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We have the following two formula about the Lie derivative of a vector field:

$$ \left.\frac{d}{dt}\right|_{t=0}T\varphi_{-t}\cdot Y_{\varphi_t(p)}=[X,Y]_p = (\mathcal{L}_XY)(p) $$

where $\varphi=\varphi^X(t,p)$ is the flow along the vector field $X$,

and equivalently,

$$ \mathcal{L}_XY=\left.\frac{d}{dt}\right|_{t=0}(\varphi_t^{-X})^*Y $$

where $(\varphi_t^{-X})^*$ is the pull-back of $\varphi_t^{-X}$.

I have a rough idea about what this formula is saying: let $Y$ is a vector field defined along a integral curve of $X$, and we "pull-back" the vector of $Y$ at point $q=\varphi_t(p)$ to its original point $p$ and measure the change rate w.r.t $t$.

But such explanation is quite forced and can not satisfy me. Can anyone provide a intuitive explanation of the Lie derivative?...

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It measures the "difference" between two flows. If you take local coordinates and expand $\Phi^X_{-t} \circ \Phi^Y_{-t} \circ \Phi^X_t \circ \Phi^Y_t$ as a power series in $t$, you should find that the $t$ term vanishes and the next term is $[X, Y] t^2$. (Equivalently, take the second derivative.) –  Zhen Lin Oct 29 '12 at 7:57
    
eh..I just cannot prove this theorem. How to expand $\Phi^X_{-t} \circ \Phi^Y_{-t} \circ \Phi^X_t \circ \Phi^Y_t$? –  hxhxhx88 Oct 29 '12 at 8:52
    
It's just a long laborious calculation, I'm afraid. –  Zhen Lin Oct 29 '12 at 11:18
    
OK...but could you just roughly show me how to expand $\Phi^X_{-t} \circ \Phi^Y_{-t} \circ \Phi^X_t \circ \Phi^Y_t$? –  hxhxhx88 Oct 29 '12 at 12:06

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