Elements of Homology Theory (Graduate Studies in Mathematics 81)

V. V. Prasolov

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

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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.

1. Simplicial Homology

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

1. Simplicial Homology

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

1. Simplicial Homology

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.

1. Simplicial Homology

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!<)-l and g. = .,p.Uf)-l. The equality (gf). = g./. is easy to verify. 0 Corollary. II simplicial complexes K and L are homeomorphic, then their homology groups are isomorphic. 5The dimension of this simplex may equal any of the numbers 0,1, ...• 2k + 1.

2. Invariance of Homology

11

Proof. If f: IKI -+ ILl and g: ILl -+ IMI are mutually inverse continuous maps, then f. and g. are mutually inverse homomorphisms of homology groups. 0 As an application of the topological invariance of homology, we give yet another proof of the nonexistence of a retraction of the ball onto its boundary. Theorem 1.9. There exists no continuous map r: nn that rex) = x for all x E sn-l.

-+

sn-l = ann such

Proof. Suppose that r: nn -+ sn-l is a continuous map for which rlsn 1 = idsn-l. Consider the inclusion i: sn-l ~ nn. The composition ri is the identity mapj therefore, r.i.: H n - 1 (sn-l j Z) -+ H n - l (sn-lj Z) is the identity map Z -+ Z (or Z ffi Z -+ Z ffi Z, if n = 1). On the other hand, the image of i.: Hn-l (sn-lj Z) -+ H n- l (nnj Z) is zero (or contained in Z, if n = 1) because Hn_1(nn j Z) = 0 for n ~ 2 and Ho(nl; Z) = Z. 0 2.3. Homotopy Invariance of Homology.

Theorem 1.10. Homotopic maps of simplicial complexes induce the same map on homology. Proof. Let /0, II: IKI -+ ILl be homotopic simplicial maps of simplicial complexes K and L. This means that there is a continuous map F: IKI x I -+ ILl for which FIIKlx{o} = fo and FIIKlx{l} = II. The Cartesian product IKI x I can be turned into a simplicial complexj moreover, we can assume that IKI x {O} and IKI x {I} are endowed with the same simplicial structure and for any simplex Ll in K, the set Ll x I is a sub complex in IKI x I (see Part I, p. 130). Consider the inclusions io: K -+ IKI x {O} ~ IK1 x I and i 1 : K -+ IKI x {1} ~ IKI x I. Clearly, fo = Fio and h = Fil. Therefore, it is sufficient to verify that io. = it. (we use the equalities fo. = F.io. and h .. = F.it.). Let Llk be a simplex in K. The chains io(Llk) and il(Ll k) have the common support Llk x I. This support is acyclic because it is homeomorphic to nk+l R:: Llk+l. 0 Corollary. If simplicial complexes K and L are homotopy equivalent, then their homology groups are isomorphic. As an application of the homotopy invariance of homology, we give yet another proof of dimension invariance (one proof was given in Part I on pp.60-61). Theorem 1.11. If n phic.

i- m,

then the spaces lRn and]Rm are not homeomor-

12

1. Simplicial Homology

Proof. For definiteness, suppose that n > 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

1. Simplicial Homology

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

3. Relative Homology

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

1. Simplicial Homology

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

3. Relative Homology

(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 <1>: 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

1. Simplicial Homology

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

3. Relative Homology

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

1. Simplicial Homology

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

1. Simplicial Homology

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.

4. Cohomology and Universal Coefficient Theorem

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.

1. Simplicial Homology

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

4. Cohomology and Universal Coefficient Theorem

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

1. Simplicial Homology

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

4. Cohomology and Universal Coefficient Theorem

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.

1. Simplicial Homology

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

1. Simplicial Homology

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,

4. Cohomology and Universal Coefficient Theorem

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

1. Simplicial Homology

=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.

4. Cohomology and Universal Coefficient Theorem

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

1. Simplicial Homology

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

1. Simplicial Homology

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).

1. Simplicial Homology

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

1. Simplicial Homology

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

1. Simplicial Homology

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

1. Simplicial Homology

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

1. Simplicial Homology

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

1. Simplicial Homology

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

1. Simplicial Homology

= 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<j
X(K x L)

= X(K)X(L).

6. The Euler Characteristic

53

Problem 40. Suppose that K is a finite simplicial complex and f: K - K is a continuous map for which fn = f 0 ••• 0 f is the identity. Prove that if

X(K)

to

(mod n), then

f

------n

has a fixed point.

The Euler characteristic of a closed manifold M n was defined in Part I as the sum of the indices of singular points of any smooth vector field (with isolated singular points) on Mn. On the other hand, we can triangulate M n and calculate the Euler characteristic of the simplicial complex thus obtained. The equivalence of these two definitions is a corollary of the following theorem. Theorem 1.32. Suppose that Mn is a triangulated closed manifold. Then there exists a smooth vector field on M n whose singular points coincide with the barycenters of the simplices in the triangulationj moreover, the barycenter of each k-simplex has index (_I)k. Proof. Let us construct a vector field on the triangulated manifold Mn by induction on the dimension of skeletons so that, under the passage from the (k - 1)-skeleton to the k-skeleton, the additional integral trajectories be incoming for the barycenters of k-simplices ("new" singular points) and outgoing for the barycenters of simplices of dimension less than k ("old" singular points). The construction of this vector field for simplices of dimensions 1, 2, and 3 is shown in Figure 13.

" \ A 'I

,

I

I

. . . ..

~/'

.

, "

II

--------~

Figure 13. Construction of a vector field

In a neighborhood of the barycenter of a k-simplex, the constructed vector field has the form vex) = Ax, where

A

= diag(-I,.~:, -~, 1, ... ,1). k

The index of such a singular point equals (_I)k. Theorem 1.33. If M2n+l is a closed manifold, then X(M 2n +l)

0

= o.

Proof. One proof of this theorem has already been given in Part I. Another proof can be obtained by using the isomorphism H k (M 2n+lj Z2) ~ H 2n+l_k(M 2n +lj Z2).

54

1. Simplicial Homology

Indeed, let bk

= dim H k (M 2n+lj Z2). 2n+l

X(M 2n+l)

=

Then bk

= b2n+1-

k , whence

n

L (-l)kbk k-O

= L((-l)kbk + (-1)2 n+l-kb2n+l_k) k=O

n

= L( -l)k(bk -

b2n+ 1-k)

= o.

0

k=O Theorem 1.34. If M 2n is a closed manifold, then we hove X(M 2n ) _ dimHn (M 2n ;Z2) (mod 2). Proof. Let bk

= dim Hk(M 2n j Z2).

X(M 2n )

=

2n L(-llbk k-O

Then bk

= b2n -k;

therefore,

n-l

= bn + L((-l)kbk + (-1)2n- kb2n _k) k-O

n-l

= bn + L(-l)k(bk

+ b2n -k) == bn

(mod 2).

0

k 0

Theorem 1.35. If a closed manifold M m is the boundary of a compact manifold Wm+l, then X(M m ) is even. Proof. It is sufficient to consider the case m = 2n because otherwise X(Mm) = o. Let us triangulate the manifold W 2n+l and consider the closed manifold W 2n+1 obtained from two copies of W 2n +l by gluing together their boundaries M2n. The manifold W 2n +l also has the structure of a simplicial complex with a sub complex M2n, and all simplices not contained in M 2n come in pairs. Therefore, X(W 2n+ 1 ) == X(AI 2n ) (mod 2). But X(W 2n+ 1 ) = O. 0 Corollary. Neither the manifold Rp2n nor cp2n is the boundary of a compact manifold. On the other hand, let us show that the manifolds IRP2n+l and cp2n+l are the boundaries of compact manifolds. For this purpose, we can use, e.g., the following assertion. Theorem 1.36. Suppose that M n is a closed manifold and u: M n ~ Mn is a fixed-point-free smooth involution, i.e., u(u(x)) = x and u(x) =I- x for all x E Mn. Then Mn is the boundary of a compact manifold. Proof. Consider the coset space

Wn+l = (Mn x [0, l])/(x, 0)

I"V

(u(x), 0).

The involution u has no fixed points; therefore, any point (x, 0) in Wn+l has a neighborhood homeomorphic to the ball obtainpd by gluing together two half-disks D~+l and u D~+l. Hence Wn+l is a manif ld with boundary Mn x {I}. 0

6. The Euler Characteristic

55

Fixed-point-free involutions on Rp2n+l and cp2n+1 can be defined as

u(xo, Xl,···, X2n, x2n+d = (-XI, xo,···, -X2n+1. X2n), u(zo, Zl,"" Z2n, Z2n+l) = (-Z1. Zo,···, -Z2n+l, Z2n). Exercise. Prove that the orientation covering manifold (Part I, p. 207) for a closed manifold is the boundary of a compact manifold. Theorem 1.37. If a closed manifold M 2n is the boundary of a compact manifold, then the dimension of the space Hn(M2n j Z2) is even. Proof. According to Theorem 1.34, we have X(M 2n ) == dim H n (M 2n j Z2) (mod 2), and Theorem 1.35 implies that X(M 2 n) is even. 0 Problem 41. Given a finite exact sequence of vector spaces ...

prove that

~

1.+1

T7

Tr

Vi+l - - Vi

E( _l)i dim ~ =

I.

~

Tr

Vi-l

~

... ,

O.

Problem 42. Prove that the Euler characteristic of the complement of any link in 8 3 is zero.

If A is a finite simplicial complex and B is its subcomplex, then we can define the Euler characteristic of the pair (A, B) by

where the Hi(A, B) are the homology groups with coefficients in some field. Problem 43. Prove that X(A, B)

= X(A) - X(B).

Problem 44*([143]). Let K be a Euclidean CW-complex for which Mn, where M n is a closed manifold. Consider the alternating sum n

IKI

~

n

W(K) = LL(-l)i+ j O!ij, i=O j=O

where each O!ij is the number of ordered pairs of disjoint (closed) faces of dimensions i and j in the complex K. (Note that D:ij = D:ji and Qii is twice the number of unordered pairs of disjoint i-faces.) Prove that w(K) = X2(Mn) - X(Mn), where X(Mn) is the Euler characteristic of the manifold

Mn. Remark. The number W(K) defined in Problem 44* can be interpreted as the Euler characteristic of the deleted product of the manifold Mn. (The definition of deleted product is given in Part I on p. 161.)

56

1. Simplicial Homology

Geometric conditions often impose various constraints on the Euler characteristic. As an example, consider transnormal embeddings (see [110]). An embedding of a manifold Mk into JRn is said to be transnormal if for any points x, y E Jv[k such that x belongs to the normal subspace at y, the normal subspaces at the points x and y coincide. Examples of transnormal embeddings are the standard embeddings of JRk and Sk into JRn. Problem 45. Prove that the number of intersection points of trans normally embedded (connected) manifolds with a normal subspace does not depend on the normal subspace. (If this number is equal to r, then the embedding is said to be r-transnormal.) Problem 46. Prove that if a closed manifold admits an r-transnormal embedding, then the number r is even. Problem 47. Prove that if a closed manifold Mk admits an r-ttansnormal embedding, then X(M k ) = 0 or r. Problem 48. Let Mk be a closed manifold admitting a transnormal embedding. Prove that if X(Mk) 0/= 0, then Hi(Mk j G) = 0 for all odd i. 6.2. The Lefschetz Fixed Point Theorem. Let K be a finite simplicial complex, and let f: IKI ---+ IKI be a continuous map. The induced map (J*)k: Hk(IKljJR) ---+ Hk(IKljJR) is a linear operator on a finite-dimensional space; therefore, we can consider its trace tr(J*)k. The Lefschetz number of the map f is defined by A(J) = Lk~O(-l)ktr(J*)k' If the complex K is path-connected, then (J*)o: JR ---+ JR is the identity map, and hence tr(J*)o = 1. Example 15. If the complex K is acyclic with respect to the coefficient group JR, then A(J) = 1 for any continuous map f: IKI ---+ IKI. Example 16. The equality A(idK) = X(K) holds. Theorem 1.38 (Lefschetz fixed point theorem, [77]). Lf..t K be a finite simplicial complex. If A(J) 0/= 0, then the map f: K ---+ K has a fixed point. Proof. Suppose that the map f has no fixed points. Since the complex K is compact, one of its subdivisions K' of K has the following property: for any point x E IKI, the distance between x and f(x) is much larger than the maximum diameter of a simplex from K'. In turn, this property implies that if K" is a subdivision of K' and cp: K" ---+ K' is a simplicial approximation of f, then, for any simplex t:.." c K", the sets t:.." and cp(t:..") are disjoint. Thus, to prove the Lefschetz theorem, it suffices to interpret the number A(cp) so that if t:.." n cp(t:..") = 0 for all simplices t:.." c K", then A(cp) = O.

57

6. The Euler Characteristic

Lemma. (a) (Hopf [57]) If K is a finite simplicial complex and CPk: Ck(K, R) --+

Ck(K, R) is a chain map, then L(-l)k trCPk = L(-l)ktr(CP.)k. k~O

k~O

(b) ( [120]) If K is a finite simplicial complex, K' is its finite subdivision, and cP: K' --+ K is a simplicial map, then

A(cp) = L(-l)kcI>k, 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

1. Simplicial Homology

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
61

1. Multiplication in Cohomology

The map cpl!>: C. (K) -+ C. (K) is the identity. The map I!>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)

1. Multiplication in Cohomology

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.

1. Multiplication in Cohomology

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
q
(-1)q&dP,..... [va, ... ,vp+q]

= (-l)q (&dP, [vq-I. ... ,vp+q]) [va, ... ,Vq-l]

L

(_1)i+l(dP, [Vq-I. ... ,Vi, ... ,vp+q])[vo, ... ,Vq].

q-l:O;i:O;p+q When we add the second sum in the first expression to the &e ond expression, all terms except for the one with i = q cancel out. It if:, o.lso clear that

2. Homology and Cohomology of Manifolds

71

adding the term with i = q to the first sum (in the first expression), we obtain 8(d? r-.. [vo, ... ,vp+q]). 0 The formula for 8( d? for (co ) homology.

r-..

Cp+q) shows that the cap product is well defined

Theorem 2.5. Cap product is natural in the sense that a continuous map, then I.U·o: r-.. (3) = 0: r-.. 1.«(3).

ill: IKI - ILl zs

Proof. The required equality holds even at the level of (co )chains:

I. (reP

r-..

[vo, ... , vp+q])

= f.( (reP, [vq, ... , vp+q]) [vo, ... , vq]) = (reP, [vq, ... ,vp+q]}/.[vo, ... ,Vq] = (eP, I. [Vq, ... , Vp+q]) I. [vo, ... , vq] = eP I. [Vq, ... ,vp+q]. r-..

o

Theorem 2.6. The cap product is dual to the cup product in the sense that

(o:q, (3P

r-..

'Yp+q)

= (o:q

~

(3P, 'Yp+q) .

Proof. The required equality holds even at the level of (co)chains:

(c q, eP

r-..

[vo, ... ,vp+q])

= (cq, (eP, [Vq, ... ,VP+q]) [vo, ... ,Vq]} = (cq, [vo, ... , vq])(eP, [vq, ... , vp+q]) = (cq ~ eP, [vo, ... , vp+q]).

o

Theorem 2.7. The cap and cup products are related by

aP

r-..

«(3q

r-..

'Yp+q+r) = (aP

~

(3q) ,.-,. 'Yp+q+r'

Proof. The required equality holds even at the level of (co) chains.

If

'Yp+q+r = [vo, .. . , vp+q+r], then both expressions are equal to (aP, [vr. ... ,vr+p]}((3q, [vr+p,'" ,vp+q+r])[vo, ... ,vr].

0

In the relative case, a similar construction of cap product yields a map

,.-,.: HP(K, Lt) x Hp+q(K, L1 U L 2 ) - Hq(K, L2)' Indeed, let [vo, ... , vp+q] be a simplex in L1 U L2. It is contained entirely either in L1 or in L 2. If it is contained in L1. then any co chain from CP(K, L 1 ) vanishes at the simplex [v q •••• , VP+q ]. and if it is contained in L 2 • then the simplex [vo •...• vq] corresponds to the zero element of the group C q (K.L2)' The important role played by the cap product in the topology of manifolds is related to the following description of the Poincare isomorphism Hn-k(Mnj R) - Hk(Mn; R) (see p. 44), where Mn is a closed manifold and R = Z2 if M n is nonorientable, in terms of the cap product.

2. Cohomology Rings

7'2

Theorem 2.8. The Poincare duality isomorphism has the form a n - k a n- k ,..... [Mn], where [MnJ is the fundamental class of the manifold Mn.

1--+

Proof. Consider a triangulation K of the manifold M n and its barycentric subdivision K'. Let us number the vertices of K' so that the least numbers are assigned to the vertices of K, next follow the barycenters of the I-simplices in K, then the barycenters of the 2-simplices are numbered, etc. For a representative of the homology class [MnJ we take the sum E ±[vo, ... , vnJ, where the summation is over all simplices K' whose vertices are writt~n in the order of increasing numbersj the signs indicate whether the orientations of the simplex and manifold are compatible. The vertices of the complex K' are numbered in such a way that a simplex [vo, ... , VkJ is contained in one of the k-simplices Ui in K. Each vertex Vk uniquely determines a k-simplex Ui, and each (n - k)-cell ui is represented as E ±[Vk, .. . , vnJj here the vertices are again written in the order of increasing numbers. At the level of (co )chains, the Poincare isomorphism associates to each chain Ck = E a,Ui a cochain cn- k for which (c n- k , = ai. Therefore,

un

cn- k ,..... L±[vo, ... ,vnJ

= L(cn-k,±[Vk, ... ,VnJ)[vo, ... ,VkJ = (c n- k , Ui}Ui = L aiUi = Ck,

o

as required.

When combined together, the two approaches to the Poincare duality isomorphism have the following important consequences. (1) The map an k 1--+ an k ,..... [MnJ is an isomorphism between the groups Hn-k(lvfnj R) and Hk(Mnj R)j (2) The correspondence between the cochains cn- k and the chains Ck, which is not invariant at the level of (co ) chains, is invariant at the (co )homology level. The invariant definition of the Poincare isomorphism allows us to give an invariant definition of the intersection number of the cycles representing the homology classes of cycles ak E Hk(Mnj R) and (3n-k E Hn_k(Mnj R). Namely, for ak = a n- k ,..... [MnJ and {3n k = {3k ,..... [Mn], we have

= (a n- k '--' (3k) ,..... [MnJ = (a n- k '--' (3k, [MnJ). Indeed, suppose that ak = E aWi and {3n-k = E bwi at the level of chains. (9)

((ak,{3n-k})

Then

(a n- k '--' (3k) ,..... [MnJ = a n- k ,..... ({3k ..-... [MnJ) = a n- k ,..... (3n-k

= (an-k,Lbwi) = Lbi(an-k u;)

= Lbiai.

2. Homology and Cohomology of Manifolds

73

Formula (9) can be rewritten so as to express the cup product in terms of the intersection number. Namely, let A be an element of Hn(Mnj R) such that A ,..... [Mn] = 1 E Ho(Mnj R) ~ R. Then

(10) Relation (9) and anticommutativity of the cup product imply (11) To see that (11) holds at the level of chains, it suffices to take chains representing the classes ak and J3n-k in the chain complexes Ck(K) and Cn-k(K*). Therefore, equality (11) follows directly from the invariance of the definition of intersection number. The relation dim(V + W) = dim V + dim W - dim(V n W), well-known in linear algebra, shows that if dim V = k, dim W = I, and the spaces V and Ware subspaces of an n-space, then dim(V n W) ~ k + I - n. If dim(V n W) = k + I - n, then the subspaces V and Ware said to be transversal. In agreement with this terminology, we say that a k- and an I-dimensional cycles in an n-manifold are transversal if the intersection of any two simplices contained in different cycles is either empty or of dimension k + I - n. In particular, cycles of complementary dimensions are transversal if and only if they intersect in finitely many points. Relation (9) has the following geometric interpretation: if Ok and J3n-k are two transversal cycles of complementary dimensions, then the cohomology class dual to their intersection is the cup product of the classes on-k and {3k dual to Ok and {3n-k. This statement is true not only for cycles of complementary dimensions but also for any transversal cycles. Let us prove this. Take cycles Ok and J3, j their dual co cycles are on-k and {3n-l. We have [ (a n-k ......... /J(.In-I ,Vo,· . . , V2n-k-1 ]) = (on-k, [vo, ... , Vn-k]) . (J3 n - l , [Vn-k, ... , V2n-k-d).

Suppose that Ok = EXia~ and (3, = EYja~. Then on-k = EXi(af)* and (3n-1 = EYj(a~)*. For transversal cycles, we have Ok n (3, = ')'k+l-n = E olJ3Ja~ n a~j therefore, ')'2n-k-1 = E oiJ3j(a~ n a~)*. Thus, we must show that if [vo, .. . , Vn-k) = (a k )* and [Vn-k, ... , V2n-k-d = (a l )*, then [vo, . .. , V2n-k-d = (a k n a ' )*. In the barycentric coordinates, the cell a k is determined by the equations Xo = Xl = ... = Xn-k, and the cell a l is determined by the equations Xn-k = ... = X2n-k-l. Hence the cell a k n a l is determined by the equations Xo = Xl = ... = X2n-k-l. Now, it is clear that (a k n a l )* coincides with the simplex [vo, ... , V2n-k-,].

74

2. Cohomology Rings

a

a

b

Figure 9. The basis cycles on a sphere with handles

Figure 10. The basis cycle on the Mobius band

Problem 58. (a) Calculate the integral cohomology ring of Sk x Sk. (b) Find the self-intersection number of the diagonal in Sk x Sk. Problem 59. Prove that for positive m and n, any map f: sn+m induces the trivial map f*: Hn+m(sn x sm) --+ Hn+m(sn+m).

--+

sn X sm

2.2. Cohomology Rings of Manifolds. The Poincare isomorphism and the relation

(12) substantially facilitate calculating cohomology rings for manifolds (here A E Hn(Mnj JR) is the element for which A ........ [Mnj = 1, i.e., (A, [Mn]) = 1). For example, the calculation performed above for closed two-dimensional surfaces becomes almost obvious, because it suffices to determine the arrangement of the dual cycles. Figure 9a shows basis cycles o:}, .•• , O:n and f31, ... ,f3n for a sphere with n handles. Two cycles intersect only if they have equal subscripts. However, it is seen from Figure 9b that each cycle 0: is homologous to a cycle 0:' disjoint from it. For the Mobius band attached to the sphere, the basis cycle 0: is represented by the diagonal (see Figure 10). It is homologous to a cycle 0:' which transversally intersects 0: in precisely one point. For the projective plane, 0: and 0:' correspond to two projective lines.

2. Homology and Cohomology of Manifolds

75

An important information about the structure of the cohomology ring of a closed manifold is provided by the following assertion, which we have essentially proved already. Theorem 2.9. Suppose that M n is a closed manifold and F is a field; if M n is nonorientable, then we assume in addition that F = Z2. Then the spaces Hn-k(Mn; F) and Hk(Mn; F) have bases 01,.·., am and /31,"" /3m such that 0i '-'" /3j = dijA, where A is a basis element of the one-dimensional space Hn(Mn; F). Proof. The Poincare isomorphism and relation (12) show that the required assertion can be stated as follows: The spaces Hk(M n ; F) and Hn k(Mn; F) have bases ai, ... , o:n and /3i, ... , /3:n such that ((a;, /3;)) = dij. This assertion coincides with the corollary of Theorem 1.28 on p. 46. 0

A similar argument, together with Theorem 1.27, proves the following assertion. Theorem 2.10. Let M" be a closed orientable manifold. Then the free Abelian groups Hn-k(Mn; Z)/T n - k and Hk(Mn; Z)/Tk, where rn- k and Tk are the torsion subgroups, have bases 01,· .. , am and /31, ... , /3m such that 0i '-'" /3j = dij A, where A is a generator of the group H n (Mn; Z) .

Using Theorems 2.9 and 2.10, we can calculate the cohomology rings H*(lRpn; Z2) and H*(cpn; Z). Theorem 2.11. For 0 ~ k ~ n, the group Hk(lRpn; Z2) contains only one nonzero element ok, and

if k if k

+ l ~ n, + l > n.

Proof. We already know that Hk(lRpn; Z2) ~ Z2 for 0 ~ k ~ n; therefore Hk(lRpn; Z2) ~ Z2 for 0 ~ k ~ n. It remains to verify that ok '-'" 0 ' i- 0 for k+l ~ n. Theorem 2.9 implies ok '-'" ol i- 0 for k+l = n, because ok and (i are the only nonzero elements of the groups Hk(lRpn; Z2) and H'(lRpn; Z2). Suppose that k + l = m < n. The homomorphism H* (lRpn) _ H* (lRpm) induced by the natural embedding is a ring isomorphism in dimensions at most m. Since the product ok '-'" ol is nonzero in the ring H*(lRpm), it follows that it is nonzero in H* (lRpn). 0 Theorem 2.12. For k = 0,2,4, ... , 2n, Hk(cpn; Z) = Z, and for all other k, the cohomology group is trivial. Moreover, there exist generators 0 2, 0 4, ... ,02n such that 02k '-" 0 21 = 02(k+l) for k + l ~ n.

2. Cohomology Rings

nj

Proof. Cohomology groups are easy to obtain from homology groups by using the universal coefficient theorem. The cohomology groups of cpn are torsion-freej therefore, Theorem 2.10 applies to the cohomology groups themselves. Let 13 2,134, ... , f32n be their generators. According to Theorem 2.10, we have f32k '-' f3 2(n-k) = ±f32n . For k + l = m < n, we have f32k '-' 13 21 = ±f32m j the proof is similar to that in the case of JRpn. Now, for generators with correct signs we can take a 2 = (32, a 4 = 132 ........ 13 2 = ±f3\ a 6 = 13 2 '-' 132 '-' 132 = ±f36, . . . . 0 Problem 60. (a) Prove that if m > 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

2. Homology and Cohomology of Manifolds

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

2. Homology and Cohomology of Manifolds

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

2. Homology and Cohomology of Manifolds

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) <.....+ (sn, M). Consider also the inclusion j: (sn \ M) <.....+ sn and the homomorphism of the zero-dimensional homology groups which it induces. The homomorphisms i* and j* enter the diagram

Hn(sn, M) ~ Hn(sn)

l~

.

l~

Ho(sn \ M) ~ Ho(sn), in which the vertical arrows are the Lefschetz and Poincare isomorphisms. This diagram is commutative up to sign; hence Ker i* ~ Ker j*. The homomorphism p*: Ho(S" \ M) --. H o(*), where * is a point from sn \ M, can be represented as the composition Ho(sn \ M) ~ Ho(sn) ~ Ho(*), whence fIo(sn \ M) = Kerp* = Kerj •. Thus, fIn-I(M) ~ ImtS· = Keri· ~ Ker j. = fIo(sn \ M). The second isomorphism is established in a similar way. o Corollary 1. If L is the image, under an embedding into S3, of a disconnected manifold that consists of n circles, then H 1 (S3 \ L) ~ HI(L) ~ and H2(S3 \ L) ~ fIO(L) ~ zn-l.

zn

Corollary 2 ([19]). If Mn-l c sn is a closed connected submanifold, then M n- l is orientable and sn \ M n 1 has two connected components. Proof. According to the Alexander duality theorem, we have Hn-1(M n- 1) ~ fIo(sn \ M n - l ). If Mn-l is a closed connected nonorientable manifold, then Hn-I(Mn-l) ~ Z2 (by Theorem 1.26 on p. 45). On the other hand, fIo(sn \ M n- 1 ) is a free Abelian group. Therefore, the manifold M n - l is orientable, and Hn-l(Mn-l) ~ Z. The isomorphism fIo(sn \ M n- 1 ) ~ Z implies that sn \ Mn-l has two path-connected components. 0

2. Homology and Cohomology of Manifolds

83

Remark. The Alexander duality holds not only for submanifolds but also for any sub complex K ~ sn (see Theorem 5.7). Problem 71. Suppose that m pairwise disjoint copies of the sphere sn-2 are embedded in sn, where n ~ 3. Let X be the complement of the embedded spheres. Calculate the homology of the space X. Problem 72. Given the sphere SP arbitrarily embedded in sn+!, prove that the space X = sn+! \ SP has the same homology as sn-p. Problem 73. Given the spheres SP and sq arbitrarily embedded in sn+!, prove that the space X = sn+! \ (SP u sq) has the same homology as sn-p V sn-q V sn. The Linking Number and Multiplication in Cohomology. The linking number can be defined not only for circles embedded in S3 but also for any closed connected oriented manifolds Mi and Mi embedded in sn, where n = p + q + 1. The definition repeats almost word for word the one given on p. 46. Namely, consider fundamental cycles [Mf] and [Mil as cycles in sn. Since the p-dimensional homology group of sn is trivial, it follows that there exists a chain Wi+! for which oWi+! = [Mil. The linking number lk( Mi, Mi) is defined to be equal to the intersection number ((Wi+ 1 , [Mil}). This number does not depend on the choice of Wi+!; the proof repeats that for circles embedded in S3. Taking a chain Wi+! for which oWi+! = [Mil and considering the intersection number (([MiL Wi+!}) , we obtain the same number (up to sign); this follows from the lemma on p. 43. The Alexander duality makes it possible to interpret linking number in terms of multiplication in cohomology. This can be performed as follows. Let X = Mi U Mi. The Alexander duality implies iIk(sn \ X) ~ iIp+q_k(X) (recall that n = p + q + 1). Therefore, there is a class in Hq(sn \ X) that corresponds to the fundamental class [Mil, and there is a class in HP(sn \ X) that corresponds to the fundamental class [Mil. The product of these cohomology classes belongs to HP+q(sn \ X) ~ Ho(X) ~ Z, i.e., it is an integer. This integer is equal to the linking number (up to sign). To prove this, note that the Alexander duality isomorphism is the composition of isomorphisms

HP+q(sn \X) __ Hl(sn,X) ~ Ho(X). The product of the cohomology classes under consideration corresponds the intersection ofthe chains Wi+ 1 and Wi+! in HI (sn , Mi U M:i) (these chains are assumed to be transversal). Indeed, the Alexander duality isomorphism is the composition of isomorphisms

Hq(sn \X) - - Hp+1(Sn,X) ~ H'D(X);

84

2. Cohomology Rings

therefore, the fundamental class [Mi] corresponds to the homology class in Hp+! (sn, X) that is represented by the chain Wi+!. Taking the boundary of the intersection Wi+! n Wi+! (which consists of arcs of curves), we obtain the union of the two intersections Wi+! n [M~] and [Mi] n Wi+!. However, we not only take the boundary but also pass to the reduced homology; hence only one of these intersections (say, the former) remains. Indeed, the contribution of each arc from Wi+! n Wi+! whose endpoints belong to the same manifold (Mi or Mi) is zero. Nonzero contributions are made only by arcs having one endpoint in Mi and the other in M~. To calculate such contributions, one of the two intersections is sufficient. 2.6. The Triple Massey Product. Take three cohomology classes laP] E HP(X), [,eq] E Hq(X), and [rr] E Hr(x) represented by three co cycles a P , ,eq, and ,r. Suppose that laP] '-"' [,eq] = 0 and [,eq] '-"' [rr] = o. Then a P '-"' ,eq = dcP+ q- 1 and ,eq '-"' , r = ddq+r - 1 for some cochains cP+ q- 1 and dq+r - 1 . Consider the cochain

It is easy to verify that this is a cocycle. Indeed, we have d,r = 0 and daP - 0, whence

dZ = dd'+q-l '-"' , r + (_l)p-l( -l)PaP '-' ddq+ r -

= a P '-"' ,eq '-"'

,r - aP '-"' ,eq '-"' ,r = O.

1

The co cycle Z corresponds to the cohomology class [z] E HP+q+r-l(x). This class is not uniquely determined because we can add coboundaries to the co cycles a P , ,eq, and and cocycles to the co chains cP+ q- 1 and dq+r - 1 . The set of all cohomology classes [z] thus obtained is called the triple Massey product of the classes laP]' [,eq], and [rr] and denoted by ([aP ]' [,eq], [,rD.

,r

Theorem 2.16. The triple Massey product of cohomology classes laP], [,eq], and [rr] is an element of the quotient group of HP+q+r-l(x) modulo laP] '-"'

Hq+r-l(x)

+ HP+q-l(X)

'-"' [rr].

Proof. Choose cochains q+q-l and ~+q-l so that a P '-' ,eq = dq+q-l = d~+q-l. Let Zl and Z2 be the co cycles constructed by using tht'se cochains. Then Zl - Z2 = (q+q-l - ~+q-l) '-"' Here q+q-l _ ~+q-l is a cocycle; therefore, [Zl] - [Z2] E HP+q-l(X) '-"' [rr]. Similarly, for different cochains d~+r-l and c4+r - 1 , we have [Zl]- [Z2] E laP] '-' Hq+r-l(x).

,r.

Now, suppose that ~ = ai + daP - 1 . If ai '-' ,eq = 6cf+ q - 1 , then ~ '-"' ,eq = dq+q-l +daP- 1 '-"' ,eq = d(q+q-l +aP- 1 '--" (Jq) becaLlse d"(Jq = o.

2. Homology and Cohomology of Manifolds

85

Figure 12. The Borromean rings

+ aP- 1 '-' {3q and obtain aP- 1 '-' {3q '-' 'Y r + (-l)Pd"aP- 1 '-' cf/+r-l aP- 1 '-' d"dq+r - l + (-l)Pd"aP-1 '-' dq +r - l

Thus, we can set ~+q-l = cr+ q- l Z2 -

Zl

= =

= (-l)Pd"(aP-1 ......... dq+r -

l ),

whence [Z2J = [ZlJ. The case where the co cycle 'Y r changes is handled similarly. Finally, suppose that {3~ = {3i + 6bq - l . In this case, we have ~+q-l = cr+ q- l + (-l)P a P '-' bq- l and r4+ r - 1 = di+ r - l + bq- l ......... 'Yr. Therefore, Z2 - Zl = 2(-1)P a P ......... bq - l '-' 'Yr. 0 The triple Massey product can be used to prove that the Borromean rings are linked (see Figure 12). It should be mentioned that for usual Borromean rings, there is a quite elementary proof, which is based on colorings of the link diagram (see [105], p. 31). However, for the multidimensional generalization of the Borromean rings, the triple product Massey must be applied. For calculations, it is convenient to specify the Borromean rings by the equations Z2 X= 0, y2+ _ = 1 (circle 8I); y=O, Z=

0,

4 x2 z2+ 4

x2+

=1

(circle 82);

2

~ = 1 (circle 83).

4 Figure 13 shows that these equations do indeed determine the Borromean rings. Instead of the space ]R3 in which the given circles are embedded, it is more convenient to consid~r its one-point compactification S3. By the Alexander duality theorem, Hk(S3 \ (81 U S2 U S3» ~ H2-k(S1 U 8 2 U 8 3 ).

86

2. Cohomology Rings

Figure 13. The transformation of the diagram

d

Figure 14. The intersections of chains

The circles 81, 8 2, and 83 are not pairwise linked; therefore, the pairwise products of the cohomology classes aI, (31, and 7 1 corresponding to the fundamental classes of 81, 82, and 8 3 under the Alexander duality are zero. Thus, the triple product (aI, (31,71) is defined. Let us show that for the coefficient group Z2, this triple Massey product is nonzero. First, note that it belongs to H2(8 3 \ (81 U 8 2 U 8 3)) ~ Ho(81 U 82 U 83) because the group a 1 ....... HI (X) + HI (X) ....... 7 1, where X = 8 3 \ (81 U 82 U 83), is trivial. Indeed, not only the pairwise products of the classes aI, (31, and 7 1 but also the products a 1 ....... aI, (31 ....... (31, and 7 1 ....... 7 1 vanish because the circles 8 1, 8 2, and 83 are unknotted and, therefore, 8 i can be transformed into 8~ by a small deformation so that the disks Di and D~ spanned by 8 i and 8! are disjoint. Let WI, w2, and W3 be the chains (relative cycles) corresponding to the disks spanned by 81, 82, and 8 3. Under the isomorphism H2(8 3 ,81 U 82 U 83) ~ HI (83 \ (81 U 8 2 U 83)), the homology classes of these relative cycles correspond to aI, (31, and 7 1. Let c and d be the half-disks such that WI n W2 = ac and W2 n W3 = ad in the group of relative chains (see Figure 14). Then the triple Massey product (al,(31.7 1) corresponds to the relative cycle en W3 + WI n d in HI (83,81 U 8 2 U 8 3); this cycle is shown by the heavy line in Figure 14. The isomorphism

2. Homology and Cohomology of Manifolds

87

takes the homology class of this cycle to a nonzero element because the cycle endpoints belong to different connected components: one of them belongs to 81 and the other to 8 3. On the other hand, if the link 81 U 8 2 U 83 C 8 3 is trivial (in particular, 8}, 8 2, and 83 are contained in three pairwise disjoint balls), then the disks WI. W2, and W3 can be chosen to be pairwise disjoint, so that the triple Massey product (0 1, {31, 'YI} vanishes. Thus, the Borromean rings form a nontrivial link. A similar argument proves the nontriviality of the following multidimensional generalization of the Borromean rings. Suppose that n = p + q + r, x = (Xl, ... , X p), Y = (YI,"" Yq), and Z = (Zl,"" zr). Consider the three spheres in ]Rn determined by the equations X=o,

lIyI12

+ IIzII2 = 1

(sphere 8i+r - I );

y=O,

IIzII2

+ IIxI12 = 1

(sphere ~+r-l);

IIxII2

+ lIyI1 2 = 1

4 4

(sphere 8K+ q - I ). 4 Any two of these spheres are separated by an (n - I)-sphere. For example, 8i+r - 1 lies outside the sphere z=O,

IIxI1 2 ~ 32 + {I/2)2

~-I

+ (3/2)2

-

,

while 8K+ q - 1 lies inside it. Thus, the spheres under consideration are not pairwise linked. Therefore, the pairwise products of the cohomology classes uP, {3q, and 'Y r corresponding to the fundamental classes of 8i+r - l , 8~+r- \ q - 1 under the Alexander duality vanish. The products aP ........ uP, and {3q ........ {3q, and 'Y r ......... 'Yr vanish as well. Let us show that for the coefficient group Z2, the triple product (uP, {3q, 'Yr} E H n - I (8n \81 U82U83) ~ Ho(81 U 8 2 U 83) does not vanish. Consider the disks WI, W2, and W3 spanned by 81, 8 2, and 8 3, that is,

SK+

WI

=

{(x,y,z) I X

= 0, lIyII 2 +

11~12 ~ I},

W2

=

{(x, y, z) I Y

= 0, IIzll2 +

11~12 ~ I},

W3

=

{(x,y,z)

I z = 0, IIxI1 2 + 1I~12 ~

I}.

Clearly, WI

n W2

=

{(x,y,z)

IX

=

O,y = 0,

IIzll

~ I}

88

2. Cohomology Rings

and

W2

n W3 = {(X,y,Z) I Y = O,Z = 0, Ilxll

~

I}.

We set

e = {(x, Y, z)

I Xl > 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.

2. Homology and Cohomology of Manifolds

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

2. Homology and Cohomology of Manifolds

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

2. Homology and Cohomology of Manifolds

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,

3. The Kiinneth Theorem

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

3. The Kiinneth Theorem

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.

3. The K iinneth Theorem

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

3. The K iinneth Theorem

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

-+

3. The K iinneth Theorem

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
X

X {3~2)

{3~2).

0

= 0: x {3.

Proof. We have Pi
108

2. Cohomology Rings

Problem 82. Prove that if at least one of the groups H*(K) and H*(L) is torsion-free, then multiplication in the ring H*(K x L) is completely determined by multiplications in H*(K) and H*(L). Problem 83. Given K = lRP2, Ll = lR'p 2 V S3, and L2 = lRp 3, prove that the rings H*(Ll) and H*(L2) are isomorphic, whereas the rings H*(K xLI) and H*(K x L2) are not. The Signature of Cartesian Products. Cohomology cross product can also be used to pr~ve the following theorem concerning the signature of the Cartesian product of two manifolds. We assume that u(Mn) = 0 if n is not divisible by 4. Theorem 2.28. The equality u(MP x Nq) = u(MP)u(Nq) holds. Proof. If p + q is not divisible by 4, then one of the numbers p and !l is not divisible by 4 either; hence u(MP x Nq) = 0 and u(MP)u(Nq) z= O. In what follows, we assume that p + q = 4k. For cohomology with coefficients in lR, the Kiinneth theorem implies

2k H 2k (MP x Nq) ~ E9(H i (MP) ®IR H 2k - i (Nq». i=O

a1

If i + j i=- p, then the intersection form vanishes at the elements ® f3~k-i and ~ ® f3~k-j. We decompose the space H2k(MP x Nq) into the sum of spaces

(19)

(Hi(MP) ®IR H 2k - i (Nq» ffi (HP-i(MP) ®IR H 2k+i-p(Nq)),

where i < p, and the space HP/2(MP) ®IR Hq/2(Nq) (if p and q are even). These spaces are mutually orthogonal with respect to the intersection form. We prove that the intersection form has zero signature on each subspace of the form (19), i.e., the signature of u(MP x Nq) equals that of the restriction of the intersection form to the subspace HP/2(MP) ®IR Hq/2(Nq). Let i < p/2. Choose a basis {as} in Hi (MP) and let {a:} be the dual basis in the space HP i(MP); then (as '--' a;, [MP]) = 6rs . Similarly, choose a basis {f3t} in H 2k -'(Nq) and the dual basis {f3;} in H2k+'-P(Nq). For a basis in the space (19) we can take the vectors a s®f3t and a;®f3:. The intersection form vanishes at all pairs of basis vectors except at the pairs (as ® f3t, a: ® f3;). According to Theorem 2.27, at such pairs of basis vectors the intersection form takes the value (_l)dim,BedimQ: = (_1)(2k-i)(p-i). Clearly, this form is symmetric. Therefore, in the chosen basis, the restriction of the intersection form to the space (19) is determined by the direct sum of matrices ( _l)d ( ~ fi), where d = (2k - i) (p - i). Each of these matr· es has signature zero.

3. The Kiinneth Theorem

109

If the numbers p and q are odd, then the space HP/2(MP) ®li Hq/2(Nq) is undefined, and u(MP x Nq) = o. If each of the numbers p and q is congruent to 2 modulo 4, then the intersection forms on the spaces HP/2(MP) and Hq/2(Nq) are antisymmetric. Therefore, these spaces have bases in which the intersection form is determined by the direct sum of matrices (J 5). The restriction of the intersection form to the tensor product of two two-dimensional spaces with such intersection matrices is given by

01 01 ( -(-1 0)®(-1 0)=

0 0 0 -1) 0010 010 0; -1 0 0 0

the minus sign appears because the number ~ . ~ is odd. It is easy to verify that this matrix has signature zero. Therefore, u(MP x Nq) = o. Finally, suppose that both numbers p and q are divisible by 4. Choose bases in the spaces HP/2(MP) and Hq/2(Nq) so that the intersection matrices in these bases are diagonal. Take the tensor product of these bases and consider the matrix A of the restriction of the intersection form to the space HP/2(MP) ® Hq/2(Nq). Let rand s be the numbers of positive diagonal elements for the intersection forms on HP/2(MP) and Hq/2(NQ), respectively. The matrix A is diagonal; its diagonal has rs+(~-r) (~-s) positive elements and r(~ - s) + s(~ - r) negative elements. The signature of this matrix equals (r - (~ - r)) (s - (~- s)) = u(MP)u(Nq). 0

Nondegenerate Bilinear Maps. A bilinear map /: lRr x ]RB --+ ]Rn is said to be nondegenemte if /(x, y) = 0 implies that x = 0 or y = o. To any nondegenerate bilinear map /: lRr x ]RB --+ lRn we associate a map F: ]Rpr-l x ]RpB-l --+ lRpn-l by setting F( {>.x}, {j.ty}) = {a/(x, y)}. This map is well defined because /(>..x, J.LY) = >"J.L/(x, y). According to the solution of Problem 81, the cohomology ring of the space ]Rpn x ]Rpm with coefficients in Z2 is isomorphic to the quotient of the polynomial ring Z2[U,V] by the relations un+! = 0 and v m +! = o. This property determines constraints on the numbers r, s, and n that ensure the existence of a nondegenerate bilinear map ]Rr x ]RB --+ lRn. These constraints have been obtained simultaneously by Stiefel [133] and Hopf [62]; they submitted their papers to the journal Commentarii Mathematici Helvetici on the same day. Theorem 2.29 (Stiefel Hopf). If there exists a nondegenemte bilinear map f: lRr x ]RB - R n , then the number (~) is even for all k such that n - s < k < r.

110

2. Cohomology Rings

Proof. Consider the homology classes a and b in HI (IRpr-l X IRps-l; Z2) that are represented by the cycles lRpl x {*} and {*} X lRpl. The cohomology classes 0 and {3 dual (in the sense of linear algebra) to a and b multiplicatively generate the cohomology ring of IRpr-l x IRps-l. Let c and 'Y be the generators of the one-dimensional homology and cohomology groups of lRpn-l. It is sufficient to prove that p.'Y = 0 + (3. Indeed, this equality implies 0 = p.(-yn) = (0 + (3)n = E~=o (~)ok{3n-k. Therefore, if k < rand n - k < s (i.e., n - s < k < r), then the number (~) is even. First, we show that P.a = c. It suffices to verify that P.a i= O. The homology class represented by a loop in IRpn-l is nonzero if and only if the lifting of this loop to sn-l is a nonclosed path. In the situation under consideration, the condition f( -x, y) = - f(x, y) implies that the liftings of loops are nonclosed. Hence P.a = c and, similarly, P.b = c. Thus, (p.'Y, a) = (-y, P.a) = (-y, c) = 1 and (P*'Y, b) = 1. Therefore, p.'Y = o +(3. 0 Problem 84. Prove that if there exists a nondegenerate bilinear map lRn x IR n --+ IRn , then n = 2k for some k. Problem 85. (a) Prove that there exists a nondegenerate bilinear multiplication IRr x IRs --+ lRr +s- l . (b) Prove that if rand s are even numbers, then there exists a nondegenerate bilinear multiplication IRr x IRS --+ lRr +s- 2 . (c) Prove that if rand s are numbers divisible by 4, then there exists a nondegenerate bilinear multiplication lRr x IRs --+ lRr +Il - 4 . (d) Prove that if rand s are numbers divisible by 8, then there exists a nondegenerate bilinear multiplication lRr x IR" --+ lRr +II - 8 .

Chapter 3

Applications of Simplicial Homology

This chapter discusses various applications of simplicial homology. They are based on studying the properties of simplicial homology and cohomology in more detail and introducing additional structures.

1. Homology and Homotopy In this section, we consider relations between homology and homotopy. First, we prove a theorem of Hurewicz, which asserts that the groups 7l"n(K) and Hn(K) are isomorphic for any (n -I)-connected simplicial complex K. Then, we construct the apparatus of obstruction theory, which is applied to interpret cohomology groups as homotopy classes of maps to K(7l", n) spaces. 1.1. The Hurewicz Theorem. Suppose that K is a simplicial complex and Xo E K. Consider the map h: 7l"n(K,xo) --+ Hn(K) defined as follows. Take a spheroid ep: (sn, so) --+ (K, xo). It induces the homomorphism ep.: Hn(sn) --+ Hn(K) of homology groups. We set h(ep) = ep.([sn]), where [sn] is the fundamental class. Homotopic maps epo and epl induce the same homomorphism in homology; therefore, h(epo) = h(epl), i.e., the map h is indeed defined for the homotopy group 7l"n(K, xo).

Let us show that h is a group homomorphism. We regard spheroids ep and 'If; as maps (D n , aDn) --+ (K, xo). We assume that the sphere sn is obtained by gluing together two balls D n with triangulations so fine that the maps ep and 1/J are homotopic to simplicial maps; we also assume that the union of the triangulations of the balls is a triangulation of the sphere. In this case, the homology classes h(ep) and h(1/J) are represented by the cycles

-

111

3. Applications of Simplicial Homology

112

~a rp(~ra.) and ~t3 rp(~2t3)' where the ~ra and ~2t3 are simplices of the

triangulations of the first and second ball and the homology class h(rp + 1/J) is represented by the cycle ~a rp(~ra.) + ~t3 rp(~2t3)· Therefore, h is a group homomorphism. It is called the Hurewicz homomorphism.

The homomorphism h, as well as the group 7I"n(K, xo) itself, depends on the choice of the base point Xo. However, this dependence is not very essential in the sense that if 'Y is any path from Xo to Xl, then the following diagram is commutative:

Theorem 3.1 (Poincare). If K is a connected simplicial complex, then the homomorphism h: 71"1 (K) --+ HI (K) is an epimorphism, and its kernel is thi! commutator subgroupl [7I"1(K),7I"1(K)] of the group 7I"l(K). Proof. We assume that the base point Xo is a vertex of the complex K. First, we prove surjectivity. Let ~! = [VOi' VIi], where VOi and Vii are vertices of K and ~ ai~! is a cycle. Then 0 = a(~ ai~n = ~ aivli - ~ aivOi. Suppose that 'Ya is a path between the base point Xo and a vertex Va going along edges of the complex K. This path corresponds to a chain 1'a E Cl(K). The path 'Yoi~hii1 is a loop based at Xo. Clearly,

h(IIboi~hiil)a,) = Lai(1'oi+~! -1'Ii). ~ aivli = ~ aivOi implies ~ aii'li = ~ ai1'oi because

The equality Va is a free generator of the group CoCK). Thus, we have constructed an element of the fundamental group which is mapped to E ai~! . The image of the homomorphism h is an Abelian group; therefore, its kernel is contained in the commutator subgroup of 71"1 (K). Let w be a loop in K based at Xo for which hew) = a(E al~n. Any loop in K based at a vertex is homotopic to a loop along edges. Therefore, we can assume that w = IT 'Y,O"I. where 'Y,O"l is an oriented edge of the complex K. For I-chains, we have

(20)

L

1',0,,1 = L

a,(1',o,il

+ 1'il,i2 + 1'i2,iO) ,

where 1'a,t3 is the oriented I-simplex [va, vt3]. Let 'Ya be a path from the base point Xo to a vertex Va (if Va = Xo, then the path 'Ya is constant). Suppose that 'Ya'Ya,t3'Yi/ is the image of the loop lThe commutator lJubgroup [G, G] of a group G is the subgroup generated by all elements of the form aba-1b- 1 , a,b E G.

