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.$\square$
A natural topology consistent with the ultraweak topology would be the ultrastrong topology.
Definition 37.1.2 (Ultrastrong Topology).label Let $H$ be a complex Hilbert space. For each $x = \seq{x_n}\in L^{2}(\natp; H)$, let
then the ultrastrong/$\sigma$-strong topology on $B(H)$ is the topology generated by the maps $\bracsn{\Phi_x|x \in L^2(\natp; H)}$.
Seeing that $B(H)$ is a dual Banach space, the following fact is immediate:
Proposition 37.1.3.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 37.1.4.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)
The composition map $(S, T) \mapsto ST$ is left-hypocontinuous with respect to the strong operator topology and strong-operator bounded subsets of $B(H)$.
Proof. (2): By Proposition 37.1.3.
(4): By the Banach-Steinhaus Theorem, every strong-operator bounded subset of $B(H)$ is equicontinuous.$\square$
Proposition 37.1.5.label Let $H$ be a complex Hilbert space, then:
- (1)
The adjoint map $T \mapsto T^{*}$ is continuous in the weak operator topology and the ultraweak topology.
- (2)
$T \mapsto T^{*}$ restricted to the normal operators is continuous in the strong operator topology.
- (3)
For any $f \in C(\complex; \complex)$, the mapping $T \mapsto f(T)$ restricted to any bounded set of normal operators is continuous in the strong operator topology.
Proof, [Section 19.1, Zhu93]. (2): Let $S, T \in B(H)$, then
Now, if $S$ and $T$ are normal, then $\normn{S^*x}_{H} = \norm{Sx}_{H}$ and $\norm{T^*x}_{H} = \norm{Tx}_{H}$, so
Therefore
and the adjoint map restricted to normal operators is continuous in the strong operator topology.
(3): Let $S, T \in B_{B(H)}(0, 1)$ and $x \in H$ and $n \in \natp$, then
so the mapping $T \mapsto T^{n}$ on $B_{B(H)}(0, 1)$ is continuous in the strong operator topology. By (2), the mapping $T \mapsto p(T, T^{*})$ is strong-operator continuous for all $p \in \complex[z, \ol z]$.
By the Stone-Weierstrass Theorem, there exist polynomials $p_{n} \in \complex[z, \ol z]$ such that $p_{n} \to f$ uniformly on $\ol{B_\complex(0, 1)}$. For any $T \in B_{B(H)}(0, 1)$, $x \in H$, and $n \in \natp$,
by the continuous functional calculus. Thus $f$ is a uniform limit of strong-operator continuous functions on $B_{B(H)}(0, 1)$, and as such also strong-operator continuous by Proposition 7.3.2.$\square$
Proposition 37.1.6.label Let $A$ be a $C^{*}$-algebra, $H_{1}, H_{2}$ be complex Hilbert spaces, $\pi_{1}: A \to B(H_{1})$ and $\pi_{2}: A \to B(H_{2})$ be injective representations of $A$, and $U: H_{1} \to H_{2}$ be a unitary equivalence, then the mapping
is strong-operator and weak-operator continuous.
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