The following problem appears to be an easy exercise on von Neumann algebra tensor products, but since I've been failing to find a rigorous proof, I'd like to make sure it's not that trivial. Suppose $M$ and $N$ are von Neuamann factors such that $L^\infty[0,1] \mathbin{\bar\otimes} M \cong L^\infty[0,1] \mathbin{\bar\otimes} N$. Does it follow $M \cong N$? It's true under the assumption of separability by the disintegration theory, but not so sure if $M$ (and hence $N$) does not have separable predual.
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asked May 25, 2017 at 10:32
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$\begingroup$ I'm late here, but I'm curious as to how the question is resolved even in the separable case. All the treatments I could find on direct integral theory always needs a spatial isomorphism between $L^\infty(0, 1) \bar{\otimes} M$ and $L^\infty(0, 1) \bar{\otimes} N$. More precisely, suppose $M$ acts on $H$ and $N$ acts on $K$, it seems to always require a unitary between $L^2(0, 1) \otimes H$ and $L^2(0, 1) \otimes K$ that conjugates $L^\infty(0, 1) \bar{\otimes} M$ and $L^\infty(0, 1) \bar{\otimes} N$ onto each other. $\endgroup$– David GaoCommented May 31, 2023 at 4:18
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$\begingroup$ This is obviously achievable if $M$ and $N$ are $\mathrm{II}_1$ factors, with $H$ and $K$ being their standard representations. In which case $L^2(0, 1) \otimes H$ and $L^2(0, 1) \otimes K$ are standard representations of $L^\infty(0, 1) \bar{\otimes} M$ and $L^\infty(0, 1) \bar{\otimes} N$, respectively, so a spatial isomorphism of the above form just comes from any isomorphism between $L^\infty(0, 1) \bar{\otimes} M$ and $L^\infty(0, 1) \bar{\otimes} N$. But without trace can this still be done? $\endgroup$– David GaoCommented May 31, 2023 at 4:22
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$\begingroup$ Obviously when $M$ or $N$ are Type $\mathrm{I}$ it can done in a different way and the result is well-known, but I'm unsure how to proceed if the factors in question are Type $\mathrm{II}_\infty$ or Type $\mathrm{III}$, even assuming separability. $\endgroup$– David GaoCommented May 31, 2023 at 4:23
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