteristic zero and the space An is finite-dimensional for any n. Then the algebra A is isomorphic (as a Hopf algebra) to the tensor product of an exterior algebra on odd-dimensional generators and a polynomial algebra on even-dimensional generators. Proof. Since the space An is finite-dimensional for any n, the algebra A has generators XI,X2, ••• such that dimxi :=:; dimxi+1 (it is assumed that


363

Xi E Adimx.). Let An be the sub algebra of A generated by Xl. ... ,Xn . It is a Hopf subalgebra, i.e., ~(An) C An ® An. Indeed, we have ~(xd = Xi ® 1 + 1 ® Xi + ~ Yr ® Ys, where dim Yr, dim Ys < dim Xi. Therefore, Yr, Ys E A i - l C AI. We assume that Xn ¢ A n - l (otherwise, the generator Xn can be eliminated).

According to Theorem 6.38, the algebra generated by Xn is the polynomial algebra F[xnl (if the dimension of Xn is even) or the exterior algebra A(xn) (if the dimension of Xn is odd). Therefore, since the algebra A is graded commutative and associative, there is a natural epimorphism A n - l ® F[xnl --+ An (if the dimension of Xn is even) or An 1 ® A(xn) --+ An (if the dimension of Xn is odd). It is sufficient to prove that this epimorphism is a monomorphism (and then apply induction on n). Consider the ideal I generated by x~ and the elements of A n - l of positive dimension. It consists of all expressions of the form ~ akx~, where ak E An 1 and the zero-dimensional components of ao and al vanish. Clearly, Xn ¢ I because any element of I of dimension dimx n must belong to A n - l (the ideal I has no elements of the form AXn , where A E F). Consider the composition An ~ An ®An ~ An ® (An/I),

where p is the natural projection. If a E A n - l is an element of positive dimension, then ~(a) = a ® 1 + 1 ® a + Lai ® a l , where aj E An 1 has positive dimension. Therefore, p~(a) = a ® 1 (this equality holds for any a E F). Moreover, ~(xn) = Xn ® 1 + 1 ®xn + ~Yr ®Ys, where Ys E A n- l is an element of positive dimension. Hence p~(xn) = Xn ® 1 + 1 ® xn, where Xn is the image of Xn in An/I. First, consider the case where the element Xn has even dimension. Suppose that there is a nontrivial relation ~ akx~ = o. Let us apply the map p~ to this relation. Taking into account the equality x~ = 0, we obtain

o= L

(ak ® l)(xn ® 1 + 1 ® Xn)k

= L akx~ ® 1 + L kakx~-l ®

xn·

By assumption, we have L akx~ = OJ hence ~ kakx~-l ® xn = o. It follows that ~ kakx~-l = o. Indeed, as mentioned above, we have Xn E I, whence xn f. o. The degree of the relation ~ kakx~-l = 0 is lower than that of the initial relation L akx~ = o. Clearly, this relation is nontrivial because the field has characteristic zero. Choosing an initial relation of minimum degree, we immediately obtain a contradiction. Now consider the case where the dimension of Xn is odd. Suppose that there is a nontrivial relation ao + alX n = O. Applying the map p~ to it, we

6. Miscellany

364

obtain 0= ao ® 1 +

al

® l(xn ® 1 + 1 ® xn)

=

+ alxn ) ® 1 + al ® Xn· al ® Xn = o. As in the preceding

By assumption, ao + alx n = 0; therefore, case, we have xn I- O. Hence al = 0 and ao

(ao

= O.

0

Corollary 1. Let G be a Lie group. Then the cohomology algebra H*(G; 1R) is an exterior algebra on generators of odd dimensions. Proof. The polynomial algebra is infinite-dimensional; hence a finite-dimen0 sional algebra cannot contain a polynomial subalgebra. Corollary 2. If G is a simply connected Lie group, then H2(G; JR.)

= O.

Proof. Since G is simply connected, Hl(G;JR) = O. Therefore, all generators in the cohomology algebra H*(G; 1R) are of dimension at least 3. 0

Hints and Solutions

1. Obviously, the required isomorphism holds even at the level of relative chains. 2. This commutative diagram can be regarded as a short exact sequence of chain complexes 0 - c~ - c. - c: - 0, where C~ is the chain complex . .. - 0 - 0 - A ~ A' - 0; the complexes C. and C: are defined similarly. The nontrivial homology groups of the complex C~ are precisely Kera and Cokera. 3. The proof of Theorem 1.13 applies to the case under consideration without any substantial changes, except that isomorphism should be replaced by monomorphism or epimorphism. 4. First, let us prove that 'P2 induces a homomorphism of the required form. Take x E Ker('P3G2). Clearly, 'P2X E Im'P2. It remains to show that 'P2X E Im,lh = Ker,82. But 0 = 'P3G2X = ,82'P2X. If x E KerG2, then x E 1m Gl; therefore, 'P2X E 1m 'P2Gl. If x E Ker 'P2, then 'P2X = o. Thus, the map is well defined. It is also clear that is a homomorphism. Now let us prove that is an epimorphism. Take y E 1m 'P2 n 1m ,81. There exists an x E A2 for which 'P2X = y. Moreover, 'P3G2X = ,82'P2x = ,82Y = 0 because Y E Im,81 = Ker,82. Hence x E Ker('P3G2). Finally, let us prove that is a monomorphism. Suppose that x E Ker( 'P3(2) and 'P2X E Im( 'P2GI) , i.e., 'P2X = 'P2GIZ for some zEAl. Then x = GIZ + w, where w E Ker 'P2. Hence x E Ker G2 + Ker 'P2. 5. (a) Consider the isomorphism / = "11 1"12. The required equality is equivalent to / + /-1 = 2 id, i.e., /2 + id = 2/, and the latter equality can be written in the form (J - id)(J - id) = o.

