# Finding the commutant of a von Neumann algebra

Suppose you have a von Neumann algebra $A$ of operators on $H$ and would like to compute its commutant. You have constructed a collection $B\subset A'$ which you suspect generates it (i.e. you think $\overline{B}=A'$). Is there a general technique for proving that $B$ actually does generate the commutant? For example, is it enough to show that bounded operators on $H$ commuting with both $A$ and $B$ must be multiples of the identity? Are there any other tricks like this that might work?

is it enough to show that bounded operators on $H$ commuting with both $A$ and $B$ must be multiples of the identity?
No. First of all, this condition only makes sense if $A$ is a factor: if $A$ has a nontrivial center, then everything in its center commutes with both $A$ and $B$. But the condition fails even for factors: you just need to find a proper von Neumann subalgebra $B$ of $A'$ such that $B' \cap A' = \mathbb{C}I$. (Anything that commutes with $A$ and $B$ lies in both $A'$ and $B'$.) There are lots of examples of this; Google "trivial relative commutant".
I guess the general technique for proving that $B$ generates the commutant of $A$ is to check that every operator that commutes with $B \cup B^*$ belongs to $A$. This would imply that the commutant of the von Neumann algebra generated by $B$ equals $A$, and hence this von Neumann algebra equals $A'$. (Maybe a disappointing answer.)
• The way I would approach this is by trying to find a good, usable characterization of the operators in $A$. Then fix an operator that commutes with $B \cup B^*$ and check that it satisfies this condition. Nov 26, 2013 at 17:13