#### Hyperbolicity vs. Amenability for Planar Graphs

Hyperbolicity vs. Amenability for Planar Graphs
Bruno Federici 0
Agelos Georgakopoulos 0
0 Mathematics Institute, University of Warwick , Coventry CV4 7AL , UK
The aim of this paper is to clarify the relationship between Gromovhyperbolicity and amenability for planar maps.
Hyperbolic graph; Non-amenable graph; Planar graph; Coarse geometry
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.
Let P denote the class of plane graphs (aka. planar maps), with no accumulation
point of vertices and with bounded vertex degrees. Let P denote the subclass of P
comprising the graphs with no unbounded face. We prove
1 See Sect. 2 for definitions.
Editor in Charge: János Pach
Theorem 1 Let G be a graph in P . Then G is hyperbolic and uniformly isoperimetric
if and only if it is non-amenable and it has bounded codegree.
Here, the length of a face is the number of edges on its boundary; a bounded face
is a face with finite length; a plane graph has bounded codegree if there is an upper
bound on the length of bounded faces. A graph is uniformly isoperimetric if satisfies
an isoperimetric inequality of the form |S| ≤ f (|∂ S|) for all non-empty finite vertex
sets S, where f : N → N is a monotone increasing, diverging function and ∂ S is the
set of vertices not in S but with a neighbour in S.
Theorem 1 is an immediate corollary of the following somewhat finer result
Theorem 2 Let G be a graph in P. 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
featuring in Theorem 2 can be weakened, and that the no accumulation point condition
is needed.
We expect that Theorem 2 remains 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 Theorem 2
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
motivation behind Theorem 2 comes 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).
In order to show the independence of the hypotheses in Theorem 3, the second
author provided a counterexample to a conjecture of Northshield [9] asking whether a
plane, accumulation-free, non-amenable graph with bounded vertex degrees must be
hyperbolic. That counterexample had unbounded codegree, and so the question came
up of whether Northshield’s conjecture would be true subject to the additional
condition of bounded codegree. The first part of Theorem 2 says that this is indeed the case.
A related problem from [6] asks whether there is a planar, hyperbolic graph with
bounded degrees, no unbounded faces, and the Liouville property. Combined with a
result of [4] showing that the Liouville property implies amenability in this context,
the third part of Theorem 2 implies that such a graph would need to have
accumulation 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 Theorem 2 implies 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
nonamenability does not imply non-amenability it suffices to consider the square grid Z2.
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 graph G of bounded codegree (G∗) and perform the following
construction on any infinite sequence {Fn} of faces of G. Enumerate the vertices of
Fn as f1, . . . fk in the order they appear along Fn starting with an arbitrary vertex.
Add a new vertex vn inside Fn, and join it to each fi by a path Pi of length n (i.e. with
n edges), so that the Pi ’s meet only at vn. Then for every 1 ≤ i ≤ k − 1, and every
1 < j < n, join the j th vertices of Pi and Pi+1 with an edge. Call G1 the resulting
graph. Then G1 has unbounded codegree, because P1, Pk and one of the edges of
Fn bound a face of length 2n − 1. Moreover G1 is not uniformly isoperimetric: the
set of vertices inside Fn is unbounded in n, while its boundary has |Fn| ≤ (G∗)
vertices. Finally, G1 is hyperbolic: it is quasi-isometric to the graph obtained from G
by attaching a path R of length n to each Fn.
