# Relative commutants of abelian von Neumann algebras

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 $R' \cap N = \mathbb C$.

• 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

I learned recently that (see here) Popa proved that every separable II$$_1$$ factor $$M$$ contains (an embedding of) the hyperfinite von Neumann algebra $$R$$ such that $$L^2M\ominus L^2R\cong _RL^2(R\overline{\otimes}R^{op})^{\oplus \infty}_R$$.

It is a standard fact that this implies $$A'\cap M\subseteq R$$ for every diffuse subalgebra $$A\subseteq R$$.

Then, as in $$R$$, there are diffuse abelian von Neumann subalgebras $$A\neq B$$ such that $$A'\cap R=B'\cap R$$. Therefore, by the above result, we also have $$A'\cap M=B'\cap M$$.

For example, write $$R=L(\mathbb{Z}\wr \mathbb{Z})=L(\langle t\rangle \wr \langle s\rangle)$$, $$A=L(\langle s\rangle)$$ and $$B=L(\langle s^2\rangle)$$. Or write $$R=L((\mathbb{Z}/2\mathbb{Z})G)\rtimes G$$, $$A=L((\mathbb{Z}/2\mathbb{Z})G)$$ and $$B=L(\{x\in (\mathbb{Z}/2\mathbb{Z})G: x_{e_G}=\bar{0}\})$$.

In both examples, we have that $$A'\cap R=B'\cap R=A\neq B$$.

(1) Following Kadison's paper, a von Neumann subalgebra $$A\subseteq M$$ is called normal if $$(A'\cap M)'\cap M=A$$. There are several old papers (see e.g. Anastasio's paper) giving concrete examples of the most extreme case of non-normal abelian subalgebras, i.e. thick subalgebras following Bures's book here. Recall $$A\subseteq M$$ is thick if $$A'\cap M$$ is a masa in $$M$$; equivalent characterizations can be found in Lemma 10.1 of this book.
(2) If $$A$$ is abelian, then since $$(A'\cap M)'\cap M=\cap_BB$$, where $$B$$ is a masa in $$M$$ containing $$A$$, we know that the above question is the same as asking whether there is an abelian von Neumann subalgebra $$A\subset M$$ such that $$A$$ is not normal, e.g. $$A\neq A'\cap M$$ and $$A$$ is thick.
(3) Let $$B\subseteq M$$ be a masa. Then every diffuse proper von Neumann subalgebra of $$B$$ is NOT normal iff $$B$$ satisfies the disjointness property, i.e. if $$C\subset M$$ is a masa with $$B\cap C$$ being diffuse, then $$B=C$$.
(4) Popa's result implies every separable II$$_1$$ factor $$M$$ contains a mixing masa, as we can take it to be a mixing masa in $$R$$, where $$R\hookrightarrow M$$ is the above mixing inclusion. Inside a mixing masa, every proper (diffuse) von Neumann subalgebra is thick. It seems not clear whether one can always find proper thick subalgebras inside every singular, equivalently weakly mixing, masa in a separable II$$_1$$ factor.