Let $M$ be a smooth ($C^\infty$) manifold. Let $\mathfrak{X}(M)$ be a set of all vector fields on $M$ and let $\mathfrak{F}(M)$ be a set of all real smooth functions on $M$. $\mathfrak{X}(M)$ is a real vector space and it is also a module over $\mathfrak{F}(M)$.
We know that partial derivatives constitute a basis for tangent space at any point $p\in M$.
Is there some sort of basis for $\mathfrak{X}(M)$ (as a vector space or as a module)? Do partial derivatives constitute a basis here as well?
I think this question Basis of vector fields on manifold is similar to mine, but because of the way it's written, I'm not really sure I understand the question and I'm not sure we're asking about the same thing.
Best Answer
The set of vector fields of a manifold $M$ is infinite dimensional. For example, over $\mathbb{R}$, it is isomorphic to $\mathcal{C}^{\infty}(\mathbb{R})$ because the tangent bundle is trivial, and the datum of a vector field $(x,V_x)$ over $\mathbb{R}$ is the same as the datum of the function $x \mapsto \langle V_x,e_1\rangle$ where $e_1$ is the constant unit vector field.
In fact, as soon as you have a non-zero vector field $X$ over $M$, the familly $\left\{fX~|~ f \in \mathcal{C}^{\infty}(M) \right\}$ spans an infinite dimensional linear subspace of $\mathfrak{X}(M)$. So one cannot hope to find a basis of it in the general case.
In addition, $\mathfrak{X}(M)$ is a $\mathcal{C}^{\infty}(M)$-module of finite rank. But this does not mean that there is a basis of $\mathfrak{X}(M)$ as a module. This can happen only if the tangent bundle is trivial, that is $TM \simeq M\times \mathbb{R}^n$. This is because if $(X_1,\ldots,X_m)$ is a basis of $\mathfrak{X}(M)$ as a module, then the map \begin{align} TM& \longrightarrow M \times \mathbb{R}^m \\ (p,v) &\longmapsto (x,v_1,\ldots, v_m) \end{align} where $v=\sum v_i X_i(p)$, would be surjective and injective, and linear in the fibers. Thus, it would be a trivilaization of $TM$ and $M$ will be a parallelizable manifold (and moreover, $m = \dim M$).