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Suppose $\Omega$ is a $\sigma$-finite measure space with measure $\mu.$ Let $\mathcal M\subseteq B(H)$ be a von Neumann algebra.

  1. Can an element of $L_\infty(\Omega)\overline{\otimes}\mathcal M$ be regarded as a measurable map from $\Omega$ to $\mathcal M$? A reference where the properties of $L_\infty(\Omega)\overline{\otimes}\mathcal M$ have been studied will be appreciated also.

  2. Is it true that there exists a canonical injective $*$-homomorphism $\pi:L_\infty(\Omega;\mathcal M)\to L_\infty(\Omega)\overline{\otimes}\mathcal M$ such that $\pi(L_\infty(\Omega;\mathcal M)) $ is dense in $L_\infty(\Omega)\overline{\otimes}\mathcal M$ in weak operator topology? In above $L_\infty(\Omega;\mathcal M)$ is all strongly measurable essentially bounded $\mathcal M$-valued functions.

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Assuming $H$ is separable, $L^\infty(\Omega)\overline{\otimes} \mathcal{M}$ can be identified with the essentially bounded weakly measurable functions from $\Omega$ into $\mathcal{M}$. (Weakly measurable = its composition with any normal state on $\mathcal{M}$ is measurable.) This is a minor variation on Theorem 6.5.8 of my book Mathematical Quantization. Here the $\mathcal{M}_x$ appearing in that theorem aren't factors, but they all equal $\mathcal{M}$.

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    $\begingroup$ Separable predual is the nice condition you want. Actually, for your question the measurability issues aren't so bad, so maybe you don't even need this hypothesis. $\endgroup$
    – Nik Weaver
    May 17 at 19:51
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    $\begingroup$ $L^\infty(\Omega; \mathcal{M})$ and $L^\infty(\Omega)\overline{\otimes}\mathcal{M}$ are the weak* closures of $L^\infty(\Omega)\otimes \mathcal{M}$ in $B(L^2(\Omega; H))$ and $B(L^2(\Omega)\otimes H)$, respectively. You see how to identify the two Hilbert spaces; now just observe that this identification respects the copies of $L^\infty(\Omega)\otimes \mathcal{M}$ in the two algebras. $\endgroup$
    – Nik Weaver
    May 18 at 21:32
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    $\begingroup$ Hi Nik, it seems the OP is using the definition from Banach-space world where $L^\infty(\Omega; E)$ means the essentially bounded strongly-measurable E-valued functions on $\Omega$. I think that for E an infinite-dimensional von Neumann algebra this space is going to be smaller than the spatial tensor product $L^\infty(\Omega)\overline{\otimes} {\mathcal M}$ (issues with the Radon-Nikodym property) $\endgroup$
    – Yemon Choi
    May 18 at 22:18
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    $\begingroup$ Oh you're right, I didn't notice that before. (Maybe added in an edit?) My answer applies to the set of weakly measurable functions. $\endgroup$
    – Nik Weaver
    May 18 at 22:41
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    $\begingroup$ That sounds like several doubts ... $f: \Omega \to \mathcal{M}$ is "essentially bounded" if the function $t \mapsto \|f(t)\|$ is bounded off of a null set. Every essentially bounded weakly measurable map operates in a natural way on $L^2(\Omega; H)$ and you can take $L^\infty(\Omega; \mathcal{M})$ to be this set of operators. Yes, this identifies maps which agree almost everywhere $\endgroup$
    – Nik Weaver
    May 19 at 19:12

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