Chapter 5. Easy Cohomology
In this short chapter we consider some cases in which the cohomology groups of a groupGhave an easy interpretation. This is for example the case in low degrees (zero, one, two). Next we consider families of groups whose cohomology groups are easy to compute with the methods we have so far at our disposal.
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.
15 Low-degree Cohomology
A. Degree-zero cohomology.
We have already proved in Proposition 12.4 that H0pG�Aq –AG, the G-fixed points of A.
In particular, ifA is a trivial ZG-module, then H0pG�Aq “A.
B. Degree-one cohomology.
Using the bar resolution to compute H1pG�Aqyields H1pG�Aq “Z1pG�Aq{B1pG�Aq.
1-cocycles: By definition, and the description of the differential maps of the bar resolution, we have Z1pG�Aq “ t� PHomSetpG1�Aq |�˚2p�q “0u
“ t� PHomSetpG1�Aq |0“�p�1r�2sq ´�pr�1�2sq `�pr�1sq @ r�1|�2s PG2u In other words, a map � :G›ÑA is a 1-cocycle if and only if it satisfies the
1-cocycle identity: �p�1¨�2q “�1¨�p�2q `�p�1q @�1� �2 PG � 57
Skript zur Vorlesung: Cohomology of Groups SS 2018 58 1-coboundaries: C0pG�Aq “HomSetptr su�Aq “ t�� :tr su›Ñ A�r s fiÑ �|�PAu –bij.ÑA. It follows that the differential map
�˚1 :C0pG�Aq›ÑC1pG�Aq
is such that �˚1p��qp�q “ ��p�r sq ´��pr sq “ ��´� for every � P G and every � P A. Therefore,
� :G›ÑAis a1-coboundary if and only if there exists�PAsuch that�p�q “��´�for every�PG.
Definition 15.1 (Derivation, principal derivation) LetA be aZG-module and let � :G›ÑA be a map.
(a) If�satisfies the1-cocycle identity, then it is called aderivationofG. We denote byDerpG�Aq the set of all derivations of G to A.
(b) If, moreover, there exists � P A such that �p�q “ ��´� for every � P G, then � is called a principal derivation (or an inner derivation) of G. We denote by InnpG�Aq the set of all inner derivations ofG to A.
Remark 15.2
It follows from the above thatH1pG�Aq –Z1pG�Aq{B1pG�Aq “DerpG�Aq{InnpG�Aq.
Example 10
LetA be a trivialZG-module. In this case, the 1-cocycle identity becomes
�p�¨�q “�p�q `�p�q� so thatZ1pG�Aq “HomGrp`
pG�¨q�pA�`q˘
. Furthermore B1pG�Aq “0. Therefore H1pG�Aq “HomGrp`
pG�¨q�pA�`q˘
�
C. Degree-two cohomology.
Again using the bar resolution to compute H2pG�Aq yields H2pG�Aq “Z2pG�Aq{B2pG�Aq.
2-cocycles: By definition, and the description of the differential maps of the bar resolution, we have Z2pG�Aq “ t� PHomSetpG2�Aq |�˚3p�q “0u
“ t� PHomSetpG2�Aq |0“�p�1r�2|�3sq ´�pr�1�2|�3sq
`�pr�1|�2�3sq ´�pr�1|�2sq @ r�1|�2|�3s PG3u In other words, a map � :GˆG ›ÑAis a 2-cocycle if and only if it satisfies the
2-cocycle identity: �1�p�2� �3q `�p�1� �2�3q “�p�1�2� �3q `�p�1� �2q @�1� �2� �3 PG � 2-coboundaries: If φPC1pG�Aq, then
�˚2pφqpr�1|�2sq “φp�1r�2sq ´φpr�1�2sq `φpr�1sq @ r�1|�2s PG2�
Skript zur Vorlesung: Cohomology of Groups SS 2018 59 Therefore a map � :GˆG›ÑA is a 2-coboundary if and only if there exists a map �:G›ÑA such that �p�1� �2q “�1�p�2q ´�p�1�2q `�p�1q @�1� �2 PG�
16 Cohomology of Cyclic Groups
Cyclic groups, finite and infinite, are a family of groups, for which cohomology is easy to compute. Of course, we could use the bar resolution, but it turns out that in this case, there is a more efficient resolution to be used, made up of free modules of rank 1.
