Date: Fri, 29 Sep 2000 22:27:07 -0500 (CDT) From: Bill Richter Subject: Re: 3 more on Grassmanians Back to Doug Ravenel's question: 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 smallest 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? I was wrong, and Tom Goodwillie was right: 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. My error was using the unit sphere rather than the unit ball. That is, the distance between the unit vectors V = (1,0) W = ( cos(t), sin(t) ) for t in [0, pi/2]. is indeed sqrt{2} sqrt{ 1 - cos(t) }. But the distance from W to the line through V is sin(t), which is also the max norm distance | pi_W - pi_V | from pi_W to pi_V. My apologies to the list for not reading Doug's definition carefully. And as Tom says, the pattern persists! For V, W in G_{n,k} = { V^k subset R^n}, the Hausdorff distance from V cap B^n to W cap B^n is the operator norm distance | pi_W - pi_V | between pi_W, pi_V in M_nxn(R). I think Tom's proof indeed works (reducing to the case of G(1,2) and then calculating), but here's another proof, following my unpublished thesis on attaching maps of Schubert bundle decompositions, which sheds some light on why the max norm is coming up: Proof: We reduce (by generalities on orthogonal complements and the max norm) to the case when W cap V^perp = 0. Then W = V_f, the graph of a function f: V ---> V^perp, i.e. the image of the map [1 f] : V ---> V + V^perp = R^n. Using the real version of the polar form, we can uniquely write f as f = u tan(theta), for an orthogonal map u and a positive (symmetric + eigenvalues positive) map theta with max norm |theta| < pi/2. Actually, u and theta are only defined the ortho complement of Ker(f). Then we have: | pi_W - pi_V | = D_Hausdorff(V, W) = | sin(theta) | i.e. the maximum eigenvalue of the positive operator sin(theta). I find that a satisfying formula, because the obvious psuedo-distance function is D( V_f, V) = | f | and D(W, V) = infty for W cap V^perp \ne 0. So we're moving from | f | = | tan(theta) | to | sin(theta) |. To see this, reduce to the case when f is injective. Then the orthogonal map u conjugates away, and we're reduced to f = tan(theta): V ---> V for a symmetric map theta: V ---> V with positive eigenvalues. Then W subset V + V is the image of the map [ cos(theta), sin(theta) ] : V ---> V + V and our actual V is V + 0 subset V + V. For any unit vector v in V, the distance from [ cos(theta) v, sin(theta) v ] to V + 0 is | sin(theta) v |, and so the Hausdorff distance from V to W is D_Hausdorff(V, W) = max_{unit vector v in V} | sin(theta) v | = | sin(theta) | by definition of the max norm (or operator norm). But we write pi_V as pi_V = 1 0 0 0 w.r.t. V + V, and as before pi_W = [ cos(theta)^2 sin(theta) cos(theta) ] [ ] [ cos(theta) sin(theta) sin(theta)^2 ] One way to see this is that we're conjugating pi_V with the exponential of the matrix [ 0 -theta ] [ ] [theta 0 ] which is [ cos(theta) -sin(theta) ] [ ] [ sin(theta) cos(theta) ] . We're used to exponentiating square matrices, but sin, cos and tan work just as well on square matrices. Then as before, pi_W - pi_V = -sin(theta)^2 sin(theta) cos(theta) cos(theta) sin(theta) sin(theta)^2 which is the product of the 2x2 matrix selfmaps of V + V: [ sin(theta) 0 ] [ -1 0 ] [ ] * [ ] * [ 0 sin(theta) ] [ 0 1 ] times [ sin(theta) -cos(theta) ] [ ] [ cos(theta) sin(theta) ] which is the exponential of the matrix pi/2 - theta : V ---> V So pi_W - pi_V = [ sin(theta) 0 ] [ ] * O [ 0 sin(theta) ] where O is an orthogonal matrix. An easy fact about the max norm is that the max norm of the map [f 0] * O [0 g] is the maximum of the max norms |f| and |g|. Therefore, | pi_W - pi_V | = | sin(theta) |. \qed Now that we have this result, we can go back to our reduction and see that if W cap V^perp \ne 0, then | pi_W - pi_V | = D_Hausdorff(V, W) = 1 -- Bill