# Cohomology of a group with a topological generator

Hello,

I've read that if $N_0$ has a topological generator $\gamma$, then its continuous cohomology on some vector space can be computed by the complex $$V \overset{\gamma-1}{\longrightarrow} V$$

If someone knows why...? Thanks !

edit : $N_0$ is isomorphic to $\mathbb{Z}_p$ and by topological generator I mean an element which generate a dense subgroup (here $1$)

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What is $N_0$?, what is a topological generator?, where did you read it? – Mauricio May 31 '12 at 1:10
I am editing the details. – Arkandias May 31 '12 at 3:32

Well, if you consider the definition of cohomology, you see that $H^0(N_0,V)$ are the fixed point, namely $v\in V$ such that $gv=v$ for all $g\in N_0$ (which, I suppose, is your group). If the group were cyclic, all $g$ would be of the form $\gamma^k$ for some $k\in\mathbb{Z}$ so $gv=v$ for all $g$ if and only if $\gamma v=v$: since you speak about continuous cohomology the action of $N_0$ on $V$ must be continuous, and the argument goes through. We then see that $H^0(N_o,V)=\mathrm{ker}(\gamma-1)$. To see that $H^1(N_0,V)=V/(\gamma-1)$ and $H^q(N_0,V)=0$ for higher $q$ is slightly more involved. The first isomorphism can be obtained by writing $$H^1(N_0,V)=\varinjlim_k H^1(N_0/N_0^k,V^{N_0^k})\cong \varinjlim_k \hat{H}^{-1}(N_0/N_0^k,V^{N_0^k})$$ where I denoted by $\hat{H}$ Tate cohomology, for which we have an isomorphism $\hat{H}^1\cong \hat{H}^{-1}$ in the case of finite cyclic groups. This can be computed explicitely: namely, if $G$ is finite cyclic generated by $\gamma$ acting on $M$, we have $$\hat{H}^{-1}(G,M)=\mathrm{ker}(\mathrm{Norm}_G)/(\gamma-1)M$$
For higher $q$, the point is that a topologically cyclic group is of cohomological dimension $1$: you can see this in Serre's book on Galois cohomology.
@Filippo: Sorry, but your formula for $H^1$ is wrong, since it has to be an inductive limit (and it should be noted that $k$ runs through $p$-powers). In general, if $G$ is profinite: $$H^n(G,M) = \varinjlim_U\; H^n(G/U;M^U)$$ where $U$ runs through the open normal subgroups of $G$. – Ralph May 31 '12 at 10:16