TITLE: Concrete example of the Lie derivative of a one-form
QUESTION [11 upvotes]: Let $\alpha$ be a one-form and $X$ a vector field. For example take:
\begin{align*}
\alpha &= y^2 dx + x^2 dy\\
&\\
X &= \frac{\partial}{\partial x}+xy \frac{\partial}{\partial y}.
\end{align*}
I'm trying to understand how the Lie derivative works on concrete examples. In particular I want to apply the formula:
$$\mathcal{L}_X\alpha = \left(X^j\frac{\partial \alpha_i}{\partial \phi^j} + \alpha_j \frac{\partial X^j}{\partial \phi^i}\right)d\phi^i.$$
If I understand the definitions correctly, it should be:
\begin{align*}
\mathcal{L}_X\alpha =&\ \left(1\frac{\partial y^2}{\partial x} + y^2 \frac{\partial 1}{\partial x}\right)dx
+\left(xy\frac{\partial y^2}{\partial y} + x^2 \frac{\partial xy}{\partial x} \right)dx\\ 
&\ +\left(1\frac{\partial x^2}{\partial x} + y^2 \frac{\partial 1}{\partial y}\right)dy + \left(xy\frac{\partial x^2}{\partial y} + x^2 \frac{\partial xy}{\partial y}\right)dy\\
=&\ (0+0)dx+(2xy^2+x^2y)dx+(2x+0)dy+(0+x^3)dy\\
=&\ (2xy^2+x^2y)dx+(2x+x^3)dy.
\end{align*}

Is this correct at all?

If not, I'd appreaciate any tip or hint as to what I'm doing wrong.

REPLY [2 votes]: I'm late for the party, but here's another solution: fixed$\renewcommand\vec[1]{{\boldsymbol #1}}$ $\vec{X} \in \mathfrak{X}(\Bbb R^2)$, the derivation $L_{\vec{X}}$ defined on functions and vector fields by $L_{\vec{X}}f \doteq \vec{X}(f)$ and $L_{\vec{X}}\vec{Y} \doteq [\vec{X},\vec{Y}]$ extends to tensor fields of any rank by imposing the Leibniz rule, and commuting with contractions. It'll boil down to Cartan's magic formula again if you think (maybe not so) hard about it, but we can use $${\rm d}\alpha(\vec{X},\vec{Y}) = \vec{X}(\alpha(\vec{Y}))-\vec{Y}(\alpha(\vec{X}))-\alpha([\vec{X},\vec{Y}])$$too. So:
$$\begin{align} (L_{\vec{X}}\alpha)(\vec{Y})&\doteq L_{\vec{X}}(\alpha(\vec{Y})) - \alpha(L_{\vec{X}}\vec{Y}) \\ &= \vec{X}(\alpha(\vec{Y})) - \alpha([\vec{X},\vec{Y}]) \\ &= {\rm d}\alpha(\vec{X},\vec{Y})+\vec{Y}(\alpha(\vec{X}))\\ &= {\rm d}\alpha(\vec{X},\vec{Y}) + {\rm d}(\alpha(\vec{X}))(\vec{Y}).\end{align}$$So we compute $$\alpha = y^2{\rm d}x+x^2{\rm d}y \implies {\rm d}\alpha = 2(x-y){\rm d}x\wedge {\rm d}y \implies {\rm d}\alpha(\vec{X},\cdot) = 2(x-y)({\rm d}y - xy\;{\rm d}x),$$and $$\alpha(\vec{X}) = y^2 + x^3y \implies {\rm d}(\alpha(\vec{X})) = 3x^2y\,{\rm d}x + (x^3+2y){\rm d}y.$$Summing: $$L_{\vec{X}}\alpha = (x^2y+2xy^2){\rm d}x + (x^3+2x){\rm d}y,$$as wanted.