Let $R$ be the hyperfinite $II\_1$ factor and let $X$ be any $R$-$R$-bimodule.



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> **Question**: Is $X$ completely reducible (i.e. a direct integral of irreducible $R$-$R$-bimodules)? 
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**Example**: If $(N \subset M)$ is an irreducible, finite depth and finite index (hyperfinite $II\_1$) subfactor, then $\_NL^{2}(M)\_N$ is completely reducible.



Are all the R-R-bimodules completely reducible?
Let $\_RM\_R:={}\_R(L^2R\otimes\_{\mathbb C} L^2R)\_R$, where the first $R$ acts on the first $L^2R$ and the second $R$ acts on the second $L^2R$.
Its algebra of $R$-$R$-bimodule endomorphisms is $R^{\mathrm{op}}\,\bar\otimes\, R$.


Using (misleading!) intuition from the representation theory of separable $C^\*$-algebras, one might guess that every MASA in $R^{\mathrm{op}}\bar\otimes R$ gives rise to a direct integral decomposition of $\_RM\_R$ into irreducible $R$-$R$-bimodules. But that is not true. There is no way of writing $\_RM\_R$ as a direct integral of irreducible bimodules.


~~Indeed, suppose that ${}\_R(M\_x)\_R$, $x\in X$, are irreducible bimodules, and that $\int^\oplus\_{x\in X} {}\_R(M\_x)\_R dx$ is an $R$-$R$-bimodule that is isomorphic to $\_RM\_R$. Then for every $M\_x$, the left and right actions of $R$ induce an action of $R\,\bar\otimes\, R^{\mathrm{op}}$ on $M\_x$. But $R\,\bar\otimes\, R^{\mathrm{op}}$ is type $II$ and does not admit irreducible representations. Contradiction.~~


My argument above is not valid.
Nevertheless, I still maintain that there is no way of writing $\_RM\_R$ as a direct integral of irreducible bimodules.




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Here is where my intuition comes from:


Let $R\_1$ and $R\_2$ be two factors such that one type II (or III) while the other is of type I.  

Let $\_{R\_1}M\_{R\_2}:={}\_{R\_1}(L^2{R\_1}\otimes\_{\mathbb C} L^2{R\_2})\_{R\_2}$, where the first $R\_1$ acts on the first $L^2R\_1$ and the second $R\_2$ acts on the second $L^2R\_2$.
Its algebra of bimodule endomorphisms is $R\_1^{\mathrm{op}}\,\bar\otimes\, R\_2$.


Using (misleading!) intuition from the representation theory of separable $C^\*$-algebras, one might guess that every MASA in $R\_1^{\mathrm{op}}\bar\otimes R\_2$ gives rise to a direct integral decomposition of $\_{R\_1}M\_{R\_2}$ into irreducible $R\_1$-$R\_2$-bimodules. But that is not true. There is no way of writing $M$ as a direct integral of irreducible bimodules as... *there are no
irreducible $R\_1$-$R\_2$-bimodules*!


