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In general what is it meant to say two Von Neumann Algebras are isomorphic? Is it an injective and surjective *-homomorphism which is also strongly continuous or one which is also normal? I would suppose the correct definition would be normal as that is an intrinsic property.

For example when it is said that any type I factor is isomorphic to $B(H)$ what exactly are the properties of the map?

sirjoe
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  • What does it mean for a map from one VN algebra to another to be "normal"? – David C. Ullrich Jun 17 '19 at 22:24
  • continuous with respect to the ultraweak topology.(i.e. the w* topology coming from it's canonical predual) – sirjoe Jun 17 '19 at 22:29
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    This does not answer the question, but might be helpful: any unit-preserving bijective *-homomorphism is automatically normal as it is an order isomorphism. – westerbaan Jun 18 '19 at 09:16
  • @westerbaan would you need unit-preserving wouldnt it necessarily be unital if it is a bijective -homomorphism between untial C-algebras? – sirjoe Jun 18 '19 at 10:16
  • Yes, that's right. Any bijective *-hom between unital C*-algs is automatically unital. – westerbaan Jun 18 '19 at 14:04

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As noted by westerbaan in the comments, an algebraic *-isomorphism between von Neumann algebras $M$ and $N$ is automatically normal. This is the typical definition for an isomorphism of von Neumann algebras, and it's the type of morphism used in the classification of Type I factors you mentioned.

You can find more info in Blackadar's Operator Algebras book here (see Section III.2.2).

But the most relevant info is that an algebraic *-isomorphism $\psi$ is always a homeomorphism for the $\sigma$-weak (i.e. ultraweak) and $\sigma$-strong topologies. If the commutant of $M$ is properly infinite, then the weak and $\sigma$-weak (resp., strong and $\sigma$-strong) topologies coincide on $M$. So if both $M'$ and $N'$ are properly infinite, then $\psi$ is obviously also a homeomorphism for the weak or strong topologies. But $\psi$ won't be strongly or weakly continuous in general (i.e. without the assumption about the commutants).

That said, $\psi$ will be a homeomorphism for the strong or weak topologies if you restrict $\psi$ to the unit balls of $M$ and $N$, no matter what their commutants are. So $\psi: B_1(M) \rightarrow B_1(N)$ is always a homeomorphism for any of the topologies we've talked about.

Josh Keneda
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