1
$\begingroup$

Question: Can the Fell topology be expressed in terms of the distributions of the the tracial states of a unitary representations, that, is $\pi_j \rightarrow \pi$ if and only if $tr\; \pi_j \rightarrow tr\;\pi$?

This makes of course only sense in a more restricted setting, say reductive groups over a local field $F$. Otherwise, it isn't clear whether irreducible unitary reps have tracial states.

Examples: This is true for $GL(n,F_v)$ for $F_v$ local field and $n=1,2$. Also works well for $SL(2, \mathbb{R})$.

$\endgroup$
4
  • 2
    $\begingroup$ It does NOT work for $SL(2,\mathbb{R})$, because of the non-Hausdorff points in the dual. Consider the trivial rep $\pi_0$, and the two discrete series rep's $\pi_+$ and $\pi_-$ which are not Hausdorff-separated from $\pi_0$. Then,as you go to the trivial along the complementary series $\pi_j$, you have $tr \pi_j\rightarrow tr\pi_0 + tr\pi_+ +tr\pi_-$. If you care only about the reduced dual, you have a similar phenomenon for the mock-discrete series at the bottom of the non-spherical principal series. $\endgroup$ Dec 19, 2013 at 15:18
  • $\begingroup$ Alright, fair enough - I didn't see that phenomena. Thank you. $\endgroup$
    – Marc Palm
    Dec 20, 2013 at 7:35
  • $\begingroup$ This is the difference between continuous trace $C^*$-algebras (which have Hausdorff spectrum) and bounded trace $C^*$-algebras (see my answer below). $\endgroup$ Dec 20, 2013 at 7:39
  • $\begingroup$ Then it doesn't work for $GL(2, F_v)$ as well, where the complementary series converge to a double point being the trivial plus Steinberg if $v$ non-archimedean. I confused the definition of the Fell topology: for each matrix coefficient in $\pi$ there exist a series of matrix coefficient in $\pi_j$, against for every sequence of converging matrix coefficients in $\pi_j$ there exists a matrix coefficient in $\pi$.... $\endgroup$
    – Marc Palm
    Dec 20, 2013 at 7:41

1 Answer 1

3
$\begingroup$

Here is a result by D. Milicic, On $C^{\ast} $-algebras with bounded trace, Glasnik Mat. Ser. III 8(28) (1973), 7–22. Say that a $C^*$-algebra $A$ has bounded trace if the linear span of $T(A^+)$ is dense in $A$, where $T(A^+)$ is the set of those positive elements such that $\pi\mapsto Tr\,\pi(x)$ is bounded on $\hat{A}$. Milicic proves that, for such a $C^*$-algebra $A$, a set $S\subset\hat{A}$ is the set of limits of a net $(\pi_j)$ in $\hat{A}$, if and only if there exists positive integers $n_\sigma$ (for $\sigma\in S$) such that, in your sense, $Tr\,\pi_j$ converges to $\sum_{\sigma\in S} n_\sigma.Tr\,\sigma$.

A result of Fell tells you that the reduced $C^*$-algebra of a semi-simple Lie group with finite centre, has bounded trace.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.