1. Homology and Homotopy

113

'Ya'Ya,rnj/ in the commutator quotient Of7r1(K)j this quotient is an Abelian

group. Equality (20) holds for the generators 'Ya,fj of the free group; hence a similar equality holds for the elements 'Ya'Ya,fj'Yj/. Let us write it in the multiplicative form:

II IjOljO,jl'Yjl-1 = II(I,Ol,O,il IiI-1 . li111l,i21i2-1 . li2'Yi2,iO'Y1_1)BI 0 . Since the path w is closed, we have

Moreover, -1

-1

1

1

liOliO,il III . li11i1,12112 . 112'Yi2,10'Y,0 = II0'Y10,11 'Yil,12'Yi2,IOI,0 =

1

,

because the loop II0,il'Yil,,21i2,iO is contractible (see Figure 1). Thus, iii = 1, as required. 0

'rIO

Figure 1. A contraction of the loop

Theorem 3.2 (Hurewicz [64]). If K is a simplicial complex and 7ro(K) = ... = 7rn -l(K) = 0 for n ~ 2, then the Hurewicz homomorphism ---+ Hn(K) is an isomorphism.

7r1(K) = h: 7rn (K)

Proof. First, we show that h is an epimorphism. Take any cycle E ai6.i E Cn(K). The equalities 7ro(K) = 7rl(K) = ... = 7rn -l(K) = 0 imply that the identity map K ---+ K is homotopic to a map f: K ---+ K taking the skeleton Kn-l to the base point Xo E K (see the proof of the Whitehead theorem in Part I). We can assume that Xo is a vertex of K. Then the restriction of f to Kn-1 is a simplicial mapj hence it has a simplicial approximation g: K(m) ---+ K whose restriction to Kn-1 coincides with f. i.e., g(Kn-1) = Xo·

Thus, g: K(m) ---+ K is a simplicial map homotopic to the identity and taking K n - 1 to Xo· Let 'Pi be the restriction of 9 to 6.ij it can be treated as a spheroid (D n ,8Dn) ---+ (K,xo). Clearly, Eaih('Pi) = g.([Eai6.i]), where [E ai6.i] is the homology class represented by the cycle E ai6.i. Since the map 9 is homotopic to the identity, it follows that 9. ([E aidi]) =

[E ai6.i] .

114

3. Applications of Simplicial Homology

Now, we show that h is a monomorphism. Consider a map cp: sn -+ K for which cp.([sn]) = O. It is required to prove that cp is the zero element of the group 7l"n(K). We can assume the map cp to be simplicial. Lemma. There exists a simplicial complex Ln+l ::J sn such that the cycle [sn] in this complex is homologous to zero and the map cp can be extended to Ln+l.

Proof. Suppose that at the level of chains, cp.([sn]) = 8(E ~~+l) (some of the simplices ~~+l C K may coincide). For the first approximation to Ln+l we take the disjoint union of the sphere sn and the simplices Li~+!, each identified with the corresponding simplex ~~+l C K; the map cp can be extended over Li~+! to the identity map. In the simplidal complex thus obtained, the cycle Zn = [sn]- E 8Li~+l is homologous to [sn], and cp.(zn) =

O. The next approximation to Ln+l is constructed so that the- cycle Zn becomes homologous to a cycle Wn = E S'J, where cp(S'J) C Kn-l. Suppose that Zn = E~:! and cp(~:!J rt. Kn 1, i.e., cp maps ~:!l homeomorphically to a simplex from K. The equality cp.(zn) = 0 implies that some other simplex ~:!2 is mapped homeomorphically onto the same simplex as ~:!l' but the orientations of ~:!l and ~:!2 are opposite. To ~:!l and ~:!2 we attach the triangulated prism ~n x I oriented so that the orientations induced on ~n X{O} = ~:!l and ~n x {I} = ~:!2 are as required. Consider an extension of cp to ~n X I with constant value on {x} x I. We have 8(~n x I) = ~:!l + ~:!2 + en, where the chain Cn contains only simplices satisfying the condition cp(~n) C K n 1. Let us replace the chain ~:!l + ~:!2 in the cycle Zn by the homologous chain -Cn. After several such operations, we obtain the desired cycle W n . Finally, to the simplicial complex obtained at the preceding step, we add the cones over all simplices ~; in the cycle W n . Let V be the union of all such simplices, and let CV be the cone over V. This cone is acyclic; therefore, the cycle Wn is now homologous to zero. It remains to extend cp over CV. A map cp: V -+ K can be extended over CV if and only if cp is null-homotopic (that is, homotopic to a constant map). By assumption, we have cp(V) C K n - 1. The equalities 7l"o(K) = 7l"l(K) = ... = 7l"77-1(K) = 0 imply that the identity map K -+ K is homotopic to a map f: K -+ K taking the skeleton Kn-l to xo. Therefore, the map cp: V -+ K n - 1 c K is homotopic to a constant map to K. 0 Suppose that cp: Ln+l -+ K is a continuous map, sn c Ln+l, and [sn] = 8en+l for some chain Cn+! E Cn+!(K). Replacing cp by a homotopic map, we can assume that cp(Ln-l) = xo. Then each simpl('x ~n c Ln+l

1. Homology and Homotopy

115

corresponds to a map (an, aan) !!.... (K, xo), and we obtain a homomorphism

cI>.: Cn(Ln+l)

-+

7I"n(K).

It remains to show that cI>.([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.

3. Applications of Simplicial Homology

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

1. Homology and Homotopy

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

3. Applications of Simplicial Homology

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

1. Homology and Homotopy

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

3. Applications of Simplicial Homology

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.)

1. Homology and Homotopy

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

3. Applications of Simplicial Homology

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.

1. Homology and Homotopy

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

3. Applications of Simplicial Homology

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.

1. Homology and Homotopy

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

3. Applications of Simplicial Homology

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.

1. Homology and Homotopy

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

3. Applications of Simplicial Homology

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

3. Applications of Simplicial Homology

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

2. Characteristic Classes

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.

3. Applications of Simplicial Homology

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

2. Characteristic Classes

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

3. Applications of Simplicial Homology

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 •

2. Characteristic Classes

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
and if k = nand n even, if k ::; nand k odd, otherwise. Proof. The boundary homomorphism 0: C k -+ Ck-l acts as multiplication by 1 + (_l)k(-l) = 1 + (_l)k+1. The coboundary homomorphism 6: C k - 1 -+ C k has the same form because it is dual to O. 0 Remark. The local system ZT is defined for any space with fundamental group Z2' Poincare Duality for Local Coefficients. Recall that the usual Poincare duality for an oriented manifold Mn with two dual cell decompositions K and K* is established by an isomorphism Ck(K; Z) -+ Cn-k(K*; Z), which

3. Applications of Simplicial Homology

138

induces the isomorphism Hk(M n ) ~ Hn-k(Mn; Z). Namely, each chain Ck - E aiUi is associated with the cochain c n k for which (c n - k , = ai. The orientation of the cell depends on that of the manifold Mn; for the same manifold with opposite orientation, is replaced by

un

u;

u;

-u:.

In a similar way, for an arbitrary manifold Mn (orient able or not), we can construct isomorphisms Ck(K; Or) --+ C n k(K*; Z) and Ck(K; Z) --+ cn-k(K*; Or), where Or = ZOrMn. The first isomorphism is constructed as follows. The elements of the groups Ck ( K; Or) are the sumb E (Eia1) Ui, where ai E Z and each Ei is the orientation of Mn at the points of the k-dimensional cell Ui. To such a sum we associate the cochain c n k E C n k(K*; Z) for which (cn k, u;) = ai. This definition makes sense because the choice of Ei determines both the number ai and the orientation of the cell Replacing E, by -E 1 , we obtain ai instead of ai and instead of

u;. u;.

-u;

The second isomorphism is constructed as follows. Each cochain in the group Cn-k(K*; Or) takes an (n-k)-cell to aiEi, where ai E Z and Ei is the orientation of M n at the points of To the chain Ck = E aWi E Ck(K; Z) we assign the cochain c n - k E cn-k(K*; Or) that takes every u; to a1Ei. This definition makes sense because replacing Ei by -Ei, we obtain the cochain taking -u; to -a1Ei.

u;.

u;

Thus, we have proved the following theorem. Theorem 3.15 (the Poincare duality isomorphism). For any closed manifold M n , the Poincare isomorphisms with local coefficients Hk(M n ; Or) ~ Hn-k(Mn; Z) and Hk(M n ; Z) ~ Hn k(Mnj Or) hold. Obstructions to Extending Sections. Consider a lorally trivial bundle p: E --+ B with fiber F. We assume that B is a simplicial complex. Suppose that on B n (the n-skeleton of the complex B), a section s: B n --+ E of the bundle p is given. We want to determine whether or not this section can be extended to Bn+l, and if not, whether it can be changed on B n \ Bn+l so that the new section does admit such an extension. This problem closely resembles the usual problem of obstruction theory. Indeed, the bundle is trivial over each simplex ,6,n+l C Bn+l, so its section over ,6,n+l is completely determined by a map to the fiber F, and it is requirpd to extend a map 8,6,n+l --+ F to a map ,6,n+l --+ F. However, this argument is valid only if a trivialization p-l(,6,n+l) --+ F x ,6,n+l is fixed. The obstructions cn+1 (s) E C n+l(Bj7rn(P-l(b))) are definpd as in the usual theory of obstructions, but the co chains have coefficients in a local system of Abelian groups {7rn (p-l(b))} on the space B. Indeed, under the assumption that the space F is n-simple (which is a stanJdrd assumption in obstruction theory), each path between bl E Band b2 E B determines an

2. Characteristic Classes

139

isomorphism 7I'"n(p-l(bI)) --+ 7I'"n(P l(b2)); this isomorphism is constructed by using the covering homotopy theorem. All assertions of Section 1.2 carryover to obstructions to extending sections without substantial changes except that cohomology with coefficients in the local system of homotopy groups of the fiber should be considered. As mentioned before, the fiber F must be simple. 2.3. The Stiefel-Whitney Characteristic Classes. Suppose a bundle p: E --+ B has fiber F such that 7I'"o(F) = '" = 7I'"n I(F) = 0 but 7I'"n(F) !- 0 for n ~ 1; for n = 1, we assume in addition that the group 71'"1 (F) is Abelian. Then the fiber F is n-simple. There are no obstructions to extending sections over the k-skeletons for k ~ n. The first nontrivial obstruction lies in the group Hn+l(B; ~), where ~ is the local system {7I'"n(P l(b))}. This obstruction does not depend on the choice of a section on the n-skeleton. Indeed, if 81 and 82 are two sections on the n-skeleton, and cn +l(8d, Cn +l(82) E zn+l(B;~) are obstructions to extending these sections over Bn+l, then there exists a co chain rtn (81, 82) E cn(B;~) for which dd n (Sl,82) = Cn +l(8I) - Cn +l(82) (see Lemma 2 on p. 118). Thus, we can associate the bundle with the first nontrivial obstruction to extending sections. For vector bundles, this construction cannot be used directly to define characteristic classes because the fiber of a vector bundle is a contractible space. But to each n-dimensional vector bundle we can assign n - 1 bundles over the same base whose fibers are sets of k orthonormal vectors, where 1 ~ k ~ n-1. Indeed, let ~ be a vector bundle over a finite simplicial complex B. The construction used in Part I, p. 204, to define a Riemannian metric on a manifold yields a Riemannian metric on~. Clearly, any two Riemannian metrics are homotopic in the class of Riemannian metrics. Indeed, if (" ')0 and (,,·h are two inner products on a vector space, then t(·,· )0+ (l-t)(., ·h is an inner product for any t E [0, 1J. Consider the bundle Pk: Ek --+ B, where Ek consists of all pairs (b E B, an ordered set of k orthonormal vectors in the fiber of ~ over b).

The space Ek is endowed with the natural topology. The fiber of the bundle ~k constructed above is known as the Stiefel manifold V(n, k); it consists of k-tuples of orthonormal vectors in JRn . For the bundle ~k, a characteristic class is defined. We start by calculating the first nontrivial homotopy group of the space V(n, k).

Problem 96. Prove that V(n, k) is a manifold of dimension nk _ k(k~H).

3. Applications of Simplicial Homology

140

Exercise. Prove that V(n + 1, 2) is the manifold of unit tangent vectors to the sphere sn.

< n - k, then 7l"i(V(n, k)) = O. The first nontrivial homotopy group 7l"n-k (V (n, k)) coincides with Z (if n - k is even or k = 1) or Z2 (ifn - k is odd and k > 1).

Theorem 3.16. If i

Proof. First, let us show (by induction on k) that 11",(V(n, k)) = 0 for i < n - k. For k = 1, the manifold V(n,l) is the sphere sn-l. Clearly, 11"i(sn-l) = 0 for i < n - 1. The induction step is as follows. Consider the map V(n, k+l) --+ V(n, k) deleting the last vectors from each (k+l)-tuple of orthonormal vectors. This map is a locally trivial bundle with fiber sn-k-l. The exact homotopy sequence for this bundle is

11",(sn-k-l) For i

<

---+

11"i(V(n, k

+ 1»

---+

11"i(V(n, k»

---+

11",_I(sn-k-l).

n - k - 1, this fragment of the exact sequence has the form. 0 --+

0 --+ 0; therefore, 7l"i(V(n, k + 1» = O. Now, let us calculate 7l"n-k(V(n, k». The case k = 1 is handled separately: V(n, 1) ~ sn-l. In what follows, we assume that 1 < k < n. Consider the map V(n + 1, k + 1) --+ V(n + 1,1) = sn assigning Vk+l to orthonormal vectors VI, •.. ,Vk+l. This map is a locally trivial bundle with fiber V(n, k). The exact homotopy sequence for this bundle is

7l",(V(n, k + 1»

--+

11"i+l(sn)

---+

7l",(V(n, k))

---+

11"i(V(n + 1, k + 1))

---+

11"i(sn).

If i ~ n - 2, then 11"i(V(n, k» ~ 11"i(V(n + 1, k + 1». Successively applying such isomorphisms, we obtain 11"n-k(V(n, k» ~ 11"n-k(V(n - k + 2,2»; the isomorphism 7l"n k(V(n - k + 3, 3» ~ 11"n-k(V(n - k + 2, 2» follows from the inequality n - k ~ (n - k + 2) - 2. It remains to verify that the group 7l"n-2(V(n, 2)) coincides with Z if n is even and with Z2 if n is odd. Consider the bundle V(n, 2) --+ V(n, 1) = sn- l with fiber V(n -1,1) = sn-2. The exact homotopy sequence for this bundle is

7l"n l(sn- 1) ~ 7l"n_2(sn-2)

a.

---+

11"i(V(n, 2))

---+

7l"n_2(sn-l)

= o.

The homomorphism maps Z to Z; therefore, it suffices to calculate a.(idsn 1). By definition, the element a.(ids n-1) E 7l"n_2(sn-2) is constructed as follows. On the sphere sn-l, consider the one-parameter family of spheres S~-2, where t E (0,1), which pass through a fixed point eo E sn-l and have a common tangent (n - 2)-plane at this point. We lift the point eo to the total space V(n, 2) of the bundle and extend the lifting to S~-2 first for small t and then for all t < 1. Letting t tend to 1 we obtain a map from the sphere S~-2 to the fiber. The homotopy class of this map is the required element of 7l"n_2(sn-2).

2. Characteristic Classes

141

To lift the point eo E sn-l (treated as a unit vector) to V(n, 2), we have to choose a unit vector el orthogonal to eo. The vector el can be regarded as a vector tangent to sn-l at eo. The lifting can be constructed using a vector field with one isolated singular point. It is required that the vectors be parallel in a half of a neighborhood of the singular point and all the rotation of the vectors take place in the second half (the first half of the neighborhood corresponds to 2 with t < 1/2 and the second, to the spheres with t > 1/2). the spheres In Figure 4, the dashed line shows the circle sf with t > 1/2 for n - 3. As

Sr-

Figure 4. The trajectories of the vector field on S2

t -+ 1, the degree of the map

Sr-

2 -+ sn-2

taking each point to the tangent vector at this point equals the index of the singular point, i.e., the Euler characteristic of sn-l. Thus, 8.(ids n-l) = X(sn-l) = 1 + (_l)n-l. This means that if n is even, then 1m 8. = 0, and if n is odd, then 1m 8. = 2Z. D Remark. Many other homotopy groups of Stiefel manifolds, which are very important for homotopy theory, were calculated in the series of papers [99]. Problem 97. Calculate the homology groups with coefficients in Z for the Stiefel manifold V(n + 2, 2), where n ~ 1. The characteristic class of the bundle Pk, which is contained in the group Hn-k+I(B, {7rn-k(V(n, k))}), is an obstruction to extending k linearly independent sections over the (n-k+l)-skeleton. Ifn-k is odd and k > 1, then this class is contained in the ordinary cohomology groups Hn-k+l(Bj Z2) because there are no nontrivial automorphisms Z2 -+ Z2 j thus, in this case, the system of local coefficients is trivial. The corresponding cohomology class W2; E H2J (B; Z2) is called the Stiefel Whitney characteristic class. This class is an obstruction to extending n + 1 - 2j linearly independent sections over the 2j-skeleton. Problem 98. A manifold Mn is said to be almost parallelizable if the manifold M n \ {x}, where x E Mn, is parallelizable. Prove that a simply connected 4-manifold M4 is almost parallelizable if and only if the class W2 for its tangent bundle vanishes.

3. Applications of Simplicial Homology

142

The odd-dimensional Stiefel Whitney classes (and the class Wn for even n) are defined by reducing the integer coefficients modulo 2. As a result, the Stiefel Whitney classes W2j+l E H2j+l(B; Z2) are obtained. We emphasize that they do not coincide with the obstructions 2 °+1 02j+lEHJ (B,{7I"2j(V(n,n-2j))}),

although they are very close to them (see Problem 100). Steenrod showed that the obstruction 02j+I. where j ~ 1, can be expressed in terms of the characteristic class W2j = 02j by means of the following construction. Consider the locally trivial bundle V (n, n - 2j + 1) S2,

1

-----t

V (n, n - 2j). The exact homotopy sequence for this bundle is

(24) 7I"2j(V(n, n - 2j)) ~

7I"2j_l(S2j -l)

~ 7I"2j-l(V(n, n - 2j + 1)) -

o.

On the right-hand side, we have 0 instead of 7I"2j I(V(n, n - 2j)) because

2j - 1 < n - (n - 2j). Thus, we have an exact sequence Z ~ Z. ~ 'Z2 .0; therefore, 1m 8* = Ker i* = 2Z, and Ker 8. = O. Adding U - to the exact sequence (24) on the left, we obtain a short exact sequence of groups. Moreover, in fact, we obtain a short exact sequence of local systems of Abelian groups (this is easy to check). As for ordinary cohomology groups, this exact sequence induces the Bockstein homomorphism (3*: H 2j (B, {7I"2j-l(V(n, n - 2j + I))}) - H2j+l(B, {7I"2j(V(n, n - 2j))}). Problem 99. Prove that

02j+l

= (3*02j.

Remark. Problem 113 shows that if WI

= 0,

then W2j+l

= Sq l W2j .

Problem 100. Prove that the obstructions Ok, where k < dim~, have order 2. Moreover, if the number n = dim~ is odd, then the class On has order 2 as well. Problem 100 shows that almost all classes Wk(~) are true obstructions, i.e., extensions of sections exist if and only if these classes vanish. The only exception is the class wn(~) with even n = dim~. Only this class is obtained by the reduction modulo 2 of a class which may be of order different from 2. 2.4. Properties of the Stiefel Whitney Classes. We have defined the classes W2, .. . , Wn for n-dimensional vector bundles. We assume that Wo = 1 and Wi = 0 for i > n. The Class WI' The class WI must be considered separately because it must correspond to an obstruction to extending n linearly independent sections from the O-skeleton to the I-skeleton. Applying the standard construction of obstruction theory, we see that the corresponding co cycle rrllst take values in 7I"o(V(n, n)), and 71"0 is not a group. We have defined V(n, k) only for

2. Characteristic Classes

143

k ~ n - 1; applying the same definition to k = n, we obtain a disconnected manifold diffeomorphic to O(n). Thus, the set 11"0 (V(n, n)) consists of two elements. This violates uniformity too because for the obstructions Oi with odd numbers, the local coefficient group is Z. For this reason, instead of 11"0 (V(n, n)), we take the reduced homology group Ho(V(n, n); Z) ~ Z and obtain the local coefficient system ac, where a E Z and c is the orientation of the fiber. Indeed, the connected components of the space Yen, n) are identified with orientations of the fiber, and hence the elements of Bo(V(n, n); Z) ~ Z have the form a Orl -a Or2; to each orientation we assign its coefficient.

The orientation of the fiber can be transferred along any path in the base; thus, we indeed obtain a local system of coefficients. We now construct obstructions to extending n linearly independent sections from the O-skeleton to the I-skeleton. The connected components of the space V(n, n) are orientations. So, we are to specify orientations in the fibers over the I-skeleton. Suppose that at each vertex Vi, an orientation Ci is given. Consider the cochain c1 E C 1 (B;{Ho(V(n,n);Z)}) for which (c l , [viVjD = Cj - Ci. This cochain is a co cycle because

(del,

[ViVjVkJ)

= (c l , [VjVkJ) - (c I , [ViVkJ)

=

+ (c l , [ViVjJ)

+ (Cj -

= O. I-skeleton if and only if c l = O.

(ck - Cj) -

(ck - ci)

E."i)

The orientations Ci extend to the Changing orientations at some vertices corresponds to adding some coboundary dc o to c 1 . Therefore, the orientations extend to the I-skeleton if and only if the cohomology class of the cocycle c1 , which we denote by 01, vanishes. The class 01 does not depend on the choice of the section over the 0skeleton. Indeed, as mentioned above, changing the orientation at vertices corresponds to adding a coboundary. This implies, in particular, that 201 = O. Indeed, if we replace the orientations Ci by -E."i at all vertices Vi, then the co cycle c l will change sign, as well as the class 01. But the class 01 does not depend on the choice of orientations; therefore, 01 = -01. i.e., 201 = O. Reducing the local coefficient system Ho(V(n, n); Z) ~ Z modulo 2, we obtain the class WI from the class 01. The equalities 01 = 0 and WI = 0 are equivalent because 201 = O. Thus, WI = 0 if and only if the fibers of the bundle over the I-skeleton admit compatible orientations. This property has the following geometric meaning. Let { be an n-dimensional vector bundle over a linearly connected base. Consider the set of unit vectors in the one-dimensional bundle An{ endowed with a Riemannian metric (these vectors correspond to orientations). If the space thus obtained is connected, then the bundle € is said to be nonorientable, and if it consists of two connected components, then € is orientable.

144

3. Applications of Simplicial Homology

Choosing one of these two components, we specify orientations of all fibers of the bundle. Example 43. A one-dimensional bundle is orient able if and only if it is trivial. Theorem 3.17. A vector bundle over a simplicial complex is orientable if and only if its restriction to the I-skeleton is orientable. Proof. The space {x E Anf,llIxll = I} doubly covers the base. This space is a CW-complex, and the connectedness of a CW-complex is equivalent to that of its I-skeleton. 0 Thus, WI (~) = 0 if and only if the bundle f, is orient able. We apply this property to calculate the class WI of the bundle 'Y~ over JRpn, which is very important for various applications of characteristic classes. The total s,Pace E(T~) of 'Y~ consists of pairs (x E JRpn, v = ,xx); it is assumeq that each point x E IRpn is represented by a nonzero vector in JRn+l, and the fiber over this point consists of all vectors proportional to this vector. Theorem 3.18. The class WI('Y~) is equal to the generator of the group HI (JRpn; Z2). Proof. The group HI (JRpn; Z2) has only one nonzero element; therefore, it suffices to verify that WI ('Y~) i= o. The bundle 'Y~ is one-dimensional; hence we must show that it is nontrivial. For every section s: JRpn --+ E(T~), consider the composition sn --+ JRpn ~ E(T~), where the first map is the canonical double covering. We obtain a map sn --+ JR, which is defined by x t-+ ,x(x). We have ,x(x)x = A( -x)( -x), i.e., ,x( -x) = -,x(x). Thus, the values of the function ,x at the endpoints of any path in sn between x and -x are of different signs; therefore, this function vanishes at some point of any such path. It follows that the bundle 'Y~ has no nowhere vanishing sections, i.e., is nontrivial. 0 The Class Wn and the Euler Class e. By definition, the element On E Hn(B; {7l"n-I(V(n, I»)}) is an obstruction to extending a single nowhere vanishing section to the n-skeleton. Note that V(n, 1) ~ sn-I. If the bundle is oriented, then each point b E B is associated with an isomorphism cp: 7l"n I (sn-I) --+ Z. (There exist two isomorphisms Z --+ Z; changing the orientation corresponds to changing the isomorphism.) Applying the isomorphism cp, we obtain an isomorphism Hn(B; {7l"n-I(V(n, I»)}) ......... Hn(B; Z). The image of the class On under this isomorphism is called the Euler class and denoted bye. The class wn is obtained from by reduction modulo 2.

2. Characteristic Classes

141

For the bundle with opposite orientation, e has the opposite signj i.e.

e( -e) = -e(e)· Theorem 3.19. If the dimension of the bundle

e is odd, then 2e(e) = O.

Proof. The map v 1---+ -v induces a fiberwise self-homeomorphism of the total space of the bundle. Therefore, if the dimension is odd, then the bundles and are isomorphic. (If the dimension is even, then the map v 1---+ -v preserves orientation.) 0

e

-e

Thus, for vector bundles of odd dimension, the classes wnCe) and e(e) do not differ substantially, although they belong to different groups. Both of them are true obstructions, i.e., sections extend to the n-skeleton if and only if wn(e) = 0 or e(e) = 0 (these conditions are equivalent). But for evendimensional orient able bundles, only the class e(e) is a true obstruction, while wn(e) is not: the equality wn(e) = 0 does not imply the existence of a nowhere vanishing section over the n-skeleton. Such a section exists only if eCe) = O. For example, W2(S2) = 0 but e(S2) = 2 i= OJ therefore, no vector field on the sphere S2 is nowhere vanishing. Let Mn be a closed oriented manifold, and let en be an oriented ndimensional bundle over Mn. Using the interpretation of the class e(e n ) E Hn(Mnj Z) as an obstruction, we calculate it as follows. Locally, the total space E of the bundle en is the product of the oriented manifold M n and another oriented manifold, the fiber of enj hence this space can be oriented. (Note that for odd n, the order of factors in the product is essentialj for transposed factors, the orientation is opposite.) Consider any section s: M n -- E transversally intersecting the zero section MO' C E in isolated points. We have (e(e n ), [Mn]) = ((MO', s(Mn»)). Indeed, for a representative of the cohomology class e(e) we can take the cocycle that is an obstruction to extending the section s from the (n -1 )-skeleton to the n-skeletonj to each n-simplex ~ n in the triangulation of the manifold M n this cocycle assigns the sum of the indices of the singular points of s in ~ n. On the other hand, this sum is equal to the intersection number of ~n with s(Mn) in MO'.

Example 44. If Mn is a closed oriented manifold, then (e(7Mn), [Mn])

=

XCMn). Note that under a change of orientation, both the Euler class e( TMn) and the fundamental class [Mn] change sign.

Example 45. The Euler characteristic of the sphere wnCTsn) = 0 for all n ~ 1.

sn is evenj therefore,

3. Applications of Simplicial Homology

140

For the sphere 8 2 , the tangent bundle is nontrivial but WI = 0 and O. Therefore, even the complete set of Stiefel Whitney classes does not distinguish between trivial and nontrivial bundles.

W2 -

Theorem 3.20. If dim (. - nand dim 11 - m, then w n +m ((. Wm

X

11) - w n ((.) x

(11)·

Proof. Take an n-cell u in the base of the bundle (.. Let us construct a (nonvanishing) section of (. over au and extend it inside u so that all zeros of the extension are nondegenerate. Let a(u) be the number of these zeros. Then the cohomology class w n ((.) is represented by the co cycle which assigns the element a(u) (mod 2) E Z2 to the cell u. Over each m-cell r of the base of 11, we similarly construct a section with b(r) nondegenerate zeros. The direct sum of these sections is a section of the bundle (. x 11 over the cell u x r. Let c( u X r) be the number of zeros of this section. Clearly, c(u x r) = a(u)b(r) because the direct sum of two sections vanishes precisely at the common zeros of these sections. Thus, to thG cell u x r the co cycle representing the class w n +m ((. X 11) assigns the product of the values of the co cycles representing w n ((.) and W m (11) for the cells u and r. According to the definition of cohomology cross product, this means that the representatives of the classes w n +m ((. X 11) and w n ((.) x W m (11) take equal values at the cell u x r. It remains to verify that if dim u' + dim r' = m + n but dim u' =/: n, then the representatives of both classes vanish at u' x r'. For the class w n ((.) x W m (11), this follows directly from the definition. Suppose that, say, dimu' < n. Then there exists a nowhere vanishing section over u'. The direct sum of this section and any section over r' is nowhere vanishmg. Therefore, the representative of the class w n +m ((. X 11) vanishes at the cell u' x r'. 0 The following theorem is a refinement of Theorem 3.20. Theorem 3.21. Ifdim(. - nand dim 11

=

m, then e((. x 11)

=

e((.)

X

e(11).

Proof. The main argument is the same as in the proof of Theorem 3.20. We must only trace the orientations. Namely, we need to prove that if vector fields v and W on the spaces V and W have nondegenerate singular points with indices (_I)k and (_I)l at the origin, then the vector field vffiw on the space V ffi W has a singular point of index (-1 )k+l at the origin. It suffices to prove this for the linear vector fields v(x) = Ax and w(y) = By. The vector field v ffi W is determined by the matrix ( ~ ~), which has determinant detAdetB. 0 Naturality Property of the Stiefel-Whitney Classes. The StiefelWhitney classes have the following naturality property: If (. is a bundle over

147

2. Characteristic Classes

Band f: B' -+ B is a continuous map, then wi(f*(~)) = r(w,(~)), i.e., the characteristic classes of the pullback r(O are obtained from those of the bundle ~ by means of the induced map H*(B; Z2) -+ H*(B'; Z2) of cohomology groups.

r:

The naturality property is fairly obvious. Indeed, let E' and E be the total spaces of the bundles r(~) and ~. By definition, we have

E' = {(b', v) Iv

E Ff(bl)}j

i.e., the fibers over b' E B' and f(b') E B are canonically isomorphic. Therefore, the map f induces a cochain map f#: en k+l(Bj{7l'n k(V(n,k»)})

-+

en k+l(B'j{7l'n k(V(n,k»}).

This cochain map takes the obstructing cocycle of the bundle the bundle (0.

r

The Whitney Formula. Let complex B. Then

(25)

Wk(~ ffi TJ) =

2:

~

~

to that of

and TJ be vector bundles over a simplicial

Wi(~) ........ Wj(TJ)

(the Whitney formula).

I+,-k Instead of the Whitney formula, we shall prove the equivalent relation (26)

Wk({ x TJ) =

2:

Wi({) x Wj(TJ),

i+j=k where { and TJ are bundles over different bases. Relation (25) is derived from (26) as follows. Suppose that { and TJ are bundles over the same base B, Pl : B x B -+ Band P2: B x B --+ B are the projections to the first and second fadors respectively, and d: B --+ B x B is the diagonal map. Then P1d = P2 d = idB and ~ x TJ = (Pi~) ffi (P2TJ). Therefore,

wk((Pi{) ffi (P2TJ»

= Wk({ x TJ) =

2:

Wi({) x Wj(TJ)·

i+j=k Applying the map d* to both parts in this equality and using naturality of the characteristic classes, we obtain d*(Wk«pi{) ffi (P2TJ») = wk(d*pi{) ffi (d*p;TJ» = Wk({ ffi TJ)· Moreover, d*(w,«() x w,(TJ» = Wi({) ........ Wj(TJ). Now, let us derive (26) from (25). First, note that ( x TJ = (px{) ffi (PyTJ), where px and py are the natural projections of X x Y to X and Y. Hence Wk«( x TJ) = Wk«Px() ffi (pyTJ» = ~i+j k Wi (P'X() ........ Wj(pyTJ) = Ei+j=kPXwd{) ........ pyWj(TJ) (the last equality follows from naturality of the characteristic classes). Finally, the corollary of Theorem 2.27 on p. 107 implies P'XWi«() ........ pyWj(TJ) = w,«() x Wj(TJ)·

3. Applications of Simplicial Homology

148

Vector bundles ~ and TJ over the same base are said to be stably equivalent if ~ ffi Em ~ TJ ffi EP , where Em and EP are the trivial bundles of dimensions m and p, respectively. We start by proving the following very special case of the Whitney formula. Lemma. Stably equivalent bundles have the same Stiefel Whitney characteristic classes.

Proof. We show that obstructions to extending the sections ofthe bundles ~ and ~ ffi Em can be chosen so as to coincide even at the level of cocycles, or, to be more precise, after the groups of co chains have been identified. The obstructing cocycle Ok(~) belongs to e n- k+1(B; {7l"n k(V(n, k))}), and the obstructing cocycle Ok(~ ffi Em), to en k+1(B; {7l"n k(V(n + m, k + m))}); here n = dim ~. The coefficient groups 7l"n k (V (n, k)) and 7l"n-k(V(n+m, k+m)) reduced modulo 2 are identified by the map V(n, k) -+ V(n + m, k + m) that takes any k vectors in ]Rn to the same k ;vectors Tn lRn ffi]Rm plus a fixed orthonormal basis of ]Rm. Take k linearly independent sections over the (n - k )-skeleton of the base of ~. They can be naturally supplemented to k + m sections of the bundle ~ ffi Em. After the identification of the groups, the coefficients of the obstructing co cycles Ok(~) and Ok(~ ffi Em) in these sections coincide. D We prove (26) by using this lemma and Theorem 3.20 on p. 146. Let ~ and TJ be bundles over simplicial complexes X and Y. Consider the inclusions ix: Xi .......... X and jy: yj .......... Y and the restrictions ~Ix' and TJly;. The first bundle has dim ~ - i linearly independent sections; therefore, it is stably equivalent to an i-dimensional bundle ~o. Similarly, the second bundle is stably equivalent to a j-dimensional bundle "10. According to Theorem 3.20, we have wl+j(~o x TJo) = WI(~O) x Wj(TJo). The naturality property of the characteristic classes implies

(ix x jy )*(Wi+j(~ x TJ)) = wi+J (~ X' x TJ YJ)

= wl+j(~o

X

"10)

= Wi(~O)

= wi(ix~) x Wj(jyTJ)

x Wj(TJo)

= (ix

X

jY)*(Wi(O x Wj(TJ))·

As is easily seen, if 0: + {J = i + j and 0: i= i, then (ix x jy)* (wa+/1(~ x TJ)) x w/1(T/O) = o. Indeed, if 0: > i, then wQ(~o) = 0 because dim~o = i < 0:, and if {J > j, then w/1(TJo) = o.

= wQ(~o)

Let Wk = Wk(~ X TJ) - LQ+/1=k wQ(~) X w/1(TJ). We have proved that Wk belongs to the kernel of the homomorphism (ix x jy)*: Hk(X x Y; Z2) -+ Hk(Xi x yj;Z2) for all dimensions i and j such that i + j = k. Let us prove that Wk = o. To simplify the formulas, we omit Z2 from the notation. According to the Kunneth theorem, we have Hk( y" x Y) ~

2. Characteristic Classes

149

Eeo+.B=k HO(X) ®H{1(y) and Hk(Xi x yi) ~ Eeo+{1=k HQ(Xi) ®HP(yi) = Hi(Xi)®Hi(yi) because HO(Xi) = 0 for 0> i and H.B(yi) = 0 for (3 > j. Therefore, the direct sum of the maps (ix x jy)* over all i and j such that i

+j

= k is a homomorphism

EB

HO(X) ® H.B(y)

a+.8=k

-+

EB

HQ(xa) ® H.B(y.B).

Q+.B=k

This homomorphism is injective because the maps HO(X) -+ HO!.(XO) and H{1(y) -+ H.8(y{1) are monomorphisms (they are dual to the epimorphisms Ho(xa) -+ Ha(X) and Hp(Y.8) -+ H.8(Y»' It is convenient to write the Whitney formula in terms ofthe total Stiefel Whitney class w({) = 1 + Wl({) + W2({) + ... + w n ({), where n = dim{. In this notation, the Whitney formula becomes w({ E9.,,) = w({) '-' w(.,,). 2.5. Applications of Stiefel-Whitney Classes. The total Stiefel Whitney class W = 1 + WI + W2 + .. , is invertible in the sense that there exists an element W = 1 + WI + W2 + .. , E H*(Bj Z2) for which ww = 1 (for brevity, instead of cups, we use the standard notation for the product of ring elements). Indeed, if a is a nilpotent element of the ring, then (1 + a)-I = 1 - a + a2 - a3 + . .. . Therefore,

(1

+ WI + W2 + ... r l = 1 - (Wl + W2 + ... ) + (WI + W2 + ... )2 = 1-WI + (W~ -W2)

+ (-W~ +2WIW2 -

(WI

W3)

+ W2 + ... )3 + ...

+ ....

All elements of the ring H*(Bj Z2) have additive order 2j hence

The classes Wi({) are called the dual Stiefel- Whitney classes. The dual Stiefel Whitney classes are important because of the following Whitney duality theorem. Theorem 3.22 (Whitney). Suppose that a manifold M n is immersed in ]RN and l/Mn is the normal bundle over Mn. Then Wi(l/Mn) = Wi(TMn). In particular, Wi (VMn ) does not depend on the immersion. Proof. The bundle l/Mn E9 TMn is isomorphic to the trivial bundle eN over Mnj the isomorphism takes each pair of vectors in the fibers of these bundles

150

3. Applications of Simplicial Homology

over a point x E M n to their sum (one of the vectors is normal to the tangent space and the other belongs to this space). Thus, W(lIMn)W(TMn) = W(lIMn ED TMn) = 1, i.e., W(lIMn) = W(TMn). D Corollary. w( Tsn)

= 1.

Proof. For the standard embedding sn

c

IR n+1, the normal bundle is triv-

W.

D

Problem 101. Given a Riemannian manifold M, prove that the normal bundle to d(M) in M x M endowed with the structure of a Riemannian manifold is isomorphic to the tangent bundle of the manifold Mj here d(x) = (x, x) is the diagonal map.

For a manifold M n , the class w( TMn) is called the Stiefel Whitney class of the manifold M n and is denoted by w(Mn). Problem 102. Calculate the classes WI and W2 for a sphere with 9 handles. Problem 103. Calculate the classes WI and W2 for the sphere with m Mobius bands attached and prove that w~ = W2. Problem 104. Prove that if a manifold M Wk for all k (here Wk = wk(Mn)).

wf

n

is immersed in IRn +1, then

The Stiefel Whitney classes of all spheres are trivial. A more interesting example is the Stiefpl Whitney classes of projective spaces IRpn. They are calculated using the following assertion. Theorem 3.23. If c 1 is the trivial one-dimensional bundle over IRpn, then 1" ,'Yn 1= =1 71Rpn = '17c '17 • • • '17'Yn·

---------n+l

Proof. The total spaces of the bundles involved in the statement of the theorem can be described as follows: T(~'pn) consists of the pairs

± (x,v),x E sn,v E IR n+1,xl.Vj

c 1 consists of the pairs (±x, A), A E IRj 'Y~ consists of the pairs (±x, A(X)), A( -x)

h~)n+1 consists of the pairs (±x, vex)), v( -x)

= -A(X)j = -vex).

The map 71Rpn ED c 1 _ h~)n+l takes pairs ±(x, v) and (±x, A) to the pair (±x, v + AX). The inverse map acts as follows. Let us represent a vector v(x) in the form vex) = Vr + Vv, where Vv = AX and vrl.x. The pair (±x, vex)) is mapped to the pairs ±(x, v r ) and ±(x, cVv), where c = 1 ifthe directions of the vectors x and Vv coincide, and c = -1 if they are oppo!!ite. D

2. Characteristic Classes

Corollary. w(JRpn)

151

= (1 + O')n+1 = 1 +

Exercise. Prove that

WI

(JRpn)

(ntl)O' + .,. + (n~l)on.

= 0 if and only if n

Theorem 3.24. The equality w(lRRn)

=1

is odd.

holds if and only if n

= 2k -

1

for some k.

Proof. Clearly, (a+b)2 == a 2 +b2 (mod 2). Therefore, (1+O')2k = 1+0'2". Suppose that n = 2k - 1. Then w(JRRn) = (1 + 0')2'" = 1 + 0'2'" = 1 because 2k > 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
The Stiefel-Whitney Numbers. Let Mn be a closed manifold (possibly disconnected), and let [Mn] E Hn(M; Z2) be its fundamental class. Then, for any cohomology class a E Hn(M; Z2), the number (a, [Mn)) E Z2 is defined. The number

(WI (TMntl '" Wn(TMn)rn, [Mn)) E Z2 obtained for the class a = W1(TMntl"'Wn(TMntn, where rI, ... ,rn are nonnegative integers for which T1 + 2r2 + ... + nrn = n, is called the Stiefel Whitney number corresponding to the cohomology class w? ... w~n. This number is denoted by w?··· w~n[Mn]. Problem 106. Prove that if n is even, then wn[~pn] =f; 0 and wr~pn] =f; O. Problem 107. Prove that if n is odd, then all the Stiefel Whitney numbers of the manifold ~pn are zero. Theorem 3.28 (Pontryagin). All the Stiefel Whitney numbers of the manifold Mn = oWn+!, where Wn+l is a compact manifold, are zero. Proof. We consider homology and cohomology with coefficients in Z2. Let [wn+l] E Hn+l (Wn+l, Mn) be the fundamental class of Wn+l. In the proof of the Lefschetz duality theorem, we showed that the homomorphism 0*: Hn+! (wn+l , Mn) __ Hn (Mn) takes [wn+!] to the fundamental class [Mn] E Hn(Mn). Hence, if a E Hn(Mn), then (a, [Mn)) = (o,o*[Wn+l)) = (c5*a, [wn+l)) , where 8*: Hn(Mn) __ Hn+l(wn+1,Mn) is t-he homomorphism dual to 0*.

2. Characteristic Classes

153

The collar theorem (for smooth manifolds) readily implies rwn+1lMn ~ rMn ffi c I , where £1 is the trivial one-dimensional bundle over Mn. Indeed, we are interested only in the part of Wn+l near aWn+l = Mn. Therefore, we can assume that Wn+l = M n x I; in this case, wj(Mn) = w,(rwn+1IMn) for all j. We have w;(rwn+1IMn) = i·wj(W n+I ), where i*: Hj(wn+l) -+ Hj(Mn) is the homomorphism induced by the inclusion i: M n '--+ Wn+l. Hence w;:l(Mn) ... w~n(Mn) = i·a, where a = w?(wn+l) ... w~n(wn+l). The exact cohomology sequence of a pair

Hn(wn+l) ~ Hn(Mn)

!:... Hn+l(wn+l,M n )

implies 6·i· = O. Consequently, 6*(wp(Mn) .. ·w~n(Mn)) = 6·i·a = 0, and therefore

. I.e., WIrl .,

- a ,wnrn [Mn] .

o

2.6. The Universal Bundle. There exists another important approach to constructing characteristic classes, which is based on the fact that any vector bundle of a given dimension is a pullback of a fixed bundle over the Grassmann manifold. Under this approach, the Stiefel Whitney characteristic classes are obtained from a certain set of multiplicative generators of the cohomology ring of the Grassmann manifold by means of the induced map in cohomology. We recall that the points of the Grassmann manifold G(n, k) are kdimensional subspaces of ]Rn. The natural embedding ]Rn '--+ ]Rn+l taking every point with coordinates (Xl, ... ,X n ) to the point with coordinates (X}, ••• , X n , 0), induces an embedding G(n, k) '--+ G(n + 1, k). Thus, we can consider the infinite Grassmann manifold G(oo, k) = Un>k G(n, k), which is endowed with the direct limit topology; a set U c G(oo,lc) is open in this topology if and only if all the sets Un G(n, k), n ;?: k, are open. In terms of the infinite Grassmann manifold, the main results of this section (Theorems 3.29, 3.30, and 3.31) can be formulated as follows: There is a one-to-one correspondence between the k-dimensional vector bundles over a compact Hausdorff space B considered up to isomorphism and the homotopy classes of maps [B, G(oo, k)]. This assertion can also be stated and proved in terms of the ordinary Grassmann manifolds; this is what we do below. There is a canonical k-dimensional vector bundle 'Y~ over the Grassmann manifold G(n, k). The space E(-y~) of this bundle consists of the pairs (a k-dimensional subspace Ilk C

]Rn,

a vector v E Ilk);

3. Applications of Simplicial Homology

154

the projection p: E(-y~) -+ G(n, k) assigns to every such pair the subspace nk viewed as a point of the manifold G(n, k). The topology of the space E(-y~) is induced from G(n, k) x jRn. The fiber over a point rrk E G(n, k) is naturally identified with the k-space rrk j thus, is has a linear space structure. It remains to verify that ,~ satisfies the local triviality condition. We use the coordinate neighborhoods Uj constructed in the proof that G(n, k) is a smooth manifold (see Part I, Chapter 5, Section 1.5). Here I denotes a set of numbers {il, ... , id satisfying the inequalities 1 S. il < ... < ik S. nj the set Uj C G(n, k) consists of k-dimensional subspaces of jRn transversal 6 to the orthogonal complement of the space jRj generated by the vectors eil' ... , e tk . The orthogonal projection of any k-subspace rrk E Uj to jRj is an isomorphism of linear spaces. There is a homeomorphism hi: Uj x jRj -+ p-l(Uj)j it maps each pair (rrk,v) E Uj x jRj to the pair (ilk, Vi) E E(-y~), where Vi E ilk is the vector with orthogonal projection v E jR1.

Theorem 3.29. Any k-dimensional vector bundle ~ over a compact Hausdorff base B is a pullback of the canonical bundle ,~ for some n. Proof. Choose open sets U}, ... , Um so that they cover B and the bundle ~ is trivial over each of them. Any compact Hausdorff space is paracompact and, therefore, normal. Hence there exists a partition of unity iI,·· ., fm subordinate to the cover U}, ... , Urn. We use the homeomorphism h t : Ui X jRk -+ p-l(Ui) to define a map 'Pi: p-l (Ut ) -+ jRk as the composition h

1

p-l (Ut ) ~ Ui x jRk and use the latter to define a map gt: E(~) g,(x)

~ { ~.(x)'I"(X)

---t

-+

jRk

jRk as

if x E Ui, if x ¢ Ui .

Finally, we set g(x) = (gl(X), ... ,gm(x». Each fiber of the map g: E(~) -+ jRmk is isomorphic to some k-dimensional subspace rrk C jRmk. Thus, we have obtained a map g: B -+ G(mk, k). It is easy to verify that g*'!.k ~~. Indeed, the total space of the bundle g*'!.k consists of all triples (ilk, v E ilk, b), where g(b) = rrk. Therefore, the fibers of the bundles ~ and g*'!.k over each point b E B are canonically isomorphic.

0

6Two linear subspaces of dimensions k and n - k in IR n are called transve al If their inter_ section consists of the zero vector only.

2. Characteristic Classes

155

For a compact Hausdorff space B, two homotopic maps fo, It: B ~ G(n, k) induce isomorphic bundles, i.e., fOI~ ~ fil~' This assertion is valid even in the following more general situation. Theorem 3.30. Suppose that B is a compact Hausdorff space, { is a bundle over X, and fo, It : B ~ X are homotopic maps. Then fo { ~ fi {. Proof. Let F: B x I ~ X be a homotopy between fo and fl' and let It be the restriction of F to B x {t}. Consider two bundles F* { and p* ft { over B x I, where p: B x I ~ B is the natural projection. The restrictions of these bundles to B x it} are isomorphic. If two bundles 6 and {2 are isomorphic, then the isomorphism between them is a section of the bundle Hom({1,6), for which all maps of fibers are isomorphisms. Lemma. Let Y be a closed subset of a compact Hausdorff space Z, and let { be a vector bundle over Z. Then any section of the bundle {Iy can be extended to a section of the entire bundle {. Proof. Let us cover Z by finitely many coordinate neighborhoods. Over each coordinate neighborhood Ui, any section of {Iy can be regarded as a map fi: Ui n Y ~ ]Rn. The set Ui n Y is closed in the topology of Ui; therefore, fi can be extended to a map F i : Ui --. lRn. Take a partition of unity {At} subordinate to the cover {Ui}. We set

Si(X) = {Ai(X)Fi(X)

o

The sum

E Si (x)

~f x E Ui, If x ¢ Ui.

is a section of { extending the given section over Y.

0

Take an isomorphism between the restrictions of the bundles F* € and p* ft € to B x {t} and consider an extension of this isomorphism to a section of the bundle Hom(F*{,p* ft{). In the space Hom(lRn , ]Rn), isomorphisms form an open subset. Since B is compact, the restrictions of the bundles P* € and p* ft{ to B x {V(t)}, where V(t) is an open subset ofthe interval I containing t, are isomorphic. Consequently, the bundles p* J:€ are isomorphic for all T E V(t), and fo€ ~ fi{ because the interval I is compact (and connected). 0 The embedding lR n <--. lR n+a induces an embedding ia: G(n, k) --. G(n + a, k). Moreover, i;(T~+a) = I~' Thus, any pullback of I~ is also a pullback of I~+a for all a E N. Theorem 3.31. Let B be a compact Hausdorff space, and let fo and II be maps from B to G(n, k). Then the bundles fo'~ and fi,~ are isomorphic if and only if the maps iafo and iaII are homotopic for some a.

156

3. Applications of Simplicial Homology

Proof. We have just proved that homotopic maps induce isomorphic bundles. Thus, we must only show that if bundles fo'Y~ and fi'Y~ are isomorphic, then the maps iafo and iah are homotopic for some a.

