Often, hairy ball theorem is mentioned as an application of degree theory derived from singular homology in algebraic topology or from de Rham cohomology in differential topology; we also mentionned it as an application of Brouwer’s degree in Brouwer’s Topological Degree (III): First Applications. Here we describe an argument due to Milnor, exposed in Gallot, Hullin and Lafontaine’s book Riemannian Geometry, using only integration of differential forms.

For convenience, we begin by exposing some basic definitions and first properties about integration on manifolds.

If $M$ is a smooth manifold, let $TM$ and $\Omega^k(M)$ denote respectively its tangent bundle and its space of differential forms of degree $k$ (ie. the vector space $\Gamma( \Lambda^k T^*M)$ of sections of the $k$-th extorior power of the cotangent bundle).  We say that $M$ is orientable if there exists an atlas $(U_i,\varphi_i)$, said oriented, such that $\varphi_i \circ \varphi_j^{-1}$ is orientation-preserving for all $i,j$.

Property: A smooth manifold $M$ is orientable if and only if there exists $\omega \in \Omega^n(M)$ satisfying $\omega_m \neq 0$ for all $m \in M$. Such a differential form is a volume form.

Sketch of proof. Let $\omega \in \Omega^n(M)$ be a volume form and let $(U_i,\varphi_i)$ be a family of charts covering $M$. Because $(\varphi_i)_* \omega_{|U_i} \in \Omega^n (\varphi_i(U_i))$ is a volume form, we can write

$(\varphi_i)_* \omega_{|U_i} = f_i ~ dx_1 \wedge \cdots \wedge dx_n$

where $f_i \in C(\varphi_i(U_i), \mathbb{R})$ does not vanish. Without loss of generality, we may suppose that $f_i > 0$ (otherwise, switch two coordinates of the chart). Then $(U_i, \varphi_i)$ is an oriented atlas.

Conversely, let $(U_i, \varphi_i)$ be an oriented atlas and $(\chi_i)$ be an associated partition of unity. Then

$\displaystyle \omega = \sum\limits_i \chi_i \varphi_i^* (dx_1 \wedge \cdots \wedge dx_n)$

is a volume form. $\square$

As an example, we may mention the differential form

$\displaystyle \omega = \sum\limits_{i=0}^n (-1)^i x^i dx_0 \wedge \cdots \wedge \widehat{dx_i} \wedge \cdots \wedge dx_n \in \Omega^n(\mathbb{R}^{n+1})$.

Then $\omega$ induces a volume form on $\mathbb{S}^n$ viewed as the unit sphere in $\mathbb{R}^{n+1}$. [In fact, $\omega$ is the canonical volum form of $\mathbb{S}^n$ as a Riemannian manifold. In particular, we define $\displaystyle \mathrm{Vol}(\mathbb{S}^n) = \int_{\mathbb{S}^n} \omega$.]

An interesting property of volum forms is that, if $\omega \in \Omega^n(M)$ is a volume form and $\alpha \in \Omega^n(M)$, then there exists $f \in C(M, \mathbb{R})$ such that $\alpha= f \omega$. It is clear locally, and then it is sufficient to extend it globally using a partition of unity.

Definition: Let $M$ be an orientable compact $n$-dimensional smooth manifold. There exists one and only one linear map $\int_{M} : \Omega^n(M) \to \mathbb{R}$ such that for every $\omega \in \Omega^n(M)$ supported in an open chart $(U,\varphi)$,

$\displaystyle \int_M \alpha = \int_{\varphi(U)} (\varphi^{-1})^* \alpha$.

[Where the right-hand side is the usual integral of a differential form defined on an open subspace in $\mathbb{R}^n$.]

