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Federici, Bruno and Georgakopoulos, Agelos. (2017) Hyperbolicity vs. amenability for planar
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DOI 10.1007/s00454-017-9859-x
Hyperbolicity vs. Amenability for Planar Graphs
Bruno Federici1 · Agelos Georgakopoulos1
Received: 4 October 2015 / Revised: 1 December 2016 / Accepted: 16 January 2017 © The Author(s) 2017. This article is published with open access at Springerlink.com
Abstract The aim of this paper is to clarify the relationship between Gromov-hyperbolicity and amenability for planar maps.
Keywords Hyperbolic graph·Non-amenable graph·Planar graph·Coarse geometry
Mathematics Subject Classification 05C10 (Primary)·05C63 (Secondary)
1 Introduction
Hyperbolicity and non-amenability1 are important and well-studied properties for groups (where the former implies the latter unless the group is 2-ended). They are also fundamental in the emerging field of coarse geometry [1]. The aim of this paper is to clarify their relationship for planar graphs that do not necessarily have many symmetries: we show that these properties become equivalent when strengthened by certain additional conditions, but not otherwise.
LetP denote the class of plane graphs (aka. planar maps), with no accumulation point of vertices and with bounded vertex degrees. LetPdenote the subclass ofP comprising the graphs with no unbounded face. We prove
1 See Sect.2for definitions.
Editor in Charge: János Pach
Bruno Federici
Agelos Georgakopoulos
Theorem 1 Let G be a graph inP. Then G is hyperbolic and uniformly isoperimetric if and only if it is non-amenable and it has bounded codegree.
Here, thelengthof a face is the number of edges on its boundary; aboundedface is a face with finite length; a plane graph hasbounded codegreeif there is an upper bound on the length of bounded faces. A graph isuniformly isoperimetricif satisfies an isoperimetric inequality of the form|S| ≤ f(|∂S|)for all non-empty finite vertex setsS, where f:N→Nis a monotone increasing, diverging function and∂Sis the set of vertices not inSbut with a neighbour inS.
Theorem1is an immediate corollary of the following somewhat finer result
Theorem 2 Let G be a graph inP. Then the following hold:
1. if G is non-amenable and has bounded codegree then it is hyperbolic; 2. if G is hyperbolic and uniformly isoperimetric then it has bounded codegree; 3. if G is hyperbolic and uniformly isoperimetric and in addition has no unbounded
face then it non-amenable.
In the next section we provide examples showing that none of the conditions fea-turing in Theorem2can be weakened, and that the no accumulation point condition is needed.
We expect that Theorem2remains true in the class of 1-ended Riemannian surfaces if we replace the bounded degree condition with the property of having bounded curvature and the bounded codegree condition with the property of having bounded length of boundary components.
1.1 Tightness of Theorem2
We remark that having bounded degrees is a standard assumption, and assuming bounded codegree is not less natural when the graph is planar. Part of the motiva-tion behind Theorem2comes from related recent work of the second author [4,6], especially the following
Theorem 3 ([6])Let G be an infinite, Gromov-hyperbolic, non-amenable, 1-ended, plane graph with bounded degrees and no infinite faces. Then the following five boundaries of G (and the corresponding compactifications of G) are canonically homeomorphic to each other: the hyperbolic boundary, the Martin boundary, the boundary of the square tiling, the Northshield circle, and the boundary∂∼=(G).
result of [4] showing that the Liouville property implies amenability in this context, the third part of Theorem2implies that such a graph would need to have accumula-tion points or satisfy no isoperimetric inequality. (Note that such a graph could have bounded codegree.)
The aforementioned example from [6] shows that non-amenability implies neither hyperbolicity nor bounded codegree, and is one of the examples needed to show that no one of the four properties that show up in Theorem2implies any of the other in
P (with the exception of non-amenability implying weak non-amenability). We now describe other examples showing the independence of those properties.
To prove that bounded codegree does not imply hyperbolicity or that weak non-amenability does not imply non-non-amenability it suffices to consider the square gridZ2. To prove that hyperbolicity does not imply weak non-amenability nor bounded codegree, we adopt an example suggested by B. Bowditch (personal communication). Start with a hyperbolic graphGof bounded codegree(G∗)and perform the following construction on any infinite sequence{Fn}of faces ofG. Enumerate the vertices of
Fnas f1, . . .fk in the order they appear along Fnstarting with an arbitrary vertex.
