[Math] Can one find the hodge number by counting points over finite fields

ag.algebraic-geometryhodge-theoryweil-conjectures

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.)

Best Answer

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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