# Intertwining Integral defined on a Weyl group?

Why does the intertwining integral such as the one defined in A. W. Knapp's paper "Intertwining operators for semisimple groups" depend only on an element w of a Weyl group?

http://www.jstor.org/discover/10.2307/1970887?uid=3739256&uid=2&uid=4&sid=56038494493

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As written, this question can only be answered by people with a copy of the book or who are sufficiently familiar with it that can remember whatever it is that Knapp does without having a copy of the book at hand. Writing it up a little bit to actually include context would probably make it answerable by lots of additional people! – Mariano Suárez-Alvarez Apr 15 '12 at 1:37
I edited the original post to provide a link to the paper. The first page of the paper is freely accessible and should provide the necessary definitions. – Joe Tarmet Apr 15 '12 at 1:52

It is not true that the intertwiner depends only upon the Weyl element but also upon the parabolic subgroup $P$. Moreover, it is better to consider the intertwiner to depend upon a set of complex parameters. Its analytic properties give alot of information about the irreducibility, growth of matrix coeffecients, and their like...

But I guess, you are more looking for an explanation, why the Weyl group enters:

1.) The Weyl group of a parabolic $P$ is the normalizer of a Levi subgroup $M$ inside $G$ modulo the inner automorphisms of the Levi subgroup, I guess it will actually be the group of outer automorphisms of $M$.

2.) Most (unitary) representations are subrepresentation/subquotient of parabolic induced representations $\pi$ of $M$. Here $\pi$ is not necessary unitary, even though the induced or subquotients/subrepresentations might be. This goes under the name Langlands classification theorem and Harish-Chandras philosophy of cusp forms.

3.) An intertwiner of the induced representation will come from an Weyl element by the Frobenius reciprocity together with Mackeys restriction prinicple and Bruhat decomposition $G//P =W$. For simplicity, assume that you are working with the group $G=GL(n,F)$. where $F$ is a finite field. For two ireducible reps $\pi, \pi_0$ of $M$, this yields that $$Hom_G( Ind_P^G \pi , Ind_P^G \pi_0) \cong \bigoplus_{w \in W} Hom_M( \pi^w , \pi_0),$$ where the later is zero, if and only if $\pi \cong \pi_0^w$ for some $w$. In principle that suggests at least, that every nontrivial intertwiner should come from Weyl elements.

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I was, as you said, looking for an explanation as to why the Weyl group enters. These three statements go a long way in clearing this up, however I am still wondering why all three of the above statements are true (do they just follow from the definitions?) – Joe Tarmet Apr 15 '12 at 14:53
I added a little bit information. Note that I have restricted myself in the last step to finite groups, to get not to messy, but even for nonfinite group, the guiding principle such as variants of the Frobenius reciprocity, the Mackey restriction formula and Schur's lemma are there, but more technical. – Marc Palm Apr 15 '12 at 15:22
Perfect, thanks. – Joe Tarmet Apr 16 '12 at 15:12
Glad I could help! – Marc Palm Apr 16 '12 at 15:44