Let $G$ be a locally compact unimodular group with center $Z$ and let $\omega$ be a unitary character of $Z$. To fix the discussion, an irreducible unitary representation $\pi$ with central character $\omega$ is

  • square integrable if for all $u,v\in V_\pi$, the matrix coefficient $g\mapsto \left< gu,v\right>$ is square integrable when viewed as a function on $G/Z$, and

  • tempered if $\pi$ is weakly contained in the right regular representation $\mathrm{L}^2(G)$. That is, for any compact $C\subseteq G$, $\varepsilon>0$ and unit vector $v\in V_\pi$, there is a unit vector $\psi\in \mathrm{L}^2(G)$ such that $|\left<gv,v\right>-\left<g\psi,\psi\right>|<\varepsilon$ for all $g\in C$.

(There are other equivalent definitions for temperedness when $G$ is a real or $p$-adic reductive Lie group.)

It is not immediately obvious from the definitions that irreducible square-integrable representations are tempered, but it is known to be true in the following cases:

  1. $Z=1$ (since then $\pi$ above can be embedded in $\mathrm{L}^2(G)$),
  2. $G$ is a real or $p$-adic reductive Lie group.
  3. $G=Z$, or $G$ is compact, or more generally $G$ is amenable (since any irreducible representation is tempered in the previous sense).

My question is whether this is true for general locally compact groups. (Assume $G$ is unimdular and all irreducible unitary representations are admissible if this simplifies things.)

The question can be generalized even further as follows: Suppose that $Z$ is just a normal amenable subgroup of $G$ and let $\mathrm{L}^2_\omega(G/Z)$ denote the set of $(Z,\omega)$-equivariant functions $\psi:G\to \mathbb{C}$ such that $|\psi|$ is square integrable when viewed as a function on $G/Z$. Is it true that any irreducible representation of $G$ that is weakly contained in $\mathrm{L}^2_\omega(G/Z)$ ($G$ acts by right translations) is weakly contained in $\mathrm{L}^2(G)$?

  • $\begingroup$ Can I assume Type I as well? (I am thinking aloud here; this may be a red herring, since there exist amenable groups which are not Type I) $\endgroup$ – Yemon Choi May 11 '16 at 2:06
  • $\begingroup$ Yes. Assume also that the group is second countable. $\endgroup$ – Uriya First May 11 '16 at 2:08
  • 1
    $\begingroup$ A trivial comment: not the matrix coefficient $g \mapsto \langle g u, v\rangle$, but only its absolute value, is defined on $G/Z$. $\endgroup$ – LSpice May 11 '16 at 2:31
  • $\begingroup$ Dear Uriya, is being "tampered" related to properties of Caylet graphs (in a similar way that property-T is related to being an expander)? (I vaguely remember that I was told so but not any detail.) $\endgroup$ – Gil Kalai Apr 14 '17 at 14:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.