# Morita-invertible C*-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 C*-algebras.

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 $$R$$. Do similar phenomena occur for C*-algebras?

For von Neumann algebras, I asked a question on Math.SE, and someone commented that the theory is basically empty: since factors of type I are Morita-trivial, and a tensor product of a factor of type II or III with another factor is again of type II or III, in the end the only way to be Morita-invertible is to be Morita-trivial.

Do C*-algebras offer more theory? I understand that in that context one has to be a little more careful: the Morita-equivalence I care about is the strong Morita equivalence (as defined by imprimitivity bimodules). Also, talking about tensor products can be awkward, so maybe I should restrict to nuclear C*-algebras (but if there are results of Morita-invertibility for well-chosen tensor products on non-nuclear algebras, I am also interested in hearing about it).

Clearly, one has to restrict to unital and central algebras, so in the end my question is the following:

If $$A$$ is a central unital C*-algebra (maybe nuclear), and there exists $$B$$ such that $$A\otimes B$$ is strongly Morita-equivalent to $$\mathbb{C}$$ (for some tensor product if $$A$$ is not nuclear), does it follow that $$A$$ itself is strongly Morita-equivalent to $$\mathbb{C}$$?

I am also interested in similar results for real C*-algebras (maybe even more so).

• There has definitely been stuff on the Cstar version of Azumaya algebras, with reference to the Brauer group, but I don't recall a precise reference right now (maybe Raeburn, or Raeburn and Taylor, or D. Williams?) - my impression was that this was more actively pursued in the 1970s and 1980s. Dec 31 '19 at 18:11
• @YemonChoi Indeed, thanks to your comment I have come across the book "Morita equivalence and continuous-trace C*-algebras" by Raeburn and Williams, which seems to address my question rather well. Jan 2 '20 at 11:39

## 1 Answer

I know my answer is coming a bit late, but the answer to your question is: yes. If $$A$$ is a $$C^\ast$$-algebra, and there exists a $$C^\ast$$-algebra $$B$$ such that $$A\otimes_\alpha B$$ is strongly Morita equivalent to $$\mathbb C$$ for some $$C^\ast$$-tensor product $$\otimes_\alpha$$, then $$A\cong \mathcal K(H)$$ is the compact operators on some Hilbert space $$H$$, and in particular $$A$$ is strongly Morita equivalent to $$\mathbb C$$. If $$A$$ is unital (as in the question), then $$H$$ is finite dimensional, so $$A$$ is a matrix algebra.

It is well-known that strong Morita equivalence preserves: (1) being a Type I $$C^\ast$$-algebra (also called GCR); and (2) the spectrum. Hence $$A\otimes_\alpha B$$ is of Type I and its spectrum is a point. In particular $$A\otimes_\alpha B$$ is simple so $$\otimes_\alpha$$ is the minimal tensor product (since the $$A\otimes_{\min{}} B$$ is a quotient of $$A\otimes_{\alpha} B$$). By Theorem 2 in Tomiyama's paper "Applications of Fubini type theorem to the tensor products of C∗-algebras. Tohoku Math. J. (2) 19 (1967), 213–226." it follows that $$A$$ and $$B$$ are Type I. By Theorem B.45 in Raeburn and William's book on Morita equivalence, it follows that the spectrum of $$A$$ is a singleton. As Naimark's problem* is true amongst Type I $$C^\ast$$-algebras it follows that $$A \cong \mathcal K(H)$$ for some Hilbert space $$H$$.

*Naimark's problem: if $$A$$ is a $$C^\ast$$-algebra for which the spectrum is a singleton, is $$A$$ isomorphic to the compact operators on some Hilbert space?

• ... in the last sentence, perhaps "is $A$ isomorphic" instead of "$A$ is isomorphic"? Jan 15 '20 at 5:41
• Thanks, Nik, it has been changed. Jan 16 '20 at 0:20