Sketch of proof. First, suppose that such a linear map exists. Let $(U_i, \varphi_i)$ be a finite family of charts covering $M$ (such a family exists by compactness) and $(\chi_i)$ be an associated partition of unity. Then necessarily:

$\displaystyle \int_M \omega= \int_M \sum\limits_{i} \chi_i \omega_{|U_i} = \sum\limits_{i} \int_M \chi_i \omega_{|U_i} = \sum\limits_{i} \int_{\varphi(U_i)} (\varphi_i^{-1})^* \chi_i \omega_{|U_i}$,

hence $\int_M$ is uniquely defined by the mentionned properties. Conversely, it is sufficient to show that the above equality does not depend on the family of charts and on the partition of unity chosen. An argument for that is to notice that if $\varphi : U \to V$ is a diffeomorphism between two open subsets $U,V \subset \mathbb{R}^n$, and $\omega \in \Omega^n(\mathbb{R}^n)$, then

$\displaystyle \int_U \varphi^* \omega= \epsilon \int_V \omega$,

where $\epsilon=1$ if $\varphi$ is orientation-preserving and $-1$ otherwise. We can write $\omega= f dx_1 \wedge \cdots \wedge dx_n$ for some $f \in C(M, \mathbb{R})$, and

$\begin{array}{lcl} \displaystyle \int_U \varphi^* \omega & = & \displaystyle \int_U f \circ \varphi ~ \varphi^* (dx_1 \wedge \cdots \wedge dx_n) \\ \\ & = & \displaystyle \int_U f \circ \varphi ~ \mathrm{Jac} (\varphi) ~ dx_1 \wedge \cdots \wedge dx_n \\ \\ & = & \displaystyle \epsilon \int_U f \circ \varphi ~ | \mathrm{Jac} ( \varphi ) | ~ dx_1 \wedge \cdots \wedge dx_n \\ \\ & = & \displaystyle \epsilon \int_{\varphi(U)} f = \epsilon \int_V \omega \hspace{1cm} \square \end{array}$

Using the proof above, we may notice:

Property: Let $M,N$ be two compact orientable $n$-dimensional smooth manifolds and $\omega \in \Omega^n(N)$. If $f : M \to N$ is a diffeomorphism then

$\displaystyle \int_N \omega= \epsilon \int_M f^*\omega$,

where $\epsilon=1$ if $f$ is orientation-preserving and $\epsilon=-1$ otherwise.

Theorem: (Hairy ball theorem) If $n$ is even, any continuous vector field on $\mathbb{S}^n$ has a zero.

Proof. Let $X$ be a continuous vector field on $\mathbb{S}^n$, that is $X : \mathbb{S}^n \to \mathbb{R}^{n+1}$ is a map satisfying $X(p) \cdot p =0$ for every $p \in \mathbb{S}^n$. By contradiction, suppose that $X$ does not vanish.

According to Stone-Weierstrass theorem, it is possible to approximate $X$ by a sequence of smooth maps $X_k : \mathbb{S}^n \to \mathbb{R}^{n+1}$. Then, if $X_k(p)^{\top}$ denotes the orthogonal projection of $X_k(p)$ on $T_p \mathbb{S}^n$, $X_k^{\top}$ is a sequence of smooth vector fields on $\mathbb{S}^n$ converging uniformely to $X$. Therefore, we may suppose without loss of generality that $X$ is smooth.

[Such an argument may be generalized to other manifolds; see Tubular neighborhood.]

Let $\omega \in \Omega^n(\mathbb{S}^n)$ be the volume form as defined above by the restriction on $\mathbb{S}^n$ of

$\displaystyle \omega = \sum\limits_{i=0}^n (-1)^i x^i \underset{ := \omega_i }{ \underbrace{ dx_0 \wedge \cdots \wedge \widehat{dx_i} \wedge \cdots \wedge dx_n }} \in \Omega^n ( \mathbb{R}^{n+1})$

and let $f_{\epsilon} : S(1) \to S( \sqrt{1+\epsilon^2}), \ x \mapsto x + \epsilon X(x)$. For every $p \in S(1)$ and $v_0, \dots, v_n \in \mathbb{R}^{n+1}$, we may write

