A vector field is differentiable if and only if the map $X: M \to TM$ is differentiable

Question

Let $X$ be a vector field defined on a manifold $M$. Then $X$ is differentiable if and only if the application $\psi:M\rightarrow TM$ such that $\psi(p)=(p,X_p)$ is differentiable. I have some problems trying to prove this. Any help is welcome.

Answer 1

I assume that what you mean by a function to be differentiable is $C^{\infty}$, and i would replaced it by the word smooth. You can learn about this proof (and the machinery to understand it) in Lee's Intro to smooth Manifolds 2nd ed, Proposition 8.14. What i have wrote here is basically detailed version of the author's proof. The tricky part is the $\Rightarrow$ part. As far as i know, we have to prove that the assumption is true locally on any open subset $U \subseteq M$ first by the help of bump function to extend local function, and then prove that $X : M \to TM$ is smooth (as map between manifolds) locally on $U$. You can see the generalization of this characterization of vector fields on general tensor fields in here or in Lee’s fantastic book cited above.

The converse is easier, so lets do this first. Assume $X : M \to TM$ is smooth as a map between manifolds. For any $f \in C^{\infty}(M)$, the map $Xf : M \to \Bbb{R}$ defined as $Xf(p) = X_pf$. For any point $p \in M$, choose a smooth charts $(U,(x^i))$ contain $p$. In this coordinates, the value of $X$ at any point $x \in U$ is $X(x) = X^i(x) \partial_i|_x$. So $$ Xf(x) = X(x)f = \bigg(X^i(x) \frac{\partial}{\partial x^i}\bigg|_x \bigg)f = X^i(x) \frac{\partial f}{\partial x^i}(x). $$ Since $X$ is smooth, then $X^i$ is smooth on $U$. It follows that $Xf$ is smooth on $U$. Since the same is true in any neighbourhood of each point, $Xf$ is smooth on $M$.

For the $\Leftarrow$ part, assume that $Xf : M \to \Bbb{R}$ smooth, whenever $f :M \to \Bbb{R}$ is smooth.

First we have to show that this assumption implies $Xf : U \to \Bbb{R}$ is smooth for any smooth function $f : U \to \Bbb{R}$ defined on open subset $U \subseteq M$.

Let any $p \in U$ and $V$ be a neighbourhood of $p$ such that $\overline{V} \subseteq U$. Choose a smooth bump function $\psi : M \to \Bbb{R}$, with $\psi \equiv 1$ on $\overline{V}$ and supp $\psi \subseteq U$. Define function $\tilde{f} = \psi f : U \to \Bbb{R}$ as $\tilde{f}(q) = \psi(q)f(q)$. By extend $\tilde{f}$ to $M$ we have $\tilde{f}$ has zero value on $M \smallsetminus \text{supp }\psi$ and $$ (X\tilde{f})|_V = X(\tilde{f}|_V) = X(\psi f)|_V = X(f|_V) = (Xf)|_V, $$ hence $Xf$ is smooth on the neighbourhood $V$ of $p$. Because $p \in U$ is arbitrary, we showed that $Xf$ is smooth on some neighbourhood of each point in $U$. Therefore $Xf$ is smooth on $U$.

Next, we will prove that

If for every open subset $U \subseteq M$ and any smooth function $f : U \to \Bbb{R}$, the function $Xf : U \to \Bbb{R}$ is smooth $\implies X $ is smooth as a map $X : M \to TM$ between manifolds.

To show this, assume that $Xf$ is smooth whenever $f$ is smooth on some open subset. Let $p$ be any point in $M$ and $(U,x^i)$ be any smooth chart contain $p$. Thinking $x^i : U \to \Bbb{R}$ as a smooth function on $U$, we have $$ Xx^i = X^j \frac{\partial }{\partial x^j} (x^i) = X^i, $$ which is smooth. Therefore $X|_U = X^i \partial_i$ is smooth. Since $X : M \to TM$ is smooth locally, then $X : M \to TM$ is smooth. This completes the proof.