Cohomology of Groups

Cohomology of Groups This is Volume 34 in PURE AND APPLIED MATHEMATICS A Series of Monographs and Textbooks Edited by...

Author: Edwin Weiss

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Cohomology of Groups

This is Volume 34 in PURE AND APPLIED MATHEMATICS A Series of Monographs and Textbooks Edited by PAULSMITHA N D SAMUEL EILENRERG A complete list of titles in this series appears at the end of this volume

Cohomology of Groups EDWIN WEISS Department of Muthemutics Boston University Boston, Mussuchusetts

A C A D E M I C P R E S S New York and London

1969

COPYRIGHT 0 1969,

BY ACADEMIC pRB8s, INC. ALL RIGHTS RESERVED. NO PART OF THIS BOOK MAY BE REPRODUCED IN ANY FORM, BY PHOTOSTAT, MICROPILM, REI'RIBVAL SYSTEM, OR ANY OTHER MEANS, WITHOUT WRITTBN PERMISSION FROM THE PUBLISHKRS.

ACADEMIC PRESS, INC. 111 Fifth Avenue, New York, New York 10003

Umted Kingdom Edition published by ACADEMIC PRESS, INC. (LONDON) LTD. Berkeley Square House, London W.l

LIBRARY OF

CONGRESS CATALOG CARD

NUMBER: 78-84239

AMS 1968 SUBJECT CLASSIFICATIONS 1820, 1068

PRINTED IN THE UNITED STATES OF AMERICA

TO M y Parents

This Page Intentionally Left Blank

Preface

T h e prerequisites for the study of the “modern version” of class field theory, as formulated by Artin and Tate, are of two kinds. On the one hand, the student needs to know certain basic arithmetic results. This subject matter, which is commonly referred to as algebraic number theory, is now readily available in several books, including one by the author. On the other hand, the student must be well versed in certain topics from homological algebra, centering upon the cohomology of groups. I t is the purpose of this book to provide a reasonably mature student of mathematics with direct access to this somewhat specialized material. Thus, in virtue of its focus and content, the most appropriate title for this book is the overly long ... Cohomology of Groups in Preparation for Class Field Theory. We have taken a fairly classical approach to cohomology theory. Among the reasons underlying this choice of approach, the following may be listed: it is historically valid; it is, hopefully, relatively easy to understand ; it probably constitutes a shorter path to the desired results than other, more abstract, approaches. The presentation of the material is intended to be detailed, expansive, and essentially self-contained. A conscious attempt has been made not to lose sight of the concrete since explicit computations are important for future class field theoretic applications. vii

...

Vlll

PREFACE

The structure of the book is apparent from the table of contents and from the introductory remarks at the beginning of chapters and sections. There is little to be gained by a detailed analysis of the structure at this point. The problems and exercises are of varying degrees of difficulty or significance, but as a rule they are not required for later use in the text. A number of topics have been omitted-for example, spectral sequences, of the Brauer group (including its connections with central simple algebras). The fact that our attitude is not encyclopedic, but rather expository, is reflected in the bibliography, which is far from complete. My sincere thanks are due to a number of individuals-especially to Sandra Spinacci for her skill and efficiency in typing the manuscript, to my family and to S. Shufro for relieving me of assorted burdens, and for assistance in both tangible and intangible form.

May, I969 Brookline, Massachusetts

E. W ~ r s s

Contents

vii

PREPACE

1.

Cohomology Groups of G in A

1-1. 1-2. 1-3. 1-4. 1-5. 1-6.

Differential Groups Herbrand’s Lemma G-Complexes (Positive Part) G-Complexes (Negative Part) Cohomology Groups of G in A Problems and Supplements

1 8 12 18 26 35

II.

Mappings of Cohomology Groups

2-1. 2-2. 2-3. 2-4. 2-5.

Homomorphism of Pairs Independence of the Complex Conjugation, Restriction, and Inflation The Transfer or Corestriction Explicit Formulas

45 55 64 70 77

111. Some Properties of Cohomology Groups 3-1. Miscellaneous Facts 3-2. Cyclic Groups and the Herbrand Quotient

ix

87 93

CONTENTS

X

3-3. 3-4. 3-5. 3-6. 3-7.

Dimension Shifting The Inflation-Restriction Sequence The Group H-*(G, Z) Cohomological Equivalence Problems and Supplements

101 108 112 119 125

IV. The Cup Product 4-1. 4-2. 4-3. 4-4. 4-5. 4-6.

Cup Products Properties of the Cup Product Cup Products for a Pairing The Duality Theorem The Nakayama Map Problems and Supplements

136 145 155 163 170 181

V. Group Extensions 5-1. 5-2. 5-3. 5-4.

VI. 6- 1. 6-2. 6-3. 6-4. 6-5. 6-6.

The Extension Problem and Ha Commutator Subgroups in Group Extensions Factor Extensions The Principal Ideal Theorem

195 206 213 218

Abstract Class Field Theory Formations Field Formations Class Formations The Main Theorem Reciprocity Law and Norm Residue Symbol The Existence Theorem

226 232 231 243 249 262

References

267

SUBJECTINDEX

27 I

Cohomology of Groups

This Page Intentionally Left Blank

CHAPTER

I

Cohomology Groups of G in A

I n this chapter, and throughout this book, G = (0, 7 , p, ...} denotes a finite multiplicative group (unless there is an explicit statement made to the contrary). T h e main objectives of this chapter include : introduction of the notion of a G-complex, detailed construction of the standard G-complex, use of an arbitrary G-complex to define the cohomology groups of G in a G-module A (the question of uniqueness of cohomology groups will be treated in Chapter 11), and the interpretation (which depends on the standard G-complex) of the lower-dimensional cohomology groups. Differential groups serve as a convenient starting point, as an introduction to some of the basic techniques, and as a tool for arriving at the infinite cohomology sequence (which is discussed in Chapter 11); in addition, they are used to give a cohomological approach to the Herbrand quotient and its properties. 1-1.

DIFFERENTIAL GROUPS

A differential group is a pair (A, d) where A is an abelian group (which we shall usually write additively) and d (the 1

I. COHOMOLOGY GROUPS

2

G

OF

IN

A

differential operator) is an endomorphism of A such that o d = 0. Since image d C kernel d, we may form the group H ( A ) = ker d/im d and call it the derived group of (A, d). If (A,, d,) and (A,, d,) are differential groups, then a homoA, is said to be admissible when morphism f : A, d, o f = f o d, . In this section, all differential operators (and, in particular, d, and d, above) will be denoted by d; this should cause d2 = d

-

no confusion.

(B,d )

Proposition. An admissible map f : (A, d ) of differential groups induces a natural homomorphism

1-1-1.

f* : H ( 4

given by

f*(a

If f , g : (A, d ) admissible and

__+

-

+w

-

H(B)

+ dB

=f ( a )

where da = 0

(B, d) are both admissible, then f

g is

-

( f h g ) * =f* + g *

If f : (A, d) (B, d) and g : (B, d ) missible, then g o f is admissible and

(C, d) are both ad-

Proof: f * is well-defined-for if a is replaced by another representative of the same class, say a da, with a, E A, then f(a da,) = f(u) +f(da,) = f ( a ) df(a,) and this belongs to f(a) dB. The remaining assertions are clear. I

+

+

1-1-2.

+

+

Corollary.

We have 0, = O

and

1,

-

=

1

-

In more detail, if 0 is the trivial map of (A, d) into (B, d) then O* is the trivial map of H ( A ) into H(B), and if 1 : A A is the identity map then 1, : H ( A ) H ( A ) is the identity.

1 - 1. 1-1-3.

Theorem.

DIFFERENTIAL GROUPS

3

Suppose that

is an exact sequence (sometimes it will be referred to as a short exact sequence) of differential groups with admissible maps, then there exists a homomorphism d* : H ( C ) --t H ( A ) such that the following triangle is exact :

H(B)

Moreover, if O - A ~ B i c - 0

0 -A'

11 gi -LB'

ci-+ o

is a commutative diagram of differential groups with exact rows and all maps admissible, then the following prism has exact triangles and commutative faces :

Proof: in [50].)

(The first formulation of this result probably appears

definition of d* :

An element y

EH(C)

is represented by

I. COHOMOLOGY GROUPS OF G IN A

4

an element c E C with dc = 0; that is, y = c + dC. Since j is onto, there exists b E B with j b = c. Becausej(db) = d(jb) = dc = 0, there exists a E A such that iu = db. Now, i is a monomorphism and i(da) = d(ia) = d(db) = 0, so that da = 0. Thus, a determines an element a! = a dA E H ( A ) , and we put d,(y) = a!. In other words, where j b = c, ia = db d,(c dC) = u d A

+

+

or symbolically : d,(c

+

+ dC) = (i-ldj-l)(c) + dA.

First let us show that if y = 0 E H ( C ) d, is well-defined: then a! = d,y = O~H(A)-that is, if c = dc,, C E C then a = da, . Now, if c = dc, there exists b, E B such that jb, = c1 so that j(b - db,) = j b -jdb, = c - djb, = 0. Hence, there exists a, E A such that ia, = b - db, . Therefore, iu

=

db

=

d(iu,

+ db,) = idu,

and

u =

du,

.

Furthermore, it is immediate from its definition that d, is additive, so it follows that d, is indeed a well-defined homomorphism of

H(C) +H(A).

The next step is to prove the six kernel-image relations which imply that the triangle is exact.

imj, Ckerj, :

j , o i , = ( j o i ) , = 0, = 0.

im j , C ker d, : An element of H ( B ) is of form b + dB with db = 0, and d,[j,(b dB)] = d,(jb dC) = a d A where iu = db = 0. Hence, a = 0 and a dA = 0 E H(A).

+

+

+

+

im d* C ker i, : i,[d,(c

+ dC)] = i,(a + dA) = iu + dB = db + dB = dB = 0 E H(B).

kerj, C im i* : db=O

and

Suppose that j,(b+dB)=jb+dC=OEH(C).

There exists c E C such that j b = dc. Choose b, E B such that jb, = c. Sincej ( b - db,) = 0 there exists a E A with ia = b - db, .

1-1 . Now, idu

=

diu = d(b - db,) and

du = 0

ker d, C im j , : dc

and

=0

5

DIFFERENTIAL GROUPS

i,(u

=

0, so that

+ dA) = b - db, + dB = b + dB.

Suppose that d,(c

+ dC) = a + dA = 0 E H(A),

so that a = du, , a, E A . Put b, = b - ia, (where c = j b , iu = db). T h e n j b , = j b = c and db, = db - idu, = 0, and we have j,(b, dB) = j b , dC = c dC.

+

+

ker i, C im d* : du = 0

+

Suppose that and

i,(u

+ d A ) = 0 E H(B).

This means that iu = db for some b E B. Putting c = jb, we have dc = 0 and d,(c dC) = a dA.

+

+

Finally, the relations g, 0 i, = i; 0 f* and h , oj, = j i 0 g, follow from (1-1-1), while g, 0 d , = d , 0 h , may be checked as follows :f,d,(c dC) = f*(u d A ) = !(a) dA’ where dc = 0, j b = c, iu = db, and d,h,(c d C ) = d,(hc dC’) =f(u) dA’ since hc = hjb = j’gb, dgb = gdb = giu = i’( fa). This completes the proof. I

+

+ +

+ +

+

An abelian group A is said to be graded when it is of the form = @ A , (weak direct sum). Suppose that the graded group A is also a differential group with differential operator d, then d is said to be compatible with the grading if for Y = + 1 or - 1 we have d : A , A,+, for each n. I n such a situation ( A , d, Y ) is said to be a differential graded group. Note that any abelian group A can be made into a differential graded group in trivial fashion by putting A , = A , A, = (0) for n # 0, d = 0. (It is clear that the notion of a differential graded group could be defined for any integer Y and the subsequent theory developed on this basis, but these matters are not of interest to us.) Consider a differential graded group ( A , d, Y ) ; then for each n, d, = d I A , is a homomorphism of A , A,,, and d,,, 0 d, = 0. Thus, ( A , d, Y) may be written in a more suggestive fashion as a chain or sequence

A

x?:

-

-

I.

6

G

COHOMOLOGY GROUPS OF

IN

A

Conversely, suppose we are given a chain (*) of abelian groups and homomorphisms d, such that dn+? 0 d, = 0 (and I = f l ) , then @A,, d = @ d, it is clear that upon writing A = (A, d, Y) is a differential graded group. These two attitudes toward a differential graded group will be used interchangeably. T h e case Y = $1 is usually called cohomology (and d is then denoted by 6 and called the coboundary operator), while the case T = -1 is called homology (and d is then denoted by a and called the boundary operator). This distinction is highly artificial in that if (A, d, + I ) is a differential graded group of cohomology type we may put A' = @ A; where A; = A_, and d' = @ dk where dk = d-, , and then (A', d', -1) is a differential graded group of homology type having all properties possessed by (A, d, $ 1)-and conversely. Because of this, we shall usually treat one of the cases I = *I and leave it to the reader to supply the details for the other case, when necessary. Consider the differential graded group (A, d, +l). For each n, the group A, (which we may view as a subgroup of A) is called the group of n-cochains. T h e operator d determines for each n the groups : %on = A, n ker d = A, n ker d, = A n ker d, (the group of n-cocycles) and

xTz

:x

:x

:x

Bn = A, n dn-lAn-l = A, n dA

=

A n d,-lA

(the group of n-coboundaries). T h e nth derived group H J A ) = 23'7$4?n is also known as the nth cohomology group of A. Since ker d = @ 3,and im d = 09,it follows that

:x

x:

-

so that we may identify, and write H ( A ) = 1 :: @ H,(A). A homomorphism f :A B is said to be an admissible map for the differential graded groups (A, d, $1) and (B, d, +1) when f 0 d = d 0 f and f maps A, into B, for every n. It is then

-

-

H ( B ) we have clear that for the induced map f* : H ( A ) : H,(A) H,(B) for every n. It may also be noted that if we put f = f I A,, then f, : A, -+ B, ,f,+lo d = d o f, , and f= f, ; conversely, if for each n we have a homomorphism

fh

xT$@

-

1- 1.

7

DIFFERENTIAL GROUPS

B, such that fn+l 0 d = d 0 f, , then f = C$ @ f, is an admissible map for the differential graded groups.

f, : A,

1-1-4. Proposition. Suppose that we are given the following commutative diagram of differential graded groups, all of cohomology type, with all maps admissible and the rows exact :

O+A&BLC-O

o

-fl

A'

O1

LB' 2C' --+o

- h*l - '*I -

Then, the following diagram is commutative and has exact rows :

% H,(A) -!LH,(B)

*..-+

Hn-l(C)

...

H,+l(C') -% H,(A') 1*_ H,(B')

-

"1

d

f.1

g.1

2H,(C) i;

H,(C')

d*

Hn+l(A)

...

H,+l(A')

...

Proof: Immediate from (1-1-3) as soon as one checks that d , raises dimension by 1. I t should be noted that the earlier discussion of differential groups with the consequent result (1-1-3) on the derived groups of an exact triangle was, in large part, a matter of technical convenience-namely, to minimize the need for subscripts in the proof of (1-1-4). I 1-1-5. Exercise. If the following diagram of differential groups and admissible maps has exact rows and columns 0 0 0

1

1

1

0

0

0

I.

8


G

IN

A

then the following diagram is anticommutative-that d , d , = -d, d , : 0

is,

0

H(C")-% H(A") d*l

(-1)

Id*

H ( C ) -% H(A') 1-2.

HERBRAND'S LEMMA

This section is in the nature of a digression from the main theme of the chapter. We apply the results of Section 1-1 on derived groups to interpret cohomologically a classical index computation (see, for example, [16, p. 3751) which is useful in class field theory. 1-2-1. Herbrand's Lemma. Suppose that A is an additive abelian group and that a and T are endomorphisms of A such that a 0 7 = T o a = 0. Suppose that B is a subgroup of A of finite index which is stable under both a and T (that is, aB C B and TBC B). Let A, denote the kernel of a on A and Au = aA denote the image of A under a. If (B, : BT) and (B, : B') are both finite, then so are ( A , : k )and (A, : A,), and then (A, : A') - (B, : B7) (A, : A,) - (B, : B")

Proof: Since the natural map of A + k / B is onto with kernel B A T ,and since BA,/B GW A,/B n A, = A,/B, it follows that ( A : B ) = ( A : BA,)(BA, : B ) = (A' : B7)(A,: B,) (A, : Bu) = (A' : B7)

(B, : Bu) (A, : Au)(Au: Bu) = (A7 : B7) (B, : Bo)

where all the factors are finite. By interchanging the roles of a and

1-2.

9

HERBRAND'S LEMMA

7 we get another formula for (A : B); comparing the two gives the desired result.

It may be noted that the proof applies when A is not abelian; of course, B must then be a normal subgroup. One may also give a cohomological treatment of Herbrand's lemma. Given A, u,T as before, we construct a differential graded group ( A = C$ @ A , , d = @ d , , +l)byputtingA, = A for all n, d, = u for even n, d, = 7 for odd n. Thus (A,d, 1) may be viewed as a sequence

:x

**.

- - - A_, = A

dl=r

A, = A

4=u

A,

=A

4-7

.**

and because this sequence has period 2, it may be expressed in closed form as

The derived groups of (A, d, 1) are

He"&q

=

ker do Ho(A)= _ -A, im dl

A*

dl A, Had(k7) = Hl(A)= ker 7 -imd, Aa

If both Ho(A)and H,(A) are finite groups we put

and call it the Herbrand quotient of A with respect to u and 7. Note that Q(A) depends on the ordering of the endomorphisms u and T . If we wish to keep track of u and 7 we write Qq,, for Q; of course, QT,u(A) = l/Qa,T(A)1-22.

Proposition.

If A is finite, then Q(A) = 1.

I.

10


G

IN

A

Since A is finite, all indices in the following diagram Proof: are finite (and, in particular, Q(A)is defined) : A A,

/ \

A,

I

A'

I

A0

\ / (0)

-

Because (A : A,) = (Au: (0)) and (A : A,) = (AT : (0)) it follows that (A, :A,) = (A, : Au)-so Q(A) = 1. I Suppose that 0 A B 6 C -+ 0 is an exact sequence of groups and that u is an endomorphism of B such that (when i is viewed as inclusion) the subgroup A is stable under o; then u induces an endomorphism of C, which we denote also by u. Since u o i = i 0 u and u 0 j = j 0 u we say that u is an admissible endomorphism for the short exact sequence. Clearly, the same terminology applies when there are three endomorphisms o, ,uB ,o, of A, B, and C, respectively such that uBo i = i o u, and u c o j =j

o , .

1-2-3. Proposition. Suppose that u and endomorphisms of the short exact sequence

r

are admissible

O-A-LBLC-O of abelian groups and that u r = r u = 0 (on B and, hence, Q(C)are defined, then on A and C too). If any two of Q(A),Q(B), 0

0

so is the third, and then

Q(B) = Q(4 Q(C)

Proof: From A, B, C, u, r we may construct the differential d, I), (C, d, 1). The maps i and j graded groups (A, d, l), (B, determine, in the obvious fashion, mappings (which we also denote by i and j) such that

O-A'-BL€-O

1-2.

HERBRAND’S LEMMA

11

is an exact sequence of differential graded groups with admissible maps. By (1-1-4) and periodicity, we have the closed exact sequence of derived groups

r H o ( A ).AHo(B)A Ho(C)2Hl(A) -L Hl(B) -%Hl(C),

If two of Q(A),Q(B), Q(C)are defined, then four of these derived groups are finite; it then follows easily that the two remaining derived groups are finite, so that the third Herbrand quotient is defined. Now, when all the derived groups are finite, let us put

Therefore, exactness yields :

so that

This completes the proof.

I

In particular, if A is a subgroup of B with (B : A) < co, then, according to (1-2-2) and (1-2-3), Q ( A ) is defined if and only if Q(B) is defined, and then Q(A) = Q(B)-which is simply another way to express the assertion of (1-2-1). 1-24

Exercise.

(i) A differential graded group

I.

12


is said to be periodic of period m d, = d,,, for all integers n. Let

>0

G

IN

A

when A, = A,+, and

denote the order of the nth derived group. If all A, are finite and we write a, = #(A,), and if m is even, then

(ii) The Herbrand quotient may be generalized as follows. Let

(A, d, +1) be a differential graded group of even period m such that all h,(A) < cx). We call

the Herbrand characteristic of A. Suppose that

is an exact sequence of differential graded groups of even period m and that the maps i a n d j are admissible; if two of Q(A),Q(B),Q ( C ) are defined, then so is the third and Q(B) = Q(A)Q(C). 1-3.

G-COMPLEXES (POSITIVE PART)

Let G = (1, u,T ,...} be a finite multiplicative group, and let F = Z[G]denote the integer group ring of G. Thus, the elements of the ring r are formal sums ~ , , e cn,u, n, E Z; addition in r is componentwise, while multiplication is determined in the natural way by the multiplication in G-in more detail,

1-3. G-COMPLEXES (POSITIVE PART)

13

and

We may view G as imbedded in r under the identification of a E G with 1 * a E In particular, 1 denotes the identity of both G and r. It will be customary for us to rename objects associated with in terms of G. For example, we say that an additive abelian group A = {a, b, c, ...} is a C-module when it is a left unitary F-module (unitary means that 1 E F is the identity operator on A). If A is a G-module, then clearly

r.

r

o(a

+ b) = oa + ob .(.a)

= (m) a

l(a) = a

Conversely, if G acts on the additive group A according to the rules (*), then A becomes a G-module when the operation of F on A is defined by noo)(a)= [no(aa)]. If the group A is multiplicative, then the action of G is often denoted by a"; note that then (ao). = aToand aConoo = n o (a%"). For every G we define the action of G on Z to be trivial-that is, an = n for all u E G and n E Z-so that 2 becomes a G-module. In the same way, Q and Q/Z (more precisely, their additive groups) will always be viewed as G-modules with trivial action. Of course, the multiplication in r always permits us to view F as a G-module. Suppose that (X = C @ X , , a = C @ a,, -1) is a differential graded group of homology type, then (X, a, -1) is said to a chain complex over r (or over G) when each X, is a F-module and a (and hence each a,) is a r-homomorphism. T h e chain complex (X, a, -1) is said to be : free if each X , is a F-free module, finite free if each X, is r-free with a finite basis over F, acyclic if its derived groups are (0) (that is, if im a, = ker an-, for all n), and augmented if there exist F-homomorphisms E :X, Z (meaning that E is onto) and p : Z 2- X-, (meaning that p is 1-1) such that a, = p E . Any finite free, acyclic, and augmented chain complex over the finite group G will be referred

xooc

(xoEG

-

0

14

I. COHOMOLOGY

GROUPS OF G IN

A

to as a C-complex and denoted by (X, a, e, p) or simply by X. The standard diagram for a G-complex is the following :

0

0

Since E is onto and p is 1-1, it follows from a, = p E that im a, = im p and ker a, = ker E , and therefore, a G-complex can be broken up into two exact G-sequences (by a C-sequence we mean a sequence of G-modules and G-homomorphisms), namely : 0

which is called the positive part of the G-complex, and which is called the negative part. Note that, conversely, if we are given a positive part and a negative part, then they can be combined (by putting 8, = p E ) to form a G-complex. 0

1-3-1. Theorem. Given a finite group G, there exists a positive part for a G-complex.

Proof:

and put

For n = 0, define a symbol

[.I,

call it the empty cell,

x, = I-[.]

so that X , is a finite free r-module with a basis consisting of a single element. For n > 0, define symbols [q,..., a,] consisting of ordered n-tuples of elements a( E G, call them n-cells, and put

xn= C O r [ u , ...u,,eG

v * - , on]

ul,.

so that X, is a finite free r-module with the set of all n-cells serving as a basis.

1-3. G-COMPLEXES (POSITIVEPART)

15

Since E and the boundary operators a, are to be G-homomorphisms, it suffices to define them on a G-basis-that is, on the cells-and extend linearly. Now define

Note that the formulas for a, and a, are indeed special cases of the formula for a,. The definition of the boundary operators a, may seem somewhat artificial, but we shall see in (1-5-13) that it arises in a rather natural way. I t remains to show that the sequence

... -x2-

a2

x,

a1

x o - L z-0

-

(*I

is exact, and one way to accomplish this is via the following preparations. Suppose that f,g : (A, a, - 1) (B, a, - 1) are admissible maps of differential graded groups, then f and g are said to be homotopic if there exists a homomorphism D : A B such B,+l and f - g = Da aD. In other words, that D :A, if we put D, = D I A,, we have the diagram

-

and for each n

+

-

I.

16


G IN A

Proposition. If f and g are homotopic maps of differential graded groups, then f* = g,

1-3-2.

.

Proof:

Let a E A be a cycle. Since aa = 0 we have

a boundary. Thus f ( a ) and g(a) represent the same homology class. I 1-3-3. Corollary. Consider the admissible mapsf = 1 and g = 0 of the differential graded group (A, a, -1) into itself. If there exists a homotopy D between f and g (such a D is called a contracting homotopy), then H ( A ) = (0).

,

Proof: From above, 0, = 1 ; but 0, maps H ( A ) to (0), while 1, is the identity map of H(A). I

This result is applied to the proof of (1-3-1)by showing that the sequence (*) (with 0's adjoined on the right to make it infinite in both directions) is a differential graded group for which there exists a contracting homotopy. For this, we need to define maps E :Z X , and D, : X , X,,, , n 2 0, such that

-

-

eE = 1

(1)

a,D,

(2)

(3)

a,,+,&

+ Ee = 1

+D,&

=

1

on

Z

(#I

on X ,

on X,, n

t1

Now, a centracting homotopy is an ordinary group homomorphism (rather than a G-homomorphism) so it suffices to define E and D, and to verify (I), (2), (3) on 2-bases of the modules in question. Thus we may put

1-3. G-COMPLEXES (POSITIVE

PART)

17

and extend linearly from these Z-bases. Then eE(1) = 4.1) = 1, which proves (I), and

(a100 + Ee)(o[.l) = al([ol) + E 4 . 1 ) = u[*] = =

+ Eo(1)

4.1 - [*I + [*I .[.I [a]

which proves (2). As for (3), we check that

Dn-lan(o[~l,*-.,on])

= [oo1 3

02

,.**,

on1

and that

Therefore, D is a contracting homotopy, provided we are dealing with a differential graded group; hence, in order to complete the proof, it remains to show that the square of the boundary is 0. For this we observe first that from (1) and (2) it follows that ~ E= E e = ea D0 .+ eEe; therefore, &$Do = 0, and since the image of Do contains the set of I-cells, we have ea, = 0. From the case tt = 1 of (3) and from (2) we have

a,

=

ala2D, + alD,a,

=

ala2o1 + a,

- Eeal ;

therefore, a1a2D1= 0, and since all 2-cells are in the image of D, , alaz = 0. Finally, suppose inductively that a,_,a, = 0, I 2 2; then multiplying the case I - 1 of (3) on the right by a, and the case Y of (3) on the left by a, we have

a,D,a,

+ ~,-,a,-~a,

=

a,

=

a,a,+,~, + ap,-,a,

I.

18

COHOMOLOGY GROUPS OF G IN

Consequently, a,a,+,D, = 0, and then a,a,+, the proof. I

A

= 0. This completes

1-34. Exercise. By a straightforward computation, show that in the positive part of a G-complex as constructed above, e o a, = 0 and an-, 0 a, = 0 for n 2 2.

14.

G-COMPLEXES (NEGATIVE PART)

141. Theorem. Any positive part for a G-complex can be completed to a full G-complex. More precisely, if there is given an exact G-sequence 9..

x, a*

+

__t

- z---

XI 4

x,A

0

in which each X , is G-free with finite basis, then there exists an exact G-sequence 0 __t

z

x-, x-, a-¶ a-1

x-3

..-

such that each X , is G-free with finite basis. Proof:

We require some preliminaries :

142. Proposition. Suppose that A, B, and C are Gmodules. Let Hom (A, B) denote the set of all homomorphisms of A into B (as additive groups), and let Hom, (A, B) denote the set of all G-homomorphisms (that is, r-homomorphisms of r-modules) of A into B. Then (1)

Hom (A, B) is an additive group which becomes a G-module when the action of G is defined by

f”

= ~ o f o u - ~ fEHom(A,B),

(2) Hom, (A, B) is a subgroup of Hom (A, B); in fact, Hornc ( A , B ) = {f E Hom (A, B ) If”

=f

Va E G }

UEG

1-4. G-COMPLEXES (NEGATIVE PART)

19

(3) If f~ Hom (A, B) and g E Hom (B, C), then g o f € Horn (A, C )

and

(g of)"

= gu of".

Proof: Straightforward. Note that to prove that Horn (A, B) is a G-module one must check that ( f, fi), = f; +f; , (fu), = f7(r, and f 1 = f. The action of r on Horn (A, B) is given in our exponential notation by

+

I 143. then :

Let A, A,, A,, B, B, ,B, be G-modules,

Proposition.

(1) If cp E Hom (A,, A) and th, E Hom (B, B,), then we may define a homomorphism (of additive groups)

(v,$1 : Horn ( A ,B )

-

Horn (A1 ,Bl)

by putting for f E Horn (A, B) (9',+)f=

(2) If in addition,

cp,

E

# o f o v

Horn (A, , A,) and

4, E Horn (B, ,B2),

(% $4= (9' O

4)

then

(v1 ! $1)

O

v1

9

41 O

(cp, 4) is additive in each variable; (cp, 0) and (0, t,h) are 0-maps; (1, 1) is the identity map. (4) If q~ and 4 are both G-homomorphisms, then (cp, t,h) is a G-homomorphism; symbolically,

(3)

b, $1 E Horn,

(Horn (4B), Horn (A, Bl))

-

9

( 5 ) If cp and I) are both G-homomorphisms, then (p), Hom, (A, , Bl); symbolically, Hom, (A, B)

(v,rl) E Horn (Horn, (4B), Horn, (4 BlN 9

Proof:

Straightforward verification.

I

4) maps

20

I.


G IN A

Proposition. Let A, B, C, and Z be H-modules, where, as usual, the action of G on 2 is trivial, and define A = Hom (A, Z); then (1) A is a G-module. (2) If A is G-free with finite basis (over F ) , then so is A. (3) If f~ Horn (A, B) and we write f = (f,I), then f E Hom (8, A). If, moreover, f E Hom, (A, B), then f E Hom, (8, A). Iff, fi ,fi E Hom (A, B) and g E Hom (B, C),then (4) n A f1+f2=fi,+f2, g o f = f o t , l=l, o=o.

144.

A

( 5 ) If f~ Horn (A, B) is an epimorphism, then morphism.

f is

a mono-

(1) follows from (1-4-2)¶ and (3) follows from (1-4-3). As for (4), it too follows from the properties of the symbol (a, $)-for example,

Proof:

n g

of

=

(gf, 1)

= (f, 1)

0

(g,1)

=

34.

To prove (5), suppose that f ( t ) = O E A for some t ~ 8Thus . (f,l)t = t of = 0, and since f ( A ) = B this implies t(B) = 0hence t = 0. It remains to prove (2). By hypothesis, A can be written in the form A = @ Fai ; therefore, A = @ Z(ua,), so that A is Z-free with basis {mai 1 i = I (...,n, u E G). For i = 1)..., n

xi,.

xF=l

define fi E

A by putting

u = 1,

.

2'1

.

otherwise and extend linearly from the Z-basis to all of A. Now, {f; I T E G, i = 1,...,n}

is a Z-basis for

A because U = T ,

.

.

otherwise


21

Therefore, { fi I i = 1,..., n} is a T-basis for A^, since i,o

I

i

i,o

145. Proposition. For any G-module A we have the following isomorphisms, which shall often be taken as identifications :

( I ) Horn, (r,A) M A (2) Horn (Z, A) A (3) 2

z

(as

additive groups)

(as

G-modules)

(as G-modules)

Proof: Any f E Hom, (T, A) is determined by its action on 1 E F , and the map f -f(l) is an isomorphism of Horn, A) onto A-so (1) is proved. Similarly, the map f+f(l) provides an isomorphism of Horn (Z, A) onto A as additive groups. It is a G-isomorphism, since f a fu(I) = ufu-'(I) = u(f(1))which proves (2). Since the action of G on 2 = Horn (Z, Z) is trivial, (3) follows immediately. I

(r,

-

Now, let us return to the proof of (1-4-1). From the exact G-sequence **-

--x, a, x,

a1

x,az-0

(*I

we get by dualization the G-sequence

By changing notation and using part (3) of (1-4-5) we may write this as 0

-z x-,-x-,-x-,P,

a-1

a4

...

(***I

According to (1-4-4), each X-, (n 2 1) is G-free with finite basis, and p is a monomorphism since E in onto. Again from (1-4-4), a_, = 0. Thus (***) (with an we have a_, p = 0, a_(,+,) infinite string of 0's added on the left) is a differential graded 0

0

I.

22


G IN A

group. T o complete the proof of (1-4-1) it suffices, therefore, to show that (***) has a contracting homotopy. For this we need : 146. Proposition. Suppose that (A, a, - 1) is a differential graded group with H ( A ) = C @ H,(A) = (0).If each A, is a free abelian group (that is, a free module over Z), then there exists a contracting homotopy.

-

Proof: For each n, aA, = a,A, is a submodule of the free 2-module An-l, so it too is Z-free (see, for example, [42, p. 441). Hence, there exists a Z-homomorphism Tn-l : aA, A, which makes the following diagram commute :

-

It follows that on A,, a, (1 - Tn-l a,) = a, - a, im (1 - T,-l o a,) C ker a, = im a, .Now define D, : A, by putting D, = T, (1 - T,-~ a,) 0

0

+

and check that on A, we have 0 a, a,+, fore, D = C @ D, is a contracting homotopy.

-

0

=

0, so A,+1

D, = 1. There-

I

-

This result applies to the differential graded group determined by (*). More precisely (using the notation of the proof of (1-3-1)), X,,and D, : X, X,+, there exist Z-homomorphisms E : Z which provide a contracting homotopy for (*) (that is, the relations (#) of the proof of (1-3-1) hold for any positive part of a G-complex). By dualization, we have 2-homomorphisms and

B:Xo-Z%Z

Dn:Xn+l-Xn

such that ,

A

(1) I = l = e E = E o , (2)

on on

l=~=Doo81+,-o~

(3) 1 = 1 = S,

0

a,+,

+

Dn-1

on

Z%Z Xo X,,, n > l

23


In other words, the differential graded group determined by (**) has a contracting homotopy. Therefore, the sequence (*w) is exact, and the proof of (1-4-1) is complete. I

A diagram which illustrates the proof of (1-4-1) and the resulting G-complex is as follows :

0

J \

0

In the case where the positive part is the one constructed in (1-3-1), the negative part just constructed may also be given explicitly. (It may be noted in passing that in this situation, (1-4-6) is not needed for the proof of (1-4-l).) The full G-complex derived from them is known as the standard C-complex. Since X, has a G-basis consisting of the empty cell -1, it follows from the proof of part (2) of (1-4-4) that X-, 7 has a canonical G-basis consisting of a single element which is denoted by (.) and is called the (-1)-cell. For any 7 E G, the action of the element (.)’ of Xo on X , is given by its action on the 2-basis (4.3I CT E G}; namely,

4,

u = r

otherwise 1, {[ol ,..., u,] I uiE G} is a G-basis for X , ; hence, For n a G-basis for X-(n+l)= X, is the set of all -(n 1)-cells,

,..., 0),.

+

(Note that a (-n)-cell is given by an ordered (n - 1)) X, is tuple of elements of G.) The action of Xn = X - G + ~ on given (see part (2) of (1-4-4)) by (u,

I.

24

COHOMOLOCY GROUPS OF G IN

As for the action of

p, and

-e,

A

-zEG

a, , note first that, by construction,

: X, +Z is the G-homomorphism that maps [*I 1 E Z, in particular, e is an element of Hom (X,, Z) = 3,= X-l, and 1. On the other hand, the element (->’ for u E G, e : u[.] of 3,also maps u[.] 1. If we denote the element 7 of e

sEG

Z [ q by S (and call it the trace element), then e =

-x,,

C (*>’

= (.>S,

*G

and by an abuse of notation we write this also as S ( . ) . T o compute w2 recall that under the identification 2 w 2, 1 E Z corresponds to the identity map 1 E 2. Consequently, p(1) = i(1) = (e, 1) o 1 = 1 o e = S ( . ) . Finally, a, is determined by a,[.] = pe[-] = p(1) = S(*). The explicit formulas for negative dimensional boundaries are given in (1-4-7), which summarizes the results of Sections 1-3 and 1-4.

p :Z

147. Theorem. A G-complex exists for any finite group G. The standard C-complex (sometimes denoted by X*) is

...

x,a,_ x,

+

a1

__f

-

x,

a0

x-1-

a-1

x-2

-

**

Z 0

/\

0

n>O n>l

1-4. G-COMPLEXES (NEGATIVE

PART)

25

and the boundaries are given by

a,([ol) = 4.1 - [.I an([o1 ,*.s

an])

= u 1 [ ~ 2v.**v on]

Proof: Only the negative dimensional boundary formulas remain to be proved, and the formula for a_, is a special case of that for a_, . (Note that in the expression for X, we abuse for (q ,..., It suffices notation and write r(ol,..., to verify that both sides of the formula for the G-homomorphism a_,, have the same action on a Z-basis of X , . Thus

I. COHOMOLOGY GROUPS OF G

26

IN

A

and when the right-hand side of the formula is evaluated at ,...,un], it too gives an expression of form

u[ul

A

+

n-1

(-1)* Bi

(-1

+ C.

