A Non-Algebraic Proof for the Hairy Ball Theorem.

Here, I want to share a peculiar analytic proof of the Hairy Ball Theorem, which states colloquially that “you cannot perfectly comb a coconut.” It may not be clear that this can never occur, but if you start going around the 2-sphere, there is no way to “smooth out” a bunch of vectors without creating either a swirl or some “clash.”

There is an absolutely standard proof using a notion of degree, which I will include at the end of this post just for completeness. Yet, there’s a relatively unknown proof  due to John Milnor, and we’ll borrow from his paper  liberally. I think it is very nice that we need virtually no machinery beyond multi-variable calculus and general topology. Somehow when different theories can establish the same theorem, the result becomes more fundamental for me. I recommend reading Milnor’s paper because he is an excellent writer, but I think that by reordering the presentation and emphasizing the main idea in the proof here, I can both clarify it for myself and share something quirky.

Main Theorem (adapted): An even dimensional n-sphere does not have a continuously differentiable field of unit tangent vectors.

While this is a stronger hypothesis on the type of vector field that the hairy ball theorem holds for, the proof can be extended with a little bit of effort for the general result.

We use the usual notation of

    \[S^{n-1}=\{\vec{u}=(u_1,...,u_n) \in \mathbb R^{n} : \|\vec{u}\|=1\}\]

for the standard unit sphere. Additionally, we say that \vec{v}(u) (some vector whose tail is at \vec{u}) is tangent to S^{n-1} if \vec{u} \cdot \vec v(u)=0.[ I know that \vec{u} is ugly symbolically, but I just want to be clear.]

With this definition of tangency, it is immediate why we require n-1 to be even, since otherwise \vec{v}(u)=(u_2-u_1,\dots,u_n-u_{n-1}) would provide the (un)desired tangent vector field.

The basic idea is that given continuously differentiable vector field \vec{x} \mapsto \vec{v}(x) on any compact region A \subseteq \mathbb R^n, we can define a corresponding function, f_t(\vec u) :=\vec{u}+t \vec{v}(u) for some parameter t \in \mathbb R. With this definition, we have the following lemma:

Lemma 1: the image of S^{n-1} under the map f_t will be a sphere of radius \sqrt{1+t^2} (let’s call this sphere S_b) for sufficiently small t.

While on the other hand,

we will take A:= \{\mathrm{the \,region\, between\, any\, two \,concenteric\, spheres}\}, and consider the area of the sphere under f_t, giving a contradiction by this

Surprising Lemma: for t sufficiently small, f_t maps some compact region A onto f_t(A) injectively and f_t(A) has a volume that can be expressed as a polynomial in t.

To be more specific, we will extend the vector field through the region between spheres and consider the volume of its image under f_t. The key will be to use lemma 1 combined with the fact that (\sqrt{1+t^2})^{n} will not be a polynomial for odd n. The idea is quite elegant, but I think that the proof is a little technical. Nevertheless here are some gory details:

proof of Lemma 1: the difficulty here is showing that f_t:S^{n-1} \to S_b is surjective, since it is clear that \|f_t(\vec{u})\|=\sqrt{1+t^2} for all \vec{u} \in S^{n-1} (we normalized in advance.)

We claim that injectivity, in this case, implies surjectivity: If f_t is injective for sufficiently small t, then f_t will have nonsingular derivatives through any compact region A. So, by the inverse function theorem, we have

    \[f_t: \mathrm{int}(A) \to \{\mathrm{ open \, sets}\}.\]

This would imply that f_t(S^{n-1}) is a relatively open subset of our sphere. Well, since S^{n-1} is compact and connected, its image is also closed. But the only clopen set in a connected space is trivial or the whole space, so we can conclude that f(S^{n-1}) is actually the full S_b.

So, we will show that f_t is indeed injective. This will follow from a more general nice Lipschitz condition that arises more generally for continuously differentiable functions on any compact region A. More carefully, we want to find a constant c so that

    \[\|\vec{v(x)}-\vec{v(y)} \| \leq c\|\vec{x}-\vec{y}\|.\]

Yet, it is a well known theorem that continuously differentiable functions are locally lipschitz. Hence, we choose U_x as a neighborhood for each x where we can find a lipschitz constant. Since the collection of U_x is an open cover, there is a finite subcover. So, we just pick the maximum bound. Well, pick t<c^{-1}, where c is the global Lipschitz constant for A. The rest of the theorem is a matter of algebraic manipulations, if f_t(\vec x)=f_t(\vec y), then \vec{x}-\vec{y}=t(\vec{v}(y)-\vec{v}(x)), which together with \|\vec{x}-\vec{y}\| \leq |t| c \|x-y\| implies that x=y. So, we are done!

Proof of the Surprising Lemma: this is actually fairly straightforward if we recall injectivity for sufficiently small t:

we take a matrix of first derivatives for f_t(\vec{u}) := \vec{u}+\vec{v}, obtaining I+t[\partial v_i/ \partial x_j] . But then, the determinant will be a polynomial in t with “coefficients” that are continuous functions of x. Since this determinant will be strictly positive, we can just integrate over the region A, providing explicit coefficients for our polynomial.

Finally, we provide the formal contradiction: we extend our vector field linearly from S^1 to the region of \vec{x} so that 1\leq \|\vec{x}\| \leq b. Then f_t(\vec{x})=\vec{x}+t \vec{v}(x) is well defined. For our A, we take the region between two concentric spheres S^1 and a sphere of radius b. But then f_t(A)=\{\vec{x} \mid \sqrt{1+t^2} \leq \|x\| \leq \sqrt{1+t^2} \cdot b\} for sufficiently small t. But, the volume of the image then can’t be a polynomial in t for odd n, since

    \[\mathrm{volume\, of}(F_t(A))=\left(\sqrt{1+t^2}\right)^{n}(\mathrm{volume\, of} (A)\]

completing our proof.

I’m not sure how one even thinks of such a proof, but there it is. With a little more effort, one can drop the continuously differentiable hypothesis in the proof to just continuous, and Milnor does so in his paper, but I don’t want to include the details here.

As promised, here is the more typical proof:

Definition: given a map f:S^n \to S^n, there is an induced map f_*:H_n(S^n) \to H_n(S^n) which is an endomorphism of of an infinite cyclic group, so it has to look like f_*(a)=ak, where k is fixed and depends only on the function f. The integer k is denoted by deg(f), and is called the degree.

From this, the functorial properties of homology immediately tell us that deg(i)=1, where i:S^n \to S^n is the identity map,  and that deg(fg)=deg(f)\cdot deg(g). Furthermore, any reflection of S^n about some axis has degree (-1), so the antipodal map, -i, has degree (-1)^{n+1}. Equipped with this, we can prove the theorem, using most essentially the fact that homology is homotopy invariant.

Alternative Proof: note that we already assumed that our tangent vector field was normalized. But then we can define \cos(t)\vec{x}+\sin(t)\vec{v}(x) for t \in [0,\pi], a painfully explicit homotopy from i to -i. Because the induced homomorphisms are the same on the level of homology , we then know that 1=deg(i)=deg(-i)=(-1)^{n+1}, which occurs only if n is odd.