Von Neumann algebras and commutants

Assume AC. Let $H$ be a complex Hilbert space and let $\mathcal{B}(H)$ denote its bounded linear operators. Applying AC to a family indexed by the natural numbers supplies Countable Choice, the hypothesis of Hilbert-adjoint identities, which provides adjoints on $\mathcal{B}(H)$. A concrete von Neumann algebra on $H$ is a unital $\ast$-subalgebra $\mathcal{A}\subseteq \mathcal{B}(H)$ that is closed in the weak operator topology (WOT). Here unital means $I_H\in \mathcal{A}$; the zero Hilbert space is allowed, with $I_H=0$ and sole unital algebra $\{0\}$. This is the only use of AC in this definition.

For any set $\mathcal{S}\subseteq \mathcal{B}(H)$, its commutant and double commutant are ${\mathcal{S}}^{\mathrm{\prime }}:=\{T\in \mathcal{B}(H):TS=ST\text{ for every }S\in \mathcal{S}\},{\mathcal{S}}^{\mathrm{\prime }\mathrm{\prime }}:=({\mathcal{S}}^{\mathrm{\prime }}{)}^{\mathrm{\prime }}\mathrm{.}$ Commutants are taken inside $\mathcal{B}(H)$. The von Neumann algebra generated by $\mathcal{S}$ is $W^{\ast }(\mathcal{S}):={\overline{{\mathrm{Alg}}^{\ast }(\mathcal{S}\cup \{I_H\})}}^{\mathrm{W}\mathrm{O}\mathrm{T}},$ the WOT closure in $\mathcal{B}(H)$ of the unital $\ast$-algebra generated by $\mathcal{S}$. No bicommutant theorem is part of these definitions.

Two elementary properties will be used. For fixed $S\in \mathcal{B}(H)$, $⟨TS\xi ,\eta ⟩=⟨T(S\xi ),\eta ⟩,⟨ST\xi ,\eta ⟩=⟨T\xi ,S^{\ast }\eta ⟩$ are WOT-continuous matrix coefficients as functions of $T$. Thus each equation $TS=ST$ defines a WOT-closed set, and so ${\mathcal{S}}^{\mathrm{\prime }}$ is WOT-closed. If $\mathcal{S}$ is self-adjoint, then ${\mathcal{S}}^{\mathrm{\prime }}$ is a unital $\ast$-subalgebra: products preserve commutation with every $S$, and taking adjoints of $TS=ST$ gives $S{T}^{\ast }=T^{\ast }S$. These facts do not identify $W^{\ast }(\mathcal{S})$ with ${\mathcal{S}}^{\mathrm{\prime }\mathrm{\prime }}$.