What is magical about Cartan's magic formula?

Question

Why is Cartan's magic formula $$\mathscr{L}_X\omega = i_Xd\omega + d(i_X\omega)$$

called "magic"?

Should it be considered a highly surprising result? Does it "magically" prove several other theorems? What is the etymology? (Why it is variously referred to as E.Cartan's formula and H.Cartan's formula?)

Well it spells out $\mathcal{L}_X$ in terms of contraction and the exterior derivative, it's very user-friendly. In particular you can establish easy results, for instance when $\omega$ is a closed symplectic form and $X$ is the Hamiltonian vector field. — Chris Gerig, Aug 30 '12 at 02:18
Either one of the Cartans, father Élie or son Henri could plausibly have invented it... See also the MO thread Is “Cartan’s magic formula” due to Élie or Henri? — t.b., Aug 30 '12 at 07:45
Yes, I've seen the MO thread, in fact. The parenthetical question at the end of the post is not mine, but was added by others. — Jesse Madnick, Aug 30 '12 at 18:16
Maybe because $\mathscr{L}_X$ is of $0$-degree (derivation) and is nicely related to a $1$-degree map $d$ and a $-1$-degree map $i_X$ — amine, Feb 21 '13 at 21:21

score 23 · Answer 1 · answered Aug 31 '15 at 19:16

I would have liked to put this as a comment but I do not have enough points to do it.

I think there is some magic in this formula because it tells us that the Lie derivative $\mathscr{L}_X$ is homotopic to zero with the homotopy $i_X$ going from top-right $\Omega^{p+1}(M)$ to bottom-left $\Omega^p(M)$ diagonally in the following diagram:

$$\require{AMScd} \begin{CD} \cdots @>{d}>> \Omega^p(M) @>{d}>> \Omega^{p+1}(M) @>{d}>> \cdots \\ \qquad @V{\mathscr{L}_X}V{0}V @V{\mathscr{L}_X}V{0}V\\ \cdots @>{d}>> \Omega^p(M) @>{d}>> \Omega^{p+1}(M) @>{d}>> \cdots \end{CD}$$

What is magical about Cartan's magic formula?

1 Answers1