Commutators and coordinate-induced basis (General Relativity) There is an exercise in MTW that asks to prove that, given two vector fields u and v, there exists a coordinate system for which
$$
\textbf{u}=\frac{\partial}{\partial x^1} \mbox{ and }\textbf{v}=\frac{\partial}{\partial x^2}
$$
if and only if u and v are linearly independent and commute.
Now, I know that given those vectors defined above, it's easy to show that they commute and are linearly independent, but I'm stuck at showing the other way around: given two vectors that commute and are linearly independent, we can write them like that. I appreciate if anyone could give me some hints.
 A: Given that this is an exercise problem from MTW (Exercise 9.9), I am trying to find a simpler solution without invoking advanced general results. Here is my take on this,
Given that $\textbf{u}$ and $\textbf{v}$ are vector fields, at any arbitrarily chosen point $\mathcal{P}_0$ they can be written as,
$$\textbf{u}=u^\alpha \frac{\partial}{\partial {y^\alpha} }$$ and $$ \textbf{v}=v^\alpha \frac{\partial}{\partial {y^\alpha} } $$
where, $\frac{\partial}{\partial {y^\alpha}}$ is some coordinate basis for the tangent space at $\mathcal{P}_0$
Since $\textbf{u}$ and $\textbf{v}$ are elements of the tangent space at $\mathcal{P}_0$, they are differential operator themselves. We can write,
$$\frac{d}{d{\lambda}} = u^\alpha \frac{\partial}{\partial {y^\alpha}} = \textbf{u}  $$
where,
$$
\frac{d{y^\alpha}}{d \lambda } = u^\alpha
$$
and similarly,
$$\frac{d}{d {\xi}} = v^\alpha \frac{\partial}{\partial {y^\alpha}} = \textbf{v}$$
where,
$$
\frac{d{y^\alpha}}{d \xi } = v^\alpha
$$
To solve the problem we have to show that there exists a coordinate basis $\frac{\partial}{\partial {x^\alpha}}$ with
$$\frac{\partial}{\partial {x^1}} = \frac{\partial}{\partial {\lambda}} = \textbf{u} $$
and, $$\frac{\partial}{\partial {x^2}} = \frac{\partial}{\partial {\xi}} = \textbf{v} $$
given that the two fields $\textbf{u}, \textbf{v}$ commute and are linearly independent.
Now we have,
$$[\textbf{u},\textbf{v}] = 0,$$
that is,
$$[ \frac{d}{d {\lambda}}, v^\alpha \frac{\partial}{\partial {y^\alpha}}  ] = 0,$$
implying,
$$ \frac{d v^\alpha}{d {\lambda}} \frac{\partial}{\partial {y^\alpha}} = 0, $$
therefore,
$$\frac{d v^\alpha}{d {\lambda}} = 0.$$
We can conclude that in a sufficiently small neighborhood of $\mathcal{P}_0$ the change of parameter $\lambda$ does not affect $\textbf{v}$.
Similarly,
$$\frac{d u^\alpha}{d {\xi}} = 0.$$
and change of $\xi$ does not affect $\textbf{u}$
Therefore, $\lambda$ and $\xi$ can be used as coordinate parameters, with $\frac{\partial}{\partial {\lambda}}$, and $\frac{\partial}{\partial {\xi}}$ as the respective coordinate basis vectors (because they are orthogonal). One can fill out the remaining basis vectors for the tangent space at $\mathcal{P}_0$ to build the required coordinate basis $\frac{\partial}{\partial x^\alpha}$.
A: If the vectors form a basis then it is fairly easy to use Poincaré's lemma to show it, namely if $e_1,...,e_n$ is a frame, and $\theta^1,...,\theta^n$ is its dual frame, and the direct frame satisfies the commutation relations $$ [e_a,e_b]=C^c_{\ ab}e_c, $$ then it is easy to show that the dual frame satisfies the differential relations $$ \mathrm d\theta ^a=-\frac{1}{2}C^a_{\ bc}\theta^b\wedge\theta^c=-\sum_{b<c}C^a_{\ bc}\theta^b\wedge \theta^c. $$
If the frame commutes, i.e. $C^c_{\ ab}=0$, then we have $$ \mathrm d\theta^a=0, $$ thus by Poincaré's lemma, each point in the frame's domain of definition has an open neighborhood $\mathscr U$ on which functions $u^1,...,u^n$ exist such that $$ \theta^a=\mathrm du^a . $$
From this point, it is rather easy to see that the $u^1,...,u^n$ form a local coordinate system and that $$ e_a=\frac{\partial}{\partial u^a}. $$

If the $e_1,...,e_n$ do not form a frame (but are linearly independent, its just there are not enough of them to be a frame), the proof is more difficult. I am going to skip the details but this can be found in for example Lee's Introduction to smooth manifolds (called "canonical form for commuting vector fields" there).
The proof itself is based on the flowout/flow box/straightening theorem. Basically, we know that if $X$ is a vector field that does not vanish in the neighborhood of a point $x$, then there exists a coordinate system $x^1,...,x^n$ such that $$ X=\frac{\partial}{\partial x^1}. $$
The sketch of the proof is that we can take a hypersurface $\Sigma$ to which $X$ is transversal (not tangential), and we may put a coordinate system $x^2,...,x^n$ on the hypersurface, then use the flow $\phi^X(t,x)$ of the vector field to construct a local foliation of hypersurfaces for which $t=x^1$ labels the individual hypersurfaces and $x^2,...,x^n$ labels a point on the $x^1$ hypersurface. Points with different $x^1$ but same $x^2,...,x^n$ lie on the same integral curve of $X$.
If one has $e_1,...,e_m$ commuting vector fields in the $n$ dimensional manifold $M$, then the basic idea of the proof is quite similar, take an $n-m$ dimensional submanifold to which the $e_1,...,e_m$ are all transversal, put a coordinate system $x^{m+1},...,x^n$ on the submanifold, define $$ \phi_{x^1,...,x^m}(x)=\phi^{e_1}_{x^1}\circ...\circ\phi^{e_m}_{x^m}(x), $$ where $\phi^{e_i}_{x^i}$ is the flow of $e_i$ for time $x^i$.
The fact that the flows of the individual vector fields commute means that $$ \phi_{x^1,...x^i,...,x^j,...,x^m}=\phi_{x^1,...,x^j,...,x^i,...,x^m}, $$ which is crucial in proving that we actually have $$ e_i=\frac{\partial}{\partial x^i} $$ once the construction is finished.
