Let $\omega$ denote the $(n-1)$ form on $\mathbb{R}^n\setminus\{0\}$ defined by:
$$\omega = \sum_{i=1}^n (-1)^{i-1} \space f_i \space dx_1\wedge... \wedge dx_{i-1} \wedge dx_{i+1} \wedge ... \wedge \space dx_n$$
where $f_i(x) = \frac{x_i}{|x|^m}$ ($m$ is a fixed positive integer).
Calculate $d\omega$.
I'm only familiar with 1-forms and 2-forms. Can anyone explain to me how (n-1) forms work in the context of this question?
I'm not sure if this is correct but
If: $\omega =\sum_{i=1}^n (-1)^{i-1} \space f_i \space dx_1\wedge... \wedge dx_{i-1} \wedge dx_{i+1} \wedge ... \wedge \space dx_n$, then would
$d\omega = \sum_{i=1}^n (-1)^{i-1}\left(\sum_{j=1}^n(\frac{\partial f_i}{\partial x_j}-\frac{\partial f_j}{\partial x_i})\right)dx_1 \wedge dx_{i-1} \wedge dx_{i+1} ...\wedge dx_n$
Where $\left(\frac{\partial f_i}{\partial x_j}-\frac{\partial f_j}{\partial x_i}\right) = 0$
because $$\frac{\partial \frac{x_i}{|x|^m}}{\partial x_j} = \frac{-2mx_ix_j}{2|x|^{m+2}} = \frac{\partial \frac{x_j}{|x|^m}}{\partial x_i}.$$
Then $d\omega = 0$.
Is this right? Or do I at least have the right idea?
Hint
$d^2 = 0,$ and $d$ obeys a product rule, so we have $$ d\omega = \sum_{i=1}^n (-1)^{i-1} df_i\wedge dx_1\wedge\ldots \wedge dx_{i-1}\wedge dx_{i+1}\wedge \ldots \wedge dx_n.$$ We also have $$ df_i = \sum_j \left(\frac{\partial}{\partial x_j} f_i\right)dx_j.$$ Then use that $\wedge$ is anticommutative and distributive.