# Explicitly exhibit that vector fields over the 2-sphere is a projective module

We know by the Serre Swan theorem that smooth vector fields over a smooth manifold form a projective module over the ring of smooth functions. We also know that the hairy ball theorem implies that the module of vector spaces on the sphere $$\mathfrak X(S^2)$$ is not a free module. Hence, it must be a "real" projective module, not a free module. So, there must exist some non-trivial module $$F$$ over $$C^\infty(S^2)$$ such that $$\mathfrak X(S^2) \oplus F \simeq \oplus_i C^\infty(S^2)$$.

1. Do we know what $$F$$ is, explicitly? Do we know what the right hand side of the equality $$\mathfrak X(S^2) \oplus F \simeq \oplus_i C^\infty(S^2)$$ should be explicitly? (what is the rank of the free module?)
2. Can we compute $$F$$ "in general" for a given (nice) smooth manifold $$M$$?

The normal bundle of $$S^2$$ embedded in $$\mathbb{R}^3$$ is trivial (see this question), so the sections of this bundle give a rank $$1$$ free $$C^{\infty}S^2$$ module $$F$$ so that $$\mathfrak X(S^2) \oplus F= \oplus_{i=1}^3 C^{\infty}S^2$$.
• Thanks! So for any smooth manifold $M$, I can first embed into $\mathbb R^n$ and then direct sum the normal space? Will this always be a free module? I suppose not; Therefore, is there some general theory on finding the explicit $F$? Mar 8, 2021 at 0:25
• Actually, I have an even more basic doubt: how does one show that the final bundle, which is the the restriction of the bundle on $\mathbb R^3$ to points on the sphere is trivial? Mar 8, 2021 at 0:33
• The final bundle is the pullback of the tangent bundle of $\mathbb{R}^3$ to $S^2$, but the pullback of a trivial bundle is trivial. Summing the tangent bundle with the normal bundle (for an embedding into $\mathbb{R}^n$) always gives a trivial bundle, but the normal bundle isn't always trivial, so this $F$ might not be a free module. One keyword is "stably trivial bundle", these are the bundles that become trivial after summing with a trivial bundle, they are the bundles that are zero in $K$-theory (this question of being stably trivial is just whether or not the complement $F$ is free). Mar 8, 2021 at 19:32