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:



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> 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 $A$.
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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.



Relative Bicommutant
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 $M$.


