The hyperbolicity constant of infinite circulant graphs

Theorem 2.1

(Invariance of hyperbolicity). Let f : X → Y 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 δ′ is 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 δ′ is 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) with 〈g〉 = Aut (G). Consider the graph G^*with V(G^*) := V(G) and E(G^*) := {[gⁿ(v⁰), g^{n + 1}(v⁰)]}_{n ∈ Z} for some fixed v⁰ ∈ V(G). We define

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 v ∈ V(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 w_n be the midpoint of the edge [gⁿ(v⁰),g^{n + 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 ∈ [w_{n − 1}w_n] ∖ {w_{n − 1}}. Note that i is the inclusion map on V(G^*) = V(G); hence, i is (1/2)-full.

Fix u, v ∈ V(G) = V(G^*). Let u₀ = u, u₁,⋯,u_{k − 1},u_k = v ∈ V(G) with dG(u,v)=∑j=1kdG(uj−1,uj) and [uj−1,uj]∈E(G) for every 1 ≤ j ≤ k; then we have

Fix u, v ∈ G^* and a geodesic [uv] in G^*. Recall that i(u),i(v) ∈ V(G) = V(G^*). We have

Fix u,v∈V(G)=V(G∗). Let v0=u,v1,⋯,vr−1,vr=v∈V(G) with dG∗(u,v)=∑j=1rdG∗(vj−1,vj) and [v_{j − 1},v_j] ∈ E(G^*); then v_j = gⁱ(v_{j − 1}) for some i ∈ {1,−1} and we have

Fix u, v ∈ G^*. We have

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, …, G_r, since G is a circulant graph; therefore, δ(G) = max{δ(G₁), … δ(G_r)} = δ(G₁). Since G₁ is connected and circulant, Theorem 2.3 gives the result. □

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, z ∈ J(G) and δ(T) = δ(G).

We also need the following technical lemmas.

Lemma 3.3

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

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

2. a₂ ≥ a_k − 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 := ⌊a_k/2⌋. If a₂ ≤ r, then r = d_G(0,r) by hypothesis and r = d_G(0,r) ≤ d_G(0,a₂) + d_G(a₂,r) ≤ 1 + r − a₂ < r, which is a contradiction. Thus we conclude a₂ > r. Therefore,

and a₂ ≥ 2r − 1. Hence, a₂ ≥ a_k − 1 if a_k is even. Since a₂ > r, if a_k is odd, then

and a₂ ≥ 2r = a_k − 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 < ⋯ < a_k with a₂ < a_k − 1, and G = C_∞(1,a₂,…,a_k) ∉ 𝓔. Then

for every u ∈ ℤ with 0 ≤ u ≤ a_j and 1 ≤ j ≤ k.

Proof

Since a₂ < a_k − 1, Lemma 3.3 gives:

1. if a_k is even, then d_G(0,⌊a_k/2⌋) < ⌊a_k/2⌋,

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

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

If a₂ ≤ ⌊a_k/2⌋, then d_G(0,⌊a_k/2⌋) < ⌊a_k/2⌋, which is a contradiction. Therefore, a₂ > ⌊a_k/2⌋ and

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

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

If j = 2, then for every 0 ≤ u ≤ a₂

If j = 3, then d_G(⌊a₃/2⌋,a₃) = d_G(0,⌊a₃/2⌋ + 1) = ⌊a₃/2⌋ − 1, and for every 0 ≤ u ≤ a₃ with u ≠ ⌊a₃/2⌋,⌊a₃/2⌋ + 1,

This finishes the proof. □

Proposition 3.5

Consider any integers k > 1 and 1 < a₂ < ⋯ < a_k. If we have either k = 2, or k = 3 and a₂ = a₃−1, or C_∞ (1, a₂, …, a_k) ∈ 𝓔, then

Proof

Define r := ⌊ a_k/2⌋ and G := C_∞ (1, a₂, …, a_k).

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)a_k given by

Lemma 3.3 gives that γ₁ and γ₂ are geodesics; then d_G(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 + a_k], then d_G(p, x) = d_G(p, y) = r + 1/2 and Lemma 3.3 gives that d_G(p, γ₂) = r + 1/2. Hence,

Assume now that C_∞ (1, a₂, …, a_k) ∈ 𝓔. Note that r + ra_k = (r+1) a_k−1, since a_k = a_k−1 + 1 is odd. Consider the curves γ₁,γ₂, γ₃ in G

joining x := −(r+1)a_k−1, y := (r+1)a_k−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 d_G(0, r) = r, since G ∈ 𝓔. Therefore, if p is the midpoint of [−r, r], then

 □

Lemma 3.6

Consider any integers k > 1 and 1 < a₂ < ⋯ < a_k. If x, y ∈ J(C_∞(1, a₂, …, a_k)) and x and y are not related, then

Furthermore, if a₂ < a_k−1 and C_∞ (1, a₂, …, a_k)∉ 𝓔, then

Proof

Let G = C_∞ (1, a₂, …, a_k). If x₁ ≤ y₁ ≤ y₂ ≤ x₂, then the cycle

has length at most 1 + a_k. Since σ is a cycle containing the points x and y, we have

If a₂ < a_k − 1 and G ∉ 𝓔, then Lemma 3.4 gives d_G(x, y₁) ≤ 1/2 + ⌊ a_k/2⌋−1=⌊a_k/2⌋−1/2 and d_G(x, y) ≤ ⌊ a_k/2⌋.

