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Let $\omega$ be a $2$-form on $\mathbb{R}^3\setminus\{0\}$ defined by $$ \omega = \frac{x\,dy\wedge dz+y\,dz\wedge dx +z\,dx\wedge dy}{(x^2+y^2+z^2)^{\frac{3}{2}}} $$ Show that $\omega$ is closed but not exact.

In order to show that $\omega$ is closed, I need to show that $d\omega=0$. I'm having some problems getting all of the calculus right and somewhere along the way I'm messing up. I started by rewriting $\omega$ as $$ \omega = (x\,dy\wedge dz+y\,dz\wedge dx +z\,dx\wedge dy)(x^2+y^2+z^2)^{-\frac{3}{2}} $$ Now I should be able to use the product rule to evaluate (I think). Then $$ d\omega = (dx\wedge dy\wedge dz+dy\wedge dz\wedge dx +dz\wedge dx\wedge dy)(x^2+y^2+z^2)^{-\frac{3}{2}} + (\ast) $$ where $$ (\ast) = (x\,dy\wedge dz+y\,dz\wedge dx +z\,dx\wedge dy)\left(-\frac{3}{2}(2x\,dx+2y\,dy+2z\,dz)\right)(x^2+y^2+z^2)^{-\frac{5}{2}} $$ Even after trying to simplify everything, I can't get it to cancel. This makes me think that perhaps I can't apply the product rule like this. What should I do to calculate $d\omega$?

If $\omega$ is a globally defined smooth form and if $d\omega=0$, then $\omega$ is exact because there is some other form $\alpha$ with $d\alpha=\omega$ and $d^2\alpha=d\omega=0$. Because $\omega$ is not defined at $(0,0,0)$, it makes sense that it isn't exact. Is there a way to show that there can't be an $\alpha$ such that $d\alpha=\omega$?

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Maybe cylindrical or spherical coordinates will help. – Pragabhava Oct 29 '12 at 20:22
To show something that a 2-form is not exact, it is sufficient to integrate it over a closed, bounndaryless region and get something non-zero. In this case, I would recommend integrating over the sphere (since you have an $x^2 + y^2 + z^2$ term). – Eric O. Korman Oct 29 '12 at 20:41
My method for integrating 2-forms has previously involved parameterizing the surface that I'm integrating over. Do I need to do the same thing and parameterize the sphere $S=\{(x,y,z)|\;x^2+y^2+z^2=1\}$ or is there a way to integrate $\omega$ just using $x^2+y^2+z^2=1$? – chris Oct 29 '12 at 22:41
By the way, this is the Solid Angle form. – diff_math Jul 24 '13 at 21:53
up vote 6 down vote accepted

Your idea is good. Define $r = (x^2 + y^2 + z^2)^\frac{1}{2}$, $f(x,y,z) = \frac{1}{r^3}$ and $\mu = x dy \wedge dz + y dz \wedge dx + z dx \wedge dy$.

Then by the product rule: $d(\omega) = d(f\mu) = df \wedge \mu + f d\mu$. Let us hold hands and calculate:

$$ d\mu = dx \wedge dy \wedge dz + dy \wedge dz \wedge dx + dz \wedge dx \wedge dy = 3 dx \wedge dy \wedge dz. $$

$$ df = \frac{-3}{r^5} (x dx + y dy + z dz)$$

$$ df \wedge \mu = \frac{-3}{r^5} (x dx + y dy + z dz) (x dy \wedge dz + y dz \wedge dz + z dx \wedge dy) = \frac{-3}{r^5} (x^2 dx \wedge dy \wedge dz + y^2 dy \wedge dz \wedge dz + z^2 dz \wedge dz \wedge dy) = \frac{-3}{r^5} (r^2 dx \wedge dy \wedge dz) = \frac{-3}{r^3} dx \wedge dy \wedge dz $$

$$ df \wedge \mu + f d\mu = \frac{-3}{r^3} dx \wedge dy \wedge dz + \frac{3}{r^3} dx \wedge dy \wedge dz= 0. $$

Phew. As you see, in the calculations, you use a lot the antisymmetrization properties of the wedge product. You just need to do everything carefully, and it will come out.

For the second question, if it would be an exact form, the result of integration of $\omega$ on every two-dimensional closed submanifold (compact, without boundary) of $\mathbb{R}^3$ would be zero by Stokes's theorem. Try to find a closed submanifold on which you can calculate the integral directly relatively easily and for which the result is non-zero.

If you are familiar with conservative vector fields, this is just like showing that the field is not conservative by showing that the work done by it along some closed loop is non-zero.

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When you're calculating $df\wedge \mu$, you start by saying $df\wedge\mu=df$. Did you mean to write that $df\wedge\mu=df\mu$? – chris Oct 29 '12 at 22:24
No, it's just a typo. $df$ is a one-form, $\mu$ is a two-form, so the wedge is a three-form. $df \mu$ doesn't make sense. I've corrected it. Thanks! – levap Oct 30 '12 at 8:46
I believe typo in the 2th line: $ dz \wedge dz $ written, however I think it should be : $dz \wedge dx$ – Imago Mar 12 at 12:56
@Imago I've corrected it, thanks for noticing! – levap Mar 13 at 14:52

Geometric calculus is slightly different in notation from differential forms, but the math is very similar, and I hope I can provide a useful insight into this problem, even with a slightly different background.

Geometric calculus replaces differentials like $dx$ by vectors $e_x$, but it still uses wedges, which are still asymmetric. So your $\omega$, in GC language, would be

$$\omega = \frac{x e_y \wedge e_z + y e_z \wedge e_x + z e_x \wedge e_y}{(x^2 + y^2 + z^2)^{3/2}}$$

Not a whole lot of difference, I'll grant. Still, GC interprets this as a bivector (field), an oriented planar subspace in 3D space.

Instead of Hodge duality, GC uses the geometric product, denoted wholly by juxtaposition: $e_i e_j = -e_j e_i$ if $i \neq j$. Otherwise, $e_i e_i = 1$ (no summation implied). Hodge duality is replaced by multiplication with the pseudoscalar, $e_x \wedge e_y \wedge e_z \equiv i$. For instance, $i(e_y \wedge e_z) = -e_x$.

Let's use this to simplify your expression for $\omega$ to:

$$\omega = \frac{i r}{|r|^3}$$

where $r = xe_x + ye_y + z e_z$.

This field, $r/|r|^3 \equiv G$, is in fact the free space Green's function for the vector derivative. $\nabla r/|r|^3 = 4\pi\delta(r)$. (You might understand this more intuitively if I say $\nabla$ is the rough equivalent of $d + \delta$, the exterior derivative plus the coderivative. We say $\nabla \wedge A$ is the exterior derivative of $A$, and $\nabla \cdot A$ is the interior derivative, or coderivative.)

Finally, it suffices in 3D to say that $i$ commutes with everything, at the cost of turning wedges into dots and dots into wedges. You have $\omega = iG$, and you want to prove that $\nabla \wedge \omega = \nabla \wedge (iG) = 0$. Pulling the $i$ out gives $i\nabla \cdot G = 0$, which is true in some places. (Question for you: where is it not true?)

As has been said, you can prove that $\omega$ is not exact (that is, there is no $\alpha$ such that $\nabla \wedge \alpha = \omega$) by seeing if $\omega$ is integrable. This is tightly coupled to the question earlier: where, if anywhere, is $\nabla \wedge \omega \neq 0$? When you integrate the exterior derivative of this field over a volume, will you get zero?

If you take nothing else from this answer (I know geometric calculus can still look and feel quite different from differential forms), I think you should see that this is problem probes at the nature of the derivative in 3d space. The 2-form you have here is just a disguise for the 3D free space Green's function.

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Use spherical coordinates. In spherical coordinates $r,\theta,\phi$, the form reads:

$$ \omega = \sin\theta\,d\theta\wedge d\phi. $$

This is closed, because the coefficient only depends on a coordinate that is already used:

$$ d\omega = d(\sin\theta)\wedge d\theta\wedge d\phi = \cos\theta\,d\theta\wedge d\theta\wedge d\phi = 0, $$

since $d\theta\wedge d\theta=0$.

This is anyway not exact, because you can integrate it on the sphere! Integrating it on the closed area $0\le\theta\le \pi, 0\le\phi\le2\pi$, that is a sphere (of radius one), you find (it's a simple calculation):

$$\oint\omega = 4\pi.$$

Therefore, the form is not exact.

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