0
$\begingroup$

Suppose $k$ is a number field and $\sigma:k \hookrightarrow \mathbb{C}$ is an embedding. If $\sim$ is an adequate equivalence relation, we can construct the category of pure motives with rational coefficients, $\textbf{M}_{\sim}(k,\mathbb{Q})$. Let the Hodge realisation functor be $R$, \begin{equation} R:\textbf{M}_{\sim}(k,\mathbb{Q}) \rightarrow \text{MHS}_{\mathbb{Q}} \end{equation}

Question 1: Suppose $M$ is a pure motive of $\textbf{M}_{\sim}(k,\mathbb{Q})$ such that its Hodge realisation is the pure Tate object $\mathbb{Q}(0)$ of $\text{MHS}_{\mathbb{Q}}$, what could $M$ be?

For example, suppose $\chi: (\mathbb{Z}/N\mathbb{Z})^\times \rightarrow \mathbb{Q}$ is a real Dirichlet character, then the Artin motive $h^0(\text{Spec}\,\mathbb{Q}(\zeta_N))$ has a direct summand $\chi$ whose etale realisation is the Galois representation associated with the Dirichlet character $\chi$. The Hodge realisation of $\chi$ is $\mathbb{Q}(0)$, and when $\chi$ is the trivial character, the associated motive is just the Tate motive $\mathbb{Q}(0)$. But are there any other possibilities? If none, how to prove it?

Question 2: Suppose $M_1$ and $M_2$ are two pure motives of $\textbf{M}_{\sim}(k,\mathbb{Q})$ that have isomorphic Hodge realisations, is it true that $M_1$ is isomorphic to $M_2 \otimes \epsilon$, where $\epsilon$ is a pure motive whose Hodge realisation is $\mathbb{Q}(0)$?

Remark: If $k$ is algebraically closed and $\sim$ is numerical equivalence relation, then these two questions are immediate consequence of Hodge conjecture. I am also wondering whether they could be answered without citing any conjecture!

$\endgroup$
  • 2
    $\begingroup$ See also here and here. $\endgroup$ – Dietrich Burde Mar 25 '18 at 10:53
  • 1
    $\begingroup$ If $\sim$ is numerical equivalence, then we do not know that there is a Hodge realisation functor. You need the “hom = num” standard conjecture for that. $\endgroup$ – jmc Mar 26 '18 at 16:56
  • $\begingroup$ One more remark: we don't presently know if the motivic Galois group is a connected group if $k$ is algebraically closed. Not in the case of Nori motives, nor for André motives, nor for motives modulo $\sim$. Even over $\CC$ we cannot currently conclude that a $1$-dimensional motive is Tate. $\endgroup$ – jmc Mar 26 '18 at 17:14
  • $\begingroup$ @jmc Thank you. The math community need to work harder. $\endgroup$ – Wenzhe Mar 26 '18 at 20:00

Your Answer

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

Browse other questions tagged or ask your own question.