Question

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?

Answer 1

I think this is not quite right.

For example, the map you write down $\bigwedge ^2 TM \to TM$ is supposed to be given by the bracket $X\wedge Y \mapsto [X,Y]$.

This is not actually a map of vector bundles, as the bracket operation is not linear over smooth functions. Moreover, this expression does not define a well defined operation on sections (in contrast to the operator $d$, say).

To see this, note that the expressions $(fX)\wedge Y$ and $X\wedge (fY)$ represent the same section of $\bigwedge ^2 TM$, but $[fX,Y] \neq [X,fY]$.

The expressions you write are (I guess) trying to be the Lie algbera homology of the space of vector fields. Vector fields do not form a Lie algebra over smooth functions which is why this doesn't make sense. It is interesting to note that the definition of Lie algebra cohomology can be modified to get something interesting (de Rham cohomology).