Let $(U,\varphi)$ be a chart of a smooth manifold $M$, and let $\omega \in \Omega^k(M)$, where $\Omega^k(M)$ is the $C^{\infty}(M)$ module of differential k-forms on $M$.
We write $\omega|_U = \sum_{I} \omega_{I} dx^{I}$ where $I$ is a multi-index such that $i_1 < \ldots < i_k$.
Then we have $d\omega|_U = \sum_{I}d\omega_I \wedge dx^I$.
In the proof, there is a step where this happens $$d\omega_I \wedge dx^I = \sum_{i_0 = 1}^{n} \partial_{i_0} \omega_{i_1,\ldots,i_k} dx^{i_0} \wedge \ldots \wedge dx^{i_n}$$ Can someone explain what happens here?
Where does the part $$\sum_{i < j} (-1)^{i+j}\omega([X_i,X_j], \ldots,\hat{X_i}, \ldots,\hat{X_j},\ldots,X_k)$$ disappear (this is one term in the coordinate-independent definition of the exterior derivative)?
I know that, for the Lie-bracket $[-,-]$, and coordinate-vector fields $\partial_i$ with respect to a chart $(U,\varphi = (x^1,\ldots,x^n))$, we have $$[\partial_i,\partial_j] = 0$$ so the first argument in $\omega$ (in the local formulation with respect to a chart) will be $0$, but does this necessarily mean that it is identically zero?
Well, to maybe answer my own question, maybe it is the case: $$\omega(0,\partial_{i_0},\ldots,\hat{\partial_{i_i}}, \ldots,\hat{\partial_{i_j}},\ldots,\partial_{i_k}) = \omega(0\partial_{i_0}+\ldots+0\partial_{i_k},\partial_{i_0},\ldots,\hat{\partial_{i_i}}, \ldots,\hat{\partial_{i_j}},\ldots,\partial_{i_k}) = \sum_{k = i_0}^{i_k} 0 \cdot \omega(\partial_k,\partial_{i_0},\ldots,\partial_{i_k}) = 0$$
Is my reasoning correct?
Edit: To be more explicit, part of my thinking here was trying to connect the coordinate-dependent definition with the coordinate-independent definition, where the coordinate-independent definition for $$d:\Omega^k(M) \to \Omega^{k+1}(M)$$ is defined by $$d\omega(X_0,\ldots,X_k) = \sum_{j = 1}^{n}(-1)^j X_j(\omega(X_0,\ldots,\hat{X_j},\ldots,X_k)+\sum_{i < j} (-1)^{i+j}\omega([X_i,X_j],\ldots,\hat{X_i},\ldots,\hat{X_j},\ldots,X_k)$$
where, for an argument $a$, $\hat{a}$ means removing that argument.
If $f$ is a function and $\alpha$ is a differential form, then $$ d(f\alpha) = df \wedge \alpha + f d\alpha. $$ Applying this to $f = \omega_I$ and $\alpha = dx^I$, and noticing that $d(dx^I) = 0$, yields $$ d(\omega_I dx^I) = d\omega_I \wedge dx^I. $$ The result now follows by writing everything in coordinates. In particular, $d\omega_I=\sum_{i_0=1}^n (\partial_{i_0} \omega_I) dx^{i_0}$, so that one has $$ d(\omega_I dx^I) = \sum_{i_0 = 1}^n \partial_{i_0}\omega_I dx^{i_0}\wedge dx^I. $$ Since the multi-index $I$ is given by $(i_1,\ldots,i_k)$, we have $\omega_I=\omega_{i_1,\ldots,i_k}$ and $dx^I = dx^{i_1}\wedge \cdots \wedge dx^{i_k}$. Hence, $$ d(\omega_I dx^I) = \sum_{i_0=1}^n \partial_{i_0} \omega_{i_1,\ldots,i_k} dx^{i_0}\wedge dx^{i_1}\wedge\cdots\wedge dx^{i_k}. $$