# How do I prove $d\omega = \frac{1}{p!}(\partial \omega_{i_1 \ldots i_p}/\partial x^j) \,dx^j \wedge dx^{i_1} \wedge \ldots \wedge dx^{i_p}$?

How do I prove $d\omega = \frac{1}{p!}(\partial \omega_{i_1 \ldots i_p}/\partial x^j) \, dx^j \wedge dx^{i_1} \wedge \ldots \wedge dx^{i_p}$ for

$$\omega = \frac{1}{p!} \omega_{i_1 \ldots i_p} dx^{i_1} \wedge \ldots \wedge dx^{i_p} \text{?}$$

I can use the following:

a) $d(\alpha + \beta) = d\alpha + d\beta$

b) $d^2 = 0$,

c) $df = \frac{\partial f}{\partial x^j} \, dx^j$,

d) $d(f\omega) = df \wedge \omega + f \, d\omega$,

e) $d(dx^{i_1} \wedge \ldots \wedge dx^{i_p}) = 0$;

I think that factorial is off. By linearity which is your property $(a)$, it suffices we show this for a form of the form (heh) $\omega = f dx^{i_1} \wedge \cdots dx^{i_n}$. If $n=0$, this is property $(c)$. So assume that $n>0$, and write $\omega = f dx^{i_1} \wedge \cdots dx^{i_n}=f\eta$. Then $(d)$ means that $d\omega = df\wedge \eta + f\wedge d\eta$ and by property $(e)$ we have that $d\eta = 0$, so that $d\omega = df\wedge \eta$. Again property $(c)$ gives you what you want.

• Really quite simple, actually ;) Thanks a lot! – jo_elk Dec 12 '17 at 19:09
• Concerning the factorial: this is the way the lecturer defined it, but he mentioned that there are different conventions. – jo_elk Dec 12 '17 at 19:11
• My point is that it is not deduced from what you have: you must artificially introduce the factorial. – Pedro Tamaroff Dec 12 '17 at 19:16