# Level vs. conductor of a supercuspidal representation

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

• 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). – LSpice May 17 '20 at 21:12
• 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. – Kimball May 18 '20 at 1:31
• @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 $m/e$? – LSpice May 18 '20 at 15:18
• @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. – Kimball May 18 '20 at 16:30
• @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). – LSpice May 18 '20 at 17:46

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.
• If we choose $c(\psi)$ to be $0$, can we say something about $\pi$? – Kiddo May 17 '20 at 22:34
• @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. – LSpice May 17 '20 at 22:48
• 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 $\sigma_1' \otimes \sigma_2$? – Kiddo May 17 '20 at 22:58
• @Kiddo, as the statement of the theorem says, it is notation (6.2.1): $m/e = \max (m_1/e_1, m_2/e_2)$. – LSpice May 17 '20 at 23:02