Weng, C., Weak-geodetically Closed Subgroups in Distance-Regular Graphs, Graphs and Combinatorics, 14(1998), 275-304 

Weak-geodetically Closed Subgraphs in Distance-Regular Graphs CHIH-WEN WENG†

Abstract. Let Γ = (X, R) denote a distance-regular graph with diameter D ≥ 2 and distance function δ. A (vertex) subgraph Ω ⊆ X is said to be weak- geodetically closed whenever for all x, y ∈ Ω and all z ∈ X,

δ(x, z) + δ(z, y) ≤ δ(x, y) + 1 −→ z ∈ Ω.

We show that if the intersection number c2 > 1 then any weak-geodetically closed subgraph of X is distance-regular. Γ is said to be i-bounded, whenever for all x, y ∈ X at distance δ(x, y) ≤ i, x, y are contained in a common weak- geodetically closed subgraph of Γ of diameter δ(x, y). By a parallelogram of length i, we mean a 4-tuple xyzw of vertices in X such that δ(x, y) = δ(z, w) = 1, δ(x, w) = i, and δ(x, z) = δ(y, z) = δ(y, w) = i − 1. We prove the following two theorems.

Theorem 1. Let Γ denote a distance-regular graph with diameter D ≥ 2, and assume the intersection numbers c₂ > 1, a₁ 6= 0. Then for each integer i (1 ≤ i ≤ D), the following (i)-(ii) are equivalent.

(i) Γ is i-bounded.

(ii) Γ contains no parallelogram of length ≤ i + 1.

Restricting attention to the Q-polynomial case, we get the following stronger result.

Theorem 2. Let Γ denote a distance-regular graph with diameter D ≥ 3, and assume the intersection numbers c₂ > 1, a₁ 6= 0. Suppose Γ is Q-polynomial.

Then the following (i)-(iii) are equivalent.

(i) Γ contains no parallelogram of length 2 or 3.

(ii) Γ is D-bounded.

(iii) Γ has classical parameters (D, b, α, β), and either b < −1, or else Γ is a dual polar graph or a Hamming graph.

1. Introduction.

Let Γ = (X, R) denote a distance-regular graph with diameter D ≥ 2, and let δ denote the distance function of Γ.

Recall a (vertex) subgraph Ω ⊆ X is geodetically closed whenever for all vertices x, y ∈ Ω, and for all vertices z ∈ X,

δ(x, z) + δ(z, y) = δ(x, y) −→ z ∈ Ω.

† Department of Mathematics, University of Wisconsin, 480 Lincoln Dr., Madison, WI 53706.

Distance-regular graphs containing many geodetically closed subgraphs have been studied by several authors. Shult and Yanushka[4], Brouwer and Wilbrink[3]

showed that if Γ is a near polygon with c2 > 1, a1 6= 0, then there exist sub 2j-gons in Γ for each integer j (2 ≤ j ≤ D). Also, Ivanov and Shpectorov[5]

showed that if Γ is a Hermitian forms graph, then Γ has geodetically closed subgraphs of any diameter j (1 ≤ j ≤ D).

In the present paper, we study the following special kind of geodetically closed subgraphs. We say a subgraph Ω ⊆ X is weak-geodetically closed whenever for all vertices x, y ∈ Ω, and for all z ∈ X,

δ(x, z) + δ(z, y) ≤ δ(x, y) + 1 −→ z ∈ Ω.

We have two main results. First, given an integer i (1 ≤ i ≤ D), we give necessary and sufficient conditions for the existence of a weak-geodetically closed subgraph of diameter δ(x, y) containing any two given vertices x, y with δ(x, y) ≤ i. Theorem 6.4 is our main result in this area.

We then tighten Theorem 6.4 in the case Γ is Q-polynomial, and obtain our second main result, Theorem 7.2.

The paper is organized as follows. In sections 2-5, we set up the necessary tools for the proof of Theorem 6.4. To do this, we study the structure theory of a weak-geodetically closed subgraph Ω of Γ.

