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Volume 15, Issue 1 (Jun 2017)

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The hyperbolicity constant of infinite circulant graphs

José M. Rodríguez
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  • Departamento de Matemáticas, Universidad Carlos III de Madrid, Avenida de la Universidad 30, 28911 Leganés, Madrid, Spain
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/ José M. Sigarreta
  • Facultad de Matemáticas, Universidad Autónoma de Guerrero, Carlos E. Adame No.54 Col. Garita, 39650 Acalpulco Gro., Mexico
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Published Online: 2017-06-09 | DOI: https://doi.org/10.1515/math-2017-0061

Abstract

If X is a geodesic metric space and x1, x2, x3X, a geodesic triangle T = {x1, x2, x3} is the union of the three geodesics [x1x2], [x2x3] and [x3x1] in X. The space X is δ-hyperbolic (in the Gromov sense) if any side of T is contained in a δ-neighborhood of the union of the two other sides, for every geodesic triangle T in X. Deciding whether or not a graph is hyperbolic is usually very difficult; therefore, it is interesting to find classes of graphs which are hyperbolic. A graph is circulant if it has a cyclic group of automorphisms that includes an automorphism taking any vertex to any other vertex. In this paper we prove that infinite circulant graphs and their complements are hyperbolic. Furthermore, we obtain several sharp inequalities for the hyperbolicity constant of a large class of infinite circulant graphs and the precise value of the hyperbolicity constant of many circulant graphs. Besides, we give sharp bounds for the hyperbolicity constant of the complement of every infinite circulant graph.

MSC 2010: 05C75; 05C12; 05A20

Keywords: Circulant graph; Gromov hyperbolicity; Geodesics; Infinite graphs

1 Introduction

Hyperbolic spaces play an important role in geometric group theory and in the geometry of negatively curved spaces (see [13]). The concept of Gromov hyperbolicity grasps the essence of negatively curved spaces like the classical hyperbolic space, simply connected Riemannian manifolds of negative sectional curvature bounded away from 0, 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 a rich general theory (see [13]).

The first works on Gromov hyperbolic spaces deal with finitely generated groups (see [3]). Initially, Gromov spaces were applied to the study of automatic groups in the science of computation (see, e.g., [4]); indeed, hyperbolic groups are strongly geodesically automatic, i.e., there is an automatic structure on the group [5].

The concept of hyperbolicity appears also in discrete mathematics, algorithms and networking. For example, it has been shown empirically in [6] that the internet topology embeds with better accuracy into a hyperbolic space than into a Euclidean space of comparable dimension (formal proofs that the distortion is related to the hyperbolicity can be found in [7]); furthermore, it is evidenced that many real networks are hyperbolic (see, e.g., [812]). A few algorithmic problems in hyperbolic spaces and hyperbolic graphs have been considered in recent papers (see [1316]). Another important application of these spaces is the study of the spread of viruses through the internet (see [17, 18]). Furthermore, hyperbolic spaces are useful in secure transmission of information on the network (see [1719]); also to traffic flow and effective resistance of networks [2022].

In [23] it was proved the equivalence of the hyperbolicity of many negatively curved surfaces and the hyperbolicity of a graph related to it; hence, it is useful to know hyperbolicity criteria for graphs from a geometrical viewpoint. In recent years, the study of mathematical properties of Gromov hyperbolic spaces has become a topic of increasing interest in graph theory and its applications (see, e.g., [11, 1719, 2340] and the references therein).

If (X; d) is a metric space and γ : [a, b] → X is a continuous function, we define the length of γ as L(γ):=sup{i=1nd(γ(ti1),γ(ti)):a=t0<t1<<tn=b}.

We say that a curve γ : [a, b] → X is a geodesic if we have L(γ|[t, s]) = d(γ(t),γ(s)) = |ts| for every s, t ∈ [a, b], where L and d denote length and distance, respectively, and γ|[t, s] is the restriction of the curve γ to the interval [t, s] (then γ is equipped with an arc-length parametrization). The metric space X is said geodesic if for every couple of points in X there exists a geodesic joining them; we denote by [xy] any geodesic joining x and y; this notation is ambiguous, since in general we do not have uniqueness of geodesics, but it is very convenient. Consequently, any geodesic metric space is connected. If the metric space X is a graph, then the edge joining the vertices u and v will be denoted by [u, v].

Along the paper we just consider graphs with every edge of length 1. In order to consider a graph G as a geodesic metric space, identify (by an isometry) any edge [u, v] ∈ E(G) with the interval [0,1] in the real line; then the edge [u, v] (considered as a graph with just one edge) is isometric to the interval [0,1]. Thus, the points in G are the vertices and, also, the points in the interior of any edge of G. In this way, any connected graph G has a natural distance defined on its points, induced by taking shortest paths in G, and we can see G as a metric graph. If x, y are in different connected components of G, we define dG(x, y) = ∞. Throughout this paper, G = (V, E) = (V(G),E(G)) denotes a simple (without loops and multiple edges) graph (not necessarily connected) such that every edge has length 1 and V ≠ ∅. These properties guarantee that any connected component of any graph is a geodesic metric space. Note that to exclude multiple edges and loops is not an important loss of generality, since [27, Theorems 8 and 10] reduce the problem of compute the hyperbolicity constant of graphs with multiple edges and/or loops to the study of simple graphs.

If X is a geodesic metric space and x1,x2,x3X, the union of three geodesics [x1x2], [x2x3] and [x3x1] is a geodesic triangle that will be denoted by T = {x1,x2,x3} and we will say that x1,x2 and x3 are the vertices of T; it is usual to write also T = {[x1x2],[x2x3],[x3x1]}. We say that T is δ-thin if any side of T is contained in the δ-neighborhood of the union of the two other sides. We denote by δ(T) the sharp thin constant of T, i.e. δ(T) := inf{δ ≥ 0 : T is δ-thin}. The space X is δ-hyperbolic (or satisfies the Rips condition with constant δ) if every geodesic triangle in X is δ-thin. We denote by δ(X) the sharp hyperbolicity constant of X, i.e. δ(X) := sup{δ(T) : T is a geodesic triangle in X}. We say that X is hyperbolic if X is δ-hyperbolic for some δ ≥ 0; then X is hyperbolic if and only if δ(X) < ∞. If X has connected components {Xi}iI, then we define δ(X) := supiIδ(Xi), and we say that X is hyperbolic if δ(X) < ∞.

