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Aug 31, 2012 at 7:01 comment added Ulrich Pennig @Mikael: Not really an argument. But there are states on $\mathcal{O}_n$ for finite $n$ which yield type $III_{1/n}$-factors. Unfortunately, those are built upon the fact, that $\mathcal{O}_n$ contains a core UHF-algebra and a conditional expectation (via the gauge action) onto it. For $\mathcal{O}_{\infty}$ you only get an AF-algebra.
Aug 31, 2012 at 6:55 vote accept Ulrich Pennig
Aug 30, 2012 at 19:22 comment added Mikael de la Salle @Jon: The state constructed in Dima's paper is just the vector state $\langle \cdot \Omega,\Omega\rangle$ restricted to the $C^*$-algebra generated by the operators $s_v + s_v^*$ for $v$ is some real subspace of $H$ (see def 2.3 and remark 2.6 in the paper). I do not believe this is so relevant if one is looking for a state on $\mathcal O_\infty$.
Aug 30, 2012 at 18:41 comment added Jon Bannon What about these free quasi-free states: msp.berkeley.edu/pjm/1997/177-2/pjm-v177-n2-p07-p.pdf? (I haven't looked carefully, this is an off the cuff comment...)
Aug 30, 2012 at 18:33 comment added Mikael de la Salle I do not know, but I am not very familiar with $\mathcal O_\infty$ (and with $C^*$-algebras in general). Do you have an argument why such a state should exist?
Aug 30, 2012 at 13:45 comment added Ulrich Pennig Do you know about any states on $\mathcal{O}_{\infty}$ which would yield a type $III$-von Neumann algebra?
Aug 30, 2012 at 12:06 comment added Ollie Snap! Nice proof.
Aug 30, 2012 at 11:55 history answered Mikael de la Salle CC BY-SA 3.0