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Given a von Neumann algebra $M$, let $$ S(M) = \{u\in M: uu^*u=u\} $$ be the set of partial isometries in $M$. Given $u,v\in S(M)$, it is well known that $uv \in S(M)$, provided $u^*u$ commutes with $vv^*$.

Now suppose that $N$ is another von Neumann algebra and that $$ \varphi :S(M)\to S(N) $$ is a bijective function such that:

  1. for every $u,v\in S(M)$, such that $u^*u$ commutes with $vv^*$, one has that $\varphi (u)^*\varphi (u)$ commutes with $\varphi (v)\varphi (v^*)$, and $\varphi (uv)=\varphi (u)\varphi (v)$.

  2. $\varphi ^{-1}$ satisfies (1).

Does it follow that $\varphi $ extends to an isomorphism from $M$ to $N$?

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No, consider the map $\phi: u \mapsto u^*$ from the partial isometries in $M$ to the partial isometries in its opposite algebra $M^{op}$. One easily checks that it has the desired properties, but it cannot extend to a linear map because already e.g. $\phi(iI) = -iI \neq iI = i\phi(I)$.

If $M$ is not isomorphic to its opposite algebra, then we have an example where one has a map of the type described between non-isomorphic von Neumann algebras.

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    $\begingroup$ Fair enough! Let me think about it. $\endgroup$
    – Nik Weaver
    Commented Nov 8, 2020 at 18:18
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    $\begingroup$ Thanks Nik. I'll accept your answer but I suppose you will not be surprised if I follow up with the additional hypothesis $\phi(\lambda I)=\lambda I$. Another sensible hypothesis one might try to use is $\phi(\sum u_i) =\sum\phi( u_i)$ for a countable set of partial isometries with pairwise orthogonal initial and final spaces. $\endgroup$
    – Ruy
    Commented Nov 8, 2020 at 18:23
  • $\begingroup$ How about using Dye'a theorem on projection lattices? Restrict $\phi$ to the projections and infer that there is a $*$-isomorphism implementing this restriction. Then we just have to show that $\phi$ is determined by its restriction to projections. $\endgroup$
    – Nik Weaver
    Commented Nov 9, 2020 at 1:23
  • $\begingroup$ But that should be easy: we only need to show that if $\phi$ is the identity on projections then it is the identity on all partial isometries. For any $p \leq u^*u$ and $q \leq uu^*$ we have $\phi(puq) = p\phi(u)q$ and that should be enough to get $\phi(u) = u$. $\endgroup$
    – Nik Weaver
    Commented Nov 9, 2020 at 1:26
  • $\begingroup$ This is just a sketch and it needs something more if $M$ has an $M_2$ direct summand (since Dye's theorem fails then). $\endgroup$
    – Nik Weaver
    Commented Nov 9, 2020 at 1:28

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