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!

  • 1
    $\begingroup$ 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. $\endgroup$ – Alex Provost Jan 16 at 12:42
  • $\begingroup$ @AlexProvost What does it mean to integrate the unit sphere? $\endgroup$ – MathOverview Jan 16 at 12:43
  • $\begingroup$ @MathOverview I said to "integrate it on the unit sphere", meaning the 2-form. $\endgroup$ – Alex Provost Jan 16 at 12:44
  • $\begingroup$ @AlexProvost But why integrate in the unitary sphere and not in $ G $? $\endgroup$ – MathOverview Jan 16 at 12:46
  • 1
    $\begingroup$ 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, $4\pi$. $\endgroup$ – Alex Provost Jan 16 at 13:42

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.


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.