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I know this may be a trivial question, but I can't find the answer on, for example, Milne's online notes and Danilov's Cohomology of Algebraic Varieties.

Suppose $K$ is a number field (say), $\overline K$ its algebraic closure, $G_K:=\text{Gal}(\overline K/K)$, $X$ is a variety over $K$, $\mathcal F$ is a locally constant (or constructible (?), don't know if it's true) abelian sheaf on $X_{\text{et}}$. It is said that for any $i\geqslant 0$, there is a continuous $G_K$ action on $H_{\text{et}}^i(X_{\overline K},\mathcal F)$.

My question is, how can one define this action? I found the answer that if $i=1$ and $\mathcal F$ is the constant sheaf associated to an abelian group $A$, then $H_{\text{et}}^1(X_{\overline K},\mathcal F)=\text{Hom}(\pi_1^{\text{et}}(X_{\overline K}),A)$, using the following exact sequence $$ 0\to\pi_1^{\text{et}}(X_{\overline K})\to \pi_1^{\text{et}}(X)\to\pi_1^{\text{et}}(\text{Spec }K)\cong G_K\to 0 $$ we get the Galois action. But what about $i\neq 1$ or $\mathcal F$ is not constant sheaf?


[EDIT] After viewing other questions on this site, I found that it is the proper base change theorem: if $f:X\to\text{Spec }K$ is proper, then there is a canonical isomorphism $H_\text{et}^i(X_{\overline K},\mathcal F_{\overline K})\xrightarrow\sim(R^if_*\mathcal F)_{\overline K}$, the latter is a $G_K$-module. (Complaint: this theorem is contained in notes I mentioned, but it relates etale cohomology to Galois representations is unmentioned, strange.)

But what about $X$ not proper, for example $X=Y_1(N)$ the open modular curve?


[EDIT2] Thanks for Roland's answer, but could anyone explain the simplest example: when $X=\text{Spec }k$, why does this definition of Galois action on $H_{\text{et}}^0$ coincides with the action given by the following category equivalence?

\begin{align*} \mathbf{Sh}\big((\text{Spec }k)_{\text{et}},\mathbf{Ab}\big) &\cong\{\text{discrete abelian group with continuous }G_k\text{-action}\} \\ \mathcal F&\mapsto\varinjlim_{k'/k\text{ finite separable extension}} \mathcal F(\text{Spec }k') \\ \big(\text{Spec }k'\mapsto A^{\text{Gal}(\overline k/k')}\big)&\leftarrow A \end{align*}

In fact, at first I thought I understand this category equivalence, but later I found that I didn't.

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    $\begingroup$ It acts by functoriality. For every $\sigma\in G_K$ you have an automorphism $\sigma:X_{\overline{K}}\to X_{\overline{K}}$ and thus you get an isomorphism $H^i(X_{\overline{K}},\sigma^{-1}\mathcal{F})\to H^i(X,\mathcal{F})$. If $\mathcal{F}$ is constant there is a canonical identification of $\sigma^{-1}\mathcal{F}$ with $\mathcal{F}$. If $\mathcal{F}$ is actually an LCC sheaf, and $X$ proper you can think of $H^i(X_{\overline{K}},\mathcal{F}_{\overline{K}})$ as being the geometric stalk of $R^if_\ast\mathcal{F}$ (where $f:X\to\mathrm{Spec}(K)$) and geometric stalks always have a $G_K$-action. $\endgroup$ – Alex Youcis Sep 23 '15 at 15:16
  • $\begingroup$ Of course, what I am hiding in the above (the second part) is that since $\mathcal{F}$ is defined over $K$, you have an isomorphism of $\sigma^{-1}\mathcal{F}_{\overline{K}}$ with $\mathcal{F}_{\overline{K}}$ (which is why there is an honest to god action opposed to just a system. $\endgroup$ – Alex Youcis Sep 23 '15 at 15:19
  • $\begingroup$ @AlexYoucis Why do you need $X$ proper and $\mathcal{F}$ LCC ? $\endgroup$ – Roland Sep 23 '15 at 15:21
  • $\begingroup$ @Roland You don't. I just wanted to make sure that $R^if_\ast\mathcal{F}$ is LCC—it makes me happier. $\endgroup$ – Alex Youcis Sep 23 '15 at 15:21
  • $\begingroup$ @Roland Perhaps I should have mentioned it. If no one answers the question, I'll be more detailed in my answer. $\endgroup$ – Alex Youcis Sep 23 '15 at 15:22
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Please correct me if I am wrong, but I really don't see where we need the proper assumption to get a Galois action.

So let $X/k$ be any scheme, $\overline{k}$ a separable (or algebraic) closure of $k$, and $\mathcal{F}$ be any sheaf on $X$. Write $X_\overline{k}$ for the base change of $X$, $p:X_\overline{k}\rightarrow X$ the induced morphism. Write also $\mathcal{F}_\overline{k}=p^*\mathcal{F}$ for the pullback of $\mathcal{F}$ to $X_\overline{k}$.

For $g\in\mathrm{Gal}(\overline{k}/k)$, we have an induce morphism $g:X_\overline{k}\rightarrow X_\overline{k}$, which induce a pullback $g^*\mathcal{F}_\overline{k}$. I claim that there is a canonical isomorphism $g^*\mathcal{F}_\overline{k}\simeq\mathcal{F}_\overline{k}$. This is because $\mathcal{F}$ comes from $X$. Indeed : $g^*\mathcal{F}_\overline{k}=g^*p^*\mathcal{F}=p^*\mathcal{F}=\mathcal{F}_\overline{k}$.

Hence we have an induced action on cohomology : $$ H^i(X_\overline{k},\mathcal{F}_\overline{k})\rightarrow H^i(X_\overline{k},g^*\mathcal{F}_\overline{k})\simeq H^i(X_\overline{k},\mathcal{F}_\overline{k})$$ which is the action you are looking for.

Now there is also the approach of Alex Youcis : if $f:X\rightarrow\operatorname{Spec}k$ is the structural morphism, $R^if_*\mathcal{F}$ is a sheaf on $\operatorname{Spec}k$ hence a set equipped with a continuous discrete $\operatorname{Gal}(\overline{k}/k)$-set. Unless I'm mistaken, I don't think we need to add any assumption for the following claim : the underlying $\operatorname{Gal}(\overline{k}/k)$-set is exactly $H^i(X_\overline{k},\mathcal{F}_\overline{k})$ with the above action. (This prove in particular that the above action is continuous).

This follows from the following continuity result $$\varinjlim_{k'}H^i(X_{k'},\mathcal{F}_{k'})=H^i(X_\overline{k},\mathcal{F}_\overline{k})$$ where $k'$ runs through the finite Galois extensions of $k$ in $\overline{k}$ and the trivial base change $(R^if_*\mathcal{F})_{k'}=R^if_*\mathcal{F}_{k'}$.


EDIT : Let me expand a bit on the last assertion. First, let us show that the stalk of $R^nf_*\mathcal{F}$ at a geometric point $\operatorname{Spec}\overline{k}$ is indeed $H^n(X_\overline{k},\mathcal{F}_\overline{k})$.

By definition, the stalk of $R^nf_*\mathcal{F}$ is $\varinjlim_{k'}R^nf_*\mathcal{F}(k')$ where the limit is taken over all the finite extension of $k'$ inside $\overline{k}$. Now recall that $R^nf_*\mathcal{F}$ is the sheaf associated to the presheaf $k'\mapsto H^n(X_{k'},\mathcal{F}_{k'})$, and since the stalk of a presheaf is the same as its associated sheaf, we get $(R^nf_*\mathcal{F})_\overline{k}=\varinjlim_{k'}H^n(X_{k'},\mathcal{F}_{k'})=H^n(X_\overline{k},\mathcal{F}_\overline{k})$, the last equality is from a limit argument (we should add $X$ quasi-compact and quasi-separated here).

Recall that the equivalence between sheaves on $(\operatorname{Spec}k)_{ét}$ and $\operatorname{Gal}(\overline{k}/k)$-sets is the following : if $\mathcal{F}$ is a sheaf on $\operatorname{Spec}k$, then $\mathcal{F}_\overline{k}=\varinjlim\mathcal{F}(k')$ is a $\operatorname{Gal}(\overline{k}/k)$-set. The action of $\sigma\in\operatorname{Gal}(\overline{k}/k)=\varprojlim\operatorname{Gal}(k'/k)$ on $\mathcal{F}_\overline{k}$ is induced by the compatibles actions on the $\mathcal{F}(k')$ (check that is indeed compatible).

Now you should convinced yourself that this is indeed the pull-back to $\overline{k}$ and that this action is the same as the induced action by functoriality (that is very first one I wrote). Do the same with $R^if_*$ and you will get the compatibility of the two actions in general.

Exercise : Take $k=\mathbb{R}$ and $\mathcal{F}=\mu_4$ the sheaf $A\mapsto\{x\in A, x^4=1\}$ and compute the action of the stalk (there is no trap here).

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  • $\begingroup$ How do you commute the pull-backto get that $R^if_\ast$ has fiber $H^i(X_{\overline{k}},\mathcal{F}_{\overline{k}})$? You need to use some sort of base change theorem. Either proper or smooth. $\endgroup$ – Alex Youcis Sep 24 '15 at 21:53
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    $\begingroup$ @AlexYoucis Yes you need a base change, but neither proper nor smooth, actually a much easier one that requires no assertions : the "étale base change", and the go to the limit by continuity. Or put differently, by definition the stalk of $R^if_*\mathcal{F}$ at $\overline{k}$ is $\varinjlim_{k'}(R^if_*\mathcal{F})(k')=\varinjlim_{k'}H^i(X_{k'},\mathcal{F}_{k'})=H^i(X_\overline{k},\mathcal{F}_\overline{k})$ (because a stalk of a presheaf is the same as the stalk of its associated sheaf). $\endgroup$ – Roland Sep 25 '15 at 11:46
  • $\begingroup$ Excuse me for being ignorant, but it seems all you've explained is how to reduce the result to finite extensions. $\endgroup$ – Alex Youcis Sep 25 '15 at 11:55
  • $\begingroup$ Oh! I see, now I understand. Good point! I didn't think to use the presheaf. :) $\endgroup$ – Alex Youcis Sep 25 '15 at 11:56
  • $\begingroup$ Thanks for your answer, but could you explain when $X=\text{Spec }k$, why does this definition of Galois action coincides with the action given by category equivalence? $\endgroup$ – Jz Pan Sep 28 '15 at 3:08

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