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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 $M(A)$?

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  • $\begingroup$ 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. $\endgroup$ – Adrián González-Pérez Jul 23 '18 at 11:05
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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.

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