The Unapologetic Mathematician

Mathematics for the interested outsider

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.

About these ads

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

No comments yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Follow

Get every new post delivered to your Inbox.

Join 391 other followers

%d bloggers like this: