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?


1 Answer 1


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?.]

  • $\begingroup$ 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, 2014 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, 2014 at 11:16
  • 2
    $\begingroup$ 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, 2014 at 11:25
  • $\begingroup$ Ahhhh of course!! Thanks!! I lost sight of that definition; silly me $\endgroup$
    – Hunter
    Apr 3, 2014 at 11:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.