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Thread: Lie derivative and commutator

  1. #1
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    [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.
    Last edited by paweld; Sep 23rd 2009 at 01:08 AM.
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  2. #2
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    Thee is mistake here and it's why I can't prive it:
    Quote Originally Posted by paweld View Post
    $\displaystyle (\mathcal{L}_X Y)_p f = \frac{d}{d t}\bigg |_{t=0} \frac{d}{d s}\bigg |_{s=0} f \circ \psi_{-t} \circ \phi_s(p)$
    ($\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.
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