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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?

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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?.]

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  • $\begingroup$ Thanks! So I need to think of $\mathrm{d} \omega$ as a $2$-form and then take the inner product with $XY$, right? I know how to take the inner product between, say, $\langle \mathrm{d} x^\mu \mathrm{d} x^\nu , e_\lambda e_\delta \rangle$. But I don't know how to do it for $\langle \mathrm{d} x^\mu \wedge \mathrm{d} x^\nu , e_\lambda e_\delta \rangle$. The book I'm reading doesn't seem to mention this, could you explain this or give a reference? $\endgroup$
    – Hunter
    Apr 3 '14 at 11:15
  • $\begingroup$ For instance, I can't seem to figure out where the minus sign comes from, although I assume this has to do with the anti-commutative behavior of the wedge product. $\endgroup$
    – Hunter
    Apr 3 '14 at 11:16
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    $\begingroup$ The definition of $dx^\mu\wedge dx^\nu$ is $$dx^\mu\wedge dx^\nu = (dx^\mu dx^\nu - dx^\nu dx^\mu).$$ $\endgroup$ Apr 3 '14 at 11:25
  • $\begingroup$ Ahhhh of course!! Thanks!! I lost sight of that definition; silly me $\endgroup$
    – Hunter
    Apr 3 '14 at 11:48

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