# Hairy Ball Theorem: homotopy from the identity to the antipodal map

I'm having trouble to prove the following statement:

If the $n$-sphere $S^n\subset\mathbb{R}^{n+1}$ admits a (continuous) nonvanishing tangent vector field, then $n$ is odd.

The idea of the proof is pretty clear on my mind: we use the fact that the antipodal map has degree $(-1)^{n+1}$ and that the Brouwer degree is invariant under homotopies.

We have a (continuous) nonvanishing tangent vector field $v\colon S^n\to\mathbb{R}^{n+1}$ that can be supposed unitary (otherwise, just consider $u=v/\|v\|$). Paraphrasing Hirsch: "a homotopy of $S^n$ from the identity map to the antipodal map is obtained moving each $p\in S^n$ to $-p$ along the great semicircle in the direction $v(p)$".

My question is: is there an explicit formula for this homotopy?

The application $H\colon S^n\times[0,1]\to S^n$ given by $$H(x,t) = \cos(\pi t)x + \sin(\pi t) v(x)$$ is what I was looking for.
• That will not stay in $S^n$. Jun 3, 2017 at 20:25
• @PVAL-inactive Well, for all $t\in[0,1]$ and all $x\in S^n$, we have that $\|\cos(\pi t)x + \sin(\pi t)v(x)\|^2 = \langle \cos(\pi t)x + \sin(\pi t)v(x), \cos(\pi t)x + \sin(\pi t)v(x) \rangle = \cos^2(\pi t)\|x\|^2 + \cos(\pi t)\sin(\pi t)\langle x,v(x)\rangle + \sin^2(\pi t)\|v(x)\|^2 = \cos^2(\pi t) + \sin^2(\pi t) = 1$, since $\|x\| = \|v(x)\| = 1$ and $\langle x,v(x) \rangle = 0$. Jun 4, 2017 at 4:54
How about this? Define $\Phi:S^n\times[0,1]\to S^n$, $n=2m-1$ by $\Phi(v,t)=A(t)v$ where $A(t)$ is the matrix $$\pmatrix{ \cos \pi t&\sin\pi t&&&&&&\\ -\sin \pi t&\cos\pi t&&&&&&\\ &&\cos \pi t&\sin\pi t&&&&\\ &&-\sin \pi t&\cos\pi t&&&&\\ &&&&\ddots&&&\\ &&&&&\cos \pi t&\sin\pi t\\ &&&&&-\sin\pi t&\cos\pi t\\ }.$$
• Nice trick! But I have to prove that $n$ is odd, haven't I? May 31, 2017 at 18:30
• @rldias Lord Shark is thinking of $S^n$ as a subset of $\Bbb R^{n+1}$. There is a reason this construction only makes sense when $n$ is odd (i.e. when $n+1$ is even). Jun 3, 2017 at 20:27