To prove that bounded codegree does not imply weak non-amenability, consider
the graph G2 obtained from the same construction as above except that we now also
introduce edges between Pk and P1: now G2 has 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. Let H be the hyperbolic graph constructed as follows. The vertex set of H is
the subset of R2 given by {( 2in , n) | i ∈ Z, n ∈ N}. Join two vertices ( 2in , n), ( 2jm , m)
with an edge whenever either n = m and i = j + 1 or n = m + 1 and i = 2 j . The
finite graph H (a) is the subgraph of H induced by those vertices contained inside the
square with corners (0, 0), (a, 0), (a, 0), (a, a). We construct the graph G by attaching
certain H (n) to H as follows. For every n ∈ N, attach a copy of H (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) of H (n), k = 0, . . . , n. Note that the resulting graph G is 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 that G is amenable and uniformly isoperimetric. It is also not hard to check
that G is hyperbolic, by noticing that the ray {(x , 0) | x ∈ Z} ⊂ G contains the only
geodesic between any two of its vertices, and using the fact that the H (n) were glued
2 B. Bowditch (personal communication) noticed that G1 is quasi-isometric to G2, showing that having
bounded codegree is not a quasi-isometric invariant in P, although he proved that having bounded codegree
is a quasi-isometric invariant among uniformly isoperimetric graphs.
onto the hyperbolic graph H along that ray; one could for example explicitly check
the thin triangles condition.
To see that Theorem 2 becomes false if we allow accumulation points of vertices,
consider the free product of the square grid Z2 with the line Z; this graph can be
embedded with bounded codegree, and it is non-amenable but not hyperbolic.
2 Definitions
The degree deg(v) of a vertex v in a graph G is the number of edges incident with v; if
(G) :=
is finite we will say that G has bounded degree.
An embedding of a graph G in the plane will always mean a topological
embedding of the corresponding 1-complex in the euclidean plane R2; in simpler words, an
embedding is a drawing in the plane with no two edges crossing. A plane graph is a
graph endowed with a fixed embedding. A plane graph is accumulation-free if its set
of vertices has no accumulation point in the plane.
A face of an embedding σ : G → R2 is a component of R2 \σ (G). The boundary of
a face F is the set of vertices and edges of G that are mapped by σ to the closure of F .
The length |F | of F is the number of edges in its boundary. A face F is bounded if
the length |F | is finite. If
is finite we will say that G has bounded codegree.
The Cheeger constant of a graph G is
(G∗) :=
c(G) :=
|∂ S| ,
where ∂ S = {v ∈ G \ S | there exists w ∈ S adjacent to v} is the boundary of S.
Graphs with strictly positive Cheeger constant are called non-amenable graphs. A
graph is uniformly isoperimetric if satisfies an isoperimetric inequality of the form
|S| ≤ f (|∂ S|) for all non-empty finite vertex sets S, where f : N → N is a monotone
increasing diverging function.
A x –y path in a graph G is called a geodesic if its length coincides with the
distance between x and y. A geodetic triangle consists of three vertices x , y, z and
three geodesics, called its sides, joining them. A geodetic triangle is δ-thin if each of
its sides is contained in the δ-neighbourhood of the union of the other two sides. A
connected graph is δ-hyperbolic if each geodetic triangle is δ-thin. The smallest such
δ ≥ 0 will be called the hyperbolicity constant of G. A graph is hyperbolic if there
exists a δ ≥ 0 such that each connected component is δ-hyperbolic.
If C is a cycle of G and x , y lie on C then they identify two arcs joining them
along C : we will write xC y and yC x for these two paths —it will not matter which is
which. Similarly, if P is a path passing through these vertices, x P y is 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
Theorem 2 separately.
We will assume throughout the text that G ∈ P, i.e. G is an accumulation-free
plane graph with bounded degrees, fixed for the rest of this paper. Theorem 2 is trivial
in the case of forests, so from now on we will assume that G has at least a cycle, or in
other words it has a bounded face.
A geodetic cycle C in a graph G is a cycle with the property that for every two points
x , y ∈ C at least one of xC y and yC x (defined in the end of Sect. 2) is a geodesic in G.
|C | < ∞.
Proof Let δ be the hyperbolicity constant of G. We will show that no geodetic cycle
has more than 6δ vertices.
Let C be a geodetic cycle, say with n vertices, and choose three points a, b, c on C
as equally spaced as possible, i.e. every pair is at least n3 apart along C . Let ab be
the arc of C joining a and b that does not contain c, and define bc and ca similarly.
We want to show that ab, bc and ca form a geodetic triangle.
