# smooth vector fields over the 2-sphere is not a free module

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?

## 1 Answer

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.

• 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? – Siddharth Bhat Mar 8 at 13:21
• @SiddharthBhat: In this context, a "non-zero section" means a "nowhere zero section", i.e. a vector field that vanishes nowhere on $S^2$. – Lee Mosher Mar 8 at 13:23
• 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? – Siddharth Bhat Mar 8 at 13:29
• @SiddharthBhat: Yeah, if $\sigma(p) \neq 0$ and $F_p \colon \mathbb{R}^n \rightarrow T_p S^2$ is a linear isomrphism then $F_p(\sigma(p)) \neq 0$. – levap Mar 8 at 19:49
• @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). – levap Mar 9 at 6:27