Differential forms and hairy ball theorem

Question

Take the $2$-form $\omega = \sin\phi \,\mathrm d \phi \wedge \mathrm d \theta$ in spherical coordinates, defined on the portion of $\mathbb R^3$ outside of the unit sphere. Clearly, $\omega$ is closed but not exact. I was wondering if there exists a clear link between this fact and the hairy ball theorem.

Answer 1

You can prove the hairy ball theorem also using forms. You can see a tangent vector field as a smooth map $F:S^2\rightarrow\mathbb{R}^3$ such that $$ \langle F(x),x \rangle=0, \, \, \mbox{for all } x\in S^2, $$ where $\langle\cdot,\cdot\rangle$ is the standard scalar product of $\mathbb{R}^3$. Suppose you have a never vanishing vector field $F$, then you can normalize it $$ \bar{F}:=\frac{F}{||F||}, $$ so you can see it as (smooth) map $\bar{F}:S^2\rightarrow S^2$. If yuo have such a map you can perform a (smooth) homotopy between the identity $id_{S^2}$ and the antipodal map $-id_{S^2}$. Condier the map $G:S^2\times [-1,1]\rightarrow S^2$ given by $$ G(x,t):=\cos(t\pi)x+\sin(t\pi) \bar{F}(x). $$ Fix a volume form $\omega$ for $S^2$ (it is a closed form), then because $G$ is an homotopy between the identity $id_{S^2}$ and the antipodal map $-id_{S^2}$ you have (note that $id_{S^2}^{\ast}(\omega)=\omega$) $$ \int_{S^2}\omega=\int_{S^2}-id_{S^2}^{\ast}(\omega) $$ Because 2 is even (as you can guess this proof works for any even dimensional sphere) the antipodal map $-id_{S^2}$ is the composition of an odd number of reflections (you have $S^2\subset \mathbb{R}^3$ so you have to do 3 reflections in order to have $-id_{S^2}$). Then you have $$ vol(S^2)=\int_{S^2}\omega=\int_{S^2}-id_{S^2}^{\ast}(\omega)=-\int_{S^2}\omega=-vol(S^2) $$ That is a contraddiction because for any volume form $\omega$ the volume of $S^2$ is non zero. So you can not have such a never vanishing tangent vector field $F$.