Group cohomology is a useful language for expressing the results of class field theory, among (many) other things. There are a few ways I could introduce this. I could define them as derived functors (i.e. as a special case of ${\mathrm{Tor},\mathrm{Ext}}$) or satellites, which would be the most general, but I try to keep my posts somewhat self-contained. I could define them additionally as cochains or coboundaries. I’ve decided to give an axiomatic definition, which will include the previous ones.

Axioms for Cohomology

First of all, here is a useful way of constructing a right ${R}$-module from an abelian group ${A}$: consider ${\hom_{\mathbb{Z}}(R, A)}$ and let ${R}$ act on the left factor. Such modules are said to be co-induced. In the case of ${R=\mathbb{Z}[G]}$, this yields a left-module too by the anti-involution ${\mathbb{Z}[G] \rightarrow \mathbb{Z}[G]}$ sending ${\sigma \rightarrow \sigma^{-1}}$. Note that each ${G}$-module can be imbedded inside a co-induced one.

Given ${G}$-modules (i.e. ${\mathbb{Z}[G]}$-modules) ${A,B,C \in \mathbf{Mod}_G}$ with an exact sequence

$\displaystyle 0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0,$

the sequence of fixed points

$\displaystyle 0 \rightarrow A^G \rightarrow B^G \rightarrow C^G$

is exact except at the end. The cohomology functors extend this into a long exact sequence.

Here are the axioms:

1. For each ${i \geq 0}$, ${H^i(G, \cdot)}$ is a covariant functor ${\mathbf{Mod}_G \rightarrow \mathbf{Ab}}$
2. ${H^0(G, A) = A^G}$ for ${A \in \mathbf{Mod}_G}$
3. Given an exact sequence ${0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0}$, there is a long exact sequence

$\displaystyle 0 \rightarrow H^0(G,A) \rightarrow H^0(G,B) \rightarrow H^0(G,C) \rightarrow H^1(G,A) \rightarrow H^1(G,B) \rightarrow \dots,$

which is functorial in the short exact sequence ${0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0}$.

4. For ${A}$ co-induced, ${H^i(G,A)=0}$ if ${i>0}$.

I think there’s a definite analogy between this approach and the Eilenberg-Steenrod axioms for homology theories, except that in this case on already has a more general theory (derived functors), so an axiomatic approach is less useful.

Anyway, the third axiom shows that the ${H^i}$ form a ${\delta}$-functor, a notion introduced by Grothendieck  which basically means a sequence of functors ${F^i}$ from one abelian category to another with a long exact sequence

$\displaystyle F^i(A) \rightarrow F^i(B) \rightarrow F^i(C) \rightarrow F^{i+1}(A) \rightarrow F^{i+1}(B) \rightarrow \dots$

associated functorially to a short exact sequence ${0 \rightarrow A \rightarrow B \rightarrow 0}$. The last condition means that the ${H^i}$ are “effaceable.” Grothendieck proved a theorem that if given two ${\delta}$-functors ${F,G}$ with ${F}$ effaceable in this way, a natural transformation ${F^0 \rightarrow G^0}$ extends uniquely to natural transformations ${F^i \rightarrow G^i}$ which make the diagram

commutative. I won’t prove this theorem, since it’s a messy sequence of diagram chases, but you can find it in books on homological algebra. For us, this will be important.

Anyway, the disadvantage of using an axiomatic approach is that we have to give an existence and uniqueness proof.

Theorem 1 Up to natural isomorphism, there is precisely one ${\delta}$-functor ${H^i}$ satisfying the above axioms. If ${P \rightarrow \mathbb{Z}}$ is a projective resolution in ${\mathbf{Mod}_G}$ (with ${\mathbb{Z}}$ the trivial ${G}$-module), then

$\displaystyle H^i(G, A) \simeq H^i(\hom_G(P,A)),$

where on the left the ${H^i}$ refers to the cohomology of a cochain complex.