-

3 65

Hints and Solutions

366

Let us show that (f id)Q2 - 0 and Q3(f - id) - O. The former equality is derived as follows: Til 1172Q2 - 171 1cp2(32 - 1711171Q2 = Q2. To prove the latter, note that CP4 Q3 = (33171, i.e., Q3 = CP4 1(33171, whence Q31711172 = cP 4 1(33171171 1172 - cP 4 1(33172 = cP 4 1CP4 Q3 = Q3· Thus, Im(f - id) c KerQ3 - ImQ2 and ImQ2 c Ker(f id); therefore, (f - id)2 - o. (b) First, suppose that such a homomorphism d exists. We set 172 171 + 171 d. The first two conditions ensure the commutativity of the diagram, and the last condition implies that 172 is a homomorphism of rings. Let us show that 172 is a monomorphism. If 172(X) - 0, then 171 (x) = 171 (-d(X)); therefore, x - -d(X), which means that d(x) -d2(x) - 0 because Imd C Ker Q3 - 1m Q2 C Ker d. Let us show that 172 is an epimorphism. Suppose that y E B3 and y - 171 (X). Then 172(X - d(X)) - 171 (X - d(x)) -171(d(x)d2 (x)) - 171 (x) - y. Now suppose that there exists a diagram with two different isomorphisms 171 and 172. Consider d 171 1172 - id. It is easy to show that d has all the required properties. (c) The commutative diagrams cp 1/J o -------t Z2 -------t Z2 X Z2 -------t Z2 ~ 0

Ihl

cp

l~

1/J

Ihl

o -------t Z2 -------t Z2 X Z2 -------t Z2 -------t 0 , where i = 1,2, 171 - id, 172 (X, y) - (y, x), cp(X) - (x, x), and tf;(x, y) = x + y, give the required example. 6. Let us represent the torus T2 as the union of simplicial complexes Ko and Kl and choose generators a and b in H1(Ko nKd (see Figure H.1). The exact sequence

Ht{KonKd

(10. 31),

H 1 (Ko )ffiH1 (K1 )

---+

HI (T2)

---+

Ho(KonKd

---+ 0

gives an exact sequence

o ---+ 1m cP ---+ HI (T2)

---+

Z ---+ 0,

where r.p = (jo, -j1). The group Hl(Ko n Kd consists of elements of the form na + mb. The homomorphism r.p takes any such element to the pair (njo(a) + mjo(b), -njl(a) - mj1(b)). But jo(a) = jo(b) and jl(a) = jl(b). Therefore, cP has the form (m, n) ~ (n + m, -n - m). Hence Ker cP = Z and 1m cP = Z. Since the exact sequence 0---+ Z ---+

splits, we have HI (T2) ~ Z ffi Z.

HI (T2)

---+ Z ---+ 0

Hints and Solutions

367

Figure H.1. The torus T'J.

From the exact sequence

o ~ H2(T2) ~ HI(Ko n Kd

(3D. il).

HI (Ko) E9 HI(Kd

we obtain an exact sequence

o ~ H2(T2) Therefore, H2(T2) ~ Kercp ~

~ Kercp ~

o.

z.

1. According to Problem 37 in Part I, we have SP x sq / SP V sq ~ Sp+q. Therefore, Hk(SP x sq, SP V sq) ~ Hk(SP x sq U C(SP V sq)) ~ Hk(Sp+q) for k ~ 1. 8. Writing the exact sequence for the pair (SP x sq, SPV sq) and applying Problem 7, we see that Hk(SP x sq) ~ Z for k = 0, p, q, p + q (if p = q, then Hp(SP x sq) ~ z E9 Z)j the remaining groups are trivial. 9. (a) Suppose that Ko is the closed E-neighborhood of the given knot K and KI is the closure of S3\Ko. Then KoUKl = S3, and KonKl = T2 is the 2-torus. We must calculate the homology of S3 \ K '" K 1 . We assume that the sphere S3 is triangulated so that its triangulation induces triangulations of Ko and K I . First, note that the space KI is homotopy equivalent to a simplicial complex containing no simplices of dimension ~ 3. Indeed, we can successively kill all of the 3-simplices in K I , starting with the boundary. Thus, Hi(Kd = 0 for i ~ 3. Clearly, Ho(Kd = Z. Let us write the Mayer Vietoris sequence:

Hi+l(Ko) E9 HHl(Kd ~ HHl(S3)

~ Hi(T2) For i

=

k... Hi(Ko) E9 Hi(K1 )

~ H i (S3).

