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 inverseis $g: Tr \rightarrow N$ given by $g((s, r)) \equiv (s, r \cdot s)$. It's easy to check that theseare 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 thismust 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 asclubbing them all together into a vector of $N$ values at each point at $S^2$.
- So we get a smoothfunction $S^2 \rightarrow \mathbb R^n$, 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 nowherevanishing vector field over $S^2$. But such an object cannot exist, and hence vector fields over thesphere cannot be free.

#### § References