Lie derivative of a differential form

Let $\varphi_t:M\to M$ be the (local) flow generated by $X$. For a $k$-form $\omega\in\Omega^k(M)$, the Lie derivative of $\omega$ along $X$ is

where $\varphi_t^*$ denotes the pullback of differential forms.

The Lie derivative can be computed without explicitly using the flow, via Cartan’s magic formula:

Here $\iota_X$ is the interior product (contraction) with $X$, and $d$ is the exterior derivative.

For $\alpha\in\Omega^k(M)$ and $\beta\in\Omega^\ell(M)$:

• Degree 0 derivation with respect to the wedge product:
  $$\mathcal{L}_X(\alpha\wedge\beta)=(\mathcal{L}_X\alpha)\wedge\beta+\alpha\wedge(\mathcal{L}_X\beta).$$
• Commutes with the exterior derivative:
  $$\mathcal{L}_X(d\omega)=d(\mathcal{L}_X\omega).$$
• On functions: for $f\in C^\infty(M)$, $\mathcal{L}_X f = X(f)$.

1. Translation on $\mathbb{R}^n$. On $\mathbb{R}^n$, let $X=\partial/\partial x^1$. For the 1-form $\omega=f(x)\,dx^2$,
   $$\mathcal{L}_X\omega = \frac{\partial f}{\partial x^1}\,dx^2,$$
   and in particular $\mathcal{L}_X(dx^2)=0$.

1. Radial vector field scales the area form on $\mathbb{R}^2$. On $\mathbb{R}^2$ with coordinates $(x,y)$, let $X=x\,\partial_x+y\,\partial_y$. For the standard 2-form $\mu=dx\wedge dy$, one computes
   $$\mathcal{L}_X\mu = 2\,\mu,$$
   reflecting that the flow of $X$ is dilation by $e^t$, which scales area by $e^{2t}$.

1. Rotations preserve the standard area form on $\mathbb{R}^2$. Let $X=-y\,\partial_x + x\,\partial_y$, whose flow is rotation. Then
   $$\mathcal{L}_X(dx\wedge dy)=0,$$
   expressing invariance of the area form under rotations.