Any set of linearly independent and commuting vector fields CAN be realized (locally) as partial derivatives of a local coordinate I am a beginner of differential geometry. I wonder if the following proposition is true:


Let $M$ be an n-dimensional manifold and
    $X_1, \dots ,X_m(m \le n)$ be m commuting and linearly independent vector fields in a neighborhood of a point $p$ in M, then there is a coordinate system $(U, x_1, . . . , x_n)$ around $p$ such that 
    $X_1 =\frac{∂}{∂x_1},...,X_m =\frac{∂}{∂x_m}$ on $U$.


Since $X_1, \dots ,X_m(m \le n)$ are commuting, we have $[X_i,X_j]=0$ for any $i,j$. 
When $m=1$, this is the proposition 1.53(Page 40) in Warner's book.
Honestly speaking, I don't even know how to prove second simplest cases when $m=2.$ If this proposition is true, then I think it should also be a theorem in some book. Every hint, solution, or reference will be appreciated!
 A: Yes, your proposition is true. This can be proven using the fact that two vector fields commute iff. their flows commute:
$$
[X,Y] = 0 \Leftrightarrow \phi^X_t\circ\phi_s^Y=\phi^Y_s\circ\phi^X_t.
$$
First of all, the proof can be found in J. M. Lee's "Introduction to Smooth Manifolds", Thm. 9.46.
To explain the idea, I will assume that $m=n$ (for the general case, see Lee). Given a point $p\in U$ (the "origin"), the flows define maps $\phi^i|_p: I_i\rightarrow U$ onto U which are your coordinate lines through $p$, and the tangent vectors are exactly the $X_i$ by definition. The flows span a coordinate grid on $U$ and the corresponding chart is then defined by the inverse of $\Phi(t^1,\dots,t^n)=\phi^1_{t^1}\circ\dots\circ\phi^n_{t^n}(p)$. To prove that this is a chart it is crucial that the flows commute, because this guarantees that a point $q\in U$ has unique coordinates $(t^1,\dots,t^n)$ and $\Phi$ is invertible.
To see this, consider the pushforward of the cartesian vector fields $\partial_{t^i}$ at some point $t_0$:
$$
(\Phi_*(\partial_{t^i}|_{t_0}))f=\partial_{t^i}|_{t_0}f(\Phi(t^1,\dots,t^n))=\partial_{t^i}|_{t_0}f(\phi^1_{t^1}\circ\dots\circ\phi^n_{t^n}(p))=\partial_{t^i}|_{t_0}f(\phi^i_{t^i}\circ\phi^1_{t^1}\circ\dots\circ\hat{\phi}^i_{t^i}\circ\dots\phi^n_{t^n}(p)=:\partial_{t^i}|_{t_0}f(\phi^i_{t^i}(q))=X_i|_{\Phi(t_0)}f.
$$
Here we used the commutivity of the flows and that $t^i\mapsto\phi^i_{t^i}(q)$ is a integral curve of $X_i$. Hence $\Phi_*$ maps the $\partial_{t^i}$ onto $X_i$. In particular $\Phi_*|_0$ maps $\partial_{t^1}|_0,\dots,\partial_{t^n}|_0$ to $X_1|_p,\dots,X_n|_p$, which are both basises of $T_0\mathbb{R}^n$ and $T_pM$, respectively. Hence, $\Phi_*|_0$ is an isomorphism and by the inverse function theorem, $\Phi$ is a local diffeomorphism, thus $\Phi$ defines a chart on some neighbourhood of $0\in\mathbb{R}^n$.
