Group Cohomology

If $G$ is a multiplicative group, the group ring $\mathbb{Z}[G]$ is the ring of all finite formal sums of elements of $G$, with multiplication defined using distributivity and extending linearly. Let $A$ be an additive group. We say that $A$ is a $G$-module if $A$ is equipped with a module structure over the group ring $\mathbb{Z}[G]$.

Let $A^G$ be the submodule of elements of $A$ that are fixed by $G$. Notice that if $A\to B$ is a homomorphism of $G$-modules, then restriction defines a homomorphism $A^G \to B^G$, so $A\mapsto A^G$ is a functor. In fact, it is a left-exact functor:

Proposition 2.1   If $0 \to A \to B \to C$ is an exact sequence of $G$ modules, then $0 \to A^G \to B^G \to C^G$ is also exact.

Definition 2.2 (Group Cohomology)   The group cohomology $H^n(G,A)$ is by definition the right derived functors of the left exact functor $A \to A^G$. These are the unique, up to canonical equivalence, functors $H^n$ such that

• The sequence
  
  $$0 \to A^G \to B^G \to C^G \xrightarrow{\delta} H^1(G,A) \t... ...G,B) \to H^n(G,C) \xrightarrow{\delta} H^{n+1}(G,A) \to \cdots$$
  
  
  is exact.
• If $A$ is coinduced, i.e., $A = \Hom (\mathbb{Z}[G], X)$ for $X$ an abelian group, then
  
  $$\H^n(G,A) = 0 \text{ for all $n\geq 1$}.$$
  
  

Remark 2.3   For those familiar with the $\Ext$ functor, we have

$$\H^n(G,A) = \Ext _{\mathbb{Z}[G]}^n(\mathbb{Z},A).$$

We construct $H^n(G,A)$ explicitly as follows. Consider $\mathbb{Z}$ as a $G$-module, equipped with the trivial $G$-action. Consider the following free resolution of $\mathbb{Z}$. Let $P_i$ be the free $\mathbb{Z}$-module with basis the set of $i+1$ tuples $(g_0,\ldots, g_i) \in G^{i+1}$, and with $G$ acting on $P_i$ componentwise:

$$s (g_0,\ldots, g_i) = (sg_0, \ldots, sg_i).$$

The homomorphism $d:P_i\to P_{i+1}$ is given by

$$d(g_0,\ldots, g_i) = \sum_{j=0}^{i} (-1)^j (g_0,\ldots,g_{j-1},g_{j+1},\ldots g_i),$$

and $P_0\to \mathbb{Z}$ is given by sending every element $(g_0)$ to $1\in\mathbb{Z}$.

The cohomology groups $\H^i(G,A)$ are then the cohomology groups of the complex $K_i = \Hom _{\mathbb{Z}[G]}(P_i, A)$. We identify an element of $K_i$ with a function $f:G^{i+1} \to A$ such that the condition

$$f(sg_0,\ldots, sg_i) = s f(g_0,\ldots, g_i)$$

holds. Notice that such an $f\in K_i$ is uniquely determined by the function (of $i$ inputs)

$$\varphi (g_1,\ldots, g_i) = f(1,g_1,g_1 g_2, \ldots, g_1\cdots g_i).$$

The boundary map $d:K_i\to K_{i+1}$ on such functions $\varphi \in K_{i}$ is then given explicitly by the formula
$$\begin{align*} (d\varphi )(g_1,\ldots, g_{i+1}) &= g_1 \varphi (g_2,\ldots, g_... ...}, \ldots, g_{i+1})\\ &+ (-1)^{i+1} \varphi (g_1,\ldots, g_{i}). \end{align*}$$
The group of $n$-cocycles is the group of $\varphi \in K_{n}$, as above are functions of $n$ variables such that $d\varphi = 0$. The subgroup of $n$-coboundaries is the image of $K_{n+1}$ under $d$. Explicitly, the cohomology group $H^n(G,A)$ is the quotient of the group group of $n$-cocycles modulo the subgroup of $n$-coboundaries.

When $n=1$, the $1$-cocycles are the maps $G\to A$ such that

$$\varphi (g g') = g\varphi (g') + \varphi (g),$$

and $\varphi$ is a coboundary if there exists $a\in A$ such that $\varphi (g) = ga - a$ for all $g \in G$. Notice that if $G$ acts trivially on $A$, then

$$\H^1(G,A) = \Hom (G,A).$$