Recall that a separable C\*-algebra $A$ is *quasi-diagonal* if there are completely positive and contractive maps $\varphi\_k \colon A \rightarrow M\_{n(k)}$ such that $||\varphi\_k(ab) - \varphi\_k(a)\varphi\_k(b)|| \rightarrow 0$ and $||\varphi\_k(a)|| \rightarrow ||a||$ for every $a, b \in A$, where $M\_{n(k)}$ denotes the algebra of $n(k) \times n(k)$ complex matrices.


The following ought to be known to experts, but I've had some issues trying to prove it.


**Question:** Let $A$ be a quasi-diagonal separable C\*-algebra. Does $A$ admit a faithful trace?


Observe we do not assume that $A$ is simple. The idea would be to use an ultralimit of the cp maps $\varphi\_k \colon A \rightarrow M\_{n(k)}$ composed with the usual traces $tr\_{n(k)} \colon M\_{n(k)} \rightarrow \mathbb{C}$. This does not necessarily yield a faithful trace, though, as the traces $tr\_{n(k)}(\varphi\_k(a^\*a))$ might be very small while $||\varphi\_k(a^\*a)||$ remains large. However, one could in some cases tweak $\varphi\_k$ by adding a factor that enlarges the trace, maintains the asymptotic multiplicativity and the norm. The actual proof requires to do this everywhere at once, that is, for every possible element $a^\*a$, and this seems impossible (or unreasonably hard).


Note, moreover, that there is no obvious K-theoretic obstruction, as every countable abelian group $K\_0(A)$ embeds into $\mathbb{R}$.


Any help is greatly appreciated.



Faithful traces on quasi-diagonal C*-algebras
No, separable (unital) quasi-diagonal $C^\ast$-algebras do not necessarily admit a faithful tracial state. For instance, the $C^\ast$-algebra
\begin{equation}
A= \{ f\in C([0,1], \mathcal O\_2) : f(0) \in \mathbb C 1\_{\mathcal O\_2}\}
\end{equation}
(where $\mathcal O\_2$ is the Cuntz algebra with two canonical generators) is homotopic to $\mathbb C$, and hence is quasi-diagonal by the homotopy invariance of quasi-diagonality due to Voiculescu. However, any trace on $A$ vanishes on the ideal $C\_0((0,1], \mathcal O\_2)$, since this is purely infinite in the sense of Kirchhberg-Rørdam [Kirchberg, Eberhard; Rørdam, Mikael Non-simple purely infinite C∗-algebras. Amer. J. Math. 122 (2000), no. 3, 637–666.]. So the only tracial state on $A$ is the one factoring through evaluation at 0.


If you consider non-unital $C^\ast$-algebras, $C\_0((0,1], \mathcal O\_2)$ is an example of a separable quasi-diagonal $C^\ast$-algebra with no tracial states.


On the other hand, if you consider separable residually finite-dimensional (RFD) $C^\ast$-algebras, then they always admit a faithful tracial state: a separable $C^\ast$-algebra is RFD if it embeds into $\prod\_{n\in \mathbb N} M\_{k(n)}(\mathbb C)$ for some sequence $k(n)$ of natural numbers. As $\prod\_{n\in \mathbb N} M\_{k(n)}(\mathbb C)$ has a faithful tracial state, e.g. $\sum\_{n\in \mathbb N} 2^{-n} \mathrm{tr}\_{k(n)}$, so does any $C^\ast$-subalgebra.


