§ Vector fields over the 2 sphere

We assume that we already know the hairy ball theorem, which states that no continuous vector field on S2S^2 exists that is nowhere vanishing. Using this, we wish to deduce (1) that the module of vector fields over S2S^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 R3\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 pS2p \in S^2, we have the vector pp as being normal to TpS2T_p S^2 at pp. So the normal bundle is of the form:
N{{s}×{λs:λinR}:sS2} \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{s}×R:sS2}Tr \equiv \{ s \} \times \mathbb R : s \in \mathbb S^2 \}.
  • We want to show an isomorphism between NN and TT.
  • Consider a map f:NTrf: N \rightarrow Tr such that f((s,n))(s,n)f((s, n)) \equiv (s, ||n||). The inverse is g:TrNg: Tr \rightarrow N given by g((s,r))(s,rs)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,FE, F over any manifold MM, a module isomorphism f:X(E)X(F)f: \mathfrak X(E) \rightarrow \mathfrak X(F) of vector fields as C(M)C^\infty(M) modules is induced by a smooth isomorphism of vector bundles F:EFF: E \rightarrow F.
  • 2. The module X(M)\mathfrak X(M) is finitely generated as a CC^\infty module over MM.
  • Now, assume that X(S2)\mathfrak X(S^2) is a free module, so we get that X(S2)iC(S2)\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 NN: mathfrakX(S2)=i=1NC(Sn)mathfrak X(S^2) = \oplus_i=1^N C^\infty(S^n).
  • But having NN different independent non-vanishing functions on S2\mathbb S^2 is the same as clubbing them all together into a vector of NN values at each point at S2S^2.
  • So we get a smooth function S2RnS^2 \rightarrow \mathbb R^n, AKA a section of the trivial bundle RnS2×Rn\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 S2S^2 were a free module.
  • Now, pick the element S2×{(1,1,1,1,)}S2×RnS^2 \times \{ (1, 1, 1, 1, \dots) \} \in S^2 \times \mathbb R^n. This is a nowhere vanishing vector field over S2S^2. But such an object cannot exist, and hence vector fields over the sphere cannot be free.

§ References