Can one find the hodge number by counting points over finite fields? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T04:30:22Z http://mathoverflow.net/feeds/question/92958 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/92958/can-one-find-the-hodge-number-by-counting-points-over-finite-fields Can one find the hodge number by counting points over finite fields? Yuhao Huang 2012-04-03T03:15:59Z 2012-04-03T17:13:19Z <p>Given a proper smooth variety $X$ of dimension $n$ over $\mathbb{C}$, assume it has a model over a DVR of mixed characteristic $(0,p)$ with residue field $\mathbb{F}_q$, and assume the closed fiber $X_0$ is smooth.</p> <p>By the Weil conjecture, one can find the Betti number of the complex manifold $X^{an}$ by counting $\mathbb{F}_{q^r}$-points of $X_0$. If by counting points we find $|X_0(\mathbb{F}_{q^r})|=\sum\pm u_j^r$ for all $r$ and $b_i$ of the $u_j$'s has absolute value (in $\mathbb{C}$) equal to $\sqrt{q}^{i}$, then the $i$-th Betti number is equal to $b_i$.</p> <p>My question is, can one find the Hodge number of $X$, which is $h^{ij}=\dim H^i(X,\Omega^j)$ by counting points of the closed fiber $X_0$? (The reason I'm asking this is, I guess both should connect to the theory of weights on the motive. So even if one cannot find the Hodge number this way, the reason must be interesting.)</p> http://mathoverflow.net/questions/92958/can-one-find-the-hodge-number-by-counting-points-over-finite-fields/92968#92968 Answer by ulrich for Can one find the hodge number by counting points over finite fields? ulrich 2012-04-03T06:28:00Z 2012-04-03T08:40:53Z <p>No, one cannot find the Hodge numbers this way. </p> <p>For an example, consider $X_0$ the Kummer surface associated to a product of supersingular elliptic curves $E_1$ and $E_2$. Recall that this is the surface give by taking the quotient of $E_1 \times E_2$ by $\{1,-1\}$ and then blowing up the 16 singular points (we assume $p \neq 2$). </p> <p>The Betti numbers of $X_0$ are $1,22,1$. By replacing $q$ by a power if necessary, we may assume that Frobenius acts on the $H^1(E_i)$ by multiplication by the positive square root of $q$. This implies that Frobenius acts on $H^2(X_0)$ by multiplication by $q$. It follows that the number of points of $X_0$ over $\mathbb{F}_{q^r}$ is the same as the number of points of $Y_0$ which is $\mathbb{P}^2$ blown up in $21$ points.</p> <p>Now let $X$ be the surface over $\mathbb{C}$ constructed in the same way as $X_0$ using lifts of $E_1$ and $E_2$ to characteristic zero. It is easy to compute the Hodge numbers of $X$ and one sees that $h^{2,0}(X) = 1$ (in fact the same is also true for $X_0$). Now $Y_0$ also lifts to a variety $Y$ in characteristic zero and $h^{2,0}(Y) = 0$. It follows that the Hodge numbers cannot be found by counting points over finte fields.</p> http://mathoverflow.net/questions/92958/can-one-find-the-hodge-number-by-counting-points-over-finite-fields/93004#93004 Answer by David Speyer for Can one find the hodge number by counting points over finite fields? David Speyer 2012-04-03T14:35:43Z 2012-04-03T17:13:19Z <p>It is worth mentioning the "Newton above Hodge" theorem. This is a way in which the point count can impose nontrivial conditions on the Hodge numbers, beyond knowing the Betti numbers. </p> <p>As I assume you know, the number of points of $X(\mathbb{F}_{q^k})$ is <code>$\sum_{r=0}^{2n} (-1)^r \sum_{i=1}^{b_r} \alpha_{i,r}^{k}$</code> where $b_i$ is the $i$-th betti number and $\alpha$'s are algebraic integers obeying $|\alpha_{i,r}| = q^{r/2}$, where the left hand side is any archimedean absolute value. Also, we can number the $\alpha$'s such that $\alpha_{i,r} \alpha_{b_r+1-i, r} = q^r$. (This is a consequence of Hard Lefschetz.). In particular, $\prod_i \alpha_{i,r} = \pm q^{r b_r/2}$.</p> <p>Fix $r$, so it will no longer appear in our notation. Let $F(x) = \prod_{i=1}^{b_r} (x-\alpha_{i,r})$; this is the characteristic polynomial of Frobenius on $H^r(X)$. Let $N$ be the $p$-adic Newton polytope of $F$. Since the constant term of $F$ is $\pm q^{r b_r/2}$, the endpoints of the Newton polytope are at $(0,0)$ and $(b_r, r b_r/2 \cdot v_p(q))$. The symmetry $\alpha_{i,r} \alpha_{b_r+1-i, r} = q^r$ means that the segments of slope $\mu$ and $r-\mu$ have the same length. In ulrich's example, all of the eigenvalues of Frob on $H^2$ have norm $q$, so $N$ is just a straight line from $(0,0)$ to $(22,22)$.</p> <p>We now define the Hodge polygon $H$. This is also a piecewise linear convex curve joining $(0,0)$ to $(b_r, r b_r v_p(q)/2)$. All of the segments have slope of the form $k r v_p(q)/2$ for some integer $k$ between $0$ and $2r$, and the horizontal length of this segment is $h^{k,r-k}$. The symmetry of the Hodge diamond tells us that this polygon, also, has the property that the segments of slope $\mu$ and $r-\mu$ have the same length.</p> <p>In Ulrich's example $X$, the Hodge polygon goes from $(0,0)$ to $(1,0)$, then to $(21,20)$, and then to $(22,22)$.</p> <blockquote> <p>The Newton above Hodge theorem says that $N$ is above $H$.</p> </blockquote> <p>I learned this from Kiran Kedlaya $p$-adic differential equations course, which he has now converted into <a href="http://books.google.com/books?id=Enrw2kCqZOsC" rel="nofollow">a book</a>, see Chapter 14. I am not familiar with the history of this result, but a quick scan of mathscinet suggests that the result was proved for hypersurfaces by <a href="http://www.ams.org/mathscinet-getitem?mr=175895" rel="nofollow">Dwork</a>, conjecturd in general by Katz (unpublished, I think), mostly proved by <a href="http://www.ams.org/mathscinet-getitem?mr=321932" rel="nofollow">Mazur</a> and the final details added by <a href="http://www.ams.org/mathscinet-getitem?mr=1859245" rel="nofollow">Ogus</a>. Mazur's paper looks very readable -- that's where I would start.</p>