In each case, it is easy to check that A=

:1

= 1, otherwise

001

Ti

= 02

,...,

Tn-l

= 0,

Exercise. How do the results of Sections 1-3 and 1-4 carry over for an infinite group G ? (Of course, the same question is appropriate for future results, too.) In particular, the positive part of a G-complex exists (in another terminology, this says that a free resolution of Z over r exists) with the understanding that the free r-modules need not have finite bases. What about the negative part of a G-complex?

1-48.

1-5.

COHOMOLOGY GROUPS OF G IN A

Suppose that G is a finite group and that A is a G-module; let (X, a, e, p) be any G-complex and consider the additive group Hom, (X, A) = C @ Hom, ( X , , A). Since a : X +X and 1 :A A are both G-homomorphisms, it follows from (1-4-3) that 6 = (a, 1) is an endomorphism of Hom,(X, A) and that a2 = (a, 1) o (a, 1) = (a2, 1) = (0, 1) = 0. There also exist homomorphisms 6, = (a,, 1) : Hom, (X,-l, A) +Hom, (X, , A) with

-

an+,

0

6,

=0

and

6

=

C @ 6, .

1-5. COHOMOLOGY GROUPS

OF

G

IN

A

27

In other words,

is a differential graded group of cohomology type which may also be written as the sequence : **.

-

A ) -% Horn,

Horn,

(x,,A )

&+l

Horn,

(x,,,,A )

-

*.*

We call V" = Vn(G, A) = Hom, (X, , A) the group of n-cochains of G in A, 2Tn = %""(G, A) = { f E $fn I Sf = 0} the group of n-cocycles of G in A, and an= Bn(G, A) = SVn-l the group of n-coboundaries of G in A. The nth derived group H,(Hom, (X, A)) = P/anis known as the nth cohomology group of C in A; it is denoted by Hn(G,A). Two n-cocycles f and g are said to be cohomologous (denoted f -g) when they differ by a coboundary-that is, when they determine the same element in the cohomology group Hn(G,A). Strictly speaking, the cohomology groups (and the other objects too) should be denoted in such a way as to indicate their dependence on the G-complex; thus, we shall write Hn(G, A, X) or H:(G, A) for Hn(G, A) when necessary. It will be shown, in Section 2-1, that the cohomology groups are independent (up to isomorphism) of the choice of G-complex, so that in the current discussion, where the objective is to describe the lower-dimensional cohomology groups, it is sufficient to make use solely of the standard G-complex. The following notations will be used : AG =

{ U E AI uu

SA

{SU 1 U E A }

As

=

={ U E A

I su

=u

VUEG)

where S

Proposition.

H2(G,A ) w

{factor sets} {splitting factor sets}.

c

U

E

~

O€C

= O}

~ = /oSC~ n , ~ EaeCr / ~ n . = /O€C o /~ = ~ 1-5-1.

=

~

1

(

~

-

i

)

I.

28


G IN A

Proof: An n-cochain (with respect to the standard complex X = X8) is completely determined by its values on the n-cells. E is For an f E V2 = Hom, ( X , ,A), its coboundary, S ~ V, a function on the 3-cells, which, in virtue of (1-4-7), is given by

m,

7 , PI

PI

=fqu,

7,

=

PI -fb, PI + f l u , .PI -fb,TI

d 7 ,

Vo, T , p E G

Thus, f is a cocycle if and only if it satisfies

Furthermore, the 2-cochain f is a coboundary if and only if it is of the form f = Sg for some g E %“-thus, it must satisfy

Functions of two variables satisfying the cocycle identity were known classically as factor sets, while those satisfying the coboundary identity were known as splitting factor sets. They arose in the study of both group extensions (which we will consider in Chapter V) and simple algebras (see, for example, [12], [4, Chapter 81). The group A was usually multiplicative, f [a,T ] was written as and the action of G on A was written exponentially; thus, in this notation, the function of two variables {u,,~}is a factor set (i.e. cocycle) when it satisfies

and {u,,~}is a splitting factor set (i.e. coboundary) when there is a function {bo} of one variable such that

1-52,

Proposition.

H’(G’ A )

{crossed homomorphisms} {principal crossed homomorphisms}

1-5. Proof:

G

A standard 1-cochain f is a cocycle 6f[u, T ]

-or

COHOMOLOGY GROUPS 07

= Uf[T] -f[UT]

IN

A

29

-e*

+ f [ u ] = 0 for all (I,T E G

what is then same, if and only if

-

Such functions f :G A are (for obvious reasons) called crossed homomorphisms. Furthermore, the 1-cochain f is a coboundary v there exists f E Wo such that f = Sg u f [ u ] = Sg[u] = ga[u] = g(a[.] -

[ a ] )

-

= u(g[.])- g [ . ] for all a E G.

Thus, if we set up the natural 1-1 correspondence 'iP A given by g t--f a = h [ . ] , then f is a coboundary if and only if there exists a E A such that f [ u ] = (u - 1) a

VUEG

Such functions f : G -+ A are called principal crossed homomorphisms. We shall write e for the character group of G (since G is finite, we have e = Horn (G, Q/Z) where Q/Z is additive) and Gc for the commutator subgroup of G. It then follows from above that : 1-5-3.

Corollary.

If G acts trivially on A, then W ( G ,A ) m Horn (G, A).

In particular, H1(G,2) = (0)


30

1-54 Theorem. (Noether's equations). If K/F is a finite Galois extension with Galois group G,then

-

H'(G, K * ) = (1)

Proof: Let f be a standard 1-cocycle of G in the multiplicative K* satisfies f(m) = (of(.))( f(u)) for all group of K ;so f : G u, T E G.We recall (see, for example, [3, p. 351) that distinct automorphisms al ,..., u, of a field K are independent-which means that if a i E K are such that aiui(/?) = 0 for all P E K , then ai = 0 for i = 1,..., n. Consequently, there exists E K* such that

xy

Y =

1f ( 4+)

f 0

7€G

Then, for any u E G

so that

(.r)f(4= 1f ( 4 ( u 4 9 = Y

-

7EG

This means that yU-l = ( q ) / y = 1/(f(u)) so that the function l/(f(u)) is a boundary; hence, f is a coboundary.

a

1-5-5. Corollary. (Hilbert's theorem 90). Suppose that K/F is a cyclic extension of degree n, and let u be a generator of G = B(K/F).If a E K has NK+Fa= 1, then there exists /? E K* such that a = /?l-".

Define a 1-cocycle f :G

Proof: f(1) 1-54.

=

1, f(u)

-

K* by putting

...,

= a,f(u2) = ~(uoL),f(un-') = a(ua)

Proposition.

a * *

(u"-*a)

I

Ho(G,A ) m AG/(SA).

Proof: The mapping f -f[.] onto A. Given f E P we have

is an isomorphism of

'30

1-5. If we putf[.]


G IN A

31

= u, then

f is a 0-cocycle t> (u - 1) u

-

=0

Vu E G

and our mapping induces an isomorphism of Toonto AG. g( .) is an isomorphism of Furthermore, the mapping g V-' onto A. Given g E V-' we have

%[*I Thus, for f E V0,

-

= g(S(.)) =

W.))

f is a 0-coboundary t> there existsg ~ 5 f - l with a =f[*]= S(g(.))

and the map onto SA. I

aESA

f-j[*]

also induces an isomorphism of 990

1-5-7. Corollary. If the group G of order n acts trivially on A, then Ho(G,A) w A/(nA).In particular,

If K/F is a Galois extension with Galois

1-54. Corollary. group G, then

HO(G, K * ) w

F* N K d *

Since F is the fixed field of G we have = F*, and Proof: because the module is multiplicative, the action of S is that of the norm from K to F. I 1-5-9.

Proposition.

H-l(G, A ) w AJIA).

Consider (as in (1-5-6)) the isomorphism f-f(-) Proof: of V-l onto A. Then forf %-l,

f is a -1 cocycle u S(f(*))= 0

32


and we get an induced isomorphism of 3-l onto A,. Furthermore, for g E we have (see (1-4-7))

%(*>

=

c .-'g(.>

TE

-

c

c

g(.> =

-

TEC

and therefore, for f E V-l, f is a -1 coboundary

c

(.-l

- l)(g(.))

TEG

there exists g E V-a with f(.) =

c (4

-

l)(g(T))

TEG

-f

(*)

EI

*

A

Thus, we have an induced isomorphism of 97-l onto IA. 1-5-10.

Corollary.

If G has order n, then

H-l(G, Z)= (0) 1-5-11.

Corollary.

I

H-l(G, Qlz) w

(5

2) /Z

If K/F is a Galois extension with Galois

group G, then

In particular, if K/F is cyclic, then H-'(G, K*) = (0) Proof: The first part is a translation of (1-5-9) to multiplicative notation. In the cyclic case, apply ( 1 - 5 4 , after observing that if u is a generator of G and A is any G-module, then (ut - 1)A C (a - l)A for all i 2 1. I 1-5-12. Remark. In proving the preceding results, we have made use of the coboundary formulas in low dimensions. For convenience and future reference, we state them for all dimensions; of course, they are immediate consequences of the boundary formulas given in (1-4-7).

1-5.


G

IN

A

33

Letf E Vn be an n-cochain of G in A with respect to the standard G-complex, then for all ul , u2 ,..., urnE G

The cases n = 0, 1, 2 may be viewed as special cases of the general formula :

The case n = -2 may be viewed as a special case of

n

< -2

Remark. The cohomology groups of G in A were introduced by Eilenberg and MacLane [22] (inspired in part by the work of Hochschild [32] on cohomology of associative algebras) according to the following procedure.

1-5-13.

I.

34

(n

COHOMOLOGY GROUPS OF G I N A

For each n 2 0 consider functions @ which map ordered 1)-tuples of elements of G into A-that is,

+

@ : Gx

* - a

x G-A.

_ I

n+l

Those which are homogeneous in the sense that

,

quao 001

,...,uu,)

, ,...,u,)

VU,

= d ( u o 01

u0

,...,on E G

are called n-dimensional homogeneous cochains; they form a group denoted by an = P ( G , A). For @ E an define 6@ E an+lby

,..*,an+,)

(S@)(uo

n+l

=

1 (-1)'

@(uo,...s

i=o

$i

,.*-,un+l)

-

where $$ indicates that this term is omitted. 6 (which is called the and it coboundary) is clearly a homomorphism of 6" (called the group satisfies 8 0 8 = 0. Denote the kernel of 6 by 3% of n-cocyctes) and the image of 6 by b"+l(called the group of (n + 1)-coboundaries). Put bo= (0). Now b" C 3", and the A) = 3"/Sn is called the nth cohomology group of group $"(G, G i n A. Consider further the group of nonhomogeneous n-cochains, n 2 1; these are the functions C$ which map ordered n-tuples of elements of G into A, with no restrictions on 4. Define a nonhomogeneous 0-cochain to be simply, an element of A; so A is the group of nonhomogeneous 0-cochains. For each n 2 0 this group may be identified with an because it is isomorphic to it under the correspondence @ ++C$ given by

,...,on) = uo+(u;lul, u;"uz ,..., +(~1 ,...,on) = @(l, , ola, UIU~

q u o , a,

U;:pn)

CT~

p..,

on)

Note that for n = 0, @ corresponds to the element C$ = @( 1) of A. When the coboundary 6 is transferred to the nonhomogeneous i - #& turns out that situation-that is, so that 6@t +--c

1-6.

PROBLEMS AND SUPPLEMENTS

35

For n = 0, this specializes to (W)(al) = a1+ - 4 where 4 E A. Then, under identification, 3" becomes the group of nonhomogeneous cocycles, dn becomes the group of nonhomogeneous coboundaries, and $"(G, A) = 3"/Sn. For n 1, these cohomology groups are clearly identical with those treated earlier in terms of the standard complex. In fact, this explains the origins of the positive part of the standard complex and its boundary operator. For n = 0, the cohomology group defined in terms of the nonhomogeneous cochains differs from the standard complex cohomology group-it is AG, since there are no 0-dimensional coboundaries. I t should be noted that the foregoing discussion (except for the comparison of H o and $O) applies when G is infinite.

>

1-6.


Let R denote a ring with 1 ; in particular, if G is a multiplicative group, finite or infinite, then R may be taken as the integer group ring Z[G] (of course, for infinite G, Z[G] consists of the elements ~ u : O E G ~n u, E ~ ,Z nu , = 0 for almost all a, which are added and multiplied in the obvious fashion) or the group algebra k[G] =

1 1a,a I a, E k UEG

of G over the field k. By any R-module, we mean a left unitary R-module. In this section, unless otherwise specified, we are concerned with R-modules and R-homomorphisms.

-

(i) Let f : A 1-6-1. coker f = B/im f, coim f have exact sequences

=

B be a homomorphism, and write A/ker f. Then coim f w im f and we

0-ker

f-A-coim

f-0

0-im

f-B-coker

f-0

0

- - ker f

A f,B

coker f -0

36

I.


(ii) Five Lemma.

G IN A

Consider the commutative diagram

with exact rows. (a) If v1 and v-, are monomorphisms, and v2 is an epimorphism, then vo is a monomorphism. (b) If v1 and v-, are epimorphisms, and rp-2 is a monomorphism, then v0is an epimorphism. (iii) Consider the commutative diagram with exact rows 0-A-B-C-0

(a) Iff and h are epimorphisms, then so is g. (b) Iff and h are monomorphisms, then so is g. (c) Iff and h are isomorphisms onto, then so is g. (iv) Four Lemma. with exact rows

Consider the commutative diagram

(a) If cp2 is an epimorphism and y-, is a monomorphism, then ker vo = fi(ker rpl) and im vl = gi'(im v0). In particular, if rp, is a monomorphism, then so is lo, and if yo is an epimorphism, then so is y l . (b) Use part (a) to prove the five lemma.

1-6.

37


(v) Consider the commutative diagram with exact rows

(a) Iffo and cpl are epimorphisms and cpo is a monomorphism, then cp-l is a monomorphism. (b) If g, and cp-, are monomorphisms and cpo is an epimorphism, then cpl is an epimorphism. (c) If g, is a monomorphism, then we have an exact sequence

- - - - - ker tpl

ker q~~

ker y o

(d) Iffo is an epimorphism then we have an exact sequence coker v1

coker v0

coker q ~ ,

(e) If g, is a monomorphism andf, is an epimorphism, then we have an exact sequence ker v1

1-6-2.

__t

ker vo

ker q~~

coker v1

coker rpo

Consider the commutative diagram of modules 0

0

0

0-

A'-

A -A"

0-

B'

B

B" -0

0

C'

c

C"

0

0

0

1 5 - 1

-

0

-1 1 1 - -- 1 - 1 1

0

coker rp-,

I.

38


G

IN

A

in which the columns are exact. If the first two rows are exact, or if the last two rows are exact, then so is the remaining row. If the first and third rows are exact, find a condition that will guarantee the exactness of the middle row. 1-63, (i) Consider a collection of modules Mu where a runs over some arbitrary indexing set U. The (external) direct product Mu is a module consisting of all mapsf :U Mu such that f(a) = mu E M u , and with the natural definitions of addition and scalar multiplication; it is often convenient to denote the generic element f of Mu by mu or {mu},so the module operations are then componentwise. There is a submodule of Mu consisting of those {mu}for which almost all mu = 0; it is called the (external) direct sum and denoted by C. @ M u or LI.. Mu For each / 3 ~ C l there l is a map i, : ME Mu which takes m, E M Eto the element with m, as the &component and 0 elsewhere, M , , the projection on the /3-coordinate. and a map nE: Mu It is clear that n, is = SUE where by SUE : ME Mu we mean the 0-map when a # /Iand the identity map when 01 = /3. It is 2 @ Mu and 7,: @ Mu M, , also clear that i, : ME and that i, nu is the identity map of C @ Mu .

nu

__t

n

nu

n

0

xu

-

0

-

-

uu

n

-n -

(ii) Suppose that we have modules M u , a E 02 and M. If for each 01 we have homomorphisms q ~ =: Mu M and t,bu : M M u , and such that $, 0 q~~ = S U E , then we say that {MuA M A Mu}is a direct family. Given such a direct family we have homomorphisms q~ : C @ Mu M and t,b : M Mugiven by

-n

The map q~ is a monomorphism; if q~ is an isomorphism onto, we say that {Mu M A Mu} is a complete representation of M as a direct sum. The direct family {Mu-% M & Mu} is a complete representation of M as a direct sum EuP ) 0~ if the identity; moreover, the a,hu’s can then be recaptured from the vu’s.

-

39

1-6. PROBLEMS AND SUPPLEMENTS

+

If is an isomorphism onto, we say that {Ma M 2Ma} is a complete representation of M as a direct product. In such a situation, the va’s can be recaptured from the +,’s. When the indexing set is finite, then direct sums and products are identical; also the direct family {Ma-% M -f% Ma} is a complete representation of M a s a direct sum it is a complete representation of M as a direct product.

-

a,

aE be a complete represent(iii) Let {Ma5 M -%Ma}, ation of M as a direct sum, and let {N, s,N -% N,}, /? E a,be a complete representation of N as a direct product. If Hom, (M, N) denotes the additive group of all R-homomorphisms of M into N , then

{HOmR ( M aN,) ,

HOmR ( M , N )

“a’”)*

HOmR (M,,, No))

with 01 E a,/? E is a complete representation of Hom, (M, N) as a direct product (of abelian groups). Symbolically, one may write

What is the situation for

and

-

1-64 (i) The monomorphism i : A B is said to be direct when iA is a direct summand of B-that is, when there exists a module C such that B = iA @ C. Another way to express this is to say that the exact sequence 0 A -L B splits; and this is true w there exists a homomorphism g : B A such that g o i is the identity map on A; furthermore, in this situation, B = im i @ kerg.

-

-

-

-

C is said to be direct when (ii) The epimorphism j : B k e r j is a direct summand of B. Another way to express this is to C 0 splits; and this is say that the exact sequence B --L

-


40

true v there exists a homomorphism f : C B such that f is the identity map of C; furthermore, in this situation, B = im f @ kerj.

j o

-- -

(iii) Consider the short exact sequence

-

B

0- A

C-0,

-

then 0 A +B splits v B +C 0 splits-and, in this situation, we say that the short exact sequence splits.

(iv) Consider the sequence 0 +Ml & M 4M , 0; it is exact and splits e+ there exist maps f2 : M2 +M and gl:MMl such that { M i & M&Mi}, i = 1,2, is a complete representation of M as a direct sum. 145.

diagram

(i) A module P is said to be projective if for any P

with the row exact there exists a homomorphism h : P +B such that g 0 h = f-that is, such that the triangle commutes. This condition may be replaced by the requirement that for any diagram P

in which f(P)C g ( B ) there exists a homomorphism h : P such that g 0 h = f. 1

(ii) If 0 +A +B is any module then

j __+

-

C is any exact sequence and

0 +Horn, (P,A ) -%Horn, (P,B)*

B

P

HornR(P, C)

is an exact sequence of abelian groups. Moreover, P is projective

1-6.

for any short exact sequence 0 the sequence t=r

0

-

Horn, (P, A )

-

41


*

A &B

Horn, (P, B )

is exact.

9 __+

C

-

Horn, (P, C )

__+

0

0

xa

@ Pa is projective v each direct (iii) A direct sum P = summand Pais projective. (iv) A module P is projective v P is a direct summand of a free module. (v) A module P is projective v every exact sequence 0 A B P 0 splits. A' A A" 0, (vi) Given exact sequences 0 0 -+ M'+ P'- A'- 0, and 0 M" P" A" 0 with P' and P" projective, then these may be imbedded in a commutative diagram

--- - -- - -__+

----).

0

0

0

-1 1 1 1 -1 1 1 -1

1

0 +M ' -

M

M"-O

0-

p'

P

P" -0

0-

A'

A

A" -0

0

0

1

0

1

1

in which the rows and columns are exact, P is projective, and the middle row splits.

---

1-66. (i) A free module is projective. Every module A is a quotient of a projective module-that is, A can be imbedded in an exact sequence 0 B P A 0 with P projective. __+

I.

42


G

IN

A

(ii) For n > 1 consider the ring Z, = Z/(nZ). For each divisor Y of n, put Y' = n / i , and consider the ideals YZ, and Y'Z, in Z, along with the exact sequence of Z, modules

This sequence splits w ( I , Y') = 1 v the Z,-module YZ, is projective. Give examples of projective modules which are not free. (iii) If R = Z (so that R-modules are the same as abelian groups), then a module is projective v it is free. (i) An additive abelian group D is said to be divisible 1-6-7. when for each x E D and every integer n # 0 there exists an other words, the map element Y E D such that ny = x-in n :D D which takes 2 nz is an epimorphism. Suppose that B is a subgroup of the abelian group A and that D is a divisible group, then any homomorphism fo : B +D can be extended D. to a homomorphism f :A (ii) If, in addition, n : D +D is not a monomorphism for any n > 1, then given any a E A with a 4 B and fo as above, it is possible to choose f so that f(a) # 0. In particular, this statement holds for D = Q/Z. (iii) Let BJ-= { f Hom ~ (A, D) If(B) = O}. Then for subgroups B, and B, of A we have, under the hypotheses of (ii), B, < B, -4 B,I < B: and Bf = B,I B, = B, (iv) A direct sum of divisible groups is divisible. A homomorphic image of a divisible group is divisible. The additive group of Q is divisible; thus the additive group of Z and any free abelian group can be imbedded in a divisible group. Finally, any abelian group can be imbedded in a divisible group. __+

-

-

-

1-64.

.

(i) A module Q is said to be injective if for any diagram

1-6.

43


with the row exact there exists a homomorphism h : B +Q such that the triangle commutes. This condition may be replaced by the requirement that for any diagram A A B

Q

--

in which ker g C kerf there exists a homomorphism h : B such that h g = f. 0

(ii) If A 2B is any module, then 0

-

C

-

Q

0 is any exact sequence and Q

a Horn, (B,8)*HOmR ( A ,Q)

HornR(c,Q)

----

is an exact sequence of abelian groups. Moreover, Q is injective i for any short exact sequence 0 A B j C 0 the sequence 0

-

a HOmR (B,Q)-@%

HOmR (c,Q)

-

HOmR (A, 8)

0

is exact. (iii) A direct product of modules is injective e+ each factor is injective. (iv) If R = Z, then our modules are simply abelian groups, and a module Q is injective v the abelian group Q is divisible. I n particular, (see 1-6-7) the 2-module Q/Z is injective. 1-6-9. for each left ideal I (i) An R-module Q is injective e=> of R (with I viewed as a left R-module) there exists, for each f E Hom, (I,Q) an element q E Q such that f(a) = aq for all a E I v for each left ideal I of R, every f E Hom, (I,Q) can be extended to an element of Hom,(R,Q) (where R is viewed as a left R-module). (ii) If X is a left R-module and Y is an abelian group (i.e. a 2-module; right or left is irrelevant), then Horn (X, Y) may be made into a right R-module by cfr)(x) = f ( r x )

f~ Horn (X, Y)

IE R

44

I. COHOMOLOGY GROUPS

OF

G IN A

If X is a right R-module, then Hom (X, Y) may be made into a left R-module by

f E Horn (X,Y )

( r f ) ( x ) =f ( m )

YE

R

(In this connection, see (3-7-8)). (iii) If the Z-module Y is injective, then the R-module Hom (R, Y) is injective (here R may be viewed as either a left or a right R-module). Every R-module can be imbedded in an injective module; in fact, the injective R-module may be taken of form Hom (R, Y) where Y is an injective 2-module. every exact sequence (iv) The module Q is injective 0 Q B C 0 splits. A' A A" +0, (v) Given exact sequences 0 0 -A' -Q' --+ N ----+. 0 and 0 + A" -Q" N" 0 with Q' and Q" injective, then these may be imbedded in a commutative diagram

----- 0-

iYY

A' - A

-A"

--

__+

-0

0-Q'-Q-Q"-O

0

0

0

in which the rows and columns are exact, Q is injective and the middle row splits. 14-10. Compute the contracting homotopies 0, in negative dimensions for the standard G-complex.

Suppose that G is a group, finite or infinite, and that A is a free G-module (with finite or infinite basis). Is A a free G-module? How does this affect the notion of a G-complex, and the definition of cohomology groups ? 1-6-11.

CHAPTER

I1

Mappings of Cohomology Groups

Having defined the cohomology groups of G in A in Chapter I we turn in this chapter to an investigation of how changes in the group or in the module (or both) induce changes in the cohomology groups. For this, the notion of homomorphism of pairs is extremely useful. I t leads, by specialization, to three important mappings of cohomology groups-namely, restriction, inflation, and conjugation. It also serves as a convenient tool for proving that cohomology groups are independent of the complex. Another mapping of cohomology groups is the corestriction, or transfer. Although it cannot be defined via homomorphism of pairs, it has many properties in common with the other mappings. With an eye to future applications, and for historical reasons too, we conclude this chapter with a discussion of explicit formulas for the various mappings of cohomology groups-all in terms of standard complexes. 2-1.

H O M O M O R P H I S M O F PAIRS

-

Let G and G' be finite groups, and let A and B be G-modules. Suppose that A : G' G is a homomorphism; then 45

2-1-1.

Proposition.

46

11.

MAPPINGS OF COHOMOLOGY GROUPS

(1) A may be made into a G'-module by putting u'EG', U E A

a'a = (ha') a

(2) Any G-homomorphism of A into B is also a G'-homomorphism; that is,

Hornc ( A , B ) C Hornc-

9a(G,A )

As >-

%"i:(G,A )

ZA

2?it(G, A )

>-

a E A, then ria E % ,! ( k a ) [ * ]= a

From

ri

and 4a E % ?.;

are given by

(4a)(.) = a

and 4 there now arise, in the natural way, epimorphisms K

: A'-

H;,(G, A )

71 : A s -++ H,:(G, A )

with kernels S A and I A , respectively, and isomorphisms

(Of course, strictly speaking, these maps should be denoted by

11. MAPPINGS

60

OF COHOMOLOGY GROUPS

K , , v S ,E , , 3, .) Note that for standard cocycles f E To and g E 9'-l their cohomology classes are K( f [*I) and q(g(*)), respectively. For an arbitrary full G-complex X we may define

KX

-

H$(G, A )

: AG

v x : As --H Hi'(G, A )

by putting KX

=

lX#J

O

Ks

rlx = lX'.x O v a

so that K~ is an epimorphism with kernel S A and qx is an epimorphism with kernel IA. It is clear that K.

= lXJ8

O

KX

rl. =

1X.X'

O

rlx

and that if X' is any other G-complex, then Kx'

= 1x.x. 0

KX

T]x' =

lx,,y*

0

r],y

In view of this, we have abstract homomorphisms (the notation for which should cause no serious confusion) K

: A'

-

Ho(G,A )

9 : As --H H-'(G, A )

which are independent of the complex and whose kernels are SA and IA, respectively. It then follows that there exist isomorphisms AG

r7 : - >-

SA

Ho(G,A )

A IA

fj : 2>-

H-'(G, A )

As a rule, we shall make statements only about K and 7 and leave it to the reader to carry them over to r7 and fj. Note that ri = ri, and 4 = 4, are not carried over for an arbitrary complex X.

-

Proposition. Let f : A B be a G-homomorphism of G-modules; then the following diagrams commute :

2-2-7.

AG

4

Ho(G,A )

BG

4

2Ho(G,B )

AS 71

H-'(G, A )

*Bs

4

H-'(G, B )

2-2.

61

INDEPENDENCE OF THE COMPLEX

Note first that f maps AG into B G , SA into SB,A, into B, , and IA into IB. We consider only the dimension 0 case (since

Proof:

the proof in dimension - 1 goes the same way) and must show that for any G-complexes X and X' the following diagram commutes : I BG AG

For this consider the diagram

-

-

Since the triangles commute, it follows that it suffices to prove that the square commutes. In order to compute (1, f ) x , , x , we may use A = 1 :X 8 X s as the map associated with 1 : G G. Now the desired commutativity is a consequence of the fact that for a E AC

((1,f)w.1= f ( 4= (W))[.l

I

O

2-28,

Proposition.

Suppose that

O-A-LB-C-O j is an exact G-sequence; then the following diagram commutes : A,&B,

4

4

AG I BG

+cs .

4G )2 4

71

H-l(G) 2H-l(B) I*cH-'(C)p*- H

*

k C G

4

Ho(B)%HO(C)

0(

Moreover, to compute 6, , we have for c E C, S,(vc)

= KU

where c = j b ,

iu

=

Sb

62

11.


Proof: Only the formula for 8, requires proof, and for this it suffices to use the standard complex and examine 8

C, & H;:(G, C )2Hi8(G,A )

AG

Let us in this proof (and, henceforth, whenever possible) omit explicit reference in the notation to the standard complex. Now, recall that 8, arises from the exact sequence

0

-

-

HOmc ( x , A) %Hom, ( x , B ) -@+ HOmc ( x , c)

0

In particular, 7jc E Hom, (X-l, C) C Horn, ( X , C) (since Hom, (X-, , C) = V-l(G, A)) and there exists u E Hom, (X-l ,B) such that (i,j)u = 7jc. Then taking w E Horn, (X, ,A) such that (1, i ) w = Su, we know that ~ * ( v c= ) t~

+ *(GI

A)

-

On the other hand, there exist a E A such that ria w and b E B such that 7jb = u. I n particular, S*(r]c) = ~ awhere , a E AG because w is a cocycle. Since the following diagrams commute : 9

B

A

*C

I*

*1

I

;I

+ B

1;

Homc(X-, , C) Homc(Xo, A) %Homc(Xo ,B )

HomG(X-l, B)

it follows that 7jc = (1,j)u = (l,j)(@) = 7j(jb), so that j b = c, and if2 = (i 0 ria)[*]= ((1, =

0

ria)[.] = (8u)[*]= (u 0

a)[*]

(qb)(S(*))= S(($)(*)) = Sb.

This completes the proof. 2-2-9.

i)

Remark.

morphism S, : C,/(lC)

-

I

This result permits us to define a homoAG/(SA)by going around the diagram

CS IC

4

A, SA p-1

H-'(G, C )3Ho(G,A)

2-2.

-+

63

INDEPENDENCE OF THE COMPLEX

+ IC

u SA where c E C, , 6, may also be defined directly, and is sometimes called the Tate linking (see, for example [75] and (3-7-7)). By applying these results to (#) we have the exact sequence

Thus, 6, = k - l o 6,o i j maps c c = j b , S b = iu. This mapping

... -+ H-Z(A) A H-Z(B)
that for a E A, wa = w , where w,( 1) = a. The restriction of w to AG is an isomorphism w : AG (Hom (z, A))G= Horn, (z, A), and for a E AG

-

K x ( 4 = wue

+ a;(G, A )

(d) Let ( A , f ) : (G, A) ( G , A') be a homomorphism of pairs with A a monomorphism; then the following diagram commutes AG

4

1

Ho(G,A ) 2-3.

(A,f)*

,A'G'

-1.

HO(G', A')

CONJUGATION, RESTRICTION, AND INFLATION

In this section we introduce certain mappings of cohomology groups that arise from homomorphisms of pairs. Henceforth, by a G-complex we shall mean, once again, a full G-complex and the notation Hn(G, A) shall be reserved for the cohomology groups computed with respect to a full complex. If a half-complex is to be considered, this hypothesis will be stated explicitly; in this situation we may still use the notation Hn(G, A) for n 2 1. Suppose that A is a G-module. We shall use exponential notation for the action of the inner automorphisms of the multiplicative group G. I n other words, for u, r E G we write 70

= mu--1

T ~ T ; and ( ~ 1 ) 0 1= TO*"'. Suppose that H is a subgroup of G, then every u E G determines a conjugate subgroup

so that (7p2)0=

H a = UHU-1

of H. Of course, A is an H-module and also an Ho-module. Suppose further that B is an H-submodule of A, then clearly uB={ubIb~B}

2-3.

CONJUGATION,

RESTRICTION, AND INFLATION

65

is an Hu-submodule of A. Now, conjugation by u-l provides an isomorphism of Hu onto H-u-l : H' >+ H-so that by using this map, B becomes, in the canonical way, an H'-module for which the action of Huis given by

-

(pa) b

Moreover, the map u : B

u(p"b)= a(@)

= (pup-' b = pb

PEH, beB

oB is an Hu-homomorphism since

= upu-lob = p"(0b)

-

PEH, beB

This means that we have a homomorphism of pairs (u-1,

u) : ( H , B )

(#I

(Hu,uB)

and because u-l is a monomorphism there arise (by (2-2-3)) mappings of cohomology groups (u-l, u)* : Hn(H,B )

-

Hn(Hu,uB)

neZ

We shall write u* = (u-l, u)* and call it conjugation by u. We have therefore : Proposition. Let A be a G-module, let H be a subgroup of G, and let B be an H-submodule of A ; then every u E G determines an isomorphism

2-3-1.

(I* = (u-', u)* : H"(H, B )

-

Hn(Hu,uB)

tl

E Z (##)

such that (m)* = u* 0 T* (that is, conjugation is transitive), 1* is the identity, and (u-')* is the isomorphism inverse to u* . Moreover, if p E H (so that HP= H and pB = B), then p* is the identity map of Hn(H, B). The first part is immediate from (2-1-8). As for the second part, let X be any G-complex. Then it is clear that X is an H-complex, and an Hu complex, too-so X may be used for the cohomology of both H and Hu, and we can be fairly explicit about the map

Proof:

66

11. MAPPINGS OF

COHOMOLOGY GROUPS

-

More precisely, in order to compute (u-1, u ) ~ we , ~need an Hahomomorphism A : X +X associated with A = u-l : Ha H (where the first X is viewed as an Hu-complex, and the second X is viewed as an H-complex which becomes an Ha-module via the map A). We assert that A = u-l will do. I n fact, &A, = E since for x E X, ,eA,x = eu-lx = u-lex = ex; also A o a = a o A because 8 is a G-homomorphism; and finally, A is an H'-homomorphism since for x E X, p E H we have A(pax) = u - ~ u ~ u - ~=xp(u-lx) and pu(Ax) = pa(u-'x) = p(u-'x) by the way in which the second X becomes an Hu-module. Thus, the homomorphism of pairs (#) leads to the cochain map (u-l,

u ) : Horn,

(X,B )

-

Horn,, (X, uB)

-

which in turn determines the mapping of cohomology groups (u-'u)X,X

:Hi(H,B)

H;(H: UB)

This means that if u E Hom, (X, B) is a cocycle, then its u ) X Xto the cohomology class cohomology class is mapped by (4, represented by the cocycle (u-l, u)u = uuu-1= u' E Hornp (X, uB). Now, if u E H, then ua = u because (as was noted in the proof of (2-1-3)) Hom,(X, B) = (Horn (X, B))S It follows that the cochain map (u-l, u) is the identity, and then so is (u-l, u ) ~ . This means that u* = 1, and the proof is complete. I 2-3-2.

Corollary. u*

-

For every u E G the map : H"(G, A )

Hn(G,A )

neZ

is the identity.

If H is a normal subgroup of G and B is a G-submodule of A,then the factor group G/H acts on Hn(H, B)that is, H"(H, B) is a G / H module.

2-3-3.

Corollary.

Another mapping of cohomology groups may be defined as follows. Let A be a G-module, and let H be any subgroup of G. If,

~

2-3.

CONJUGATION,

RESTRICTION, AND INFLATION

67

as usual, we let 1 denote the identity map of A and i = iH4 denote the inclusion map from H into G then clearly

is a homomorphism of pairs. Since i is a monomorphism there arise mappings of cohomology groups

(i, l), : H"(G,A ) We shall write res from G to H.

=

res,+,

=

-

H"(H, A )

neZ

(i, l), and call it the restriction

Remark. T o deal with restriction concretely one takes any G-complex X and any H-complex X' and examines the map

2-34

As an illustration, let X be any G-complex. Then, as observed in the proof of (2-3-1), X is also an H-complex, and it is not hard to describe res,,, . In fact, in order to compute res,,, we need an H-homomorphism A : X X associated with X = a' : H G (here the first X is an H-complex and the second X is a G-complex which becomes an H-module via the map A). I t is clear that A = 1 will do. Thus, the homomorphism of pairs (i, 1) leads to the cochain map

-

( 1 , l ) : Horn, ( X , A )

-

-

HornH( X , A )

(note that this map is simply inclusion) and this determines the mapping of cohomology groups res,,, : H;(G, A )

-

HF(H, A )

nEZ

This means that if u E Hom, (X, A) is a cocycle, then its cohomology class is mapped by res,,, to the cohomology class represented by the cocycle u E Hom, (X, A).

