I am trying to follow a proof and I am not so sure with a step there.
So, let $X\in\Gamma(TM)$ be a vector field with compact support on a $\partial$-manifold $M$ and $\psi^X_t$, $t\in[-1,1]$ the flow generated by $X$.
Let $\Sigma^X_t$ be the approximate Lie derivative, i.e. the difference quotient $$\Sigma^X_t:L^2\Omega^k(M)\to L^2\Omega^k(M),\quad \Sigma^X_t\omega = \frac{1}{t}\big((\psi^X_t)^*\omega - \omega\big).$$
Then $$\lim\limits_{t\to 0}\Sigma^X_t\omega = \mathcal{L}_X\omega.$$
Now in the proof I am trying to follow, they say that we can write $$\Sigma^X_t\omega = \int_0^1\mathcal{L}_X\bigg((\psi^X_{ht})^*\omega\bigg)dh.$$
I don’t quite understand that. Is it just that the fundamental theorem of calculus is applied?
Does applying the chain rule yield the following? $$\frac{d}{dh}\bigg((\psi^X_{ht})^*\omega\bigg) = \frac{d}{dh}(ht) \cdot \mathcal{L}_X\bigg((\psi^X_{ht})^*\omega\bigg) = t \cdot \mathcal{L}_X\bigg((\psi^X_{ht})^*\omega\bigg) $$ Why the Lie derivative for the outer derivative?
And then from that it follows the “claim” with the fundamental theorem of calculus?
$$\int_0^1\mathcal{L}_X\bigg((\psi^X_{ht})^*\omega\bigg)dh = \frac{1}{t}\int_0^1 \frac{d}{dh}\bigg((\psi^X_{h\cdot t})^*\omega \bigg)dh = \frac{1}{t}\big((\psi^X_{1\cdot t})^*\omega -(\psi^X_{0\cdot t})^*\omega\big) = \frac{1}{t}\big((\psi^X_{t})^*\omega -\omega\big).$$
Thanks in advance!
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EDIT: I guess I figured it out:
$$\mathcal{L}_X((\psi^X_h)^*\omega)= \lim\limits_{t\to 0}\Sigma^X_t\big((\psi^X_h)^*\omega\big) = \lim\limits_{t\to 0}\frac{1}{t}\big((\psi^X_t)^*\big((\psi^X_h)^*\omega\big) - (\psi^X_h)^*\omega\big) = \lim\limits_{t\to 0}\frac{1}{t}\big((\psi^X_{t+h})^*\omega- (\psi^X_h)^*\omega\big) = \frac{d}{dh} \big((\psi^X_h)^*\omega\big). $$
Is that correct?