Remember that exercise I mentioned, about the path components of a certain subset of the orthogonal group O(n)?
Consider the subspace Xn of orthogonal matrices with square the identity: in other words, symmetric orthogonal matrices. Let O(n) act on these by conjugation: certainly any orthogonal conjugate of a matrix in Xb is still in Xn. Consider the conjugacy classes. Since symmetric matrices are orthogonally diagonalisable, each conjugacy class contains a diagonal matrix; since it's orthogonal, its diagonal entries are +1 and -1 with certain multiplicities. Two such matrices are similar iff they have the same number of +1 entries; thus we have distinct classes Ck, k=0...n, where Ck is the conjugacy class of the diagonal matrix Dk with k +1 entries followed by n-k -1 entries. The elements of this conjugacy class are uniquely determined by their 1-eigenspaces, which are k-planes in Rn: in fact, Ck is the Grassmannian GkRn.
Note that each conjugacy class is path-connected as a subspace of Xn, since the special orthogonal group SO(n) is path-connected and acts transitively on each class. Note also that the trace function takes on the constant value 2k-n on each Ck. Therefore, the Ck are exactly the path components of Xn.
Since O(n) acts transitively on each Ck with the isotropy group of Dk a closed Lie subgroup--the same as the isotropy group of the standard k-plane in the Grassmannian; you can find it explicitly without much fuss--we can give Ck a unique smooth manifold structure so that the O(n)-action is smooth; with this structure, Ck is diffeomorphic to O(n) mod the isotropy group, which is also diffeomorphic to the Grassmannian; so Ck is diffeomorphic to the Grassmannian of k-planes (equivariantly, even). It's not too hard to go on and show that with this smooth structure, Ck is an embedded submanifold of O(n).
So there.
Posted by aloysius at March 02, 2004 04:35 PM | TrackBack |