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How would you prove the identity

$\displaystyle \frac{\partial}{\partial s}\Psi_{s^*} \mathbb{X} = (-1)L_{\mathbb{Y}}\Psi_{s^*}\mathbb{X}$

where $\Psi_{s}$ is the flow of $\mathbb{Y}$ and $\Psi_{s^*}$ is its push forward. $\mathbb{X}$ and $\mathbb{Y}$ are vector fields.

The Lie derivative with respect to $\mathbb{X}$ is defined by $L_{\mathbb{X}}\mathbb{Y} = [\mathbb{X},\mathbb{Y}]$

The Lie bracket is defined by $\displaystyle \frac{\partial}{\partial s}\Bigg|_{s=0} \Psi_{s^*}\mathbb{X} = [\mathbb{X}, \mathbb{Y}]$

I think one of main issues is how to rid of the $\Bigg|_{s=0}$ and make it more general.

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    $\begingroup$ How are you defining the Lie derivative? $\endgroup$ – Jez Sep 4 '14 at 9:06
  • $\begingroup$ Have edited my question to accomodate this but shouldnt all definitions be equivalent? $\endgroup$ – Permian Sep 4 '14 at 9:11
  • $\begingroup$ How are you defining the Lie bracket? The equality you are trying to prove is essentially one definition of the Lie derivative. $\endgroup$ – Jez Sep 4 '14 at 9:19
  • $\begingroup$ Edited again... $\endgroup$ – Permian Sep 4 '14 at 9:24
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We want to show that

$\frac{\partial}{\partial s} \Psi_{s^*} X = [\Psi_{s^*}X, Y]$

(using the definition of $L$ and antisymmetry of $[,]$), i.e. that

$\frac{\partial}{\partial s} \Psi_{s^*} X = \left. \frac{\partial}{\partial t} \right|_{t=0} \Psi_{t^*} \left( \Psi_{s^*} X\right)$.

But we can write the RHS as

$\left. \frac{\partial}{\partial t} \right|_{t=0} \Psi_{t+s^*} X$

and this is just

$\left. \frac{\partial}{\partial t} \right|_{t=s} \Psi_{t^*} X$,

which is precisely what we mean on the LHS.

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