0
$\begingroup$

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?

$\endgroup$
  • 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.