Three more responses on Grassmanians (to see the previous ones, go to the archive (www.lehigh.edu/~dmd1/postings.html))...........DMD _______________________________________________________ Date: Thu, 28 Sep 2000 13:06:30 -0600 (MDT) From: "Peter D. Zvengrowski" Subject: Re: another on Grassmanians Dear Doug, The approach mentioned by Goodwillie and Korbas is more or less the same as Milnor-Stasheff, see an exercise at the end of the chapter on Grassmann manifolds. I recall seeing a paper (while I visited Bochum in 1998) that gave an explicit metric for Grassmann manifolds, in terms of what one would think of as the angle between two k-planes. Unfortunately I can't recall either the journal or the authors, but it did come out in 98. Regards, Peter Zvengrowski _________________________________________________ Subject: Re: 2 more on Grassmanians Date: Thu, 28 Sep 2000 23:13:27 -0400 (EDT) From: "Tom Goodwillie,304 Kassar,863-2590,617-926-3565" Doug, at first I thought you were seeking to make the Grassmannian into a Riemannian manifold. My suggestion was (meant to be) to embed it in the matrices and let it inherit a Riemannian metric from them. Now I see that you wanted to make it a metric space, so we are thinking of the straight-line distance between two projection matrices. (Norm of the difference.) But, come to think of it, there are various norms available. I was thinking that the square of the norm of a matrix is the sum of the squares of its entries, so that for a symmetric matrix the square of the norm is also the sum of the squares of the eigenvalues. With this norm, the metric is not proportional to the Hausdorff metric. But what if we use the operator norm, so that the norm of a symmetric matrix is the maximum of the absolute values of an eigenvalues? it seems to me that this makes the metric exactly equal to the Hausdorff metric. The only proof I've thought of (I haven't written it down, but I think it's right) involves reducing to the case of G(1,2) and then calculating. Tom Goodwillie ____________________________________________________ Date: Thu, 28 Sep 2000 22:39:58 -0500 (CDT) From: Bill Richter Subject: Re: 2 more on Grassmanians 2. I can think of another way to define a metric on G(m,n). Each subspace gives us a compact subset of the unit ball in R^n. One has the Hausdorf metric on the set of all nonempty compact subsets of R^n defined as follows. The distance from A to B to be the samllest number r such that each point in A is within r of some point in B and vice versa. Is this metric on G(m, n) a scalar multiple of the one below? No, not if I understand you. Take the simplest case, lines in R^2, and the lines through the points V = (1,0) W = ( cos(t), sin(t) ) for t in [0, pi/2]. Then your metric is the distance between V & W, which is sqrt{2} sqrt{ 1 - cos(t) } But Tom's metric is the norm of the 2x2 matrix pi_W - pi_V the difference between the orthogonal projections onto these lines. pi_V = 1 0 0 0 pi_W = c^2 sc cs s^2 pi_W - pi_V = -s^2 sc cs s^2 which is sin(t) times a reflection times a rotation matrix, so the max norm of pi_W - pi_V is | pi_W - pi_V | = sin(t) The Euclidean norm (which I don't like as much) just gives you a scalar multiple | pi_W - pi_V | = sqrt{2} sin(t) But that's not a scalar multiple of sqrt{ 1 - cos(t) } -- Bill