Categories, abelian categories and cohomologies.

Yoshifumi Tsuchimoto

Let $\mathcal{C}$ be an abelian category. For any object $M$ of $\mathcal{C}$ , the extension group $Ext^j_\mathcal{C}(M,N)$ is defined to be the derived functor of the ``hom'' functor

$$N\mapsto \operatorname{Hom}_\mathcal{C}(M,N).$$

Let $G$ be a group. Let us consider a functor

$$F^G:M\mapsto M^G=\{ m\in M ; \quad g.m=m(\forall g\in G)\}$$

The functor is left-exact. The derived functor of this functor

$$H^j(G,M)=R^j F^G(M)$$

is called the $j$ -th cohomology of $G$ with coefficients in $M$ . Let us consider $\mathbb{Z}$ as a $G$ -module with trivial $G$ -action. Then we may easily verify that

$$F^G(M)=M^G\cong \operatorname{Hom}_G(\mathbb{Z},M).$$

Thus we have

$$H^j(G,M)=\operatorname{Ext}^j_G(\mathbb{Z},M).$$

The extension group $\operatorname{Ext}^\bullet_\mathcal{C}(M,N)$ may be calculated by using either an injective resolution of the second variable $N$ or a projective resoltuion of the first variable $M$ .

EXAMPLE 08.1   Let us compute the extension groups $\operatorname{Ext}^j_\mathbb{Z}(\mathbb{Z}/36\mathbb{Z}, \mathbb{Z}/108\mathbb{Z})$ .

1. We may compute them by using an injective resolution
   $$0 \to \mathbb{Z}/108\mathbb{Z}\to \mathbb{Q}/108\mathbb{Z}\to \mathbb{Q}/\mathbb{Z}\to 0$$
   of $\mathbb{Z}/108\mathbb{Z}$ .
2. We may compute them by using a free resolution
   $$0 \leftarrow \mathbb{Z}/36\mathbb{Z}\leftarrow \mathbb{Z}\leftarrow 36 \mathbb{Z}\leftarrow 0$$
   of $\mathbb{Z}/36 \mathbb{Z}$ .

EXERCISE 08.1   Compute an extension group $\operatorname{Ext}^j(M,N)$ for modules $M,N$ of your choice. (Please choose a non-trivial example).

To compute cohomologies of $G$ , it is useful to use $\mathbb{Z}[G]$ -resolution of $\mathbb{Z}$ . For any tuples $g_0,g_1,g_2,\dots,g_t$ of $G$ , we introduce a symbol

$$[g_0,g_1,g_2,\dots, g_t]$$

and we consider the following sequence

$$*_G 0\leftarrow \mathbb{Z} \overset{d}{\leftarrow} \bigoplus_{g_0\in ... ..._{g_0,g_1,g_2\in G} \mathbb{Z}\cdot [g_0,g_1,g_2] \overset{d}{\leftarrow} \dots$$

where $\epsilon,d$ are determined by the following rules.

$$d ([g_0])=1$$

$$d([g_0,g_1])=[g_1]-[g_0]$$

$$d([g_0,g_1,g_2])=[g_1,g_2]-[g_0,g_2]+[g_0,g_1]$$

$$d([g_0,g_1,g_2,g_3])=[g_1,g_2,g_3]-[g_0,g_2,g_3]+[g_0,g_1,g_3]-[g_0,g_1,g_2]$$

$$\dots$$

To see that the sequence $*_G$ is acyclic, we consider a homotopy

$$h([g_0,g_1,\dots,g_t]) =[1,g_0,g_1,\dots,g_t]$$

EXERCISE 08.2   Show that $h\circ d+d \circ h=\operatorname{id}$

LEMMA 08.2  

1. Each of the modules that appears in the sequence $*_G$ admits an action of $G$ determined by
   $$g.[g_0,g_1,g_2,\dots,g_t]= [g\cdot g_0,g\cdot g_1,g\cdot g_2,\dots,g\cdot g_t]$$

2. $$C_t=\bigoplus_{g_0,g_1,g_2,\dots g_t\in G} \mathbb{Z}\cdot [g_0,g_1,g_2,\dots ,g_t]$$
   is $\mathbb{Z}[G]$ -free

There are several choices for the $\mathbb{Z}[G]$ -basis of $C_t$ . One such is clearly

$$\{ [1,g_1,g_2,g_3,\dots,g_t] ; g_1,g_2,\dots, g_t\in G\}.$$

It is traditional (and probably useful) to use another basis

$$\{ \langle g_1,g_2,g_3,\dots,g_t\rangle ; g_1,g_2,\dots, g_t\in G\}.$$

where

$$\langle g_1,g_2,g_3\dots g_t\rangle = [1, g_1, g_1 g_2, g_1 g_2 g_3 , \dots, g_1 g_2 g_3 \dots g_t].$$

Conversely we have

$$[1,a_1,a_2,\dots,a_t] = \langle a_1, a_1^{-1}a_2,a_2^{-1}a_3,\dots, a_{t-1}^{-1} a_t \rangle.$$