2, we obtain

0----. Z ~ Z ----. H2(Kd ~

o.

The homomorphism 8* is as follows. Any generator 0: of the group H3(S3) can be represented as a sum of simplices with compatible orientations into which the sphere S3 is decomposed. These simplices are divided into two

Hints and Solutions

368

parts according to whether they belong to Ko or K 1 . We define 8.a to be the common boundary of these parts. Clearly, 8. is an isomorphism. Therefore, H2(K1) - o. For i

=

1, we obtain 0---- Z ffi Z

~ Z ffi HI(Kd ---- O.

The map j. takes a parallel on the torus to a generator of the group HI(Ko) = Zj therefore, a meridian is mapped to a generator of HI(Kd. Hence H1(K1) = Z, and a generator of HI (K1) is represented by a small circle in a plane transversally intersecting the knot (see Figure H.2).

Figure H.2. A generator of the homology group of the complement of a knot

(b) As in (a), we obtain exact sequences 0---- Z ~ Zn ---- H2(K1) ____ 0

and 0---- z2n ~ Zn ffi HI(Kd ____ O.

We have 8.(1) = (1, ... , l)j hence H2(Kd = zn-l. The map j. takes parallels of n tori to generators of the group HI (Ko) = zn j therefore, meridians of these n tori are mapped to generators of H1(KI). Thus, HI(Kd = zn, and generators of this group are represented by small circles put on the link components. 10. Let us write the Mayer Vietoris sequence for Ko = M n \ int Dn and KI = D n : Hk(sn-l) ____ Hk(M n \ int Dn) ffi Hk(D n ) ____ Hk(M n ) - + Hk_l(sn-l). If k ~ 1, then Hk(Dn) = 0, and if 2 ~ k ~ n - 2, then Hk(sn-l) = 0 and Hk-I (sn-I) = OJ for k = 1, we consider reduced homology groups. 11. Let K' be the barycentric subdivision of the simplicial complex K. Consider the simplicial map f: K' -+ N(e) defined as follows. Each simplex ~ in the complex K is covered by one of the sub complexes L i . To the barycenter v of ~ we assign the minimum index i. We must show that if K' contains a simplex with vertices vo, VI, ... , Vk which correspond to

Hints and Solutions

369

sub complexes LiD, Lip ... , L ik , then N(£) contains a simplex with vertices LID' ... ' L ik , i.e., LID n ... n Lik =1= 0. We can assume that in the initial simplicial complex K, Vo is a vertex, VI is the midpoint of an edge, V2 is the center of a 2-face, etc. Then Vm E Lim implies Vo, .• ·, Vm 1 E Lim. Therefore, Vio E LiD n ... n L'k. Let us prove that f induces an isomorphism of homology groups. We use induction on n. For n - 1, the complex K = Ll is acyclic, and N(£) = * consists of one point. Thus, the basis of induction is obvious. The induction step is as follows. Suppose that £1 = {Ll. ... , L n }, Kl = Ll U ... U L n , and K2 - Ln+l. Consider the sub complex Nl = N(£l) of N(£) and let N2 be the closed star of the vertex (n + 1). By construction, f takes K~ to N l , and by the induction hypothesis, the restriction of f to K~ induces an isomorphism in homology. The restriction of f to K~ also induces an isomorphism in homology because the complex K2 - Ln+l is acyclic and N2 is a cone with vertex (n + 1). Let M, = Li U Ln+l. Then M = {Ml , ... , Mn} is a cover of the set Kl n K 2, to which the induction hypothesis applies. Therefore, f induces an isomorphism between the homology groups of the complexes Kl n K2 and N(M) = Nl n N 2. Let us write the Mayer Vietoris exact sequence: Hk(Kl

n K2) ------+ Hk(K1 ) E9 Hk(K2) ------+ Hk(K)

l~ Hk(Nl

n N2)

1

l~ ~ Hk(Nd E9 Hk(N2)

------+ Hk(N)

------+ Hk-l(Kl n K2)

----t

Hk-l(Kl ) E9 Hk-l(K2)

l~ ~ Hk-l (Nl

l~

n N2) ------+ Hk-l (Nd E9 Hk-l (N2) .

The pairs of the left and right maps are isomorphisms; hence, by the five lemma, the middle map is an isomorphism as well. 12. In case (b), consider the diagram

O~A

'P

>B

'I/J

)C----tO

1 1 1 idA

EB..p

ide

o~ A ~ A E9 C ------+ C ------+ 0 , where (
'n), then >.f+·· .+>.~ can be expressed as a polynomial with integer coefficients in the coefficients of the polynomial f, and xf + ... +:?n can be expressed bimilarly in terms of the coefficients of f. 0

Hints and Solutions

383

This lemma immediately implies the required assertion. 50. This is so because the definition of cup product for relative cohomology coincides with that for absolute cohomology. The only point to be noted is that L U 0 = L U L. 51. (a) The exact cohomology sequence

shows that the map HP' (K, L,) -+ HP' (K) is an epimorphism, that is, HP' (K, L,) has an element {3, mapped to 0,. Moreover, {31 ....., ... ....., {3n is mapped to 01 ....., ... ....., On. But {31 ....., ... ....., (3n E H·(K, U~-l L,) = o. (b) The required assertion follows from (a) because the suspension can be represented as the union of two contractible spaces (cones).

