Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\mathcal{H}$ be a Hilbert space and $B(\mathcal{H})$ the algebra of bounded linear operators on $\mathcal{H}$. A MASA $\mathcal{M}$ is a subalgebra of $B(\mathcal{H})$ that is abelian and self-adjoint, and is maximal with respect to these conditions.

MASAs are quite standard materials in operator theory, but I wonder whether we have studied MAAs, that is, we drop the condition that the algebra has to be closed under involution.

Just like for MASAs, MAA exists if we assume Zorn's lemma. But MASAs can be constructed by hand (for instance, $\mathcal{H}=\mathcal{L}^2(X,\mu)$, a MASA is the multiplication algebra). I am not sure whether we have such kind of explicit examples of MAA.

SO, do we have a nice example of MAA (on $\mathcal{L}^2$ maybe)? Do we know something about MAAs?

Thanks!

share|improve this question
2  
This might be appropriate for mathoverflow, if you don't receive answers here. –  Potato Jan 2 '13 at 6:09
    
@Potato Is there a convenient way to migrate questions to MO? –  Hui Yu Jan 14 '13 at 16:32
    
I think you could just repost it with a link to this question and a note saying you are reposting it because you received no answers. –  Potato Jan 14 '13 at 17:54

1 Answer 1

up vote 2 down vote accepted

The usual proof that $L^\infty$ is a MASA in actuality proves that it's an MAA in your sense; that is, to prove that a given operator $T$ is a multiplication operator all you need to know is that it commutes with all the other multiplication operators (and oftentimes all you need to know is that it commutes with one particular multiplication operator; see e.g. Arveson's "A short course on spectral theory", section 4.1).

On the other hand, there are MAA's which are not MASAS. This is easiest to see in the finite-dimensional case. Let $S$ be the $n\times n$ matrix with $1$'s along the diagonal just below the main diagonal, and $0$'s elsewhere (so $S$ is just the unilateral shift). It's not hard to prove that any matrix which commutes with $S$ must be lower-triangular and constant along all the diagonals, and these are exactly the matrices given by polynomials in $S$; hence they form an MAA. But obviously this algebra is not self-adjoint. A similar argument also works in the infinite-dimensional case.

share|improve this answer

Your Answer

 
discard

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.