Notation: If Ais a ZG-module and � PZG, then we let �� :A›ÑA� � fiÑ�¨�denote the left action of � onA (or left external multiplication by� inA).
Proposition 16.1 (Free resolution of finite cyclic groups)
LetC� be a finite cyclic group of order�PZ°0 generated by�, and let�:“∞�´1
�“0 ��PZC�. Then
¨ ¨ ¨ �� //ZC� ��´1
//ZC� �� //ZC� ��´1
//ZC��
is a freeZC�-resolution of the trivial ZC�-module.
Proof : SetG:“C�. By Lemma 11.3,
IG“ xt��´1|1§�§�´1uyZ“ x�´1yZG�
Therefore, the image of��´1 is equal toIG, which is the kernel of the augmentation mapε:ZG›ÑZ.
Now, let�“∞�´1
�“0 λ��� PZG. Then, ��“∞�´1
�“0 λ��. Hence kerp��q “ �´1ÿ
�“0λ���|�´1ÿ
�“0λ� “0(
and we claim that this is equal to the image of��´1. Indeed, the inclusionImp��´1qÑkerp��qis clear, and conversely, if �“ ∞�´1
�“0 λ��� P kerp��q, then ∞�´1
�“0 λ� “0, so that � P kerpεq “ IG “Imp��´1q, whencekerp��qÑImp��´1q. Finally, we claim thatkerp��´1q “Imp��q. We have
�´1ÿ
�“0λ���Pkerp��´1q ñ p�´1q
˜�´1 ÿ
�“0λ���
¸
“0
ñ �´1ÿ
�“0λ���`1´�´1ÿ
�“0λ���“0
ñ �´1ÿ
�“0λ�´1��´�´1ÿ
�“0λ���“0� whereλ1:“λ�´1
ñ �´1ÿ
�“0pλ�´1´λ�q��“0
ñ @0§�§�´1� λ�´1 “λ�“:λ
ñ
�´1ÿ
�“0λ���“λ� ñ
�´1ÿ
�“0λ���PImp��q�
Skript zur Vorlesung: Cohomology of Groups SS 2018 60 Theorem 16.2 (Cohomology of finite cyclic groups)
LetC� “ x�|��“1y be a finite cyclic group of order�PZ°0 and letA be a ZC�-module. Then H�pC��Aq –
$&
%
AC� if�“0�
AC�{Imp��q if�•2� �even�
kerp��q{Imp��´1q if�•1� �odd�
where �“∞�´1
�“0 ��PZC� and for� P t�� �´1u,�� denotes left external multiplication by� in A.
Proof : By Proposition 16.1 the trivialZG-moduleZadmits the projective resolution
� � � �� //ZC� ��´1
//ZC� �� //ZC� ��´1
//ZC��
For�“0, we already know thatH0pC��Aq “AC�. For�°0, applying the functorHomZC�p´�Aqyields the cochain complex
HomZC�pZC��Aq �˚�´1 //HomZC�pZC��Aq � //
˚� //HomZC�pZC��Aq �˚�´1 //¨ ¨ ¨�
where in each degree there is an isomorphismHomZC�pZC��Aq›–ÑA� � fiÑ�p1q. Hence for�P t�´1� �u, there are commutative diagrams of the form
HomZC�pZC��Aq �˚� //
–
✏✏
ö
HomZC�pZC��Aq
–
✏✏A �� //A�
Hence, the initial cochain complex is isomorphic to the cochain complex A ��´1 //A �� //A ��´1 //A �� //¨ ¨ ¨
(degree) 0 1 2 3
and the claim follows.
For infinite cyclic groups the situation is even simpler:
Theorem 16.3 (Cohomology of infinite cyclic groups)
If G “ x�y is an infinite cyclic group, then 0 //ZG ��´1//ZG is a free resolution of the trivial ZG-module, and
H�pG�Aq “
$&
%
AG if �“0�
A{Imp��´1q if �“1�
0 if �•2�
where ��´1 denotes the left external multiplication by �´1inA.
Proof : Exercise 1, Exercise Sheet 9.