How to proof the Lie derivative of a one-form with respect to Lie bracket, equals to the Lie bracket of the Lie derivative of the one-form, namely
$$
\def\LL{\mathcal{L}}
\LL_{[X,Y]} \omega
= [\LL_X, \LL_Y] \omega
= \LL_X \LL_Y \omega – \LL_Y \LL_X \omega ?
$$
Where $X,Y$ are vector fields, while $\omega\in\Omega^1(M)$ is a one-form.
Best Answer
Here is maybe a roundabout way to do it. As Andreas said, it is true for functions $f$, since $\mathcal L_X f = Xf$, $$\mathcal L_{[X,Y]}f = [X,Y]f = XYf - YXf = \mathcal L_X\mathcal L_Y f - \mathcal L_Y\mathcal L_X f.$$
It is also true for vector fields, using $\mathcal L_X Y = [X,Y]$ and the Jacobi identity of the Lie bracket (which you can rearrange to look like a Leibniz rule to remember easier): \begin{align*} \mathcal L_{[X,Y]} Z = [[X,Y],Z] &= [[X,Z],Y] + [X,[Y,Z]]\\ &= -[Y,[X,Z]] + [X,[Y,Z]]\\ &= -\mathcal L_Y \mathcal L_X Z + \mathcal L_X \mathcal L_Y Z\\ &= \mathcal L_X \mathcal L_Y Z - \mathcal L_Y \mathcal L_X Z. \end{align*} Then it follows from \begin{align*} (\mathcal L_{[X,Y]} \omega)(Z) &= \mathcal L_{[X,Y]} (\omega (Z)) - \omega(\mathcal L_{[X,Y]} Z) \\ &= [\mathcal L_X,\mathcal L_Y](\omega(Z)) - \omega([\mathcal L_X,\mathcal L_Y]Z)\\ &= ([\mathcal L_X,\mathcal L_Y]\omega)(Z) \end{align*}