If we have a triangle with two identical vertices, we call it a bigon; note that since this is a special case of the definition, every geodesic bigon in a δ-hyperbolic space is δ-thin.

In the classical references on this subject (see, e.g., [1, 2, 41]) appear several different definitions of Gromov hyperbolicity, which are equivalent in the sense that if X is δ-hyperbolic with respect to one definition, then it is δ′-hyperbolic with respect to another definition (for some δ′ related to δ). The definition that we have chosen has a deep geometric meaning (see, e.g., [2]).

Trivially, any bounded metric space X is ((diam X)/2)-hyperbolic. A normed linear space is hyperbolic if and only if it has dimension one. A geodesic space is 0-hyperbolic if and only if it is a metric tree. If a complete Riemannian manifold is simply connected and its sectional curvatures satisfy Kc for some negative constant c, then it is hyperbolic. See the classical references [1, 2, 41] in order to find further results.

We want to remark that the main examples of hyperbolic graphs are the trees. In fact, the hyperbolicity constant of a geodesic metric space can be viewed as a measure of how “tree-like” the space is, since those spaces X with δ(X) = 0 are precisely the metric trees. This is an interesting subject since, in many applications, one finds that the borderline between tractable and intractable cases may be the tree-like degree of the structure to be dealt with (see, e.g., [42]).

A graph is circulant if it has a cyclic group of automorphisms that includes an automorphism taking any vertex to any other vertex. There are large classes of circulant graphs. For instance, every cycle graph, complete graph, crown graph and Möbius ladder is a circulant graph. A complete bipartite graph is a circulant graph if and only if it has the same number of vertices on both sides of its bipartition. A connected finite graph is circulant if and only if it is the Cayley graph of a cyclic group, see [43]. Every circulant graph is a vertex transitive graph and a Cayley graph [44]. It should be noted that Paley graphs is an important class of circulant graph, which is attracting great interest in recent years (see, e.g., [45]).

The circulant is a natural generalization of the double loop network and was first considered by Wong and Coppersmith [46]. Circulant graphs are interesting by the role they play in the design of networks. In the area of computer networks, the standard topology is that of a ring network; that is, a cycle in graph theoretic terms. However, cycles have relatively large diameter, and in an attempt to reduce the diameter by adding edges, we wish to retain certain properties. In particular, we would like to retain maximum connectivity and vertex-transitivity. Hence, most of the earlier research concentrated on using the circulant graphs to build interconnection networks for distributed and parallel systems [47], [48]. The term circulant comes from the nature of its adjacency matrix. A matrix is circulant if all its rows are periodic rotations of the first one. Circulant matrices have been employed for designing binary codes [49]. Theoretical properties of circulant graphs have been studied extensively and surveyed [47].

For a finite graph with n vertices it is possible to compute δ(G) in time O(n3.69) [50] (this is improved in [10, 51]). Given a Cayley graph (of a presentation with solvable word problem) there is an algorithm which allows to decide if it is hyperbolic [52]. However, deciding whether or not a general infinite graph is hyperbolic is usually very difficult. Therefore, it is interesting to relate hyperbolicity with other properties of graphs. The papers [24, 29, 40] prove, respectively, that chordal, k-chordal and edge-chordal are hyperbolic. Moreover, in [24] it is shown that hyperbolic graphs are path-chordal graphs. These results relating chordality and hyperbolicity are improved in [33]. In the same line, many researches have studied the hyperbolicity of other classes of graphs: complement of graphs [53], vertex-symmetric graphs [54], line graphs [55], bipartite and intersection graphs [56], bridged graphs [32], expanders [22], Cartesian product graphs [57], cubic graphs [58], and random graphs [3739].

In this paper we prove that infinite circulant graphs and their complements are hyperbolic (see Theorems 2.3, 2.4 and 3.15). We obtain in Theorems 3.7 and 3.8 several sharp inequalities for the hyperbolicity constant of a large class of infinite circulant graphs, and the precise value of the hyperbolicity constant of many circulant graphs. Besides, Theorem 3.15 provides sharp bounds for the hyperbolicity constant of the complement of every infinite circulant graph.

2 Every circulant graph is hyperbolic

Let (X, dX) and (Y, dY) be two metric spaces. A map f : XY is said to be an (α, β)-quasi-isometric embedding, with constants α ≥ 1, β ≥ 0 if for every x, yX: α1dX(x,y)βdY(f(x),f(y))αdX(x,y)+β.

The function f is ε-full if for each yY there exists xX with dY(f(x),y) ≤ ε.

A map f : XY is said to be a quasi-isometry, if there exist constants α ≥ 1, β, ε ≥ 0 such that f is an ε-full (α, β)-quasi-isometric embedding.

A fundamental property of hyperbolic spaces is the following:

Theorem 2.1

(Invariance of hyperbolicity). Let f : XY be an (α, β)-quasi-isometric embedding between the geodesic metric spaces X and Y. If Y is δ-hyperbolic, then X is δ′-hyperbolic, where δis a constant which just depends on α, β, δ.

Besides, if f is ε-full for some ε ≥ 0 (a quasi-isometry) and X is δ-hyperbolic, then Y is δ′-hyperbolic, where δis a constant which just depends on α, β, δ, ε.

We denote by Aut(G) the set of automorphisms of the graph G (isomorphisms of G onto itself). If g ∈ Aut(G) we will denote by 〈g〉 the cyclic subgroup of Aut(G) generated by g.