We identify the isomorphic bundles fo'Y~ and fi'Y~ and assume that the commutative diagrams E ~ E('Y~) C G(n, k) x;Rn

lp

lp'

B~G(n,k) are given for i = 0,1. Instead of the map 'PI' we consider its composition -+ lRn with the natural projection G(n, k) x]Rn -+ lRn. Note that h(b) is a subspace of .,pi(P l(b)) E G(n, k), i.e., the map.,pi uniquely determines Ii-

.,pi: E

Consider the embeddings in,O,in,}: lRn -+ lR n E9 lRn , where in,o is an isomorphism between lRn and the first summand and in,l is an is(}fil9rphism between lRn and the second summand. They induce embeddings in,o, in,l: G(n, k) -+ G(2n, k), where in,o = in and in,} ~ in (for n = 1, the homotopy between in,o and in,l is constructed by rotating the coordinate axes; for n > 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

2. Characteristic Classes

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),

3. Applications of Simplicial Homology

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

2. Characteristic Classes

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).

3. Applications of Simplicial Homology

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)).

2. Characteristic Classes

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

3. Applications of Simplicial Homology

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)).

2. Characteristic Classes

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

3. Applications of Simplicial Homology

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».

2. Characteristic Classes

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

3. Applications of Simplicial Homology

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.

2. Characteristic Classes

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

3. Applications of Simplicial Homology

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 .

2. Characteristic Classes

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

3. Applications of Simplicial Homology

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

2. Characteristic Classes

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
3. Group Actions

175

joined by an edge Al (or coincide). Clearly, v E Al n gAl, whence gv = v. Suppose that (ii) holds. Let v E An gAo Then the vertices v and gv belong to the same simplex A, and therefore v = gv. 0

Theorem 3.48. The second barycentric subdivision of any simplicial Gcomplex is a regular G-complex. Proof. First, we prove that the barycentric subdivision K' of a simplicial G-complex K has property (A). Each vertex v of the complex K' determines a simplex A(v) in K for which this point is interior. Moreover, vertices VI and V2 belong to the same simplex of K' if and only if one of the simplices A(VI) and A(V2) is a face of the other. We have A(gv) - gA(v)j in particular, the simplices A(v) and A(gv) have the same dimension. Therefore, if vertices v and gv belong to the same simplex of K', then one of the simplices A(v) and A(gv) is a face of the other, and the dimensions of these simplices are equal. Therefore, A(v) = A(gv), and v = gv. Now we prove that if a simplicial G-complex K has property (A), then its barycentric subdivision K' has property (R). The proof is by induction on n. Suppose that [vo, ... , vnJ and [hovo, ... , hnvnJ are simplices of K' and all (n - 1)-simplices of K', where n ~ 1, have property (R) (zero-dimensional simplices always have property (R)). Renumbering the vertices if necessary, we can assume that A(vo) c A(VI) c ... C A(vn ), i.e., vo is a vertex of a simplex of K, VI is the midpoint of an edge, and so on. By the induction hypothesis, there exists an element hE H such that hVi = hiVi for i = 0,1, ... , n - 1. Let us show that hnVi = hiVi for all i. This equality is obvious for i = nj it remains to verify it for i = 0,1, ... ,n - 1. Under the action of h- l , the simplex [hovo, ... ,hnvnJ is mapped to [vo, ... , Vn-l, h-1hnvnJ. Moreover,

A(vo) C A(VI) c ... C A(Vn-l) C A(h-1hnvn ) = h-1hnA(vn ). Therefore, A(Vn-l) C A(vn ) n h-1hnA(vn ). Property (A) implies that all points of the (n - 1)-simplex A(Vn-l) are fixed under the action of h-1hn . Hence h-1hnVi = Vi for i = 0,1, ... , n - 1. Therefore, hnVi = hVi = hiVi, as 0 required. Thus, after refining the triangulation of the complex K, we can identify the sets IKI/G and IK/G. The natural topologies on the identified sets coincide, i.e., the map IKI/G -+ IK/GI is a homeomorphism. Indeed, since the simplicial map K -+ K/G is surjective, it follows that a subset of IK/GI is open if and only if its preimage is open in IKI. On the other hand, a subset of the orbit space IKI/G is open if and only if its preimage is open

110

J. AppllcatlOns ot :::iimpllClal 110molOgy

in IKI. It remains to note that the composition coincides with the map IKI ~ IKIGI.

IKI

~

IKI/G ~ IKIGI

3.2. Equivariant Simplicial Approximation. We say that a simplicial G-complex L has property (I) if for an arbitrary element 9 E G and an arbitrary simplex ~ in L, the equality g(~) = ~ implies that the restriction of the action of 9 to ~ is the identity map. The barycentric subdivision K' of any simplicial G-complex K has property (I). This is a general property of simplicial maps (see Part I, Theorem 3.14).

Theorem 3.49. Suppose that K and L are simplicial G-complexes and L has property (I). Then for any continuous map f: IKI ~ ILl, there exists an equivariant simplicial approximation cp: K(n) ~ L. Proof. Let 'I/J: K(n) ~ L be a simplicial approximation of the map f .. We define an equivariant simplicial map cp: K(n) ~ L as follows. For each vertex v E K(n), consider its orbit {g(v) Ig E G}. The orbits of two vertices either coincide or are disjoint. In each orbit, we choose some vertex v and put cp(v) = 'I/J(v); for all other vertices from the same orbit, we set cp(g(v)) = g(cp(v)) = g('I/J(v)). First, we must verify that the map cp is well defined, i.e., gl('I/J(V)) = g2('I/J(V)) whenever gl(V) = g2(V), or, equivalently, g('I/J(v)) = 'I/J(v) whenever g(v) = v; here 9 = g1 1 g2. To show this, we use property (I). Let ~ be the bimplex of L that contains f(v) as an interior point. The point g(J(v)) = f(g(v)) = f(v) is also an interior point of the simplex ~; therefore, g(~) =~. According to property (I), we have gl~ = id~. But 'I/J is a simplicial approximation of f; hence 'I/J(v) E ~, which means that

g('I/J(v)) = 'I/J(v). It remains to show that cp is a simplicial approximation of the map f, i.e., f(st v) cst cp(v) for any vertex v E K(n). If v is the distinguished point of its orbit, then cp(v) = 'I/J(v) and f(st v) cst cp(v) because'I/J is a simplicial approximation of f. For any element 9 E G, the simplicial map 9 is oneto-one; hence stg(v) = g(stv), where v is a vertex of K(n) or L. Thus, if f(stv) c stcp(v), then f(stg(v)) = f(g(stv)) = g(J(stv)) C g(stcp(v)) = st(gcp(v)) = stcp(g(v)). 0

Remark. The map cp constructed in the proof of Theorem 3.49 is equivariantly homotopic to f; i.e., these maps are homotopic in the class of equivariant maps. The homotopy is constructed in a standard way (for any x, the points cp(x) and f(x) belong to the same simplex and can therefore be joined by a line segment). 3.3. Fixed Points and Fixed Simplices. For a G-space X, lpt XC be the set of fixed points of the action of G, i.e., XC = {x E X I gx = x Vg E G}.

3. Group Actions

17;

For a simplicial G-complex K, by KG we denote the sub complex of K formed by the simplices that are pointwise fixed under the action of G. Clearly, IKGI C IKIG. A point x belongs to the set IKIG \ IKGI if and only if it is interior for a simplex .6. = [vo, . .. ,vnl and gx = x for all 9 E G but gVi =I- VI for some i and some 9 E G. If an interior point of a simplex .6. is fixed under the action of an element g, then g.6. = .6.. In particular, if a simplicial G-complex K has property (A), then IKGI = IKIG. Recall that the barycentric subdivision of any simplicial G-complex has property (A). Therefore, after refining the subdivision, we can assume that IKGI = IKIG.

Theorem 3.50. Let K be a simplicial Zn-complex with property (A)7, and let 9 be a generator of the group Zn. Then X(KG) = A(g), where A(g) is the Lefschetz number of the map g: IKI --+ IKI. Proof (see [148]). We represent Ck(K; lR) in the form Vk EB Wk, where Vk = Ck(K G; lR) and W k is the space spanned by the simplices not belonging to KG (such simplices form a basis in this space). By definition, X(KG) = ~) _1)k dim Vk and A(g) = E( _1)k tr gk, where gk: Ck(K; JR) --+ Ck(K; lR) is the map induced by the action of 9 on k-simplices. Property (A) implies that if g.6. = .6., then .6. E KG. Therefore, the first dim Vk diagonal elements in the matrix of gk in the natural basis of the space Vk EB Wk are equal to 1 and the remaining diagonal elements are zeros. Hence tr gk = dim Vk. 0 The action of a group G on the space X is said to be effective if for each nonidentity element 9 E G, there exists a point x E X such that gx -=I- x.

Theorem 3.51 (Minkowski). Suppose that M is a triangulated sphere with n handles (n ~ 2) on which a finite group G acts effectively (and simplicially) and the G-complex M has property (A). Then for any nonidentity element 9 E G, the map g*: H 1 (M;lR) --+ Hl(M;lR) is not the identity. Proof (see [148]). We show that the set of points of the manifold M that are fixed under the action of the subgroup Zm C G generated by 9 consists of several isolated points and disjoint circles. First, an effective action on M cannot have fixed 2-simplices because any 2-simplex sharing an edge with a fixed simplex must be fixed as well (and a free vertex can be mapped only to itself). Second, if v is a fixed vertex, then it is either isolated in the set of fixed points or incident with precisely two fixed edges. Indeed, let VI, ... ,Vm be the vertices neighboring V and numbered in some order (say, clockwise). Suppose that gVl = Vk. Then gV2 = Vk+l or gV2 = Vk-l' We have gVi = Vk-l+i in the former case and gVi = Vk+l-i in the latter. In the former case, the only fixed point inside the polygon VI ... Vm is V; in 7Property (A) was defined on p. 174. Recall that the barycentric subdivision of any simplicial G-complex has property (A).

:1. Appllcatwns ot :;implicial Homology

178

the latter case, there are two fixed line segments joining v either to some vertex Vj or to the midpoint of some edge VjVj+l' Property (A) eliminates the second possibility. Suppose that g. is the identity map. Then A(g) = X(M) = 2 - 2n < 0 because n > 1. On the other hand, according to Theorem 3.50, we have A(g) = X(M z",) and X(MZm) ~ 0 because MZ'" consists of several isolated points and pairwise disjoint circles. D 3.4. Transfer Homomorphisms. Suppose that G is a finite group, K is a regular simplicial G-complex, and C.(K) is a simplicial chain complex. We define an action of the group G on C.(K) by setting g[vo, ... , vnl = [gvo, ... ,gvnl. Consider the homomorphism u: C.(K) --+ C.(K) that takes every chain e to LgEG ge. If a subcomplex L C K is invariant with respect to the action of G , then we can transfer this action to C.(K, L) = C.(K)/C.(L) and define a sim11ar homomorphism u: C.(K, L) --+ C.(K, L).

Let p: K --+ K/G be the canonical projection (which is a simplicial map). The map p induces the chain map P#: C.(K, L) --+ C.(K/G, L/G). Theorem 3.52. If a subeomplex L C K is G-invariant, then Ker u Kerp#.

=

Proof. First, we show that the regularity of the complex K implies property (A). Suppose that vertices v and gv belong to the same simplex. Then [v, vl and [v, gvl are simplices in K. By the regularity condition, v = g'v and gv = g'v for some g' E G. Thus, v = g'v = gv, as required. Consider a simplex f::..k in K/G. Let f::..~, ... , f::..~ be the simplices over f::..k in K. Since K is regular, each of these simplices admits the orientation compatible with that of f::..k, i.e., such that the projection Pi: f::..f --+ f::..k is orientation-preserving. Indeed, the regularity condition implies property (A). Therefore, if g[vo, .. . , vnl = [vo, ... , v n ], then all points of the simplex [vo, ... , vnl are fixed under the action of the element gj in particular, 9 preserves the orientation of the simplex [vo, ... , vnl. Let e

and

= L~=l aif::..~. Then

3. Group Actions

179

the second equality holds because

To calculate Ker (1' and Ker P#' we decompose every chain into summands of the form L~ 1 a,~f; a chain belongs to the kernel of (1' or P# if and only if so does each of the summands. The conditions for such a summand to belong to the kernels of (1' and P# almost coincide: they are ~ (L~ 1 ai) = o and L~=l ai = O. For integer coefficients ai, these conditions coincide

D

~~

Remark. Theorem 3.52 is valid not only for Z but also for an arbitrary coefficient group A provided that A has no element a for which Gla - o. Since P#: C.(K, L) --+ C.(K/G, L/G) is an epimorphism, it follows that C .. (K/G,L/G) ~ C.(K,L)/Kerp# = C.(K,L)/Ker(1' ~ (1'C.(K,L); thus, there is a canonical chain isomorphism J.L: C.(K/G, L/G) --+ (1'C.(K, L). The composition of this homomorphism and the inclusion (1'C.(K, L) c C .. (K, L) induces a homomorphism J.L.: H.(K/G, L/G) --+ H.(K, L), which is called the transfer homomorphism. Clearly, the map p induces a homomorphism P.. : H .. (K, L) --+ H.(K/G, L/G). Consider the compositions P.J.L .. and J.L.P ... It follows directly from the definition that J.LP#c = (1'C; therefore, J.L.P .. = LgEG g., where g.: H.(K, L) --+ H.(K, L) is the map induced by the action of g. Moreover, any element c E C.(K/G, L/G) can be represented in the form c = P#c, where c E C.(K, L). Hence J.LC = J.LP#c = (1'C, which means that P#J.Lc = P#(1'C = IGlp#c = IGIC. Thus, p .. J.L. is multiplication by IGI·

Any chain (1'C = L9EG gc is G-invariant. Hence, 1m J.L C C. (K, L)G and ImJ.L. C H.(K, L)G. Clearly, the restriction of the map J.L ..P. to H.(K, L)G is the sum of IGI identity maps; thus, this restriction is multiplication by IGI as well. Using the remark after Theorem 3.52, we obtain the following result.

Theorem 3.53. If F is a field whose characteristic is either coprime to IGI or equal to zero, then the maps p.: H.(K, L; F)G

--+

H.(K/G, L/G; F)

and J.L .. : H .. (K/G, L/G; F)

--+

H*(K, L; F)G

are isomorphisms. Also, the kernel of the homomorphism p .. : H.(K, L; F) --+ H*(K/G, L/G; F) coincides with that of (1'•.

180

3. Applications of Simplicial Homology

We now show that in Theorem 3.53, the assumption that the characteristic of the coefficient field F is coprime to IGI cannot be removed. Consider the lens space L(p, q). Let G = Zp. For K = 8 3 , we have L(p, q) = K/G, and K is a regular G-complex. It is easy to derive from the universal coefficient theorem that if p' is a prime, then, for k = 1 and 2, we have if p is divisible by p', if p is not divisible by p'. Thus, if p = GI is divisible by p', then the groups Hk(L(P,q)jZp') and Hk(8 3 j Zp')G - 0 are not isomorphic. Theorem 3.53 can be carried over to cohomology groups in a natural way. 3.5. Smith Theory. In this section, we consider G = Zp for a pEime p. (Note that, for applications, even the simplest case p = 2 is interesting.) Moreover, we consider homology and cohomology only with coefficients in Zp. Let K be a regular G-complex, and let L be an invariant subcomplex of K. The chains from C.(K, Lj Zp) can be multiplied by elements of the group ring ZpG, which consists of the formal sums L aigi, where ai E Zp and g, E G = Zp. Let 9 be a generator (that is, any nonzero element) of the group G. We use mUltiplicative notation for the group G, that is, assume that G consists of the elements I, g, ... , gP-l. Consider the elements U = 1 + 9 + ... + gP 1 and T = 1 - 9 in the group ring ZpG. The equality gP = 1 implies UT = o. Let us show that T P- 1 = u, i.e., (_I)k(P k 1) == 1 (mod p). Indeed, (1 p)(2 - p) ... (k p) = 1 ·2··· k (mod p)j therefore, 1·2···(p-k 1)·(-I)k(p-k) ... (p-l) =1 1 . 2· .. (p - k - 1) . 1 ·2· .. k -

(mod p).

For each element of the form p = Tk, we set p = TP-k. In particular, T = U and U - T. For every p = Tk, where 1 ~ k ~ p - 1, consider the chain subcomplex pC.(K, Lj Zp) C C.(K, Lj Zp). In what follows, we omit the coefficient group Zp from the notation. Theorem 3.54. For any p chain complexes

=

Tk, where 1 ~ k ~ p - 1, the sequence of

0---+ pC.(K, L) ffi C.(K G , L G) ~ C.(K, L) px, pC.(K, L)

---+

0

is exact. (Here i is the sum of 'Ilatural embeddings and px i~ multiplication by p.)

3. Group Actions

181

Proof. First, note that the embedding C.(KG, L G) defined because LG = L n KG.

--+

C.(K, L) is well

We are interested only in simplices in K \ L. Suppose that 6 c KG. Then r 6 = 0; therefore, p6 = 0 and p6 = O. This means, in particular, that the map i takes pC.(K, L) EB 0 and 0 EB C.(KG, L G) to subspaces with trivial intersection. Moreover, px (multiplication by p) maps the subspace C.(KG, L G) c C.(K, L) to zero. Hence it is sufficient to verify exactness only for chains of the form 2: aigi6, where 6 ct KG. For fixed 6, every chain of this form is uniquely determined by an element 2: a,gi E A, where A = ZpG is the group ring. It remains to prove the exactness of the sequence i

px

O--pA --A--pA--O. Clearly, i is a monomorphism, and px is an epimorphism. Thus, we have to verify that dim(pA) + dim(pA) = dimA = p (the dimension is over the field Zp). Recall that p = rk, where 1 ~ k ~ p - 1, and p = r P - k . We show that dime rk A) = p - k. The kernel of r consists of the elements 2: aig i such that 2: aig i = 2: aigi+l. This means that ai = ai+l = ... , i.e., the kernel is generated by the element u = 1 + 9 + ... + gP-l. Therefore, dim(rA) = p - 1. Moreover, u = r P - 1 = rkrP - k- 1 , whence Kerr = (u) erA. Thus, dim(r k+1A) = dimr(rkA) = dim(rkA) -1. 0 The homology groups Hf(K,LjZp) = H.(pC.(K,Lj'Lp» are known as the Smith homology groups. The short exact sequence from Theorem 3.54 induces the exact sequence of homology groups

... __ Hg(K,L) EBHq(KG,LG) ~ Hq(K,L) ~ Hg(K,L)

~ H:_ 1 (K,L) EBHq_ 1 (K G,LG) - - .... It is called the Smith exact sequence. As a rule, the Smith exact sequence is used for p = u and p = r.

For the Smith homology groups, yet another set of exact sequences can be constructed (for k = 1, ... ,p - 2). The argument at the end of the proof of Theorem 3.54 shows that the sequence (27) is exact. It follows that u = r P - 1 and dim(rkA) = p - k. Note that, for k = p-l, the sequence (27) degenerates: 0 --+ uA --+ uA --+ 0 --+ O. From the short exact sequence (27) we obtain the exact sequence of homology groups

(28)

... - - H;(KG,L G) ~ Ht(K,L)

~ Ht+

1

(K, L)

~ H;-l (KG, L G) - - ....

oJ. J1.jJjJllGCLLlUIlS

.LO~

or

;:)lmpllClCLI

nUIllUlU~y

Suppose that K is a finite simplicial complex and L is a sub complex of K. We define the Euler characteristic of the pair (K, L) by X(K, L) = E(-I)idimHi(K,LjF), where F is a field. Like the ordinary Euler characteristic, the number X(K, L) does not depend on the choice of F. Theorem 3.55. (a) For an integer n ~ 0,

dimH~(K,L)

+ LdimHn(KG,LG) ~

LdimHn(K,L)j

(b) x.(KG, L G) - X(K, L) (mod p). Proof. (a) Suppose that a, - dimH,(K G, L G), b, - dimH,(K, L), dimH:(K,L), and Ci = dimHf(K,L). The Smith exact sequence

Ci -

H:+ 1 (K, L) ----. H:(K, L) EEl Hi(K G, L G) ----. Hz(K, L) implies Ci + ai ~ c;+! + b,. Similarly, writing the Smith sequence for p, we obtain Ci + ai ~ Ci+! + b,. Consider the sequence of inequalities

cn + an Cn+! Cn+2

~

Con+! + bn,

+ an+! ~ Cn+2 + bn+!, + an+2 ~ Cn+3 + bn+2,

The number of inequalities is finite (that is, the inequalities with large numbers have the form 0 ~ 0) because the simplicial complex K is assumed to be finite. Summing up these inequalities, we obtain the desired relation. (b) Any exact sequence of spaces 0 - Va - VI - ... - Vn - 0 can be represented as the direct sum of the short exact sequences 0 - . .. - 0 ~~l - ~Q - ~~l - 0 - ... - O. For each of these short exact sequences, we have E( -I)' dim ~Q = OJ therefore, E( _I)i dim Vi = o. Thus, if spaces AI! Bi, and C i are such that an exact sequence 0 - Ao - Bo - Co Al - BI - ... - Cn - 0 holds, then E(-I)idimAi + E(-I)idimCi = E(-I)'dimB,. Hence the Smith exact sequence implies

(29)

x(p)

+ X(K G, L G) + X(p)

=

X(K, L),

=

1, ... ,p - 2.

where X(p) = E( _I)i dim Hf(K, L). Similarly, the exact sequence (28) gives (30)

x(O")

+ x(rk+!) = x(rk),

k

Summing the equalities (30) over k = 1, ... ,p - 2, we obtf'·n (p - 2)X(0") x(rP - 1 ) = X(T). But 0" = rP-1j therefore, x(r) = (p - l)A(u).

+

3. Group Actions

Setting p

=r

183

in (29) (so that p - r P

X(O') Thus, X(K, L) = PX(O')

0'), we obtain

1

+ x(r) + X(K G, L G)

+ X(KG, LG)

X(K, L). X(KG, L G) (mod p).

o

Theorem 3.55 gives strong constraints on the structure of the sets of fixed points for actions of the group Zp on spheres and disks. Its proof uses only homological information; so, the same argument applies to homology spheres and disks. A simplicial complex K is called a Zp-homology r-sphere if H,(K; Zp) ~ H,(ST; Zp) for all i. A paIr (K, L) is called a Zp-homology r-disk if H,(K,L;Zp) rv H,(DT,ST l;Zp) for all i. Theorem 3.56 (Smith [122]). Suppose that G

Zp (Jor prime p), K is a regular G-complex, and L is an invariant subcomplex of K.

(a) If K is a Zp-homology n-sphere, then KG is a Zp-homology r-sphere, where -1 < r S n. (It is assumed that the sphere of dimension -1 is the empty set.) If p > 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.

3. Applications of Simplicial Homology

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
O(cp(Wk ® a»

=
The chain o( cp( Wk ® a» is a cycle, and it is contained in the acyclic chain complex G.(a x a). Therefore, G.(a x a) = .c(Wk ® oa) contains a chain c for which O(
+ (-1)kTDk_l = oDk + (-1)k+1D ko.

Proof. Since cP is a chain map, we have

ODk(C)

= O(cp(Wk ® e» = cp(O(Wk ® c» = cp(OWk ® C + (_1)kwk ® oe).

Taking into account the fact that Dkoe = cp(Wk ® oe), we obtain

oDk(e)

+ (_1)k+1 Dkoe = =

as required.

cp(OWk ® c)

= cp«Wk-l + (-1)k Twk _ 1 ) ® c)

Dk-le+ (-l)kTDk_l e,

o

3. Applications of Simplicial Homology

186

Corollary. Do-TDo = 8DI +D I 8. Thus, Dl is a chain homotopy between Do and TDo·

For k ~ 0, we define the '""'k-product cP '""'k cq of cP E CP(K) and q c E cq (K) to be the cochain that assigns

to every chain e E C p+q k(K). It is convenient to set rJ' '""' 1 cq = O. Note that '""'k-pwduct depends on the choice of the map cpo Thus, we have to verify that the operation Sqk' which is defined in terms of '""'k, does not depend on cpo Theorem 3.58 (the coboundary formula). The following equality holds:

5(rJ' '""'k eq) - (_1)k5cP '""'k eq + (-l)P+kcP '""'k 5eq (-l)kcP '""'k-l eq - (-l)pqc q '""'k-l cPo Proof. Let c E Cp +q k+l(K). Then

(5(cP '""'k eq), e) By definition, 8(Wk ® e)

= (cP '""'k eq, 8e) = (cP ® cq, CP(Wk ® c)).

= 8Wk ® c + (_l)kwk ® 8e.

Therefore,

(5(cP '""'k eq), e)

= (-l)k(cP ® eq, cp(8(Wk ® e))) - (_l)k(cP ® cq, CP(8Wk ® c)) = (-l)k(cP

cq, 8(cp(Wk ® c))) - (_l)k(cP ® cq, CP(Wk

1

® e))

- (_1)2k cP

cq,cp(TWk-l ® c)) = (-1)k(5(cP ® cq), cp(Wk ® c)) - (-l)k(cP ® cq, cp(Wk-l ® e)) - (cP ® cq, Tcp(Wk 1 ® e))

= (_l)k 5(cP ® eq), cp(Wk ® c)) - (_l)k(cP ® cq, cp(Wk-I ® e)) - (-l)PQ eq ® cP, CP(Wk-l ® e)). By definition, 5(cP ® eq) = (5cP) ® cq + (-l)PcP ® 5cq. Hence the first term in the obtained expression is equal to

(-1)k(5cP ® cq), cp(Wk ® e))

+ (-l)P+k(cP ® 5eq), cp(Wk ® c)).

This completes the proof of the coboundary formula.

0

Corollary. If a cochain cP E CP(K) is a cocycle modulo 2, i.e., bcP = 2cP+ 1 , where cP+! E CP+ 1 (K), then the cochain cP '""' k cP E C 2p-k (K) is a cocycle modulo 2 as well.

4. Steenrod Squares

187

Proof. According to the coboundary formula, we have

5(cP '-'k cP)

= (-1)k2cP+1 '-'k cP + (-1)p+kcP '-'k 2cP+1 - (_1)kcP '-'k

It remains to note that (_1)k

+ (_1)p2

1

cP - (_1) p2 cP

'-'k-l

cPo

- ±1 ± 1 - 0 (mod 2).

D

Thus, the formula Sqk cP = cP '-'k cP defines a map Sqk: ZP(Kj Z2) --+ Z2 p-k(Kj Z2). The natural projection Z2p k(Kj Z2) --+ H 2p k(Kj Z2) then gives a map Sqk: ZP(KjZ2) --+ H2p k(KjZ2). Theorem 3.59. For any k ~ 0, the map cP 1-+ Sqk cP induces a group homomorphism Sqk: HP(Kj Z2) --+ H 2p k(Kj Z2). Proof. First, we verify that if cP 5c (mod 2), then Sqk cP is a coboundary modulo 2. Indeed, cP '-'k cP - 5(c '-'k 5c + c '-'k 1 c) (mod 2) because

5(e '-'k 5c + e

'-'k-l

e)

= 5e '-'k 5e + e '-'k 1 5e + 5e '-'k-l e + 5e '-'k 1 e + e '-'k-l 5c + e '-'k 2 e + c '-'k - 5e '-'k 5e = cP '-'k cP (mod 2).

2

e

Now, we show that the map Sqk is a homomorphism. Clearly,

Sqk(cP

+ cfP) = Sqk cP + SQk cfP + cP '-'k cfP + cfP '-'k cPo

== 0 (mod 2) and 5dP == 0 (mod 2), then 5(cP '-'k+1 cfP) == 5cP '-'k+1 cfP + cP '-'k+1 5cfP + cP '--'k cfP + cfP '--'k cP == cP '-'k cfP + cfP '-'k cP (mod 2).

Moreover, if 5cP

Theorem 3.60. The homomorphism f*: H*(Lj Z2) mutes with SQk for any simplicial map f: K --+ L.

--+

D

H*(Kj Z2) com-

Proof. By definition, we have

= (e®c,'Pdwk®f(e'))) = (e®e,'PL(1®f)(wk®e')), (SQk(j*e), c') = (j*e ® j*e, 'PK(Wk ® c')) = (e ® e, (J ® f)'PK(Wk ® e')).

(j*(Sqkc),e')

The equivariant chain maps 'Pd1 ® f) and (J ® f)'PK have the common support £(Wk ® u) = C*(Ju x fu). This support is equivariant in the sense that T£(Wk®U) = £(Wk®U). The cyclicity and equivariance ofthe support allow us to apply a standard construction (see pp. 185 and 7) and obtain a chain homotopy D between the maps 'Pd1 ® f) and (J ® f)'PK for which DT = D (if D(1llk ® u) = c, then we set D(T1llk ® u) = e). Thus,

+ Da)(1llk ® c')) (5(e ® e), D(1llk ® e')) + (e ® e, D(a1llk ® c' ± 1llk ® ae')).

((J* Sqk - Sqk j*)c, c') = (c ® e, (aD =

3. Applications of Simplicial Homology

If dc = 0 (mod 2), then 6(c ® c)

== 0 (mod 2). Moreover,

D(8wk ® c') = D((Wk-l ± TWk-l) ® c') = D(Wk-l ® c') ± D(Wk-l ® c')

== 0 (mod 2).

Therefore,

U· Sqk -

S~ f*)c

== 6d (mod 2),

o

where (d, c") = (c ® c, D(Wk ® c')).

Corollary. The operation Sqk does not depend on the choice variant chain map cpo

0/ the

equi-

Proof. For K = Land / = id K , there are two possible choices of 'P, 'PK and 'PL. If S~ and Sqk are the operations constructed for cp K and 'P L, then Sqk - Sqk = id· S~ - S~ id· = O. 0 For certain reasons, it is more convenient to use the notation Sqk = Sqp k: HP(K; Z2) - HP+k(K; Z2). It is assumed that 0 ~ k ~ p. For k > p, Sqk vanishes identically on HP(K; Z2). It follows directly from the definition that for the k-dimensional cohomology group, Sqk = Sqo is the cohomology square map. For this reason, the operations Sqk, which were introduced by Steenrod in [125, 128, 129J, are called the Steenrod squares. The Steenrod squares can also be defined for relative cohomology. Let L c K be a simplicial subcomplex. Then we have the short exact sequence 0-- C·(K,L)

~ C·(K) ~ C·(L) - - O.

By construction, 'PK(Wk ®u) E C.(u x u) c C.(L x L) for all u c L. Therefore, we can assume that 'PL = 'PKlw®c.(L)· For cochains c, c' E C·(K), we have i·(c '--'k c') = i·c '--"k i·c'; in particular, i·(p·c '--"k p.c') = i·p·c '--"k i·p·c' = 0 for any two cochains c, c' E C·(K, L) because i·p· = O. Since the sequence is exact, p·c '--"k p.c' = p·d for some cochain d E C·(K, L); this cochain d is determined uniquely because p. is a monomorphism. We set c '--'k c' = d. By definition, p·Cc '--"k c') = p·c '-'k p.c'. It easy to show that the coboundary formula remains valid for the relative '--'k-product. Thus, we can define a map Sqk: HP(K, L; Z2) - HP+kCK, L; Z2) similar to that in the absolute case. It can be seen from the definition that p. Sqk = Sqk p •. 4.2. Properties of the Steenrod Squares. We have already proved the following properties of the Steenrod squares. 1. The Steenrod squares satisfy the naturaIity condition J*Sqk = Sqk for any map f: (K, L) - (KI, Ld of simplicial pairs. 2. If k > p, then Sqka = 0 for all a E HP(K, L; Z2)'

r

189

4. Steenrod Squares

3. For any a E Hk(K, L; Z2), Sqka

=a

......... a.

Other properties of the Steenrod squares are given in the following theorem. This section is devoted to proving it; for brevity, we omit the coefficient group Z2 from the notation. Theorem 3.61. (a) The operation Sqk commutes with the connecting homomorphism d: H*(L) ---+ H*(K,L). (b) The operation Sqk commutes with the suspension isomorphism E* : HP(K) --+ HP+l(EK).

(c) The operation SqO is the identity map. (d) The operation Sql is the Bockstein homomorphism for the exact sequence 0 --+ Z2 ---+ Z4 ---+ Z2 --+ O. (e) The Cartan formula Sqk(a ......... (3) - E,(Sqia) ......... (Sqk i(3) holds. The Steenrod squares have yet another very important property; namely, they satisfy the so-callpd Adem relations. The formulation and the proof of the Adem relations is beyond the scope of this book. Commutation with 6. The connecting homomorphism d: H*(L) ---+ H*(K, L) can be described as follows. Take cq E Zq(L) and let [&] be the corresponding cohomology class. The homomorphism i*: C*(K) ---+ C*(L) is an epimorphism; hence there exists a cochain cq E C·(K) for which i·cq = cq. But i·(dcq) = dc q = 0; thus, there exists a cochain dq+1 E C q+1(K, L) for which dc q = p·dq+1. The homomorphism p.: C·(K, L) ---+ C·(K) is a monomorphism; therefore, dd q+1 = 0 because p·ddq+1 = d2 cq = O. For the cohomology class d· [c q ] we take the class [d q+1]. We must prove that the co cycles d·Sqkcq and Sq k d q +1 are cohomologous. By definition, p·(Sqk~+1) = Sqk(p·d q+1 ) = Sqk(d(iQ)

= d(iQ ......... q+l-k dcq = d((iQ '-"q+l-k dcq + cq ......... q-k cq). i·Ccq ......... q+1-k dc q + CQ ......... q-k cq) = cq ......... q-k cq because i·(dCQ) =

Moreover, a and i·CCQ) = cq. Consider d' = Sqk dq+1, d = cq ......... q-k cq, and e' = cq .........q+1-k dc q + cq .........q-k cq. We have p·d' = de' and i·C' = c'. This means that d·[d] = [d']. But d = cq ......... q-k cq ; this implies the required assertion.

Commutation with the Suspension Isomorphism. The proof of Ca) does not depend on any special properties of the short exact sequence used to construct the connecting homomorphism d·. The same argument shows that the operation Sqk commutes with the connecting homomorphism from the Mayer-Vietoris cohomology sequence. The suspension isomorphism is

3. Applications of Simplicial Homology

HIU

the connecting homomorphism from the Mayer Vietoris sequence for K o and Kl, where Ko and Kl are copies of the cone over K and K o U Kl = y:,K.

The Operations SqO and Sq1 • Let {3: HP (K, L; Z) -+ HP+l (K, L; Z) be the Bockstein homomorphism for the short exact sequence 0 -+ Z2 -+ Z4 -+ Z2 -+ O. To calculate SqO and Sql, we need the following auxiliary assertion. Lemma. (3Sqk

=

{o

k+l

Sq

if k is odd, if k is even.

Proof. Suppose that c E HP(K, L; Z2) and c is an integer cochain such that reducing it modulo 2 yields a co cycle representing c. By definition, Sqkc is the cohomology class of the cochain c '-"P k c reduced modulo 2. By assumption, c - 2d for some integer cochain d; therefore, setting q = p k, we obtain a(c '-"P k c) - (-1)q2d '-"q c + (-l)P c '-"q 2c' - (l)1fc q I c - ( l)P c '-'q I C. Thus, the cohomology class (3(Sq k c) is determined by the co cycle

c + c '-'q c' + c:(c '-"q-I c) (mod 2), where c: - 0 for odd k - q - p and c = 1 for even k. But d '-"q c+ C '-"q c' a(c '-"q+l d) (mod 2) because ac - 2d == 0 (mod 2) and 2ad = a 2 c = 0, Le., ad - O. Thus, (3(Sq k c) - 0 for odd k and (3(Sq k c) = [c '-'q-l c] = Sqk+ l c c'

for even k.

'-"q

D

The lemma shows that (d) follows from (c), because if SqO = id, then SqI _ {3SqO = (3. In turn, (c) can be proved by using the lemma. To be more precise, the lemma is used to prove (c) for K = ~p2, from which (c) is deduced for all simplicial pairs (K, L). Thus, we show that (c) holds for ~p2. Let a be the generator of the group HI(~p2; Z2)' According to the lemma, we have {3SqOa = Sq1 a = a '-" a i= 0; hence SqOa i= O. Therefore, SqOa - a because HI (~p2; Z2) has only one nonzero element. The natural embedding f: ~pI -+ ~p2 induces the homomorphism f*: Hl(~p2; Z2) -+ HI(~pl; Z2) that takes a to the generator a' of the group HI (~pI; Z2). The naturality condition implies SqOa' = Sq(f*a) f*(SqOa) = f*a = a'. Thus, (c) holds also for ~pI ~ SI. Since the operation SqO commutes with the suspension isomorphism, (c) holds for sn. Let K be an n-dimensional simplicial complex. According to Theorem 3.6, any element of the group Hn(K; Z) can be realized by means of a map K -+ sn. This is also true for the group Hn(K; Z2) because the reduction of coefficients modulo 2 induces an epimorphism Hn(K; Z) -+ Hn(K; Z2)' Note that only the cohomology groups have th;:, property; it

4. Steenrod Squares

191

does not hold for homology. The map H2(~'p2; Z) -- H 2(Rp2; Z2), say, cannot be an epimorphism because H2(Rp2; Z) = a whereas H 2(Rp2; Z2) =IO. The surjectivity of Hn(K; Z) __ Hn(K; Z2) follows from the fact that the group Hn(K) is torsion-free for each n-dimensional simplicial complex; it is proved as follows. According to the universal coefficient theorem, the group Hn(K; G) is the direct sum of the groups Hom(Hn(K), G) and Ext(Hn l(K), G). The map Hom(Hn(K), Z) -- Hom(Hn(K), Z2) is an epimorphism because the group Hn(K) is torsion-free, and Ext(Hn l(K), Z) -Ext(Hn - 1 (K), Z2) is an epimorphism in any case (this follows from the calculation of Ext(A, B) for cyclic groups A and B). The naturality of the operation SqO implies (c) for K. If K is an infinite-dimensional simplicial complex, then the natural embedding K n -- K induces a monomorphism Hn(K; Z2) __ Hn(Kn; Z2). Therefore, the naturality of SqO implies (c) for an arbitrary K.

Now, consider the relative cohomology groups Hn(K, L; Z2). Let CL be the cone over L. Then Hn(K, L) rv Hn(K U CL, CL) ~ Hn(K U CL). The first isomorphism is obvious at the level of (co)chains, and the second follows from the exact sequence for the pair (K U C L, C L) because the space C L is contractible. Both isomorphisms commute with SqO; therefore, (c) holds for any pair (K, L) of simplicial complexes. The Cart an Formula. It is sufficient to show that Sqk (a x.B) = Ei Sqia x Sqk i{3. Indeed, we have a '-'" (3 = d*(a x (3), where d: IKI -- IK x KI is the diagonal map. Therefore, Sqkd*(a x (3) = d*Sqk(a x (3) = d* Ei Sqia x Sqk-t {3 = El Sq1 a '-'" Sqk-i (3. Let r: W -- W ® W be the linear extension of the map defined by r(wk) = E( _1)i(k-i)Wi ® TiWk_i. This is a chain map. Indeed, we have r(8wk)

= r(wk-l + (_1)kTwk_d = L( _1)i(k-l-i)Wi ® rWk-l-i

+ L ( _1)i(k

l-i)+kTwi ® rWk-l-i

and 8r(wk)

= 2:( -1)i(k-i) 8wi ® rWk-i + L ( _1)i(k-i)+i wi ® r8Wk-i = L(-1)i(k-i)Wi_l ®Tiwk_i

+ L(-1)i(k-i)+iTwi_l ®rWk-i

+ 2:( -1)i(k-i)+i wi ® Tiwk_i_l + L( _1)i(k-i)+k Wi ® Ti+l wk _ i _ 1 . In the expression for 8r(wk), the first and fourth sums cancel each other because i(k - i) is replaced by i(k - i) + k - 1 when i is replaced with i + 1.

3. Applications of Simplicial Homology

Now it is easy to see that 8r(wk) = r(8wk) (the first sum in the expression for r(8wk) coincides with the third sum in the expression for 8r(wk), and the coincidence of the second sums in these expressions is proved by replacing i with i + 1). We apply r to define the map 'PKxL as the composition

W ® C.(K) ® C.(L) ~ W ® W ® C.(K) ® C.(L)

~ W ® C.(K) ® W ® C.(L)

'PK®'PL,

C.(K) ® C.(K) ® C.(L) ® C.(L)

~ C.(K) ® C.(L) ® C.(K) ® C.(L), where T transposes the second and third factors in both cases. The map 'PKxL is equivariant, and is supported on the image of the corresponding map C; hence we can use it to calculate the ........ k-product in the chain complex C.(K x L) ~ C.(K) ®C.(L), and therefore, to calculate the operati'OInJ.Sqk. Suppose that 0 E ZP(K; Z2), /3 E ZP(L; Z2), a E C.(K; Z2), and b E C.(L; Z2). We are interested in the action of the operation Sqk = Sqp+q-k on the {p+q)-cocycle oX/3. For brevity, we set p+q-k = n. Using the definition of 'PKxL and discarding signs (because we deal with Z2 coefficients), we obtain (Sqk(O ® /3), a ® b)

= ((0 ® /3) ........ n (0 ® /3), a ® b) = (0 ® /3 ® 0 ® /3, 'PKxL(Wn ® a ® b))

(0 ® 0® /3 ® /3, L 'PK(Wi ® a) ® Ti'PL(Wn-i ® b)) = L (0 ........ 0, a) (/3 ........ n-i /3, b) = L (SqP-i o , a)(Sqq-n+i/3, b) =

i

=

L(SqP-i o X Sqq-n+i/3, a ® b),

where the summation is over all i from 0 to n. Thus, Sqk to x /3) = E~-o SqP-io X Sqq-n+i/3 = E~=p-n Sqio X Sqk-i /3. It remains to show that the summation limits are as required. The terms in Sqk(o x /3) = Ei Sqio x Sqk-i/3 are nonzero only for 0 ~ i ~ p and 0 ~ k - i ~ q, i.e., for k - q ~ i ~ k. Therefore, the summation is over i from k - q = p - n to p. The Cartan formula has the following interpretation. For each element a E H· (K; Z2), we set Sq 0 = Ek Sqk 0 (this sum contains only finitely many nonzero terms). The map Sq: H·(K; Z2) -+ H·(Kj Z2) is a ring homomorphism. Indeed, if dim 0 = p and dim /3 = q, then the componen~ of dimension (p + q + k) of the element Sq 0 ........ Sq /3 = (Ei Sq a) ........ (E, Sq' /3) has the form E Sqio ........ Sqi /3, where the summation is over all i and j for which (i + p) + (j + q) = P + q + k, i.e., i + j = k. Therefore, it has the form

4. Steenrod Squares

19~

Ei Sqio '-"' Sqk-i{3.

According to the Cartan formula, this sum is equal tc Sqk(o ........ {3); thus, it is equal to the (p + q + k)-dimensional component oj Sq(o ........ {3). Using the homomorphism Sq, we can calculate Sqk(on) for any 0 E H 1 (K;Z2). Theorem 3.62. If 0 E Hl(K; Z2), then Sqk(on) = (~)on+k. Proof. The properties of the Steenrod squares imply that Sqko = 0 for k ~ 2, SqOo = 0, and Sq10 = 0 2. Therefore, Sqo = 0 + 0 2. Taking into account the fact that Sq is a homomorphism, we obtain Sq(on) = (Sqo)n = (0 + (2)n = 0 Ek (~)ok. Hence Sqk(on) = (~)on+k. 0 Problem 112. Prove that if the cohomology class 0 is one-dimensional, then Sq(l + 0 + 0 2 + 0 4 + ... ) = 1 + o. Problem 113 (Wu formula). Prove that

Sqk Wm =

WkWm

+

(m-k) 1

+

Wk-l Wm+l

(m- 2k + 1)

Wk-2 W m+2

+ ... +

(m k- 1)

WOWm+k·

Chapter

4

Singular Homology

1. Basic Definitions and Properties Simplicial homology groups are defined only for simplicial complexes, and their topological invariance is far from being obvious. There is an approach to define homology for an arbitrary topological space X, and the topological invariance of homology thus defined is obvious. This approach is as follows. Consider all continuous maps I: !:l. k _ X, where !:l. k is the simplex [0,1, ... , k]j such maps are called singular k-simplices. By Ck(Xj G) we denote the group of all finite sums '2: ai/i, where ai E G and Ii is a singular k-simplex. To define the boundary homomorphism a, we introduce maps Ej = Ej: !:l.k-l _ !:l.k for j = 0,1, ... , k by setting Ej[i] = [i] for i < j and Ej[i] = [i + 1] for i 2: jj these maps are extended to the entire simplex by linearity. For a singular simplex I: !:l. k - X, we set aI = '2:;=0 (-l)j I Ej E Ck-l(Xj G). For a zero-dimensional simplex I, we set al = O. Theorem 4.1. {)a

= O.

Proof. First, note that if i < j, then Ej+1Ef = E7+1Ej_l. Indeed, both maps Ej+1Ef and E7+1Ej_l take [a] to [a] (if a < i), to [a + 1] (if i ~ a < j - 1), or to [a + 2] (if a 2: j - 1). Therefore,

aal = a(:L (-l)j IEj) J

=

2: (-l)i+j IEjEi 'J

= :L (-l)i+j IEjEi + :L (-l)i+j IEiEj-l. i~j

i<j

-

195

196

4. Singular Homology

Let us replace i by j and j by i + 1 in the second sum. After such a change, the second sum cancels the first one. 0 Again, consider the groups Zk = Ker8k and Bk = Im8k+1. According to Theorem 4.1, we have Bk C Zk; therefore, the quotient Hk(X;G) = Zk/Bk is defined. This group is called the singular homology group of the space X with coefficients in G. We shall use the same notation for singular and simplicial homology; we shall show in Section 1.4 that the singular homology of any simplicial complex is isomorphic to its simplicial homology.

Theorem 4.2. If a space X has n path-connected components, then we have Ho (X; G) = G EB ••• EB G.

----------n

Proof. Choose points Xl, ... , X n , one in each path-connected component of X. If a point Yi belongs to the same component as Xi, then there exists a path from Xi to Yi. This path is a singular simplex f: ~l -- X; we have 8(aJ) = a[YiJ - a[x,J. Thus, the singular chain a[YiJ coincides with a[xiJ up to a boundary. Now suppose that a[xiJ = 8c, where c = Eadi and fi is a path from Zi to Wi. Then 8c = E a,[wiJ - E adziJ. In this expression, the sum of all coefficients vanishes; comparing the sum of coefficients on the leftand right-hand sides of the equality a[xiJ = E adwiJ - E ai[zi], we obtain a=O. 0 ConsiderCk = Ck(X) andC~ = Ck(Y). Any continuous map If': X -+ Y induces the map If'k: Ck -+ C~ that assigns to a singular simplex f: ~k -+ X the singular simplex If'f: ~k -+ Y. Moreover, If'k-18kf = E (-l)ilf'fcj = i4lf'kf. The equality If'k-18k = i4lf'k implies If'k(Ker8k ) C Keri4 and If'k(Im8k +l) C Imi4+l. We obtain a homomorphism If'.: Hk(X) -+ Hk(Y). The identity map X -+ X induces the identity map Hk(X) -+ Hk(X); hence a homeomorphism X -+ Y induces an isomorphism Hk(X) -+ Hk(Y). This assertion can be generalized as follows: any homotopy equivalence X -+ Y induces an isomorphism Hk(X) -+ Hk(Y). The following theorem is even more general.

Theorem 4.3. If maps If', 1/;: X -+ Y are homotopic, then the homomorphisms If'., 1/;.: Hk(X) -+ Hk(Y) coincide. Proof. As in the case of simplicial homology, we define a chain homotopy as a family of homomorphisms D k : Ck(X) -+ Ck+I(Y) for which

(31) Dk-1 8k + a:c+lDk = If'k -1/;k. The proof of the equality If'. = 1/;. remains the same. Let H: T x I -+ Y be a homotopy between If' and 1/;. From tius homotopy we construct a chain homotopy Dk: Ck(X) -+ Ck+l(Y) as follows. Let

197

1. Basic Definitions and Properties

I:

fj.k --+ X be a singular simplex. Consider the map F: fj.k x I --+ Y defined by F(x, t) = H(f(x), t). We set Dkl = E~=0(-I)iF67+1, where 67+1 is the linear map from the simplex [0,1, ... , k + 1] to the simplex [00, ... , io, i 1, ... ,k1] C fj.k X Ij here 00,' .. , io are the vertices of fj.k x {O} and i 1, ... , k 1 are the vertices of fj.k x {I}. The singular chains Dk-18kl and ~+1Dkl are linear combinations of singular simplices of the form Fo.: fj. k --+ Y, where 0. is a map from [0, ... , k] to a simplex in 15 k x I. To prove (31), is suffices to determine which simplices in 15 k x I are obtained in this way (with signs taken into account). First, consider the case k = 2. The map D2 takes the simplex [0,1,2] to +000 11121 - 00101121 + 00 102021 , and the map

8fJ takes these simplices to +0 11121

-10 1121

-00 1121 +000 121

+00 1121

-002021

-001021 +00 1011

+00 1021

-00 0111

-00 1020'

Note that the simplices underlined likewise cancel each other. For an arbitrary k, the picture is similar: for i i- 0, k, the map ~+1 deletes the vertex i + 2 from the simplices (-I)i[Oo, ... , io, i\ (i + 1)1, ... , k 1] and (_1)i+l[00, ... , io, (i + 1)0, (i + IF, ... , k 1], and the simplex [00,"" io, (i + 1)1, ... , k 1 ] occurs twice with opposite signs. The map fh takes the simplex [0,1,2] to [1,2] - [0,2] + [0,1]' and Dl takes these simplices to +10 1121

-000 121

+00 0111

In the sum 8':JD2 + D 1 eh, all simplices except +0 11121 and -01h21 are canceled. The remaining simplices correspond to the maps cpI and 1/;/. It is required to show that, for any k, each of the simplices [... 3... ioi 1 ... ] (or [... ioi 1 ··.3 ... ]) is encountered precisely once in ~+1 Dk and in Dk-18k with opposite signs. This is easy. 0 Example 50. Suppose X is a contractible space. Then Ho(Xj G) Hk(XjG) = 0 for k > O.

