exterior derivative in local coordinates

Question

Given that the exterior derivative d acting on n-form w is defined as:

$dw(X_1, \ldots, X_{n+1}) = \sum_{i=1}^{n+1} (-1)^{i+1}X_i(w(X_1, \ldots, \hat X_i, \ldots, X_{n+1})) + \sum_{i<j}(-1)^{i+j}w([X_i, X_j], X_1, \ldots, \hat X_i, \ldots, \hat X_j, \ldots, X_{n+1})$

where the $X_i$ are members of the tangent space and the hat denotes omission.

My goal is to prove that dw can be written in local coordinates as

$dw = \partial_b w_{a_1, \ldots,a_n} dx^b \wedge dx^{a_1} \wedge , \ldots, \wedge dx^{a_n}$

I started by writing $w = w_{a_1, \ldots,a_n} dx^{a_1} \wedge , \ldots, \wedge dx^{a_n}$ as an expansion of w in some chosen coordinate system, which I believe we can do because since we are only working locally, the relevant topolgy is that of $\mathbb{R}^m$ for some m, and therefore the n-forms have a basis in the covector space.

Then I started with the first term in the def. of dw, and wrote:

$\sum_{i=1}^{n+1} (-1)^{i+1}X_i((w_{a_1, \ldots,a_n} dx^{a_1} \wedge , \ldots, \wedge dx^{a_n})(X_1, \ldots, \hat X_i, \ldots, X_{n+1}))$

applying product rule(since $X_i$ is a derivation)

$\sum_{i=1}^{n+1} (-1)^{i+1}X_i(w_{a_1,\ldots,a_n}) (dx^{a_1}\wedge , \ldots, \wedge dx^{a_n})(X_1, \ldots, \hat X_i, \ldots, X_{n+1}) + \sum_{i=1}^{n+1} (-1)^{i+1}w_{a_1,\ldots,a_n}X_i( (dx^{a_1}\wedge , \ldots, \wedge dx^{a_n})(X_1, \ldots, \hat X_i, \ldots, X_{n+1}))$

And from here I'm stuck, im not sure how to complete this proof as the defintion of the exterior derivative involves its action on a number of vectors, while the statement we want to prove seems to be divorced from this notion(or rather implicitly, the vectors seem to ordered without omissions)

Answer 1

After consulting a bit more with Spivak's "Introduction to Differential Geometry" and with help from Ted I believe I have solved it.

I forgot to recongize that as an n-form, dw could also expanded in a basis as

$dw = dw(\frac{\partial}{\partial x^{a_1}},\ldots,\frac{\partial}{\partial x^{a_{k+1}}}) dx^{a_{1}}\wedge ,\ldots,\wedge dx^{a_{k+1}}$

Then

$dw(\frac{\partial}{\partial x^{a_1}},\ldots,\frac{\partial}{\partial x^{a_{k+1}}}) = \sum_{i=1}^{k+1} (-1)^{i+1}\frac{\partial}{\partial x^{a_i}}(w_{b_1,\ldots,b_n})(dx^{b_1}\wedge\ldots,\wedge dx^{b_n})(\frac{\partial}{\partial x^{a_1}},\ldots,\hat{\frac{\partial}{\partial x^{a_{k+1}}}}\ldots, \frac{\partial}{\partial x^{a_{k+1}}})$

where the Lie bracket in the second sum causing that term to go to zero since partial derivatives commute

$ = \sum_{i=1}^{k+1} (-1)^{i+1} \frac{\partial}{\partial x^{a_i}}w_{a_1,\ldots, \hat a_i,\ldots,a_n}$

so

$dw = \sum_{i=1}^{k+1} (-1)^{i+1} \frac{\partial}{\partial x^{a_i}}w_{a_1,\ldots, \hat a_i,\ldots,a_n} dx^{a_{1}}\wedge ,\ldots,\wedge dx^{a_{k+1}}$

upon rearranging of indicies for each term gives

$dw = (k+1)\frac{\partial}{\partial x^{a_{k+1}}}w_{a_1,\ldots,a_n} dx^{a_{k+1}}\wedge dx^{a_1} ,\ldots,\wedge dx^{a_{k}}$

The only thing that seems to be wrong is the k+1 factor which should be due to convention.