The Unapologetic Mathematician

Mathematics for the interested outsider

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 ith coordinate vector field by saying it’s the unique vector field on U that is x-related to the ith 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.


June 22, 2011 - Posted by | Differential Topology, Topology

1 Comment »

  1. […] we can find a patch and an everywhere-nonzero vector field on so that spans for every . Then we know we can find a chart around such that on . Then the curve with coordinates and for all other […]

    Pingback by Integrable Distributions Have Integral Submanifolds « The Unapologetic Mathematician | June 30, 2011 | Reply

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: