Scholarly article on topic 'Gromov hyperbolicity of planar graphs'

Gromov hyperbolicity of planar graphs Academic research paper on "Mathematics"

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Academic research paper on topic "Gromov hyperbolicity of planar graphs"

Cent. Eur. J. Math. • 11(10) • 2013 • DOI: 10.2478/s11533-013-0286-9

1817-1830

VERS ITA

Central European Journal of Mathematics

Gromov hyperbolicity of planar graphs

Research Article

Alicia Cantón1*, Ana Granados2^, Domingo Pestana3^, José M. Rodríguez3^

1 Universidad Politecnica de Madrid, Avenida Arco de la Victoria, s/n, Ciudad Universitaria, 28040 Madrid, Spain

2 St. Louis University (Madrid Campus), Mathematics Department, Avenida del Valle 34, 28003 Madrid, Spain

3 Departamento de Matemáticas, Universidad Carlos III de Madrid, Avenida de la Universidad 30, 28911 Leganés, Madrid, Spain

Received IS July SOIS; accepted S9 November SOIS

Abstract: We prove that under appropriate assumptions adding or removing an infinite amount of edges to a given planar graph preserves its non-hyperbolicity, a result which is shown to be false in general. In particular, we make a conjecture that every tessellation graph of R2 with convex tiles is non-hyperbolic; it is shown that in order to prove this conjecture it suffices to consider tessellation graphs of R2 such that every tile is a triangle and a partial answer to this question is given. A weaker version of this conjecture stating that every tessellation graph of R2 with rectangular tiles is non-hyperbolic is given and partially answered. If this conjecture were true, many tessellation graphs of R2 with tiles which are parallelograms would be non-hyperbolic.

MSG: 05C10, 05C63, 05C75, 05A20

Keywords: Planar Graphs • Gromov Hyperbolicity • Infinite Graphs • Geodesics • Tessellation © Versita Sp. z o.o.

1. Introduction

Hyperbolic spaces play an Important role In geometric group theory and In the geometry of negatively curved spaces. The concept of Gromov hyperbolicity grasps the essence of both negatively curved spaces like the classical hyperbolic space or Riemannian manifolds of negative sectional curvature, and of discrete spaces like trees and the Cayley graphs of many finitely generated groups. It is remarkable that a simple concept leads to such rich general theory [1, 16, 17].

* E-mail: alicia.canton@upm.es î E-mail: agranado@slu.edu

* E-mail: dompes@math.uc3m.es § E-mail: jomaro@math.uc3m.es

Springer

The theory of Gromov spaces was used Initially for the study of finitely generated groups (see [17] and the references therein), where its practical importance was discussed. This theory was mainly applied to the study of automatic groups [31], which appear in computational science. The concept of hyperbolicity appears also in discrete mathematics, in particular, a number of algorithmic problems in hyperbolic spaces and hyperbolic graphs has been considered in recent papers [12, 13, 15, 28]. Another application of these spaces is secure transmission of information on the internet [21-23], playing a significant role in the spread of viruses through the network [21, 23]. It has been shown empirically in [42] that the internet topology embeds with better accuracy into a hyperbolic space than into an Euclidean space of comparable dimension. Hyperbolicity is also useful in the study of DNA data [7].

The study of mathematical properties of Gromov hyperbolic spaces and its applications is a topic of recent and increasing interest in graph theory;see, for instance [3-5, 7, 8, 10, 14, 21-25, 27, 29, 30, 33-36, 39-41, 43, 44].

In recent years several researchers have been interested in showing that metrics used in geometric function theory are Gromov hyperbolic. For instance, the Gehring-Osgood j-metric is Gromov hyperbolic;and the Vuorinen j-metric is not Gromov hyperbolic except in the punctured space [18]. The study of Gromov hyperbolicity of the quasihyperbolic and the Poincaré metrics is the subject of [2, 6, 19, 20, 36-38, 40, 41]. In particular, in [36, 40, 41, 43] it is proved the equivalence of the hyperbolicity of many negatively curved surfaces and the hyperbolicity of a very simple graph. Deciding whether a space is hyperbolic is a difficult problem since the location of geodesics is unknown, and hence, it is useful to know hyperbolicity criteria for graphs. This will be the topic of our further discussion.

One of the main questions in the study of any mathematical property is to find transformations which preserve that property. In [8, Theorem 3.15] the authors prove that adding or removing any finite amount of edges of a graph preserves its non-hyperbolicity (or hyperbolicity). It is thus natural to consider what happens if the amount of edges is infinite. Theorem 3.1 below gives a positive answer to this question under some appropriate hypotheses for planar graphs; Theorem 3.6 shows that the general answer is negative, even for planar graphs.

The papers [9, 35] study the hyperbolicity of some type of planar graphs. In particular, in [9], the authors conjectured that every tessellation graph of R2 with convex tiles is non-hyperbolic. Sections 4 and 5 deal with this open problem. Theorem 5.1 shows that in order to prove this conjecture, it suffices to consider tessellation graphs of R2 such that every tile is a triangle. A weaker conjecture is stated here, namely that every tessellation graph of R2 with rectangular tiles is non-hyperbolic. Theorem 4.6 gives a partial answer to this question. Finally, Theorem 4.2 shows that if this weaker conjecture is true, then many tessellation graphs of R2 with tiles which are parallelograms are non-hyperbolic.

2. Background on Gromov hyperbolic spaces

Let (X,d) be a metric space and let y: [a,b] —> X be a continuous function. The curve y is a geodesic if L(y|"[ts]) = d(y(t),y(s)) = \t — s| for every s,t e [a, b], where L denotes the length of a curve;a geodesic line is a geodesic with domain R, and a geodesic ray is a geodesic with domain [0, to). X is a geodesic metric space if for every x, y e X there exists a geodesic joining x and y; denote by [xy] any of such geodesics (since uniqueness of geodesics is not required, this notation is ambiguous, but convenient). It is clear that every geodesic metric space is path-connected. If the metric space X is a graph, [u, v] denotes the edge joining the vertices u and v.

In order to consider a graph G as a geodesic metric space, one must identify any edge [u,v] e E(G) with the real interval [0,1] (if I = L([u, v]));therefore, any point in the interior of any edge is a point of G and, if the edge [u, v] is considered as a graph with just one edge, then it is isometric to [0, /]. A connected graph G is naturally equipped with a distance defined on its points, induced by taking shortest paths in G, inducing in G the structure of a metric graph. Note that edges can have arbitrary lengths.

Throughout the paper only simple, connected and locally finite graphs are considered (i.e., graphs without loops or multiple edges and so that each ball contains a finite number of edges); these properties guarantee graphs are geodesic metric spaces. The study of the hyperbolicity of graphs with loops and multiple edges can be reduced to the study of the hyperbolicity of simple graphs [5, Theorems 8 and 10].

If X is a geodesic metric space and J = {J1,J2,... ,Jn} is a polygon with sides Jj C X, then J is said to be 5-thin if for every x e J¡ one has that < 5. The sharp thin constant of J, 5(J), is then 5(J) = inf {5 > 0 : J is 5-thin}. If

x1,x2,x3 are points in X, a geodesic triangle T = {x1,x2,x3} is the union of the three geodesics [xix2], [x2x3] and [x3x-i].

The space X is 5-hyperbolic (or satisfies the Rips condition with constant 5) if every geodesic triangle in X is 5-thin. Denote by 5(X) the sharp hyperbolicity constant of X, i.e., 5(X) = sup {5(T) : T is a geodesic triangle in X}. The space X is hyperbolic if X is 5-hyperbolic for some 5 > 0; in this case, 5(X) = inf {5 > 0 : X is 5-hyperbolic}.

Trivially, every bounded metric space X is (diamX)-hyperbolic. The real line R is 0-hyperbolic whereas the Euclidean plane R2 is not. In general, a normed vector space E is hyperbolic if and only if dim E = 1. Every metric tree with arbitrary length edges is 0-hyperbolic; every simply connected complete Riemannian manifold with sectional curvature verifying K < —c2 < 0 is hyperbolic. More background and further results are given in, e.g., [1, 16]. The spaces X with 5(X) = 0 are precisely the metric trees, and the hyperbolicity constant of a geodesic metric space can be viewed as a measure of how "tree-like" the space is.

There are several definitions of Gromov hyperbolicity, all equivalent in the sense that if X is 5-hyperbolic with respect to definition A, then it is 5'-hyperbolic with respect to definition B for some 5', see, e.g., [1, 16].

Let (X, dX) and (Y, dY) be two metric spaces. A map f: X ^ Y is said to be an (a, fi)-quasi-isometric embedding, with constants a > 1, fi > 0 if for every x, y e X:

a—1dX(x,y) — fi < dY(f(x),f(y)) < adX(x,y)+ fi.

The function f is e-full if for each y e Y there exists x e X with dY(f (x),y) < e. A map f: X ^ Y is said to be a quasi-isometry, if there exist constants a > 1, fi,e > 0 such that f is an e-full (a, fi)-quasi-isometric embedding. In that case we say that X and Y are quasi-isometric. Note that a quasi-isometric embedding, in general, is not continuous. Let X be a metric space, Y a non-empty subset of X and e a positive number. The e-neighborhood of Y in X, denoted by Ve(Y) is defined as the set {x e X : dX(x, Y) < e}.

A fundamental property of hyperbolic spaces is the following

Theorem 2.1 (invariance of hyperbolicity).

Let f: X ^ Y be an (a, fi)-quasi-isometric embedding between the geodesic metric spaces X and Y. If Y is hyperbolic, then X is hyperbolic. Besides, if f is e-full for some e > 0 (a quasi-isometry), then X is hyperbolic if and only if Y is hyperbolic.

If D is a closed subset of X, the inner metric considered in D is defined as

dD(z, w) = inf {LX(y) : Y C D is a continuous curve joining z and w} > dX(z, w). Consequently, LD(y) = LX(y) for every curve y C D.

In an informal way, a tessellation, T, on a complete Riemannian surface, X, is a partition of X by geometric shapes (called tiles) with no overlaps and no gaps. The tessellation graph associated to T is the union of the boundaries of the tiles. More precisely, for n > 1, an n-cell is a topological space homeomorphic to the open ball in Rn. A 0-cell is a singleton space. A tesselation on a complete Riemannian surface, X, is a CW 2-complex on X such that every point on X is contained in some n-cell of the complex for some n e {0,1, 2}. A tessellation graph is the 1-skeleton (the set of 0-cells and 1 -cells). The edges (1 -cells) of a tessellation graph are just rectifiable paths in X and have the length induced by the metric on X (these paths may or may not be geodesics in X). Throughout the paper X = R2 with the exceptions of Theorem 3.4 and the proof of Theorem 3.6, where X will stand for the hyperbolic plane.

