I am the Manifolds King! I make all manifolds my bitches! (Yea, even space-time itself.)
The problem was this: let Xn={A in O(n) : A2=I}. That is, it's the set of all symmetric orthogonal nxn matrices. Your mission, if you choose to accept it, is to show that all the path components of Xn (as a subspace of O(n)) are embedded submanifolds of O(n), and O(n)-homogeneous spaces, and in fact figure out which familiar homogeneous spaces they are.
I exhibited the path components in what I thought was an awfully nice way...And as soon as I saw them, I had a sudden flash of intuition revealing exactly what these manifolds had to be. It just seemed right, even though there didn't seem to be any geometric or topological reason for it...But I could see how I might build a mapping between them. When I worked through the details, it turned out I was right. Not only did I have intuition, I had correct intuition! That is just about the coolest thing ever.
I'm not going to tell you what they turned out to be. The problem's not due until Monday, and I don't want to give anything away if, by some bizarre twist of Fate or Chance someone else in my Manifolds class should read this before then. I do not want to spoil anyone's fun. For it was fun! Yes, it was fun when I finished writing it down...Heady, even.
There were a lot of details to wade through, it's true, the Equivariant Rank Theorem spraying left and right, remembering that all symmetric matrices are orthogonally diagonalisable, exhibiting the path components as different level sets for the trace map, a few commutative diagrams...
But the payoff was pretty.
Also, I think the set of all Borel subgroups of a complex semisimple Lie group, since it carries a transitive action of the group by conjugation, and the isotropy group of any Borel is the Borel itself (since it's its own normaliser), should have the structure of a smooth manifold, via identifying it with the coset space of any handy Borel.
Posted by aloysius at February 28, 2004 12:02 AM | TrackBack |