I am looking for a reference/idea, how this passage from Labesse's Snowbird Lecture "Introduction to endoscopy" pg.5 can be explained:

"We shall denote by $f_\pi$ a pseudo-coefficient for $\pi$, although it is highly non unique. But as regards invariant harmonic analysis this plays no role. In particular the orbital integrals are independent of the choice of the pseudocoefficient; they are also independent of the choice of the Haar measure on $G(F)$ but one has to use the canonical measure on the compact torus $G (F)$. The orbital integrals of $f_\pi$ are easily computed for regular semisimple: $$O_\gamma(f_\pi) =\begin{cases} \Theta_\pi(\gamma), & \gamma \mathrm{ elliptic}, \newline 0, &\mathrm{else},\end{cases} $$ where $\Theta_\pi(\gamma)$ is the character of $\pi$."

Here $G$ is a reductive group over a local field $F$ and $\pi$ a squareintegrable representation.


Existence of pseudo-coefficients for square-integrable representations (and the link with character values of the representations) is stated and proved in

D. Kazhdan, Cuspidal geometry of $p$-adic groups. J. Analyse Math. 47 (1986), 1–36.

| cite | improve this answer | |
  • 1
    $\begingroup$ I have seen this reference, in fact Labesse gives it himself. The existence is not the issue. What I want to understand/reference is the statement about the orbital integral. Thanks nevertheless. $\endgroup$ – Marc Palm May 14 '12 at 10:56
  • $\begingroup$ You also have the exact statement on the link bewteen orbital integral of a pseudo-coefficient and the character of the representation in Kazhdan's paper. $\endgroup$ – Paul Broussous May 14 '12 at 13:32
  • $\begingroup$ Okay, I was searching the document for keywords, probably not the correct ones;) I will have a more careful look. Thank you. $\endgroup$ – Marc Palm May 14 '12 at 16:16
  • 1
    $\begingroup$ I find Kazhdan's paper difficult to read ! $\endgroup$ – Paul Broussous May 14 '12 at 20:27
  • $\begingroup$ I went back again. It is theorem K on page 7. $\endgroup$ – Marc Palm Jun 12 '12 at 15:48

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.