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I need help with the following problem:

"Let $X$ be a vector field defined on surface $S$, and $p \in S$ such that $X(p) \neq 0$. Prove that there exists a local parametrization $\phi \colon U \to S$ with $U$ an open set of $\mathbb{R}^2$ such that $X|_{\phi(U)} = \phi_1$ where $\phi_1$ is the derivative of the parametrization with respect to it's first parameter."

I intuitively imagine I need to use the vector field in a creative way to satisfy the condition requiered. In a way I think I need something like

$$ \phi(s,t)=\int_0^s \int_0^t X(s,t) ds dt $$

But that doesn't really work. I was wondering if anyone could guide me to the good parametrization so I can complete the rest...

Jarana
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  • I got a comment from someone that the problem can be solved by using the inverse function theorem or by describing the problem through an apropiate ODE. Still I don't know from where to start – Jarana Jun 12 '13 at 00:39

1 Answers1

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Here is a sketch of a solution. A few significant details are omitted.

By ODE theory: There exists an open neighborhood $W \subset S$ of $p$ and an $\epsilon > 0$ such that for every point $q \in W$, there exists a smooth curve $\gamma^q\colon (-\epsilon, \epsilon) \to S$ solving the initial-value problem $$\begin{cases} (\gamma^q)'(t) = X(\gamma^q(t))\\ \gamma^q(0) = q. \end{cases}$$ Such curves are called integral curves of the vector field $X$.

Let $\psi\colon V \to S$ be any local parametrization with $\psi(0,0) = p.$

Define a local parametrization $\phi\colon (-\epsilon, \epsilon) \times (-\delta,\delta) \to S$ by $$\phi(t,y) = \gamma^{\psi(0,y)}(t).$$ Let $U = (-\epsilon, \epsilon) \times (-\delta,\delta)$. For all $(t_0, y_0) \in U$: $$\left.\frac{\partial}{\partial t}\right|_{(t_0, y_0)}\phi(t,y) = \left.\frac{d}{d t}\right|_{t = t_0} \gamma^{\psi(0,y_0)}(t) = X(\gamma^{\psi(0,y_0)}(t_0)) = X(\phi(t_0, y_0)).$$ In other words, $\phi_1 = X|_{\phi(U)}$.

Jesse Madnick
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  • Isn't there a problem with the fact that the width of the interval (-\epsilon, \epsilon) changes with each curve \gamma^q? Or is it independant of the point and only depends on the original parametrization? – Jarana Jun 15 '13 at 21:46
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    That's probably the main "significant detail" that I omitted: There exists an open neighborhood $W \subset S$ of $p$ and an $\epsilon > 0$ such that every integral curve $\gamma^q$ starting at $q \in W$ is defined on $(-\epsilon, \epsilon)$. – Jesse Madnick Jun 16 '13 at 05:01
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    The other omitted details are: (1) $\delta > 0$ has to be chosen small enough so that $\psi(0,y)$ lies in $W$ for all $y \in (-\delta, \delta)$, and (2) that $\phi$ is a local parametrization (homeomorphism onto its image) -- possibly after shrinking its domain -- follows from the Inverse Function Theorem: to this end, one should check that the total derivative (Jacobian matrix) of $\phi$ at $(0,0)$ is invertible. – Jesse Madnick Jun 16 '13 at 05:15
  • One last doubt: How can we use the inverse function theorem on \phi if it goes from R^3 to R^2? – Jarana Jun 17 '13 at 04:40