A couple of weeks ago, when Miodrag Milenkovic posed an interesting general problem in connection with POW-7, I was reminded of the “hairy ball theorem” (obviously a phrase invented during a more innocent era!), and a surprisingly easy proof of same that John Baez once told me over beers in an English pub. John is quite a good story-teller, and I wasn’t able to guess the punch line of the proof before it came out of his mouth –when it came, I was so surprised that I nearly fell off my stool! Well, I happened to run across the original source of this proof recently, and though it may be “old news” for some, it’s such a nice proof that I thought it was worth sharing.

The hairy ball theorem says: every continuous tangent vector field on a sphere of even dimension must vanish somewhere (at some point of the sphere, the tangent vector is zero). In the case of an ordinary 2-dimensional sphere, if you think of a vector at a point as a little “hair” emanating from that point, then the theorem says that you can’t comb the hairs of a sphere so that they all lie flat against the sphere: there will be a cowlick sticking straight out somewhere.

The classical proofs usually make some kind of appeal to homology theory: for example, a deep result is that the Euler characteristic of a compact manifold can be computed in terms of any continuously differentiable tangent vector field, by adding up the so-called “indices” of the vector field in the neighborhoods of critical points, where the vector field vanishes (a technical result shows there is no loss of generality in assuming the vector field is continuously differentiable). If the vector field vanishes nowhere, then the Euler characteristic is the empty sum 0; this cannot be in the case of an even-dimensional sphere, because its Euler characteristic is 2. The hairy ball theorem follows.

Some of these homology-based proofs are quite slick, but generally speaking, homology theory requires some heavy infrastructure; the question is whether a more elementary proof exists. The following “analytic” proof is due to John Milnor and uses very little machinery, basically nothing beyond advanced calculus. I will follow his exposition (American Mathematical Monthly, July 1978, pp. 521-524) pretty closely.

For the first step, suppose that we have a continuously differentiable vector field $v$ defined in a compact region of space $\mathbb{R}^n$, $A$. For any real $t$ and for $x \in A$, define a function

$f_t := (x \mapsto x + tv(x))$

The matrix of first partial derivatives of $f_t$ is $\displaystyle I + t[\frac{\partial v_i}{\partial x_j}]_{ij}$, where $I$ denotes the $n \times n$ identity matrix. For $|t|$ sufficiently small, the determinant of this matrix is strictly positive over all of $A$.

Lemma 1: If $|t|$ is sufficiently small, then $f_t$ is a one-to-one function of $A$ onto its image, and $\mbox{vol}(f_t(A))$ is a polynomial function of $t$.

Proof: Since $v$ is continuously differentiable over a compact region, there is [using e.g. the mean value theorem, evidently a red rag for some people 😉 ] a Lipschitz constant $c > 0$ so that

$|v(x) - v(y)| \leq c|x-y|, \qquad x, y \in A$

We have $f_t(x) = f_t(y)$ only if $t(v(x) - v(y)) = y - x$; if we assume $|t| < c^{-1}$, this can happen only if $x = y$. So $f_t$ is one-to-one for such $t$.

The determinant of the matrix of first partials above is of the form

$1 + t\sigma_1(x) + t^2\sigma_2(x) + \ldots + t^n \sigma_n(x)$

where the $\sigma_k$ are continuous functions in $x$. By the first part of the lemma, we may take $|t|$ so small that $f_t$ is a continuously differentiable embedding, and then by a change-of-variables formula in multivariate calculus we have that

$\displaystyle \mbox{vol}(f_t(A)) = a_0 + a_1 t + \ldots + a_n t^n$

where $\displaystyle a_k = \int_A \sigma_k(x) dx_1 \ldots dx_n.$ This completes the proof. $\Box$

Next, suppose that on $S^{n-1}$ we have a continuously differentiable non-vanishing field $v$ of tangent vectors. Applying the continuously differentiable map $v \mapsto v/|v|$, we assume the vector field $v$ consists of unit tangent vectors. For each $u \in S^{n-1}$, the vector $u + tv(u)$ is thus of length $(1 + t^2)^{1/2}$, hence $f_t$ maps the unit sphere $S^{n-1}$ to the sphere of radius $(1 + t^2)^{1/2}$.

