# Adelic open image for modular forms?

There's a famous theorem of Serre that if $E$ is a non-CM elliptic curve over $\mathbf{Q}$, and $\rho_{E, \ell} : Gal(\overline{\mathbf{Q}}/{\mathbf{Q}}) \to GL_2(\mathbf{Z}_\ell)$ is its $\ell$-adic Galois representation, then the product $\rho = \prod_{\ell} \rho_\ell: Gal(\overline{\mathbf{Q}}/{\mathbf{Q}}) \to GL_2(\widehat{\mathbf{Z}})$ has open image. In particular, this implies that $\rho_{E, \ell}$ is surjective for almost all $\ell$; but is much stronger than this, as it shows that the $\rho_{E, \ell}$ for different $\ell$ are "independent" in some sense.

If one works instead with general non-CM modular forms $f$ of weight $k \ge 2$, then I know of theorems (due to Ribet and Momose) describing the images of the $\rho_{f, \ell}$, showing that they are "as large as possible" for almost all $\ell$ (cf. this earlier question of mine). (The notion of "as large as possible" is much more delicate in this generality, because the coefficient field can be nontrivial, and there can be "inner twists".)

Are there analogues of Serre's adelic open image results for higher-weight modular forms?

EDIT. I'll just put up a guess of mine, just to show that there is a reasonable conjectural formulation which is compatible with the known results for individual $\ell$. Momose has shown that there is a subfield $F$ of the coefficient field $E = \mathbf{Q}(f)$, a quaternion algebra $B$ over $F$, and an open subgroup $H$ of $Gal(\overline{\mathbf{Q}}/{\mathbf{Q}})$, all independent of $\ell$, such that for any $\ell$ the representation $\rho_{f, \ell}: Gal(\overline{\mathbf{Q}}/{\mathbf{Q}}) \to GL_2(E \otimes \mathbf{Q}_\ell)$ sends $H$ to an open subgroup of the group $\{ x \in B(F \otimes \mathbf{Q}_{\ell}) : \operatorname{norm}(x) \in \mathbf{Z}_{\ell}^{\times(k-1)}\}$. For all $\ell$ coprime to the discriminant of $B$, we have $B(F \otimes \mathbf{Q}_{\ell}) = GL_2(F \otimes \mathbf{Q}_{\ell})$, and Ribet has shown that for all but finitely many such $\ell$ the image of $H$ is all of $\{ x \in GL_2(O_F \otimes \mathbf{Z}_{\ell}) : \operatorname{det}(x) \in \mathbf{Z}_{\ell}^{\times(k-1)}\}$.

Conjecture: The image of $H$ in $GL_2(L \otimes \mathbf{A})$ contains an open subgroup of $\{ x \in B(O_F \otimes \widehat{\mathbf{Z}}) : \operatorname{norm}(x) \in \widehat{\mathbf{Z}}^{\times(k-1)}\}$.

This is visibly consistent with (and implies) Momose and Ribet's results.

• There is an immediate obstruction to the naïve generalization: the determinant of $\rho_{\ell}$ has values in a smaller ring than the ring of coefficients of $\rho_{\ell}$ for an infinite number of $\ell$ in general so $\rho_{\ell}$ is not surjective for an infinite number of $\ell$ and so the products of the $\rho_{\ell}$ cannot have open image. But you are aware of this in your question, so you seem to be thinking about a more sophisticated generalization. Sep 23, 2014 at 7:42
• @David: as far as I understand, you are looking for a generalization both in the weight and in letting the coefficient field be bigger than $\mathbb{Q}$. Why not just sticking to $k\geq 2$ but $E_f=\mathbb{Q}$? Sep 23, 2014 at 11:35
• @Olivier: Indeed I am aware of these issues -- that is exactly what I meant when I said "the notion of as large as possible is more delicate in this generality". Sep 23, 2014 at 14:31
• David: what you want is certainly conjectured, at least. The standard reference is Serre's "Proprietes conjecturales des groups de Galois Motiviques..." in volume 1 of the book Motives, PSPM 55. Specifically, section 11 is about your problem in the more general setting of systems of representations attached to motives, where conjectural statement 11.4? page 390 is a very general conjecture which should specialize to yours in the case of a modular form (I have not checked). Now, as for the proof, I don't know. Serre might say something about the status on page 392 which I can't access now.
– Joël
Sep 23, 2014 at 15:41
• Serre's open image theorem follows in a purely group-theoretic way from the fact the the $\ell$-adic representations have open image for all $\ell$ and are surjective for all but finitely many, and the knowledge that the determinant is the cyclotomic character. It seems to me that Aaron Greicius's result about maximal open subgroups of direct products of profinite groups (see Proposition 2.5 of this paper) should help determine the group theoretic obstruction (if any) to an open image result of the type you seek. Sep 23, 2014 at 17:55

## 1 Answer

As I mentioned in my comment, an adelic open image theorem should follow in a purely group-theoretic way from the knowledge that the $\ell$-adic representations are surjective (for an appropriately specified codomain) for all sufficiently large $\ell$ and have open image for all $\ell$.

The group theory necessary is Aaron Greicius's classification of maximal closed subgroups of a direct product of profinite groups (see Proposition 2.5 of this paper - the link is correct this time). Greicius's result is the following:

Assume that $G_{\alpha}$, for $\alpha \in \Lambda$ are profinite groups with the property that there is no non-abelian finite simple group $M$ that is a quotient of $G_{\alpha}$ and $G_{\alpha'}$ for $\alpha \ne \alpha'$. Then, every maximal closed subgroup of $G = \prod_{\alpha} G_{\alpha}$ comes from a maximal closed subgroup $H_{\alpha} \subseteq G_{\alpha}$, or from a maximal subgroup of $G/G'$.

In the case at hand, $\Lambda$ can be taken to be the set of primes $\ell \ne 2, 5$, and $G_{\ell} = {\rm im}~\rho_{f,\ell}$. The only possible non-abelian finite simple quotients of $GL_{2}(O_{F} \otimes \mathbf{Z}_{\ell})$ are the $PSL_{2}(\mathbf{F}_{\ell^{r}})$, and it's not possible to have $PSL_{2}(\mathbf{F}_{\ell_{1}^{r_{1}}})$ isomorphic to $PSL_{2}(\mathbf{F}_{\ell_{2}^{r_{2}}})$ for odd primes $\ell_{1} \ne \ell_{2}$. The proof of Greicius's proposition implies that the adelic image contains $G'$. Moreover, $SL_{2}(O_{F} \otimes \mathbf{Z}_{\ell})$ has no abelian quotients for $\ell \geq 5$, and the commutator subgroup of $SL_{2}(O_{F} \otimes \mathbf{Z}_{\ell})$ has finite index for $\ell = 3$.

Greicius's result doesn't quite apply if you include $\ell = 2$ and $\ell = 5$ because $PSL_{2}(\mathbf{F}_{4}) \cong PSL_{2}(\mathbf{F}_{5})$, but it is easy to see that if $K_{2}$ is the fixed field of $\rho_{f,2}$ and $K_{5}$ is the fixed field of $\rho_{f,5}$, then $K_{2} \cap K_{5}/\mathbb{Q}$ is a finite extension.