Intersections of maximal abelian von Neumann algebras - MathOverflow 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 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 Answer by Chris Heunen for Intersections of maximal abelian von Neumann algebras 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>