Definition 2.2

Let G be any circulant connected infinite graph and g ∈ Aut(G) withg〉 = Aut (G). Consider the graph G*with V(G*) := V(G) and E(G*) := {[gn(v0), gn + 1(v0)]}nZ for some fixed v0V(G). We define n(G,g):=dG(v0,g(v0)),N(G,g):=max{dG(v0,w)|[v0,w]E(G)}.

Note that the definition of n(G, g) and N(G, g) do not depend on the choice of v0, since G is a circulant graph. Hence, for every vV(G), n(G,g)=dG(v,g(v))=dG(v,g1(v)),N(G,g)=max{dG(v,w)|[v,w]E(G)}.

Theorem 2.3

Any circulant connected infinite graph G satisfies the inequality δ(G) ≤ c, where c is a constant which just depends on n(G, g) and N(G, g).

Proof

For each n ∈ ℤ, let wn be the midpoint of the edge [gn(v0),gn + 1(v0)] ∈ E(G*). Let us define a map i : G*G as follows: for each n ∈ ℤ, let i(u) := gn(v0) for every u ∈ [wn − 1wn] ∖ {wn − 1}. Note that i is the inclusion map on V(G*) = V(G); hence, i is (1/2)-full.

Fix u, vV(G) = V(G*). Let u0 = u, u1,⋯,uk − 1,uk = vV(G) with dG(u,v)=j=1kdG(uj1,uj) and [uj1,uj]E(G) for every 1 ≤ jk; then we have dG(u,v)j=1kdG(uj1,uj)j=1kN(G,g)dG(uj1,uj)=N(G,g)dG(u,v).

Fix u, vG* and a geodesic [uv] in G*. Recall that i(u),i(v) ∈ V(G) = V(G*). We have dG(u,v)dG(u,i(u))+dG(i(u),i(v))+dG(i(v),v)1/2+N(G,g)dG(i(u),i(v))+1/2=N(G,g)dG(i(u),i(v))+1,1N(G,g)dG(u,v)dG(i(u),i(v))+1N(G,g)dG(i(u),i(v))+1.

Fix u,vV(G)=V(G). Let v0=u,v1,,vr1,vr=vV(G) with dG(u,v)=j=1rdG(vj1,vj) and [vj − 1,vj] ∈ E(G*); then vj = gi(vj − 1) for some i ∈ {1,−1} and we have dG(u,v)j=1rdG(vj1,vj)=j=1rn(G,g)dG(vj1,vj)=n(G,g)dG(u,v).

Fix u, vG*. We have dG(i(u),i(v))n(G,g)dG(i(u),i(v))n(G,g)(dG(i(u),u)+dG(u,v)+dG(v,i(v)))n(G,g)(1/2+dG(u,v)+1/2)=n(G,g)dG(u,v)+n(G,g).

We conclude that i is a (1/2)-full (max{N(G, g),n(G, g)},n(G, g))-quasi-isometry and Theorem 2.1 gives the result, since G* is 0-hyperbolic. □

Theorem 2.4

Every circulant graph is hyperbolic.

Proof

Let us consider any fixed circulant graph G. If G is a finite graph, then it is (diam G)-hyperbolic. Assume now that G is an infinite graph.

If G is connected, then Theorem 2.3 gives that it is hyperbolic.

If G is not connected, then it has just a finite number of isomorphic connected components G1, …, Gr, since G is a circulant graph; therefore, δ(G) = max{δ(G1), … δ(Gr)} = δ(G1). Since G1 is connected and circulant, Theorem 2.3 gives the result. □

3 Bounds for the hyperbolicity constant of infinite circulant graphs

Let {a1,a2,…,ak} be a set of integers such that 0 < a1 < ⋯ < ak. We define the circulant graph C(a1,…,ak) as the infinite graph with vertices ℤ and such that N(j)={j±ai}i=1k is the set of neighbors of each vertex j ∈ ℤ. Then C(a1,…,ak) is a regular graph of degree 2k. If k = 1, then C(1) is isometric to the Cayley graph of ℤ, which is 0-hyperbolic. Hence, in what follows we just consider circulant graphs with k > 1.

Denote by J(G) the set of vertices and midpoints of edges in G, and by ⌊t⌋ the lower integer part of t.

The following results in [25] will be useful.

Theorem 3.1

([25, Theorem 2.6]). For every hyperbolic graph G, δ(G) is a multiple of 1/4.

As usual, by cycle we mean a simple closed curve, i.e., a path with different vertices, unless the last one, which is equal to the first vertex.

Theorem 3.2

([25, Theorem 2.7]). For any hyperbolic graph G, there exists a geodesic triangle T = {x, y, z} that is a cycle with x, y, zJ(G) and δ(T) = δ(G).

We also need the following technical lemmas.

Lemma 3.3

For any integers k > 1 and 1 < a2 < ⋯ < ak, consider G = C(1,a2,…,ak). Then the following statements are equivalent:

  1. dG(0,⌊ak/2⌋) = ⌊ak/2⌋ and, if ak is odd, then dG(0,⌊ak/2⌋+1) ≥ ⌊ak/2⌋.

  2. a2ak − 1.

  3. We have either k = 2 or k = 3 and a2 = a3 − 1.

Proof

Assume that (1) holds. If k = 2, then (2) holds; hence, we can assume k ≥ 3. Define r := ⌊ak/2⌋. If a2r, then r = dG(0,r) by hypothesis and r = dG(0,r) ≤ dG(0,a2) + dG(a2,r) ≤ 1 + ra2 < r, which is a contradiction. Thus we conclude a2 > r. Therefore, r=dG(0,r)dG(0,a2)+dG(a2,r)1+a2r, and a2 ≥ 2r − 1. Hence, a2ak − 1 if ak is even. Since a2 > r, if ak is odd, then rdG(0,r+1)dG(0,a2)+dG(a2,r+1)1+a2(r+1), and a2 ≥ 2r = ak − 1. Then (2) holds.

A simple computation provides the converse implication.

The equivalence of (2) and (3) is elementary.□

Let us define the subset 𝓔 of infinite circulant graphs as 𝓔:= {C(1,2m + 1,2m + 2,2m + 3)}m ≥ 1.