More precisely, in section 2, we define the notion of a subgraph being weak- geodetically closed with respect to a vertex. We find necessary and sufficient conditions for a subgraph to be weak-geodetically closed with respect to some vertex.

In section 3, we get some inequalities involving the intersection numbers of Γ, when we assume the existence of certain weak-geodetically closed subgraphs.

Proposition 3.2 is the main result in this section.

In section 4, we consider a regular connected subgraph Ω of Γ. First, we find a lower bound for

¯¯

¯Ω

¯¯

¯ and necessary and sufficient conditions for this bound to be met(Lemma 4.4). These conditions involve the notion of weak-geodetic closure.

In our main result of this section, Theorem 4.6, we show Ω is weak-geodetically closed if and only if Ω is weak-geodetically closed with respect to at least one vertex.

In section 5, we restrict to the case c2 > 1, and prove a weak-geodetically closed subgraph Ω of Γ is distance-regular.

We prove the two main theorems in section 6 and section 7.

For the rest of this section, we give some definitions.

Let Γ = (X, R) be a finite undirected graph without loops or multiple edges, with vertex set X and edge set R. We say vertices x, y are adjacent if xy ∈ R.

Pick any integer i (0 ≤ i ≤ D) and any vertices x, y ∈ X. By a path of length i from x to y, we mean a sequence x = x0, x1, · · · , xi = y of vertices from x such that x_j, x_j+1are adjacent for all j (0 ≤ j ≤ i − 1). Being joined by a path is an equivalence relation. Its equivalence classes are called the connected components of Γ. Γ is said to be connected whenever Γ has a unique connected component.

From now on, assume Γ is connected. The distance δ(x, y) between two vertices x, y ∈ X is the length of a shortest (geodesic) path from x to y. By the diameter of Γ, we mean the scalar

D := max{δ(x, y)|x, y ∈ X}.

Sometimes we write diam(Γ) to denote the diameter of Γ. By a clique in Γ, we mean a set of mutually adjacent vertices in X.

Let Γ = (X, R) be a graph with diameter D. By a subgraph of Γ, we mean a graph (Ω, Ξ), where Ω is a nonempty subset of X and Ξ = { xy |x, y ∈ Ω, xy ∈ R}.

We refer to (Ω, Ξ) as the subgraph induced on Ω and by abuse of notation, we refer to this subgraph as Ω. For any x ∈ X and any integer i, set

Γi(x) := {y|y ∈ X, δ(x, y) = i}, and for y ∈ Γi(x), set

B(x, y) := Γ1(x) ∩ Γi+1(y), (1.1) A(x, y) := Γ1(x) ∩ Γi(y), (1.2) C(x, y) := Γ₁(x) ∩ Γ_i−1(y). (1.3) Note that for all x, y ∈ Γ and for all z ∈ C(y, x), we have

C(x, z) ⊆ C(x, y), (1.4)

B(x, z) ⊇ B(x, y). (1.5)

The valency k(x) of a vertex x ∈ X is the cardinality of Γ1(x). The graph Γ is called regular (with valency k) if each vertex in X has valency k. Γ is said to be distance-regular whenever for all integers i (0 ≤ i ≤ D), and for all x, y ∈ X with δ(x, y) = i, the numbers

ci :=

¯¯

¯C(x, y)

¯¯

¯, (1.6)

a_i :=

¯¯

¯A(x, y)

¯¯

¯, (1.7)

bi :=

¯¯

¯B(x, y)

¯¯

¯ (1.8)

are independent of x, y. The constants ci, ai, bi (0 ≤ i ≤ D) are known as the intersection numbers of Γ. The sequence

{b0, b1, · · · , bD−1; c1, c2, · · · , cD}

is called the intersection array of Γ. Note that the valency k = b0, c0 = 0, c1 = 1, bD = 0, and

k = c_i+ a_i+ b_i (0 ≤ i ≤ D) (1.9) [2, 126].

