# The Unapologetic Mathematician

## 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 $i$th coordinate and $t$ to the $j$th coordinate can be done in either order. That is, their flows commute.

June 18, 2011