= G and

Proof. Any contractible space is homotopy equivalent to a singleton *. Therefore, it suffices to show that HO(*j G) = G and Hk(*j G) = 0 for k > O. For any k ~ 0, there exists a unique map fj.k --+ *j let us denote it by A. Clearly, 8k A = E~=o(-l)iA-l for k ~ 1 and 80 /0 = O. Hence

4. Singular Homology

198

~k d2k I

0 and 8 2k hk - 12k I (k ~ 1). Thus, we have Zo Bo - 0, Z2k 1 = G and B 2k-l = G, and Z2k - 0 and B2k - o.

=

G and

D

1.1. Excision Theorem and the Mayer Vietoris Exact Sequence. The definitions of relative and reduced singular homology groups, as well as of singular cohomology groups, are the same as in the simplicial case. The construction of the exact homology sequence of a pair does not require any changes either. But when constructing the Mayer Vietoris exact sequence, one encounters problems because not every singular chain in X U Y can be represented as the sum of two singular chains with supports in X and Y. This problem is essential; the Mayer Vietoris exact sequence for singular homology exists only under certain constraints. Its existence is closely related to the following theorem.

Theorem 4.4 (on excision). Suppose U cAe X and U C int.4....,., Then the inclusion of pairs (X \ U, A \ U) c (X, A) induces an isomorphism H*(X \ U, A \ U) --+ H*(X, A). Proof. Consider Xl - A and X 2 = X \ U. We have int(X \ U) X \ int A; therefore, int Xl U int X2 = X.

=X

\ U:::J

Lemma 1. Let X - int Xl U int X 2. Then, for any singular simplex f: t:,.k --+ X, there exists a positive integer m = m(J) such that the image of any simplex from the mth barycentric subdivision of t:,.k is contained entirely either in Xl or in X 2 . Proof. The sets f-I(intXt} and f- l (intX2 ) form an open cover of the compact set t:,.k C IRk. Let d be the Lebesgue number (see Part I, p. 59) of this cover. If d is the diameter of the simplex t:,.k, then the diameter of any simplex in its mth barycentric subdivision is at most (k!l)m d . It remains to choose m so that (k!I)m d < d. D To each singular simplex f: t:,. k --+ X we can assign its baryct'ntric subdivision sd f = E fa, where fa is the restriction of f to a simplex t:,.~ from the barycentric subdivision of t:,. k. Formally, the singular chain sd f is defined as follows. On p. 8, we defined a chain map ik: Ck(K) --+ Ck(K') for which ik(t:,.k) = [b,ik_ I 8t:,.k] = E±[b,b',b", ... ,Vi], where b is thf' barycenter of the simplex t:,.k, b' is the barycenter of its (k - I)-face t:,.' c t:,.k, b" is the barycenter of a (k - 2)-face t:,." c t:,.' c t:,. k, ... , and Vi is a vertex of t:,. k. We set sd f = E ±go., where go. is the composition t:,. k

=

[vo,

VI, ... , Vk] --+

[b, b', b", . .. , Vi]

1) X

(here the first arrow is a homeomorphism of simplices treated as ordered sets of vertices).

1. Basic Definitions and Properties

199

The map sd can be extended to a homomorphism sd: G.(X) - G.(X) by linearity. This is a chain map: the equality 8 sd = sd 8 follows from

i8 = 8i. Lemma 2. For each positive integer m, there exists a family of homomorphisms D k : Gk(X) - Gk+l(X) such that

(i) 8 k+lDk c + Dk-18kC = sdm C - Cj (ii) the support1 of the chain DkC is contained in that of the chain c. Proof. We set Do = 0 and construct the homomorphism Dk assuming that Do, ... , Dk-l are already constructed. Consider the singular simplex idk: tlk - tlk in Gk(tl k ). It is easy to show that the chain Zk = sdm idk - idk -Dk-18k idk is a cycle. Indeed, by assumption, for the chain ik-l = 8 k id k , we have

8k D k-Iik-l

+ Dk-28k

Iik

1 -

sdm ik

1 -

ik-l,

= Dk 28k-18k idk = O. Therefore, = 8 k sdm idk -8k idk - sdm 8k idk +8k idk = 0

and Dk-28k-Iik-l

8kZk because 8 k sd m

= sdm 8k .

We already know that Hk(tl k ) = 0 for k ~ 1 because the simplex tlk is a contractible space. Hence there exists a singular chain bk+l E Gk+l(tl k ) for which Zk = 8b k+l' Suppose that bk+l = Eadi, where fi: tlk+l _ tlk. For any singular simplex f: tl k - X, we set Dkf = E ai (J 0 fd and extend Dk by linearity. The required equality

8 k+lDkf + Dk-18kf = sdm f - f

E

follows from adi E ak8k+l(J 0 fi).

+ Dk-18k

=

sdm idk - idk in view of 8 k+lDk f

= 0

We continue the proof of Theorem 4.4. According to Lemma 2, any relative cycle Zk E Gk(X, A) is homologous to the relative cycle sdm Zk because Dk-18kZk E Ck(A). Any cycle contains only finitely many singular simplicesj therefore, by Lemma 1, we can choose a number m for each cycle Zk so that the support of any singular simplex included in sdm Zk is contained entirely in Xl = A or X2 = X\U. Thus, the map H(X\U,A \U) - H.(X, A) is an epimorphism. It remains to show that if Zk E C k (X \ U) and Zk - 8b E C k (A) for some b E Ck+l(X), then Zk - 8b' E Ck(A \ U) for some b' E Ck(X \ U). Let C!Xl,X2}(X) be the group of sums of chains from Ck(Xd and Gk(X2). By assumption, 8b E C~Xl,X2} (X). Choose a number m so that 1 By the support of a chain we mean here the union of the images of all singular simplices in this chain.

4. Singular Homology

200 sdmb E Cl~t,X2}(X). We have b + 8Db

Let sdmb - D8b

= b' + b",

Zk - 8b' - 8b"

= sdm b -

D8b E Cl~t,X2}(X).

where b' E Ck+1(X2) and b" E Ck+1(Xd. Then

= Zk -

8(b + 8Db)

= Zk -

8b E Ck(Xd·

Moreover, zk,8b' E Ck(X2) and 8b" E Ck(Xd. Hence Zk - 8b' E Ck(Xl)

n Ck(X2) = Ck(A) n Ck(X \

U)

= Ck(A \

U),

o

as required. Now, consider the Mayer Vietoris exact sequence. Xl, X 2 C X, we have the short exact sequence 0-+ C.(XlnX2)

(31,-32),

For any two sets

C.(XI)EBC.(X2) ~ C!X1,X2} (X l UX2'} ~ 0,

where C!X1,X2} (Xl UX2) = C.(Xd +C.(X2) is the group of sums of chains. This short exact sequence induces the exact sequence ... - +

Hk(Xl n X2) ~ Hk(Xd EB H k (X 2)

~ Hk(C!X1,X2} (Xl

U X2))

~ Hk-l(X l n X 2 )

- + ....

If the embedding C!X1,X2}(Xl UX2) ---+ C.(Xl UX2) induces an isomorphism of homology groups, then we say that the pair {Xl, X 2} satisfies the excision axiom. In this case, the group H k (C!Xl,X2}(X l U X 2)) can be replaced by H. (X 1 U X 2) ; the exact homology sequence is then called the Mayer Vietoris sequence. Example 51. Suppose that Xl and X2 are two semicircles, Xl contains both boundary points, and X 2 contains no boundary points (see Figure 1). Then the Mayer Vietoris exact sequence does not hold for the pair {Xl, X2}.

0 2

XI

Figure 1. The semicircles Xl and X 2

Proof. By assumption, Xl U X 2 = 8 1 and Xl n X 2 = 0. Therefore, the fragment HI (XI) EB HI (X2) ---+ Hl(X l U X2) ---+ HI (Xl n X2) has the form o ---+ z ---+ o. 0

1. Basic Definitions and Properties

201

In the proof of the excision theorem, we showed that if Xl UX2 = int Xl U int X2, then the pair {Xl, X 2 } satisfies the excision axiom. The surjectivity of the homomorphism H*(C!Xl,X2}(X)) ~ H*(X), where X = Xl U X2, was proved explicitly, and its injectivity follows from the existence of a chain d = Db E Ck+2(X) such that b + ad E C~::,X2}(X) for any b E Ck+1(X) with ab E C~Xl,X2}(X); such a chain exists because the chain b + ad.

-

Example 52. Hk(sn)

=

{Zo

if k if k

ab is

the boundary of

= n, ~

n.

Proof. Suppose that Xl and X2 are two different points on the sphere sn, Xl = sn\{xI}, andX2 = sn\{X2}. The spaces Xl andX2 arecontractiblej therefore, Hk(Xi ) = 0 for all k. The pair {X I, X 2} satisfies the excision axiom because the sets X I and X 2 are open in sn. The Mayer Vietoris exact sequence gives the isomorphism Hk(sn) ~ Hk_l(sn \ {XI,X2}). Moreover, sn \ {XI,X2} '" sn-l. Thus, Hk(sn) ~ Hk_l(sn-l). It remains to note that the required assertion is obviously true for n = O. o Example 53. Hk(Dn, sn- I ) =

{Zo

if . k If k

= n, ~

n.

Proof. The exact sequence of the pair (Dn, sn-l) shows that

o

Hk(D n , Sn-l) ~ Hk_l(sn-l).

Exercise. Prove the suspension isomorphism HkCL.X) ~ Hk-I(X) for all k~1.

Problem 114. Prove that Hi (X, Y)

~

Hi(X U CY, CY).

Problem 115. Prove that Hi (X, Y)

~

Hi(X U CY) for i

~

l.

Problem 116. Given a connected CW-complex X and its sub complex Y, prove that Hi (X, Y) ~ Hi(X/Y). Suppose that (Xl, AI) and (X2' A 2) are two pairs oftopological spaces, Ai C Xi C X for i = 1,2, and the pairs of sets {Xl, X 2 } and {AI, A 2} satisfy the excision axiom. Then we have the exact sequence ---- Hk(XI

n X2, Al n A 2) ~ Hk(XI, Ad EEl Hk (X2 , A2)

~ Hk(XI

U X2, Al U A 2)

~ Hk-l (Xl n X2, Al n A 2) ---- j

4. Singular Homology

202

it is called the relative Mayer Vietoris sequence. To construct this sequence, note that the map induced in relative homology by the embedding

C.(XI) C.(AI)

+ C.(X2) + C.(A2)

_--':---'----'---'-

-

C.(XI C.(AI

u X 2) u A 2)

_-':--_--'-

is an isomorphism. Indeed, consider the exact homology sequence of the pair (Xl U X2, Al U A 2) and the corresponding exact sequence for homology of the form H.(C!Xl,X2} (Xl UX2», which arises from the obvious short exact sequence. The natural embeddings of chain complexes induce a map from the latter exact sequence to the former, and the maps of absolute homology groups are isomorphisms. It follows by the five lemma that the maps of relative homology groups are isomorphisms as well. It remains to construct a short exact sequence

n X 2) --. C.(XI ) EB C.(X2 ) --. C.(XI) + C.(X2) C.(AI n A2) C.(AI) C.(A2) C.(AI) +9.(A2)

O --. C.(XI

~

0

,

i.e., to prove that the quotient of a short exact sequence by an exact subsequence is an exact sequence (this is true for any, not necessarily short, exact sequences). Consider the exact sequence as a chain complex C and the subsequence as a chain sub complex C'. The exactness of these sequences means that H.(C) = 0 and H.(C') = O. Let us write the exact sequence induced in homology by the short exact sequence 0 - C' - C - C/C' - 0:

Hk(C) ~ Hk(C/C') ~ Hk-I(C'). We obtain Hk(C/C') = 0, as required.

Singular Cohomology. Singular cohomology groups are defined in a natural way. Let Ck(X) be the group of singular chains with coefficients in Z. Then Ck(Xj G) = Hom(Ck(X), G) is the group of singular cochains with coefficients in G. The coboundary operator d: Ck(XjG) _ Ck+I(XjG) is the dual of the boundary operator 8: Ck+l(X) - Ck(X). The group Zk of cocycles, the group Bk of coboundaries, and the cohomology group Zk / Bk are defined in a standard way. For singular cohomology, a theorem similar to the excision theorem for singular homology is valid. Indeed, according to the universal coefficient theorem, any chain map inducing an isomorphism in integral homology induces also an isomorphism in cohomology.

1.2. Axioms of (Co)homology Theory. Singular (co)homology theory is only one of the possible theories of (co ) homology defined for all topological spaces. Another example is Cech cohomology. For simplirial complexes, these theories coincide, but for more general topological spaces, they may give different results. There are also other (co ) homology theolles.

1. Basic Definitions and Properties

203

Steenrod and Eilenberg [131] suggested axioms which any (co)homology theory must satisfy and proved that if two (co)homology theories satisfy these axioms and have the same coefficient group, then they coincide for simplicial complexes. We say that a class A of pairs (X, A) of topological spaces is admissible if the following conditions hold: (i) if a pair (X, A) belongs to A, then the pairs (X, X), (X,0'), (A, A), and (A, 0') belong to Aj

(ii) if a pair (X, A) belongs to A, then the pair (X x I, A x I) belongs to Aj (iii) the pair (*,0'), where * is a singleton, belongs to A. Any homology theory is defined for an admissible class of pairs. To every pair (X, A) from the given admissible class and any integer n it assigns an Abelian group Hn(X, A). To any map of pairs I: (X, A) - (Y, B) and any integer n a homomorphism f.: Hn(X, A) - Hn(Y, B) is assigned. Moreover, boundary homomorphisms 8.: Hn(X,A) - Hn-1(A), where H n - 1 (A) = H n - 1 (A, 0'), are defined. Every homology theory must satisfy the following axioms (the Steenrod Eilenberg axioms). 1. If I is the identity map, then so is I. j ·2. (gl). = g.l.j

3. 8.1. = (JIA).8., i.e., the following diagram is commutative:

4 (the exactness axiom). If i: A - X and j: X embeddings, then the following sequence is exact:

(X, A) are natural

5 (the homotopy axiom). If maps I, g: (X, A) - (Y, B) are homotopic, then the homomorphisms I. and g. coincide for all nj 6 (the excision axiom). If U c X is an open set whose closure is contained in int A and the natural embedding eX \ u, A \ U) _ (X, A) is an admissible map, then, for any n, this embedding induces an isomorphism Hn(X \ U, A \ U) - Hn(X, A)j 7 (the dimension axiom). H n (*) = 0 for n =1= o. The group Ho(*) is called the coefficient group.

4. :}mgUlar nomolOgy

Any cohomology theory is defined for an admissible class of pairs. To every pair (X, A) from the given admissible class and any integer n it assigns an Abelian group Hn(x, A). Any map of pairs /: (X, A) - (Y, B) and integer n are assigned a homomorphism /*: Hn(y, B) - Hn(x, A). Moreover, coboundary homomorphisms c5*: Hn-l(A) _ Hn(x, A), where Hn-l(A) = Hn-l(A, 0), are defined. Any cohomology theory must satisfy the following axioms (the Steenrod Eilenbery axioms): 1. If / is the identity map, then so is

/*;

2. (gf)* = /*g*; 3. c5* /* = (JIA)*c5*, i.e., the following diagram is commutative:

Hn(x, A) ~ Hn(y, B)

1

1 6•

6•

Hn-l(A) ~ Hn-l(B); 4 (the exactness axiom). If i: A - X and j: X embeddings, then the following sequence is exact:

... ~ Hn-l(A) ~ Hn(x, A)

L

(X, A) are natural

Hn(x) ~ Hn(A) ~ ... ;

5 (the homotopy axiom). If maps /,g: (X, A) - (Y,B) are homotopic, then the homomorphisms /* and g* coincide for all n; 6 (the excision axiom). If U c X is an open set whose closure is contained in int A and the embedding (X \ U, A \ U) - (X, A) is admissible, then, for any n, this embedding induces an isomorphism Hn(x, A) -

Hn(x \ U,A \ U); 7 (the dimension axiom). Hn(*) = 0 for n i- O. The group HO(*) is called the coefficient group. Shortly after Steenrod and Eilenberg published their book. it turned out that the dimension axiom plays a special role; some naturally arising theories satisfy all Steenrod Eilenberg axioms except the dimension axiom. Such (co ) homology theories are said to be extraordinary, or generalized. They include, e.g., cobordism theory and K-theory. 1.3. The Jordan-Brouwer Theorem. Brouwer [14 18] proved that if a subspace B C lRn is homeomorphic to sn-l, then lRn \ B has two connected components. In the special case of n = 2, this theorem was proved earlier by Jordan. 2 Generalizing the proof of the Jordan-Brouwer theorem, Alexander obtained the following result. 2The proof of Jordan was not quite rigorous. The first complete proof of Jordan's theorem was suggested by Veblen in 1905.

1. Basic Definitions and Properties

20~

Theorem 4.5 (Alexander [5]). If A, BeRn are homeomorphic closeQ subspaces, then Hk(R n \ A) ~ Hk(R n \ B) for all k ~ O. Proof (see [30]). Let R 2n = R n x Rn. Suppose that A C R n x {O} and B C {O} x Rn. Then (R2n \ A) :::::: (R2n \ B) (see Part I, Theorem 2.5). If A =I- R n and B =I- R n , then the required assertion is easily derived from the following lemma. Lemma. If A c Rn is a closed set and A =I- Rn, then Hk+t(Rn+i \ A) ~ Hk(R n \ A) for any i ~ 0 (it is assumed that JR.n = R n x {O} C Rn+i). Proof. We first consider the case i X = R n +1 \A,

=

1 and then use induction. We set

> 0 or x ! t < 0 or x

X+ = {(x, t) ERn x JR.! t

E JR.n \ A},

X_ = {(x, t) ERn x R

E JR.n \

A}.

The sets X+ and X_ are open in X, X+ U X_ = X, and X+ n X_ (JR.n \ A) x R. Let us write the exact Mayer Vietoris sequence for reduced homology: ... - - Hk+l(X+) EI1 Hk+1(X-) - - H k+1(X+ U X_) - - Hk(X+ n X_) - - Hk(X+) EI1 Hk+l(X-) - - ....

The spaces X+ and X_ are contractiblei hence they have trivial reduced homology groups, and therefore Hk+l(X+ u X_) ~ Hk(X+ n X_), i.e., Hk+1(Rn+1 \ A) ~ Hk(JR.n \ A) x JR.) ~ Hk(IR n \ A). 0 If A =I- IR n , then, according to the lemma, we have Hk(JR. n \ A) c:,; Hk+n(JR. 2n \ A)i if A = IRn, then we obtain the isomorphism Hk(Rn+l \ A) ~ Hk+n_l(R 2n \ A). Thus, if A =I- JR.n and B =I- IRn, then Hk(JR.n \ A) ~ Hk(JR.n\B) and Hk(Rn\A) ~ Hk(JR.n\B). If A = ]Rn, then Hk(Rn+l \A) ~ H k (Rn+1 \ B). In particular, Ho(]Rn+1 \ A) ~ Ho(JR.n+1 \ B). Since the set ]Rn+1 \ A = ]Rn+1 \ ]Rn has precisely two connected components, it follows that so does ]Rn+1 \ B. But this is possible only if B = JR.n . 0

The Alexander theorem readily implies the following assertion, which was proved by Brouwer in [17].

Theorem 4.6 (invariance of domain). If U and V are homeomorphic subspaces of]Rn and U is open, then V is open as well. Proof. Take a point Vo E V. We must prove that V has a subset that is open in ]Rn and contains the point Vo. Consider a homeomorphism h: U ~ V and let Uo = h- 1 (vo). Choose c > 0 so that Dn = {x E ]Rn ! IIx - uoll ~ c} C U. According to Alexander's theorem, the set ]Rn \ h(Dn) is connected, while

4. Singular Homology

206

the set an \ h(sn-l), where sn-l = aDn, has two connected components. The sets jRn \ h(Dn) and h(Dn) \ h(sn-l) = h(int Dn) are connected, and their union is jRn\h(sn-l). Therefore, it is these sets that are the connected components. In particular, they are open subsets of the space jRn \ h(sn-l), which, in turn, is an open subset in IRn. Hence the set h(int Dn) 3 h( uo) = Vo is open in IRn. 0 Corollary (invariance of boundary). A homeomorphism Dn _ Dn cannot take a point x E aDn to an interior point. Example 54. Suppose that X = (a x {O}) U ({O} x a) c 1R 2 , U = {(O, t) I 2 < t < 3}, and V = {(O, t) I -1 < t < I} (see Figure 2). Then the sets U

u

v -1

2

3

Figure 2. Noninvariance of domains

and V are homeomorphic, U is open in X, but V is not open in X. Problem 117. Suppose that U and V are homeomorphic subspaccs of an and U is closed. Is it true that V must be closed? 1.4. Isomorphism between Simplicial Homology and Singular Homology. For any simplicial complex K, both simplicial and singular homology groups are defined. In this section, we prove that these groups are isomorphic. It is more convenient to construct a chain map that induces this isomorphism by using the total (ordered) chain complex C.(K), whose homology groups are isomorphic to the simplicial homology groups (see Theorem 2.1 on p. 60), rather than the simplicial chain complex. To any ordered k-simplex (vo, ... , Vk) there corresponds the singular simplex f: [0, ... , k] - [vo, ... , Vk], where f is the linear map of simplices taking each point i to Vi. This correspondence, when extended by linearity, determines a homomorphism jk: Ck(K) _ Ck(K), where Ck(K) is the singular chain complex. It follows directly from the definition of boundary homomorphisms in C. and C. that j is a chain map. Therefore, j induces a homomorphism j.: Hk(C.(K)) - Hk(K) of homology groups. Theorem 4.7. The homomorphism j. is an isomorphism.

1. Basic Definitions and Properties

20~

Proof. First, suppose that the simplicial complex K is finite. In this case the theorem is proved by induction on the number n of all simplices of a1: dimensions in K. If n = 1, then K consists of one point. For a one-point space, the isomorphism is obvious even at the level of chains. Now suppose that i", is an isomorphism for any simplicial complex containing fewer than n simplices. Let K be a simplicial complex containing precisely n simplices, and let 6 be a simplex of maximal dimension in K. The simplex 6 is not a face; therefore, KI = K \ 6 is a sub complex of K containing fewer than n simplices. To simplify notation, we set Hk(K) = Hk(C", (K)). For singular homology, we consider the Mayer Vietoris sequence for 6 and K{, where K{ = KI U (int ll. \ {xo}) for Xo E int ll. because the pair ll., KI does not satisfy the excision axiom. As a result, we obtain the following commutative diagram with exact rows: --~~ Hk+l(all.) ---+~ Hk(K) ~ Hk(ll.) ffi Hk(Kd ---+ ...

l~o Hk+l(all.)

1~o

11 Hk(K)

l~o

l~o Hk(ll.) ffi Hk(KI)

lisO

The upper vertical arrows coincide with i"" and the lower vertical arrows are the isomorphisms induced by homotopy equivalences (or the identity maps). It follows from the five lemma that the map i",: Hk(K) -+ Hk(K) is an isomorphism (to apply this lemma, we must add two terms, on the left and on the right, to the above fragment of the diagram). Now suppose that the simplicial complex K is infinite. For any

0:

E

Hk(K), there exists a finite sub complex L of K such that 0: belongs to the image of Hk(L) under the homomorphism induced by the embedding L -+ K. Therefore, i",: Hk(K) -+ Hk(K) is an epimorphism. Suppose that it E Hk(K) and i .. it = O. At the level of chains, this means that a singular chain Ck E Ck(K) corresponding to a representative of the homology class it is the boundary of some singular chain O:k+l E Ck+I(K). Choose a finite sub complex L in K so that it is a support for the chain O:k+l. The chain Ck represents the zero homology class in Hk(L); therefore, it = 0, i.e., i", is a monomorphism. D

Theorem 4.8. If U is an open subset ofIRn, then Hk(U)

= a for

k ~ n.

Proof. Consider a singular cycle Z E Ck(U), where k ~ n. We must prove that z = aw, where w is a singular chain in U. Let X be the union of all

4. Singular Homology

208

images of singular simplices in z. The set X C U is compact; therefore, the distance c between X and lR n \ U is positive. Let K be a simplex in ]Rn which contains U and is triangulated so that the diameter of any simplex in its triangulation is less than c. Consider the subcomplex L of K that consists of all simplices intersecting X. We have ILl c U because the triangulation is sufficiently fine. Let us write the exact sequence of simplicial homology groups for the pair (K, L): ... ---+

lh(K, L)

---+

Hk(L)

---+

Hk(K)

---+ ....

Rere Hk(K) = 0 because K ~ An, and Hk+I(K, L) = 0 because k + 1 ~ n + 1> 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.

1. Basic Definitions and Properties

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

4. Singular Homology

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.

1. Basic Definitions and Properties

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

4. Singular Homology

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.

1. Basic Definitions and Properties

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

4. Singular 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

1. Basic Definitions and Properties

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.

4. Singular Homology

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

1. Basic Definitions and Properties

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

4. Singular Homology

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.

1. Basic Definitions and Properties

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

4. Singular Homology

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.)

1. Basic Definitions and Properties

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

4. Singular Homology

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

4. Singular Homology

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

., . .

4. Singular Homology

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.

2. The Poincare and Lefschetz Isomorphisms

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).

4. Singular Homology

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

2. The Poincare and Lefschetz Isomorphisms

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



:0:0) Hn(Mn,Mn \ {x}jZm).

Here 'PI and 'P2 are the monomorphisms from the universal coefficient formulas. The homomorphism p. ® id is a monomorphism as well. On the other hand, if 0: is a generator of the group Hn(Mn), then (p. ®id)(o:® 1) = p.o: ® 1 = m ® 1 = O. 0 For every point x E Mn, consider the groups Tx = Hn(Mn, Mn \ {x}) ~ Z and T x (Z2) = Hn(Mn,Mn \ {X}jZ2) ~ Z2. Let us introduce topologies on the sets T = UxEMn Tx ar,d T(Z2) = UXEMn T x (Z2) so that the natural projections of T and T(Z2) onto Mn be coverings. Note that such topologizations are equivalent to defining local systems of coefficient groups isomorphic to Z and Z2 on Mn. For coefficients in Z2, the local system is trivial because there exists precisely one isomorphism Z2 --+ Z2. The space T(Z2) consists of two copies of the manifold Mnj one of them corresponds to the zero element of the group Z2, and the other corresponds to the nonzero element. We want to construct a local system of coefficient groups isomorphic to Z that generalizes the system Or M n for smooth manifolds. We say that a set V is admissible if V is the interior of a closed disk contained in some chart on Mn. To construct a topology on T, we use the observation that if a set V is admissible and x E V, then the inclusion

4. Singular Homology

232

Mn \ V ~ M n \ {x} is a homotopy equivalence. It follows that the homomorphism iv,x: Hn(Mn,M n \ V)

---+

Hn(Mn,M n \ {x})

= Tx

induced by this inclusion is an isomorphism (the proof is precisely the same as for simplicial homology; see Theorem 1.14 on p. 16). We declare the sets {iv,xa I x E V} to be open for all a E Hn(M n , Mn \ V). These open sets form a base for the topology of T. Note that the admissible sets V form a base for the topology of Mn. An orientation of a topological manifold M n is, by definition, a continuous section s of the covering T ---+ Mn such that every element sex) E Tz is a generator of the group Tx ~ Z. Clearly, any connected topological manifold without boundary has either none or two orientations. A topological manifold admitting an orientation is said to be orientable, and a topo'ttlgical manifold with a given orientation is oriented. An orientation of a topological manifold Mn can also be defined for the coefficient group Z2 as a continuous section s of the covering T(Z2) ---+ M n such that every element sex) E T x (Z2) is the generator of the group T x (Z2) ~ Z2. Clearly, such a section always exists and is unique because Z2 has only one generator. A homology class a E Hn(Mn) is said to be fundamental if for any x E M n , the map p.: Hn (Mn) ---+ Hn (Mn, Mn \ {x}) = Tx takes a to a generator of Tx. We shall see that there is a close relation between fundamental classes and orientations; namely, each orientation of a closed topological manifold determines a fundamental class. In particular, for a closed connected orient able manifold, we have Hn(Mn) ~ Z. But before proceeding to orient able manifolds, we consider the nonorientable case. Theorem 4.20. If Mn is a connected nonorientable topological manifold without boundary, then Hn(Mn) = o. Proof. Let a E Hn(Mn). For each point x E Mn, consider the image ax E Tx of a under the homomorphismp.: Hn(Mn) ---+ Hn(Mn, Mn\{x}) = Tx. The nonorientability of M n implies the existence of a path 'Y such that traversing it replaces ax by -ax. Therefore, ax = O. According to Corollary 1 of Theorem 4.19, a = O. 0

We use the notation ax introduced in the proof of Theorem 4.20 in the formulation and the proof of the following theorem. Theorem 4.21. Let M n be a closed connected orientable ma ifold endowed with an orientation s: M n ---+ T. Then there exists a untr.jU fundamental class a such that ax = sex) for all x E Mn.

233

2. The Poincare and Lefschetz Isomorphisms

Proof. The most important and difficult part of the theorem is the existence of a fundamental class. We start with this part, postponing the proof of uniqueness. We recall that if V is an admissible 6 set, then the map iv.% : Hn(M n , M n \ V) -+ T% is an isomorphism. This enables us to construct the required homology class Ov in Hn(Mn, M n \ V). Our goal is to glue such classes together by using the relative Mayer Vietoris sequence and obtain a homology class in Hn(M n , 0) = Hn(M n ). In dealing with Mayer Vietoris sequences, open sets are more convenient to handle because they automatically satisfy the excision axiom. Thus, instead of the admissible set V, we take its closure Vi the map iv.% is an isomorphism as well, and we can construct the class 0v E Hn(M n , M n \ V). Since the manifold Mn is compact, it can be covered by finitely many closures of admissible sets V 11 ••• , V m. Suppose that we have already constructed an element OI •...• k E

Hn(M n , M n \ (VI

u··· u Vk))

such that ixoI .... ,k = s(x) for all x E VI U··· U Vki here ix is the homomorphism of homology groups induced by the map of pairs

(Mn,M n \ (VI u··· u Vk))

-+

(Mn,M n \ {x}).

Consider the relative Mayer Vietoris sequence

Hn(M n , M n \ (VI u··· u Vk+d) ---+ Hn(M n , M n \ (VI u··· u Vk)) ffi Hn(M n , M n \ Vk+d ---+ Hn(M n , M n \ ((VI u··· u Vk) n Vk+d). Take an element of the form (OI •...• k, -ok+d in the direct sum. Let us prove that it is the image of some element OI •...• k+! E

Hn(M n , M n \ (VI u··· u Vk+!)),

or, equivalently, that the image of

(OI •...• k,

-Ok+!) in

Hn(M n , M n \ ((VI u··· u Vk) n Vk+d) vanishes. We denote this image by fJ. By assumption, for any x E C = (VI u··· u Vk) n V k+!, the homomorphism Hn(Mn, Mn \ C) -+ Tx takes fJ to zero. Lemma. Let U C M n be an open set, and let fJ E Hn(Mn, U) be an element such that the homomorphism Hn(Mn, U) -+ Tx takes fJ to zero for any x E U. Then fJ = o. 6Admissible sets were defined on p. 231.

4. Singular Homology

234

Proof. The proof of the lemma follows a scheme which has already been used many times in this book. It consists in proving the required assertion for simple domains and extending it by using the Mayer Vietoris sequence. First, suppose that the set Mn \ U is contained in a coordinate neighborhood W ~ ~n. Consider the commutative diagram

Here the vertical arrows are excision isomorphisms. The equality p*(3 - 0 is equivalent to i*(3 - 0, where iJ is the preimage of (3 under the excision isomorphism. Lemma 4.2 implies (3 - 0 and, therefore, (3 - O. Now we consider the general case. Let Vi be an open eoordinate neighborhood in M n homeomorphic to ~n. The compact set M n \ U can be covered by finitely many open sets Vi. Let U, = (Mn \ U) n Vi. Then M n \ U - U (Mn \ Ut ), i.e., U Ui . Thus, it is sufficient to show that if the assertion of the lemma holds for open sets U' and U", then it holds for U' n U". Consider the relative Mayer Vietoris sequence

n

Hn+I(M n , U'

u U")

- - Hn(M n , U'

It is easy to see that sequence of a pair

n U") - - Hn(M n , U') EB Hn(M n , U"). H n+l(Mn, U' u U") = o. Indeed, consider the exact

H n+l(M n ) - - Hn+l(M n , U' u U") - - Hn(U' U U"). Here Hn+l(M n ) 0 and Hn(U'uU") - 0 if U'UU" i- M n (if U'UU" _ M n , then the equality H n +l(Mn , U' U U") = 0 is obvious). Thus, we have a monomorphism Hn(M n , U'

n U")

-+

Hn(M n , U') EB Hn(M n , U").

Suppose that it takes (3 to ((3', (3"). Let (3x, (3~, and (3~ be the images of (3, (3', and (3" under the map to Tx· Then (3~ = (3x and (3~ = -(3x. Therefore, if (3x - 0, then (3~ - 0 and (3~ = o. By assumption, (3x = 0 for all x E Mn \ (U' n U"). Therefore, (3~ = 0 for all x E M n \ U' and (3~ = 0 for all x E Mn \ U". It follows that (3' = 0 and (3" = O. SincE' the map under consideration is a monomorphism, we have (3 = O. 0 The element 01, ... ,k+l that is mapped to (Ol, ... ,k. -Ok+r) has the required property, i.e., ixaI, ... ,k+l = s(x) for all x E VI U··· U Vk+l. The uniqueness of the fundamental class is very easy to prove. If 0 and a' are two fundamental classes corresponding to the same orientation s, then, for each x E Mn, the homomorphism Hn(Mn) _ Hn(M n , M n \ {x})

2. The Poincare and Lefschetz Isomorphisms

235

takes the cla..<;s a - a' to s(x) - s(x) - O. But this homomorphism is a monomorphism (see Corollary 1 of Theorem 4.19). D Any manifold admits a unique orientation with respect to the coefficient group Z2. The argument from the proof of Theorem 4.21 applies to the group Z2 without any changes; using it, we can construct the fundamental class in Z2-homology for any closed connected topological manifold. 2.2. The Thorn Isomorphism. The Thom isomorphism relates the kdimensional cohomology of a closed manifold M n to the (n + k )-dimensional cohomology of the pair (Mn x M n , M n x M n d( M n )), where d: Mn ---. M n x M n is the diagonal map. The construction of this isomorphism is based on Lemmas 4.3 and 4.4 below. We start by introducing some notation. Let M n be a closed oriented topological manifold with an orientation s: M n ---. T; by PI, P2: Mn x M n ---. M n we denote the projections onto the first and second factors. For every point x E M n , consider the map lx: M n ---. Mn x M n defined by lx(Y) - (x, y). This map induces a map of pairs

which we denote by the same symbollx . Lemma 4.3. Suppose that V c M n is an open neighborhood contained in a disk (in local coordinates), which, in turn, is contained in a disk D n of larger radius centered at the same point Xo E V. Then there exists a homeomorphism (): pil (V) = V x Mn ---. PI I (V) with the following properties:

(i) ()(x, y) = (x, y') for all x E V and y E M n (here y' is some point of the manifold Mn); (ii) ()(x, x) = (x, xo) for all x E V; (iii) (P2()lx) .. (S(x)) = s(xo) for all x E V (here we mean maps of pairs because s(x) is the homology class of a pair). The statement of (iii) uses (ii): (p2()lx)(x)

= P2()(X, x) = P2(X, xo) = Xo.

Proof. Consider the self-homeomorphism of V x Dn defined by the condition that its restriction to each line segment with endpoints (x, x) and (x, y), where x E V and y E aDn, is a linear map onto the segment with endpoints (x, xo) and (x, y). Figure 4 shows the restriction ofthis self-homeomorphism to {x} x D n for n = 2. On V x aDn, this map is the identity; therefore, it can be extended over V x Mn by the identity map outside V x Dn, The composition P2()lx coincides with the map shown in Figure 4; clearly, it is orientation preserving. D

4. Singular Homology

Figure 4. The construction of the homeomorphism

For any open set V

c

M

n,

the pair

(p-l(V),p-l(V) \ d(Mn))

= (V x M n , V x

M

n \

d(Mn))

is defined; we denote it by V x . In what follows, an important role is played by the pair M nx = (Mn x Mn, M n x Mn \ d(Mn)). Note that the map l:z: can be regarded as a map of pairs (M n , M n \ {x}) -+ M n x . Lemma 4.4. (a) For i < n, Hi(Mnx) = O. (b) The isomorphism Ho(Mn) ~ Hn(Mnx) holds; it is induced by the map taking the O-chain corresponding to a point x E M n to the relative

n-cycle representing the class (l:z:).(s(x)). Proof. We apply Lemma 4.3. Suppose that a set V satisfies the conditions of the lemma and () is the corresponding homeomorphism. Then () takes d(V) to V x {xo}; therefore, it induces a homeomorphism of pairs VX

= (V x M n , V x

Mn

\

d(Mn))

-+

V x (M n , M n

\

{xo}).

Thus, H.(V X) ~ H.(V x M n , V x (Mn \ {xo})). Applying the Kiinneth theorem for relative homology, we see that the group Hk(VX) is isomorphic to the direct sum of the groups

EB

Hp(V) ® Hq(Mn,Mn \ {xo})

= Ho(V) ® Hk(Mn,M n \ {xo})

p+q=k and

EB

Tor(Hp(V), Hq(Mn, Mn\{xo})) =Tor(Ho(V), Hk_l(M n , Mn\{xo} »). p+q=k-l But Hi(Mn, M n \ {xo}») = 0 for i < n. Therefore, Hk(V..c:) = a for i < n, and Hk(VX) ~ Ho(V) ® Hk(Mn, Mn \ {xo}).

2. The Poincare and Lefschetz Isomorphisms

237

Consider the diagram

Ho(V) ® Hn(M n , Mn \ {xo})

Hn(V, V \ {x})

(I z ). ~

Hn(VX)

6. ~

r~ 1

Hn(V x M n , V x (Mn \ {xo})) (P2).

Hn(M n , M n \ {xo}). We know that (P20lx) .. (s(x)) = s(xo). Hence the map Ho(V) - Ho(V) ® Hn(Mn, M n \ {xo}) induced by the map [xl - [xl ®s(xo) is an isomorphism. Returning to Hn(V X), we obtain the map [xl - (lx) .. (s(x)) instead of this isomorphism. Thus, the assertion of the lemma is true for the set V. Note that we have not used the connectedness of Vj the isomorphism Hn(V X) ~ Ho(V) holds even for disconnected open sets V. It remains to show that if the assertion is true for open sets U1, U2, and Ul nu2, then it is also true for U1 UU2. We take two segments of the Mayer Vietoris sequences and map them to one another by the homomorphism described in the statement of the lemma:

HO(UI n U2)

--~)

HO(Ul) ffi H O(U2)

---~)

Ho(Ul U U2)

111

Hn((Ul n U2)X)

~

Hn-l(U{") ffi H n-l(U2X) ~ Hn-1((Ul

U

U2)X).

To this diagram we can add ------to)

0

-------~)

1

0

1

on the right without violating commutativity. The homology groups in the lower row are trivial. Applying the five lemma, we obtain the required result. 0 By Lemma 4.4 and the universal coefficient theorem, Hi(Mnx) = 0 for i < nand Hn(Mnx) ~ Hom(Hn(MnX),Z) ~ Hom(Ho(Mn),Z)j both isomorphisms are canonical. The class U E H n (Mn X) corresponding to the homomorphism Ho(Mn) - Z induced by the augmentation is called the Thorn class of the topological manifold Mn. The Thorn class is uniquely determined by the condition that (U, (lx) .. (s(x)) = 1 for all x E M n , i.e., (l;U, s(x)} = 1. This means, in particular, that l;U is a generator of the group Hn(Mn,M n \ {x}).

238

4. Singular Homology

For X M n x M n and A - M n x Mn\d(Mn), we have U E Hn(x, A). For a cohomology class a k E Hk(Mn), the class pi(a k ) belongs to Hk(X)j therefore, we can consider the relative cup product Hk(X) ® Hn(x, A) --+ Hn+k(x, A). We set <J>·(a k ) - pi(a k ) '-' U E Hn+k(Mnx).

Theorem 4.22 (Thorn isomorphism). For any k, the map <1>.: Hk(Mn) Hn+k(Mnx) is an isomorphism.

--+

Proof. For every open set V c M n , let Uv E Hn(vx) be the restriction of the cohomology class U to V. Consider the cap product Hn(vx) ® Hk(VX) --+ Hk n(V x Mn). For (3k E Hk(V X), we set

CP.({3k)

(pd.(Uv ,..... (3k) E Hk n(V),

Let V be an open set satisfying the conditions of Lemma 4.3. Then the homeomorphism 0 from this lemma induces an isomorphism

0·: Hn(v x M n , V x (Mn \ {x}))

--+

Hn(vx).

Using Theorem 4.15 on p. 217, we can identify the element O· I(UV) with 1 ®,n E HO (V) ® H n (Mn , M n \ {x} ), where ,n is the generator of the group Hn(M n , Mn\ {x}) such that (,n, s(x)) - 1 (in the notation of Theorem 4.15, ,n = en: we identify the groups Hn(M n , M n \ {x}) and Hn(lRn,lR n \ {a}) by means of the excision isomorphism). The homomorphism 0 has the property PI = piO (on the set V). Therefore, the naturality of the cap product implies

(pt).(Uv ,..... (3k) - (pt).O.(O·(O·-l(UV)) ,..... (3k) - (pd.(O· l(UV)""" (}.({3k)). The element (}.({3k) belongs to Hk(V x Mn, V x (Mn \ {x})). According to the Kiinneth theorem, we can identify it with bk n ® s(x), where bk n E

Hk n(V). The homomorphism (pt). takes the cap product of 1 ®,n E HO(V) ® Hn(M n , Mn\{x}) andbk n s(x) E Hk n(V)®Hn(Mn,Mn\{x}) tobk-n. This means that CP.({3k) - bk n' The map (3k 1-+ bk n is the composition of isomorphisms

Hk(VX) ~ Hk(V x M n , V x (Mn \ {x})) ~ Hk n(V) ® Hn(Mn,M n \ {x}) ~ Hk n(V), Thus, cp.: Hk(VX)

--+

Hk n(V) is an isomorphism.

Using the standard argument involving the Mayer Vietoris sequence, we obtain an isomorphism cp.: Hk(Mnx) --+ Hk_n(M n ) for each k > n. We pass to the level of chains and construct the dual map of cochains for cp •. Then, from the map of cochains we pass to the map cp.: Hk(Mn) --+

2. The Poincare and Lefschetz Isomorphisms

239

Hn+k(Mnx) of cohomology groups. Let us show that cp* pi(a k ) '-' U. Indeed,

= ~*, i.e., cp*(a k )

(cp*(ak),/h+n) - (ak,cp*(.Bk+n)) - (ak,(pt)*{U .--....Bk+n)) - (Pi (a k ), (U .--... .Bk+n)) (pi(a k ) '-' U, .Bk+n); the last equality follows from (aq,.B'P .--... l'p+q) - (a q .B'P, l'p+q) (see Theorem 2.6 on p. 71). It remains to verify that cp* is an isomorphism. The isomorphism 'P* induces the commutative diagram

o ~ Ext(Hk

l(Mn),Z)

l~·l 0------+ Ext(Hk+n

1 (Mnx),

--4)

Hk(Mn)

--4)

Hom(Hk(Mn),z)

l~·

~

0

1~·2

Z) ------+ Hk+n(Mnx) ------+ Hom(Hk+n(Mnx), Z) ------+ o.

Here 'P*l and CP*2 are isomorphisms becaufle Cp* is an isomorphism. According to the five lemma, cp* is an isomorphism as well. D

2.3. The Poincare Isomorphism. To prove the Poincare isomorphism theorem for topological manifolds, we need the following two properties of closed topological manifolds. Theorem 4.23. (a) Any closed topological manifold M n can be embedded in IRk for some k.

(b) If I: Mn --+ jRk is an embedding of a closed topological manifold Mn, then there exists an open set U ::J J(Mn) such that J(Mn) is its retract. Proof. (a) Let Ul , ... , Um be open sets homeomorphic to IRn and covering Mn. Take homeomorphisms hi: Ui --+ jRn ~ sn \ {yo} and consider the map Ii: M n --+ sn defined by if x E Ui , if x ¢ Ui. The map Ii is continuous because limx--+8u. I(x) = yo. The maps /I, ... , 1m can be treated as a single map I: M n --+ Sf C jRm(n+l). Clearly, if x =I- x', then I{x) =I- I(x'). Since Mn is compact and jRm(n+l) is Hausdorff, it follows that I is a homeomorphism from Mn to J(Mn) C jRm(n+l).

n:l

(b) For brevity, we denote the image of the manifold Mn in IRk by the same symbol Mn. Choose a sufficiently large simplex t:1 k in jRk so that the manifold M n is contained strictly inside t:1 k. Let us construct an (infinite) triangulation of the space t:1 k \ Mn as follows. First, we take the barycentric subdivision of t:1 k and choose the simplices intersecting M n in this subdivision. Then, we take the barycentric subdivisions of the chosen

4. Singular Homology

240

simplices and leave the remaining simplices intact. Among the new simplices we chose those intersecting M n , and so on. Since M n is compact, it follows that the distance from any point x E ~k \ M n to M n is positive; therefore, each point x has a neighborhood intersecting only finitely many simplices constructed above. This means that what we have constructed is indeed a triangulation K of the space ~k\Mn.

Now, let us construct a family of subsets No C Nl C ... of ~k and continuous maps ri' N, --+ Mn such that M n C N, and the restriction of each ri to M n is the identity map. For No we take the union of M n and the vertices of the simplicial complex K. The map ro takes each point x E No to a point of M n nearest1 to x. In particular, ro (x) - x for all x E Mn. The sets N, and maps r, are constructed by induction on i. Pas~n,.,from Ni to Ni+l' we add some (i+l)-simplices from the complexfo; these are the simplices ~~+l C K such that their boundaries are contained in Ni and the maps a~~+l ~ Mn can be extended to continuous maps ~~+l --+ Mn. The map ri+l on ~~+l is defined as follows. Consider all continuous extensions cp: ~~+l --+ Mn ofri. Let 8(cp) be the diameter of the set cp(~~+l). Clearly, 8(cp) is not smaller than the diameter of ri(a~~+l). Let a be the greatest lower bound of the set of all numbers 8(cp). If a = 0, then ri takes a~~+l to one point Xa E Mn. In this case, we set ri+l(~~+l) = Xa' If a> 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

2. The Poincare and Lefschetz Isomorphisms

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,

4. Singular Homology

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),

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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.

4. Singular Homology

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)

2. The Poincare and Lefschetz Isomorphisms

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.

4. Singular Homology

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

4. Singular Homology

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

2. The Poincare and Lefschetz Isomorphisms

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

4. Singular Homology

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

4. Singular Homology

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

4. Singular Homology

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..<;ses: Continuation

for which there exists a homeomorphism h: U p'lu,h - pu·

--t

U' such that hi

=

i' and

Theorem 4.30. The tangent bundle of a closed smooth manifold M is equivalent to its tangent micro bundle. Proof. Consider a Riemannian metric on the tangent bundle. For each tangent vector v E TxM, there exists a unique geodesic 'Yv(t) such that 'Yv(O) = x and ct'Yv(O) - v. Consider the map expx: TxM --t M that assigns the point 'Yv(l) to each tangent vector v. Since M is compact, we can choose a number c > 0 such that for each x EM, the restriction of the map expx to the open ball in TxM that consists of vectors of length less than c is a diffeomorphism from this ball to some open neighborhood of x in M.

Let U be an open neighborhood of the zero section in T M consisting of vectors of length less than c. The map h: U --t M x M defined by hex, v) = (x, expx(v)) is a diffeomorphism from U to a neighborhood U' of the diagonal d(M) in MxM. This map h has all the required properties. 0 The Thom cla..<;s U E Hn(M x M, M x M \ d(M)) is characterized by the condition that for any point x E M, the map of pairs

Ix: (M,M\ {x})

--t

(M x M,M x M\d(M))

defined by Ix(Y) = (x, y) takes U to a generator ofthe group Hn(M, M\ {x}), i.e., I;U generates Hn(M, M \ {x}). By analogy, the Thom class of an ndimensional microbundle p: E --t B is defined as a cohomology class U E Hn(E, E \ i(B)) with the property that for any point b E B, the natural embedding

ib: (P-l(b),p-l(b) \ i(b))

--t

(E, E \ i(B))

takes U to a generator of the group Hn(p-l(b),p l(b) \ i(b)). Now we consider the Thom classes of real vector bundles. The following theorem is valid. Theorem 4.31. Let ~ be an oriented real n-dimensional vector bundle over a compact CW -complex B. Suppose that E = E(~) is the total space of this bundle and Eo is an open subset of E consisting of all nonzero vectors. Then (a) there is a unique integral cohomology class Ut;, E Hn(E, Eo) such that for any bE B, the cohomology class ibUt;, is the generator of H n (lRn , JRn\ {O}) determining a given orientation of the fiber JRn j (b) Hi(E, Eo) = 0 for i < n;

4. Singular Homology

(c) the map Hi(B) --+ Hi+n(E, Eo) defined by 8 a isomorphism for each i.

