For a smooth manifold $M$, what is the relation between $\mathfrak X(M)$ and $C^{\infty}(M) $

Question

Let $M$ be a smooth manifold. We denote $C^\infty(M)$ is the space of all smooth functions $f:\ M\longrightarrow\mathbb R$ and $\mathfrak X(M)$ is the space of all smooth vector fields $X:\ M\longrightarrow T(M)$. It is known that if $X$ is a smooth vector field, then for any local chart $x:\ U\subset M\to\mathbb R^n$, there exists the smooth functions $X_1,\dots,X_n:\ U\to\mathbb R$ such that \begin{align} X_p\,=\,\sum_{i=1}^nX_i(p)\,\frac{\partial}{\partial x_i}\bigg|_p \end{align} or we can write $X=\sum X_i\,\displaystyle\frac{\partial}{\partial x_i} $ as a operator $C^\infty(M)\to C^\infty(M) $ such that \begin{align} X(f)\,=\,\sum X_i\,\displaystyle\frac{\partial f}{\partial x_i}. \end{align} From here, I wonder if the spaces $\mathfrak X(M)$ and $C^\infty(M)$ have any relationshio ?

Answer 1

If we identify $X \in \mathfrak X(M)$ with the map $$C^{\infty}(M) \to C^{\infty}(M), \qquad f \mapsto X(f),$$ as in the question statement, checking directly shows that $\mathfrak X(M)$ is a module over $C^{\infty}(M)$. The product rule $X(fg) = X(f) g + f X(g)$ precisely makes the elements of $\mathfrak X(M)$ derivations, and all derivations of $C^\infty(M)$ turn out to arise this way, giving a (natural) isomorphism $$\mathfrak X(M) \cong \operatorname{Der}(C^\infty(M)) .$$ Indeed, sometimes we define a vector field on $M$ to be a derivation of $C^\infty(M)$.