6
$\begingroup$

Lance's theorem asserts that a discrete group $G$ is amenable if and only if the reduced and full groups C*-algebras coincide. The group von Neumann algebra is the weak closure of the reduced group C*-algebra concretely represented on $\ell_2(G)$. The definition of the full group C*-algebra is abstract as we range through all possible repsresntations of the algebraic group algebra so we can't mimick the definition of the group von Neumann algebra in this case.

Does the full group C*-algebra canonically embed into some finite von Neumann algebra like the reduced group C*-algebra does?

If so, what is the relation (if any) to the group von Neumann algebra of $G$?

$\endgroup$

1 Answer 1

3
$\begingroup$

I think the answer is no. Glimm proved that every group that has a unitary representation whose image generates a type II factor, also has a unitary representation that generates a type III factor. When you look at the full group C$^*$-algebra all representations are on equal footing, so you cannot embed it canonically in an environment where there is a trace.

$\endgroup$
3
  • $\begingroup$ Thanks Martin. What worries me a little is that you make no reference to amenability whatsoever. In this case of course, the reduced and full C*-algebras are the same, yet by Glimm's result there are still representations which generate type III factors. $\endgroup$ Jun 16, 2015 at 13:33
  • $\begingroup$ Yes, I actually waited for a while to post my answer because of that, and my last sentence vaguely tried to address it. I think that the point is simply that when the group is amenable all its faithful representations can be made to live in the same algebra and are thus isomorphic (weather they live in a II $_1$ factor or a III factor), which is not the case for non-amenable. $\endgroup$ Jun 16, 2015 at 15:59
  • $\begingroup$ I think I don't get it. There are still type III factors lurking behind amenable groups. The point is that they contribute nothing to the norm(s) on the algebraic group algebra. $\endgroup$ Jun 16, 2015 at 22:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .