Suppose $X/\mathbb F_q$ is a smooth projective variety. Katz-Messing (eudml) shows that the characteristic polynomial of the Frobenius on $H^i_{et}(\overline{X},\mathbb Q_\ell)$ and $H^i_{crys}(X)$ are equal. Suppose now that $G \subset Aut(X)$ is a finite group. Can we identify the two cohomologies as $G$-representations (over an algebraically closed field $K$ containing both $\mathbb Q_\ell$ and $W(\mathbb F_q)$)?
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1$\begingroup$ For $\mathbb{Q}_\ell$ vs $\mathbb{Q}_{\ell'}$ where $\ell\neq\ell'\neq p$, an analogous result was proved by Olsson in "Independence of $\ell$ and traces of cohomology" (and independently by Bondarko). Maybe the motivic methods used there would give you the crystalline version. $\endgroup$– Piotr AchingerCommented Jun 7, 2022 at 6:31
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1$\begingroup$ From reading that paper, it seems that Olsson only talks about the virtual character corresponding to the alternating sum over all cohomologies but I am really interested in the individual cohomology groups. When we consider the virtual character, then I think it might be easier in the smooth projective case over a finite field by appealing to Lefschetz. $\endgroup$– AsvinCommented Jun 7, 2022 at 7:58
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2$\begingroup$ Maybe one can adapt the arguments of Olsson / Bondarko to compute the alternating sum of traces of something like $\gamma \circ \operatorname{Fr}^m$, for $\gamma \in G$ and large $m$. If these traces are equal for all $m \gg 0$, then from RH over finite fields it follows that the traces on each individual cohomology group are equal. (This is basically how the Katz-Messing argument works). $\endgroup$– David LoefflerCommented Jun 8, 2022 at 15:03
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This is OK when $X$ is projective and smooth, at least when the Kunneth components of the diagonal are algebraic. Let $k$ be the algebraic closure of a finite field. Define $M_{1}$ and $M_{2}$ to be the categories of motives over $k$ for homological equivalence (w.r.t. some Weil cohomology $H$) and numerical equivalence respectively. They are rigid tensor categories, and we have a tensor functor $M_{1}\rightarrow M_{2}$ (which preserves traces). An automorphism $a$ of $X$ defines automorphisms $h_{1}^{r}X$ and $h_{2}^{r}X$, and $Tr(a|H^{r}(X))=Tr(a|h_{1}^{r}X)=Tr(a|h_{2}^{r}X)$, which is a rational number (not depending on $H$). When applied to the powers of $a$, this shows that the characteristic polynomial of $a$ on $H^{r}(X)$ has rational coefficients independent of $H$ (by the Newton identities).
The following argument may prove the general case (or may not). Let $M(k)$ be a (good) triangulated category of motives with $\mathbb{Q}{}$-coefficients, and let $f\colon M(k)\rightarrow P$ be the universal determinant functor of $M(k)$ (Breuning 2011, 3.1). The determinant of an automorphism $a$ of an object $X$ of $M(k)$ is an element of $\pi _{1}(P)$, which contains $\mathbb{Q}^{\times}$. The result of Bondarko/Olsson should show that, for all integers $n$, $\det(1-na)\in \mathbb{Q}^{\times}\subset\pi_{1}(P)$. If so, then it follows that there exists a polynomial in $\mathbb{Q}[T]$ that is equal to characteristic polynomial of $a$ on the $l$-adic realization of $X$ for all $l$ (including $l=p$) (see Proposition 3.1 of arXiv:1311.3166).
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$\begingroup$ If I understand right, the argument is that automorphisms $a$ are really "motivic" and the traces of $a$ can be computed at the level of cycles. Is it known that the tensor functor $M_1 \to M_2$ is faithful? If I am remeber right, this is automatic for tensor functors between rigid tensor categories? $\endgroup$– AsvinCommented Jun 9, 2022 at 10:07