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A supercuspidal representation is an admissible representation with trivial Jacquet module for any parabolic.This implies that the matrix coeff. are compactly supported mod center.

Cartier (in Corivalis) defines super cuspidal rep as admissible rep. with compactly supported matrix coeff. mod center.

I am wondering if this (Cartier's) definition is equivalent to triviality of Jacquet module? If yes, is there a reference for such a proof?

Thanks in advance.

flag	asked Oct 13 2011 at 0:38 isildur 214●1●8

It seems a little odd to me to include the adjective "admissible" in the definition of "supercuspidal". I would just say a supercuspidal representation is a smooth representation with trivial Jacquet module for any parabolic. Eventually, you prove that every irreducible smooth representation is admissible, by using parabolic induction and proving the result for irreducible supercuspidals. – Marty Oct 13 2011 at 16:41

Joël · Answer 1 · 2011-10-13 00:49:48Z UTC

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The answer is yes (and this is a classical result). A good reference with complete self-contained proofs is Bill Casselman's book (http://www.math.ubc.ca/~cass/research/pdf/p-adic-book.pdf). cf Theorem 5.3.1

answered Oct 13 2011 at 0:49

Joël
9,499●19●52

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It is funny how, when you read papers from the 1970`s or 1980`s on p-adic groups, they always quote this book as "the soon to be published book", and yet, we are in 2011 and this book is still unpublished! – M Turgeon Oct 13 2011 at 0:59

Supercuspidals : clarification

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