I am trying to compute the Lie derivative $\mathcal{L}_X\omega$ of the 2-form
$$\omega:= xdy\wedge dz + (z-y)dx\wedge dy$$
against the vector field
$$X:=-y \frac{\partial}{\partial x} + x \frac{\partial}{\partial y}$$
on the 2 sphere $S^2 \subset \mathbb{R}^3$. I believe that the flow $F$ of $X$ is given by
$$F\left(t, \left[\begin{matrix} x \\ y \\ z \end{matrix} \right]\right) = \left[\begin{matrix} \cos(t) & -\sin(t) & 0\\ \sin(t) & \cos(t) & 0 \\ 0 & 0 & 1 \end{matrix}\right] \cdot \left[\begin{matrix} x \\ y \\ z \end{matrix}\right].$$
Using $\mathcal{L}_X \omega := \frac{d}{dt}|_{t=0} F^*\omega$, I get $$\mathcal{L}_X \omega = xdx\wedge dz – ydy\wedge dz – xdx\wedge dy$$
from $F^*\omega = \sum_i F^*(a_i)d(F^*x_{i_1})\wedge d(F^*x_{i_2})$. For Cartan's homotopy formula, I get
$d\omega = 2dx\wedge dy \wedge dz, \hspace{1cm}\text{ and }\\
\iota_X \omega = x^2 dz + (z-y)(-ydy-xdx)$
so that
$d \iota_X\omega = 2xdx\wedge dz – y dz\wedge dy – xdz\wedge dx + x dy\wedge dx \\ \hspace{1cm} = 3xdx\wedge dz – ydz\wedge dy + x dy\wedge dx$
and
$\iota_X(d\omega) =d\omega(X,-,-)\\
\hspace{1.3cm} = 2dx\wedge dy \wedge dz\left(\begin{matrix} -y & – & – \\ x & – & – \\ 0 & – & – \end{matrix}\right)\\
\hspace{1.3cm} = 2(-ydy\wedge dz – x dx\wedge dz) \\
\hspace{1.3cm} = -2y dy\wedge dz – 2x dx\wedge dz.$
This gives $\mathcal{L}_X\omega = d \iota_X\omega + \iota_X(d\omega)\\
= (3xdx\wedge dz – ydz\wedge dy + x dy\wedge dx) – (2y dy\wedge dz + 2x dx\wedge dz)\\
= x dx \wedge dz – y dy\wedge dz -x dx\wedge dy.$
This now agrees with what I got from $\frac{d}{dt}|_{t=0}F_t^* \omega$, though my question is: How to restrict this to $S^2$? For instance, in the upper hemisphere $U_{z>0}$ of $S^2$, we have coordinates $(x,y,\sqrt{1-x^2-y^2})$ so that $d(i^*z) = \frac{-x}{\sqrt{1-x^2-y^2}}dx – \frac{y}{\sqrt{1-x^2-y^2}}dy$ gives
$i^*(\mathcal{L}_X\omega):= xdx\wedge d(i^*z) – ydy\wedge d(i^*z) – xdx\wedge dy\\
= (\frac{2xy}{\sqrt{1-x^2-y^2}} – x)dx\wedge dy \\
= \frac{-2xy – x\sqrt{1-x^2-y^2}}{\sqrt{1-x^2-y^2}} dx\wedge dy.$
Is this correct? I suppose one could also use spherical coordinates, rather than worry about charts.
Best Answer
Here's what I got, although it's not 100% certain that I got the correct results. You should check your calculations and mine again and compare them.
So $$L_X \omega = d i_X \omega + i_X d \omega.$$
First, we calculate $$i_X \omega = \omega(X, \cdot) = x \cdot \begin{vmatrix} x & 0 \\ dy & dz \end{vmatrix} + (z-y)\cdot\begin{vmatrix} 0 & x \\ dz & dy \end{vmatrix} = x^2 dz - (z-y)xdz = x(x+y-z)dz,$$and $d i_X \omega = (2x+y-z)dx \wedge dz + x dy \wedge dz.$
Since $\omega$ is a two-form on the sphere $S^2$, its differential is zero, so the $i_X d \omega$ term vanishes. Note that, as it was written, $\omega$ is a form on $\mathbb{R}^3$, so in this calculation, whenever we write $\omega$, we mean $i_{S^2}^* \omega$, where $i_{S^2}$ is the inclusion $S^2 \hookrightarrow \mathbb{R}^3$. But since the pullback commutes with the interior and exterior derivatives, we can do the calculation in $\mathbb{R}^3$ and then pull everything back.