From: rusin@vesuvius.math.niu.edu (Dave Rusin) Newsgroups: sci.physics,sci.math Subject: Re: S^1, S^3, and S^7 are parallelizable. Date: 9 Feb 1998 21:18:37 GMT I had tried to outline an easy proof that S^{n-1} parallelizable implies n is a power of two. The proof used the assumption that RP^{n-1} was parallelizable to draw the conclusion. John Baez objected: >Actually, why is this true? I can see this if we have a trivialization >of the tangent bundle of S^{n-1} which is invariant under the map sending >each point to its antipode, but I don't see why the parallelizability >of S^{n-1} implies the existence of a trivialization with this property. Indeed, the implications of parallelizability do go the other way but only go this way if the vector fields are nice enough (although since when the smoke clears it's the same n=1, 2, 4, 8 in both cases, I guess there is an implication both ways after the fact.) Let me then offer a comparable proof for spheres: see Steenrod and JHC Whitehead Proc Nat Acad Sci 37 (1951) 58-63. If you had n-1 tangent vector fields, you could perform a Gram-Schmidt reduction to them (and the normal vector) to get an orthogonal matrix at each point, i.e., a map S^{n-1} -> O(n) which splits the projection O(n) -> S^{n-1} onto the first column. But this would give corresponding maps in mod-2 cohomology H*(O(n)) <-> H*(S^{n-1}). It turns out that on the ring level there is no obstruction here, but each side is an algebra over the mod-2 Steenrod algebra A_2, and the map of A_2 modules does not split. This Steenrod/Whitehead proof has the added feature that it gives a more general upper bound on the number of independent vector fields on a sphere: if n = s*2^r with s odd, there cannot be as many as 2^r vector fields on S^{n-1}. This includes the hairy ball theorem (r=0) and the parallelizable case (s=1) as two extremes. dave