Why $L_X \circ d = d \circ L_X$?

Question

How to understand that the Lie derivative $L_X$ commutes with exterior differentiation $d$?

I can follow the proof step by step, but I do not think I understand it.

A similar question concerns $F^* \circ d = d \circ F^*$, namely, the pull-back commutes with exterior differentiation.

There should be some simple reason for two things to commute, right?

Answer 1

Let $M$ be a smooth manifold, let $F:M\to N$ be a smooth map, and let $\omega \in \Omega^k(N)$ be a smooth $k$ form on $N$.

Then for each $p\in M$ and $v_1,\dots,v_k \in T_pM$ we have $(F^*\omega)_p(v_1,\dots,v_k):=w_{F(p)}(dF_p(v_1),\dots,dF_p(v_k))$.

If $(V,\psi=(y^j))$ is a chart on $N$, then $\omega_{|V}=\sum'_Jw_J \;dy^{j1}\wedge\dots\wedge dy^{j_k}$.

Then $(F^*\omega)_{|F^{-1}(V)}=\sum'_J (w_J\circ F) \;d(y^{j1}\circ F)\wedge\dots\wedge d(y^{j_k}\circ F)$

While $(dw)_{|V}=\sum'_J dw_J \wedge dy^{j1}\wedge\dots\wedge dy^{j_k}$

And $[d(F^*\omega)]_{|F^{-1}(V)}=\sum'_J d(w_J\circ F) \wedge d(y^{j1}\circ F)\wedge\dots\wedge d(y^{j_k}\circ F)$

And $[F^*(dw)]_{|F^{-1}(V)}=\sum'_J d(w_J\circ F) \wedge d(y^{j1}\circ F)\wedge\dots\wedge d(y^{j_k}\circ F)$

Answer 2

If $f(t)^\ast \alpha = b_I(x,t)dx_I$, then \begin{align*} dL_X\alpha &=d\frac{d}{dt} f^\ast\alpha\\ &= d (b_I)_tdx_I = (b_I)_{tx_i}dx_i\wedge dx_I \\&= \frac{d}{dt} d(b_Idx_I) = \frac{d}{dt} d f^\ast\alpha =\frac{d}{dt} f^\ast d\alpha \\&= L_Xd\alpha\end{align*}