Let $\mathcal M$ and $\mathcal N$ be two von Neumann algebras. A linear map $J:\mathcal M\to\mathcal N$ is said to be a Jordan isomorphism if $J$ is bijective, $*$-preserving and $J(xy+yx)=J(x)J(y)+J(y)J(x)$ for all $x,y \in \mathcal M.$

Is there any nice classification of type I von Neumann algebras up to Jordan isomorphism? Also is there is any classification of type I factors up to Jordan isomorphism?

  • $\begingroup$ I saw in the literature two notion of Jordan isomorphism: one is the condition $J(xy+yx)=J(x)J(y)+J(y)J(x)$. The second is the condition $J(x^2)=J(x)^2$ (which is equivalent to the other one by polarization, as soon as 2 is invertible). This doesn't fit your definition, which forces $J(xy-yx)$ to vanish and hence is excluded if $\mathcal{M}$ is non-commutative. You seem to confuse with the trace condition. $\endgroup$ – YCor Apr 30 '20 at 10:35
  • $\begingroup$ @ YCor!Sorry! Edited! $\endgroup$ – A beginner mathmatician Apr 30 '20 at 11:48

I have not read through all the details carefully, but I think your questions can probably be answered using the results in Kadison's 1951 paper "On isometries of operator algebras". That paper actually has a discussion of Jordan $\ast$-isomorphisms between ${\rm C}^\ast$-algebras (confusingly, Kadison calls these "${\rm C}^\ast$-isomorphisms") and in particular:

Theorem 10 (with terminology updated) a Jordan $\ast$-isomorphism from a von Neumann algebra onto a ${\rm C}^\ast$-algebra is the direct sum of a $\ast$-isomorphism and a $\ast$-anti-isomorphism.

As for your question about the factor case, this is handled by

Corollary 11: any Jordan $\ast$-isomorphism of a factor is either a $\ast$-isomorphism or a $\ast$-anti-isomorphism.

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    $\begingroup$ Yes. And because every type I von Neumann algebra is $*$-isomorphic to its opposite algebra, this means that Jordan isomorphic $\Rightarrow$ $*$-isomorphic, $\endgroup$ – Nik Weaver Apr 30 '20 at 15:03
  • $\begingroup$ @ Ni and Choi. Actually, I was looking for a more concrete kind of classification. For example can it be seen to be a von Neumann algebra by gluing concretely known type von Neumann algebras e.g. $B(\mathcal H)$ ($\mathcal H$ is a Hilbert space) etc.? $\endgroup$ – A beginner mathmatician Apr 30 '20 at 15:31
  • $\begingroup$ I can't make any sense of this comment. $\endgroup$ – Nik Weaver Apr 30 '20 at 15:36
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    $\begingroup$ Are you just asking for a classification of type I von Neumann algebras? That is achieved through direct integral decomposition. $\endgroup$ – Nik Weaver Apr 30 '20 at 15:37
  • $\begingroup$ Alright! I now understand. $\endgroup$ – A beginner mathmatician Apr 30 '20 at 17:22

I believe two type I von neumann algebras are Jordan isomorphic if and only if they are $*$-isomorphic. Jordan isomorphism preserve order, so if two von Neumann algebras are Jordan isomorphic they are order isomorphic and hence have the same normal state space. And one should be able to recover any type I von Neumann algebra from its normal state space. (I don't have the reference handy, but this is probably in State Spaces of Operator Algebras by Alfsen and Schulz.)

BTW, this will not be true outside the type I setting. There is a von Neumann algebra (even a factor) that is not $*$-isomorphic to its opposite algebra. But the two are Jordan isomorphic via the identity map.


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