How to define a vector field passing through a given point.

Question

Given a manifold $M$ we can consider its tangent bundle $TM$. Fix $m \in M$ and $v \in TM$. Is it always possible to define a vector field $F$ such that $F(p)=v$?

Answer 1

Consider any chart of $M$ $\psi:\mathbb{R}^n \rightarrow M $ of $M$ with image $U=\psi(\mathbb{R}^n)$ such that $\psi(0) =p$. Then $L:=d\psi(0)^{-1}$ is an isomorphism $T_pM \rightarrow \mathbb{R}^n$.

Get yourself a smooth cutoff function $\Phi:x:\mathbb{R}^n \rightarrow \mathbb{R}$ such that $\Phi(0)=1$, such that $0\le\Phi\le 1$ and such that $\Phi(x)=0$ whenever $|x|\ge1$.

Define $w:= Lv$ and define the vectorfield $W$ on $\mathbb{R}^n$ by $W(x) := \Phi(x)w$. (Here, the tangent space of $\mathbb{R}^n$ is identified with $\mathbb{R}^n$)

Then, finally, transfer that vectorfield to $M$ by letting $V(\psi(x)):= d\psi(x)W(x)$

That vectorfield will do what you asked for.

(Notation may differ from what you are used to.)

Answer 2

I assume you meant $F(m) = v$.

As $v \in TM$, $v \in T_pM$ for some $p \in M$. Suppose now that we have a map $F : M \to TM$ with $F(m) = v$ and let $\pi$ denote the projection $TM \to M$. Note that

$$(\pi\circ F)(m) = \pi(F(m)) = \pi(v) = p.$$

Hence, if $F$ to be a section of $TM$ (i.e. a vector field on $M$), we must have $p = m$, i.e. $v \in T_mM$. We will proceed under this assumption.

As per Thomas' suggestion, let $(U, (x^1, \dots, x^n))$ be a chart about $m$. Note that $\frac{\partial}{\partial x^1}, \dots, \frac{\partial}{\partial x^n}$ is a basis of local sections, so $TM|_U \cong U\times \mathbb{R}^n$ given by the map $\Phi(w) \mapsto (q , [w])$ where

• $q = \pi(w)$ (i.e. $w \in T_qM$), and
• $[w]$ is the coordinate vector of $w$ with respect to the basis $\{\frac{\partial}{\partial x^1}|_q, \dots, \frac{\partial}{\partial x^n}|_q\}$ of $T_qM$.

Now let $V \subset U$ be open and $K \subset V$ compact with $m \in K$, then there is a smooth bump function $g : U \to \mathbb{R}$ such that $g|_{U\setminus V} = 0$ and $g|_K = 1$. Setting $F(p) = \Phi^{-1}((p, g(p)[v]))$, we obtain a map $F : U \to TM|_U$. If $p \in U\setminus V$, then

$$F(p) = \Phi^{-1}((p, g(p)[v])) = \Phi^{-1}((p, 0[v])) = \Phi^{-1}((p, 0)) = 0 \in T_pM.$$

We also have

$$F(m) = \Phi^{-1}((m, g(m)[v])) = \Phi^{-1}((p, 1[v])) = \Phi^{-1}((p, [v])) = v \in T_mM.$$

By defining $F(p) = 0 \in T_pM$ for $p \in M\setminus U$, we get a smooth extension of $F$ to a map $M \to TM$ which satisfies $\pi\circ F = \operatorname{id}$ and $F(m) = v$. That is, $F$ is a vector field on $M$ with $F(m) = v$.

Answer 3

Preparation

Any rough section has a local representation: $$S=\sum_k S^k\frac{\partial}{\partial x^k}$$ A rough section is smooth iff its coordinates are so: $$T\varphi\circ S=(S^1,\ldots,S^n)$$ Any smooth function with closed support extends trivially: $$F_E:=\chi_{\overline{\mathrm{supp}}F}F$$

Construction

Consider the rough section: $$S:=\sum_k v^k\frac{\partial}{\partial x^k}$$ Patch a smooth bump on it: $$S_0:=\chi_K S$$ Extend it trivially: $$S_E:=\chi_{\overline{\mathrm{supp}}S_0}S_0$$

Caution

The support must be closed in your manifold: $$K\subseteq U\text{ compact}\implies K\subseteq M\text{ compact}\implies K\subseteq M\text{ closed}$$