What's going on with these identities involving $d$, $\mathcal L_X$, and $\iota_X$?

Question

Let $\Omega^k$ denote the smooth $k$-forms on a given smooth manifold. Then we have the following operators:

1. Exterior derivative: $d:\Omega^k\to\Omega^{k+1}$ (takes you to the right in the de Rham complex)
2. Interior multiplication: $\iota_X:\Omega^k\to\Omega^{k-1}$ (takes you to the left)
3. Lie derivative: $\mathcal L_X:\Omega^k\to\Omega^k$ (stays put)

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On wedge products, the Lie derivative behaves a little differently than the other two:

1. $d(\omega\wedge\eta)=d\omega\wedge\eta+(-1)^{|\omega|}\omega\wedge d\eta$
2. $\iota_X(\omega\wedge\eta)=\iota_X\omega\wedge\eta+(-1)^{|\omega|}\omega\wedge \iota_X\eta$
3. $\mathcal L_X(\omega\wedge\eta)=\mathcal L_X\omega\wedge\eta+\omega\wedge \mathcal L_X\eta$

Is there an obvious reason for this?

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For $f\in\Omega^0$, we have some sort of "grand unification": $\mathcal L_Xf=Xf=df(X)=\iota_X(df)$. When we start composing them,

1. $dd=0$.
2. $\iota_X\iota_X=0$.
3. $\mathcal L_X\mathcal L_X=d\iota_Xd\iota_X+\iota_Xd\iota_Xd$. (Does this have any interpretation?)
4. Playing with $d\iota_X$ and $\iota_Xd$ gives $d\iota_X+\iota_Xd=\mathcal L_X$. (Cartan's magic formula)
5. Playing with $d\mathcal L_X$ and $\mathcal L_Xd$ gives $d\mathcal L_X-\mathcal L_Xd=0$. (Exterior and Lie derivatives commute)
6. Playing with $\iota_X\mathcal L_Y$ and $\mathcal L_Y\iota_X$ gives $\iota_X\mathcal L_Y-\mathcal L_Y\iota_X=\iota_{\mathcal L_XY}=-\iota_{\mathcal L_YX}=\mathcal L_X\iota_Y-\iota_Y\mathcal L_X$. (What is this?!)
7. In particular, we have $\iota_X\mathcal L_X=\mathcal L_X\iota_X$.

Edit: I suppose we should also add these to the list for completeness:

8. $\iota_X\iota_Y+\iota_Y\iota_X=0$.
9. $\mathcal L_X\mathcal L_Y-\mathcal L_Y\mathcal L_X=\mathcal L_{[X,Y]}=\mathcal L_{\mathcal L_XY}$.

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Cartan's magic formula says that, with $\iota_X$ as the homotopy operator, $\mathcal L_X$ is the zero map in cohomology. Is there a geometric interpretation that makes this obvious?

In general, all these formulas are obtained through algebraic manipulation and I have no geometric intuition for why they should be true. Is there a convenient mnemonic device that makes all of this "obvious"?

Answer 1

OP might find the following rephrasing useful. Consider a graded associative algebra $({\cal A},\circ,+)$ containing the exterior derivative $d$ of form degree $|d|:=+1$, and the contraction $i_X$ of form degree $|i_X|:=-1$, where $X$ is a vector field. The graded Lie superbracket is the supercommutator

$$[a,b]~:=~a\circ b -(-1)^{|a||b|}b\circ a~=~-(-1)^{|a||b|}[b,a], \tag{1}$$

which is graded skewsymmetric, and which satisfies a graded Jacobi identity

$$ 0~=~\sum_{\text{cycl. }a,b,c} (-1)^{|a||c|} [a,[b,c]]. \tag{2} $$

Note that the following supercommutators

$$ [d,d]~=~2d^2~=~0 \quad\text{and}\quad [i_X,i_Y]~=~0\tag{3}$$

vanish. Cartan's magic formula is the supercommutator

$${\cal L}_X ~=~ [i_X,d] .\tag{4}$$

Contraction with the Lie bracket $[X,Y]$ of vector fields is again a supercommutator

$$ i_{[X,Y]}~=~[{\cal L}_X, i_Y] .\tag{5}$$

Several of OP's relations are consequences of the above relations, e.g.

$$ [d, {\cal L}_X]~\stackrel{(2)+(3)+(4)}{=}~0, \tag{6}$$

$$ {\cal L}_{[X,Y]}~\stackrel{(2)+(4)+(5)+(6)}{=}~[{\cal L}_X,{\cal L}_Y],\tag{7}$$

and so forth.