52. Let olsn be the image of 0 under the natural homomorphism H*(sn,Mk) -+ H*(sn). Then 0 ....., {3 = (olsn) ....., {3 = 0 ....., ({3lsn). If o < dimo < n, then 0 Sn = 0, and if 0 < dim{3 < n, then {3lsn = o. It remains to consider the case of dim 0 = dim {3 = n. In this case, we have 0""" (3 E H2n(sn, Mk) = O. 53. (a) Suppose that G and H are Abelian groups and cp: G -+ H and 1f;: H -+ G are homomorphisms such that 1/;cp = ide. Then H = 1m cp ffi Ker 1/;. Indeed, any element h E H can be represented as h = cp1/;h + (h - cp1f;h), where cp1/;h E Imcp and h - cp1/;h E Ker1/;. Moreover, if x = cpg and 1/;x = 0, then 9 = 1/;cpg = 1/;x = 0; therefore, 9 = 0, and hence x=O.

Applying this assertion to the homomorphisms i.: H.(A) -+ H.(X), r.: H.(X) -+ H.(A), r*: H·(A) -+ H*(X), and i*: H·(X) -+ H*(A), we obtain the required result. (b) The equality r*o ....., r·{3 = r*(o ....., (3) implies that 1m r* is a subring, and the equality i· (0 ....., (3) = i· 0 ....., i· {3 implies that if i* 0 = 0 or i· (3 = 0, then i*(o ....., (3) = O. 54. First, note that HP(X V Y) = HP(X) E9 HP(Y) for p ~ 1. The existence of natural retractions X V Y -+ X and X V Y -+ Y implies that H*(X) and H*(Y) are ideals in H·(X V Y) (see Problem 53). Therefore, (x + y) ....., (x' + y') = x ....., x' + y ....., y' for any cohomology classes x, x' E H·(X) and y, y' E H·(Y) of positive dimension. Indeed, X""" y' E H·(X) n H·(Y), and therefore x ....., y' = O. 55. Suppose that there exists a retraction r: SP x sq -+ SP V sq. Then, according to Problem 53, we have H*(SP x sq) = Ker i· ffi 1m r*. Let 1, oP, {3q, and oP ....., (3q be generators of the nontrivial cohomology groups of Sp x sq. It is easy to see that Keri· = Hp+q(SP x sq). Therefore, Imr·

Hints and Solutions

384

consists of all linear combinations of 1, o.P, and f3 q • Thus, o.P '-" f3 q ¢ 1m r* , and 1m r* is not a subring. We have obtained a contradiction. 56. (a) Let o.(n) and o.(m) be the generators of the groups HI (JRpn; Z2) and H1(JRpm;Z2). Suppose that there exists a retraction r: JRpn ---t JRpm. Then, according to Problem 53, the group H*(JRpn; Z2) -is the direct sum of Keri* and Imr*. Clearly, i*(o.(n») = o.(m); therefore, Ker1* consists of O,o.~rl, ... ,o.~)' and Imr* consists of 0, l,o.(n),o.fn), ... ,o.(n)" This is not a subring, while 1m r* must be a subring by Problem 53. (b) The solution is similar to that of (a). 57. (a) Take the same bases o.l, ... ,o.n,f31, ... ,f3n and o.~, ... ,o.~,f3~, ... ,f3:n of H1(M~) and Hl(Ml) as in Theorem 2.3. Let A = 0.1 '-" 131 and A' = o.~ '-" f3~ be generators of the groups H2(Mi) and H2(Ml). If R(o.I)

=

alo.~

+ ... + amo.~ + b1f3~ + ... + bm f3:n

and

then

I*(A) =

1*(0.1 '-"

f3d = (ald l

+ ... + amdm -

b1Cl - ... - bmCm)A'.

(b) The homomorphism h that takes 0.1 to o.~ and 132 to f3~ is not induced by any continuous map f: M? ---t Mi because 1*(0.1) '-" 1*(132) =

1*(0.1 '-" 132) =

o.

58. (a) Clearly, HO(Sk x Sk) ~ H2k(Sk X Sk) ~ Z and Hk(Sk x Sk) ~ Z EB Z, and all of the remaining cohomology groups are trivial. Let 0.,13 E Hk(Sk X Sk) be the cohomology classes dual (in the sense of linear algebra) to the homology classes of the cycles Sk x {xo} and {Yo} x Sk. Then 0. '-" 13 = A, where A is a generator of H2k(Sk x Sk) because Sk x {xo} and {yo} x Sk intersect transversally in only one point, namely, (Yo, xo). Let us show that 0. '-" 0. = 0 and 13 '-" 13 = O. The cycle Sk x {xo} is homologous to Sk x {x~}; to see this, it suffices to consider the chain Sk x I, where I is a path joining xo and x~. The cycles Sk x {xo} and Sk x {x~} are disjoint; therefore, 0. '-" 0. = o. Similarly, 13 '-" 13 = O. (b) First, note that the co cycle 0. + 13 is dual to the diagonal and that (0. + 13) '-" (0. + 13) = 0. '-" 13 + 13 '-" o.. For k odd, 0. '-" 13 = -13 '--' 0.; in this case, the self-intersection number of the diagonal vanishes. If k is even, then 0. '-" 13 + 13 '-" 0. = 20. '-" 13 = 2A, and therefore the diagonal has self-intersection number ±2. The sign depends on the orientations of the diagonal and Sk x Sk. The orientation of the diagonal is canonically determined by that of Sk. For even k, the orientation of the sphere Sk x Sk is canonically determined as well because the bases el, ... , ek, cI, ... , Ek