1-+

p*(a) ........, Ue is an

Proof. If the bundle is trivial, i.e., E = B x R.n, Eo = B x (R.n \ {O}), and p: E --+ B is the projection to the first factor, then all the required assertions are in fact contained in Theorem 4.15 on p. 217 (in the notation of this theorem, X = B, A = 0, and en = Ue). Thus, it suffices to show that if the required assertions hold over open sets U, V, and U n V, then they also hold over U U V. We set El = p-l(U), E2 = p-l(V), E3 = p-l(U n V), and E4 = p 1 (U U V). Let us write the Mayer Vietoris sequence: Hi 1(E3, E3) --+ H'(E 4 , E~) --+ H'(El, EJ) Ee H'(E 2, E3) --+ H'(E 3,E3). For i < n, the groups on the left and right of Hi(E4, E~) are trivial;-otherefore, Hi(E4, E~) = O. This proves (b). Uniqueness implies tl)at any classes Ul E Hn(El,EJ) and U2 E Hn(E 2,E3) determine the same element of the group H n (E 3, E3). Hence there exists a class U4 E Hn(E4, E~) whose projections are Ul and U2. The equality H n - 1(E 3, E3) = 0 implies the uniqueness of such a class. This proves (a). To prove (c), consider a similar Mayer Vietoris sequence for the bases of bundles and the map from this sequence to that written above that takes each element a to p*(a) ........, U. Applying the five lemma, we obtain the required result. 0 For coefficients in Z2, an analog of Theorem 4.31 for nonorientable bundles ~ is valid. The Gysin Sequence. Let ~ be an oriented vector bundle of dimension n with projection p: E --+ B. Consider the exact cohomology sequence of the pair (E, Eo): ..• --+

Hi+n(E, Eo)

L

Hi+n(E) ~ H'+n(Eo) ~ Hi+n+1(E, Eo)

--+ • . . .

Since the map p is a homotopy equivalence, it follows that p*: Hi+n(B) --+ H'+n(E) is an isomorphism. We also have the Thom isomorphism cp*: Hi(B) --+ Hi+n(E, Eo). Using these isomorphisms, we obtain the exact sequence ••• --+

H'(B)

p. IJ.'P,

Hi+n(B) ~ Hi+n(Eo) ~ Hi+1(B)

--+ ••.•

This exact sequence is called the Gysin sequence [46]. In the Gysin sequence, the map i*p* is induced by the restriction of p to Eo. Moreover, the map p*-lj*cp can be expressed in terms of the Euler class e(~) as p*-lj*cp(a) = a '-" e(~). Indeed, we have p 1 j*cp(a) = 8The relative cup product of the classes p. (0) E H'(E) and UE E Hn(E, Eu) belongs to

Hi+n(E, Eo).

3. Characteristic Classes: Continuation

257

p* Ij*(p*(a:) '-' U~) = p* l(p*(a:) '-' j*(U~)) = a '-' (p* Ij*(U~)). It remains to verify that p*-lj*(Ud = e({), i.e., j*(U{) = p*e({). Let]Rn be the fiber over some point of the base. Consider the co chain in E which assigns 0 to any simplex Lln c ]Rn \ {O} and ±1 to any simplex Lln c ]Rn containing the origin (the sign depends on whether the orientation of the simplex is compatible with that of the fiber). The cohomology class of such a chain is determined uniquely, and it coincides with both j*(U{) and p*e({); this follows directly from the definitions.

In the case of a nonorientable vector bundle, the Gysin exact sequence holds for cohomology with coefficients in Z2; in this case, the Euler class e({) is reduced modulo 2, i.e., replaced by w n ({). The exact Gysin sequence holds also for any oriented9 locally trivial Sn-l-bundle p: E ---. B because any such bundle can be associated with an oriented n-dimensional vector bundle. Problem

120. Calculate the cohomology groups of the manifold

G+(2n + 2,2) using the bundle V(2n

+ 2, 2)

81

~

G+(2n + 2, 2).

3.2. The Thorn and Wu Formulas. In this section, we consider cohomology with coefficients in Z2. Suppose that { is a real bundle over a compact simplicial complex B, U{ is its Thorn class, and cp: Hi(B) ---. Hi+n(E, Eo) is the Thorn isomorphism defined by cp(a:) = p*(a) '-' U~.

Theorem 4.32 (Thorn [137]). The Stiefel Whitney class Wi({) is equal to cp l(SqiU{), where Sqi is the Steenrod square. Proof. Let us show that cp-l(SqiUd satisfies all of the conditions determining the ith Stiefel Whitney class. Clearly, cp-l(SqOU{) = cp-l(Ud = l. Moreover, if i > 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.

4. Singular Homology

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

4. Singular Homology

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).

4. Singular Homology

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

5. Cech and de Rham Cohomology

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
Theorem 5.5. Let X be a normal topological space, and let Y be its compact subspace. Then Jik(y; G) ~ lim Jik(U; G), where U is the cover ofY by open sets in X (not in Y, as in the definition of Cech cohomology). ~

Proof. Let U be a cover of Y by open sets in X. Take its finite sub cover {Ut, ... , Un} and consider the cover of Y by the sets U, nY, which are open in Y. Any subspace of a normal space is Hausdorff, and any compact Hausdorff space is normal. Therefore, the space Y is normal, and the cover {UinY} has an open refinement {~} such that Vi c UinY. Here Vi is the closure of Yi in Y; it is also the closure of Yi in X because Y is closed in X. Let us construct open sets WI,"" Wn in X such that Vi c Wi, Wi CUi, and the nerves of the covers {V t, ... , V n} and {W 1. ... , W n} coincide, i.e., Vii n ... n Vip = 0 if and only if W il n .. , n Wi p = 0. Let CI be the union of all sets of the form Vii n ... n Vip disjoint from V 1. The sets V 1. Ct, and X \ UI are closed in X, and the sets VI and C I U (X \ UI) are disjoint. Therefore, we can choose an open set WI in X so that Vi C Wi and WIn (CI U (X \ Ud) = 0; the latter condition means that WIn CI = 0 and WI CUI. Let us show that the nerves of the covers {WI, V2, ... , V n} and {VI, ... , V n } coincide. Clearly, if WI n Vii n ... n Vip = 0, then VI n Vii n ... n Vip = 0 because VI C WI C WI. Now suppose that VinViIn···nVip = 0. Then Vi1n· ··nVip c CI. But WInCl = 0; hence WI n Vii n··· n Vip = 0. Applying the same procedure to WI, V2,.··, V n, we construct W 2 , and so on. Let us show that the nerve of the cover of Y by the sets WI,.'" W n , which are open in X, coincides with that of the cover by the sets WI n Y, ... , Wn n Y, which are open in Y. Clearly, if Wi l n ... n Wi p = 0, then W il n··· n Wip n Y = 0. Suppose that W il n··· n W ip n Y = 0. Then Vii n .. , n Vip = 0, or, equivalently, W il n ... n Wi p = 0. Therefore, W il n ... n Wi p = 0. Thus, we have shown that the cover {WI, ... , W n } of the space Y by open sets in X, whose nerve coincides with that of {WInY, ... , WnnY}, is cofinal in the family of covers of Y by open sets in X. According to Theorem 5.2, we have ~iIk(U;G) ~ ~Hk(W;G). By construction, iIk(W;G) ~ iIk(W n Y; G); here W n Y = {WI nY, ... , Wn n Y}. Moreover, the cover

5. tech and de Rham Cohomology

270

W n Y is cofinal in the set of all open covers of Y. This implies the required isomorphism. 0 In calculating Cech cohomology, the following assertion is often useful. Theorem 5.6. Let X be a compact triangulable 1 space, and let Y n be a subcomplex of some triangulation of X. Suppose that IY11 ::J IY2 1 ::J ..• and n~=llYnl = Y. Then iIk(Y;G) ~ ~Hk(Yn;G). Proof. Let Xn be a triangulation of X for which some2 subdivision Yn is a full subcomplex (Le., any simplex in Xn spanned by some vertices of Yn is a simplex in Yn). We can choose this triangulation to be so fine that (i) the diameter of any simplex in Xn is less than lin (we assume that X is embedded in Euclidean space) and (ii) the star of each vertex of Xn is contained in the star of some vertex of Xn-l. We endow Yn"tlit.h the triangulation induced by X n .

Consider two covers Un and U~ of Yn ; Un is the cover by the stars in Yn of all vertices of the complex Yn , and U~ is the cover by the stars in Xn of all vertices of the same complex Yn . The first cover is open in Yn , and the second is open in X n . Condition (ii) allows us to construct a simplicial approximation h n : Yn - Yn- 1 of the inclusion IYnl C IYn-ll. The nerve of the cover Un can be identified with Yn ; thus, we obtain a simplicial map h n : N(Un ) - N(Un-d and the induced homomorphism (which is not necessarily an isomorphism) (h n )*: iIk(Un_d - iIk(Un ); we omit the coefficient group G in thE' notation. Thus, ~Hk(Yn) ~ ~iIk(Un)' For the covers U~, condition (ii) implies U~ < U~_l; thus, a homomorphism (h~)*: iIk(U~_l) _ iIk(U~) arises. Let us show that the nerves of the covers Un and U~ coincide, and therefore the homomorphisms (h n )* and (h~)* coincide as well (after the nerves are identified). Clearly, st(v, Yn ) C st(v, Xn); therefore, if n st(Vi, Yn ) =I 0, then also n St(Vi' Xn) =1o. Suppose that n st(Vi, Xn) =1o. This means that the vertices Vi of a simplex in Xn span a simplex in Yn . Moreover, the Vi are vertices of the complex Yn , and the simplicial subcomplex Yn is full by construction. Therefore, the vertices Vi span a simplex in Yn , i.e., nst(Vi, Yn ) =10. Thus, ~iIk(Un) ~ ~iIk(U~). To apply Theorem 5.2, we must show that the covers U~ > 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.

5. Cech and de Rham Cohomology

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

5. tech and de Rham Cohomology

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.

5. Cech and de Rham Cohomology

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
+ w([cp~o, ~i], ... , ti, ... , ~k) = cp~i(W(cpeO, . .. , ti, ... , ek)) + ~i(cp)W(cpeo, . .. , {, ... , ~k) + cpw([eo, ei], ... ' ti, . .. , ek) - ei(cp)W(cpeo,· .. e,,···, ~k) = cp(ei(W(eO, ... , ti, ... , ek)) + w([eo, eil,···, ti,.··' ek))·

2. De Rham Cohomology

277

Let us introduce local coordinates Xl, ... , Xn and calculate the differential of the form W = cp(x) dXil 1\ ... 1\ dXile' where il < ... < ik. We set {o = ~, ... , {k = ~ for jo < ... < jk· The vector fields {i commute, X'Q ]Ie i.e., [{i, {j] = O. Therefore, k

dw({O,· .. ,{k)

a

= L(-l)P ax. p-O

~

(W({O, ... ,{p, ... ,{k)).

3"

e

Here W({O, . .. , p, ... , {k) = cp(X) if il = jo, ... , ip = jp-l, ip+1 = jp, ... , ik = jkj in all the other cases, we obtain O. We have jp-l < jp < jp-tl. Hence

a

~

(-l)P-a (w({o, ... ,{p, ... ,{k)) xi"

Using this expression, we can easily verify that dod = OJ moreover, if WI is a k-form, then d(WI 1\ W2) = dwl 1\ W2 + (_I)kwl 1\ dw2. Note also that the operation d is local in the sense that, at each point x, the form dw is completely determined by the restriction of W to an arbitrarily small neighborhood of x. The commutator of two vector fields is also a local operation in this sense. A differential form W is said to be closed if dw = o. The differential of a form W is exact if W = dw' for some form w'. The coset space of the set of closed k-forms on the manifold M n modulo the exact k-forms is called the kth de Rham cohomology space and denoted by H~R(Mn). Example 70. HgR(JR) ~ JR and H6R(JR)

= o.

Proof. For JR, there is only one nontrivial differential d: nO(JR) --+ nl(JR), which is defined by cp ~ ~ dx. Clearly, Ker d consists of constant functions. Therefore, HgR(JR) = Kerd ~ JR. It is easy to show that Imd = nl(l~). Indeed, any I-form on JR can be represented as {dx, where { E CCXl(JR). We set cp(x) = {(t) dt for X ~ 0 and cp(x) = I~ {(t) dt for X ~ o. We have dcp = {dx. Thus, H6R(JR) = nl(JR)/Imd = n l (JR)/nl (JR) = o. 0

Ie:

Example 71. If M n is a connected manifold, then HgR(Mn) ~ JR.

5. Cech and de Rham Cohomology

278

Proof. The kernel of the homomorphism d: nO(Mn) --+ nl(Mn) consists of the functions satisfying the condition .EY!.... = ... = .!!£.. = 0 in each coordinate aXl OXn neighborhood. Such functions are constant on each connected component of the manifold. 0

r:

It is easy to see that the map nk(Nm) --+ nk(Mn) induced by a smooth map I: M n --+ N m is a chain map, that is, drw = r dw. Indeed, at vector fields ~o, ... , ~k' both forms take the same value

L

(_l)i~iW(J",~O, ... , t,,···, I"'~k)

+ L (-1)i+;w(J"'[~i'~3],··· ,t,,·.· ,t;, ... ,I"'~k)' 1<3

Theorem 5.9. II M n = UI U U2, where Ul and U2 are open sets, then the Mayer Vietoris exact sequence --+

H~R(Mn) -----. H~R(Ul)ffiH~R(U2) -----. H~R(UlnU2) -----. H~1tI(Mn)

--+

holds.

Proof. It is sufficient to construct an exact sequence of groups of co chains 0-----. nk(Mn) ~ nk(Ul) ffi n k (U2) ~ nk(Ul

n U2) -----. O.

Let W E nk(Mn). We set i"'(w) = (iiw, i 2w), where iiw and i 2w are the restrictions of w to Ul and U2, respectively. Take WI E nk(ut) and W2 E n k (U2). We set j"'(Wl,W2) = jiwl - j2w2, where jiwi is the restriction of Wi to Ui. The injectivity of the homomorphism i'" is obvious because any form on Mn is uniquely determined by its restrictions to U1 and U2. Let us show that j'" is an epimorphism. Let {~1' ~2} be a smooth partition of unity subordinate to the cover {U}, U2}. Then the form w E nk (UI n U2) can be represented as ii ~2W + iPIWj here ~2W is a well-defined form on Ul because ~2 = 0 on UI \ (UI n U2). Finally, let us verify that Ker j'" = 1m i"'. The equality J"'i'" = 0 implies Ker j'" ::J 1m i"'. Suppose that (WI, W2) E Ker j"', i.e., jiWl = j2W2. This means that Wl(X) = W2(X) for all x E U1 n U2. Hence there exists a form W on M n such that it is equal to WI on U1 and to W2 on U2. Clearly,

i"'w

=

(WI, W2).

0

For a noncompact manifold M n , it is often useful to consider, in addition to the algebra n"'(Mn) of differential forms, the algebra n~(Mn) of compactly supported differential forms (that is, forms identically vanishing

2. De Rham Cohomology

279

outside some compact sets). The definition of the differential d for compactly supported forms is the same as for general forms, and dod = O. Therefore, for compactly supported forms, cohomology can be constructed; it is denoted by H~(Mn). If the manifold M n is compact, then its cohomology H~ coincides with H But if Mn is noncompact, then this cohomology is different.

rJR .

Example 72. If Mn is a connected noncompact manifold, then H~(Mn) =

O. Proof. The kernel of the homomorphism d: n~ (Mn) -+ n~ (Mn) consists of compactly supported constant functions. But since Mn is noncompact (and connected), there exists only one such function (identical zero). 0 Example 73. The isomorphism

HJ (JR) ~ JR holds.

Proof. A form .,p(x) dx (with compact support) belongs to the image of the homomorphism d: n~(JR) -+ n~(JR) if and only if the function 'P(x) = J~oo .,p(t) dt has compact support, i.e., J~oo .,p(t) dt = O. Consider the map n~(JR) -+ JR defined by .,p(x) dx 1--+ J~CXl.,p(t) dt, which clearly is an epimorphism. Its kernel coincides with 1m d. Therefore, n~(JR)/Imd ~ JR. 0 There is a fundamental difference between cohomology with compact supports and the ordinary de Rham cohomology; it manifests itself even in the case of compact manifolds. Namely, in the Mayer Vietoris exact sequence for compactly supported cohomology, the maps act in the reverse direction, that is, they behave as in the case of homology rather than cohomology. This is so because the operation of restriction to an open set is not defined for compactly supported forms. Indeed, if the compact support of a form contains an open set, then the restriction of this form to this set is not compactly supported. In return, for compactly supported forms, the operation of extension to a set containing a given open set U is defined. Since any point outside U has a neighborhood disjoint from the compact support K c U, we can extend any form with support contained in U by setting it to zero outside U.

Theorem 5.10. If Mn = U1 U U2 , where U1 and U2 are open sets, then the following Mayer Vietoris exact sequence holds: ••. ----+

H:(M n )

----+

H:(UI n U2) ----+

H:(Ud EEl H k (U2)

----+

H:+1(Mn)

----+ .•••

5. Cech and de Rham Cohomology

280 Proof. Consider the exact sequence

o ~ n~(UI n U2)

£

n~(Ud ffi n~(U2)

.!L n~(Mn) ~ 0

defined as follows. Let W E n~(UI n U2). We set j*w = (jiw, -j2w), where j;w is an extension of w to Up. For two forms WI E n~(Ud and W2 E n~(U2)' we set i*(Wb W2) = iiwi + i;W2, where i;wp is an extension of wp from Up to the entire manifold. The injectivity of j~ and exactness at the middle term are obvious. Let us verify the surjectivity of i~. Take W E n~(Mn) and let PI, >'2} be a smooth partition of unity subordinate to the cover {UI, U2}. The support of the form >'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<. If {V.a} is another orientation atlas and {JL.a} is a partition of

281

2. De Rham Cohomology

unity subordinate to it, then the equality L(j J.L(j = 1 implies La IUa AaW = La,(j AaJ.L(jW. Since the support of AaJ.L(jW is contained in Ua n V(j, we have IUa AaJ.L(jW = IVIJ AaJ.L(jW. Hence La IUa AaW = La,(j IVIJ AaJ.L(jW = L(j IVIJ J.L(jw because La Aa = 1. This means that the integral of the form W over the manifold Mn is well defined.

Iua.

The Stokes Theorem. The Stokes theorem says that IMn dMJ = ± IaMn W for any form W E n~(Mn). Formally, in the second integral we should write i*w instead of w, where i: aM n --+ M n is the natural embedding; in other words, i*w is the restriction of W to aM n . The sign ± before the second integral can be removed by imposing the requirement that the orientation aM n must be induced by that of Mn. Let us prove the Stokes theorem. Case 1. M n

= lRn.

Let W = cpdXl /\ ... /\ dXn-l E n~-l(lRn). Then dMJ = ... /\ dXn-l = (-1)n- 1 dXl/\'" /\ dXn. Therefore,

1t

Ln

dMJ

= (_l)n-l

J([: :::

It dX n /\ dXl /\

dXn) dXl' ··dXn-l·

But I~oo /!; dXn = cp(XI, ... , Xn-I. 00) - CP(Xl,"" Xn-l, -00) the function cp is compactly supported. Case 2. M n

Li

= lR+. = {(Xl, ... , Xn)

E lR n

I Xn

because

o}.

Let W = Li CPi dXI /\ ... /\ dxi /\ ... /\ dX n E n~-l(lRn). Then dMJ (_l)n-l~ dXI/\"'/\ dx n . Therefore,

{n dMJ =

llR+

L

l~i
(_l)i-l

(1

00

-00

t=

n, then I~oo ~ dXi integral

=

acp~ dXi) dXl ... dxi ... dX n ax,

+ (_l)n-l If i

~

=0

(1

= 0, as in Case

00

:~: dXn)

1. For i

dXl ... dXn-l.

= n, we obtain the

282

5. Cech and de Rham Cohomology

the last equality holds because the form dX n vanishes on aIR+., and therefore the restriction of w to aIR+. contains only one term, r.pn dXl/\ ... /\ dXn-l. There is no sign ( _1)n ifthe orientation of aIR+. is determined by the form (_I)n dXl/\· .. /\ dXn-l. This condition can be stated differently as follows. Suppose that a basis el, ... , en-l determines the positive orientation of the boundary and E is the outward normal to the boundary. Then the basis E, el. .. . , en-l determines the positive orientation of the manifold. Indeed, we have E = -en, and hence the basis E, el, ... , en-l has the same orientation as (-I)nel. ... ,en-l.en. In what follows, we assume that the orientation of aIR+. satisfies this condition and omit the sign from the Stokes formula. Note that on p. 96, we were led to the same condition on the orientation of the boundary by different considerations. Case 3. M n is an arbitrary oriented manifold. The passage from IRn and IR+. to an arbitrary manifold is performed by means of partitions of unity. Let {Ua } be an orientation atlas for Mn in which all sets Ua are diffeomorphic to IR n or IR+., and let Pa} be a smooth partition of unity subordinate to it. We represent w E n~-I(Mn) as w = E ,xaw. It is sufficient to prove the Stokes theorem for each form ,xaw separately. The support of such a form is a closed subset of the compact set supp,xa c Ua ; therefore, it is compact. Homotopy Invariance of the de Rham Cohomology. Theorem 5.11. Let io, il: IRn --+ IRn x IR = IR n+! be the embeddings defined by x 1--+ (x,O) and x 1--+ (x, 1). Then, for any k ~ 0, there exists a linear map D: nk(IRn+!) --+ nk-l(IRn) such that dD + Dd = ii - i o. Proof. For r.p E no (IRn+!) = Coo (IRn+! ), we set Dr.p = O. The basis forms in nk(IRn+!), where k ~ 1, are of two types. Namely, let t be a coordinate on the line JR, and let Xl, ... ,Xn be coordinates on IRn. We denote the forms not containing dt by a = a dXil /\ ... /\ dXilc and the forms containing dt by (3 = b dt /\ dXjl /\ ... /\ dXjlc_l. Any form is a sum of such forms; so it is sufficient to prove the required assertion only for forms of types a and (3.

U:

We set Da = 0 and D(3 = b dt) dXjI /\ ... /\ dXjlc_l. Let us prove the equality dD+Dd = ii -io for (i) the function r.p, (ii) the form a, and (iii) the form (3.

= 0 and Ddr.p = (ii) dDa = 0 and Dda = (i) dDr.p

J: ~ (Jo

l

dt

= r.p(1) -

r.p(0) = (~:

o)r.p·

~~) dXil /\ ... /\ dXilc = (ii - io)a.

2. De Rham Cohomology

283

(iii) Note that io{3 = ii{3 = 0 because the restrictions of dt to io(l~n) and to il(lRn) vanish identically (the t-coordinate of any tangent vector is zero). We have

and

Therefore dD{3 + Dd{3

o

= o.

Corollary (the Poincare lemma). In lRn , any closed differential k-form (with k > 0) is exact; i.e., H~R(l~n) = 0 for k > o. Proof. Consider the map H: ]Rn x lR -+ lRn defined by (x, t) 1-+ tx. We have Hil = id and Hio = * (* denotes the constant map to 0). Suppose that w E nk(lRn) and dw = O. Then w = (ii - io)H*w = dD(H*w) + Dd(H*w). But D d(H*w) = DH* dwj therefore, w = dD(H*w), i.e., the form w is exact. 0 Theorem 5.11 remains valid for maps io, il: Mn x lR -+ Mnj no substantial changes in the proof are required. This implies that homotopic maps induce the same map of the de Rham cohomology.

Example 74. Hk(lRn) c

~ Hk

DR

(sn)

~ {lR

Proof. Using the diffeomorphism lRn sequence

0

~

if k if k

= n, i- n.

sn\ {xo}, we can construct an exact

o -+ n~(lRn) -+ nk(sn) -+ nk(sn)/n~(lRn) -+ o. Here nk(sn)/n~(lRn) is the space of germs of k-forms at the point Xo. The cohomology of this space is easy to calculate by using the Poincare lemma. Indeed, any sufficiently small neighborhood of Xo can be assumed to be contained in lRnj therefore, for a k-form with k ~ 0, closedness in a small neighborhood of Xo implies exactness. Using a short exact sequence of forms, we can construct the exact sequence of cohomology. As a result, we obtain H:(lR n ) ~ H~R(sn) for

284 k

5. Cech and de Rham Cohomology

~

2. Moreover, we obtain an exact sequence

Therefore, H:(I~.n) ~ Hf,R(sn). In particular, Hf,R(Sl) ~ R. It remains to calculate the de Rham cohomology of the sphere sn for sn as the union of open sets Ul = sn \ {Xl} and U2 = sn \ {X2} diffeomorphic to Rn. We have UI n U2 ~ sn-l X R; hence H8R(UI n U2) ~ H8R(sn I). Therefore, for k > 0, the Mayer Vietoris sequence implies H1R(sn-l) ~ H~1iI(sn). For k = 0 and n ~ 2, we obtain an exact sequence n ~ 2. Let us replesent

Exactness at the term R ED R implies that Ker ~ R and 1m ~ R in this term; hence the kernel ofthe map HgR(sn-l) - Hf,R(sn) is R. Therefore, Hf,R(sn) = o. 0 Any n-form W E nn(sn) is closed. If this form is exact, then it belongs to the kernel of the map nn(sn) - R defined by W 1-+ Isn:.l. Indeed, Isn do. = O. The equality HOR(sn) = R shows that the space of exact n-forms coincides with the kernel of this map. Therefore, the cohomology class of any n-form w for which Isn w # 0 is a nonzero element of the group HOR(sn) = R. To obtain a nonzero cohomology class in H:;(Rn), we take the form cp(XI, .. . , xn) dXl 1\ ... 1\ dx n , where cp is a compactly supported function such that cp(XI, ... , xn) dXI ... dX n # O.

Inln

2.2. The Poincare Isomorphism for de Rham Cohomology. Suppose that Mn is a compact oriented manifold, o. k E n k (M"n), and (3n-k E nn-k(Mn). Then o. k 1\ (3n-k E nn(Mn); thus, we can consider the integral IMn o. k 1\ ~-k. For a noncompact manifold mn, this integral may be divergent. To prevent this, we assume that (3n-k E n~-k(Mn).

If Mn is a manifold without boundary, then the map nk(Mn) x

n~-k(Mn)

_ R

thus constructed carries over to the de Rham cohomology Indeed, suppose that do. k = a and d{3n-k = o. Then d(w k - 1 1\ (3n-k) = dw k - l 1\ {3n-k and

285

2. De Rham Cohomology

d( o:k " w n- k- l ) = ±o:k "dw n- k 1; therefore, the forms dw k 1" {3n k and o:k "dw n- k 1 are exact. But the integral of any exact (compactly supported) form over a manifold without boundary vanishes. This means that the map carries over to the coset space of the set of closed forms modulo the exact forms. As a result, we obtain a bilinear map H8R(M n ) x H~-k(Mn) --+ JR, i.e., a linear map D: H8R(Mn) --+ (H~-k(Mn)t.

Theorem 5.12 (the Poincare isomorphism). For any oriented manifold M n without boundary, the map

is an isomorphism. Proof (see [42]). First, let us show that the required assertion is true for M n = JRn. We have already calculated the groups Hk(JR n ) and H~(JRn) (see the Poincare lemma and Example 74). The only nontrivial groups among these are HD(JRn) and H~(JRn), and both of them are isomorphic to JR. Therefore, it suffices to verify that the map

is nonzero. Let 1 E HD(JR n ) be the cohomology class of the function identically equal to 1 on JRn , and let 'P(XI,"" xn) dXI""'" dX n E n~(JRn) be a form whose cohomology class is nonzero. Then

and therefore D(l) i= O. We have proved the required assertion for Mn = JRn , and thereby for all open subsets of any manifold M n diffeomorphic to JRn. To transfer it to open subsets of Mn (and, thereby, to the entire manifold Mn), we apply the standard technique based on maps of Mayer Vietoris sequences. Consider the Mayer Vietoris sequence for compactly supported cohomology:

... _

H-:-k(M n ) _

H~-k(UI) EB H~-k(U2)

- - H-:-k(U1 n U2) _

H~-k-I(Mn)

__ ....

Let us dualize it, i.e., take the sequence of dual spaces and dual maps. The Mayer-Vietoris sequence for the de Rham cohomology can be mapped to

286

5. Cech and de Rham Cohomology

the dualized sequence by means of D: Hk(Mn)

1

DMn

H~ k(Mn)*

) Hk(Ul) E9 Hk(U2)

1

DUI EBDUI

---+ H~-k(Ul)* E9 H~-k(U2)* -~) Hk(Ul

n U2)

lDu

1

U:I

) Hk+l(Mn)

lDMn

-----+ H~-k(Ul n U2)* -----+ H~-k-l(Mn)*. Let us show that this diagram is commutative up to sign. The commutativity of the left square and of the central square, in which only the restriction and extension operations are used, is obvious. Consider the right square. The upper map in the right square is as follows. Suppose that a k E nk(Ul n U2) and da k = O. Let us represent the form a k as a k = at - a~, where a~ E nk(Ui). The upper map takes a k to ak+l E nk+1(Mn), which coincides with da~ on Ui. The map H~-k-l(Mn) -+ H~-k(Ul n U2) dual to the lower map in the square is as follows. Suppose that f3 n - k- 1 E n::-k-1(Mn) and df3 n - k - 1 = O. Let us represent the form f3 n - k - 1 as f3 n - k - 1 = f3~-k-l + f3~-k-l, where f3f- k - 1 E nc(Ui). The form f3 n - k - 1 is taken to the form f3 n - k E n::-k(Ul nU2), which coincides with df3~-k-l = -df3~-k-l (on UlnU2). We must prove that lUlnu:I a k /\f3n-k = ± lMn ak+l/\ f3 n - k - 1 • Clearly,

The form a~ I\f3f- k - 1 has compact support in Ui; consequently, the integral lUI d(a~ 1\ f3f- k - 1 ) vanishes. Therefore,

The support of the form df3~-k-l = -df3~-k-l is contained in the intersection of the supports of the forms f3~-k-l and f3~-k-l; hence it is contained

2. De Rham Cohomology

28'j

in Ul n U2' Thus, this form can be regarded as an extension of fjn-k, and (_l)k+l [

ci+l" fjn-k-l

=

[

at" fjn-k

}Ulnu2

) Mn

=

[

-

[

a~ "fjn-k

}Ul nu2

(at - a~) "fjn-k

}ulnu2

=

1

a k "fjn-k,

UlnU2

as required. According to the five lemma, if the Poincare isomorphism holds for the manifolds Ulo U2, and U1 n U2, then it holds also for the manifold Ul U U2 = Mn. The next (and most important) step is proving the Poincare isomorphism for any open set U C jRn. Let U be the base for the topology of R n which consists of all open rectangles given by ai < Xi < bi for i = 1, ... , n. This base has two properties important for our purposes; namely, (i) the Poincare isomorphism theorem is valid for any set from U, and (ii) the intersection of any two sets from U belongs to U. Lemma 5.1. Let U be a base for the topology of Mn with properties (i) and (ii). Then, for any finite union of sets from U, the Poincare isomorphism theorem is valid. Proof. Take Ulo "" Uk E U. We shall prove that the Poincare isomorphism holds for UI U ... U Uk by induction on k. For k = 1, this follows from (i). For k > 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

5. Cech and de Rham Cohomology

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

5. Cech and de Rham Cohomology

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

3. The de Rham Theorem

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

5. Cech and de Rham Cohomology

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.

3. The de Rham Theorem

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

5. tech and de Rham Cohomology

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

3. The de Rham Theorem

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

5. Cech and de Rham Cohomology

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

3. The de Rham Theorem

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

5. Cech and de Rham Cohomology

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:

3. The de Rham Theorem

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

5. Cech and de Rham Cohomology

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)).

1. The Alexander Polynomial

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

.

1. The Alexander Polynomial

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. The Alexander Polynomial

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'--~><-- ~ a

(§)-

b Figure 7. Twistings of a handle

is even. Two twistings in opposite directions cancel each other, and two twistings in the same direction are replaced by a loop (see Figure 7b).

Problem 125. Calculate the Alexander polynomial for the knot shown in Figure 8.

Figure 8. The knot diagram for Problem 125

Problem 126. (a) Given odd numbers p, q, and r, we can construct a knot P(p, q, r) as shown in Figure 9 (the numbers p, q, and r may be negative; for each negative number, the types of crosses change). Calculate the Alexander polynomial for the knot P(p, q, r). (b) Prove that the Alexander polynomial of the knot P( -3, 5, 7) is t; i.e., it coincides with the Alexander polynomial of the trivial knot. (This knot is called the Seifert knot.) Now we prove some general properties of the Alexander polynomial. It follows from the properties of determinants that ~L(t) ~ ~L(Cl) for any oriented link L. Indeed, ~L(t) ~ det(tA - AT) = (_t)n det(C I A - AT) ~ ~L(t 1).

Theorem 6.3. (a) If K is a knot, then ~K(l) ~ l. (b) If L is a link with at least two connected components, then ~d1) ~

o.

Proof. Let L be an oriented knot or a link. Take the standard generating cycles ft, .. . , !2g+n-l on its Seifert surface (a knot is characterized by n = 1). By definition, ~d1) ~ detB, where B = A - AT, i.e., btl = al} - aji = lk(h ft) -lk(fj, f/) = lk(fi-' Ii) -lk(f/, Ii). To calculate this difference

1. The Alexander Polynomial

311

p

Figure 9. The knot diagram for Problem 126

of linking numbers, consider the strip F, spanned by the curves I j - and li+ and orient it so that its boundary is the cycle It 1,+. The required difference of linking numbers is equal to the intersection number ((Fi, IJ)) = ±((J" fJ)), where the last intersection number is calculated on the Seifert surface. It is easy to show that under the sign convention which we use, the sign in the last expression is minus. When two I-dimensional cycles are interchanged, their intersection number changes sign. Therefore, the matrix B consists of 9 diagonal blocks 6) and n - 1 zeros on the diagonal. If n = 1, then det B = 1, and if n > 1, then det B = o. 0

(J

The Multivariable Alexander Polynomial. In constructing the Alexander polynomial ~L(t), we used the infinite cyclic covering Xoo -+ 8 3 \ L. It is constructed by using the subgroup of 71"1(83 \ L) consisting of the loops for which the linking number with L is zero. For a knot, this is the commutator subgroup [71"1,71"1]. For a link, we can also consider the covering constructed modulo the subgroup [71"1,71"1]. This subgroup is normal; therefore, we obtain a regular covering X -+ 8 3 \ L with automorphism group 71"1/[71"1,71"1] ~ HI (83 \L) ~ Zt, where r is the number of components in the link L. The automorphism group of this covering is the free Abelian group with generators tlo· .. , t r , which correspond to small circles put on the components of L. Thus, H 1(X; Z) is a module over the ring Z[tt1, ... , f;-l], and we can use this module to construct the Alexander polynomial ~dt1' ... ' t r ) in the variables t1, . .. , t r · 1.5. The Conway Polynomial. The Alexander polynomial is determined up to multiplication by ±tr ; actually, this is a whole class of polynomials rather than a single polynomial. The Conway polynomial is a uniquely determined polynomial in the class of Alexander polynomials, which is obtained

312

O. Mlscel1any

by a change of variables. The Conway polynomials of different links can be summed, as opposed to the Alexander polynomials (sums of different representatives of two Alexander polynomials may belong to different Alexander polynomials). Moreover, there is a remarkable relation between the Conway polynomials of links depending on each other in a certain way, which makes it possible to calculate the Conway polynomial of any link directly from the diagram of this link. To prove that the Conway polynomials are well defined, we need an auxiliary assertion concerning relationships among different Seifert surfaces of the same oriented link L. This assertion will also be used to prove that the Arf invariant of a link is well defined. Let F be a Seifert surface of an oriented link L C S3. Suppose that the cylinder I x D2 is embedded in S3 so that its intersection with F is precisely the bases of the cylinder, i.e., F n (I x D2) = 81 X D2; moreover, in a small neighborhood of the bases, the cylinder lies on one side of the surface F. Let us remove the bases of the cylinder from F and attach I x 8D2 instead of them. As a result, we obtain another Seifert surface F' of the link L. We say that the Seifert surface F' is obtained from the Seifert surface F by attaching a handle. Theorem 6.4. Let F1 and F2 be Seifert surfaces of an oriented link L. Then there exists a sequence of Seifert surfaces S1,'" ,Sk, where S1 = F1 and Sk = F2, such that, for any i = 1, ... , n - 1, one of the surfaces Si and Si+1 is obtained from the other by either attaching a handle or applying an isotopy (fixed on L). Proof. Applying a small deformation (fixed on L), if necessary, we can assume that the surfaces F1 and F2 transversally intersect each other in finitely many closed self-avoiding curves (the intersection along L is also transversal). Let M3 be the closure of one of the domains into which the surfaces F1 and F2 divide S3. Suppose that M3 lies on one side of each of the surfaces F1 and F2; this means that if different parts of M3 are adjacent to F1 (or F 2), then all of them lie on one side of F1 (or F2); below, we shall show that such a manifold M3 always exists. Let 8M3 = 8 1 M 3 U 8 2M 3, where 8~M3 = 8M3 n F~. Any triangulation of thE' sphere S3 for which F1 and F2 are subcomplexes can be restricted to M3. Let A 3 be the union of all simplices from the second barycentric subdivision of this triangulation of M3 that intersect 8 1 M3 or some I-simplices in the initial triangulation. Informally, A3 is the union of the c-neighborhood (collar) of the boundary 8 1 M3 in M3 and the c-neighborhood of the I-skeleton of the initial triangulation. Let F{ be the surface obtained from Ft by removing 8 1 M 3 and attaching the closure of the manifold 8A3 \ 81A1 instead. This operation reduces to a sequence of isotopies and attachments of handles.

1. The Alexander Polynomial

313

Indeed, the transformation in question is as follows. We have a surface to which several edges of a graph are attached so that all of the edges are on one side of the surface. Consider the boundary of the c-neighborhood of the graph. To the initial surface we attach the part of this boundary that lies on the same side of the surface as the graph; the fragments of the initial surface that are contained in the c-neighborhood of the graph are removed. This procedure reduces to successive isotopies and attachments of handles. The surface F2 is changed as follows. Let B3 be the closure of M3 \

A 3 , i.e., the union of the cones whose vertices are the centers of the 3simplices from the initial triangulation and bases are the stars (in the second barycentric subdivision) of the centers of those 2-faces of these simplices which are not contained in 81M3. Removing B3 n F2 from the surface F2 and attaching the closure of 8B3 \ (B 3 n F 2) instead, we obtain a surface F'. This transformation is of the same form. Indeed, if some 3-simplex has n 2-faces not contained in 8 1 M 3 , then this simplex corresponds to the attachment of the boundary of the c-neighborhood of a graph consisting of n edges going from one vertex. The surfaces F{ and F~ coincide inside M 3, and F{ n F~ = (F1 n F 2 ) U (8A3 \ 8 1M 3). But, translating Fl along the collar 81M3, we might violate the necessary condition 8F1 = L. This must be fixed. Each connected component of the set Fl n F2 is either contained entirely in the boundary of M3 or does not intersect it. If some component K of the link L is contained in the boundary of M 3, then we change F{ by an isotopy that moves 8F{ back along the collar until 8F{ coincides with K. These transformations of the Seifert surfaces Fl and F2 decrease the number of domains into which these surfaces divide the sphere 8 3 . Therefore, we can achieve the situation that Fl and F2 do not divide 8 3 , i.e., coincide. It remains to prove that there always exists a manifold M3 such that it lies on one side of both Fl and F2. For this purpose, we use the infinite cyclic covering p: Xoo -+ 8 3 \ L constructed in Section 1.2. The space Xoo can be constructed by using either of the surfaces Fl and F2. Therefore, Xoo contains the two families of surfaces t i Fi and ti F2 (these are the boundaries of the 3-manifolds constituting Xoo). We are interested in the case where Fl and F2 intersect not only in L. The idea is to consider the maximum integer n for which F2 n t n F1 -:f:. 0 (see Figure 10). For such an integer to exist, we assume that the set L is removed from 8 3 together with its open c-neighborhood, so that the space obtained is compact. Now we deal with compact sets, and the required number n does exist. The surface F2 cuts Xoo into two parts; let Y be the part that contains the surfaces t r F2 for r < o. The surface t n PI also cuts Xoo; let Z be the ;>art

6. Miscellany

314

y

Figure 10. The choice of the manifold M3

of Xoo containing the t n +r Fi for r p(Y n Z) is the required manifold.

> O. It is easy to see that the projection 0

Two square matrices are said to be S-equivalent if one of them can be obtained from the other by applying several operations of the following two types: (1) A 1---+ PApT , where P is an integer matrix with determinant ±1; (2) A

~ (~8~) 000

or A

~ (~8 8), 010

where

Q

is a column and f3 is a row.

Theorem 6.5. Any two Seifert matrices of the same oriented link are Sequivalent. Proof. A basis change in HI (Fi Z) leads to a type (1) transformation of the Seifert matrix. According to Theorem 6.4, it remains to determine what happens to the Seifert matrix when a handle is attached. We can assume that, under this operation, the basis of HI (F; Z) is supplemented by the generators fn+! and fn+2 shown in Figure 11. If i ~ n, then IkU':;+2' fd = O. Choosing

Figure 11. The additional generators

appropriate orientations of the curves fn+l and fn+2, we an assume that either IkU;t+l,/n+2) = 0 and IkU;+!,fn+2) = 1 or IkU;tt-1 fn+2) = 1 and

.. (AOO) 0 0"6 .

IkU;+!, fn+2) = O. In the former case, the new Seifert matriX

IS

315

1. The Alexander Polynomial

Transformations of type (1) reduce this matrix to the form (~o ~). Simio 00 A 0 0) larly, in the latter case, the matrix reduces to the form ( {3 0 0 • o o lO Let us introduce the formal variable t l / 2 , for which (t l / 2 )2 = t, and define the Alexander polynomial in Conway's normalization ~LCt) E Z[C l / 2 , t l / 2 ] by ~L(t) = det(t l / 2 A - C l / 2 AT), where A is the Seifert matrix of the link L. If A is a matrix of order r, then ~L(t) = C r / 2 det(tA - AT); therefore, the Alexander polynomial in Conway's normalization coincides with the Alexander polynomial up to multiplication by a unit of the ring Z[C l / 2 , t 1/ 2 ]. Theorem 6.6. For any oriented link L, the Alexander polynomial in Conway's normalization is determined uniquely; i.e., it does not depend on the choice of the Seifert matrix A. Proof. First, note that

det(t 1 / 2 pT AP _ C

1/ 2 pT AT P)

= det(pT(t l / 2 A _ t

1/2 AT)P)

= (det p)2 det(t 1 / 2A _ C

Let B

0) = ( Ao0 00 0 1 . Q

l / 2 AT).

Then

Multiplying the last row of this matrix by suitable elements of the ring Z[C 1 , t] and adding the result to the remaining rows, we kill the column t 1/ 2 o. The row _C 1/ 2 o T can be killed in a similar way. Therefore, det(t 1 / 2 B - C

1 / 2 BT)

= det(t 1/ 2A -

C

1 / 2 AT)

det ( _C0 1/ 2

t l / 2) O

= det(t l / 2A _ C l / 2AT) because det similar.

(-t 01 2 tl~2) =

1. For a transformation of type (2), the proof is 0

The proof of Theorem 6.3 implies the following properties of the Alexander polynomial in Conway's normalization. (i) If K is a knot, then ~K(t) E Z[C 1 , t], ~K(I)

=

1, and ~K(t)

=

~K(Cl).

(ii) If L is a link with at least two connected components, then ~LCl) = O.

6. Miscellany

316

Theorem 6.7. Suppose that L+, L_, and La are oriented links whose diagrams coincide everywhere except in a small disk, in which they are as shown in Figure 12. Then their Alexander polynomials in Conway's normalization are related by (41)

~L+ -~L

= (C 1/ 2 _

t 1 2)~Lo·

X X )( L+

L

La

Figure 12. The three links

Proof. Let us construct a Seifert surface Fa for the link La so that, in the disk under consideration, it has the form shown in Figure 13. Attaching

F

Fa

Figure 13. The three Seifert surfaces

a twisted strip to La, we obtain Seifert surfaces F+ and F_ for the links L+ and L_. Take the basis in H 1 (F±) that consists of the class of a closed curve h going along the strip and the basis elements of Hl(Fo). If Ao is the Seifert matrix for La, then the Seifert matrices for L_ and L+ are (a Ao) and (n/i 1 Xo )' where n is an integer. Considering the determinants of the matrices t 1 2 A - C 1/2 AT for the three Seifert matrices and taking into account the fact that det (~ Ao) = a det Ao + c, where c does not depend on a, we obtain the required result. 0 Relation (41), together with the condition that the Alexander polynomial in Conway's normalization for the trivial knot equals 1, makes it possible to calculate ~L for any oriented link L (see [108] for details). Therefore, ~L is a polynomial in z = C 1/ 2 - t 1/ 2 . The polynomial VL(Z) such that V dt- 1 / 2 - t 1 / 2 ) = ~dt), where ~L is the Alexander polynomial in Conway's normalization, is called the Conway polynom· I. The Conway polynomial satisfies the skein relation

(42)

2. The ArE Invariant

317

Problem 127. Prove that if a link L lies on both sides of a plane disjoint from this link, then V'L(Z) = O. Let us write the Conway polynomial in the form V'L(Z) al(L)z + a2(L)z2 + ....

=

ao(L)

+

Problem 128. Let n(L) denote the number of components in the link L. Prove that aieL) = 0 for i == n(L) (mod 2). Problem 129. (a) Prove that if K is a knot, then ao(K) = 1. (b) Prove that if L is a link with more than one component, then ao(L)

=

O. Problem 130. Let L be a link with n(L) components. Prove that al(L) for i < n(L) - 1.

=0

Problem 131. Prove that if L is a link with precisely two components Ll and L2, then al(L) = Ik(LI, L2)' Problem 132. Suppose that L+ and L_ are knots and Lo is the link with components Ll and L2. Prove that a2(L+) - a2(L_) = Ik(LI, L2)'

2. The Arf Invariant 2.1. The Arf Invariant of Quadratic Forms. Associated with each real quadratic form q(x) is the symmetric bilinear form B(x, y) for which B(x, x) = q(x). The identity q(x + y) - q(x) - q(y) = 2B(x, y) makes it possible to reconstruct B from q. For the field Z2, the definition of a quadratic form and the associated bilinear form changes as follows. Let V be a finitedimensional space over Z2. A map q: V - Z2 is called a quadratic form if there exists a bilinear form B(x, y) for which q(x+y)+q(x)+q(y) = B(x, y). Of course, the signs do not matter; it only matters that 2B(x, y) is replaced by B(x, y), because division by 2 is not defined over the field Z2' Note that, in contrast with the real case, q(x) is not equal to B(x, x). The quadratic form q is said to be nondegenerate if the bilinear form B is nondegenerate. It is seen from the definition that B(x, y) = B(y, x) and B(x, x) = q(2x)+2q(x) = O. Therefore, the same argument as in the case ofreal skewsymmetric matrices shows that if q is a nondegenerate quadratic form, then the space V has a basis el, II, ... , en, fn such that B(ei' ej) = B(Ji, Ii) = 0 and B(ei' fJ) = dij; such a basis is called a symplectic basis. Let q be a nondegenerate quadratic form, and let el, .. " fn be a symplectic basis. The element c(q) = E~=l q(ei)q(Ji) E Z2 of the group Z2 is called the Arf invariant of q.

6. Miscellany

318

Theorem 6.8 (Arf [8]). Any nondegenerate quadratic form q can be reduced to the form XIYI + ... + XnYn + c(q)(x~ + Y~), where c(q) is the Arf invariant. Moreover, the Arf invariant c(q) does not depend on the choice of a symplectic basis. Proof. First, we consider the case n = 1. The equality q(el) + q(JI) + q(el + h) - 1 implies that either one of the elements q(eI), q(h), and q(el + fI) is equal to 1 and the two others are 0 or all of these elements are equal to 1. The elements el, h, and el + h are equivalent in the sense that any two of them can be taken for a symplectic basis because B(el + h, el + h) - 0 and B(el + h, et) - 1. Therefore, we can assume that q(el + fd = 1. Thus, for n = 1, we obtain two nondegenerate quadratic forms, qo and ql; here qo(el) = qo(Jd - 0 and ql(el) - ql(Jd = 1. In coordinates, these forms can be written as qo = XIYl and ql - XIYl +x~+Y~. Clearly, c(qo) - 0 and C(ql) = 1. To prove the independence of the Arf invariant on the choice of a symplectic basis, we must show that the forms qo and ql are not equivalent, that is, cannot be obtained from each other by a basis change. Indeed, the form qo takes the value 1 precisely once, while qi takes this value three times. Now consider an arbitrary nondegenerate quadratic form q. Choose a symplectic basis el, ... , fn. Let 'Pi be the restriction of q to the subspace spanned by the vectors ei and fl' Then q = 'PI ffi ... ffi 'Pn, where each form 'Pi is equivalent to qo or to qi.

Lemma 1. The quadratic forms 'l/Jo

= qo ffiqo

and 'l/JI

= ql ffiql

are equivalent.

Proof. Let eI, h, e2, 12 be a symplectic basis for which 'l/Jo(ei) = 'l/Jo(J,) = 0 and 'l/Jl(ei) = 'l/Jl(f,) = 1. Consider the basis e~ - el + e2, ff = el + 12, e~ = el + h + e2 + 12, f~ = el + h + h· It is easy to show that this basis is symplectic. Let us show that 'l/Jl(e~) = 'l/Jl(JI) = 0, i.e., 'l/J1(eD = 'l/Jo(ei) and 'l/JI (JI) - 'l/Jo (Jl)' Indeed, we have 'l/Jl (el + e2) = 'l/Jl (ed + 'l/Jl (e2) + B( eI, e2) = 1 + 1 + 0 = 0, 'l/J1(el + h) - 'l/Jl(ed + 'l/J1(h) + B(el' h) = 1 + 1 + 0 = 0, 'l/J1(el + h + e2 + h) = 'l/J1(el + e2) + 'l/Jl(Jl + h) + B(el + e2, h + h) = 0+ 0 + 0 = 0, and 'l/Jl (el + h + h) = 'l/Jl (el + h) + 'l/Jl (JI) + B(el + 12, fI) = 0+ 1 + 1 = O. 0 We introduce the notation q ffi ... ffi q

-------

= nq.

n times

Lemma 1 implies that the form q is equivalent to either nqo or (n-l)qoffiql. In coordinates, these forms are written as XIYl + ... + XnYn and XIYl + ... + XnYn + x~ + Y~, respectively. Clearly, c(nqo) = 0 and c((n - l)qo ffi qt} = 1.