If y₁ ≤ x₁ ≤ x₂ ≤ y₂, then the same argument gives these inequalities.

If x₁ ≤ y₁ < x₂ ≤ y₂, then x₂ − y₁+y₁ − x₁ ≤ a_k and min{x₂−y₁, y₁ − x₁} ≤ ⌊ a_k/2⌋. If x₂ − y₁ ≤ ⌊ a_k/2⌋, then d_G(x, y) ≤ d_G(x, x₂)+x₂ − y₁+d_G(y₁, y) ≤ 1+⌊ a_k/2⌋. If y₁ − x₁ ≤ ⌊ a_k/2⌋, then d_G(x, y) ≤ d_G(x, x₁)+y₁ − x₁+d_G(y₁, y) ≤ 1+⌊ a_k/2⌋. If a₂ < a_k−1 and C_∞(1, a₂, …, a_k)∉ 𝓔, then Lemma 3.4 also gives d_G(x, y₁) ≤ ⌊ a_k/2⌋−1/2 and d_G(x, y) ≤ ⌊ a_k/2⌋.

If y₁ ≤ x₁ < y₂ ≤ x₂, then the same argument gives these inequalities. □

Theorem 3.7

For any integers k > 1 and 1 < a₂ < ⋯ < a_k,

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, …, a_k), 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, z ∈ J(G). We consider several cases.

1. If x and y are not related, then Lemma 3.6 gives

   dG(p,[xz]∪[yz])≤dG(p,{x,y})≤12dG(x,y)≤12+12⌊ak2⌋. (2)

2. Assume that xR y. Without loss of generality we can assume that xL y. Denote by w₁, …, w_m the vertices in [xy] such that w₁ ∈ {x₁, x₂}, w_m ∈ {y₁, y₂} and d_G(w_j, w_j+1) = 1 for every 1 ≤ j < m; we define

   i0:=min{1≤i≤m|wj≥x2∀i≤j≤m},x0:=wi0.

(B.1) Assume that either x ∈ V(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

has length at most 1 + a_k. Since x ∈ V(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

(B.2) Assume now that x is the midpoint of an edge [r, r + a_j]∈ E(G) for some r ∈ ℤ and 1 < j ≤ k. If x₀ = x₂, then L([xx₀]) = 1/2. If w_i0−1 = x₁, then L([xx₀]) = 3/2. If x₀ > x₂ and w_i0−1 ≠ x₁, then we have either w_i0−1 < x₁ < x₂ < w_i0 or x₁ < w_i0−1 < x₂ < w_i0. In the first case the cycle

has length at most a_k. Since σ is a cycle containing the points x and x₀, we have

Assume that x₁ < w_i0−1 < x₂ < w_i0. Then

Therefore, in Case (B), if p ∈[xx₀]\ B(x₀, 3/4), then

Define

A similar argument to the previous one shows that if p ∈[y₀y]\ B(y₀, 3/4), then

Since T is a continuous curve, we have

(a) Assume that y₀ ∈ [xx₀]\ {x₀}.

(a.1) If d_G(x₀, y₀)≥2, then

(a.2) If d_G(x₀, y₀) = 1, then w_j0 = y₀ = w_i0−1, w_i0 = x₀ = w_j0+1 and the definitions of x₀ and y₀ give y₀ < x₂ ≤ x₀ and y₀ ≤ y₁ < x₀. Since [x₀, y₀] ∈ E(G),

(b) Assume that y₀∉[xx₀]\ {x₀}. Since x₂ ≤ x₀, y₀ ≤ y₁ and [xz] ∪ [yz] is a continuous curve joining x and y, if p ∈ V(G)∩[x₀y₀]⊂[xy], then there exist u, v ∈ V(G)∩([xz]∪[yz]) with [u, v] ∈ E(G) and u ≤ p ≤ v. Since T is a cycle, we have

Therefore, if p ∈[x₀y₀]∪ B(x₀, 3/4)∪ B(y₀, 3/4), then

This inequality, (2), (3) and (4) give

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

where p is the point in [xx₀] with d_G(p, x₀) = 3/4 or the point in [y₀y] with d_G(p, y₀) = 3/4. By symmetry, without loss of generality we can assume that p is the point in [xx₀] with d_G(p, x₀) = 3/4; thus d_G(p, w_i0−1) = 1/4. Therefore, we are in Case (B.2) with x₁ < w_i0−1 < x₂ < w_i0 and

Then

and we conclude

(I) Assume that x₂∈[xx₀]. Since T is a cycle, then x₁∈[xz]∪[yz] and x₁ < w_i0−1. Hence, since [xz]∪[yz] is a continuous curve joining x and y, and d_G(p, w_i0−1) = 1/4, we obtain as above

which is a contradiction.

(II) Assume that x₁∈[xx₀].

