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It seems that most people are talking about $*$-isomorphisms of von Neumann algebras. However, I can not find any references for the Banach isomorphisms, i.e., let $A,B$ be two different von Neumann algebras (consider them as Banach spaces equipped with uniform norms), are there any bijective mapping from $A$ onto $B$? In particular, I would like to consider the case when $A$ is a finite von Neumann algebra and $B=B(H)$.

I have posted this question in stackexchange but I think it fits mathoverflow more. https://math.stackexchange.com/questions/4717820/banach-isomorphisms-between-von-neumann-algebras

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    $\begingroup$ To clarify: is the question whether or not there is a bounded linear bijective function $f:A\to B$? $\endgroup$
    – Ben Johnsrude
    Commented Jun 13, 2023 at 6:52
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    $\begingroup$ All non-subhomogeneous injective von Neumann algebras with separable predual (including the hyperfinite II_1 factor and $B(\ell_2)$) are Banach isomorphic by Pelczynski's decomposition method. On the other hand, I'm not sure if there is any non-injective von Neumann algebra that is Banach isomorphic to $B(\ell_2)$. See Rosenthal's survey for a relevant literature. ams.org/books/conm/321/5650/conm321-5650.pdf $\endgroup$
    – Narutaka OZAWA
    Commented Jun 13, 2023 at 7:27
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    $\begingroup$ @NarutakaOZAWA The paper arxiv.org/abs/2010.13743 of Pisier says that certain non-injective vN algebras are isomorphic as Banach spaces to $B(\ell_2)$, via some non-cb complementation maps and the usual Pelczynski decomposition machine $\endgroup$
    – Yemon Choi
    Commented Jun 21, 2023 at 4:37
  • $\begingroup$ @Yemon Choi: Thanks! I should have remembered it. $\endgroup$
    – Narutaka OZAWA
    Commented Jun 21, 2023 at 6:49
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    $\begingroup$ Not exactly the question itself but related: Theorem 7 in doi.org/10.2307/1969534 implies that if two $C^*$-algebras are isometrically isomorphic as Banach spaces, then they are Jordan $*$-isomorphic. In particular, for von Neumann factors this implies that they are either $*$-isomorphic or $*$-anti-isomorphic. So if you are asking about isometric isomorphisms, then no, $B(l^2)$ is not isometrically isomorphic to any finite vNa as Banach spaces. $\endgroup$
    – David Gao
    Commented Aug 18, 2023 at 12:04

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