Isoperimetric Clusters in Homogeneous Spaces via Concentration Compactness

Abstract

We show the existence of generalized clusters of a finite or even infinite number of sets, with minimal total perimeter and given total masses, in metric measure spaces homogeneous with respect to a group acting by measure preserving homeomorphisms, for a quite wide range of perimeter functionals. Such generalized clusters are a natural “relaxed” version of a cluster and can be thought of as “albums” with possibly infinite pages, having a minimal cluster drawn on each page, the total perimeter and the vector of masses being calculated by summation over all pages, the total perimeter being minimal among all generalized clusters with the same masses. The examples include any anisotropic perimeter in a Euclidean space, as well as a hyperbolic plane with the Riemannian perimeter and Heisenberg groups with a canonical left invariant perimeter or its equivalent versions.

Appendices

Appendix A: Useful Facts About Group Actions

We collect here the following lemma, useful in applications of our results to verify condition (ii) of Theorem 3.3, and some easy remarks related to it.

Lemma A.1

Let $(X,d,\mu )$ be a complete locally compact metric space with nonnegative $\sigma$-finite Radon measure. Suppose G is a topological group acting on X by homeomorphisms preserving $\mu$-nullsets, and

• (i) the action of G is proper discontinuous,

• (ii) the set S of $x\in X$ having nontrivial stabilizer subgroup

  $$\begin{aligned}G_x:=\{g\in G:gx=x\},\end{aligned}$$

  i.e. $S:=\{x\in X:G_x\ne \{1\}\}$, is $\mu$-negligible, that is, $\mu (S)=0$,

• (iii) and X is compactly generated, i.e. there exists a compact $K\subset X$ such that $GK=X$.

Then there is a precompact Borel set $B\subset X$ (a fundamental domain for the action of G) such that $GB=X$, $\mu (\partial B)=0$, and $\mu (gB \cap B)=0$ for all $g\in G$, $g\ne 1$. Moreover, if the action of G on X is free, then one can choose B so as to have $gB \cap B=\emptyset$ for all $g\in G$, $g\ne 1$.

Remark A.2

Condition (iii) of Lemma A.1 for locally compact metric space X is equivalent to cocompactness of the action of G, that is, to compactness of the quotient space X/G. In fact, since the natural projection map $\pi :X\rightarrow X/G$ is continuous, then condition (iii) implies that $X/G=\pi (K)$ is compact. Vice versa, note that $\pi$ is also an open map (as a projection map under group action, namely, because for every open $U\subset X$ one has $\pi ^{-1}(\pi (U)))=\cup _{g\in G} gU$ is open since so is each gU). Therefore, taking a cover $\{U_\lambda \}$ of X by precompact open sets, we have that $\{\pi (U_\lambda )\}$ is an open cover of X/G, and if X/G is compact, extracting a finite subcover $\{\pi (U_{\lambda _j})\}_{j=1}^m$, $m\in \mathbb {N}$ of $\{\pi (U_\lambda )\}$, we get that

$$\begin{aligned}K:=\bigcup _{j=1}^m {\bar{U}}_{\lambda _j}\end{aligned}$$

is a compact set satisfying $GK=X$.

Remark A.3

Another immediate observation worth mentioning is that for the condition (iii) of Lemma A.1 to hold, it is necessary that G be infinite, unless, of course, X is compact.

Proof

We divide the proof in two steps.

Step 1. For all $x\in X\setminus S$ we show that there exists a $\rho >0$ (which can be taken arbitrarily small) such that $\bar{B}_\rho (x)$ is compact, $B_\rho (x)\cap S = \emptyset$, $\mu (\partial B_\rho (x)) = 0$ and $gB_\rho (x)\cap B_\rho (x) = \emptyset$ for all $g\in G$, $g\ne 1$. In particular this proves that S is closed.

