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Let $A$ be a separable $C^*$-algebra and $S\subseteq A$ a norm-dense, countable set in $A$. Assume that there are two faithful (meaning that the corresponding GNS-construction gives a faithful representation) tracial states $\tau$, $\rho$ on $A$. Does this already imply that $\tau = \rho$?

Background and own thoughts: The GNS-construction gives us two separable Hilbert spaces $\cal H_1$ $\cal H_2$ which are (by the faithfulness) the closure of $S$ by the norms coming from $\tau$ and $\rho$. As the Hilbert spaces are separable, there exists an isometric isomorphism $U: \cal H_1 \rightarrow \cal H_2$, so for any $x \in S$ we have $\rho(x^*x)=\left\Vert x\right\Vert _{\rho}^{2}=\left\Vert Ux\right\Vert _{\tau}^{2} = \tau\left(\left(Ux\right)^{*}\left(Ux\right)\right)$. Now, if $U$ is implemented by a unitary (or isometry) in $A$, this implies that $\rho$ and $\tau$ coincide on positive elements of $S$, hence are equal. The question is: Does such a unitary/isometry exist? If not, does uniqueness hold nevertheless?

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  • $\begingroup$ This is false even in Abelian $C^\ast$-algebras. Just take two different probability measures with full support in $\mathbb R$. $\endgroup$ – Adrián González-Pérez Dec 3 '18 at 11:46
  • $\begingroup$ There are some simple $C^\ast$-algebras that have the unique trace property , like $C^*_r ( \mathbb F_2)$, but that is far from the norm. $\endgroup$ – Adrián González-Pérez Dec 3 '18 at 11:49
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This fails almost always, unless you have unique trace. For instance let $A=\mathbb C\oplus\mathbb C$. Let $\tau(x,y)=(x+y)/2$, $\rho(x,y)=x/3+2y/3$.

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