Let $k$ be a field with separable closure $\overline k$ and absolute Galois group $G$. We know that sheaves of abelian groups on the etale site of $k$ correspond to abelian groups with a $G$ action (obtained by taking stalks of the sheaf at a geometric point).

Let $l/k$ be a finite separable extension of fields of dimension $n$ and let $\pi: \operatorname{Spec} l \to \operatorname{Spec} k$ be the map on schemes. Given a sheaf $\mathscr A$ on the etale site over $l$, we can define it's pushforward $\pi_*\mathscr A$.

Is there a simple description of this in terms of the module $\mathscr A$ along with what the Galois action of $G$ looks like in terms of the Galois action of $\operatorname{Gal}(\overline k/l)?$

At least when $\mathscr A$ is the constant sheaf, I think the pushforward is $\mathscr A\times {\operatorname{Gal}(l/k)}$ and the action is trivial on the first factor and on the left on the second factor. I suspect that the answer in general is something like $\mathscr A\times {\operatorname{Gal}(l/k)}$ but I am not sure what the action on the first factor should be...


Your description $\mathscr A\times\operatorname{Gal}(l/k)$ does not works, in fact $\mathscr A\times\operatorname{Gal}(l/k)$ might not be an abelian group.

You know that the pull-back corresponds to forgetting the action. More precisely, if $\mathcal{F}$ is an étale sheaf on $k$, corresponding to a Galois representation on $\mathcal{F}_{\overline{k}}$, then the pull-back $\pi^*\mathcal{F}$ correspond to the restriction of the representation to the group $\operatorname{Gal}(\overline{k}/l)\subset\operatorname{Gal}(\overline{k}/k)$.

The push-forward is the right-adjoint of this functor. You can explicity compute this right-adjoint using the adjunction isomorphism with $\pi^*\mathbb{Z}[\operatorname{Gal}(\overline{k}/k)]$. You will find that this corresponds to $\mathscr{A}\mapsto \operatorname{Hom}_{\operatorname{Gal}(\overline{k}/l)}(\mathbb{Z}[\operatorname{Gal}(\overline{k}/k)],\mathscr{A})$ with the action $(\sigma.f)(x)=f(x\sigma)$.

When $\mathscr{A}$ is constant and if $l/k$ is Galois, another description is as follow : $\pi_*\mathscr{A}=\prod_{\sigma\in\operatorname{Gal}(l/k)}\mathscr{A}$ where the action on the right-hand side is given by $\tau(a_\sigma)_\sigma=(a_{\sigma\tau})_\sigma$. (If $l/k$ is not Galois, you will have to replace $\operatorname{Gal}(l/k)$ by $G_l\backslash G_k$ the set of right-cosets). Using representative of right-cosets, you can have a similar description when $\mathscr{A}$ is not constant.

Note that this also works if $l/k$ is not-finite but you will need to be careful with the topology as $\pi_*\mathscr{A}$ will not be discrete anymore.

  • $\begingroup$ Thanks for answering - the adjoint perspective clears everything up immediately and I really should have seen that. Also, I was using non standard notation, by $\mathscr A\times G$, I meant $\prod_{g\in G}\mathscr A$ and the action is by permuting the factors. Sorry for that. $\endgroup$ – Asvin Jun 20 '17 at 15:15
  • $\begingroup$ Yes I was confident that it was what you meant though it is not clear in the context of groups ! (And just to be pedantic, $\mathscr{A}\times G=\coprod_{g\in G}\mathscr{A}$ in categories where $\coprod\neq\prod$, this distinction is also relevant in the case of infinite extension). $\endgroup$ – Roland Jun 20 '17 at 15:24

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