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?

  • 1
    $\begingroup$ do you know a good way to get from one forms to two forms? $\endgroup$ – Andres Mejia Feb 23 '18 at 0:02
  • $\begingroup$ No not really.. $\endgroup$ – WannaBeRealAnalysist Feb 23 '18 at 0:38


$d^2 = 0,$ 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.

  • $\begingroup$ Thank you. You're very kind. $\endgroup$ – WannaBeRealAnalysist Feb 23 '18 at 4:15
  • $\begingroup$ I posted an answer, is it correct? $\endgroup$ – WannaBeRealAnalysist Feb 23 '18 at 22:36
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    $\begingroup$ @WannaBeRealAnalysist No. For one, $\omega$ is an $(n-1)$-form so $d\omega$ is an $n$-form. Second, even if your formula were right $\frac{\partial f_i}{\partial x_j}-\frac{\partial f_j}{\partial x_i} \ne 0.$ $f_i$ and $f_j$ are different functions (as far as I can tell) and these are first derivatives so mixed partial derivatives commuting has no relevance. (But your formula isn't right so really this doesn't need to be explored further.) $\endgroup$ – spaceisdarkgreen Feb 24 '18 at 0:12
  • $\begingroup$ but given the $f$ is defined, wouldn't $\left(\frac{\partial f_i}{\partial x_j}-\frac{\partial f_j}{\partial x_i}\right) = 0$? because the partial of $f_i$ with respect to $f_j$ would equal the partial of $f_j$ with respect to $f_i$ $\endgroup$ – WannaBeRealAnalysist Feb 24 '18 at 0:56
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    $\begingroup$ @WannaBeRealAnalysist Yes, actually it does (I didn't seen the explicit form) but it has nothing to do with mixed partials. But it's a moot point I think since that combination does not actually appear in the answer. $\endgroup$ – spaceisdarkgreen Feb 24 '18 at 1:23

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