Hints and Solutions

385

and El, . .. , Ek, el, .•. , ek have the same orientation. Under this choice of orientations, the sign is plus. 59. Let 0 and fJ be the cohomology classes dual (in the sense of linear algebra) to the homology classes of the cycles sm x {xo} and {Yo} xsn. Then 0'-" fJ is a generator of Hn+m(sn x sm). Thus, it is sufficient to prove that /*(0) = 0 and /*(fJ) - O. But /*(0) E Hm(sn+m) - 0 and /*(fJ) E

Hn(sn+m) _ O. 69. Any contractible manifold is orientablej therefore, we have the Lefschetz isomorphism Hk(M n , aM n ) !:!:!. Hn k(Mn). Hence Hn(Mn, aM n ) Z and Hk(M n , aMn) - 0 for k i- n. It follows from the exact sequence for the pair (Mn,aM n ) that Hk(aM n ) '" Hk(M n ) for k < n-l and H n _ 1 (aMn) !:!:! Hn(Mn,aM n ) = Z. 70. Consider the following commutative (up to sign) diagram for homology and cohomology groups with coefficients in the field Q:

H 2 (M 3, aM 3) ~ HI (aM3)

r

~ ~[M3l H 1 (M 3)

""

,.

i.

r~[aM3l

)

HI (M 3 ) ""

r~[M3l

) HI (aM3) ~ H2(M3, aM3).

Since we consider homology and cohomology with coefficients in a field, the maps i* and i* are dual to each other in the sense of linear algebra. Any linear map A: U ---+ V determines the exact sequence

o -----+ Ker A

-----+

U

A

-----+

V

-----+

Coker A

-----+

o.

Passing to dual spaces, we obtain the exact sequence A·

o -- (Ker A)* - - rr -- V* - -

(Coker A)* - -

o.

Thus, Coker(A*) = (Ker A)*. Therefore, dim Coker i* = dim Ker i*, which means that dim 1m a = dim Ker i. = dim Coker i* = dim Coker a. As a result, we obtain dimlma = dim(HI(aM3)/lma) = dimH1 (aM 3) dimlma, i.e., dimlma = ~ dim HI (aM 3 ), as required. 71. According to the Alexander duality theorem, we have if k = n - 2, if k = 0, otherwise.

386

Hints and Solutions

Therefore, Z

H,(X)

zm = zm-l 0

if i = 0, if i = 1, if i = n - 1, otherwise.

72. According to the Alexander duality theorem, we have fI,(X) ~ fIn '(SP) for 0 < i s: n. Moreover, H n+1(X) = 0 because the space X is homotopy equivalent to an (n+ I)-manifold with boundary. Thus, fIi(X) = Z for i - n - p, and all the remaining reduced homology groups of X are trivial.

73. According to the Alexander duality theorem, we have fI,(X) ~ fIn-'(sP u sq) for 0 ~ i ~ n. Moreover, H,(X) = 0 for i ~ n + l. 60. (a) Let o(n) and o(m) be the generators of the groups HI (Rpn; Z2) and Hl(Rpm; Z2). If /*(O(n») = 0, then /*(otn») = 0, where otn) = O(n) '-" ... '-" o(n) is the generator of Hk(Rpn; Z2). Suppose that /*(O(n») =1= 0, i.e.,

/*(O(n») = o(m). Then /*(0(:;1) = 0(~)1 =1= 0 because n + 1 ~ m. On the other hand, 0(:;1 = O. We have obtained a contradiction. (b) The map f. is dual to /*. 61. Suppose that such a map g exists. Then the diagram

where Pm and Pn are double coverings, is commutative. According to Problem 60(a), the map f!H): HI (Rpm; Z2) --+ HI (Rpn; Z2) is zero; therefore, so is the map f!7f): 71"1 (Rpm; Z2) --+ 71"1 (Rpn; Z2). Indeed, if n = 1, then the homomorphism f.: Z2 --+ Z is zero for obvious algebraic reasons, and if n > 1, then we can identify the generators of the groups HI (Rpn; Z2) and 71"1 (Rpn) and the generators ofthe groups HI (Rpm; Z2) and 7I"} (Rpm). After such an identification, we obtain f!7f) = f!H). The map f induces the zero homomorphism of fundamental groups; hence it has a lifting