68

11.


Let H be a subgroup of G and let A be Proposition. a G-module, then there exists a map

2-3-5.

resG+H: Hn(G, A)

-

H"(H, A )

nEZ

other words, if H' is a sub-

Moreover, restriction is transitive-in group of H , then

resG+H'= resH-,H, 0 resG+H

Let X, X and X be G, .H and H'-complexes, respecProof: tively. Then, from (2-1-8) we have

which proves transitivity.

I

Still another mapping of cohomology groups may be defined in the following manner. Let A be a G-module and suppose that H is a normal subgroup of G. If a E AH and u E G, then ua E AHbecause, given p E H there exists p' E H such that pa = up', so that p(ua) = up'a = ua. Furthermore, (up)a = ua, so that each coset uH has a well-defined action on AH.In other words, AHmay be viewed as a G/H-module. Let 7r : G G/H be the canonical A be the inclusion map (we also write ~ ( u )= 6)and let i : AH map. When AH is made into a G-module via the map T , then i is a G-homomorphism, and consequently

--

-

(r,i) : ( G / H , A H )

(G, A )

is a homomorphism of pairs. Since i is not 1-1, there arise mappings of cohomology groups

-

(r,i)* : H"(G/H, A H )

H"(G, A )

n>l

We shall write inf = inf(t/H)+ = (r,i)* and call it the inflation map. The inflation map is not considered for n = 0 because of our convention about the meaning of Hn(G, A). In virtue of the transitivity principle of (2-1 -8) we have therefore :

2-3.

69

CONJUGATION, RESTRICTION, AND INFLATION

Proposition. Let H be a normal subgroup of G and let A be a G-module; then there exists a map

2-36.

Moreover, inflation is transitive; more precisely, if H , C Hl C G and H I , H, are normal subgroups of G then the following diagram commutes : H"(G/H, ,AH')2H " ( G / H 2 ,~

~

2

)

We may also note that the homomorphisms of pairs (u-l, o),(i, I), and ( x , i) commute with each other, so that application of the transitivity principle shows that conjugation, restriction, and inflation commute with each other. In detail, we have : 2-3-7. Proposition. Consider the groups H , C H I C G and the G-module A, and let B be an H,-submodule of A (so that B is also an H,-submodule of A); then for u E G the following diagram commutes :

H"(Hl , B ) 2%H"(H2 ,B) O.1

O*

1

nEZ

H"(H,", UB)z H"(Hz", UB)

If, in addition, H , is normal in Hl , then the following diagram commutes :

11. MAPPINGS

70


If, in addition, H , is normal in G, then the following diagram commutes :

H"(G/H2,AH')-!% H"(G, A ) re8

1

re81

n>l

H"(Hl ,A )

Hn(Hl/H2, AH2)

Remark. It is clear that (2-1-9) and (2-2-5) carry over to u* ,res, and inf; it is only necessary to restate the hypotheses carefully. We shall express this by saying that conjugation, restriction, and inflation commute with the coboundaries arising from short exact sequences.

2-3-8.

2-4.

THE TRANSFER O R CORESTRICTION

I n this section we define still another mapping of cohomology groups and discuss some of its elementary properties. Suppose that H is a subgroup of G, and let G = UE'=,u,Hbe a left coset decomposition-so (G : H) = m. For any G-module A, we define a function m

=

S,,,(a)

1uia

acAH

1

This function is called the trace from H to G, since it generalizes . Of course, S,,, may be viewed as our previous trace S = S{l)+c multiplication by the element cya, of Z[G]. A few of the basic properties of the trace are given in : Proposition. Let H be a subgroup of G, and let A, B, C be G-modules, then :

241.

(1) S,, is a well-defined (that is, it is independent of the choice of representatives for the cosets) homomorphism of

AH +A,.

(2) The trace is transitive; that is, if H ' C H C G, then sH+G

S&+H

=

sH'+G

m

2-4.

THE TRANSFER OR CORESTRICTION

71

-

= SI1)+, : A -+ AG is a G-homomorphism. (4) SH+,is a homomorphism of HOm,(A, B) Horn, (A, B). ( 5 ) Iff E Hom, (A, B), then on AHwe have S, of =f o S,, . (6) I f f E Hom, (A, B), then S,, f = (G : H)f. (7) If f E Hom,(A, B) and g E Horn, (B, C), then

(3) S

SH+G(g

(8) Iff

E

O f )

=g

SH+G(f)*

Hom, (A, B) and g E HOm, (B, C), then SH+Ck

o f )

= (SH+GR)

of.

Entirely straightforward; for example, (4) is immediate from (1) and the fact that Hom (A, B) is a G-module with

Proof:

Horn, (A, B) = (Horn (A,B)), and HOm,(A, B ) = (Horn (A,B)),.

I

Now, let X be any (full) G-complex. Then X is also an Hcomplex and we have a homomorphism SH+G: Horn,

(x, A ) +HOmc (x, A)

For u E Hom, ( X , , A) we note that 8

SH+G# = ( S H + & )

a

SH-+G(ua) =

SH+G(8u),

so that 6 o SH+, = SH+,o 6. This means that S,, is an admissible map of differential graded groups, and determines, therefore, a homomorphism, called the transfer or corestriction, of cohomology groups (corH+&.x

= corx,x : fG(H,

A)

-

For an arbitrary H-complex X we define

as the composite map

H , W , A)

nEZ

12

11.

-


In order to define an abstract map cor,, we make use of the following result :

: Hn(H,A)

Hn(G,A)

Lemma. Let ( h , f ) : (G, A ) +( G , A') be a homomorphism of pairs, and consider subgroups H C G and H' C G'. Suppose that X(H') C H (so that ( h , f ) : (H, A ) +(H, A') is a homomorphism of pairs), that X(G)H = G, and that ( G :H )= (G : H) = m. Then for any full G, G', H, H' complexes X, X', X,X',respectively, the following diagram commutes for all n E Z when A is a monomorphism, and for n 2 1 otherwise :

242.

Using X and X' as H and Proof: let us consider the diagram H&H,

4

(A,f)x,xi t H&(H', A')

p:,.,

15.1 H W , A)

(I,f)X.X'

t

.OW,XJ.

H W , A)

H complexes, respectively,

Hi,(H', A')

1

CO'X,X'

0,f)x.x'

+ Hi#(G',A')

-

The top rectangle commutes by (2-1-8), so it remains to show that X be the bottom rectangle commutes. For this, let A : X' any G-homomorphism suitable for h in the computation of the lower (A, f ) x x * . This same A may then be used as the H'-homoX in the computation of the upper (A, f morphism of X Thus, it suffices to show that we have commutativity in the cochain diagram t HornH' (X', A') Horn, ( X , A )

SH-0

)xxn

1

HornG( X , A )

1S.L.

F

HornG' (XI, A')

.

2-4. Let G'

=


uy a;H'

73

be a left coset decomposition; so h(G') =

u X(U;) h(H'), m

1

and if we put A(ui) = q , then G = X(G')H = uy qH. Since (G : H) = m, this is a left coset decomposition. Now, for u E Hom, (X, A) we have (S,.,c*

0

( l l , f ) u)

=


m

m

i=l

i=l

1 (full)"; = 1 a;fu/la:-l

I

Proposition. Let H be a subgroup of G, then for any G-module A there exists a homomorphism

243.

which is independent of the choice of complexes. Proof: 2-44.

Immediate from (2-4-2). Proposition.

if H' C H C G, then

I

The transfer map is transitive; that is,

We must show that if X, X,X are complexes for G, H, H', respectively, then

Proof:

74

11. MAPPINGS


= (corH&.X

and the proof is complete.

I

In virtue of (2-4-2), it is easy to see that, under suitable hypotheses, the transfer map commutes with conjugation, restriction, and inflation; more precisely, we have : 245. Proposition. Consider the groups H2C Hl C G and the G-module A. If B is an HI-submodule of A and t~ E G, then the following diagram commutes :

H"(Hz, B ) 5 H"(Hl ,B ) a*

1

a*

1

H"(H,",uB)S H"(H,", uB)

If, in addition, H2 is normal in G, then the following diagram commutes : H"(Hl/H2 ,AHa)aHn(G/Hz,AH') inti

H"(H,, A ) 246.

Proposition.

-

11.1

n>,l

Hn(G, A )

Let Hl and H2be subgroups of G such

2-4.

75


that H I H , is a group; then for any G-module A, the following diagram commutes :

1

H n P 2

res

,A )

cor

~"(HlH,9 4

1

n E Z.

res

H ~ ( Hn , H ~A ,)

H ~ ( H ,A, )

-

Consider the homomorphism of pairs

Proof:

0 9 1 ) : (HlH, A ) 9

Then i(Hl n H,) C H , ,

(H, A )

i(H,) H ,

9

=

H,H,

,

and

( H I H z : H2) = (Hl : Hl n H2), so that (2-4-2) applies.

I

247. Proposition. T h e transfer commutes with the coboundary arising from short exact sequences; that is, if

is an exact G-sequence and H C G then the following diagram commutes :

*..

...

H "(H , A ) i t - H n(H , B )

-

__f

co.1

Hn(G, A )

cor1

Hn(H, c)-%Hn+'(H, A)-**.

..*J.

.1

Hn(G, B ) -% Hn(G,C ) A Hn+'(G,A)-..* 8

Proof: Let X be any G-complex. From the commutative diagram with exact rows 0 --+ Horn, ( X , A ) 0

-

sH+C1

Horn, ( X , A )

a Horn, sH&l

( X ,B )

a Horn, ( X , B )

uHorn, ( X , C ) SH&1

HOm,

-

(x,c)

0

0

76

11.


we derive the desired commutativity when the cohomology groups of both rows are computed with respect to X. If X' is any Hcomplex, then the desired commutativity (when the cohomology groups of H are with respect to X and those of G are with respect to X)is gotten by use of If.,x. I Proposition. Consider H C G and the G-modules A and B. Iff E Hom, (A, B), then the following diagram commutes :

2-48.

f)*

Hn(G,A ) A Hn(G,B ) CSH

re*

1

Hn(H, A )

* f

tort

nEZ

Hn(H, B )

Proof: According to (2-4-1), SH+,f E Hom, (A,B). Let X be any G-complex; it suffices to prove our result when all cohomology groups are computed with respect to X. From (2-3-4), we know X may be used for the computation of that A = 1 : X (res,+H)x+. Then the proof is complete as soon as one checks the commutativity of the cochain diagram Horn, ( X , A )

(1 *SH,Gf)

Horn, ( X , B )

(I,1)/,

HomH 249.

Corollary.

any ar E Hn( G, A)

pH+G

(x, A)

(1.f)

I

If H C G and A is a G-module, then for

corH+, res,+,, a

Proof:

Horn, ( X , B )

=

(G : H)a

In (2-4-8) take A = B and f = 1.

Exercise. Let A be a G-module, and let H be a subgroup of G; then the following diagram commutes :

2410.

AH

4

sH.rc

+

AC

4

Ho(H, A ) 5Ho(G,A )

2-5. 2-5.

EXPLICIT FORMULAS

77

EXPLICIT FORMULAS

The mappings of cohomology groups defined in the preceding sections have concrete realization when standard complexes are used, and this section is devoted to the derivation of explicit formulas for these maps. First of all, let us consider a framework within which inflation, restriction, and conjugation are included. Thus, suppose that ( h , f ) : (G, A) (G', A') is a homomorphism of pairs. Let X and X denote the standard complexes for G and G', respectively. Then knowledge of an explicit formula for a G'-homomorphism (1 : X' X suitable for h determines an explicit formula for the map of cochains

-

-

(A,f): HornG( X , A )

-

Horn,, ( X ,A')

which, in turn, describes

Putting I' = Z[G] and r' = Z[G'] we recall (see Section 2-1) that (1 must make the following diagram commutative :

\z/

"0

...

- -x, *l

4

=

T o define A it suffices to do so on the G'-cells and then extend G'-linearly. If we put

A,([.;

[,I

Ao([&l)

=

,...,.A])

= [hu;

,..., AUA]

nZ1

78

11.


then e o A, = e1 and, as in easily checked, A, o a&, = a,+, o /I,+, on the GI-cells; thus we have the positive part of A. For arbitrary A, the negative dimensional A's do not exist, in general. However, if h is a monomorphism, then (as in (2-1-6)) the negative dimensional A's can be found; the details may be left to the reader, because we shall later find such A's, in the cases of interest to us, by another method. Now, consider the inflation map inf

=

-

inf(G,H)+G: Hn(G/H,AH)

Hn(G,A )

-

It arises from the homomorphism of pairs

-

i) : (G/H,A H )

(7,

(G, A )

where ~ ( u = ) 6. Let XG and XGIHbe the standard complexes; if XGIHsuitable for h = T constructed according to the A : XG the foregoing is denoted by Ainf, then

4TGl) = n;'([U,

[CjHI

,...,an]) = [6,,...,On]

n>l

We shall be somewhat careless and write inf = (Pf, i)-so that this cochain map inf induces the cohomological map (inf(GIH)+G)XGIH,KG , This means that the inflation map is given in terms of the standard inf u where u is a standard complexes by the cochain map u cochain of GIH in AHand

-

(inf u)([ul

,...,an] = u([6, ,...,Cn])

-

n>l

I n the same way, we may consider the restriction map res = reG+,, : P ( G , A )

-

Hn(H, A )

It arises from the homomorphism of pairs (i, 1) : (G, A) --t (H, A), and if we let Ares : XH XG be the map of standard complexes suitable for h = i described above, then

2-5.

79

EXPLICIT FORMULAS

We write res = (Are*, 1) so that this cochain map res induces (resC+H)Xc,XH . Since exists (see (2-1-6)), although we have not found it explicitly, we see that the restriction map in dimensions 2 0 is given in terms of the standard complexes by the cochain map u --+ res u where u is a standard cochain of G in A and

(res u)([P~*..*,

pnl)

=

-

u([P~*...,~

n l )

tl

Z 1,

~1

,*.*i pn

EH

Finally, in order to describe conjugation by a, we consider the (W, aB), let homomorphism of pairs (a-l, a) : ( H , B ) A,, : X H " X H denote the map of standard complexes suitable for h = a-l described earlier, write con,, = (Au,0 ) for the cochain map, and leave the details to the reader. T o summarize these results, we have : For positive dimensions, inflation, restriction, and conjugation are determined by standard complex cochain maps of form : 2-5-1.

Proposition.

(inf u)([ul ,..., on])= u([G1 ,...,4)

tl

2 1, a1 ,...,on E G

(res u)([,l) = u([;;l)

(res u)([pl **-.* pnl)

=

u ( [ Pv~* * , pnl)

Z 1,

~1 ,..s

2 1,

p1 v.., pn E

pn

EH

(conC8 U > ( [ + l ) = Mil)

(con,,u)([pl ,.*.,pn])

= a([u-'p1Ui**-i

u-lpnu])

Ha

The transfer map does not come from a homomorphism of pairs, so that determination of a standard complex cochain map which induces it involves a return to first principles. We recall that if X H and XG are the standard complexes, then the cohomological map we seek to make explicit is (CorH+G)XH,XG

= (CorH-rG)XC,XG

H 'XH,XC

Consequently, the desired standard complex cochain map, which we denote by cor, is the composite Horn, ( X H ,A )

HOmH (XG,A )

Horn, (Xc, A )

11.

80

-


where Acor : Xc XH is an H-homomorphism suitable for A = 1 in the computation of (1, l ) X ~ , YIn~ .other words, for a standard n-cochain u of H in A, we have cor u = sH-,,(u 0 A?)

n E Z (#)

Now, let us turn to the definition of Akor : X z for n 2 0. Let C = {c} denote the set of right cosets of H in G, and for each coset c choose, once and for all, a representative Z E c. There is no harm in requiring that for c = H, E = 1. We may write __+

G= UC=UHZ osc

C

and every element u of G has a unique expression u = pE with p E H. If we write rc = Z[G] and rH = Z[H], then X,G = M G l ) =

c 0 Gf(Z[;;I)

C€C

so that the set {E[$ I c E C} constitutes an H-basis for Xf.In the same way, {E[ul ,...,a,] I c E C, ul ,...,an E G} is an H-basis for Xg ,n 2 1. Because Agar must satisfy 8 0 A p = 8, and because it suffices to define A y on an H-basis, it follows from

eH{Ar(Z[;;])} = ec(t[;;]) = Eec([;;]) = E * 1 = 1 = e H ( [ J )

that &Or

should be defined by

~,cor(~[Gl)= [I;] From the requirement that i3f A?, = Apro i3: and the fact that for c E C, u 5 G the elements and Tu belong to the same coset cu, (so ZuZ- E H ) we have 0

cor c -

~ i Y ~ q " ' ( ~ [ ~= l I A0 )}

a1 (C[%l)

cor-

G

= A0 cal([Ull)

= A,co'{EU1[G] - E[G]} cor - --I

= A,

-

{CuIq (C4;;l)

- Z[;;l}

--1

= ZU1CU1 [h] - [i]

(since A, is an H-homomorphism) =

a;[r.&j

2-5. so that

EXPLICIT FORMULAS

81

AF should be defined by /ly(E[ol]) = [Eu17G;l]

In the next step, we must have

--1

--1

= ~ulG;1~lu2culu2 ] - [Eu1u2cu1u2 ] =

H - --l-

+ [Eo17G;'1

--1

a, [culcul ,cu1u2cu1u,3

so that A;'' should be defined by

It may now be left to the reader to show inductively that Akor, for

n 2 1, may be taken as AF(E[ul ,...,u,]) --1

-

--1

,...,cu1u2

= [EUlCU1 ,cu1u2cu1u2

--I

***

u"-lu,cul

*..

u,

3

From (#) and the fact that G = U P H is a left coset we conclude that : decomposition, so that SH+G=

zEct?-l,

Proposition. For positive dimensions, the transfer (or corestriction) is determined by a standard complex cochain map of form

2-5-2.

82

11.


It remains to consider negative dimensions. Inflation does not exist for negative dimensions, so it does not enter into the discussion. As for restriction and corestriction we have already defined, for n 2 0, H-homomorphisms

which are associated with the commutative diagrams

2

and

Using the notation of Section 1-4 we have the dual mappings A

A

A p : XE

-

A

Xff

n

A?

and

A

-

: Xff

A

Xf.

Now putting A

= /Icor -(n+l)

A?

n

=

Ares -(n+l)

n, 2, p1 ,..., pndl E H

As for it is sufficient to give its action on an H-basis of X Zn. We assert first that if c = H

otherwise In fact, for p

EH

and the assertion follows immediately. In the same way, one may show that for n >, 2, u1 ,..., un-l E G

From all this we conclude

2-5.

85

EXPLICIT FORMULAS

24-3. Proposition. For negative dimensions, restriction and corestriction are determined by standard complex cochain maps of form :

(COT U)((Ol

,...,on-1))

=

I

u((al

,..., on-,)) if all uiE H

0 otherwise

I

n22

all ui E G

2-5-4. Proposition. For negative dimensions, conjugation is determined by a standard complex cochain map of form

-

Proof: In connection with the homomorphism of pairs (u-1, U) : ( H , B) (H',uB) we have for n 2 0 the maps A; : X:u +X z used for the computation of con,. They are given by the formulas (see (2-5-1))

-

-

In the same way, for (u,u-l) : (Hu,uB) (H, B) we have A;-' : X f X f " for n 0. According to the dualization procedure used earlier we may define

n (1,n = A u-(n+l) -1

.x y + q *

n t O

11.

86


It follows easily that &,((,.">)

=

so that, in negative dimensions, con, = (A",u) is as described.

I

In virtue of these explicit formulas our mappings of cohomology groups can, for dimensions 0 and -1, be described in terms of mappings of modules-more precisely, 2-55.

The following diagrams commute :

Proposition.

AG

4

HO(G, A )

i

CE *AH

~~BG-.H

1."

HO(H, A )

A,

4

H-'(G, A )

'

reBG+H

bn

H-'(H, A )

CHAPTER

111

Some Properties of Cohomology Groups

This chapter is devoted to several special and unrelated topics about cohomology groups. The only thread connecting them is their applicability in the development of class field theory. As usual, G will refer to a finite group; however, we shall be careless about the notation P ( G , A). It will be clear from the context whether we are dealing with concrete cohomology groups computed with respect to some G-complex or with abstract ones. 3-1.

MISCELLANEOUS FACTS

This section deals with some simple properties of cohomology groups-for example, special conditions under which the cohomology groups vanish, and also the relations between the cohomology of a group and the cohomology of its Sylow subgroups.

-

Lemma. Let f : A B be a G-homomorphism of G-modules. If f is a trace (meaning that f = Sg for some g E Hom (A, B)), thenf, = 0.

3-1-1.

Proof:

Let X be a G-complex, and let D be a contracting 87

88

111. SOME

PROPERTIES OF COHOMOLOGY GROUPS

+

homotopy for it-so Da aD = 1. For any n, let u be an n-cocycle of G in A with respect to X ; in other words, u E Hom, (X, ,A) with Su = 0. Therefore, u = u 1 = uDa uaD = uDa. Now, making use of (2-4-1) we have

-

(1, f) u

= fu =

+

S(g) u = S(gu)= S(guDa) = S(guD) a

= G[S(guD)]

*

so that the image of an n-cocycle is a coboundary, and f = 0.

I

If the identity map of A, 1 :A -+ A is the trace of an endomorphism of A, then P ( G , A) = (0) for all n. 3-1-2.

Corollary.

From above 1* = 0; but 1* : H"(G, A) Proof: is the identity. I

-

W(G, A)

3-1-3. Corollary. If the module A is G-regular, then W(G, A) = (0) for all n.

Proof:

By (2-1-4) the identity map of A is a trace.

I

Corollary. Suppose that K/F is a finite Galois extension with Galois group G, then

3-14

neZ

W G , K+) = (0)

According to the normal basis theorem-see Proof: or [41, pp. 57,611-there exists a E K such that

I

in other words, K+ is G-regular. 3-1-5.

Proof:

Corollary.

If G

=

[3, p. 661

{l},then P ( G , A)

The identity is a trace.

I

=

(0) for all n.

3- 1.

89

MISCELLANEOUS FACTS

Proposition. If G has order m, then for any n every element of Hn(G, A) has m as a period-that is

3-16.

--

-

m . Hn(G,A ) = (0)

neZ

Proof: Let m : A A be the map which takes u mu. Since m is the trace of the identity, we have m , = 0. On the other hand, m , : Hn(G, A ) Hn(G, A) is the map which takes a+ ma. I 3-1-7. Corollary. Suppose that G has order m and that I is an integer for which (m, I ) = 1 and r A = (0). Then Hn(H, A) = (0) for all n and all subgroups H of G.

Proof: xm : A

-

Let x , y E Z be such that xm A is the identity map, and xm

= (xm), : Hn(G,A )

-

+ y = 1.

Then

H*(G, A )

is the identity. Because mHn(G, A) = (0), we have Hn(G, A) = (0). The same argument applies to any H C G because its order is prime to r. [ Corollary. Let V be a vector space over the field F of characteristic p # 0. If V is a G-module and ( p , m) = 1, where m = #(G), then Hn(H, A) = (0) for all n and all subgroups H of G.

3-18.

Proof:

Apply (3-1-7).

[

3-1-9. Proposition. If A is a finitely generated G-module, then Hfl(G, A) is finite for every n.

Proof: The result holds if A is finite, because then, for any G-complex X,there are only a finite number of cochains of G in A. In the general case, consider the exact G-sequence O - - - + A * A + - + On

A mA

m

=

#(G)

90

111. SOME


where T is the canonical map. For each n, it determines an exact sequence m

--))I

H"(G, A ) *-_ H"(G, A )

H" (G,

A

a)

Now, T* is a monomorphism because m , = 0, and Hn(G, A/(mA)) is finite because A/(mA) is finite. This completes the proof. I 34-10.

Corollary.

Hn(G, Z)is a finite group for every n.

The G-module A is said to be uniquely divisible by t when A is a G-isomorphism onto. The inverse map then map Y : A is denoted by l / r , and we have ( 1 / ~ o) ( t ) = ( Y ) ( I / t ) = 1, , the identity map of A. __+

0

34-11. Proposition. If A is uniquely divisible by the order m of G, then P ( G , A) = (0) for all n.

Proof:

We have m , = 0 and also (m)*(l/m)*= (lA)*= 1.

I

Suppose that G acts trivially on a field F of characteristic 0 (viewed additively), then 3-1-12.

Proposition.

Proof:

Consider the exact additive G-sequence

where the action of G is trivial. Since F is uniquely divisible by m = #(G), the desired isomorphism follows from the exactness of

3- I. MISCELLANEOUS FACTS

91

For all n, we have

3-1-13.

Corollary.

3-1-14.

Proposition.

If G is a group of automorphisms of a

finite field K, then

Ho(G,K*) = (1) Proof:

Let F = KG (the fixed field of G),

q = #(F),

m = #(G)

=

-

[K :4,

and

qm = #(K).

Of course, G is cyclic with generator u : a &, a t z K , and according to (1-5-8), Ho(G, K*) M F*/(NK+FK*). Now,

NK+F K*

__+

F*

is a homomorphism given by N K-bF).(

= OLl+Q+"*+Qm-l = a(Qm-l)/(Q-l)

--H

H*-'(G, A )

nEZ

The location of the augmentation in X is obviously not significant. It is easy to see that A, has the "proper" cohomological behavior. In particular, it commutes with induced homomorphisms and with coboundaries arising from exact sequences. Therefore, it follows Ai , BAC 0 is an exact G-sequence, that if 0 then we have an exact hexagon

-

A)

H'G

4

and can treat the Herbrand quotient by the method used in Section 1-2. However, we prefer to do so by a slightly different procedure.

96

111. SOME


For any finite group G and any G-module A we denote the order of the nth cohomology group (in a manner consistent with the notation introduced in (1-2-4)) by M A ) = Q(G, 4 = #(H"(G, 4)

If G is cyclic, we define

provided both terms on the right side are finite, and call it the Herbrand quotient of the G-module A. Now, for any u E G, A is an abelian group with endomorphisms u - 1 and S such that (u - 1) o S = S 0 ( 0 - 1) = 0. In particular, if u is a generator of the cyclic group G, then

(*I

44 = Q~-I.s(A)

where Q+JA) is the Herbrand quotient of A with respect to the endomorphisms o - 1 and S as defined in Section 1-2. Consequently, (1-2-2) and (1-2-3) can be carried over to h, and we have : 3-23. Proposition. Suppose that G is a cyclic group and that 0 A +B C4 is an exact G-sequence such that two of the three quotients h(A), h(B), h(C) are defined, then the third one is defined and __+

__+

h(B) = h(A)h(C)

Moreover, if the G-module A is finite then h(A) = 1

-

3-24. Corollary. If G is cyclic and f :A +B is a G-homomorphism of G-modules such that the kernel f -1(O) and h(B) the cokernel B / ( f ( A ) )are both finite, then h(A)is defined is defined, and in this situation h(A) = h(B)

3-2.

CYCLIC GROUPS AND THE HERBRAND QUOTIENT

- - - - --

97

Proof: Apply the preceding result to the exact G-sequences 0 kerf A A/kerf0,O A/ker f --* imf0, and 0imf +B B/imf 0. I

Any G-module A may also be viewed as a G-module for which the action of G is trivial. If G = (1, u,..., urn-'} is cyclic, then there is a Herbrand quotient defined in this case too; we denote it by ho(A) = ho(G, A). Because the action of G is trivial, u - 1 is the 0-map on A and S is multiplication by m ; so according to (*) we have W A ) = Qo,m(A)

In other words,

-

provided both terms on the right are finite, and where A,,, is the mu. It is surprising, perhaps, that for kernel of the map u certain G there is a connection between h(A) and ho(A). This connection is given by our next result, which is an extension of a result that appears in [17, Theorem 10.31, and whose primary application is to give an efficient proof of the first inequality of global class field theory. 3-2-5.

(Tate)

Proposition.

G

Suppose that

= (1, u,...,

is a cyclic group of order p, p prime, and that A is a G-module. If hO(A) is defined, then so are ho(Ac)and h(A), and then

Proof:

3-26.

We prepare for the proof by proving three lemmas. Lemma.

Let B C A C E, C C E, D be abelian groups,

111.

98

SOME PROPERTIES OF COHOMOLOGY GROUPS

let Q : A D be a homomorphism and write, as usual, A, for the image of A under Q and A, for the kernel of Q on A; then __+

( A : B) = (Am :B")(A, : B,)

and

Proof: Start from (A :B) = ( A : BA,)(BA, : B), and observe that the natural map A -A@/Bw has kernel BA,, while (BA,)/B w A"/P n A,) = A,/& * For the second part, consider the natural map Q : A(AC)/C, and apply the fist part. I

Let the abelian group A, then 3-27.

Lemma.

Q,#

Qo.&)

be commuting endomorphisms of

= Qo.@)

Q 0 . M

with the understanding that if the left side is defined then so is the right side, and conversely.

Proof:

The assertion is that

T o prove it, note first that (A@)S= AS@= A*$ c A,,

(AS), = {flu) I q$(u) = 0) = (A&)*

and VedS

Now consider ( A : A",)

From (3-2-6)we have

= AS '

= ( A :A'XA'

: A'S)

3-2.

CYCLIC GROUPS AND THE HERBRAND QUOTIENT

99

and also

( A : A'J) - ( A : A') 1 .

( 4 4 : (0))

(A : AS)

1

(4:(42) ((Ad)*:(wo, :(0)) *

which is the desired result. The finiteness conditions are easily checked. I

3-24 Lemma. Let G = (1, u,...,u9-l) be a cyclic group of prime order p, and let A be a G-module. Then, on A, the maps (u - 1)P-l and p (that is, multiplication by p) have the same image and the same kernel. Proof: We work in the ring Z[G]. Since u is essentially a primitive pth root of unity, it follows that

X*'+

*..+ x + 1 =

so that

p

n (1

n (X- d),

P-1

1

0-1

=

- u')

1

+ + +

For any i = 1,...,p - 1, (1 - ui) = (1 - u)(l u - * u*-.-l) so that im (1 - u f )C im (1 - u) and ker (1 - u) C ker (1 - uf). But the roles of u and ui may be interchanged, because u* is also a generator of G; consequently, for i = 1 p - 1

,...,

im(l -u*)

=

im(l -u)

ker(1 -u*) = ker(1 - 0 )

These relations hold on A and on any G-submodule of A; in particular, they hold on and A(o--lp. Therefore,

and similarly ker ( p) = ker ( ( 0 - 1)P-l).

I

100

111.

SOMB PROPERTIES OF COHOMOLOGY GROUPS

Now let us turn to the proof of (3-2-5). Consider the exact G-sequence

The modules and homomorphisms are fixed, and the sequence is exact for each of the actions of G. If @A) = (A :mA)/(A, :(0)) is defined, then so is P(A"') = (Au-' : dU-')/((Au-')(.: (0)), since ( A : d )= (A*' : (mA)"-l)(Au-l: (mA)u-l),

(mA)o-' = mAu-1, and (AU-l),C A,,, . Then, according to (3-2-3), hO(A3 is defined, and P ( A ) = ho(AG)h0(Au-l).Furthermore, h ( M ) = &(A3 < 00, and making use of (3-2-8)and (3-2-7)we have for B = AU-l

In particular, h(Au-l) is defined; so according to (3-2-3),h(A) is defined and h(A) = h(AG)h(AU-'). Therefore


a

3-3. 3-3.

101

DIMENSION SHIFTING

--

DIMENSION SHIFTING

The G-module P is said to be projective if whenever we have G-homomorphisms of G-modules f : P C and g : B -H C, there exists a G-homomorphism h : P B such that f = g o h. This is usually expressed by the requirement that every diagram P

,?'I

rc B-C-0 of G-modules and G-homomorphisms in which the row is exact can be completed to a commutative diagram. It is clear that P is C and g : B C projective if and only if whenever f : P are G-homomorphisms of G-modules with f(P)Cg(B), then there B with f = g o h. exists a G-homomorphism h : P __+

__+

__+

3-3-1.

A G-free module is projective.

Proposition.

Let {x.} be a I' = Z[G] basis for the G-free module X. Proof: For each a, choose b, E B such that g(b,) = f(x.), and then define h by putting h(x,) = b.. I

3-3-2. (1)

Let X be an arbitrary G-module, then : B A C +0 is an exact G-sequence, then If A the following sequence of abelian groups is exact : Proposition.

0

H o q (C, X)-!% Homo (B,

(2) If 0 +A & B

x)aHOmG (A, x)

I

C is an exact G-sequence, then the following sequence of abelian groups is exact :

0

H o q ( X , A)

(3) Let 0 - A sequence, then 0

__t

__+

HOmG (X, B)

B-

C-0

aHomo (x,c) be an exact G-

-

Homo (X, A) _* HOmG (X, B) +Homo (X, C)

is exact for all such short exact sequences projective.

O

e=. X

is

102

111.

Proof:


(1) To show that (j,1) is a monomorphism, consider

f E Horn, (C, X), then (j,l ) f = f o j

=0

==-fjB) = 0 - f ( C )

=0

-f=

0.

Since

(i, 1) o ( j , 1) = (ji,1) = (0, 1) = 0, we have im ( j , 1) C ker (i, 1). To prove the reverse inclusion, suppose fis Horn, (B, X ) satisfies (i, I)f = f 0 i = 0. Then f (iA)= 0 and we may define a G-homomorphism3 :B/(iA)+X by putting j ( b iA) =f(b). Furthermore, j induces a G-isomorphism j : B/(iA) C and the map g =30 j-l E Horn, (C, X) satisfies (j,1) g = 30 f 1 0 j = f. This completes the proof of (1). The proof of (2)involves similar arguments. As for (3), exactness C) is precisely equivalent to the assertion that X is at Horn, (X, projective. I

+

-

3-34. Remark. To every ordered pair of G-modules A and B, we may assign the Z-module (that is, abelian group) Hom, (A, B). Furthermore, given G-homomorphisms of Gmodulesf : B +B' and g : A' +A, there exist homomorphisms of groups

We summarize these facts by saying, loosely, that the functor Horn, is contravariant in the first variable and covariant in the second variable. (A precise formulation of these terms is given in [15, Chapter 21.) We also say that the functor Horn, is left exact which is, by definition, the assertion that parts (1) and (2)of (3-3-2) hold. An exact G-sequence 0 +A i B i C 0 is said to split (with respect to G, or over G) when iA is a direct summand of B. In other words, there exists a G-submodule B of B such that B = iA @ B'; note that then B' is an isomorphic copy of C, since j :B' >C.

---

3-3.

103

DIMENSION SHIFTING

3-34 Exercise. The following conditions on a G-module P are equivalent : (I) P is projective. (2) P is a direct summand of a free G-module. (3) Every exact G-sequence 0 A B +P 0 splits. (4) P is Z-free and the identity map of P is a trace.

-

__t

__+

-j

Let 0 4 A B C an exact G-sequence, and let X be any G-module, then 3-3-5.

-

(1) If

0

6

Proposition.

__+

0 be

iA is a direct summand of B as a G-module, the sequence

Horn,

*

(c,x)

HornG (A, x ) - o

HOmG (B, x)

of abelian groups is exact and splits. (2) If iA is a direct summand of B as a Z-module, then Horn (C,X)

0

Horn (B, X) %Horn (A, X)

-

0

is an exact G-sequence which splits as a Z-sequence. Proof: HOmG (B,

(1)

Using the notation introduced in (3-3-3), we have

x)= Horn, (iA@ B’,x)

Horn, (;A,

x)@ HOmG ( E ,x)

and when the appropriate identifications are made the result follows immediately. (2) Apply part (1) and (1-4-3). Note that we do not assert that the sequence splits as a G-sequence. I

3-36. Proposition. If A is G-regular, then, for any Gmodule X , so are Hom (X, A) and Hom (A, X). Proof:

Let B be a Z-module such that A = LEG @ uB.Then

Horn (X, A)

=

Horn ( X ,

1@ uB)w 0

@ Horn (X, uB) 0

w

1 @ (Horn (X,B))o. 0

A similar proof applies for Hom (A, X).

I

111. SOME

104


If A is G-regular and H is a normal 3-3-7. Proposition. subgroup of G, then AHis (G/H)-regular. Proof: Write A = LEG @ aB and let G = Uca,H = Ut Hag be a coset decomposition. Any element of A has a unique expression a = LEG ab, , b, E B. Thus, a E AH u for each a E G, b, = b, for all p E H. In other words, if S, = h p is the H-trace, then a E AH w a is of the form a = &(SHat)bu,, bu,E B. Now, SHB C AH and because the action of G/H on AHis given completely by the action of the a, ,it follows that

AH= C @ (Sei) B = i

C @ u ~ S H B=) C

&GIH)

i

0 Il(SHB)

which completes the proof.

3-34.

F

Proposition.

