Lie derivative formula for time dependent differential forms. I am new to the topic and I am trying to show the Lie derivative formula for time dependent differential forms, i.e.,
$\frac{d}{dt} \phi_t^* \alpha_t = \phi_t^* \mathcal L_X \alpha_t + \phi_t^* \frac{d}{dt}\alpha_t$
with $\phi_t$ the flow of the vector field $X$ and $\alpha_t$ some time-dependent differential form. To do this I take the derivative on the left-hand side
$\frac{d}{dt} \phi_t^* \alpha_t = \lim_{h\to0}\frac 1 h ( \phi_{t+h}^* \alpha_{t+h} - \phi_t \alpha_t)=$
$\lim_{h\to0}\frac 1 h ( \phi_{t+h}^* \alpha_{t+h} - \phi_{t+h}^* \alpha_{t} + \phi_{t+h}^* \alpha_{t} - \phi_t \alpha_t) = $
$ \phi_t^* \mathcal L_X \alpha_t + \phi^*_t( \lim_{h\to0} \frac 1 h \phi^*_h [ \alpha_{t+h} - \alpha_t ])$
I am stuck here. It seems that you can just apply the limit at $\lim_{h\to 0}\frac 1 h\phi_h=id$ separately and write the rest as the derivative of $\alpha_t$. How is this possible?!
 A: I think it's easiest to think about this as a "diagonal derivative": if we define $$\omega(s,t) = \phi_s^* \alpha_t$$ then $$\frac{d}{dt}\phi_t^* \alpha_t|_p = \frac{d}{dt}\omega(t,t)|_p$$ is just the directional derivative of $\omega(s,t)|_p$ (considered as a function $\mathbb R^2 \to \Lambda^k T^*_p M$) in the direction $(1,1)$ evaluated at the point $(t,t).$ Thus, assuming everything is nicely behaved, we can use the formula for the directional derivative in terms of the partial derivatives:
$$\frac{d}{dt} \omega(t,t)|_{p} = \partial_s\omega(t,t)|_p + \partial_t\omega(t,t)|_p. \tag{1}$$ These partial derivatives are easy to compute: for fixed $t$ we find $$\partial_s\omega(s,t) = \partial_s \phi_s^* \alpha_t = \phi_s^* \mathcal L_X \alpha_t$$ by the definition of the Lie derivative, while for fixed $s$ we find $$\partial_t \omega(s,t) = \partial_t \phi_s^* \alpha_t = \phi_s^* \partial_t \alpha_t=\phi_s^* \frac d{dt} \alpha_t$$ by the linearity of the pullback $\phi_s^*.$ Evaluating these both at $s=t$ and plugging in to $(1)$ yields the formula you're looking for.
We still need to show things are actually nicely behaved; i.e. that $\omega(s,t)|_p$ is fully (not just partially) differentiable. It suffices to show that this function has continuous partial derivatives. We computed these above; and looking at them, we see that (assuming smoothness of $X$) it suffices for the family $\alpha_t$ to be continuously differentiable on its domain $M \times \mathbb R.$
