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A *-algebra of operators on a given Hilbert space is a von Neumann algebra if it's equal to its double-commutant. That's a nice purely algebraic way of characterizing von Neumann algebras on a given Hilbert space.

Now suppose that $M_1$ and $M_2$ are two von Neumann algebras on the same separable Hilbert space ${\cal H}$ over $\mathbb{C}$. Is there a purely algebraic way of expressing the condition that $M_1$ and $M_2$ are isomorphic to each other as von Neumann algebras, without explicitly referring to topology?

Here are examples of algebraic conditions that don't work (as far as I know), but they illustrate what I mean by "algebraic":

  • Suppose that $M_1=U^{-1}M_2 U$ for some unitary operator $U$ on ${\cal H}$. That's an algebraic condition, but it's not general enough, because two von Neumann algebras can be isomorphic to each other without being unitarily equivalent to each other.

  • Suppose that the two von Neumann algebras $M_1$ and $M_2$ are isomorphic as *-algebras. That's an algebraic condition, but I doubt that it's specific enough, because I don't see any reason why *-isomorphism would imply isomorphism as von Neumann algebras.

Clarification: A comment pointed out that there are two notions of isomorphisms of von Neumann algebras: spatial and abstract. I am interested in abstract isomorphisms.

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    $\begingroup$ There are two notions of isomorphisms of von Neumann algebras: spatial isomorphisms (your first bullet point) and abstract ones (your second bullet point). If neither of these is what you want, you should be say what it is instead. $\endgroup$
    – MaoWao
    Jul 8, 2020 at 19:52
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    $\begingroup$ In any case, $\ast$-isomorphisms between von Neumann algebras are necessarily isometric (preserve the norm) and normal (preserve the weak$^\ast$ topology). Maybe that was the piece you were missing when you discarded them in your second bullet point. $\endgroup$
    – MaoWao
    Jul 8, 2020 at 19:53
  • $\begingroup$ @MaoWao I edited the question to clarify that I'm interested in abstract isomorphism, and I think your second comment answers my question. I didn't realize that *-isomorphisms automatically preserved those other properties. I'll look into that and try to understand why it's true. Thank you! $\endgroup$ Jul 8, 2020 at 21:16

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Based on hints in MaoWao's comment, I found https://math.vanderbilt.edu/peters10/teaching/spring2013, which includes the notes vonNeumannAlgebras.pdf. Corollary 1.3.2 in those notes confirms that a *-isomorphism between two von Neumann algebras is automatically isometric, and corollary 4.2.4 confirms that it automatically preserves the relevant topology. (See the top of page 32 for the definition of a key term used in the latter corollary.)

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