If x , y, z are distinct points in C then let x zy be the arc in C from x to y that
passes through z. Then we know that one of ab, acb is a geodesic joining a and b, and
|acb| ≥ 2 n3 > |ab|, so ab is a geodetic arc. Similarly, bc and ca are geodetic arcs.
Consider now the point p on ab at distance n6 from a along C . Since G is a
δ-hyperbolic graph, we know that there is a vertex q on bc or ca which is at distance
at most δ from p. But as C is a geodetic cycle, the choice of a, b, c implies that
d( p, q) ≥ min{d( p, a), d( p, b)} =
By the Jordan curve theorem, we can say that a point of R2 is strictly inside a given
cycle C of G if it belongs to the bounded component of R2 \ C and is inside C if it
belongs to C or is strictly inside. We say that a subset of R2 is inside (resp. strictly
inside) C if all of its points are inside (resp. strictly inside) C . A subset of R2 is outside
C if it is not strictly inside C .
Recall that we are assuming G to have no accumulation point, so inside each cycle
we can only have finitely many vertices.
Corollary 1 Suppose G ∈ P is hyperbolic and uniformly isoperimetric. If every face
of G is contained inside a geodesic cycle, then (G∗) < ∞.
Proof Consider a face F contained inside a geodetic cycle C ; by Lemma 1 we know
that |C | ≤ 6δ, where δ is the hyperbolicity constant of G. Let S be the set of all vertices
inside the geodetic cycle C 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 of C
sends less than (G) edges to ∂ S, implying that |∂ S| < (G)|C |. Let f : N → N be
a monotone increasing diverging function witnessing the weak non-amenability of G.
Then, since F ⊆ S,
which is uniformly bounded for every face F of G.
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 Corollary 1 whenever 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 paths P, Q, R sharing
the same endpoints, if P ∪ R is a cycle and Q lies inside it, we will say that Q is
between P, R.
Now, suppose we are given a cycle C and two points x , y ∈ C such that there exists
a geodesic γ joining x and y lying outside C . Consider the set S = S(x , y) of x –y
geodesics that lie outside C . This set can be divided into two classes:
S1 := { ∈ S | xC y is between yC x , },
S2 := { ∈ S | yC x is between xC y, }.
These two subsets of S cannot be both empty because one of them must contain γ .
Let us assume, without loss of generality, that S1 = ∅. For the proof of Theorem 2,
we will make use of the notion of ‘the closest geodesic’ to a given cycle; let us make
this more precise. Consider the above cycle C in a plane graph, two points x and y on
C and a choice of an arc on C joining them, say xC y. Let us define a partial order on
the set S1 defined above: for any two geodesics , ∈ S1 we declare if is
between xC y, .
Lemma 2 With notation as above, (S1, ) has a least element.
Proof The set S1 is a subset of all paths from x to y of length d(x , y). These paths
are contained in the ball of center x and radius d(x , y). As G is locally finite, this ball
is finite and so is S1. Therefore, it suffices to produce for every couple of elements a
(greatest) lower bound.3
Pick two geodesics ,
to y) of maximal subpaths of
in S1; let P1, . . . , Ph be the collection (ordered from x
lying inside the cycle xC y ∪ and Q1, . . . , Qk the
3 In a similar fashion we can produce a least upper bound, showing that S1 is a finite lattice.
collection (ordered from x to y) of maximal subpaths of lying inside the cycle
xC y ∪ (note that h − k ∈ {−1, 0, 1}). Without loss of generality, we can assume
that x belongs to P1, so h − k ∈ {0, 1}.