Lemma 3.4

Consider any integers k > 1 and 1 < a2 < ⋯ < ak with a2 < ak − 1, and G = C(1,a2,…,ak) ∉ 𝓔. Then min{dG(0,u),dG(aj,u)}ak21,(1) for every u ∈ ℤ with 0 ≤ uaj and 1 ≤ jk.

Proof

Since a2 < ak − 1, Lemma 3.3 gives:

  1. if ak is even, then dG(0,⌊ak/2⌋) < ⌊ak/2⌋,

  2. if ak is odd, then dG(0,⌊ak/2⌋) < ⌊ak/2⌋ or dG(0,⌊ak/2⌋ + 1) < ⌊ak/2⌋.

If dG(0,⌊ak/2⌋) < ⌊ak/2⌋, then inequality (1) trivially holds. Hence, we can assume that ak is odd, dG(0,⌊ak/2⌋) = ⌊ak/2⌋ and dG(0,⌊ak/2⌋ + 1) < ⌊ak/2⌋. These facts imply that dG(0,⌊ak/2⌋ + 1) = ⌊ak/2⌋ − 1.

If a2 ≤ ⌊ak/2⌋, then dG(0,⌊ak/2⌋) < ⌊ak/2⌋, which is a contradiction. Therefore, a2 > ⌊ak/2⌋ and ak21=dG(0,ak2+1)1+a2ak21,ak2a2<ak1.

We conclude a2 = ak − 2, and k = 3 or k = 4. If k = 4, then a2 = a4 − 2, a3 = a4 − 1 and G ∈ 𝓔, since ak is odd. This is a contradiction, and we conclude k = 3 and a2 = a3 − 2.

If j = 1, then min{dG(0,u),dG(1,u)} = 0 for every 0 ≤ u ≤ 1.

If j = 2, then for every 0 ≤ ua2 min{dG(0,u),dG(a2,u)}min{u,a2u}a22=a321.

If j = 3, then dG(⌊a3/2⌋,a3) = dG(0,⌊a3/2⌋ + 1) = ⌊a3/2⌋ − 1, and for every 0 ≤ ua3 with u ≠ ⌊a3/2⌋,⌊a3/2⌋ + 1, min{dG(0,u),dG(a3,u)}min{u,a3u}a321.

This finishes the proof. □

Proposition 3.5

Consider any integers k > 1 and 1 < a2 < ⋯ < ak. If we have either k = 2, or k = 3 and a2 = a3−1, or C (1, a2, …, ak) ∈ 𝓔, then δ(C(1,a2,,ak))12+ak2.

Proof

Define r := ⌊ ak/2⌋ and G := C (1, a2, …, ak).

Assume first that we have either k = 2, or k = 3 and a2 = a3−1. Consider the curves γ1γ2 in G joining x := 0 and y := r + (r + 1)ak given by γ1:=[0,1][1,2][r1,r][r,r+ak][r+ak,r+2ak][r+rak,r+(r+1)ak],γ2:=[0,ak][ak,2ak][rak,(r+1)ak][(r+1)ak,(r+1)ak+1][(r+1)ak+1,(r+1)ak+2][r+(r+1)ak1,r+(r+1)ak].

Lemma 3.3 gives that γ1 and γ2 are geodesics; then dG(x, y) = L(γ1) = L(γ2) = 2r+1. Let T be the geodesic bigon T = {γ1, γ2} in G. If p is the midpoint of [r, r + ak], then dG(p, x) = dG(p, y) = r + 1/2 and Lemma 3.3 gives that dG(p, γ2) = r + 1/2. Hence, δ(C(1,a2,,ak))dG(p,γ2)=12+ak2. Assume now that C (1, a2, …, ak) ∈ 𝓔. Note that r + rak = (r+1) ak−1, since ak = ak−1 + 1 is odd. Consider the curves γ1,γ2, γ3 in G γ1:=[rrak,r(r1)ak][r2ak,rak][rak,r][r,r][r,r+ak][r+ak,r+2ak][r+(r1)ak,r+rak],γ2:=[(r+1)ak1,rak1][2ak1,ak1][ak1,0],γ3:=[0,ak1][ak1,2ak1][rak1,(r+1)ak1], joining x := −(r+1)ak−1, y := (r+1)ak−1 and z := 0. One can check that γ1, γ2 and γ3 are geodesics in G. Let T be the geodesic triangle T = {γ1, γ2, γ3}. Note that dG(0, r) = r, since G ∈ 𝓔. Therefore, if p is the midpoint of [−r, r], then δ(G)dG(p,γ2γ3)=12+dG(r,γ3)=12+dG(r,{0,2r})=12+r=12+ak2. □

Let G = C (1, a2, …, ak). If x∈ V(G), we define x1 := x2 := x; if xG \ V(G), we define x1 and x2 as the endpoints of the edge containing x with x1 < x2. Therefore, 1 ≤ x2x1ak if xG \ V(G). If x, yJ(G), we say that xL y if x = y or x2y1, and x and y are related (and we write xR y) if xL y or yL x. Note that L is an order relation on J(G).

Lemma 3.6

Consider any integers k > 1 and 1 < a2 < ⋯ < ak. If x, yJ(C(1, a2, …, ak)) and x and y are not related, then dG(x,y)1+ak2. Furthermore, if a2 < ak−1 and C (1, a2, …, ak)∉ 𝓔, then dG(x,y)ak2.

Proof

Let G = C (1, a2, …, ak). If x1y1y2x2, then the cycle σ:=[x1,x1+1][x1+1,x1+2][y11,y1][y1,y2][y2,y2+1][x22,x21][x21,x2][x2,x1] has length at most 1 + ak. Since σ is a cycle containing the points x and y, we have dG(x,y)12L(σ)1+ak21+ak2. If a2 < ak − 1 and G ∉ 𝓔, then Lemma 3.4 gives dG(x, y1) ≤ 1/2 + ⌊ ak/2⌋−1=⌊ak/2⌋−1/2 and dG(x, y) ≤ ⌊ ak/2⌋.

