next up previous
Next: Bibliography

Categories, abelian categories and cohomologies.

Yoshifumi Tsuchimoto

\fbox{Examples of derived functors}

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

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

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

% latex2html id marker 732
$\displaystyle 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

$\displaystyle 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

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

Thus we have

$\displaystyle 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

    $\displaystyle 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

    $\displaystyle 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

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

and we consider the following sequence

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

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

      $\displaystyle d ([g_0])=1$
      $\displaystyle d([g_0,g_1])=[g_1]-[g_0]$
      $\displaystyle d([g_0,g_1,g_2])=[g_1,g_2]-[g_0,g_2]+[g_0,g_1]$
      $\displaystyle d([g_0,g_1,g_2,g_3])=[g_1,g_2,g_3]-[g_0,g_2,g_3]+[g_0,g_1,g_3]-[g_0,g_1,g_2]$
      $\displaystyle \dots$

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

$\displaystyle h([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

    $\displaystyle 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. $\displaystyle 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

$\displaystyle \{ [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

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


$\displaystyle \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

$\displaystyle [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.

next up previous
Next: Bibliography