4
$\begingroup$

Suppose $M$ is an object of the (conjectured) abelian category of mixed motives over $\mathbb{Q}$, $\textbf{MM}_{\mathbb{Q}}$, Scholl defines that $M$ is defined over $\mathbb{Z}$ if the following condition is satisfied.

Let $M_{\ell}$ be the $\ell$-adic realisation of $M$, which has a weight filtration $W_*$. For every $ p \neq \ell$, when we consider $M_{\ell}$ as a representation of the inertia group $I_p$, the weight filtration $W_*$ splits.

Suppose $X$ is a hypersurface of the projective space $\mathbb{P}^n_{\mathbb{Z}}$, and let $h^i(X)$ be the mixed motive associated to its $i$-th cohomology, is $h^i(X)$ (expected to be) a mixed motive defined over $\mathbb{Z}$?

If we further assume the singularity of $Y$ consists of finitely many $\mathbb{Z}$-valued points, then is $h^i(X)$ (expected to be) a mixed motive defined over $\mathbb{Z}$?

Here let's assume the existence of the category of mixed motives and its expected properties!

Improve this question
$\endgroup$
1
$\begingroup$

In your first question, what do you mean by $h^i(X)$? Assuming you mean $h^i(X_\mathbb{Q})$, then its $\ell$-adic realization is the étale cohomology of $X_{\overline{\mathbb{Q}}}$, which is described (after restriction to a decomposition group at $p$) by a spectral sequence involving the étale cohomology of the special fiber of $X$ at $p$. It is certainly not true in general that the weight filtration splits even after retricting to inertia (see e.g. the weight-monodromy conjecture). On the other hand if you assume $X_{\mathbb{Q}}$ to be smooth projective over $\mathbb{Q}$ then $h^i(X)$ is defined over $\mathbb{Z}$ because the motive is pure of weight $i$, so the weight filtration Scholl refers to is trivial.

Improve this answer
$\endgroup$
7
  • $\begingroup$ Yes I mean $h^i(X_{\mathbb{Q}})$. For a smooth variety $X$ over $\mathbb{Q}$, $h^i(X_{\mathbb{Q}})$ is pure of weight $i$, so from the definition it is defined over $\mathbb{Z}$. If $X$ has very mild singularity, e.g. only one rational singular point, let's say it has weight filtration $W_{i-1} \subset W_i$, do you think in general it does not split into $W_{i-1} \oplus W_i/W_{i-1}$ considered as representation of inertia group? Do you know any references that might be helpful? $\endgroup$ – Wenzhe Apr 4 '18 at 14:13
  • $\begingroup$ If $E$ is an elliptic curve over $\mathbb{Q}_p$ with split multiplicative reduction then its Tate module sits in an exact sequence $0 \to \mathbb{Q}_\ell(1) \to V_\ell(E) \to \mathbb{Q}_\ell \to 0$ so there is a subspace of weight -2 and a quotient of weight 0. The action of inertia is trivial on $\mathbb{Q}_\ell(1)$ and $\mathbb{Q}_\ell$, but is unipotent on $V_\ell(E)$, hence the action of inertia does not split. $\endgroup$ – François Brunault Apr 4 '18 at 15:53
  • $\begingroup$ The following lecture notes by V. Dochitser could be helpful to you: arxiv.org/pdf/1410.1039.pdf $\endgroup$ – François Brunault Apr 4 '18 at 15:55
  • $\begingroup$ Feel sorry for bothering your again with a naive question. In section 4.5 of Nekovar's note Beilinson's conjecture (also in Scholl's note Remarks on special values of $L$-functions), it says that the category of pure motives form a full subcategory of category of mixed motives defined over $\mathbb{Z}$, does this mean every pure motive is defined over $\mathbb{Z}$? But the elliptic curve counter-example is against this statement. (I feel there is something I have not understood.) $\endgroup$ – Wenzhe Apr 4 '18 at 18:22
  • $\begingroup$ @Wenzhe You're welcome. I have to say I'm a bit confused by the definition of Scholl. In Groupes fondamentaux motiviques de Tate mixte, Deligne and Goncharov define mixed Tate motives over $\mathbb{Z}$ as those mixed Tate motives over $\mathbb{Q}$ which are unramified, in the sense that the associated $\ell$-adic representation is unramified at all primes $\neq \ell$ (see Proposition 1.7). It's not clear to me what is the relation between these two definitions. $\endgroup$ – François Brunault Apr 4 '18 at 18:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.