# Thread: Lie derivative and commutator

1. ## [SOLVED] Lie derivative and commutator

Does anyone know an elegant proof of equality: $\mathcal{L}_X Y=[X,Y]$ ( $\mathcal{L}$ is Lie derivative).

I know the following definition of Lie derivative:
$(\mathcal{L}_X Y)_p = \frac{d}{d t}\bigg |_{t=0} (\psi_{-t})_* Y_{\psi_t(p)}$
( $\psi_{t}$ is the flow of vector filed X and ( $\psi_{-t})_*$ is appropriate pushforward)

Using the fact that $X f = \frac{d}{d t}\bigg |_{t=0} f \circ \psi_t$ one may write Lie derivative accordingly:
$(\mathcal{L}_X Y)_p f = \frac{d}{d t}\bigg |_{t=0} \frac{d}{d s}\bigg |_{s=0} f \circ \psi_{-t} \circ \phi_s(p)$
( $\phi_s$ is the flow of Y)

Commutator may be written in form:
$
[X,Y]f=XYf-YXf=
\frac{d}{d t}\bigg |_{t=0} \left( \frac{d}{d s}\bigg |_{s=0} f \circ \phi_s \right) \circ \psi_t(p) -
\frac{d}{d t}\bigg |_{t=0} \left( \frac{d}{d s}\bigg |_{s=0} f \circ \psi_s \right) \circ \phi_t(p)
$

I'm stuck here.

2. Thee is mistake here and it's why I can't prive it:
Originally Posted by paweld
$(\mathcal{L}_X Y)_p f = \frac{d}{d t}\bigg |_{t=0} \frac{d}{d s}\bigg |_{s=0} f \circ \psi_{-t} \circ \phi_s(p)$
( $\phi_s$ is the flow of Y)
There should be:
$(\mathcal{L}_X Y)_p f = \frac{d}{d t}\bigg |_{t=0} \frac{d}{d s}\bigg |_{s=0} f \circ \psi_{-t} \circ \phi_s \circ \psi_t (p)$

Now the proof is strightforward:

\begin{aligned}
(\mathcal{L}_X Y)_p f = & \frac{d}{d t}\bigg |_{t=0,u=0} \frac{d}{d s}\bigg |_{s=0} f \circ \psi_{-t} \circ \phi_s \circ \psi_u (p)+\\
&\frac{d}{d u}\bigg |_{u=0,t=0} \frac{d}{d s}\bigg |_{s=0} f \circ \psi_{-t} \circ \phi_s \circ \psi_u (p)=
\end{aligned}


\begin{aligned}
(\mathcal{L}_X Y)_p f = & - \frac{d}{d t}\bigg |_{t=0} \frac{d}{d s}\bigg |_{s=0} f \circ \psi_{t} \circ \phi_s (p)+\\
& \frac{d}{d u}\bigg |_{u=0} \frac{d}{d s}\bigg |_{s=0} f \circ \phi_s \circ \psi_u (p)
\end{aligned}

The right hand side is equal to commutator of X and Y.