When are tracial states continuous in the strict topology?

Question

Let $A$ be a separable (non-unital) C*-algebra. Let $\tau: A\to\mathbb{C}$ a tracial state. Consider the strict topology on $A$, i.e. $a_n\to a$ iff $a_nx\to ax$ and $xa_n\to xa$ for all $x\in A$. It is clear that this is the same as the usual norm-topology when $A$ has a unit.

My question is:

When is $\tau$, restricted on the unit ball of $A$, continuous in the strict topology?

In other words: Under what conditions is it true that every norm-bounded sequence $a_n\in A$, with $a_nx~,~ xa_n\to 0$ for all $x\in A$, satisfies $\tau(a_n^*a_n)\to 0$?

Is this always possible? If not, I would be very glad to see a counterexample. If it turns out to be wrong in general, I would also like to know about the case that $A$ has a unique tracial state and is algebraically simple. Looking at that case would be sufficient for my purposes.

score 3 · Accepted Answer · answered Oct 18 '13 at 17:33

3

Go to the GNS representation. There is a cyclic vector $v$ such that $\tau(x) = \langle \pi(x)v,v\rangle$ for all $x \in A$. Now if $\|a_nx\| \to 0$ in $A$ for every $x \in A$ then $\|\pi(a_nx)v\|\to 0$ for all $x \in A$. Since $v$ is cyclic and $(a_n)$ is bounded, this implies that $\pi(a_n)v \to 0$, so $\tau(a_n^*a_n) = \|\pi(a_n)v\|^2 \to 0$.

answered Oct 18 '13 at 17:33

Nik Weaver

42,041

Thank you! We actually pondered about this with some collegues for the whole day and somehow failed to see this. Now I'm almost embarassed about asking such an easy question, but whatever...thanks for clearing up our headaches! ;) – Gabor Szabo Oct 18 '13 at 23:46
No problem, the GNS representation always helps with this kind of question. – Nik Weaver Oct 19 '13 at 00:02
This is really nice! – Ulrich Pennig Oct 29 '13 at 08:52

When are tracial states continuous in the strict topology?

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