Meaning of $H^*(X)$ Being Hodge-Tate in Algebraic Geometry

ag.algebraic-geometry

Let $X$ be a (let us say smooth to obscure any confusions I have between $H(X)$ and $H_c(X)$) algebraic variety defined over some subfield of $\mathbb{C}$. I have occasionally overheard the expression "$H^*(X)$ is Hodge-Tate" used to mean something which, as far as I could tell from context, resembled one of the following:

(1) $H^*(X)$ is generated by $(p,p)$ classes, i.e. those in some intersection $W_{2p} H^i(X,\mathbb{Q}) \cap F^p H^i(X,\mathbb{C})$, where $W$ and $F$ are the weight and Hodge filtrations from the mixed Hodge structure. In particular were $X$ smooth and proper, $H^*(X) = \bigoplus H^{p,p}(X)$.

(2) Spread $X$ out as appropriate and reduce mod a good prime, then it is `polynomial count', i.e. the number of points over $\mathbb{F}_{p^n}$ is a polynomial in $p^n$.

(3) Spread $X$ out as appropriate and reduce mod a good prime, then all the eigenvalues of Frobenius are powers of $p$.

(4) The class of $X$ in the Grothendieck group of varieties is in $\mathbb{Z}[\mathbb{A}^1]$

But when I searched for "Hodge-Tate" on google, I arrived at some description of "Hodge-Tate numbers" etc which seemed to have something to do with p-adic Hodge theory and apply to any variety. Anyway my question is as in the title,

What does it mean for $H^*(X)$ to be Hodge-Tate?

Also I guess (4) => (3) => (2) and I vaguely recall from some appendix of N. Katz that => (1) can be tacked on the end (?) I would also like to know

Which of the reverse implications is false, and what are some counterexamples?

Best Answer

2 does imply 1 (for smooth projective varieties) via $p$-adic Hodge theory and perhaps a simpler argument.

1 does not imply 2. Indeed, blow up $\mathbb P^2$ at the Galois orbit of some point that is not $\mathbb Q$-rational but is rational over some quadratic field extension, say $(1: \sqrt{-1} : 0)$ . Mod a prime $p$ where that point is not $\mathbb F_p$-rational, there are $p^2+p+1$ $\mathbb F_p$-rational points. Mod a prime $p$ where that point is rational, there are $p^2+3p+1$ points. Obviously, this cannot be explained by any polynomial.

2 does imply 3 for smooth projective varieties. Using the polynomial for the number of points, one can compute the Weil zeta function as a product of terms of the form $\left( \frac{1}{1 -p^n t} \right)$. Using the Lefschetz trace formula, this is a product of factors corresponding to the eigenvalues of Frobenius in the etale cohomology. By the Riemann hypothesis, none of these terms cancel, so all eigenvalues are powers of $p^n$.

Not sure about 3 and 4.