Timeline for Irreducibility of the $n$th symetric power of the reduction of the Galois representation of a non-CM newform
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 10, 2022 at 16:32 | comment | added | Fouad Fahmi | Many thanks for the clarifications. | |
| Mar 10, 2022 at 16:12 | comment | added | Kevin Ventullo | As Kimball indicated, question (ii) as stated doesn’t make sense because ${\rm Sym}^n$ is not 2-dimensional for $n>1$. I don’t think it is true that ${\rm Sym}^n\overline{\rho}_{f,\ell}(G_{\mathbb{Q}}))$ is absolutely irreducible in general, since the dimension grows with $n$ but there are only finitely many isomorphism classes of characteristic $\ell$ irreducible reps of the finite group ${\rm Image}(\overline{\rho}_{f,\ell})$; in particular there is one of maximal dimension. | |
| Mar 10, 2022 at 15:43 | comment | added | Kevin Ventullo | If a 2-dim’l representation is reducible, it is an extension of a 1-dim’l representation by another 1-dim’l representation. That is, it sits in the middle of an exact seq. where the sub and quotient are both 1-dim’l. Since 1-dim’l reps are abelian, this corresponds group theoretically to an exact sequence where the sub and quotient are both abelian. This is basically the definition of solvable. | |
| Mar 10, 2022 at 9:49 | comment | added | Fouad Fahmi | Thanks for your answer, could you please provide a reference for the fact about the solvability of the image of a reducible Galois representation? Do you have any thoughts about (ii)? Or how we can deduce directly from $(*)$ that ${\rm Sym}^n\bar{\rho}_{f,\ell|\mathbb{Q}(\zeta_\ell})$ is absolutely irreducible? | |
| Mar 10, 2022 at 9:14 | history | answered | Kevin Ventullo | CC BY-SA 4.0 |