# Uniqueness of faithful (!) tracial states on separable $C^*$-algebra

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?

• This is false even in Abelian $C^\ast$-algebras. Just take two different probability measures with full support in $\mathbb R$. – Adrián González-Pérez Dec 3 '18 at 11:46
• 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. – Adrián González-Pérez Dec 3 '18 at 11:49

This fails almost always, unless you have unique trace. For instance let $$A=\mathbb C\oplus\mathbb C$$. Let $$\tau(x,y)=(x+y)/2$$, $$\rho(x,y)=x/3+2y/3$$.