I am familiar with the Morita theory of rings, and the hermitian Morita theory of rings with involution, and I am trying to understand some parallels and differences with the Morita theory of von Neumann algebras (or more generally C*-algebras, but let us start with the von Neumann case).

In the algebraic version, we are interested in the monoid structure of the Morita equivalence classes of $R$-algebras (where $R$ is a commutative ring), given by the tensor product over $R$. In particular, the invertible elements of this monoid are given by the Azumaya algebras over $R$, and they form the Brauer group of $R$.

For von Neumann algebras, is there a characterization of Morita-invertible algebras, that is algebras $M$ such that there exists $N$ with $M\otimes N$ Morita-equivalent to $\mathbb{C}$? Same question for real von Neumann algebras. Unless I'm mistaken such algebras should be factors, but is there a restriction on the type? If I understand correctly, type I algebras are already Morita-trivial, so I would be looking for restrictions in type II and III.

I'm not sure what the analytic equivalent of the sandwich map $A\otimes_R A^{op}\to \operatorname{End}_{R-mod}(A)$ should be, so I don't even have likely candidates for the would-be inverses. Maybe something like $M\otimes M^{op}\to L^2(M)$? If that makes sense.

  • $\begingroup$ This feels more appropriate on MathOverflow $\endgroup$ Dec 28, 2019 at 15:02
  • $\begingroup$ @Iwassuspendedfortalking Ok, I did wonder about that before asking, I'll wait to see if someone has an answer, and otherwise I'll ask again on MO. $\endgroup$ Dec 28, 2019 at 15:28
  • $\begingroup$ @CaptainLama: The tensor product of type II or III factors is again a type II or III factor. So only type I factors are invertible. $\endgroup$
    – Dmitri P.
    Dec 29, 2019 at 4:19
  • $\begingroup$ @DmitriPavlov I see, so there's nothing to see here, then, the only way to be invertible is to be trivial... $\endgroup$ Dec 29, 2019 at 16:51


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