Maximal localizations of von Neumann algebras - MathOverflow most recent 30 from http://mathoverflow.net2013-05-21T17:46:59Zhttp://mathoverflow.net/feeds/question/1940http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/1940/maximal-localizations-of-von-neumann-algebrasMaximal localizations of von Neumann algebrasDmitri Pavlov2009-10-22T20:30:14Z2010-08-05T14:47:27Z
<p>Suppose M is a von Neumann algebra.
Denote by L its maximal noncommutative localization,
i.e., the Ore localization with respect to the set of all left and right regular elements,
i.e., elements whose left and right support equals 1.</p>
<p>Denote by A the set of all closed unbounded operators with dense domain
affiliated with the standard representation of M on a Hilbert space, i.e., L^2(M),
also known as the standard form of M.</p>
<p>Von Neumann proved that if M is finite, then L and A are canonically isomorphic.</p>
<p>What can we say about the relationship of L and A when M has type III?</p>
<p>I am also interested in the properly infinite semifinite case.</p>
http://mathoverflow.net/questions/1940/maximal-localizations-of-von-neumann-algebras/34635#34635Answer by Andreas Thom for Maximal localizations of von Neumann algebrasAndreas Thom2010-08-05T14:47:27Z2010-08-05T14:47:27Z<p>I think the question is not well-posed or has a negative answer.</p>
<p>One first has to deal with the question whether the left-right-regular elements satisfy the Ore condition, or equivalently, we have to ask: Can we find common denominators? If one is not in the finite case, this is not possible. </p>
<p>For $B(H)$, let us take injective bounded operators $T$ and $S$ such that the images are dense but intersect only in the zero vector. In order to find a common denominator, we need to find an operator $R$ (bounded, injective, dense image) and bounded operators $X$ and $Y$ such that $R = TX$ and $R= SY$. This cannot possibly work since $T$ and $S$ have disjoint image.</p>
<p>Since $B(H)$ sits inside any type $III$-factor, no Ore localization in the above sense exists.</p>