Cartan's formula says: $\mathcal{L}_X \omega = (di_X + i_Xd) \omega $

For a 1-form, this is equivalent to:

$d\omega(X,Y) = X(\omega(Y)) - Y(\omega(X)) - \omega([X,Y]) $

For a 2-form, you can get:

$d\omega(X,Y,Z) = X(\omega(Y,Z)) - Y(\omega(X,Z) + Z(w(X,Y)) - \omega([X,Y],Z) + \omega([X,Z], Y) - \omega([Y,Z] X) $

Now let $X$ be a an element of the $n+1$ exterior power of the space of vector fields defined on an open set, $\omega$ be a $n$-form and define:

$X_1 \wedge \dots \wedge X_{n+1} (\omega) = X_1(\omega(X_2, \dots, X_{n+1})) - X_2(\omega(X_1, \dots, X_{n+1})) + \dots$

also define a map $\[\] : \Lambda^{n+1} \rightarrow \Lambda^n$:

$\[\] (X_1 \wedge \dots \wedge X_{n+1} ) = ([X_1,X_2]\wedge X_3 \dots \wedge X_{n+1}) - ([X_1,X_3]\wedge X_2 \dots \wedge X_{n+1}) + \dots$

Then you can keep iterating Cartan's formula and get:

$ d\omega(X) = X(\omega) - \omega([]X) $

$\[\]^2 = 0$ by the Jacobi identity, so that makes $\Lambda^{\bullet}$ into a chain complex, So my question is: what do you get when you take the homology?