From: ikastan@uranus.uucp (Ilias Kastanas) Subject: Re: countable compactness / finite intersection Date: 1 Apr 2000 02:03:50 GMT Newsgroups: sci.math Summary: [missing] In article <8bsi1e$ads$1@nnrp1.deja.com>, Robin Chapman wrote: ... @Something slightly less elementary might be Hocking and Young's "proof" @of the nontrivialilty of the Hopf fibration (p. 186). I'm not convinced @by this. For a start I'm sure their equations for x_1, x_3 and x_4 @in terms of alpha and beta are wrong, and in any case this would @not extend to a continuous function y on S^2 (which they need). @The function y would fail to be continuous at one of the "poles" @of the sphere. Can *this* proof be rescued? Indeed those equations are wrong; flip the sign of cos(beta). The north pole is supposed to correspond to the 2-plane x_1 = x_2 = 0, and the south pole to x_3 = x_4 = 0. Ilias ============================================================================== From: Robin Chapman Subject: Re: countable compactness / finite intersection Date: Fri, 31 Mar 2000 23:17:24 -0800 Newsgroups: sci.math Summary: [missing] In article <8c3le6$8a4m$1@hades.csu.net>, ikastan@uranus.uucp (Ilias Kastanas) wrote: >In article <8bsi1e$ads$1@nnrp1.deja.com>, >Robin Chapman wrote: >.... >@Something slightly less elementary might be Hocking and Young's "proof" >@of the nontrivialilty of the Hopf fibration (p. 186). I'm not convinced >@by this. For a start I'm sure their equations for x_1, x_3 and x_4 >@in terms of alpha and beta are wrong, and in any case this would >@not extend to a continuous function y on S^2 (which they need). >@The function y would fail to be continuous at one of the "poles" >@of the sphere. Can *this* proof be rescued? > > > > Indeed those equations are wrong; flip the sign of cos(beta). >The north pole is supposed to correspond to the 2-plane x_1 = x_2 = 0, >and the south pole to x_3 = x_4 = 0. But even with the corrected formula surely the function y (defined on the 2-sphere sans poles) will not extend to a continuous function on all of S^2? Robin Chapman * Sent from RemarQ http://www.remarq.com The Internet's Discussion Network * The fastest and easiest way to search and participate in Usenet - Free! ============================================================================== From: ikastan@uranus.uucp (Ilias Kastanas) Subject: Re: countable compactness / finite intersection Date: 3 Apr 2000 14:51:43 GMT Newsgroups: sci.math Summary: [missing] In article <01741414.22cdbbdd@usw-ex0105-036.remarq.com>, Robin Chapman wrote: [quote of previous message deleted --djr] I gave a telegraphic (if not Delphic) answer, not addressing the full question... a nice question too -- thanks for bringing it up, Robin! Such a proof would be remarkably simple; no Hopf invariant, cohomology of differential forms or whatnot; just homotopy. For readers who don't have H&Y handy: the idea is to build a continuous y "inverting" the Hopf map h: S^3 -> S^2. They want to pick y(alpha,beta) in S^1(alpha,beta) by intersecting the latter with E, the 3-dim subspace x_1=x_2... in fact, just with E+, the x_1=x_2 >= 0 half of E. [For a "clean" version of h, think of S^3 as the unit sphere in C^2 rather than R^4. For each z in C - {0} identify all (w,zw) in C^2, thus obtaining the complex projective line, PC^1. Then h is the natural map from S^3 to PC^1. It is easy to view PC^1 as S^2: each equivalence class contains an (r_1 e^i(phi), r_2), r_1^2 + r_2^2 = 1, r_i >= 0; the r_1 e^i(phi) fill the unit disk D^2, while its boundary r_1=1 corresponds to a single class; this amounts to S^2. (In effect, we are using stereo- graphic projection). Note that points of S^3 "differing by a factor of e^i(theta)" have the same image under h -- hence the circles S^1 above, h^-1 of singletons.] Well, as Robin pointed out, the construction via E+ doesn't work; the plane x_1=x_2=0 (beta=0, alpha immaterial) is a subspace of E, i.e. an entire S^1 is contained in E; intersection with E+ yields a semi- circle, not a single point. No way to patch it up either. Now this is not accidental; such a construction _has_ to fail, for a simple, general reason: the envisioned y is 1-to-1, and so a homeomorphism (S^2 being compact). Its range would be inside the intersection of E+ and S^3, that is, a 2-hemisphere -- which is homeo- morphic to a subset of R^2. Thus y would be a homeomorphism of S^2 and a subset of the plane; that's impossible. One has to look for a different method then. Clearly, using all of E (or some other 3-dim E) does not help; E meets S^3 in a 2-sphere, but y cannot be onto it (each S^1 has at least two points on the 2-sphere), and we are in the same boat as before. Is there a more involved approach... a range sitting inside S^3 in some funny way?! No; no amount of tinkering can work. A y as desired does not exist. Suppose a y existed. Its range would be a subset of S^3 homeomorphic to S^2, as seen above; call it Sigma^2. The composition f of h and y would map S^3 to Sigma^2, and each one of h's S^1's would have all its points mapped by f to a single x in S^1. In particular, f(x) = x. Since the set of x's is just Sigma^2, f would be a retraction of S^3 to Sigma^2. But there is no such thing. Ilias