I was reading Sean Carrol's GR book, when on page 85 he introduces the Leibniz rule analogue for exterior derivatives:
$$\text d(\omega\wedge\eta) = (\text d\omega)\wedge\eta + (-1)^p\omega\wedge(\text d\eta)$$ where $\omega$ is set to be a p-form and $\eta$ a q-form. I was trying to prove it by expanding the RHS and showing that it could be rewritten as the LHS:
$$\text d(\omega\wedge\eta)= (p+q+1)\partial_{[\mu_1}X_{\mu_2...\mu_{p+q+1}]}=\frac{p+q+1}{(p+q+1)!}\sum_{\pm\circlearrowleft}\partial_{\mu_1}X_{\mu_2...\mu_{p+q+1}}$$
where $X=\omega\wedge\eta$, and $\sum_{\pm\circlearrowleft}$ denotes the sum over all permutations of the indices with a positive sign if it's an even permutation and a negative sign if it's an odd permutation. Then
$$X_{\mu_1...\mu_{p+q}} = \omega\wedge\eta=\frac{(p+q)!}{p!q!}\frac{1}{(p+q)!}\sum_{\pm\circlearrowleft}\omega_{\mu_1...\mu_p}\eta_{\mu_{p+1}...\mu_{p+q}}$$ So we get $$\text d(\omega\wedge\eta) = \frac{1}{(p+q)!p!q!}\sum_{\pm\circlearrowleft}\partial_{\mu_1}\bigg(\sum_{\pm\circlearrowleft}\omega_{\mu_2...\mu_{p+1}}\eta_{\mu_{p+2}...\mu_{p+q+1}}\bigg)=\frac{1}{p!q!}\sum_{\pm\circlearrowleft}\partial_{\mu_1}\omega_{\mu_2...\mu_{p+1}}\eta_{\mu_{p+2}...\mu_{p+q+1}}$$ However, when I expand $(\text d\omega)\wedge\eta$ I get the exact same thing
$$\text d\omega = (p+1)\partial_{[\mu_1}\omega_{\mu_2...\mu_{p+1}]}=\frac{p+1}{(p+1)!}\sum_{\pm\circlearrowleft}\partial_{\mu_1}\omega_{\mu_2...\mu_{p+1}}$$
$$(\text d\omega)\wedge\eta = \frac{1}{p!}\frac{(p+q+1)!}{(p+1)!q!}\frac{1}{(p+q+1)!}\sum_{\pm\circlearrowleft}\bigg(\sum_{\pm\circlearrowleft}\partial_{\mu_1}\omega_{\mu_2...\mu_{p+1}}\bigg)\eta_{\mu_{p+2}...\mu_{p+q+1}}$$
$$(\text d\omega)\wedge\eta = \frac{1}{p!q!}\sum_{\pm\circlearrowleft}\partial_{\mu_1}\omega_{\mu_2...\mu_p+1}\eta_{\mu_{p+2}...\mu_{p+q+1}}$$ I can explain more about the notation and how I'm simplifying the double sums. I can't find my mistake, and I would appreciate it if someone could tell me if I'm treating these derivatives wrong. Thanks.

  • 1
    $\begingroup$ I suggest you be explicit with your sum over permutations, and you reinstate the wedge products. Also, I suggest you prove things in the following simpler cases: first prove it when $\omega$ is a 0-form (i.e a real-valued function). Next, prove it when $\omega=dx^{i_1}\wedge\cdots\wedge dx^{i_p}$ and $\eta=dx^{j_1}\wedge\cdots\wedge dx^{j_q}$ (both sides vanish; make sure you know why). Now, using linearity of $d$, you should be able to deduce the general case. If in doubt, just open up a textbook on differential geometry or something similar (eg Spivak’s calculus on manifolds). $\endgroup$
    – peek-a-boo
    Jul 31, 2023 at 15:10
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    $\begingroup$ actually I think I see your error. You didn’t apply the product rule on $\partial_{\mu_1}(\omega_{\cdots}\eta_{\cdots})$. You really have to be careful with notation here (your weird circle notation and the lack of the wedge products will almost definitely confuse you, especially when dealing with the $(-1)^p\omega\wedge d\eta$ term… and notice you have a typo, it should be $(-1)^p$, not $(-1)^q$ as you wrote). $\endgroup$
    – peek-a-boo
    Jul 31, 2023 at 15:14
  • $\begingroup$ @peek-a-boo Thanks I just corrected the exponent and you are right about not applying the product rule, once I did I got the extra term I needed to make both sides the same. I got the $(-1)^p$ from the second term of the product rule, by moving the indices so $\omega_{\mu_1...\mu_{p}\partial_{\mu_{p+1}}}(\eta_{\mu_{p+2}...\mu_{p+q+1}})$ turned into $(-1)^p\omega_{\mu_2...\mu_{p+1}\partial_{\mu_{1}}}(\eta_{\mu_{p+2}...\mu_{p+q+1}})$. I think I'm still not being careful enough so I will check a book on differential geometry like you said. Thanks again. $\endgroup$
    – JS30
    Jul 31, 2023 at 15:48

1 Answer 1


The easiest way to derive these types of relations is to write the forms as

$$ \omega = \omega_I dx^I \hspace{2cm } \eta = \eta_Jdx^J$$

where $I$ and $J$ are some permutation of $\{ \mu_1, ..., \mu_p \}$ and $\{ \mu_1, ..., \mu_q \}$ respectively. Throughout, I am going to ignore the combinatoric factors. We can therefore write

$$ d(\omega \wedge \eta) = d( \omega_{I} \eta_{J}dx^I \wedge dx^J) = \partial_\mu(\omega_{I} \eta_{J}) dx^{\mu} \wedge dx^I \wedge dx^J = \partial_\mu(\omega_{I}) \eta_{J} dx^{\mu} \wedge dx^I \wedge dx^J + \omega_{I} \partial_\mu(\eta_{J}) dx^{\mu} \wedge dx^I \wedge dx^J$$

We can identify the first term as $d\omega\wedge\eta$. Note for the second term, we can pull $dx^\mu$ through $dx^I$ to get

$$ dx^\mu \wedge dx^I = (-1)^P dx^I\wedge dx^\mu$$

Then, we shall have

$$ (-1)^P(\omega_{I} dx^I )\wedge(\partial_\mu\eta_{J} dx^{\mu} \wedge dx^J) = (-1)^P\omega\wedge d\eta$$

Putting all of this together, we get the desired relation

$$ d(\omega \wedge \eta) = d\omega\wedge\eta + (-1)^P\omega\wedge d\eta$$


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