If y1x1x2y2, then the same argument gives these inequalities.

If x1y1 < x2y2, then x2y1+y1x1ak and min{x2y1, y1x1} ≤ ⌊ ak/2⌋. If x2y1 ≤ ⌊ ak/2⌋, then dG(x, y) ≤ dG(x, x2)+x2y1+dG(y1, y) ≤ 1+⌊ ak/2⌋. If y1x1 ≤ ⌊ ak/2⌋, then dG(x, y) ≤ dG(x, x1)+y1x1+dG(y1, y) ≤ 1+⌊ ak/2⌋. If a2 < ak−1 and C(1, a2, …, ak)∉ 𝓔, then Lemma 3.4 also gives dG(x, y1) ≤ ⌊ ak/2⌋−1/2 and dG(x, y) ≤ ⌊ ak/2⌋.

If y1x1 < y2x2, then the same argument gives these inequalities. □

We can state now the main result of this section, which provides a sharp upper bound for the hyperbolicity constant of infinite circulant graphs.

Theorem 3.7

For any integers k > 1 and 1 < a2 < ⋯ < ak, δ(C(1,a2,,ak))12+ak2, and the equality is attained if and only if we have either k = 2, or k = 3 and a2 = a3−1, or k = 4, a2 = a4−2, a3 = a4−1 and a4 is odd.

Proof

In order to bound δ(G) with G = C(1, a2, …, ak), let us consider a geodesic triangle T = {x, y, z} in G and p ∈ [xy]; by Theorem 3.2 we can assume that T is a cycle with x, y, zJ(G). We consider several cases.

  • (A)

    If x and y are not related, then Lemma 3.6 gives dG(p,[xz][yz])dG(p,{x,y})12dG(x,y)12+12ak2.(2)

  • (B)

    Assume that xR y. Without loss of generality we can assume that xL y. Denote by w1, …, wm the vertices in [xy] such that w1 ∈ {x1, x2}, wm ∈ {y1, y2} and dG(wj, wj+1) = 1 for every 1 ≤ j < m; we define i0:=min{1im|wjx2ijm},x0:=wi0.

(B.1) Assume that either xV(G) or x is the midpoint of an edge [r, r +1]∈ E(G) for some r ∈ ℤ. If x0 = x2, then L([xx0]) ≤ 1/2. If x0>x2, then the cycle σ:=[wi0,wi01][wi01,wi01+1][wi01+1,wi01+2][wi02,wi01][wi01,wi0] has length at most 1 + ak. Since xV(G) or x is the midpoint of an edge [r, r + 1], thus σ is a cycle containing the points x and x0, and we have L([xx0])=dG(x,x0)12L(σ)1+ak2<32+ak2.

(B.2) Assume now that x is the midpoint of an edge [r, r + aj]∈ E(G) for some r ∈ ℤ and 1 < jk. If x0 = x2, then L([xx0]) = 1/2. If wi0−1 = x1, then L([xx0]) = 3/2. If x0 > x2 and wi0−1x1, then we have either wi0−1 < x1 < x2 < wi0 or x1 < wi0−1 < x2 < wi0. In the first case the cycle σ:=[wi0,wi01][wi01,wi01+1][x11,x1][x1,x2][x2,x2+1][wi01,wi0] has length at most ak. Since σ is a cycle containing the points x and x0, we have L([xx0])=dG(x,x0)12L(σ)ak2. Assume that x1 < wi0−1 < x2 < wi0. Then L([xx0])=dG(x,x0)12+min{dG(x2,wi01),dG(wi01,x1)}+132+min{x2wi01,wi01x1}32+ak2. Therefore, in Case (B), if p ∈[xx0]\ B(x0, 3/4), then dG(p,[xz][yz])dG(p,x)L([xx0]B(x0,3/4))=L([xx0])3434+ak2.(3)

Define j0:=max{1im|wjy11ji},y0:=wj0. A similar argument to the previous one shows that if p ∈[y0y]\ B(y0, 3/4), then dG(p,[xz][yz])34+ak2.(4)

Since T is a continuous curve, we have ([xz][yz]){x1,x2},([xz][yz]){y1,y2}.

(a) Assume that y0 ∈ [xx0]\ {x0}.

(a.1) If dG(x0, y0)≥2, then L([xy])=L([xx0])+L([y0y])L([y0x0])32+ak2+32+ak22=1+2ak2,dG(p,[xz][yz])dG(p,{x,y})12L([xy])12+ak2.

(a.2) If dG(x0, y0) = 1, then wj0 = y0 = wi0−1, wi0 = x0 = wj0+1 and the definitions of x0 and y0 give y0 < x2x0 and y0y1 < x0. Since [x0, y0] ∈ E(G), dG(x2,y1)1+ak2,L([xy])=dG(x,y)dG(x,x2)+dG(x2,y1)+dG(y1,y)12+1+ak2+12=3+ak2,dG(p,[xz][yz])dG(p,{x,y})12L([xy])3+ak412+ak2.

(b) Assume that y0∉[xx0]\ {x0}. Since x2x0, y0y1 and [xz] ∪ [yz] is a continuous curve joining x and y, if pV(G)∩[x0y0]⊂[xy], then there exist u, vV(G)∩([xz]∪[yz]) with [u, v] ∈ E(G) and upv. Since T is a cycle, we have dG(p,[xz][yz])dG(p,{u,v})min{pu,vp}ak2. Therefore, if p ∈[x0y0]∪ B(x0, 3/4)∪ B(y0, 3/4), then dG(p,[xz][yz])<34+ak2.