Along the paper, given a set E contained in a Riemannian surface X, we denote by AX(E) its area and by E its closure.

3. Hyperbolicity of tessellation graphs

If G0 is a non-hyperbolic tessellation graph of R2, a natural question is whether this non-hyperbolicity will be preserved when adding to it any number (possibly infinite) of vertices and edges. Theorem 3.1 gives an affirmative answer to this question under some regularity hypotheses on G0 (Theorem 3.6 below will show this result to be false in general). It will also show a connection between the continuous and discrete frames, see e.g., [26].

Theorem 3.1.

Let G0 be the 1 -skeleton of a tessellation of R2 with tiles {Fn}neI. Assume that there exists a partition I = A-j U A2 of the set of indices and positive constants c1, c2, verifying the following properties:

(I) diamGodFn < c1 and AR2(Fn) > c2 for every n e A1,

(II) ddFn (x, y) < c1dR2(x, y) for every x,y e dFn and for every n e A2.

Then G0 is not hyperbolic. Moreover, any 1-skeleton G of a tessellation of R2 which contains G0 as a subgraph is not hyperbolic.

Proof. It will be proven that the Inclusion i: G0 —> R2 Is a quasl-lsometrlc embedding; In fact, It Is shown that

dR2(x,y) < dCo(x,y) < {2cjc-1 + c-j) dK2(x, y) + nc3:c-1 (1)

for every x, y e G0.

First of all, it is clear that dR2(x, y) = dR2(i(x), i(y)) < dGo(x, y) for every x, y e G0. Fix now x, y e G0 and let a be the Euclidean segment joining x and y in R2. If n e A1 and Fn n a = 0, then Fn C Vdiam Fn (a) C Vcj(a). Since Vcj(a) is the union of two half-disks and a rectangle, clearly, AR2(Vcj(a)) = 2c1L(a) + nc2. Let N(a) denote the number of Fn with n e A1 that crosses a, then

c2N(a) < AR2(Vci(a)) = 2c-L(a) + nc2v

Therefore,

N(a) < c^(2cidvi(x,y) + ncj).

Consider a as an oriented segment from x to y. A finite set of points will be inductively defined as follows: let y1 be the first point on a with y1 e UneA, Fn; then y1 e Frj for some r1 e A1; take y2 to be the last point on a n Frj. Proceeding this way, assume that {y1,..., y2j} have been defined with y2s-1 the first point and y2s the last on anFrs for s = 1,... ,j.

If a \ [y1y2j] does not intersect UneAA-^.....0} Fn, then this process is stopped. If a \ \y1y2j] intersects (JneA^.....ri} Fn,

then define y2j+1 as the first point in (a\[y^2j]) U y} with y2j+1 e LUaa-^.....0} Fn; then yv+1 e F^ for some

rj+1 e A1\ and define y2j+2 as the last point in a n Frj+1. Eventually this process will finish and a finite set {y1,..., y2N} will be obtained.

Let [x1x2], [x3x4],..., [x2m-1x2m] be the Euclidean segments contained in the closure of a \UN=1[y2i-1y2i]. Notice that [x2j-1x2j] C a n UneA2 Fn, and thus

dG0(x,y) < sup {diamG0 dFn}N(a)^^ d G0(x2j-1, x2j) < c1N(a) + q ^ dR2(x2j-1, x2j) ^

neA1 j=1 j=1

< v{2c\c- + c^dM2(x, y) + nc]c-\ which completes the proof of (1).

We shall show next that G0 is not hyperbolic. To this end it will be first proven that

5(G0) > = sup diamj.2 dFn. (2)

For any fixed n, let us consider the set An of closed curves in G0 freely homotopic to dFn in R2 \ Fn. Choose a closed curve an e An with L(an) = min {L(a) : a e An}; it is clear that L(an) > 2diamR2 dFn, and that dan (x, y) = dG0(x, y) for every x,y e an since an is a shortest curve in An. Let xn,yn be points in an with dG0(xn,yn) = dan (xn,yn) = L(an)/2. Then there are two different geodesics aI, a^ in G0 joining xn and yn with a^n U a2 = an. Therefore the set Bn = {altf} is a geodesic bigon (a geodesic triangle having two of its vertices to be the same point). If un is the midpoint of an, then 5(Bn) > dG0(un, af,) = dG0(un, {xn, yn}) = L(an)/4 > (1/2) diamR2 dFn. Taking the supremum on n, (2) follows.

If supn diamR2 dFn = oo, by (2), G0 is not hyperbolic. If supn diamR2 dFn = c^ < to, then the inclusion i: G0 —> R2 is a c\-full (a0, b0/a0)-quasi-isometry, with a0 = 2c1c—1 + c^ > c^ > 1 and b0 = ncijc-1, since diamR2 Fn < diamG0 dFn. In this case, Theorem 2.1 implies G0 is not hyperbolic.

Let us finally show that any 1-skeleton G of a tessellation of R2 which contains G0 as a subgraph will not be hyperbolic. Clearly, dG(x, y) < dG0(x, y) for every x, y e G0, and also dR2(x, y) < dG(x, y) for every x, y e G,. By (1),

do(x, y) < dC0(x, y) < a0dR2(x, y) + b0 < a0dc(x, y) + b0

or, equivalently,

— dG0(x, y) — — < dG(x, y) < dG0(x, y), a0 a0

for every x,y e G0. Thus the inclusion i0: G0 —> G is an (a0, b0/a0)-quasi-isometric embedding. Therefore, since G0 is not hyperbolic, by Theorem 2.1 one obtains that G is not hyperbolic. □

The arguments just given in the proof of Theorem 3.1 have the following consequences.

Theorem 3.2.

Let G0 be the 1-skeleton of a tessellation of R2 such that there exist non-negative constants a0,b0 so that dG0(x, y) < a0dR2 (x, y) + b0 for every x, y e G0. Then any 1 -skeleton G of a tessellation of R2 which contains G0 as a subgraph is not hyperbolic.

Theorem 3.3.

Let G be the 1 -skeleton of a tessellation ofR2 with tiles {Fn}. Then 5(G) > (1/2) sup diamR2 dFn.

A direct consequence of Theorem 3.3 is that, if supn diamR2 dFn = to, then G is not hyperbolic. It will be shown in Theorem 3.6, one of the main results of this section, that if supn diamGdFn = oo, this is false. The following results on hyperbolicity will be needed there.

Theorem 3.4 ([35, Theorem 3.1 and Remark 3.2]).

Let G be the 1 -skeleton of a tessellation of the hyperbolic plane H with tiles {Fn }. If for some positive constants c, c2, one has diamGdFn < c and AH(Fn) > c2 for every n, then G is hyperbolic.

Let us denote by G \ {v} the metric space obtained by removing the point {v} from the metric space G. A vertex v of a graph G is a cut vertex if G \ {v} is not connected. Note that in a tree, any vertex with degree greater than one is a cut vertex. Finally, let us denote by {Gr}r the closures in G of the connected components of the set

G \ {v G V(G) : v is a cut vertex of

The set {Gr}r is the canonical T-decomposition of G.

Example.

Let us consider two cycle graphs H, r2, and x^ G V(H), x2 G V(r2). Define the graph G as the graph with V(G) = V(r1) U V(r2) and E(G) = E(r:) U E(r2) U [x:,x2]. Then r2, [x^x-^} is the canonical T-decomposition of G.

Theorem 3.5 ([5, Theorem 5]).

If {Gr}r is the canonical T-decomposition of G, then 5(G) = sup 5(Gr).

The next result will deal with periodic graphs. The tessellation graph G of R2 Is periodic If there exists (u,v) e R2 \ {(0,0)} such that T(G) = G, where T: R2 ^ R2 is defined as T(x, y) = (x, y) + (u, v). Recall that a geodesic line is a geodesic with domain R. By Euclidean line we mean an straight line in R2, i.e., a geodesic line in the Euclidean plane.

Theorem 3.6.

There exists a periodic hyperbolic 1 -skeleton G of a tessellation of R2 with tiles {Fn} verifying supn dlamG dFn = oo and containing infinitely many Euclidean lines. Furthermore, there exists a periodic non-hyperbolic subgraph G0 of G which is also a tessellation graph of R2.

Remark 3.7.

The main idea in the construction of such a tessellation is to include in R2 a tessellation graph quasi-isometric to a periodic model of the hyperbolic plane. The example given in Theorem 3.6 shows that it is not possible to replace supn dlamR2 dFn by supn dlamGdFn in Theorem 3.3. Theorem 4.6 shows a large class of non-hyperbolic tessellation graphs containing infinitely many Euclidean lines.

Proof. The structure of the proof is as follows: first, a hyperbolic graph G3, which is a tessellation of H, will be defined;based on G3, define a new hyperbolic graph G6 which is a tessellation of R2; finally, the graph G satisfying all conditions in the statement will be defined from G6.

Let us consider the hyperbolic plane H with its Fermi coordinates, see, e.g., [11, p.247], i.e., the plane R2 with the Riemannian metric ds2 = cosh2y dx2 + dy2 (thus dA = cosh y dxdy).

Let [t] stand for the integer part of t. Consider the segments Im>n in H given by Im>n = {(x, y) e H : n/[cosh m] < x < (n + 1)/[cosh m], y = m} for m = 0,1, 2,... and 0 < n < [cosh m] — 1,and Jm>n = {(x, y) e H : x = n/[cosh m], m < y < m + 1} for m > 0 and 0 < n < [cosh m].

Let G1 = Um.nVmn U Jmn}, S0(x,y) = (x, —y) and G2 = G1 U SvG). Also, let Rk(x,y) = (x + k,y) for k e Z. The graph G3 is now defined as G3 = |Jk Rk(G2). Clearly, G3 is a tessellation graph of H. Let us check that it verifies the hypotheses in Theorem 3.4. To this end, let {F*} be the tiles of the tessellation G3. Since S0 and Rk are isometries of H, diamG3 dF* < LH(dF*). A standard computation gives

, ,, > f(n+1)/lcoshm , , coshm „ , ,, , fm+1 , Lrn(Im,n)= I cosh mdx = f-—- < 2, LB{Jmsn) = dy = 1,

Jn/[cosh m] [cosh m] Jm

m cosh(m + 1) cosh m em+1 „

cosh(m +1) dx = -^-1 • -,—T < --• 2 = 4e,

Jn/[coshm] cosh m [cosh m] em/2

(x, y) G H : ;—7—T < x < ,n + ^ 1 , m < y < m + 1 i Icosh m] Icosh m] J

r(n+1)/[coshm] rm+1 r(n+1)/[coshm] cosh m

I I cosh ydydx > I cosh mdx = ---—- > 1.

