Let $M$ be a smooth manifold. Consider the set of derivations $\mathcal{X}(M)$ and the set of derivations $\mathcal{D}:= \{D: C^\infty(M, \mathbb{R}) \to C^\infty(M, \mathbb{R})\mid D \mathrm{ \ linear \ map \ and \ D \ satisfies \ the \ Leibniz \ rule}\}$
Note that I define vector field as a map $X: M \to \bigcup_{p \in M}T_pM$ such that for every $p \in M$ $X_p \in T_pM$ and given a chart $(U, x^1, \dots, x^n)$ one can write
$$X_p = \sum_i a_i(p) \frac{\partial }{\partial x^i}\vert_p, \quad p \in U$$
I have shown that the map
$$\Phi: \mathcal{X}(M) \to \mathcal{D}: X \mapsto (f \mapsto(f_*)(X))$$
where $(f_*)(X)(p) = \sigma((f_*)_p(X_p))$ where $\sigma: T_{f(p)}\mathbb{R} \to \mathbb{R}$ is the canonical map given by$\sigma([\gamma]) = \gamma'(0)$
is a well-defined linear map.
I'm trying to show it is surjective.
Here is my idea:
Let $D$ be a derivation. Let $p \in M$. Choose a chart $(U, x^1, \dots, x^n)$ near $p$ and define $$X_p = \sum_{i} D(x^i)(p) \frac{\partial{}}{\partial{x^i}}\vert_p$$
It is clear to me that this is a vector field. Now, I want to show that $\Phi(X) = D$
My calculations lead to:
$$\Phi(X)(f)(p) = \sum_i D(x^i)(p) \frac{\partial{(f \circ \phi^{-1})}}{\partial{x_i}}(\phi(p))$$
However, I'm not even sure if this is the right vector field to show surjectivity and I can't complete my calculation.
Any help will be appreciated.
Best Answer
I don't think this is as easy as it looks. I did the proof when I first started self-studying this subject, and it was a slog. Here is a sketch of how I did it: following your idea,
$1).\ $ it's easier to do this locally first. So, take an open $U\in M$. Basically, you want to show that if $D$ is a derivation, then there are smooth coefficient functions $a_i$ in some chart $(U,\phi)$ such that $D=\sum^n_{i=1}a^i\frac{\partial }{\partial x^i}.$ The obvious candidate for the $a^i$ is $D(x^i).$ So,
$2).\ $ If $f$ is any smooth function on $M$ by composing with $\phi^{-1}$ we may as well assume that $M=\mathbb R^n$, and wlog that $U$ contains the origin.
$3).\ $ Loring proves that $f(x)=f(0)+\sum^n_{i=1}x^i(x)g_i(x)$ for $g_i(0)=\frac{\partial f}{\partial x^i}(0).$ It's just Taylor's formula. Apply $D$ to get $Df(x)=\sum^n_{i=1}x^i(x)D(g_i)(x)+g_i(0)D(x^i)(x).$ Let $x\to 0$ so that $Df(0)=\sum^n_{i=1}g_i(0)D(x_i)(0)=\sum^n_{i=1}D(x_i)(0)\frac{\partial f}{\partial x^i}(0).$ Translating this to any $z\in U$, we have $Df(z)=\sum^n_{i=1}D(x_i)(z)\frac{\partial f}{\partial x^i}(z),$ as desired.
$4).\ $ To globalize this result, cover all of $\mathbb R^n$ by a countable collection $\{U_i\}$ and let $\{\psi_i\}$ be a partition of unity subordinate to this cover. Consider the derivations $D_i=\psi_iD$ initially defined on $C^{\infty}(\mathbb R^n)$ and show how they descend uniquely to $C^{\infty}(U_i).$
$5).\ $ From $3).\ $ we have that each $D_i$ corresponds uniquely to a vector field $X_i$, such that $X_i(p)=0$ whenever $p\notin \text{supp}\ \psi_i$, so it can be extended by zero to a global vector field which we still call $X_i.$
$6).\ $ Define $X=\sum_iX_i$ show that this is well-defined and that $X(f)=\sum_iX_i(f)=D(f).$