In connection with the G-module

= Z[Gl we have exact G-sequences

o-z-r'-z-o and

o-z+r-j-o

where 41) = 1, p(1) = S,and n

Z=C@Z(u-l) a€G

1

] = -zs

Moreover, I and J are direct 2-summands of r (that is, the exact sequences split as 2-sequences) and we have isomorphisms of G-modules Horn (Z,Z)M

1

Horn (J,Z)M Z

As usual, the action of G on 2 is trivial. Since the element 1 is a G-basis for F we may put 41) = 1 E Z and extend n,,a E F to CuEG nu E Z. G-linearly. Thus, e maps an element y = Leo Clearly, e is onto, and its kernel is

Proof:

z = { y E r l ~ n u = o } = ~ ~ z ( u c- i@)~=( USC

aeG

u#l

~ - i )

3-3.

105

DIMENSION SHIFTING

r

As for the second sequence, define p(1) = S E and extend , p is a G-homomorphism. linearly-so p(n) = nS for ~ E Z and Consequently, if we put J = r/(ZS), then the second sequence is exact. Note that if we define a G-homomorphism S : + by uE and extending G-linearly from putting S(1) = S = EGG the r-basis {l}, then we have a commutative triangle

r

r

r

S

r

r

Z which is essentially the augmentation of the standard G-complex. Because Z M r/I and J M r/(p(Z)) are Z-free, it follows from (3-3-4) that our sequences split over Z. However, these assertions may also be proved directly. Thus, it is clear that

r=igz.imIBz In fact, y = LEG n,u E r is of form

As for J = r/(ZS),let y It is clear that

-

p denote the natural map of r--

T = -Ezl6 and that J = Cozl @ Z6; so

We note in passing that the map y =

c

a&

n,a-+p

=

r +J is given by

c n,6 c (n, - %)6 =

a&

a+l

The uniqueness of this last expression implies that if

then n, - n1 = ni - n; for all u # 1.

J.

106

-

111. SOME


To prove the final assertion, let us first define a map Horn (I, Z) J by

(and the right side equals z , f ( u - 1)6, sincef(0) = 0). This is clearly an isomorphism of abelian groups-and the inverse map takes no6 E J to thefE Horn (I,Z)for whichf(u - 1) = n o , u # 1. To prove that this is a G-isomorphism, we must show that for T E G

zzl

c f'@

u#l

-but

zzlf'(u

- 1)6 = &lf(T-lu

=

c

f(7-b

- ~-l)6,while

- 1) 6 =

U€G

[f(T-'U

- 1) -f(T-'

- I)]

6

o#1

-

Finally, define a map Hom (J,Z)

I by

(and the right side equals z , f ( 6 ) ( ~- l), where, of course, f ( T ) = -ztlf(6)). This is clearly an isomorphism of abelian nu(u - 1) to the groups-and the inverse map takes CUzl f E Hom (J,Z)for which f ( 6 ) = nu,u # 1. To prove that this is a G-isomorphism, we must show that for T E G

3-3.

107

DIMENSION SHIFTING

c f ( a 7 . - 1)

U€C

This completes the proof. I For any G-module A we get, from (3-3-8) and (3-3-5) the exact G-sequences

0 0

-

Horn (Z, A)

Horn ( J , A )

-

aHorn (r,A ) Horn (r,A )

-

Horn (I, A )

Horn (Z, A)

-

__t

0 0

and these sequences split as 2-sequences. According to (1-4-5), Hom (2, A) m A as G-modules, and according to (3-3-6), Hom (I',A) is G-regular; thus we have proved : 3-3-9. Proposition. exact G-seqeunces

Given a G-module A, there exist two

O-A-M-A--O

0-

A+-

M-

A-0

which split as 2-sequences, and where M is G-regular. In fact, M = Hom (F,A), A+ = Hom (I,A), A- = Horn (I, A ) . 34-10. Theorem. (Dimension Shifting). Given a Gmodule A, there exist G-modules A+ and A- such that for any subgroup H of G and any integer n we have

H"-l(H, A-) w H"(H, A ) w H"+l(H, A+)

Proof:

For any H C G the sequences of (3-3-9) are exact

111. SOME PROPERTIES

108


H-sequences, and it is easy to check that M is H-regular. From (2-2-4) and (3-1-3) we arrive at exact sequences

-

(0)= H"-l(H, M ) (0)= H"(H, M )

H"-'(H, A-) -L H"(H, A )

-

H"(H, M )=(O)

H "(H , A ) *_ H"+'(H, A+)--+ H"+'(H, M )=(O) 8

and the proof is complete. I Because of these isomorphisms, the exact sequences of (3-3-9) are known as dimension shifters. Corollary. Given a G-module A and an integer d, there exists a G-module B such that for any subgroup H of G and any integer n we have H"(H, A ) M H"+d(H, B )

3-3-11.

Iterate the preceding result, using A+ or A- according Proof: as d is positive or negative.

3-4.

THE INFLAT10N-RESTRICT10 N SEQUENCE

In this section, we suppose that A is a G-module and that H is a normal subgroup of G. Then, using the results of Section2-3 we have for each n 2 1 the sequence 0

-

H"(G/H,A H )

Hfl(G,A ) 5Hn(H, A )

which is known as the inflation-restriction sequence in dimension n. 341.

For each n

Proposition.

2 1, res inf 0

= 0.

Proof: Applying (2-3-7) with Hl = H, = H gives the commutative diagram

H"(G/H,AH) -%H"(G, A ) re8

1

Imp

H"(H/H, A H ) -%H"(H, A )

-and

by (3-1-5), Hn(H/H,AH)= (0).

I

3-4. THE INFLATION-RESTRICTION SEQUENCE

342.

Proposition.

sequence

I n dimension 1,the inflation-restriction

0 +H1(G/H,A") %H1(G,A )

is exact.

109

-

H1(H,A )

Proof: Let u G denote the canonical map of G --tf G/H, and for each of the groups G/H, G, H , use its standard complex for computation of the cohomology groups. As in Section2-5 we use inf and res to denote the cochain maps and are thus in a position to use the appropriate formulas in (2-5-1). Let u be a I-cocycle of G/H in exactness at H'(C/H, A"): AH such that the cohomology class of inf u is 0. Since infu is a coboundary and (inf u)[u] = 2451, u E G, there exists a E A such that VUEG u[b] = ( 0 - 1) a

For p E H we have ( p - 1)a = u[/3] = 0-as may be seen by substituting p for T in the cocycle identity &[?I - u[*] u[b] = 0, a,? E GI€€.This means that a E A" and that

+

u[5] = (5 - l ) a

Vb E G / H

Thus, u is a coboundary, and its cohomology class is 0. exactness at H'(C,A): In virtue of (3-4-1)) it suffices to show that if u is a 1-cocycle of G in A for which res u is a coboundary, then there exists a I-cocycle v of G/H in A H such that inf w is cohomologous to u. According to (2-5-1) and the hypothesis on u there exists a E A such that Now put

-

V U EG

u"u1 = u[u] - (u - 1) a

so that u' is a I-cocycle of G in A and u' u. Then u'b] = 0 for all p E H , and from the cocycle identity m ' [ ~-] ~'[uT] u'[u] = 0 we get (by taking T = p E H ) u'[up] = u'[u] Vu E G, p E H and (by

+

111. SOME

110

PROPERTIES OF COHOMOLOCY CROUPS

taking o = p E H ) p u ' [ ~= ] u'[p7] V7 E G, p E H . Then, upon defining a I-cochain v of G/H in AH by v[tj] = u"u]

Vtj E G/H

it is immediate that v is a cocycle with infv = u' completes the proof.

N

u. This

343. Theorem. Let A be a G-module, let H be a normal subgroup of G, and suppose that for an integer n 2 1 we have

H'(H, A ) = (0)

-

Y

= 1, 2,

...,n - 1

Then, in dimension n, the inflation-restriction sequence 0

Hn(G/H,AH)%Hn(G,A )

Hn(H, A )

is exact. Proof: By induction on n. For n = 1, the hypotheses are vacuous and the result is given by (3-4-2). Suppose then that the result is true for n - 1 2 1. According to (3-3-9), A can be imbedded in an exact G-sequence

(*I

0-A-B-C-0

in which B is G-regular. Because this is also an exact H-sequence and H1(H, A) = (0) it follows from (2-2-4) (or, more precisely, from the comments following ( 2 - 2 4 , in the case where cohomology is done with respect to the half-complex) that the sequence 0

-

AH-

BH

__t

CH

__t

Q

is exact. Moreover, this is clearly an exact (G/H)-sequence, and according to (3-3-7), BH is (G/H)-regular. Therefore, in the commutative diagram 0 0

-

Hn-I(G/H, CH)%Hn-l(G, C )5 Hn-l(H, C )

.*I &

Hn(G/H,AH)

Hn(G, A ) r88 Hn(H, A )

3-4.

111

THE INFLATION-RESTRICTION SEQUENCE

the three coboundary maps are isomorphisms onto. Thus, to prove our result, it suffices to show that the top row is exact. Since B is also H-regular, we conclude from the exact H-sequence (*) that Nr(H, C) = Hr+'(H, A) for all Y. Now, the hypothesis on the cohomology of H in A (for Y = 2, ..., n - 1) implies that Hr(H, C) = (0) for Y = 1,2, ...,n - 2. Consequently, the induction hypothesis applies for C, and the top row is exact. This completes the proof. I The case n = 2 of this result will play an important role in abstract class field theory (see Chapter VI) when we discuss the Brauer group. This case n = 2 was first proved in [34] for Galois cohomology and applied to local class field theory; its formulation for global class field theory was given in [54]. Our general result, considered from the point of view of spectral sequences, appears in [38].

344. Remark. It may be noted that it is possible to define a deflation map (see [79])in negative dimensions, which is dual to inflation, and for which an analog of (3-4-3) holds (with inf replaced by defl and res replaced by cor). The sole fact of interest to us (since it will be used in Section 3-6) is that if H is a normal subgroup of G and A is any G-module, then we may define a deflation map in dimension 0 by the requirement that the diagram

Ho(G,A )

be commutative-so

aHo(G/H,AH)

that defl is defined by

Moreover, the sequence HO(H, A ) %HO(G, A )

-

Ho(G/H,AH)

0

112

111.


is exact. To see this consider the diagram

4

.1

Ho(H, A )

-

HO(G, A ) %Ho(G/H,A H )

Here, the top row ie induced from AH (which is not exact). It is exact since

Sn.,

-

0

AG5 AG

-

0

and

By (2-5-5) and the definition of defl the diagram commutes; and because the I? maps are isomorphisms onto, the bottom row is exact. 3-5.

THE GROUP W ( G , Z )

In this section, we examine the group WZ(G,Z), which will play a key role in our formulation of the reciprocity law (see Section 6-5).

3-54.

For any group G we have an isomorph-

Proposition.

ism

G

given by UGC

-

I

(u - 1)

+ IS

UEG

Proof: Here, as usual, Gc is the commutator subgroup of G, while le (which is viewed additively) is the square of the ideal I= @ Z(0 - 1) of r =

z[q.

3-5.

Let us map G 1/12by the rule u is a homomorphism since __+

(UT -

1) f 1'

-

THE GROUP H-2(G, z )

= (U

+ - 1) + 12}+

- 1)(T

= {(U

-

1)

-

(u -

(U -

{(T

113

1)

- 1)

+

(T

1)

+ 12. This

- 1) + 1'

+ 1').

Because 1/12is commutative, it follows that Gc is in the kernel, so that we have an induced homomorphism of G/Gc +1/12given (0 - 1) 12. T o show that this is an explicitly by : uGC isomorphism, we construct its inverse. Define a map I G/GC uGc, u E G and extending linearly from by mapping (u - 1) the Z-basis. This is clearly a homomorphism; moreover, I 2 is in the kernel because {(u - I)(T - 1) I cr, T E G} is a set of Z-generators for Z2 and

-

+

__+.

(U

-

1)(T -

1) = (UT - 1) - (T - 1) - (U - 1) _+

(UTGC)(T-~G~)(U-~G~) = Gc

Thus, we have an induced homomorphism of 1/12+G/GCgiven by : (cr - 1) + 1 2 +uGc. This completes the proof. I

-- - -

Remark. Let us show how fP2(G,Z) can be interpreted as the group G/Gc. From the exact G-sequence 0 I r Z 0 we conclude (since r is G-regular) H-'(G, I) is an isothat the coboundary S* : W2(G, Z) morphism onto. Furthermore, according to (2-2-6) we have H-l(G, I) with kernel I - I = 12, (since I , = I ) a map 7 :I which determines the isomorphism ;i: Z/12>H-'(G, I). Combining these with (3-5-1) leads to the chain of isomorphisms

3-5-2.

__+

-

and provides a canonical procedure for identifying H-2(G, 2)with G/Gc. In particular, if we denote by 5, the element of IF2(G, Z) corresponding to uGC (so that u 5, maps G onto W2(G, Z) with kernel Gc), then

S*L

= V(U

- 1)

114

111. SOME PROPERTIESOF

COHOMOLOGY GROUPS

Our next objective is then to describe the usual cohomological maps of H-g(G, Z) as reflected in G/GC.

3-53. Proposition. Let H be a subgroup of G. Then the map cor :H-B(H, 2) +H-Z(G, Z) corresponds to the natural map i :H/Hc +G/GC which is induced by the inclusion map i : H d G.

rE

Let r G = Z[q, I, = &;lsG 0 Z(u - l), = Z[rrJ, = CpZleE @ Z ( p - 1) and consider the commutative diagram

Proof: IH

o -zE+

a z -0 rH-

where i denotes inclusion and 8,.8 are the respective augmentations (as in (3-3-8)). The top row is an exact H-sequence, the middle row is an exact G-sequence which we view as an exact H-sequence, and the bottom row is an exact G-sequence. This leads to the diagram

r,,

r,

is H-regular, and is The 8, maps are isomorphisms because both G-regular and H-regular. The inclusion map i : I, maps IH I, H I t and induces the map i of the factor groups. All the squares commute-by (2-2-5), (2-2-7), (2-4-7)) and (2-5-5).

-

3-5.

THE CROUP H-2(G,

115

z)

In virtue of this, we may compute cor 0 1* (which is indeed the map we seek for standard complexes) by using the third column. Consequently, for p E H pHC+

(p

- 1)

+ 1;-

+ 1,

*IG

1) +Z:-pGC

-(p-

and this completes the proof.

3-54.

(p - 1)

I

As preparation for our next result, we need

Remark.

to know the definition of the group theoretical transfer (or

verlagerung)

-

V=

VG+H

G/GC +H/HC

The definition, along with the basic properties of this map, may be found in [83, Chapter 51 or [31, p. 2021; however, we shall arrive at the definition through cohomological considerations. Let denote the character group of G-that is, is the group of all continuous homomorphisms of G into Q/Z. Because G is finite and the action of G on Q/Z is trivial, we may identify A

Hl(G, Q/Z) = Horn (G, Q/Z) = e = (G/Gc)

-

If H is a normal subgroup of G the identification enables us to carry the map inf : Hl(G/H, Q/Z) H1(G, Q/Z) over to a map A

-e

inf : (G/H)

which, in virtue of (2-5-1), is given by n (inf x)(u)= ~ ( 5 ) y, E (GIH),

uE

G

In other words, inflation of a (standard) cocycle corresponds to inflation (or lifting) of a character of G / H to G. Similarly, if H is any subgroup of G, then, according to (2-5-1), restriction of a cocycle corresponds to restriction of a character of G to H-in other words, the map res : € - A

116

111.

-


corresponding to res : H'(G, Q/Z)

H1(H,Q/Z) is given by

x

(resx)(p)= X ( P )

4

PEH

As for the corestriction or transfer; here we have a map

-

cor: A + €

corresponding to cor : H'(H, Q/Z) H'(G, Q/Z). According to (2-5-2), if G = (J c = HZ is a right coset decomposition, this map is given by

u

A

A

Since E? = (H/Hc) and since ( H / H c ) may be identified with H/Hc, there exists a dual map -

V

-

= V C + H : GIGc

H/HC

called the reduced group theoretical transfer, which satisfies the relation X{V(;~GC)} = (cor x ) ( u ~=~ x) tu~-hc) E ( Hn I H C ) , (I E G

(n c

n Because this is true for all x E (H/Hc),it follows that

which is indeed the customary formula for the (reduced) group theoretical transfer. Of course, the usual group theoretical transfer is defined by this formula when G is infinite, provided that ( G : H) *” 8

H-’(H, 1,)

,*=I1

ct +l-
tnere exists an integer I such that for all H C G we have f(r-l) : Hr-l(H, A ) -+

Hr-l(H, B )

(isomorphism)

j-(r) : H'(H, A ) >--HH'(H, B ) f(r+l)

: Hr+l(H, A ) >-

(epimorphism)

Hr+l(H, B )

(monomorphism)

By (3-6-2) we have (with obvious choice of notation) Proof: an exact sequence

-

Hn(H, A )%H n(H , B ) %Hn(H, C )%Hn+l(H, A )

-

.

Now, f(r-l) is an epimorphism

= (0)

im QJ(?) = ker

-

--

imS(,) = (0)

=

H*(H, C)

Hr(H, C ) = (0)

-

since Hr(H, B) = imf(r) = ker CJ+) . Therefore, H+'(H, C) = Hr(H, C) = (0) for all H C G, and by (3-6-1) C is cohomologically B is trivial. Returning to our exact sequence, we see thatf : A indeed a cohomological equivalence-which completes the proof. I The same result and proof are valid if the exact sequence (*) is

3-6.

123

COHOMOLOGICAL EQUIVALENCE

used, because one can only work to the right (left) from an epimorphism f(r-l) (monomorphism f(r+l)). It is also known (see [26]) that this result holds when f(r) and f(r+l) are isomorphisms onto.

If the G-module A has the properties

3-6-4. Theorem.

(1) H-l(H, A) = (0) for all H C G. (2) HO(H, A) is cyclic of order # ( H ) for all H C G. then Z and A are cohomologically equivalent. Let a be a generator of Ho(G,A), so a has order #(G). Proof: Fix any subgroup H of G. Take any positive integer s < #(H). Then corH+G(s resC+Ha) = s corH+GresC-rHa

= s(G : H ) a

s(G:H)~#OEHO(G,A)

Therefore, resG+, a E HO(H, A) has order 2 # ( H ) ; so from (2) resG+Ha has order # ( H ) , and it is a generator of Ho(H,A). Now choose a E AG such that ."(a) = a, and using this Q define the G-homomorphism f : 2 A by f(1) = a. We show that f is a cohomological equivalence (with Y = 0). Clearly,

-

-

Z) : H l ( H , Z)>-

f(-') : H-'(H. f(+l)

H-'(H, A )

since H-'(H, A ) = (0) since

H'(H, A )

H'(H, Z)

=

(0)

and it remains to show that f(o) : H"H,

Z)

'----H

HO(H, A )

(isomorphism)

In virtue of (2-5-5), K ~ ( u )= resG,,a is a generator of Ho(H, A). On the other hand, Ho(H, Z) is cyclic of order # ( H ) and K H ( ~ is ) a generator. By (2-2-7), we have the commutative diagram

z = ZH

f

b

AH

4 Z)-Ho(H,

Ho(H,

f*=f(o)

A)

124

111.


which says thatf(o) maps the generator ~ ~ ( to1 the ) generator ."(a). Thus, f(o) is an isomorphism onto, and the proof is complete. I 3-6-5.

If the G-module A has the

(Tate).

Corollary.

properties

(1) H1(H,A) = (0) for all H C G, (2) H2(H,A) is cyclic of order # ( H ) for all H C G, then, for all H C G and all n we have Hn(H, Z)w Hn+2(H,A )

By dimension shifting, there exists a G-module C with Hfl(H, C) M Hfl+*(H,A)Vn,V H C G, and (3-6-4) may then be applied to C. I

Proof:

This result is a weak form of Tate's theorem (see [74]). The full theorem also gives an explicit form for the isomorphisms Hn(H, 2) M Hn+2(H,A), and this is one of the major objectives of the next chapter. Roughly speaking, the arithmetic part of class field theory consists of the verification of the hypotheses of (3-6-5), and then the case n = -2 of Tate's theorem gives the reciprocity law isomorphism.

3-66, Remark. The proofs of (3-6-3), (3-6-4), and (3-6-5) depend on (3-6-1) and (3-6-2). Furthermore, it is easy to see that given the stronger version of (3-6-1) (mentioned at the end of that proof) the following versions of (3-6-3), (3-6-4), and (3-6-5) hold.

-

(i) Let f : A B be a G-homomorphism of G-modules. Suppose that for each prime p dividing #(G) we have a p-Sylow subgroup Gp and an integer rp such that f(rp-l)

f(r,)

--

: HrP-'(G, , A ) : HrP(Gpi A )

f(',+,) : H'p+'(G,, A )

H'.-'(G,, B )

Hs(G, B ) 9

H'p+'(G,, B )

-then f is a cohomological equivalence.

is an epimorphism is an isomorphism is a monomorphism

3-7.

125


(ii) Consider the G-module A, (Y E Ho(G, A), and above. If the following properties hold

(1)

H-’(G,

3

A ) = (0)VP,

(2) Ho(Gp ,A) is cyclic of order #(G,), a generator,

-then

and resG+G,(a)is

2 and A are cohomologically equivalent.

(iii) Suppose that

(1)

p, Gp as

(Y

E H2(G,A) and

H’(Gp 9 A ) = (0)VP,

(2) H2(Gp,A) is cyclic of order #(Gp), and res,, a generator,

-then

Hn(H,2)

3-7.

w

(a) is

Hn+2(H,A ) V H C G, Vn.

-

PROBLEMS A N D SUPPLEMENTS

-

3-7-1. Given a homomorphism of groups h : G‘ G, we have constructed (at the beginning of Section 2-5) an explicit X, suitable for A, in dimensions G’-homomorphism A : X’ n 2 0 where X and X are the standard GI- and G-complexes, respectively. If h is a monomorphism, find the negative dimensional A’s explicitly.

-

3-7-2. Suppose that ( X , f ) : (G, A) (G’, A’) is a homomorphism of pairs. If t : G A is a crossed homomorphism is a crossed (see (I-5-2)), then ( h , f ) t = fth : u‘ -ft[ho’] homomorphism of G‘ in A’ whose cohomology class is the image of the class o f t under the map ( h , f ) , : H’(G, A) H1(G’, A’). Thus, inflation of crossed homomorphisms is analogous to lifting of homomorphisms from factor group to group, while restriction of crossed homomorphisms is analogous to restricting homomorphisms to a subgroup. What about corestriction of crossed homomorphisms. __t

-

126

111.

-


Suppose that ( A , f ) : (G, A) (G‘, A’) is a homomorphism of pairs. What can be said about the diagram

3-7-3.

H-1(

G,A )

(A f )

H-l( G’, A’)

Induced homomorphisms f* commute with in-Aon, restriction, corestriction, and conjugation; of course, the hypotheses need to be formulated accurately.

3-74

-- - ---

3-7-5. Suppose that G is a multiplicative group, finite or A B C 0 is an exact G-sequence infinite. If 0 show that 0 AG BC Cc is an exact sequence of abelian groups. Can the hypothesis be weakened? __+

(i) Suppose that H is a subgroup of G. If the G-module 3-7-6. A is G-free, then it is also H-free. If A is G-regular, then it is also

H-regular.

(ii) Let us call a G-module A C-special when its identity map is a trace; so G-regular implies G-special. Show that the converse is false. Discuss the properties of G-special modules; in particular, show that a module which is either G-projective or G-injective is G-special. I n general, the notion of G-specialness is easily seen to be sufficient for our theory-for example, for dimension shifting. 3-7-7.

Consider the exact G-sequence

For c E C, , choose b E B and a E A such that j b = c and ia = Sb. Then a E AG and the map c +u + SA of Cs AG/(SA)is well-defined (that is, independent of the choice of 6 ) and a homo__t

3-7. morphism. Call it homomorphism

127

PROBLEMS AND SUPPLEMENTS 9).

Since I C C k e r v , we have an induced

Consider the sequence

where z a n d j are the natural induced maps, and show that it is exact. In virtue of the interpretation of cohomology groups in dimensions 0 and - 1, this provides a direct proof of the exactness of H-l(A)

- H-'(B)

H - (C )A Ho(A)

- Ho(B)

HO(C)

where 6, = tjj, the Tate linking (see (2-2-8) and (2-2-9)).

3-7-8. Suppose that A is a left G-module, and r = Z[G], then the additive group M = Hom (F,A) = Hom, (F,A) may be made into a left G-module in several ways. First of all, r may be viewed as a left G-module, and as in (1 -4-2), M then becomes a left G-module (which we denote by Ml) in which the action of G on Ml is given by (fu)l = u 0 f 0 u-l, that is, ( f ")1(~)= Uf(U-lT), 0 , T E G, f E M . Let A denote the additive group of A viewed as a G-module with trivial action. Then M = Hom (F,A) = Hom (r,A) can be made into a G-module (which we denote by M,) in the same way. Thus, the action of G on M , is here given by (fu), = u o f o 0-1, that is, (f"),(T) = f(u-%). On the other hand, F may be viewed, in a natural way, as a right G-module, and A may be made into a right G-module by putting au = u- k. Of course, A is also a right G-module with trivial action. For f E M , 0 E G let us now define "f = f s-l = u - l o f 0 u-which means, act first by u (on the right!), then byf, and then u-l. Then M = Hom (r,A) becomes a left G-module M 3 , with the action of G on M3 given by pf)3(~) = (f(7u))o-l = u( f (7.)). Finally M = Hom (r,A ) becomes a left G-module M4 with action given by (of)4(~)= f(Tu). Show that the four G-modules Ml ,M , ,M a , M4 are G-isomorphic.

111. SOME

128


(i) Let G be a multiplicative group, finite or infinite, 3-7-9. and let = Z[GJ If A and B are G-modules then A @ B (the usual tensor product over Z-that is, the tensor product as abelian groups) may be made into a G-module in which u(a @ b) = ua @ ob. In particular, I' @ B becomes a G-module in this way. If X is any abelian group, then X may be viewed as a G-module with trivial action, and we may speak of the G-module A @ X. Let B denote B viewed solely as an abelian group-so that B is a G-module with trivial action. Then I' @ B is a G-module with u(y @ b) = my @ b, y E r and it is isomorphic to the G-module r @ B-in fact, the desired isomorphism of G-modules is given by u b +u @ub. (ii) The G-module A is said to be induced when there exists @ uX-thus, induced a subgroup X of A such that A = LEG means G-regular. The G-module A is induced e- there exists an abelian group X such that A M @ X. Any G-module A can be expressed as a quotient of an induced module-more precisely, there is a canonical epimorphism of @ d A which maps

r

r

r

a @ a +aa.

-

(iii) For any abelian group X we may consider the G-module Hom (r,X)-namely, X is. viewed as a G-module with trivial is viewed ag..'a (left) G-module, and (see (1-4-2)) action, Hom X) becomes a G-module with f* = u o f 0 0-1 and consequently fo(y) =f(u-v). The G-module A is said to be co-induced when there exists an abelian group X such that A M Hom(F, X). Any G-module A can be imbedded in a coinduced module-more precisely, there is a canonical monomorphism of A +Hom (F,A ) which maps a fa, where

r (r,

-

fa(u) = u-la.

(iv) If G is finite, then for any abelian group X the G-modules Hom (r,X) and @ X are isomorphic; in particular, the notions of regular, induced, and co-induced modules then coincide.

r

-

3-7-10, Suppose that H is a subgroup of G and that G = (J c = HE is a right coset decomposition. Is the map y +C,c?y a homomorphism of IG IH ?

u,

3-7-11.

For a cyclic group G, we have defined a special G-

3-7.


129

complex in (3-2-1). Using this complex, describe (for any Gmodule A) the cochains, cocycles, and coboundaries in dimensions 0 and -1, and the maps K and 7. Suppose that H is a subgroup of G; show, directly and cohomologically, that the reduced transfer has the property 3-7-12.

(i) If H and K are subgroups of G, then we have a disjoint decomposition into double cosets G = H q K . If we write KO#= uIKu~',then (G : K) = C6,( H : H n KO[); in fact, if H = U, P,~F,,pjl E H , Fg = H n KO[ is a left coset decomposi(P,~U,)K is a left coset decomposition. tion for each i, then G = (ii) For any G-module A and any integer n, the map : Hn(K, A) Hn(H, A) is given by resG+H cor,, 3-7-13.

ui=,

u,,*

-

0

If, in addition, H is a normal subgroup of G, then

and for

01

E Hn(H, A )

(see (2-3-3) concerning the action of G/H on Hn(H, A)). (iii) An element 01 E Hn(H, A) is said to be stable when reSHo+HnHO

= reSH-,HnHO a

for every u E G. (In particular, if H is normal, then a is stable, v U*(Y = 01 for every u E G.) If /I E Hn(G, A) and (Y = res,+,/I then (Y is stable. Finally, for any stable (Y E Hn(H, A) we have resC+HcorH+c (a)= (G : H ) a

130

111. SOME PROPERTIES OF

COHOMOLOGY GROUPS

(i) Suppose that H is a subgroup of G and that B is an H-module. In the natural way, we may view I' = Z [ q as a left G-module, and hence as an H-module, too. According to our standard procedure Hom (F,B) is a left H-module, with f p = p 0 f 0 p-1, p E H. Let us put 3-7-14.

B*

= (Hom (F, B)), =

-

Hom,(F, B)

Since there is a 1-1 correspondence between Hom (I',B) and the B, it follows that under this idenset of all mappings f : G tification B* = {f:G - B

V p E H , VUEG}

If(pu) = pf(0)

We make B* into a G-module as follows : for T E G, f E B* define B* by Y(4 = f ( 4 UEG

YE

(ii) Define O : B*

-

B by

-

Of = f V )

Then O is an H-homomorphism, and if i : H G is the inclusion map, then (i, 0) : (G, B*) +(H, B) is a homomorphism of pairs. In order to compute (i,el, : H"(G, B*) -+

-

-

H ~ HB,)

nEZ

choose a G-complex X and view it as an H-complex also. The map A :X X associated with i : H G may be taken as A = 1 (see (2-3-4)), and (1,d): HOmG(X,B*)+Hom,(X,B)

then commutes with coboundaries and induces (i, O), Now, for any G-module A, the map (1,e) : HOmG ( A , B*)

-

.

Hom, ( A , B)

is an isomorphism onto. In fact, the inverse mapping is the following : for v E Hom, (A, B ) define Q. E Hom, (A, B*) by

V " ( 4 ( 4= 4 . 4

Thus, (1, O)-l : v

+py

.

UEG,U E A

3-7.

131


It follows then that (i, O), is an isomorphism onto in all dimensions-that is, H"(G, B*) w H"(H, B ) neZ (This result is known as Shapiro's lemma.) Moreover, we have the commutative diagram (i

I),=ree

Hn(G, B*) L Hn(H, B*)

so that

(i, 8),

= 8,

o

res

(iii) Suppose further that B is a G-module. Define

Then p is a G-homomorphism, (1, p ) : (G,B ) (G, B*) is a homomorphism of pairs, and O p = 1 (the identity map on B). Therefore, the following diagram commutes __+

0

H"(G, B*)

so that

(i, O),

0

p* = res

(iv) Suppose again that B is a G-module, and let G = be a left coset decomposition. Define

u, T H

132

111.


7rf

=

c

7f(7-1)

7

Then T is independent of the choice of coset representatives, and is a G-homomorphism. Furthermore, 7~ is an epimorphism; in fact, if we define T

h:B-B*

by putting

then h is an H-homomorphism with T 0 A = 1 (the identity map on B). For any G-module A, the following diagram commutes :

-

It follows, therefore, that we have a commutative diagram H"(H, B)

(i,G'

H*(G,B*)

and the relation

-

(v) The mappings p and 7r arise in a natural way; more precisely, suppose that B is a G-module-then the mapping of B Hom, (r,B) given by 6 +fa where fb(l) = 6, is an

3-7.


133

isomorphism onto (of G-modules), and we may identify the two G-modules. Now, p and 7r are defined in order to make the following diagrams commute : B

21 B

(i) Suppose that H is a subgroup of G, and let 3-7-15. G = (JYTiH be a left coset decomposition (and for convenience, take 7, E H). Suppose that A is a G-module and A, is an H-sub@ 7J1. This kind of situation module of G such that A = occurs naturally and often in class field theory, and leads us to say here that {G, H, A, A,} is semilocal. Note that the translates TiA, are independent of the choice of coset representatives rS and that H={uEG~uA,=A~}. Suppose that A is a G-module and A; is an additive subgroup (i.e., 2-submodule) of A such that A is the direct sum of some of the translated subgroups u A ; , u E G-that is, A = @ u,A;, for example. Consider the subgroup H = {u E G I u A ; = A;} of G; are any additional hypotheses needed to guarantee that (G, H, A, A;} is semilocal ? (ii) Let X be any G-module and {G, H, A, A,} be semilocal; we seek to set up inverse isomorphisms

xy

HOmG

(x, A)

B

Horn,

(x,A,)

134

111. SOME


-

Let 7~ : A A, = T,A, be the projection map (there is no loss of generality in taking T , = l), so 7 is an H-homomorphism, as is the inclusion map i : A, A. Consider the diagram __+

(1.1)

(1 1

SH-G

(1.f)

Horn, (X, A)7 Horn, (X, A)

Horn, (X, A,)

If we put a = (1, T ) o (1, 1) and fi = S , o (1, i) then a and fi are the desired isomorphisms. Now, suppose that X is a G-complex, and use it to compute the cohomology of both G and H. We arrive at a diagram rB8

Hn(G,A ) 1_Hn(H, A ) 001

c*

Hn(H,A,)

'I*

neZ

and the maps provide inverse isomorphisms

3-7-16. We show that the hypotheses of Shapiro's lemma (see (3-7-1qii))) and those of the semilocal situation (see (3-7-15)) are essentially equivalent-hence, each of the conclusions about the isomorphism of cohomology groups may be used to prove the other. (i) Let B be an H-module, and form the G-module B* = H o m , ( F , B ) = { f : G - B I f ( g u ) = p f ( u ) p E H , u r z G } . Fix a left coset decomposition G = T*H and define homomorphisms

u1

...,m

at:B*+B*

i = 1,

by

If we put Bl = { f e B) *. ( f I

=0

Vu#H}

3-7.

135


1) then B, is an H-module isomorphic to B; in fact, the mapf+f( is an H-isomorphism of B, >B. Moreover, every f E B* can be expressed uniquely in the form

-it follows that B* = cy @ ‘*(B,)and that {G, H,B*, B,} is semilocal. (ii) Suppose that {G, H, A, A,} is semilocal; thus, for a left coset decomposition G = (Jy riH we have A = Cy @ rsA1 and H = {u E G I uA, = Al}. Then the G-module A,*= Hom,(r, A , ) = { f : G -

A,

If@)

= pf(u)

pE

H , u E G}

is isomorphic to A; in fact, the mapping

is a G-isomorphism of isomorphism ?)

A? >-A.

(What is the inverse

Give an example of a G-module with Hn(G, A) = (0) 3-7-17. for all n, but where A is not cohomologically trivial.

CHAPTER

IV

The Cup Product

The main objective of this chapter is to prove a more refined version of Tate’s theorem (see (3-6-5))and in this way to provide a hold on the reciprocity law isomorphism and the Nakayama map of class field theory. This will be accomplished by exploitation of the notion of cup product.

41.

CUP PRODUCTS

In order to define the cup product it is necessary to prepare some background material about tensor products. As usual, let G denote a finite group, and let r = Z[G]. If A and B are G-modules, we denote by A @ B the tensor product of A and as Z-modules; in general, all our tensor products will be taken over Z. (The basic facts about tensor products, which we shall take for granted, may be found in [I I] or in [18, $121.) Now, the additive group A @ B may be viewed as a G-module in the following way. For each u E G, consider the bilinear map 0, : A x B A @ B given by 0,(u, b) = uu @ ab. Then, by a standard property of the tensor product, there exists a unique linear

-

136

-

4-1. CUP PRODUCTS

137

map (which we denote by u) of A @ B B,(a, b) = u(a @ b). In other words,

A @ B such that

u ( u @ ~= ) u~@ob

~ E G ,~ E A b, e B

and A @ B is clearly a G-module. Thus, the operator u is an element of Horn (A @ B, A @ B) and is sometimes denoted by u @ u. Of course, this u @ u must be distinguished from u @ u, belonging to the tensor product of G-modules Hom (A, A) @ Hom (B, B). As a matter of fact, the first of these maps u @ u is the image of the second under the natural homomorphism which is a special case of the one given in the following statement. Proposition. Let A, A', B, B' be G-modules. Then the mapping of G-modules

4-1-1.

Horn (A,A') @ Horn (B, B')

-

Horn ( A 0 B, A' @ B')

which takes

f Og-f

Og

is a G-homomorphism. Furthermore, if f~ Hom, (A, A') and g E Hom, (B, B'), thenf @ g E Hom, (A 6B, A' @ B').