2. Weak-geodetically closed subgraphs with respect to a vertex.

Let Γ = (X, R) denote a graph, and let Ω denote a subgraph of Γ. In this section, we define what it means for Ω to be weak-geodetically closed with respect to a vertex. We find some necessary and sufficient conditions for Ω to have this property.

We begin with a definition.

Definition 2.1. Let Γ = (X, R) denote a graph with distance function δ. Fix a subgraph Ω of Γ, and pick any vertex x ∈ Ω. Ω is said to be geodetically closed with respect to x (resp. weak-geodetically closed with respect to x), whenever for all y ∈ Ω and for all z ∈ X,

δ(x, z) + δ(z, y) = δ(x, y) −→ z ∈ Ω (resp. δ(x, z) + δ(z, y) ≤ δ(x, y) + 1 −→ z ∈ Ω).

Lemma 2.2. Let Γ = (X, R) denote a graph with distance function δ. Fix a subgraph Ω of Γ, and pick any vertex x ∈ Ω. Then with the notation of (1.3), the following (i)-(iii) are equivalent.

(i) Ω is geodetically closed with respect to x.

(ii) C(y, x) ⊆ Ω for all y ∈ Ω.

(iii) For all y ∈ Ω, and for all w ∈ Γ1(y) \ Ω, δ(x, w) ≥ δ(x, y).

Proof. This is immediate from Definition 2.1.

Lemma 2.3. Let Γ = (X, R) denote a graph with distance function δ. Fix a subgraph Ω of Γ, and pick any vertex x ∈ Ω. Then with the notation of (1.2), (1.3), the following (i)-(iii) are equivalent.

(i) Ω is weak-geodetically closed with respect to x.

(ii) C(y, x) ⊆ Ω and A(y, x) ⊆ Ω for all y ∈ Ω.

(iii) For all y ∈ Ω, and for all w ∈ Γ₁(y) \ Ω,

δ(x, w) = δ(x, y) + 1. (2.1)

Proof. (i)−→(ii). Let the vertex y ∈ Ω be given, and pick any z ∈ A(y, x) ∪ C(y, x). Then δ(x, z) ≤ δ(x, y), and of course δ(z, y) = 1, so

δ(x, z) + δ(z, y) ≤ δ(x, y) + 1.

Hence z ∈ Ω by Definition 2.1.

(ii)−→(iii). Let y, w be given. Observe

w ∈ Γ1(y) \ Ω

⊆ B(y, x) by (ii), and (2.1) follows from (1.1).

(iii)−→(i). Suppose Ω is not weak-geodetically closed with respect to x. Then by Definition 2.1, there exists a vertex y ∈ Ω and a vertex z 6∈ Ω such that

δ(x, z) + δ(z, y) ≤ δ(x, y) + 1. (2.2) Of all such pairs y, z, pick one with δ(z, y) minimal. Note that z 6= y by the construction, and δ(z, y) 6= 1 by (2.1)-(2.2), so there exists a vertex z⁰ ∈ C(z, y).

Observe

δ(z⁰, y) = δ(z, y) − 1 (2.3)

by the construction, and

δ(x, z⁰) ≤ δ(x, z) + 1 (2.4)

by the triangular inequality. Adding (2.2)-(2.4), we obtain

δ(x, z⁰) + δ(z⁰, y) ≤ δ(x, y) + 1. (2.5) Observe z⁰ ∈ Ω by (2.3), (2.5) and the construction. Now by (iii) (with y :=

z⁰, w := z), we find

δ(x, z) = δ(x, z⁰) + 1. (2.6)

By the triangular inequality,

δ(x, y) ≤ δ(x, z⁰) + δ(z⁰, y). (2.7)

Adding (2.2), (2.3), and (2.7), we obtain

δ(x, z) ≤ δ(x, z⁰),

contradicting (2.6). We conclude Ω is weak-geodetically closed with respect to x.

