# Are there smooth and proper schemes over $\mathbb Z$ whose cohomology is not of Tate type

Is there an example of smooth and proper scheme $X \to \mathrm{Spec}(\mathbb Z)$, and an integer $i$ such that $H^i(X, \mathbb Q)$ is not a Hodge structure of Tate type?
Alternatively: such that $H^i(X_{\bar{\mathbb{Q}}}, \mathbb{Q}_\ell)$ is not a Galois representation of Tate type?

• A result of Fontaine says that $H^q(X,\Omega^p) = 0$ if $p \ne q$, and $p+q \le 3$. So we need $\dim(X) > 3$.
• If we allow stacks, then examples come from the theory of modular forms: $H^{11}(\bar{M}_{1,11})$ is associated with the Ramanujan $\Delta$ function. So this question is explicitly about schemes.

An explicit example would be wonderful. An inexplicit proof that such an $X$ exists is fine as well.

Related questions:

• Doesn't the result of Fontaine need $p \neq q$? – user19475 Sep 25 '17 at 12:22
• @TimoKeller: Oo, yes of course. – user114562 Sep 25 '17 at 12:28
• I think I calculated once that under GRH, Fontaine's argument goes through for $p+q \leq 4$. So one needs dimension at least 5. However, there is no real hope I see of doing this in any dimension $<11$. One approach would be to try to resolve the singularities of the coarse moduli space of $\overline{\mathcal M}_{1,11}$. Taking the Hilbert scheme might be a good first step, as this resolves some singularities in coarse moduli spaces of quotient stacks. – Will Sawin Sep 25 '17 at 14:51
• This is a notorious open problem. As Will says, there are natural examples of smooth proper DM stacks with interesting cohomology (e.g. $\overline{\mathcal{M}_{g,n}}$). But there is no known scheme example. – Daniel Litt Sep 25 '17 at 15:17
• @DanielLitt: Yes, I am aware of those natural examples (see the question, where I mention $\overline{M}_{1,11}$). One line of attack that I was thinking about for a proof of existence was to use general results about categories of motives built from smooth projective varieties versus motives built from smooth proper DM stacks. I think that over fields there are theorem that say they give the same category. I don't know if something like this also works over $\mathbb Z$. – user114562 Sep 25 '17 at 17:01