# 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?

• An $\mathbb{R}$-basis of $\Gamma(TS^1)$ will contain infinitely many elements. Jun 10, 2021 at 7:37

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.
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.