By proper dicontinuity of the action of G for a precompact neighborhood W of x the set $\{g\in G:g{\bar{W}}\cap {\bar{W}} \ne \emptyset , g\ne 1\}$ is finite; let $g_1, \dots , g_n$ be its elements. For each $k=1,\dots , n$ since $x\not \in S$ we have $x\ne g_k x$ and hence we can find $U_k$ neighbourhood of x and $V_k$ neighbourhood of $g_k x$ which are disjoint, i.e. $U_k\cap v_k=\emptyset$. Since $W, U_1,\dots , U_n$ and $g_1^{-1} V_1, \dots , g_n^{-1} V_n$ are all neighbourhoods of x (because $g_k$ are homeomorphisms), their interesection U is also a neighbourhood of x. Since $U\subset W$ we have $gU\cap U=\emptyset$ unless $g=g_k$ for some $k=1,\dots , n$. But in the latter case, i.e. when $g=g_k$, since $U\subset U_k$ and $gU=g_kU\subset V_k$ we find that $gU\cap U \subset U_k\cap V_k = \emptyset$. Clearly $U\cap S=\emptyset$ since otherwise there would exist an $x\in U\cap S$ and $g\ne 1$ such that $gx=x\in U$.

Now since U is a neighbourhood of x, there exists an $\varepsilon >0$ such that for all $\rho \le \varepsilon$ the ball $B_\rho (x)$ is precompact and $\bar{B}_\rho (x)\subset U$. To complete the proof of the claim of this step notice that all but a countable number of $\rho <\varepsilon$ have the property $\mu (\partial B_\rho (x))=0$, because otherwise $\mu (B_\varepsilon (x))$ would be infinite, contrary to the fact that is should be finite since $B_\varepsilon (x)$ is precompact and $\mu$ is a Radon measure.

Step 2. Let ${\tilde{K}}\supset K$ be a cover of K by a finite number of open balls so small to be precompact, so that in particular, ${\tilde{K}}$ is open and precompact. Let $S_k$ be a sequence of open sets satisfying

$$\begin{aligned}S\cap K \subset S_k \subset {\bar{S}}_k \subset {\tilde{K}}\end{aligned}$$

and such that $\mu (\bar{S}_k)\rightarrow 0$ as $k\rightarrow \infty$ (one can take for instance $S_k:=\{x\in X:\mathrm {dist}(x, S\cap K)< t_k\}$ for a sequence of numbers $t_k\searrow 0$). By compactness we can take a finite cover $U_1,\dots , U_{n_1}$ of $K\setminus S_1$ with open balls $U_i$ satisfying the properties stated in Step 1 and such that ${\bar{U}}_i \subset {\tilde{K}}$. Inductively we can take a finite cover $U_{n_k+1},\dots , U_{n_{k+1}}$ of the compact set $(K\setminus S_{k+1})\setminus \bigcup _{i=1}^{n_k} U_i$ by open balls $U_i$ satisfying the properties stated in Step 1 which are all contained in $S_k$ so that we have

$$\begin{aligned}K\setminus S \subset \bigcup _{i=1}^{+\infty } U_i, \qquad \bigcup _{i=n_k}^{+\infty } U_i \subset S_k.\end{aligned}$$

Define inductively the disjoint sets

$$\begin{aligned} V_1&:= U_1,\\ V_{i+1}&:= U_{i+1}\setminus \bigcup _{j=1}^i G U_j. \end{aligned}$$

Note that since $U_j$ are all precompact and G acts properly discontinuously, then each $G U_j$ in the definition of $V_{i+1}$ can be substituted by a finite union of $g U_j$ over a finite subset of $g\in G$ (depending of course on i and j): in fact,

$$\begin{aligned}g U_j\cap U_{i+1} \subset g \overline{(U_j\cup U_{i+1})}\cap \overline{(U_j\cup U_{i+1})}\ne \emptyset \end{aligned}$$

for an at most finite set of $g\in G$. Therefore, recalling also that $\partial (g U_j)= g\partial U_j$ (since the action of g is a homeomorphism), and hence $\mu (\partial (g U_j))= \mu (g\partial U_j)=0$ (since the action preserves $\mu$-nullsets), we get $\mu (\partial V_i)=0$ for all i.

Let

$$\begin{aligned}{\tilde{B}}:= \bigcup _{i=1}^{+\infty } V_i, \qquad B:= (S\cap {\tilde{K}})\cup {\tilde{B}},\end{aligned}$$

so that $B\subset {\tilde{K}}$ is a precompact Borel set.

