Lemma 43.4.10.label Let $M$ be a von Neumann algebra, $\varphi \in S(M)$ be a state, and $(H, \pi, \xi)$ be the GNS triple associated with $\varphi$, then $\pi$ is normal if and only if $\varphi$ is normal. In which case, $\pi(M) \subset B(H)$ is a von Neumann algebra.
Proof. (1): Suppose that $\varphi$ is normal. Since $\pi(M)\xi$ is dense in $H$, the Arzelà-Ascoli Theorem implies that the weak operator topology on $\ol{B_{\pi(M)}(0, 1)}$ coincides with the topology of pointwise convergence on $\bracsn{\eta \otimes \theta|\eta, \theta \in \pi(M)\xi}$. For each $x, y, z \in M$,
By separate continuity of composition in the weak operator topology, the functional $x \mapsto \dpn{\pi(x)\pi(y)\xi, \pi(z)\xi}{H}$ is weak-operator continuous. As such, $\pi|_{\ol{B_M(0, 1)}}: \ol{B_M(0, 1)}\to \ol{B_{\pi(M)}(0, 1)}$ is weak-operator continuous.
(2): By Proposition 43.4.9.$\square$
Post a Comment