# Extension of tracial state to the multiplier algebra

If $$A$$ is a $$C^*$$ algebra, I know the fact: If $$\tau$$ is a state on $$A$$, we can extend it to a state on $$M(A)$$, by continuity of $$\tau$$ and where $$M(A)$$ is the multiplier algebra of $$A$$.

If $$\tau$$ is a tracial state on $$A$$, can we extend $$\tau$$ to get a tracial state on $$M(A)$$?

• I haven't checked the details but I will proceed as follows: First $M(A)$ has a coarser topology that the norm topology, that is the strong topology given by seeing $M(A)$ as the space of adjointable operators over $A$ as a Hilbert $C^*$-module. Then, extend $\tau$ by strong continuity and use the fact that the tracial identity is preserved. – Adrián González-Pérez Jul 23 '18 at 11:05

## 1 Answer

Let $$\tau$$ be a state on $$M(A)$$ such that its restriction to $$A$$ satisfies $$\tau(ab) = \tau(ba)$$ for all $$a,b \in A$$. For $$x,y \in M(A)$$ we then have $$\tau(xy) = s-\lim\tau((xe_\lambda)(e_\lambda y)) = s-\lim \tau(e_\lambda yx e_\lambda) = \tau(yx),$$ where $$(e_\lambda)_\lambda$$ is an approximate unit for $$A$$. Note that $$e_\lambda \to 1$$ strictly and hence $$e_\lambda x \to x$$ strictly. Since $$\tau$$ is strictly continuous, the result follows.

Note that in the case $$A$$ is separable, $$A$$ is a hereditary C*-subalgebra of $$M(A)$$, namely $$A = \overline{hM(A)h}$$, where $$h \in A$$ is strictly positive. In that case every state on $$A$$ has a unique extension to $$M(A)$$ given by $$\tau(x) = \lim \tau(e_\lambda x e_\lambda) \qquad (x \in M(A)).$$

Edit: For the argumet to work one needs that $$\tau$$ is strictly continuous on $$M(A)$$. I will check if this is automatic and then update the answer.