Suppose $G$ is a smooth and abelian $k$-group scheme, for $k$ a field.


Is it possible to get back galois cohomology groups $H^\*(k,G)$ studying the cohomology of the classifying stack $BG=[\*/G]$ ?



Galois cohomology out of the classifying stack
Hmm, a priori Galois cohomology is cohomology of $Spec(k)$ with values in $G$. For $H^1$, for instance, this amounts to considering the set (in fact group) of isomorphism classes of maps of stacks $Spec(k) \to BG$. Whereas the degree-one cohomology of $BG$ with values in some $A$ (some sort of abelian group scheme over $k$, say) is like considering the set of isomorphism classes of maps of stacks $BG \to BA$. One is measuring maps out of $BG$ and one is measuring maps into $BG$ (in particular, $H^1(k,G)$ is the group of $k$-points of $BG$). 


For higher-degree cohomology $H^n(k,G)$ one can think of this as maps of higher stacks $Spec(k) \to B^nG$, so you're not even using $BG$ at this point. There may be some way to do something tricky and use $H^\ast(BG,\mathcal{F})$ to inform you what the Galois cohomology is, but there's no formal reason it could be true, from the point of view I just outlined.