Hints and Solutions

387

We have PnjPm = fPm = Png. For a point x E 8 m , either g(x) = jPm(X) or g( -x) - jPm(X) = jPm( -x). Since the maps 9 and jPm coincide at one point, they coincide everywhere because the map of covering spaces that covers a given map of the bases of coverings is completely determined by the image of one point. This is impossible because 9 takes x and -x to different points, whereas Pm takes them to the same point. 62. (a) This problem is solved by using Theorems 2.11 and 2.12, which describe multiplication in the cohomology rings of the spaces Rpn and cpn. In addition, Problem 51 should be applied. (b) The sets {(xo : Xl : ••. : x n ), X$ '# O} are contractible (they are homeomorphic to open disks) and cover Rpn (cpn). 63. Let a be a generator of the group HI(cpn). Then r(a) = '\a, where ,\ E Z. Therefore, f*(a k ) = (f*a)k = ,\kak • In calculating A(f), we can use maps of both homology and cohomology groups. Hence A(f) 1 +,\ +,\2 + ... + ,\n. 64. According to Problem 63, we have A(f) = 1 +,\ +,\2 + ... +,\n = A~+lll. Therefore, if A(f) = 0, then ,\n+! = 1 and ,\ '# 1. Thus, ,\ = -1, and the number n is odd. 65. As in the solution of Problem 63, we have ran = ,\nan. Hence f*[cpnJ = ,\n[cpnJ. 66. (a) If n is even, then deg f = ,\n ~ 0 by Problem 65. (b) Let us show that the diffeomorphism cpm -+ cpm specified by (zo : ... : zm) 1-+ (zo : ... : zm) in homogeneous coordinates is orientationreversing for odd m. Indeed, in a neighborhood of the fixed point (1 : 0 : ... : 0), this diffeomorphism coincides with the map C m -+ C m defined by (ZI,"" zm) 1-+ (Zl,"" zm). This map considered as a map 1R2m -+ R 2m has determinant (-I)m. 67. The cohomology groups (with coefficients in Z) of the given spaces are isomorphic, but their cohomology rings are different: multiplication in the ring H*(Cp2) is nontrivial, whereas in the ring H*(8 2 V 8 4 ) it is trivial. The space CP2 is obtained by attaching D4 to 8 2 via the map 8D4 -+ 8 2 , which is a Hopf fibration. If this map were null-homotopic, then the space thus obtained would be homotopy equivalent to 8 2 V 8 4 • 68. The cohomology groups of the given spaces are Z in dimensions 0, m, n, and m + n (if m = n, then the cohomology group in dimension m = n is ZEBZ); the remaining cohomology groups are trivial. For the space 8 m X 8 n , the product of generators of the m- and n-dimensional cohomology groups

388

Hints and Solutions

is a generator of the (m + n)-dimensional cohomology group. For the space sm V sn V sn+m, multiplication in cohomology is trivial.

74. Let us write the Mayer Vietoris sequence for K 1 and K2 = M~n \ int D 4n :

= Mt n \

int D 4n

Here H 2n (S4n-l) = 0 and Il2n l(s4n-l) = OJ therefore, H2n(Kt)ffiH2n (Kd ~ H 2n (Mtn#Mr). Moreover, according to Problem 10, we have H2n(K,) ~ H2n(M~n). Clearly, there are 2n-cycles that generate H2n(M~n) and are disjoint from D4n (the disk D4n can be chosen to be arbitrarily small). Therefore, the intersection form of the manifold Mt n # M~n is the direct sum of the intersection forms of the manifolds Mt n and M~n.

75. Let zP and w q be cochains with values in G which represent the cohomology classes aP and bq. Then 8 (zP '-" w q), 8 ( zP) '-" w q, and zP '-' 8 (w q) is a cochain representing the classes [3*(a P '-'" bq), [3*(a P) '-" bq, and aP '-" [3*(b q). Clearly, 8(zP '-" w q) = 8(zP) '-" w q + (-l)PzP '-" 8(wq). 76. Obviously, Tor(Hp(MP), Hq_ 1(Nq» = 0 because Hp(MP) = 0 or Z. Therefore, according to the Kiinneth theorem, we have Hp+q(MP x Nq) ~ Hp(MP) ®Hq(Nq). This group is isomorphic to Z if and only if Hp(MP) = Z and Hq(Nq) = Z. 77. Suppose that sn = MP x Nq, where MP and Nq are manifolds of dimensions p, q ~ 1. Clearly, these manifolds must be closed. Moreover, according to Problem 76, they must be orientable. Therefore, by the Kiinneth theorem, the group Hp(sn) contains the subgroup Z = Hp(MP) ® Ho(Nq), and 0 < p < n. This is impossible. 78. The fundamental groups of both spaces coincide with Z2, and all of the remaining homotopy groups for the sphere and projective space of the same dimension are isomorphic. For homology, it suffices to prove that the homology groups with coefficients in Z2 are different. By the Kiinneth theorem, the dimension of the space Ef),>o Hi(X x Yj Z2) is equal to the product of the dimensions of Ef),>o Hi (X;Z2) and Ef),>o Hi(Yj Z2)' Hence the dimensions of Ef)i>O H.(sn x !Rpmj Z2) and Ef)i>O H~(sm x !Rpnj Z2) are equal to 2m + 2 and 2n + 2, respectively. 79. Clearly, 7I"1(S2 x !RpOO) ~ Z2 ~ 7I"1(!Rp2) and 7I"k(S2 x !RpOO) ~ 7I"k(!Rp 2) for k ~ 2 because 7I"k(POO) = 0 and 7I"k(S2) ~ 7I"k(!Rp2) for k ~ 2.

