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Math Help - Lie derivative and commutator

  1. #1
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    [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:
    <br />
[X,Y]f=XYf-YXf=<br />
\frac{d}{d t}\bigg |_{t=0} \left( \frac{d}{d s}\bigg |_{s=0} f \circ \phi_s \right) \circ \psi_t(p) - <br />
\frac{d}{d t}\bigg |_{t=0} \left( \frac{d}{d s}\bigg |_{s=0} f \circ \psi_s \right) \circ \phi_t(p)<br />

    I'm stuck here.
    Last edited by paweld; September 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
     (\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:
    <br />
 \begin{aligned}<br />
(\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)+\\<br />
&\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)=<br />
\end{aligned}<br />
    <br />
 \begin{aligned}<br />
(\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)+\\<br />
& \frac{d}{d u}\bigg |_{u=0} \frac{d}{d s}\bigg |_{s=0} f \circ \phi_s \circ \psi_u (p)<br />
\end{aligned}<br />
    The right hand side is equal to commutator of X and Y.
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