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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 $SL(2, \mathbb{R})$.

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    $\begingroup$ 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. $\endgroup$ – Alain Valette Dec 19 '13 at 15:18
  • $\begingroup$ Alright, fair enough - I didn't see that phenomena. Thank you. $\endgroup$ – Marc Palm Dec 20 '13 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). $\endgroup$ – Alain Valette Dec 20 '13 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 '13 at 7:41
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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.

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