Relative Bicommutant

Question

If $A \subseteq \mathcal B(\mathcal H)$ is an algebra of operators that is closed under adjoint, then its bicommutant $A''$ is a von Neumann algebra, and is the ultraweak closure of $A$; this is one version of von Neumann's bicommutant theorem. Does the theorem hold relative to an arbitrary von Neumann algebra $\mathcal M$? Concretely, what is the truth value of the following statement:

Let $\mathcal M$ be a W*-algebra, and $A\subseteq \mathcal M$ be a subalgebra closed under adjoint. Then the relative bicommutant $A'' = \{ m \in \mathcal M | \forall x \in \mathcal M. (\forall a \in A. ax = xa) \implies mx=xm \}$ is a W*-algebra, and is the ultraweak closure of $A$.

A W*-algebra is C*-algebra that is isomorphic to a von Neumann algebra. I use the term W*-algebra to emphasize that the bicommutant is being computed relative to $\mathcal M$ itself rather than relative to a Hilbert space on which $\mathcal M$ is represented.

Edit: As Matthew points out below, $A$ should contain the unit of the ambient algebra, i.e., of $\mathcal B(\mathcal H)$ in the bicommutant theorem, and of $\mathcal M$ in the statement in question.

Answer 1

Let $A\subseteq B(H)$ be a subset, and let $\text{alg}(A)$ be the algebra generated by $A$. Then it's easy to see that $$A' = \text{alg}(A)'.$$ A similarly easy check shows that if $A$ and $B$ are subsets, then $$A' \cap B' = (A\cup B)' = \text{alg}(A\cup B)'.$$

So, for your question, pick some normal representation $M\subseteq B(H)$ (so that $M''=M$), and let $A\subseteq M$ be a subset. Set $$X=\{x\in M:ax=xa \ (a\in A)\} = A'\cap M = A'\cap M'' = (A\cup M')',$$ so your relative commmutatant is \[\{m\in M:xm=mx \ (x\in X) \} = X' \cap M = X' \cap M'' = (X\cup M')'.\] So, yes, this is a von Neumann algebra.

In fact, as $A\subseteq M$, clearly $M'\subseteq A'$ and so $A''\cap M = (A'\cup M')' = A''$. So as $X\subseteq A'$, thus $A''\subseteq X'\cap M$.

Edit: There is probably an easier example than this... But, let $M=VN(\mathbb F_2)$, say with canonical generators $a$ and $b$. Let $A$ be a star-algebra generated by $b$ and $a^{-1}ba$: so $A$ is just linear combinations of $b^n$ and $a^{-1}b^na$ for $n\in\mathbb Z$. A bit of combinatorics shows that $X=\mathbb C1$ and so the the relative bicommutant is all of $M$. However, $A$ is not ultraweakly dense in $M$, because we cannot approximate the generator $a$.

Conclude: So, if I haven't messed up, this shows that the relative bicommutant is always a W*-subalgebra of $M$, but that it might be larger than the ultraweak closure of $A$ in $M$.

Answer 2

In the theory of subfactors, it frequently happens that a subfactor of finite index $N \subset M$ satisfies $N' \cap M = {\mathbb C}$. Those subfactors are called irreducible. In this case the relative bicommutant of $N$ would be $M$.

Answer 3

To be true in any generality, the original assertion requires ${\mathcal{M}}$ to be a type I factor (for non-type I factors, the existence of subfactors with trivial relative commutant will always give a counterexample, as Andreas mentioned above). Otherwise, it fails even in finite dimension: let $\mathcal{M}=\mathbb{C}\oplus M_2(\mathbb{C})$, and let $A=\mathbb{C} I\subset\mathcal{M}$. Then $A'\cap\mathcal{M}=\mathcal{M}$, and so the relative bicommutant of $A$ is the centre of $\mathcal{M}$, which is strictly larger that $A$.

Notice that this example will work with any non-factor $\mathcal{M}$, and any $A$ which is a proper subalgebra of the centre.