$\begin{array}{lcl} ( f_{\epsilon}^* \omega)_p(v_0, \dots, v_n) & = & \sum\limits_{i=0}^n (-1)^i (p^i+ \epsilon X(p)^i) \omega_i (v_0+ \epsilon dX(p) \cdot v_0, \dots, v_n+ \epsilon dX(p) \cdot v_n) \\ & = & \sum\limits_{i=0}^n \sum\limits_{k=0}^{n+1} (-1)^i (p^i+ \epsilon X(p)^i) \epsilon^k \omega_{ik} (v_0, \dots, v_n) \end{array}$

where $\omega_{ik} \in \Omega^n(\mathbb{R}^{n+1})$ is such that $\omega_{i0}=\omega_i$ and $\omega_{ik}(v_0, \dots, v_n)= \omega_i (\tilde{v_0}, \dots, \tilde{v_n})$ with $\tilde{v_j} = v_j$ or $dX(p) \cdot v_j$. Therefore, there exist $\alpha_k \in \Omega^n (\mathbb{R}^{n+1})$ such that

$\displaystyle f_{\epsilon}^* \omega= \omega+ \sum\limits_{k=1}^{n+2} \epsilon^k \alpha_k \hspace{1cm} (1)$.

First, let us show how conclude the proof if $f_{\epsilon}$ is a diffeomorphism:

Let $r= \sqrt{1+ \epsilon^2}$ for convenience, $\psi : x \mapsto rx$ be a diffeomorphism from $S(1)$ to $S(r)$ and  as mentioned above $\displaystyle \mathrm{Vol}(\mathbb{S}^n)= \int_{\mathbb{S}^n} \omega$. Then

$\displaystyle \mathrm{Vol}(\mathbb{S}^n) (1+ \epsilon^2)^{\frac{n+1}{2}} = r^{n+1} \int_{\mathbb{S}^n} \omega = \int_{S(1)} \psi^* \omega= \int_{S(r)} \omega= \int_{S(1)} f_{\epsilon}^* \omega$,

so according to $(1)$, the expression $\displaystyle (1+\epsilon^2)^{\frac{n+1}{2}}$ is a polynomial in $\epsilon$. Of course, it is only possible when $n$ is odd.

To conclude the proof, we shall see that $f_{\epsilon}$ is a diffeomorphism when $\epsilon$ is sufficiently small. From $(1)$, if $g_k \in C(\mathbb{S}^n,\mathbb{R})$ is such that $\alpha_k = g_k \omega$,

$\displaystyle f_{\epsilon}^* \omega= \left( 1+ \sum\limits_{k=1}^{n+1} \epsilon^k g_k \right) \omega$.

By compactness, the $g_k$ are bounded so the quantity $1+ \sum\limits_{k=1}^{n+1} \epsilon^k g_k$ does not vanish when $\epsilon$ is small enough. Therefore, $f_{\epsilon}^*\omega$ is a volume form and $f_{\epsilon}$ is an immersion.

Moreover, if $f_{\epsilon}$ were not injective for $\epsilon$ small enough, there would exist sequences $(\epsilon_k)$, $(x_k)$ and $(y_k)$ such that $\epsilon_k \to 0$, $x_k \neq y_k$ and $f_{\epsilon_k}(x_k)= f_{\epsilon_k}(y_k)$, hence

$\displaystyle \frac{x_k-y_k}{\| x_k-y_k \|} = - \epsilon_k \frac{X(x_k)-X(y_k)}{ \| x_k-y_k \| }$.

The first term has norm one, $\epsilon_k \to 0$ and the last term is bounded according to mean value theorem: a contradiction.

Therefore, for $\epsilon$ small enough, $f_{\epsilon}$ is an injective immersion, and consequently a local diffeomorphism according to inverse function theorem. In particular, $\mathrm{Im}(f_{\epsilon})$ is open but also closed by compactness of $S(1)$; by connectedness, $\mathrm{Im}(f_{\epsilon})= S(r)$ and $f_{\epsilon}$ is a diffeomorphism. $\square$