# The Unapologetic Mathematician

## Building Charts from Vector Fields

Sorry for the delay; I’ve been swamped at my actual job the last couple days.

Anyway, I left off last time by pointing out that coordinate vector fields commute:

$\displaystyle\left[\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}\right]=0$

Today I want to show a certain converse: if we have $k$ vector fields $X_1,\dots,X_k$ on some open region $V\subseteq M^n$ that are linearly independent at some $p\in V$ and which commute — $[X_i,X_j]=0$ for all $i$ and $j$ — then we can find some coordinate chart $(U,x)$ around $p$ so that the $X_i$ are the first $k$ coordinate vector fields. That is,

$\displaystyle\frac{\partial}{\partial x^i}=X_i\vert_U$

In particular, if $X$ is a vector field with $X_p\neq0$ then there is a coordinate chart $(U,x)$ around $p$ with $\frac{\partial}{\partial x^1}=X\vert_U$.

If $(U,x)$ is any chart, then we can describe the $i$th coordinate vector field by saying it’s the unique vector field on $U$ that is $x$-related to the $i$th partial derivative in $\mathbb{R}^n$. That is, we’re trying to prove that: $x_*\circ X_i\circ x^{-1}=D_i$ on $x(U)$.

In fact, we can further simplify our claim by assuming that $M=\mathbb{R}^n$, $p=0$, and $X_{i0}=D_{i0}$ — that the vector fields agree at the point $0\in\mathbb{R}^n$. Indeed, if $z$ is any coordinate map taking $p$ to $0$ then we can define the vector fields $Y_i=z_*\circ X_i\circ z^{-1}$ on $\mathbb{R}^n$. These must have vanishing brackets because we can calculate:

\displaystyle\begin{aligned}{}[Y_i,Y_j]&=[z_*\circ X_i\circ z^{-1},z_*\circ X_j\circ z^{-1}]\\&=z_*\circ[X_i,X_j]\circ z^{-1}\\&=z_*\circ0\circ z^{-1}=0\end{aligned}

What’s more, if $y$ is a local diffeomorphism of $\mathbb{R}^n$ with $y_*\circ Y_i\circ y^{-1}=D_i$, then $x=y\circ z$ is a coordinate map satisfying our assertion.

Now, let $\Phi^i_t$ be the flow of $X_i$, and let $W$ be a small enough neighborhood of $0$ that we can define $f:W\to\mathbb{R}^n$ by

$\displaystyle f(a_1,\dots,a_n)=\left[\Phi^1_{a_1}\circ\dots\circ\Phi^k_{a_k}\right](0,\dots,0,a_{k+1},\dots,a_n)$

The order that the flows come in here doesn’t matter, since we’re assuming that the $X_i$ — and thus their flows — commute. Anyway, given any smooth test function $\phi$ on $\mathbb{R}^n$ we can check

\displaystyle\begin{aligned}\left(f_*D_1\right)_a\phi&=\left(D_1\right)_a(\phi\circ f)\\&=\lim\limits_{h\to0}\frac{1}{h}\left[\left[\phi\circ f\right](a_1+h,a_2,\dots,a_n)-\left[\phi\circ f\right](a_1,a_2,\dots,a_n)\right]\\&=\lim\limits_{h\to0}\frac{1}{h}\left[\left[\phi\circ\Phi^1_{a_1+h}\circ\dots\circ\Phi^k_{a_k}\right](0,\dots,0,a_{k+1},\dots,a_n)-\left[\phi\circ f\right](a)\right]\\&=\lim\limits_{h\to0}\frac{1}{h}\left[\left[\phi\circ\Phi_h\circ\Phi^1_{a_1}\circ\dots\circ\Phi^k_{a_k}\right](0,\dots,0,a_{k+1},\dots,a_n)-\left[\phi\circ f\right](a)\right]\\&=\lim\limits_{h\to0}\frac{1}{h}\left[\left[\phi\circ\Phi_h\right](a)-\left[\phi\circ f\right](a)\right]\\&=\left(X_1\right)_{f(a)}(\phi)\end{aligned}

That is, $f_*D_1=X_1\circ f$. To see this for any other $X_i$, simply swap around the flows to bring $\Phi^i_t$ to the front.

We can also check that

\displaystyle\begin{aligned}\left(f_*D_{k+i}\right)_0\phi&=\left(D_{k+i}\right)_0(\phi\circ f)\\&=\lim\limits_{h\to0}\frac{1}{h}\left[\left[\phi\circ f\right](0,\dots,0,h,0,\dots,0)-\left[\phi\circ f\right](0,\dots,0)\right]\\&=\lim\limits_{h\to0}\frac{1}{h}\left[\phi(0,\dots,0,h,0,\dots,0)-\phi(0,\dots,0)\right]\\&=\left(D_{k+i}\right)_0\phi\end{aligned}

Thus $f_{*0}$ is the identity transformation on $\mathcal{T}_0\mathbb{R}^n$. The inverse function theorem now tells us that there is a chart $(U,x)$ around $0$ with $x=f^{-1}$, which will then satisfy our assertions.