This inequality, (2), (3) and (4) give δ(C(1,a2,,ak))34+ak2.(5)

By Theorem 3.1, in order to finish the proof it suffices to show that the equality in (5) is not attained. Seeking for a contradiction, assume that the equality is attained. The proof of (5) gives that we have dG(p,[xz][yz])=34+ak2, where p is the point in [xx0] with dG(p, x0) = 3/4 or the point in [y0y] with dG(p, y0) = 3/4. By symmetry, without loss of generality we can assume that p is the point in [xx0] with dG(p, x0) = 3/4; thus dG(p, wi0−1) = 1/4. Therefore, we are in Case (B.2) with x1 < wi0−1 < x2 < wi0 and L([xx0])=32+min{dG(x2,wi01),dG(wi01,x1)}=32+min{x2wi01,wi01x1}=32+ak2. Then min{dG(x2,wi01),dG(wi01,x1)}=min{x2wi01,wi01x1}=ak2, and we conclude ak2x2wi01,wi01x1ak2+1.(6)

(I) Assume that x2∈[xx0]. Since T is a cycle, then x1∈[xz]∪[yz] and x1 < wi0−1. Hence, since [xz]∪[yz] is a continuous curve joining x and y, and dG(p, wi0−1) = 1/4, we obtain as above 34+ak2=dG(p,[xz][yz])dG(p,wi01)+dG(wi01,[xz][yz])14+ak2, which is a contradiction.

(II) Assume that x1∈[xx0].

(II. 1) If x2x1 = x0wi0−1 = ak, then wi0−1x1 = x0x2, dG(x1, wi0−1) = dG(x2, x0) and dG(x,x0)=dG(x,x1)+dG(x1,wi01)+dG(wi01,x0)>dG(x,x1)+dG(x1,wi01)=12+dG(x2,x0)=dG(x,x2)+dG(x2,x0)dG(x,x0), which is a contradiction.

(II.2) If x2x1 < ak, then (6) gives x2wi01=wi01x1=ak2, and 34+ak2=dG(p,[xz][yz])dG(p,x2)dG(p,wi01)+dG(wi01,x2)14+ak2, which is a contradiction.

(II.3) Assume x2x1 = ak and x0wi0−1 < ak. Since x1∈[xx0], we have wi0−1x1 = ⌊ ak/ 2 ⌋.

(II.3.1) If x2wi0−1 = ⌊ak/2⌋, then 34+ak2=dG(p,[xz][yz])dG(p,x2)dG(p,wi01)+dG(wi01,x2)14+ak2, which is a contradiction.

(II.3.2) If x2wi0−1 = ⌊ ak/2⌋+1, then ak = x2x1 = 2⌊ ak/2⌋+1 and dG(x0,x2)x0x2=x0wi01(x2wi01)<akak21=ak2,34+ak2=dG(p,[xz][yz])dG(p,x2)dG(p,x0)+dG(x0,x2)<34+ak2, which is a contradiction.

This finishes the proof of the inequality.

If we have either k = 2, or k = 3 and a2 = a3−1, or k = 4, a2 = a4−2, a3 = a4−1 and a4 is odd, then Proposition 3.5 gives δ(C(1,a2,,ak))12+ak2. Since we have the converse inequality, we conclude that the equality holds.

Assume now that the equality holds. Denote by G the circulant graph C (1, a2, …, ak). By Theorem 3.2 there exist a geodesic triangle T = {x, y, z} in G that is a cycle with x, y, zJ(G), and p ∈ [xy] with dG(p, [xz] ∪ [yz]) = 1/2+⌊ ak/2⌋. Hence, pJ(G).

Seeking for a contradiction, assume that a2 < ak−1 and that G ∉ 𝓔.

If x and y are not related, then Lemma 3.6 gives dG(x, y) ≤ ⌊ ak/2⌋. Since dG(x, y) = dG(x, p) + dG(p, y)≥ 1 + 2 ⌊ ak/2⌋, thus x and y are related, and without loss of generality we can assume xL y. Since dG(p, x), dG(p, y)≥1/2+⌊ ak/2⌋, the previous argument gives that x and p are related, and p and y are related.

(i) Assume first xL p and pL y.

Since p1x1p1x2dG(x2,p1)≥⌊ ak/2⌋−1/2, we have p1x1p1x2≥⌊ ak/2⌋, and there is some point u ∈([xz]∪[yz])∩ V(G) with up1. A similar argument gives that there is some point v ∈ ([xz]∪[yz])∩ V(G) with p2v. Denote by u and v any couple of vertices satisfying these properties. We are going to prove that vu > ak.

If pV(G), then min{pu,vp}min{dG(u,p),dG(p,v)}12+ak2,min{pu,vp}1+ak2,vu2+2ak2>ak. If p is the midpoint of some edge [m, m+aj] with m ∈ ℤ and 2 ≤ jk, then Lemma 3.4 gives dG(p, w) ≤ ⌊ ak/2⌋−1/2 < 1/2 + ⌊ ak/2⌋ for every w ∈ ℤ with mwm + aj, since a2 < ak−1 and G ∉ 𝓔. Furthermore, min{mu,vmaj}min{dG(u,m),dG(m+aj,v)}ak2,vuaj+2ak22+2ak2>ak.

Finally, assume that p is the midpoint of some edge [m, m+1] with m ∈ ℤ. By Lemma 3.3, we have dG(0, ⌊ ak/2⌋) ≤ ⌊ ak/2⌋−1 or dG(0, ⌊ ak/2⌋+1) ≤ ⌊ ak/2⌋−1. Therefore, dG(p,p±(ak2+12))ak212,min{pu,vp}32+ak2,vu3+2ak22+2ak2>ak.

Hence, we have proved vu >ak in every case.

Since if p is the midpoint of some edge [m, m +aj] with m ∈ ℤ and 2 ≤ jk then dG(p, w) < 1/2 + ⌊ ak/2 ⌋ for every w ∈ ℤ with mwm + aj, and [xz]∪[yz] is a continuous curve joining u and v, there exist u0, v0 ∈([xz]∪[yz])∩ V(G) with u0p1, p2v0 and [u0, v0]∈ E(G). Hence, akv0u0 > ak, which is a contradiction.