Lemma 2.4. Let Γ = (X, R) denote a graph with distance function δ. Fix a subgraph Ω of Γ, and pick a vertex x ∈ Ω. Suppose Ω is weak-geodetically closed with respect to x, and suppose there exists a vertex z ∈ Γ1(x) \ Ω. Then the following (i)-(ii) hold.

(i) For any vertex y ∈ Ω,

δ(z, y) = δ(x, y) + 1.

(ii) x is the unique vertex in Ω adjacent to z.

Proof. (i). By Definition 2.1 and since z 6∈ Ω, we have δ(x, z) + δ(z, y) >

δ(x, y) + 1. Of course δ(x, z) = 1, so δ(z, y) > δ(x, y). Also by the triangular inequality,

δ(z, y) ≤ δ(z, x) + δ(x, y)

= 1 + δ(x, y).

Hence δ(z, y) = δ(x, y) + 1.

(ii). This is immediate from (i).

Definition 2.5. Let Γ = (X, R) denote a graph with distance function δ, and let Ω be any subgraph of Γ. For all vertices x ∈ Ω, define

diamx(Ω) := max{δ(x, y)|y ∈ Ω}.

Lemma 2.6. Let Γ = (X, R) denote a distance-regular graph. Fix a subgraph Ω of Γ, and pick a vertex x ∈ Ω. Suppose Ω is weak-geodetically closed with respect to x. Set d :=diamx(Ω). Then the following (i)-(iv) hold.

(i) For all y ∈ Ω ∩ Γ_d(x), ¯

¯¯Ω ∩ Γ1(y)

¯¯

¯ = cd+ ad. (ii) For all y ∈ Ω ∩ Γd(x),

C(x, y) ∪ A(x, y) ⊆ Ω ∩ Γ₁(x). (2.8)

(iii) ¯¯

¯Ω ∩ Γ1(x)

¯¯

¯ ≥ cd + ad.

(iv) Equality holds in (iii) if and only if equality holds in (2.8) for at least one y ∈ Ω ∩ Γd(x), if and only if equality holds in (2.8) for all y ∈ Ω ∩ Γd(x).

Proof. (i). Note that Γ₁(y) = C(y, x) ∪ A(y, x) ∪ B(y, x). Observe that Ω ∩ B(y, x) = ∅ since δ(x, y) =diamx(Ω). Now Ω ∩ Γ1(y) = C(y, x) ∪ A(y, x) by Lemma 2.3(ii), and (i) follows by (1.6), (1.7).

(ii). Pick z ∈ C(x, y) ∪ A(x, y). Then certainly z ∈ Γ1(x) and δ(x, z) + δ(z, y) ≤ δ(x, y) + 1,

so z ∈ Ω by Definition 2.1.

(iii), (iv). These are immediate from (ii).

3. Weak-geodetically closed subgraphs.

Let Γ = (X, R) be any graph. In this section, we study a subgraph Ω that is weak-geodetically closed with respect to all vertices in Ω. We prove that when Γ is distance-regular, the existence of Ω forces certain inequalities involving the intersection numbers of Γ.

Definition 3.1. Let Γ = (X, R) be a graph. A subgraph Ω of Γ is said to be geodetically closed (resp. weak-geodetically closed) , whenever Ω is geodetically closed (resp. weak-geodetically closed) with respect to all x ∈ Ω.

Note. A weak-geodetically closed subgraph Ω of Γ is geodetically closed in Γ.

In particular, Ω is connected, and the distances as measured in Ω are the same as distances as measured in Γ.

Proposition 3.2. Let Γ = (X, R) denote a distance-regular graph with diameter D ≥ 2. Fix an integer d (1 ≤ d < D), and suppose there exists a weak- geodetically closed subgraph Ω of Γ that has diameter d. Then the intersection numbers b_i, a_i, c_i of Γ satisfy the following inequalities.

