## Dembélé on Abelian Surfaces with good reduction everywhere

New paper by Dembélé (friend of the blog) on abelian surfaces with good reduction everywhere (or rather, the lack of them for many real quadratic fields of small discriminant). I have nothing profound to say about the question of which fields admit non-trivial abelian varieties with everywhere good reduction, but looking at the paper somehow dislodged the following question from my brain, which I often like to ask and don’t mind repeating here:

Question: Fix a prime p. Does there exist a non-solvable extension of $\mathbf{Q}$ unramified everywhere except for p?

There is a (very) related question of Gross, who (and I can’t track down the precise reference) was generous and allowed ramification at infinity. That makes the question easy to answer for big enough p just by taking the mod-p Galois representations associated to either the weight 12 or weight 16 cusp form of level 1. But what if you impose the condition that the extension has to be unramified at the infinite prime as well (so totally real) then you are completely out of luck as far as Galois representations from algebraic automorphic forms go, because for those, complex conjugation will always be non-trivial. (Things don’t get any easier if you even allow regular algebraic automorphic representations, as Caraiani and Le Hung showed). Except, that is, for the case when p = 2. There is a different paper by Lassina on this topic, which solved Gross’ question for p=2 by finding a level one Hilbert modular form over the totally real field $\mathbf{Q}(\zeta + \zeta^{-1})$ for a 32nd root of unity with a non-solvable mod-2 representation. But (as he shows) this extension is ramified at infinity — in fact, the Odlyzko discriminant bounds show that to get a totally real extension (assuming GRH) one would have to take the totally real field to be at least as large as $\mathbf{Q}(\zeta + \zeta^{-1})$ for a 128th root of unity. Is it even possible to compute Hilbert Modular Forms for a field this big?

Leaving aside the computational question, there is also a theoretical one as well, even for classical modular forms. Given a Hilbert modular form, or even a classical modular form, is there any easy way to compute the image of complex conjugation modulo 2? One reason this is subtle is that the answer depends on the lattice so it really only makes sense for a residually absolutely irreducible representation. For example:

Question: For every n, does there exist a (modular) surjective Galois representation

$\mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q}) \rightarrow \mathrm{SL}_2(\mathbf{F})$

for a finite field of order divisible by $2^n$ which is also totally real? Compare this to Corollary 1.3 of this paper of Wiese.

I don’t even have a guess as to the answer for the first question, but the second one certainly should have a positive answer, at least assuming the inverse Galois problem. As usual, an Aperol Spritz is on offer to both the second question and to the first in the special case of p=2.

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### 4 Responses to Dembélé on Abelian Surfaces with good reduction everywhere

1. Emmanuel Lecouturier says:

The paper of Merel “Formes modulaires modulo 2 et composantes réelles de jacobiennes modulaires” gives conditions for the image of the complex conjugation to be trivial (cf. in particular Prop. 4.1 and Cor. 4.2).

• Thanks, that’s nice! I didn’t know this paper, although I was aware of earlier results of Merel on a related problem. It’s not obviously such a useful characterization for answering the second question, however (or for computations with Hilbert modular forms).

2. AP says:

It is probably unrealistic to compute Hilbert modular forms over the totally real subfield of the $128$th cyclotomic field: the order of magnitude of the dimension of the spaces of forms over $F$ is $\Delta_F^{3/2}$, which in this case is more than $10^{86}$.

• AP says:

Even if you take into account all the terms of the mass formula, you get more than 10^35. Even for the 64th real cyclotomic field the order of magnitude is 10^10.