# Derivations on Manifold are Vector Fields

How can I show that every derivation of $C^\infty(M)$ on a smooth manifold can be represented by a vector field? I want to show that the space of vector fields is isomorphic to the space of derivations of $C^\infty(M)$. I know the proof when $M = \mathbb{R^n}$, but would like to do it for a general smooth manifold.

If you know the proof on $\mathbb{R}^n$, it is the same in every chart, of $(U_i)_{i\in I}$ and if $X_i$ is the vector field defined on $U_i$, the restriction of $X_i$ and $X_j$ on $U_i\cap U_j$ are equal.
• vectorfields can be defined locally, you know to proof on $\mathbb{R}^n$ a chart is an open subset of $\mathbb{R}^n$, so the derivation on $M$ induces a derivation $U_i$ which is a derivation on an open subset of $\mathbb{R}^n$, you can just copy the construction that you know for $\mathbb{R}^n$. – Tsemo Aristide Dec 18 '17 at 22:57
• You consider a derivation $D$ defined on $U\subset \mathbb{R}^n$ and consider $D(x_i)$ where $x_i$ are the restriction of the coordinates on $U$,... – Tsemo Aristide Dec 18 '17 at 22:59
• What do you mean by $D(x_i)$? The proof I know uses Taylor's formula. – Soup Dec 18 '17 at 23:09
• Do you mean take $X = \sum Dx^i \partial{f}/\partial{x^i}$? – Soup Dec 18 '17 at 23:12