Let $f\colon X \to S$ be a smooth proper morphism of schemes. If $S$ is of characteristic zero (i.e., $S$ is a $\mathbb Q$-scheme), then Deligne has shown:

  1. $R^af_*\Omega^b_{X/S}$ is locally free for all $a,b \geq 0$.

  2. The Hodge-De Rham spectral sequence $E^{ab}_1 = R^af_*\Omega^b(X/S) \Rightarrow H_{\rm DR}^{a+b}(X/S)$ degenerates in $E_1$.

This is known to fail in positive characteristic. Mumford gave examples of smooth projective surfaces over algebraically closed fields. Nevertheless there are several interesting cases of schemes $X \to S$ in characteristic $p > 0$ where I know this to be true:

a. $X$ is an abelian scheme, a relative curve, a global complete intersection in projective space, or a K3-surface over $S$.

b. $X$ is a smooth projective toric variety over a field.

c. There is also a criterion of Deligne and Illusie which in particular shows 1. und 2. to hold if $\dim(X/S) < p$ and $X$ can be lifted to $W_2(S)$.

Question: What are other examples in positive characteristic, where 1. and 2. hold?

There is also a variant of the result of Deligne for logarithmic schemes. In particular I would be also interested for examples where the logarithmic analogue of 1. and 2. hold.

ADDITION: I am taking the risk to name two examples of smooth projective schemes over a field, where I would not be too surprised if (1. and) 2. hold, but where I know of no results:

d. $X$ is Calabi-Yau (i.e., its canonical bundle is trivial).

Edit: As Torsten Ekedahl pointed out below this definition of "Calabi-Yau" is not the "right" one (not even in char. $> 2$ as I wrote in an earlier edit) and does not imply in general that the Hodge-De Rham spectral sequence degenerates.

e. $X$ is $G$-spherical for a reductive group $G$ (i.e., $X$ carries a $G$-action such that there exists a dense $B$-orbit, where $B$ is a Borel subgroup of $G$).

Edit: Again one might to have exclude some small primes depending on the Dynkin type of $G$.

  • 4
    $\begingroup$ This is a good question. I don't know many examples beyond the ones you listed. Things like Grassmanian bundles can be added to b. More generally, any time the Hodge numbers are concentrated along the diagonal ($h^{pq}=0$ for $p\not= q$), the spectral sequence trivially degenerates. $\endgroup$ – Donu Arapura Mar 18 '11 at 14:41
  • 2
    $\begingroup$ d) needs some extra condition as is shown by a s.s. Enriques surface in char 2. (It should probably not qualify as Calabi-Yau somehow.) $\endgroup$ – Torsten Ekedahl Mar 21 '11 at 19:22
  • $\begingroup$ @Torsten Ekedahl: Yes, thank you. I had not thought that the triviality of the canonical bundle is not a good definition for CY in characteristic 2. I edited d). $\endgroup$ – Torsten Wedhorn Mar 23 '11 at 8:22
  • $\begingroup$ It is not clear that this is related to char 2. In arXiv:math/0306435 I compute the Hodge s.s. of an example of Schröer in char 3 which has trivial canonical bundle and non-degenerate Hodge s.s. $\endgroup$ – Torsten Ekedahl Mar 23 '11 at 11:42
  • $\begingroup$ @Torsten Ekedahl Since this is an old thread, no one will probably look at this, but I wonder if you impose the condition that the CY variety be of finite height (i.e. not supersingular) if you get degeneration of the HdRSS. The example in your paper has infinite height. $\endgroup$ – Matt May 14 '11 at 18:26

[I misunderstood Torsten Ekedahl's earlier comment. I'm reverting the lemma to its original form which was a bit stronger.]

Since the question seemed to resonate with me, I've been thinking about this on and off (but mostly off) for a couple of days now. Here's what I've come up with.

What seems to make the example of complete intersections work is the fact that the Hodge numbers can be computed by formulas independent of the characteristic (a standard generating function can be found in SGA7, exp XI). Here's the underlying principle.

Lemma. Suppose that $D$ is the spectrum of a mixed characteristic DVR with closed point $0$ and generic point $\eta$. Let $\mathcal{X}\to D$ be a smooth projective family such that $$\dim H^q(\mathcal{X}_0,\Omega_{\mathcal X_0}^p)= \dim H^q(\mathcal{X}_\eta,\Omega_{\mathcal X_\eta}^p)$$ for all $p,q$. Then Hodge to De Rham degenerates on the closed fibre.

Proof. It degenerates on $\mathcal{X_\eta}$ by Hodge theory. This plus semicontinuity implies $$\dim E_1(\mathcal{X}_0)\ge \dim E_\infty(\mathcal{X}_0)\ge \dim E_\infty(\mathcal{X}_\eta) =\dim E_1(\mathcal{X}_\eta)=\dim E_1(\mathcal{X}_0)$$

This can be used to check degeneration for the following cases:

Ex 1. Complete intersections in projective spaces as noted already.

Ex 2. Products of smooth projective curves and complete intersections. Use Kuenneth and the fact that curves lift into characteristic $0$ (the obstruction lies in $H^2$ of the tangent sheaf; or see Oort, Compositio 1971).

Ex 3. Certain cyclic branched covers of projective space, and more generally certain hypersurfaces in weighted projective space. I'm too lazy to say what "certain" means exactly. But a careful reading of Dolgachev's notes on weighted projective spaces ought to yield something more precise.

  • $\begingroup$ This is subtle, because the cases are new only for $p\leq\dim X$, otherwise we can use Deligne-Illusie... $\endgroup$ – Piotr Achinger Mar 21 '11 at 16:51
  • 1
    $\begingroup$ Not quite, the lifting could be over a highly ramified DVR. $\endgroup$ – Torsten Ekedahl Mar 21 '11 at 19:19
  • $\begingroup$ Torsten, by "not quite", you're referring to lifting curves? Then yes, not quite. $\endgroup$ – Donu Arapura Mar 21 '11 at 19:37
  • $\begingroup$ I guess I didn't understand the argument of Deligne-Illusie, that we can reduce to positive characteristic in such a way that the special fiber lifts to $W_2$. I thought that if we have a lift to char. 0 then we have a lift to $W_2$ by some universal property of the Witt vectors :). Now I read it more carefully... $\endgroup$ – Piotr Achinger Mar 21 '11 at 20:38
  • $\begingroup$ You are of course right. In fact, all the examples I gave in a) and b) can be lifted to characteristic zero - although not in general over an absolutely unramified dvr. But I think your argument does not use this. $\endgroup$ – Torsten Wedhorn Mar 23 '11 at 8:51

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.