7
$\begingroup$

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?

$\endgroup$

2 Answers 2

9
$\begingroup$

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.)

$\endgroup$
2
  • $\begingroup$ Thank you for your answer! Indeed, the algebras I am dealing with are Type II factors acting on B(L^2(X)) where X is a topological space. Is there some classification of the operators on B(L^2(X)) that would allow me to systematically check that the operators in B are the only ones commuting withoperators in A (or vice versa)? $\endgroup$
    – mkreisel
    Nov 26, 2013 at 14:04
  • $\begingroup$ 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. $\endgroup$
    – Nik Weaver
    Nov 26, 2013 at 17:13
2
$\begingroup$

I was pointed to the paper "A framework to study commutation problems" https://eudml.org/doc/87328 which may help. There are other papers listed in the references there which also deal with the same problem.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.