We now verify the claimed properties of B. Let us prove first that $GB\supset K$. For $x\in K$ we have either $x\in S$ or $x\in U_i$ for some $i \in \mathbb {N}$. If $x\in S$ we have $x\in B$ because $x\in S \cap K \subset S\cap {\tilde{K}}$. If $x\in U_i$ then either $x\in V_i$ and hence $x\in B$ or there exists a $g\in G$ and $j<i$ such that $x \in g U_j$. If j is the minimum possibile index with this property, then $V_j=U_j$ and we obtain that $g^{-1} x\in V_j\subset B$. Hence $GB\supset K$ and since $GK=X$, we conclude that $GB=X$.

To prove $\mu (\partial B)=0$, note that

$$\begin{aligned}\partial {\tilde{B}} \subset \bigcup _{i=1}^{n_k} \partial V_i \cup \partial \left( \bigcup _{i=n_k+1}^{\infty } \!\!\!V_i\right) \subset \bigcup _{i=1}^{n_k} \partial V_i \cup {\bar{S}}_k, \end{aligned}$$

the latter inclusion being due to the fact that $V_i\subset S_k$ for $i\ge n_k$. Thus from $\mu (\partial V_i)=0$ we get

$$\begin{aligned}\mu (\partial {\tilde{B}}) \le \mu (\bar{S}_k)\rightarrow 0\end{aligned}$$

as $k\rightarrow \infty$, which gives $\mu ({\tilde{B}})=0$. Recalling that $\mu (S)=0$ we get the claim.

It remains to prove that $\mu (gB \cap B)=0$ for all $g\in G$, $g\ne 1$. To this aim first notice that $g {\tilde{B}} \cap {\tilde{B}} = \emptyset$ because $V_i \cap g V_j = \emptyset$ for all i, j and $g\ne 1$. Now

$$\begin{aligned}gB\cap B \subset gS \cup S \cup (g{\tilde{B}}\cap {\tilde{B}}) = gS \cup S\end{aligned}$$

hence $\mu (gB\cap B)\le \mu (gS)+\mu (S)=0$ for all $g \ne 1$. Finally note that when $S=\emptyset$, then $B\subset {\tilde{B}}$, hence $gB\cap B\subset g{\tilde{B}}\cap {\tilde{B}}=\emptyset$ concluding the proof. $\square$

Remark A.4

It is worth remarking that in many applications already the Dirichlet-Voronoi fundamental domain

$$\begin{aligned}D_x:=\{ y\in X:d(y,x)< d(gy, x)\quad \text{ for } \text{ all } g\in G\text{, } g\ne 1\}\end{aligned}$$

for some $x\in X$ satisfies the statement of Lemma A.1. Note in fact that

$$\begin{aligned}g D_x\cap D_x=\emptyset \quad \text{ for } \text{ all } g\in G, g\ne 1,\end{aligned}$$

because if $y\in g D_x$, that is, $gy\in D_x$, then $y:=g^{-1}gy \not \in D_x$. Thus one can take $B:= D_x$ when $\mu (\partial B)=0$. For instance, when $X=\mathbb {R}^n$, and $G=\mathbb {Z}^n$ acting by translations (say, by integer valued vectors), then one can take for B the Dirichlet-Voronoi fundamental domain for this action which is the cube containing part of its boundary, once $\mu$ is absolutely continuous with respect to the Lebesgue measure.

The following easy technical lemma is also used in the paper.

Lemma A.5

Let G be a group acting on a set X and $V\subset X$. If

$$\begin{aligned}\nu :=\#\{g\in G:gV\cap V\ne \emptyset \} <\infty ,\end{aligned}$$

then there is a partition of G into $\nu$ pairwise disjoint subsets $G_1, \dots , G_\nu$, such that $hV\cap gV =\emptyset$ for every $h,g\in G_j$, $h\ne j$, $j=1,\dots , \nu$.

