Today I’ve found myself mainly thinking about the interaction between Frobenius and the filtration on crystalline cohomology, referring to Mazur’s classic paper http://projecteuclid.org/download/pdf_1/euclid.bams/1183533965 as well as Ogus’ paper on “Griffiths Transversality in Crystalline Cohomology” and thought I would record a summary here before I forget everything. Of course, both of these papers are from the 1970s, so if there has been any significant advance or later examples of interest that any readers know of I’d be extremely interested to hear.

Firstly the setup. Let $X_0/k$ be a smooth projective variety over a perfect field of characteristic $p$, and let us also suppose we have a smooth proper lift $X/W(k)$. We can form the crystalline cohomology and have comparisons $M:= H^i_{crys}(X_0/W(k)) =H^i_{crys}(X/W(k)) = H^i_{dR}(X/W(k))$. Let us assume these are free modules and in fact also that the Hodge cohomology groups $H^q(X, \Omega^p_{X/W(k)})$ are free. Then we have the further relation that $M/pM = H^i_{dR}(X_0/k)$.

What structures are in play? Algebraic de Rham cohomology comes with a Hodge filtration $F^k \subset M$ (which depends on the lift $X$, though of course mod $p$ it does not), and crystalline cohomology is equipped with a semilinear Frobenius $\Phi: \sigma^*M \rightarrow M$ (which does not depend on the lift).

The fundamental relationship between these two structures is given by a theorem of Mazur, which implies under the freeness assumptions we have made that the Frobenius determines the mod $p$ Hodge filtration.

Theorem (Mazur): The reduction mod $p$ of $\Phi^{-1}(p^jM)$ is precisely the reduction of $F^j$.

(Note for the statement that Frobenius induces an isomorphism $\Phi^*M/p\Phi^*M = H^i_{dR}(X_0^{(p)}/k) \rightarrow H^i_{dR}(X_0/k)=M/pM$ preserving the Hodge filtration.)

One immediate consequence of this is the conjecture of Katz relating the slopes of Frobenius to the shape of the Hodge filtration.

Corollary: For the crystalline cohomology of $X_0/W(k)$, the Newton polygon lies above the Hodge polygon.

We do not explain what this means here, except to remark that Mazur’s theorem allows you to find bases in which the matrix for $\Phi$ has columns divisible by powers of $p$ of widths given by the Hodge numbers. The statement, which is about relating these numbers to valuations of eigenvalues, is then just an easy result in linear algebra.

The next obvious question is what can we say about the Hodge filtration $F^k$ coming from our smooth lift? The above result tells us that

$\Phi(\sigma^*F^k) \subset p^kM + p\Phi(\sigma^*M)$.

Mazur was also able to prove that $\Phi(\sigma^*F^k) \subset p^{v(k)} M$, where $v(k) := \text{sup}_{l \geq k} v_p(p^l/l!)$.

Question (Mazur): Is it in fact the case that $\Phi(\sigma^*F^k) \subset p^k M$? If so, we say $M$ is strongly divisible, and this statement is equivalent to Frobenius inducing an isomorphism $\sum_i p^{-i} \sigma^*F^i \rightarrow M$.

Note that whenever the Hodge filtration has length shorter than $p$, this is immediate from Mazur’s second inequality, since $v(k)=k$ in this case for all nonempty pieces of filtration. For example, for the cohomology of a curve or $H^1$ of an abelian variety we do get something strongly divisible.

The reason for the distance between Mazur’s estimate and the notion of strong divisibility is the phenomenon of Griffiths transversality, which was investigated by Ogus in the crystalline context following Griffiths’ work in classical Hodge theory.

Classically suppose you have a variety $X/\mathbb{C}[[t]]$ and you want to study its de Rham cohomology. Using the Gauss-Manin connection, one can identify the cohomology group itself with the cohomology of the constant family $X'$ defined by the fibre at $t=0$. However, the Hodge filtrations will not agree, but are allowed to vary within the confines imposed by Griffiths transversality. Explicitly, one can show that

$F_{X}^k \subset \sum_{i \geq 0} t^i/i! F_{X'}^{k-i}$.

In the crystalline situation, given two smooth lifts $X,X'$, one obtains a similar formula,

$F_{X}^k \subset \sum_{i \geq 0} p^i/i! F_{X'}^{k-i}$.

However, note that the denominators start to cancel off the powers of $p$, once $i \geq p$, and this is exactly why Mazur can only get the estimate involving $v(k)$. For a more general statement in the same vein see Ogus’ “corollary 2.5.”

These are all inequalities, and it is natural to ask if they are “strict”. In other words, does Griffiths transversality really happen (do these filtrations vary), and to what extent? One obvious example to bear in mind is abelian varieties, where every possible lift of the Hodge filtration corresponds to a lift (and determines it uniquely: this is Grothendieck-Messing theory). If, as the filtration lengths increase, there is enough freedom for the lifts to vary widely within the constraints imposed by transversality, then one would expect a counterexample to strong divisibility.

Ogus manages precisely this, and his example is as follows. Suppose $p>2$, and consider the hypersurface $X'/W(k)$ given by

$X': X_0^{p+2} + \dots X_{p+1}^{p+2} + pX_0\dots X_{p+1} = 0$.

Then $\Phi(\sigma^*(F^p_{X'}(H^p_{crys}(X_0/W(k)))))$ fails to lie in $p^p H^p_{crys}(X_0/W(k))$.

This is achieved by comparing to the Fermat hypersurface which also visibly lifts the special fibre. One can show that in fact the cohomology of this is strongly divisible, exploiting the fact the group actions in play give an explicit decomposition of the cohomology, making it easy to control. Comparing the filtration coming from this to the filtration coming from $X'$, using a careful deformation-theoretic study of these differences, Ogus deduces that the latter fails to be strongly divisible.