Jn/[cosh m] Jm Jn/[cosh m] [cosh m]

Therefore,

diamG3 dF* < LB(dF*) < 4e + 4, AB(F*) > 1, for every r, and Theorem 3.4 allows to conclude that G3 Is hyperbolic.

Consider now the graph G3 embedded in the Euclidean plane R2. Let us define Ko,o = /0,0; for m > 1 and 0 < n < [cosh m] — 1, let Km n be a polygonal curve joining the endpoints of /mn7 which is contained in the rectangle {(x, y) G R2 : n/[cosh m] < x < (n + 1)/[cosh m], m —1/6 < y < m + 1 /6}, where LR2(Km n) = LH(/m n) and

[cosh(m—1)]—1 . . [cosh m]—1

U [cosh(nm — 1)],m) C U Km,n.

n=0 [ ( )] n=0

Set G4 = |Jmn{Kmn U Jmn}, G5 = G4 U S0(G4) and define the next graph G6 as G6 = |Jk Rk(G5). Clearly, G6 is a tessellation graph of R2. Since the graphs G3 (in H) and G6 (in R2) are isometric, the graph G6 is hyperbolic.

Finally, let us define G. For m > 0, 0 < n < [cosh m] — 1 and 0 < s < m, let Mmns be the cycle graph which is the union of the four Euclidean segments joining the points

n + s/(4m + 4) m + 1 \ / n + (2s + 1)/(8m +8) ^ 2

cosh m] 2 [cosh m] 3

n + (s + 1)/(4m + 4) m + 1 \ / n + (2s + 1)/(8m +8) ^ + 1

[cosh m] 2 [cosh m] 3

Set G7 = |JmnsMmns, G8 = G7 U S0(G7), G9 = |Jk Rk(G8) and G = G6 U G9. Clearly, G is a tessellation graph of R2. Note that the sets Mmns, their images by S0 and Rk, and G6, are the canonical T-decomposition of G; hence, Theorem 3.5 gives that 5(G) = max{5(Ge), supmns 5(Mm n s)}. One can check that 5(Mmns) = LR2(Mmns)/4 < 1; since 5(G) < max{5(G6), 1} < oo, the graph G is hyperbolic.

Let us check the condition on the tiles of this graph. Denote by {Fr} the tiles of G; if dFr contains

Mmsns0, Mm,n,1.....Mm,n,m, then

diamG dFr > ^Y- Lr2(M m,n,s) >

m + 1 2 m + 1

2L_ R V m,n,s/ c- 2 3 3 '

and one concludes that supr diamG dFr = oo. Furthermore, the graph G is periodic and contains infinitely many Euclidean lines by construction.

Finally, let us construct a periodic non-hyperbolic subgraph G0 of G which is also a tessellation graph of R2. Let us define Km = ^=0 m— Km,n, G™ ^ Um{Km u Jm,0 U Jm\wshm]}, Gn = G^ U S0(G,0) and G0 = Uk Rk(Gn). It is clear that G0 is a tessellation graph of R2 and a subgraph of G. For each m > 0, consider the midpoint pm of Km, i.e., the point with dGti(pm, (0, m)) = dG0(pm, (1, m)) = LR2(Km)/2, and the geodesic bigon Bm in G0 with two different geodesics yI,, Ym, joining Pm and pm+1, then Y1m U yI = Km U Km+1 U Jm,0 U Jm,[coshm]. If qm is the midpoint of Ym, then

1 1 I 1 1 \

5(Bm) > dG0(qm,Y2m) = d G0(qm , {Pm,Pm+1}) = ^ lR2(yI,) = 2(2 LR2(Km) + LR2(Jm,0) + 2 LR2(Km+1)\

111 cosh m „1 cosh(m + 1) \ 1 , „.

= - --—- [cosh m] + 1 +-----{:[cosh(m +1)U = - (cosh m +cosh(m + 1) + 2),

2\2 [coshm] 2 [cosh(m + 1)^^ J/ 4V '

and one concludes 5(G0) > supm 5(Bm) = to. □

A corollary for 2-quasiperiodic graphs follows. Recall that the tessellation graph G of R2 is 2-periodic if there exist two linearly independent vectors (u^, v-\), (u2, v2) e R2 such that Tj(G) = G, j = 1,2, where Tj: R2 ^ R2 are defined as

Tj(x,y) = (x,y) + (uj,vj^ j = 1, 2

The graph G is 2-quasiperiodic if there exists a 2-periodic subgraph G0 of G.

Corollary 3.8.

If G is 2-quasiperiodic then G is not hyperbolic.

Proof. If G0 is a 2-periodic subgraph with tiles {Fn}nei, then one can take the partition A-i = I, A2 = 0 of the set of indices in the statement of Theorem 3.1. Therefore, G is not hyperbolic. □

4. Tessellations with parallelograms and rectangles

In this section It Is shown that the hyperbolicity of certain tessellations with parallelograms Is equivalent to the hyperbolicity of tessellations with rectangles. It is also shown that under some hypotheses rectangular tessellations are not hyperbolic.

4.1. Tessellations with parallelograms

Next it will be shown that considering tessellations of parallelograms with bounded inclinations is equivalent to considering rectangular tessellations with sides parallel to the axis in order to study hyperbolicity.

Consider the standard basis in R2 defined by ,e2} and, given a,ß, let Ua and Vß be the vectors defined by Ua = (cos a, sin a), Vß = (sin ß, cos ß). Fix real numbers a and b with a < b < n + a. A tessellation T is a p-tessellation of R2 if its tiles satisfy the following conditions:

• F is a parallelogram, for all F e T.

• For each F e T there exist a pair of angles a, ß such that a e (a, b), ß e (—n/2 — a, n/2 — b), and the sides of F are parallel to Ua and Vß, respectively.

Notice that the second condition above implies that if two adjacent tiles in T partially share a side, then (for both of them) there is a side that is either parallel to Ua (for some a) or to Vß (for some ß).

Theorem 4.1.

Given a p-tessellation T of R2, there exist a tessellation T of R2 with rectangular tiles and a bijective continuous function f: T —> T such that f [g is an isometry from the 1-skeleton 9 of T to the 1-skeleton G of T.

Proof. Applying a rotation, without loss of generality, we may assume a e (—c,c) where c = (b — a)/2. Then the vectors Ua and Vß give, respectively, the "almost-horizontal" and "almost-vertical" directions of a tile. In fact, if P is a vertex of a tile, F e T, the lines with directions Uc and U—c through P divide the plane in four sectors. Since the sides of every F e T are parallel to Ua and Vß with a e (—c, c) and ß e (—n/2 + c, n/2 — c), no more than four tiles can share a vertex and therefore T has the structure of a rectangular tessellation.

For a given Ua and Vß, let S(a,ß) = |F : F e T is a parallelogram of angles a,ß}. The tessellation T induces a "partition" of R2, S, given by the connected components of S(a, ß), with a e (—c, c), ß e (—n/2 + c, n/2 — c). If B = 0 is a connected component of S(a, ß) then B is a union of closures of tiles of T. If B = R2, then there exist a and ß such that all the tiles in T have sides parallel to Ua and Vß. In this situation define f = faß where faß: R2 —> R2 is the linear map such that faß(Ua) = e1 and faß(Vß) = e2. Clearly, f is an isometry from the 1-skeleton 9 of T to the 1-skeleton G of T.

In what follows it is assumed that B = R2. Since each B = 0 is the union of parallelograms whose sides have a fixed inclination, then its boundary components are polygonal lines with two possible angles. Moreover, B is a convex set and therefore it is either a parallelogram (if B bounded) or otherwise, it is a generalized parallelogram with a side at infinity, that is, a half-strip (if there is one side at infinity), a strip or a sector (if there are two) or half-plane (if there are three). Indeed, if it is not convex, there is a tile F e T, F e B, that shares two sides with B and therefore F is a parallelogram with sides parallel to those of B, thus F e B. The same argument implies that if B' = B are two connected components of S(a,ß) and S(a', ß') such that B n B' = 0, then B and B' share a whole side or a vertex.

The function f: T ^ T will be a piecewise linear function defined on the sets B inductively and so that if B e S(a, ß), then

f(x) — f(y) = faß(x) — faß(y) = faß(x — y) for all X,y e B, (3)

where faß: R2 ^ R2 is the linear map such that faß(Ua) = e1 and faß(Vß) = e2. To start the induction, let O = (0, 0) and define f(O) = O. Denote by BO one of the sets which contains O, and let a0 and ß0 be such that BO e S(a0, ß0).

If x G BO then

f (x) = fa0ß0(x] = f (O) + faoßo(x — O).

Notice that, for all x, y G BO, relation (3) trivially holds by the linearity of fa0ß0. Let Co = BO. Assume now that f Is defined and continuous on a connected set Cn which is a finite union of blocks B G S defined as Cn = {B G S : B n Cn—1 = 0} and that (3) holds for every set B G Cn. Extend f from Cn onto Cn+1 = {B G S : B n Cn = 0} in the following way: for B G Cn+1 \ Cn, B G S(a, ß), take any point P G dB n Cn, and define

f (X)= f (P) + faß(X - P), X G B.

Notice that (3) holds for points x, y G B by the linearity of faß. We are left to show that the extension is well defined. Indeed, since no more than four tiles of T can meet at a vertex, and since different B's share a whole side, at each vertex exactly four different sets B G S meet. The function f straightens the sides of each B and places it adjacent to the images of its neighbors. Concretely, if B G Cn+1 \ Cn and x G dB n Cn then considering x as a point on B G Cn+1 \ Cn, fB(x) = f(P) + faß(x — P), for a point P G dB n Cn where f(P) was already defined. If x = P there is nothing to prove. If x = P, then there exists B' G Cn such that x G dB', thus B and B' share a side the one with P and x. By (3), fB,(x) = f(P) + fa'ß'(x — P). Since both x,P G dB' n dB then, faß(x — P) = fa'ß'(x — P) and therefore f is well defined on B n Cn. To see that it is well defined on Cn+1, consider now B, B' G Cn+1 \ Cn such that B n B' = 0. Then, there exists a point Q G B n B' n Cn and by (3),

fB(x) = f (Q) + faß(x — Q); fB' (x) = f (Q) + fa'ß' (x — Q).

Since f is well defined on Cn, f (Q) is the same in both definitions, and since x,Q G Bn B' then faß(x — Q) = fa'ß'(x — Q). Thus fB(x) = fB'(x) = f(x) and f is well defined on Cn+1. An induction argument gives that f is continuous in R2.