-

Proof: Application of the standard procedure for defining a linear map of a tensor product via a bilinear map shows that f @g f @ g determines an additive map of G-modules. As for the action of G, if u E G, f E Hom (A, A'), g E Hom (B, B'), then

(fOg)"=f"@gO-f"Og"= (.@o)(f Og)(.-'Oo-')

= (fOg)"

so that the map is a G-homomorphism. Furthermore, i f f and g are both G-homomorphisms, so that f c = f a n d g u = g f o r a l l u E G , then(f@g)o = f @ g a n d f @ g is a G-homomorphism. It may be noted that this statement differs from the assertion that there is a homomorphism of

-

Horn, ( A , A') @ Horn, (B, B')

Horn, ( A @ B, A'

0B'). I

Iv.

138

THE CUP PRODUCT

The behavior of exact sequences under tensoring is given in part by the results which follow. 4-1-2.

Proposition.

If

A A B A C -and O A‘L L CB ’ - O‘ are exact G-sequences, then so is

Proof: According to (4-1-1) this is indeed a G-sequence. Since j and j’ are onto and C @ C’ is generated by the elements of the form c @ c’, it follows that j @ j’ is onto. In order to prove exactness at B @ B’, we observe that the image D of (i @ 1) @ (1 @ i’) is contained in ker ( jBY), so that j @j’ induces a linear map p : (B @ B‘)/D C @ C’. On the other hand, the map y’ : C x C‘ -+ (B @ B‘)/Dgiven by

-

y‘(c, c’) = fl(c)

+

c E C,

@j‘-’(~’) D

C’ E

C‘

(where j-l(c) and j’-l(c‘) denote any preimages of c and c‘, respectively) is well-defined (since both iA @ B’ and B @ ?A‘ are contained in D) and bilinear. Therefore, y‘ induces a linear map y : C @ C‘ __c (B 6B’)/D. It is clear that /3 y = identity and y p = identity-consequently, D = ker ( j BY). I 0

0

- -

4-13. Corollary. The functor @ is right exact; this means A +B C __t 0 is any exact G-sequence that if 0 and B’ is any G-module, then the sequences A @ B’ B’ @ A

- B @ B’

C @ B +0

B’ @ B

B’ @ C -

0

are exact G-sequences. The first sequence arises from (4-1-2) with A‘ = (0), Proof: C’ = B’. A switch of notation leads to the second sequence. I

4-1.

-

CUP PRODUCTS

139

41-4. Corollary. If 0A A B i C-0 is an exact G-sequence which splits as a Z-sequence, then, for any G-module B'.

is an exact G-sequence which splits as a Z-sequence.

Proof: By hypothesis, there exists an additive subgroup B" of B with B = iA @ B" as Z-modules, and then B 8 B'

=

(iA@ B') @ (B" @ B')

as Z-modules. (Note that we are somewhat careless about distinguishing between = and w . ) This provides a Z-splitting and shows that i 0 1 is a monomorphism. I 41-5. Proposition. If the G-module A is G-regular, then so is A @ A' for any G-module A'.

Proof:

If A, is a subgroup of A with A

=

zOsG @ oA, , then

-

-

Remark. (Dimension Shifting). If A is a Gmodule, the bilinear map (n,a ) --+ nu of Z x A A leads A which takes n @ a nu. This to a linear map of Z @ A is an isomorphism; the inverse map is given by a -+ I 0a, so that nu --3 1 @ nu = n @ a. Moreover, this is a G-isomorphism, since for u E G,

4-1-6.

u(n

0a) = un @ ua

n @ uu

-

n(ua) = +a)

We shall often identify the G-modules Z @ A w A. I n particular, we have the identification of G-modules Z 0Z w Z.

Iv.

140

THE CUP PRODUCT

Now consider the exact G-sequences (see (3-3-8)) o - + I - r ~ z - + o 0 - Z A r - 1 - 0

Since these split as 2-sequences, we have, by (4-1-4), exact G-sequences (which split as 2-sequences) O - I @ A - - i+81r @ A s Z @ A & O - Z @ A % A ~ I ' @ A - +

A-0 ]@A-0

for any G-module A. By (4-1-5), r @ A is G-regular, and hence H-regular for every subgroup H of G. Therefore, for all n and all H C G, we have

0A )

Hn-l(H, J @ A ) m Hn(H, A ) w Hn+'(H, I

Of course, tensoring by A on the left leads to an analogous result. Remark. Consider the G-complex (X, a, E , p); for the construction of the cup product it is necessary to look at X @ X . We recall that for all n there exist 2-homomorphisms D :X , X,+, (see (1-4-6)) and T : X , X , (see (2-1-4)) aD = 1 and the G-trace S ( T )= 1. Then, for all such that Da p and q we may define G-homomorphisms a', a", D', D" in the diagram

41-7.

-+

-

x, 0xu+, a" tDm

XV-l0x,

a'

-

a* 7x, 0x,7 X,+l 0x, a*/

tDN

xv 0XI?-1

that a'2 = 8"' = 0, a' o a" = a" o a', a'D' + D'a' = 1, a"D" D"a" = 1. In fact, a' = a @ 1 and a" = 1 6 a are commuting G-homomorphisms of "square" zero. Furthermore, X,,, 6 Xq from the 2-homomorphisms D @ T : X , 0 Xq so

+

-

-

4-1. CUP PRODUCTS

141

and r @ D : X , @ X , X , @ X,+, we construct the Ghomomorphisms D' = S(D @ r) and D" = S(r @ D). Finally, a'D'

+ D'a' = (a @ 1)

0

+ S(D @ T )

S(D @r)

+ s(Da @

=

s(aD 0

=

S(1 07r) = C(1 @ P ) O

7r)

o

(a @ 1)

s((aD + Da) @ r)) = C(1 = 1 @-p= 1, =

T)

@ T O )

0

0

0

+

and similarly a"D" D"a" = 1. The cup product is to be an operation which, for any G-modules A and B, provides a pairing of Hp(G, A ) x HQ(G,B )

-

H*q(G, A @ B )

for all p and q. I n order to define it, one naturally goes back to cochains. Consider then f E Hom, (X, , A) and g E Horn, (X, , B); we seek to define f w g E Hom, (X,,, , A @ B). An obvious procedure is to construct G-homomorphisms y,,, in

x,,, 3xv 8xu

A 0B

in order to put f u g = (f@g) vP,,. If this cup product of cochains is to lead to a bilinear map of cohomology classes, it is necessary that the cup of two cocycles be a cocycle and that the cup of a cocycle and a coboundary be a coboundary. Since it is also desirable that the coboundary relation be symmetric, this suggests 6(f w g) = 6f w g f w Sg. In other words, we want (f0g) 9)P .q a = (Sf 0g) V p + 1 , , (f0 Vp,q+1 which requires rpP,, a = a'~,+,,, aN~,,,+, . I n view of all this we prove: 0

+

+

41-8.

For any G-complex X there exist G-

Lemma.

homomorphisms

VV.,

such that

I: 11,A

+

8

:XV,,

-

= (& @ &)

: V Q . 2 = a'?b+1.u

xv

0

0xu

all

P,q

all

P,4

po.0

+ (-I),

a"Pv.U+l

Iv.

142

THE CUP PRODUCT

Condition I is a normalization which expresses the Proof: commutativity of the diagram

xo

'pg.0

xo 0xo

Z%Z@Z

It excludes, in particular, the choice of all 'pP,, = 0. The introduction of the factor (- 1)p in 11,. is a purely technical device. At the outset, we observe that operating on 11,. on the left by a' gives 1IIP& : a'qlP,oa= (-1)P

a'a"P)D,,O+l

The first step is then to construct qonfor all q such that I holds and III,, holds for all q. Choose ( E X , with e(5) = 1 and for each q define a 2-homomorphism $o,q : X , +X , 6X , by #ob.0(4 =

Then

s 0m4x

Yo.0 = S(4fO.Q)=

X € X ,

c 4fL

O€C

is a G-homomorphism of X , +X o 6 X , whose action is given by Po,&) =

c (05 0mF14

X€X,

O€G

In particular, for x E X , we have (.

0l)pO&)

=

Consequently, for x E X , (e

0ebo.o(x)

c (1 Om%)

=

OEC

1 0X €

z@ X u

,

= (1 04 ( e

01)Po,o(4 = 1 04 4 9

which upon identification of 2 6 Z with 2 equals holds. As for III,, , for x E X,,, , we have a'a"PO.u+l(x) = a"~PO.u+1(x)= a"@ 01bo*u+d4 =

e(x).

Thus I

a"b 0 1)(e 0lb0,,+1(x= ) 4) 0 ax

which is also equal to a'qo,, a(x).

4-1.CUP

143

PRODUCTS

The next step is to prove the existence of all F,,~with negative p by showing that if, for a given p and all q there exist G-homomorphisms F , , ~which satisfy III,,*, then we can define G-homomorphisms P ) , - ~ , ~ for all q such that IIp-l holds. Since IIp-l,q implies 4 IIIp-l the proof may proceed inductively. To do this, put

v,-l,aa

=

(-1)po"[(-i)pa'a"~,,,1

=

= a'Pv.0 - a"((-l)'v,-i.a+i)

D"a"a'v,,, = a'vs.s

=

(1 - a"~")a'~,,,

+ (--l)P-la"~p-i.r+i

and IIp-l holds. with positive p, we Finally, to prove the existence of all show that if for a given p and all q there exist G-homomorphisms F, which satisfy IIIp,Q, then we can define G-homomorphisms vp+l,Qfor all q such that II,,Q holds. Since, as one verifies easily, II,,Q and 11, Q+l together imply III,+l,Q, the proof may then proceed inductively. Now, using IIIp,Q, we have ,Q

vg,qa = ( a w + D'a'k,,,a =

a'D'p,,,a

=

a"D'cp,,,a

=

awv,,,a

+ (-i)p~'a'a"~,,,+~

+ (-l)P(l a'D')a"p,,,+l + (- l)P+'D'a"p,v.a+lI + (-l)pa"vp.a+i -

so that we should put V,+l,a = D'vp,aa

+ (-l)'+lD'a"vp.a+l

all q

This completes the proof. I A slightly different version of the construction of such tpP,*'s is given in [15, Chapter 141. Suppose that A and B are G-modules. Then for every pair of integers p and q there exists a bilinear map, called the cup product (of dimension (p, q)), of

4-1-9.

Proposition.

Hp(G, A ) x H$G, B )

-

Hp+Q(G,A @ B )

Iv.

144

THE CUP PRODUCT

and a homomorphism of HP(

Proof:

G, A ) @ HQ(G, B )

-

H*Q( G, A 0 B)

Let

X be any G-complex and consider

fsf'

E

HomG

(x,A),

g , 8'

9

HomG

(xq

9

B)*

From what has gone before, it is clear that

( f + f ' )u g = f u g + f ' u g ,

(g +g') = f u g

f

and

+ (-1)YU

S ( f u g ) = Sf u g

+ f ug',

447

Therefore, cupping two cocycles gives a cocycle, and cupping a cocycle with a coboundary gives a coboundary-and it follows that we have an induced bilinear cup product of cohomology classes. This, in turn, leads to a homomorphism of the tensor product. I Strictly speaking, the cup product we have constructed depends on the G-complex X and maps HP(G, A, X )

+ HQ(G,B, X )

-

HHQ(G,A @ B, X )

However, we shall continue to be careless with the notation; eventually we shall see that the cup product is indeed independent of the complex.

-

Exercise. Suppose that X is the standard Gcomplex; then the G-homomorphisms p,,, : X,,, X, @ X, are completely determined by their action on the G-cells. Show that the pp,p)scan be constructed in such a way that

4-1-10.

= [.I 0[*I )'..)4 )= t.1 0 to1 Uq-1)) = 1.1 0 (01

~o.ll([~l) rpO,U([01

rpO.,(CrJl

*.P

P-1.1([*1)=

=

%,-l((.))

%I

*'**I

c ((.>" 0+-'I) ,.**9

[.I 0(*> 9>0

q>l

00-1)

oSG

(?'i,d[U,

71) = 0[1]

= %,-l([-I)

=

0

m1 -

c

oEG

- u[o-'] @ u[T] [ 0 ' 1 )

0 [TI>

{(u[lI - 0[u-ll)

0")

4-2. 42.

PROPERTIES OF THE CUP PRODUCT

145

PROPERTIES OF T H E C U P P R O D U C T

In this section we describe the cup product in dimension (0,O) and its behavior with respect to the coboundary arising from exact sequences. These properties serve to characterize the cup product completely, and the uniqueness of the cup product then implies its independence of the choice of a complex. Proposition. (CPl). Let A and B be G-modules; then for a E AC and b E BG we have a @ b E (A @ B)Cand

42-1.

KU

Kb

=

K(U

@ b)

Proof: Of course, we must compute with respect to some G-complex ( X , a, e, p). According to (2-2-10),

-

A is the unique G-homomorphism such that where wa : 2 ua(1) = a. Similarly, we have ~ , ( b ) = Wb& B!(G, B). I t is clear that a @ b E ( A @ B)G,so that .,(a 6b) = Wa@be B%(G,A @B); and under the identification

+

+

This result provides an explicit way of computing the cup product in dimension (0,0)-namely, it is determined by the canonical map K . More precisely, for a E Ho(G, A), fl E Ho(G. B)

Iv.

146

THE CUP PRODUCT

we compute a w B (all with respect to some G-complex X) by choosing a E AG,b E BG such that K a = a, K b = /?and noting that W = K(U @ b). 42-2.

(CP2).

Proposition.

Let A‘ be a G-module.

(i) If the G-sequences O-A’.B

o --+ A

A’%

4 - c - 0

B

c

A’%

A‘ - 0

are both exact, then, for all p and q, the following diagram commutes: Hp(G, C ) x W ( G ,A’) a*

HP+l(G,A ) x H *(G, A’)

W

Hp+*(G,C @ A’)

W

HWq+l(G, A @ A ’ )

a*

1

In other words, for y E W(G, C) and a’ E HQ(G,A’) we have S*(y

u a’)= s*y w

a’

(ii) If the G-sequences

o-A-LB-Lc-o 0-

A’

A%

A’

0B %A’

c-o

are both exact, then, for all p and q, the following diagram has commutativity character (- 1)p: Hp(G, A’) x Hq(G, C ) -.%

W

1.8.1

Hp(G, A’) x H*+l(G,A )

Hp+g(G,A’ @ C )

Id*

H**+l(G, A’ @ A )

In other words, for a’ E Hp(G, A‘) and y E H*(G, C)we have 8*(a‘ w y ) = (-l)P(a‘ w S*y)

4-2.

147


Proof: We prove (ii) in order to show how the factor (-l)p arises; the proof of (i) goes the same way. Let X be any G-complex. We observe that the coboundary maps arise (see (1-1-4) and (2-1-9)) from the exact sequence of differential graded groups 0 0

-

Horn, Horn,

--

-

(x,A ) Horn, (x, B ) %Horn, (x, c) (x, A’ @ A ) HOm, (x, A’ @ B ) (1.1 @A HOmG (x, A’ @ c) 0 (1.i)

-

(1,1@i)

0

Let f‘ E Hom, ( X , , A’) and h E Hom, ( X , , C) be cocycles representing the cohomology classes a’ and y, respectively. Choose g E Horn, ( X , , B) such that h = (1, j ) g = jg and f E Hom, (X,,, , A) such that Sg = (1, i) f = q. Then 6,y is represented by f and 01’ w 6,y is represented by f ’ w f. On the other hand, a’ w y is represented by the cocycle f’w h E Hom, (X,,, , A’ @ C). We have, therefore,

f’u h = (f‘0h)PP.* = (1 @i)(f‘ @g)P,.* =

(1

O i ) ( f ’ U E )= (1, 1 0 i ) ( f ‘ U B )

and S(f’wg)

=

S f ’ v g +(-l)”(f’wSg)

=

(-l)”(l @ i ) ( f ’w f )= (I, 1

Oq((-l)p(f’ Wf)),

so that 6,(a’ w y ) is represented by (- l)p (f‘ w.f).This completes the proof. I 42-3. Remark. In virtue of the foregoing, the cup product w is really a collection {u,,,~ I p, q E Z} of bilinear maps wp,* : HI’(G, A ) x H*(G,B )

-

Hp+*(G, A @ B )

which are related to the coboundaries arising from exact sequences by the properties CPI and CP2. It follows that for each pair ( p , q ) and any G-modules A and B we have a homomorphism of groups Q D , q : H p ( G , A ) 8 He(G, B) Hp+q(G, A @ B)

-

148

Iv.

THE CUP PRODUCT

given by %,,(a

0B) = a

=a

U P , B

”B

and these Q,, satisfy CP1 and CP2 when the appropriate notational changes are made. It is clear that statements made about u = {up,} can be transferred to Q = {Qp,p}, and conversely. Theorem. The cup product is characterized by the fact that it is a bilinear map satisfying CPl and CP2. More precisely, if @ = {OP,Jis such that for any G-modules A and B

4-24

@J,,,~ : HP(G, A ) x HQ(G,B )

-

HP+Q(G,A @ B )

is a bilinear map, and the following conditions are satisfied:

CPI: If the hypotheses of CPl hold, then @o,o(KU, K b )

=

K(Q

@ b).

CP2: If both hypotheses of CP2 hold, then we have a*@P*QtY,

a’)= @ P + I . Q ( a * Y ,

a’)

and 8*@P.&’,

-then

@=u

Y) = (-l)p@D,Q+l(a’,

S*Y)

.

Proof: We work with respect to some fixed G-complex X (since the cup product is defined in terms of X-see (4-1-8) and (4-1-9)) even though it will not appear explicitly in the proof. T o. prove that According to CPl, we have @o,o = w ~ , ~ @p,p = up,* for all (p, q) we proceed inductively via the use of dimension shifters. Thus, suppose that for given (p, q) = up,* . Then it follows from (4-1-4) and (4-1-6) that for any fixed G-modules A and B the following are exact G-sequences which split as Z-sequences:

0-2-T-J-0 O-B---+B@r--+B@]-O O-A@B-A@B@r-A@B@

1-0

4-2.

149


Since B @ I' and A @ B @ r are both G-regular, the maps 6, in the following diagram are isomorphisms onto:

-

Hp(G, A ) x HQ(G,B @ 1) 1.8.1

HP( G, A ) x H*+'(G,B)

H*@(G,A @ B @ J)

b*

Hf)+*+l(G, A @ B )

From CP2 we know that this diagram has commutativity character (-l)P when both horizontal maps are given by w or by @. Consequently, = wp,qimplies that @p,q+l = . The proof that @ p , q = up,q implies @p,p-l = wp,Q--l is based on the exact G-sequences 0-B@I

-B@r

-B-0

O-A@B@I-A@B@I'-A@B-O

and the diagram Hp(G, A ) x HQ-'(G,B)

___t

IX8*1

Hp(G, A ) x HQ(G,B @ I )

while the proof that @ p , q = based on the exact sequences

-

HP+Q-l(G,A @ B )

18*

(-W

Hp+g(G,A @ B @ I )

implies

@p-l,q

=

up-l,q is

and the diagram HP-l(G, A ) x H@(G, B) 8* X 1 1

HP( G, Z @ A ) x HQ(G, B )

-

Hf'+Q-'(G,A @ B )

18*

HP+@( G, I @ A @ B )

Finally, the details of the proof that op,* = up,* implies @p+l,q = wP+',* are left to the reader. I

Iv.

150

THE CUP PRODUCT

It is clear that this result can be generalized to a formal statement about conditions under which multilinear maps of cohomology groups are identical; however, we shall not do so. Instead, when the occasion arises, we shall, at best, provide a sketch of a proof based upon the principles used in the proof of (4-2-4). 42-5.

Corollary.

The cup product is independent of the

choice of G-complex. Proof: The statement means that if X and X' are G-complexes with associated functions {p7p,q} and {p7k,g} in terms of which the cup products for cohomology groups w and a,respectively, are defined, then

-

is a commutative diagram for all p, q, A, B. T o prove this, consider H$+g(G,A @ B) given the maps @p,q : H$(G, A) x H:(G, B) It is easy to verify that x ao by @p,q = 1;t'o @ = { @ p , g } is a bilinear map which satisfies CPl and CP2; therefore, @ = w. I I t may also be noted that independence of the complex applies to the behavior of cup products with respect to coboundaries. More precisely, under the hypotheses of part (i) of CP2, if y E H;(G, C) and a' E H$(G, A'), then 1x,x4*(y w a') = 6*(lx,x,y a lx,xa')and an analogous statement is valid in case (ii) of CP2. 4-26.

Proposition.

words, if a E Hp(G, A ) ,

The cup product is associative. In other

E H*(G, B), y E Hr(G, C),then

4-2. Proof:


Define maps

0,Y : Hp(G, A )

x HQ(G,B ) x H'(G, C )

151

-

Hp+Q++'(G, A @B @ C)

by @(a, 8, y ) = (a u 8) u y and Y(a,8, y ) = a w (8 w y). These maps are multilinear-that is, linear in each variable. For a E A G , b E B G , c E C G we have @(Ku, ~ bKC) , = K ( U @ b @ c), so that @ satisfies the analog of CPI. Of course, Y satisfies the same relation, and, in particular, Y = @ in dimension (0, 0,O). Furthermore, Qi satisfies the analog of CP2 which would have three parts in this situation. T o be explicit, part (iii) would read as follows: if the G-sequences

0

-

O-A-B-C-0

A' @ B' @ A -A'

@ B' @ B

-

A' @ B' @ C -0

are both exact, then the diagram Hp(G,A') x HQ(G,B') x Hr(G, C ) 0Hp+*+'(G,A' @ B' @ C ) I

X I

xs*l

Id*

(-l)P+Q

Hp(G, A') x HQ(G,B')

x Hrf1(G,A ) 0HP+Q+'+l ( G , A ' @ B ' @ A )

has commutativity character (- l)P* for all (p, q, r)-in words, with obvious choice of notation, @(a',8', 6,y) = ( -1)P+Q8*@(af,

other

/9', y).

As for parts (i) and (ii), their conclusions would be

respectively. Again, Y satisfies the same relations. Therefore, by the method used to prove (4-2-4), we conclude that @ = Y. For example, the proof that Qip,q,r = Yp,q,rimplies @= q+l,r = !Pp,q+l,r is based on the exact sequences

Iv.

152

THE CUP PRODUCT

and the diagram

This completes the proof-with reader. I

the missing details left to the

Let X be a G-complex and for every (p, q, I) consider the trilinear map

-

Horn, (X,,A ) x Horn, (Xu, B ) x Horn, (X,,C)

Horn, (Xs+a+r A 9

O B O C)

given by (f,g, h) +(fw g) w h. For notational convenience put ( f w g ) w h = f w g w h. Then S ( f v g uh) = Sf ug v h

+ (-1)PfW

Sg v h

+ ( - l ) ” + ” f ~ g v Sh

and this mapping induces the map of cohomology classes (a w j3) w y. In the same way, we have

(a,j3,y )

__+

with corresponding coboundary formula and induced map of cohomology classes (a,fi, y ) +a u (fi w y). Of course, (a wfi) w y = a w (j3 w y ) so that we may write a wfi w y without ambiguity. The question of equality of (fug) w h and f w ( g w h) is left to the reader. Bx A For any sets A and B there is a pairing T : A x B called the twist, which is given by T ( a , b) = (b, a). In particular, if A and B are G-modules, there is a canonical G-isomorphism (which we shall also call the twist and denote also by T) T :A @ B B @ A for which T ( a @ 6 ) = b @ a . Then, for every integer n, T induces an isomorphism of cohomology groups

-

_ +

T* :Hn(G, A

B ) >-

H”(G, B

A)

4-2.


153

If the roles of A and B are reversed, we still use the same notation, and this will cause no confusion. 4-2-7.

Proposition.

any (p, q) and any

Proof:

Let A and B be G-modules. Then for E H*(G, B) we have

a E Hp(G, A),

The map

given by

that @ = (-1)m T* 0 u o T) is bilinear and satisfies CPl and CP2. The verification of CP1 makes use of the commutative diagram (so

(B @A)-

-

Ho(G,B @ A )

T

( A @ B),

T*.Ho(C,A@B)

while the verification of CP2 makes use of T* 0 6, = 6 , o

-

T* .

I

Suppose that Q : A +A' and B' are G-homomorphisms of G-modules, so that Q I,4 :A @ B A' @ B' is a G-homomorphism. Then for a E H*( G, A) and fl E H*(G, B) we have

4-24 I,4 : B

Proposition. __t

Fix a G-complex X, and let f E Hom, ( X , , A) and Proof: g E Hom, (X,,B) be any cochains. Then

Iv.

154

THE CUP PRODUCT

since Vf u *g = (d0 +g) 0 P)P.U = (TO *)(f 0 8 ) V9.u = (F 0 #)(fu g) 0

In other words, the following diagram commutes:

and the desired result follows easily.

I

Let X be a G-complex and let A denote the cochain cup product. In other words,

4-2-9.

Exercise.

is the bilinear map d(f,g) = f ug. Then A , which induces the cup product u of cohomology classes, satisfies the properties:

-

Moreover, if d' is any bilinear map (more precisely, A' = {A;,*} is a set of such maps) satisfying (CCPl), (CCPZ), and (CCP3), then HP+q(G, A €3 B) the bilinear mapping H*(G, A) x H*(G, B) which it induces on cohomology classes is precisely the cup product. Strictly speaking, the symbol kx which appears in (CCPl) has never been defined. What is meant is kxu = wae, a E AG (compare (2-2-10)), so that kx : A'

-

Px(G, A )

is a homomorphism whose image contains cocycle representatives of all the cohomology classes in Ho(G, A).

4-3. 43.

155

CUP PRODUCTS FOR A PAIRING


The cup product discussed in the preceding section is not quite what we want; a slight extension is needed. Suppose that A, B, C are G-modules and consider a C-pairing 8 :A x B C. By this we mean that 8 is bilinear and that

-

- -

qua, ub) = ue(a, b)

E

G,

a E A,

b EB

This latter condition may be expressed as 8 (u x u) = o o 8. If we let 7r : (a,b) a @ b denote the canonical map of A x B A @ B, then there exists a unique homomorphism 0

~ : A @ B - c

such that 8 o 7r = 8-or what is the same, 8(a @ b) = 8(a, 6). Moreover, on the generators a @ b of A @ B we have, for u E G, B+ 0 b) = B ( @~ ub)~ = qua,ub) = ue(a, b) = .8(. 8 b), so that 8 is a G-homomorphism. As natural examples of G-pairings we have:

-

(1) If A is any G-module, define 8 : Z x A A by d(n, a) = nu. This G-pairing will often be denoted by 6’”. (2) If A and B are any G-modules, let O:Hom(A,B)x A - B

be given by O(f, a) = f ( a ) . Note that this is a G-pairing since e(p, ua) = j q u a ) = .f(a) = ue(f, a).

-

Of course, to every G-pairing 8 : A x B C there is assoC which is the G-pairing given by ciated its twist OT : B x A qb,

= qa,b)

__+

-

-so that OT = 8 T and OT = 8 0 T. It is also worth noting that a G-pairing 6’ : A x B C gives rise to G-homomorphisms of G-modules. Thus, for a E AC,define a 0 : B-Cby aqb) = e(a, b) 0

Iv.

156

THE CUP PRODUCT

-

Since me(&) = 8(a, ~ b =) 8(ua, ~ b =) &(a, b) = A ( b ) , a8 is indeed a G-homomorphism. I n the same way, for brz BC, 8b :A C given by @(a) = O(a, b)

-

is a G-homomorphism. If 8 : A x B C is a G-pairing we define the cup product with respect to 8 (or simply the @-cupproduct) : H'(G, A )

x H'(G, B ) +H"'(G, C )

for all ( p , q) by putting (with 8, = (8),) a we

-in

other words,

=

Hp(G, A ) x HQ(G,B )

B = o*(a

-

8,o

v

w; that is,

B) is the composite map

Hptu(G,A 0 B ) 5 H*'(G, C)

The properties of w carry over to W e , as will be seen in the next few results. Of course, the &cup product W e on cohomology classes is induced from a &cup product (also denoted by we) on cochains. In more detail, let X be a G-complex, and suppose that f~ Hom, ( X p, A), g E Hom, (X,,B ) are cochains. Then define f we g E Horn, (Xp+q 9 C ) by f w e g =Jo(.fwg)=

8o(fOg)ope,q

so that on cochains

= 8 0 u,and it follows easily that it induces we on cohomology classes. It should be noted that we is a generalization of w; in more A @ B given by O(a, b) = a 6b detail, the map 8 : A x B is clearly a G-pairing for which 8 is the identity, so that we = u on both cochains and cohomology classes.

-

43-1.

Consider the diagram

Proposition.

B'

B

C'

4-3.

157


-

where u, o, w are G-homomorphisms and 8 is a G-pairing. Then 8' : A' x B' C' given by 8' = w 0 8 0 (u x w ) is a G-pairing, and for a' E Hp(G, A'), 8' E H*(G, B') we have a'

Wg'

8'

= w*(u*(Y'

w g

w*8')

It is clear that 8' is a G-pairing, and that

Proof:

8'=

W O ~ O ( U @ W )

and

8; = w , o ~ , ~ ( u @ w ) ,

Hence, wg'

=

8;

0

w

=

w*

= W* = W*

which is the desired result.

8*0 (u @ w)* w o 8, o uo (u, x w*) (by 0

0

0

V g

0

(u*

(4-2-8))

x w*)

I

Let 0 : A x B +C be a G-pairing, Proposition. then for a E Hp(G, A), /IE Hq(G, B) we have 43-2.

Proof:

In virtue of (4-2-7),

which is the desired result.

I

Proposition. The cup product with respect G-pairings is associative; more precisely, if

43-3.

to

158

Iv.

THE CUP PRODUCT

for a, E A , ,a, E A , ,a, E A3 then for a1 E W(G, A J , a2 E H’J(G,A2), a3 E Hr(G, A,) we have

4 U,O,,,

( a 1 VO,,

Proof:

~g =

4 wel,la

(.z w e a , a 4

The hypotheses imply that B12.3[B1,2(a1

@

@ as] = Bl.23[‘l @ B2,3(a2 @ %)I

-

Define the G-pairings d&,3 : ( A , @ A,) x A , A123 and %,23 : x @ A3) by %2,3 = e12,3 (d1,2 x and d ; , s 3 = d1,23 0 (1 x B2,3), respectively. We have, therefore, Oi2.3

so that

= B12,3

=

(4.2

(8;,23)*.

__+

@I)

= O1.23

(l @ e2,3) = &,23

Now, by (4-3-1) and (4-2-8), it follows

that

Proposition. Let 6 : A x B -+ C be a G-pairing; then for a E AG,b E BC we have O(a, b) E Cc and

4-34.

KU

Kb

= K(e(U, b))

It is clear that O(a, b) E Cc and then from (4-2-1) and Proof: (2-2-10) it follows that

4-3-5.

Exercise.

The preceding result is a generalization of

CP1 (see (4-2-1)) to &cup products. State and prove the appro-

4-3. CUP PRODUCTS FOR

159

A PAIRING

priate generalization of CP2 (see (4-2-2)) for the case of cup products with respect to G-pairings. Proposition. Cupping with respect to a G-pairing by a 0-dimensional cohomology class is the same as applying an induced map of cohomology groups. More precisely, if 0 :A x B C is a G-pairing, then

4-36.

-

(i) for a E AG,8 E W(G, B) we have KO W e

(ii) for

a E Hp(G,

S = (ae)*(P)

A), b E BG we have a

~6 = (eb)*(a)

Choose a G-complex (X, a, e, p ) and letg E Hom, (Xq.,B) be a cocycle representing 8. According to (2-2-3), (1, “0)g is a cocycle representing (“d), (8), and in virtue of (2-2-10) w,e is a cocycle representing KU. Thus, U,E weg is a cocycle representing ~a w e 8. Now, for x E X , we have

Proof:

(W&

weg)(x)

Ogboa(x) = &Ja

=

OR)(.0Ibo.*(x)

and ((1, ae)g)(x) = “@M) = %4 g(x)) = =4

4 l ) Og(.))

=

40 0g(x))

4% Og)(l 0 4

But, as seen in the proof of (4-1-8), (e @ 1) y0,Xx) = 1 @ x for the yo,, defined there. Therefore, w,e Wag = (1, “0)g and part (i) is proved. The proof of (ii) is now immediate: a

~b = (-1)O~b U ~ T a

-

= (b((eT)),(a) = (&)*(a).

I

Now let us turn to the cohomology of an arbitrary subgroup H of G. Let X be a G-complex with associated G-homomorphisms yPpn: X,,, X , @ X, in terms of which the cup product is defined. Then X is also an H-complex and the functions yp,qare H-homomorphisms which may be used to define the cup product u for the cohomology of H. Of course, all tensor products are

160

Iv.

THE CUP PRODUCT

-

over 2, so they are not affected by a change from G to H. FurtherC is a G-pairing, then it is also an more, if 8 : A x B H-pairing, and we can define We (for both H-cochains and H-cbhomology classes) in precisely the same way that was used for G.

-

Proposition. Let H be any subgroup of G, and C is a G-pairing. Then, writing suppose that 0 : A x B res = resG+, and cor = corH+G we have:

43-7.

(i) res (a

j?) = res a

res j?

(ii) cor (res a uep) = a we cor /3 (iii) cor (a

res p) = cor a

p

a E Hp(G,

A), j? E H*(G, B)

E Hp(G,

A), j? E Hg(H, B)

01

aEW(H, A),

j? E H*(G, B )

Proof: (i) Choose a G-complex X with tppA's, and use X for the cohomology of H, too. Let f ~ H o m , ( X , , A ) and g E Hom, (X,,B) be cocycles representing a and /I, respectively. According to (2-3-4), f represents res a and g represents res j?. It follows that a /I, res (a we j?), and res a wgres j? are all represented by the cocycle f g = 8 0 (f@ g) 0 vpA .

(ii) Let f E Hom, (X,,A) represent a and g E Hom, (X,,B) represent j?. Thenfrepresents res a and 8 (f @ g) 0 tppA represents j?. According to the definition of cor (see Section 2-4), res a SH+,g represents corj?, so that a carp is represented by 8 o (f 6Sg,) o tpP . On the other hand, if we write G = otH and note that f, and tppA are G-homomorphisms, then cor (res 01 we j?) is represented by Q

u

e"

This proves (ii). The proof of (iii) goes the same way.

1

4-3.

-


161

Theorem. Let 8 : A x B C be a G-pairing and fix an element a E Hp(G, A). For each subgroup H of G and each integer q, define a map

4-34

by &cupping on the left by resG+Hol-in other words,

Suppose there exists an integer qo such that for all subgroups H of G we know that (i) OH: Hqo-l(H, B )

HP*o-l(H, C ) is an epimorphism,

-

__t

(ii) OH: Hqo(H, B) +HP*o(H, C) is an isomorphism onto, (iii) OH: Hqo+'(H, B )

HP*o+'(H, C ) is a monomorphism.

Then G j H is an isomorphism onto in all dimensions and for all subgroups H of G. Proof: By induction on p. Consider first p = 0. If we choose a E AC such that K ~ ( u )= a, then, according to (2-5-5), #(a) = resG+Ha. Now, 8 is an H-pairing, and application of (4-3-6) to it yields OH = ("e), . Furthermore, "8 : B +C is a G-homomorphism which, in virtue of (3-6-3), is a cohomological equivalence. This proves our assertion for p = 0. T o pass from p to p 1 we use dimension shifting techniques. Suppose the result holds for p. Consider the exact G-sequence (see (4-1-6))

+

and the commutative diagram

o-c-rgc-

J@C-O

Iv. THE

162

CUP PRODUCT

whose rows are exact G-sequences. Then for any H C G we have the diagram Hp(H, J 0A ) x H*(H, B ) a*

X I

1

H*l(H, A ) x H *(H , B)

W

H**(H, J

v

0A @ B )

18*

-

H**+'(H, A @ B ) L

The rectangles commute by (4-2-2) and (2-2-S), and the coboundary maps 8, are isomorphisms onto since F @ A, F @ C and F @ A @ B are H-regular. Now, define the G-pairing 8' : ( J @ A) x B J @ C by 0'(j @ (I,b) = j @ e(a, b). It is clear that 8' = 1 @ 8 so that ( p ) , = (1 @J 8), and the composite map in the top row is (1 @ 8), 0 w = we. . To prove our result for p 1, let a E Hp+'(G, A) satisfy the hypotheses. Let a' E Hp(G, J @ A) be the unique element for which &a' = a. Then for any subgroup H of G,

-

+

and the following diagram commutes for all q,

V

ll

HqH, B)

*Ii

V

f

8*

H**+l(H, C )

where @&?) = res, jH a p and @&@) = res, j H a' u e , p. Since the hypotheses are satisfied for @ , , (using a,p 1, qo , 0) they are satisfied for 0; (using a', p , qo ,el). By the induction hypothesis, @ is an isomorphism I onto in all dimensions and for all H, hence 1. the same is true for Q H , and our result holds for p The passage from p to p - 1 proceeds in the same manner. I

+

+

4-4. THE

-

163

DUALITY THEOREM

-

Exercise. Let O : A x B C be a G-pairing of G-modules. If H is a normal subgroup of G then 6 induces a CH.Consider (G/H)-pairing of (G/H)-modules 8' : AH x BR a E (A*)"/" = AC and /3 E HQ(G/H,BH) with q 2 1; then

43-9.

inf,,,,

(K"/"(u)

we,8) = K

infcIH4

~ U we )

8

A similar relation holds for the other variable. Furthermore, if a E Hp(G/H, A H )with p 2 1, then infGIH-1G

44.

