RETRACTION: wrong word used.
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Bokuan Li
2026-09-24 21:31:39 -04:00
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The complexity of $\Omega$, that is, the number of copies of $\Omega(A)$ that it contains, depends on two factors:
\begin{enumerate}
\item 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 \autoref{theorem:hilbert-measures-dual}.
\item The size of the Hilbert space $H$ relative to $B$: 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 $H$.
\item The size of the Hilbert space $H$ relative to $B$: If $H$ is extremally 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 $H$.
\end{enumerate}
More concretely, (1) manifests as the size of the space $\mathscr{E}$, and (2) manifests as the size of the kernel of the mapping $L^1(\Omega) \to \mathscr{E}$.