most recent 30 from http://mathoverflow.net 2013-05-25T16:52:35Z http://mathoverflow.net/feeds/question/100576 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/100576/intersections-of-maximal-abelian-von-neumann-algebras blackpeacock 2012-06-25T09:13:50Z 2012-07-09T12:59:04Z

<p>Let $H$ be separable Hilbert space. Let $A$ be a maximal abelian von Neumann subalgebra of $B(H)$, and $B$ an abelian von Neumann algebra with $A\cap B={\mathbb C}I$, where $I$ is the indentity element of $B(H)$. Does there exist another maximal abelian von Neumann subalgebra of $B(H)$, say $C$, such that $C\supseteq B$ and $A\cap C={\mathbb C}I$?</p>

http://mathoverflow.net/questions/100576/intersections-of-maximal-abelian-von-neumann-algebras/100583#100583 Chris Heunen 2012-06-25T10:48:16Z 2012-06-25T10:48:16Z

<p>This is only a partial answer, but it didn't fit in the comment box.</p> <p>In finite dimension, say $\dim(H)=n$, a maximal abelian von Neumann algebra $A \subseteq B(H) \cong M_n(\mathbb{C})$ just comes down to (the set of matrices that are diagonal in) a choice of basis for $H$. Similarly, $B$ consists of diagonal matrices in some (second) basis, possibly with repeated eigenvalues. So maximality forces $C$ to consist of all diagonal matrices in some (third) basis that spans the eigenspaces of the second one. The question is whether this third basis can be chosen while respecting $A \cap C=\mathbb{C}I$. If each eigenspace of $B$ has dimension an integer power of a prime number, then <a href="http://en.wikipedia.org/wiki/Mutually_unbiased_bases" rel="nofollow">mutually unbiased bases</a> are known to exist, and the answer is affirmative.</p>