From now on we assume that we are given a group pG�¨q (in multiplicative notation) and consider modules over the group algebraKGof Gover a commutative ringK. The main aim of this chapter is to introduce the cohomology groups of G and describe concrete projective resolutions which shall allow use to compute them in some cases.
References:
[Bro94] K. S. B����,Cohomology of groups, Graduate Texts in Mathematics, vol. 87, Springer-Verlag, New York, 1994.
[Eve91] L. E����,The cohomology of groups, Oxford Mathematical Monographs, The Clarendon Press, Oxford University Press, New York, 1991.
[Rot09] J. J.R�����,An introduction to the theory of groups. Fourth ed., Graduate Texts in Mathema- tics, vol. 148, Springer-Verlag, New York, 1995.
[Wei94] C. A.W�����,An introduction to homological algebra, Cambridge Studies in Advanced Mathe- matics, vol. 38, Cambridge University Press, Cambridge, 1994.
11 Modules over the Group Algebra
Lemma-Definition 11.1 (Group algebra)
IfG is a group andK is a commutative ring, we may form thegroup ringKG whose elements are the formal linear combinations∞
�PGλ��withλ� PK, and addition and multiplication are given by ÿ
�PGλ��` ÿ
�PGµ��“ ÿ
�PGpλ�`µ�q� and ` ÿ
�PGλ��˘
¨` ÿ
�PGµ��˘
“ ÿ
���PGp��q�� � ThusKG is aK-algebra, which as aK-module is free with basisG. Hence we usually callKG the group algebra ofG over K rather than simplygroup ring.
Proof : By definition KG is a free K-module with basisG, and the multiplication in G is extended by K- bilinearity to the multiplication KGˆKG ›ÑKG. It is then straightforward to check that this makes KGinto aK-algebra.
In this lecture, we will mainly work with the following commutative rings: K “Z the ring of integers, and fields.
48
Remark 11.2
(a) KG-modules and representations:
If K is a field, then specifying a KG-module V is the same thing as specifying a K-vector spaceV together with a K-linear action ofG onV, i.e. a group homomorphism
G ›ÑAutKpVq “:GLpVq� or in other words aK-representation of G.
Similarly, if K “ Z, then specifying a ZG-module M is the same thing as specifying a Z-moduleM together with a Z-linear action ofG onM, i.e. a group homomorphism
G݄AutZpMq also called an integral representation ofG.
(b) Left and right KG-modules:
Since G is a group, the map KG ›Ñ KG such that � fiÑ �´1 for each � P G is an anti- automorphism. It follows that anyleft KG-moduleM may be regarded as aright KG-module via the right G-action �¨� :“�´1�. Thus the sidedness of KG-modules is not usually an issue.
(c) The trivialKG-module:
The commutative ringK itself can be seen as a KG-module via the G-action
¨ : GˆK ›ÑK
p�� λqfi›Ñ�¨λ:“λ
extended by K-linearity to the whole of KG. This module is called the trivial KG-module.
An arbitraryKG-module A on whichG acts trivially is also calleda trivial KG-module.
(d) Tensor products ofKG-modules:
If M and N are two KG-modules, then the tensor product MbK N of M and N balanced overK can be made into aKG-module via the diagonal action ofG, i.e.
�¨ p�b�q:“��b�� @�PG�@�PM�@�PN� (e) Morphisms ofKG-modules:
If M and N are twoKG-modules, then the abelian group HomKpM�Nq can be made into a KG-module via the conjugation action of G, i.e.
p�¨�qp�q:“�¨�p�´1¨�q @�PG�@�PM� (f) The augmentation map and the augmentation ideal:
The map ε : KG ›Ñ K defined by εp�q :“ 1for every �PG and extended by K-linearity to the whole of KG is called the augmentation map. This is a surjective homomorphism of K-algebras whose kernel
kerpεq “:IG
is called the augmentation ideal of KG, and KG{IG – K. (Notice that ε is hence also a homomorphism of KG-modules, so that we can also see IG as aKG-submodule ofKG.)