(ii) Assume now that pL x or yL p. By symmetry, we can assume that pL x and thus p ∈[xx0] and dG(p, x0)≥1. Since dG(x, x0) ≤ 3/2 + ⌊ ak/2⌋, we have 1/2+ ⌊ ak/2⌋=dG(p, [xz]∪[yz]) ≤ dG(x, p) ≤ 1/2 + ⌊ ak/2⌋. Therefore, dG(x, p) = 1/2 + ⌊ ak/2⌋, dG(p, x0) = 1, pV(G) and x is the midpoint of [x1, x2]∈ E(G). Thus dG(p, x1)≥⌊ ak/2⌋, dG(p, x2)≥⌊ ak/2⌋ and px1x2x0. By Lemma 3.3, we have dG(0, ⌊ ak/2⌋) ≤ ⌊ ak/2⌋−1 or dG(0, ⌊ ak/2⌋+1) ≤ ⌊ ak/2⌋−1.

If dG(0, ⌊ ak/2⌋) ≤ ⌊ ak/2⌋−1, then x1≥⌊ ak/2⌋+1, x2ak−(⌊ ak/2⌋−1) and 1x2x1ak(ak/21)ak/21=ak2ak/20, which is a contradiction.

If dG(0, ⌊ ak/2 ⌋ + 1) ≤ ⌊ ak/2⌋−1, then x2 ≤ ⌊ ak/2⌋, x1≥⌊ ak/2⌋ and 1 ≤ x2x1 ≤ ⌊ ak/2⌋ − ⌊ ak/2⌋ ≤ 0, which is a contradiction.

Therefore, we conclude in every case that a2ak−1 or G ∈ 𝓔. Hence, Lemma 3.3 gives k = 2, or k = 3 and a2 = a3−1, or k = 4, a2 = a4 − 2, a3 = a4 − 1 and a4 is odd. □

We also have a sharp lower bound for the hyperbolicity constant.

Theorem 3.8

For any integers k > 1 and 1 < a2 < ⋯ < ak we have δ(C(1,a2,,ak))32,

and the equality is attained if ak = k.

Proof

Define G = C (1, a2, …, ak), and consider the geodesics γ1, γ2 in G given by γ1:=[0,ak][ak,2ak][2ak,2ak+1],γ2:=[0,1][1,1+ak][1+ak,1+2ak].

Let T be the geodesic bigon T = {γ1, γ2} in G. If p is the midpoint of [ak, 2ak], then δ(G)dG(p,γ2)=min{dG(p,ak)+dG(ak,1+ak),dG(p,2ak)+dG(2ak,2ak+1),12L(γ1)}=32.

Assume now that ak = k, i.e., G = C (1, a2, …, ak) = C(1, 2, ⋯, k). Therefore, dG(m, m + w) = 1 for every m, w ∈ ℤ with |w| ≤ k. Note that if x, yJ(G) and x and y are not related, then |x1y1| < k, dG(x1, y1) = 1 and dG(x,y)dG(x,x1)+dG(x1,y1)+dG(y1,y)=2.(7)

Let us consider a geodesic triangle T = {x, y, z} in G and p ∈ [xy]; by Theorem 3.2 we can assume that T is a cycle with x, y, zJ(G). We consider several cases.

  • (A)

    If x and y are not related, then (7) gives dG(p,[xz][yz])dG(p,{x,y})12dG(x,y)1.

  • (B)

    Assume that xR y. Without loss of generality we can assume that xL y. Define wj, i0, x0 and y0 as in the proof of Theorem 3.7. Then wi0−1 < x2wi0 = x0.

If x0 = x2, then L([xx0]) ≤ 1/2. If x0 > x2, then wi0−1 < x2 < wi0. Since x2wi0wi0wi0 −1k, dG(x2, x0) = 1 and L([xx0])=dG(x,x0)dG(x,x2)+dG(x2,x0)12+1=32.

Therefore, in both cases, if p ∈ [xx0], then dG(p,[xz][yz])dG(p,x)L([xx0])32.

A similar argument to the previous one shows that if p ∈ [y0y], then dG(p, [xz] ∪ [yz]) ≤ 3/2. If y0 ∈ [xx0], then dG(p, [xz] ∪ [yz]) ≤ 3/2 holds for every p ∈ [xy]. Consider now the case y0 ∉ [xx0].

Since x2x0 and y0y1, every pV(G) ∩ [x0y0] ⊂ [xy] verifies x2py1. Since T is a continuous curve, we obtain ([xz][yz]){x1,x2},([xz][yz]){y1,y2}.

Since [xz] ∪ [yz] is a continuous curve joining x and y, if pV(G) ∩ [x0y0], then there exist u, vV(G) ∩ ([xz] ∪ [yz]) with [u, v] ∈ E(G) and upv. Hence, puvuk, vpvuk and dG(p,[xz][yz])dG(p,{u,v})=1.

Therefore, if p ∈ [x0y0], then dG(p,[xz][yz])32.

These inequalities give δ(T) ≤ 3/2 and, hence, δ(C(1,2,,k))32.

Since we have proved the converse inequality, we conclude that the equality holds. □

As usual, the complement G of the graph G is defined as the graph with V(G) = V(G) and such that eE(G) if and only if eE(G). We are going to bound the hyperbolicity constant of the complement of every infinite circulant graph. In order to do it, we need some preliminaries.

For any graph G, we define, diam V(G):=sup{dG(v,w)|v,wV(G)},diam G:=sup{dG(x,y)|x,yG}.

We need the following well-known result (see a proof, e.g., in [36, Theorem 8]).

Theorem 3.9

In any graph G the inequality δ(G) ≤ (diam G) /2 holds.

We have the following direct consequence.

Corollary 3.10

In any graph G the inequality δ(G) ≤ (diam V(G) + 1)/2 holds.

From [34, Proposition 5 and Theorem 7] we deduce the following result.

Lemma 3.11

Let G be any graph with a cycle g. If L(g) ≥ 3, then δ(G) ≥ 3/4. If L(g) ≥ 4, then δ(G) ≥ 1.

