Deriving Cartan formula I have trouble deriving Cartan formula of the form:
$$
\mathrm{d} \omega (X,Y) = X[\omega(Y)] - Y[\omega(X)] - \omega([X,Y]) \tag{1}
$$
where $\mathrm{d}$ is the exterior derivative, $\omega$ is a one-form, $X$ and $Y$ are tangent vectors in the coordinate basis $\{e_\mu \}$, and $[ \cdot,\cdot]$ denotes the Lie bracket. I can show that the right-hand side equals:
\begin{equation}
X[\omega(Y)] - Y[\omega(X)] - \omega([X,Y]) = (X^\nu  Y^\mu - Y^\nu X^\mu ) \partial_\nu \omega_\mu \tag{2}
\end{equation}
but I have difficulties evaluating the left-hand side of equation $(1)$.
I have tried this:
$$
\begin{aligned}
\omega(X,Y) & = \omega_\mu \mathrm{d} x^\mu X^\nu e_\nu Y^\lambda e_\lambda \\&
= \omega_\mu \delta^\mu_\lambda X^\nu Y^\lambda e_\nu \\&
= \omega_\mu X^\nu Y^\mu e_\nu
\end{aligned}
$$
This is not any kind of form anymore, so I'm not sure if the exterior derivative is well-defined to act on it. If I would try it, then I get:
$$
\begin{aligned}
\mathrm{d} \omega & = \left( \partial_\lambda \omega_\mu X^\nu Y^\mu e_\nu \right) \mathrm{d} x^{\lambda} \\&
= \partial_\lambda \omega_\mu X^\nu Y^\mu \delta_\nu^\lambda \\&
= \partial_\nu \omega_\mu X^\nu Y^\mu 
\end{aligned}
$$
which does not equal the right-hand side equation $(2)$. And so it does not obey equation $(1)$.
Does anybody where I went wrong in my derivation?
 A: When you write $\omega(X,Y)$ this gives the impression that $\omega$ is a 2-form! It is $d\omega$ that is a 2-form, if $\omega$ is a 1-form. The action of the 1-form $\omega =\omega_\mu dx^\mu$ on the vector field $X = X^\mu \partial_\mu$ is simply $$\omega(X) = \omega_\mu X^\mu.$$
(Since you use index notation, you probably know about relativity and this should be familiar: the contraction of a contravariant and a covariant vector is exactly a 1-form eating a vector field!)
Now you can always write $\omega = \omega_\mu dx^\mu$. Since $d(dx) = 0$ (this is part of the definition of $d$), it must be that $$d\omega = (d\omega_\mu) \wedge dx^\mu.$$
But on a function, $$d\omega_\mu = \frac{\partial \omega_\mu}{\partial x^\nu} dx^\nu.$$
(This formula is what they should teach you in multivariate calculus but don't...)
Now you can see that when you let $X$ act on $\omega(Y)$ and $Y$ on $\omega(X)$, you will get derivatives also on the components $X^\nu$ and $Y^\nu$. But you will get that from $\omega([X,Y])$ too, and they will cancel. You should be able to work this out yourself.
[There is actually a theorem that if you want to know if a tensor identity is true, it is sufficient to check it under the assumption that any Lie brackets involved are 0. Can you think of why?.]
