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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 $M$?
Siddharth Bhat
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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 $\mathfrak X(S^2) \oplus F= \oplus_{i=1}^3 C^{\infty}S^2$.

J Cameron
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    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 $F$? – Siddharth Bhat Mar 08 '21 at 00:25
  • 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? – Siddharth Bhat Mar 08 '21 at 00:33
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    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). – J Cameron Mar 08 '21 at 19:32