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Let $\omega$ be a closed $k$-form on $\mathbb{R}^n$ and $c:I^k\rightarrow\mathbb{R}^n$ a $k$-cube on $\mathbb{R}^n$. Let $\mathbb{X}$ be a vector field on $\mathbb{R}^n$ with flow $\Phi_t$. Show that $$\frac{d}{dt}\int_c\Phi_t^*\omega=\int_{\Phi_t\circ\partial c}i_{\mathbb{X}}\omega$$ where for a ($k-1$)-chain $\mathcal{C}=\sum_ra_rc_r$, the ($k-1$)-chain $\Phi_t\circ \mathcal{C}$ is defined to be $\sum_ra_r(\Phi_t\circ c_r)$

I don't know where to start, I cannot see the first step I should take.

Any help would be greatly appreciated.

Formulas to use:

$L_\mathbb{X}\omega=\left.\frac{\partial}{\partial t}\right|_{t=0}\Phi_t^*\omega=i_\mathbb{X}d\omega+di_\mathbb{X}\omega$

$\int_cd\omega=\int_{I^k}c^*d\omega$

I still havn't been able to make much progress, could someone verify: is it correct to say that the right will have one more integral from $0$ to $1$ than the left or the other way around? $c$ a $k$-cube, whereas $\partial c$ is a $(k+1)$-cube or $(k-1)$-cube, which one?

Progress: I have realized that the question mentions some subtle information which I had been previously overlooking.

"$\omega$ a closed $k$-form on $\mathbb{R}^n$"

A differential $k$-form $\omega$ is closed if $d\omega=0$

This means that in the equation of the Lie derivative $i_\mathbb{X}d\omega=0$

Therefore we have

$\left.\frac{\partial}{\partial t}\right|_{t=0}\Phi_t^*\omega=di_\mathbb{X}\omega$

This looks remarkably similar to what I am trying to prove, any hints as to what I should do next would be greatly appreciated.

Sam Houston
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    What you need is the Cartan's formula: http://math.stackexchange.com/questions/507753/cartans-magic-formula –  Dec 31 '14 at 15:29
  • @John: I am familiar with this formula, I have tried rearranging $L_\mathbb{X}\omega$ formula to make some progress, but cant see the steps to simplify either side to the equation. If you were to have another look it would be greatly appreciated. – Sam Houston Jan 10 '15 at 12:05

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We have

$$ \begin{aligned}{d\over dt}\int_c \Phi_t^* \omega &= \int_{\Phi_t\circ c} \left.{\partial \over \partial t}\right|_{t=0}\Phi_t^* \omega \\ &= \int_{\Phi_t\circ c} \mathcal{L}_\mathbb{X}\omega \\ &= \int_{\Phi_t\circ c} \iota_\mathbb{X} \underbrace{d \omega}_{=0} + d \iota_\mathbb{X} \omega~\text{(by Cartan's Formula)} \\ &= \int_{\underbrace{\partial(\Phi_t\circ c)}_{=\Phi_t\circ \partial c}} \iota_\mathbb{X} \omega~\text{(by Stokes' Theorem)} \end{aligned} $$

Joel Ong
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