Let $B$ be a unital C*-algebra and let $A⊆B$ be a closed *-subalgebra containing the unit of $B$. I am mostly interested in the case that $A$ is abelian but, for the strict purpose of stating my question, this does not seem to matter much.
Let us say that the inclusion "$A⊆B$" has property (EPS) (for Extended Pure states Separate) provided the set of all state extensions of pure states on $A$ separates points of $B$ in the sense that, if $b∈ B$, and $\psi(b^*b)=0$, for all such states $\psi$, then $b=0$.
An example failing this property is $A=C([0,1])$ (represented on $L^2[0, 1]$ as multiplication operators), and $B=A+K$ (compact operators), but I suspect the failure is due to the fact that $A$ is not a regular subalgebra of $B$ (normalizers, in the sense of Kumjian, do not span $B$). Nevertheless there are regular counter-examples as well.
For obvious reasons any inclusion of the form "$ℂ⊆B$" satisfies (EPS) and so does "$ℂ^n⊆B$".
Other situations in which (EPS) holds are:
- $B$ is abelian,
- $A$ is abelian, $B=A\rtimes G$, and $G$ is amenable.
My questions are:
Questions. Has property (EPS) been studied before? Can it be characterized in some sensible way?
I suspect that this might have something to do with nuclearity, so here is another:
Question. If $A$ is abelian and regular and $B$ is nuclear, can one show that (EPS) holds?