Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

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.

share|improve this answer
    
(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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.