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


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.

  • $\begingroup$ Thanks! But I focus on infinite dimensional Hilbert space $H$. $\endgroup$ Jun 27 '12 at 6:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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