37.1 Topologies on $B(H)$
Let $H$ be a complex Hilbert space. Thanks to its self-duality, there is a natural dual pairing
Depending on the topology placed on $H \otimes H$, and the corresponding completion, a handful of different topologies arise on $B(H)$. In fact, the above duality produces a predual for $B(H)$, being the trace class operators:
Definition 37.1.1 (Ultraweak Topology).label Let $H$ be a complex Hilbert space, then the dual of $H \wh \otimes_{\pi} H$ is $B(H)$, and the $\sigma(B(H), H \wh \otimes_{\pi} H)$-topology is the ultraweak/$\sigma$-weak topology on $B(H)$.
Proof. By Proposition 13.11.2 and the Riesz Representation Theorem.$\square$
Seeing that $B(H)$ is a dual Banach space, the following fact is immediate:
Proposition 37.1.2.label Let $H$ be a complex Hilbert space, then every bounded subset of $B(H)$ is relatively compact with respect to the ultraweak topology.
Proof. By the Banach-Alaoglu Theorem.$\square$
Now, a few facts about the more familiar operator topologies:
Proposition 37.1.3.label Let $H$ be a complex Hilbert space, then:
- (1)
The dual of $B(H)$ with respect to its strong and weak operator topologies is $H \otimes H$.
- (2)
Every bounded subset of $B(H)$ is relatively compact in the weak operator topology.
- (3)
The composition map $(S, T) \mapsto ST$ is separately continuous in the strong and weak operator topologies.
- (4)
For any bounded subset $B \subset B(H)$, the composition map $(S, T) \mapsto ST$ restricted to $B \times B(H)$ is continuous in the strong operator topology.
- (5)
The adjoint map $T \mapsto T^{*}$ is continuous in the weak operator topology and the ultraweak topology.
Proof. (2): By Proposition 37.1.2.
(4): Let $\angles{S_\alpha}_{\alpha \in A}\subset B$, $\angles{T_\alpha}_{\alpha \in A}\subset B(H)$, and $(S, T) \in B \times B(H)$ such that $S_{\alpha} \to S$ and $T_{\alpha} \to T$ in the strong operator topology. Since $\{S_{\alpha}| \alpha \in A\}$ is equicontinuous, for any $x \in H$,
$\square$
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