According to the Kiinneth theorem, the group H.(S2 x lRPOOj Z2) contains a subgroup isomorphic to HO(S2j Z2) ® Hi (lRpoo j Z2) ~ Z2j in particular, this group is nontrivial for any i. On the other hand, II. (lRp 2 j Z2) = 0 for i ~ 3.

Hints and :iolutions

80. Solution 1. The space A * B is the quotient of A x [0,1] x B by the following equivalence relation: (al, tI, bl ) (a2, t2, b2) if either tl = t2 = a and al = a2 or tl = t2 = 1 and bl = b2. Consider the set A c A * B of all points for which t ~ 1/2 and the set B c A * B of all points for which t ~ 1/2. The spaces A and B can be identified with the subsets of A * B consisting of the points for which t = a and t = 1. Under this identification, A and B become deformation retracts of A and B. Clearly, An B = A x B. Thus, we obtain the Mayer Vietoris exact sequence f'V

... ~ H r+1(A*B)

-+

Hr(AxB) ~ Hr(A)ffiHr(B) ~ Hr(A*B)

-+ . . . .

The map cp is induced by the inclusions A c A * Band B C A * B. It is easy to see that both inclusions are null-homotopic. For example, for the inclusion A C A * B, the homotopy is defined by !t(a) = (a, t, bo), where bo is a point of B. Thus, the above exact sequence has the form

0---+ H r+1(A * B)

---+

Hr(A x B) ~ Hr(A) ffi Hr(B)

---+

o.

The group Hr(A x B) is found using the Kiinneth theorem. It remains to calculate the kernel of tP. This map is the direct sum of the maps tP A and tP B, and the composition

is induced by the projection onto the first factor. For q > 0, the map Hr_q(A)®Hq(B) -+ Hr(A) is zero, and Hr(A)®Ho(B) -+ Hr(A) ~ Hr(A)® Ho(*) is induced by the map B -+ *, which has kernel Hr(A) ® Ho(B). For the map tPB, the argument is similar.

E(A" B), where A" B Solution 2. Recall that A * B (see Part I, Problem 49). Therefore, f'V

=A

x B/A V B

H r+1(A * B) ~ Hr(A" B) ~ Hr(A x B, A V B). Let us write the exact sequence for the pair (A x B, A V B): ... ---+

Hr(A V B)

---+

Hr(A x B)

---+

Hr(A x B, A V B)

---+

Hr-I(A V B)

---+ ....

Here

Hr(A x B)

=

E9 (Hi(A) ® Hj(B» i+j=r

and Hr(A V B)

-+

ffi

E9

Tor (Hi (A), Hj(B»,

i+j=r-l

Hr(A x B) is an isomorphism onto the direct term (Hr(A) ® Ho(B» ffi (Ho(A) ® Hr(B».

In particular, this map (and a similar map of the (r - 1)-dimensional homology groups) is a monomorphism. Therefore, the group Hr(A x B, A V B) is

Hints and Solutions

390

isomorphic to the quotient group Hr(A x B)/ Hr(A V B). Taking the quotient modulo the subgroup Hr(A V B) is equivalent to passing to reduced homology. 81. The Kiinneth theorem implies that the graded cohomology groups of the given spaces are isomorphic to the quotients of thp additive groups of the polynomial rings Z2[XI, ... , Xk] and Z[Xl, .. . , Xk] modulo the relations X~l +l = ... = X~k+l _ O. These groups are generated by elements of the form a;nl x ... X a;k, a ~ m, ~ ni. Here a, is a generator of the onedimensional cohomology group in the real case and of the two-dimensional cohomology group in the complex case. By Theorem 2.27 not only the groups but also the rings are isomorphic. In the complex case, the sign (_1)Q1P2 can be ignored because all nonzero elements have even dimension. 82. Theorem 2.27 expresses multiplication in the image of the natural monomorphism H* (K) ® H* (L) --+ H* (K xL) in terms of those in the rings H*(K) and H*(L). But if one of the groups H*(K) and H*(L) is torsion-free, then this monomorphism is an isomorphism. 84. If there exists such a nondegenerate bilinear map, then the number a < k < n. Therefore, n (a + b)n == an + b (mod 2) for all integers a and b. (~) is even for all k satisfying the inequalities

Clearly, (a + b)2 == a 2 + b2 (mod 2). Therefore, (a + b)2k == a 2k + b2k (mod 2). Suppose that n = 2km, where m is an odd number larger than 1. Then

(a + b)n

== (a21t + b21t )m

== an + ma(m-l) 2k b21t + m(m -

1) a(m

2)2Itb2.2k

2

Here m ¢.

+...

(mod 2).

a (mod 2).

