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I'm interested in the following result (chapter 5, theorem 7 in volume 1 of Spivak's Differential Geometry):

Let $X$ be a smooth vector field on an $n$-dimensional manifold M with $X(p)\neq0$ for some point $p\in M$. Then there exists a coordinate system $x^1,\ldots,x^n$ for $U$ (an open subset of $M$ containing $p$) in which $X=\frac{\partial}{\partial x^1}$.

Could someone please explain, in words, how to prove this (or, if you have the book, how Spivak proves this)?

I've read Spivak's proof, and have a few questions about it:

1) How is he using the assumption $X(p)\neq0$?

2) Why can we assume $X(0)=\frac{\partial}{\partial t^1}|_0$ (where $t^1,\ldots,t^n$ is the standard coordinate system for $\mathbb{R}^n$ and WLOG $p=0\in\mathbb{R}^n$)?

3) How do we know that in a neighborhood of the origin in $\mathbb{R}^n$, there's a unique integral curve through each point $(0,a^2,\ldots,a^n)$?

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To complete Adam's answer, the assumption $X(p) \neq 0$ is used in (2). You can assume by translation your coordinate chart is such that $p$ corresponds to $0$ in $\mathbb{R}^n$. Then, in that coordinate chart, $X$ is represented as some $\sum X^i \frac{\partial}{\partial t^i}|_0$. Because $X(p) \neq 0$, not all $X^i$ are zero, and so you can complete the vector $(X_1, ..., X_n)$ to a basis of $\mathbb{R}^n$ and construct a linear isomorphism that maps $(1, 0, ..., 0)$ to $(X_1, ..., X_n)$. Composing it with the chart, you get (2). –  levap Nov 22 '12 at 20:26

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As a partial answer: uniqueness of integral curves I believe boils down to the existence and uniqueness theorem from ordinary differential equations.

Furthermore, any coordinate chart can be translated so that 0 maps to the point $p$ on the manifold, whatever point you're interested in. You're just doing a composition. If you have $\varphi:\mathbb{R}^n\rightarrow M$ and $\varphi(x_0) = p$ then $\phi(x) = \varphi(x + x_0)$ is the new map. It's easily seen to be smooth and satisfy all the properties you need. Because you can always do this, we often just assume it is done and take $\varphi$ to be the map taking 0 to $p$ without further comment.

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ah yes, thanks! Picard-Lindelof. –  Kelly Nov 22 '12 at 19:51

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