Now consider the subgraph
P1 ∪ Q1 ∪ . . . ∪ Ph ∪ Qk ,
P1 ∪ Q1 ∪ . . . ∪ Ph−1 ∪ Qk ∪ Ph , if h = k + 1.
if h = k;
Note that each Pi shares one endvertex with Qi−1 and the other with Qi , and similarly
Q j shares the endvertices with Pj and Pj−1. We want to prove to be an element of
S1 and specifically the greatest lower bound of and . Note that and intersect
in some points x = x1, x2, . . . , xn = y (the endvertices of all Pi and Q j ) and, being
geodesics, xi xi+1 is as long as xi xi+1. This implies | | = | | = | | = d(x , y),
i.e. is a geodesic (in particular, it is a path). The fact that ∈ S1 follows from
having put together only sub-paths of elements from S1. Lastly, we need to show that
both and hold. But all paths Pi and Q j are inside both xC y ∪ and
xC y ∪ , therefore so is .
Of course, there is nothing special with S1 and thus S2 has a similar partial order
and admits a least element too, provided it is non-empty.
Let us say that in a plane graph a path P crosses a cycle C if the endpoints of P
are outside C but there is at least one edge of P that lies strictly inside C .
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 geodesic in 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 set S1 given above, and B
satisfies that condition. By Lemma 2, there exists a least x –y geodesic with respect
to . Let us show that this is the required geodesic. Suppose there is a geodesic
that crosses the cycle xC y ∪ and lies outside C , so the endpoints of are outside
yC x ∪ and has an edge e strictly inside xC y ∪ . Let a b the longest subpath
of containing e and lying inside xC y ∪ . Then a, b are on and the geodesic
:= x a ∪ a b ∪ b y
, contradicting the minimality of . This contradiction proves our
Note that Corollary 2 does not claim uniqueness for the geodesic: if satisfies the
claim and then satisfies it as well. However, the unique least element of
S1 satisfies the statement of (2) thus such a geodesic will be referred to as the closest
geodesic to the cycle C in S1. We conclude that, given a pair of points x , y on a cycle
C , there are exactly one or two x –y geodesics closest to C , depending on how many of
S1, S2 are non-empty. If these two geodesics both exist, they can intersect but cannot
cross each other.
Theorem 4 If G ∈ P is hyperbolic and uniformly isoperimetric then
(G∗) < ∞.
Proof We want to show that if F is a face of G, then it is contained in a geodetic
cycle and then apply Corollary 1. The idea of the proof is to construct a sequence of
cycles C0, C1, . . . each containing F , 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 cycle C0 coinciding with the boundary of the face F . If C0 is
geodetic we are done, otherwise there are two points x , y such that both xC0 y and
yC0x are not geodesics. Consider a geodesic 1 joining them: since F is a face, 1
must lie outside the cycle C0. Therefore, we have three paths xC0 y, yC0x and 1
between x and y. Assume without loss of generality that xC0 y is between yC0x , 1.
Then the union of 1 with yC0x yields a new cycle C1 with the following properties:
• |C1| = |yC0x | + | 1| < |yC0x | + |xC0 y| = |C0|, since xC0 y is not a geodesic
while 1 is;
• the face F is inside the cycle C1 since it was inside (or rather, equal to) C0 which
in turn is inside C1.
Using Lemma 2 we can require the geodesic 1 to be the closest to the cycle C0 with
respect to the arc xC0 y. Note that the cycle C1 cannot 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
Corollary 2.
We can iterate this procedure: assume by induction that after n steps, we are left with
a cycle Cn such that the face F is still inside Cn and Cn cannot be crossed by geodesics.
If Cn is a geodetic cycle we are done, otherwise there are two points x , y ∈ Cn that
prevent that, and we can find a closest geodesic n+1 as before, creating a new cycle
Cn+1. We conclude that the face F is inside Cn+1 and |Cn+1| < |Cn|. Since these
lengths are strictly decreasing, the process halts after finitely many steps, yielding the
desired geodetic cycle. Note that Cn+1 still has the property that it cannot be crossed
by a geodesic: indeed, if a geodesic crosses Cn+1, then since it cannot cross Cn by
the induction hypothesis, it would have to cross the cycle xCn y ∪ n+1, which would
contradict condition (2) of Corollary 2.
