3
$\begingroup$

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 $BG=[*/G]$ ?

$\endgroup$
1
  • $\begingroup$ Cohomology of a stack is contravariant in the stack, but Galois cohomology is covariant in $G$. $\endgroup$ Mar 25, 2015 at 20:09

1 Answer 1

3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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