The theory of Clifford algebras gives us an explicit lower bound for the number of linearly independent vector fields on the $n$-sphere, and Adams proved that this is actually always the best possible: there are never more linearly independent vector fields.
More precisely, this gives the following number: if $n+1 = 16^a 2^b c$ with $c$ odd, $0 \leqslant b \leqslant 3$, we get $\rho(n) = 2^b + 8a$ and there are exactly $\rho(n) - 1$ linearly independent vector fields on $S^n$. This lower bound comes by construction of vector fields from Clifford module structures on $\mathbb{R}^{n+1}$, and figuring these out isn't too hard, it follows from the classification of real Clifford algebras with negative definite quadratic form. This is detailed for example in Fibre Bundles by Husemöller; the material comes from the paper Clifford Modules by Atiyah, Bott, Shapiro. This classification hinges on a particular mod 8 periodicity for real Clifford algebras.

Question: How does this description of vector fields on spheres relate to Bott Periodicity in the real case (either for real $K$-Theory, in the form $KO^{n+8} \cong KO^{n}$, or for the homotopy groups of the infinite orthogonal group, $\pi_{n+8}(O) \cong \pi_n(O)$)?

In particular, I'm inclined to think there should be a rather direct relationship: after all, $K$-theory is talking about vector bundles, sections of which are vector fields! Surely the formula for the number of vector fields on spheres should have a concrete interpretation in terms of $K$-theory? The (underlying) mod $8$ periodicities must be linked!
In addition, the result of periodicity mod 8 for Clifford algebras is also often called Bott Periodicity; what is the deeper relationship here? This other post mentions that the periodicity for Clifford algebras relates to the periodicity for complex K-Theory and so it mentions BU and not BO.

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    $\begingroup$ Husemoller's Fiber Bundles has details, if I recall correctly. $\endgroup$ Mar 14, 2010 at 20:24
  • $\begingroup$ I was looking at that, but I didn't find any information about Bott Periodicity in the real case. I think the answer will probably involve something about the J-homomorphism and the Adams conjecture, but I don't really know much about that. $\endgroup$ Mar 14, 2010 at 20:48
  • $\begingroup$ Have you looked at the paper by Atiyah, Bott, and Shapiro, "Clifford Modules" (Topology v.3 supp.1)? This talks about both the real and complex cases, and explicitly relates modules over clifford algebras to the homotopy of the infinite orthogonal group and infinite unitary group. In the last section, they discuss the relation between Clifford algebras and the vector field problem. $\endgroup$ Mar 15, 2010 at 20:14
  • $\begingroup$ I have been trying to read this paper, but I find it quite technical at times. Also, I did not find an explicit explanation of the relation between KO and the vector field problem, but maybe someone who has a better understanding of the material can find this connection implicitly described. $\endgroup$ Mar 28, 2010 at 2:00
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    $\begingroup$ I take it you are unsatisfied with Adams' account in his Annals paper "Vector Fields on Spheres", but if you haven't read that, you should. He is a master of exposition. $\endgroup$ Feb 27, 2012 at 14:25

3 Answers 3


This is not really answering your question. But it's worth pointing out that periodicity of Clifford algebras (closely tied to Bott periodicity) already gives you the "periodicity" in the explicit lower bound. If you want a sphere with $8a-1$ independent vector fields, you can take an irreducible representation $M(8a)$ of the real Clifford algebra $C(8a)$; the Clifford module structure provides that many vector fields on the unit sphere in $M$. The algebras $C(8a)$ are all Morita equivalent, and since $\dim_{\mathbb{R}}C(8(a+1)) = 16^2\, \dim_{\mathbb{R}}C(8a)$, you must have $\dim M(8(a+1))=16\, \dim M(8a)$, so we learn that $\rho(16^a-1)\geq 8a-1$.

As for the upper bound, this reduces to something about the real K-theory of truncated projective spaces, which Adams calculated, and which clearly has something to do with Bott periodicity.

But I would agree that this all sounds like a big coincidence. I wish I knew something better to say about it.

  • $\begingroup$ This is more or less what I had in mind about the lower bound: you get the vector fields on spheres from Clifford algebras, and the Clifford algebras of interest do satisfy a form of Bott Periodicity. It is this relation that motivated my curiosity about how this form of Bott periodicity (for Clifford algebras) and the form for K-Theory tie in; and especially how this becomes apparent in the situation of vector fields on spheres. $\endgroup$ Mar 15, 2010 at 0:57

Another source for this material: Matt Ando's notes from a course Haynes Miller gave on this subject are available from Miller's webpage: Part I, Part II.

(I couldn't get the links to appear correctly in the comment box, but I copied the exact same html code here and it works just fine. Anyone know why?)


This is not really different from Charles's answer, but you might want to look at two papers by Beno Eckmann. They construct Hurwitz-Radon matrices, and point out that this is effectively saying that a linearized form of the homotopy groups of the stable classical groups is the same as their homotopy groups, and that if this could be established directly by some form of approximation, you would have a very transparent proof of Bott periodicity.

The papers are Beno Eckmann, "Hurwitz-Radon Matrices and Periodicity Modulo 8", L'Enseignement Mathematique Vol 35 (1989) pp 77-91 (which you can get on SEALS)

and Beno Eckmann," Hurwitz-Radon Matrices Revisited: From Effective Solution of the Hurwitz Matrix Equations to Bott Periodicity" pp 23-35 in "the Hilton Symposium 1993" CRM Proceedings and Lecture Notes Vol 6

  • $\begingroup$ That's a cute paper. It looks to me that Eckmann is doing something similar to what Atiyah, Bott, & Shapiro talk about in their famous paper, only he's using a finite group $G_s$ in place of the clifford algebra $C(s)$. $\endgroup$ Mar 15, 2010 at 13:53

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