A Universe of Sorts

§ Vector fields over the 2 sphere

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

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

p

. So the normal bundle is of the form:

\mathfrak N \equiv \{ \{ s \} \times \{ \lambda s : \lambda in \mathbb R \} : s \in \mathbb S^2 \}

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. 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.