Ok, first of all, the last part of the theorem gives us a clean way to prove existence: just fix the projective resolution ${P}$, and check the axioms. The first is evident. Next, let ${P_1 \rightarrow P_0 \rightarrow \mathbb{Z} \rightarrow 0}$ be the end of the resolution; then

$\displaystyle 0 \rightarrow \hom_G(\mathbb{Z}, A) \rightarrow \hom_G(P_0, A) \rightarrow \hom_G(P_1, A)$

is exact, which proves 2. For 3, given an exact sequence ${0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0}$, there is a natural short exact sequence of complexes

$\displaystyle 0 \rightarrow \hom_G(P,A) \rightarrow \hom_G(P,B) \rightarrow \hom_G(P,C) \rightarrow 0$

by projectivity, so taking the associated long exact sequence gives a natural long exact sequence of ${H^i(G,\cdot)}$ of the right form. Finally, if ${A = \hom_{\mathbb{Z}}(\mathbb{Z}[G], B)}$ for ${B}$ an abelian group, we have

$\displaystyle \hom_G(P,A) = \hom_{\mathbb{Z}}(P,B) \ \mathrm{(given \ } \phi:P \rightarrow A \mathrm{ \ consider \ } \phi(1): P \rightarrow B) ,$

and for ${B}$ abelian the latter cochain complex has trivial cohomology after the first index, because ${P \rightarrow \mathbb{Z}}$ splits as ${\mathbb{Z}}$-complexes.

Uniqueness follows because given two such families of functors ${F^i, G^i}$, we have transformations ${F^0 \rightarrow G^0, G^0 \rightarrow F^0}$ which is the identity, and which by Grothendieck’s theorem yields natural transformations ${F^i \rightarrow G^i, G^i \rightarrow F^i}$; the composition of these extend the identity transformations ${F^0 \rightarrow F^0, G^0 \rightarrow G^0}$ and must be the identity.

Alternatively, one often defines ${H^i(G, \cdot)}$ as the derived functors of the left-exact functor ${A \rightarrow A^G = \hom_G(\mathbb{Z}, A)}$; the uniqueness in the above theorem says that deriving the ${\hom}$ functor in either of the two variables gives the same result. This fact can be extended to ${\hom}$ or ${\otimes}$ over a ring.

Homology

Group homology is dual to cohomology, except that one starts with the right exact functor ${A \rightarrow A_G:=A/I_GA}$, where ${I_G}$ is the augmentation ideal of ${\mathbb{Z}[G]}$ generated by ${\sigma -1, \sigma \in G}$. The axioms are as follows.

1. For each ${i \geq 0}$, ${H_i(G, \cdot)}$ is a covariant functor ${\mathbf{Mod}_G \rightarrow \mathbf{Ab}}$
2. ${H_i(G, A) = A_G}$ for ${A \in \mathbf{Mod}_G}$
3. Given an exact sequence ${0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0}$, there is a long exact sequence

$\displaystyle \dots \rightarrow H_1(G,B) \rightarrow H_1(G,C) \rightarrow H_0(G,A) \rightarrow H_0(G,B) \rightarrow H_0(G,C) \rightarrow 0 ,$

which is functorial in the short exact sequence ${0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0}$.

4. For ${A}$ induced (i.e. ${A = \mathbb{Z}[G] \otimes_{\mathbb{Z}} B}$),  ${H_i(G,A)=0}$ if ${i>0}$.

The ${H_i}$ form a ${\delta}$-functor in the opposite direction.

Theorem 2 Up to natural isomorphism, there is precisely one ${\delta}$-functor ${H^i}$ satisfying the above axioms. If ${P \rightarrow \mathbb{Z}}$ is a projective resolution in ${\mathbf{Mod}_G}$ (with ${\mathbb{Z}}$ the trivial ${G}$-module), then

$\displaystyle H^i(G, A) \simeq H^i(P \otimes_{\mathbb{Z}G} A),$

where on the left the ${H^i}$ refers to the cohomology of a cochain complex.

In the tensor product, ${A}$ is considered as a right ${\mathbb{Z}[G]}$-module via its left-module structure and the anti-involution on ${\mathbb{Z}[G]}$.The rest of the proof is similar, since

$\displaystyle \mathbb{Z} \otimes_{\mathbb{Z}[G]} A = (\mathbb{Z}[G]/I_G) \otimes_{\mathbb{Z}[G]} A = A_G.$

Next, I’ll discuss the canonical resolution of ${\mathbb{Z}}$ and how one can compute cohomology and homology.