I keep seeing that the hairy ball theorem implies that smooth vector fields over $S^2$ are not a free module is implied by the hairy ball theorem; I don't understand how. How do I fill in the gaps?

The hairy ball theorem tells us that we cannot have a smooth nonvanishing vector field over $S^2$. For contradiction, assume that $\mathfrak X(S^2)$ is a free module over the ring $C^\infty (S^2)$. Thus, $\mathfrak X(S^2) \simeq \oplus_i C^\infty(S^2)$, since a free module is isomorphic to a direct sum of copies the ring.

I am unsure how to continue. I have some ideas:

  1. First, see that we need at least two copies of $C^\infty(S^2)$, because the sphere locally looks like $\mathbb R^2$.
  2. If we have exactly two basis vector fields $V_1, V_2$, then these must both vanish at points $p_1$, $p_2$ by the hairy ball theorem. Consider the neighbourhood of $p_1$: since $v_1$ vanishes, we have only one vector field $V_2$ which is not sufficient to locally span $\mathbb R^2$.
  3. If we have three vector fields, we can have $V_{1, 2, 3}$ vanish at distinct $p_{1 2, 3}$ thereby leaving two vector fields even when one of them vanishes. However, now we can pick a point where none of $V_{1, 2, 3}$ vanish. (Why does such a point exist?). At this point, locally, we have three degrees of freedom $V_{1, 2, 3}$, but the vector space looks like $\mathbb R^2$ so they cannot be linearly independent.

Unfortunately, as is obvious from the above, I have no idea how to make this rigorous. I would appreciate whether this proof is repairable, and if now, how does one actually prove that smooth vector fields over $S^2$ are not free module?


Recall the following two basic facts:

  1. Given two vectors bundles $E,F$ over $M$, any homomorphism $f \colon \Gamma(E) \rightarrow \Gamma(F)$ of $C^{\infty}(M)$-modules is induced by a unique smooth homomorphism of vector bundles $F \colon E \rightarrow F$ (i.e a "smoothly varing" family of linear maps $F_p \colon E_p \rightarrow F_p$). This is sometimes called the "tensor characterization lemma" when the bundles involved are tensor bundles. For a proof, see for example Lemma 10.29 in Lee's Introduction to Smooth Manifolds. If $f$ is an isomorphism of $C^{\infty}(M)$-modules then $F$ is also an isomorphism of smooth vector bundles.
  2. Given a vector bundle $E$ over $M$, the module $\Gamma(E)$ is finitely generated as a $C^{\infty}(M)$-module. For a proof, see here.

Now, assume that $\Gamma(TS^2)$ is free so that $\Gamma(TS^2) \cong \oplus_{i} C^{\infty}(S^2)$. By $(2)$, the right hand side must be a finite direct sum $\oplus_{i=1}^n C^{\infty}(S^2) \cong \Gamma(\underline{\mathbb{R}^n})$ where $\underline{\mathbb{R}^n}$ is the trivial bundle $S^2 \times \mathbb{R}^n$. Now, by $(1)$, we have an vector bundle isomorphism $F \colon \underline{\mathbb{R}^n} \rightarrow TS^2$ (which in particular implies that if such an $F$ exist, $n = 2$). Choose any nonwhere zero section of $\sigma$ of $\underline{\mathbb{R}^n}$ and consider the section $F_{*}(\sigma)$ of $TS^2$ given by $F_{*}(p) = F_p(\sigma(p))$. Since $F_{p} \colon \mathbb{R}^n \rightarrow T_p S^2$ is a linear isomorphism, $F_{*}(\sigma)$ is a nonwhere zero section of $TS^2$, i.e, a non-vanishing vector field on $S^2$, a contradiction.

  • $\begingroup$ Why does $F_p$ being a linear isomorphism imply that $F_*(\sigma)$ is nonvanishing? As I understand it, a zero section is the zero vector field: a vector field that vanishes everywhere on $S^2$. Why can't $F_*$ map each $\sigma$ to a vector field on $S^2$ that vanishes at a single point? $\endgroup$ – Siddharth Bhat Mar 8 at 13:21
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    $\begingroup$ @SiddharthBhat: In this context, a "non-zero section" means a "nowhere zero section", i.e. a vector field that vanishes nowhere on $S^2$. $\endgroup$ – Lee Mosher Mar 8 at 13:23
  • $\begingroup$ I'd like to then understand why $F_*$ maps nowhere vanishing vector fields to nowhere vanishing vector fields. If I had to guess, is the idea that since $F_p$ is locally an isomorphism of vector spaces, it can't annhilate vectors? And thus a nowhere zero section gets mapped to a nowhere zero section? $\endgroup$ – Siddharth Bhat Mar 8 at 13:29
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    $\begingroup$ @SiddharthBhat: Yeah, if $\sigma(p) \neq 0$ and $F_p \colon \mathbb{R}^n \rightarrow T_p S^2$ is a linear isomrphism then $F_p(\sigma(p)) \neq 0$. $\endgroup$ – levap Mar 8 at 19:49
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    $\begingroup$ @SiddharthBhat: Yeah, exactly. A section of $\underline{\mathbb{R}^n}$ is just an $n$-tuple $(f_1,\dots,f_n)$ of scalar valued smooth functions on $M$ which is the same as an element of $\oplus_{i=1}^n C^{\infty}(M)$ (and the module structures also coincide by definition). $\endgroup$ – levap Mar 9 at 6:27

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