5
$\begingroup$

Let $f$ a newform of weight $2$ on $\Gamma_0(Np^r)$, $N$ coprime to $p$, and consider its $p$-adic Galois representation $$ \rho:G_{\mathbb Q}\longrightarrow GL_2(\bar{\mathbb Q}_p) $$ It's a theorem of Carayol that the prime-to-$p$ conductor $N(\rho)$ of $\rho$ equals $N$. Hence, one can recover $N$ from $\{\rho\vert_{I_q}\}_{q\mid N}$.

The question is:

Can $r$ be read in $\rho\vert_{I_p}$?

Thanks for your time!

$\endgroup$
4
  • 1
    $\begingroup$ Could $p^r$ be the conductor of the Weil-Deligne representation attached to $\rho|G_{\mathbb{Q_p}}$ by Fontaine? $\endgroup$
    – Aurel
    Nov 11, 2016 at 0:59
  • $\begingroup$ What Aurel says is true -- a theorem of Takeshi Saito. $\endgroup$
    – wrigley
    Nov 11, 2016 at 17:19
  • $\begingroup$ @wrigley It was a guess; thanks for confirming it. Do you have a precise reference? $\endgroup$
    – Aurel
    Nov 14, 2016 at 18:51
  • $\begingroup$ "modular forms and p-adic Hodge theory" 1997 Inventiones. $\endgroup$
    – wrigley
    Nov 25, 2016 at 12:04

1 Answer 1

1
$\begingroup$

The answer to the question in the title is yes, as explained in the last paragraph below.

However, under a literal interpretation of "can" (implying actual feasibility), I believe the answer to the question in the body of the text is no.

Assume for instance that $f$ and $g$ are two is $p$-ordinary eigencuspforms ($p$-ordinary means that under a fixed embedding of $\bar{\mathbb Q}$ into $\bar{\mathbb Q}_{p}$, the $p$-adic valuation of $a_{p}(f)$ and of $a_{p}(g)$ is zero), that $\pi(f)_p$ (the automorphic representation of $\operatorname{GL}_{2}(\mathbb Q_{p})$ attached to $f$) is unramified principal series and that $\pi(g)_{p}$ (same notation) is unramified Steinberg.

Then the conductor of $f$ at $p$ is trivial ($r=0$) whereas the conductor at $p$ of $g$ is $p$ ($r=1$). However, after restriction to $I_{p}$, both $\rho_f$ and $\rho_g$ are equivalent to \begin{equation} \begin{pmatrix} 1&*\\ 0&\chi^{-1} \end{pmatrix} \end{equation} where $\chi$ is the cyclotomic character. I don't know how to distinguish between them using the class of the extension of $\chi^{-1}$ by $1$ (the $*$, so to speak) and it seems hard to me though I admit I also don't know that it is definitely not possible.

One can construct many such examples of ambiguous $I_{p}$-representation, so I doubt one can reconstruct $p^{r}$ in general. As more generally the representation $\rho_f|G_{\mathbb Q_{p}}$ is the representation $V_{2,a_p}$ in the notation of C.Breuil Sur quelques représentations modulaires et $p$-adiques de $\operatorname{GL}_{2}(\mathbb Q_{p})$ II (Journal de l'IMJ, 2003) it might be a good idea to have a look at this article if you want a definite answer.

As Aurel points out, $p^{r}$ is the conductor of the Weil-Deligne representation attached to $D_{\operatorname{pst}}(\rho_f|G_{\mathbb Q_{p}})$ so you certainly can reconstruct $r$ from $\rho_{f}|G_{\mathbb Q_{p}}$ and what you are missing in your setting are the eigenvalues of the image of $\operatorname{Fr}(p)$ through $\rho_f$. In the case above for instance, both eigenvalues would have the same $p$-adic valuations in the first case and different valuations in the second.

$\endgroup$
3
  • 1
    $\begingroup$ That's really not that hard. Such a class is an element of $H^1(\mathbb Q_p, \chi) $, i.e., by Kummer theory, an element of $\mathbb Q_p^ \times \otimes \mathbb Q_p = \mathbb Q_p \times \mathbb Q_p$ where the first coordinate comes from the valuation and the second comes from the logarithm. The unramified extensions necessarily form a one-dimensional subspace. I claim the image of the logarithm map is the unramified one. $\endgroup$
    – Will Sawin
    Nov 11, 2016 at 3:55
  • $\begingroup$ A really silly way to check this is to note that every other one-dimensional subspace contains the image of an element $q$ in $\mathbb Q_p$ with positive valuation, and thus is the $p$-adic Tate module of the elliptic curve with uniformization $\mathbb Q_p^\times / q$, which has multiplicative reduction at $p$ and hence has ramified Weil-Deligne representation. $\endgroup$
    – Will Sawin
    Nov 11, 2016 at 3:56
  • $\begingroup$ Oh, I see. I'll let my example stand for the moment though, and maybe someone can give a definite answer. $\endgroup$
    – Olivier
    Nov 11, 2016 at 4:55

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.