(i)

ci ≥ ci−1(c2− 1) + 1 (1 ≤ i ≤ d + 1).

(ii)

ai ≥ ai−1(c2− 1) + a1 (1 ≤ i ≤ d + 1).

(iii)

b_i ≤ (b_i−1− k)(c₂− 1) + b₁ (1 ≤ i ≤ d + 1).

(iv) Suppose c2 > 1. Then

bi < bi−1 (1 ≤ i ≤ d + 1).

Proof. Let the integer i be given. Our result is clear if i = 1, since c1 = 1, c0 = 0, a0 = 0, b0 = k. Hence we may assume i ≥ 2. First we claim there exist vertices x, y ∈ Ω and z ∈ X \ Ω such that

δ(x, y) = i − 1, δ(y, z) = 1, δ(x, z) = i. (3.1) Indeed, since diam(Ω) = d, we can pick vertices x⁰, y ∈ Ω with δ(x⁰, y) = d.

Observe B(y, x⁰) 6= ∅ since d < D, so pick a vertex z ∈ B(y, x⁰). Note that z 6∈ Ω, since

δ(x⁰, z) = d + 1

> diam(Ω).

Now pick a vertex x in a geodesic path from x⁰ to y with δ(x, y) = i − 1. Clearly, x ∈ Ω, and x, y, z satisfy (3.1). This proves our claim. Recall by Lemma 2.4(ii),

Γ1(z) ∩ Ω = {y}. (3.2)

Now we consider the four parts of the proposition.

(i). Observe each vertex in C(y, x) is adjacent to c2− 1 vertices in C(z, x) \ {y}.

Next observe each vertex in C(z, x) \ {y} is adjacent to at most 1 vertex in C(y, x). To see this, pick any w ∈ C(z, x) \ {y}. Then w 6∈ Ω by (3.2). Note that C(y, x) ⊆ Ω by Lemma 2.3(ii), so w is adjacent to at most one vertex in C(y, x) by Lemma 2.4(ii). Now by counting the edges between C(z, x) \ {y} and C(y, x), we find

ci− 1 =

¯¯

¯C(z, x) \ {y}

¯¯

¯

≥

¯¯

¯C(y, x)

¯¯

¯(c2− 1)

= c_i−1(c₂− 1), as desired.

(ii). We first prove

A(z, y) ⊆ A(z, x), (3.3)

and then count the edges between A(z, x) \ A(z, y) and A(y, x) to establish the inequality.

Note that

A(y, x) ⊆ Ω (3.4)

by Lemma 2.3(ii),

A(z, y) ∩ Ω = ∅ (3.5)

by (3.2), and

A(z, y) ⊆ A(y, x) ∪ A(z, x) (3.6)

by construction. Now (3.3) follows from (3.4)-(3.6). We now count the edges between A(z, x) \ A(z, y) and A(y, x).

Claim 1. Each vertex in A(z, x) \ A(z, y) is adjacent to at most one vertex in A(y, x).

Proof of Claim 1. Observe that by (3.2),

A(z, x) ∩ Ω = ∅, so Claim 1 follows from (3.4) and Lemma 2.4(ii).

Claim 2. Each vertex in A(y, x) is adjacent to c₂− 1 vertices in A(z, x) \ A(z, y).

Proof of Claim 2. Pick w ∈ A(y, x). Observe

w ∈ Ω (3.7)

by (3.4), so w is not adjacent to z by (3.2); in particular δ(w, z) = 2. It now suffices to show

