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

  • $\begingroup$ As a naive comment, the condition implies that $N$ is a factor. $\endgroup$ Jan 9, 2012 at 10:06
  • $\begingroup$ That's correct. I mentioned it above. $\endgroup$ Jan 9, 2012 at 12:01
  • $\begingroup$ 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). $\endgroup$ Jan 9, 2012 at 20:49
  • $\begingroup$ 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. $\endgroup$ Jan 9, 2012 at 22:23
  • 2
    $\begingroup$ 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. $\endgroup$
    – Jiang
    Jan 16, 2012 at 15:47

1 Answer 1


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 $A'\cap R=B'\cap R=A\neq B$.

A few comments:

(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.

  • $\begingroup$ This is great! Thanks. $\endgroup$ Feb 4, 2019 at 17:50

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