Proof

We consider the family ${\mathcal {F}}$ of all subsets $F\subset G$ with the property that $hV\cap gV = \emptyset$ for all $h,g\in F$, $h\ne g$. Thanks to Zorn’s lemma we can find $G_1$ in ${\mathcal {F}}$ which is maximal with respect to inclusion. Then define inductively $G_{n+1}$ by taking any maximal subset of $G \setminus \bigcup _{j=1}^{n} G_j$ in the family ${\mathcal {F}}$. We claim that $G_n=\emptyset$ for all $n> \nu$ because if $g\in G_n$ then for all $j<n$ by the maximality of $G_j$ there exists $h_j \in G_j$ such that $h_j V \cap g V \ne \emptyset$. This means $g^{-1} h_j V \cap V \ne \emptyset$ for $j=1,\dots , \nu$ and, of course, also $V\cap V\ne \emptyset$ (if V is empty the lemma is trivial). So we have found $\nu +1$ distinct elements of G in the set $\{g\in G:gV \cap V \ne \emptyset \}$ against the hypothesis. $\square$

Appendix B: Auxiliary Lemmata

We collect here, mainly for the sake of completeness and readers’ convenience, some auxiliary lemmata of more or less folkloric nature.

Lemma B.1

(Equisummability) Suppose that $m_{k,i}\ge 0$, and:

$$\begin{aligned}&\lim _{k\rightarrow +\infty } \sum _i m_{k,i} = m, \\&\lim _k m_{k,i} = m_i\\&\lim _n \left( \sup _k \sum _{i=n}^{+\infty } m_{k,i}\right) = 0. \end{aligned}$$

Then

$$\begin{aligned}\sum _i m_i = m.\end{aligned}$$

Proof

For every $\varepsilon >0$ there is an $n_\varepsilon \in \mathbb {N}$ such that for all $k\in \mathbb {N}$ and $n\ge n_\varepsilon$ one has

$$\begin{aligned}\sum _{i=n}^{+\infty } m_{k,i}< \varepsilon , \quad \text {and}\quad \sum _{i<n} m_{k,i}\le \sum _i m_{k,i} \le \sum _{i<n} m_{k,i}+\varepsilon .\end{aligned}$$

Hence either $m=+\infty$ and $m_i=+\infty$ for some $i<n_\varepsilon$, or $m\in \mathbb {R}$. In the first case the thesis follows. In the second letting $k\rightarrow +\infty$ we obtain for $n\ge n_\varepsilon$ the estimate

$$\begin{aligned} m-\varepsilon \le \sum _{i=1}^n m_i \le m.\end{aligned}$$

Letting now $n\rightarrow +\infty$, one gets

$$\begin{aligned}m-\varepsilon \le \sum _{i} m_i \le m\end{aligned}$$

for all $\varepsilon >0$ implying the thesis. $\square$

Lemma B.2

Suppose that for all $k\in \mathbb {N}$, $j=1,\ldots , N$, where $N\in \mathbb {N}\cup \{\infty \}$ the sets $A_k\subset X$, $A_k^j\subset X$ be $\mu$-measurable and

$$\begin{aligned}\mu \left( A_k \triangle \bigcup _{j=1}^N A_k^j\right) =0, \qquad \mu \left( A_k^j \cap A_k^i\right) =0 \quad \text{ whenever } i\ne j.\end{aligned}$$

If $A_k\rightarrow A$ and $A_k^j\rightarrow A^j$ in $L^1_{\mathrm {loc}}(\mu )$ as $k\rightarrow \infty$, then

$$\begin{aligned}\mu \left( A\triangle \bigcup _{j=1}^N A^j\right) =0, \qquad \mu \left( A^j \cap A^i\right) =0 \quad \text{ whenever } i\ne j.\end{aligned}$$

Proof

It suffices to write for every compact $K\subset X$ the relationships

$$\begin{aligned} \mu \left( \left( A_k \triangle \bigcup _{j=1}^N A_k^j\right) \cap K\right)&=\int _K \left| {\mathbf {1}}_{A_k}(x)-\left( \sum _{j=1}^N {\mathbf {1}}_{A_k^j}(x)\right) \right| \,\mathrm{d}\mu (x),\\ \mu \left( \left( A_k^j \cap A_k^i\right) \cap K\right)&=\int _K {\mathbf {1}}_{A_k^j}(x){\mathbf {1}}_{A_k^i}(x) \,\mathrm{d}\mu (x), \end{aligned}$$

and pass to the limit as $k\rightarrow \infty$. $\square$