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.


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


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

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  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 […]

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