Hodge numbers of reduction mod $p$ - MathOverflow

Hodge numbers of reduction mod $p$

2012-12-17T19:52:42Z

Let $X$ be a projective variety defined over a number field $K$, and $p \in \textrm{Spec }\mathcal{O}_K$ a maximal ideal, so that reduction mod $p$ makes sense, and the resulting scheme (mod $p$) $\bar{X}$ is smooth over the relevant finite field. Assume that $X$ smooth over $K$.

1.) In the case that $X$ is a curve, is there a short argument to show that the geometric genus of $X$ and of $\bar{X}$ are the same? Certainly if $X$ is a plane curve this is clear.

2.) The hodge numbers $h^{p,q}_X = \dim H^p(X, \Omega^q)$ make sense in all characteristics. Are the hodge numbers preserved under reduction mod $p$, that is, $h^{p,q}_X = h^{p,q}_\bar{X}$ ?

3.) The Weil conjectures tell us that we can recover the Betti numbers of $X$ (considered as a complex manifold) from the zeta function of $\bar{X}$. There are many smooth projective varieties that have reduction $\bar{X}$ mod $p$ and the Weil conjectures tell us that all of them have the same Betti numbers. Can one prove this without using the Weil conjectures, perhaps with Etale cohomology?

4.) More generally, if $\mathcal{L}$ is a locally free sheaf on $X$, and $\bar{\mathcal{L}}$ denotes the reduction mod $p$, I would guess that the numbers $\dim H^p(X, L)$ and $\dim H^p(\bar{X}, \bar{L})$ don't match up - but I don't have a good example.

I am interested in proofs (not using the Weil conjectures if possible).

Answer by wccanard for Hodge numbers of reduction mod $p$

2012-12-17T21:29:13Z

This is only a comment but I'm a new user and can't make comments.

For (4), the dimensions of the $H^i$ might not even be the same if $i=0$. let $X$ be an elliptic curve and choose two distinct points $P$ and $Q$ on the generic fibre whose reductions mod $p$ are the same. Let $L$ be the line bundle $P-Q$. This has no global sections on the generic fibre, because any function with a simple pole at $P$ and a simple zero at $Q$ would be an isomorphism between the curve and projective 1-space. However mod $p$ the line bundle becomes trivial, so has global sections.

Answer by Donu Arapura for Hodge numbers of reduction mod $p$

2012-12-17T21:53:27Z

For some explicit counterexamples to (2) and therefore also (4) again, see J. Suh, Plurigenera of general type surfaces in mixed characteristic, Compositio (2008). The only thing that you can say for sure is that $h^{pq}(\bar X)\ge h^{pq}(X)$ by upper semicontinuity of cohomology.

Answer by Daniel Litt for Hodge numbers of reduction mod $p$

2012-12-17T22:23:30Z

Since people have addressed (2-4), I'll address (1). There is indeed a fast argument.

Namely, you have a flat family of curves $X\to \operatorname{Spec}(\mathcal{O}_{K, p})$. You are interested in comparing the geometric genus of the generic fiber $X_K$ to that of the special fiber $X_{\mathbb{F}_q}$ . Note that by constancy of Euler characteristic in flat families, $$1-p_a({X_K})=\chi(\mathcal{O}_{X_K})=\chi(\mathcal{O}_{X_{\mathbb{F}_q}})=1-p_a(X_{\mathbb{F}_q}).$$ where $p_a$ denotes arithmetic genus. So the arithmetic genera of the fibers are equal. Since you've required that the fibers $X_K$ and $X_{\mathbb{F}_q}$ are smooth, their geometric genera equal their arithmetic genera, which completes the argument.

ADDED (12/19/2012): I'd also like to add a comment about the Hodge decomposition for curves, to complement Christian's excellent answer. Namely, one can see that the Hodge-theoretic predictions one would make from the situation over $\mathbb{C}$ are true for curves over any base. Here's what I mean by this:

Let $\pi: C\to S$ be a smooth projective relative curve, with geometrically connected fibers. Let $\Omega^{dR}_{C/S}$ be the relative de Rham complex $$\Omega^{dR}_{C/S}: 0\to \mathcal{O}_C\overset{d}{\to} \Omega^1_{C/S}\to 0.$$ Then $$R^2\pi_*(\Omega^{dR}_{C/S})=R^1\pi_*(\Omega^1_{C/S})$$ $$R^0\pi_*(\Omega^{dR}_{C/S})=R^0\pi_*(\mathcal{O}_C)$$ and there is a natural short exact sequence $$0\to R^0\pi_*(\Omega^1_{C/S})\to R^1\pi_*(\Omega^{dR}_{C/S})\to R^1\pi_*(\mathcal{O}_C)\to 0.$$

These statements are equivalent to the claim that the Frolicher spectral sequence degenerates at $E_2$. But this is clear, since all of the differentials have $0$ as either their target or source (if you write out the spectral sequence, you'll see that all the non-zero entries are in a little $2\times 2$ box, but all the differentials have bi-degree $(1, 2)$).

So one cannot hope to have any "pathological" behavior in $h^{p,q}$'s for smooth relative curves. One can see this from Christian's answer (since the cohomology of a curve has no torsion); I wanted to observe that this is true over any base for essentially formal reasons.

Answer by Christian Liedtke for Hodge numbers of reduction mod $p$

2012-12-19T10:01:58Z

As already pointed out, the Hodge numbers may go up under reduction modulo $p$. On the other hand, let me also point out that the situation can be controlled:

1.) For all $p$, where $\overline{X}_p$ is smooth, the $\ell$-adic Betti numbers of $X$ and $\overline{X}_p$ are the same.

2.) Now, by the universal coefficient formula relating crystalline and deRham cohomology, we have for all $i$ short exact sequences

$$ 0 \to H^i_{cris}(\overline{X}_p/W)/p\to $$

$$ H^i_{dR}(\overline{X}_p/k_p)\to {\rm Tor}_1^{W(k_p)}(H_{cris}^{i+1}(\overline{X}_p/W),k_p)\to 0 $$

where $k_p={\cal O}_K/p$. Now, if $\overline{X}_p$ has torsion-free crystalline cohomology, then the term on the right is zero, and the term on the left is a $k_p$-vector space of dimension equal to the $i$.th $\ell$-adic Betti number. Then, the Fr\"olicher spectral sequence relating Hodge- and deRham-cohomology degenerates at $E_2$, we have that $\sum_{p+q=i}h^{p,q}$ is equal to the $i$.th $\ell$-adic Betti number. Thus, simply for dimension reasons, the Hodge numbers of $X$ and $\overline{X}_p$.

The upshot is that torsion in crystalline cohomology of $\overline{X}_p$ detects and controls the differences in Hodge numbers of $X$ and $\overline{X}_p$. For almost all $p\in {\rm Spec} {\cal O}_K$, the reduction $\overline{X}_p$ will be smooth and will have torsion-free crystalline cohomology.