# differential forms- $\omega$ closed but not exact

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!

• 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. Commented Jan 16, 2019 at 12:42
• @AlexProvost What does it mean to integrate the unit sphere? Commented Jan 16, 2019 at 12:43
• @MathOverview I said to "integrate it on the unit sphere", meaning the 2-form. Commented Jan 16, 2019 at 12:44
• @AlexProvost But why integrate in the unitary sphere and not in $G$? Commented Jan 16, 2019 at 12:46
• 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, $4\pi$. Commented Jan 16, 2019 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.