Notice that, by construction, f maps each B to a rectangle with sides parallel to the axes, and each F G B to a rectangle inside f (B) also with sides parallel to the axes. Also if B1 and B2 are adjacent to B on opposite sides (that is, Bn Bj = 0, j = 1,2, and B1 n B2 = 0), then f(B1) and f (B2) are also adjacent to f(B) on opposite sides. Therefore, the function f is both injective and surjective. Finally, since f is linear on B each tile F G T is mapped to a rectangle and its side lengths are preserved. That is, when restricted to the 1-skeleton G of T the function f is an isometry. □

The next result is a consequence of the previous theorem.

Theorem 4.2.

All p-tessellation graphs of R2 are non-hyperbolic if and only if all tessellation graphs of R2 whose tiles are rectangles are non-hyperbolic.

4.2. Tessellations with infinitely many parallel rays Lemma 4.3.

Let T be a rectangular tessellation in R2 ^ C with tiles parallel to the coordinate axes and with infinitely many vertical rays in the upper-half plane. If F is any tile on the tessellation, L and I are the lengths of its longest and shortest sides respectively, consider the following two conditions:

(i) There exists an increasing function g: R+ ^ R+ such that for every tile F, L/l < g(dR2(F, iR)), where iR denotes the imaginary axis.

(ii) There exists a constant C so that for every tile F, ldG(F, 0) > C.

If (i) or (ii) hold, then for any point x lying on a vertical ray and any other vertical ray y with Re x < Re y, there is a geodesic advancing always rightwards and upwards which joins x to y.

Remark.

Note that any curve advancing always rightwards and upwards is a geodesic.

Proof. Let y0 and Y1 be any two of these vertical rays, and without loss of generality suppose y0 lies on the left of Yi. Let D be defined as D = d(Y0, iR) > 0. Let a be the geodesic ray starting in x defined as follows: a(t) = x + it for t e [0, t0], where t0 = max{0, infzeY1 Imz — Imx}; after that a advances rightwards when it is possible and otherwise upwards. Denote by {Fk} a choice of (ordered) tiles with a C UkdFk.

For any tile Fk, let hk denote the length of its horizontal side, and vk the length of its vertical side. The goal is to show that, in any case, there exists N such that

y hk > d(Y0, Y1).

Suppose not. Then, there exists C1 such that ^k=1 hk < C1. It will be shown that this implies that there exists C2 such that ^kL1 vk < C2, contradicting the fact that T is a tessellation.

Assume (i) holds. Without loss of generality, we can assume that g(t) > 1 for every t > 0; then vk < hkg(dR2(Fk, iR)) for all k. Therefore,

to to to I k—1 \ to

^ vk < ^ hkg(d(Fk,iR)) < ^ hkgi ^ hn + D < ^ hkg(Q + D) < C g(Q + D) = C2.

k=1 k=1 k=1 n=1 k=1

Assume (ii) holds. Without loss of generality, by Theorem 3.3 we can assume that supkL(dFk) = C0 < to. Then,

k—1 k—1 dG(Fk, 0) < dG(F1 , 0) + ^(Vj + hj) < dG(F1, 0) + ^ C0 < dG(F1, 0) + C0(k — 1).

j=1 j=1

Thus, by hypothesis,

k > dG (Fk, 0) > dGF:oTc0(k-^),

and therefore one concludes ^ hk = to. □

As it was mentioned above, there are several equivalent definitions of hyperbolicity. For the proof of the next result, the one involving uniformity in the divergence of the geodesics is used. Namely

Definition 4.4.

A function e : R+ ^ R+ is a divergence function for the geodesic metric space X if for all x G X, all R G R+ and all geodesics y = [xy], y' = [xz], e satisfies the following condition: if r > 0, R + r < min {d(x, y), d(x,z)}, d(y(R),y'(R)) > e(0) > 0 and a is a path in X \ B(x, R + r) from y(R + r) to y'(R + r), then L(a) > e(r). Let X be a geodesic metric space. The geodesics diverge in X if there is a divergence function e(r) such that lim e(r) = x>.

In [1, 32] the following result was proved.

Theorem 4.5.

A geodesic metric space X is hyperbolic if and only if geodesics diverge in X.

Theorem 4.6.

Let T be a rectangular tessellation of R2 ^ C with tiles parallel to the coordinate axes and with infinitely many vertical rays. For F any tile on the tessellation, if (i) or (ii) of Lemma 4.3 hold, then the 1 -skeleton of T is not hyperbolic.

Remark.

Theorem 3.6 shows that the existence of infinitely many vertical rays (or even infinitely many vertical lines) does not guarantee the non-hyperbolicity of a tessellation graph.

Proof. Seeking for a contradiction assume that the 1-skeleton G of the tessellation T is hyperbolic. Then, there exists a divergence function, e: R+ ^ R+.

Denote by {yk}k the vertical rays in G. Without loss of generality we can assume that Re yk increases with k and that limt^^ Im yk(t) = oo for every k. Let x e y0. Let n be a geodesic starting at x which is the union of horizontal and vertical displacements and such that n H yk = 0 for every k > 0 (recall Lemma 4.3). Denote by nk the segment of n which starts at the point x and finishes at the first point zk of yk. Fix n to be so that dG(y0, yn) > e(0). Let R be such that nn(R) = zn; then dc(Y0(R),nn(R)) > dG(Y0,nn) > e(0).

