Let $\pi$ be an automorphic cuspidal representation of $\operatorname{GL}_2(\mathbb A_{\mathbb Q})$, and let $\Pi = \operatorname{sym}^2(\mathbb\pi)$, which is a representation of $\operatorname{GL}_3(\mathbb A_{\mathbb Q})$. Let $\pi= \bigotimes'\pi_p$ and $\Pi= \bigotimes'\Pi_p$. Let $c(\Pi_p)$ and $c(\pi_p)$ be the conductors for $\Pi_p$ and $\pi_p$ respectively. Can we get some relation between the conductors $c(\Pi_p)$ and $c(\pi_p)$ using theorem 6.5 of Bushnell - Local Rankin–Selberg convolutions for $\operatorname{GL}_n$: Explicit conductor formula or any other way?

Thank you in advance.

  • $\begingroup$ You had some $\mathbb\pi$ $\mathbb\pi$ and some $\pi$ $\pi$; since I think the latter is more usual, I edited for uniformity. If you do restore \mathbb, note that it shouldn't apply to everything; so, for example, $\mathbb\pi_p$ $\mathbb\pi_p$, not $\mathbb{\pi_p}$ $\mathbb{\pi_p}$. $\endgroup$ – LSpice Nov 26 '19 at 18:09
  • $\begingroup$ Also, the Bushnell paper wouldn't load for me, so I changed the reference to the official JAMS page. $\endgroup$ – LSpice Nov 26 '19 at 18:09
  • 3
    $\begingroup$ The conductor of the symmetric square should be the conductor of the Rankin-Selberg convolution $\pi \otimes \pi$ minus the conductor of the central character, and I think this should get you the formula you want. $\endgroup$ – Will Sawin Nov 26 '19 at 18:10
  • $\begingroup$ @WillSawin Thanks for the comment. But will you explain it little more? How are you saying that the conductor of symmetric square will be like this? please explain!! $\endgroup$ – Kiddo Nov 26 '19 at 18:25
  • 2
    $\begingroup$ Will's answer is spot-on. See also my answer here: mathoverflow.net/questions/285135/…. For references, try Gelbart-Jacquet: doi.org/10.24033/asens.1355 $\endgroup$ – Peter Humphries Nov 26 '19 at 19:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.