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This question arose from the discussion over at the question Centralizers in $C^*$-algebras.

Which von Neumann algebras $N$ satisfy the property that $A' \cap N = B' \cap N \implies A = B$, for all commutative von Neumann subalgebras $A, B \subset N$?

Note that $N = \mathcal B(\mathcal H)$ has this property by von Neumann's double commutant theorem, and perhaps this property characterizes $\mathcal B(\mathcal H)$. It is clear that $N$ must be a factor by considering $A = \mathbb C$ and $B = \mathcal Z(N)$. If $\mathbb F_2 = \langle a, b \rangle$ is the free group on two generators, then by considering the Fourier expansion of elements in $L\mathbb F_2$ it is not hard to see that for $A = L\langle a \rangle$ and $B = L\langle a^2 \rangle$ we have $A' \cap L\mathbb F_2 = B' \cap L\mathbb F_2$ thus $L\mathbb F_2$ does not have this property.

Also note that if we were to consider the case when $A$ and $B$ are allowed to be non-commutative then relevant is Corollary 4.1 in Popa's paper On a Problem of R.V. Kadison on Maximal Abelian $*$-Subalgebras in Factors which shows that every type $II$ factor $N$ contains a hyperfinite subfactor $R$ such that $R' \cap N = \mathbb C$.

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As a naive comment, the condition implies that $N$ is a factor. – Martin Argerami Jan 9 '12 at 10:06
That's correct. I mentioned it above. – Jesse Peterson Jan 9 '12 at 12:01
Sorry, I should read with more care! And your proof is better than mine, as my argument used non-unital subalgebras (which you are probably not considering). – Martin Argerami Jan 9 '12 at 20:49
Yes, it is important that $A$ and $B$ contain the same unit as $N$ (I think in most books this is part of the definition of a von Neumann subalgebra). Otherwise this property is not even satisfied for $\mathcal B(\mathcal H)$, since if we consider $A \subset \mathcal B(\mathcal H)$ any self-adjoint subalgebra, then we have $A' = (A'')'$ and $A''$ always contains the identity operator. – Jesse Peterson Jan 9 '12 at 22:23
One comment, if we could choose one abelian von Neumann subalgebra $A$ such that there exists one unitary $u$ in the normalizer of $A'\cap N$ but not in the normalizer of $A$, then set $B=uAu^*$, this would give us one counterexample. – Jiang Jan 16 '12 at 15:47

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