# Exterior derivatives Leibniz rule

I want to prove Sean Carroll's "spacetime and geometry"'s eq.(2.78): $$\mathrm{d}(\omega \wedge \eta)=(\mathrm{d} \omega) \wedge \eta+(-1)^p \omega \wedge(\mathrm{d} \eta) \tag{2.78}$$ where $$\omega$$ is a $$p-$$form, $$\eta$$ is a $$q-$$form. There are many versions of proof online, but most of them use directly the wedge product $$\wedge$$, which is more mathematics. But I want to use the component version to prove it. Start with $$(A \wedge B)_{\mu_1 \cdots \mu_{p+q}}=\frac{(p+q) !}{p ! q !} A_{\left[\mu_1 \cdots \mu_p\right.} B_{\left.\mu_{p+1} \cdots \mu_{p+q}\right]} \tag{2.73}$$ and $$(\mathrm{d} A)_{\mu_1 \cdots \mu_{p+1}}=(p+1) \partial_{\left[\mu_1\right.} A_{\left.\mu_2 \cdots \mu_{p+1}\right]} \tag{2.76}$$ The L.H.S of (2.78) is \begin{aligned} \mathrm{d}(\omega \wedge \eta)&=\mathrm{d}(\frac{(p+q) !}{p ! q !} \omega_{\left[\mu_2 \cdots \mu_{p+1}\right.} \eta_{\left.\mu_{p+2} \cdots \mu_{p+q+1}\right]}) \\ &=(p+q+1)\frac{(p+q) !}{p ! q !}\partial_{\left[\mu_1 \right.}\omega_{ \left[\mu_2 \cdots \mu_{p+1}\right.} \eta_{\left.\left.\mu_{p+2} \cdots \mu_{p+q+1}\right]\right] } \\ &\stackrel{?}{=}\frac{(p+q+1) !}{p ! q !}\partial_{\left[\mu_1 \right.}\frac{1}{(p+q)!}(\omega_{ \mu_2 \cdots \mu_{p+1}} \eta_{\mu_{p+2} \cdots \mu_{p+q+1} }+ \text{alternating sign sums })_{\left. \right]}\\ &=\frac{(p+q+1) !}{p ! q !}\partial_{\left[\mu_1 \right.}(\omega_{ \mu_2 \cdots \mu_{p+1}} \eta_{\mu_{p+2} \cdots \mu_{p+q+1} })_{\left. \right]} \\ &=\frac{(p+q+1) !}{p ! q !}(\partial_{\mu_1 }\omega_{ \mu_2 \cdots \mu_{p+1}} \eta_{\mu_{p+2} \cdots \mu_{p+q+1} }) \end{aligned} where in the $$\stackrel{?}{=}$$ step, I use the fact that all the $$(p+q)!-1$$ terms can equal to the first term under suitable change of signs.

Now, for the R.H.S of (2.78) \begin{aligned}(\mathrm{d} \omega) \wedge \eta&=((p+1)\partial_{\left[\mu_1 \right.}\omega_{\left. \mu_2 \cdots \mu_{p+1}\right]})\wedge \eta \\ &=(p+1)\frac{(p+q+1)!}{(p+1)!q!}(\partial_{\left[\left[\mu_1 \right. \right.}\omega_{\left. \mu_2 \cdots \mu_{p+1}\right]} \eta_{\left. \mu_{p+2} \cdots \mu_{p+q+1} \right]} ) \\ &=\frac{(p+q+1)!}{(p)!q!}(\partial_{\mu_1 }\omega_{ \mu_2 \cdots \mu_{p+1}} \eta_{\mu_{p+2} \cdots \mu_{p+q+1} } )\end{aligned} where in the last step I use a similar method with $$\stackrel{?}{=}$$. But, this already equal to the L.H.S of (2.78). So what's wrong with my derivation?

• I think you weren't careful enough about the derivatives. On the LHS, you are taking derivative of the product of $\omega$ times $\eta$, whereas the RHS term you computed has only a derivative of $\omega$. If you expand the derivative of the product on the LHS, you get a second term which you can only capture with the $\omega \wedge (d \eta)$ term Feb 19, 2023 at 15:50
• Would Mathematics be a better home for this question? Feb 19, 2023 at 17:56

I think your derivation is correct, except that $$d(\omega \wedge \eta)_{\mu_1 ... \mu_{p+q+1}} = \frac{(p+q+1)!}{p! q!} \partial_{\mu_1}(\omega_{\mu_2 ... \mu_{p+1}} \eta_{\mu_{p+2} ... \mu_{p+q+1}}) \neq \frac{(p+q+1)!}{p! q!} (\partial_{\mu_1}\omega_{\mu_2 ... \mu_{p+1}}) \eta_{\mu_{p+2} ... \mu_{p+q+1}}$$
Instead $$d(\omega \wedge \eta)_{\mu_1 ... \mu_{p+q+1}} = \frac{(p+q+1)!}{p! q!} (\partial_{\mu_1}\omega_{\mu_2 ... \mu_{p+1}}) \eta_{\mu_{p+2} ... \mu_{p+q+1}} + \frac{(p+q+1)!}{p! q!} \omega_{\mu_2 ... \mu_{p+1}} (\partial_{\mu_1} \eta_{\mu_{p+2} ... \mu_{p+q+1}})$$ The first term, as you've already shown, is equal to $$d\omega \wedge \eta$$ and for the second term: $$(\omega \wedge d \eta)_{\mu_1 ... \mu_{p+q+1}} = \frac{(p+q+1)!}{p! q!} \omega_{\mu_1 ... \mu_p} (\partial_{\mu_{p+1}} \eta_{\mu_{p+2} ... \mu_{p+q+1}})$$ which differs from the previous second term by a cyclic permutation of $$\mu_1 ... \mu_{p+1}$$ with sign $$(-1)^p$$