8) = i n f G / H 4

OL " 9

8

infGIH-G

T H E DUALITY THEOREM

Let us consider any G-modules B and C. As was observed in the preceding section there is a canonical G-pairing 8:Hom(B,C) x B - C

such that for 4 E Hom (B, C) and 6 E B we have If B' is another G-module, we shall also write

O(4,

b)

= 4(b).

8:Hom(B',C) x B ' - C

with O(+', 6') = +'(6') for the canonical G-pairing; this should cause no confusion. For any $ E Hom (B, B') we have an adjoint map, which is a homomorphism of groups,

$ = ($, 1) : Hom (B', C)

-

Hom (B, C)

Since for 4' E Horn (B', C) we have $4' = ($, 1) 4' = 4'$, it follows that (with 6 E B) O(&', 6 ) = O(t$', $6). Suppose further that $ is a G-homomorphism, $ E Horn, (B, B'); then (see (1-4-3)) 1+8 is a G-homomorphism, and we may define a G-pairing 8, : Horn (B', C) x B C

-

by putting Ob(4', 6) = O(+', $6) = O(&', 6 ) = #'($b) = (&')(6); in other words, 8, = e o ( i x $1 = e o ( $ x I )

Iv.

164

THE CUP PRODUCT

G, O , ( p , ob) = p ( + ~ b = ) +'%(+b) = ~ + ' ( + b )= of&(+', b), 0, is indeed a G-pairing. Consequently, 0, determines a cup product For

E

so that

vgs : Hp(G, Horn (B', C ) ) x Hq(G, B )

-

HH'J(G,C)

which is induced by the corresponding cup product on cochains, and can be computed according to the following rule. 441.

Proposition.

If and

.f E HP(G, Horn (B', C ) )

/3

E H*(G, B )

then

. f ~ , , B = J * . f v e B = 5'u,**B Proof: Fix a G-complex X with associated functions vPa. It clearly suffices to prove that i f f ' E Hom, ( X , , Hom (B', C)) and g E Horn, ( X , , B) are p - and q-cochains, respectively, then

f' we, g = $f' v

9

g=

f' We *g

Now, let us observe that

f' w9,g

=

Uf'Ogb,.,?

Jf' v 9 g = 4 J @ l ) ( f ' Ogb,.,

andf' we+g = &l @ +)(f'@g) tpP,*, so it suffices to show that

8,=Bo(*@1)=00(1@*) -in

other words, that the following diagram commutes: Horn (B', C )

B'

7 99

Horn (B', C ) @ B

Horn (B, C )

b

C

B

But this is immediate from the definition of 0,. This completes the proof. I

4-4.

165

THE DUALITY THEOREM

For each pair of integers ( p , q) let us consider the homomorphism

-

Yp,q: HP(G, Horn (B, C ) )

Horn (Hq(G, B), Hp+q(G,C ) )

which is defined by taking the image of ,$ E Hp(G, Horn (B, C)) to be the homomorphism whose action on Hg(G, B) is given by 8-cupping on the left by ,$-in other words, (YpA(B) =

442.

6 we B

B E Hq@, B )

Suppose that

Proposition.

0 - B ' L B

L B " - O

is an exact G-sequence and that C is a G-module for which the adjoint G-sequence

0

-

-

Horn (B", C ) s-0.1) Horn (B, C )

Mi.1)

Horn (B', C)

__t

0

is exact. Then the following diagram has commutativity character (- 1)p-1: Horn (H'J(G,B'), H*+$G, C ) )

Hp(G, Horn (B', C ) ) 6*

1

(-

I@*

1)P-1

.1)

Horn (Hq-l(G, B"),Hp+q(G, C ) )

HP+l(G, Horn (B",C ) )

We must show that for Proof: j?"E Hq-'(G, B") we have S,cf

u

g

fl"

=

5' E Hp(G, Horn (B', C)) and

(-1)P-lcf

W g

8.J"

Choose a G-complex X. Let g" E Horn, (X,-, ,B") be a cocycle representing 8". Then there exist g E Horn, (X,-l ,B) such that g" = (1, j )g = jg and g' E Horn, ( X , , B') such that Sg = (1, i)g' = %', and g' is a cocycle representing S,fl". Similarly, iff ' E Horn, (X,, Horn (B', C)) is a cocycle representing .f, then there exist f E Horn, ( X , , Horn (B, C)) such that f ' = ( 1 , i ) f = if

and f " E HornG(X,+,, Horn (B", C)) such that Sf = ( 1 , f ) f "

=r,

Iv.

166

THE CUP PRODUCT

and f“ is a cocycle representing S,f. It now follows that S,f‘ (- 1)P f‘ we = 0 since it is represented by

+

S,r

f” we g” + (-

1)”f’ we g‘ jg (-1)P;fVeg‘ = jf ” g (-1)”f We Q’ = af w e g (-l)”fu, = W we 8)

r

+

=fa

+

(see proof of (4-4-1))

+

which is a coboundary. This completes the proof. 443.

-0

I

Suppose that for a fixed pair of integers

Proposition.

-

( p o , qo) and a fixed G-module C the mapping Y,,o.Qo:H”o(G, Horn ( B , C ) )

Horn (Hqo(G,B), H”o*o(G, C ) )

is an isomorphism onto for every G-module B. Then for all (p, q) with p q = po qo , and all B, the map YP,*is an isomorphism onto.

+

+

Proof: If the result is true for (p, q) then, using dimension shifting, we show that it is true for ( p 1, q - I). For any B,

+

O-I@B-r@B-B-O

is an exact G-sequence which splits as a Z-sequence and has G-regular middle term (see (3-3-8), (4-1-4), and (4-1-5)). By (3-3-5) and (3-3-6), for any C 0

-

Horn (B, C )

-

Hom(I‘@ B , C )

-

Horn(Z @ B , C ) -0

is an exact G-sequence with G-regular middle term. Thus the vertical maps in the following diagram (which, by (4-4-2), has commutativity character (- 1)P-l) are isomorphisms onto: Hp(G,

Horn ( I @ B, C ) ) 8*

HP+l(C,

I

Horn ( B , C ) )

Horn (Hq(G,I @ B), H*(G, C ) )

(-

1)P-1

1 ~ )

Horn (Hq-l(G, B), Hp+q(G, C ) )

4-4.

167

THE DUALITY THEOREM

Consequently, if Y p , is an isomorphism onto, then so is Y,+l,q-l . The details of the passage from ( p , q) to (p - I, q 1) are left to the reader. 1

+

444.

so that

Proposition.

K ~ Ho(G, E Hom

Suppose that

Yo,u(d)= .f* : Hq(G,B ) Proof:

If

-

(B, C)). Then for all q, Hq(G,C )

b E Hq(G, B), then

Suppose that C is a G-module on which the action of G is trivial and whose additive group is divisible (see (1-6-7)); then for all q and all G-modules B

U S .

Theorem.

Y-D,D-l : H-p(G, Horn (B, C ) )

-

Horn (Hp-l(G, B), H-l(G, C ) )

is an isomorphism onto. Proof: By (4-4-3), it suffices to prove that Yo,-l is an isomorphism onto. For this it is useful to observe that any 5~ Ho(G, Hom (B, C)) can be written as 5 = Kf for some f E (Hom (B, C))"= Hom, (B, C) and B E H-'(G, B) can be b some ) b E B, , so that the action of Yo,-l written as fl = ~ ~ (for is described by (YO,--l(Kf))(?Bb) = f*(?Bb)

-

=

V d f (4)

(by (4-4-4)) (by (2-2-7))

Moreover, because the action of G on C is trivial, the map : C, H-l(G, C) is an isomorphism onto. Suppose that Yo,-l(~f) = 0. Then f(b) = 0 for every b E B, ,

r),

Iv.

168

THE CUP PRODUCT

and we may define an ordinary homomorphism go : SB -+ C by putting go(S4 =f@)

brzB

Since C is divisible, go may be extended (see (1-6-7)) to g E Hom (B, C). Now, for b E B we have (Sg)(b)= Cg"b = Cgu-1b = g(Sb) = g,(Sb) = f ( b ) . U

0

-

Therefore, Kf = K(Sg) = 0 and Yo,-'is a monomorphism. Finally, if v E Hom (H-'(G, B), H-'(G, C)) is given, then we may define a homomorphism fo :B, C, by fo = '7; o v o y B and extend fo to f E Hom (B, C). Then f E Hom, (B, C ) since f(ub) - uf(b) =f(ub) -f(b) = f ( ( u- 1)b) = 0, and Yo,-l(~f) =v = rlc(f(b)) = rlC(fO(b)) = Y(rlBb). since for b E B, (YO,--I(Kfl)(rlBb) Thus, Yo,-' is an epimorphism, and the proof is complete. 1 The preceding result may be applied to C = Q/Z. For any G-module B, we denote its dual by B = Hom (B, Q/Z), and call the elements of characters of B in Q/Z. If G has order n, then TO/Z

:

(;1 z)/z = (Q/Z)s

H-'(G, Q/Z)

+

-

is an isomorphism onto. Therefore, the map (1, .I$~) Hom (Hp-l(G, B), H-'(G, Q/Z)) into Horn (Hp-'(G, B),

;(1 Z)/Z)

= Horn (H*l(G,

of

B), Q/Z) = H*'(G, B )

is also an isomorphism onto. We have therefore,

446. Corollary. (Duality Theorem). B and any integer p the mapping

For any G-module

is an isomorphism onto. This isomorphism associates with [ E H-p(G, E)) the character of HP-'(G, B) in Q/Z which takes

-

169

4-4. THE DUALITY THEOREM &z(f G-pairing. __t

447.

w e b)

Corollary.

where 8 : I? x B

Q/Z is the canonical

(Integral Duality Theorem).

p, we have an isomorphism

For every

H-p(G, Z ) = W ( G ,2)

More precisely, to 6 E H-p(G, Z) this isomorphism associates the character of Hp(G, Z) in Q/Z whose action on 4 E Hp(G, Z) is given by

5

-

-1 8-1 ?QlZ

* (6"Oz 5 )

where Bz : Z x Z -+ Z is the natural G-pairing &(m, n) = mn and 6, :H-p-l(G, Q/Z) -+ H-p(G, Z) is the coboundary arising from the exact G-sequence (with trivial action of G)

Proof: According to (3-1-13), 6, is an isomorphism onto. The identification of 2 and Q/Z permits us to identify H-P-l(G, Q/Z) and H-P-l(G, 2). Therefore, by applying (4-4-6) with B = Z we conclude that

(I, ?&)

0

Y-p-l.98,' 0

-

: H-'(G, 2) -+ HqG, 2)

is an isomorphism onto, and the element of Hp(G, Z) corresponding to 5 E H-P(G, Z) maps

-

where 0 denotes both the canonical G-pairing

e : Horn (Z, Q l Z ) x Z

QlZ

B(f, a) = f(a), and the natural G-pairing 8 : Q/Z x Z

__+

B(a, n) = an. Note that under the correspondence f-f(l)

Q/Z

Hom (Z, Q/Z) and Q/Z the two 0's correspond to each other.

of

Iv.

170

THE CUP PRODUCT

Thus it remains to show that

8 2 5 we 5

=8

3 5 we2 f)

- o r , since 6, is an isomorphism, that 6,(6;’[ Now, tensoring the exact G-sequence

0-

2 -Q-QQlZ

5) = I wez 5.

-0

with Z yields the same sequence (after identification), so that application of (4-3-5) gives the desired result. This completes the proof. I

448.

If S1 , S2 E H-p(G, Z) then

Corollary.

= 5 2 e* 51 we2

51

5

=

52

5

for all 5 E H*(G, Z). 45.

THE NAKAYAMA MAP

In this section, we shall use the standard G-complex X exclusively (therefore, we shall omit all reference to it) and carry out some explicit computations. These revolve about the Nakayama map (which is sometimes referred to as the “Japanese homomorphism”) and its connection with a certain cup product. 4-54.

Proposition.

which takes f

-

-

For any G-module A, the map of

Horn, (I, A )

3Y1(G, A )

ut where

is an isomorphism onto. Proof: ForfE Hom (I,A), let us define the standard 1-cochain ut by the given formula. Note that because f is a homomorphism

4-5. THE

171

NAKAYAMA MAP

u,([I]) = f ( 0 ) = 0. Since {(u - I)[ u # I> is a 2-basis for I, it is clear that f *--+ u, is an isomorphism between Hom (I,A) and the group of standard I-cochains u for which u([l]) = 0. It should also be noted that if a I-cochain u is a cocycle, then it is automatic that u([l]) = 0; this is immediate from the coboundary formula 0U[T] - U [ U T ] u[u] = 0. Moreover,

+

is a cocycle

u,

--

- U f [ U T ] + u,[7] = 0 for all u,7 E G uf(7 - 1) -f(m - I) +f(u - 1) = 0 e j ( u ( 7 - 1)) = uj(7 - 1) UU,[T]

f is a G-homomorphism

tj

w

I

f E Horn, (I,A).

According to (3-3-9) we have, for any G-module A, the dimension shifting exact G-sequence 0

-

A

%

Horn (Z, A)

1- (. ,1)

Horn (F, A)

3Horn (I, A )

-

0

(#)

-

with middle term G-regular. It follows, in particular, that 6 * : Ho(G, Horn ( I , A))

H1(G, A ) % H1(G, Horn (Z,A))

is an isomorphism onto. We are now in a position to describe S* explicitly. 45-2.

Proposition.

Consider

f E (Horn (I,A)), = Hom, (I,A)

and the corresponding U, E

P ( G , A) C Horn, (XI , A) % Horn, ( X , , Horn (2, A))

If a E H1(G,A) is the cohomology class of u, , then ~f and correspond to each other under 6, ; in other words, S*(Kf)

= -a

-a

Iv.

172

THE CUP PRODUCT

T o compute 6, we must go back to the exact sequence Proof: of differential graded groups 0

__+

Horn,

(x, A ) !S Horn, (x, Horn (z, A))

4

-

2Horn, ( X , Horn (P, A)) -@+ Horn, ( X , Horn (I, A ) )

0

Let us write &f = c E Horn, ( X , ,Horn (I,A)), so that c[.] = f . Then there exist b E Horn, (X, , Hom (F, A)) such that (I, i ) b = c and a E Horn, (X, ,Horn (Z, A)) such that (1, El) a = Gb-so that G * ( K ~ ) is the class of the 1-cocycle a. Now, let us put b[.] = g E Hom (I', A), so that i'g

= i(b[.])= c[.]

=f

-

Thus, g i =f,andf is the restriction of g to I. Since I' = I @ Z 1, we may reverse matters and define g to be the unique extension off to r for which g(1) = 0; then define b by 4.1 = g, so that b and a satisfy all the conditions mentioned above. Next, we note that El 0 a = 6b E Horn, ( X , , Hom (r,A)) is determined by its action on the I-cells, and 0

66([u])= b(u[.] - [.I) =

- bC.3 = gu - g E Horn (P,A )

But g" - g vanishes on I since g(. - 1) -so

that

=f(7

- 1) =f"(.- 1)

= g"(7 - 1)

(by (#)) there exists, for each u E G, h, E Horn (Z, A) such = g" - g. It follows that t(a[u])= (Sb)([u])= a,

all

(I

EG

and then a[.] = h, E Horn (Z, A). It remains to determine which cocycle in Hom,(X,, A) corresponds to a E Hom, (XI , Hom (Z, A)). For this we compute, ha(1) = h,(e( 1)) = (&,)( 1) = r(l) - g( 1) = ugu-'( 1) - ug( 1) (since g( I) = 0)= ug(u-' - 1) = uf (u-l - 1) = f(I - u) = -ur([u]). In other words, a corresponds to the cocycle - u r , so that &(Kf) = -01, and the proof is complete. I

4-5.

173

THE NAKAYAMA MAP

I n virtue of (4-5-2), given a E H1(G, A) % H1(G, Hom (Z, A)) there exists f E Horn, (I,A) such that -6*(~f) = a, and by (4-5-1) there is an associated 1-cocycle ut E Horn, (XI, A) which represents a. Conversely, any cocycle u E Horn, (XI,A) representing OL is of form u = ut for some f E Horn, (I,A). We wish to examine the cup product of a with any element of W2(G, 2)-which element can be expressed (according to (3-5-2)) as 5, = 8_*lq(u- 1) for some u E G. We shall also make use of the formula given in the proof of (4-4-2), and apply it to the case B' = I, B = r, B" = C = Z.

-

Proposition. A be the natural Let 8 : A x Z G-pairing, O(a, n) = na. Then the cup product

45-3.

we: H1(G,A )

-

x H-2(G,Z)

H-'(G, A )

is given explicitly by the formula

where u Proof:

= ut

is any cocycle representing

a, and f E

Hom, (I,A).

For the purposes of this proof, let

O':Hom(Z,A) X I - A

and

B":Hom(Z,A)x Z - A

denote the canonical G-pairings; in particular, 8" m 8. We have then, for any u E G (from the identification) (by (4-4-2), with p = 0) (by (3-5-2)) (by (4-3-5)) (ffl= f ) (by (2-2-7)) (by definition of uf)

Iv.

174

THE CUP PRODUCT

Suppose that 6 : 2 x A -A is the natural G-pairing, d(n, a) = nu. Then we have the cup product we: H-Z(G, Z)x W ( G ,A )

-

i P Z ( G ,A )

all q

and the relation

This indicates that in studying cupping by lo it does not matter very much whether we cup on the left or the right. Of course, dT is then the G-pairing used in (4-5-3), so that, in particular, for q = 1, we have

-

-

Given any G-module A, we have then for each u E G and each 5, wea of H*(G, A) H,w2(G, A). integer q a mapping a Our objective is to give an explicit cochain description of these 1, a homomormaps for q 2 1. Let us define, for u E G and q phism u#

: Horn, ( X , , A )

-

Horn,

A)

of standard cochain groups. Consider I( E Hom, (X, ,A) and define O#U on the (q - 2)-cells (and then extend linearly to get U#U E Hom, (X,-, ,A)) by the formulas: (.#U)( * * * I or)

00102

a * *

o r g(or+l i - * -or+,) i

How is this cup product related to the one we have defined (in Section 4-1) via the ppq)s ? Can it be extended to all dimensions Y and s ? 4-6-13.

then for

Letf : A

-

A' be a G-homomorphism of G-modules,

5 E H*(G, Z) and OL E H'J(G,A ) f*(t " e A

-in

a) =

5 weAf4.1

other words, the following diagram commutes: Hfl(G,Z)

Hp(G, Z)

We also havef,(ar

x HQ(G,A )

f*l

2HP+q(G,A ) U

If*

U

x Hq(G, A') -% W+q(G,A')

wo; 5) = ~ * ( o LuB; ) 5.

Suppose that G is a G-module and 0 : A x Z the natural G-pairing. In view of the identification

4-6-14.

e=Horn (G, Q/Z) w H1(G,Q/Z) we may define a pairing of AC x

e

by putting (a,

x>

a* >-

-

A is

W(G, 2)

-

H2(G,A )

=

K a "0

s*x

aEAG,

XEG

4-6.

189


If H is a subgroup of G, then (with notation as in (3-5-4)) res (a, x)

= (a,

res x)

= (a,

inf x'>

aeAG, XE&

and if H is normal in G, then

-

inf (a, x')

a E AG, X'

n

E (G/H)

What about cor ? Given X E c, let be a representative of x in Q ;by this we Q is a function with ~ ( u (mod ) 2)equal to mean that R : G x(u) for every ~ E G Then . ~,xEH~(G Z), is represented by the standard 2-cocycle u = 62 for which u[u,71 = g(u)

+ n(.)

-

n(4

Furthermore, ~a wg6,x is represented by the standard 2-cocycle w for which

40, 71 = (40, 7I)a 4-6-15.

If the G-module A is a ring which satisfies the condition UEG, a , b ~ A

u(ab) = (ua)(ab)

we say that A is a C-ring. I n this situation, ring multiplication given by 8(a, b) = ab) in A (that is, the map 8 : A x A -A is a G-pairing. The abelian group H(G, A) = C_'," @ Hr(G, A) becomes a ring when multiplication is defined in terms of cupping with respect to 8; it is called the cohomology ring of A. If A is commutative, H(G, A) is skew-commutative. This means, in particular, that for (Y EHP(G,A)/?E H*(G, A) av

g p = ( - 1 1 ~ 9 ~wg

If A has an identity 1, then 1 E AG and ~ ( 1E) Ho(G, A) is an identity of H(G, A). Suppose further that M is a G-module that is also a (left) A-module and such that the action of G is compatible with the module operation of A-this means that u(am) = (ua)(um)

U E G,

U E

A, m E M

Iv.

190

THE CUP PRODUCT

Thus, the map 8' : A x M --+ M given by 8'(a, m) = am is @ Hr(G, M) a G-pairing. The abelian group H(G, M) = becomes an H(G, A)-module under the operation defined via cupping with respect to 8'. If A has an identity 1, this module is unitary. An important example of a cohomology ring is H(G, Z). This ring is known to be finitely generated-see [25]. Note that for any G-module M, H(G, M) becomes a unitary H(G, Z)-module.

xTz

Suppose that G is a finite group of order n > 1. If H*(G, 2) is a cyclic group of order n then the integer q is said to be a period for G, and an element 6 E Hq(G, Z) which generates this cyclic group is called a maximal generator. Consider the coho@ Hr(G, 2) mology ring with identity (see (4-6-15)) H(G, Z) = with multiplication (which we denote by u)arising from the Z. natural G-pairing of 2 x Z

4-6-16.

-

(i) For 6 E Hq(G, Z) the following conditions are equivalent: (a) 6 is a maximal generator. (b) $, has order n. (c) There exists 5' E H-Q(G,Z) such that t' u 5 = 1. 5 is an isomorphism of Hr(G, A) a (d) The map a onto H"J(G,A) for all Y E Z and all G-modules A, where 0 :A x Z A is the natural G-pairing. (ii) If

E

-

H*(G, Z) is a maximal generator then

5' is unique,

so we denote it by t-l, and 6-' E H-q(G, 2) is a maximal generator. If 5 E Hp(G, Z) is also a maximal generator, then so is 5 v 5 E HpW(G, Z). The periods of G form a subgroup of Z, and

all the periods are even. If q is a period of G then it is a period of every subgroup H of G; in fact, if 6 E Hq(G, Z) is a maximal generator, then so is resC+H6 E H*(H, Z).

(iii) Suppose that H is a p-Sylow subgroup of G, and that H*(H, Z) is a maximal generator. Let n be an integer such that ~n = 1 (mod #(H)) for every integer Y prime to p. Then 6" E Hm(H, Z) is stable (see (3-7-13)) and corH+G5" is of order #(H). E

4-6.

191


(iv) G has a period >O v every Sylow group Gp has a period >O. Furthermore, a p-group has a period >O v it is cyclic or a generalized quaternion group.

(A generalized quaternion group is one with two generators u,7 with relations U'

= T 2 = (UT)2

where t is a power of 2. Such a group then has order 4t, and every element can be expressed uniquely in the form U'T

*

s=O,l

OQrO, then it has a cyclic subgroup Hl of order p contained in the center; if H2is another cyclic subgroup of G of order p, then Hl + H2is a H , , 2) group of order p2 which has no period >O since H*(Hl is not cyclic of order p2-so G has only one cyclic subgroup of order p.)

+

Let B be a multiplicative group, finite or infinite, and consider a free resolution of 2 over

4-6-17.

Z'

=

Z[G]=

I1

n,u

1 almost all n,

=0

osG

By this is meant an exact G-sequence

where each X,, is G-free. (Actually, it suffices to deal with a projective resolution-that is, one in which each X , is G-projective.) Thus, a free resolution is simply the positive part of a G-complex (see Section 1-3) in which the requirement that each X,, have a finite G-basis is dropped.

Iv. THE

192

-

CUP PRODUCT

-

Let A be a G-module and consider the differential graded groups

0

*..

HOQ (Xo,A ) A

**.

-- -

Horn,

(Xn-1,

A) --%

-..

Hom, (X,, A )

aa X, & A A Xn-l

A

Xo @ p A

**.

(*I

0

(**I

where 8, = (a,, 1). The derived groups of (*), Xn(C, A) = ker S,+,/im 8,

are called the cohomology groups of G in A. The derived groups of (**), H%(G,A) = ker (a, @ l)/im (a,,, @ l), are called the homology groups of G in A. Both the homology and cohomology groups are independent of the choice of resolution (the proof makes use of homotopies). Moreover, *(G, A) w A,, and Z0(G,A) = (X,@r A)/im (a, @ 1) w A/(IA) = A,. Given the A B C 0 there short exact G-sequence 0 arise the exact homology and cohomology sequences

----

...

- - - - - -__t

&yG, B )

X (G, C )

2X"+l(G,A ) .#"+'(G, B)

-.*

K+,(G,C ) 2% K ( G , A ) Xn(G,B )

&+l(G, B )

__t

-..

n>O

**.

If G is finite these sequences may be joined to give an exact sequence which is infinite in both directions. This may be done as follows. We have the homomorphism S* : Xo(G,A )

-

P ( G , A)

induced by the trace (see (4-6-3)).Put &(C, A)

=

ker S*

&(G, A)

Thus, and

=

coker S*

4-6.

193


Now, we have a commutative diagram with exact rows

0

- - P( G, A )

,#"( G, B )

3P(G, C )

_+

X1(G,A )

-

From this there arises a homomorphism 8 :ker S s

@(G, B )

coker S,*

- -

S ( G , A ) - -P**

@(G, C )

Now put H"(G, A ) = Hn(G,A)

n>l AG/SA

HO(G, A ) = @(G, A )

H-'(G, A ) = &(G, A )

As/IA

H-n(G, A ) = Hn-i(G, A )

n 2 2

These are our customary cohomology groups of G in A. In fact, this is clear in dimensions 2 - 1. As for the remaining dimensions, using the resolution given by the positive part of the standard complex, we recall (see (4-6-4)) that X, & A is isomorphic to Hom, (2,, A). Moreover, under this isomorphism the boundary and coboundary correspond-that is, the following diagram commutes: Xn

Or G

001

1

-

Xn-l Or A

It follows that for n group.

> 2,

Home (Xl,A ) )*(a,

I)

HOW ( - L i ,A )

H-"(G, A) is our usual cohomology

194

Iv.

THE CUP PRODUCT

Historically, the procedure we have outlined for defining the cohomology groups in all dimensions via cohomology and homology and then connecting these by the Tate linking (see (3-7-7), (2-2-8), and (2-2-9))comes first. The use of a full $-complex is a matter of technical convenience. Note that since X, % X , and 8, & A M Horn, (X,,A), we could get all the cohomology groups by tensoring the full G-complex rather than taking Horn. We leave it to the reader to find the explicit formulas (using the standard complex) for the various cohomologicalmaps when the cohomology groups are given in terms of the tensor product.

CHAPTER

V

Group Extensions

In this chapter we study some of the basic properties of group extensions and their connection with cohomology. The main objective is the group theoretical version of the principal ideal theorem. 5-1.

THE EXTENSION PROBLEM AND Hz

Consider the exact sequence of multiplicative groups 1 - A L U i - G - 1

(54

where G may be finite or infinite. There is no loss of generality in assuming that i is the inclusion map; thus, A is a normal subgroup of U and UIA rn G. For each u E G, let us choose a representative u, E U-so that j(u,) = u. We shall refer to {uU I u E G} as a complete system of representatives for G in U, or more briefly, as a section of G in U . (Of course, it is really the mapping u u, of G U which should be called a section.) Since j(u,uJ = UT = j(uu7)for all u,7 E G, there exist elements u , , ~E A such that

- -

w

7

= au,7%7

195

Vu, T E G (5.2)

196

v. GROUP EXTENSIONS

Clearly, I u, T E G} may be viewed as a standard 2-cochain of G in A. For purposes of simplicity and, even more, because it suffices for our needs, we shall always assume that A is abelian. (A good deal of the theory can be carried over when A is not abelian-see, for example, [31, Chapter 151 or [45, Chapter 121). Now, U acts on A by inner automorphisms and, because A is abelian, the elements of A act trivially on A-so there is induced a "natural" action of G on A. In more detail, for a E A , u E G define an exponential action of u on a by

(5.3)

au = u,au;'

Then A is indeed a G-module; in fact, (ub)" = u,ubu;' = u,,au,-'u,bu;'

= a%",

a1 = ulauyl = a since u1 E A, and (uTy = u,(u,uu;')u~' = u , , , ( u , ~ u ~ = ~ )u,,uu;~ u ~ ~ = auT

Moreover, this action of G on A is independent of the choice of representatives {u,}; in fact, if {w, I u E G} is another section of G in U , then for each u E G there exists c, E A such that w, = c,u,, and then V , U V ; ~ = C , ~ , L I U ; ~ C ; = ~ u,au;'. Once the section {uo I u E G} is fixed, every element u E U has a unique expression of form u = au,

aEA, O E G

which is given by u = j ( u ) and a = uu;'. Because Eq. (5.3) can be written as UOU,

= u,u

U E A , U E G (5.4)

it follows that multiplication in U can be described in terms of the multiplications in A and in G, the action of G on A, and the 2-cochain {a,,,}. Thus, if u, w E U are of form u = au, , w = bu, , then uv

=

abuil,u,

= ab"~,,,u,,

(5.5)

5-1.

THE EXTENSION PROBLEM AND

H2

197

Furthermore, associativity in U leads to the equality (for all u, 7 , p E G) of (

~

~

=b a o ,pT ~ u T u p= ao,TaoT.p*,Tp

and u u ( ~ p = ) ~ T , p u T p= 4 . p ~ T = p a:,pa,,T~,Tp

Therefore, for all

IJ,

ao,TaoT,p

=

%,p E G, a:,oau,Tp

or

-1

=1

a:,~~:t~o,Tpau,T

(5.6)

This formula is precisely (see (1-5-12)) the multiplicative form of the coboundary formula for standard cochains of G in A in dimension 2; consequently, the 2-cochain { u , , ~ } is a standard 2-cocycle of G in A (or, in classical terminology, {u,,~}is a factor set), and we may express this by ( S U ) , , ~ , = ~ 1. Of course, if {w,, = c,u,} is any other section and we write w,w, = b , p , , , then {bu,T} is also a standard 2-cocycle of G in A. From vuwT

=

(cA)(cTuT)

=c

o~~c~~ao,T~uT

it follows that bu+z;; = c,c:c;., and the right side is the coboundary (6~),,~ = c&c;, of the standard 1-cochain {c,} of G in A. This means that the cocycles {u,~,} and { l ~ , , ~belong } to the same cohomology class in H2(G, A). It should be noted that the action of G on A used for cohomology is derived from the sequence (5.1) and is expressed by the formula (5.3). Suppose now that G is a multiplicative group, finite or infinite, and that A (also written multiplicatively) is a G-module with the action of G given exponentially. By a solution of the extension problem for the pair (C, A) we shall mean an exact sequence of the form (5.1) (which shall also be denoted simply as a triple {U, i , j } ) such that the action of G on A determined by (5.1) and expressed by (5.3) coincides with the given action of G on A. We shall then say that { U , i, j } is an extension of A by G. Thus, the upshot of the foregoing discussion is that if {U, i , j } is an extension of A by G, then it determines an element of H2(G, A). Our next objective is to show that, conversely, an element (Y E H2(G, A) determines a solution { U, i, j } of the extension problem, and such that the associated cohomology class in H2(G, A)

v. GROUP

198

EXTENSIONS

is precisely a. For this, let {a,,,} be a standard 2-cocycle of G in A which. belongs to the class 01. In particular, Eq. (5.6) holds. By . substituting u = T = 1 or T = p = 1 or T = u-l = p-1, respectively, one arrives at the formulas (for all u E G)

Now, for each

(I

E

G,choose a formal symbol U

= {(a, v,)

1 u E A,

uE

u, and let

G}

(5.8)

Define multiplication in U according to the rule (a, u,)(b, u,) = (ab"~,,,,v,,)

a, b E A

u, T E G (5.9)

From the coboundary relation (6u),,,, = 1 and commutativity of A, it follows that multiplication in U is associative; in fact,

and this is equal to

T o exhibit a left identity in U , one seeks an element (x, up) such that (x, #,)(a,u,) = (a, u,) for all a E A, u E G-that is, (xuvz,,,, ,u,) = (a, u,). Consequently, p = 1, and, making use of the first part of (5.7), it is clear that (ay,?,,ul) is a left identity. Finally, given (a, u,) E U, we see that 1 u -1

( ~ L l aa,-i.,

3

1

u,-iMa, u,) = (aL1, u,)

-

-

so that (Ui,?,tfU;!i,,, u , - ~ ) is a left inverse for (a, u,). This shows U and j : U G are that U is a group. The maps i : A now defined by

i(u) = (.a,:

,ul)

j(a, u,) = u

(5.10)

5-1. THE EXTENSION PROBLEM

AND

H2

199

It is easy to check that i is a monomorphism, that j is an epimorphism, and that im i = {(a,u,)l a E A} = kerj. This means that

is an exact sequence. We must also verify that the 2-cocycle of G in A determined by this exact sequence coincides with the one from which the construction started. Consider then the section ((1, u,)( u E G} of G in U ; we must check that

But this is straightforward, both sides being equal to uUT). Finally, it remains to verify that the action of G on A determined by the exact sequence coincides with the original action-that is,

Now, the left side is (auu:,\, ul) while the right side is

1= ';.,.(

9

q

(in virtue of (5.7))

The foregoing discussion shows (among other things) that a 2-cocycle {a,,,} belonging to the cohomology class 01 E HYG, A) leads to an extension {U, i,j} of A by G. Since any other cocycle {bu,T}belonging to a also leads to a solution of the extension problem, it is desirable.to compare solutions of the same extension problem, and we are led to the following definitions. Suppose that A is a G-module with { U , i , j } an extension of A by G and that A' is a G'-module with {U',C , f } an extension of A' by G'. By a homomorphism (or translation) of {U, i , j } into { U', i', j ' } we shall mean a triple (f,v, A) of homomorphisms

200

- - -

v. GROUP EXTENSIONS

A', A : G diagram commutes:

f :A

G , (p : U

U' such that the following

l-A'-U&G-l

4 4 4

1 --+A'&

(5.1 1)

U'L G ' - 1

Of course, since i and i' are inclusions, f is the restriction of (p to A. When f,(p, A are all isomorphisms onto, we shall say that the translation (f,(p, A) is an isomorphism. I n the special case where A = A', G = G , f = 1, A = 1, we say that the extensions{U, i , j } and {U', i',j'} of A by G are equivalent if there exists a homo{ U', i', j'}.This defimorphism (f,(p, A) = (1, (p, 1) : { U,i, j } nition can be restated as the requirement that there exist a homomorphism (p : U +U' such that

-

U

1-A

G-1

(5.12)

U'

is a commutative diagram. It should be noted that, as is easily checked, such a is automatically an isomorphism onto, so we are dealing with an equivalence relation and (p may be referred to as an equivalence of {U,i, j } and {U',i',j'}. If {U',i',j'} = {U, i, j } , we say that (p is an automorphism of { U, i, j } . It is easy to see that equivalent extensions of A by G determine the same cohomology class a E NZ(G,A); in fact, matters may be arranged so that they determine the same cocycle. To see this, suppose that { U, i, j } and { U',i',j'} are equivalent extensions with U',and let {u.} be a section of G in U. equivalence (p : U Then {u,,~},which is defined via uuuT= a,,Tuu,, is a 2-cocycle belonging to a. If we put u: = (p(u.), then j'(u:) = 0. Therefore, {u: I u E G}is a section of G in U' and u:u: = u,,,u;-which says i',j'}. It is also true that {au,T}is a cocycle associated with {U',

-

5-1.

