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What is the relation between level and conductor of a supercuspidal representation of $\operatorname{GL}_2(\mathbb{Q}_p)$ for some prime $p$?

Proposition 3.4 in Loeffler and Weinstein - On the computation of local components of a newform refers to Breuil and Mézard - Multiplicités modulaires et représentations de $\operatorname{GL}_2(\mathbb Z_p)$ et de $\operatorname{Gal}(\overline{\mathbb Q}_p/\mathbb Q_p)$ en $\ell = p$. Appendice par Guy Henniart. Sur l'unicité des types pour $\operatorname{GL}_2$, which I am unable to understand because of the language.

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    $\begingroup$ I don't know if it's just me, but the PDFs to which you linked won't load. I changed the arXiv PDF link to an abstract link, as is the usual convention, but I had to guess at the Breuil paper, which I think is Breuil and Mézard - Multiplicités modulaires et représentations de $\operatorname{GL}_2(\mathbb Z_p)$ et de $\operatorname{Gal}(\overline{\mathbb Q_p}/\mathbb Q_p)$ en $\ell = p$ (MSN). $\endgroup$
    – LSpice
    Commented May 17, 2020 at 21:12
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    $\begingroup$ I explained how to get this from standard references in Section 2.2 of my basis problem paper. You get that the level is one less than one half of the conductor. $\endgroup$
    – Kimball
    Commented May 18, 2020 at 1:31
  • $\begingroup$ @Kimball, it may be necessary to be cautious here: I think that there are notions of normalised and of un-normalised level. According to Bushnell, Henniart, and Kutzko - Local Rankin–Selberg convolutions for $\operatorname{GL}_n$, the level is an integer $m$, and $\frac m e = \frac1 2(f - 1)$, where $e = 1$ if $\pi$ is unramified and $e = 2$ if $\pi$ is unramified. Probably you are using a normalised notion, where the level is the rational number $m/e$? $\endgroup$
    – LSpice
    Commented May 18, 2020 at 15:18
  • $\begingroup$ @LSpice Maybe I should have clarified, but the question was about supercuspidal $\pi$, and I am only stating what you get when $\pi$ is discrete series, so $e=2$ in your notation, using the normalization of level as in Bushnell-Henniart's book. I haven't looked at Bushnell-Henniart-Kutzko, at least not in detail, but I thought the references I use (which are just for GL(2)) might be a candidate for "a more elementary reference" that you mentioned in your answer. $\endgroup$
    – Kimball
    Commented May 18, 2020 at 16:30
  • $\begingroup$ @Kimball, definitely, and I think you should post it as an answer. However, I'm pretty sure that, even for supercuspidals, $e$ can be $1$ or $2$ (or else I'm totally misunderstanding their notation, which is entirely possible). $\endgroup$
    – LSpice
    Commented May 18, 2020 at 17:46

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There's probably a more elementary reference, but, according to Bushnell, Henniart, and Kutzko - Local Rankin–Selberg convolutions for $\operatorname{GL}_n$, (6.1.2), if $m$ is the level of $\pi$, then the conductor of $\pi$ depends on a choice of additive character $\psi$, which will be trivial on $\mathfrak p^{c(\psi)}$ but not on $\mathfrak p^{c(\psi) - 1}$ for some integer $c(\psi)$, and is given by $$ f(\pi) = 2(1 + c(\psi) + m/e), $$ where $e$ is $1$ if $\pi$ is unramified and $2$ if $\pi$ is ramified.

EDIT: I'll leave this answer since it's been accepted, but @Kimball's comment provides a better, as more elementary, reference in Section 2.2 of his paper Kimball - The basis problem revisited.

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  • $\begingroup$ If we choose $c(\psi)$ to be $0$, can we say something about $\pi$? $\endgroup$
    – user15243
    Commented May 17, 2020 at 22:34
  • $\begingroup$ @Kiddo, there is no "the conductor of $\pi$"; it depends on a choice of $\psi$, so that (as the reference indicates) one should really write $f(\pi, \psi)$. We may certainly choose $\psi$ as you say, and it simplifies the formula, but of course doesn't affect $\pi$ at all. $\endgroup$
    – LSpice
    Commented May 17, 2020 at 22:48
  • $\begingroup$ Thank you for the reply. $\endgroup$
    – user15243
    Commented May 17, 2020 at 22:55
  • $\begingroup$ Theorem 6.5,(ii) says $f(\sigma_1' \otimes \sigma_2)= n_1n_2(1+m/e)$. What is this $m$? It is not mentioned in the theorem. Is it the level of $\sigma_1' \otimes \sigma_2$? $\endgroup$
    – user15243
    Commented May 17, 2020 at 22:58
  • $\begingroup$ @Kiddo, as the statement of the theorem says, it is notation (6.2.1): $m/e = \max (m_1/e_1, m_2/e_2)$. $\endgroup$
    – LSpice
    Commented May 17, 2020 at 23:02

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