Let $c:I\to M$ be a $C^{\infty}$ curve on smooth manifold $M$ of dimension $n$ and $X:I\to TM$ be a vector field along $c$, $$\forall t\in I\qquad X(t)\in T_{c(t)}M.$$ does there exist a vector field $\bar{X}:M \to TM$ such that $X=\bar{X}\circ c$?


2 Answers 2


As @Bebop showed, globally it is not always possible to find the extension of a vector field along $c$, but (as you might guess) $\textit{locally}$ we can. We can do this by the help of Constant Rank Theorem provided that the curve is regular ($c'(t) \neq 0$ so that the curve $c$ is an immersion). By this theorem, we can find an adapted coordinates for the curve. And using this coordinate chart we can build the local extension for $X$.

Construction :

By constant rank theorem, $\forall t_0 \in I$ we can always find a small interval $J$ contain $t_0$ and a chart $(U,\varphi)$ contain $c(t_0)$ s.t the curve $c : I \rightarrow M$ has the following representation

$$ c(t) = (t,0,\dots,0) \qquad \forall t \in J$$

The vector field along $c$, $X : I \rightarrow TM$ has representation on $J\subset I$ $$ X(t) = X^j(t) \frac{\partial}{\partial x^j} \Bigg|_{(t,0,\dots,0)}$$ where $c(J)\subset U$. Because $c(J)$ is a closed subset contained in $U$ then by Extension Lemma for Smooth Function (also in Lee's Smooth Manifold) we can define the component functions $\tilde{X}^j : U \rightarrow \mathbb{R}$ of a local vector field $\tilde{X} : U \rightarrow TM$ $$ \tilde{X} (x^1,\dots,x^m) = \tilde{X}^j (x^1,\dots,x^m) \frac{\partial}{\partial x^j} \Bigg|_{(x^1,\dots,x^m)} $$ such that the restriction to $c(J)$ in the chart $(U,\varphi)$ is

$$ \tilde{X}^j |_{c(J)} (x^1,\dots,x^m) = \tilde{X}^j (t,0,\dots,0) := X^j(t)$$

equal to the component functions of the vector field along curve $X^j(t)$. Therefore by construction $\tilde{X} \circ c = X$ on $J\subset I$.


An arbitrary smooth curve $c:I\rightarrow M$ may have self-intersections or being closed so that not every vector field along $c$ is the restriction of a vector field on $M$.

As an example, let $M=\mathbb R^2$ and $$\begin{array}{rccl}c:&[0,2\pi]& \rightarrow & M\\ & t&\mapsto & (\cos(t),\sin(2t))\end{array}$$

Drawing of $c$

This is a smooth curve with one self-intersection point at $t_1=\dfrac{\pi}{2}$ and $t_2=\dfrac{3\pi}{2}$.

Consider the smooth vector field along $c$ to be $X(t)=c'(t)$ and observe that $$X(t_1)=\begin{pmatrix}-1\\-2\end{pmatrix} \qquad \text{and} \qquad X(t_2)=\begin{pmatrix}1\\-2\end{pmatrix}.$$

Since $X(t_1)\neq X(t_2)$, you can't have a vector field $Y:M\rightarrow TM$ satisfying $$Y(c(t))=X(t)$$ for any $t$.


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