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I am reading Milne's lecture notes on etale cohomology and I'm hoping someone could help me clear up some minor confusion.

Let $X$ be a nonsingular variety over an algebraically closed field $k,$ say $k=\overline{\mathbb{F}_p}$ and let $\varphi:X\to X$ be any regular map. Let $\ell$ be a prime that is coprime to $p.$

In the course of proving the Lefschetz fixed-point formula, at the end of page $148,$ Milne states the following:

"Although in the above discussion, we have identified $\mathbb{Q}_\ell$ with $\mathbb{Q}_\ell(1),$ the above theorem holds as stated without this identification. The point is that $$ H^r(X,\mathbb{Q}_\ell(s)) = H^r(X,\mathbb{Q}_\ell)\otimes \mathbb{Q}_\ell(s) $$ and $\varphi$ acts through $H^r(X,\mathbb{Q}_\ell).$ Tensoring with the one-dimensional $\mathbb{Q}_\ell$ vector space $\mathbb{Q}_\ell(s)$ doesn't change the trace."

I am confused about why the last part of this statement is true. For example, if $X = X_0 \times_{\text{Spec }\mathbb{F}_p}\text{Spec }\overline{\mathbb{F}_p}$ and $F$ is the corresponding Frobenius map, shouldn't $F$ act on $\mathbb{Q}_l(s)$ by multiplying by a factor of $p^s?$

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    $\begingroup$ The Frobenius in the Galois group acts nontrivially on $\mathbb Q_\ell(s)$, but the Frobenius endomorphism of a variety acts trivially on $\mathbb Q_\ell(s)$, as does every other map from the variety to itself. $\endgroup$
    – Will Sawin
    Commented Jun 11 at 16:22
  • $\begingroup$ @WillSawin Thank you. It's true that when I thought of the nontrivial action, I was thinking about the Frobenius in the etale fundamental group (Is this correct or also wrong?). Can you please expand in an answer why a map from $X$ to itself will induce a trivial action on $\mathbb{Q}_\ell(s)?$ If $\varphi: X\to X$ is an etale map, is it because the action is simply the restriction map from the definition of an etale sheaf? $\endgroup$
    – Hasan Saad
    Commented Jun 11 at 17:04
  • $\begingroup$ It is correct in the sense that for $X/\mathbb F_p$ you may use the Lefschez trace formula to calculate the action of the Frobenius in the étale fundamental group of $\mathbb F_p$ on the cohomology of $X_{\overline{\mathbb F}_p}$ (not the Frobenius in the étale fundamental group of $X$, which doesn't act on cohomology) by comparing the action of the Frobenius in the fundamental group with the Frobenius morphism. $\endgroup$
    – Will Sawin
    Commented Jun 12 at 9:59
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    $\begingroup$ For your last question, yes. To take the trace we need a natural map $\phi ^* \mathbb Q_\ell(s) \to \mathbb Q_\ell(s)$. There is a natural map, which can be constructed in several ways. One is to write $\mathbb Q_\ell(s)$ as the pullback along a projection $\pi$ to a point and observe $\pi \circ \phi= \pi$ so $\phi^* \pi^* = \pi^*$. This construction is compatible with restriction to $\phi$-invariant subschemes thus in particular to $\phi$-fixed points, but at $\phi$-fixed points $\phi$ is trivial so this constructs the identity map, showing the action is trivial. $\endgroup$
    – Will Sawin
    Commented Jun 12 at 10:03
  • $\begingroup$ @WillSawin Thank you! This definitely answers my question and it cleared up some misconceptions I have made. Can you please post this as an answer so I can accept it? $\endgroup$
    – Hasan Saad
    Commented Jun 12 at 14:11

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The Frobenius in the Galois group acts nontrivially on $\mathbb Q_\ell(s)$, but the Frobenius endomorphism of a variety acts trivially on $\mathbb Q_\ell(s)$, as does every other map from the variety to itself.

More precisely, to define the trace of a map $\phi \colon X\ to X$ on the cohomology of $\mathbb Q_\ell(s)$ we need a natural map $\phi^* \mathbb Q_\ell(s) \to \mathbb Q_\ell(s)$.

There is a natural map, which can be constructed in several ways. One is to write $\mathbb Q_\ell(s)$ as the pullback along a projection $\pi$ to a point and observe $\pi \circ \phi = \pi $ so $\phi^* \pi^* =\pi^*$.

This construction is compatible with restriction to $\phi$-invariant subschemes thus in particular to $\phi$-fixed points, but at $\phi$-fixed points $\phi$ is trivial so this constructs the identity map, showing the action is trivial.

When we use the Lefschetz trace formula to compute for a variety $X/\mathbb F_p$ the trace of Frobenius (in the Galois group of $\mathbb F_p$) on the cohomology of $X_{\overline{\mathbb F_p}}$, we are doing this by comparing the action of Frobenius in the Galois group to the action of the Frobenius endomorphism. This is because the composition of the two Frobeniuses is the absolute Frobenius which acts as the identity on cohomology, though one may need to throw an inverse on there as I can never keep track of the signs.

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  • $\begingroup$ a question about the construction of $\phi^* \mathbb Q_\ell(s) \to \mathbb Q_\ell(s)$ why once having observed that it is compatible with restrictions to $\phi$-invar subschemes and that it is especially trivial restricted to the fixed points of $\phi$, consequently the construction gives identity map everywhere (that's what we want to have to conclude $\phi$ acts trivially on $Q_\ell(s) $)? $\endgroup$
    – user267839
    Commented Jun 16 at 1:39
  • $\begingroup$ Or are you invoking here an axiomatic approach for the existence of the trace in the sense that any candidate for $ \phi^* \mathbb Q_\ell(s) \to \mathbb Q_\ell(s)$ "we can find so far" satisfying a bunch of properties - eg restricts to identity of fixed points - is already the candidate we are looking for? $\endgroup$
    – user267839
    Commented Jun 16 at 21:26
  • $\begingroup$ @user267839 I guess this is just for checking that for one side of the trace formula the map is trivial. The same idea shows that the construction gives the identity map everywhere (in a sense this construction is the definition of "the identity map" in this context - how else would you define it?) $\endgroup$
    – Will Sawin
    Commented Jun 16 at 23:25
  • $\begingroup$ maybe following clarifies the issue: Before, you wrote that the Frobenius endom, or even more generally every endomorphism of $X$ acts trivially on $\Bbb Q(s)$. Could you explain how in more general setting a general endom on $X$ induce a selfmap on $\Bbb Q_\ell(s)$? Saying plainly, this means any fixed $F: X \to X$ should induce a map of sheaves $\Bbb Q_\ell(s) \to \Bbb Q_\ell(s)$. May I assume that more generally such induced action by an endo on $X$ on a sheaf $\mathcal{F}$ here only possible when $\mathcal{F}$ arise as pullback of a sheaf under structre map $\pi: X \to \text{Spec}(k)$? $\endgroup$
    – user267839
    Commented Jun 17 at 18:34
  • $\begingroup$ @user267839 Saying "such induced action is only possible" seems extreme. For example a space $X$ may have no nontrivial endomorphisms in which case every endomorphism acts naturally on every sheaf. But certainly when $\mathcal F$ arises as the pullback of a sheaf under a structure map, every endomorphism of $X$ acts on $\mathcal F$. $\endgroup$
    – Will Sawin
    Commented Jun 17 at 18:35

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