MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $A\cap C={\mathbb C}I$?

share|cite|improve this question

This is only a partial answer, but it didn't fit in the comment box.

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 mutually unbiased bases are known to exist, and the answer is affirmative.

share|cite|improve this answer
Thanks! But I focus on infinite dimensional Hilbert space $H$. – blackpeacock Jun 27 '12 at 6:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.