§ Vector fields over the 2 sphere
We assume that we already know the hairy ball theorem, which states that
no continuous vector field on exists that is nowhere vanishing. Using
this, we wish to deduce (1) that the module of vector fields over 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 . So at each point, we have
a tangent plane, and a normal vector that is perpendicular to the sphere:
for the point , we have the vector as being normal to at .
So the normal bundle is of the form:
- If we think of the trivial bundle, that is of the form .
- We want to show an isomorphism between and .
- Consider a map such that . The inverseis given by . 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 over any manifold , a module isomorphism of vector fields as modules is induced by a smooth isomorphism of vector bundles .
- 2. The module is finitely generated as a module over .
- Now, assume that is a free module, so we get that.
- By (2), we know that thismust be a finite direct sum for some finite : .
- But having different independent non-vanishing functions on is the same asclubbing them all together into a vector of values at each point at .
- So we get a smoothfunction , AKA a section of the trivial bundle.
- This means that we have managed to trivialize the vector bundle over the sphere if vector fields over were a free module.
- Now, pick the element . This is a nowherevanishing vector field over . But such an object cannot exist, and hence vector fields over thesphere cannot be free.