2. The ArE Invariant

319

To prove that the Arf invariant does not depend on the choice of a symplectic basis, we must prove that the forms nqo and (n - l)qo ffi qi are not equivalent.

Lemma 2. The form nqo takes the value 1 precisely 2 2n - 1 - 2n I times, and the form (n - l)qo ffi ql takes this value precisely 22n - 1 + 2n - 1 times. Proof. We prove the lemma by induction on n. For n = 1, the assertion is easy to check. Suppose that we have already proved that the form nqo takes the value 1 precisely 22n - 1 - 2n 1 times. The form qo takes the value 1 precisely once and the value 0 precisely three times. Therefore, the form (n + l)qo takes the value 1 precisely 3(2 2n - 1 - 2n - 1 ) + 22n I + 2n - 1 = 22n+l - 2n times (because nqo takes the value 0 precisely 22n - 1 + 2n 1 times). The form ql takes the value 1 precisely three times and the value o precisely once. Therefore, the form nqo ffi ql takes the value 1 precisely 3(2 2n - 1 + 2n - I ) + 22n I _ 2n 1 times. 0 This completes the proof of the theorem.

o

2.2. The Arf Invariant of Oriented Links. Let L C S3 be an oriented link, and let F be its Seifert surface. Consider the map q: HI (Fj Z2) --+ Z2 defined by q(x) = a2(x, x), where a2 is the Seifert form a reduced modulo 2. For example, if the homology class of x is represented by a closed self-avoiding curve on the surface F, then q(x) is the number of full turns (modulo 2) of the c-neighborhood of this curve in the surface F. Clearly, q(x + y) = a2(x, x) + a2(x, y) + a2(Y, x) + a2(Y, Y)j therefore, q(x + y) + q(x) + q(y) = a2(x, y) + a2(Y, x). Let us show that the righthand side of this equality equals f(x, y), where f(x, y) is the intersection form reduced modulo 2. Indeed, a(x,y) + a(y,x) = lk(x-,y) + lk(y-,x) = lk(x-, y) + lk(y, x+). It is seen from the definition of linking number that the parity of lk(x-, y) + lk(y, x+) coincides with that of the number of intersection points of curves representing y and x. Thus, q is the quadratic form corresponding to the bilinear form f. Look at Figure 2 on p. 302. It shows that the intersection form is nondegenerate if and only if the boundary of the Seifert surface is connected, i.e., when we deal with a knot. For this reason, the definition of the Arf invariant for a knot is simpler than for links with several components: the Arf invariant of a knot is defined as the Arf invariant of the quadratic form q. For links with several components, the definition of the Arf invariant is not so simplej moreover, the Arf invariant is not defined for some links. To obtain a nondegenerate form from the intersection form, we must take the quotient of the space HI(Fj Z2) modulo the subspace generated

6. Miscellany

320

by the cycles h g+1. ... , hg+n-l corresponding to n - 1 boundary components. The cycle corresponding to the remaining boundary component is homologous to the sum of these cycles. Thus, consider the quotient space H1(Fj 71.2)/i.H1 (8Fj 71. 2 ), where i: 8F - F is the natural embedding. For the form q to be well defined on this quotient space, each component L, must satisfy the condition q(Li) = 0, or, equivalently, lk(Li, Li) = O. The Seifert surface does not intersect Li j therefore, the linking coefficient of Li and Li is congruent modulo 2 to that of Li and L - L" which, in turn, is equal to the linking coefficient of Li and L - L i . Thus, we can assume that the link L has the property (43)

lk(L;, L,) = 0

(mod 2)

for all i.

Then the form q is well defined on the quotient space. The Arf invariant A(L) of a link L with property (43) is defined as the Arf invariant of the quadratic form q on the quotient space H1(Fj71.2)/i.H1(8Fj71.2). Theorem 6.9. The Arf invariant A(L) of an oriented link L is well defined, i.e., it does not depend on the choice of the surface F. Proof. According to Theorem 6.4, it suffices to show that the Arf invariant does not change when a handle is attached to F. The first 2g curves in Figure 2 constitute a symplectic basis el, h, ... , eg, fg for the form q in the space HI (F)/i.H1 (8F). After a handle is attached, we complete it by the curves eg+1 and f g+1 chosen as shown in Figure 11 on p. 314. If eg+l is the curve fn+2 in this figure, then q(eg+1) = O. Therefore, E~=l q(ek)q(fk) = E~!~ q(ek)q(/k). 0

Clearly, if K is the unknot, then A(K) = o. It is also easy to verify that if K is the trefoil, then A(K) = 1. Indeed, in calculating the Alexander polynomial for the trefoil (see p. 309), we constructed a basis h, h. By the definition of the form !(x, y), we have f(x, x) = O. Moreover, f(h, h) = 1 because hand 12 intersect in one point. Hence A(K) = q(h)q(h) = 1. Theorem 6.10. Suppose that K+ and K_ are two knots and L is a twocomponent link, and their diagrams coincide everywhere except for a small disk, in which they are as shown in Figure 14. Then

where Ll and L2 are the components of the link L.

Proof. For K+ and K_, we can construct Seifert surfaces F+ and F_ which are arranged as shown in Figure 15 over the small disk undt:!r consideration and coincide otherwise. Using F+ and F _, we construct a surface Fo for the

2. The ArE Invariant

321

xx

L

K

Figure 14. The diagrams of the two knots and the link

link L. If this surface is disconnected, then, to obtain a Seifert surface for L, we attach a handle to all of the three surfaces, thus rendering Fo connected. Let ell h, ... , P n , fn be a symplectic basis for the form q on the space H 1(Fo; 2:2 ). Let us complete it to a symplectic basis by adding the cycles en+1 and fn+1,± shown in Figure 15; the cycle e n+1 is represented by the

Fa Figure 15. The three Seifert surfaces

curve L 1. The strips corresponding to the curves fn+1,+ and fn+1,- differ by precisely one full turn; therefore, q(Jn+1,+) - q(Jn+1,-) = 1. Thus, A(K+) - A(K-)

= q(en+I) (q(Jn+1,+) - q(Jn+1,-)) = q(en+d = Ik(L 1 , L 1) = Ik(LI, L - Ld = Ik(LI, L2).

0

Theorem 6.11. (a) For any knot K, A(K) == a2(K) (mod 2), where a2(K) is the coefficient of z2 in the Conway polynomial VK(Z). (b) For any knot K, A(K)

=

{a 1

i!

AK(-l) if AK(-l)

== ±1 == ±3

(mod 8), (mod 8),

where AK(t) is the Alexander polynomial in Conway's normalization.

Proof. (a) If L+ and L_ are knots, then Lo is a two-component link. According to Problem 132,we have a2(L+) - a2(L_) = Ik(L1. L2), where L1

6. Miscellany

322

and L2 are the components of the link L. Theorem 6.10 implies A(L+) == A(L_) + lk(Ll' L2) (mod 2). Moreover, if K is the unknot, then a2(K) = 0= A(K). Therefore, A(K) == a2(K) (mod 2) for any knot K. (b) By definition, ~K(t) = V K(C 1/ 2 - t 1 2); thf'refore, ~K( -1) = VK(-2i). Problem 128 implies VK(Z) = 1 + a2(K)z2 + a4(K)z4 + .... Hence VK(-2i) == 1-4a2(K) (mod 8). If a2(K) is even, then V'K(-2i) == 1 (mod 8), and if a2(K) is odd, then V' K( -2i) == -3 (mod 8). D 2.3. Knotting of Embeddings of the Graph K7. Using properties of the Arf invariant, Conway and Gordon [26] proved the following theorem.

Theorem 6.12. Any embedding of K7 (the complete graph on 7 vertices) in IR3 contains a nontrivial knot. Proof. For every embedding of K 7 in IR3 , we define (7 E Z2 by (7 = E A( C), where the summation is over all of the 360 = ~ cycles C of length 7 passing through all vertices of the graph K7. Any embedding of the graph K7 in IR3 can be transformed into a given embedding by applying the transformations of edges shown in Figure 16. Let us prove that (7 does not change under such transformations. Generally, we must consider three types of transformations of crosses, namely, a self-intersection of an edge, an intersection of two adjacent edges,

x-x Figure 16. The transformations of edges

and an intersection of two nonadjacent edges. But it is seen from Figure 17 that in K7 one intersection of an edge with itself can be replaced by five intersections of different edges; therefore, it suffices to consider only intersections of different edges (adjacent or nonadjacent).

r

~-----b Figure 17. A self-intersection of an edge

323

2. The ArE Invariant

Dragging the vertices of the intersecting edges l to the cross under consideration, we obtain a diagram of the form shown in Figure 18. Figure 18a refers to an intersection of adjacent edges, and Figure 1Sb, to an intersection of nonadjacent edges (of course, the cross itself may be of the opposite type).

,

,,

\

X X

,, \

\

, ,, I

, .... __

....

b

a

Figure 18. Simplification of a cross

Let us see what happens to A(C) under a change of the type of a cross. We denote the variation of A(C) bye(C). If the cycle C does not contain at least one of the two edges in the cross under consideration, then e( C) = o. First, suppose that the cycle C contains both of these edges. Then, according to Theorem 6.10, we have e(C) == lk(Ll, L 2) (mod 2), where Ll and L2 are the components of the link L constructed as shown in Figure 14 (p. 321). Suppose that the intersecting edges are adjacent. The construction of the link L = LIUL2 in this case is shown in Figure 19a. Note that the component

~LI ,,

I

,

....

_-_

~LI ,

I

I

....

a

b

Figure 19. The two components of the link

L1 is the same for all cycles C containing the edges under consideration. Therefore, Ik(L 1 ,L2) = Lew(Ll,e), where the summation is over all edges e of the cycle C different from the two edges in question, and w(Ll, e) is the number of crosses at which L1 passes over e with signs taken into account. The change of u equals La Le W(L1' e) (mod 2), where the first sum is 1 Together

with the edges incident to them.

6. Miscellany

324

over the cycles C containing both edges under consideration. Thus, it is sufficient to prove that for each edge e of K 7 different from the two edges in question, the number of cycles C containing e is even. The following cases are possible. 1. The edge e and the two edges under consideration are either incident with the same vertex or form a cycle. Then no cycles C contain e. 2. The edge e shares a vertex with precisely one of the edges under consideration. Then precisely six cycles C contain e. 3. The edge e has no common vertices with the edges under consideration. Then precisely 12 cycles C contain e. Now, suppose that the intersecting edges are nonadjacent. The construction of the link L = Ll U L2 in this case is shown in Figure 19b. We have Ik(L1, L 2) = E w(el' e2), where the summation is over all pairs of edges el C L1, e2 C L 2. We number the vertices of the graph K7 starting with the endvertices of the two edges under consideration as shown in Figure 20; the arrows indicate

,"

" ----- ..... ,

\L2 ,

4~1

3~2 Figure 20. The numbering of graph vertices

the direction of traversing the cycle C. The curves Ll and L2 may contain the following edges el and e2 (up to a renumbering of vertices): (a) (23), (45); (b) (23), (56); (c) (27), (45); (d) (27), (56). It is easy to see that in these cases, the cycles C are as follows: (a) (1234567), (1234576); (b) (1234567), (1234657), (1234756), (1234765); (c) (1273456), (1276345); (d) (1273456), (1273465). In every case, an even number of cycles C is obtained; therefore, u does not change. A simple verification shows that for the embedding of K7 1O]R3 shown in Figure 21, all but one of the cycles Care unknotted. Moreover, the knotted

3. Embeddings and Immersions

325

Figure 21. An example of an embedding

cycle is the trefoil, and the Arf invariant of the trefoil is 1. Thus, we have u = 1 for any embedding, so there always exists a knotted cycle C. 0 Remark. In Part I, we proved that for any embedding of the graph Ka in ]R3, there exists a pair of linked cycles (Theorem 1.10). The proof us much simpler than that of Theorem 6.12 about knotting of cycles. It is known that for any link, there exists a positive integer N such that this link can be inscribed in any generic set of N points in]R3 (i.e., the line segments joining some points from this set form a link equivalent to the given one); a simple proof of this fact is given in [107].

3. Embeddings and Immersions 3.1. The Strong Whitney Embedding Theorem. In Part I, we proved the weak Whitney theorem about embeddings and immersions, which says that, for n ~ 2, any n-manifold without boundary can be immersed in ]R2n and embedded in ]R2n+l. In this section, we prove the strong Whitney embedding theorem that any closed manifold Mn can be embedded in ]R2n.

The idea of the proof is to first construct a regular (generic) immersion f: M n -+ ]R2n, that is, an immersion such that it has no triple points (in other words, no three different points are mapped to one point) and, at each double point (self-intersection) a = f(x) = f(y), where x 1= y, the equality df(TzM n ) E9 df(TyMn) = Ta]R2n holds. The so-called Whitney trick makes it possible to kill all pairs of double points. If the manifold M n is orient able and n is even, then the points in each pair to be killed must match in a certain sense. Thus, to kill all double points, it is sometimes required to first add several double points. Regular Immersions. Theorem 6.13. Suppose that Mn be a closed manifold and f: Mn -+ ]R2n is an immersion. Then for any c > 0, there exists a regular immersion g: Mn -+ 1R2n such that IIf(x) - g(x)11 < c for all x E Mn.

326

6. Miscellany

Proof. We use the same construction as in the proof of the existence of immersions (see Part I, Theorem 5.17), but now, instead of the change f,(Y) - fi I(Y) + Ay, we make the change f,(y) - fl I(Y) + v, where v is a sufficiently small constant vector. First, let us see what happens in a small neighborhood of the image of a point x E Mn. We can assume that f(x) - 0 and the image of a neighborhood UI,l 3 x in ]R2n is locally (i.e., in some neighborhood V) determined by the equations Yl - 0, ... , Yn - O. For the manifold f I(V), consider the composition f-I(V)

L

V ~

]Rn,

where p(YI, . .. , Y2n) - (Yl, ... , Yn). Let v' - (VI, ... , v n ) be a regular value of this composition different from the images of the double points belonging to V. The n-dimensional subspace nn given by YI - VI, ... , Yn - Vn does not pass through the double points of the image of M n that belong to V, and at each point a E rrn n f(Mn) n V, the tangent space df(Tf l(a)Mn) is complementary to nn, i.e., the direct sum of this space and nn has dimension 2n. This means, in particular, that all intersection points of nn and f(Mn)n V are isolated. Thus, adding the vector (VI, ... , V n , 0, ... ,0) to each point from the image of Ui,1 and leaving all the other parts of the intersection f(Mn) n V intact, we obtain an extension of the regular immersion over Ui,l. We described what is going on in Ui,l. The global map that remains the same outside Ui ,2, is constructed by using the bell-shaped function A, as in the proof of the weak Whitney theorem. The only thing to care about is the vector v', which should be sufficiently small for the map obtained to be an immersion. 0 If a closed manifold Mn is regularly immersed in R2n, then the number of

double points is finite. Indeed, suppose that there are infinitely many double points. Let Xo be a limit point of their preimages. It has a neighborhood U with self-avoiding image. Let {Xi} be a sequence of preimages of double points from U converging to Xo. The corresponding points Yi do not belong to U. Let Yo be a limit point for {Yi}. Choosing a neighborhood of Yo whose image intersects the image of U in at most one point, we obtain a contradiction.

Double Points. Suppose that f: M n _ ]R2n b a regular immersion and a = f(x) = f(y) is a double point. If the manifold M n is orientable and n is even, then, for the point a, the self-intersection number is defined. Namely, choose an orientation of Mn and take positively orIented bases of df(TxMn) and df(TyMn). From these bases, we compose a basis of space ]R2n by consecutively writing the clements of the first and second bas.es. If

3. Embeddings and Immersions

327

the orientation of the basis thus obtained is positive (negative), then the self-intersection number is 1 (-1). The self-intersection number does not depend on the choice of orientation of Mn because a simultaneous change of the orientations of the bases of df(TxMn) and df(TyM n ) does not affect the orientation of the basis of ]R2n composed of them. The self-intersection number does not change under transpositions of the points x and y either. Indeed, interchanging the first n vectors and the last n vectors requires n 2 transpositions. The self-intersection number of a regular immersion f is defined as the sum of the self-intersection numbers of all double points. If n is odd or the manifold M n is nonorientable, then the self-intersection numbers of double points have indefinite signs; in these cases, the self-intersection number of the immersion f is defined to be the remainder of the total number of double points on dividing by 2. We denote the self-intersection number of a regular immersion f by If·

Theorem 6.14. Let M n be a clo,~ed manifold. Then there exists a regular immersion f: M n - 4 ]R2n for which If takes any given value (it is assumed that If E Z if M n is orientable and n is even, and If E Z2 otherwise). Proof. First, we construct a regular immersion intersection. For n = 1, we set

2x

Yl =x- - - I +x 2

(see Figure 22). The points x

=

]Rn

-4

]R2n

with one self-

1

and

Y2

= 1 + x2

±1 are mapped to (0,1/2); at the other

Figure 22. A self-intersection

points, no self-intersection occurs, The Jacobian matrix of this map is (1 2 ~~~~, -1!~2); it is nowhere vanishing. For x = ±l, we obtain the linearly independent matrices (1, =fl). For n ~ 2, consider the map defined by Yl = Xl - ~, Yi = Xi for i = 2, ... , n, where X = (1 + x~) ... (1 + x~), Yn+! = and Yn+i = x)t, for i = 2, ... , n. Let us show that this map is a regular immersion with a

1-,

6. Miscellany

unique self-intersection. Its Jacobian matrix is

4X1X p

4X1X2

X(l+x~)

X(l+xa)

o

1 0

0

-2xl X(I+xV x2(I xJ) X(l+x 1 )

1

2x~

2xn X(I+xa)

X(l+x~) Xl(l-X~)

2xlx~xn

X(I+x 2 )

X(I+xa)

-2X1X2Xn

X(l+x~)

The last n elements of the first column vanish if and only if Xl O. In this case, we have

= ... = Xn

=

Therefore, the Jacobian matrix has rank n at each point. Now let us find all pairs of points (xi, ... ,X~) = (Xl, ... ,xn ) such that Y! = Yi for all i. The equalities Yi = Xi, where i = 2 .... ,n, imply x~ = Xi for i = 2, ... ,no We have Y~+1 = Yn+1; therefore, X' = X, and (xi? = x~, i.e., xi = ±XI. Thus, xi = -Xl =1= o. The equalities Yn+i = x~', where i = 2, ... , n, imply X2 = ... = Xn = O. Therefore, X = 1 +x~. Consequently, Yl = Xl - l~;i and = -Yl· Hence YI = 0 and x~ = 1. Thus, only the points (±1, 0, ... ,0) are mapped to the same point.

yi

The Jacobian matrices at the points (±l, 0, ... ,0) have the form

1

o o

o 1

o

1

o

±1/2

o o

o

o

±1/2

=f 1 / 2

o

o o

3. Embeddings and Immersions

329

These two matrices constitute the matrix

1 o o 1 o o 1 o o o o 1 ..............................................

0 0 -1/2 0 0 1/2

o

o

1 0 0 1/2

1 0 0

0 0 0 1/2 0 -1/2 0

o

-1/2

1 0

o o

1

o

0 0 0 0 0 0

1 0 0 1

0 0

0 0

-1

o 1 o 0 o 1 ............................... 1 0 0

............................... 0 0 ... 0 0 0 . .. -1 This matrix is nonsingular, and hence the immersion is regular. It follows directly from the definition of the self-intersection number (provided that n is even and the manifold Mn is orientable) that it changes sign when the orientation of ]R2n is changed. Therefore, using the regular immersion constructed above and its composition with the symmetry of 1R2n about a hyperplane, we can construct two regular immersions with one selfintersection each and self-intersection number +1 for one immersion and -1 for the other. Suppose that f: M n --+ ]R2n is a regular immersion and U c M n is a sufficiently small neighborhood homeomorphic to lRn such that the restriction of f to U is an embedding. It is easy is construct a regular immersion 9 coinciding with f outside U and such that the sets g(int U) and f(M n \ int U) are disjoint, the image g(U) has a unique self-intersection, and the self-intersection number for this point has a given sign (provided that n is even and Mn is orientable). Compared to f, the immersion 9 has precisely one additional self-intersection point. This construction proves the required assertion. 0

The Whitney Tt-ick. To prove the strong Whitney embedding theorem, it remains to learn how to kill pairs of self-intersection points of the manifold M n in lR2n (if M n is orient able and n is even, then it is assumed that the self-intersection numbers of the points in these pairs have opposite signs). Note that the Whitney trick only works for n ~ 3. For n = 2, the Whitney

6. Miscellany

330

embedding theorem is proved by a different method (see p. 334). In what follows, we assume that n ~ 3. Let I: M n " be a assume only at

p and q be two self-intersection points for a regular immersion ]R2n. Consider 1 I(p) - {PI,P2} and 1 I(q) - {qI,q2}. Let path joining p, and q, in M n (i - 1,2). We set Ii - Ib,). We can that the curves II and 12 are self-avoiding and intersect each other the points P and q.

First, we show that II and 12 span a disk transversal to f(Mn). To be more precise, let r be a closed convex set in ]R2 with two corners (see Figure 23), and let r' be its open e-neighborhood. Then there exists an

Figure 23. The domain

T

embedding 'Ij;: r' - ]R2n with the following properties: (1) the corners are mapped to p and q;

(2) 'Ij;(r') n f(Mn) = II U 12; (3) at any point of II U 12 different from p and q, the tangent space Tx'lj;(r') is not contained in Txl(Mn). The required map 'Ij; is constructed as follows. First, we construct a vector field on II transversal 2 to I(Mr) in ]R2n; here 11vfr is a small neighborhood of the curve II in Mn. For the vectors at p and q we take the tangent vectors to the curve 12 directed toward each other (see Figure 24). They are transversal to I(Mr) because I(M!]) is. Consider the bundle over II whose fiber over each point x is the space ]R2n \ TxMn '" sn-l. This bundle is trivial because its base space is an interval. The sections of this bundle at the endpoints of II are given. Since the space sn-l is path-connected, these sections can be extended over the entire curve II. As a result, we obtain the required vector field. In a similar way, we construct a vector field on the curve 12. Outside small neighborhoods of the points p and q, we take the endpoints ofthe vectors tv(x), where x E ,lU,2, v(x) is the vpctor field at x, and t :=:; e (e is a fixed sufficiently small positive number). In small neighborhoods of p and q, we take parts of the planes spanned by the tangent vectors to the curves II and 12. In constructing the vector fields at p and q, we used 2 Transversal here means that, at each point of 1'1, the vector does not belung to the tangent space to I(Mf') at this point.

331

3. Embeddings and Immersions

1'1

q

p

Figure 24. The vector field on 1'1

tangent vectors to ,1 and ,2; therefore, slightly moving the surfaces under consideration, we can glue them together so as to obtain an embedding of a small neighborhood of the boundary of the disk r (see Figure 25). Transversality is not violated.

Figure 25. The embedding of a neighborhood of the boundary of

T

To the interior of the obtained strip (this is the curve J in Figure 25) we can attach the disk n 2 because any continuous map 8 1 ~ ]R2n extends to a continuous map n2 ~ ]R2n. Since 2n > 5, we can approximate the constructed map D2 ~ ]R2n by a smooth embedding without changing it near U Then, using the inequality 2 + n < 2n, we can remove all intersections of J(Mn) with the image of D2 that arise in the approximation. This completes the construction of the required map 1/J.

,1 ,2.

The next step is to construct an orthonormal system of vector fields on the disk 1/J (r) that are orthogonal to 1/J (r). First, we suppose that n is even, Mn is orient able, and the self-intersection numbers of the points p and q have opposite signs.

W3, ..• , W2n

,1

Consider the restriction to of the tangent bundle of M n and take the orthogonal complement of the tangent bundle Of,1 in this restriction. As a result, we obtain an (n -l)-dimensional bundle. This bundle is trivial because its base is an interval. Therefore, it has linearly independent sections W3, .•. , W n +1· In a similar way, we construct vector fields W n +2, ... , W2n on the curve At the points p and q, all of the 2n - 2 vectors W3, •. . , W2n are defined, but they determine opposite orientations of the normal bundle to the disk 1/J(r) at these points. Indeed, transferring vectors tangent to the curves ,1 and along these curves, we obtain bases of the tangent spaces Tp 1/J(r) and T q 1/J{r) at p and q with opposite orientations (see Figure 26).

,2·

,2

6. Miscellany

332

p

Figure 26. Change of orientation

We shall assume for convenience that the curves 1'1 and 1'2 are not only transversal but also orthogonal to T M!] and T Mf . On the curve 1'1, the vector fields W3, ... , Wn+1 are defined. Consider the bundle over 1'1 whose fiber over each point x is the (n -I)-dimensional subspace orthogonal to the vectors W3(X), ... , Wn+l(x) and tangent to the space T:z;'!/J(T). At the points p and q, the vectors Wn+2, .. . , W2n are defined; these are orthonormal sections of this bundle. We need to extend them over the entire curve 1'1. First, we extend the section wn +2. We can do this because the bundle is trivial and the fiber jRn-1 \ {O} '" sn 2 is connected. Then, we extend a section of the orthogonal complement of the bundle determined by the extended section, and so on. The bundle used to extend the section W2n-1 has fiber jR2 \ {O} '" Sl. When we try to extend the section W2n, the fiber will be the disconnected set ]R1 \ {O}. To extend this section, we need the orientations to be compatible; but they are compatible in the case under consideration. Now we have the vectors W3, ... , W2n on the curve 1'1 and the vectors Wn+2, ... , W2n on the curve 1'2. By Theorem 3.16, "Tr1(V(2n-2, n-I)) = 0 for 1 < (2n - 2) - (n -I), i.e., n > 2. Therefore, the vector fields Wn+2, ... , W2n can be extended to the entire disk '!/J(T). Consider the orthogonal complement of the vectors Wn+2, ... , W2n and of the tangent space T:z;'!/J(T) at each point x E '!/J (T). On the curve 1'1. the sections W3, ... ,Wn+l of this bundle are given. Each point of the disk '!/J(T) can be represented as 1't(s), where 1 ~ s, t ~ 2. Let us extend the sections W3, ... , Wn+l, which are given at 1'1 (s), to constant sections of the trivial bundle over the curve 1't (s), where s is fixed and t varies. As a result, we obtain the required vector fields W3, ... , Wn+l on the disk '!/J(T). Now, we can prove the strong Whitney embedding theorem. We assume that T C ]R2 C ]R2n. Let us continue the embedding '!/J: T _ ]R2n to a map from a small neighborhood of T to ]R2n as follows. For x = (Xl, ... , X2n), we set 2n

+L

XiWi ('!/J(x)) , where x = (x}, X2)· i=3 Since the vectors W3, .. . ,W2n form a basis of the orthogonal complement of the tangent space T:z;'!/J(T) at X E '!/J(T), the Jacobian of the map t/J is '!/J(X) = '!/J(x)

333

3. Embeddings and Immersions

nonzero at x. Therefore, 'ljJ is a diffeomorphism from some neighborhood U of the disk 'ljJ(r) onto a domain in ]R2n. We set Ni = 'ljJ-l(f(Mi)) and N; = 'ljJ-l(f(M;». Our immediate goal is to deform N; in the domain U so as to kill two intersection points with Ni while preventing the appearance of new intersection points. To describe the required deformation, it is convenient to assume that the preimage of the curve 1'1 under 'ljJ lies on the xl-axis (see Figure 27). Moreover, we assume that the curves 1'1 and 1'2 are naturally extended so

,pI

Figure 27. The location of the disk

T

as to exit the small neighborhood of the disk 'ljJ(r); on their extensions, the vector fields wa, ... , W2n are constructed in the same way as on the curves themselves. At each point X E 'ljJ-l bI), the tangent space TxNi is the plane (Xl, Xa, ..• , Xn+l); therefore, Ni differs little from this plane. Performing a small deformation inside a small neighborhood of the curve 'ljJ-lbl), we can assume that Ni coincides with a part of this plane (possibly, we need to replace U with a smaller neighborhood of'ljJ-lbJ). Similarly, we can assume that N; coincides with a part of the manifold obtained by applying all vectors of the space (X n +2, ... , X2n) to each point of the curve 'ljJ-l('Y2) and its extension. Consider the projection 71':

(Xl, X2,

Xa, ..• , X2n) ~

(Xl, 0, Xa,···,

X2n)'

Clearly, 7l'(Ni) = Ni, and the set 7l'(Nr) is contained in the space (Xl, X n+2, Therefore, the set Ni n 7l'(N;) lies on the xl-axis. Let us subject Ni to the following deformation, which is the identity outside the c-neighborhood of the disk r. First, all points of N; are translated parallel to the x2-axis; thus, the set 7l'(Ni) does not change. The curve 'ljJ-I (2) and its extension are moved to the domain X2 < 0 (see Figure 28). For any other point (Xl, X2, •.• , X2n) E Ni, the displacement vector is obtained upon multiplying the displacement vector of (Xl, X2, 0, ... , 0) by >'(x~ + ... + x~n), where >'(0) = 1 and >'(t) = 0 for t > c 2. The projection 71' takes the intersection points of Ni and Nr to those of Ni and 7l'(Nr); hence if the plane (Xl, X2) contains no intersection points of .•• , X2n)'

6. Miscellany

334

Figure 28. The Whitney trick

the manifold Nf and the deformed manifold N!], then these manifolds do not intersect at all. Clearly, if c is sufficiently small, then the image of the c-neighborhood of the disk T under the map 1j; contains no other parts of the manifold J(Mn). Therefore, no new self-intersections arise. We have proved the Whitney theorem in the case where the self-intersection numbers for the points p and q are of opposite signs. In the cases where the manifold Mn is nonorientable and where it is orient able and n is odd, the main construction is the same. But to apply it, we may need to change the curves 'Yl and 'Y2. It is required that under the translation of the pair of normal spaces to Mf and M'2 from p to q along the curves 'Yl and 'Y2, the orientation of JR2n determined by this ordered pair of spaces change. If it does not change and the manifold M n is nonorientable, then we can extend one of the curves 'Yl and 'Y2 so that the translation along the added part of the curve will change orientation (a change of the orientation of the tangent space entails a change of the orientation of the normal space). If the manifold is orient able but has odd dimension n > 1, we replace the curves 'Yl and 'Y2 with 'Yi and 'Y~, where 'Yi starts from the point p along the curve 'YI and arrives at the point q along 'Y2, and 'Y~ starts along 'Y2 and ends along 'YI. As a result, the spaces in the ordered pair are interchanged, which leads to a change of orientation for odd n.

Remark. A similar argument proves that for n > 2, any continuous map 8 n --+ M 2n, where M2n is a simply connected manifold, is homotopic to an embedding. But it turns out that there exist maps 8 2 --+ M 4, where M4 is a simply connected manifold, that are not homotopic to embeddings. An example is given in [70]. Now, let us prove that any closed 2-manifold can be embedded in ]R4. Clearly, a sphere with 9 handles can be embedried even in IR3. It is also clear that if MJ and M1 can be embedded in IR 4 , then the connected sum MJ # MJ can be embedded in ]R4 as well. Therefore, it is sufficient to prove that JRp2 embeds in ]R4. An embedding of ~'p2 into ]R4 can be constructed ru. follows. Consider the map from the unit sphere 8 2 C ]R3 to ]R4 defined by (x, y, z) '-

3. Embeddings and Immersions

335

(xy, yz, zx, x 2 - y2). Let us show that this map induces an embedding of JRp2 into JR4, i.e., that it is an immersion and the preimage of each point consists of precisely two antipodal points of the sphere. First, we prove the assertion concerning the preimage. If abc i- 0, then the equalities xy - a, yz - b, and zx - e imply x = ±Jae/b, y - a/x, and z = e/x. If a - b - e - 0, then two of the numbers x, y, and z are zero and the third is equal to ±1. It remains to consider the case where two of the numbers a, b, and e are zero and the third is nonzero. In this case, one of the numbers x, y, and z is zero and the two others are not. Now, we employ the equality x 2 - y2 = d (or the equivalent equality 2x2 + z2 = 1 + d). If x - 0, then y - ± R and z = b/y. If y - 0, then x = ±Vii and z - e/x. If z = 0, then x = ±~v'f+d and y = a/x. Let us prove that the map f: JRp2 ~ JR4 is an immersion, i.e., its Jacobian matrix has rank 2 at each point (x, y, z) E 8 2 . The projective plane can be covered by three charts with coordinates (x, y), (x, z), and (y, z), in which z i- 0, y i- 0, and x i- 0, respectively. In the first chart, the Jacobian matrix of f is (

y

-~

x

z-~

2X) _11.

-2y

z

The matrix formed by the first and last columns is singular if and only if x 2 +y2 = 0, i.e., x = y = and z = ±1. But in this case, the matrix formed by the second and third columns is nonsingular. In the second chart, the Jacobian matrix of f has the form

°

-.! Y

z

4X).

x

2z

We are interested only in the points with z = 0. At these points, the matrix formed by the two last columns is nonsingular if x i- 0, and if z = and x = 0, then the matrix formed by the two first columns is nonsingular. The nonsingularity of the Jacobian matrix at the point (±1 : 0) in the third chart is proved similarly.

°

°:

3.2. The Normal Degree of an Immersion. Let f: Mn ~ JRn+l be an immersion of a closed connected oriented manifold. Since Mn is oriented, at each point x E j(Mn), the unit normal vector N(x) is determined uniquely. Therefore, the map f induces a Gaussian map N: Mn ~ sn. The degree of this map is called the normal degree of the immersion f. The question of what normal degree an immersion f can be, was considered, in particular, by Hopf [55, 56] and Milnor [87]. In this section, we present some of their results.

6. Miscellany

336

First, note that if an immersion /: M n -+ Rn+1 has normal degree d, then the immersion s /, where s: IRn +1 -+ IRn + 1 is the symmetry with respect to a hyperplane, has degree (-I)nd. Indeed, if vectors el,"" en, N(x) form a positively oriented basis, then sel, .. . , sen, sN(x) form a negatively oriented basis because the map s is orientation-reversing. Therefore, for the map sf, the new map Mn -+ sn is the composition of the old map N: M n -+ sn and the map -siSTi. The restriction siSTi has degree -1, and the antipodal map sn -+ sn has degree (_I)n 1. Two oriented manifolds Mf and M; determine an oriented manifold Mf # M;, which is called their connected sum. It is constructed as follows. Let Df C Mf and D~ C M; be unit disks embedded in these manifolds. We remove disks of radius 1/2 from these disks and glue together the remaining cylinders I x s~ 1 and I x ~-l so as to preserve orientation (see Figure 29). More formally, the construction is as follows. Suppose that il: D n -+ Mf is

Figure 29. Connected sum

an orientation-preserving embedding, and i2: D n -+ M; is an orientationreversing embedding. In Mf \ il(O) and M; \ i2(O), we identify il(tU) with i2((1 - t)u) for every unit vector u and each t, 0 < t < 1. The map il(tU) 1---+ i2((1 - t)u) is orientation-preserving; therefore, on the obtained manifold Mf # M;, we can choose an orientation compatible with those of the manifolds Mf and M;. Clearly, Mn # sn ~ Mn; in particular, sn#sn~sn.

Theorem 6.15. 1/ Mf and M; admit immersions into R n +1 with normal degrees d l and d 2, then Mf # M; admits an immersion into ]Rn+1 with normal degree d l + d 2 - 1. Proof. Take any points Xl E Mf and X2 E Mr. We apply to the manifold Mr an orientation-preserving Euclidean motion of R n +1 so as to render the

3. Embeddings and Immersions

337

normal vectors at Xl and X2 directed toward each other (see Figure 30); this is needed for the orientations to be compatible. Then, we perform a

cc( M~#M~

Figure 30. The immersion of a connected sum

surgery and obtain an immersion of the manifold Mf # Mr. To calculate the normal degree of this immersion, it suffices to trace how many times the map Mf # Mr --+ sn takes an arbitrary given value. The most convenient value is N(XI). Slightly moving the manifold Mf if necessary, we can assume that the contribution of the value N(XI) taken by the map Mf --+ sn at a neighborhood of Xl is +1. This means that in this neighborhood of Xl, the manifold Mf looks as a sphere, and N(XI) is the outward normal to this sphere. In this situation, no new points taken to N(xd arise under the surgery. Thus, the maps Mf --+ sn and Mr --+ sn take to N(XI) not only Xl but also some other points. Consequently, the normal degree of the constructed immersion is equal to d l + d 2 - 1. 0 Theorem 6.16 (Hopf). For any odd n, the sphere with any odd normal degree.

sn

can be immersed into

]Rn+l

Proof. Applying the symmetry about a hyperplane, we construct an immersion of into ]Rn+l with normal degree -1. We take two copies of the sphere thus immersed and apply Theorem 6.15. As a result, we obtain an immersion of the manifold # ~ with normal degree (-1) + (-1) -1 = -3. Then, we construct an immersion of with normal degree (-3) + (-1) -1 = -5, and so on. into lRn+l with all negative odd We have constructed immersions of degrees. Applying symmetries about hyperplanes, we obtain immersions with all positive odd degrees. 0

sn

sn sn

sn

sn

sn

Corollary. Suppose that n is odd and a manifold Mn is immersed into with normal degree d. Then Mn can be immersed into ]Rn+l with any normal degree of the same parity as d.

lRn+l

6. Miscellany

338

Proof. We take an immersion of sn with normal degree 2k+ 1 and construct an immersion of M n ~ M n # sn with normal degree d + 2k. D Theorem 6.17. If a closed oriented manifold Mn can be immersed into ]Rn+1 with normal degree zero, then M n is parallelizable. Proof. It is easy to see that TMn = N.(TS n ), i.e., the tangent bundle of the manifold M n is a pullback of the tangent bundle of thE:' bphere sn by means of a map N: Mn ---+ sn. Indeed, N.(Tsn) consists of the pairs (x,v) with x E M n and v 1.. N(x).

If the map N: Mn ---+ sn has degree zero, then, by the Hopf theorem, it is null-homotopic and, therefore, induces the trivial bundle. D Corollary. Suppose that a manifold Mn zs not parallelizable and n is odd. Then, if M n can be immersed into lRn +1, then it can be immersed into lR n +1 with any odd normal degree, but it admits no immersions into ]R.n+ 1 of even normal degree. Proof. Apply Theorem 6.17 and the corollary of Theorem 6.16.

D

Problem 133. Prove that if the sphere sn is parallelizable, then so is any closed orient able manifold M n that can be immersed into ]R.n+1. Every embedding of a manifold Mn into ]R.n+1 is an immersion. Therefore, for any embedding, its normal degree is defined. The normal degree of an embedding satisfies fairly strong conditions. To formulate them, we introduce the quantity E(Mn) = Ei>O dimH,(Mnj F), where F is the additive group of a field. Theorem 6.18 (Milnor [87]). If d is the normal degree of an embedding of Mn into ]R.n+1, then d == ~E(Mn) (mod 2) and Idl ~ ~E(Mn). Proof. The set ]R.n+1 \ Mn has two connected components, one of which is boundedj we denote it by A. We can assume that the manifold Mn is oriented so that the normal vector is directed outside of A. The normal degree d is equal to X(A). This is proved in the same way as the main lemma in the proof of the Poincare-Hopftheorem (Theorem 5.28 in Part I). It is seen from the definitions that X(A) == E(A) (mod 2) and -E(A) ~ X(A) ~ E(A). Moreover, dim Ho(Aj F) = 1j therefore, 2 - E(A) ~ X(A). Now let us prove that E(A) = !E(M). Consider the one-point compactification sn+1 of lRn+1. We set B = sn+1 \ A. According to the Alexander duality theorem, we have Hk(Mn) ~ H n _k(sn+1 \ Mn) for 0 ~ k :::; n. If 1 :::; k ~ n - 1, then dimHn_k(M n )

= dim Hk(Mn) = dim H n _ k (sn+1

= dim Hn-k(A) + dim Hn-k(B).

\ Mn)

4. Complex Manifolds

339

The equality dimHO(Mn) = 0 implies dim Hn(A) - dim Hn(B) = O. Therefore, ~(M) = ~(A) + ~(B) because dimHo(Mn) + dim Hn(Mn) = 2 = dim Ho(A) + dim Ho(B). We can apply the Alexander duality theorem not only to M n but also to A. As a result, we obtain Hk(A) '" fIn k(sn+1 \A) = hn k(B) for 0 ::; k ::; n, whence ~(A) = ~(B). Clearly, ~(A) = ~(A) because the spaces A and A are homotopy equivalent. Thus, 2~(Mn) = E(A). 0

4. Complex Manifolds 4.1. Complete Intersections. Let Y be a complex manifold of dimension n, and let X c Y be its complex submanifold of dimension n - 1. This means that X can be covered by open subsets Ui of Y so that in each Ui, X is determined by the equation It - 0, where I, is a holomorphic function such that grad It (x) =1= 0 for all x E Ui n X. We can assume that the sets U, cover Y (if Ui n X - 0, then we set Ii - const -I 0). We associate with the submanifold X C Y a one-dimensional complex bundle3 ~(X) over Y with transition functions gij(X) = 11(x)/1i(x). If x E X, then gij (x) = liIDyf/.x,y ...... x 11 (x) /IJ (x); this limit is equal to the proportionality coefficient of the vectors grad 11 (x) and grad 1i (x). The bundle ~(X) does not depend on the choice of the functions IiIndeed, suppose that in Ui , the set X is determined by the equations Ii = 0 and Pi = O. Then the function hi = Id Pi does not vanish on Ui , and .. g- ZJ

-!!:J.. . h. g1J'

-

The bundle ~(X) has a holomorphic section that vanishes precisely on X. On Ui, this section is determined by k Such a section is well defined because Ii = gij Ij· Example 75. Let Xd be a nonsingular hypersurface of degree d in cpn given by a homogeneous equation F(zo, ... , zn) = 0 of degree d. Then (~(Xd)* ~ '"Y~ ® ... ® '"Y~. ~

d

Proof. In the chart Ui , the hypersurface Xd is defined by the equation li(wo, ... , Wi,··· ,Wn ) = 0, where F J'

(ZO Zi

1~ "'."

... ,

Zn) _ F(zo,· d.. , Zn) . -

Zi

Zi

3In algebraic geometry, this bundle is called the line bundle associated with the divisor X and denoted by [Xl· We do not use this notation because in topology, [Xl usually denotes the fundamental class of the manifold X.

6. Miscellany

340

Therefore, the bundle ~(Xd) is determined by the transition functions 9ij = = (.;.-) d, while the canonical bundle 'Y~ over CPR is determined by the transition functions 9ij = ~ (see Example 67 on p. 273). 0

l

J

In particular, for a hypersurface CP"-l C CPR specified by a linear equation, the bundle ~(CP") is isomorphic to the bundle defined in Example 69 on p. 274. It is easy to show that the restriction of the bundle ~(X) to X is the normal (in the Hermitian metric) bundle to X in Y. Indeed, consider the bundle v over X whose fibers over Ui are vectors proportional to grad fi. If A, grad fi = Aj grad f" then Ai = Aj. Therefore, v is determined by the transition functions 9,j = f,/ fl. The bundle v is complex conjugate to the normal bundle v; hence v is determined by the transition functions 9,j = f,/ h, which coincide with those for ~(X).

f.

Example 76. Let Xd be a nonsingular hypersurface of degree d in CP".

Then

C(Xd) = (1 + a)"+1(1 - da + d2 a 2 - d3 a 3 + ... ), where a is the restriction to Xd of the canonical4 generator of the group H2(cpn; Z) (the sum is finite because an = 0). Proof. Let j: Xd --+ cpn be the natural embedding. Then a = j*n, and j*(~(Xd)) = VXd is the normal bundle to Xd in CpR. Therefore, j*(Tcpn) = TXd ffiVXd = TXd ffij*(~(Xd)). Taking into account the equalities C(~(Xd)) = -dch~) = dn and c(cpn) = (1+n)n+1, we see that (1+a)n+1 = c(Xd)(1 + da), and hence

C(Xd) = (1 + a)"+1(1 + da)-l

o Example 77. Let Xd be a nonsingular hypersurface of degree d in CPR.

Then x(Xd )

= ~ (_1)kdk+1 ( ~

k=O

n

+1

n-1-k

).

Proof. In Example 76, we calculated the total Chern class of the manifold X d • In particular, we obtained

C.-l(X.) To calculate X(Xd)

~ ~ (-l)kd"(n:; ~ k};n-l

= (Cn-l(Xd), [Xd]),

4That is, taking the value +1 on

Cp2

it remains to find (l"i n -

with canonical complex orientation.

l ,

[Xd]).

4. Complex Manifolds

341

The number (ii n- l , [XdD is equal to the number of the intersection points in cpn of the cycle dual to o:n-l and Xd (with signs taken into account). The cycle dual to o:n-l is the complex line Cpl. In general position, it intersects a surface of degree d at d points because the restriction to this line of a homogeneous polynomial F in n + 1 variables is a homogeneous polynomial of degree d in two variables, which corresponds to an inhomogeneous polynomial of degree d in one variable. All intersection numbers are positive because we deal with complex manifolds. Thus, the homology class [Xd] is d[cpn-l]. 0 In particular, for a smooth algebraic curve Xd of degree din C 2 , we have X(Xd) = (~)d - (~)d2 = 3d - cPo Therefore, Xd is a sphere with 9 handles, where 2 - 2g = 3d - d 2 , i.e., 9 = (d-l~d 2). In a similar way, we can calculate the total Chern class and the Euler characteristic not only for a nonsingular hypersurface but also for an intersection of generic nonsingular hypersurfaces, as in [22]. A complete intersection is the complex n-manifold M = MI n ... n Mp c cpn+p , where M I , ... , Mp are the hypersurfaces defined by the homogeneous equations fI = O,···,/p = 0 such that the vectors gradfI, ... ,grad/p are linearly independent at each point x EM. Example 78. Suppose M = MI n· .. n Mp c cpn+p is a complete intersection, and the hypersurfaces Ml, ... , Mp have degrees dl, ... , dp, respectively. Suppose also that the hypersurfaces M I , ... , Mp are nonsingular. Then

X(M)

~ (t.(-I)t:~: 1)"0) P'd;,

Proof. Suppose that j: M - cpn+p is the natural embedding, and 0: E H2(cpn+p; Z) is the canonical generator. We set a = j*o:. Then c(Cpn+P) = (1 + o:)n+p+l and j*T(cpn+p ) ~ TM E9 lI, where II is the normal bundle to M in Cpn+p. Clearly, II is the direct sum of the bundles {(Mi); therefore, C(lI) = (1 + dla) ... (1 + dpa). Thus, c(M)

=

(1

+ a)n+p+l(1 + dla)-l ... (1 + dpa)-l.

In particular,

It remains to prove that (an, [M])

dimensional subspace of

CP

=

d l ... dp , i.e., that any generic pintersects M in d l ... dp points. This can be

6. Miscellany

342

done in different ways. First, we can employ the fact that the cycle Mi is homologous to d, [cpn+p 1]. Indeed, consider the intersection of p generic copies of cpn+p I in cpn+p (this intersection is cpn). The intersection number of [CP P ] and [M] is equal to that of [CP P ] and [cpn] multiplied by dl ... dp . Secondly, we can use the method of elimination of variables, which is based on calculating the resultant of two polynomials. Applying this method to the restrictions of homogeneous forms of degrees dl,"" dp to a p-dimensional subspace, we obtain a polynomial of degree dl ... dp • D

Remark. A more careful argument shows that we can dispense with the assumption that the hypersurfaces MI, ... , Mp are nonsingular; see [9].

+ ... +

4.2. The Homology of the Hypersurface ZOO zBn = 1. Let ao, ... , an be positive integers, and let a - (ao, ... ,an), Consider the hypersurface Va in C n +! defined by the equation zao

+ ... + zan

_ 1.

In [101]' Pham calculated the homology groups of the space Va by finding a deformation retract Ua of Va whose homology groups are easy to calculate. In [53], it was shown that Ua is homotopy equivalent to the wedge of (ao - 1) ... (an - 1) n-spheres. Thus,

-

Hk(Va )

-

{o

z(ao-l) ... (an-l)

if k if k

i- n, = n.

The subspace Ua C Va is given by Ua - {x E Va

n ]Rn+! I x~·

~ 0 for i

= 0, ... , n},

and the deformation retraction from Va to Ua is constructed as follows. Consider the map C n +! --+ C n +! defined by

The image of Va under this map is the hyperplane X in en+! determined by the equation Uo + ... + Un = 1. The map p: Va --+ X is an (ao ... an)-fold covering branched over the intersections of X with the hyperplanes Ui = O. The standard simplex l:!,.n in ]Rn+! C C n +!, which is given by the inequalities to ~ 0, ... , tn ~ 0 and the equation to + ... +tn = 1, is a deformation retract of the hyperplane X. Indeed, a deformation retraction can be obtained from a deformation retraction of C n+ l onto ]Rn+l, a deformation retraction of ]Rn+! onto the hyperplane X n ]Rn+!, and a deformation retraction of the hyperplane onto the simplex l:!,.n, which is shown in Figure 31 for n = 2. This deformation retraction does not affect the type of branching over any moving point; therefore, the deformation retraction of X onto the simplex l:!,. n can be lifted to a deformation retraction of Va onto the preimage

4. Complex Manifolds

343

Figure 31. The deformation retraction

of ~ n under the branched covering p: Va coincides with Ua. Example 79. For the hypersurface z~ with the graph Kn,m (see Part I, p. 7).

~

X.

