This is based on a comment here: https://mathoverflow.net/questions/67485/can-proper-smooth-base-change-be-used-to-show-that-varieties-cannot-be-lifted-to

I felt funny about the comment and I tried to prove it, but instead came up with a way to construct a counterexample. Since I'm still learning this stuff, it is possible I've made some glaring error, though.

The comment: If $X$ is a smooth proper variety over a field of positive characteristic and $h^n(X, \mathbb{Q}_\ell)\neq \sum_{p+q=n} h^{pq}(X)$ for some $n$, then $X$ cannot be lifted to characteristic $0$. Or in other words, if the $\ell$-adic and Hodge Betti numbers do not match for some $n$, then it cannot be lifted.

Here is my proposed counterexample. Find a smooth proper $Y\to S$ where $S=\mathrm{Spec}(R)$ for $R$ a DVR with mixed characteristic ($R/m=k$ of characteristic $p>0$, and $\mathrm{Frac}(R)=K$ of characteristic $0$) with the following property: Some Hodge number satisfies $h^{st}(Y_0)> h^{st}(\overline{Y}_\eta)$ where $\overline{Y}_\eta$ means the geometric generic fiber. So we are looking for an example where some Hodge number jumps up at the special fiber. I don't see any immediate reason why this can't happen, since they should be merely upper semicontinuous.

Here is my attempted proof of why this gives a counterexample. Well, first $Y_0/k$ is a characteristic $p$ variety that can be lifted to characteristic $0$ just by construction. Suppose $n=s+t$ from the example. We have the sequence of (in)equalities $h^n(Y_0, \mathbb{Q}_\ell)= h^n(\overline{Y}_\eta, \mathbb{Q}_\ell)=\sum h^{pq}(\overline{Y}_\eta)<\sum h^{pq}(Y_0)$ which comes respectively from smooth base change for etale cohomology, and then the fact that $\overline{Y}_\eta$ is a smooth variety over an algebraically closed field of characteristic $0$ (by Deligne-Illusie the Hodge-de Rham SS degenerates at $E_1$ but maybe the error is that $\ell$-adic and de Rham don't match up?), and then the assumption about Hodge numbers jumping.

In other words, such a variety $Y_0$ would be liftable with $\ell$-adic and Hodge Betti numbers being different.

Question: What is a good example of where this happens? Or, if the original comment was correct what is an actual proof and where does mine go wrong? Thanks!

P.S. It is very annoying that you can't reply to comments on MO, so the person never learns of a reply. This has happened several times to me now.


1 Answer 1


Suppose we find a smooth projective surface $X$ over $W(k)$ (the ring of Witt vectors of an algebraically closed field $k$ of char. $p > 0$) whose geometric generic fibre has $p$-torsion in its middle dimensional cohomology. (This geometric generic fibre is over an algebraically closed field of char. $0$, so its middle cohomology is the same whether I compute it as etale cohomology, or by converting it --- via the Lefshcetz principle --- into a surface over the complex numbers and computing the cohomology as singular or even simplicial cohomology.)

For safety, I want to assume that $p$ is odd (so that crystalline cohomology will work optimally well, and also so that $p$ is greater than the relative dimension of $X$).

Because $p$ is greater than the relative dimension of $X$, the integral comparison theorem of Fontaine--Messing and Faltings should show that $p$-torsion in the $2$nd cohomology of the geometric generic fibre of $X$ is the same as $p$-torsion in the $2$nd crystalline cohomology of $X$ over $W(k)$.

Now universal coefficients relates $p$-torsion in the crystalline cohomology of $X$ over $W(k)$ to the de Rham cohomology of the special fibre of $X$. More precisely, non-trivial $p$-torsion in crystalline $H^2$ of $X$ implies that the de Rham $H^1$ of the special fibre of $X$ has rank greater than the de Rham $H^1$ of the generic fibre of $X$.

If I'm not confused, a consideration of the Hodge-to-de Rham spectral sequence should then imply that the sum of the $h^{pq}$s of the special fibre is greater then the $1$st Betti number of the generic fibre.

So this should give an example of the kind that you want.

Of course, for this to work, you have to produce smooth projectivce surfaces with $p$-torsion in their $H^2$ and good reduction at $p$. One can find such examples as Picard modular surfaces, I believe. There may well be easier (and more explicit examples); unfortunately, I don't know them.

Summary: The basic idea here is that, in circumstances of good reduction over an absolutely unramified $R$ with $p$ large enough, the difference between the rank of de Rham cohomology in the generic and special fibres comes from $p$-torsion in the cohomology of the generic fibre. Since this $p$-torsion can't be detected by Betti numbers, you can't expect equality of ($\ell$-adic) Betti numbers between the generic and special fibres to force an equality of dimensions of de Rham cohomology.

[Note: This is related to one of Grothendieck's original motivations for introducing crystalline cohomology, which was to find a way to study $p$-torsion in cohomology of varieties in char. $p$.]

  • $\begingroup$ Thanks. This is really fantastic. It might take me a bit to absorb as I'm still struggling through the basics of crystalline cohomology. I like the idea of looking at $W$. I was making it too hard by picking an arbitrary DVR. As you point out, this makes some calculations easier since $p>\dim$ and lifting to $W_2$ allows you to conclude Hodge-de Rham degenerates on the special fiber. $\endgroup$
    – Matt
    Jul 13, 2011 at 16:10

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