Consider a new geodesic j which starts at y0(R) and which is the union of horizontal and vertical displacements, and such that j H Yn = 0; let us fix wn e j H Yn and let jn be the segment of j which finishes at wn e Yn. Denote by nn1 and jn1 the vertical rays starting at zn and wn, respectively. The curve r given by the geodesic segments [xy0(R)]U jn U jn1 is a geodesic;similarly, the curve r2 defined as r2 = nn U nn1 is also a geodesic. Note that if wn = r1 (t0), then r (t) = r2(t) for every t > t0, and dG(H(R), r2(R)) > e(0). This contradicts the hyperbolicity assumption. □

5. Tessellations with convex tiles

In [9], the authors conjectured that every tessellation gra ph of R2 with convex tiles is non-hyperbolic. Our next result shows that in order to prove this conjecture, it suffices to consider tessellation graphs of R2 with triangular tiles.

Theorem 5.1.

All tessellation graphs of R2 whose tiles are convex polygons are non-hyperbolic if and only if all tessellation graphs of R2 whose tiles are triangles are non-hyperbolic.

Proof. Let G be a tessellation graph of R2 whose tiles are convex polygons, and consider its tiles Fn. If supn diamR2 dFn = oo, then G is non-hyperbolic and the conclusion holds. Therefore, assume that c = supn diamR2 dFn < to. For each n, let Pn1 and Pn2 be two vertices of Fn accomplishing the maximum Euclidean distance between the vertices of Fn. Let us consider a new tessellation graph of R2, G', obtained from G by adding in each tile Fn new edges which join each vertex of Fn with Pn1 by the Euclidean segment between them. That is, all the tiles of G' are triangles and therefore, by hypothesis, G' is non-hyperbolic. We shall show that the inclusion i: G —> G' is a c-full (1 + n/2, 0)-quasi-isometry and, therefore, by Theorem 2.1, G will also be non-hyperbolic.

Let us consider a tile Fn and its corresponding vertices Pn1 and Pn2. Then Fn is contained in the closure of the Euclidean circle with center Pn2 and radius equal to the Euclidean distance between Pn1 and Pn2. Without loss of generality one can assume that Pn 2 is the origin of coordinates and Pn1 is the point with coordinates (1, 0). Let Pn be a point of dFn; since Fn is a convex polygon, Pn is contained in the right half-plane, i.e., if (r, 9) are the polar coordinates of Pn, then 0 < r < 1 and —n/2 < 9 < n/2. Let P'n be the projection of Pn over the circumference {x2 + y2 = 1}. The goal is to compare the Euclidean distance between Pn and Pn1 and the sum of the Euclidean distance between Pn and P'n plus the length of the arc of the circumference {x2 + y2 = 1} between P'n and Pn1. To this end, one needs to bound the function

r, ™ 9 + 1 — r 9 + 1 — r n „ n

f (r,9) = --- = , == , 0 < r < 1, — < 9 <- .

( ) \1 — re91 V1+r2-2Fcose ~ ~ 2~ ~2

Let us consider the functions

92 (1 — r)2

91 (r, 9) = 1+ r2 — 2r cos 9 , g2(r, 9) = 1+ r? — 2rcos 9

For fixed 9, the function g(r) = 1 + r2 — 2r cos 9 attains its minimum value when r = cos 9, therefore g^ (r, 9) < 92/sin2 9. Since the ratio 9/sin 9 increases for 9 e [0,n/2], g1(r,9) < tt2/4 for 0 < r < 1, —n/2 < 9 < n/2. Also, since —2r < —2r cos 9, g2(r, 9) < 1, 0 < r < 1, —n/2 < 9 < n/2. Therefore it follows

sup f(r,9) < 1 + n.

0<r<1 2

—n/2<9<n/2

The tile Fn is convex, thus

ddp„ (Pn.Pni) < & + 1 - r < (1+^1 Vi + r2 - 2r cos (

1 + 2 dR2(Pn, Pnl) = 1 + 2 I dc,(Pn, Pn,i)- (4)

For any points P, Q on the graph G, let us consider a geodesic y In G' joining P and Q. Let y„ = y n F'n, where F'n Is the subgraph of G' obtained by adding to dFn the new edges joining the corresponding point Pn1 with the other vertices of dFn. If yn is contained in dFn, then the length of yn in G' coincides with its length in G. If yn is not contained in dFn, then yn = y'n U y" where y'n = yn n dFn, y"n = yn \ y'n. Note that the closure of y" is connected and its endpoints are vertices in dFn n V(G). Let an be a geodesic in G joining the endpoints of y"n; since Fn is convex, an is contained in dFn. From (4) one gets

de (P, Q) = L(y) = £ L(Yn) = £ [L(Y'n ) + L№)\ > £

L(Y'n ) + 1 +

1 + 2 I dc (P, Q).

In any case one concludes that (1 + n/2)-1dG (P, Q) < de (P, Q) < de (P, Q), which means that the Inclusion i : G —> G' Is a (1 + n/2, 0)-quasi-isometric embedding. It is clear that i is c-full, with c = supn diamR2 dFn < to. By hypothesis the graph G' is non-hyperbolic, and by Theorem 2.1 it follows that G is also non-hyperbolic. □

Acknowledgements

Research is supported in part by two grants from Ministerio de Ciencia e Innovación (MTM 2009-07800 and MTM 2008-02829-E), Spain. J. Rodríguez is also supported in part by a grant from CONACYT (CONACYT-UAG I0110/62/10), Mexico.

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