Let $X_0$ be a smooth projective variety over $\mathbf{F}_q$ and ${X}$ its base change to an algebraic closure $k$ of $\mathbf{F}_q$.

Crystalline cohomology $H^*_{\rm cris}(X) := H^*((X/W(k))_{\rm cris},\mathcal{O}_{(X/W)_{\rm cris}})[1/p]$ is known to be a Weil cohomology.

It is also known to be computed as the Zariski hypercohomology of the de Rham-Witt complex $W\Omega^*_X$, and the Hodge to de Rham spectral sequence:

$$E_2^{i,j} := H^i(X_{\rm Zar},W\Omega^j_X)[1/p]\Rightarrow H^{i+j}_{\rm cris}(X)$$

degenerates at the second page.

The questions below are asked with Zariski hypercohomologies replaced by étale hypercohomologies, if necessary.

  • Do we therefore have a “Hodge decomposition” $$\bigoplus_{i+j=n}H^i(X_{\rm Zar},W\Omega^j_X)[1/p]=H^n_{\rm cris}(X)\ \ ?$$
  • When $n$ is even and $i=j=n/2$, for each $x\in H^i(X_{\rm Zar},W\Omega^i_X)[1/p]$, is there an integer power of the geometric Frobenius $F^N$ on $H^i(X_{\rm Zar},W\Omega^i_X)[1/p]$ such that $F^N(x) = (q^i)^N$? In other words, is $H^i(X_{\rm Zar},W\Omega^i_X)[1/p]$ the subspace of Tate classes of $H^{2i}_{\rm cris}(X)$? It is already known that $H^i(X_{\rm ét}, W\Omega^i_{X,{\rm log}})[1/p]$ is $H^{2i}_{\rm cris}(X)^{F = q^i}$, where $W\Omega^*_{X,{\rm log}}$ is the log de Rham-Witt complex.
  • Are the Frobenius eigenvalues on $H^i(X_{\rm Zar},W\Omega^j_X)[1/p]$ (base changed to an algebraic closure of $\text{Frac}(W)$) not integers when $i\neq j$?
  • And on a more naive level, where do we crucially use that $k$ is algebraically closed? We know there is no Weil cohomology with coefficients in $\mathbf{Q}_p$, so when $p = q$ we know crystalline cohomology with $W(\mathbf{F}_p)[1/p]$ coefficients can't work. Where do the proofs of the axioms break down when $k$ is not algebraically closed?

I will content myself of some short comments with only references: these things should be well present in the literature.

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    $\begingroup$ Have you looked at Brinon and Conrad's CMI notes, or Ouyang and Fontaine's notes? $\endgroup$ Jan 20 '19 at 2:08

1)Yes, such decomposition follows from the fact that Frobenius on the de Rham-Witt differential forms acts in a way that slopes on $H^i(X, W\Omega^j)[1/p]$ are in the interval $[j,j+1)$. This forces the spectral sequence coming from the stupid filtration on the de Rham-Witt complex to degenerate at the first page and, moreover, provides a canonical splitting of the resulting filtration on $H^n_{cris}(X)$ which identifies $H^i(X, W\Omega^j)[1/p]$ with the direct summand of $H^n_{cris}(X)$ where Frobenius has slopes in the interval $[j,j+1)$(such direct summand is provided by the Dieudonne-Manin classification). See part II.3.A of Illusie's "Complexe de de Rham-Witt et cohomologie cristalline". Though, note that the spectral sequence you wrote down is not the Hodge-to de Rham, but rather the one coming from the conjugate filtration on the de Rham-Witt complex combined with Illusie-Raynaud's Cartier isomorphism. Anyway, the same argument shows that this spectral sequence degenerates and leads to a decomposition which is, a posteriori, the same by the characterization using slopes.

2)Even though there is no clear analogy between the above decomposition and the usual Hodge theory(e.g. if our variety admits a lift over $W(k)$, the dimensions $\dim_{W[1/p]}H^i(X,W\Omega^j)[1/p]$ need not coincide with the Hodge numbers of the generic fiber), it might be useful to evaluate this statement from the point of view of classical Hodge theory. There the condition of being a Tate class is stronger than just sitting in the middle piece of the Hodge filtration, and the same is actually happening in our situation. Let $X$ be a product of a supersingular elliptic curve $E_1$ with an ordinary elliptic curve $E_2$. We have an isomorphism of Frobenius-modules $H^2_{cris}(X)=\Lambda^2(H^1_{cris}(E_1)\oplus H^1_{cris}(E_2))$. The slopes of $H^1(E_1)\oplus H^1(E_2)$ are $0,1/2,1/2,1$ so after taking the exterior square we get the slopes $1/2,1/2,1,1,3/2,3/2$. There are four slopes from the interval $[1,2)$ so the space $H^1(W\Omega^1)[1/p]$ is four-dimensional, but no power of Frobenius acts as a power of $q$ on it because that would force all the slopes to be equal to $1$.

3)It follows from the Weil conjectures on purity of Frobenius eigenvalues on etale cohomology and the Lefschetz formula that any Weil cohomology theory yields the same list of eigenvalues of Frobenius on the $n$-th cohomology, see Katz and Messing's beautiful proof https://eudml.org/doc/142251 So, indeed, if $\alpha$ is an eigenvalue of Frobenius on $H^i(W\Omega^j)[1/p]$ which is an integer, then by the Weil conjectures $\alpha=\pm q^{\frac{i+j}{2}}$ so it has slope $\frac{i+j}{2}$ which means that $j\leq \frac{i+j}{2}<j+1$. This can happen only if $i=j$ or $i=j+1$, that is, your guess is almost correct, eigenvalues which are integers can only occur in the pieces $H^i(W\Omega^i)$ and $H^i(W\Omega^{i-1})$. The second case can actully happen, e.g. consider $i=1$ for a supersingular elliptic curve over $\mathbb{F}_{p^2}$ with vanishing trace of Frobenius(any supersingular elliptic curve for $p>3$ will work), it will have eigenvalues $\pm p$ on $H^1(W\mathcal{O})$.