Add a new vertexvninsideFn, and join it to each fiby a pathPiof lengthn(i.e. with n edges), so that the Pi’s meet only atvn. Then for every 1≤ i ≤ k−1, and every
1 < j <n, join the jth vertices of Pi andPi+1with an edge. CallG1the resulting
graph. ThenG1 has unbounded codegree, because P1,Pk and one of the edges of Fnbound a face of length 2n−1. MoreoverG1is not uniformly isoperimetric: the
set of vertices inside Fn is unbounded inn, while its boundary has|Fn| ≤ (G∗)
vertices. Finally,G1is hyperbolic: it is quasi-isometric to the graph obtained fromG
by attaching a pathRof lengthnto eachFn.
To prove that bounded codegree does not imply weak non-amenability, consider the graphG2obtained from the same construction as above except that we now also
introduce edges between Pk and P1: nowG2has bounded codegree while still not
being uniformly isoperimetric.2
To prove that hyperbolicity and weak non-amenability together do not imply non-amenability without the condition of no unbounded face consider the following example. LetHbe the hyperbolic graph constructed as follows. The vertex set ofHis the subset ofR2given by{(2in,n)|i ∈Z,n∈N}. Join two vertices(2in,n), (
j
2m,m)
with an edge whenever eithern =mandi = j+1 orn =m+1 andi =2j. The finite graphH(a)is the subgraph ofHinduced by those vertices contained inside the square with corners(0,0), (a,0), (a,0), (a,a). We construct the graphGby attaching certainH(n)toHas follows. For everyn ∈N, attach a copy ofH(n)along the path
{(n2,0), . . . , (n2+n,0)}of H by identifying the vertex(n2+k,0)of H with the vertex(k,0)ofH(n),k=0, . . . ,n. Note that the resulting graphGis planar because
n2+n < (n+1)2, and so the H(n)’s we attached to H do not overlap. It is easy to prove thatGis amenable and uniformly isoperimetric. It is also not hard to check thatGis hyperbolic, by noticing that the ray{(x,0)|x ∈Z} ⊂Gcontains the only geodesic between any two of its vertices, and using the fact that theH(n)were glued
2 B. Bowditch (personal communication) noticed thatG
onto the hyperbolic graph Halong that ray; one could for example explicitly check the thin triangles condition.
To see that Theorem2becomes false if we allow accumulation points of vertices, consider the free product of the square grid Z2 with the lineZ; this graph can be
embedded with bounded codegree, and it is non-amenable but not hyperbolic.
2 Definitions
Thedegreedeg(v)of a vertexvin a graphGis the number of edges incident withv; if
(G):= sup
v∈V(G)
deg(v)
is finite we will say thatGhasbounded degree.
Anembeddingof a graphGin the plane will always mean a topological embed-ding of the corresponembed-ding 1-complex in the euclidean planeR2; in simpler words, an embedding is a drawing in the plane with no two edges crossing. Aplane graphis a graph endowed with a fixed embedding. A plane graph isaccumulation-freeif its set of vertices has no accumulation point in the plane.
Afaceof an embeddingσ :G→R2is a component ofR2\σ (G). Theboundaryof a faceFis the set of vertices and edges ofGthat are mapped byσto the closure ofF. Thelength|F|of Fis the number of edges in its boundary. A faceF isbounded if the length|F|is finite. If
(G∗):= sup
Fbounded face ofG |F|
is finite we will say thatGhasbounded codegree. TheCheeger constantof a graphGis
c(G):= inf
∅=S⊂Gfinite
|∂S| |S| ,
where ∂S = {v ∈ G\S | there existsw ∈ Sadjacent tov}is theboundaryof S. Graphs with strictly positive Cheeger constant are callednon-amenable graphs. A graph isuniformly isoperimetricif satisfies an isoperimetric inequality of the form
|S| ≤ f(|∂S|)for all non-empty finite vertex setsS, where f:N→Nis a monotone increasing diverging function.
IfC is a cycle ofG andx,y lie onC then they identify two arcs joining them alongC: we will writexC yandyC xfor these two paths —it will not matter which is which. Similarly, ifPis a path passing through these vertices,x P yis the sub-path of
P joining them.
3 Hyperbolicity and Weak Non-amenability Imply Bounded Codegree
In this and the following sections we will prove each of the three implications of Theorem2separately.
