Let $M$ be a smooth manifold, $f: M \to \Bbb{R}$ a smooth function and let $X$ be a smooth vector field on $M$, with flow $(t,p) \mapsto\phi(t,p) \equiv \phi_t(p)$. I'm trying to prove that the Lie derivative commutes with exterior derivative, i.e \begin{align} L_X(df) = d(L_Xf) \end{align}
The definition of Lie derivative I'm working with is that for any tensor field $T$ on $M$, \begin{align} L_XT &:= \dfrac{d}{dt}\bigg|_{t=0} \phi_t^*T \end{align}
For my particular case where $T=df$, by using the properties of $d$, we have that \begin{align} \dfrac{\phi_t^*(df) - \phi_0^*(df)}{t} &= d \left( \dfrac{\phi_t^*f - f}{t} \right) \end{align} I'd now like to take the limit $t \to 0$ on both sides. However, if on the RHS I can justify why \begin{align} \lim_{t \to 0}d \left( \dfrac{\phi_t^*f - f}{t} \right) = d \left( \lim_{t \to 0} \dfrac{\phi_t^*f - f}{t} \right) \end{align} then of course my proof will be complete. But this is where I'm having trouble.
I've tried the following. For any $p \in M$, we have that \begin{align} (\phi_t^*f)(p) &= (f \circ \phi)(t,p) \\ &= f(p) + t \dfrac{\partial(f \circ \phi)}{\partial t} \bigg|_{(0,p)} + R_t(p) \\ &= f(p) + t (L_Xf)(p) + R_t(p) \end{align} for some smooth $(t,p) \mapsto R_t(p)$ which satisfies $\lim_{t \to 0} \dfrac{R_t(p)}{t} = 0$
But then I realized this doesn't simplify much because \begin{align} d \left( \dfrac{\phi_t^*f - f}{t} \right) &= d(L_Xf) + d \left( \dfrac{R_t}{t} \right) \end{align} so I still have to justify why interchanging the limit $t \to 0$ with the exterior derivative on second term is valid.
I also tried to write it out in coordinates, and it seems to have something to do with equality of mixed partials, but I just can't figure out how (if that is even the case). SO, any help is appreciated in justifying the exchange of limits and $d$.
Yes, it's precisely equality of mixed partials. Working in local coordinates $(x^1,\dots,x^n)$ on $M$, let's set $F(x,t) = f(\phi_t(x)) = (\phi_t^*f)(x)$. Then $F$ is smooth and $$\frac{\partial}{\partial t}\Big|_{t=0}\frac{\partial F}{\partial x^i} = \frac{\partial}{\partial x^i} \frac{\partial F}{\partial t}\Big|_{t=0}.$$ Therefore $$\sum \frac{\partial}{\partial t}\Big|_{t=0}\frac{\partial F}{\partial x^i}dx^i = \sum \frac{\partial}{\partial x^i} \frac{\partial F}{\partial t}\Big|_{t=0} dx^i.$$ This means that $\mathscr L_X(df) = d(\mathscr L_Xf)$, as desired.