+ zf =

Clearly, this preimage

1, the space Ua coincides

Proof. The corresponding branched covering over the I-simplex ~l for n = 2 and m = 3 is shown in Figure 32. 0

Zo

Figure 32. The branched covering

Now, let us prove that the space Ua is homotopy equivalent to the wedge of (ao - 1) .. , (an - 1) n-spheres. We identify Ua with the space of n-tuples (uoto, ... ,untn ), where each Ui is an aith root of unity, ti ~ 0, and E ti = 1. This means that Ua coincides with the join Zao * Zal * ... * Zan' where Zm is the m-point discrete space. Since Zm = ~, we have m-l Zao

* Zal * ... * Za

rv

sn V ... V sn

n~'

(ao-l) .. ·(an- 1)

6. Miscellany

344

because SO

* .,. * SO =

~

sn; we have also used the relation

n+l

Problem 134. Consider Zm as the group of mth roots of unity. Then Ua = Zao * ... *Zan is an n-dimensional simplicial complex, and there is a natural one-to-one correspondence between its n-simplices and the elements of the group Za = Zao E9 ... E9 Zan: each n-simplex is obtained from the simplex e corresponding to the identity element of Za under the action of an element of Za. Let WO, ••• , Wn be the elements of Za corresponding to the identity elements of the groups Zao' ... , Zan' (a) Prove that e = (1 - wo) ... (1 - wn)e is a cycle. (b) Prove that the kernel Ia of the self-map of the group algebra Z[Za] defined by W t-+ w(1 - wo) ... (1 - wn ) is generated by the elements 1 + Wi

+ w? + ... + w~·-l,

where i

= 0,1, ... , n.

(c) Prove that the elements W~OW~l

•••

w~ne, where 0 ~ ki ~

ai -

2,

form a basis in the integral homology of the space Ua .

5. Lie Groups and H-Spaces 5.1. Some Properties of Lie Groups. A Lie group is a manifold G that is also a group for which the maps G x G - G and G - G defined by (g, h) t-+ gh and 9 t-+ glare smooth. To avoid delving deeply into the details of Lie group theory not related to algebraic topology, we consider only matrix Lie groups, which are submanifolds in the space of nonsingular square matrices over the field 1R; the multiplication in this group is the usual multiplication of matrices, and the identity element is the identity matrix I. To each square matrix A we assign the matrix exp A = I + A + ~ + 1~ + . .. (the proofs of the convergence of this series and of the basic properties of the map expA are given, e.g., in [104]). The map expA is a diffeomorphism from some neighborhood of the zero matrix to a neighborhood of the identity matrix. The tangent vector to the curve exp(tA) at t = 0 equals A. Therefore, the tangent space to the (matrix) Lie group at the identity is the linear space spanned by the preimage of a neighborhood of the identity under the exp map; this linear space is called the Lie algebra of the given Lie group. Any connected Lie group is completely determined by a neighborhood of the identity element (which we regard as a subset of the space of matrices); therefore, it is determined by its Lie algebra (which we also regard as a subset of the space of matrices).

5. Lie Groups and H -Spaces

345

A Lie group is commutative if and only if all matrices in its Lie algebra pairwise commute. Problem 135. Prove that the image of a linear subspace V of the space of matrices under the map exp is a Lie group if and only if, for any two matrices A, BE V, the matrix [A, B] = AB - BA belongs to V. The Lie group T" =

21

X •••

'"

x

S:

= ]Rn /zn is called the toros.

n

Theorem 6.19. Any connected commutative Lie group G is isomorphic to T" x jRm. Proof. For a commutative Lie group G, the map exp: TeG -+ G is a group homomorphism because if AB = BA, then eAe B = eA +B . For a connected Lie group, this homomorphism is an epimorphism. Therefore, G ~ TeG/K, where K = Ker expo The group K is discrete because exp is a diffeomorphism in a neighborhood of zero. Hence this group is a discrete subgroup in Te G ~ jRn+m. There is a basis in jRn+m in which the elements of K have the form (k 1 , ••• , kn, 0, ... ,0), where k, E Z. Thus, jRn+m/K ~ Tn X ]Rm. 0 Corollary. Any compact connected commutative Lie group is the torus. Theorem 6.20. Any closed subgroup H of a Lie group G is a submanifold. Proof. Take a norm on the space of matrices, say IIAII = max laijl for A = (aij). Suppose that in the space TeG, a matrix A and a sequence of matrices An such that expAn E H, II An II -+ 0, and An/IIAnll -+ A are given. Let us show that etA E H for any real number t. Since IIAnli -+ 0, we can choose integers mn such that mnllAnli converges to any given number t. We have mnAn -+ tA and emnAn -+ etA. But e mnAn = (eAn)mn E H, and the group H is closed. Therefore, etA E H. Let W be the space of all matrices of the form tA, where A is obtained by using the passage to the limit specified above and t is an arbitrary real number. Let us show that W is a linear space. To this end, it suffices to check that if A, B E W, then A + B E W. In a small neighborhood of the identity matrix, the map exp is invertible; therefore, for small t, we have etAetB = eF(t), where F(t) is a smooth curve in TeG and F(O) = O. Taking into account the relation etAe tB = I + t(A + B) + ... , we see that F(t)/t -+ A + Bast -+ O. Consider the sequence of matrices C n = F(l/n). Clearly, IICnll -+ 0, eC " E H (because etAetB E H for all t), and Cn/IICnll -+ oX(A + B), where oX = J~~ ~IF(~) Hence A + BE W.

1-1.

Let us prove that the set exp W contains some neighborhood of the identity element in H. We decompose TeG into the direct sum W ffi W' and I W d W I E W.I On some neighborhood of the put cp(w, w) = e w eWi £or wEan

346

6. Miscellany

zero matrix in TeG, the map r.p acts as a diffeomorphism to a neighborhood of the identity matrix in G because r.p(TeG) c G and the differential of r.p at zero coincides with that of exp, since eWe w' ~ eW+w' for small wand w'. Suppose that there exists a sequence of points (w n , w~) such that r.p(wn, w~) E H, (wn,w~) -.. 0, and w~ f. o. Then eW~ E H because eWnew~ E Hand eWn E H. Choose a subsequence W~k such that the sequence w~lk/llw~k II has a limit w'. On one hand, w' E W' and w' f. o. On the other hand, w' E W. We have arrived at a contradiction. Thus, the identity element of G has a neighborhood U whose intersection with H is a sub manifold in U. For an arbitrary point h E H, we take the neighborhood hU. 0 Theorem 6.21. For an arbitrary action of the torus Tn on a vector space V, there exists an invariant inner product in V (that is, an inner product such that (u, v) = (tu, tv) for any tErn and u, v E V). Proof. Take any inner product (u, vh in V and consider the function f(s) = (su,svh on the torus. Let (u,v) = ITnf(s)dJ1., where dJ1. = dXl···dxn is the canonical measure on Tn. This measure is translation-invariantj therefore, if g(s) = f(s + t), then ITn g(s) dJ1. = ITn f(s) dJ1.. Thus, (u, v) = ITn(su, svh dJ1. = ITn g(s) dJ1. = ITn f(s) dJ1. = ITn«t + s)u, (t + s)vh dJ1. = (tu, tv). 0 5.2. Cohomology of Lie Algebras. The de Rham cohomology of a compact connected Lie group G can be expressed in terms of the Lie algebra g of G, which reduces calculating the real cohomology of G to a purely algebraic problem. In this section, we describe this reduction. As in Subsection 5.1, we consider only matrix Lie groups and algebras. The Lie Algebra of a Lie Group. As mentioned on p. 344, the Lie algebra g of a Lie group G is the tangent space at the identity element e E G. But this definition refers only to the linear structure of g. To define an algebra structure, we must define the product of any two matrices A, BEg in gj this product is usually denoted by [A, B] and called the commutator of A and B. The commutator is defined as follows. Each matrix A ~ g determines the left-invariant vector field VA(9) = (Lg).A, where Lg(x) = gx is the left translation by 9 on the group G. Since the translation is by a constant matrix g, it follows that, in the matrix form, this left-invariant vector field can be written as VA(g) = gAo A left-invariant vector field can be defined as a vector field on G invariant with respect to left tran~latlOns. Then any left-invariant vector field is uniquely determined by the vector at the point ej this vector is spread over the entire group by left translations. For two

347

5. Lie Groups and H -Spaces

left-invariant vector fields VA and VB, their commutator [VA, VB] is defined. This is again a left-invariant vector field vc. The matrix C, i.e., the vector [VA, VB] at e E G, is defined to be the commutator of the matrices A and B. Theorem 6.22. For matrix Lie algebras, [A, B] = AB - BA is the usual commutator of matrices. Proof. Any matrix Lie group G is a subgroup in the group GL(n, R) of nonsingular matrices of order n. The commutator of two vector fields on G is the same as the commutator of these vector fields considered on GL(n, JR). Therefore, is suffices to prove the required assertion for GL(n, R). In this case, the elements x) of the matrix Ilx)ll? can be taken for local coordinates. We are interested in the commutator of vector fields with coordinates VA(X)) = x~A~ and VB(X)) = x~Bf (with summation over repeated indices). By definition, [VA(X), VB(X)] = Vc{X) , where

k _ i AP 8 (x:Bj) _ i BP 8 (x:A7) vc (x )I - xp j 8 i xp j 8 i Xj

Xj

q j J>k ~j 6k - xpi BPj AqI U -- xpi APj B I 6qUi q i - xpk(APqBqI Thus, vc{x)t

= x!Cr,

where C

-

BPqAq) I .

= AB - BA.

0

The Maps Ad and ad. Consider the transformation x 1-+ gxg- 1 on the Lie group G. In a neighborhood of the identity, it can be represented as etX 1-+ ge tX g-l, where X is an element of the Lie algebra g. It is easy to see that the derivatives with respect to t of etX and ge tX g-1 at t = a are X and gXg- 1 , respectively. Therefore, the map X 1-+ gXg-l, where 9 E G, is a self-map of the Lie algebra g. It is denoted by Adg • Moreover, any fixed matrix A E 9 determines the map X 1-+ [A, X] = AX - XA of the Lie algebra g. It is denoted by adA. The maps Ad and ad are related as follows: adA X is the derivative of Adexp(tA) X with respect to t at t = O. Indeed,

etA X e- tA

= (I + tA + ... )X(I - tA + ... ) = X + t(AX -

X A)

+ ....

Theorem 6.23. If G is a compact connected Lie group, then det Ad g = 1 for any 9 E G. Proof. It is easy to see that the map 9 1-+ Adg is a homomorphism from the group G to the automorphism group of the space g. Therefore, the map 9 1-+ det Ad g is a homomorphism from the group G to the multiplicative group JR* of nonzero real numbers. Clearly, this homomorphism is continuous; hence its image is a compact connected subgroup in JR*. But such a subgroup can contain only the element 1. 0

6. Miscellany

348

Integration over Compact Lie Groups. A differential k-form W on a Lie group G is said to be left-invariant if L;w = w for any 9 E G. Similarly, a right-invariant k-form is a form for which R;w = w, where Rg(x) = xg is the right translation by g. A differential form is bi-invariant if it is both left- and right-invariant. Theorem 6.24. On any compact connected Lie group G of dimension n, there exists a nonzero bi-invariant n-form, and it is unique up to proportionality. Proof. Let We be a nonzero n-form on the tangent space at the identity element e E G. It uniquely determines a left-invariant form w L and a right-invariant form w R by w;(X1, ... , Xn) = we(gX1 , ... , gXn) and by w:(XI, ... ,Xn ) = we(X1g, ... ,Xng) at every 9 E G. Up to multiplication by a nonzero constant, we(X1 , ... , Xn) is the determinant of the matrix composed of the coordinates of the vectors XI, ... , X n . Hence w: = (det A)w;, where A is the determinant of the transformation taking each vector Y = X 9 to gX = gYg-1j i.e., A = Adg • Thus, on any Lie group of dimension n, there exists at most one (up to proportionality) nonzero bi-invariant n-form, and such an n-form exists if and only if det Ad g = 1 for all 9 E G. According to Theorem 6.23, any compact connected Lie group does satisfy this condition. D Theorem 6.24 implies that, on any compact connected Lie group G of dimension n, there exists a unique bi-invariant n-form w for which JG w = l. Using this n-form, we can define the integral JG f = JXEG f(x)w for any continuous function f: G -+ JR. This integral has the properties JG 1 = 1 and

1

~G

f(xg)

=

1

~G

f(gx)

=

1

f(x)

for any 9 E G.

~G

Exercise. Prove that Theorem 6.21 is valid not only for the torus T" but also for any compact connected Lie group. To define averaging, which is an analog of the arithmetic mean of functions on the group Zn, for functions on a compact Lie group G, we need integration over G. Left-Invariant Forms. Any left-invariant differential form is determined by its values on the sets of vectors of the tangent space at the identitYj therefore, it is also determined by its values on the sets of left-invariant vector fields. If w is a left-invariant differential k-form and 6, .. · ,t;.k are leftinvariant vector fields, then w(t;.I, ... ,t;.k) = constj hence T/W(t;.l,··· ,t;.k) = a

5. Lie Groups and H -Spaces

349

for any vector field TJ. As a consequence, the formula for the differential of a form becomes simpler, namely,

(44)

dw({o, ... ,{k) =

L

(-l)i+i w ([{IO{J],···,€,,···,€i,···,{k).

O$I
For the Lie algebra 9 of a Lie group G, we define the space of co chains Ck(g; JR) as the space of skew-symmetric k-linear maps 9 x ... x 9 --+ JR; the map d: C k --+ CHI is defined by (44). The cohomology of the cochain complex thus obtained is called the cohomology of the Lie algebra 9 with coefficients in JR. This is the same as the cohomology of the complex of left-invariant differential forms. Generally, it does not coincide with the cohomology of the complex of all differential forms, i.e., with the de Rham cohomology of the Lie group G. But these cohomologies do coincide for compact (and connected) Lie groups G.

Theorem 6.25. If a Lie group G is compact and connected, then the cohomology of its Lie algebra 9 is isomorphic to the de Rham cohomology of the manifold G. Proof. The proof uses the existence of the integral IgEG f(g) on a compact connected Lie group G and its properties specified on p. 348. First, note that if forms w and w are left-invariant and w - w = do. for some form a, then w-w = da L for some left-invariant form a L . Indeed, since the form do. is left-invariant, it follows that do. = IgEG L;(da) = d IgEG L;a. Clearly, the form a L = IgEG L;a is left-invariant as well. Now we prove that if w is an arbitrary closed form, then there exists a left-invariant form w L such that w = w L + do. for some form a. Since the group G is path-connected, for each 9 E G there exists a path g(t) from g(O) = e to g(l) = g. Using this path, we obtain a family of forms Wt = L;Ct)w which join Wo = w to WI = L;w. Clearly, the form L;w is closed because so is w. Therefore, the homotopy invariance of the de Rham cohomology (see p. 282) implies that w - L;w = dag for some form a g. The dependence of a g on g is not necessarily continuous, but, in any case, it can be assumed to be integrable (piecewise continuous) in g. Consider the left-invariant form w L = IgEG L;w. Clearly,

w - wL = where

0.=

IgEG a g.

1

gEG

(w - L;w) =

{

19EG

dag = drag = do., 19EG D

Corollary. If GI and G 2 are compact connected Lie groups one of which covers the other, then H* (G 1; JR) ~ H* (G 2 ; lR) .

6. Miscellany

350

Example. For the torus rn, all differentials in the complex of left-invariant differential forms are zero; therefore, Hk(Tn; JR) ~ Ak(JRn ) is the space of skew-symmetric k-forms on JRn . Remark. For noncompact Lie groups, Theorem 6.25 is false. For example, rn and JRn have isomorphic Lie algebras, but the spaces H*(rn; 1R) and H*(JRn ; 1R) are not isomorphic. Bi-Invariant Forms. Theorem 6.26. If G is a compact connected Lie group, then the de Rham cohomology algebra of the manifold G is isomorphic to the algebra of biinvariant differential forms. Proof. According to Theorem 6.25, the de Rham cohomology algebra of the manifold G is isomorphic to the cohomology algebra of the complex of left-invariant differential forms. For the latter, we can repeat the argument from the proof of Theorem 6.25 with left translations replaced by right ones. As a result, we obtain an isomorphism between the de Rham cohomology algebra of the manifold G and the cohomology algebra of the complex of bi-invariant differential forms. It remains to prove that the differential on this complex is zero, i.e., dJJJ = 0 for any bi-invariant form w. Consider the map i: G -+ G defined by i(g) = g-l. It takes the curve ge tX to e- tX g-I; therefore, the tangent vector Y = gX at 9 is taken to -Xg-I = _g-IYg- l . Thus, ifw is a bi-invariant k-form, then

(i*wg)(Xl. ... ) = Wg

= Wg 1 (_g-1 XIg-l, ... ) = wee -XIg- I , ... ) = Wg( -Xl' ... ) = (_I)kWg(Xl' ... )'

i.e., i*w

= (_I)kw.

1

(i*Xl, ... )

Hence

(_I)k+1 dJJJ = i* dJJJ = di*w = (_I)k dJJJ, which means that dJJJ = O.

o

5.3. Maximal Tori. Let G be a topological group. An element 9 EGis called a topological generator of G if the closure of the subgroup generated by 9 coincides with the entire group G. Theorem 6.27. The torus Tn has a topological generator. Proof. Consider a countable open base UI, U2, ... for the topology of T"'. By a cube in Tn we mean the image of the cube given by IXi - ail ~ e: in JRn under the canonical projection. Clearly, any cube is a compact subset of the torus. We refer to the number min(I,2e:) as the edge length of the cube. If a cube has edge length 1, then it coincides with Tn.

5. Lie Groups and H -Spaces

351

Take any cube Co in Tn. Suppose that we have constructed a sequence of cubes Co ::J Cl ... ::J Ck-l' Let us construct a cube C k as follows. Suppose that the edge length of C k - 1 equals c. Choose a number N(k) so that 2N(k)c > 1. The N(k)th power of the cube Ck-l coincides with rn. Therefore, we can choose a cube Ck C Ck-l so that C~(k) C Uk. Take 9 E n~ 0 Ck. We have gN(k) E Uk; therefore, 9 is a topological generator of the torus rn. 0 Let G be a compact connected Lie group. The term maximal torus is used for a subgroup T C G that is a torus which is maximal in the sense that if T c T' C G, where T' is a torus, then T = T'. Any torus T eGis contained in some maximal torus. Indeed, if tori T C c G form a strictly increasing sequence, then their dimensions strictly increase as well; indeed, if a closed n-manifold is a submanifold in an n-dimensional manifold, then it is both open and closed. Therefore, any strictly increasing sequence of tori is finite.

Tl C T2 C ...

Let 8 C G be an arbitrary subset of a group G. The normalizer of 8 is the group N(8) = {g E G I g8 = 8g}, and the centralizer of 8 is the group

Z(8)

= {g

E G

I g8 = sg Vs E 8}.

If 8 is a subgroup of G, then 8 is a normal subgroup of N(8). If G is a Lie group and S is its subgroup, then N(8) and Z(8) are closed subgroups of G, and therefore, according to Theorem 6.20, they are Lie groups. For any element 9 E N(8), the inner automorphism x to 8; thus, N(8) acts on 8.

1-+

gxg- 1 maps 8

Theorem 6.28. Let T C G be a maximal torus. Then

(a) the quotient group N(T)/T is finite; (b) the dimension of the spa"e G/N(T) is even, X(G/N(T)) X(G/T) = IN(T)/TI·

= 1, and

Proof. (a) Consider the map N(T) ~ Aut(T) that assigns the automorphism x 1-+ gxg- 1 to each element 9 E N(T). The group Aut(T) is discrete because any automorphism of the torus is determined by an integer matrix with determinant ±1 (any such automorphism leaves the integer lattice invariant). Therefore, the connected component N(T)e of the identity element is mapped to the identity automorphism, i.e., N(T)e C Z(T). Let A be a tangent vector to Z(T) at e. Consider the subgroup H generated by the torus T and the elements eAt, where t E JR. The subgroup H is closed; therefore, it is a Lie subgroup. Moreover, the group 1I is

6. Miscellany

352

connected and commutative. Therefore, it is a torus containing T. The maximality of T implies H = Tj in particular, eAt E T. Hence Z(T)e = T. But T C N(T)e C Z(T)e, whence Z(T)e = N(T)e = T. Thus, N(T)/T = N(T)/N(T)e is a discrete compact group, and it must be finite. (b) Consider the action of the torus T on the quotient space G/N(T) by left translations. Under the action of t E T, the class gN(T) is mapped to tgN(T). The class N(T) remains fixed because T C N(T). Consider the action of T on G by left translations. This action can be transferred to the tangent space TeG. Take a T-invariant inner product in TeG. Let (TeN(T».L be the orthogonal complement to TeN(T) in TeG. We have TNCT)(G/N(T» ~ TeG/TeN(T) ~ (TeN(T».L j each element ofT acts on TNCT) (G/N(T» as an orthogonal transformation. Using the restriction ofthe map exp to the space (TeN(T».L, we construct a disk D m in G/N(T), where m = dim(G/N(T», such that it is centered at the point N(T) and the three closed sets Dm, sm-l = aDm, and M = G / N (T) \ int D m are invariant with respect to the action of T. Let us show that the coset N(T) is the unique fixed point of the action of T on G / N (T) j moreover, this is so not only for the action of the entire torus T but also for the action of its topological generator t. Indeed, tgN(T) = gN(T) if and only if g-ltg E N(T), or, equivalently, g-ltng E N(T) for n = 1,2, ... j the last condition, in turn, is equivalent to g-ITg C N(T) because the set {tn} is everywhere dense in T. Finally, the condition g-ITg C N(T) is equivalent to g-ITg = T, i.e., 9 E N(T). Indeed, N(T)e = T, and the set g-ITg is connected and contains the identity element c. The torus T contains a path from t to ej therefore, the actions of t on G / N (T) and on G are homotopic to the identity map, and the actions on TeG during the homotopy respect the inner product in TeG chosen above. Thus, we have a homotopy in the class of maps leaving invariant D m , sm-l, and M. Consequently, there exist fixed-point-free maps sm-l --+ sm-l and M --+ M homotopic to the identity. By the Lefschetz theorem, we have X(M) = 0 and X(sm-l) = OJ hence m is even. The formula for the Euler characteristic of the union of two CW -complexes implies

X(G/N(T»

= X(D m ) + X(M)

- X(sm-l)

= X(Dm) = 1.

The natural projection G/T --+ G/N(T) is an IN(T)/TI-fold coveringj 0 therefore, X{G/T) = IN{T)/Tlx{G/N(T» = IN{T)/TI. Theorem 6.28 and the Lefschetz theorem readily imply the follOwing assertion.

5. Lie Groups and H -Spaces

353

Theorem 6.29. Let T c G be a maximal torus. Then, lor any element 9 E G, there exists an x E G such that 9 E xTx- l . Proof. Consider the left coset space GIT = {xT} and the map l(xT) = gxT (this map is well defined because the coset gxT, when x is replaced by xt, where t E T, does not change, gxtT = gxT). Suppose that a coset xT is a fixed point of I, i.e., gxT = xT. Then 9 E xTx-l. Therefore, it suffices to prove that the map I: GIT --+ GIT has a fixed point. For this purpose, we employ the Lefschetz theorem. To calculate the Lefschetz number A(f), we can take any map 10 homotopic to I. This means that we can replace 9 by any other element go E G because the group G is connected. Take the identity element e for go. Then 10 = id c / T and A(fo) = x(GIT). According to Theorem 6.28, we have x(GIT) = IN(T)ITI -::/: O. Therefore, A(f) = A(fo) -::/: 0, and hence the map I~a~d~~.

0

Corollary 1. All maxtmal tori are conjugate; i.e., ilT and T' are maximal tori 01 a compact Lie group G, then T' = gTg-l lor some 9 E G. Proof. Let t' be a topological generator of the torus T'. Take 9 E G for which t' E gTg-l. We have T' C gTg- l . Clearly, gTg- 1 is a torus. Therefore, the maximality of T' implies T' = gTg-l. 0 Corollary 2. For any compact connected Lie group G, the image 01 the map exp: TeG --+ G is the entire group G. Proof. Let T be a maximal torus of the Lie group G. For each x E G, we can choose 9 E G so that x belongs to the torus gTg-l. In the proof of Theorem 6.19, we showed that the map exp is an epimorphism for a torus. 0 The dimension of a maximal torus of a compact connected Lie group G is called the rank of G. Problem 136. Given a compact connected Lie group G and a positive integer n, prove that the image of the map I: G --+ G defined by f (g) = gn coincides with the entire group G. 5.4. Regular Elements. Let G be a compact connected Lie group. An element 9 EGis said to be regular if it is contained in precisely one maximal torus; if g is contained in more than one of the maximal tori, then it is called singular. Regular elements can also be defined as follows: An element 9 E G is regular if and only if the dimension of its normalizer NCg) is equal to that of a maximal torus. To prove the equivalence of these definitions, we need two lemmas.

354

o. 1Vllscellany

Lemma 1. Suppose that S is a connected Abelian subgroup of a compact connected Lie group G and 9 E G commutes with all elements of S. Then there exists a torus T containing 9 and S.

Proof. Consider the subgroup H generated by 9 and S in G. Since H is Abelian, it follows that its closure H is a compact Abelian group; hence the connected component (H)e of the identity is a torus. The compactness of H implies the finiteness of the quotient group H /(H)e. This quotient group is generated by one coset g(H)e because the connectedness of S implies S c (H)e. Thus, H/(H)e ~ Zk for some positive integer k. Let t be a topological generator of the torus (H)e. Clearly, gk E (H)e and tg- k E (H)e. The group (H)e is divisible; therefore, tg- k - sk for some s E (H)e. Consider the element h - gs E H. Clearly, hk = (gs)k - t; hence the powers of h are everywhere dense in (H)e. Applying the translation by h r , we see that the powers of h are also everywhere dense in the coset of gr in (H)e. In this way, all cosets are obtained; therefore, h is a topological generator of the group H. The element h belongs to some maximal torus T, and {g} USc H c H c T. D Lemma 2. Let N(g)e be the connected component of the identity in the normalizer of some element 9 E G. Then N(g)e is the union of all maximal tori of G containing g.

Proof. First, suppose that 9 belongs to some maximal torus T. Then N(g) => 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 .

5. Lie Groups and H -Spaces

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.

5. Lie Groups and H -Spaces

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.

5. Lie Groups and H -Spaces

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

5. Lie Groups and H -Spaces

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 (

Hints and Solutions

370

In case (c), consider the diagram

where (IP + W)(a, c) - IP(a) .,pIP = 0 and .,pw = ide.

+ W(c).

This diagram is commutative because

13. (a) The map 1f'k(B) ~ 1f'k(E) ~ 1f'k(B) is the identity; therefore, P. is an epimorphism. For k = n+ 1, the map 8.: 1f'n+I(B) --+ 1f'n(F) is zero. Thus, we obtain an exact sequence 0 --+ 1f'n(F) ..!..4 1f'n(E)~1f'n(B) --+ 0, and s. is a right inverse of the homomorphism P•.

(b) The map 1f'k(F) ~ 1f'k(E) ~ 1f'k(F) is the identity; therefore, i. is a monomorphism. For k - n - 1, the map 8.: 1f'n(B) --+ 1f'n-I(F) is zero. Thus, we obtain an exact sequence 0 --+ 1f'n(F) ~ 1f'n(E) ~ 1f'n(B) and T. is a left inverse of the homomorphism i •.

--+

0,

(c) Since the space F is contractible in E, any spheroid in F is contractible in E. Therefore, the map 1f'n(F) --+ 1f'n(E) is zero. Thus, we obtain an exact sequence 0 --+ 1f'n(E) ~ 1f'n(B) ~ 1f'n-I(F) that the homomorphism 8. has a right inverse.

--+

O. Let us show

Take a homotopy It: E --+ E for which lolF = idF and !l(E) = eo is a singleton; we can assume that eo E F. Let IP: sn-l --+ F be a spheroid to F. Then ItIP can be considered as an n-spheroid in B. Homotopic spheroids in F correspond to homotopic spheroids in B; thus, we obtain a homomorphism 1f'n-I(F) --+ 1f'n(B). Clearly, this homomorphism is a right inverse for 8 •. 14. Each homomorphism IP E Hom(A, Z) determines the homomorphism rj; E Hom(mA, Z) defined by rj;(ma) = IP(a); the same formula defines the inverse correspondence. We only need to verify that if ma = 0, then IP(a) = O. Suppose that IP(a) = k. Then 0 = IP{ma) = mk, and k = O. 15. Let us construct a sequence aI, a2, a3, ... E QjZ as follows. The element al is arbitrary; the element a2 is such that 2a2 = al; the element a3 is such that 3a3 = a2, etc. After the element an-l is chosen, there are n possible choices of an. Therefore, the set of such spquences is uncountable. Indeed, even if there were only two choices at each step, we would obtain an uncountable set of sequences (the set of all dyadic fractions is uncountable).

For each sequence aI, a2, a3, ... E QjZ, we define a homomorphism Q --+ QjZ by setting I(~) = an. The equality n! I(~!) = 1(1) holds because n! an = al. Clearly, if I(~) is given and I is a homomorphism,

I:

Hints and Solutions

then f (~) integer m.

=

(n - I)! f

371

(;h)

is defined; therefore,

f ('::) is defined for any

16. For the group T, there is a free resolution of the form

where i(ek) = mkek for the standard generators of zr. The group Ext (A, 71..) is the quotient of Hom(Zr, 71..) ~ zr modulo the image of Hom(Zk EEl zr, 71..) under the map induced by (0, i), i.e., modulo the image of Hom(r, 71..) under the map i. induced by i. Let Ck E Hom(zr, 71..) be the element dual to ek, i.e., such that ck(eJ ) = 8kj. Then i.(ck) = mkCk. Therefore, the quotient of Hom(Zr,Z) by the image of i. is isomorphic to T. 17. Using free resolutions

and

O-R'-+F'-+B-O and the canonical isomorphism

we obtain the commutative diagram

o

o

0

1

Tor(B, A)

111 111 111 1 1 1o. o o

o

) R® R' ----+ F ® R' ---+) A ® R' ---+) 0

o

)R®F'----+F®F'

) A®F'---+) 0

o~Tor(A,B) ----?R®B~F®B

) A®B---+) 0

We construct an isomorphism Tor(A, B) -+ Tor(B, A) in the same way as the connecting homomorphism in the exact sequence for a pair. For convenience,

Hints and Solutions

372

we introduce a uniform notation for groups and homomorphisms:

o

o.

o

We must construct an isomorphism Ker a13

-

Ker 1331.

Take X13 E Ker a13. Using the surjectivity of 1312, we choose an element X12 E X 12 for which J312(X12) - X13. We have J322aI2(X12) = a13J312(XI2) = a13(x13) = 0; therefore, for the chosen element X12, there exists a unique X21 such that J321(X21) - a12(x12). The element X12 itself is not uniquely determined. But if x~2 is another element for which J312(X12) = X13, then J312(X12 - X~2) = 0; therefore, X12 - x~2 = J3u(xu) for some Xu E Xu. We pass from X21 to X31 - a21(x2!) E X31. This element is determined uniquely. Indeed, J321(X21 - X~I) = a12(x12 - X~2) = a12J311(xU) = J321aU(Xl1). But 1321 is a monomorphism; therefore, X21 - x~l = all(xu), whence a21 (X21 - x~l) = a21aU (Xll) = O. Let us show that X31 E Ker 1331. By construction, 1331 (X3!) = 1331 a21 (x2d = a22J321 (X2t) = a22aI2(X12) = O. The map Ker 1331 - Ker a13 is constructed is a similar way. Clearly, the element X31 = a21 (X2!) is mapped to X13; thus, the constructed maps are mutually inverse. For the periodic group T, there exists a resolution of the form o - F ~ F~T - 0, where F is a free Abelian group. Indeed, suppose that the epimorphism p: F - T takes each generator JO/. to an element of order nO/.. Then we set i(fO/.) = nO/.JO/.. If Ext(T, Z) = 0, then the map 18.

Hom(F, Z) ~ Hom(F, Z) is an isomorphism. Suppose that cp E Hom(F, Z) and cp(fO/.) = kO/.. Then icp(fO/.) = nO/.kO/.. Since i is an isomorphism, we have nO/. = ±l. Thus, i is an isomorphism, and hence T = O. 19. Consider homomorphisms a: G _ A and 13: G _ B. If cpa = 0, then a = 0 because cp is a monomorphism. Therefore, r:p is a monomorphism. Let us show that 1m r:p = Ker'l~, i.e., 13 = cpa for some a if and only if

Hints and Solutions

373

If (3 = CPO!, then 'I/J{3 = 'l/JCPO! = 0 because 'l/Jcp = o. If 'I/J{3(g) = 0, then Ker'I/J = Imcp, i.e., (3(g) = cp(a), and the element a E A is determined uniquely because cp is a monomorphism. Thus, we can set a(g) = cp l{3(g).

'I/J{3

= O.

(3(g)

E

20. (a) According to Problem 19, it suffices to prove that -J; is an epimorphism. In other words, for any homomorphism -y: F -+ C, there exists a homomorphism (3: F -+ B such that -y = 'I/J{3. Let {JQ} be a basis in the group F. We set (3(fQ) = b where bQ is any element of B for which 'I/J(b = -y(fQ); such an element exists because 'I/J is an epimorphism. Any homomorphism of a free group is determined by its values on generators. Q ,

Q )

(b) According to Theorem 1.19, it suffices to prove that rj; is an epimorphism. In other words, if A c B is a subgroup, then any homomorphism a: A -+ G c~ be extended to a homomorphism (3: B -+ G. Take x E B\A. Let us extend the homomorphism a to the group generated by x and A as follows: (1) if mx ¢ A for all mEN, then we set (3(x)

= 0;

(2) if mx E A for some mEN, then we take the minimum number n E N with this property and set (3(x) = g, where 9 is the element of G for which ng = a(nx). Now, the required homomorphism can be constructed by induction (if the group B/A is not finitely generated, then the induction is transfinite). 21. We shall repeatedly use Problem 4; so, for convenience, we introduce the following notation. Let E be a commutative diagram

We set ImE = (Imcp2 n 1m (3)/lm(cp2a) and KerE = Ker(CP2a)/(Kera + Ker cpd. In this notation, the assertion of Problem 4 is that Ker El ~ 1m E 2, where El is the left square and E2 is the right square. Note that writing the same commutative diagram as

we obtain the same 1m E and Ker E because {3CPl

= CP2a.

Hints and Solutions

374

Consider now the following commutative diagram with exact rows and columns:

Hom(A, B)

~

1

Hom(A, G)

1

~

E,

Hom(A, H)

1

~Ext(A,B)

E,

1

Hom(F, B) -----+ Hom(F, G) -----+ Hom(F, H) -----+) 0

1 Hom(R, B)

1 Ext(A, B)

E. ~

E,

1 Hom(R, G)

E3 ~

1 Hom(R, H)

1 ) o.

The two zeros in this diagram appear because the group F is free and G is divisible. It is easy to show that Ker El = Ext(A, B) and Ker E5 Indeed, in both cases we deal with diagrams of the form

= Ext(A, B).

where
Hints and Solutions

375

we construct the commutative diagrams

o ~ Hom(F,A) ~ Hom(F,B) ~ Hom(F,C) ~ 0

111

o ~ Hom(R, A) ~ Hom(R, B) ~ Hom(R, C) ~ 0 and

o ~ Hom(A, G) f---- Hom(B, G) f---- Hom(C, G) +--- 0

111

o ~ Hom(A, H)

~

Hom(B, H) ~ Hom(C, H) ~ 0

with exact rows. The exact sequences

o --+ Hom(X, A) --+ Hom(F, A) --+ Hom(R, A)

--+

Ext(X, A)

--+

0

and

o --+ Hom(A, X)

--+

Hom(A, G)

--+

Hom(A, H)

--+

Ext(A, X)

--+

0

imply that the kernel and cokernel of the map Hom(F, A) --+ Hom(R, A) are Hom(X, A) and Ext(X, A), respectively, and the kernel and cokernel of the map Hom(A, G) --+ Hom(A, H) are Hom(A, X) and Ext(A, X). Applying Problem 2, we obtain the required result. 23. According to the universal coefficient theorem, we have

HI (K; Z) ~ Hom(HI (K), Z) ffi Ext(Ho(K), Z).

But the group Ho(K) is torsion-free; therefore Ext(Ho(K), Z) = O. Clearly, if Tl is a finite group, then Hom(TI, Z) = 0 and Hom(TI ffi zr, Z) ~ zr. 24. According to the universal coefficient theorem, we have

Hi(K; Z) ~ Hom(zn. ffi 1i, Z) ffi Ext(zn,

because Ext(zn,

1

ffi 1i-I, Z) ~ Ti

1

ffi ~-l, Z) ~ Zn, ffi 1i-l

I by Problem 16.

25. First, let us prove the assertion about torsions. Consider the two sequences of homomorphisms

Ci+I

8,+1

~

C

8.

i --+

C

i-I

and

where Ck is a free group and C k = Hom(Ck , Z) ~ Ck. By a theorem of Smith (see, e.g., [104]), multiplying the matrix of the map 0i+I by integer

376

Hints and Solutions

matrices with determinants ±1 on the left and on the right, we can reduce this matrix to a matrix in which the diagonal contains 1, ... ,1, dl, ... , dl (d] ~ 2) and all of the remaining elements (both diagonal and off-diagonal) are zero. It is easy to see that Ti ~ Zdl EEl ... EEl Zdl . To calculate T, we must reduce the matrix of the map 6i - 1 = to the Smith normal form. It is obtained from the matrix of ai by transposition; therefore, the group T, which is calculated by using the operator is isomorphic to the group Ti-l, which is calculated by using the operator i .

a:

a:,

a

Now let us prove the assertion about the ranks of groups. Note that the torsion subgroup T, is completely determined by the operator a,+l; however, to calculate the rank of the ith homology group, we need both operators a,+l and a,. Suppose that the diagonal of the Smith normal form of the matrix of contains ai nonzero elements. Then

a,

rkKera, rk Ker 6'

= rkC, = rk Ci -

ai, a,+l.

rkIma,+l = ai+I. rkIm6 i - 1 = ai·

Therefore,

Indeed, in the quotient by the image, every 1 on the diagonal kills one direct summand Z, and every d > 1 replaces one summand Z by Zd, which makes no contribution to the rank. 26. Since the groups Ho(X) and Ho(Y) are torsion-free, Ext(Ho(X), Z) o. Thus, the universal coefficient theorem gives natural isomorphisms Hl(X) -+ Hom(Hl(X), Z) and Hl(y) -+ Hom(Hl(Y)' Z). The zero homomorphism f.: H1(X) -+ H1(Y) induces the zero homomorphism Hom(Hl(Y), Z) -+ Hom(HI(X), Z); therefore, after natural identifications we obtain the zero homomorphism Hl(y) -+ Hl(X). = Ext(Ho(Y), Z) =

27. Consider the same Mayer Vietoris sequence as in the solution of Problem 10. If Mn is a closed orientable manifold, then the connecting homomorphism Hn(Mn) -+ Hn_1(sn-l) takes the fundamental class [Mn] to the fundamental class [sn-l]; therefore, it is an isomorphism. It follows that the map Hn_1(M n \ int Dn) -+ Hn_1(M n ) is an isomorphism. The closedness of the manifold M n is essential. For example, suppose M n = Di is the n-disk containing a smaller disk Dn. Then M n \ int Dn '" sn-l, and therefore H n_ l (M n \ int Dn) = Z i- 0 = H n_l(M n ). The orientability of Mn is essential as well. For example, if M n = lRp2, then lRp 2 \ int D2 '" SI, and HI (lRp2 \ int D2) = HI (SI) = Z i- Z2 = HI (lRP2).

377

Hints and Solutions

28. We have the commutative diagram

"" Hn(sn) H n+1(ESn ) -=---+

1

(E/).

1/.

"" Hn(sn) , H n+1(ESn ) ---=---t where the horizontal arrows are isomorphisms of suspensions. Therefore, if f. is multiplication by d, then so is (Ef) •. 29. Clearly, Ho = Zp and H2 = Zp (in the orient able case) or H2 (in the nonorientable case). We have

HI(Kj Zp)

~

(HI(K) ® Zp) EB Tor(Ho(K) , Zp)

~

=

H1(K) ® Zp

because Ho(K) = Z. Therefore, Hl(nT2jZp) = z;n and HI(mp2jZp) '7Im

ILJp

I

0

=

•

30. Clearly, H O = Z. We have

H 1 (Kj Z) ~ Hom(H1 (K), Z) EB Ext(Ho(K), Z) ~ Hom(H1(K), Z) because Ho(K) = Z. Thus, HI(nT2jZ) = z2n and H 1(mP2 j Z) = zm-l. Finally,

H2(KjZ) ~ Hom(H2(K),Z) EB Ext(Hl(K),Z). Thus, H2(nT2j Z) ~ Hom(Z, Z) ~ Z and H 2(mP2 j Z) ~ Ext(H1(mP2), Z) ~

Ext(Z2' Z) ~ Z2. 31. Obviously, HUX = S3 and HnX = 8H is a sphere with 9 handles. Therefore, the Mayer Vietoris sequence for X and H has the form

... - - H2(S3) - - HI(8H) - - HI (X) EB H1(H) - - HI (S3) - - .... Here H2(S3) = H 1(S3) = OJ hence H 1(8H) ~ H1(X) EB HI(H), i.e., Z2g ~ HI (X) EB zg. Thus, HI(X) ~ zg. The groups HI(H) and H1(X) are isomorphic to the kernel and the image of the homomorphism i.: H 1(8H) --+ HI(H), respectively. The kernel of i. is generated by the meridians of the boundary of the handlebody, and the image is generates by its parallels. Therefore, the group HI (H) is generated by circles put on the handles of H. 32. The torus rn can be represented as the n-cube with identified respective points of opposite faces. In the n-torus, each k-face is determined by n - k equations of the form Xi = ±l. Therefore, the number of k-faces is equal to 2n - k (~). After identifications in the cube, we obtain (~) k-cells in the torus. Simlarly to the case of the 2-torus, all boundary homomorphisms are zero. Therefore, the homology of the n-torus coincides with that of the chain complex Z - - Zn

--+

Z(;) __ ... __ Z(;) __ Zn

with zero homomorphisms. Hence Hdrn) = Z(~).

--+

Z

Hints and Solutions

378

33. The integral cohomology groups are easy to calculate if the integral homology groups are known (see Problem 24). Example 10 shows that

for k - 0 and k - n, where n oddj for even k, 0 < k < nj otherwise. 34. Taking I-cells a and b and attaching them to two 2-cells along the words a 5 b 3 and b3 (ab) 2, we obtain a CW-complex X. The chain complex

for calculating the cell homology of X has the form Z2 ~ 'Z} - 0, where the homomorphism is determined by the matrix ( 3 ~ ). The determinant of this matrix equals -lj therefore, is an isomorphism. Thus, the CWcomplex X is acyclic.

a

a

To prove that X is noncontractible, it suffices to show that the group is nontrivial. The group 71"1 (X) is defined by the generators a and b and the relations a 5 - b3 - (ab)2. It admits a nontrivial homomorphism to the group of proper motions of the dodecahedron. Namely, let A be a rotation of the dodecahedron through 271"/5 about an axis passing through the center of a face, and let B be a rotation through 271"/3 about an axis passing through a vertex (the particular axes and directions of rotation are specified in Figure H.3). Then AB is a rotation through 71" about an axis

71"1 (X)

Figure H.3. The rotations of the dodecahedron.

passing through C (the midpoint of an edge). Thus, A5 = B3 = (AB)2 is the identity transformation. Therefore, the map which takes a to A and b to B can be extended to a homomorphism of the groups. 35. The universal coefficient theorem implies Zbk ffi Tk ~ Zak ffi T k - 1 , i.e., bk = ak and Tk ~ Tk-1. In particular, Tl ~ To = O. By the Poincare

379

Hints and Solutions

duality theorem, we have ak = bn- k and Tk ~ Tn k. Therefore, Tk ~ T n- k ~ Tn k-l and Tk ~ Tn-k ~ Tn k+1. 36. Let Hk(Mn) = Hk ffiTk, where Hk is a free Abelian group and Tk is the torsion subgroup. Applying Problem 35 to the manifold M n , we obtain Hk = H n- k for k = 1,2, ... , n - 1 and Tk = Tn k 1 for k - 1,2, ... , n - 2. Applying Problem 35 and the suspension isomorphism to the manifold EM n , we obtain Hn-I = 0, Hk-I = Hn k for k = 2, ... ,n - 1, Tn 2 = 0, and Tk-I = Tn-k-l for k = 2, ... , n - 2. Therefore, HI - ... = Hn I = 0 and Tl = ... = T n-2 = o. Moreover, T n- 1 = 0 for any closed orientable manifold Mn.

37. Consider the unit vector perpendicular to the plane of the diagram and directed upward; we treat it as a point of the sphere 52. The points of the torus mapped to this point are in one-to-one correspondence with the crosses at which the curve J passes under K. It is seen from Figure H.4 that different types of crosses correspond to different signs of the Jacobian of /.

~ ~ ~ K

--y7J K e= -1

e= +1 Figure H.4. The map

f

To ensure that deg / = lk(J, K), we choose the orientations as follows. The basis eI, e2, where el and e2 are the tangent vectors to J and K, determines the positive orientation of the torus. A basis Cl, C2 determines the positive orientation of the sphere if the basis CI, C2, C3, where C3 is an outward normal, has positive orientation. 38. Refining the triangulation of L if necessary, we can assume that each simplex from L is completely covered by the neighborhood U involved in the definition of the covering. For the triangulation of K we can take the preimages of the simplices from L. Let us denote the number of i-simplices in these triangulations of K and in L by ai (K) and ai (L), respectively. Then ai(K) = nai(L); therefore, X(K) = nx(L).

Hints and Solutions

380

39. Let b~ and b~ be the number of k-simplices in K and L, respectively, and let bk be the number of k-cells in K x L. Then (_1)kbk = L p +q k(-1)Pb~(-1)qb~ because bk = L p +q kb~b~. Therefore,

X(K x L) = L

(_1)kbk = (L

(-1)Pb~) (L (-1)qb;) = X(K)X(L).

40. If In = idK and the map I has no fixed points, then we can construct an n-fold covering p: K - K/"" where x '" I(x) '" P(x) '" ... '" In(x) = x. If the triangulation of the complex K is sufficiently fine, then the simplices ~ and I(~) do not intersect. Such a triangulation induces a triangulation of K/",. According to Problem 38, we have X(K) - nx(K/",) 0 (mod n).

=

41. Clearly, 1m I,

VJ./ Ker I,; therefore, dim VJ. = dim 1m II + dim Ker Ii~

Thus,

L

( _1)i dim VJ. = L

=L

+L (-1)'dimlml, + L

( _1)i dim 1m Ii

( _1)i dim Ker Ii (-1)idimlmll+1

= o.

42. Let Ko be the c-neighborhood of the given link, and let Kl be the closure of 8 2 \ Ko. Then X(8 3 ) = X(Ko) + X(KI) - X(Ko n Kl). Here X(8 3 ) = 0, X(Ko) = 0, and X(Ko n K 1 ) = 0 because Ko n Kl consists of one or several tori. Hence X(Kl) = 0, as required. 43. Let us write the exact sequence of a pair for homology with coefficients in some field:

This sequence contains only finitely many nonzero terms, so we can use Problem 41. As a result, we obtain

L

(_1)i dimH,(B)

+L

(_1)1+1 dim Hi(A)

+L

(_1)i dimH,(A, B) =

o.

44. * Let us enumerate the i-faces of the complex K by 1, 2, ... , I" where I, is the number of i-faces, and consider the number pij(k) of j-faces that intersect the kth i-face F~. The number of j-faces disjoint from this face is Ii - Pi) (k); therefore, I.

(1)

O!ij

I.

= L (Ij - pij(k)) = Iilj - LPij(k). k=l

Lemma. For each lace F~,

k=l

Lj=o( -1)jPij(k) = (_1)n.

Proof. Let us endow the manifold M n with a Riemannian metric and consider the set Qe of all points of Mn at distance at most c from F~. If c is

Hints and Solutions

381

sufficiently small, then Qe is homeomorphic to the n-disk and Qe contains no vertices of K different from those of the face F~. The intersections of the faces of K with 8Qe ~ sn-l determine a CWcomplex structure on sn-l. The number of (j -I)-cells in the CW-complex thus obtained is equal to the number of j-faces of K intersecting the face F~ and not contained in F~. Therefore, (2) where qj-l is the number of (j - I)-cells in the CW-complex 8Qe (it is assumed that q-l = 0) and Tj is the number of j-faces in the convex polyhedron F~ (it is assumed that Ti = 1 and Tj = 0 for j > i). Clearly, E~=o(-I)jTj = X(Dk) = 1 and E~:~(-I)jqj = X(sn-l) = 1 + (_I)n-l. Using (2), we obtain n

n

n

L(-I)jpij(k) = L(-I)jq,-l+ L(-I)jTj = -1-(-I)n-l+1 = (_I)n, 3=0

j-O

j 0 D

as required. Formula (1) implies n

\lI(K)

=

n

L L (-I)i+

jQij

i=O 3=0 n n

n

n

h

= LL(-I)i+ j fdj - LLL(-I)i+ j pij(k). i=O j=O

i-O j=O k=l

The first sum can be calculated directly:

To calculate the second sum, we use the lemma: n n h n h n LLL(-I)i+ j pij(k) = L(-I)iLL(-I)jpi3(k) 1=03=0 k=l

i=O n

Ii

k=l

j=O n

= L(-I)i L (-I)n = (-I)n L (-I)ifi = (-I)nX(Mn). i=O k=l As a result, we obtain \lI(K) = X2(Mn) - (-I)nX(Mn). But if n is odd, then X(M n ) = o. Therefore, \lI(K) = X2(Mn) - X(Mn). 45. Consider the following equivalence relation on Mk: x "..- y if the normal subspaces at x and y coincide (Le., x and y belong to the same i=O

382

Hints and Solutions

normal subspace). The map Mk ---. N k /'" is a covering because each normal plane intersects Mk transversally. 46. It is sufficient to prove that for coefficients in Z2, the intersection number of Mk and every normal subspace is O. Any (n - k)-dimensional subspace in IRn can be transformed into any other (n - k)-dimensional subspace by a motionj therefore, the intersection number of Mk and an (n - k)dimensional subspace does not depend on the subspace. But for any compact manifold, there exists a subspace disjoint from it. 47. Choose a puint a E Mk such that the function f(x) - Ilx - a1 2 , x E Mk, has no degenerate critical points. The critical points of this function are the intersection points of Mk with the normal subspace at a. Let ri be the number of critical points with intersection number i. Then 0 r, = r k and o( -l)ir, - X(M ). Clearly, ro > O. Therefore, -r < X(Mk) ~ r. But X(M k ) - rx(N k ), where N k is the manifold introduced in the solution of Problem 45. Hence -1 < X(N k ) ~ 1. Thus, X(N k ) is equal to 0 or 1, i.e., X(M k ) is equal to 0 or r. 48. Solving Problem 47, we proved that if Mk is a closed manifold admitting a transnormal embpdding and X(M k ) -:f:. 0, then ri = 0 for all odd i. Therefore, Mk is homotopy equivalent to a CW-complex containing no cells of odd dimension. 49. It is sufficient to prove the required assertion in the case where f is a simplicial map. In this case, the chain map ik: Ck(Kj IR) ---. Ck(Kj IR) is determined by a matrix A with integer elements, and the chain map Jk is determined by the matrix AP.

