## Subspaces of a Subfactor

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Is the following true?

Let $\mathcal N \subset \mathcal M$ be a subfactor. There is a bijective correspondence between the ultraweakly closed subspaces of $\mathcal M$ that are bimodules over $\mathcal N'\cap \mathcal M$, and the ultraweakly closed subspaces of $\mathcal N$.

If the statement is false, is there a simple way to modify it to make it true? I am particularly interested in the case that $\mathcal M$ is of type $\mathrm I$.

If $\mathcal V \subseteq \mathcal N$ is ultraweakly closed, then $\mathcal V (\mathcal N' \cap \mathcal M)$ is a bimodule over $\mathcal N'\cap \mathcal M$. If the subfactor admits a conditional expectation, then this function is injective. (Edit. Jesse points out that the conditional expectation should be ultrweakly continuous.)

Edit. Steven points out that the statement as written is trivially true by a counting argument. Of course, I'm asking about the function $\mathcal V \mapsto \mathcal V \mathcal (\mathcal N'\cap \mathcal M)$, or something similarly natural. He also notes that irreducible subfactors are a counterexample to the bijectivity of $\mathcal V \mapsto \mathcal V(\mathcal N' \cap\mathcal M )$ in general. This leaves a single concrete question:

Let $\mathcal N \subseteq \mathcal B (\mathcal H)$ be a factor. Is the function $\mathcal V \mapsto \mathcal V \mathcal N'$ a bijection between the ultraweakly closed subspaces of $\mathcal N$ and the ultraweakly closed subspaces of $\mathcal B(\mathcal H)$ that are $\mathcal N'$ bimodules?

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What is $\mathcal V (\mathcal N' \cap \mathcal M)$? Is this the algebraic span of $\mathcal V$ and $(\mathcal N' \cap \mathcal M)$ or are you taking the closure in some topology? – Jesse Peterson Apr 5 2011 at 12:31

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This is a nice question, Andre. Thanks for asking! If N is a type I factor, we should be able to decompose B(H) as N⊗N′. Of course N′ is also type I. We can get the identity as a strong-operator limit of finite rank operators in N′, should it be infinite dimensional. The compact operators feel like the only obstruction for doing what you want in this case. Your map should be bijective in the type I case... – Jon Bannon Apr 6 2011 at 1:40

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Thanks, and why is the the function $\mathcal V \mapsto \mathcal V (\mathcal N' \cap \mathcal M)$ 1-1 if there is a conditional expectation from $\mathcal M$ to $\mathcal N$? For instance, what if $\mathcal N = R$ is the amenable $II_1$ factor and $\mathcal M = \mathcal B(L^2 R)$? If $\mathcal V$ is any irreducible subfactor in $R$ then doesn't it follow from von Neumann's bicommutant theorem that $\mathcal V ( R' \cap \mathcal B(L^2R) ) = \mathcal B(L^2R)$? – Jesse Peterson Apr 6 2011 at 6:03

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@ Jesse: Let $E: \mathcal M \rightarrow \mathcal N$ be a conditional expectation. For all $m \in \mathcal N' \cap \mathcal M$ and $n \in \mathcal N$, $nE(m)= E(nm) = E(mn)= E(m)n$, so $E(m) \in \mathcal N' \cap \mathcal N = \mathbb C$. It follows that $E(\mathcal V(\mathcal N' \cap \mathcal M)) = \mathcal V E(\mathcal N'\cap \mathcal M) = \mathcal V$. Thus, the function $\mathcal V \mapsto \mathcal V (\mathcal N' \cap \mathcal M)$ is injective. I may be missing something. – Andre Apr 6 2011 at 20:51

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A good reference might be Section IV.2 of Blackadar's "Operator algebras" (ams.org/mathscinet-getitem?mr=2188261). Of particular interest to you might be Theorem IV.2.2.3 which shows that if there is an ultraweakly continuous conditional expectation from $\mathcal B(\mathcal H)$ to $\mathcal N$, then $\mathcal N$ is type I. – Jesse Peterson Apr 7 2011 at 5:45
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## 1 Answer

I suppose the statement is true because the set of ultraweakly-closed subspaces of $N$ has the same cardinality as the set of $N^\prime\cap M$-bimodular subsets of $M$. You probably want an explicit description of the correspondence, preferably such that the $N^\prime\cap M$-bimodule that corresponds to $V\subset N$ is exactely $V(N^\prime\cap M)$.

This is certainly not true in general. For an irreducible subfactor $N\subset M$, i.e. $N^\prime\cap M=\C$, the condition that $V\subset M$ is an $N^\prime\cap M$-bimodule is empty, while there are strictly more subsets of $M$.

Of course, the bicommutant theorem shows that there are no irreducible subfactors of type I factors. I do not immediately see a counterexample in the type I case.

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Thank you for the answer. Irreducible subfactors came up as a counterexample to one of my previous questions, but I managed to completely forget about them! Since I'm interested primarily in the type $\mathrm I$ case, I'd like to wait until that part of the question is resolved. – Andre Apr 5 2011 at 9:12
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