# Isoperimetric inequality for exterior domains on $\mathbb{H}^{n}$

Fix $$n\geq 2$$ and let $$\mathbb{H}^{n}=\mathbb{R}_{+}\times \mathbb{S}^{n-1}$$ be the hyperbolic space be defined as a Riemannian manifold equipped with the Riemannian metric $$g=dr^{2}+\sinh^{2}rd\theta^{2},$$ where $$dr^{2}$$ is the standard metric on $$\mathbb{R}_{+}$$ and $$d\theta^{2}$$ is the standard metric on the sphere $$\mathbb{S}^{n-1}$$. Denote with $$\mu$$ the Riemannian measure on $$\mathbb{H}^{n}$$ and with $$\sigma$$ the Riemannian measure of co-dimension $$1$$ on hypersurfaces on $$\mathbb{H}^{n}$$. It is a known fact that $$\mathbb{H}^{n}$$ admits the Isoperimetric inequality $$\sigma(\partial \Omega)\geq f(\mu(\Omega)),$$ for all precompact open sets $$\Omega\subset \mathbb{H}^{n}$$ with smooth boundary, where $$f$$ is defined by $$f(v)=\left\{\begin{array}{cl}v, &v\geq 1\\ v^{\frac{n-1}{n}}, &v\leq 1.\end{array}\right.$$

Let us fix some compact set $$K\subset \mathbb{H}^{n}$$ and consider $$\mathbb{H}^{n}\setminus K$$ as a manifold. Does the same Isoperimetric inequality now hold for precompact open sets $$\Omega\subset \mathbb{H}^{n}\setminus K$$ with smooth boundary and if yes, where can I find a reference for this statement? If this is not known for the hyperbolic space $$\mathbb{H}^{n}$$, is a similar statement known for other Riemannian manifolds, for example $$\mathbb{R}^{n}$$?

I might be missing something, but if you require $$\Omega$$ to be precompact in $$\mathbb{H}^n \setminus K$$, then $$K$$ makes no difference for the purpose of determining the measures and you can use the original inequality. The same if you only require pre-compactness in $$\mathbb{H}^n$$ but include the measure of $$\partial \Omega \cap K$$.

But if you do neither and only consider bounded $$\Omega$$ and their relative boundary in $$\mathbb{H}^n \setminus K$$, then generally there is no isoperimetric inequality: Pick any set $$\Omega$$ of positive measure and $$K := \partial \Omega$$, then $$\sigma(\partial \Omega \setminus K) = 0 < f(\mu(\Omega)).$$

• I changed something. I consider $\mathbb{H}^{n}\setminus K$ as a manifold with boundary, that means $\partial \Omega\cap K=\emptyset$. Jul 2, 2021 at 10:58
• @Shaq155 In that case the measures of $\Omega$ and $\partial \Omega$ do not change if you consider them as subsets of $\mathbb{H}^n$ instead, so the inequality trivially holds.
– mlk
Jul 2, 2021 at 11:09
• Maybe my question was misleading, I consider domains $\Omega \subset \mathbb{H}^{n}\setminus K$. Jul 2, 2021 at 11:14
• @Shaq155 That part was the one thing that was clear. The detail that was initially missing was if $\Omega$ can come close to $K$. But since $\mathbb{H}^n \setminus K \subset \mathbb{H}^n$, if $\Omega$ is precompact in the former, it has some distance to $K$ and thus $\partial \Omega$ does not depend on which topology you use and the inequality then follows trivially.
– mlk
Jul 2, 2021 at 11:24
• Could you explain why it has some distance, even though I guess this is trivial? Jul 2, 2021 at 11:28