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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quick scan of mathscinet suggests that the result was proved for hypersurfaces by Dwork, conjecturd in general by Katz (unpublished, I think), mostly proved by Mazur and the final details added by Ogus.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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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.