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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?

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  • $\begingroup$ The paper "Eigenvalues of Froebenius and Hodge numbers" from Kisin and Lehrer discusses the relations between 1), 2) and 3), using p-adic Hodge theory. $\endgroup$
    – Simon Pepin Lehalleur
    Feb 4, 2013 at 6:28
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    $\begingroup$ Also, there is a potential confusion with the notion of "Hodge-Tate representation" in p-adic Hodge theory. According to Faltings' theorem, the cohomology of any smooth proper variety over a p-adic field is Hodge-Tate (see definition 2.3.4 and theorem 2.2.3 in the Brinon-Conrad lecture notes, math.stanford.edu/~conrad/papers/notes.pdf) so this does not quite match the notions 1-4) (which are closer to "the motive of X is a mixed Tate motive", I guess) $\endgroup$
    – Simon Pepin Lehalleur
    Feb 4, 2013 at 6:36
  • $\begingroup$ The appendix of Katz is in this paper: arxiv.org/pdf/math/0612668.pdf It does indeed seem to say (2)=>(1). $\endgroup$
    – Jim Bryan
    Feb 4, 2013 at 7:31
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    $\begingroup$ The Tate conjecture implies that every Galois-invariant map between the etale cohmology of two varieties comes from an $\mathbb Q_l$-linear combination of correspondences. Thus every Galois-invariant map between the etale cohomology of two motives comes from a $\mathbb Q_l$-linear combination of morphisms in the category of motives. In particular, by linear algebra, an isomorphism of Galois representations comes from an isomorphism of motives. (3) gives an isomorphism of Galois representations between $[X]$ and a sum of powers of the Tate motive (one may also have to assume semisimplicity) $\endgroup$
    – Will Sawin
    Feb 4, 2013 at 23:39
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    $\begingroup$ thus we have an isomorphism in the category of motives, so an equality in the Grothendieck group of motives. I think the problem is that an isomorphism in the category of motives may come from a correspondence in the category of algebraic varieties, like an isogeny between two elliptic curves, that is not an isomorphism, so does not induce an equality in the Grothendieck group. $\endgroup$
    – Will Sawin
    Feb 4, 2013 at 23:40

1 Answer 1

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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.

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  • $\begingroup$ Do you know if (1) and "$X$ is defined over $\mathbf Q$" implies (2)? $\endgroup$
    – Dan Petersen
    Feb 4, 2013 at 8:01
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    $\begingroup$ I don't think that (2) implies (1) in general. Balazs mentions in mathoverflow.net/questions/92657 an example due to Nick Katz (that I haven't been able to locate) of a variety with polynomial count but whose cohomology is not all of $(p,p)$ type. I guess the result is true if you make some purity assumption (like $X$ smooth proper) to avoid cancellation in the Hodge-Deligne polynomial. $\endgroup$
    – Dan Petersen
    Feb 4, 2013 at 9:14
  • $\begingroup$ Oh - the fact that it's true when $X$ satisfies purity is exactly what is proven in the appendix by Katz linked by Jim Bryan above. $\endgroup$
    – Dan Petersen
    Feb 4, 2013 at 9:29
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    $\begingroup$ The variety I described is defined over $\mathbb Q$, although I said it in a strange way. I meant to blow up at the closed point on $\mathbb P^2_\mathbb Q$, thus a Galois orbit of $\mathbb P^2_{\bar{\ mathbb Q}}$-points, which is why I added $2p$ and not $1p$. This is defined over $\mathbb Q$. $\endgroup$
    – Will Sawin
    Feb 4, 2013 at 17:35

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