This is problem 7 on page 172 of Spivak's Differential Geometry pt. 1. Given a smooth manifold $M$ and a smooth vector field $X$ on $M$,
Check that if the coordinate system $x$ is $x = \chi^{-1}$ for $\chi: \mathbb{R}^n \to M$, then $X = \partial / \partial x^1$ is equivalent to $\chi_*(\partial / \partial t^1) = X \circ \chi$.
This is the last step towards to the proof on page 148 that, for $p \in M$ with $X(p) \neq 0$, there is a coordinate system $(x,U)$ around $p$ such that $X = \partial / \partial x^1$. Namely, Spivak proves that $\chi_*(\partial / \partial t^1) = X \circ \chi$ and says that this is equivalent to $X = \partial / \partial x^1$.
This seems like a very simple problem (and Spivak says as much) but I seem to be having trouble unravelling the definitions. A couple methods I have tried:
- Assuming $\chi_*(\partial / \partial t^1) = X \circ \chi$, rewrite this as $X = \chi_*(\partial / \partial t^1) \circ \chi^{-1} = \chi_*(\partial / \partial t^1) \circ x$. I then tried to evaluate this on a smooth function $f: M \to \mathbb{R}$ and show that it is equal to $\partial f/\partial x^1$ but the precomposition by $x$ throws me off as to how to proceed.
- Let $p \in M$, and $x(p) \in \mathbb{R}^n$. Let $\gamma: (-\epsilon, \epsilon) \to \mathbb{R}^n$ be such that $\gamma(0) = x(p)$ and $\gamma^\prime(0) = \partial/ \partial t^1$. Then define a path $\tilde{\gamma} = \chi \circ \gamma$ in $M$. I think that $\tilde{\gamma}^\prime(0)$ should be $\partial / \partial x^1 |_p$ but had trouble proving it.
- Since $x(U) \cong \mathbb{R}^n$, I think that I can use $x$ as a coordinate for $\mathbb{R}^n$. Then I have the change of coordinates formula
$$\frac{\partial}{\partial t^1} = \sum_{i=1}^n \frac{\partial x^i}{\partial t^1} \frac{\partial}{\partial x^i}$$
I feel that this should be useful somehow, and I'd like to combine it with the first bullet point, but again precomposing by $x$ makes me unable to parse what is actually happening. Thanks.
You're getting a bit lost in the notation. By definition, to say $X=\dfrac{\partial}{\partial x^1}$ is to say that, for any smooth function $f$ on $M$ defined in a neighborhood of $p$, we have $$(Xf)(p) = \dfrac{\partial (f\circ x^{-1})}{\partial t^1}\Big|_{x(p)},$$ where $(t^1,\dots,t^n)$ are the standard coordinates on $\Bbb R^n$. Setting $x^{-1} = \chi$, we have $Xf(p) = \dfrac{\partial (f\circ\chi)}{\partial t^1}\Big|_{x(p)}$. Now, by definition, $$\left(\chi_*\Big(\dfrac{\partial}{\partial t^1}\Big)\right)_p(f) = \dfrac{\partial (f\circ\chi)}{\partial t^1}\Big|_{x(p)},$$ and so the two vector fields unravel to the same thing.