Let a locally compact group $G$ act on a probability space $(X,\mu)$. Define the centralizer by $C(G)=\{\Delta\in Aut(X,\mu)\mid \Delta(gx)=g\Delta(x)\text{ almost everywhere}\}$. $Aut(X,\mu)$ denotes the set of $\mu$-preserving transformations of $X$. I am searching for actions $G\curvearrowright X$ such that the second centralizer $C(C(G))=G$. In the following, i tacitly assume that all actions are faithful and i identify the acting group $G$ with the corresponding subgroup of $Aut(X,\mu)$.

Of course, for compact groups $G$ such actions exist: let $G$ act on itself (with normalized Haar measure) by left translation. The centralizer is the right translation action of $G$, and the second centralizer is again the left translation action of $G$.

I found a paper by Daniel Rudolph that asserts that the second centralizer of the Bernoulli action of $\mathbb{Z}$ is just $\mathbb{Z}$ itself.

Now I was wondering if the similar result holds for the Gaussian action associated to the left-regular representation of an arbitrary second countable locally compact group. Has this been studied?

  • $\begingroup$ sounds kind of like the bicommutant characterization of von Neumann algebras. $\endgroup$ – Ian Agol Feb 14 '12 at 17:30
  • 2
    $\begingroup$ Do you know if you have $C(C(C(G))) = C(G)$? If so this might be a good way to find such actions. $\endgroup$ – Jesse Peterson Feb 14 '12 at 23:30
  • $\begingroup$ Let me simply note that your condition cannot be satisfied very often - since centralizers are obviously closed it can only be satisfied if $G$ is closed, so for a countable $G$ the image of $G$ in $Aut(X,\mu)$ must be discrete. Hence, for some fixed countable $G$, a generic (in the sense of Baire category) $G$-action fails to satisfy your condition. However it often happens for abelian groups that the double commutant is the closure of $G$. Of course that does not say anything about the case of Gaussian actions and it seems reasonable to expect that they might behave like Bernoulli shifts... $\endgroup$ – Julien Melleray Feb 14 '12 at 23:43
  • $\begingroup$ @agol: If we replace $Aut(X,\mu)$ by the infinite unitary group, we see that every group $G$ with this property must be the unitary group of some von Neumann algebra. Hence if it is locally compact, it is a direct product of finite unitary groups. (and hence compact) $\endgroup$ – Steven Deprez Feb 15 '12 at 9:02
  • $\begingroup$ @Jesse: For any group/algebra, the triple centralizer is again just the centralizer. In fact, i am searching for groups $G$ that are the centralizer of some group action, which is a equivalent to asking that $C(C(G))=G$. $\endgroup$ – Steven Deprez Feb 15 '12 at 9:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.