7
$\begingroup$

For elliptic curves over $\mathbb{Q}$ the Mumford-Tate group is either $\mathrm{GL}_2$ or $\mathrm{Res}_\mathbb{Q}^F (\mathbb{G}_m)$ if it has CM with the imaginary quadratic field $F$. In this case the $\mathbb{Q}$-endomorphism algebra completely determines the Mumford-Tate group. On the automorphic side, having CM can be seen as the associated modular form $f$ satisfying $f=f\otimes \chi$ for some quadratic character $\chi$.

For a simple abelian surface $A/\mathbb{Q}$, again the endomorphism algebra determines the Mumford-Tate group. Here one can find the classification of the Hodge group (special Mumford-Tate group) for low dimensional abelian varieties. In particular there are four cases for abelian surfaces. One expects to be able to associate an automorphic representation $\pi$ of $\mathrm{GSp}_4$ to the abelian surface $A$.

My question is that if the four cases of Mumford-Tate groups (or the endomorphism algebra if you like) can be translated to properties (hopefully some kind of symmetries) of the automorphic representation $\pi$. For instance, I think the CM case (Type IV(2,1) here) should be similar to the case of elliptic curves, namely $\pi=\pi \otimes \chi$ for some character.

My main concern is the two dimensional case but any comments on the higher dimensional cases where the Mumford-Tate group is not determined by the endomorphism algebra is also highly appreciated.

$\endgroup$

1 Answer 1

4
$\begingroup$

Yes, knowing the endomorphism algebra of $A$ (conjecturally) translates to certain properties of an associated automorphic representation $\pi$.

First, you should look at the Galois type, which is labelled (A)-(F) in FKRS's paper on Sato-Tate groups. (These labels also include non-simple $A$, which correspond to products of elliptic curves, and thus products of rational elliptic newforms.)

If the $\mathbb Q$-endomorphism algebra contains a real quadratic field $K$, i.e., $A$ has real multiplication, then $A$ is of GL(2) type as defined by Ribet. By Ribet + Serre's conjectures, this means $A$ corresponds to weight 2 newform of GL(2) with rationality field $K$. If further $A$ has quartic CM, then $f$ will have CM, and have inner twists.

Otherwise, then the (generalized) paramodular conjecture predicts $A$ corresponds to a weight 2 automorphic representation $\pi$ of GSp(4). See Brumer and Kramer's formulation and/or BCGP's potential modularity paper. In the typical case (trivial endomorphisms), the Galois representations are strongly irreducible, but the other Galois types (B)-(F) correspond to representations that are reducible upon restriction to some finite extension $K/\mathbb Q$, with details depending on the type---see Section 9.2 of [BCGP]. Consequently this means that $\pi_K$ is an isobaric sum, which is one way of stating the automorphic property corresponding to the endomorphism algebra of $A$.

There should be similar stories in higher dimensions. Certainly the GL(2)-type situation is understood in higher dimensions.

$\endgroup$
2
  • $\begingroup$ Thanks, this is very helpful. But, even though the description as isobaric sums is very concrete, I was hoping for some kind of symmetry on the automorphic representation like the CM case for modular forms. I guess in the $\mathrm{GL}_2$ case it's not hard to go from the isobaric decomposition of $\pi_K$ to $\pi = \pi \otimes \chi$ since the summands should be Hecke characters. But I'm not sure if something similar to this can be said for $\mathrm{GSp}_4$. $\endgroup$ Commented May 22, 2023 at 9:45
  • 1
    $\begingroup$ @AlirezaShavali From this perspective, the CM case is special because $K$ is quadratic, so you can say $\pi$ is induced from a character of $K$, which means it is Galois invariant, i.e., satisfies that twisting property. In general you only know that $\pi$ is an isobaric component of an automorphic induction from $K$. However, if you're lucky (e.g., if it is the full automorphic induction and $K$ is solvable Galois), you can also conclude $\pi$ is Galois invariant, which is the type of symmetry you want. My guess is if you work out the details for these cases, you're lucky sometimes. $\endgroup$
    – Kimball
    Commented May 22, 2023 at 15:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .