Higher-dimensional analogue of $d^2\phi=2\pi\delta(r)dx \wedge dy$?

Question

In two dimensions, the following relation exists:

$$d^2\phi=2\pi\delta(r)dx \wedge dy.$$

See, for example, this question for proof and discussion.

My question is whether there is a higher-dimensional analogue of this relation. In other words, is there some 1-form $\omega$ such that

$$d^2\omega=\delta(r)dx \wedge dy \wedge dz,$$

up to a numerical factor?

Answer 1

Let

• $\omega=\mathrm{d}x^1\wedge\mathrm{d}x^2\wedge\cdots\wedge \mathrm{d}x^n$ be the usual volume form on $\mathbb{R}^n$;
• $\vartheta=x^1\frac{\partial}{\partial x^1}+x^2\frac{\partial}{\partial x^2}+\cdots+x^n\frac{\partial}{\partial x^n}$, the generator of dilations about the origin,
• $\iota_v$ the interior derivative operator for a vector field $v$.

Then the $(n-1)$-form $$\delta\Omega=\frac{\iota_{\vartheta}\omega}{r^n}$$ is the element of solid angle about the origin. Away from the origin, the exterior derivative of $\delta \Omega$ $$\begin{split}\mathrm{d}(\delta\Omega)&=-n r^{-n-2}\vartheta^{\sharp}\wedge\iota_{\vartheta}\omega+r^{-n}\mathcal{L}_{\vartheta}\omega\\ &=-n r^{-n}\omega+nr^{-n}\omega\\ &=0 \end{split}$$ where $\mathcal{L}_{v}$ denotes the Lie derivative with respect to $v$. But the integral of $\delta\Omega$ on the sphere of radius $a$ centered about the origin is equal to the content of the unit $(n-1)$-sphere: $$\int_{aS^{n-1}}\delta \Omega=\lvert S^{n-1}\rvert$$ Heuristically, a physicist might use Stokes' theorem to write what's been shown so far as $$\begin{align}\mathrm{d}(\delta \Omega)&=0\quad (r>0)\\ \int_{aB^n}\mathrm{d}(\delta\Omega) &=\lvert S^{n-1}\rvert \end{align}$$ where $aB^n$ is the ball of radius $a$ about the origin. On these grounds, the physicist would write $$\boxed{\mathrm{d}(\delta\Omega)=\lvert S^{n-1}\rvert \delta^{(n)}(x)\omega}\text{.}$$