Lemma 11.3
(a) IG is a freeK-module withK-basist�´1|�PGzt1uu.
(b) IfX is a set of generators for the group G, then IG is generated as aKG-module by the set t�´1|� PXu�
(c) If M is a KG-module, then IG¨M“ x�¨�´�|�PG� �PMyK.
Proof : (a) First of all, the set S:“ t�´1|�PGzt1uu is clearly contained in kerε by definition of ε.
The setS isK-linearly independent since 0“ ÿ
�PGzt1u
λ�p�´1q pλ�PKq
“ ÿ
�PGzt1u
λ�� ´ ÿ
�PGzt1Gu
λ�
implies that�0for every�PGzt1u, becauseG isK-linearly independent inKG.
To prove thatS spansIG, let∞
�PGλ��(λ�PK) be an element ofIG“kerε. Hence 0“εpÿ
�PGλ��q “ ÿ
�PGλ�
and it follows that ÿ
�PGλ��“ ÿ
�PGλ��´0“ ÿ
�PGλ��´ÿ
�PGλ� “ÿ
�PGλ�p�´1q “ ÿ
�PGzt1u
λ�p�´1q�
(b) Clearly, for every elements�1� �2PG we have:
�1�2´1“�1p�2´1q ` p�1´1q and �´11 ´1“ ´�´11 p�1´1q Thereforet�´1|�PGzt1uuÑxt�´1|�PXuyKG, which implies that
IG“ xt�´1|�PGzt1uyK Ñxt�´1|�PXuyKGÑIG and hence equality holds.
(c) Follows from (a).
Definition 11.4 (G-fixed points and G-cofixed points) LetM be a KG-module.
(a) TheG-fixed points of M are by definitionMG :“ �PM|�¨�“�@�PG( . (b) TheG-cofixed points of M are by definitionMG:“M{pIG¨Mq.
Exercise [Exercise 1, Exercise Sheet 8]
LetM and N beKG-modules. Prove that:
(a) MG is the largest submodule ofM on whichG acts trivially;
(b) MG is the largest quotient ofM on whichG acts trivially;
(c) HomKpM�NqG “HomKGpM�Nq;
(d) pMbK NqG –MbKGN;
(e) If G is finite, then pKGqG “ x∞
�PG�yK and ifG is infinite, then pKGqG“0.
12 (Co)homology of Groups
We can eventually define the homology and cohomology groups of a given groupG.
Definition 12.1 (Homology and cohomology of a group) LetA be aKG-module and let �PZ•0. Define:
(a) H�pG�Aq:“TorKG� pK�Aq, the�-th homology group ofG with coefficients in A; and (b) H�pG�Aq:“Ext�KGpK�Aq, the�-th cohomology group ofG with coefficients in A. Remark 12.2
A priori the definition of the homology and cohomology groups H�pG�Aq and H�pG�Aq seem to depend on the base ring K, but in fact it is not the case. Indeed, it can be proven that there are group isomorphisms
TorKG� pK�Aq –TorZG� pZ�Aq and Ext�KGpK�Aq –Ext�ZGpZ�Aq
for each �PZ•0 and everyKG-module A, which can also be seen as aZG-module via the unique ring homomorphism Z›ÑK, mapping1Z to1K. See Exercise 2, Exercise Sheet 8.
In view of the above remark, from now on, unless otherwise stated, we specify the ring K to Z.
Proposition 12.3 (Long exact sequences)
Let 0 //A � //B ψ //C //0 be a short exact sequence of ZG-modules. Then there are long exact sequences of abelian groups in homology and cohomology:
(a)
¨ ¨ ¨ //H�`1pG�Cqδ�`1//H�pG�Aq �˚ //H�pG�Bq ψ˚ //H�pG�Cq //¨ ¨ ¨
¨ ¨ ¨ //H1pG�Cq δ1 //H0pG�Aq �˚ //H0pG�Bq ψ˚ //H0pG�Cq //0 (b)
0 //H0pG�Aq �˚ //H0pG�Bq ψ˚ //H0pG�Cq δ0 //H1pG�Aq //¨ ¨ ¨
¨ ¨ ¨ //H�pG�Aq �˚ //H�pG�Bq ψ˚ //H�pG�Cq δ� //H�`1pG�Aq //¨ ¨ ¨
Proof : By definition of the homology and cohomology groups ofG, (a) is a special case of Proposition 10.5(a) and (b) is a special case of Theorem 10.10(i).
