Parallelization of a Sphere gives Division Algebra

Question

Is there an elementary proof of the fact, that a parallelization of $S^n$ can turn $\mathbb{R}^{n+1}$ into a division algebra?

My guess was something like this:

Let $v_1(x),\dots, v_{n}(x)$ denote the sections of my parallelization. I'm assuming that these are pairwise orthogonal. Now $v_1((1,0,\dots,0)),\dots, v_{n}((1,0,\dots,0))$together with $(1,0,\dots,0)$ defines a basis for $\mathbb{R}^{n+1}$. I would like to define multiplication $v_i((1,0,\dots,0))\times x$ as applying the rotation by 90° in the plane spanned by $v_i(x)$ and $x$ and multiplication by $(1,0,\dots,0)$ as the identity. The next step would be to extend this linearly. But this doesn't work since my parallelization is not invariant under the right things for this to be correct.

PS: With an elementary proof I mean, that it does not involve Milnor and Kervaires result.

Answer 1

Consider the statements:

(A1) There is a parallelization of $\textrm{S}^{n}$.

(A2) There is a structure of a division algebra on $\mathbb{R}^{n+1}$.

Then it is known that these statements are isomorphic. But now consider the sets:

$S1 = \{ \textrm{parallelizations of } \textrm{S}^n \}$

$S2 = \{ \mathbb{R}\textrm{-linear division algebra structures on } \mathbb{R}^{n+1} \}$

There is a standard map $S2 \to S1$ transporting the basis at $(1, 0 \ldots,0)$ via multiplication. But this map is in general not an equivalence in any sensible way. For example if $n =3$, there are up to homotopy two structures of division algebras on $\mathbb{R}^4$ (one can reverse the orientation), hence $\vert \pi_0 ( S2 ) \vert = 2$, but $\pi_0(S1) \,\tilde{=}\, \pi_3(O(3)) \,\tilde{=}\, \mathbb{Z}$.

Thus it seems there is in general no construction of a map in the other direction. (Of course, by the classification, we know that $S1$ is empty exactly if $S2$ is, so there is always a map.)