Vector fields on $(4n+1)$-spheres

Question

If $n$ is odd then $S^{n-1}$ doesn't admit a nowhere-vanishing vector field, and if $n$ is even then there does exist one (Hairy Ball Theorem). We can then ask, on $S^{n-1}$, what is the maximum number $k(n)$ of linearly independent vector fields? Rewriting $n=2^{4a+b}(2s+1)$, Adams computes $k(n)=2^b+8a-1$.

In particular, on the $(4n+1)$-spheres, there is only one nowhere-vanishing vector field up to linear-independence, whereas in every other (odd) dimension there are more.
Example: on the circle $S^1$ there are the vector fields generated by (counter)clockwise rotation, but these are the same up to a scalar. This makes sense: I start flowing along this single dimension and then I have to continue flowing in that direction until I come back to my starting point.
I tried considering the difference between $S^3$ and $S^5$, which fiber over $\mathbb{C}P^1$ and $\mathbb{C}P^2$ respectively. A nowhere-vanishing vector field in both cases is given by taking the standard nowhere-vanishing vector field on the $S^1$-fiber. But for $S^3$ there are three linearly-independent fields (the $i,j,k$-directions when representing $S^3$ as the unit quaternions -- is there a way to see this using the fibration picture?), whereas for some reason $S^5$ can only admit the one.

What can be the differential/topological reasoning behind this? I.e. is there a down-to-earth way to deduce this result on $S^{4n+1}$, or for starters, $S^5$?

Could there possibly be an analogous index theorem going on here, in the same way that the Poincare-Hopf theorem provides us the Hairy Ball result?

Answer 1

The Radon-Hurwitz number $k(n)$ is the largest $k$ such that there exists an orthogonal multiplication $\mathbb R^k\times \mathbb R^n\to \mathbb R^n$; so for an ONB $x_1,\dots, x_k$ of $\mathbb R^k$ and a unit vector $y\in \mathbb R^n$ the vectors $y, x_1.y, x_2.y,\dots x_k.y$ are orthogonal in $\mathbb R^n$. This describes vector fields on $S^{n-1}$. The orthogonal multiplications were constructed by Radon [Lineare Scharen orthogonaler Matrizen, Abh. Math. Sem. Univ. Hamburg 1, 1-14, 1921] who extended the construction of Hurwitz for $k=n=1,2,4,8$. They extend to representations of Clifford algebras $C(\mathbb R^k, -\langle\quad,\quad\rangle)$ which explains the "periodicity" in $k(n)$ with respect to 8 and 2.

Radon writes: "For real matrices the solution offered itself to me by a particular reduction method using matrices whose elements are complex numbers or quaternions." (My translation) Maybe, there is an inkling of the fiber method you are looking at.

Adams showed that there are not more linearly independent vector fields on the sphere $S^{n-1}$.

Answer 2

Thanks Misha for the reference! (Just rewriting it here to complete this thread).
It seems that an off-the-cuff calculation (what I refer to as "down-to-earth") probably isn't going to suffice; there is some intricate stuff going on in the proofs involving the homotopy groups of the rotation groups of the spheres and orthogonal frames.

Homotopy Properties of the Real Orthogonal Groups (Whitehead), 1941, which focuses precisely on our case in question. The proof comes down to the obstruction of certain maps induced from rotation groups and their fixed subspaces on spheres.

Vector Fields on the n-Sphere (Steenrod, Whitehead) 1950, which studies particular fibrations of Stiefel manifolds over the spheres. It proves that for $n=2^k(2m+1)-1$ we have $k(n)<2^k$, which includes our case in question.