# Morita-invertible von Neumann algebras

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.

• This feels more appropriate on MathOverflow Dec 28, 2019 at 15:02
• @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. Dec 28, 2019 at 15:28
• @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. Dec 29, 2019 at 4:19
• @DmitriPavlov I see, so there's nothing to see here, then, the only way to be invertible is to be trivial... Dec 29, 2019 at 16:51