Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Background: I've been studying some flows on vector spaces associated with linear actions of Lie groups, with the hope of using these flows to prove things about quotient spaces. The most successful tool so far has been a certain kind of inequality for functions associated with the action. I'm trying to generalize this inequality, and my question concerns a key step in the proof. To avoid unnecessary baggage and notation, I'll state the problem purely in terms of linear algebra.

Let $V$ be a vector subspace of the space of $n\times n$ skew-hermitian matrices (I do not assume that $V$ is closed under the matrix commutator). Call a subspace $V'$ of $V$ commuting if $[A,B] = 0$ for all $A,B \in V'$, and maximal commuting if $V'$ is not properly contained in any other commuting subspace.

My question is: is there a nice description of the maximal commuting subspaces of $V$?

Motivating example: if $V$ is itself the Lie algebra of some closed subgroup $G$ of $U(n)$, then a maximal commuting subspace is just a Cartan subalgebra, and Lie theory tells us that any two such subspaces are conjugate by some $g \in G \subset U(n)$.

Since my motivating example is already a nontrivial result in Lie theory, I suspect that the general case is hard. I thought about mimicking this example as follows: define $G_V$ to be the group of $g \in U(n)$ such that $g V g^{-1} \subset V$. Call two maximal commuting subspaces equivalent if there are conjugate via some $g \in G_V$. Is there any hope of proving a result like "there are only finitely many equivalence classes of maximal commuting subspaces"? If not, are there any hypotheses that we can put on $V$ to guarantee such a result?

share|improve this question
    
Related question: mathoverflow.net/questions/48764/… –  jvp Jul 6 '11 at 21:43

1 Answer 1

up vote 6 down vote accepted

There is no hope of a general result. Look at the vector space

$$ V = \left \lbrace t \begin{pmatrix} i & 0 \newline 0 & 0\end{pmatrix} + s\begin{pmatrix} 0 & 1 \newline -1 & 0 \end{pmatrix} \right \rbrace . $$

So $V$ is two dimensional and not closed under the commutator. Any one-dimensional subspace is maximal commuting.

The eigenvalues of a matrix $M(t,s) \in V$ are by explicit computation:

$$\lambda_\pm(t,s) = i t \left [ 1 \pm \sqrt{1+4 \frac{s^2}{t^2}} \right ].$$

A maximal commuting subspace is characterized by the ratio $\alpha =s/t$ and the ratio

$$\frac{\lambda_+(t,s)}{\lambda_-(t,s)} = \frac{1 + \sqrt{1 +4 \alpha^2}}{1-\sqrt{1+4 \alpha^2}}$$

is constant on the subspace. Since this ratio is a unitary invariant and all positive values of $\alpha$ occur, there are uncountably many equivalence classes.

share|improve this answer
    
Thanks for this, it's extremely helpful to have such a simple and explicit example to work with. –  Jonathan Jul 6 '11 at 22:45

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.