Let $E$ be a motive over $\mathbb Q$. (I should precise, that by a motive I mean here a pure motive over $\mathbb Q$, with coefficients in $\mathbb Q$, that I see here as a conjectural object which exists in a world where standard conjectures and anything else you can expect is true, in the spirit of Grothendieck, and like in Serre's paper in Motives, PSPM 55, volume 1 - a paper which unfortunately does not seem to be online except partially on Amazon).

Recall that the motivic galois group of $E$, denoted $G_{M(E)}$, is defined as the tensor-automorphic group of the functor "Betti realization" from the category $M(E)$, defined as the smallest Tannakian sub-category of the category of motives containing $E$, to the category of $\mathbb Q$-vector spaces.

An equivalent (under the conjectures of Tate and Hodge) and more concrete (at least for me) definition is as follows (see the same paper of Serre): let $H_b(E)$ be the Betti realization of $E$ (a finite-dimensional $\mathbb Q$-vector space), and let $\rho_\ell$ be the $\ell$-adic realization of $E$, which is a continuous semi-simple representation $\rho_\ell$ of $\Gamma_{\mathbb Q}$ (the absolute Galois group of $\mathbb Q$) over $H_b(E) \otimes \mathbb Q_\ell$. Then $G_{M(E)}$ is the algebraic subgroup $Gl(H_b(E))$ such that $G_{M(E)} \otimes \mathbb Q_\ell$ is the Zariski-closure of the image of $\rho_\ell$, for every prime number $\ell$.

In any case, $G_{M(E)}$ is a reductive group. My question is

What reductive groups over $\mathbb Q$ arise as $G_{M(E)}$ for some pure motive $E$?

This question is asked by Serre in the section 8 of his paper. At that time there was not much known on it apparently (it was 20 years ago): Serre notices that such a group $G$ must have an element $\gamma \in G(\mathbb R)$ such that $\gamma^2=1$ and whose centralizer is a maximal compact subgroup of $G(\mathbb R)$. This necessary conditions implies that $Sl_2$ is not a motivic Galois group. On the positive side, I believe that symplectic group $GSp_{2n}$ are known to be motivic Galois group (attached to $E=$ an abelian variety over $\mathbb Q$ of dimension $n$, sufficiently generic), and I know a few other examples like those attached to elliptic curves with complex multiplication.

There has been a lot of work and progresses on motives since then, and I am asking to know what progresses have been made in the direction of this question. I would be very surprised if it had been solved completely (even assuming the standard conjectures + Hodge conjecture + Tate conjecture), but I would be interested in any partial results:

For example, what if we just try to classify the connected reductive group which arise as neutral component of a $G_{M(E)}$? those groups up to isogeny ? those groups after extension to the algebraic closure of $\mathbb Q$? Conjectural answers (conjectural in the sense that they are not proved even assuming the conjectures assumed above) are also more than welcome.

Even a list of example of groups that are provably or possibly motivic Galois group which goes beyond the small list I have given would be very useful. And of course, any references more recent than Serre's could be very useful.

  • 6
    $\begingroup$ It's rather explicitly known which reductive groups arise as the motivic Galois groups of abelian motives (i.e., those in the Tannakian category generated by abelian varieties). See Deligne's Corvallis talk and Milne's second Seattle talk (same conference as Serre's article). Beyond that, not much more is known (see Rabelais's answer). It's hoped that all the groups defining Shimura varieties with rational weight arise as motivic Galois groups, but this is a (very interesting) open question. $\endgroup$ – anon Apr 15 '13 at 18:16

The first question (applied to $\mathrm{GL}(2)$-abelian varieties over $\mathbf{Q}$) seems to include the following problem: what totally real fields $F$ occur as the field of coefficients of a classical weight $2$ modular form? This seems a totally impossible question to answer. For example, it includes the question of which Hilbert modular surfaces $X_F$ have rational points; since $X_F$ is of general type for $F$ of suitable large discriminant, the answer seems hard to predict in advance (especially because fields $F$ of arbitrarily large degree do actually occur).

For the second, surely the work of Zhiwei Yun (http://arxiv.org/pdf/1112.2434v1.pdf) is relevant here.


I realise that I'm a bit late to this party, but maybe it is still interesting to you:

Under the Hodge conjecture the motivic Galois group coincides with the Mumford–Tate group. These have been classified by S. Patrikis: http://arxiv.org/abs/1302.1803.

  • $\begingroup$ What is the dependence on the $\mathbb{Q}$ structure of the variety? I think Hodge conjecture concerns about existence of cycles defined over $\mathbb{C}$. What is wrong? $\endgroup$ – Mostafa Feb 25 '15 at 17:17
  • 4
    $\begingroup$ @Mostafa — The group of algebraic cycles modulo algebraic equivalence does not depend on the ground field. The Hodge conjecture claims that the functor $H_{\text{B}} \colon \mathrm{Mot}_{K} \to \mathbb{Q}\text{-HS}$ is fully faithful. As a consequence, it induces for every motive an equivalence between the Tannakian categories $\langle M \rangle$ and $\langle H_{\text{B}}(M) \rangle$. Therefore the motivic Galois group coincides with the Mumford–Tate group. $\endgroup$ – jmc Feb 25 '15 at 17:53
  • $\begingroup$ @Mostafa — My first statement was there as a reply to your first comment. Modulo algebraic equivalence there is no difference between cycles defined over $\mathbb{Q}$ or $\mathbb{C}$. $\endgroup$ – jmc Feb 25 '15 at 20:18
  • $\begingroup$ I want the idea of proof or a reference. $\endgroup$ – Mostafa Feb 25 '15 at 20:26
  • 1
    $\begingroup$ @Mostafa — Ok, sorry, I misunderstood you. For example it is claimed in Kleiman's ICM paper (1970). However, I don't know a reference for a proof from the top of my head. $\endgroup$ – jmc Feb 26 '15 at 7:31

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.