We will assume throughout the text thatG ∈ P, i.e.G is an accumulation-free plane graph with bounded degrees, fixed for the rest of this paper. Theorem2is trivial in the case of forests, so from now on we will assume thatGhas at least a cycle, or in other words it has a bounded face.
Ageodetic cycle Cin a graphGis a cycle with the property that for every two points
x,y∈Cat least one ofxC yandyC x(defined in the end of Sect.2) is a geodesic inG.
Lemma 1 If G∈Pis hyperbolic, then the lengths of its geodetic cycles are bounded, i.e.
sup
Cgeodetic cycle ofG
|C|<∞.
Proof Letδbe the hyperbolicity constant ofG. We will show that no geodetic cycle has more than 6δvertices.
LetCbe a geodetic cycle, say withnvertices, and choose three pointsa,b,conC
as equally spaced as possible, i.e. every pair is at leastn3apart alongC. Letabbe the arc ofCjoiningaandbthat does not containc, and definebcandcasimilarly. We want to show thatab,bcandcaform a geodetic triangle.
If x,y,z are distinct points inC then let x zybe the arc in C from x to y that passes throughz. Then we know that one ofab,acbis a geodesic joiningaandb, and
|acb| ≥2n3>|ab|, soabis a geodetic arc. Similarly,bcandcaare geodetic arcs. Consider now the point p onabat distancen6froma alongC. SinceG is a δ-hyperbolic graph, we know that there is a vertexqonbcorcawhich is at distance at mostδfrom p. But asCis a geodetic cycle, the choice ofa,b,cimplies that
d(p,q)≥min{d(p,a),d(p,b)} =
n
6
,
from which we deduce thatn≤6δ.
By the Jordan curve theorem, we can say that a point ofR2isstrictly insidea given cycleCofGif it belongs to the bounded component ofR2\Cand isinside Cif it belongs toC or is strictly inside. We say that a subset ofR2is inside (resp. strictly inside)Cif all of its points are inside (resp. strictly inside)C. A subset ofR2isoutside Cif it is not strictly insideC.
Corollary 1 Suppose G∈Pis hyperbolic and uniformly isoperimetric. If every face of G is contained inside a geodesic cycle, then(G∗) <∞.
Proof Consider a faceF contained inside a geodetic cycleC; by Lemma1we know that|C| ≤6δ, whereδis the hyperbolicity constant ofG. LetSbe the set of all vertices inside the geodetic cycleC so|S| <∞as there is no accumulation point. Then the vertices of S sending edges to the boundary∂S belong to C and each vertex ofC
sends less than(G)edges to∂S, implying that|∂S|< (G)|C|. Let f:N→Nbe a monotone increasing diverging function witnessing the weak non-amenability ofG. Then, sinceF⊆S,
|F| ≤ |S| ≤ f(|∂S|)≤ f(6δ(G)),
which is uniformly bounded for every faceFofG.
In what follows we will exhibit a construction showing that in any graph each face is contained inside a geodetic cycle, which allows us to apply Corollary1whenever the graph is hyperbolic and uniformly isoperimetric.
We remarked above that by the Jordan curve theorem we can make sense of the notion of being contained inside a cycle. Similarly, given three pathsP,Q,Rsharing the same endpoints, if P ∪R is a cycle and Q lies inside it, we will say that Qis
between P,R.
Now, suppose we are given a cycleCand two pointsx,y∈Csuch that there exists a geodesicγ joiningxandylying outsideC. Consider the setS =S(x,y)ofx–y
geodesics that lie outsideC. This set can be divided into two classes:
S1:= {∈S |xC yis betweenyC x, },
S2:= {∈S|yC xis betweenxC y, }.
These two subsets ofS cannot be both empty because one of them must containγ. Let us assume, without loss of generality, thatS1= ∅. For the proof of Theorem2,
we will make use of the notion of ‘the closest geodesic’ to a given cycle; let us make this more precise. Consider the above cycleCin a plane graph, two pointsxandyon
Cand a choice of an arc onCjoining them, sayxC y. Let us define a partial order on the setS1defined above: for any two geodesics, ∈S1we declareifis
betweenxC y, .
Lemma 2 With notation as above,(S1,)has a least element.