Γ₁(w) ∩ (A(z, x) \ A(z, y)) = C(z, w) \ {y}, (3.8) since

¯¯

¯C(z, w) \ {y}

¯¯

¯ = c2− 1. The inclusion

Γ1(w) ∩ (A(z, x) \ A(z, y)) ⊆ C(z, w) \ {y}

is clear by construction. To prove

C(z, w) \ {y} ⊆ Γ1(w) ∩ (A(z, x) \ A(z, y)), pick u ∈ C(z, w) \ {y}. Of course u ∈ Γ1(w) and u ∈ Γ1(z), so

u 6∈ Ω (3.9)

by (3.2), and

u ∈ A(z, x) ∪ A(w, x) (3.10)

by construction. Note that

A(w, x) ⊆ Ω (3.11)

by (3.7) and Lemma 2.3(ii). Hence u ∈ A(z, x) by (3.9)-(3.11). Also u 6∈ A(z, y) by (3.7) and (3.9), otherwise u is adjacent to y, w ∈ Ω, contradicting Lemma 2.4(ii). Hence we have (3.8). This proves Claim 2.

Now using Claim 1, Claim 2, we count the edges between A(z, x) \ A(z, y) and A(y, x), obtaining

a_i− a₁ =

¯¯

¯A(z, x) \ A(z, y)

¯¯

¯

≥

¯¯

¯A(y, x)

¯¯

¯(c2− 1)

= ai−1(c2− 1), as desired.

(iii). By (i), (ii) and (1.9),

bi = k − ai− ci

≤ k − (a_i−1+ c_i−1)(c₂− 1) − a₁− 1

= (b_i−1− k)(c₂− 1) + b₁, as desired.

(iv). Observe bi−1 − k ≤ 0, c2− 1 ≥ 1 and b1 < k, so by (iii), bi ≤ (bi−1 − k)(c2− 1) + b1

≤ bi−1 − k + b1

< bi−1, as desired. This proves Proposition 3.2.

4. Regular subgraphs of distance-regular graphs.

In this section, we study basic properties of a regular connected subgraph Ω in a distance-regular graph, and get a lower bound of

¯¯

¯Ω

¯¯

¯. We find necessary and sufficient conditions for

¯¯

¯Ω

¯¯

¯ to meet this lower bound. These conditions are related to the weak-geodetically closed property. Theorem 4.6 is the main result of this section.

We begin with a definition.

Definition 4.1. Let Γ = (X, R) denote a distance-regular graph with diameter D ≥ 2, and let Ω denote a regular connected subgraph of Γ. We define

(i)

βi(Ω) := γ − ci− ai (0 ≤ i ≤ D), where γ denotes the valency of Ω.

(ii)

ki(Ω) := β0(Ω)β1(Ω) · · · βi−1(Ω)

c1c2· · · ci (1 ≤ i ≤ D), k₀(Ω) := 1.

(iii)

d(Ω) := min{i|0 ≤ i ≤ D, βi(Ω) ≤ 0}. (4.1) (We observe β_D(Ω) ≤ 0, so (4.1) makes sense).

Lemma 4.2. Let Γ = (X, R) denote a distance-regular graph with diameter D ≥ 2. Let Ω denote a regular connected subgraph of Γ, and write d := d(Ω).

Then the following (i)-(iii) hold.

(i) βi(Ω) > 0 (0 ≤ i < d).

(ii) k_i(Ω) > 0 (0 ≤ i ≤ d).

(iii) γ ≤ ad+ cd, where γ denotes the valency of Ω.

Proof. (i). This is immediate from Definition 4.1(iii).

(ii). This is immediate from (i) and Definition 4.1(ii).

(iii). This is immediate from Definition 4.1(i), (iii).

Lemma 4.3. Let Γ = (X, R) denote a distance-regular graph with diameter D ≥ 2. Let Ω denote a regular connected subgraph of Γ, and pick any x ∈ Ω.

Pick an integer i (0 ≤ i ≤ d(Ω)). Then the following (i)-(iii) hold.

