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

Question

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?

Answer 1

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.

Answer 2

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...