Proof The setS1is a subset of all paths from xtoyof lengthd(x,y). These paths
are contained in the ball of centerxand radiusd(x,y). AsGis locally finite, this ball is finite and so isS1. Therefore, it suffices to produce for every couple of elements a
(greatest) lower bound.3
Pick two geodesics, inS1; letP1, . . . ,Phbe the collection (ordered fromx
toy) of maximal subpaths oflying inside the cyclexC y∪andQ1, . . . ,Qk the
3 In a similar fashion we can produce a least upper bound, showing thatS
collection (ordered from x to y) of maximal subpaths of lying inside the cycle
xC y∪(note thath−k ∈ {−1,0,1}). Without loss of generality, we can assume thatxbelongs toP1, soh−k∈ {0,1}.
Now consider the subgraph
:=
P1∪Q1∪. . .∪Ph∪Qk, ifh =k;
P1∪Q1∪. . .∪Ph−1∪Qk∪Ph, ifh =k+1.
Note that eachPishares one endvertex withQi−1and the other withQi, and similarly Qjshares the endvertices withPjandPj−1. We want to proveto be an element of
S1and specifically the greatest lower bound ofand. Note thatandintersect
in some pointsx=x1,x2, . . . ,xn=y(the endvertices of allPi andQj) and, being
geodesics,xixi+1is as long asxixi+1. This implies|| = || = || =d(x,y),
i.e.is a geodesic (in particular, it is a path). The fact that ∈ S1follows from
having put together only sub-paths of elements fromS1. Lastly, we need to show that
bothandhold. But all pathsPiandQjare inside bothxC y∪and
xC y∪, therefore so is.
Of course, there is nothing special withS1and thusS2has a similar partial order
and admits a least element too, provided it is non-empty.
Let us say that in a plane graph a pathP crossesa cycleC if the endpoints of P
are outsideCbut there is at least one edge ofPthat lies strictly insideC.
Corollary 2 Let C be a cycle in a plane graph G, and let B be a geodesic between two points x and y of C such that B lies outside C and xC y lies between yC x and B. Then there exists a x–y geodesicin G satisfying the following:
(1) xC y lies between yC x and;
(2) there is no geodesic outside C crossing the cycle xC y∪.
Proof Note that condition (1) is exactly the definition of the setS1given above, andB
satisfies that condition. By Lemma2, there exists a leastx–ygeodesicwith respect to. Let us show that this is the required geodesic. Suppose there is a geodesic that crosses the cyclexC y∪and lies outsideC, so the endpoints ofare outside
yC x∪andhas an edgeestrictly insidexC y∪. Letabthe longest subpath ofcontainingeand lying insidexC y∪. Thena,bare onand the geodesic
:=xa∪ab∪by
satisfies ≺ , contradicting the minimality of. This contradiction proves our
claim.
Note that Corollary2does not claim uniqueness for the geodesic: ifsatisfies the claim and thensatisfies it as well. However, the unique least element of
S1satisfies the statement of (2) thus such a geodesic will be referred to as theclosest
geodesic to the cycleCinS1. We conclude that, given a pair of pointsx,yon a cycle
C, there are exactly one or twox–ygeodesics closest toC, depending on how many of
S1,S2are non-empty. If these two geodesics both exist, they can intersect but cannot
Theorem 4 If G∈P is hyperbolic and uniformly isoperimetric then(G∗) <∞.
Proof We want to show that if F is a face ofG, then it is contained in a geodetic cycle and then apply Corollary1. The idea of the proof is to construct a sequence of cyclesC0,C1, . . .each containingF, with the lengths|Ci|strictly decreasing, so that
the sequence is finite and the last cycle is a geodetic cycle.
Let us start with the cycleC0coinciding with the boundary of the faceF. IfC0is
geodetic we are done, otherwise there are two pointsx,ysuch that bothxC0y and
yC0xare not geodesics. Consider a geodesic1joining them: sinceF is a face,1
must lie outside the cycleC0. Therefore, we have three paths xC0y, yC0x and1
betweenxandy. Assume without loss of generality thatxC0yis betweenyC0x, 1.
Then the union of1withyC0xyields a new cycleC1with the following properties:
• |C1| = |yC0x| + |1|<|yC0x| + |xC0y| = |C0|, sincexC0yis not a geodesic
while1is;
• the faceFis inside the cycleC1since it was inside (or rather, equal to)C0which
in turn is insideC1.