4 Non-amenability and Bounded Codegree Imply Hyperbolicity
One of the assertions of Theorem 2 was 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 as linear isoperimetric inequality. For our interests,
which are planar graphs of bounded degree, that condition has been rephrased as in
Theorem 5 below. 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 of H are bounded by a cycle; let us call ∂top H
the cycle bounding the unbounded face of H .
Definition 1 A combinatorial disk H satisfies a (k, D)-linear isoperimetric inequality
(LII) if |F | ≤ D for all bounded faces F of H and the number of bounded faces of H
is bounded above by k|∂top H |.
Definition 2 An infinite, connected, plane graph G satisfies a LII if there exist k, D ∈
N such that the following holds: for every cycle C ⊂ G there is a combinatorial disk H
satisfying a (k, D)-LII and a map ϕ : H → G which is a graph-theoretic isomorphism
onto its image (so that ϕ does not have to respect the embeddings of H, G into the
plane), such that ϕ(∂top H ) = C .
Bowditch’s criterion is the following:
Theorem 5 ([2]) A plane graph G of minimum degree at least 3 and 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
linear isoperimetric inequality, i.e. having linear 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
general path-metric spaces, by proving that having a subquadratic isoperimetric function
implies hyperbolicity. Our Theorem 6 shows that for planar graphs, non-amenability
and the boundedness of the codegree together are sufficient to imply a linear
isoperimetric inequality.
An immediate corollary is the following:
Corollary 3 Let G be a plane graph of minimum degree at least 3, 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 cycle C , let H be the subgraph of G induced by all vertices inside C .
Then H is a finite plane graph of codegree bounded above by (G∗). By assumption,
the number of bounded faces of H is bounded above by k|C | = k|∂top H |. Thus G
satisfies a LII, and G is hyperbolic by Theorem 5.
We will see a partial converse of this statement in Lemma 4.
We would like to apply this criterion to our non-amenable, bounded codegree graph
G, but G might have minimum degree less than 3. Therefore, we will perform on G
the following construction in order to obtain a graph G of minimum degree 3 without
affecting any of the other properties of G we are interested in.
Define a decoration of a non-amenable graph G to be a finite connected induced
subgraph H with 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 since G is non-amenable, the size of any of its decorations
is bounded above.
Perform the following procedure on each decoration H of the graph G: if |∂ H | = 1
delete H , while if ∂ H = {v, w} delete H and add the edge {v, w} if not already there.
Call the resulting graph G . Note that the minimum degree of G is at least 3: any
vertex of G of degree at most 2 belongs to a decoration, and if H is a decoration and
x ∈ ∂ H then by maximality x sends at least 3 edges to G \ (H ∪ ∂ H ) when |∂ H | = 1
and at least 2 edges when |∂ H | = 2. Note also that the maximum degree of G is at
most (G).
Now assume G is non-amenable with Cheeger constant c(G); then the size of
2
decorations is bounded above by c(G) and thus the size of any face of G is reduced by
at most c(2G) after the procedure, so (G ∗) is finite if (G∗) is. Consider the identity
map I : V (G ) → V (G). Then
and every vertex in G is within c(2G) from a vertex of f (V (G )), hence I is a
quasiisometry between G and G . Thus G is non-amenable, since non-amenability is a
quasiisometric 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 that G is hyperbolic then so is G.
Proof Starting from G, perform the construction of the auxiliary graph G as above:
the resulting graph G is non-amenable, has bounded codegree and has minimum
degree at least 3.
Let C be a cycle and S ⊂ G the (finite but possibly empty) subset of vertices lying
strictly inside C ; by non-amenability we have
|C | ≥ |∂ S| ≥ c(G )|S|.
Let us focus on the finite planar graph G [C ∪ S] induced by C ∪ S and let F 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 )
which is equivalent to F ≤ (1+c(cG(G))) (G) |C |. Since (G ∗) is finite, by Corollary 3
G is hyperbolic. By the remark above, G is hyperbolic too.
5 Hyperbolicity and Weak Non-amenability Imply Non-amenability
Let us prepare the last step of the proof of Theorem 2 with a lemma.
