# Computing the Lie Derivative using a flow

I struggle with computing the Lie derivative of a form using the flow of a vector field. For example, let $$X=x \frac {\partial }{\partial x}+\frac {\partial }{\partial y}$$ be a vector field with flow $$\rho _{t}:(x,y)\to (xe^{t},t+y)$$, and let $$\omega = x^2y \operatorname{d}x\wedge \operatorname{d}y$$ be a two form. Let's say $$\mathbb{R}^{3}$$ is the ambient manifold with global chart $$\operatorname{id}_{\mathbb{R}^{3}}$$. I know how to compute the Lie derivative using Cartan's formula, but I have trouble applying the flow definition to this simple example. I thought to just apply $$\rho _{t}$$ to $$x$$ and $$y$$ and take the derivative \begin{align*} \mathcal{L}_{X}(\omega )&=\frac {d}{dt}\mid_{t=0}(x e^{t})^{2}\cdot (t+y)\operatorname{d}x\wedge\operatorname{d}y=\left(2 x^2 e^{2t}\cdot (t+y)+x^2 e^{2t}\right)\mid_{t=0}\operatorname{d}x\wedge\operatorname{d}y\\&=\left(2x^2y+x^2\right)\operatorname{d}x\wedge\operatorname{d}y \end{align*} But that doesen't seem to be correct if I compare with Cartan's formula. I also know that I need to "push forward" the tangent vectors onto which I apply this derivative, but I am not sure how to do that. I hope my problem is not too trivial.

\begin{align*} \mathcal{L}_{X}(\omega )&=\frac {d}{dt}\bigg|_{t=0} \rho_t^* \omega \\ &=\frac {d}{dt}\bigg|_{t=0}(x e^{t})^{2}\cdot (t+y)\operatorname{d}(x e^{t})\wedge\operatorname{d}(t+y) \\ &=\frac {d}{dt}\bigg|_{t=0}x^2 e^{3t}\cdot (t+y)\operatorname{d}x \wedge\operatorname{d} y \\ &=\left(3 x^2 e^{3t}\cdot (t+y)+x^2 e^{3t}\right)\mid_{t=0}\operatorname{d}x\wedge\operatorname{d}y\\&=\left(3x^2y+x^2\right)\operatorname{d}x\wedge\operatorname{d}y \end{align*}