Existence of basis on a hairy ball I am watching, lecture 6 of "The Mathematics and Physics of Gravity and Light" where the lecturer brings up the case of hairy ball to show that $Γ(TS^2)$ $C^∞$-module (a vector space of smooth sections over a $C^∞$ ring) does not have a basis. I have read a couple other posts regarding this, but I am still confused. Specifically, in order for the above cases to prove that $C^∞$-module does not have a basis, then it implicitly assumes that there exists a basis in $Γ(TS^2)$ $\mathbb{R}$-vector space which is a vector space over a field of real numbers.
Now, if I've understood correctly, an element in $Γ(TS^2)$ $C^∞$-module allows scaling of a vector field at individual points. Thus linear combination of them can give a vector field that has curl. Thus $Γ(TS^2)$ $C^∞$-module having no basis in hairy ball seems to amount to saying that a hairy ball cannot be combed in a single smooth stroke.
However, it also seems impossible that the elements of $Γ(TS^2)$ $\mathbb{R}$-vector which as I understood should be a vector field with vectors at each point pointing in a same direction scaled equally by $r\epsilon\mathbb{R}$, to have a basis. So am I misunderstanding something? If not, if the hairy ball example seems to show non-existence of basis for both a vector space and a module, how does it illustrate the non-existence of basis on a module?
 A: You have to make clear to yourself what the satements on existence of a basis would actually mean in this case. A basis of $\Gamma(TS^2)$ as a module over $C^\infty(M,\mathbb R)$ would mean a family of vector fields $\xi_i$ (indexed by some set $I$) such that any vector field on $S^2$ can be uniquely written as a finite sum $f_1\xi_{i_1}+\dots+f_k\xi_{i_k}$ for smooth functions $f$. It is elementary to see that the uniqueness condition implies that this family can consist of at most two elements. But writing this as $\{\xi_1,\xi_2\}$, the hairy ball theorem tells you that there is a point $x\in S^2$ such that $\xi_1(x)=0$. But then for arbitrary functions $f_1$ and $f_2$, the value of vector field $f_1\xi_1+f_2\xi_2$ in $x$ is proportinal to $\xi_2(x)$. Thus it is clear that not any vector field can be written in this form.
Indeed, modules over rings usually do not admit a basis, this is only true for so called free modules.
In contrast, a basis for the vector space $\Gamma(TS^2)$ is a family $\{\xi_i:i\in I\}$ as above such that any vector field on $S^2$ can be written uniquely as a finite sum $a_1\xi_{i_1}+\dots+a_k\xi_{i_k}$ for constants $a_i$. As remarked by @Kajelad , this evidently implies that the family must be infinite (and indeed uncountable). There is a general theorem that ensures the existence of a basis for any vector space. However, this uses the axiom of choice, so by general principles, it is impossible to write down such a basis explicitly.
