Lie derivative of a differential form (${\mathcal{L}}_X$)

Let $X$ be a smooth vector field with local flow ${\varphi }_t$. The Lie derivative of a differential form isPulling back before taking the derivative compares forms in the same cotangent space. The Lie derivative of a differential form preserves degree and obeys the Leibniz rule for the wedge product of differential forms.

Cartan's magic formula expresses it using two intrinsic operations:where $d$ is the exterior derivative and ${\iota }_X$ is the interior product. In particular, it commutes with the exterior derivative.

On the real line, $X=x{\partial }_x$ has local flow ${\varphi }_t(x)=e^{t}x$. Hence ${\varphi }_t^{\ast }dx=e^{t}dx$ and ${\mathcal{L}}_{X}dx=dx$. The vector field vanishes at zero, but its differential still stretches the one-form there. For a volume form $\omega$, the equation ${\mathcal{L}}_X\omega =0$ defines a volume-preserving vector field.

The Lie derivative of a differential form satisfies ${\mathcal{L}}_X=d{\iota }_X+{\iota }_{X}d$. Both sides agree on functions. On a differential 1-form $\eta$, evaluation at a vector field $Y$ gives $X(\eta (Y))-\eta ([X,Y])$ on both sides, by the definition of the exterior derivative and the Lie bracket of vector fields. Both operators are degree-zero derivations of an algebra on the exterior algebra. Agreement on functions and coordinate differential 1-forms therefore proves agreement on every differential form. Since $d^2=0$, the formula also gives ${\mathcal{L}}_{X}d=d{\mathcal{L}}_X$, so the Lie derivative of a differential form commutes with the exterior derivative.