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I'd like to construct a closed but not exact $n-1$-form $\omega$ on $\mathbb{R}^n\setminus\{0\}$ in analogy to the winding form: $$\frac{x~dy-y~dx}{x^2+y^2}$$

I think something like $$\omega=\frac{\sum_{i=1}^n x_i(\star dx_i)}{(x_1^2+\dots+x_n^2)^{n/2}}$$ should probably work. $\star$ is the hodge star operator, for example $\star dx_1=dx_2\wedge\dots\wedge dx_n$ and $\star dx_2=-dx_1\wedge dx_3\wedge\dots\wedge dx_n$. $\omega$ is closed since $$d\omega=\sum_{i=1}^nd\left(\frac{ x_i}{(x_1^2+\dots+x_n^2)^{n/2}}\right)\wedge\star dx_i=\sum_{i=1}^n \frac{(\sum x_j^2)^{n/2}-n x_i^2 (\sum x_j^2)^{n/2-1}}{(\sum x_j^2)^{n}} dx_1\wedge\dots\wedge dx_n=0$$ Now it remains to show that $\omega$ is not exact. I guess a direct computation is not the way to do it. I thought about assuming that there exists $d\eta=\omega$ and then using Stoke's theorem to get the contradiction $0\not=\int_{S^{n-1}}\omega=\int_\emptyset \eta=0$. Two questions about this:

  1. Is the general argument correct?
  2. How to prove $0\not=\int_{S^{n-1}}\omega$? I guess a direct computation using generalized sphere coordinates is rather tedious.
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  • $\begingroup$ Another way (not saying a better way) to see $\int_{S^{n-1}}\omega=0$ is the following.. we have already assumed $\omega=d\omega’$... so, we have $$\int_{S^{n-1}}\omega=\int_{S^{n-1}}d\omega’=\int_{Disk}d(d(omega))=0$$ as $d^2=0$... this is one way to get used to the property that $d^2=0$.. $\endgroup$
    – user537667
    Dec 27, 2018 at 18:22

1 Answer 1

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  1. Yes, this is a reasonable way to go about things.

  2. Let $\psi=\sum_{i=1}^n x_i (\star dx_i)$. Then $\left.\psi\right|_{S^{n-1}}=\left.\omega\right|_{S^{n-1}}$, and so $\int_{S^{n-1}} \omega = \int_{S^{n-1}} \psi$. Moreover, $d\psi$ is a constant multiple of the volume form on $\Bbb{R}^n$. Now, apply Stokes' theorem to $B^n$ and $S^{n-1}$...

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  • $\begingroup$ What do you mean by $d\psi$ is a constant multiple of the volume form on $\mathbb{R}^n$? What is a volume form? $\endgroup$
    – Julian
    Jan 24, 2013 at 21:42
  • $\begingroup$ The volume form is $dx_1 \wedge dx_2 \wedge \dots \wedge dx_n$ (that is, the form $\Omega$ such that $\int_{D} \Omega$ gives the volume of any domain $D$ in $\Bbb{R}^n$). $\endgroup$
    – Micah
    Jan 24, 2013 at 21:44
  • $\begingroup$ Isn't $\psi$ a $n-1$-form and this volume form a $n$-form? $\endgroup$
    – Julian
    Jan 24, 2013 at 21:46
  • $\begingroup$ Yes. It's $d\psi$ which coincides with the volume form. $\endgroup$
    – Micah
    Jan 24, 2013 at 21:46
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    $\begingroup$ So $d\psi=n dx_1\wedge\dots\wedge dx_n$ and therefore $0\not=\operatorname{vol} B_n=\int_{B_n} \frac{d\psi}{n}=\frac{1}{n}\int_{S^{n-1}} \psi$? $\endgroup$
    – Julian
    Jan 24, 2013 at 21:51

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