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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.

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 $b_i$.

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.)

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What is a DVR of mixed characteristic? –  Spice the Bird Apr 3 '12 at 4:14
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The fraction field is of char. $0$ and the quotient by maximal ideal is of char. $p$. –  temp Apr 3 '12 at 5:46
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A variety $V$ over $C$ is defined over a fin. gen. subfield $K$ over $Q$; $K$ is the function field of a smooth proper scheme $Y$ over $Z[1/N]$; a smooth proper $V$ arises from base change from a smooth proper morphism $f:X\to Y$. Given $V_1$, $V_2$, this gives us $X_1, X_2$ and now we can look at their fibers over points of $Y$ . But Ito (page 16) front.math.ucdavis.edu/0211.5378 shows that if two varieties over $Z[1/N]$ have the same zeta function mod $p$ for almost all $p$, then their Hodge numbers are equal. Ulrich's answer shows that one prime is insufficient. –  SGP Apr 3 '12 at 13:43
    
Thank you Temp. –  Spice the Bird Apr 3 '12 at 15:36
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2 Answers

No, one cannot find the Hodge numbers this way.

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 $p \neq 2$).

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.

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.

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Do you mean $\mathbb{P}^2$ blown up in 21 points? Why is the action of Frobenius on $H^2$ given by multiplication by $q$ if $q$ is a square? –  temp Apr 3 '12 at 7:39
    
Yes, thanks. I have edited my answer. –  ulrich Apr 3 '12 at 8:41
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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.

As I assume you know, the number of points of $X(\mathbb{F}_{q^k})$ is $\sum_{r=0}^{2n} (-1)^r \sum_{i=1}^{b_r} \alpha_{i,r}^{k}$ 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, $\prod_i \alpha_{i,r} = \pm q^{r b_r/2}$.

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 $(22,22)$.

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.

In Ulrich's example $X$, the Hodge polygon goes from $(0,0)$ to $(1,0)$, then to $(21,20)$, and then to $(22,22)$.

The Newton above Hodge theorem says that $N$ is above $H$.

I learned this from Kiran Kedlaya $p$-adic differential equations course, which he has now converted into a book, 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 Dwork, conjecturd in general by Katz (unpublished, I think), mostly proved by Mazur and the final details added by Ogus. Mazur's paper looks very readable -- that's where I would start.

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The link to Mazur's paper goes to the Dwork paper. –  Andy Putman Apr 3 '12 at 15:33
    
Fixed now, thanks. –  David Speyer Apr 3 '12 at 17:13
    
Wow, this is a nice example of Newton above Hodge. I guess it also explains why in Newton above Hodge people keep talking about supersingular things. Illusie said in his survey on Crystalline cohomology for very general varieties the two polygon match, and the Newton polygon goes up under specialization. –  temp Apr 4 '12 at 1:29
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