Using Lemma2we can require the geodesic1to be the closest to the cycleC0with
respect to the arcxC0y. Note that the cycleC1cannot be crossed by any geodesic:
a side of the cycle is made by a face, which does not contain any strictly inner edge, and the other side is bounded by the closest geodesic, which cannot be crossed by Corollary2.
We can iterate this procedure: assume by induction that afternsteps, we are left with a cycleCnsuch that the faceFis still insideCnandCncannot be crossed by geodesics.
IfCnis a geodetic cycle we are done, otherwise there are two pointsx,y ∈Cnthat
prevent that, and we can find a closest geodesicn+1as before, creating a new cycle
Cn+1. We conclude that the face F is insideCn+1and|Cn+1| < |Cn|. Since these
lengths are strictly decreasing, the process halts after finitely many steps, yielding the desired geodetic cycle. Note thatCn+1still has the property that it cannot be crossed
by a geodesic: indeed, if a geodesic crossesCn+1, then since it cannot crossCnby
the induction hypothesis, it would have to cross the cyclexCny∪n+1, which would
contradict condition (2) of Corollary2.
4 Non-amenability and Bounded Codegree Imply Hyperbolicity
One of the assertions of Theorem2was proved in [9] using random walks. In this section we provide a purely geometric proof of that statement.
Bowditch proved in [2] many equivalent conditions for hyperbolicity of metric spaces, one of which is known aslinear isoperimetric inequality. For our interests, which are planar graphs of bounded degree, that condition has been rephrased as in Theorem5below. Before stating it we need some definitions.
Let us call a finite, connected, plane graph H with minimum degree at least 2 a
combinatorial disk. Note that all faces ofHare bounded by a cycle; let us call∂topH
Definition 1 A combinatorial diskHsatisfies a(k,D)-linear isoperimetric inequality
(LII) if|F| ≤Dfor all bounded facesFofHand the number of bounded faces ofH
is bounded above byk|∂topH|.
Definition 2 An infinite, connected, plane graphGsatisfies a LII if there existk,D∈ Nsuch that the following holds: for every cycleC⊂Gthere is a combinatorial diskH
satisfying a(k,D)-LII and a mapϕ: H →Gwhich is a graph-theoretic isomorphism onto its image (so thatϕ does not have to respect the embeddings of H,Ginto the plane), such thatϕ(∂topH)=C.
Bowditch’s criterion is the following:
Theorem 5 ([2])A plane graph G of minimum degree at least3and bounded degree is hyperbolic if and only if G satisfies a LII.
Remark This LII condition traslates for Cayley graphs to the usual definition of lin-ear isoperimetric inequality, i.e. having linlin-ear Dehn function. Gromov proved in his monograph [7] that for a Cayley graph having linear Dehn funcion is equivalent to being hyperbolic. It is worth mentioning that Bowditch [3] extended this result to gen-eral path-metric spaces, by proving that having a subquadratic isoperimetric function implies hyperbolicity. Our Theorem6shows that for planar graphs, non-amenability and the boundedness of the codegree together are sufficient to imply a linear isoperi-metric inequality.
An immediate corollary is the following:
Corollary 3 Let G be a plane graph of minimum degree at least3, bounded degree and codegree. Suppose there exists k such that for all cycles C ⊂ G the number of faces of G inside C is bounded above by k|C|. Then G is hyperbolic.
Proof For every cycleC, letHbe the subgraph ofGinduced by all vertices insideC. ThenHis a finite plane graph of codegree bounded above by(G∗). By assumption, the number of bounded faces of H is bounded above byk|C| = k|∂topH|. ThusG
satisfies a LII, andGis hyperbolic by Theorem5.
We will see a partial converse of this statement in Lemma4.
We would like to apply this criterion to our non-amenable, bounded codegree graph
G, butGmight have minimum degree less than 3. Therefore, we will perform onG
the following construction in order to obtain a graphGof minimum degree 3 without affecting any of the other properties ofGwe are interested in.
Define adecorationof a non-amenable graphGto be a finite connected induced subgraphHwith at most 2 vertices in the boundary∂H that is maximal with respect to the supergraph relation among all subgraphs of G having these properties. For example, we can create a decoration by attaching a path to two vertices of a graph of degree at least two. Note that sinceGis non-amenable, the size of any of its decorations is bounded above.
