When proving identities about traces of functions on representations of $p$-adic groups, Kazhdan's density theorem indicates one only has to check equalities of traces on tempered representations. More precisely, let $F$ be a nonarchimedean local field, let $G$ be a reductive group over $F$, let $\mu$ be a Haar measure on $G(F)$, and let $f:G(F)\rightarrow\mathbb{C}$ be a locally constant function with compact support. For any admissible representation $\pi$ of $G(F)$, we have the trace of $f$:

$$ \operatorname{tr}(f|\pi) = \operatorname{tr}\left(v\mapsto\int_{G(F)}f(g)\pi(g)v\cdot d\mu(g)\right). $$

(One aspect of) Kazhdan's density theorem says that if $\operatorname{tr}(f|\pi)=0$ for all admissible tempered representations $\pi$ of $G(F)$, then $\operatorname{tr}(f|\pi)=0$ for all admissible representations $\pi$ of $G(F)$.

This is precisely direction (d)$\implies$(c) in Theorem 0 of Kazhdan's "Cuspidal Geometry of $p$-adic Groups." Kazhdan's paper simply says that it follows from Theorem XI.2.11 in Borel and Wallach's Continuous cohomology, discrete subgroups, and representations of reductive groups, which is some seemingly impenetrable statement about the Langlands classification for nonarchimedean local fields.

At this point, I have a few questions:

  1. How does one prove (this part of) Kazhdan's density theorem? What is this part of Borel and Wallach's book really saying, and how does their result imply Kazhdan's?

  2. This part of Borel and Wallach's book considers any nonarchimedean local field $F$, whereas Kazhdan's paper only considers $F$ with characteristic zero. Does (this part of) Kazhdan's density theorem hold over $F$ with positive characteristic too? Of course, this might just be an immediate generalization of the answer to 1.

Thank you in advance!

  • 1
    $\begingroup$ The Langlands classification says the standard representations form a basis of the Grothendieck group of representations. If $\mathrm{tr}(f)$ vanishes on all temprered representations, then it vanishes on all standard representations. Since the trace is a linear functional on the Grothendieck group, I think this is 1. $\endgroup$ – Mike B Feb 7 '17 at 17:22
  • $\begingroup$ Thank you for the reply! Why would the vanishing of $\operatorname{tr}(f|-)$ on tempered representations of $G(F)$ imply its vanishing on all standard representations? $\endgroup$ – Charles Denis Feb 7 '17 at 20:27
  • $\begingroup$ The standard representations are essentially induced from tempered. There is a descent formula (I think originally appearing in Harish-Chandra?) which says that if $\pi$ is induced from $\pi_M$ then $\mathrm{tr}\pi(f)=\mathrm{tr}\pi_M(f_M)$, where $f_M$ is the constant term. I don't have it in front of me, but try van Dijk's paper on computing induced characters. $\endgroup$ – Mike B Feb 8 '17 at 15:10
  • $\begingroup$ The standard representations are indeed essentially induced from tempered ones, but the tempered ones could be parabolically induced from proper parabolic subgroups. Why would the vanishing of $\operatorname{tr}(f|-)$ on tempered representations of $G(F)$ imply the vanishing of $\operatorname{tr}(f_{M(F)}|-)$ for the Levi subgroups $M$ of $G$? $\endgroup$ – Charles Denis Feb 8 '17 at 21:59
  • $\begingroup$ Good point. I'll try and think about that. $\endgroup$ – Mike B Feb 8 '17 at 23:36

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