The Unapologetic Mathematician

Mathematics for the interested outsider

Brackets and Flows

Now, what does the Lie bracket of two vector fields really measure? We’ve gone through all this time defining and manipulating a bunch of algebraic expressions, but this is supposed to be geometry! What does the bracket actually mean? It turns out that the bracket of two vector fields measures the extent to which their flows fail to commute.

We won’t work this all out today, but we’ll start with an important first step: the bracket of two vector fields vanishes if and only if their flows commute. That is, if X and Y are vector fields with flows \Phi_s and \Psi_t, respectively, then [X,Y]=0 if and only if \Phi_s\circ\Psi_t=\Psi_t\circ\Phi_s for all s and t.

First we assume that the flows commute. As we just saw last time, the fact that \Phi_s\circ\Psi_t=\Psi_t\circ\Phi_s for all t means that Y is \Phi_s-invariant. That is, \Phi_{-t*}Y\circ\Phi_t=Y. But this implies that the Lie derivative L_XY vanishes, and we know that L_XY=[X,Y].

Conversely, let’s assume that [X,Y]=0. For any p\in M we can define the curve c_p in the tangent space \mathcal{T}_pM by c_p(s)=\Phi_{-s*}Y\circ\Phi_s(p). Since the Lie derivative vanishes, we know that c_p'(0)=0, and I say that c_p'(s)=0 for all s, or (equivalently) that c_p(s)=Y_p.

Fixing any s we can set q=\Phi_s(p). Then we calculate

\displaystyle\begin{aligned}c_p'(s)&=\lim\limits_{\Delta s\to0}\frac{c_p(s+\Delta s)-c_p(s)}{\Delta s}\\&=\lim\limits_{\Delta s\to0}\frac{1}{\Delta s}\left[\Phi_{-(s+\Delta s)*}\circ Y\circ\Phi_{s+\Delta s}(p)-\Phi_{-s*}\circ Y\circ\Phi_s(p)\right]\\&=\lim\limits_{\Delta s\to0}\frac{1}{\Delta s}\Phi_{-s*}\left[\left[\Phi_{-\Delta s*}\circ Y\circ\Phi_{\Delta s}\right]\left(\Phi_s(p)\right)-Y\left(\Phi_s(p)\right)\right]\\&=\Phi_{-s*}\lim\limits_{\Delta s\to0}\frac{1}{\Delta s}\left[\left[\Phi_{-\Delta s*}\circ Y\circ\Phi_{\Delta s}\right](q)-Y(q)\right]\\&=\Phi_{-s*}c_q'(0)=\Phi_{-s*}0=0\end{aligned}

Now this means that Y is \Phi_s-invariant for all t, meaning that \Phi_s and \Psi_t commute for all s and t, as asserted.

As a special case, if (U,x) is a coordinate patch then we have the coordinate vector fields \frac{\partial}{\partial x^i}. The fact that partial derivatives commute means that the brackets disappear:

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

This corresponds to the fact that adding s to the ith coordinate and t to the jth coordinate can be done in either order. That is, their flows commute.

About these ads

June 18, 2011 - Posted by | Differential Topology, Topology

3 Comments »

  1. […] I left off last time by pointing out that coordinate vector fields […]

    Pingback by Building Charts from Vector Fields « The Unapologetic Mathematician | June 22, 2011 | Reply

  2. […] again pick up the question we posed earlier about what the bracket measures. Clearly it should have something to do with flows and their […]

    Pingback by What Does the Bracket Measure? (part 1) « The Unapologetic Mathematician | June 23, 2011 | Reply

  3. […] John Armstrong: The Lie Derivative, Brackets and Flows […]

    Pingback by Eighth Linkfest | June 25, 2011 | Reply


Leave a Reply

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

WordPress.com Logo

You are commenting using your WordPress.com 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 )

Google+ photo

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

Connecting to %s

Follow

Get every new post delivered to your Inbox.

Join 392 other followers

%d bloggers like this: