Suppose $A$ is a nonunital residually finitedimensional (RFD) $C^*$algebra, then the multiplier algebra $M(A)$ is also RFD. I wonder whether there exists a trace on the corona algebra $M(A)/A$?
1 Answer
By "trace" I assume you mean tracial state, and in that case the answer is "not necessarily". A counter example is produced below.
The goal is this: we construct an unital RFD $C^\ast$algebra $B$ and a $\ast$epimorphism $\pi \colon B \to \mathcal O_2$ (Cuntz algebra). Then $A := \ker \pi$ does the trick. In fact, it is RFD since this property is preserved by passing to $C^\ast$subalgebra. The canonical $\ast$homomorphism $B \to M(A)$ which extends the identity map on $A$, induces a unital $\ast$homomorphism $\mathcal O_2 \cong B/A \to M(A)/A$ (called the Busby map of the extension $0 \to A \to B \to \mathcal O_2 \to 0$). Hence $M(A)/A$ is properly infinite and does therefor not admit any tracial states.
Consider the unitisation of its cone \begin{equation} C = \{ f \in C([0,1], \mathcal O_2) : f(1) \in \mathbb C 1_{\mathcal O_2} \}. \end{equation} There is a surjection $ev_0 \colon C \to \mathcal O_2$ which is evaluation at zero. As $C$ is homotopy equivalent to $\mathbb C$, it is quasidiagonal by Voiculescu's quasidiagonality theorem. Hence there exists a sequence of integers $k(n)$ and a unital embedding \begin{equation} C \hookrightarrow \prod_{n\in \mathbb N} M_{k(n)}(\mathbb C) / \bigoplus_{n\in \mathbb N} M_{k(n)}(\mathbb C). \end{equation} Let $q \colon \prod_{n\in \mathbb N} M_{k(n)}(\mathbb C) \to \prod_{n\in \mathbb N} M_{k(n)}(\mathbb C) / \bigoplus_{n\in \mathbb N} M_{k(n)}(\mathbb C)$ be the quotient map, and $B:= q^{1}(C)$. Then $B$ is unital, RFD, and it surjects onto $C$ which surjects onto $\mathcal O_2$. This is what we wanted to show.

$\begingroup$ I wonder why $M(A)/A$ is properly infinte. $\endgroup$ Commented May 20, 2019 at 2:46

$\begingroup$ Because the Cuntz algebra $\mathcal O_2$ is properly infinite, and there is a unital $\ast$homomorphism $\mathcal O_2 \to M(A)/A$. $\endgroup$ Commented May 20, 2019 at 6:09