To start our investigation we characterise the (co)homology of groups in degree zero:
Proposition 12.4
LetA be aZG-module. Then
(a) H0pG�Aq –ZbZGA–AG, and (b) H0pG�Aq –HomZGpZ�Aq –AG as abelian groups.
Proof :
(a) By Proposition 10.10(a),H0pG�Aq “TorZG0 pZ�Aq –ZbZGA. Moreover by Exercise 1(d), Exercise Sheet 8, we haveZbZGA– pZbZAqG–AG.
(b) We already know that H0pG�Aq “Ext0ZGpZ�Aq –HomZGpZ�Aq. Moreover by Exercise 1(c), Exer- cise Sheet 8, we haveHomZGpZ�Aq –HomZpZ�AqG–AG.
The degree-one (co)homology groups with coefficients in a trivialZG-module can also be characterised using the augmentation ideal. For this we letG��:“G{rG�Gsdenote the abelianization of the groupG.
Exercise [Exercise 4, Exercise Sheet 8]
Prove that:
(a) There is an isomorphism of abelian groupspIG{pIGq2�`q – pG���¨q.
(b) IfA is a trivialZG-module, then:
¨ H1pG�Aq –IGbZGA–IG{pIGq2bZGA–IG{pIGq2bZA–G��bZA;
¨ H1pG�Aq –HomZGpIG�Aq –HomZGpIG{pIGq2�Aq
–HomZpIG{pIGq2�Aq –HomZpG���Aq –HomGrppG�Aq.
Corollary 12.5
IfZdenotes the trivial ZG-module, thenH1pG�Zq –IG{pIGq2 –G��. Proof : This is straightforward from Exercise 4, Exercise Sheet 8.
13 The Bar Resolution
In order to compute the (co)homology of groups, we need concrete projective resolutions of Z as a ZG-module.
Notation 13.1
Let�PZ•0 be a non-negative integer. Let F� be the freeZ-module with Z-basis consisting of all p�`1q-tuplesp�0� �1� � � � � ��q of elements ofG. Then the group G acts onF� via
�¨ p�0� �1� � � � � ��q “ p��0� ��1� � � � � ���q�
and it follows that F� is a freeZG-module with ZG-basis �� :“ p1� �1� � � � � ��q |�� PG( . First, for each 0§�§�, define maps
B�: G�`1 ›Ñ G�
p�0� � � � � ��q fiÑ p�0� � � � ��q�� � � � � ��q,
where the check notation means that�� is deleted from the initialp�`1q-tuple in order to produce an�-tuple, and extend them byZ-linearity to the whole ofF�. If�•1, define
�� :F�›ÑF�´1
� fi›Ñ ÿ�
�“0p´1q�B�p�q�
Since the mapsB� are ZG-linear by definition, so is��. Finally consider the augmentation map ε:F0 “ZG›ÑZ
�fi›Ñ1 @�PG�
Proposition 13.2
The sequence ¨ ¨ ¨ ��`1//F� �� //F�´1 ��´1
//¨ ¨ ¨ �1 //F0 is a freeZG-resolution of the trivialZG- module.
Proof : Set F´1 :“ Z and �0 :“ε (note that ε “�0 is consistent with the definition of ��). We have to prove that the resulting sequence
pF‚� �‚q⇣ε Z:“ˆ
¨ ¨ ¨ ��`1 //F� �� //F�´1 ��´1
//¨ ¨ ¨ �1 //F0 �0 //F´1
˙
is an exact complex.
¨ Claim 1: ��´1˝��“0for every�•1.