4)Crystalline cohomology is a Weil cohomology theory even if you work over any finite field, and really not much happens when you are changing the base field: if $k\subset k'$ is an extension of perfect fields then the canonical map $H_{cris}^i(X/W(k))\otimes_{W(k)}W(k')\to H^i_{cris}(X_{k'}/W(k'))$ is an isomorphism. It follows from the corresponding fact about crystalline cohomology complexes $Ru_*\mathcal{O}_{X/W}$, which, in turn, can be proven by devissage from the analogous fact about the de Rham complexes which is just the base change for Kahler differentials(a detailed proof should be in Berthelot's book). And, actually, in all of the above we were working with crystalline cohomology over $W(\mathbb{F}_q)$ in order to be able to view the Frobenius as a linear endomorphism of crystalline cohomology.

Serre's observation about the non-existence of a $\mathbb{Q}_p$-cohomology theory is in a slightly different setup -- it shows that there cannot be a reasonable cohomology theory for varieties over the algebraic closure of a finite field $\bar{\mathbb{F}}_p$ with coefficients in $\mathbb{Q}_p$-vector spaces. But given a variety $X$ over $\mathbb{F}_p$ the crystalline cohomology $H^n_{cris}(X/\mathbb{Z}_p)$ is only functorial under the maps of varieties over $\mathbb{F}_p$, and the Serre's argument doesn't go through here because a supersingular curve defined over $\mathbb{F}_p$ can't have all its endomorphism defined over the same finite field $\mathbb{F}_p$(crystalline cohomology actually provides a very sophisticated proof of this fact, but this can also be shown via more elementary methods -- $\mathrm{End}_{\mathbb{F}_p}(E)$ contains a $2$-dimensional subalgebra generated by the $p$-th power Frobenius and any endomorphism defined over $\mathbb{F}_p$ commutes with Frobenius but the center of division algebra $\mathrm{End}_{\bar{\mathbb{F}}_p}(E)$ is only one-dimensional)

  • $\begingroup$ I’m with you on all the answers: there are only some doubts I still have about them. (1) Is there a Hodge-to-de Rham ss for crystalline cohomology? There is a crystalline de Rham complex on $(X/W)_{\rm cris}$: do we have $E_2^{r,s}:= H^r(X_{\rm Zar}, \Omega^s_{X/W,cris})\Rightarrow H^{r+s}_{\rm cris}(X)$? (2) What do you mean when you say that in classical Hodge Theory being a Tate class is stronger than being a Hodge class? A conjecture of Deligne should imply that the two notions are equivalent, i.e. Hodge classes should match Tate classes via $\ell$-adic Artin comparison $\endgroup$
    – user134756
    Jan 20 '19 at 4:53
  • $\begingroup$ I think the spectral sequence in (1) that I’m asking about in my comment is the hypercohomology-to cohomology spectral sequence applied to $\Omega^*_{(X/W)_{cris}}$, using that $\mathbb{H}^*((X/W)_{\rm cris},\Omega^*_{(X/W)_{cris}}) = H^*_{cris}(X)$. Am I right? $\endgroup$
    – user134756
    Jan 20 '19 at 4:59
  • $\begingroup$ Erratum: when writing $E_2^{r,s}:= H^r(X_{\rm Zar},\Omega^s_{X/W,cris})$ I really meant $H^r((X/W)_{\rm cris}, \Omega_{(X/W)_{cris}}^s)$. $\endgroup$
    – user134756
    Jan 20 '19 at 5:26
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    $\begingroup$ (1)i’m not sure I understand what is $\Omega^*_{(X/W)_{cris}}. As i said, for any complex of sheaves there is a 1st(sic) page spectral sequence from cohomology of the terms of the complex to its hypercohomology, here we apply it to the de Rham-Witt complex on the Zariaki site. (2)i meant that being a Hodge class is stronger than being an element of the middle piece of the Hodge decomposition — it should also lie in the integral lattice. i mentioned this just to explain why i didn’t expect the statement you asked for to hold, this might be a bad analogy and the actual proof doesn’t rely on it. $\endgroup$
    – SashaP
    Jan 20 '19 at 5:34
  • $\begingroup$ OK. Now I see. What I meant by $\Omega^*_{(X/W),cris}$ is the complex of sheaves assigned on $(X/W)_{cris}$ by $(U,T,\delta) \mapsto \Omega^*_{\mathcal{O}_T(T),\delta}$. Denoting by $u : (X/W)_{cris}\to X_{Zar}$ the projection to the Zariski topos, we have $Ru_*\mathcal{O}_{(X/W)_{cris}} = \mathbb{R}u_*\Omega^*_{(X/W)_{cris}}$, so we do get: $$H^*((X/W)_{cris},\mathcal{O}_{X/W}) = \mathbb{H}^*((X/W)_{cris},\Omega^*_{(X/W)_{cris}})$$ so as you say we have a ss $H^r((X/W)_{cris},\Omega^s_{(X/W)_{cris}})\Rightarrow H^{r+s}((X/W)_{cris},\mathcal{O})$. $\endgroup$
    – user134756
    Jan 20 '19 at 5:40

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