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

Suppose $$A$$ is a non-unital residually finite-dimensional (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.

• I wonder why $M(A)/A$ is properly infinte. – math112358 May 20 at 2:46
• Because the Cuntz algebra $\mathcal O_2$ is properly infinite, and there is a unital $\ast$-homomorphism $\mathcal O_2 \to M(A)/A$. – Jamie Gabe May 20 at 6:09