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The setting is as follows: $(R,m)$ is a local ring (assume noetherian, complete, if you need) and $\rho\colon G\to \operatorname{Aut}(M)$ is a group representation on the free, finite-rank $R/m^n$-module $M$, for some $n\geq 1$.

Let $\operatorname{End}(M)$ be the adjoint representation on which $G$ acts by conjugation.

To a class $c\in H^1(G,\operatorname{End}(M))$ one can associate (via an infinitesimal deformation) an exact sequence $0\to M\to E\to M\to 0$ of $(R/m^n)[G]$-modules. The class of this extension in $\operatorname{Ext}^1(M,M)$ is well defined (does not depend on the cocycle representing $c$).

Question: What is a map in the inverse direction?

Note, that if $n=1$, every sequence $0\to M\to E\to M\to 0$ splits as $k:=R/m$-vector spaces. And any such splitting $s$ yields a 1-cocycle $g\mapsto g \cdot s \cdot g^{-1}-s$.

But, e.g. for $n>1$ the sequence of $\mathbb{Z}_2/(4)$-modules $0 \to \mathbb{Z}_2/(2)\to \mathbb{Z}_2/(4)\to \mathbb{Z}_2/(2)\to 0$ does not split.

Or am I missing something?

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up vote 1 down vote accepted

Given an extension $$ 0 \to M \to N \to M \to 0 $$ apply the functor $\text{Hom}_{R/\mathfrak{m}^n}(M, -)$ to get $$ 0 \to \text{End}(M) \to \text{Hom}(M, N) \to \text{End}(M) \to 0 $$ (using freeness to get exactness on the right). Now take $G$-invariants to get a map $$ \text{End}_G(M) \to H^1(G, \text{End}(M)). $$ The image of the identity endomorphism under this map is your desired cocycle.

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(I should add that this is a guess! I didn't check the constructions both ways are the identity, but what else could they be? (well, minus the identity, for one...)) – user29743 Aug 7 '12 at 18:18
recipe for explicitly getting the $H^0$ to $H^1$ map in terms of cocyles: given $0 \to M' \to M \to M'' \to 0$ exact sequence of $G$-modules, to map $M''^G$ to $H^1(G, M')$, send $m$ to the cocycle mapping $g$ to $g\widetilde{m} - \widetilde{m} \in M'$ where $\widetilde{m} \in M$ is any lift of $m$. – user29743 Aug 7 '12 at 18:20
Chosing a $R/(\mathfrak{m}^n)$-linear inverse of the identity morphism of $M$ exactly amounts to a splitting of the sequence in the first place. But I see my error now: Of course the modules in my example are not free and a basis can always be liftet to give a splitting. So I indeed missed something rather childish. Your answer gives a nice funtorial viewpoint to this, though. Thanks! – clueless Aug 7 '12 at 18:45

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