Hodge numbers of reduction mod $p$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T01:41:13Z http://mathoverflow.net/feeds/question/116639 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/116639/hodge-numbers-of-reduction-mod-p Hodge numbers of reduction mod $p$ LMN 2012-12-17T19:52:42Z 2012-12-19T21:29:26Z <p>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 $K$.</p> <p>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.</p> <p>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, <code>$h^{p,q}_X = h^{p,q}_\bar{X}$</code>?</p> <p>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?</p> <p>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.</p> <p>I am interested in proofs (not using the Weil conjectures if possible).</p> http://mathoverflow.net/questions/116639/hodge-numbers-of-reduction-mod-p/116647#116647 Answer by wccanard for Hodge numbers of reduction mod $p$ wccanard 2012-12-17T21:29:13Z 2012-12-17T21:29:13Z <p>This is only a comment but I'm a new user and can't make comments.</p> <p>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.</p> http://mathoverflow.net/questions/116639/hodge-numbers-of-reduction-mod-p/116652#116652 Answer by Donu Arapura for Hodge numbers of reduction mod $p$ Donu Arapura 2012-12-17T21:53:27Z 2012-12-17T21:53:27Z <p>For some explicit counterexamples to (2) and therefore also (4) again, see J. Suh, <em>Plurigenera of general type surfaces in mixed characteristic,</em> 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.</p> http://mathoverflow.net/questions/116639/hodge-numbers-of-reduction-mod-p/116658#116658 Answer by Daniel Litt for Hodge numbers of reduction mod $p$ Daniel Litt 2012-12-17T22:23:30Z 2012-12-19T21:29:26Z <p>Since people have addressed (2-4), I'll address (1). There is indeed a fast argument. </p> <p>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 <code>$X_{\mathbb{F}_q}$</code>. Note that by constancy of Euler characteristic in flat families, <code>$$1-p_a({X_K})=\chi(\mathcal{O}_{X_K})=\chi(\mathcal{O}_{X_{\mathbb{F}_q}})=1-p_a(X_{\mathbb{F}_q}).$$</code> 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.</p> <hr> <p>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 <strong><em>curves over any base</em></strong>. Here's what I mean by this:</p> <p>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 <code>$$\Omega^{dR}_{C/S}: 0\to \mathcal{O}_C\overset{d}{\to} \Omega^1_{C/S}\to 0.$$</code> Then <code>$$R^2\pi_*(\Omega^{dR}_{C/S})=R^1\pi_*(\Omega^1_{C/S})$$</code> <code>$$R^0\pi_*(\Omega^{dR}_{C/S})=R^0\pi_*(\mathcal{O}_C)$$</code> and there is a natural short exact sequence <code>$$0\to R^0\pi_*(\Omega^1_{C/S})\to R^1\pi_*(\Omega^{dR}_{C/S})\to R^1\pi_*(\mathcal{O}_C)\to 0.$$</code></p> <p>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 $(1, 2)$). </p> <p>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. </p> http://mathoverflow.net/questions/116639/hodge-numbers-of-reduction-mod-p/116773#116773 Answer by Christian Liedtke for Hodge numbers of reduction mod $p$ Christian Liedtke 2012-12-19T10:01:58Z 2012-12-19T10:01:58Z <p>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: </p> <p>1.) For all $p$, where $\overline{X}_p$ is smooth, the $\ell$-adic Betti numbers of $X$ and $\overline{X}_p$ are the same.</p> <p>2.) Now, by the universal coefficient formula relating crystalline and deRham cohomology, we have for all $i$ short exact sequences</p> <p>$$0 \to H^i_{cris}(\overline{X}_p/W)/p\to$$</p> <p>$$H^i_{dR}(\overline{X}_p/k_p)\to {\rm Tor}_1^{W(k_p)}(H_{cris}^{i+1}(\overline{X}_p/W),k_p)\to 0$$</p> <p>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 $\overline{X}_p$.</p> <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. </p>