2 added 131 characters in body

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 $1,22,1$ and 1,22,1$. By replacing$q$by a power if necessary, we may assume that Frobenius acts on the action$H^1(E_i)$by multiplication by the positive square root of$q$. This implies that Frobenius acts on$H^2$is given H^2(X_0)$ by multiplication by $q$ if $q$ is a square. 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$15$21$ 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.

1

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 $1,22,1$ and the action of Frobenius on $H^2$ is given by multiplication by $q$ if $q$ is a square. 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 $15$ 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.