3
$\begingroup$

I am reading the paper [Margulis, G. A.; Nevo, A.; Stein, E. M., Analogs of Wiener's ergodic theorems for semisimple Lie groups. II. Duke Math. J. 103 (2000), no. 2, 233–259] (MSN).

I want to understand the following argument. Let $(G,K)$ be a Gelfand pair. Then $M(G,K)$ i.e. the set of all bi-$K$-invariant measures on $G$ is a commutative Banach algebra. Suppose $\tau:G\to B(\mathcal H_\tau)$ be a strongly continuous unitary representation of $G.$ Denote $A_{\tau}:=\overline{\tau(M(G,K)}.$ Then spectrum of $A_\tau$ can be identified with all positive definite spherical functions on $G.$ How do you see that?

Secondly given a probability measure $\mu\in M(G,K)$ define $$\|\mu\|_T:=\sup\{|\phi_\lambda(\mu)|:\text{$\phi_\lambda$ positive-definite spherical function}\}.$$ Then $\|\mu\|_T\geq \tau(\mu).$

I do not understand the identification of spectrum of $A_\tau$ with positive-definite spherical functions. Secondly, what does one mean by $\phi_\lambda(\mu)$ since $\phi_\lambda$ is supposed to be a function on $G$?

Improve this question
$\endgroup$
9
  • 1
    $\begingroup$ Did you check the four references given for the identification? $\endgroup$
    – LSpice
    Apr 21, 2020 at 17:58
  • 1
    $\begingroup$ Also, $\phi_\lambda(\mu)$ means that $\phi_\lambda$ is viewed as a character of $\overline{\tau(M(G, K))}$, then evaluated at $\tau(\mu)$. $\endgroup$
    – LSpice
    Apr 21, 2020 at 18:07
  • 1
    $\begingroup$ Anyway, (1) notice that the spectrum is not identified with all pdsf's, but only a subset, and (2) I think that the identification is the simple one: given a character $\chi$ of $\overline{\tau(M(G, K))}$, send $g \in G$ to the value of $\chi \circ \tau$ at the unit mass on $K g K$. $\endgroup$
    – LSpice
    Apr 21, 2020 at 18:17
  • 1
    $\begingroup$ The closed span of the various unit $K g K$'s is $M(G, K)$, so injectivity follows. (Also, you can delete comments.) $\endgroup$
    – LSpice
    Apr 21, 2020 at 18:41
  • 1
    $\begingroup$ $G$ is a (finite-dimensional) Lie group, so it is $\sigma$-compact. $\endgroup$
    – LSpice
    Apr 21, 2020 at 19:04

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.