MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am looking at a von Neumann algebra constructed from a discrete group and a 2-cocylce. Does someone know some good references (article, book)? It would be very helpful for me. To be more precise, consider a countable group $G$ and a 2-cocycle $\phi :G^2\rightarrow S^1$ where $S^1$ is the group of complex number of modulus 1. You get a representation $\pi$ of the group $G$ in the Hilbert space $l^2(G)$ defined as follow: $$\pi(g)(e_t)=\phi(g,t).e_{gt}$$, where $e_t$ is the canonical hilbert basis of $l^2(G)$. I consider $L_\phi(G)$, the von Neumann algebra generated by $\pi(G)$. I am looking for reference on those kinds of algebra. Thanks, Arnaud

share|cite|improve this question
It might help if you give a bit more detail about how the cocycle is used to construct this vN algebra... – Yemon Choi May 11 '10 at 10:01
How exactly are you looking at the von Neumann algebra of interest? Is it introduced in a book, a paper, by a person talking, or something else? – S. Carnahan May 11 '10 at 13:53

You can consult the article

Bédos, Erik On Følner nets, Szegő's theorem and other eigenvalue distribution theorems. Exposition. Math. 15 (1997), no. 3, 193--228.

and this recent article:

Finally, I believe that the construction is first appeared in

Zeller-Meier, G. Produits croisés d’une C*-algèbre par un groupe d’automorphismes. (French) J. Math. Pures Appl. (9) 47 1968 101–239

share|cite|improve this answer
thanks a lot, it is nice to have an article in french. – Arnaud Brot Sep 28 '10 at 16:31

You might find these articles helpful (although it deals with the more general case): Sutherland, Colin E. Cohomology and extensions of von Neumann algebras. I, II. Publ. Res. Inst. Math. Sci. 16 (1980), no. 1, 105--133, 135--174.

share|cite|improve this answer

Your Answer


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.