Strong closure and cyclic-separating duality

Fix $T\in R^{\mathrm{\prime }\mathrm{\prime }}$, an integer $n\ge 1$, and a tuple $\xi =({\xi }_1,\dots ,{\xi }_n)\in H^{\oplus n}$, and define $D_r=r^{\oplus n}$ for $r\in R$. Let $C=\overline{\{D_r\xi :r\in R\}}$. This is a closed linear subspace because $R$ is linear and $r\mapsto D_r\xi$ is linear. For each $r\in R$, $D_{r}C\subseteq C$: it holds on the orbit because $D_r{D}_s\xi =D_{rs}\xi$, and extends to its closure because $D_r$ is bounded. Applying this to $r^{\ast }\in R$ gives $D_r^{\ast }C\subseteq C$. If $\eta \in C^{\perp }$ and $\zeta \in C$, then [F5] gives $⟨D_r\eta ,\zeta ⟩=⟨\eta ,D_r^{\ast }\zeta ⟩=0$, so $D_r{C}^{\perp }\subseteq C^{\perp }$. The orthogonal projection $P_C$ from [F4] therefore commutes with every $D_r$.

Suppose $\mathrm{\Omega }$ is cyclic for $M$, and let $y\in M^{\mathrm{\prime }}$ satisfy $y\mathrm{\Omega }=0$. For every $x\in M$, commutation gives $yx\mathrm{\Omega }=xy\mathrm{\Omega }=0$. Thus $y$ vanishes on the dense subspace $M\mathrm{\Omega }$ and, being bounded, is zero on $H$. This proves that $\mathrm{\Omega }$ is separating for $M^{\mathrm{\prime }}$.

Suppose $\mathrm{\Omega }$ is separating for $M^{\mathrm{\prime }}$. Let $C=\overline{M\mathrm{\Omega }}$ and let $P_C$ be its orthogonal projection from [F4]. For $a\in M$, both $a$ and $a^{\ast }$ preserve $C$, because $M$ is an algebra closed under adjoints. If $\eta \in C^{\perp }$ and $\xi \in C$, then [F5] gives $⟨a\eta ,\xi ⟩=⟨\eta ,a^{\ast }\xi ⟩=0$; hence $a$ also preserves $C^{\perp }$. Thus $a{P}_C=P_{C}a$ for all $a\in M$, so $P_C\in M^{\mathrm{\prime }}$ by [F1]. Since $I\in M$, $\mathrm{\Omega }\in C$ and $(I-P_C)\mathrm{\Omega }=0$. Separatingness for $M^{\mathrm{\prime }}$ forces $I-P_C=0$, whence $C=H$ and $\mathrm{\Omega }$ is cyclic for $M$.

Write $P_C=(P_{ij}{)}_{i,j=1}^n$ in coordinate blocks. The identity $P_C{D}_r=D_r{P}_C$ says $P_{ij}r=r{P}_{ij}$ for all $r\in R$, so $P_{ij}\in R^{\mathrm{\prime }}$ by [F1]. For $T\in R^{\mathrm{\prime }\mathrm{\prime }}$, it follows that $T$ commutes with every $P_{ij}$; thus $D_T{P}_C=P_C{D}_T$. Since $I\in R$, $\xi =D_I\xi \in C$ and $P_C\xi =\xi$. Consequently $D_T\xi =P_C{D}_T\xi \in C$.

By the definition of $C$, for every $\epsilon >0$ there is $r\in R$ with $\mathrm{\parallel }{D}_T\xi -D_r\xi {\mathrm{\parallel }}_{H^{\oplus n}}<\epsilon$. The direct-sum norm is the square root of the sum of the squared coordinate norms, so $\mathrm{\parallel }T{\xi }_j-r{\xi }_j\mathrm{\parallel }<\epsilon$ for each $j$. These are exactly the finite strong-operator neighborhood tests. The empty tuple imposes no conditions. Therefore $T\in {\bar{R}}^{\mathrm{S}\mathrm{O}\mathrm{T}}$.

Let $W=W^{\ast }(R)={\bar{R}}^{\mathrm{W}\mathrm{O}\mathrm{T}}$. For each $x\in R^{\mathrm{\prime }}$, the set of operators commuting with $x$ is weak-operator closed and contains $R$; therefore it contains $W$. This proves $R^{\mathrm{\prime }}\subseteq W^{\mathrm{\prime }}$, while $R\subseteq W$ gives $W^{\mathrm{\prime }}\subseteq R^{\mathrm{\prime }}$. Hence $R^{\mathrm{\prime }\mathrm{\prime }}=W^{\mathrm{\prime }\mathrm{\prime }}=W$ by [F7]. Strong convergence implies weak convergence, so ${\bar{R}}^{\mathrm{S}\mathrm{O}\mathrm{T}}\subseteq W=R^{\mathrm{\prime }\mathrm{\prime }}$. Step 3.1 gives the reverse inclusion, proving the strong-closure identity.

Steps 1.2 and 1.3 prove both directions of cyclic-separating duality, and step 4.1 proves the strong-closure identity.