# Smooth vector fields form a $C^\infty (M)$-module

I am watching these lectures and am confused at this point about the counterexample to the module of smooth vector fields having a basis. The speaker invokes the Hairy Ball Theorem to say that every vector field on $$S^2$$ must vanish, but it is not clear to me why that prevents any vector field from being a part of a basis.

• Do we know that if we are given a manifold $M$ such that $\Gamma(TM)$ has a basis then this basis has dimension $dim(M)$? This would help but I don't know if it is true in general. Commented Jun 8, 2019 at 14:48

This argument proves that if any vector field necessarily vanishes at some point, then there is no basis for the module of smooth vector fields with only $$d$$ elements, where $$d$$ is the dimension of $$M$$. In fact, if one of the vector fields vanishes at $$P$$, then the other $$d-1$$ supposed members of the basis would span only a $$d-1$$ dimensional vector space inside $$T_p M$$. So any vector field $$X$$ with $$X(P)$$ outside this $$d-1$$-dimensional space could not be written as a linear combination of the $$d$$ vector fields.
However, this argument does not prove that there is no basis, there could be a basis with more than $$d$$ elements. In order to have a complete proof, you must prove that the dimension cannot be greater than $$d$$. This is not so difficult either: Assume that there are more than $$d$$ smooth vector fields $$X_1,\dots,X_d,X_{d+1},\dots$$ in the basis. Then at a given point $$P$$, they span all of $$T_p M$$ (otherwise the same argument as above gives a contradiction). Then there is a subset of $$d$$ of the given vector fields, which we can assume (by renumbering) to be $$X_1,\dots,X_d$$, for which $$\{X_1(P),\dots,X_d(P)\}$$ is a base (because $$T_p M$$ is a vector space of dimension $$d$$). So $$X_1(P),\dots,X_d(P)$$ are linearly independent, and it is also true that there is a neighborhood $$U$$ of $$P$$ such that $$X_1(Q),\dots,X_d(Q)$$ are linearly independent for each $$Q\in U$$. Take a bump function $$g:M\to \Bbb{R}$$, i.e. such that $$g(m)=0$$ for $$m\notin U$$ but $$g(P)\ne 0$$. Then $$g X_{d+1}$$ can be written as a $$C^{\infty}(M)$$-linear combination of $$X_1,\dots,X_d$$, so $$X_1,\dots,X_d,X_{d+1}$$ are not $$C^{\infty}(M)$$-linearly independent.