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).

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    $\begingroup$ 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. $\endgroup$ – Yemon Choi Dec 31 '19 at 18:11
  • $\begingroup$ @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. $\endgroup$ – Captain Lama Jan 2 at 11:39

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?

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  • $\begingroup$ ... in the last sentence, perhaps "is $A$ isomorphic" instead of "$A$ is isomorphic"? $\endgroup$ – Nik Weaver Jan 15 at 5:41
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    $\begingroup$ Thanks, Nik, it has been changed. $\endgroup$ – Jamie Gabe Jan 16 at 0:20

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