Subalgebras of von Neumann algebras - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T04:44:20Z http://mathoverflow.net/feeds/question/34692 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/34692/subalgebras-of-von-neumann-algebras Subalgebras of von Neumann algebras Andreas Thom 2010-08-05T21:40:37Z 2010-08-09T05:04:40Z <p>In the late 70s, Cuntz and Behncke had a paper </p> <p>H. Behncke and J. Cuntz, <em>Local Completeness of Operator Algebras</em>, Proceedings of the American Mathematical Society, Vol. 62, No. 1 (Jan., 1977), pp. 95- 100</p> <p>the about the following question. Let \$A\$ be a \$C^\star\$-algebra and let \$B \subset A\$ be some \$*\$-subalgebra. Let \$B\$ be dense and such that every maximal abelian subalgebra of \$B\$ is norm-closed. Is it true that \$B=A\$? They proved this in various cases.</p> <p>I want to ask a related question in the von-Neumann-setting. More precisely: Let \$A\$ be a von Neumann algebra and \$B \subset A\$ be some ultra-weakly dense \$\star\$-subalgebra such every MASA of \$B\$ is ultra-weakly closed in \$A\$. Is it true that \$B=A\$? A result of Gert Pedersen implies that once \$B\$ is a \$C^\star\$-algebra, then \$B=A\$. Hence, the two questions are closely related. Taking the work of Behncke-Cuntz and Pedersen together, it is known that \$B=A\$ if \$A\$ has no \$II_1\$-part.</p> <blockquote> <p><strong>Question:</strong> Is it true for every von Neumann algebra?</p> </blockquote> <p>or a little more modest</p> <blockquote> <p><strong>Question:</strong> Is it true for the hyperfinite \$II_1\$-factor?</p> </blockquote> <p>Another strenghtening of the the assumption (which could help) would be the following:</p> <blockquote> <p><strong>Question:</strong> Let \$A\$ be a \$II_1\$-factor and \$B \subset A\$ be a ultra-weakly dense \$\star\$-subalgebra such that for every hyperfinite subalgebra \$R \subset A\$ one has that \$R \cap B\$ is ultra-weakly closed in \$A\$. Is it true that \$B=A\$?</p> </blockquote>