3
$\begingroup$

I am currently aiming at estimating orbital integrals. Maybe surprizingly, I hope for some help in the compact case (ramified places), in proving the usual formula

$$O_\gamma(f) = \int_G f(x^{-1}\gamma x)dx = \Theta_\pi(\gamma) dim(\pi)^{-1}$$

where $f$ is a suitably normalized matrix coefficient for $\pi$. There are general proofs of this fact (for instance in Arthur or Kottwitz, using Peter-Weyl theorem or properties of supercuspidal representations), yet I would rather find an easiest way to do so in the compact case.

For instance, it is quite easy to prove

$$O_\gamma(f) = \Theta_\pi (f) \Theta_\pi (\gamma)$$

so is there any way to see that $\Theta_\pi(f) = dim(\pi)^{-1}$ (if true) ?

Thank you in advance for any clue or reference.

Improve this question
$\endgroup$
6
  • 3
    $\begingroup$ Could you give make your notation and assumption more precise ? What is your group ? What sort of representation are you considering ? What are the assumptions on the group and representation ? What is ${\rm dim}(\pi )$ if $G$ is not compact ? $\endgroup$
    – Paul Broussous
    Commented Apr 10, 2016 at 9:18
  • 1
    $\begingroup$ @PaulBroussous I am in a compact setting, more precisely: I consider $B$ a division quaternion algebra and $G = Z \backslash B^\times$; I am trying to do the work locally for the (compact) unit group $G_p$ in a ramified place. (Otherwise $\dim(\pi)$ should probably be the formal degree of $\pi$ ?) $\endgroup$
    – Desiderius Severus
    Commented Apr 10, 2016 at 11:04
  • 1
    $\begingroup$ Of course 'supercuspidal' is vacuous in this setting! If you don't like to use Peter–Weyl, I guess that you don't want to use orthogonality of matrix coefficients to prove that $\Theta_\pi(f) = \dim(\pi)^{-1}$? $\endgroup$
    – LSpice
    Commented Apr 10, 2016 at 16:29
  • $\begingroup$ @LSpice I am not at ease with Peter-Weyl and what really hides behind, hence I prefer to avoid its use, never knowing if it would be using a huge machinery for such a simple result. I am more familiar with orthogonality of matrix coefficients, but you seems to say they are not so different, hence it would be ok ;) $\endgroup$
    – Desiderius Severus
    Commented Apr 10, 2016 at 17:41
  • 2
    $\begingroup$ I think the issue is not so much Peter-Weyl but some form of Schur's lemma (which seems to be the key mechanism behind all orthogonality relations I've come across in the setting of group reps) $\endgroup$
    – Yemon Choi
    Commented Apr 10, 2016 at 17:53

1 Answer 1

Reset to default
5
$\begingroup$

You said in comments that you are OK with using orthogonality of matrix coefficients, but let's suppose you changed your mind. Let $\langle\cdot, \cdot\rangle$ be a $G$-invariant pairing on the space $V$ of $\pi$, and let $\mathrm dg$ be the Haar measure on $G$ normalised to give it total mass $1$. What follows is all very standard, so I assume it's not what you want, but let's have something out here so that you can clarify where it falls short.

Note that, for each $b, w \in V$, we have that $v \otimes v' \mapsto {\displaystyle\int} \langle g b, v'\rangle\cdot\overline{\langle g w, v\rangle}\mathrm dg$ is an element of $\operatorname{Hom}_G(V \otimes V, \mathbb C)$, hence, by Schur's lemma, a scalar multiple of $\langle\cdot, \cdot\rangle$; and that the map sending $b \otimes w$ to this scalar is itself is an element of $\operatorname{Hom}_G(V \otimes V, \mathbb C)$, hence again a scalar multiple of $\langle\cdot, \cdot\rangle$. Call this scalar $c$. (Note the illustration of Yemon Choi's claim that Schur's lemma lies at the bottom of everything!)

Now $$ \Theta_\pi(g) = \sum_{b \in B} \langle g b, b\rangle, $$ where $B$ is an orthonormal basis of the space $V$ of $\pi$; so, if we write $f_{v, w}$ for the matrix coefficient $g \mapsto \langle v, g w\rangle$, then \begin{align*} \Theta_\pi(f_{v, w}) & {}= \sum_{b \in B} \int \langle g b, b\rangle\cdot\langle v, g w\rangle\mathrm dg \\ & {}= \sum_{b \in B} \int \langle g b, b\rangle\cdot\overline{\langle g w, v\rangle}\mathrm dg \\ & {}= \sum_{b \in B} c\langle b, w\rangle\cdot\overline{\langle b, v\rangle} \\ & {}= c\langle v, w\rangle = c f_{v, w}(1). \end{align*} On the other hand, \begin{align*} f_{v, w}(1) & {}= \int \langle g v, g w\rangle\mathrm dg \\ & {}= \int \sum_{b \in B} \langle g v, b\rangle\cdot\overline{\langle g w, b\rangle}\mathrm dg \\ & {}= \sum_{b \in B} c\langle v, w\rangle\cdot\langle b, b\rangle \\ & {}= c\cdot\lvert B\rvert\cdot f_{v, w}(1) = c\cdot\dim(\pi)f_{v, w}(1). \end{align*} Since this is true for all $v, w \in V$, we have that $c = \dim(\pi)^{-1}$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .