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Do all $*$-isomorphisms between von Neumann algebras preserve the strong operator topology?

Seems clearly true for $*$-isomorphisms with a unitary implementation, but I don't see the answer for other cases ... perhaps there is an easy argument from the fact that von Neumann algebras are closed in this topology, but I've spent a while looking for one and don't see it.

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No. Take $M$ to be any II$_1$-factor. Let $\pi:M\to B(H)$ be an irreducible representation (it exists because you can do GNS of a pure state). As $M$ is simple (as a C$^*$-algebra!), $\pi$ is injective. And $\pi(M)$ is dense in $B(H)$, but it cannot be everything.

So $\pi:M\to\pi(M)$ is a $*$-isomorphism that does not preserve the sot/wot/ultrasot/ultrawot topologies.

As mentioned in the comments, if $M\subset B(H)$ and $N\subset B(K)$ are von Neumann algebras (in the usual "double commutant" sense) then a $*$-isomorphism $\pi:M\to N$ is sot-continuous on bounded sets by passing through normality.

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  • $\begingroup$ But in this case $\pi$ is not a $\ast$-isomorphism since it is not surjective. $\endgroup$
    – Adrián González Pérez
    Nov 29, 2018 at 12:17
  • $\begingroup$ I am also not convinced since $\ast$-homomorphism must preserve the suprema of ascending families of projections and that will imply normality for $\pi$. Perhaps I am confused about that. $\endgroup$
    – Adrián González Pérez
    Nov 29, 2018 at 12:18
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    $\begingroup$ @Adrián: of course, I can take $\pi(M)$ to be the codomain. And you seem to have a misunderstanding of the equivalence normal $\iff$ sot-continuous (and you need selfadjoints, not just projections). The equivalence is true, if you are doing it in a von Neumann algebra. Unless you claim that every monotone-complete C$^*$-algebra is a von Neumann algebra. $\endgroup$
    – Martin Argerami
    Nov 29, 2018 at 15:09
  • $\begingroup$ Thanks! I didn't realize that. $\endgroup$
    – Adrián González Pérez
    Nov 29, 2018 at 15:25
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    $\begingroup$ No problem. I have mixed feelings about the fact that von Neumann algebras are always considered represented. On the one hand, it makes a lot of sense because it is how you would usually use them. But, on the other hand, because this is usually not really considered in textbooks, together with AW$^*$-algebras becoming unfashionable a few decades ago, there is little knowledge about all this (including me). $\endgroup$
    – Martin Argerami
    Nov 29, 2018 at 15:33

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