Lemma 2: For sufficiently small $|t|$, the map $f_t$ maps $S^{n-1}$ onto the sphere of radius $(1 + t^2)^{1/2}$.

Proof: Extend the vector field $v$ on $S^{n-1}$ (and therefore also $f_t$) to the compact region between two concentric spheres, $A = \{v \in \mathbb{R}^n: 1/2 \leq |v| \leq 3/2\}$, by homothety, i.e., put $v(ru) = rv(u)$, $f_t(ru) = rf_t(u)$ for $1/2 \leq r \leq 3/2$ and $u \in S^{n-1}$. There is a Lipschitz constant $c > 0$ such that

$|v(x) - v(y)| \leq c|x - y|$ for $x, y \in A.$

Take $|t| < \mbox{min}(c^{-1}, 1/3)$, and let $u_0$ be any unit vector. The function

$h(x) := u_0 - tv(x), \qquad x \in A$

maps the complete metric space $A$ to itself (because $|t| < 1/3$ and $|v(x)| \leq 3/2$ — just use triangle inequalities), and $h$ satisfies a Lipschitz condition $|h(x) - h(y)| \leq k|x - y|$ where $k = c|t| < 1$. By a classical fixed-point theorem, $h$ has under these conditions a (unique) fixed point $x$, so that $x + tv(x) = u_0$. Rescaling $u_0$ and $x$ by the factor $(1 + t^2)^{1/2}$, the statement of lemma 2 follows. $\Box$

Now we prove the hairy ball theorem. If $w$ is a continuous non-vanishing vector field of tangent vectors on $S^{n-1}$, let $\varepsilon > 0$ be the absolute minimum of $|w|$. By standard techniques (e.g., using the Stone-Weierstrass approximation theorem), there is a continuously differentiable vector field $v$ such that $|v - w| < \varepsilon/2$, and then $|v| > \varepsilon/2$ so that $v$ is also non-vanishing. As above, we may then substitute $v/|v|$ for $v$, i.e., assume that $v$ consists of unit vectors.

Given $b > 1 > a > 0$, we extend $v$ to the region $A = \{x \in R^n: a \leq |x| \leq b\}$ by homothety. For sufficiently small $|t|$, the map $f_t$ defined above maps a spherical shell $|x| = r$ in this region bijectively onto the shell $|x| = r(1 + t^2)^{1/2}$. Hence $f_t$ maps $A$ to the region $\{x: a(1 + t^2)^{1/2} \leq |x| \leq b(1 + t^2)^{1/2}\}$, and we get a dilation factor:

$\mbox{vol}(f_t(A)) = (1 + t^2)^{n/2}\mbox{vol}(A).$

By lemma 1, $(1 + t^2)^{n/2}$ is polynomial in $t$. So $n$ must be even; therefore $S^{n-1}$ admits a non-vanishing vector field only if $n-1$ is odd. This gives the hairy ball theorem. $\Box$

Man, what an awesome proof. That John Milnor is just a master of technique.

Just a quick note on how any of this bears on Milenkovic’s problem. He asked whether for any topological embedding of $S^{n-1}$ in $\mathbb{R}^n$ and any point $P$ in the region $I$ interior to the embedding, there exists a hyperplane $H$ through $P$ such that the barycenter of the $(n-1)$-dimensional region $H \cap I$ coincides with $P$.

Under the further simplifying assumption that the barycenter varies continuously with $H$, the answer is ‘yes’ for even-dimensional spheres. For (taking $P$ to be the origin) we can define a tangent vector field whose value at $u \in S^{n-1}$ is the vector from $P$ to the barycenter of $u^\perp \cap I$. For $n-1$ even, this vector vanishes for some $u$, hence $P$ coincides with the barycenter for that particular $u$.