Is the Lie derivative of a smooth vector field equal to the Lie bracket field?

[Euge]

Here is this week's POTW:

-----
Let $M$ be a smooth manifold, $X, Y$ smooth vector fields on $M$, and $\phi_t$ the flow of $X$. The Lie derivative of $Y$ along $X$, $\mathcal{L}_XY$, is given by

$$\mathcal{L}_XY:= \frac{d}{dt}\bigg|_{t=0} \phi_{-t*}Y.$$

Show that $\mathcal{L}_XY$ is equal to the Lie bracket field $[X,Y]$.
-----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Euge]

No one answered this week's problem. You can read my solution below.

Spoiler

Let $f$ be a smooth function on $M$. For each $p\in M$,

$$\mathcal{L}_XYf(p) = \frac{d}{dt}\bigg|_{t = 0} \phi_{-t*}Yf(p) $$
$$= \frac{d}{dt}\bigg|_{t = 0} Y(f\circ \phi_{-t})(\phi_t(p))$$
$$=\lim_{t \to 0} \frac{Y(f\circ \phi_{-t})(\phi_t(p)) - Yf(p)}{t}$$
$$=\lim_{t\to 0} \frac{Y(f\circ \phi_{-t})(\phi_t(p)) - Y(f\circ \phi_{-t})(p)}{t} + \lim_{t\to 0} \frac{Y(f\circ \phi_{-t})(p) - Yf(p)}{t}$$
$$= \lim_{t\to 0} \frac{\phi_t^*Yf - Yf}{t}(\phi_{-t}(p)) + Y\left(\lim_{t\to 0} \frac{(f\circ \phi_{-t})(p) - f(p)}{t}\right)$$
$$=X(Yf)(p) + Y(-Xf)(p)$$
$$= [X,Y]f(p).$$

Since $f$ and $p$ were arbitrary, $\mathcal{L}_XY = [X,Y]$.