8
$\begingroup$

Let $G$ be a connected split reductive group over $\mathbb{Z}$. Let $F$ be a local non-Archimedean field. Let $\rho$ be an irreducible smooth representation of $G(F)$. How does one define the conductor of $\rho$?

The conductor should be an integer canonically associated with the representation $\rho$ so that it matches up with the conductor of the corresponding Galois representation on the Langlands' dual side.

An example of an answer that I'm looking for is given in http://www2.imperial.ac.uk/~buzzard/maths/research/notes/old_introductory_notes_on_local_langlands.pdf Here, Kevin Buzzard gives a nice simple algebraic definition of the conductor for representations $\mathrm{GL}_2$ using the mirabolic subgroup (see middle of page six of his notes). Does his definition generalize to $\mathrm{GL}_n$? Is there a simple algebraic definition for the conductor of representations of general $G$? What is a good reference for this subject?

$\endgroup$

1 Answer 1

7
$\begingroup$

for GL_n you could look at:

H.Jacquet,I.I.Piatetski-Shapiro,J.Shalika:Conducteur des representations du groupe lineaire, Math. Annalen 256, 199–214 (1981).

$\endgroup$
4
  • 3
    $\begingroup$ Yes, but/and there was a correction, linked-to (top link) on Jacquet's homepage, at math.columbia.edu/~hj $\endgroup$ Dec 3, 2013 at 0:41
  • 1
    $\begingroup$ The case of $\mathrm{GL}_2$ was first dealt with Casselman in his paper "On some results of Atkin and Lehner". However, it seems that he uses a different subgroup from Jacquet, Piatetski-Shapiro, Shalika [JP-SS]. Namely, Casselman uses the subgroup of $\mathrm{GL_2}(\mathcal{O})$ whose reduction modulo $t^n$ is the Borel subgroup, where as, [JP-SS] use the subgroup whose reduction modulo $t^n$ is the mirabolic subgroup. Can someone explain the discrepancy? $\endgroup$
    – Dr. Evil
    Dec 5, 2013 at 3:59
  • 2
    $\begingroup$ mirabolic * centre= parabolic he probably assumes something about the conductor of the central character. $\endgroup$
    – schaffner
    Dec 6, 2013 at 18:16
  • 2
    $\begingroup$ @Paul Garret. To make the story more precise ... In fact the mistake was pointed out to H. Jacquet by Nadir Matringe (University of Poitiers, France). The error was fixed independently by Jacquet and Matringe. Both published (at least submitted) their (different) proofs. $\endgroup$ Dec 11, 2013 at 19:12

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.