Equivalence relation of differential forms

Question

My notes claim that

$\displaystyle d\omega (x) = \frac{1}{k!} d\omega_{i_1\cdots i_k} \wedge f^{(i_1)}\wedge\cdots\wedge f^{(i_k)}$

is equivalent to

$\displaystyle d\omega(x) = \frac{1}{k!} \frac{\partial\omega_{i_1\cdots i_k}}{\partial x^j}dx^j \wedge f^{(i_1)}\wedge\cdots\wedge f^{(i_k)}$.

I am not sure you could prove this and I am not sure how $d\omega_{i_1\cdots i_k}$ is related to $\displaystyle \frac{\partial\omega_{i_1\cdots i_k}}{\partial x^j}dx^j$. I assume it may come from the definition of derivative of a 0-form?

Also what is $d\omega_{i_1\cdots i_k}$?

Answer 1

It is the definition. Basically, $d$ is degree $1$ anti-derivation satisfying:

1.$df$ is the usual differential for $f$ a $0$-form

2.$d(\alpha\wedge\beta)=d\alpha\wedge\beta+(-1)^k\alpha\wedge d\beta$ for $\alpha$ a $k$-form (antiderivation).

3.$d^2=0$

It can be proved that $d$ is uniquely determined by these axioms and the local coordinate version fits 1,2,3, so it has to be the exterior differentiation that turns $f$ into $df$. For more information, you can inquire any advanced differential geometry book (I learned it from Kobayashi & Nomizu), and also the wiki page.