Irreducibility of the $n$th symetric power of the reduction of the Galois representation of a non-CM newform

Question

In "On $\ell$-adic representations attached to modular forms II", Ribet proved that the $\ell$-adic representation $\rho_{f,\ell}$ attached to a non-CM newform form $f$ satisfies $${\rm SL}_2(\mathbb{F}_\ell)\subset \overline{\rho}_{f,\ell}(G_\mathbb{Q}).\qquad\qquad(*)$$

I am wondering:

(i) Why does $(*)$ imply that $\overline{\rho}_{f,\ell|G_{\mathbb{Q}(\zeta_\ell)}}$ is absolutely irreducible?

(ii) Let $n$ be a positive integer. Does $(*)$ imply that ${\rm Sym}^n\overline{\rho}_{f,\ell|G_{\mathbb{Q}(\zeta_\ell)}}$ is irreducible? If the answer is negative, under what condition does this hold?

Answer 1

Here's an answer that's more work than Kevin Ventullo's, but works for $\ell=2$ and $\ell=3$, and gives you a method that will work in various more general cases.

There is a surjection $\pi_1\colon G_\mathbb Q \to \overline{\rho}_{f,l} (G_{\mathbb Q})$ and a surjection $\pi_2 \colon G_\mathbb Q \to \operatorname{Gal}(\mathbb Q(\zeta_\ell) / \mathbb Q)$. The group $\overline{\rho}_{f,l} (G_{\mathbb Q(\zeta_\ell)})$ is the image under $\pi_1$ of the kernel of $\pi_2$. In other words, it is the intersection of the image of the product map $(\pi_1,\pi_2) \colon G_\mathbb Q \to \overline{\rho}_{f,l} (G_{\mathbb Q}) \times \operatorname{Gal}(\mathbb Q(\zeta_\ell) / \mathbb Q)$ with $G_{\mathbb Q}) \times 1$.

By Goursat's lemma, the image of the product map is $ \{ (a,b) \mid f_1(a) =f_2 (b) \}$ where $f_1 \colon \overline{\rho}_{f,l} (G_{\mathbb Q}) \to Q$ and $f_2 \colon \operatorname{Gal}(\mathbb Q(\zeta_\ell) / \mathbb Q) \to Q$ are surjections for some finite group $Q$. Expressed in this language, $\overline{\rho}_{f,l} (G_{\mathbb Q(\zeta_\ell)})$ is the kernel of $f_1$.

Now because $f_2$ is surjective, $Q$ is a quotient of a cyclic group of order $\ell-1$, and thus is a cyclic group of order dividing $\ell-1$.

So $\overline{\rho}_{f,l} (G_{\mathbb Q(\zeta_\ell)})$ is the kernel of a map from $\overline{\rho}_{f,l} (G_{\mathbb Q(\zeta_\ell)})$ to a cyclic group of order dividing $\ell-1$. Thus it contains the kernel of a map from $SL_2(\mathbb F_\ell)$ to the kernel of a cyclic group of order dividing $\ell-1$.

Since the abelianization of $SL_2(\mathbb F_\ell)$ is trivial (for $\ell>3$) or has order relatively prime to $\ell-1$ (for $\ell=2,3$, since $2$ is relatively prime to $1$ and $3$ is relatively prime to $2$), $\overline{\rho}_{f,l} (G_{\mathbb Q(\zeta_\ell)})$ contains $SL_2(\mathbb F_\ell)$.

Since $SL_2(\mathbb F_\ell$ acts absolutely irreducibly on the standard representation for all $\ell$, it follows that $\overline{\rho}_{f,l} (G_{\mathbb Q(\zeta_\ell)})$ acts absolutely irreducibly on the standard representation, as desired.

It immediately follows that the image of $\operatorname{Sym}^n \overline{\rho}_{f,\ell} \mid_{ G_{\mathbb Q(\zeta_\ell)}}$ containsthe image of $SL_2(\mathbb F_\ell)$ under the $n$th symmetric power representation. Since the $n$th symmetric power representation of $SL_2(\mathbb F_\ell)$ is absolutely irreducible for $n \leq \ell-1$, the representation $\operatorname{Sym}^n \overline{\rho}_{f,\ell} \mid_{ G_{\mathbb Q(\zeta_\ell)}}$ is absolutely irreducible for $n\leq \ell-1$.

Answer 2

For (i), if $\overline{\rho}_{f,\ell|G_{\mathbb{Q}(\zeta_\ell)}}$ was absolutely reducible, then in particular the image of $G_{\mathbb{Q}(\zeta_\ell)}$ under $\overline{\rho}$ would be solvable. However, since $G_{\mathbb{Q}}/G_{\mathbb{Q}(\zeta_\ell)}$ is abelian, that would imply the image of all of $G_{\mathbb{Q}}$ under $\overline{\rho}$ would be solvable. This contradicts $(*)$ for $\ell > 3$.

I am not sure what happens for $\ell = 2,3$.