# The Unapologetic Mathematician

## What Does the Bracket Measure? (part 2)

Today we’ll prove the assertions we made last time: if $X$ and $Y$ are vector fields with flows $\Phi$ and $\Psi$, respectively in some neighborhood $U$ of some point $p\in M$, we define the curve

$\displaystyle c(t)=\left[\Phi_{-t}\circ\Phi_{-t}\circ\Psi_t\circ\Phi_t\right](p)$

We assert that for any smooth $f\in\mathcal{O}U$ we have

\displaystyle\begin{aligned}(f\circ c)'(0)&=0\\\frac{1}{2}(f\circ c)''(0)&=[X,Y]_p(f)\end{aligned}

To show the first, we define the three “rectangles”

\displaystyle\begin{aligned}V_1(s,t)&=\left[\Psi_s\circ\Phi_t\right](p)\\V_2(s,t)&=\left[\Phi_{-s}\circ\Psi_t\circ\Phi_t\right](p)\\V_3(s,t)&=\left[\Psi_{-s}\circ\Phi_{-t}\circ\Psi_t\circ\Phi_t\right](p)\end{aligned}

Notice that $c(t)=V_3(t,t)$, $V_3(0,t)=V_2(t,t)$, and $V_2(0,t)=V_1(t,t)$. The chain rule lets us then calculate:

\displaystyle\begin{aligned}(f\circ c)'(0)=&\frac{d}{dt}f(c(t))\Big\vert_{t=0}\\=&\frac{d}{dt}f(V_3(t,t))\Big\vert_{t=0}\\=&\frac{\partial}{\partial s}f(V_3(s,t))\Big\vert_{s=t=0}+\frac{\partial}{\partial t}f(V_3(s,t))\Big\vert_{s=t=0}\\=&\frac{\partial}{\partial s}f(V_3(s,t))\Big\vert_{s=t=0}+\frac{d}{dt}f(V_3(0,t))\Big\vert_{t=0}\\=&\frac{\partial}{\partial s}f(V_3(s,t))\Big\vert_{s=t=0}+\frac{d}{dt}f(V_2(t,t))\Big\vert_{t=0}\\=&\frac{\partial}{\partial s}f(V_3(s,t))\Big\vert_{s=t=0}+\frac{\partial}{\partial s}f(V_2(t,t))\Big\vert_{s=t=0}+\frac{\partial}{\partial t}f(V_2(t,t))\Big\vert_{s=t=0}\\=&\frac{\partial}{\partial s}f(V_3(s,t))\Big\vert_{s=t=0}+\frac{\partial}{\partial s}f(V_2(t,t))\Big\vert_{s=t=0}+\frac{\partial}{\partial t}f(V_2(t,t))\Big\vert_{s=t=0}\\=&\frac{\partial}{\partial s}f(V_3(s,t))\Big\vert_{s=t=0}+\frac{\partial}{\partial s}f(V_2(t,t))\Big\vert_{s=t=0}\\&+\frac{\partial}{\partial s}f(V_1(t,t))\Big\vert_{s=t=0}+\frac{\partial}{\partial t}f(V_1(t,t))\Big\vert_{s=t=0}\\=&-Y_pf-X_pf+Y_pf+X_pf=0\end{aligned}

as asserted. As for the other assertion, we start by observing

$\displaystyle(f\circ c)''(0)=\frac{\partial^2(f\circ V_3)}{\partial s^2}(0,0)+2\frac{\partial^2(f\circ V_3)}{\partial s\partial t}(0,0)+\frac{\partial^2(f\circ V_3)}{\partial t^2}(0,0)$

Using the fact that $\frac{\partial(f\circ V_3)}{\partial s}=-(Yf)\circ V_3$ we can turn the first term on the right into

$\displaystyle\frac{\partial^2(f\circ V_3)}{\partial s^2}(0,0)=\frac{\partial(-(Yf)\circ V_3)}{\partial s}(0,0)=Y_pYf$

Now, similar tedious calculations that make the big one above look like idle doodling give us two more identities:

\displaystyle\begin{aligned}\frac{\partial^2(f\circ V_3)}{\partial s\partial t}(0,0)&=-Y_pYf\\\frac{\partial^2(f\circ V_3)}{\partial t^2}(0,0)&=Y_pYf+2[X,Y]_pf\end{aligned}

and we conclude that

$\displaystyle(f\circ c)''(0)=2[X,Y]_pf$

as asserted.

June 27, 2011 - Posted by | Differential Topology, Topology