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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.

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You forgot to substitute also for the differential. It should be

\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*}

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