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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 $\mathbb{R}^{n}$?

Thanks in advance for your help!

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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 $$\sigma(\partial \Omega \setminus K) = 0 < f(\mu(\Omega)).$$

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  • $\begingroup$ I changed something. I consider $\mathbb{H}^{n}\setminus K$ as a manifold with boundary, that means $\partial \Omega\cap K=\emptyset$. $\endgroup$
    – Shaq155
    Jul 2 at 10:58
  • $\begingroup$ @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. $\endgroup$
    – mlk
    Jul 2 at 11:09
  • $\begingroup$ Maybe my question was misleading, I consider domains $\Omega \subset \mathbb{H}^{n}\setminus K$. $\endgroup$
    – Shaq155
    Jul 2 at 11:14
  • $\begingroup$ @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. $\endgroup$
    – mlk
    Jul 2 at 11:24
  • $\begingroup$ Could you explain why it has some distance, even though I guess this is trivial? $\endgroup$
    – Shaq155
    Jul 2 at 11:28

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