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 $1$)



Cohomology of a group with a topological generator
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.