E:

E7

Lemma. Suppose that A is a square matrix with integer elements and p is a prime. Then tr(AP) = tr A (mod p). Proof. Let f(x) - det(xI -A), where I is the identity matrix. Any polynomial f with integer coefficients determines a polynomial f over the field IFP (that is, the finite field with p elements). Adding the roots Xl, •• . ,Xn of f to lFp, we obtain the field lFpk with some k.

For any elements x, y E IFpk, we have (x + y)P = x P+ yP because (mod p) for m - 1,2, ... ,p - 1. Therefore,

U:J == 0

xi + ... + x~ = (Xl + ... + xn)P = Xl + ... + Xn (the second equality follows from Xl + .. ·+xn E lFp). But Xl + .. ·+X n = tr A (mod p) and xi + .. .+:?n = tr(AP) (mod p). The former equality is obvious, and the latter holds because if f(x) = (x->'d··· (x->'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) " , . .. ,

~<(J
m

0< p

~ 1,

IWI12 = 1 _ p2;

0< p

~

1,

IWII = 1 -

0< p

~

1,

IWII2 + IW212 = 1- p2;

2

0< p ~ 1, IWll2

2

p,

k

m

k+l

< (J < - - ; m

+ IW212 = 1- p2,

~<(J
m

This cell decomposition is compatible with the equivalence relation on Soo; therefore, it gives a cell decomposition of the lens space L'::. containing one cell of each dimension. The boundary homomorphisms are calculated in precisely the same way as for the three-dimensional lens spaces. As a result, we obtain the chain complex

· .. ~Z~Z~Z~Z~Z. The homology of this complex is as required. 89. Let M3 be the universal covering of the manifold M3. Then 7l"I(M 3 ) = 0 and 7l"2(M3) = 7l"2(M) = o. The group 7l"I(M3) is infinite; therefore, the manifold M3 is noncompact. Hence H3(M3) = o. Clearly,

Hints and tiolutions

392

~ 3. Applying the Hurewicz theorem, we obtain 71'i(M3) = o for all i. According to the Whitehead theorem, the manifold M3 is contractible, which means that 71'i(M3) = 71'i(M3) = 0 for i ~ 2.

Hk(M3)

= 0 for k

90. Suppose that the group 71' contains an element of finite order. Then it contains the cyclic subgroup Zm generated by this element. Let X -+ X be the covering that corresponds to the subgroup Zm C 71'. Then X is a K(Zm,l) space; therefore, it is homotopy equivalent to Lr:;. This is a contradiction because, on one hand, X is a finite-dimensional simplicial complex, and on the other hand, according to Problem 88, X has nontrivial homology of arbitrarily high dimensions. 91. Obviously, the required map 71' exists for the trivial bundle. Now, suppose that such a map 71' exists for the bundle E. Consider the map f: E -+ B x IRn defined by f(e) = (p(e), 71'(e)). According to Theorem 3.13, this map determines an isomorphism from E to the trivial bundle B x lin. 92. Introducing a Riemannian metric, we can assume the trivializations to be orthonormal. Take some trivialization of the bundle T S3. Any trivialization ofTS3 is uniquely determined by a map S3 -+ SO(3). A homotopy of trivializations is a homotopy of such maps. It remains to note that SO(3) R:: Rp3 (see the solution of Problem 91 in Part I) and 71'3(IRp3) = 71'3(S3) = Z. 93. Take a trivialization of the bundle T S3 . Any vector field on S3 without singular points is uniquely determined by a map S3 -+ IR3\ {O} '" S2. A homotopy of vector fields without singular points is a homotopy of such maps. Clearly, 71'3(S2) = Z. 94. We prove the required assertion by induction on k. For k = 1, this assertion is obvious. Suppose that it is true for the products snl X .•• X sn/e 1. We assume the sphere sn/e to be embedded in lRn/e+1. The normal bundle to sn/e in lin /e+1 is trivial; hence so is the normal bundle to sn/e in lR n1 +,,+n/e+1. The fiber of this trivial bundle is IRnl +·+n/e-l +1. By the induction hypothesis, snl X ••• X sn/e 1 can be embedded in this space. As a result, we obtain the required embedding. 95 [124]. We use Examples 38 and 39. If all numbers

ni

are even, then

Therefore, any vector field on the manifold snl X ••• X sn/e has a singular point, and hence this manifold is not parallelizable. Now let us prove that the manifold snl X ... X sn/e wlth k ~ 2 is parallelizable provided that one of the numbers ni is odd. For k = 2, this assertion is an obvious corollary of the following lemma.

Hints and Solutions

393

Lemma. Let Ml and M2 be closed manifolds such that the Whitney sum of the tangent bundle TM. and the one-dimensional trivial bundle is trivial and x(Md = O. Then the manifold Ml x M2 is parallelizable.

Proof. Let 177:J be the trivial n-dimensional bundle over B. By assumption, x(Md = 0; hence there exists a vector field without singular points on MI. lt determines two bundles over Ml, the trivial one-dimensional bundle 1111 and its orthogonal complement, which we denote by a. We have dima = dim Ml - 1 = ml - 1 and TMI = a EB 1711 . For any vector bundles 6 and 6 over B and any continuous map f: Bl -+ B, the bundle 1*(6 EB 6) is isomorphic to (/"'6) EB (/"'6). Moreover, 1*177:J = 177:J1 • Therefore, TMIXM2

~ 7ri(TMl) EB 7ri(TM2) ~ 7ri(a EB 1711) EB 7ri(TM2)

~ 7ri(a) EB 1711 xM2 EB 7r2(TM2) ~ 7ri(a) EB 7ri(1111 EB TM2) '" m2+1 '" "'( IT'I 1 ) m2 m2 = 7rl"'( a )1'1CD"1 17MlxM2 = 7rl a CD 17Ml l1Ml xM2 '" = 7rl"'( TMI )IT'I CD 17MlxM2

'"= 7rI'" (TMI = 17Ml 1 ) = m2-1 '" ml +1 = 17MlxM2 m2-I '" ml +m2 l1MlxM2 = 17MlxM2 = 17Ml xM2' o CD

CD

CD

For k > 2, we use induction on k. Suppose that ni is odd. Then we can consider Ml = snl X snk-l and M2 = snk. By the induction hypothesis, MI is parallelizable; therefore, TMI EB 1111 is the trivial bundle. 96. We shall not check that V(n, k) is a manifold; we shall only calculate its dimension. The map V(n, k) -+ Yen, k -1) deleting the last vector from each k-tuple v!, ... ,Vk is a locally trivial bundle with fiber sn-k. Indeed, the unit vector Vk belongs to the (n - k + I)-space that is the orthogonal complement of the subspace generated by VI,"" vk-I' Clearly, V(n, 1) ~ sn-I. We obtain the sequence of bundles

yen, k)

sn-k

----+

Yen, k - 1)

sn-k-l

sn-2 ----+

I '"

Yen, 1)

= gn-l.

Thus, the dimension of the manifold Yen, k) is equal to

(n - 1) + (n - 2)

+ ... + (n -

k)

= nk -

k(k

97. The manifold V(n + 2,2) has dimension 2n Theorem 3.16, it is (n - I)-connected and 7rn (V(n

+ 2, 2)) = {

z Z2

+ 1) .

2

+ 1.

According to

if n is even, if n is odd.

By the Hurewicz theorem, we have Hn(V(n + 2, 2)) ~ 7rn (V(n + 2,2)) and Hk(V(n + 2,2)) = 0 for 0 < k < n. We have calculated half of -:;he

Hints and Solutions

394

homology groups. The remaining groups can be found using Poincare duality. As a result, we obtain

H.(V(n + 2,2)) = {:

if k - 0, n, n otherwise,

+ 1,

2n + 1,

and if n is odd, then if k - 0, 2n + 1, if k = n, otherwise. 98. We represent M4 as a CW -complex and construct three independent sections of the tangent bundle by induction on the skeletons. The first nontrivial obstruction occurs when we extend these sections over the 2-skeleton. This obstruction is the class W2. Therefore, an extension over the 2-skeleton exists if and only if W2 = o. The obstruction to extending sections over the 3-skeleton belongs to the cohomology with coefficients in 71'"2 (V(4, 3)). But V(4, 3) ~ SO(4) and 7I'"2(SO(4)) = 0 (this it is seen from the exact sequence of the bundle SO(4) --+ 8 3 with fiber SO(3) because 7I'"2(SO(3)) = 7I'"2(lRp3) = 0). Therefore, this obstruction vanishes.

Obviously, three independent sections can be extended from the boundary of the 4-disk to this disk with the center removed. If independent sections over Mn with several punctures Xl, ... ,X n are given, then we construct independent sections over Mn with one puncture Xo. To this end, we connect the point Xo ith Xl, ... , Xn by disjoint paths and contract these paths to Xo. 101. Let 6 = (u}, vd and ~2 = (U2' V2) be vectors in the space TxM x TxM ~ T(x,x)(M x M). We set (6'~2) = (UI,U2) + (V},V2). Take u, v E TxM. The vector ~ = (u, v) E T(x,x) (M x M) is tangent to d(M) if and only if u = v, i.e., ~ = (u, u). The vector.,., = (v, w) is orthogonal to all vectors (u, u) if and only if v = -w, i.e., .,., = (v, -v). The map 7M --+ lId(M) taking each vector v E TxM to (v, -v) E T(x,x)(M x M) is an isomorphism of vector bundles. 102. The sphere with 9 handles is orientable; therefore, WI = o. Its Euler characteristic is even; therefore, W2 = o. The equalities WI = 0 and W2 = 0 can also be proved by using the triviality of the normal bundle for the standard embedding of the sphere with 9 handles in lR3 • 103. For the sphere to which m Mobius bands are attaLhed, the parity of the Euler characteristic coincides with that of the number m. Therefore,

Hints and Solutions

395

W2 = mA, where A is the generator of the two-dimensional cohomology group, i.e., W2 = 0 for even m and W2 = A for odd m. For the coefficient group Z2, the one-dimensional cohomology group of the sphere with m Mobius bands attached has a basis 01, ... ,am, where the cohomology classes 01, ... ,am are dual to cycles such that traverses along them reverse orientation; therefore, WI = 01 + ... + am. Finally, OiOj

= dijA,

whence w~

= mA = W2.

104. Let VM'" be the normal bundle, and let a = Wl(VMn). Then w(Mn)(l+o) = 1; therefore, WI +0 = 1, W2+WIO = 0, ... ,Wn+Wn-lO = O. Hence WI = 0,W2 = 0 2 , ... ,Wn = an. 105. Let Wk = Wk(JRpn) = (ntl)ok for k = 1, ... , n. According to Problem 104, we have Wk = wt. Therefore, the numbers (ntl) , where k = 1, ... ,n, are either all even or all odd. Over the field Z2, we have (l+x)2Tn = 1+x2Tn. Let aO+al ·2+·· ·+am ·2m be the binary decomposition of n + 1. Then

(1

+

x)m

= (1 +

x)a o (1

+

x 2)a 1

•••

(1

+

x 2Tn )a

Tn •

If all of the numbers (ntl) are even, then ao = ... = am-l = a and am = 1; this means that n + 1 = 2m . If all of these numbers are odd, then ao = ... = am = 1; this means that n + 1 = 2 m +! - 1.

106. If n is even, then the cohomology class wn(lRpn) = (n + l)on, where a is the generator of HI (IRpn; Z2), is nonzero. Therefore, the StiefelWhitney number wn[JRpnj is nonzero. The cohomology class Wl(JRpn) = (n + 1)0 = a is nonzero as well and an f= O. Therefore, the Stiefel Whitney number wf[l[~'pnj is nonzero. 107. If n = 2k - 1, then w(JRpn) = (1 + a)2k = (1 + a 2)k; hence Wi(IRpn) = a for any odd i. But any monomial 1 ••• w~n, where rl +2r2 + ... + nrn = 2k - 1, contains a factor Wi with odd i.

wr

108. The only if part is obvious; let us prove that if the bundle t;,k ffi e 1 is trivial, then so is t;,k. Since both bundles t;,k ffi e 1 and e 1 are orientable, so is t;,k. We assume that all fibers of t;,k ffi e 1 and t;,k are endowed with orientations. Each fiber of t;,k ffi e 1 is identified with IRk+! , and each fiber of t;,k is an oriented k-dimensional subspace in IRk+!. The space of oriented kdimensional subspaces in IRk+! is homeomorphic to the sphere Sk. Consider the bundle over Sk whose fiber over each point x E Sk is the oriented kdimensional subspace of JRk+l corresponding to this point. Let f: B ---+ Sk be the map taking every point b E B to the fiber of t;,k over this point. Clearly, t;,k = f*'Yt. Since the dimension of the complex B is less than k, it follows that the map f is null-homotopic. Therefore, the bundle t;,k is trivial. 109. (a) This is an obvious corollary of Problem 108.

'Yt

Hints and Solutions

396

be the normal bundle for an embedding M n c ]Rn+k. Then TMn ffi v k ~ c n +k+ l . We consider the situation in which k ~ n + 1. Suppose that the manifold M n is stably parallelizable. Then the bundle c l ffi TMn is trivial. Hence, according to Problem 108, the bundle v k is trivial as well. Conversely, if v k is trivial, then so is c l ffi TMn. 110. The space E(P'Y~) consists of all pairs (I, rrk), where I c rrk. The space E(p'Y~-I) consists of all pairs (I, rr k- l ), where I -.L rrk-l. The map E(P'Y~) --+ E(p'Y~-I) takes each pair (I, rrk) to the pair (I, rr k- l ), where rr k- l is the orthogonal complement to I in rrk. The inverse map takes each pair (I, rr k- l ) to the pair (I, rrk), where rrk is the space spanned by I and rrk-l. (b) Let

TMn

vk

ffi v k ~ on+kj therefore, c l ffi

111. Suppose that some finite group acts freely on ]Rn. Any finite group contains the subgroup G = Zp with some prime p. The group G acts freely on lRn. This action induces an action of G on sn such that it has the unique fixed point 00, i.e., (sn)G = {oo}. On the other hand, Theorem 3.56 implies that the space (sn)G is a homology sphere, while the topological space consisting of one point is not a homology sphere. 112. According to Theorem 3.62, we have Sqa 2m = (a + ( 2)2m = 2m a + a 2m +l because e~) == 0 (mod 2) for 1 ~ i ~ 2m - 1. Therefore, Sq(a + a 2 + a 4 + ... ) = (a + ( 2 ) + (a 2 + ( 4 ) + (a 4 + ( 8 ) + ... = a. 113. Solution 1. Theorem 3.32 implies the existence of a monomorphism

where f*w m = O"m(aI, ... , an) is the mth elementary symmetric function. Moreover, SqO ai = aI, Sql ai = a~, and Sqr ai = 0 for r > 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).

<

Hints and Solutions

397

Let us apply induction on the number n of variables. For n = 1, we have the obvious identities al = 0'00'1 and at = O'~. The induction step is as follows. If k = m, then congruence (3) takes the form ~ (ail·· . aik)2 == (O'k)2 (mod 2); obviously, this is true. Suppose that k < m. Consider a similar sum ~* a~l ... a;k aiHl ... aim for the n - 1 variables aI, ... ,an 1. Then the left-hand side of (3) takes the form

Let 17m

O'~

be the mth elementary function of the variables al, ... , an-I. Then + O'~; therefore, the right-hand side of (3) takes the form

= anO'~_I

( m k-

1) ( *

anO'm+k_I

*) + O'm+k

z:: . + z:: ((m - + i-I) + (m - . + i -2)) 2 k-I (

= an

i=O

an

m-k+i-l )

*

*

O'k-i-IO'm+i-I

'l

k

i=O

k

.

'l

k

'l-

1

*

*

O'k-iO'm+i-I

here we set (m-=.;-2) = o. To complete the proof, it remains to use the induction hypothesis. For the coefficients of a~ and a~, the required relation follows at once. For the coefficient of an, note that

because

Solution 2. It is easy to obtain the required relation for the bundle 'l over lRpoo. Indeed, WI (,1) = a is the generator of HI (lRPoo; Z2). Hence the proof reduces to the obvious equality Sql a = a 2 • Suppose that the required relation is valid for the bundles ~ and Wi = Wi(~). According to the Whitney formula, we have wm(~ x ,1) = Wm x 1 + Wm-I X a. The Cartan formula

Hints and Solutions

398

implies Sq(Wm-1 x Sqk (Wm(~

0)

x, )) 1

= (Sqk wm-I) x = WkWm

X

1+

0

+ (Sqk-l Wm-l)

(m - k) Wk-lWm+1 1

+ ... + ( m-1) k WOwm+k

+ WkWm-l

X 0

+

lWm

+ ... +

1 X 02

X

Therefore,

1

x 1

(m -: -1)Wk_1Wm

+ ... + (m;2)WoWm+k

+ Wk

X 0 2.

+

1

x

0

(m ~ k)Wk_2Wm

(7 -=- :)WOWm+k

2

x

X 0

X 0 2

0 2.

On the othet" hand, w~(~ x ,1)w](~ X ,1) = (Wi X 1 + Wi-l X o)(Wj x 1 + Wj-l X 0) = W,Wj x 1 + (Wi lWj + WiWj-l) x 0 + Wi-lWj-1 X 0 2 . Now it is easy to check the required relation for the bundle ~ x For the terms of the form ... x 1 and ... X 0 2 , it is seen directly. For the terms of the form· .. X 0, we must show that, modulo 2, 1 + (m~k) = (m-~-I), (m1 k) + (m-;+1) = (m2k), etc. These equalities follow from the equalities -1 + (m~k) = (m-~-l), _(m~k) + (m-;+1) = (m2k), etc., because the sign is inessential when residues modulo 2 are considered. Thus, the required formula is valid for the bundle ,1 X· .. x,l. Therefore, it is valid for the bundle because the natural map H"(G(n, k); Z2) -+ H" (lRpoo x ... x lRpoo ; Z2) is a monomorphism. Any n-dimensional bundle { is a pullback of ,n; therefore, the required relation holds for { as well. 114. From the cone CY, we remove the cone whose lateral side is one half of that of CY. Then we apply the excision theorem, aftf'l" which we can use the homotopy invariance of the singular homology of a pair (see Theorem 1.14 on p. 16; for singular homology, the proof is precisely the same). 115. Let us write the exact sequence for the pair (X U C'Y, CY):

,1.

,n

Hi(CY) --- Hi(X U CY) --- Hi(X U CY, CY) --- H,_I(CY).

Here Hi(CY) = H i - 1 (CY) = 0 because the cone CY is contractible. Therefore, Hi(X U CY) ~ Hi(X U CY, CY) for i ~ 1. It remains to apply Problem 114. 116. For i = 0, the required assertion is obvious. Suppose that i ;::: 1. Then Hi(X, Y) ~ Hi(X U CY) by Problem 115. Since the complex CY is contractible, we have Xu CY (X U CY)/CY = X/Yo r.J

Hints and Solutions

399

117. No, this is not true The set U c lR n determined by the equation closed. It is homeomorphic to the set V c lRn determined by the equation Xn = 0 and the inequality x~ + ... + Xn-l < 1. This set is not closed. Xn

= 0 is

119. Let us represent sn as 8,6.n+l, where ,6.n+1 = [vo, V!, ... , Vn+IJ, and consider the sets st Vi for i = 0,1, ... , n + 1. Any r ~ n + 1 vertices ViI,···, Vir span a simplex [Vil' ... ' VirJ. Moreover, U~ I st Vile is the star of this simplex; this is a contractible set. 120. We calculated the cohomology groups of the manifold V(2n + 2,2) in the solution of Problem 97. Let us write the Gysin sequence for the bundle

2, 2) ~ G+(2n + 2,2). Starting from the end, we consequtively obtain H 4n = Z, H 4n-1 = 0, H 4n-2 = Z, H 4n - 3 = 0, ... , H2n+1 = 0, H 2n = ZEeZ. After this, we can apply Poincare duality. As a result for 0 ~ k ~ 4n, we obtain V(2n +

z { Hk(G+(2n + 2,2)) = Z Ee Z o

if k is even, k

i- 2n,

if k = 2n, if k odd.

121. (a) According to the Wu theorem, we have WI = VI, W2 = V2 + Sq VI, ... , Wk+I = Vk+I + SqI Vk + Sq2 Vk-I + .... Therefore, if WI = ... = Wk = 0, then VI = ... = Vk = 0 and Vk+I = Wk+l. (b) If WI = ... = Wk = 0, then VI = ... = Vk = 0, and if n = 2k or 2k + 1, then Vi = 0 for i > k. Therefore, Wi = 0 for i > k. 122. In addition to the dimensions 0 and 4k, the manifold M4k has nontrivial homology and cohomology groups only in dimension 2k, and it has no torsion.

The Wu class has only one component of positive dimension, namely, V2k. According to Problem 121, we have W2k = V2k. Clearly, (Sqa 2k , [M4k]) = (a 2k '-"' a 2k , [M4k]) and (a 2k '-"' V, [M4k]) = (a 2k '-"' V2k, [M 4k ]). Therefore, the class V2k is completely determined by the equality a 2k '-"' a 2k = a 2k '-"' V2k (for all a 2k ). In particular, V2k = 0 if and only if a 2k '-"' a 2k = 0 for all a 2k E H2k(M4k; Z2). Now we consider a cohomology class a 2k over Z. It can be naturally associated with the cohomology class a 2k over Z2 (by reducing the value assigned to each chain modulo 2). The number (a 2k '-"' a 2k , [M4k]) is even if and only if a 2k '-"' a 2k = O. 123. It is seen from the proof of Theorem 3.26 that if n = 2k, then = 1 + a+a 2 + ... +a n - l ; therefore, Wn_I(lRpn) i- o. On the other hand, if a manifold M n is embedded in lRn+k, then wk(M n ) = o. Thus, if n = 2k, then lRpn cannot be embedded in lR 2n - i .

w(lRpn)

Hints and Solutions

400

124. Let G = Znl EB· .. EB Znk' For a presentation matrix of this module we can take the diagonal matrix A = diag(nI, ... , nk). The ideal £1 is generated by its determinant, which equals nl" ·nk = IGI. 125. For the knot under consideration, we can choose hand 12 as shown in Figure 4 (with the only difference that the lowest cross in this figure should be replaced by three crosses). Only the linking number lk(h, tt) changes and becomes equal to 2. Therefore, A = and the Alexander polynomial is 2(t 2 - 2t + 1) + t = 2t2 - 3t + 2.

(5 -l ),

126. (a) We choose hand 12 as shown in Figure H.5. It is easy to verify that, for these hand 12, the Seifert matrix is ~ (:~i Hence the Alexander polynomial is l((pq + qr + rp)(t 2 - 2t + 1) + t 2 + 2t + 1).

:t;).

p

Figure H.5. The choice of

h

and

h

(b) If p = -3, q = 5, and r = 7, then pq + qr + rp this case, the Alexander polynomial equals t.

=

-1. Therefore, in

127. The diagram of the link La = L can be represented as shown in Figure H.6. The links L+ and L_ are isotopic; hence V'L+ = V'L_, and therefore ZV'L = V'L+ - V'L_ = O.

00 00 [][] L

Lo

Figure H.6. The diagrams of the three links

128. It is easy to show that n(L+) = n(L_) = n(Lo)±l (see Figure H.7). Thus, if the required assertion is true for two of the three links L+, L_, and La, then it is also true for the third link. The assertion If> true for the trivial m-component link (if m > 1, then the Conway polynomial is zero); hence it is true for any link.

Hints and Solutions

401

,. ----,. .... ....

"

"\

: X"'" I

I

"

,

"

"" I

\

~ ("

",'

... -:;:-::::- ....

"\

\

I

\

,-",-

....

\

)()

" 'X' ",\ :" , ,,:' \

J

\"

",

I

'''''::<-'

Figure H.7. The proof of the equality n(L+)

= n(L

)

= n(Lo) ± 1

129. Clearly, ao(L) = VdO); therefore, the relation VL+(Z)-VL (z) = zV Lo(Z) implies ao(L+) = ao(L_). Changing types of crosses, we can transform any link into the trivial link (with the same number of components). Hence ao(L) = ao(L'), where L' is the trivial link with the same number of components as in L. For the unknot, we have ao = 1, whereas for the trivial m-component link with m > 1, we have ao - O. 130. We prove the required assertion by induction on n(L). For n(L) = 1, there is nothing to prove. Clearly, n(Lo) ~ n(L±) - 1; therefore, the induction hypothesis implies ai(Lo) = 0 for i < n(L±) - 2. Thus, ai(L+) = a,(L ) for i < n(L±) - 1. Changing types of crosses, we can transform any link into the trivial link with the same number of components. Such changes do not affect ai(L) with i < n(L) - 1; therefore, ai(L) = 0 for i < n(L) - 1. 131. Suppose that L+ and L_ are two-component links and both components are present at the distinguished cross. Then Lo is a knot; therefore, ao(Lo) = 1 according to Problem 129(a). Thus, al(L+) - al(L_) = ao(Lo) = 1. Clearly, Ik(Ll+' L 2+) - Ik(Ll ,L2-) = 1. Changing types of crosses, we can transform any two-component link into a two-component link with unlinked components. For such a link, we have al(L) = Ik(Ll' L 2 ) because al (L) = 0 and Ik(Ll' L 2 ) = O. Therefore, this equality is true for any two-component link.

The relation VL+(Z) - VL (z) = ZVLo(z) implies a2(L+) If Lo is a two-component link, then al(Lo) = Ik(Ll' L 2) by Problem 131. 132.

a2(L_)

= al(Lo).

133. In the proof of Theorem 6.17, it was shown that TM n If the bundle T sn is trivial, then so is any of its pullbacks. 134. (a) The action of each

Wi

= N.(TS n ).

on Ua is defined by

Under such an action, the ith face aie, which is determined by the equation ti = 0, is fixed. Therefore, (1 - w,)aie = 0, which means that

Hints and Solutions

402 (b) First, suppose that n = O. Let w where no, . .. , nao are integers. Then w(l - wo)

=

=

no + nlWO + ... + naow ao - 1 ,

(no - nao) + (nl - no)wo + ... + (n ao - nao_dw ao - l ;

therefore, an element w belongs to the kernel of the given map if and only if no = nl = ... = nao-l = nao· Now, suppose that n = 1. Again, let w = no + nlwO + ., r+ naowao-l; 1 - 1 with however, this time, the ni are linear combinations of 1,WI, ... integer coefficients rather than simply integers. Consider w' = w(l - wd = n~+n~wo+·· .+n~owao 1, where n~ = ni-n,W. The equality w'(l-wo) = 0 is again equivalent to n~ = n~ = ... = n~, i.e.,

,wr

w(l - wI)

= n'(l +

Wo + w~ + ... + w~ I).

It follows that w = a(l + Wo + w~ + ... + w~O-I) + b, where b(l - wI} = 0, 1 - 1 ). The further argument is similar. i.e., b = c(l + WI + w~ + ... +

wr

(c) The simplicial complex Ua has no (n+1)-cells; hence its n-dimensional homology group Ua coincides with the n-cycle group. Therefore, Hn(Ua ) contains the subgroup Z[ZaJc ~ Z[ZaJ/la. The elements specified in the statement of the problem form a basis in this subgroup. Clearly, the rank of the subgroup equals (ao - 1) ... (an - 1), i.e., coincides with the rank of the entire group Hn(Ua ). Hence Hn(Ua ) ~ Z[ZaJc. 136. Let x be an arbitrary element of G. According to Corollary 2 of Theorem 6.29, we have x = expA for some matrix A. We set 9 = exp(~A). Then gn = x.

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-

403

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Index

G-complex regular, 174 simplicial, 173 G-space, 173 H -space, 358 K('II", n) space, 122 S-equivalent matrices, 314 r-transnormal embedding, 56 action effective, 177 free, 183 simplicial, 173 acyclic functor, 103 model,104 theorem, 104 simplicial complex, 6 support theorem, 7 admissible set, 231 Alexander duality, 81 ideal,308 polynomial, 308 in Conway's normalization, 315 theorem, 205 Alexander Pontryagin duality, 271 Alexander Whitney diagonal approximation, 105, 214 algebra Hopf,361 connected, 361 Lie, 344 cohomology, 349 algebraically trivial map, 120 almost parallelizable manifold, 141

anticommutativity of cup product, 63 Arf invariant of a knot, 319 of a link, 320 of a quadratic form, 317 theorem, 318 associated sheaf, 265 attaching a handle, 312 augmentation, 6, 17 axiom dimension, 203, 204 exactness, 203, 204 excision, 200, 203, 204 noncommutative, 224 homotopy, 203, 204 axiomatic approach to Stiefel Whitney classes, 159

basis of a free Abelian group, 4 of a module, 305 symplectic, 317 Betti numbers, 3 bi-invariant form, 348 bilinear map nondegenerate, 109 of Abelian groups, 27 Bockstein homomorphism, 14, 92, 142 Borromean rings, 85 multidimensional, 87 Borsuk Ulam theorem, 76 boundary, 2 homomorphism, 2 of a simplex, 1

-

411

412

braid colored, 122 group, 123 bundle associated with a divisor, 339 conjugate, 170 dual, 171 induced, 133 nonorientable, 143 orientable, 143 vector, 131 equivalent, 132 isomorphic, 132 smooth, 131 stably equivalent, 148 with structure group, 272 canonical map, 9 vector bundle, 153 cap product, 70 Cartan formula, 189 Cartesian product of Abelian groups, 21 category, 103 with models, 103 Cech cohomology, 267, 268 cellular homology, 37, 210 centralizer, 351 chain, 2 complex, 4 free, 4 nonnegative, 4 ordered,60 total, 60 homotopy, 5, 196 map, 4 relative, 12 with closed supports, 48 characteristic class Chern, 163 Chern of a complex manifold, 171 Euler, 144 Pontryagin, 172 Stiefel Whitney, 141 Euler, 51, 90 of a pair, 55, 182 Chern characteristic class, 163 of a complex manifold, 171 class Chern, 163 of a complex manifold, 171 Euler, 144 fundamental, 36 cohomology, 125 Pontryagin, 172 primitive homology, 49

Index

Stiefel Whitney, 141 of a manifold, 150 Thorn, 237, 255 Wu,259 closed form, 277 coboundary, 22 formula, 186 homomorphism, 267 cochain, 21, 266 difference, 117 relative, 23 with compact supports, 48 cocycle,22 noncommutative, 222 cohomologous, 222 coefficient group, 203, 204 cofinal set, 266 cohomology Cech, 267, 268 cross product, 106 de Rham, 277 fundamental class, 125 group, 22 Lie algebra, 349 noncommutative, 222, 274 operation, 127 reduced,23 relative, 23 singular, 202 with compact supports, 48 with local coefficients, 136 cokernel, 15 collar theorem, 78, 247 colored braid group, 122 commutator, 346 of vector fields, 276 subgroup, 112 complete intersection, 341 complex chain, 4 Stiefel manifold, 162 vector bundle, 162 complexification, 172 conjugate bundle, 170 connected Hopf algebra, 361 sum of manifolds, 336 connecting homomorphism, 12, 14 consistent family, 264 constant presheaf, 264 contravariant functor, 103 Conway polynomial, 316 coproduct, 360 covariant functor, 103 cross product, 106 cup product, 59, 62

413

Index

cycle, 2 homologous, 3 degree of a map, 36 Dehn twist, 50 de Rham cohomology, 277 theorem, 289 diagonal approximation, 105, 184, 214 Alexander Whitney, 105,214 difference cochain, 117 differential fonn closed,277 exact, 277 polynomial on a complex, 296 on a simplex, 296 smooth,296 dimension axiom, 203, 204 direct limit, 265 product of vector bundles, 134 sum of Abelian groups, 21 of bundles, 134 directed set, 264 of Abelian groups, 265 divisible group, 32 domain invariance theorem 205 double point, 325 ' dual bundle, 171 Stiefel Whitney class, 149 duality Alexander, 81 Alexander Pontryagin, 271 Poincare, 44 dunce hat, 115 effective action, 177 Eilenberg theorem, 117 Eilenberg MacLane space, 122 Eilenberg Zilber theorem, 213 element regular, 353 singular, 353 elementary ideal, 307 embedding r-transnonnal, 56 transnormal, 56 equivalent microbundles, 254 vector bundles, 132 equivariant map, 173 Euler characteristic, 51, 90 of a pair, 55, 182

class, 144 exact fonn, 277 sequence Mayer Vietoris, 18 of a pair, 12 Smith,181 split, 24 exactness axiom, 203, 204 excision axiom, 200, 203, 204 noncommutative, 224 isomorphism, 12 theorem, 198 extraordinary (co)homology theory, 204 five lemma, 15 form bi-invariant, 348 closed,277 exact, 277 intersection, 88, 260 left-invariant, 348 polynomial on a complex, 296 on a simplex, 296 quadratic over 2':2, 317 right-invariant, 348 Seifert, 304 smooth,296 fonnula Cartan, 189 coboundary, 186 of universal coefficients, 33 Thorn, 257 Whitney, 147 WU,193 free action, 183 chain complex, 4 functor, 103 module, 305 resolution of an Abelian group, 28 functor acyclic, 103 contravariant, 103 covariant, 103 free, 103 fundamental class, 36 cohomology, 125 of a topological manifold, 232 with boundary, 249 generalized (co)homology theory, 204 germ, 265 Gromov norm, 221

414

group cohomology, 22 reduced,23 relative, 23 singular, 202 colored braid, 122 divisible, 32 free Abelian, 4 homology of a chain complex, 4 singular, 196 Lie, 344 of braids, 123 of coefficients, 203, 204 periodic, 31 ring, 180 simplicial homology, 3 Smith homology, 181 Gysin sequence, 256 Helly's theorem, 208, 209 homologous cycles, 3 homology cellular, 37, 210 disk,183 group of a chain complex, 4 simplicial, 3 Smith,181 primitive class, 49 reduced, 17 relative, 12 sequence for a triple, 15 of a pair, 12 singular, 196 sphere, 45, 81, 183 with closed supports, 48 homomorphism Bockstein, 14, 92, 142 houndary,2 connecting, 12, 14 Hurewicz, 112 of presheaves, 263 restriction, 263 transfer, 179 homotopic trivializations, 133 vector fields, 133 homotopy axiom, 203, 204 chain, 5, 196 Hopf algebra, 361 fibration, 219 invariant, 219 theorem, 337, 362

Index

Hopf Whitney theorem, 119 Hurewicz homomorphism, 112 theorem, 113 hyperbolic manifold, 222 ideal Alexander, 308 elementary, 307 induced bundle, 133 infinite cyclic covering, 304 Grassmann manifold, 153 infinite-dimensional lens space, 123 injective resolution, 32 integral form, 280 intersection complete, 341 form, 88, 260 number, 42 invariance of a boundary, 206 of a domain, 205 invariant Arf of a knot, 319 of a link, 320 of a quadratic form, 317 Hopf,219 isomorphic vector bundles, 132 isomorphism excision, 12 Lefschetz, 81 for topological manifolds, 250 of bundles with structure group, 273 Poincare, 44 for de Rham cohomology, 285 for topological manifolds, 243 with local coefficients, 138 suspension, 20, 201 Thom, 238 Kiinneth theorem, 99 for relative homology, 215 for singular homology, 213 relative, 215 Kan Whitehead theorem, 129 Kolmogorov Alexander multiplication, 59 Lefschetz fixed point theorem, 56 isomorphism, 81 for topological manifolds, 250 number, 56 left-invariant form, 348 vector field, 346

Index

lemma on extension, 297 Poincare, 283 lens space, 93 infinite-dimensional, 123 Leray Hirsh theorem, 168 Lie algebra, 344 group, 344 line bundle associated with a divisor, 339 linking number, 46, 83 local system of groups, 136 manifold almost parallelizable, 141 Grassmann infinite, 153 hyperbolic, 222 parallelizable, 132, 260, 338 Schubert, 157 stably parallelizable, 156 Stiefel, 139 complex, 162 topological with boundary orientable, 248 map algebraically trivial, 120 chain, 4 equivariant, 173 splitting, 167 Massey triple product, 84 matrix presentation, 305 Seifert, 304 maximal torus, 351 Mayer Vietoris sequence, 18, 200 for de Rham cohomology, 278 for de Rham cohomology with compact supports, 279 noncommutative, 224 relative, 20, 202 microbundle, 254 equivalent, 254 tangent, 254 Milnor theorem, 338 Minkowski theorem, 177 model, 103 acyclic, 104 module finitely generated, 305 free, 305 Moore space, 128 morphism, 103 Morse inequality, 210 multidimensional Borromean rings, 87 multiplication, 358 Kolmogorov Alexander, 59

415

natural transformation, 103 naturality of cap product, 71 of Stiefel Whitney classes, 146 noncommutative cocycle, 222 cohomology, 222, 274 excision axiom, 224 Mayer Vietoris sequence, 224 nondegenerate bilinear map, 109 quadratic form over Z2, 317 nonnegative chain complex, 4 nonorientable bundle, 143 normal degree of an immersion, 335 normalizer, 351 number Betti,3 intersection, 42 Lefschetz, 56 linking, 46, 83 self-intersection, 326, 327 Stiefel Whitney, 152 object of a category, 103 obstruction, 116 to extending sections, 138 ordered chain complex, 60 orientable bundle, 143 topological manifold, 232 with boundary, 248 orientation of a topological manifold, 232 with boundary, 248 system of groups, 136 oriented topological manifold, 232 parallelizable manifold, 132, 260, 338 partition of an integer, 158 periodic group, 31 Poincare duality, 44 isomorphism, 44 for de Rharn cohomology, 285 for topological manifolds, 243 with local coefficients, 138 lemma, 283 theorem, 112 point double, 325 self-intersection, 325 polynomial Alexander, 308 in Conway's normalization, 315 Conway, 316

416

differential form on a complex, 296 on a simplex, 296 Pontryagin characteristic class, 172 theorem, 152 presentation matrix, 305 presheaf, 263 constant, 264 primitive homology class, 49 product Massey triple, 84 of Abelian groups Cartesian, 21 tensor, 27 tensor of chain complexes, 97 vector bundle, 132 projective resolution, 32 projectivization of a vector bundle, 167 pullback, 133 quadratic form over Z2, 317 nondegenerate, 317 rank of a Lie group, 353 reduced cohomology, 23 homology, 17 regular G-complex, 174 element, 353 immersion, 325 relative chain, 12 cochain,23 cohomology, 23 homology, 12 Kiinneth theorem, 215 Mayer Vietoris sequence, 20, 202 resolution injective, 32 projective, 32 restriction homomorphism, 263 right-invariant form, 348 ring Borromean, 85 group, 180 roots, 355 Schubert manifold, 157 265 of a bundle, 131 zero, 132 Seifert form, 304 knot, 310

~ction,

Index

matrix, 304 surface, 305 self-intersection number, 326, 327 point, 325 sequence exact of a pair, 12 Gysin, 256 Mayer Vietoris, 18, 200 for de Rham cohomology, 278 for de Rham cohomology with compact supports, 279 noncommutative, 224 relative, 20, 202 Smith exact, 181 set admissible, 231 cofinal, 266 directed, 264 of Abelian groups, 265 sheaf,264 associated with a presheaf, 265 generated by a presheaf, 265 signature of a manifold, 90 of a product, 108 Thom theorem, 91 simplex boundary, 1 singular, 195 simplicial G-complex, 173 action, 173 complex acyclic, 6 homology group, 3 volume, 221 singular cohomology, 202 element, 353 homology, 196 simplex, 195 skein relation, 316 skew-commutativity of cup product, 63 Smith exact sequence, 181 homology group, 181 theorem, 183 smooth differential form on a compl:lx, 296 on a simplex, 296 triangulation, 289 vector bundle, 131 space K(7I",n),122

Eilenberg MacLane, 122 lens, 93

Index

Moore, 128 split exact sequence, 24 splitting map, 167 stably equivalent vector bundles, 148 parallelizable manifold, 156 Steenrod square, 188 Steenrod's five lemma, 15 Steenrod Eilenberg axioms, 203, 204 Stiefel manifold, 139 complex, 162 theorem, 260 Stiefel Hopf theorem, 109 Stiefel Whitney class characteristic, 141 dual,149 of a manifold, 150 total, 149 number, 152 Stokes theorem, 281 strong Whitney embedding theorem, 325 sum connected of manifolds, 336 direct of Abelian groups, 21 of bundles, 134 Whitney, 134 support of a chain, 6 suspension isomorphism, 20, 201 symplectic basis, 317 tangent microbundle, 254 tensor product of Abelian groups, 27 of chain complexes, 97 theorem acyclic model, 104 acyclic support, 7 Alexander, 205 Arf,318 Borsuk Ulam, 76 chain homotopy, 5 collar for smooth manifolds, 78 de Rham, 289 simplicial, 299 domain invariance, 205 Eilenberg, 117 Eilenberg Zilber, 213 excision, 198 Helly's, 208, 209 Hopf, 337, 362 Hopf Whitney, 119 Hurewicz, 113 K iinneth, 99 for relative homology, 215

417

for singular homology, 213 relative, 215 Kan Whitehead, 129 Lefschetz fixed point, 56 isomorphism, 81 Leray Hirsh, 168 Milnor, 338 Minkowski, 177 on a collar, 247 on acyclic models, 104 supports, 7 on domain invariance, 205 Poincare, 112 Pontryagin, 152 Smith,183 Stiefel, 260 Stiefel Hopf, 109 Stokes, 281 Thorn, 257 signature, 91 Whitney duality, 149 strong embedding, 325 Wu,259 Thorn class, 237, 255 formula, 257 isomorphism, 238 theorem, 257 signature, 91 topological generator, 350 manifold orientable, 232 oriented, 232 with boundary orientable, 248 torsion subgroup, 44 torus, 345 maximal,351 total chain complex, 60 Stiefel Whitney class, 149 transfer homomorphism, 179 transition function, 272 transnormal embedding, 56 transversality, 73 triple homology sequence, 15 Massey product, 84 trivial vector bundle, 132 universal coefficient formulas, 33 theorem, 33

index

41ti

vector bundle, 131 canonical, 153 complex, 162 equivalent, 132 isomorphic, 132 product, 132 smooth, 131 stably equivalent, 148 trivial, 132 field homotopic, 133 left-invariant, 346

Whitney formula, 147 sum, 134 theorem duality, 149 strong embedding, 325 trick, 329 Wu class, 259 formula, 193 theorem, 259 zero sertion, 132

Titles in This Series 84 Charalambos D. Aliprantis and Rabee Tourky, Cones and order, 2007 83 Wolfgang Ebeling, Functions of several complex variables and their singularities (translated by Philip G. Spain), 2007 82 Serge Alinhac and Patrick Gerard, Pseudo-differential operators and the Nash Moser theorem (translated by Stephen S. Wilson), 2007 81 V. V. PrBSolov, Elements of homology theory, 2007 80 Davar Khoshnevisan, Probability, 2007 79 WllIilllIl Stein, Modular forms, a computational approach (with an appendix by Paul E. Gunnells), 2007 78 Harry Dym, Linear algebra in action, 2007 77 Bennett Chow, Peng Lu, and Lei Ni, Hamilton's Ricci flow, 2006 76 Michael E. Taylor, Measure theory and integration, 2006 75 Peter D. Miller, Applied asymptotic analysis, 2006 74 V. V. PrBSolov, Elements of combinatorial and differential topology, 2006 73 Louis Halle Rowen, Graduate algebra: Commutative view, 2006 72 R. J. Williams, Introduction the the mathematics of finance, 2006 71 S. P. Novikov and I. A. Taimanov, Modern geometric structures and fields, 2006 70 Sean Dineen, Probability theory in finance, 200.') 69 Sebastian Montiel and Antonio Ros, Curves and surfaces, 2005 68 Luis Caffarelli and Sandro Salsa, A geometric approach to free boundary problems, 2005 67 T.Y. LIlIIl, Introduction to quadratic forms over fields, 2004 66 Yuli Eidelman, Vitali Milman, and Antonis Tsolomitis, Functional analysis, An introduction, 2004 65 S. RIlIIlanan, Global calculus, 2004 64 A. A. Kirillov, Lectures on the orbit method, 2004 63 Steven Dale Cutkosky, Resolution of singularities, 2004 62 T. W. Korner, A companion to analysis: A second first and first second course in analysis, 2004 61 Thomas A. Ivey and J. M. Landsberg, Cartan for beginners: Differential geometry via moving frames and exterior differential systems, 2003 60 Alberto Candel and Lawrence Conlon, Foliations II, 2003 59 Steven H. Weintraub, Representation theory of finite groups: algebra and arithmetic, 2003 58 Cedric Villani, Topics in optimal transportation, 2003 57 Robert Plato, Concise numerical mathematics, 2003 56 E. B. Vinberg, A course in algebra 2003 55 C. Herbert Clemens, A scrapbook of complex curve theory, second edition, 2003 54 Alexander Barvinok, A course in convexity, 2002 53 Henryk Iwaniec, Spectral methods of automorphic forms, 2002 52 llka Agricola and Thomas Friedrich, Global analysis: Differential forms in analysis, geometry and physics, 2002 51 Y. A. Abramovich and C. D. Aliprantis, Problems in operator theory, 2002 50 Y. A. Abramovich and C. D. Aliprantis, An invitation to operator theory, 2002 49 John R. Harper, Secondary cohomology operations, 2002 48 Y. Eliashberg and N. Mishachev, Introduction to the h-principle, 2002 47 A. Yu. Kitaev, A. H. Shen, and M. N. Vyalyi, Classical and quantum computation, 2002

TITLES IN THIS SERIES

46 Joseph L. Taylor, Several complex variables with connections to algebraic geometry and Lie groups, 2002 45 Inder K. Rana, An introduction to measure and integration, second edition, 2002 44 Jim Agler and John E. MCCarthy, Pick interpolation and Hilbert function spaces, 2002 43 N. V. Krylov, Introduction to the theory of random processes, 2002 42 Jin Hong and Seok-Jin Kang, Introduction to quantum groups and crystal bases, 2002 41 Georgi V. Smirnov, Introduction to the theory of differential inclusions, 2002 40 Robert E. Greene and Steven G. Krantz, Function theory of one complex variable, third edition, 2006 39 Larry C. Grove, Classical groups and geometric algebra, 2002 38 Elton P. Hsu, Stochastic analysis on manifolds, 2002 37 Hershel M. Farkas and Irwm Kra, Theta constants, Riemann surfaces and the modular group, 2001 36 Martin Schechter, Prmciples of functional analysis, second edition, 2002 35 James F. Davis and Paul Kirk, Lecture notes in algebraic topology, 2001 34 Sigurdur Helgason, Differential geometry, Lie groups, and symmetric spaces, 2001 33 Dmitri Burago, Yuri Burago, and Sergei Ivanov, A course in metric geometry, 2001 32 Robert G. Bartle, A modern theory of integration, 2001 31 Ralf Korn and Elke Korn, Option pricing and portfolio optimization: Modern methods of financial mathematics, 2001 30 J. C. McConnell and J. C. Robson, Noncommutative Noetherian rings, 2001 29 Javier Duoandikoetxea, Fourier analysis, 2001 28 Liviu I. Nicolaescu, Notes on Seiberg-Witten theory, 2000 27 Thierry Aubin, A course in differential geometry, 2001 26 Rolf Berndt, An introduction to symplectic geometry, 2001 25 Thomas Friedrich, Dirac operators in Riemannian geometry, 2000 24 Helmut Koch, Number theory: Algebraic numbers and functions, 2000 23 Alberto Candel and Lawrence Conlon, Foliations I, 2000 22 Gunter R. Krause and Thomas H. Lenagan, Growth of algebras and Gelfand-Kiri1lov dimension, 2000 21 John B. Conway, A course in operator theory, 2000 20 Robert E. Gompf and Andre I. Stipsicz, 4-manifolds and Kirby calculus, 1999 19 Lawrence C. Evans, Partial differential equations, 1998 18 Winfried Just and Martin Weese, Discovering modern set theory. II: Set-theoretic tools for every mathematician, 1997 17 Henryk Iwaniec, Topics in classical automorphic forms, 1997 16 Richard V. Kadison and John R. Ringrose, Fundamentals of the theory of operator algebras. Volume II: Advanced theory, 1997 15 Richard V. Kadison and John R. Ringrose, Fundamentals of the theory of operator algebras. Volume I: Elementary theory, 1997 14 Elliott H. Lieb and Michael Loss, Analysis, 1997 13 Paul C. Shields, The ergodic theory of discrete sample paths, 1996 12 N. V. Krylov, Lectures on elliptic and parabolic equations in HOlder spaces, 1996 11 Jacques Dixmier, Enveloping algebras, 1996 Printing

For a complete list of titles in this series, visit the AMS Bookstore at www.ams.org/bookstore/.

The book is a continuation of the previous book by the author (Elements of Combinatorial and Differential Topology, Graduate Studies in Mathematics, Volume 74, American Mathematical Society, 2006). It starts with the definition of simplicial homology and cohomology, with many examples and applications. Then the Kolmogorov-Alexander multiplication in cohomology is introduced. A significant part of the book is devoted to applications of simplicial homology and cohomology to obstruction theory, in particular, to characteristic classes of vector bundles. The later chapters are concerned with singular homology and cohomology, and Cech and de Rham cohomology. The book ends with various applications of homology to the topology of manifolds, some of which might be of interest to experts in the area. The book contains many problems; almost all of them are provided with hints or complete solutions.

ISBN 978 - 0 - 8218 - 3812 - 9

917 8 0821838 129

GSM/81