Let $f$ be a newform of weight $k$ and level $N$ with integer coefficients. Deligne-Serre theorem theorem says there exist a nice associated representation $\rho_{f}^{(\ell)}:\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})\to \text{GL}_2(\mathbb{Q}_{\ell}),$ for any prime $\ell$ not dividing $N$. In particular, this induces a representation $\rho_{f,\ell}:\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})\to \text{GL}_2(\mathbb{F}_{\ell}).$ Due to Serre's open image theorem, both adelic and mod $\ell$ representations are surjective for all but finitely many $\ell$, when $f$ has weight $2$ and comes from an elliptic curve. For the general case (especially when the weight is more than $2$ ?) is the mod $\ell$ representation coincides with the matrices having determinant a $(k-1)^{th}$ power in $\mathbb{F}_{\ell}^*$, for all but finitely many $\ell$ ? If not, what's stopping it to be ? and what we can say about the images ?


1 Answer 1


In the setting that $f$ has integer Hecke eigenvalues, then the result is true, so long as $f$ has weight $k\ge 2$ and does not have complex multiplication. See Theorem 3.1 of this paper by Ribet: in this setting, $R=\mathbb Z$.

In general, the Hecke eigenvalues of $f$ will all lie in a finite extension $E$ of $\mathbb Q$, and $\rho_f$ will be valued in $\mathrm{GL}_2(E_{\lambda})$ for a prime $\lambda$ above $\ell$. Let $\mathbb F_\lambda$ be the residue field of $E_\lambda$. If $k\ge 2$ and $f$ does not have CM, then, for all but finitely many primes, the image of the mod $\lambda$ representation $\overline{\rho}_f$ in $\mathrm{GL}_2(\mathbb F_\lambda)$ will contain a subgroup conjugate to $\mathrm{SL}_2(\mathbb F_\ell)$ (but not necessarily $\mathrm{SL}_2(\mathbb F_\lambda)$).

To say anything further, we have to be a lot more careful: I'd suggest looking at the answers to this question.

  • $\begingroup$ Yes, I am happy when eigenvalues are integer.s In the Ribet's paper, are you talking about (2) at page 186 ? but that's only said for level 1, no ? $\endgroup$
    – dragoboy
    May 20, 2020 at 12:42
  • 2
    $\begingroup$ @dragoboy I'm referring to Theorem 3.1 on p191. You're right that the result on p186 only applies to level $1$ forms. $\endgroup$ May 20, 2020 at 12:44
  • $\begingroup$ Great! thanks... $\endgroup$
    – dragoboy
    May 20, 2020 at 13:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.