Call the resulting graph G. Note that the minimum degree ofG is at least 3: any vertex ofGof degree at most 2 belongs to a decoration, and ifHis a decoration and
x∈∂Hthen by maximalityxsends at least 3 edges toG\(H∪∂H)when|∂H| =1 and at least 2 edges when|∂H| =2. Note also that the maximum degree ofGis at most(G).
Now assume G is non-amenable with Cheeger constant c(G); then the size of decorations is bounded above byc(2G) and thus the size of any face ofGis reduced by at mostc(2G)after the procedure, so(G∗)is finite if(G∗)is. Consider the identity mapI: V(G) →V(G). Then
dG(x,y)≤dG(I(x),I(y))≤
2
c(G)dG(x,y)
and every vertex inGis within c(2G) from a vertex of f(V(G)), henceI is a quasi-isometry betweenGandG. ThusGis non-amenable, since non-amenability is a quasi-isometric invariant for graphs of bounded degree (see for instance [5, Thm. 11.10] or [8, Sect. 4]). For the same reason, if we can prove thatGis hyperbolic then so isG.
Theorem 6 If G∈Pis non-amenable and it has bounded codegree then G is hyper-bolic.
Proof Starting fromG, perform the construction of the auxiliary graphGas above: the resulting graph G is non-amenable, has bounded codegree and has minimum degree at least 3.
LetCbe a cycle andS⊂Gthe (finite but possibly empty) subset of vertices lying strictly insideC; by non-amenability we have
|C| ≥ |∂S| ≥c(G)|S|.
Let us focus on the finite planar graphG[C∪S]induced byC∪Sand letF be the number of faces inside it. Since each vertex is incident with at most(G)faces, we have|C∪S|(G)≥ F. Thus
(1+c(G))|C| ≥c(G)(|S| + |C|)≥c(G) 1
(G)F,
which is equivalent toF ≤ (1+c(cG(G))() G)|C|. Since(G∗)is finite, by Corollary3
Gis hyperbolic. By the remark above,Gis hyperbolic too.
5 Hyperbolicity and Weak Non-amenability Imply Non-amenability
Let us prepare the last step of the proof of Theorem2with a lemma.
Proof LetHbe the subgraph ofGspanned bySand all its incident edges. Note that
H contains all vertices in∂S, but no edges joining two vertices of∂S. Then H is a finite plane graph by definition. We letCbe the closed walk bounding the unbounded face ofH. We claim thatChas the desired properties.
To see this, let x1, . . . ,xn be an enumeration of the vertices of∂S in the order
they are visited byC. Then the subwalkxiC xi+1is contained in some face ofG: all
interior vertices ofxiC xi+1lie inSby our definitions,and so all edges incident with
those vertices are inH; therefore, sincexiC xi+1is a facial walk inH, it is also a facial
walk inG. SincexiC xi+1is contained in some face of GandGhas only bounded
faces, we have|xiC xi+1|< (G∗). Applying this to alli, we obtain
|C∩∂S| =n > n
i=1
|xiC xi+1|
(G∗) ≥
|C|
(G∗).
We need a result which is almost a converse of Corollary3.
Lemma 4 Let G∈Pbe hyperbolic and uniformly isoperimetric. Then there exists k such that for all cycles C ⊂G the number of faces of G inside C is bounded above by k|C|.
Proof Using the auxiliary graphGfrom the previous section, we may assume that
G has minimum degree at least 3. LetC be a cycle of G. Since G is hyperbolic, by Theorem5 there exists a combinatorial disk H satisfying a(k,D)-LII with an isomorphismϕfromHto a subgraph ofGsuch thatϕ(∂topH)=C. The boundaries
Fi,i ∈ I,of bounded faces ofH are sent byϕ to cyclesCi :=ϕ(Fi)ofG so that |I| ≤k|C|, and the boundDon the length of bounded faces ofHis an upper bound to the length of those cycles Ci. Let Si be the (finite) set of vertices ofG strictly
insideCi, so that∂Si ⊆Ci. Let f:N→Na monotone increasing diverging function
witnessing the weak non-amenability ofG, i.e.|S| ≤ f(|∂S|)for all finite non-empty
S⊂G. Then
|Si| ≤ f(|∂Si|)≤ f(|Ci|)≤ f(D),
for all non-empty Si. Let F(Ci)be the number of faces of GinsideCi; then for a
non-empty Si we have F(Ci) ≤ (G)|Si| because no vertex can meet more than
(G)faces. In conclusion
|{faces insideC}|= i∈I
F(Ci)≤
i:Si=∅
1+
i:Si=∅
(G)|Si| ≤ |I| +(G)f(D)|I|,
from which by settingk:=k+(G)f(D)kthe assertion follows.
