# Thread: Lie derivative and commutator

1. ## [SOLVED] Lie derivative and commutator

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

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

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

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

I'm stuck here.

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

Now the proof is strightforward:
\displaystyle \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}
\displaystyle \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.