Indeed: Letp�0� � � � � ��q PG�`1 be a basis element. Then
p��´1˝��qp�0� � � � � ��q “
�´1ÿ
�“0
ÿ�
�“0p´1q�p´1q�pB�˝ B�qp�0� � � � � ��q� Now let0§�0 †�0§�. If we remove��0 first and then��0, we get
pB�0˝ B�0qp�0� � � � � ��q “ p´1q�0`�0`
�0� � � � �q��0� � � � �q��0� � � � � ��˘
�
On the other hand, if we remove��0 first, then��0 is shifted to position�0´1and must be removed with signp´1q�0´1. So both terms cancel and it follows that��´1˝�� is the zero map.
¨ Claim 2: F‚ ε
⇣Zis an exact complex.
Indeed: by definition of the modulesF�(�•0), we may viewpF‚� �‚qas a complex ofZ-modules.
By Exercise 3, Exercise Sheet 5, it suffices to prove that there exists a homotopy betweenIdF‚ and the zero chain map. For�•0, define
��: F� ›Ñ F�`1
p�0� � � � � ��q fiÑ p1� �0� � � � � ��q,
let �´1 :Z“F´1 ›ÑF0 –ZG be the Z-homomorphism sending 1to p1q, and let�� :“0for all
�§´2. For�•0andp�0� � � � � ��q PG�`1 compute
`��`1˝��`��´1˝��˘
p�0� � � � � ��q “ p�0� � � � � ��q `
�`1ÿ
�“1p´1q�`
1� �0� � � � �q��´1� � � � � ��q
`ÿ�
�“0p´1q�`
1� �0� � � � ��q�� � � � � ��q
“ p�0� � � � � ��q�
and it is clear that��`1˝��`��´1˝��“IdF� for every�§´1, as required.
Notation 13.3 (Bar notation) Given�PZ•0, set
r�1|�2|� � �|��s:“`
1� �1� �1�2� �1�2�3� � � � � �1¨� � �¨��˘
PG�`1� With this notation, we have
p1� �1� � � � � ��q ““
�1|�´11 �2|�´12 �3|� � �|�´1�´1��‰
� Hence F� becomes a free ZG-module with basis r�1|� � �|��s |�� PG(
“: G�, which as a set is in bijection with G�. In particular F0 is the freeZG-module with basistr su(empty symbol). With this notation, for every�•1and every0§�§�, we have
B�r�1|� � �|��s “
$’
&
’%
�1¨“
�2|� � �|��‰
�“0�
“�1|� � �|��´1|����`1|��`2|� � �|��‰
1§�§�´1�
“�1|� � �|��´1‰
�“��
Because of this notation the resolution of Proposition 13.2 is known as thebar resolution.
In fact, it is possible to render computations easier, by considering a slight alteration of the bar reso- lution called the normalised bar resolution.
Notation 13.4 (The normalised bar notation)
Let � P Z•0, and let F� be as above and let D� be the ZG-submodule of F� generated by all elementsr�1|� � �|��sofF�such that at least one of the coefficients��is equal to1. In other words, ifp1� �1� � � � � ��q PF�, then:
p1� �1� � � � � ��q PD� ñ D1§�§�´1 such that ��“��`1�
Lemma 13.5
(a) D‚ is a subcomplex ofF‚. (b) ��pD�qÄD�`1 for all�•0.
Proof :
(a) Let�•1. We have to prove that��pD�qÑD�´1. So let p1� �1� � � � � ��q PD�, so that there is an index1§�§�´1such that��“��`1. Then, clearly
B�p1� �1� � � � � ��q PD�´1 for each0§� §�such that �‰�� �`1�
On the other hand, we have the equalityB�p1� �1� � � � � ��q “ B�`1p1� �1� � � � � ��q and in the alter- nating sum��p1� �1� � � � � ��q “∞�
�“0p´1q�B�p1� �1� � � � � ��q, the signs ofB� andB�`1are opposite to each other. Therefore, we are left with a sum over� ‰�� �`1.
(b) Obvious by definition of ��. Corollary 13.6
SetF� :“F�{D� for every �•0. Then F‚ is a free ZG-resolution of the trivial module.
