Proposition 45.1.4.label Let $M$ be a unital $C^{*}$-algebra, $\tau: M \to \complex$ be a faithful tracial state, and $(L^{2}(M, \tau), \pi, 1_{M})$ be its GNS triple, then:
- (1)
The uniform structure of pointwise convergence[1] on $\ol{B_M(0, 1)}\subset B(L^{2}(M, \tau))$ coincides with the uniform structure of $\ol{B_M(0, 1)}\subset L^{2}(M, \tau)$ induced by the $L^{2}(M, \tau)$ norm.
- (2)
$\pi(M) \subset B(L^{2}(M, \tau))$ is a von Neumann algebra if and only if $\ol{B_M(0, 1)}$ is complete in $L^{2}(M, \tau)$. In which case, $\tau$ is normal.
Proof, [Proposition 2.6.4, AP]. (1): For any $x, y \in \ol{B_M(0, 1)}$,
so the uniform structure of pointwise convergence is stronger than the uniformity induced by the $L^{2}(M, \tau)$ norm.
On the other hand, by the Arzelà-Ascoli Theorem, the uniform structure of pointwise convergence on $\ol{B_M(0, 1)}$ coincides with the uniform structure of pointwise convergence on any dense subset of $L^{2}(M, \tau)$. In particular, this includes $M \subset L^{2}(M, \tau)$. For each $x, y \in \ol{B_M(0, 1)}$ and $t \in M$, since $\tau$ is tracial,
so the uniformity induced by the $L^{2}(M, \tau)$ norm is stronger than the uniform structure of pointwise convergence.
(2): Using (1), it is sufficient to show that $\ol{B_M(0, 1)}\subset B(L^{2}(M, \tau))$ is complete if and only if it is strong-operator closed. By Proposition 6.7.3, $\ol{B_M(0, 1)}$ is complete if and only if it is closed in $L^{2}(M, \tau)^{L^2(M, \tau)}$ with respect to the product topology. Finally, the Arzelà-Ascoli Theorem implies that $\ol{B_M(0, 1)}$ is closed in $L^{2}(M, \tau)^{L^2(M, \tau)}$ if and only if it is strong-operator closed in $B(L^{2}(M, \tau))$.$\square$
- Strong operator topology.keyboard_return
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