(II. 1) If x₂ − x₁ = x₀ − w_i0−1 = a_k, then w_i0−1 − x₁ = x₀ − x₂, d_G(x₁, w_i0−1) = d_G(x₂, x₀) and

which is a contradiction.

(II.2) If x₂ − x₁ < a_k, then (6) gives

and

which is a contradiction.

(II.3) Assume x₂ − x₁ = a_k and x₀ − w_i0−1 < a_k. Since x₁∈[xx₀], we have w_i0−1 − x₁ = ⌊ a_k/ 2 ⌋.

(II.3.1) If x₂ − w_i0−1 = ⌊a_k/2⌋, then

which is a contradiction.

(II.3.2) If x₂ − w_i0−1 = ⌊ a_k/2⌋+1, then a_k = x₂ − x₁ = 2⌊ a_k/2⌋+1 and

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

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, …, a_k). By Theorem 3.2 there exist a geodesic triangle T = {x, y, z} in G that is a cycle with x, y, z ∈ J(G), and p ∈ [xy] with d_G(p, [xz] ∪ [yz]) = 1/2+⌊ a_k/2⌋. Hence, p ∈ J(G).

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

If x and y are not related, then Lemma 3.6 gives d_G(x, y) ≤ ⌊ a_k/2⌋. Since d_G(x, y) = d_G(x, p) + d_G(p, y)≥ 1 + 2 ⌊ a_k/2⌋, thus x and y are related, and without loss of generality we can assume xL y. Since d_G(p, x), d_G(p, y)≥1/2+⌊ a_k/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 p₁ − x₁≥p₁ − x₂≥d_G(x₂,p₁)≥⌊ a_k/2⌋−1/2, we have p₁ − x₁≥p₁ − x₂≥⌊ a_k/2⌋, and there is some point u ∈([xz]∪[yz])∩ V(G) with u ≤ p₁. A similar argument gives that there is some point v ∈ ([xz]∪[yz])∩ V(G) with p₂ ≤ v. Denote by u and v any couple of vertices satisfying these properties. We are going to prove that v − u > a_k.

If p ∈ V(G), then

If p is the midpoint of some edge [m, m+a_j] with m ∈ ℤ and 2 ≤ j ≤ k, then Lemma 3.4 gives d_G(p, w) ≤ ⌊ a_k/2⌋−1/2 < 1/2 + ⌊ a_k/2⌋ for every w ∈ ℤ with m ≤ w ≤ m + a_j, since a₂ < a_k−1 and G ∉ 𝓔. Furthermore,

Finally, assume that p is the midpoint of some edge [m, m+1] with m ∈ ℤ. By Lemma 3.3, we have d_G(0, ⌊ a_k/2⌋) ≤ ⌊ a_k/2⌋−1 or d_G(0, ⌊ a_k/2⌋+1) ≤ ⌊ a_k/2⌋−1. Therefore,

Hence, we have proved v − u >a_k in every case.

Since if p is the midpoint of some edge [m, m +a_j] with m ∈ ℤ and 2 ≤ j ≤ k then d_G(p, w) < 1/2 + ⌊ a_k/2 ⌋ for every w ∈ ℤ with m ≤ w ≤ m + a_j, and [xz]∪[yz] is a continuous curve joining u and v, there exist u₀, v₀ ∈([xz]∪[yz])∩ V(G) with u₀ ≤ p₁, p₂ ≤ v₀ and [u₀, v₀]∈ E(G). Hence, a_k≥v₀ − u₀ > a_k, 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 d_G(p, x₀)≥1. Since d_G(x, x₀) ≤ 3/2 + ⌊ a_k/2⌋, we have 1/2+ ⌊ a_k/2⌋=d_G(p, [xz]∪[yz]) ≤ d_G(x, p) ≤ 1/2 + ⌊ a_k/2⌋. Therefore, d_G(x, p) = 1/2 + ⌊ a_k/2⌋, d_G(p, x₀) = 1, p ∈ V(G) and x is the midpoint of [x₁, x₂]∈ E(G). Thus d_G(p, x₁)≥⌊ a_k/2⌋, d_G(p, x₂)≥⌊ a_k/2⌋ and p ≤ x₁ ≤ x₂ ≤ x₀. By Lemma 3.3, we have d_G(0, ⌊ a_k/2⌋) ≤ ⌊ a_k/2⌋−1 or d_G(0, ⌊ a_k/2⌋+1) ≤ ⌊ a_k/2⌋−1.

If d_G(0, ⌊ a_k/2⌋) ≤ ⌊ a_k/2⌋−1, then x₁≥⌊ a_k/2⌋+1, x₂ ≤ a_k−(⌊ a_k/2⌋−1) and

which is a contradiction.

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

Therefore, we conclude in every case that a₂≥a_k−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. □

Theorem 3.8

For any integers k > 1 and 1 < a₂ < ⋯ < a_k we have

and the equality is attained if a_k = k.

Proof

Define G = C_∞ (1, a₂, …, a_k), and consider the geodesics γ₁, γ₂ in G given by

Let T be the geodesic bigon T = {γ₁, γ₂} in G. If p is the midpoint of [a_k, 2a_k], then