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Two rings are said to be Morita equivalent if their categories of (left) modules are equivalent. The notion is also used in more general contexts when certain categories of representations are equivalent.
2
votes
C*-bimodules: the mess with definitions
As far as I'm concerned $C^*$ bimodules generally denote those you attributed to A.Connes. Such a bi-module define (by tensorization over B) a functor from $B$ $C^*$-modules to $A$ $C^*$-modules. Any …
29
votes
Why is the bicategory viewpoint useful?
You believe the 1-category is interesting but somehow not the bicategory, yet one could say the exact same thing one dimension below: Why is this category even interesing while you could just consider …
10
votes
Strong Morita Equivalence and Morphisms Between $ C^{*} $-Algebras
You cannot says anything about morphisms between $A$ and $B$ in general, but from the two bi-modules you can construct a third algebra $C$, such that both $A$ and $B$ embeds into $C$ with the embeddin …
8
votes
Accepted
Can a finite von Neumann algebra be strongly morita equivalent to a properly infinite von ne...
Assume $A$ and $B$ are two unital $C^*$-algebras which are strongly Morita equivalent.
This means that there exists a Hilbert $A$-module $H$ such that $B$ is the algebra of compact operators on $H$. …