THE EXTENSION PROBLEM AND

He

20 I

that cocycles belonging to the same class a: E H2(G, A) determine equivalent extensions of A by G; this will be a consequence of the more general discussion which follows. Suppose that { U, i, j } is an extension of A by G, that { U', i',j'} is an extension of A' by G , and that f : A A', A : G +G are homomorphisms. We would like to decide if there exists a U' such that (f,q ~ A) , is a homomorhomomorphism g~ : U phism of group extensions (see the diagram (5.11)), and if so to count them. To study this question, choose actions {u, I u E G} C U be the correand {u:. I u' E G'} C U', and let { u ~ , and ~} sponding 2-cocycles. Suppose that we have such a v. Then for any u = au, E U we have

-

-

d4 = f(4 duo)

(5.13)

so that v is determined completely as soon as the values ~ ( u , E) U' are prescribed. Because j'(cpu,) = A&) = A(u) = j'(ui(,J, there exists for each u E G an element c: E A' such that

In the usual way (see (2-1-1)), we may define (a'y = ( u ' ) ~ (and ~) thus view A' as a G-module. When this is done, {c: I u E G} is a standard 1-cochain of G in A' which, in virtue of (5.13) and (5.14), serves to describe rp. Let us observe further that f : A A' is a G-homomorphism -that is:

-

f ( 4 =f(@

U E A , U E G (5.15)

In more detail, we have

-

(G, A') is a This implies, in particular, that (1,f) : (G, A) homomorphism of pairs (see Section 2-1). Of course, (A, 1) : (G', A')

-

(G A')

v. GROUP EXTENSIONS

202

is also a homomorphism of pairs. We assert that if {u,,,~}belongs to a E HB(G, A) and {~i.,~.} belongs to a' E H%(G',A'), then f*(a) = ( l , f ) * a = (A,

l),

a' E HTG, A')

(5.16)

To see this, it suffices (see the beginning of Section 2-5) to show ( ~G ) } in A' differ by the that the 2-cocycles df(a,,,)} and { u ; ( ~ , , ~of coboundary of the 1-cochain {ci}. More precisely, we have

This proves half of the following result. 5-14. Proposition. Let { U,i, j } be an extension of A by G which determines a E HB(G,A), and let { U',i',j'} be an extension of A' by G' which determines a' E HTG', A'). Suppose that f :A A' and h : G G' are homomorphisms and that A' is also viewed as a G-module by putting (u'p = (a')"'").Then there U' such that exists a homomorphism p : U

-

-

__t

-

is a commutative diagram (that is, there exists a translation (f,p, A) :{ U,i, j } { U',i',j'}) if and only if (i) f is a G-homomorphism. (ii) f&) = (l,f)* a = (A, 11, a'. Suppose that (i) and (ii) hold, and choose sections Proof: {u,} C U and {ui.} C U'.If we write

5-1.

THE EXTENSION PROBLEM A N D

H2

203

then the cocycles and {a:.,,.} belong to a: and a', respectively. Condition (i) says that Eq. (5.15) holds, while (ii) implies that there exists a 1-cochain {ci} of G in A' such that (5.17) holds. Define p : U U' by

-

(5.18)

so that q is a homomorphism. Because u l e A , we have 1 = p( 1) = q(ullul) = f(ul)-' clu; , and therefore f(ul) = ciu; . Then 'i 0 f = p 0 i, since for a E A, p(a) = p(au;'u,) = f(au;')c;u; = f ( u )

Finally, h 0 j = j ' 0 cp since (A 0 j)(auu) = h(a) and

(i' d(%) = i ' ( f ( a ) c X o ) = 44 This completes the proof. I O

Theorem. Suppose that the abelian group A is a G-module (G is finite or infinite). Then there is a natural 1-1 correspondence between the equivalence classes of extensions of A by G and the elements of H2(G, A).

5-1-2.

I t remains only to show that if Proof: and {bm,T}are cocycles belonging to a E H2(G,A) and { U , i, j } , { U', i',j'} are the extensions they determine according to our construction, then these extensions are equivalent. T o do this, simply apply (5-1-1) with A' = A, G' = G, f = 1, h = 1, a' = a. Since conditions (i) and (ii) are trivially satisfied, the existence of p : U U' is assured, and the theorem is proved. I

-

5-1-3. Exercise. In virtue of (5-1-2), the equivalence classes of extensions of A by G can be made into an abelian group by

204

v.

GROUP EXTENSIONS

carrying over the multiplication from H2(G, A). This multiplication of extensions was first defined in a direct fashion in [8]; it is consequently known as Baer multiplication. The details may be organized as follows: i A U G 1 is a group extenSuppose that 1 sion corresponding to a E H2(G,A), so that for a section (u,} the cocycle {u,,,} given by u,u;, = u,,,u,, belongs to a. Suppose, further, f' that 1 A U' G 1 is a group extension corresponding to a' E H2(G,A), so that the cocycle (u:,,} determined by the section {u:} belongs to a'. Form the direct sum

----

-- -u 0U' = ((24,

u') I u E

u, u' E U'}

and consider the subgroup (U,U') = {(u, #')I j(u) = j'(u')}. This leads immediately to an exact sequence 1+A@A-(U,

U)-G-l

Now consider the subgroup D = ((a, a-l)( a E A} of A @ A . When A @ A is viewed as a subgroup of (U,ul), D becomes a normal subgroup of (U,U'). If we put U" = (U,U')/D, then A is isomorphic with (A @ A)/D, and with suitable definitions of i" andj" I-A''-U"

L G - 1

is a group extension of A by G. Moreover, the cocycle {a:,,} associated with this extension (for the section {uz} = {(u, ,u:)D}) is precisely { u , , , ~ ~ , , }This . enables us to define multiplication in the set of equivalence classes of extensions of A by G, and it becomes a group isomorphic to HZ(G,A ) . I Now, let us place ourselves once again in the situation of (5-1-1) (where conditions (i) and (ii) hold, and the same notation applies) and investigate the number of v's that exist (for fixed f and A). Let 9 denote the set of all homomorphisms p : U U' such { U',i', j'} is a translation. According that (f,p, A) : { U,i, j } to (5-1-1), 9' # 4, and we may fix a '9 e 9'.Then, by (5.14), there is associated with q a 1-cochain c' = (c:} E W(G, A') given by c i = q(u,) u i i ; . For any y5 E 9, let us denote its associated 1-cochain by b' = {b:} E %"(G,A'), where b: = $(urn)u:;: Since

-

-

.

5-1. THE EXTENSION PROBLEM AND H2

205

Eq. (5.17) applies for both q~ and #, it follows that c’ and b’ have the same coboundary (namely, f(a,.) Consequently, their quotient d‘ = b‘c‘-l, where = b’c’-l a a =w)P(~a)-l

(5.19)

is a cocycle-that is, d‘ = {di} E 8 l ( G , A’). Furthermore, d’ is independent of the choice of section of G in U ; in fact, if {va = aaua}, a, E A, is any other section then the cocycle determined by this procedure is +(%) d%)-l= f(4 J N U a ) (P(.a’)f(a)-’

=

f ( 4daf(4-l

=

4

With each # E Y we have associated a d‘E 9”(G, A’). Conversely, given d’ E Z1(G, A‘), the steps in the discussion are reversible, so there is defined a # E 9’ which in turn determines d’. It follows that there is a 1-1 correspondence h,t

-

d‘

between Y and ZT”(G, A‘). This enables us to make 9 into a group isomorphic with P ( G , A’),but, of course, this isomorphism depends on the choice of ‘pE 9’. On the other hand, we may define namely, if e’ E b l ( G , A‘) a “natural” action of 81(G, A‘) on 9; and # E 9’ with q5 d’, then e’d’ E P ( G , A), and we define el# to be the element of Y associated with e‘d’. It follows that

-

(e‘4)(%7)= e:+(%)

UE

G (5.20)

Since this relation suffices to determine e’#, we see that the action of S 1 ( G , A’) on Y is independent of the choice of q~.It is clear that the action of 9 ( G , A’) on Y is transitive and without fixed points, and that e”(e’#) = (e”e’)# for e”, e’ E P ( G , A). Next, let us see which elements of Y correspond to coboundaries. Suppose then that # t--t d, d € B 1 ( G ,A’). There exists a’ E A’ such that di = a’”-’ for all E G, so that (J

~ ( u u , )= f(u) d;c;#;(,, = f(u) c;u‘-la’%;(a) = u’-l(f (u) c;u;(a))a’

If, for a‘ E A‘ we let I,, denote the inner automorphism of U’given by l a d = a‘-lu’a‘, then # = I,, o q ~ . This means that cp and # are equivalent (v M #) under the equivalence relation defined by: th ,l = if and only if there exists a‘ E A‘ such that t,hz = I,, 0 .

v. GROUP EXTENSIONS

206

On the other hand, if F w # (where v, # are the ones appearing in (5.19)) so that I,!I = 1,. o v, then U

-

) = a#-l t t a’ = (p-1 Caua(d

t

I

Cuua(d

But #(u,,) = b&, ; hence 41, d’ where d: = afU-’-that is d’ is a coboundary. We have shown that # w g~ ts # corresponds to a d‘ E W ( G ,A’). (We leave it to the reader to state the necessary and sufficient condition for m .) It follows that H1(G,A’) acts on 9,the set of equivalence and we have proved the following result: classes of elements of 9, 5-14, Proposition. Let the hypotheses be as in (5-1-1), and suppose that conditions (i) and (ii) are satisfied. Then H1(G,A’) and 9 have the same number of elements; in fact, H1(G,A’) operates transitively and without fixed points on 9.In particular, if H1(G,A‘) = (1) then all the elements of Y are equivalentin other words, any two elements of Y differ (with respect to composition of maps) by an inner automorphism of U’ by an element of A’. An easy consequence of (5-1-4) is the following: Exercise. Let {U,i, j } be a group extension of A by G.Then the automorphisms of { U,i, j}form a group Y which is canonically isomorphic to T1(G, A). The inner automorphisms I, : u u-luu of { U,i, j}by an element of A form a subgroup 9” isomorphic to @(G, A). Therefore, the equivalence classes of automorphisms of {U,i, j} form a group 9 = Y / Y isomorphic to H1(G,A). In particular, if H1(G,A) = (1) then every automorphism of { U,i, j}is an inner automorphism determined by an element of A.

5-1-5.

-

5-2.

COMMUTATOR SUBGROUPS IN GROUP

EXTE NSlONS

5-2-1. Remark. It is convenient to recall some facts and notation from (1-6-7). For an arbitrary multiplicative group U, we write 0 = Horn (U,Q/Z). This is an additive abelian group

5-2.

COMMUTATOR SUBGROUPS IN GROUP EXTENSIONS

207

(since Q/Z is additive) called the group of characters of U in Q/Z. If A is an abelian group and B is a subgroup of A, then Q/Z can be extended to a character any character f : B g : A -+ Q/Z. (In other words, if __t

-e

(1)

BAA Ac

-

(1)

is an exact sequence of abelian groups, then

0

b.1)

~

A^

(i.1)

B

0

is exact.) In fact, if a E A, u .$ B, then g may be chosen so that U, with A abelian and U not abelian, then a character of A need not be extendible to a character of U, a condition for extendibility is discussed in (5-2-2). Let B* = {g E A’ I g(B) = 0} = ker (i, l), and call it the annihilator of B in A’. If B,, B, are subgroups of A, then (B, + B,)* = Bt n Bf and B, < B, =a Bfr < BP. Consequently, for any subgroups B, , B, , B+ = B+ B, = B, .

g(a) # 0. If A C

-

5-2-2.

Consider the group extension

Lemma.

1-A-U-G-1

and a character f E A’, and let Uc denote the commutator subgroup of U. Then f can be extended to a character g E 0 u f( UGn A ) = 0

In other words, we must show that if ( Uc n A)* denotes Proof: the annihilator of Uc n A in A’, then (Uc n A)I

= {f E

a If

can be extended to g E #}

(5.21)

Suppose first that f E A^ can be extended to g E Hom (U, Q/Z). Then UcC ker g because Q/Z is abelian-so f( Uc n A) = 0. For the converse, consider f E ( Uc n A)* C Horn (A, Q/Z). Then f determines a character j of A/( Uc n A) given by j [ a ( Uc n A)] =f(u). Since Uc is a normal subgroup of U, A/( Uc n A ) fi: ( UcA)/Uc and

v. GROUP

208

EXTENSIONS

we have a character g of ( UcA)/Uc given by g(a Uc) = f[a(Uc n A)]. Now (UcA)/Ucis a subgroup of the abelian group U/Uc,so g can and then g' determines a be extended to a character g' of U/Uc; character g of U given by g(u) = g'(uUc).This g E f) is the desired extension off. 5-2-3.

Let G be a finite group and suppose that

Proposition.

is a group extension associated with

where 8, : Z x A

a! E

HYG,A). Then

-

A is the canonical G-pairing and

is the usual map in dimension -1. Suppose that f~ A (5-2-1),it suffices to show that

Proof:

= Hom

(A, Q/Z). In virtue of

To accomplish this, we note first that by (5-2-2),f E (Ucn A)* eaf can be extended to E f). Therefore, let us consider the diagram

where, in the bottom row (which is additive), Q/Z is viewed as an extension of Q/Z by the (0) group. Now, apply (5-1-1) (note

5-2.

209


that the action of G on Q/Z defined via A = 0 is precisely the customary trivial action, and that a' = 0 E H2((0),Q/Z) = (0)) which says that f can be extended to

E

I

0 ts (i) f is a G-homomorphism (ii) (l,f)*. = (4 l)*a' = 0 E H2(G, Q/Z)

Condition (i) says that f(u') = uf(u) = f(u) for all a E A, u E Gso it is equivalent to the condition that f vanishes on IA. T o

- -

rephrase condition (ii), we apply the duality theorem (4-4-6) with p = -2, B = Z, B = Hom (Z, Q/Z) % Q/Z. Thus, f * ( a ) = 0 E H2(G,Q/Z) the character of W S ( G Z) , in Q/Z 5) is the trivial character given by 5 ~O:~(f*(a)

* r]O:z(f*(a)WebIz 5 ) = 0 for all 5 E H-'(G, 2) v &jZlzf*(a u e i

5 ) = 0 for all 5 E H-s(G, 2) see (4-6-13))

In other words,

Now, let us consider the diagram

If (1) holds, then f is a G-homomorphism and by (2-2-7)the diagram commutes. Then, since qOlz is an isomorphism onto, f 7 i 1 = 7 ~ f* : ,~and (2) implies that 0

0

f7l1(H-'(G, Z) W e A a) = 0

(5.25)

since for 5 E H-3(G, Z), 5 weAa = a U e i 5. Conversely, suppose that (5.25) holds. Then f is a G-homomorphism because

210

v.

GROUP EXTENSIONS

q ~ ' ( 0= ) IA, so that (5.24) commutes and r)& of* = f o 7 ) ~ ' . By applying (5.23) we conclude that f E (Ucn A)*. Thus (5.22) holds, and the proof is complete. I 5-24. Remark. It is an incidental by-product of the preceding proof that

(5.26)

IACUCnACA,

These inclusions may also be proved in a more direct fashion. For ~ E AU,E G we have a"-' = u,uu;'a-l E Uc n A, and the first inclusion holds. T o show that Ucn A C A, , one makes use of the group theoretical transfer (see (3-5-4))

The section {u,} provides a complete system of representatives for the cosets of A in U, U = UosG Au, . Fix u E U, and for u E G let us write j ( u a ) = a'; thus a' is the unique element of G such that u,uu;' E A. It then follows from the transfer formula (see (3-5-4)) that V(u) =

n

(I' = ~ ( u , u ) (5.27)

u,uu21

esC

In particular, for u = Q E A we have a' = j(u,p) = u, and V(a) =

n

U&U,-'

O€G

=

n a

-

au = N(a)

Since the homomorphism V maps Uc UCnACA,. 5-24,

Proposition.

1 it follows that

Let G be a finite group and suppose that

is a group extension and that {u,} is a section of G in U with

5-2.

21 1


associated 2-cocycle {a,,,}. Then the transfer V = Vu-tA: U is given by the formulas

__+

A

(i) V(a) = N ( u ) EAC UEA. (ii) V(u,) = aU,,E AC 7 E G. (iii)

nuoc V(au,) = N(a) nua,,, E

AC.

Proof: Part (i) has been proved in (5-2-4). As for part (ii), if (T E G, then (according to (5.27)) u' = j(u,u,) = m, so that V(u,) = u,u,u;~ = a U , , .Moreover, for p E G we have

nu

nu

(n

%,r)P

=

n

=

(%.u%7,P;:T)

U

U

n 0

=w,7 =

n

%,T

U

Consequently, the homomorphism V is given by (iii) and takes values in A G . I

-nu -

It should be observed that, in virtue of (ii), V leads to a mapping a,, of G AC. This mapping need not be a homomorphism, but when the right side is reduced modulo the group of norms NA we have precisely the Nakayama map (see (4-5-7)).

7

Proposition. Let the hypotheses be those of (5-2-4) and (5-2-5). Then the following diagram is commutative and has exact rows:

5-24.

1

-

NA

A Ho(G,A )

AC

-

1

Here, i, j , are induced from i, j, N in the obvious fashion, is the reduced transfer (see (3-5-4), (3-5-9,(3-5-6)), and for each Hn+2(G,A) is the map given by integer n, a, : Hn(G,2) __+

an : p

-

fl voA p E H"(G, Z) (Y

(5.28)

Proof: The bottom row is clearly exact. As for the top row, i is given by i[a(Uc n A ) ] = aUc, so it is a monomorphism with

v. GROUP

212

EXTENSIONS

image (AUc)/Uc; and f is given by f(uUc)= j ( u ) Gc, so it is an epimorphism. Since j maps U onto G with kernel A and j ( Uc)= Gc, it follows that kerf = (AUc)/Uc.Therefore, the top row is exact. The square on the left commutes because, according to (5-2-S), u'i[a(Ucn A)] = Y(aUC)= V(a) = N(a) = N[a(Ucn A)]. AS for the square on the right, consider u = a u , ~U.Then, making use of (4-5-7), we have

5-2-7.

Corollary.

(5-2-6) and consider (i) (ii)

-

Let us place ourselves in the situation of AG.Then

Y : U/Uc

Y is an epimorphism

es a-2

V is a monomorphism

es

is an epimorphism.

and

are monomorphisms.

Proof: Viewing all groups additively, we complete the diagram of (5-2-6) as follows:

1 1 1 1 1 1

0-0-0-0-0

o-*-*-*-o

Jq

v1

I.-*

O-*-*-*-O

1 1 1

0-0-0-0-0

1 1 1

5-3. FACTOR

213

EXTENSIONS

The columns may be viewed as differential graded groups, so by (1-1-4) we have the exact sequence of derived groups: ..*

-

(0)

_+

ker

m

- ker

= ___ AN

Uc n A

ker

7

-

Now, V is an epimorphism e;=> im = Ho(G, A) a_$ is an epimorphism, thus proving (i). Furthermore, B is a monomorUc n A) = (0). From (5-2-3) it phism c- ker OLQ = (0) and AN/( follows that im % ( Uc n A)/(IA),so that H-'(G,A) ima-,

and (ii) is immediate.

5-3.

%-

A, U C nA

I FACTOR EXTENSIONS

Let A be a multiplicative G-module (where the group G is finite or infinite) and let { U , i, j ) be a group extension of A by G associated with OL E H2(G,A); in other words,

is exact, and if {u. I u E G} is a section of G in U with u,u, = a.,,u,, , then {a,,,} is a standard 2-cocycle belonging to a. Suppose that H is a subgroup of G . If we put W = j - l ( H ) , then the sequence (1)-AA

W A H - ( l )

(5.29)

is exact. Even more, { W, i, j } is an extension of A by H and may be referred to as a subextension of { U,i, j}-in fact, {up I p E H} is a section of H in W and for a € A, P E H , ap = u,au;'. The cohomology class in H2(H,A) associated with the subextension { W, i, j } is precisely resG-,Ha; this is immediate because

v.

214

GROUP EXTENSIONS

{ao,, I u, T E H } is the restriction of the cocycle { u ~I u, , ~T E is also the cocycle determined by the section {ua I u E H ) .

G } and

Now, let us turn to the situation where H is a normal subgroup of G. Here W = j - l ( H ) = {au, I a E A , p E H } = {.,a I ~ E Ap E, H } is a normal subgroup of U, and U / W w G / H . If we let 7r : G G / H denote the natural map (and i = inclusion, as usual), then we have the exact sequence

-

(5.30) Unfortunately, this cannot be considered within our framework for group extensions-W is not abelian and there is no way to define an action of G / H on W by conjugation. To overcome this, let us consider (5.31) wherei'(wWc) = i(w) Wc = wWCandj'(uWC)= j(u) H = n(j(u)). First of all, it should be noted that Wc is normal in U-for a generator of Wcmay be taken of form w1w,Wi1w,' where wul,w2 E W and then for U E U, uw1w2w;1w;1u-1 =

(uwlu-')(uw2~-')(uw;'u-')(uw,'u-')

which is the commutator of the elements uw1u-l and uw2u-l of W. From the definitions of 'i and j',it is immediate that the sequence (5.31) is exact. Because A C W there is no natural way to derive an action of G / H on W/Wcfrom the action of G on A; however, we shall view W/Wcas a G / H module according to the standard procedure arising from the exact sequence (5.31)-that is, through conjugation by representatives of G / H in U/Wc. In this way, an element LY E H2(G,A ) leads to an element a' E H2(G/H, W/We) associated with the factor extension { U /Wc,i',j'}. This is not quite satisfactory cohomologically, because A is the basic module. For this reason, let us consider the group theoretical transfer (see (3-5-6)) Vw-lA: WA/Ac = A. Note that for this we must have ( W : A) < ao; in *other words, H must be a finite group. From (5-2-5), applied to the subextension

5-3. FACTOR

215

EXTENSIONS

(5.30),it follows that VwjA maps W into AH,so that the reduced transfer is a homomorphism -

v = VWj"

:W-AH

WC

Moreover, V is a (G/H)-homomorphism (as we shall next verify) and we have the induced homomorphism of cohomology V,

=

(BW+J* : H2(-G H

-)W

Wc

-

H 2 (TI G A")

V is a (G/H)-homomorphism, note (see (5.27)) that V(wWc)= V(w) = uPwup.lE AHwhere p' = j(upw), and that for a E A", u E G we have = uu = u,uu~'. Since, for any U E G, uHu-'= H and j(~,,u,,u;~)= up@, we may also use { u , ~ ~ uI p; ~E H} as a section of H in W for the subextension (5.29). T o show that

nPEH

Consequently,

=

n

(u,u,u~')(uuw.,')(uuuD~ll~~)-'

PEH

and because j(u,uPu;'u,wu;') = ap'o-l, this expression is simply V,,,(u,wu;') computed with respect to the section { U , U ~ U ; ~1 p E H}; but V(u,wuu;') = Y{(wWC)'""'}, thus proving the assertion. 5-3-1. Exercise. The foregoing discussion shows that if A is a G-module and H is a finite normal subgroup of G, then there exists a mapping Z)GjG,H

: H2(G,A )

-

(:,

1

H 2 - AH

defined in the following manner. Give a E H2(G, A), let { U, i, j } be an extension of A by G associated with a; putting W = fl(H) there arises the factor extension (5.31) which is associated with a' E H2(G/H, W/Wc); then v G + G / H ( ~ ) = VI*(a')= (VW+")* (a'). Show that ~ ~ + ~ , , , (isa )independent of the choice of the extension { U , i, j ) belonging to the equivalence class which corresponds to a.

v. GROUP EXTENSIONS

216

Proposition. Let H be a normal subgroup of the finite group G ; then the map

5-3-2.

HZ(G,A )

W G ~ / : H

_+

HP

is a homomorphism. Moreover, if we let T E G denote the representative of the coset Y = TH, then w ~ . + is~ induced / ~ by the map where of cocycles {au,,}+{(era),,$} (“)r,a

= [NH(u~,~uitT)I

[n

‘p,,j]

*H

and X = &,$= F G - ’ E H . Proof: The fact that a,/ is a homomorphism will follow immediately from the formula for ( W U ) , , ~since both NH and are multiplicative. Starting from the section {uo I cr E G } of G in U (which determines the cocycle {Q,,~} via up, = a, 7uoT)we know that {up I p E H } is a section of H in W.If for Y E G / H we put u, = u, ,then {u, I Y E G / H } is a full set of representatives for G / H in U (see (5.30)), and {u,Wc I Y E G / H }is a section of G / H in U/Wc(see (5.31)). It is clear that u,us = a,,,~,,, and writing X = FfSrs-’ (so that X is a function of r and s in G / H , and X E H ) we have U,U, = a A , , p F J . Suppose the factor extension (5.31) is described by the 2-cocycle {w,,,Wc)-this means that

npsH

(urWC)(u8Wc)= w r . ~ m W ~

and that if { u ~ ,belongs ~} to the class a E H2(G,A), then the cocycle { W ~ , ~belongs W ~ } to the class a‘ E H2(G/H,AH).Thus W ~ + ~ / ~ ( O is L) represented by rw+JwV,

8

wc)= ~ ( u F ,~ 4 i : ~ uWA e )


I

5-3. 5-34.

Let the hypotheses be as in (5-3-2);

Proposition.

then for

01

217

FACTOR EXTENSIONS

E H2(G,A )

inf(G,H)+GvG+G/H&!

=

&!‘H:l’

Proof: For u E G, denote the representative of the coset to It will also be convenient to which u belongs by 5-so 6 = let h denote a generic element of H. When all cohomology groups are computed with respect to the standard complex, infoa! is represented by the 2-cocycle inf va for which

a.

(inf~a),,,~ = (wa),H.TH

[ IIaa.c7&-1]

= [~H(ac.~ik-1,&)1 =

n

h€H

h€H

--1) (a” a-h 6 . T 6Ta; 1,&ah.67ar

Since {aa,T}is a 2-cocycle we have the coboundary relation

-

LI

(Sa)a,T.p

--1

= aa,T.apa,$p,u~t =

1

-

0, 7 , P E

G

6,T f----t f , p h and h, which enables us to write

In this formula we substitute first u ----l OTUT , 7 ++ UT, p then u

----l

because H6 = uH,and HUTUT = H since 6 f G - l H. ~ To simplify this expression consider the 1-cochain of G in A {ca = ah,6};its coboundary is

nhsH

(sc)a,,

=

c~c,;’ca

=

n ‘;,faib%,i

h€H

and dividing by this coboundary gives

Since ( 8 ~ ) i , i ,= ~ 1 the right side is equal to

n

h€H

‘u.hT‘o:lh

n

= h€H ( a u , r h a ~ ~ )

v. GROUP EXTENSIONS

218

and since (&z),,~,, = 1 this is equal to

where d is the 1-cochain {d, = (inf 4 and the proof is complete.

-

54.

nheH~,,h}. Therefore,

U . T

-

(H;1)

au.7

I

T H E PRINCIPAL IDEAL THEOREM

--

Suppose that A is a G-module, and consider a E H2(G,A). Let 1 A +U G 1 be a group extension associated with a. Thus, if {u,} is a section of G in U and we write up, = a,,,u,, , then {a,,,} is a standard 2-cocycle belonging to a. Of course, in choosing the section {uo} we may always take = 1 E U; and when this is done u,,l

= a,

,= 1

all

UE G

(5.32)

In this section, when dealing with a cocycle {a,,,} we shall always assume that (5.32) is satisfied. If B is a G-module containing A such that for the inclusion map i :A B we have ;*(a) = 1 E H2(G, B), the B is said to be a splitting module for a. I n other words, a cocycle belonging to 01 becomes a coboundary when viewed in the larger module B.

-

541. 01

E

Proposition.

H2(G, A).

A splitting module exists for each

We construct a G-module B containing A (which we view additively) for which there exists a standard 1-cochain {xu} such that a,,, = (ax),,, = ox, - xu, xu for all u, T E G. T o do this, take a formal symbol x, for each T # 1 E G. It is also convenient to put x1 = 0. Now, put

Proof:

+

B=A@C@Zx, 7#1

5-4. THE

PRINCIPAL IDEAL THEOREM

-so B is an abelian group. Define the action of keeping the given action of u on A and putting

-

+

= ' U T - U'

u'T

219 IJ

E

7

'U,T

G on B by # 1 (5.33)

Of course, this suffices for the definition of the homomorphism B. Note that Eq. (5.33) is also true for T = 1. It is clear that 1 E G acts as the identity on B. Finally,

u :B

P(.'T)

dxOT

-

+

'UJ)

= xPUT - '

P

-

+

- '(0U)T - 'PU


'P.UT

+

-'PU

'0U.T

+

-'P.0

+ fa,,,

= (pU)'T

I

542. Exercise. Suppose that A is an additive G-module, and fix n 1. For each i = 0, I , ..., n let us say that the standard n-cochain €59, = Vn(G,A) is i-normalized (or i-normal) if f [ul,..., u,] = 0 whenever one (or more) of u1 ,..., ai is 1. I n particular, the 0-normal cochains are simply the cochains of Vn. T h e n-normal cochains are referred to simply as normal cochains.

>

(i) T h e normal cochains form a subgroup gn of P;we also put go = Vo.T h e cocycles which are normal form a subgroup 3, of 9"". I n virtue of the coboundary formula, the coboundary of an i-normal cochain is i-normal. I n particular, gn= SQn-l is a subgroup of gn,and we may form the normalized cohomology At the beginning of this section (see groups Hn(G,A ) = dnl.@. formula (5.32)) it was observed that every element of H2(G,A) may be represented by a normal cocycle. We shall see in (ii) that, even more, Hn( G, A) rn Hn( G, A) for all n 1. (ii) Given any cochain f E V n let us construct, inductively, cochains f o , fl ,...,f, E V n and g, ,..., g, E 5P-l in the following manner: i = 1 , ...,n f i = f i - 1 - sgi f o =f

>

ga[q I . . . , on_,]= (-

l ) y + 1 [ q

,...,

(Ti-1

, 1 , ui ,...,U,-J

It is clear that Sf = Sfo = Sfi . Furthermore, if Sf is normal, then fc is i-normal for i = 0, 1,..., n. (To prove this, observe that the case i = 0 is trivial, and then proceed inductively. By the induction

v. GROUP

220

EXTENSIONS

hypothesis, fi is i-normal; hence, so are gi+l ,Sgi+, and fi+l . Thus, it remains to verify that fi+l[ul

,.*.,oi 1, ui+2 9

,*.as

ow]

=0

For i >, 1, one may verify that fi+l[uI

j...,

ui 1, ui+2,...s

un]

=

(-1)'

afi[ul s-.,

ui 1, 1,

ui+2 ,-.*s

on]

+

which is 0 since Sf is (i 1)-normal. A similar argument takes care of the case i = 0.) It follows that every normal cochain which is a coboundary is the coboundary of a normal cochain-so that an = 6@b-1 = qn n 6 P - 1 = enn 9% = 3% n 9%and also that every cocycle is cohomologous to a normal cocycle. Consequently, Rn is isomorphic to Hn for n 2 1. (iii) For 01 E Hn(G, A) define the notion of a splitting module for 01 in the same way as was done for n = 2. Then for every a (and n 2 l), there exists a splitting module. in the sequel. We shall not make use of (5-4-2)

5-4-3. Proposition. Let B be the splitting module for 01 E H2(G, A) constructed in (5-4-1); then the factor module B / A is G-isomorphic to I; more precisely, the following G-sequence is exact (0) A &B I -+ (0)

-

where i is inclusion and j is given by j ( A ) = 0, j ( x 7 ) = T - 1 for T # 1. Proof: Since {T - 1 I T # 1) is a Z-basis for I, it is clear that j is a Z-homomorphism of B onto I with kernel A. Note that j x 7 = T - 1 even for T = 1. Moreover, j is a G-homomorphism, because j(ux7) =j(xu7 - xu

5-44,

+ q,J

Lemma. y =

= (m - 1) - (u -

and

S =

1 u E Z[G] U€G

1

=u(j~)~).

Let

C moo E Z[G] O€G

1) = u(T - 1)

5-4. then

yBCA

Proof:

22 1

THE PRINCIPAL IDEAL THEOREM

-

y =

mS

forsome m e Z

In virtue of (5-4-3) we have

yB C A

v t>

(0) v mu are equal.

yI

=

y(7

- 1)

= 0 for all T E G

I

-

We observe that (5-4-4) is the first place in the discussion where G is required to be finite, and in this case the trace provides a G-homomorphism S : B A which induces a G-homomorphism S : B/(IB) A. In the following, we shall view A as a multiplicative subgroup of U,U / A M G, and as an additive subgroup of B. ----f

545.

Lemma.

We have an isomorphism of abelian groups

under the correspondence a

+ + IB X,

-

We define a map log : U

Proof:

log (uu,) = a

-

au,UC

B/(IB) by

+ + IB X,

then log is a homomorphism because the difference of and log (au,)

is (a6 - b )

log (au,bu,)

=

log ( U ~ ~ U , , , + U , , ) = a

+ log (bu,)

=

a

+ (xu+

- X,

+ ub + a,,,+ + x,, + IB

+ + b + x, + ZB

(where

X,

+ a,,,) - + IB = (u- 1)b + X,

-

(U

a, b E A )

- l)~+ , IB = IB.

Since B/(IB)is abelian, there is an induced homomorphism

-log u :UC

B IB

-

222

v. GROUP EXTENSIONS

-

On the other hand, we may define a homomorphism UIUc by exp a = aUc for a E A and exp x, = u,Uc exp : B for 7 # 1, and then extending to all of B. Note that for 7 = 1, x, = 0, and u, = 1, so that exp x, = u,Uc holds in this case, too. Moreover, exp vanishes on IB because exp (u - 1).

= u0-'UC =

u,uu~'u-'UC = U c

and exp (u - l)x,

= exp (x,,

- x,

+ u,,,

- x,)

=

-

u , , u ~ ' a , , ~ ~ ;= 'UU ~c

Thus, we have an induced homomorphism K p : B/(IB) which is clearly the inverse of @. I 5-46.

Lemma.

U/ Uc

The following diagram commutes:

B IB

Proof:

For a E A, FUJaUc) = N(a) while Slog(aUC) = S(u + I B )

=

S(a)

and because we are viewing A both multiplicatively and additively, these are equal. Furthermore, V(uTUc)= aU,,while

nUsG

SIOg(u,Uc) = S(X, +ZB) = S(X,)

thus completing the proof. I We are now in a position to prove the group theoretic formulation of the theorem, first stated by Hilbert (see [2], [27], [39], [49]) that every ideal of an algebraic number field becomes principal in its absolute class field.

5-4.

THE PRINCIPAL IDEAL THEOREM

223

547. Theorem. Let U be a multiplicative group such that Uc is finitely generated and ( U : Uc) < co. Suppose that A is an abelian subgroup of U which contains Uc. If (A : Uc) = e, then

is the trivial map.

Proof: For a E A, u E U we have uau-'a-' E Uc C A ; so uuu-l E A and A is normal in U. The factor group G = U/A is then a finite abelian group, and upon writing #(G) = ( U : A) = n we have ( U : Uc) = ne. In the exact sequence (I)-A-

U-G-(l)

A may be made into a G-module in the usual way (that is, by conjugation via a section of G in U). Let 01 E H2(G,A) be the cohomology class associated with this group extension, and let B be the splitting module for a as constructed in (5-4-1). Since

Vt+Ameans apply VU+"and then raise to the eth power, it follows from (5-4-5) and (5-4-6) that it suffices to verify that eS:B-A

is the trivial map. (Here, again, A is viewed additively as a subgroup of B and multiplicatively as a subgroup of U . ) Since B / A w I is Z-free on (n - 1) generators, A/Uc is a finite abelian group and Uc is finitely generated, it follows that B is a finitely generated Z-module. In addition, B / ( I B ) is finite abelian with ( B : IB) = ( U : Uc) = ne. Let b, ,..., b,, E B be elements which are representatives of a basis of B/(IB),and if e, ,i = 1,..., m, is the order of b, (modIB), then ne = e, . Now IB, being a subgroup of the finitely generated abelian group B, is itself finitely generated-say by b,n+l ,..., b, E B . If we put e,,,, = -..= e, = 1, then the elements 6 , ,..., b, generate B , e,b, E I B for i = 1,..., s, and ne = e, . Ib, , there exist kj E I such Since B = Zb, and IB = that eibi = h,,b, , i , j = 1,...,s. In other words,

ny

n:

x;=, x:=,

s

1 (eisij - hij)bj = O

j=l

i = I , ...,s

v. GROUP EXTENSIONS

224

Because T = Z[G] is a commutative ring, we may write

A = det (e& - A,)

E

Z[Gl

and Cramer’s rule says that j = 1,

Abj = 0

...,s

Therefore, AB = (O), and by (5-4-4) there exists t E Z such that

A = tS.

It suffices then to show that t = e, and this follows from a simple application of the map E : Z[G] +Z. In fact, e(A) = te(S) = tn and since e(rL,) = 0, e(d) = e{det (etai, - A,)} = det (eta,,) =

n a

ed = ne

1

Thus, t = e and the proof is complete. Principal Ideal Theorem. Let U be a multiplicative group for which U/Ucis finite and Uc/(Uc)” is finitely generated; then U UC

54-8.

