We assume that we already know the hairy ball theorem, which states that no continuous vector field on $S^2$ exists that is nowhere vanishing. Using this, we wish to deduce (1) that the module of vector fields over $S^2$ is not free, and an explicit version of what the Serre Swan theorem tells us, that this module is projective

§ 1. Vector fields over the 2-sphere is projective

Embed the 2-sphere as a subset of $\mathbb R^3$. So at each point, we have a tangent plane, and a normal vector that is perpendicular to the sphere: for the point $p \in S^2$, we have the vector $p$ as being normal to $T_p S^2$ at $p$. So the normal bundle is of the form:

• If we think of the trivial bundle, that is of the form $Tr \equiv \{ s \} \times \mathbb R : s \in \mathbb S^2 \}$.
• We want to show an isomorphism between $N$ and $T$.
• Consider a map $f: N \rightarrow Tr$ such that $f((s, n)) \equiv (s, ||n||)$. The inverse is $g: Tr \rightarrow N$ given by $g((s, r)) \equiv (s, r \cdot s)$. It's easy to check that these are inverses, so we at least have a bijection.
• To show that it's a vector bundle morphism, TODO.
• (This is hopelessly broken, I can't treat the bundle as a product. I can locally I guess by taking charts; I'm not sure how I ought to treat it globally!)

§ 1. Vector fields over the sphere is not free

• 1. Given two bundles $E, F$ over any manifold $M$, a module isomorphism $f: \mathfrak X(E) \rightarrow \mathfrak X(F)$ of vector fields as $C^\infty(M)$ modules is induced by a smooth isomorphism of vector bundles $F: E \rightarrow F$.
• 2. The module $\mathfrak X(M)$ is finitely generated as a $C^\infty$ module over $M$.
• Now, assume that $\mathfrak X(S^2)$ is a free module, so we get that $\mathfrak X(S^2) \simeq \oplus_i C^\infty(S^2)$.
• By (2), we know that this must be a finite direct sum for some finite $N$: $mathfrak X(S^2) = \oplus_i=1^N C^\infty(S^n)$.
• But having $N$ different independent non-vanishing functions on $\mathbb S^2$ is the same as clubbing them all together into a vector of $N$ values at each point at $S^2$.
• So we get a smooth function $S^2 \rightarrow \mathbb R^n$, AKA a section of the trivial bundle $\underline{\mathbb R^n} \equiv S^2 \times \mathbb R^n$.
• This means that we have managed to trivialize the vector bundle over the sphere if vector fields over $S^2$ were a free module.
• Now, pick the element $S^2 \times \{ (1, 1, 1, 1, \dots) \} \in S^2 \times \mathbb R^n$. This is a nowhere vanishing vector field over $S^2$. But such an object cannot exist, and hence vector fields over the sphere cannot be free.

§ References

• Smooth vector fields over $S^2$ is not a free module
• Smooth vector fields over $S^2$ is a projective module