(i) ¯

¯¯Ω ∩ Γi(x)

¯¯

¯ ≥ ki(Ω) (4.2)

with equality if and only if

C(y, x) ⊆ Ω (∀y ∈ Ω, δ(x, y) ≤ i) (4.3) and

A(y, x) ⊆ Ω (∀y ∈ Ω, δ(x, y) ≤ i − 1). (4.4) (ii)

Ω ∩ Γ_i(x) 6= ∅. (4.5)

(iii)

diamx(Ω) ≥ d(Ω). (4.6)

Proof. (i). We prove this by induction on the integer i. First assume i = 0.

Then (4.2)-(4.4) hold at i; indeed both sides in (4.2) equal 1. Next assume i ≥ 1.

Then by Definition 4.1(i), a counting argument, the induction hypothesis and Definition 4.1(ii),

ci

¯¯

¯Ω ∩ Γi(x)

¯¯

¯ ≥ number of edges between Ω ∩ Γi(x) and Ω ∩ Γi−1(x) (4.7)

≥ βi−1(Ω)

¯¯

¯Ω ∩ Γi−1(x)

¯¯

¯ (4.8)

≥ βi−1(Ω)ki−1(Ω) (4.9)

= β0(Ω)β1(Ω) · · · βi−1(Ω)

c₁c₂· · · c_i−1 (4.10)

= ciki(Ω), (4.11)

and equalities hold in (4.7)-(4.9) if and only if (4.3)-(4.4) hold. Now (4.2) follows since c_i > 0.

(ii). This is immediate from (i) above and Lemma 4.2(ii).

(iii). This is immediate from (ii) and Definition 2.5.

Lemma 4.4. Let Γ = (X, R) denote a distance-regular graph with diameter D ≥ 2, and let Ω denote a regular connected subgraph of Γ. Then

¯¯

¯Ω

¯¯

¯ ≥ k0(Ω) + k1(Ω) + · · · + kd(Ω), (4.12) where d := d(Ω) is from Definition 4.1(iii). Furthermore, equality holds in (4.12) if and only if for any x ∈ Ω (and for some x ∈ Ω),

d = diam_x(Ω), (4.13)

C(y, x) ⊆ Ω (∀y ∈ Ω, δ(x, y) ≤ d) (4.14) and

A(y, x) ⊆ Ω (∀y ∈ Ω, δ(x, y) ≤ d − 1). (4.15) Proof. Pick x ∈ Ω. Then by Lemma 4.3,

¯¯

¯Ω

¯¯

¯ =

diamXx(Ω) i=0

¯¯

¯Ω ∩ Γi(x)

¯¯

¯

≥ Xd i=0

¯¯

¯Ω ∩ Γi(x)

¯¯

¯ (4.16)

≥ Xd i=0

ki(Ω), (4.17)

and equalities in (4.16)-(4.17) hold if (4.13)-(4.15) hold. Hence we have the lemma.

Theorem 4.5. Let Γ = (X, R) denote a distance-regular graph with diameter D ≥ 2. Let Ω denote a regular connected subgraph of Γ, and let d := d(Ω) be as in Definition 4.1(iii). Then the following (i)-(iii) are equivalent.

(i) Equality is obtained in (4.12).

(ii) Ω is geodetically closed, and for all x, y ∈ Ω,

A(y, x) ⊆ Ω if δ(x, y) < diamx(Ω). (4.18) (iii) There exists a vertex x ∈ Ω such that

Ω is geodetically closed with respect to x, (4.19) and for all y ∈ Ω,

A(y, x) ⊆ Ω if δ(x, y) < diam_x(Ω). (4.20)

If (i)-(iii) hold, then Ω is distance-regular, with diameter d, and intersection numbers

c_i(Ω) = c_i (0 ≤ i ≤ d), (4.21)

a_i(Ω) = a_i (0 ≤ i < d). (4.22)

Proof. (i)→(ii) is immediate from Lemma 4.4, Lemma 2.2(ii), Definition 3.1 and (4.6). (ii)→(iii) is clear. To prove (iii)→(i), by Lemma 4.4, Lemma 2.2(ii), we only need to prove (4.13). Observe by a counting argument,

¯¯

¯Ω ∩ Γi(x)

¯¯

¯ci =

¯¯

¯Ω ∩ Γi−1(x)

¯¯

¯βi−1(Ω) (1 ≤ i ≤ diam_x(Ω)), forcing

β_i > 0 (0 ≤ i < diam_x(Ω)).

