Adelic open image for modular forms?

Question

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.

Answer 1

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.