# Are two $W^*$-algebras $A$ and $B$ that are Morita equivalent (in the sense of Rieffel) also Morita equivalent in the algebraic sense (as rings)?

Let $A$ and $B$ be $C^*$-algebras, I had a question about Morita equivalence for $C^*$-algebras and $W^*$-algebras, I've come across 3 concepts:

• Morita equivalence for $C^*$-algebras: Equivalence of categories of Hermitian modules of $A$ and $B$.
• Strong Morita equivalence for $C^*$-algebras: This is given by the existence of an "$A$-$B$-imprimitivity bimodule" ($A$-$B$-bimodule satisfying certaing conditions).
• Morita equivalence for $W^*$-algebras (as defined by Rieffel in "Morita equivalence for $C^*$-algebras and $W^*$-algebras", this in the case $A$ and $B$ are $W^*$-algebras): This is given by the existence of an "$A$-$B$-equivalence bimodule".

$W^*$-algebras are $C^*$-algebras so what's the difference there? I read somewhere that two $C^*$-algebras $A$ and $B$ are strong Morita equivalent if they are Morita equivalent in the traditional algebraic sense (as rings), but I can't find the proof, so I was wondering:

Are two $W^*$-algebras $A$ and $B$ that are Morita equivalent (due to the existence of an equivalence bimodule), also algebraically Morita equivalent (as rings)?

Anyone have references for this? Are equivalence bimodules and imprimitivity bimodules the same thing? I've seen different definitions of them, some very similar.

• If I remember correctly it is enough to consider the example $\mathbb C$ and $B(H)$, which are Morita equivalent as W* algebras, but not in the sense of C* algebras?!? – Marcel Bischoff Aug 1 '14 at 1:12
• The point is $B(H)$ has non-trivial closed ideals, but the existence of non-trivial closed intervals is invariant under strong Morita equivalence link.springer.com/article/10.1007%2FBF01899246#page-1 therefore $B(H)$ and $\mathbb C$ cannot be Strongly Morita equivalent as C*-algebras – Marcel Bischoff Aug 1 '14 at 1:16
• Meyer says (in "Morita equivalence in algebra and geometry", definition 5) that two $C^*$-algebras $A$ and $B$ are strongly Morita equivalent if there's an $A$-$B$-equivalence bimodule, Rieffel says (in "Morita equivalence for $C^*$-algebras and $W^*$-algebras", definition 7.5) that two $W^*$-algebras $A$ and $B$ are Morita equivalent if there's an $A$-$B$-equivalence bimodule and the definition of equivalence bimodules in both papers are exactly identical. – Richard Jennings Aug 1 '14 at 2:44
• For $W^*$-algebras, the assumptions are a little different: you have an additional normality ($\sigma$-weak continuity) assumption and, on the other hand, only assume weak density. – Rasmus Aug 1 '14 at 9:09
• In order to turn Marcel's comment into an answer to the question, observe that $B(H)$ having a non-trivial ideal (closed or not) implies that $\mathbb C$ cannot be algebraically Morita equivalent to $B(H)$. – Rasmus Aug 1 '14 at 9:11