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.


1 Answer 1


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.

  • $\begingroup$ Very nice answer indeed! Well done! $\endgroup$
    – Ruy
    Dec 14, 2020 at 12:51

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