37.4 Commutative von Neumann Algebras

Proof. ($\Leftarrow$): Let $B \supset A$ be a commutative von Neumann algebra, then $B \subset A' = A$.

($\Rightarrow$): For each $T \in (A')_{sa}$, the von Neumann algebra generated by $A$ and $T$ is commutative. As such, $T \in A$. As this holds for all $T \in (A')_{sa}$, $A' = (A')_{sa}+ i(A')_{sa}\subset A$.$\square$

Proof, [Proposition 22.1, Zhu93]. ($\Rightarrow$): Let $T \in A'$ with $Tx = 0$, then $TSx = STx = 0$ for all $S \in A$. In which case, $T(H) \subset \ol{T(Ax)}= \bracs{0}$ by Proposition 5.5.3.

($\Leftarrow$): Let $P \in B(H)$ be the orthogonal projection from $H$ onto $\ol{Ax}$, then as $I \in A$, $x \in \ol{Ax}$. Since $\ol{Ax}$ is a reducing subspace for $A$, $P \in A'$. Thus $I, P \in A'$ and $(I - P)x = 0$. Given that $x$ is separating for $A'$, $I = P$, so $\ol{Ax}= H$.$\square$

Proof. Since $A$ is commutative, $A \subset A'$. As $x$ is separating for $A'$ by Proposition 37.4.3, it is also separating for $A$.$\square$

Proof, [Theorem 22.3, Zhu93]. Let $\seqj{x}\subset H$ be a maximal collection of non-zero vectors such that the spaces $\bracsn{Ax_j|j \in I}$ are mutually orthogonal. Such a collection exists by Zorn’s lemma, and must be at most countable given that $H$ is separable.

Let $\seq{x_n}$ be an enumeration of such a set, padding by zeroes if necessary, and $x = \sum_{n \in \natp}x_{n}/2^{n}$. For any $T \in A$, if $Tx = 0$, then as $\bracsn{Ax_n|n \in \natp}$ are mutually orthogonal, $Tx_{n} = 0$ for all $n \in \natp$. Since $A$ is commutative, $Ax_{n} \subset \ker(T)$ for all $n \in \natp$. By maximality of $\seq{x_n}$, $H = [l^{2}(\natp); \ol{Ax_n}]$, $H \subset \ker(T)$, and $T = 0$. Therefore $x$ is a separating vector.$\square$

Proof, [Corollary 22.4, Zhu93]. By Proposition 37.4.2, $A = A'$. By Theorem 37.4.5, $A$ admits a separating vector. By Proposition 37.4.3, this separating vector for $A$ is a cyclic vector for $A' = A$.$\square$