Remark 37.4.1.label Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a commutative $C^{*}$-subalgebra with $I \in A$, $B$ be the von Neumann algebra generated by $A$, $E: \cb_{\Omega(A)}\to B(H)$ be the spectral measure 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 II, there exists a decomposable measure space $\Omega$, corresponding to a number of copies of $\Omega(A)$, such that $\mathscr{E}$ is a quotient of its $L^{1}$ space, $\mathscr{E}^{*}$ is a subspace of its $L^{\infty}$ space, and $H$ is isomorphic to its $L^{2}$ space. The preceding isomorphisms are all linked by a unitary equvalence between $B$ acting on $H$, and $\mathscr{E}^{*}$ acting on the $l^{2}$ direct sum.
The complexity of $\Omega$, that is, the number of copies of $\Omega(A)$ that it contains, depends on two factors:
- (1)
The complexity of the von Neumann algebra $B$: If $B$ is sufficiently complex, then $\mathscr{E}$ cannot be expressed as the $L^{1}$ space of a single measure on $\Omega(A)$. Instead, multiple copies of $\Omega(A)$ are needed to handle mutually singular measures with overlapping supports. For more details on this phenomenon, see Theorem 24.7.2.
- (2)
The size of the Hilbert space $H$ in comparision with $B$: If $H$ is extremely large, then a large number of vectors are required for $B$ to cover it. As such, many copies of $\Omega(A)$ are required to handle the complexity of $H$.
More concretely, (1) manifests concretely as the size of the space $\mathscr{E}$, and (2) manifests as the size of the kernel of the mapping $L^{1}(\Omega) \to \mathscr{E}$.
By limiting these two sources of complexity, it is possible to remove the need of multiple copies of $\Omega(A)$. In particular,
- (1)
If $B$ admits a cyclic vector, then only one copy of $\Omega(A)$ is required for the construction in the Spectral Theorem [Theorem 23.1, Zhu93].
- (2)
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