[math-fun] Bott periodicity
The NY Times just printed an obituary for Raoul Bott, Emeritus of the Harvard math dept. who passed away Dec. 20: <http://www.nytimes.com/2006/01/08/national/08bott.html>. It mentions one of his best-known results, "Bott periodicity", which blew me away when I first learned of it as a grad student. So in case anyone's interested, I thought I'd describe it here for those not already familiar with it. (Some technical details are omitted.) Given a topological space X, there are algebraic invariants which describe its holes, which can come in various dimensions. These include the groups pi_n(X) for each n = 1,2,3,.... We consider only spaces X for which any two points can be connected by a continuous path. As a set, pi_n(X) is all continuous maps of the unit n-sphere S^n into X where a preassigned point of S^n is required to map to a preassigned point (call it *) of X . . . and THEN any two such maps f,g: S^n -> X are considered equivalent if there is a continuous *family* of such maps {h_t, 0 <= t <= 1}, with h_0 = f and h_1 = g. This set is given a group structure, so each pi_n(X) is a group. (The "fundamental" group, pi_1(X), is the only one that can be (and often is) non-abelian; any pi_n(X) for n > 1 is abelian.) For some examples, pi_1(figure-8) is the free group on 2 generators. For any n, pi_n(S^n) is the group Z of integers. ------------------------------------------------------------------------------ A convergent thread starts with the fact that any Euclidean space R^n has its own group of rotations, which is also a topological space such that the group operations are continuous. If we also toss in the isometries of R^n preserving the origin which *reverse* orientation, we get the set of all invertible nxn matrices M such that M-transpose = M-inverse. This topological ("Lie") group is denoted by O(n), the orthogonal group of R^n. (O(n) corresponds to the set of all n-tuples of orthonormal vectors in R^n.) O(n) is locally Euclidean -- i.e., a manifold -- of dimension (n^2 - n)/2. Just as R^n naturally includes in R^(n+1) as the first n coordinates, we may form the matural inclusions O(1) in O(2) in . . . O(n) in O(n+1) in . . .. Using these inclusions we can take the union of all the O(n)'s for n >= 1 to get an all-inclusive group that we call O(oo). ------------------------------------------------------------------------------ Embodying all the symmetry of all Euclidean spaces R^n, O(oo) is a very interesting object. Raoul Bott showed that pi_n(O(oo)) depends only on n mod 8. Here is the table: n mod 8 0 1 2 3 4 5 6 7 ---------------------------------------------------------------------- pi_n(O(oo) Z_2 Z_2 0 Z 0 0 0 Z It is not a coincidence that the dimensions 0,1,3,7 of the non-zero groups mod 8 are the same as the dimensions of the unit spheres in the only normed real division algebras: the reals, complexes, quaternions, octonions; these four division algebras give rise to the existence of non-zero homotopy classes in pi_n(O(oo)). There is much more to Bott periodicity, but this will do for now. --Dan
On 1/8/06, Daniel Asimov <dasimov@earthlink.net> wrote:
The NY Times just printed an obituary for Raoul Bott, Emeritus of the Harvard math dept. who passed away Dec. 20: <http://www.nytimes.com/2006/01/08/national/08bott.html>. It mentions one of his best-known results, "Bott periodicity", which blew me away when I first learned of it as a grad student. [etc etc]
I have to admit I'd kinda assumed that "Bott periodicity" referred to the fact that the real Clifford algebra Cl(n+8,0) generated by n+8 square roots of +1 is isomorphic to 16x16 matrices over Cl(n,0), on no better grounds than that the relevant ur-text is M.F.Atiyah, R.Bott, A.Shapiro "Clifford Modules" Topology vol.3 (1964). Was this conclusion entirely unjustified, or is there some connection? Which reminds me that I recently acquired a copy of the august tome J.H.Conway, D.A Smith "On Quaternions and Octonions" and promptly became mired at Lemma 3 on page 6. I can prove this result easily, but I'm hanged if I can follow their (3 line) proof --- can another reader clarify? Fred Lunnon
On Sunday 08 January 2006 14:41, Fred Lunnon wrote:
Which reminds me that I recently acquired a copy of the august tome J.H.Conway, D.A Smith "On Quaternions and Octonions" and promptly became mired at Lemma 3 on page 6. I can prove this result easily, but I'm hanged if I can follow their (3 line) proof --- can another reader clarify? Fred Lunnon
I'll have a go. Let a be any element of GO(n) fixing a k-dimensional subspace. If it's the identity, we're done. Otherwise, pick something it doesn't fix; say it takes v to w. Let b be reflection in v-w. Then ba is still in GO(n); it fixes all the same things a did; and it also fixes v. So it fixes (at least) a (k+1)-dimensional subspace, so by induction on n-k we're done. How do we know that ba fixes all the same things a did? Because anything a fixes is orthogonal to v-w (Why? Because an isometry preserves inner products; so if au=u and av=w then (u,v) = (au,av) = (u,w), so (u,v-w)=0.) and is therefore fixed by b as well as by a. Is that any better? -- g
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Daniel Asimov -
Fred lunnon -
Gareth McCaughan