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.

About these ads

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

No comments yet.

About this weblog

This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).

I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.

RSS Feeds

RSS - Posts
RSS - Comments
Feedback

Got something to say? Anonymous questions, comments, and suggestions at Formspring.me!
Subjects
Archives

June 2011

M T W T F S S

« May Jul »

1 2 3 4 5

6 7 8 9 10 11 12

13 14 15 16 17 18 19

20 21 22 23 24 25 26

27 28 29 30
Search for:

The Unapologetic Mathematician

Mathematics for the interested outsider