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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.

Doug McLellan
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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.

Martin Argerami
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  • But in this case $\pi$ is not a $\ast$-isomorphism since it is not surjective. – Adrián González Pérez Nov 29 '18 at 12:17
  • 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. – Adrián González Pérez Nov 29 '18 at 12:18
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    @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. – Martin Argerami Nov 29 '18 at 15:09
  • Thanks! I didn't realize that. – Adrián González Pérez Nov 29 '18 at 15:25
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    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). – Martin Argerami Nov 29 '18 at 15:33
  • Thanks for this & also for answer to an older question, which just led me to 7.1.16 of Kadison/Ringrose: "$$ isomorphisms between von Neumann algebras are weak- and strong-operator homeomorphisms on bounded sets*," which is the main result I was looking for. I am probably missing some key knowledge to understand your counterexample though. If $\pi(M)$ is a vN algebra that's dense in $B(H)$, musn't it be all of $B(H)$, since vN algebras are sot-closed? – Doug McLellan Nov 29 '18 at 19:28
  • The algebra $\pi(M)$ is C$^$-isomorphic to a von Neumann algebra. The usual definition of von Neumann algebra is not intrinsic (as my example shows), but depends on the environment. But yes, if $M\subset B(H)$ and $N\subset B(K)$ are von Neumann algebras (in the usual "double commutant" sense) then a $$-isomorphism $\pi:M\to N$ is sot continuous by passing through normality. – Martin Argerami Nov 29 '18 at 21:15
  • I know this is an old answer, but I just happened to stumble over it. Even if $M$ and $N$ are concrete von Neumann algebras, it is not true that an $\ast$-isomorphism is SOT-continuous. For example, the SOT on $B(H)\subset B(S_2(H))$ equalso the $\sigma$-SOT (as for any von Neumann algebra in standard form), but SOT and $\sigma$-SOT are different on $B(H)\subset B(H)$. What is true in this setting is that $\ast$-isomorphisms (or more generally normal cp maps) are always SOT-continuous on norm bounded sets. – MaoWao Nov 19 '21 at 11:23
  • Thanks for noticing! Edited. – Martin Argerami Nov 19 '21 at 12:40