Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I should be tarred and feathered for not knowing at least the status of the following question.

Question: Let $\Gamma$ be a discrete amenable group. If $\pi:\Gamma \rightarrow B(\mathcal{H})$ is a unitary representation of $\Gamma$ on a separable Hilbert space $\mathcal{H}$, is the von Neumann algebra $\pi(\Gamma)''$ necessarily injective?

Flippantly one imagines that the answer to this question is yes, by Theorem 2.2 of Bekka's paper on amenable representations. But this result only says that the images of group elements are in the centralizer of a non-normal state...it isn't immediately clear why the entire von Neumann algebra should lie in the centralizer of such a state. If one tries to sidestep this by looking at a proof using almost invariant vectors, one is busted by the fact that a representation that is "$H$-amenable" isn't necessarily amenable in Bekka's sense.

EDIT: Makoto's nice answer below provided me with some closure. I'm still worried that I can't see a more or less direct way to this result from Connes's '76 paper on the classification of injective factors. If this paper can, in a more or less direct and self-contained way, be used to resolve the question, please feel free to include another answer.

share|improve this question
    
Jon, what definition of injective vN alg are you using? Norm-one projection onto the algebra? Norm-one projection onto the commutant? Quasi-expectation onto the commutant? etc. –  Yemon Choi Dec 2 '12 at 21:32
    
Good question, Yemon. I mean norm-one (Banach space) projection onto the algebra. What I originally wanted to do was to show that there was a hypertrace...i.e. that the von Neumann algebra was the centralizer of a state on B(H). That was probably clear from the comments following the question. I'd be happy if it were possible to prove this result using Connes's paper and very little else. I'm actually quite satisfied with the existing answers, though, if I'm asking for too much. – –  Jon Bannon Dec 3 '12 at 0:28

1 Answer 1

up vote 8 down vote accepted

This is because you get a representation of the maximal(=reduced by amenability) C*-algebra $C^\ast(\Gamma)$ on $H$, and the image has to be a nuclear algebra because it's a quotient of nuclear one. Then, $\pi(\Gamma)''$, being a weak*-closure of a nuclear C*-algebra, has to be injective.

share|improve this answer
3  
Nice answer. I'd just like to add, for those of us from the "amenability side", that it is (a) straightforward to show that the $C^\ast$-algebra of an amenable group is an amenable Banach algebra (b) easy to show directly that the quotient of any amenable Banach algebra is amenable (c) using results of Bunce-Paschke (IIRC) one can show that the weak-star-closure of any amenable $C^\ast$-algebra is injective. So nuclearity per se, and "nuclear implies amenable", don't have to enter the picture –  Yemon Choi Nov 30 '12 at 23:30

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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