Let $G$ be a profinite group. Let $\Lambda$ be a discrete valuation ring over $\mathbb{Z}\_p$. Let $A\_0,A\_1$ be artinian ring over $\Lambda$. Let $\alpha : A\_1 \rightarrow A\_0$ is a small map that is $\mathfrak{m}\_{A\_1}.ker(\alpha) = 0$. Assume that a representation $\rho : G \rightarrow GL\_n(A\_0)$ is given. I would like to know when can a representation be lifted to $GL\_n(A\_1)$. 


I vaguely(and perhaps incorrect) know that the obstruction lies in $H^2(G, Ad\rho)$. What is the cocycle class associated to $\rho$ which must vanish so that $\rho$ can be lifted?


Any references/hint/solution are welcome.


Thanks!



Obstructions to lift of representations
Something like what you want is true, though it needs to be formulated slightly differently. Let $k$ be the residue field of $A\_0$, and let $\overline{\rho}$ denote the ``residual'' representation $\rho \otimes\_{A\_0} k$.


Let $I$ denote the kernel of the map from $A\_1$ to $A\_0$; then $I$ is a $k$-vector space. We can associate to $\rho$ a natural ``obstruction class'' that lives in $H^2(G, Ad \overline{\rho}) \otimes\_k I$. This class vanishes if, and only if, $\rho$ lifts to $GL\_n(A\_1)$. (Moreover, if this is the case, then the set of such lifts is a torsor over $H^1(G, Ad \overline{\rho}) \otimes\_k I$.


Here's a brief sketch of the construction of this obstruction class (ignoring some issues with continuity that are routine but slightly annoying): 


For each $g \in G$, choose a lift $x\_g$ of $\rho(g)$ to $GL\_n(A\_1)$. Of course the map sending $g$ to $x\_g$ need not be a group homomorphism, but for any $g,h$ in $G$, we have:


$$x\_{gh}^{-1} x\_g x\_h = 1 + y\_{g,h},$$
where $y\_{g,h}$ lies in $M\_n(I)$. The group $GL\_n(A\_1)$ acts on $M\_n(I)$ by conjugation; this action factors through the quotient $GL\_n(k)$. We can thus regard $G$ as acting on $M\_n(I)$ via $\overline{\rho}$; this identifies $M\_n(I)$ with $(Ad \overline{\rho}) \otimes\_k I$.


With these identifications it is easy to see that $y\_{g,h}$ is a $2$-cocycle with values in $(Ad \overline{\rho}) \otimes\_k I$. This cocycle vanishes if, and only if, our chosen set of lifts $x\_g$ define a homomorphism of $G$ into $GL\_n(A\_1)$ lifting $\rho$. Moreover if we replace our chosen lifts $x\_g$ with another set of lifts $x'\_g$, then this changes $y\_{g,h}$ by a coboundary.


Thus there exists *some* lift of $\rho$ if, and only if, $y\_{g,h}$ is a coboundary, completing the argument.