Lemma 3 Suppose G has bounded codegree and no unbounded faces. Then for every
finite connected induced subgraph S of G, there exists a closed walk C such that S is
inside C and at least |C |/ (G∗) vertices of C are in the boundary of S.
Proof Let H be the subgraph of G spanned by S and 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 let C be the closed walk bounding the unbounded
face of H . We claim that C has the desired properties.
To see this, let x1, . . . , xn be an enumeration of the vertices of ∂ S in the order
they are visited by C . Then the subwalk xi C xi+1 is contained in some face of G: all
interior vertices of xi C xi+1 lie in S by our definitions,and so all edges incident with
those vertices are in H ; therefore, since xi C xi+1 is a facial walk in H , it is also a facial
walk in G. Since xi C xi+1 is contained in some face of G and G has only bounded
faces, we have |xi C xi+1| < (G∗). Applying this to all i , we obtain
|C ∩ ∂ S| = n >
i=1
We need a result which is almost a converse of Corollary 3.
Lemma 4 Let G ∈ P be 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 graph G from the previous section, we may assume that
G has minimum degree at least 3. Let C be a cycle of G. Since G is hyperbolic,
by Theorem 5 there exists a combinatorial disk H satisfying a (k , D)-LII with an
isomorphism ϕ from H to a subgraph of G such that ϕ(∂top H ) = C . The boundaries
Fi , i ∈ I, of bounded faces of H are sent by ϕ to cycles Ci := ϕ(Fi ) of G so that
|I | ≤ k |C |, and the bound D on the length of bounded faces of H is an upper bound
to the length of those cycles Ci . Let Si be the (finite) set of vertices of G strictly
inside Ci , so that ∂ Si ⊆ Ci . Let f : N → N a monotone increasing diverging function
witnessing the weak non-amenability of G, 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 G inside Ci ; then for a
non-empty Si we have F (Ci ) ≤ (G)|Si | because no vertex can meet more than
(G) faces. In conclusion
i∈I
i : Si =∅
i : Si =∅
|{faces inside C }|=
F (Ci ) ≤
(G)|Si | ≤ |I | +
from which by setting k := k +
(G) f (D)k the assertion follows.
Note that in order to prove the non-amenability of a graph G it suffices to check
that |∂ S| ≥ c|S| for some constant c > 0 and all finite induced connected subgraphs
S, instead of all finite subsets. Indeed, if we assume so and if S is a finite induced
subgraph with components S1, . . . , Sn , then
|∂ S| =
∂ Si ≥
i=1
i=1
|∂ Si | ≥
i=1
|Si | =
where the first equality follows from ∂ Si ∩ S j = ∅ for all i = j (S is induced) and
the first inequality holds because the boundaries ∂ Si can overlap, but no vertex of ∂ S
belongs to more than (G) of them.
Proof By the above consideration it is enough to check the non-amenability only
on connected induced subgraphs of G. By Theorem 4 we know that G has bounded
codegree. Let S be such a subgraph and C as in Lemma 3. Then
|∂ S| ≥ |∂ S ∩ C | ≥
(G∗)
Let k > 0 be as in Lemma 4; if T denotes the set of all vertices inside C and F the
set of all faces inside C , we have
(G∗) vertices. Combining the last two
6 Graphs with Unbounded Degrees
We provided enough examples to show that Theorem 2 is 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 now P∗ denote the class of plane
graphs with no accumulation point of vertices; so that P is the subclass of bounded
degree graphs in P∗.
Problem 1 Is there a hyperbolic, amenable, uniformly isoperimetric plane graph of
bounded codegree and no unbounded face in P∗?
Problem 2 Is every non-amenable bounded codegree graph in P∗ hyperbolic?
Acknowledgements B. Federici was supported by an EPSRC Grant EP/L505110/1, and A.
Georgakopoulos 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
Mathematical Sciences, Cambridge, for support and hospitality during the programme ‘Random Geometry’ where
work on this paper was undertaken.
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0
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