Welcome to MathOverflow

Question

3

Some years ago, I heard about the following problem:

Find two ICC, i.e., infinite conjugacy class groups $G$ and $H$, such that $L(G)\cong L(H)$, but $G$ and $H$ are not measure equivalent, where $L(G)$ denotes the group von Neumann algebra associated to the group $G$.

Recall that $G$ and $H$ are measure equivalent iff they admit stably orbit equivalent actions.

What is the status on this problem now?

edited Feb 19 at 22:24

asked Feb 19 at 21:50

Jiang
577●3●7

	The way you stated it, you can take any group $G$ which admits a finite index subgroup $H$ and so that $L(G)$ has some restrictions on the fundamental group. I believe such groups have been constructed. If you ask only for stable equivalence of the von Neumann algebras, I do not think that there are examples. – Andreas Thom Feb 19 at 22:07
	sorry, I do not quite understand the meaning you say "so that L(G) has some restrictions on the fundamental group". The problem asked for constructing $G, H$ such that $L(G)\cong L(H)$ and $G, H$ are not measure equivalent or show such $G, H$ do not exist. Is the expression misleading or am I miss some point? – Jiang Feb 19 at 22:36
	This problem is still open as far as I know. – Jesse Peterson Feb 21 at 1:08
	If this type question is sensitive to the separable argument, then a naive approach would be to find ICC groups $\{G_{\alpha}\}_{\alpha\in A}$, where $A$ is a uncountable set, and require $\{L(G_{\alpha})\}$ has at most countable different equivalent classes (maybe we can achieve this by considering the invariant associated to the $L(G_{\alpha})$, dimension, etc.) and at the same time require $G_{\alpha}$ has uncountable many different classes module the ME relation. Of course, these $\{G_{\alpha}\}$ should not have property $(T)$, and it may be not practical. – Jiang Feb 21 at 18:13
	We can further ask a weak version of the big one: Does $L(G)\cong L(H)$ and $G$ satisfies property $(T)$ imply that $G, H$ are measure equivalent? We can also replace the property $(T)$ by any other known invariant properties under Measure equivalent. Still, much work to be done on both sides, ME and von Neumann algebra equivalence. – Jiang Feb 21 at 18:17

Not measure equivalent ICC groups $G$ and $H$, but $L(G)\cong L(H)$

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.

Your Answer

Browse other questions tagged von-neumann-algebras ergodic-theory or ask your own question.

Welcome to MathOverflow

How to write math

Not measure equivalent ICC groups $G$ and $H$, but $L(G)\cong L(H)$

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.