Suppose $A$ is a non-unital residually finite-dimensional (RFD) $C^\*$-algebra, then the [multiplier algebra](https://en.wikipedia.org/wiki/Multiplier_algebra) $M(A)$ is also RFD. I wonder whether there exists a trace on the corona algebra $M(A)/A$?



Residually finite-dimensional $C^*$-algebra
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.


