Worked examples of Lie derivatives

Question

I'm trying to find the Lie derivative of a 2-form $\sin(\theta)d\theta \wedge d\phi$ with respect to a vector field given in a differential basis $a \partial/ \partial \phi$ and I think the way to go here is to use Cartan's formula but I'm not sure how to do that. I can't find any worked examples in any differential textbook or online, everything seems to focus on proving identities not performing actual calculations. Can somebody point me to a useful resource with some exercises or step-by-step examples?

Answer 1

I understand that you want to compute $$\mathcal{L}_{a \partial_\theta}(\sin\theta\,{\rm d}\theta\wedge {\rm d}\phi).$$In this case it is easy enough to use the definition via flows. The flow of the vector field $a\partial_\theta$ is just $$\Phi_{t,a\partial_\theta}(\theta,\phi) = (\theta+at, \phi),$$so $$\begin{align}\mathcal{L}_{a\partial_\theta}(\sin\theta\,{\rm d}\theta \wedge {\rm d}\theta) &= \frac{\rm d}{{\rm d}t}\bigg|_{t=0} (\Phi_{t,a\partial_\theta})^* (\sin\theta\,{\rm d}\theta \wedge {\rm d}\phi) \\ &= \frac{\rm d}{{\rm d}t}\bigg|_{t=0} \sin(\theta+at)\,{\rm d}(\theta+at)\wedge{\rm d}\phi \\ &= \frac{{\rm d}}{{\rm d}t}\bigg|_{t=0} \sin(\theta+at)\,{\rm d}\theta\wedge {\rm d}\phi \\ &= a\cos(\theta+at)\bigg|_{t=0}\,{\rm d}\theta\wedge{\rm d}\phi \\ &= a \cos\theta\,{\rm d}\theta\wedge{\rm d}\phi.\end{align}$$If you want to double-check using Cartan's Homotopy Formula $\mathcal{L}_{a\partial_\theta} = \iota_{a\partial_\theta}\circ {\rm d}+{\rm d}\circ \iota_{a\partial_\theta}$, you can just use that $${\rm d}(\sin\theta\,{\rm d}\theta\wedge {\rm d}\phi) = \cos\theta \,{\rm d}\theta\wedge{\rm d}\theta\wedge {\rm d}\phi = 0$$because ${\rm d}\theta$ repeats, and compute $$\begin{align} \iota_{a\partial_\theta}(\sin\theta\,{\rm d}\theta\wedge{\rm d}\phi) &= \sin\theta\,({\rm d}\theta \wedge {\rm d}\phi)(a\partial_\theta, \cdot) \\ &= \sin\theta \begin{vmatrix} {\rm d}\theta(a\partial_\theta) & {\rm d}\theta \\ {\rm d}\phi(a\partial_\theta) & {\rm d}\phi\end{vmatrix} \\ &= \sin \theta \begin{vmatrix} a & {\rm d}\theta \\ 0 & {\rm d}\phi\end{vmatrix} \\ &= a \sin {\rm d}\phi,\end{align}$$leading again to $$\mathcal{L}_{a\partial_\theta} = {\rm d}(a\sin\theta\,{\rm d}\phi) = a\cos\theta\,{\rm d}\theta \wedge {\rm d}\phi.$$Anyway, you can look for more examples like this in the book Analysis and Algebra in Differentiable Manifolds - A Workbook for Students and Teachers by Gadea, et al.

Answer 2

$L_X=i_Xd+di_X$, you have $d(sin(\theta)d\theta\wedge d\phi)=cos(\theta)d\theta\wedge d\theta\wedge d\phi=0$

Let $(u,v)$ be a vector of $T_{\theta,\phi}\mathbb{R}^2$ $i_{a{\partial}\over{\partial\phi}}sin(\theta)d\theta\wedge d\phi=sin(\theta)det\pmatrix{0&u\cr a&v}=-asin(\theta)u=-asin(\theta)d\theta(u,v)$. This implies that $d(i_{a{\partial}\over{\partial\phi}}sin(\theta)d\theta\wedge d\phi)=0$