most recent 30 from http://mathoverflow.net 2013-06-20T08:03:13Z http://mathoverflow.net/feeds/question/114925 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/114925/number-of-ii-1-factors Bojan Kwitek 2012-11-29T20:16:52Z 2012-11-30T15:47:58Z

<p>McDuff proved that there exist continuum many non-isomorphic (separable) II${}_1$ factors. I would like to politely ask whether it is known/open if one can find $2^{\mathfrak{c}}$ (or at least $\mathfrak{c}^+$) many such factors.</p> <p>My feeling is that this is not possible to construct more than $\mathfrak{c}$ separable von Neumann algebras by a simple cardinality argument. The ball of $B(H)$ for $H$ separable, is metrisable under the ultraweak topology, so it has at most $\mathfrak{c}$ ultraweakly closed subsets. So we cannot have more than $\mathfrak{c}$ different balls, and consequently, have more than $\mathfrak{c}$ non-isomorphic algebras. Is this correct?</p>

http://mathoverflow.net/questions/114925/number-of-ii-1-factors/114931#114931 Eric Wofsey 2012-11-29T21:19:01Z 2012-11-30T15:47:58Z

<p>Your argument is correct. An alternate and more "intrinsic" argument is to look at the predual, which is a separable Banach space. There are only continuum many separable Banach spaces, since they are determined by the metric on a countable dense subset that is a $\mathbb Q[i]$-vector space. Thus there are only continuum many separable von Neumann algebras, when considered just with their structure of dual Banach spaces. As Nik points out in the comments, the dual Banach space structure may not determine the algebra structure. However, given any such dual Banach space, you can fix a countable weak*-dense subset, and then any algebra structure will be determined by what it does on that countable set, so you again have only continuum many possibilities.</p>