The reduced $C^\*$-algebra of a non-abelian free group $G$ has a unique trace. Hence, there is no chance to separate conjugacy classes of group elements using traces on $C^\star\_{red} G$. On the other side, for the group ${\mathbb Z}$, separation is clearly possible.



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> **Question:** Let $G$ be an amenable group. Does the reduced group $C^\star$-algebra of $G$ support sufficiently many traces to distinguish between conjugacy classes of group elements?
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**EDIT:** The question seems already interesting for $S\_{\infty} = \cup\_n S\_n$. Let's get explicit and pick $g \in S\_n$ (for some $n$) and consider the canonical trace $\tau\_{g,n}$ which sends every conjugate of $g$ to $1$ and all other elements to zero. (This can be done for any $n' \geq n$ since $S\_n \subset S\_{n'}$.) The function $\tau\_{g,n} \colon {\mathbb C}S\_n \to {\mathbb C}$ is a conjugation invariant function and hence, it must be a linear combination of the normalized traces of irreducible representations of $S\_n$. 



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> **Question:** What is the sum of the absolute values of the coefficients that come up in this linear combination of traces? 
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This is (as one can check) the norm of $\tau\_{g,n}$, call it $c(g,n)$. So, we see that the compatible family of maps $\tau\_{g,n}$ extends from ${\mathbb C}S\_{\infty}$ to $C^\* S\_{\infty}$ if and only if $c(g,n)$ remains bounded.



Conjugacy classes and reduced group $C^*$-algebra of an amenable group
There is an easy example, namely $SL\_3(F)$ where $F$ is the algebraic closure of some finite field. This group does not admit non-trivial characters (a result of Kirillov) and is locally finite, hence amenable. This gives a negative answer to the first question.


Sorry, I deleted my previous answer since it contained a mistake.