We say that a vertex v of a graph G is a cut-vertex if G ∖ {v} is not connected. A graph is two-connected if it does not contain cut-vertices.

We need the following result in [26, Proposition 4.5 and Theorem 4.14].

Theorem 3.12

Assume that G is a two-connected graph. Then G verifies δ(G) = 1 if and only if diam G = 2. Furthermore, δ(G) ≤ 1 if and only if diam G ≤ 2.

Definition 3.13

Let us consider integers k ≥ 1 and 1 ≤ a1 < a2 < ⋯ <ak. We say that {a1, a2, …, ak} is a 1-modulated sequence if for every x, y ∈ ℤ with |y| ∉ {a1, a2, …, ak} we have |x| ∉ {a1, a2, …, ak} or |xy| ∉ {a1, a2; ⋯,ak}.

We have the following sharp bounds for the hyperbolicity constant of the complement of every infinite circulant graph. In particular, they show that the complement of infinite circulant graphs are hyperbolic.

Theorem 3.14

For any integers k ≥ 1 and 1 ≤ a1 < a2 < ⋯ <ak we have 1δ(C(a1,a2,,ak)¯)32.

Furthermore, δ(C(a1,a2,,ak)¯)=1 if and only if {a1, a2, …, ak} is 1-modulated. If there is 1 ≤ j < ak/5 with j,5j ∉ {a1, a2, …, ak} and 2j,3j,4j ∈ {a1, a2, …, ak}, then δ(C(a1,a2,,ak)¯)=3/2.

Proof

Define G : = C (a1, a2, …, ak). Given u, v ∈ ℤ, consider w ∈ ℤ with w > u + ak and w > v + ak. Since [u, w],[v, w] ∉ E(G), we have [u, w],[v, w] ∈ E(G) and dG(u, v) ≤ dG(u, w) + dG(w, v) = 2. Hence, diam V(G) ≤ 2 and Corollary 3.10 gives δ(G) ≤ 3/2.

Since [0, ak + 1], [ak + 1, 2ak + 2], [2ak + 2, 3ak + 3], [3ak + 3, 0] ∉ E(G), we have [0, ak + 1], [ak + 1, 2ak + 2], [2ak + 2, 3ak + 3], [3ak + 3, 0]∈ E(G). Since the cycle C := {0, ak + 1, 2ak + 2, 3ak + 3, 0} in G has length 4, Lemma 3.11 gives δ (G) ≥ 1.

Since G is a circulant graph, the sequence {a1, a2, …, ak} is 1-modulated if and only if for every x, y1, y2 ∈ ℤ with |y2y1|∉{a1, a2, …, ak}, we have |xy1| ∉ {a1, a2, …, ak} or |xy2| ∉ {a1, a2, …, ak}. This happens if and only if dG(x, [y1, y2]) ≤ 1 for every xV(G) and [y1, y2]∈ E(G), and this condition is equivalent to diam V(G) ≤ 2. Since G is a two-connected graph, Theorem 3.12 that diam G ≤ 2 if and only if δ (G) ≤ 1. Since δ (G)≥1, we conclude that δ (G) = 1 if and only if {a1, a2, …, ak} is 1-modulated.

Assume that there is 1 ≤ j < ak/5 with j, 5j ∉ {a1, a2, …, ak} and 2j, 3j, 4j ∈ {a1, a2, …, ak}, and consider the cycle C := {0, j, 2j, 3j, 4j, 5j, 0} in G with length 6. Let x and y be the midpoints of the edges [2j, 3j] and [5j, 0], respectively. Since dG([2j, 3j], [5j, 0]) = 2, dG(x, y) = 3 and C contains two geodesics g1, g2 joining x and y, with g1V(G) = {0, j, 2j} and g2V(G) = {3j, 4j, 5j}. Since dG(j, {3j, 4j, 5j})≥ 2, we have δ (G)≥dG(j, g2) = dG(j, {x, y}) = 3/2, and we conclude δ (G) = 3/2. □

Since Paley graphs is an important class of circulant graph, which is attracting great interest in recent years (see, e.g., [45]), we finish this paper with a result on the hyperbolicity constant of Paley graphs.

Recall that the Paley graph of order q with q a prime power is a graph on q nodes, where two nodes are adjacent if their difference is a square in the finite field GF(q). This graph is circulant when q ≡ 1(mod 4). Paley graphs are self-complementary, strongly regular, conference graphs, and Hamiltonian.

Proposition 3.15

For any Paley graph G we have 1δ(G)32.

Proof

Let us denote by n the cardinality of V(G).

Since Paley graphs are self-complementary (the complement of any Paley graph is isomorphic to it), the degree of any vertex is (n − 1)/2. Hence, given u, vV(G) with [u, v] ∉ E(G), there exists a vertex wV(G) with dG(u, w) = dG(u, v) = 1, and we conclude that diam V(G) ≤ 2. Therefore, Corollary 3.10 gives δ(G) ≤ 3/2.

Since G is a Hamiltonian graph, there exists a Hamiltonian cycle g. Since L(g) = n ≥ 5, Lemma 3.11 gives δ(G) ≥ 1. □

Acknowledgement

The authors thank the referees for their deep revision of the manuscript. Their comments and suggestions have contributed to improve substantially the presentation of this work. This work is supported in part by two grants from Ministerio de Economía y Competititvidad (MTM2013-46374-P and MTM2015-69323-REDT), Spain, and a grant from CONACYT (FOMIX-CONACyT-UAGro 249818), México.

The first author is supported in part by two grants from Ministerio de Economía y Competititvidad (MTM2013-46374-P and MTM2015-69323-REDT), Spain, and a grant from CONACYT (FOMIX-CONACyT-UAGro 249818), México.

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About the article

Received: 2016-06-21

Accepted: 2017-03-30

Published Online: 2017-06-09


Citation Information: Open Mathematics, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2017-0061.

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© 2017 Rodríguez and Sigarreta. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License. BY-NC-ND 3.0

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