85. We set f(x, y)

= (Zl," ., Zr+s 1), where

rt1 1= (t 1) (t Xi ti

Zktk

k-l

,

1

Yiti-1) .

i=l

If the product of two polynomials is the zero polynomial, then one of these polynomials must be zero. Therefore, the bilinear map f is nondegenerate. We have constructed a nondegenerate bilinear map ]Rr x lRs -+ lRr +s - l . Similarly, we can construct a nondegenerate bilinear map C r xes -+ C r +s - 1 , which is a nondegenerate bilinear map ]R2r X ]R2s --+ jR2r+2s-2. Moreover, we can use not only the complex numbers C, but also the quaternions ]HI and the octonions (Cayley numbers) O. 86. (a) Let X be an acyclic CW-complex. Then the space X is pathconnected. Therefore, according to Problem 50 in Part I, the space l:X is

391

Hints and Solutions

simply connected, and hence HI('EX) = O. We have Hi+I('EX) = Hi(X) = o for i ~ 1. By the Hurewicz theorem, all homotopy groups of 'EX are trivial. Thus, the space 'EX is contractible by the Whitehead theorem. (b) An example of a noncontractible acyclic CW -complex was given in the solution of Problem 34.

sq

87. The space SP x can be represented as a CW-complex with cells of dimensions 0, p, q, and p + q. Therefore, SP V is the k-skeleton of SOP x sq if max{p, q} ~ k ~ p + q - 1. The map f corresponds to attaching the cell D'P+q to the (p + q - I)-skeleton. The homotopy class of

sq

the map Sp+q 1 L SP V sq is the obstruction to extending the identity map SP V sq ---+ SP V sq from the (p + q - I)-skeleton of SP x sq to the (p + q)skeleton. Suppose that such an extension g: SOP x sq ---+ sP V sq exists. Let a P, {3q and a'P, (3q be the cohomology classes in H*(S'PV sq) and H*(SP x sq) corresponding to SP and sq. Then a P = g*a P and {3q = g*{3q. Therefore, a'P ......... {3q = g*(a P ......... (3q). But a P ......... {3q i= 0, whence a'P ......... {3q = O. 88. Consider the following cell decomposition of the sphere SOO, which contains m cells of each dimension:

(e 27rik/m , 0 , 0 , ... )., (e27!"iO,0,0, ... ),

(WI, pe 27!"ik/m, 0, 0, ... ),

0 0, ... ) , (WI,W2,pe 27!"ik/m " (WI. W2, pe 27!"i OOO) " , . .. ,

~ 1. Therefore, by the Cartan formula, Sqk(ail ... aim)

= ail Sqk(ai:z ... aim) + a~l Sqk-l(ai:z ... ai••.) = ... =

L

a~l ... a~/oaj/O+l ... ajm'

where the summation is over all permutations of il, ... , im for which il ... < ik and ik+l < ... < im. Therefore, Sqk(/*W m )

=L

a~l ... a~/oaill:+l ... aim·

Since f* is a monomorphism, it suffices to prove that (3)

~ 2 2 ~ ~ ail··· ai/oai/O+l ... aim == ~ 1=0

(m - k+ i-I) i

O"k-iO"m+i

(mod 2).

Elements of Homology Theory (Graduate Studies in Mathematics 81)

Elements of Homology Theory (Graduate Studies in Mathematics 81)

Elements of Homology Theory (Graduate Studies in Mathematics)

Mapping Degree Theory (Graduate Studies in Mathematics)

Finite Group Theory (Graduate Studies in Mathematics)

Elements of Homotopy Theory (Graduate Texts in Mathematics)

Probability (Graduate Studies in Mathematics)

Analysis (Graduate Studies in Mathematics)

A Modern Theory of Integration (Graduate Studies in Mathematics)

Resolution of Singularities (Graduate Studies in Mathematics)

Homology (Classics in Mathematics)

Problems in Operator Theory (Graduate Studies in Mathematics, V. 51)

Homology in Group Theory

Homology in Group Theory

Elements of Functional Analysis (Graduate Texts in Mathematics) (v. 192)

Fourier Analysis (Graduate Studies in Mathematics)

Global Calculus (Graduate Studies in Mathematics)

Hamilton's Ricci Flow (Graduate Studies in Mathematics)

Measure Theory and Integration (Graduate Studies in Mathematics)

Finite Group Theory (Graduate Studies in Mathematics 92)

Mapping Degree Theory (Graduate Studies in Mathematics 108)

Galois Theory (Graduate Texts in Mathematics)

Deformation Theory (Graduate Texts in Mathematics)

Galois Theory (Graduate Texts in Mathematics)

Graph Theory (Graduate Texts in Mathematics)

Deformation Theory (Graduate Texts in Mathematics, 257)

Complex Made Simple (Graduate Studies in Mathematics)

Algebra: Chapter 0 (Graduate Studies in Mathematics)

Foliations I (Graduate Studies in Mathematics)

Geometric Relativity, Graduate Studies in Mathematics 201

Graph Theory (Graduate Texts in Mathematics)

Elements of Homology Theory (Graduate Studies in Mathematics 81)