Recall the following corollary to the proper and smooth base change theorems:

Let $\pi: X \to S$ be a proper, smooth morphism. Then the direct images $R^i \pi_* \mathcal{F}$ are locally constant constructible for any l.c.c. sheaf $\mathcal{F}$ (with torsion prime to the characteristic of the residue fields) on $X_{et}$.

It follows in particular that if $S$ is the Spec of a DVR (say with an algebraically closed residue field), then the cohomology of the generic fiber (at least base-changed to a separable closure) is the same as that of the specific fiber. Consider the case where $S$ has unequal characteristic --- then essentially, this means that the special fiber $X_0$ admits a (smooth) lifting to characteristic zero. Then the above observations says something about what the cohomology of $X_0$ has to be (by comparing to the cohomology of the generic fiber, which is also the complex cohomology).

Can this be used to show that schemes in characteristic $p$ aren't liftable to characteristic zero? I don't know if this is easy, because etale cohomology groups seem to satisfy many of the same properties (e.g. Poincare duality, dimensional vanishing) of complex cohomology. Milne's book seems to list this as an application of proper-smooth base change but does not actually give an example.

  • 2
    $\begingroup$ What you state is not quite correct: for example, if $\mathcal{F}$ is the constant sheaf $\mathbb{Z}/l$ then the residue characteristic should be prime to $l$. To see this is necessary consider a family of elliptic curves and $R^1\pi_*$. $\endgroup$
    – naf
    Jun 11 '11 at 6:20
  • 2
    $\begingroup$ Here are two instances: if Hodge symmetry fails, then the proper smooth variety cannot be lifted (e.g. Serre's example of a proj. smooth surface); if $h^n(X,\mathbb Q_l)\ne\sum_{p+q=n}h^{pq},$ then it cannot be lifted. $\endgroup$
    – shenghao
    Jun 11 '11 at 15:46
  • $\begingroup$ @shenghao Can you expand on that second point at all? That isn't obvious to me. In fact, several counterintuitive examples have been constructed with respect to liftings. Essentially what you've claimed is that any liftable variety has equal Hodge and $\ell$-adic Betti numbers. William Lang constructed an example of a liftable surface for which the de Rham and Hodge Betti numbers are not equal, and I've thought a bit about this comparison. I haven't actually put thought into the $\ell$-adic vs Hodge comparison which is why this surprises me a little. $\endgroup$
    – Matt
    Jul 3 '11 at 23:37
  • $\begingroup$ Ah. (Almost) figured it out. $Y\to S$ a lifting of $X$ as in question, then by the Theorem in the question $h^n(X, \mathbb{Q}_\ell)=h^n(\overline{Y}_\eta, \mathbb{Q}_\ell)$. But now we have a smooth variety over an algebraically closed field of char $0$, so $\ell$-adic is the same as de Rham and Hodge-de Rham SS degenerates to give $h^n(\overline{Y}_\eta, \mathbb{Q}_\ell)=h^n_{dR}(\overline{Y}_\eta)=\sum_{p+q=n} h^{pq}$. Actually, to finish we need a reason $\sum h^{pq}$ stays constant in the family as well... $\endgroup$
    – Matt
    Jul 5 '11 at 18:48
  • $\begingroup$ I'm pretty sure this second statement of shenghao is false. The Hodge numbers don't have to be constant in this family, they are only upper semicontinuous which means they could jump at the special fiber. Take any $Y\to \mathrm{Spec}(R)$ where $R$ a complete DVR of mixed characteristic with res field $k$ in which a Hodge number $h^{st}$ with $s+t=n$ jumps on the special fiber $Y_0$. By construction $Y_0/k$ is liftable and by the previous argument $h^n(X, \mathbb{Q}_\ell)\neq \sum h^{pq}$. This is a counterexample, but I don't know of a concrete realization of this off the top of my head. $\endgroup$
    – Matt
    Jul 12 '11 at 19:08

There is an example due to Hirokado of a Calabi-Yau threefold in characteristic 3 with third Betti number zero which implies that it cannot be lifted to characteristic zero. See: Hirokado, Masayuki - A non-liftable Calabi-Yau threefold in characteristic 3. Tohoku Math. J. (2) 51 (1999), no. 4, 479–487.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.