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Let $R$ be the hyperfinite $II_1$ factor and let $X$ be any $R$-$R$-bimodule.

Question: Is $X$ completely reducible (i.e. a direct integral of irreducible $R$-$R$-bimodules)?

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.

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    $\begingroup$ You will need direct intgerals. $\endgroup$
    – Marc Palm
    Commented Nov 25, 2013 at 10:22
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    $\begingroup$ Your example $_NL^2M_N$ has finite statistical dimension and is therefore completely reducible. $\endgroup$
    – André Henriques
    Commented Nov 25, 2013 at 16:46
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    $\begingroup$ It's certainly well known. I don't know the original reference, but it appears as Lemma 4.10 of my paper arxiv.org/pdf/1110.5671v1.pdf $\endgroup$
    – André Henriques
    Commented Nov 25, 2013 at 19:29
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    $\begingroup$ Yes, this is proven in my paper. The relevant definitions and propositions are: Definition 5.1, Definition 5.10, Warning 5.11, Lemma 5.16, Proposition 7.5, and Corollary 7.14. $\endgroup$
    – André Henriques
    Commented Nov 25, 2013 at 20:39
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    $\begingroup$ $_NL^2M_N = _NL^2M\boxtimes_M L^2M_N$. $\endgroup$
    – André Henriques
    Commented Nov 25, 2013 at 22:29

1 Answer 1

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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.

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!

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  • $\begingroup$ Andre, I trust you about $_RM_R$, but if your argument is not valid, how do you know that $_RM_R$ is not completely reducible? Is it intuition or do you read this result somewhere ? $\endgroup$
    – Sebastien Palcoux
    Commented Nov 25, 2013 at 21:41
  • $\begingroup$ @André: I think you're correct. If you look at the coarse correspondence as an $M=R\otimes R^{op}$ module, then sub $M$-modules correspond to projections in the commutant $M'\cong R\otimes R$, which is type II$_1$. Irreducible sub $M$-modules should correspond to minimal projections, of which there are none. Or am I missing something here? $\endgroup$
    – Dave Penneys
    Commented Nov 25, 2013 at 22:28
  • $\begingroup$ @Dave: That's not how things work. The "summands" in a direct integral decomposition are not sub(bi)modules. $\endgroup$
    – André Henriques
    Commented Nov 25, 2013 at 22:32
  • $\begingroup$ @AndréHenriques: can you complete your answer? Anyway I will accept it thanks to this meta-post: Acceptation of partial answers $\endgroup$
    – Sebastien Palcoux
    Commented Sep 4, 2014 at 12:17

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