# The Unapologetic Mathematician

## The “Hairy Ball Theorem”

We can use the concept of degree to prove the (in)famous “hairy ball theorem”. That is, any smooth vector field defined on the whole sphere $S^n$ for $n$ even must vanish at some point. The name comes from thinking of a “hair” growing out of each point on the sphere, and trying to “comb” them all to lay flat — tangent — against the sphere. If this can be done, then the direction each hair points defines a unit vector at its point, and combing smoothly means we have a smooth vector field. The assertion is that such a “combing” is impossible.

First, I say that if $n$ is even then the antipodal map sending a point $p\in S^n$ to $-p$ is orientation-reversing. Indeed, we first extend to the larger space $\mathbb{R}^{n+1}$; the antipodal map $-1:\mathbb{R}^{n+1}\to\mathbb{R}^{n+1}$ is clearly orientation-reversing, as we can see by taking its Jacobian. This matrix has $n+1$ eigenvalues, all equal to $-1$, so the determinant is $(-1)^{n+1}=-1$.

If $p$ is the position vector of a point in $S^n$, then we consider the canonical identification $\mathcal{I}_p$ of $\mathbb{R}^{n+1}$ with $\mathcal{T}_p\mathbb{R}^{n+1}$ and calculate

$\displaystyle(-1)_*\mathcal{I}_pu=\mathcal{I}_{-p}(-u)=-\mathcal{I}_{-p}u$

So if $\{\mathcal{I}_pu_i\}$ is a positively-oriented basis of $\mathcal{T}_p\mathbb{R}^{n+1}$, then $\{-\mathcal{I}_{-p}u_i\}$ is a negatively-oriented basis of $\mathcal{T}_{-p}\mathbb{R}^{n+1}$.

Now, the position vector field $p\mapsto\mathcal{I}_pp$ is $-1$-related to itself; if $\mathcal{I}_pp$ is used as the first vector in a positively-oriented basis of $\mathcal{T}_p$, then $-\mathcal{I}_{-p}p=\mathcal{I}_{-p}(-p)$ is the first vector in the antipodal basis, which we know is negatively-oriented. But this means that the positively-oriented basis of $\mathcal{T}_pS^n$ must be flipped to a negatively-oriented basis of $\mathcal{T}_{-p}S^n$.

So what does this mean? The identity map has degree $1$, while we can calculate that the antipodal map has degree $-1$. Since these are different, the two maps must not act identically on homology, and therefore cannot be homotopic.

But now let’s assume that $X$ is an everywhere-nonzero vector field which, without loss of generality, we may assume to have constant length $1$ — use the inner product we get by taking $\mathcal{T}_pS^n\subseteq\mathcal{T}_p\mathbb{R}^{n+1}$ and divide by the original length to normalize. For each point $p$ we can define the great circle

$\displaystyle c_p(t)=(\cos(\pi t))p+(\sin(\pi t))\mathcal{I}_p^{-1}X(p)$

This is a curve that lies on the sphere for every $p$. Indeed, we can check that

\displaystyle\begin{aligned}\langle c_p(t),c_p(t)\rangle=&\langle(\cos(\pi t))p+(\sin(\pi t))\mathcal{I}_p^{-1}X(p),(\cos(\pi t))p+(\sin(\pi t))\mathcal{I}_p^{-1}X(p)\rangle\\=&\langle(\cos(\pi t))p,(\cos(\pi t))p\rangle\\&+\langle(\cos(\pi t))p,(\sin(\pi t))\mathcal{I}_p^{-1}X(p)\rangle\\&+\langle(\sin(\pi t))\mathcal{I}_p^{-1}X(p),(\cos(\pi t))p\rangle\\&+\langle(\sin(\pi t))\mathcal{I}_p^{-1}X(p),(\sin(\pi t))\mathcal{I}_p^{-1}X(p)\rangle\\=&(\cos(\pi t))^2\langle p,p\rangle\\&+\cos(\pi t)\sin(\pi t)\langle p,\mathcal{I}_p^{-1}X(p)\rangle\\&+\sin(\pi t)\cos(\pi t)\langle\mathcal{I}_p^{-1}X(p),p\rangle\\&+(\sin(\pi t))^2\langle\mathcal{I}_p^{-1}X(p),\mathcal{I}_p^{-1}X(p)\rangle\\=&(\cos(\pi t))^2+(\sin(\pi t))^2=1\end{aligned}

So if we define $H:S^n\times[0,1]\to S^n$ by $H(p,t)=c_p(t)$, then we have a homotopy from the identity map on $S^n$ to the antipodal map, which is exactly what we just showed could not exist. Thus we conclude that no such non-vanishing vector field can exist on an even-dimensional sphere.

December 13, 2011 - Posted by | Differential Topology, Topology