oa.operator algebras - Relative Bicommutant

Let $Asubseteq B(H)$ 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 $$A' cap B' = (Acup B)' = text{alg}(Acup B)'.$$

So, for your question, pick some normal representation $Msubseteq B(H)$ (so that $M''=M$), and let $Asubseteq M$ be a subset. Set

$$X={xin M:ax=xa (ain A)} = A'cap M = A'cap M'' = (Acup M')',$$ so your relative commmutatant is [{min M:xm=mx (xin X) } = X' cap M = X' cap M'' = (Xcup M')'.] So, yes, this is a von Neumann algebra.

In fact, as $Asubseteq M$, clearly $M'subseteq A'$ and so $A''cap M = (A'cup M')' = A''$. So as $Xsubseteq A'$, 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 $ninmathbb Z$. 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 $M$.

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