Vu-,uo :UC

-

(UCY

is the trivial map. Proof: Let W = U/( U“)”; then Wc = Uc/(Uc)”, ( Wc)” = (l), W/Wc rn U/Uc,Wc/(Wc)ern Uc/(UCY, and W satisfies the hypotheses of the theorem. The following diagram commutes:

Therefore, it suffices to prove our result for W-in

other words,

5-4. THE

PRINCIPAL IDEAL THEOREM

225

we may assume at the start that (U")" = (l), and prove the result for such U only. In this situation, Uc is abelian, and (5-4-7) may be applied with A = Uc. Since (A : Uc) = e = 1, the proof is complete. I Example. Let us exhibit an example, due to Witt [80], which shows that in the principal ideal theorem the hypothesis that Uc/(Uc)"is finitely generated cannot be discarded. Consider the quaternions Q over the reals, R; thus, Q is a 4-dimensional space over R with basis (1, i,j , A}, and multiplication in Q is determined by the relations i2 = j 2 = R2 = -1, .. .. . ij = R = -JZ, jk = i = -Rj, Ri = j = -iR. Of course Q is a division ring which contains C = R * 1 @ R i. In the multiplicative group of Q consider

549.

U = {jmeffirI m E Z,

Y E

Q}

Because eni(*+U= -emir and j 2 = - 1, we need only take m = 0 or 1. Now, U is a group. Closure under multiplication follows from evirj = (cos m + i sin m)j = j cos m - ji sin m = jeffi(-r). Associativity and the existence of an identity are clear. The inverse of emir is ed-r), and the inverse of j&*is -j&*= jed*+l). Let G = {+1, - l} be the cyclic group of order 2, and map U onto G by jmenir (- l)m. This is a homomorphism with kernel A = {emir I Y E Q}. Thus, U / A rn G is abelian and Uc C A. Furthermore, the commutator in U of j and emiris

-

(j)(evir)( -j)(prir)

= e+Zvi(-t)

-and since {e2ni(-r)I Y E Q} = {ezmir1 I E Q}, it follows that A = Uc. Therefore, U/Ucis finite, and Uc/(Uc)"= Uc (since Uc is abelian) is not finitely generated. It remains to show that Vujuc is not the trivial map. In the extension (1) Uc = A U G (l), take as representatives of G in U, u1 = 1 and = j . Consequently, q 1= uL-l = u-l,l = 1 and = j 2 = -1; and then using ua,-l = - 1, so that Vujuc is not (5-2-5) V,,juc(u-lUc) = trivial and the conclusion of the principal ideal theorem does not hold.

- --naeG

CHAPTER

VI

Abstract Class Field Theory

This chapter is concerned with the purely cohomological aspects of class field theory (both local and global), and its contents, which are due to Artin and Tate, are rather standard (see, for example, [6],[43], [46]). The central feature is the somewhat formal notion of a class formation. Its function is to provide a set of axioms from which one is able to prove, by cohomological techniques, the basic theorems of class field theory--especially, the reciprocity law with its many consequences. It follows then that, from this point of view, the verification of the axioms for a class formation constitutes the arithmetic part of class field theory. Needless to say, this arithmetic part, which is a deep and complicated story in itself, is beyond the scope of this book. 61.

FORMATIONS

Definition. A formation is a composite object {G, {GF};A} where G is a group (usually infinite), A is a G-module, {GF}FEzis a nonempty collection of subgroups G, of G which are indexed by some set Z,and such that the following conditions hold: where, as usual, AGF denotes the set of (1) A = (JFEzAG~

614.

226

6-1.

FORMATIONS

227

elements of A invariant under the action of every element of G,. In other words, this condition says that every element of A is left fixed by some member of the family {G,}. For simplicity, we denote ACP by A,. With a view to the applications, the indices F E Z are called fields, A is called the formation module, and A, is called the F-level of the formation.

(2) G is a topological group (this includes Hausdorf€)such that (a) {G,} is a fundamental system of neighborhoods of the identity in G. (b) Every open subgroup of G is a member of {G,}. (c) Every G, is of finite index in G. The group G is called the Galois group of the formation. 6-1-2. Remark. Condition (2) says that the open subgroups of G form a fundamental system of neighborhoods of the identity, and that they are all of finite index in G. Thus, G being totally disconnected but not quite compact, is very much like a Galois group in the usual sense. Actually, the topological properties of G do not play a role until much later, and it is convenient to formulate condition (2) in the following completely equivalent, but entirely algebraic, fashion:

(2) G is a group such that (a') The intersection of two members of {G,} is an element of {G,}. (Of course, the intersection of two subgroups of finite index is of finite index.) Since we assume that for F, F ' E C , G, = G,, t> F = F', this condition says that given G,, , GFt in {GF}there exists a unique element of C, called the composite of the fields Fl and F, , denoted by FIF, , such that GF1n GFa = G,,. (b') Any conjugate of an element of {G,} is again an element of {G,}. In other words, for any G, E {G,} and o E G, G; = oG,o-l is an element of {G,}-so there exists an F' E Z (Fo is said to be a conjugate field of F) such that G p = G;. (We may also write F' = OF.) (c') G, = (11. Every element of {G,} is of finite index in G. (d')

npEP

228

VI. ABSTRACT CLASS FIELD

THEORY

(el) Any subgroup of G which contains an element of {G,} is itself an element of {G,}. It is well-known (see, for example, [59, p. 551 or [lo, p. 61) that conditions (af), (bf), (cf) together say that G can be made into a topological group with {G,} as a fundamental system of neighborhoods of the identity. It is, therefore, immediate that the two versions of (2) are equivalent. Example. The canonical example of a formation, which provides motivation and suggests terminology, is that of ordinary Galois theory. Let k be any field and let $2 be a Galois extension (this means algebraic, normal, and separable) of k. In particular, $2 is often taken as a separable closure of k. Put G = 9(sZ/k), the Galois group of the extension. Let further Z = {F, E, K, L,...} denote the set of all finite extensions of k which are contained in Q. For every F E Z , let G, =. S(Q/F). As usual, G may be made into a topological group (which is compact and totally disconnected) by taking {G,},F E Z, as a fundamental system of neighborhoods of the identity. Each G, is of finite index in G; in fact, (G : G,) = [F : k]. The fundamental theorem of Galois theory for infinite extensions says, in particular, that there is a 1-1 correspondence between the set of all intermediate fields between k and $2 and the set of all closed subgroups of G; consequently, every open subgroup of G is a G, for some F E Z. Let A = Q* be the multiplicative group of $2; this becomes a G-module in a natural way. By Galois theory, A, = A G F is precisely the multiplicative group of F-that is, A, = F*. It is A} is a formation. now immediate that {G, {G,},

6-1-3.

Remarks. In accord with the canonical example, we introduce the following notations and definitions for an arbitrary formation. If F, E E E . we write E 3 F whenever GE C G , , and then say that F is a subfield of E, or that E is an extension field of F. Note that, in this situation, A, C A, since AGF C AGE. A pair of fields F C E determines a layer of the formation; denote it by E/F. A , is called the ground level and A, the top level of the layer.

6-14

6- 1.

FORMATIONS

229

Define the degree [E :F ] of the layer EIF by [E :F ] = (G, : GE); note that it is finite. T h e layer K / F is said to be normal when G, is a normal subgroup of G, (we shall try to reserve the notations KIF, LIF, LIE for normal layers and E/F for ordinary ones); the factor group G F / G Kwhich , we denote by G,/F or G ( K / F )and call the Galois group of the normal layer, is finite and acts in a natural way on the top layer A, = AG'. Namely, for u E G,, let 6 be the corresponding element of GKIFand for a E A, put 6a = ua. This is well-defined and oa E A,, so that A, is a G,/,-module. Moreover, a E A,, so that 6a = a V 6 E GKIF=+ ua = a Vo E G, A ~ K /=F A,-or in other words, the ground level of a normal layer consists precisely of all elements of the top level which are invariant under the action of the Galois group of the layer. (A symbolic proof of this statement looks like ( A K ) G ~=I ~ (ACK)CF/GK = AGF = A, .) T h e normal layer KIF is said to be abelian, cyclic, solvable, etc., when these properties hold for GKIF. I n virtue of all this, the cohomology groups of a normal layer are defined as Y € Z

and the order of H'(K/F) will be denoted by h,(K/F).Our standard cohomological maps occur in this context in the following forms: If F C E C K with KIF normal and a E G, then

= G,/GKu, ~(AGK) = AG; = A G K ~ and that (Note that (GF/GK)(I u* here is not exactly the conjugation map discussed in Section 2-3 because GKIFand G K o pare not subgroups of the same group.

230

VI.

ABSTRACT CLASS FIELD THEORY

Instead, u* is gotten from the homomorphism of pairs (see Section 2-3) (u-l,

: (GKIF

> AK)

-

(Gp/p 9 dAK))

-and all the usual properties of conjugation hold.) If F C K C L, with L/F and K/F both normal, then

It is clearly notationally convenient and helpful to write these maps as: Y€Z

Y€Z Y€Z Y>l

The discussion in this section has been based, in large part, upon carrying over the statements of ordinary Galois theory to formations; however, this is not always valid. For example, the analog of the fundamental theorem of Galois theory does not apply to formations (if it were really needed, we would have included it among the axioms)-that is, the correspondence GE A, need not be 1-1. In fact, it is perfectly possible that G acts trivially on the formation module A, so that all the levels A, = A. On the other hand, among the results which do carry over, the following is quite useful.

-

6-1-5.

Proposition.

(i) Given a finite number of layers

Ei/F i = 1, 2, ...,s over the same ground field, there exists a normal layer K/F containing them all (that is, F C Ei C K ) . (ii) Every subgroup of the Galois group GKjF of a normal layer is of form GKIEfor some intermediate field E.

6- 1.

FORMATIONS

23 1

Proof: (i) For each i, GEi is of finite index in GF ; so as u runs over G , the number of distinct conjugates Ggi = G,; is finite. Consequently, GEP= GK for some K E Z,and it is clear that G, is normal in G , and Ei C K. (ii) Every subgroup of GK/F = GF/GKis of form RIG, where H is a group between GKand GF. Thus, H = G, for some E E Z; so F C E C K and the subgroup is G,/G, = GKIE. I 6-14. Remark. Given any two fields Fl and F, in a formation we have defined (see (6-1-2)) their compositum FIF, by G , n G,, = GF,,, . Since GF,F, is the biggest subgroup of G contained in both GF, and GF2, it follows that FIF, is the smallest field containing both Fl and F, . We may further define the intersection Fl nF, by GFlnF, = [GF, U GFO] (the subgroup of G generated by GF, and GF,). Since [GFIU G,,] is the smallest subgroup of G containing GF,and GF, , it follows that Fl nF, is the largest field contained in both Fl and F, . Because M/F abelian means that GM/F = G,/G, is abelian, we G M3 G : . From this criterion it see that M/F is abelian c:=> follows that if Ml/F and M2/F are abelian then so is (MIM,)/F. Also, any extension E/F contains a unique maximal abelian subextension M/F. T o see this, note that the smallest subgroup H of GFwhich contains G, and for which G,/H is abelian is precisely H = [G, u Gg]. If we define M by G , = [GEU G;], M / F is the desired extension. We leave it to the reader to check that if M/F is abelian and E/F is an arbitrary extension, then ( M E ) / E is abelian. 6-1-7. Exercise. Let E/F be an arbitrary layer of degree n, say. If GF = ULl uiG, is a coset decomposition, and for a E A, It n uia = a‘i, then NEjF: A, -+ A, is a we put NEIFa= homomorphism, called the norm map, which is independent of the choice of coset representatives, and satisfies the usual properties of the norm in field theory. Among these we have:

nl

nl

n.PEGK,F n

pa, a E A,-that (1) If K/F is normal then NKIFa= NK,Fis the “customary” norm. (2) If F C E C E‘, then N E * / F = N E / ,0 N,p/, .

is,

232

VI.


(3) If u E A,, then NE/pa= u [ E : F l . (4) If F C E C E’, then N E # / , A E p C N E / , A E . ( 5 ) If F C E C K with K/F normal, then N K / , A K C N K / E A K . (Note that some of the properties of the norm have already been treated in (2-4-l).) 62.

FIELD FORMATIONS

A formation is said to be a field formation if it satisfies Axiom 1.

H1(K/F)= (1) for every normal layer K/F.

According to (1-5-4), our canonical example of a formation is a field formation. However, for certain formations (namely, those of global class field theory) Axiom I is not accessible directly but rather through an ostensibly weaker formulation. 62-1. Axiom 1’.

degree.

Proposition.

I n a formation, Axiom I is equivalent to

H1(K/F)= (1) for every cyclic layer K/F of prime

Proof: This is a special case (with Y = 1, u = 0) of the following lemma which shall be used later in the case Y = 2, v=l. I Lemma. Suppose that for the formation {G, {G,}, A) there exists a positive integer Y such that for all F C K C L with KIF, L/F both normal, the sequence

6-2-2.

is exact. Suppose, further, that for a fixed integer v 2 0, h,(K/F)I[K:F]” for all cyclic layers K/F of prime degree; then h,(K/F)I[K: F l yfor all normal layers K/F.

6-2. FIELD

233

FORMATIONS

Note first that since the inflation-restriction sequence is always exact in dimension 1 (see (3-4-2)), (6-2-1) does indeed follow from the case T = 1, v = 0 of the lemma. As for the proof of the lemma, we show first that the desired divisibility relation holds for [ K :F] = p", p prime. (Of course, a statement of divisibility includes the finiteness of h,(K/F).) For n = 1, this is asserted by the hypothesis of the lemma. Suppose then, inductively, n, we must show that it holds that divisibility holds for all m for n 1-that is for p n + l . The group GKtFis then of order pn+l and (by standard properties of p-groups) contains a proper normal where F C K' C K. Therefore, subgroup-which has form GKIK~ p [ K : K'] pn and p [K' :F ] \< pn, so that by the induction hypothesis h,(K/K')I[K : K']" and h,(K'/F)J[K' :F]. Now, it follows from the exact sequence Proof:

(0 +I">

Proof: Since Axioms I and I' are equivalent, the first equivalence is trivial. T o prove the second equivalence, note that according to Axiom 0, h,(K/F) = [ K :F ] h,(K/F) for cyclic layers of prime degree-and then, Axiom I" holds w h,(K/F) = [K:F] -G=- h,(K/F) = 1 e=-Axiom I' holds. I

If a formation satisfies (0 + I"}, then the Proposition. second inequality h,(K/F)([K:F ] holds for all normal layers K/F.

6-3-5.

VI.

242


Since Axiom I holds we know (see (6-2-5)) that for any Proof: F C K C L with K / F , L/F normal, the sequence

is exact. It remains only to apply (6-2-2) in the case

t

I

v=l.

= 2,

We turn next to a weaker version of Axiom 11.

To each field F E C there is associated a subgroup H2(*IF)of the Brauer group Ha(*/F)with a monomorphism

Axiom -- 11'.

-

invp : ~2

-

(+)

such that

Q

(a) If there exists a normal layer K / F of degree n, then H2(*/F) contains a cyclic group of order n. (b) For any layer E/F the symbolic restriction resFAEmaps H2(*/F)into H2(*/E), and on Ha(*/F)

inv,

6-34. Proposition. satisfies Axioms (0 I"

-

0

rap+, = [ E :F]invF

A formation is a class formation

+ + II'}.

Proof: t=

:

t*

it

=*: It is clear that {I + 11) implies (0+ I" + II'}. It suffices to show that for every F,H2(*/F)= H2(*IF)-

ixF.

as we may then put invp = Consider any normal layer K/F with [K:&'I = n, say. By (6-3-4), Axiom I holds, so according to (6-2-5) the sequence

is exact. Now, 11' says that H2(*/F)contains a cyclic group order n, and that inv, re+-,, T, = n G F T, = invF (nT,)= invF (0) = 0

T, of

6-4.

THE MAIN THEOREM

243

-

Since inv, is a monomorphism, we have res,+, T , = (0), and then (as observed in (6-2-6)) T , C H 2 ( K / F ) . On the other hand, by (6-3-5), h,(K/F) I [ K :FJ= n = #(T,). Therefore, T , = H2(K/F), and it follows that H2(*/F)= H2(*/F). Putting inv, = inv, , we note that inv, {H2(K/F)}= ((l/n)Z)/Z since ((l/n)Z)/Z is the unique cyclic subgroup of order n in Q/Z. This completes the proof. I T h e importance of Axiom 11' is then that the whole Brauer group H2(*/F) can be captured by using only certain normal layers K/F which, in practice, may be taken of a simple type. (In local theory one uses the unramified extensions; in global theory one uses the cyclotomic extensions.) A field formation is a class formation cyclic of order [K :F] for every normal layer K/F, and forF C E C K resK/,+,IE : H 2 ( K / F --+ ) H 2 ( K / E )is an epimorphism.

6-3-7.

Exercise.

e=> H2(K/F)is

Hint: Let F, be the base field and consider a chain of fields F , C K , C K 2 C ..- such that (i) K,/F, is a normal layer for T = 1, 2, ..., and (ii) if E/F, is any layer, there exists an T for which CE :FOI", :FoI6-3-8. Exercise. Let G be an infinite cyclic multiplicative group, and consider the G-module A = Z with trivial action. If {G,} is the set of all subgroups of G (excluding (l)), then {G, {G,}, A} is a class formation.

64.

T H E MAIN T H E O R E M

We are now in a position to state the theorem which is the focal point of substantially all of the work in this and earlier chapters; the proof consists essentially of the collation of previous results. 641. Main Theorem. (Tate). Let K/F be any normal AK) be layer in a class formation, let ax/,E H2(K/F)= H2(GKlF,

244

VI.

its canonical class, and let 0 = 0; : A, x natural G,/,-pairing. Then the map @"IF

-


z)

: Hn(GK/F,

---+

Z

H*+'(GKIF, A,)

A, denote the

=

K

Hn+*(7)

whose action on [ E Hn(GK/F, Z) is given by

is an isomorphism onto for all integers n. According to (6-1-5) every subgroup of GKIFis of form Proof: GKIEwhere F C E C K,and by (6-3-1), r e S G ~ p + G ~=KIF I~

=

reSK/F-rKIEOrKIF = reSF-rE ''KIF

=

OlK/E E

K (r)

HZ

Therefore, it suffices to verify the hypotheses of (4-3-8) solely for GKIFand the fundamental class aKIF.More precisely, the proof of the main theorem is complete as soon as it has been shown that (i)

is an epimorphism. (ii) @OKIF is an isomorphism onto. (iii) @i,F is a monomorphism. @;;F

Now, (i) is clear because H1(K/J')= (0), and (iii) is clear because (see (1-5-3)) H1(GK/F,Z) = (0).As for (ii), if

then aKIFis a generator of the group H2(K/F)which is cyclic of order m. On the other hand, H0(GKlF,Z) M Z / ( m Z ) (see (1-5-7)), so that HO(G,/, , Z) is a cyclic group of order m which is generated by ~ ( 1 ) Ir . suffices, therefore, to show that aKIFwe~ ( 1 = ) From (4-3-6), we know that ( Y K / F v, ~ ( 1 = ) (el), and since 81 : A, A, is the identity map this equals 1*cuKlF = This completes the proof. I

-

For any normal layer K/F of degree m in 642. Corollary. a class formation, we have isomorphisms:

6-4.

Proof:

THE MAIN THEOREM

245

The last ispmorphism is immediate from (4-4-7).

I

643. Proposition. T h e isomorphisms of the main theorem commute with restriction and corestriction; that is, if F C E C K with K/F normal, then, for all n

Proof:

We must show that the following diagram commutes:

re8

-which

says that for

It

re81 tcor

cor

6 E Hn(GKIF,Z ) ,

reSKIF+KIE (aKIF

4) = & K / E

reSKIF+KIE

6,

5)

CorKIE+KIF

5.

and for 5 E Ha( G K / E , Z), cOrKIE+KIF (OLKIE " 0

= (yKIF

246

VI.


But, since aKtE is the restriction of aKIF, these assertions are immediate from (4-3-7). I

644.

Proposition.

then, for all n

Proof:

>1

If F C K C L with L/F and K/F normal,

We show that the following diagram commutes:

[ L : K ] !in!

Corollary. Suppose that F C K C L with L/F and KIF normal. If [K : F]I[L : K ] , then, for all n 2 3,

645.

infK/F+L/F

-

:Hn(K/F)

H"(L/F)

is the trivial map. Proof: In the preceding diagram the horizontal maps are isomorphisms onto, so it suffices to show that [L : K ] infCKIF.+GLIF is the zero map. But this is trivial since [ K :F ] Hn(G,lF , 2)= (0) (by (3- 1-6))- I This result says that infKIF+*lKis a weak map, in general. If there are sufficiently many fields-for example, if the field F has extensions of degree divisible by every integer m-then infKtF+,., is the zero map and Hfl(*/F)= (0) for n 2 3.

6-4.

247

THE MAIN THEOREM

646. Exercise. The isomorphisms of the main theorem commute with conjugation; in other words, if K/F is a normal layer in a class formation, then the following diagram commutes for all u E G and all integers n:

lo*

.1D*

Remark. For a normal layer K/F of degree m in a class formation, the main theorem asserts the existence of isomorphisms of Hn(G,/F, Z) onto Hnf2(K/F)-where SzlF is &cupping by the canonical class aKIFE H2(K/F) on the left (where f3 = 0 ; : A , x Z A , is the natural GKIF pairing). Given 5 E Hn(GK/F, Z) and /I E Hn+2(K/F),we should like to decide when 5 and /I correspond to each other under For this we need to recall and fix some notation. There are natural G,IF-pairings 0, : Z x Z Z and &Iz: Z x Q/Z Q/Z; the exact GKlFZ Q Q/Z 0 leads to the coboundsequence 0 H-"(G,/, , Z) which is an ary map 6, : Hpn-l(GK/F, Z) isomorphism onto; the canonical map in dimension -1, 7 = r]Q/z : (Q/Z), = ((l/m)Z)/Z H-l(GK/F, Q/Z) is an isomorphism onto. Now,

647.

-

- - - --

fl

= @:IF

5

= aK/F

fl

5

6,x

x E H - ~ - ~ ( G , I ~Q/Z) , (in view of ( a K i F we 5 ) wo &*x = ~ K / w F e (t; we, **XI for all

= (-l)naK/F

we

5)

=

(4-4-8)), and

weQ/z

(by (4-3-3))

x)

(by (4-3-5))

{8*dv-1(5 weQ/Z x )I1 IY~KIF we K [ T ' ( ~ weQlz x)I (by

= ( -1)naKlF =

(-

In the proof of the main theorem we have seen that %/F

K(1)

= LYK/F;

(4-5-9))

VI.

248


therefore, taking invariants, it follows that

** invF(fl V for all x E H-"-l( G K / p ,Q/Z). fl = a K / F V

O

s * X ) = (-l)n?-l(c VOo,z

O

x)

Exercise. Suppose that K / F is a normal layer in a class formation; then according to (6-4-2), H4(K/F)is isomorphic to the character group GKlF-rnore precisely: the main theorem (6-4-1) implies that : H a ( G K / F , 2) __* H4(K/F) is an isomorphism onto; furthermore, the coboundary map

6-44

841 :H'

(GKIF

9

%)

-

H8(GKIF

I

z)

is an isomorphism onto; finally, G K / p = Hom ( G K / F , Q/Z) may be identified with H ' ( G K / p , Q/Z) by associating with a character x E cKiF the standard 1-cocycle, also denoted by x, such that ~([u]) = x(a) V a E G K / p ; thus, the isomorphism between eKlF and HYK/F) is given explicitly by the correspondence

x

-

aKIF

s*x

where 0 : AK x Z A,. Consider the situation F C E C K C L with KIF and L/F both normal, and let M/F denote the maximal abelian sublayer of K/F. __+

-

(I) The restriction map res :H4(K/F)

H4(K/E) corresponds to the map of G K / F - - - + G K / E which takes x +x I GKIE.If M C E, then res is the trivial map. (2) The corestriction map cor :H 4 ( K / E ) H4(K/F)corresponds to the map of

which takes x

__+

x V where 0

6-5.

249

RECIPROCITY LAW AND NORM RESIDUE SYMBOL

- --

is the reduced group theoretical transfer (see (3-5-4)). If

M C E, then cor is the trivial map. (3) The inflation map inf : H4(K/F)

H4(L/F)corresponds G L / F given by x +f f L : W ,where to the map of G K / F f E GLIFis defined as follows: let o 6 be the natural GL/F/GL/K G K i F , then R(U) = x ( 9 . map of G L / F (4) Suppose that x E GKlFand /3 E H*(K/F);if x and /?? correspond, then (see (4-5-4))

X(.)

Hint:

= invF

(('#)*8)

VUEG

It is convenient to use (6-5-1) and (6-5-2).

62.

RECIPROCITY LAW A N D NORM RESIDUE S Y M B O L

In this section we derive the key consequences of the main theorem; these are the basic results of class field theory. Consider the normal layer K/F of degree m. (It is understood here, and henceforth, that we are dealing with a class formation.) The main theorem, in dimension -2, provides an isomorphism onto: =

: H-'(GK/F

9

-

Z)

K

H O(7 =) ~ ( G K

/ F9 A K )

which is given by a cup product. Since

(in order to conform to the applications, we write the formation module A multiplicatively) we have

%!EN-

A F

NKjFAK

Moreover, this isomorphism can be given explicitly.

250

-


THEORY

T o do this, recall first that K : A, Ho(K/F)is a homomorphism onto with kernel NKIFAKwhich induces the isomorphism I? of AF/(NKlFAK) onto Ho(K/F). On the other hand, the map a 5, = 6;'q(u - 1) is a homomorphism of G K / F onto H-a(GKIF,2) with kernel G;/, (see (3-5-2)) which induces an isomorphism of G K / F / P K / F onto H-2(G,/F, 2). (Here q is the canonical map of I onto H - ' ( G , I ~ I), , which is an isomorphism; 6, : H-2(GK/F, 2) H-l(GK/F, I) is the coboundary arising from the exact sequence 0 I Z &it is an isomorphism onto.) Combining these isomorphisms with @,lF gives the desired isomorphism in (*). If, for a E A, and v E GKIF, the elements aNKIFA, and uG&, correspond to each other under this isomorphism, we write aN,/+4, uG;/F and also a u (even though the latter is not a 1-1 correspondence). We have proved:

-

- -- - r

-

6-5-1.

(Reciprocity Law).

Theorem.

layer K/F we have

-

For any normal

In fact, if a E A , and u E GKIF,then

where 8 = 0:: A ,

x Z+

A,.

It is also possible to give a dual description of the reciprocity law isomorphism. 6-5-2.

Corollary.

If a E A , and u E G K / F , then

-

(Here 6, : H1(GK/=, Q/Z) H2(GK/F,Z), and by the standard identification x E &,lF = Hom ( G K / F , Q/Z) is viewed

Proof:

6-5.

251

RECIPROCITY LAW AND NORM RESIDUE SYMBOL

also as an element of H1(GKIF, Q/Z).) Using the case n = -2 of (6-4-7) we have: Q

-

(I

for all x

e=.KQ = a K / F w E

g

5, * invF (KQ W g 6,~)

=

v-'([,

WgOlz

X)

H1(GK/F,Q/Z), and according to (4-5-8)

6-5-3. Remark. The isomorphism of the reciprocity law provides maps in two directions. One way there is a homomorphism of GKlF onto AF/(NK/FA E # 0, it sufficesto verify the finite intersection property-but this is trivial, because from K C K' =- X K 3XKt it follows that XKln XKI3 X K I K s # 0 . If we take b E X, then b E NK/EAK = DE and NEl& = a. I

x,

x,

nmE

nmE

Proposition. For every natural number m and every field F in-a topological formation satisfying Axiom III(a), (b) we have

6-6-3.

DF = DF

n A% OD

=

n-1

Proof: T o show that D, = DF (i.e., that the group D, is infinitely divisible) it suffices to prove that D , = D; for every

6-6.

265

THE EXISTENCE THEOREM

prime p. Consider any a E D,. For sufficiently large E-that for E 3 FFp-we have (using III(a) and (b))

is,

By {allp} we mean {b E A , I b p = a}; it is a closed set in A , , and a subset of {b E A , I bp = a} which is compact in A , (in virtue of Axiom III(b)). Thus, {&P} is compact in A, and NE/,AE is closed in A,+o that X, is compact and closed in A , , and in {ul/p}. Now, the X , have the finite intersection property since X,,n X E ,3 XEIE,# 0,and by compactness X = nnFF,X , # 0. For EX we have bP =aED,and be

n

NEIFAE

E3FF9

=

n

NEIFAE

E X

= D,

This implies that D, C D$ , and then D, = 0:. Turning to the remaining part of Wthe proof, we have A: . On the other D, = DF C A% for n = 1,2,..., so D, C hand, for each E 3 F we have

n A; c A$,:“]= m

NEIFA,

c NEIFAE

n-1

Hence,

n:-l

A; C D , and the proof is complete.

I

6-6-4. Abstract Existence Theorem. In a topological class formation satisfying Axiom 111, a subgroup of any level A , is a norm subgroup -=ait is an open subgroup of finite index in A , . =*: A norm subgroup N E l F A E of A, is closed (by Axiom III(a)) and of finite index (by (6-5-8)), so it is also open in A,. e=: Let C be an open subgroup of finite index in A , , and let 93 = {B}denote the set of all norm subgroups of A , . From (6-6-3), we have D, = DkApAF:c) C A F( ” F : C ~ )C-so B C C and then

Proof:

nBEa

266


THEORY

( A F - C)C U B ( A F - B). Since A,- C is closed in A,, is an open covering of the compact set {(A, - B)I B (A, - C) n U,. Extract a finite subcovering C

u, n ( A , - C )C U ( A , - Bd. 1

By (6-5-7), B' = n ; B S is an element of a,and it is clear that (A, - C) n U,C A, - B'. Therefore, (A, - C) n U, n B' = 0 and U , n B' C C. Because B' and C are open subgroups of finite index, so is B' n C, and then (B' n C) U, is an open subgroup of finite index containing U , . By Axiom III(c), (B' n C) U , is a norm subgroup. Hence B n (B' n C) U, is a norm subgroup. Since B' n U,C C it is easy to see that B' n (B' n C)U,C(B' n CXB' n V,)C C

Thus, C contains a norm subgroup, and is therefore itself a norm subgroup.

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Subject Index A

Cohomology ring, 189 Coini, 35 Conjugate fields, 227 Conjugation, symbolic, 235 Coker, 35 Compatible operator, 5 Complete representation, 38, 39 Complete system of representatives,

.4belian group, divisible, 42 Abstract existence theorem, 265 Acyclic chain complex, 13 Adjoint map, 163 .4dmissible endomorphism, 10 Admissible homomorphism, 2 .Admissible map, 6 .innihilator, 207 h g m e n t e d chain complex, 13 Automorphism of extensions, 200

195

Composite of fields, 227 Conjugation, 65 Contracting homotopy, 16 Corestriction, 71 Crossed homomorphism, 25 Cup product, 143 with respect to e, 156

B Baer multiplication, 204 Boundary operator, 6 Brauer group, 236

D 1)eflation map, 111 Degree, 229 Derived group, 2 Differential group, 1 Differential operator, -3 Differential graded group, 5 1)imension shifters, 108, 139 Direct epimorphism, 39 Direct family, 38 Direct product, 38 Direct sum, 38 Divisible abelian group, 42 Ihality theorem, 168

C Canonical class, 238 Canonical isomorphism, 55 Cell, 14 -1, 23 -(N 1),23 Chain complex, 13 Character group, 115, 168 Class field, 257 Class formation, 237 Cohoundary operator, 6 Cochain, i-normalized, 219 Cohomology, 6 Cohomologically trivial, 119 Cohomologous, 27 Cohomological equivalence, 122 Cohomology group, 6, 27, 229

+

E Empty cell, 14 Endomorphism, admissible, 10 27 1

272

SUBJECT INDEX

Equivalence, cohomological, 122 Equivalent extensions, 200 Exact functor, left, 102 Exact sequence short, 3 split, 102 Existence theorem, 265 Extension, 199, 228 Extension problem, 197 External direct product, 38 External direct sum, 38

F F-level, 227 Factor extension, 214 Factor set, 28, 197 Field, 227 Field formation, 232 Finite free chain complex, 13 First inequality, 241 Five lemma, 36 Formation, 226 Formation module, 227 Four lemma, 36 Free chain complex, 13 Functor, 102 Fundamental class, 238 G

G-complex, 14 G-module, 13 uniquely divisible, 90 G-pairing, 155 G-regular module, 47 G-ring, 189 G-sequence, 14 G-special module, 126 Galois group, 227, 229 Ground level, 228 Group of characters, 207 derived, 2 differential, 1 graded, 5

of n-coboundaries, 6, 27 of n-cochains, 6, 27 of n-cocycles, 6, 27 nth cohomology, 6, 27 nth-derived, 6 Group theoretical transfer, 115, 118 reduced, 116

H Herbrand characteristic, 12 Herbrand quotient, 9, 96 Herbrand’s lemma, 8 Hilbert’s theorem 90, 30 Homology, 6 Homomorphism admissible, 2 crossed, 29 of extensions, 199 of pairs, 46 Homotopy, contracting, 16

I Independence of complex, 55, 56 Induced G-module, 128 Inflation, 68 symbolic, 234 Inflation-restriction sequence, 108 Injective module, 42 Integer group ring, 12 Integral duality theorem, 169 Invariant, 237

J Japanese homomorphism, 170

L Layer, 228 Level. 228

M Maximal generator, 190

273

SUBJECT INDEX

Module G, 13 G-regular, 47 G-special, 126 induced, 128 injective, 42 projective, 40,101 splitting, 218

N n-cell, 14 n-coboundary, 6 n-cochain, 6 n-cocycle, 6 nth-cohomology group, 6 nth-derived group, 6 Nakayama map, 170, 178 Negative part of a G-complex, 14 Norm group, 255 Norm index, 255 Norm map, 231 Norm residue map, 251 Norm residue symbol, 251 Normal basis theorem, 88 Normal cochain, 219 Normal layer, 229

0 Operator boundary, 6 coboundary, 6 differential, 2

P )-primary part, 92 p-Sylow subgroup, 91 Pair, 46 Pairing, G-, 155 Period, 190 Periodic, differential graded group, 12 Positive part of a G-complex, 14 Principal crossed homomorphism, 29

Principal ideal theorem, 224 Product, cup, 143 Projective module, 40,101

Q Quaternion group, generalized, 191

R Reciprocity law, 250 Reduced group theoretical transfer, 116, 118 Regular, G-, 47 Restriction, 67 symbolic, 235 Ring, cohomology, 189 G-, 189

S Second inequality, 241 Section, 195 Shapiro’s lemma, 131 Short exact sequence, 3 Special, G-, 126 Split exact sequence, 39, 40, 102 Splitting factor set, 28 Splitting module, 218 Standard G-complex, 23,24 Subextension, 213 Subfield, 228 Sylow group, 91 Symbolic conjugation, 235 Symbolic corestriction, 235 Symbolic inflation, 234 Symbolic restriction, 235

T Tate’s theorem, 243 Tensor product, 136 Top level, 228 Topological formation, 262

274 Torsion group, 92 Trace element, 24 Trace function, 70 Transfer, 71, 118 group theoretical, 115, 118 Translation, 199 Translation theorem, 258 Trivial, cohomologically, 119 Twist, 152, 155

SUBJECT INDEX

U Uniquely divisible G-module, 90 Unitary G-module, 13 Universal norm, 263

V Verlagerung, 115, 116

Pure and Applied Mathematics A Series of Monographs and Textbooks Edited by

Paul A. Smith and Samuel Ellenberg Columbia University, New York

1 : ARNOLD SOMMERFELD. Partial Differential Equations in Physics. 1949 (Lectures on Theoretical Physics, Volume V I ) 2: REINHOLD BAER.Linear Algebra and Projective Geometry. 1952 3 : HERBERT BUSEMANN A N D PAUL KELLY.Projective Geometry and Projective Metrics. 1953 4 : STEFAN BERGMAN A N D M. SCHIFFER. Kernel Functions and Elliptic Differential Equations in Mathematical Physics. 1953 BOAS,JR. Entire Functions. 1954 5 : RALPHPHILIP BUSEMANN. The Geometry of Geodesics. 1955 6 : HERBERT 7 : CLAUDE CHEVALLEY. Fundamental Concepts of Algebra. 1956 8: SZE-TSENHu. Homotopy Theory. 1959 Solution of Equations and Systems of Equations. Second 9 : A. M. OSTROWSKI. Edition. 1966 N~. on Analysis. Volume I, Foundations of Modern Analy10: J. D I E U K J N Treatise sis, enlarged and corrected printing, 1969 11 : S. I. GOLDBERG. Curvature arid Homology. 1962 12 : SICURDUR HELGASON. Differential Geometry and Symmetric Spaces. 1962 13 : T. H. HILIIEBRANDT. Introduction to the Theory of Integration. 1963 ABHYANKAR. Local Analytic Geometry. 1964 14 : SHREERAM L. BISHOPA N D RICHARD J. CRITTENDEN. Geometry of Manifolds. 1964 15 : RICHARD 16: STEVEN A. GAAL.Point Set Topology. 1964 17: BARRY MITCHELL. Theory of Categories. 1965 18: ANTHONY P. MORSE.A Theory of Sets. 1965

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Cohomology of Groups

Cohomology of groups