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!

  • 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


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.

  • 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

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