Proof : Since D‚ is a subcomplex of F‚, we can form the quotient complex pF‚� �‚q, which consists of free ZG-modules. Now by the Lemma, ��pD�qÄ D�`1 for each �•0, thereforeD� is in the kernel of ��
post-composed with the quotient map F�`1 ›Ñ F�`1{D�`1 and each Z-linear map �� : F� ›Ñ F�`1
induces aZ-linear maps�� :F� ›ÑF�`1 via the Universal Property of the quotient. Hence, similarly to the proof of Proposition 13.2, we get a homotopy ��|�PZ(
and we conclude that the sequence
¨ ¨ ¨ //F� �� //F�´1 //¨ ¨ ¨ �1 //F0 ε“�0 //Z //0�
is exact, as required.
Definition 13.7 (Normalised bar resolution)
The chain complex pF‚� �‚q is called thenormalised bar resolutionof Zas a ZG-module.
Example 9 (Bar resolution in low degrees) In low degrees the bar resolution has the form
¨ ¨ ¨ //F2 �2 //F1 �1 //F0 ε //Z //0 bases elts: r�1|�2s r�s r s
with
¨ εpr sq “1;
¨ �1pr�sq “ B0pr�sq ´ B1pr�sq “�r s ´ r s;
¨ �2r�1|�2s “ B0pr�1|�2sq ´ B1pr�1|�2sq ` B2pr�1|�2sq “�1r�2s ´ r�1�2s ` r�1s.
Similar formulae hold forF‚. (Exercise!)
14 Cocycles and Coboundaries
We now use the (normalised) bar resolution in order to compute the cohomology groups H�pG�Aq (� • 0), where A is an arbitrary ZG-module. To this end, we need to consider the cochain complex HomZGpF‚�Aq. Define
C�pG�Aq:“HomSetpG��Aq
to be the set of all maps from G� to A, so that clearly there is an isomorphism of Z-modules HomZGpF��Aq›–ÑC�pG�Aq
mapping � fi�|G�. Using this isomorphism, we see that the corresponding differential maps are:
�˚�:C�´1pG�Aq›ÑC�pG�Aq
� fi›Ñ�˚�p�q where
�˚�p�q`
r�1|� � �|��s˘
“�`
�1r�2|� � �|��s˘
`�´1ÿ
�“1p´1q��`
r�1|� � �|����`1|� � �|��s˘
` p´1q��`
r�1|� � �|��´1s˘
� Definition 14.1 (�-cochains, �-cocycles,�-coboundaries)
With the above notation:
(a) The elements ofC�pG�Aq are called the �-cochainsof G.
(b) If� PC�pG�Aqis such that�˚�`1� “0, then� is called an�-cocycle ofG, and the the set of all�-cocycles is denotedZ�pG�Aq.
(c) If � P C�pG�Aq is in the image of �˚� : C�´1pG�Aq ›Ñ C�pG�Aq, then � is called an �- coboundaryof G. We denote by B�pG�Aq the set of all�-coboundaries.
Proposition 14.2
LetA be aZG-module and�•0. ThenH�pG�Aq –Z�pG�Aq{B�pG�Aq.
Proof : Compute cohomology via the bar resolution and replace theZ-moduleHomZGpF��Aqby its isomorphic Z-moduleC�pG�Aq. The claim follows.
Remark 14.3
If we used the normalised bar resolution instead, the �-cochains are replaced by the �-cochains vanishing on�-tuplesr�1|� � �|��shaving (at least) one of coefficient�� equal to1. (This is because HomZGpF��Aq Ä HomZGpF��Aq). We denote these by C�pG�Aq, and thus C�pG�Aq Ñ C�pG�Aq.
The set of resulting normalised �-cocycles is denoted by Z�pG�Aq, and the set of resulting nor- malised�-coboundaries byB�pG�Aq. It follows that
H�pG�Aq –Z�pG�Aq{B�pG�Aq –Z�pG�Aq{B�pG�Aq�
