Economically computing $d\beta$

Question

$\displaystyle \beta = z\frac{x dy \wedge dz + y dz \wedge dx + z dx \wedge dy}{(x^2+y^2+z^2)^{2}}$

Show that $d\beta=0$.

So, let $r=x^2+y^2+z^2$,

$\begin{align} \displaystyle d\beta &= d(z\frac{x dy \wedge dz + y dz \wedge dx + z dx \wedge dy}{(x^2+y^2+z^2)^{2}}) \\ &= d(\frac{zx dy \wedge dz + zy dz \wedge dx + z^2 dx \wedge dy}{r^{2}}) \\ &= d(\frac{zx dy \wedge dz + zy dz \wedge dx + z^2 dx \wedge dy}{r^{2}}) \\ &= d(\frac{zx}{r^2}) \wedge dy \wedge dz + d(\frac{zy}{r^2}) \wedge dz \wedge dx + d(\frac{z^2}{r^2}) \wedge dx \wedge dy \\ \end{align} $

Is there a better of doing this as I dont really want to have to crunch $d(\frac{zx}{r^2})$ with the product rule?

Answer 1

Note that, in each term, we've written for instance $df \wedge dy \wedge dz$. Because $df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y}dy + \frac{\partial f}{\partial z} dz$, and $dy \wedge dy = 0, dz \wedge dz = 0$, we see that $$df \wedge dy \wedge dz = \frac{\partial f}{\partial x}dx \wedge dy \wedge dz.$$ So ultimately for this computation you only need to take three partials, not nine.

The key here is that whenever you've written $df \wedge dx_i$ for some $dx_i$, you don't need to compute $\frac{\partial f}{\partial x_i}$, since $\frac{\partial f}{\partial x_i}dx_i \wedge dx_i = 0$.

Answer 2

Let $X=z(x,y,z)/r^2$. Then $\beta=i(X)\Omega$ where $\Omega=dx\wedge dy \wedge dz$ is the volume element of $\mathbb{R}^3$ and $i$ is the interior product. Then the definition of divergence operator says $$d(i(X)\Omega)=\operatorname{div}(X)\Omega$$ So to show $d\beta=0$ is equivalent to show $\operatorname{div}(X)=0$ and we have $$\operatorname{div}(X)=\operatorname{grad}(z)\cdot (x,y,z)/r^2+z \operatorname{div}((x,y,z)/r^2)$$ $$=(0,0,1)\cdot\frac{(x,y,z)}{r^2}+z\sum_{x,y,z}\frac{r^2-x\cdot 2r\cdot 2x}{r^4}$$ $$=\frac{z}{r^2}+z\frac{3r^2-4r\sum_{x,y,z}x^2}{r^4}$$ $$=\frac{z}{r^2}+z\frac{3r^2-4r^2}{r^4}$$ $$=0$$

Answer 3

Perhaps a more understandable way to see this is to parametrize $\Bbb R^3-\{0\}$ by spherical coordinates: $$(x,y,z) = f(r,\phi,\theta)= (r\sin\phi\cos\theta,r\sin\phi\sin\theta,r\cos\phi).$$ Then $f^*\beta = (r\cos\phi)\dfrac{r^3\sin\phi\, d\phi\wedge d\theta}{r^4} = \cos\phi\sin\phi\,d\phi\wedge d\theta$. So it's clear, then, that $f^*(d\beta) = d(f^*\beta) = 0$, and, since $f$ is a diffeomorphism almost everywhere in space, we must have $d\beta = 0$.