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let be $$ \omega= |x|^{-3} \left(x_1 dx_2 \wedge dx_3+x_2dx_3 \wedge dx_1 + x_3dx_1 \wedge dx_2\right) $$ and $G:= \mathbb{R}^3 \backslash \{ 0 \} $

I want to prove, that $ \omega$ is closed, but not exact

That $ \omega $ is closed, I can prove it by looking if $ d\omega =0 $

But how can I prove it's not exact?

I know that a continous 2-form is exact in $G$ if there exists a 1-Form $y$ so, that $\omega = dy $ how can I show there doesn't exist such $y$?


Edit: Showing $\int_S \omega \neq 0 $

Where $S$ is the unitsphere Set $r=1$ and set $$x_1= \sin \phi \cos \theta $$ $$x_2= \sin \phi \sin \theta$$ $$x_3= \cos \phi $$

$ \theta=[0, 2\pi], \phi=[0, \pi]$

then $ dx_1 \wedge dx_2 = -\sin \phi \cos \phi d\theta \wedge d\phi $

$ dx_2 \wedge dx_3 = -\sin^2 \phi \cos \theta d\theta \wedge d\phi $

$ dx_3 \wedge dx_1 = -\sin^2 \phi \sin \theta d \theta \wedge d\phi $

Putting in the equation: $$ \int_0^{2 \pi} \int_0^{ \pi} |x| ( \sin \phi \cos \theta - \sin^2 \phi \cos \theta ~d \theta \wedge d\ \phi \\+ \sin \phi \sin \theta -\sin^2 \phi \sin \theta ~d\theta \wedge d\phi + \cos \phi -\sin \phi \cos \phi~ d \theta \wedge d\phi $$

what do I put for $x$ ? I don't know how to solve the integral thank you for any help!

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    $\begingroup$ The first thing to try would be to integrate it on the unit sphere; if the result is not zero, then $\omega$ is not exact. $\endgroup$ Commented Jan 16, 2019 at 12:42
  • $\begingroup$ @AlexProvost What does it mean to integrate the unit sphere? $\endgroup$ Commented Jan 16, 2019 at 12:43
  • $\begingroup$ @MathOverview I said to "integrate it on the unit sphere", meaning the 2-form. $\endgroup$ Commented Jan 16, 2019 at 12:44
  • $\begingroup$ @AlexProvost But why integrate in the unitary sphere and not in $ G $? $\endgroup$ Commented Jan 16, 2019 at 12:46
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    $\begingroup$ On the unit sphere, $|x| = 1$. By the way, this reduces to the standard volume form on $S^2$, so you expect the integral to yield the surface area, $4\pi$. $\endgroup$ Commented Jan 16, 2019 at 13:42

1 Answer 1

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Switch to spherical coordinates and integrate $\omega$ on the unit sphere $r = 1$. Let $\theta,\phi$ denote the azimuth and polar angles, respectively; then $$dx = \cos \theta \sin \phi \, dr - \sin \theta \sin \phi \, d\theta + \cos\theta \cos \phi \, d\phi$$ $$ dy = \sin \theta \sin \phi \, dr + \cos \theta \sin\phi \, d\theta + \sin \theta \cos \phi \, d\phi $$ $$dz = \cos \phi \, dr - \sin\phi \, d\phi$$

and after some algebra we find that $\omega = \sin\phi \, d\phi \wedge d\theta$. Hence $$\int_{S^2} \omega = \int_0^{2\pi} \int_0^\pi \sin \phi \, d\phi \, d\theta = 2\cdot 2\pi = 4\pi.$$

This is the expected answer, namely the surface area of the sphere, as $\omega$ restricted to $S^2$ is precisely the volume form induced by that of $\mathbb R^3$. And since the integral is not zero, $\omega$ cannot be exact, lest Stoke's theorem be violated.

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