Note that in order to prove the non-amenability of a graphGit suffices to check that|∂S| ≥c|S|for some constantc>0 and all finite induced connected subgraphs
subgraph with componentsS1, . . . ,Sn, then
|∂S| = n
i=1
∂Si ≥ 1
(G)
n
i=1
|∂Si| ≥ c
(G)
n
i=1
|Si| = c (G)|S|,
where the first equality follows from∂Si ∩Sj = ∅for alli = j (S is induced) and
the first inequality holds because the boundaries∂Sican overlap, but no vertex of∂S
belongs to more than(G)of them.
Theorem 7 If G ∈ P is hyperbolic and uniformly isoperimetric then G is non-amenable.
Proof By the above consideration it is enough to check the non-amenability only on connected induced subgraphs ofG. By Theorem4we know thatGhas bounded codegree. LetSbe such a subgraph andCas in Lemma3. Then
|∂S| ≥ |∂S∩C| ≥ |C| (G∗)
and thus
|∂S| |S| ≥
1 (G∗)
|C| |S|.
Letk >0 be as in Lemma4; ifT denotes the set of all vertices insideCandF the set of all faces insideC, we have
|C| |T| =
|C| |F| ·
|F| |T| ≥
1
k
1 (G∗),
since each face is incident with at most (G∗)vertices. Combining the last two inequalities, we have
|∂S| |S| ≥
1 (G∗)
|C| |S| ≥
1 (G∗)
|C| |T| ≥
1
k((G∗))2.
6 Graphs with Unbounded Degrees
We provided enough examples to show that Theorem2is best possible, except that we do not yet know to what extent the bounded degree condition is necessary. Solutions to the following problems would clarify this. Let nowP∗denote the class of plane graphs with no accumulation point of vertices; so thatP is the subclass of bounded degree graphs inP∗.
Problem 1 Is there a hyperbolic, amenable, uniformly isoperimetric plane graph of bounded codegree and no unbounded face inP∗?
Acknowledgements B. Federici was supported by an EPSRC Grant EP/L505110/1, and A. Georgakopou-los was supported by EPSRC Grant EP/L002787/1. This project has received funding from the European Research Council (ERC) under the European Union Horizon 2020 research and innovation programme (Grant agreement No 639046). The second author would like to thank the Isaac Newton Institute for Mathe-matical Sciences, Cambridge, for support and hospitality during the programme ‘Random Geometry’ where work on this paper was undertaken.
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References
1. Benjamini, I.: Coarse Geometry and Randomness. Lecture Notes in Mathematics, vol. 2100. Springer, Cham (2013)
2. Bowditch, B.H.: Notes on Gromov’s hyperbolicity criterion for path-metric spaces. In: Ghys, E., Hae-fliger, A. (eds.) Group Theory from a Geometrical Viewpoint, pp. 64–167. World Scientific, River Edge (1991)
3. Bowditch, B.H.: A short proof that a subquadratic isoperimetric inequality implies a linear one. Mich. Math. J.42(1), 103–107 (1995)
4. Carmesin, J., Georgakopoulos, A.: Every planar graph with the Liouville property is amenable. arXiv:1502.02542(2015)
5. Dru¸tu, C., Kapovich, M.: Lectures on Geometric Group Theory.http://people.maths.ox.ac.uk/drutu/ tcc2/ChaptersBook(2013)
6. Georgakopoulos, A.: The boundary of a square tiling of a graph coincides with the Poisson boundary. Invent. Math.203(3), 773–821 (2016)
7. Gromov, M.: Hyperbolic Groups. In: Gersten, S.M. (ed.) Essays in Group Theory. Mathematical Sciences Research Institute Publications, vol. 8, pp. 75–263. Springer, New York (1987)
8. Kanai, M.: Rough isometries, and combinatorial approximations of geometries of non-compact Rie-mannian manifolds. J. Math. Soc. Jpn.37(3), 391–413 (1985)
