Subalgebras of von Neumann algebras - MathOverflow most recent 30 from http://mathoverflow.net2013-05-25T04:44:20Zhttp://mathoverflow.net/feeds/question/34692http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/34692/subalgebras-of-von-neumann-algebrasSubalgebras of von Neumann algebrasAndreas Thom2010-08-05T21:40:37Z2010-08-09T05:04:40Z
<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>
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<p><strong>Question:</strong> Is it true for every von Neumann algebra?</p>
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<p>or a little more modest</p>
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<p><strong>Question:</strong> Is it true for the hyperfinite $II_1$-factor?</p>
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<p>Another strenghtening of the the assumption (which could help) would be the following:</p>
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<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>
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