# Semi-commutative von Neumann algebras

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.

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 $$\mathcal{M}$$.
• $L^\infty(\Omega; \mathcal{M})$ 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. May 18, 2022 at 21:32
• 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) May 18, 2022 at 22:18
• 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 May 19, 2022 at 19:12