Let $M$ be a smooth real manifold. I want to show that we have an isomorphism of real vector space $\Gamma(TM)$ of all smooth sections of $TM$ (i.e. of vector fields on $M$) and of real vector space $\mathsf{Der}_{\mathbb{R}} C^{\infty}(M)$ of all $\mathbb{R}$-derivations of $\mathbb{R}$-algebra $C^{\infty}(M)$. More precisely, I want to show that the map that sends vector field $X$ to derivation $f \mapsto X(f)$ will be an isomorphism.

This map is $\mathbb{R}$-linear, then is it sufficient to show that it is injective and surjective. Let's start from injectivity. Suppose there are two different vector fields $X$ and $Y$ such that for any $f \in C^{\infty}(M)$ we have $X(f) = Y(f)$ or $(X-Y)(f)=0$. Let's take any point $x \in M$ and a chart $(U,\varphi_{U})$, $x \in U$. By definition $$ (X-Y)(f)(x) = \langle \nabla (f\circ \varphi^{-1}_{U})(\varphi_{U}(x)),\xi-\eta\rangle, $$ where $\xi,\eta$ are vectors of coordinates of $X_x,Y_x$ in the local coordinate system $(U,\varphi_{U})$. It is sufficient to show that $\xi = \eta$ to prove injectivity. To do this it is sufficient to find such $f$ that $\nabla ( f \circ \varphi_{U}^{-1}) (\varphi_{U}(x)) = \xi - \eta$. I can't just take $f(p) = \langle \xi - \eta, \varphi_{U}(p) \rangle$ because it is defined only in the chart $(U,\varphi_{U})$. What can I do with it? I would like to multiply $\langle \xi - \eta, \varphi_{U}(p) \rangle$ by some function $\chi$ with support in $U$ and equal to $1$ in some neigborhood of $x$. But I'm not sure that one can find such neighborhood in general.

To prove surjectivity I have to find explicitely a vector field such that a given derivation is generated by such vector field. Here I have absolutely no ideas. How can I prove this?

  • $\begingroup$ The existence of $\chi$ is key in differential geometry. See for instance en.wikipedia.org/wiki/Bump_function. $\endgroup$ – Sam Lisi Mar 22 '13 at 15:44
  • $\begingroup$ My hint would be to think about what a derivation has to be, if you look at it in a coordinate chart. Another hint is to remember that you have cut-off functions and partitions of unity. $\endgroup$ – Sam Lisi Mar 22 '13 at 15:47
  • $\begingroup$ I'm struggling with exactly the same problem. Did you find the solution, OP? $\endgroup$ – hjhjhj57 Mar 26 '15 at 4:08

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