Show convergence of net associated to GNS-triplet associated to state on a $C^*$-algebra

Question

Let $A\subseteq B \subseteq B(H)$ be an inclusion of $C^*$-algebras where $H$ is some Hilbert space. We have the following conditions:

• B is a von Neumann algebra with $A'' = B$.
• The inclusion $A \subseteq B(H)$ is non-degenerate.
• $B$ contains $\operatorname{id}_H$, but $A$ does not.

Further, let $\omega \in B^*$ be a normal state and let $(H_\omega, \pi_\omega, \xi_\omega)$ the associated GNS-triplet. If $\{u_\lambda\}_{\lambda\in \Lambda}$ is an approximate unit for $A$, is it true that $$\lim_{\lambda \in \Lambda}\|\pi_\omega(u_\lambda)\xi_\omega-\xi_\omega\| = 0?$$

I know that the representation $\pi_\omega$ is a normal $*$-morphism. Maybe that's useful here.

Answer 1

We need the following fact: if a net of positive operators $(a_{\lambda})$ is increasing and bounded then it converges in the strong operator topology. It is not difficult to check that it converges in the weak operator topology to some element $a$. Therefore $(a-a_{\lambda})$ is a bounded net of positive operators that converges to zero. It follows that the net formed by the square roots $(\sqrt{a-a_{\lambda}})$ converges to zero in the strong operator topology. Recall now that multiplication is continuous with respect to the strong operator topology on bounded subsets. As the net $(\sqrt{a-a_{\lambda}})$ is bounded, the net formed by squares, i.e. $(a-a_{\lambda})$ converges to zero in the strong operator topology as well. It means that $(a_{\lambda})$ converges to $a$ in the strong operator topology.

By your assumptions, the approximate unit $(u_{\lambda})$ converges to the identity in the weak* topology. From normality of $\pi_{\omega}$ it follows that $(\pi_{\omega}(u_{\lambda}))$ converges to the unit in the weak* topology. But $(\pi_{\omega}(u_{\lambda}))$ is a bounded, increasing net, thus it converges in the strong operator topology. The limits in the two topologies have to agree, therefore $(\pi_{\omega}(u_{\lambda}))$ converges to the unit in the strong operator topology, in particular $\lim_{\lambda \in \Lambda} \|\pi_{\omega}(u_{\lambda})\xi_{\omega} - \xi_{\omega}\|=0$.

Answer 2

This is actually true in a somewhat greater generality:

Claim: Let $A$ be a $C^*$-algebra with approximate identity $(u_\lambda)$. Let $\pi:A\rightarrow\mathcal B(H)$ be a non-degenerate representation. Then $\pi(u_\lambda)\rightarrow 1_H$ strongly.

This even works for Banach algebras with a bounded approximate identity, acting on Banach spaces.

Non-degeneracy means that a typical element of $H$ is of the form $\xi = \pi(a)\eta$, and so $\lim_\lambda \pi(u_\lambda)\xi = \lim_\lambda \pi(u_\lambda a)\eta = \pi(a)\eta = \xi$. Linearity and density (and using that $(u_\lambda)$ is a bounded net) shows that $\lim_\lambda \pi(u_\lambda)\xi = \xi$ for any $\xi\in H$. But this shows exactly that $\pi(u_\lambda)\rightarrow 1$ strongly.

In your case, $\pi_\omega$, restricted to $A$, is non-degenerate. We need to equivalently show that if $\eta\in H_\omega$ with $\pi_\omega(a)\eta=0$ for all $a\in A$, then $\eta=0$. The assumption gives that $$ 0 = (\pi_\omega(a)\eta|\pi_\omega(x)\xi_\omega) = (\eta|\pi_\omega(a^*x)\xi_\omega) \qquad (x\in M). $$ As $A''=B$, we know that $A$ is weak$^*$-dense in $B$ and so $(u_\lambda)\rightarrow 1$ weak$^*$ in $B$. Thus $\pi_\omega(u_\lambda x) \rightarrow \pi_\omega(x)$ weak$^*$ (hence weakly) as $\pi_\omega$ is normal. So $$ (\eta|\pi(x)\xi_\omega) = \lim_\lambda (\eta|\pi(u_\lambda x)\xi_\omega) = 0 \qquad (x\in M). $$ By (the GNS) construction, $\{ \pi(x)\xi_\omega : x\in M \}$ is dense in $H_\omega$, so we conclude that $\eta=0$, as required.

Final comment: The Cohen-Hewitt factorisation theorem in fact shows that every $\xi\in H$ is already equal to $\pi(a)\eta$ for some $a\in A,\eta\in H$.