Corollary 37.5.6 (Representation of Commutative von Neumann Algebras).label Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a commutative von Neumann algebra with $I \in A$, then:
- (1)
There exists a LCH space $\Omega$ and a decomposable Radon measure $\mu$ on $\Omega$ such that $A$ is *-isomorphic to $L^{\infty}(\mu; \complex)$.
- (2)
If $A$ admits a cyclic vector, then $\Omega$ may be taken to be compact.
- (3)
If $H$ is separable, then $\Omega$ may be taken to be compact.
Proof. Let $E$ be the spectral measure on $\Omega(A)$ associated with $A$, and $\mathscr{E}\subset M_{R}(\Omega(A); \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$. By Spectral Theorem I, $A$ is *-isomorphic to $\mathscr{E}^{*}$.
(1): By (3) of Lemma 37.5.2 and Theorem 24.7.2, $A$ is *-isomorphic to $[l^{\infty}(I); L^{\infty}(\mu_{i}; \complex)]$, where $\seqi{\mu}\subset \mathscr{E}$ is a maximal mutually singular family. Let $\Omega = \bigsqcup_{i \in I}\Omega(A)$, then $\Omega$ is a LCH space. For each $i \in I$, let $\Omega_{i}$ denote the $i$-th copy of $\Omega(A)$, then
is the desired decomposable Radon measure.
(2): By Spectral Theorem II, there exists a single positive Radon measure $\mu \in \mathscr{E}$ on $\Omega(A)$ such that $\mathscr{E}$ is absolutely continuous with respect to it. Therefore the index set in (1) can be taken to be a singleton.
(3): If $H$ is separable, then so is $\mathscr{E}$. As such, there exists a single positive Radon measure $\mu \in \mathscr{E}$ on $\Omega(A)$ such that $\mathscr{E}$ is absolutely continuous with respect to it. Therefore the index set in (1) can be taken to be a singleton.$\square$
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