# 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?

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? Mar 8, 2021 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 $S^2$. Mar 8, 2021 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? Mar 8, 2021 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 $F_p(\sigma(p)) \neq 0$. Mar 8, 2021 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). Mar 9, 2021 at 6:27

Let $$M$$ be a smooth manifold inside some $$\mathbb{R}^N$$ given as the $$M = f^{-1}(0)$$, where $$f =(f_1, \ldots, f_{N-n})\colon \mathbb{R}^N \to \mathbb{R}$$, a submersion at every point of $$M$$. We have an orthogonal decomposition at every point $$p$$ of $$M$$

$$T_p(\mathbb{R}^N) = T_p(M) \oplus T(M)_p^{\perp}$$

We get for the tangent bundles $$i^{*}(T(\mathbb{R}^N)) = T(M) \oplus T(M)^{\perp}$$

and so for the modules of global sections

$$\mathcal{O}(M)^N = \mathcal{X}(M) \oplus \mathcal{O}(M)^{N-n}$$

(the module of global sections of the normal bundle is freely generated by \$\grad f_i).

(If $$M$$ was just imbedded in $$\mathbb{R^N}$$ but not globally the zero set of a submersion then the normal bundle might not be trivial. In any case, we still get that $$\mathcal{X}(M)$$ is a projective module).

Now, if $$\mathcal{X}(M)$$ is a free $$\mathcal{O}(M)$$ module, it has to be free of rank $$n= \dim M$$. So consider $$X_1$$, $$\ldots$$, $$X_n$$ such a basis.

Let $$m\in M$$ arbitrary. Let $$v$$ be a tangent vector to $$M$$ at $$p$$. There exists a vector field $$X$$ $$M$$ with $$X(p) = v$$. Write $$X= \sum \phi_i X_i$$. Taking values at $$p$$ gets us $$v= X(p) = \sum \phi_i(p) X_i(p)$$. We conclude that at every point $$p$$ the vectors $$X_i(p)$$ form a system of generators of $$T(M)_p$$, and so, a basis. Conversely, if we have $$n$$ vector fields $$X_i$$, such that at every $$p$$ the vectors $$X_i(p)$$ form a basis of $$T(M)_p$$, then $$X_i$$ are a basis of $$\mathcal{X}(M)$$ over $$\mathcal{0}(M)$$.

Now we see clearly that if $$X_i$$ ($$1\le i \le n$$) form a basis then $$X_i(m)\ne 0$$ for all $$i$$ and all $$m\in M$$.

Note: the argument above uses the fact that we are dealing with smooth manifolds. There exist compact complex manifolds without non-zero global vector fields. So that module ( over $$\mathcal{O}(M) \simeq \mathbb{C}$$) is $$0$$, so free...