Hence (4.13) holds by (4.6) and Definition 4.1(iii).

Now suppose (i)-(iii) hold. (4.21)-(4.22) follow from (4.13) and (ii) above. We now have Theorem 4.5.

Theorem 4.6. Let Γ = (X, R) denote a distance-regular graph with diameter D ≥ 2. Let Ω denote a regular connected subgraph of Γ, and let d := d(Ω) be as in Definition 4.1(iii). Then the following (i)-(iii) are equivalent.

(i) Equality holds in (4.12), and Ω has valency cd+ ad. (ii) Ω is weak-geodetically closed.

(iii) Ω is weak-geodetically closed with respect to at least one vertex in Ω.

Suppose (i)-(iii) hold. Then Ω is distance-regular, with diameter d, and inter- section numbers

c_i(Ω) = c_i (0 ≤ i ≤ d), (4.23)

a_i(Ω) = a_i (0 ≤ i ≤ d). (4.24)

Proof. Observe each of the three statements (i)-(iii) in the present theorem implies the corresponding statement in Theorem 4.5. Without loss of generality, we may assume Theorem 4.5(i)-(iii) hold. In particular, we may assume Ω is distance-regular with diameter d.

(i)→(ii). Since Theorem 4.5(ii) holds by assumption, it remains to show

A(y, x) ⊆ Ω (4.25)

for all x, y ∈ Ω such that δ(x, y) = d. To obtain (4.25), observe by (4.21) that

¯¯

¯A(y, x) \ Ω

¯¯

¯ = ad− ad(Ω)

= ad− (

¯¯

¯Ω ∩ Γ1(y)

¯¯

¯ − cd)

= 0, and (4.25) follows.

(ii)→(iii). This is clear.

(iii)→(i). Since Theorem 4.5(i) holds by assumption, it remains to show Ω has valency c_d+ a_d. Pick any x, y ∈ Ω such that δ(x, y) = d. Then

¯¯

¯Ω ∩ Γ1(y)

¯¯

¯ = cd + ad

by Lemma 2.6(i).

Now assume (i)-(iii) hold. Then (4.23)-(4.24) hold by (4.21)-(4.22), and since Ω has valency cd+ ad. This proves Theorem 4.6.

5. Distance-regular graphs with c2 > 1.

In this section, we restrict our attention to the case Γ = (X, R) is distance- regular with intersection number c₂ > 1. We first prove that a weak-geodetically

closed subgraph Ω of Γ is regular (and consequently distance-regular by Theorem 4.6). We then give a precise description of Ω.

Lemma 5.1. Let Γ = (X, R) be a distance-regular graph with diameter D ≥ 2, and assume the intersection number c2 > 1. Fix a subgraph Ω of Γ, and pick any vertex x ∈ Ω. Suppose Ω is weak-geodetically closed with respect to x. Then for all y ∈ Ω ∩ Γ1(x), ¯

¯¯Ω ∩ Γ1(x)

¯¯

¯ ≥

¯¯

¯Ω ∩ Γ1(y)

¯¯

¯.

Proof. Since Γ is regular, it suffices to prove

¯¯

¯Γ1(x) \ Ω

¯¯

¯ ≤

¯¯

¯Γ1(y) \ Ω

¯¯

¯. (5.1)

Observe that each vertex in Γ₁(x) \ Ω is adjacent to c₂− 1 vertices in Γ₁(y) \ Ω.

Indeed pick z ∈ Γ1(x) \ Ω. Then by Lemma 2.4(i), δ(z, y) = 2.

Note that z is adjacent to c2 vertices in Γ1(y), and x is the unique one of such vertices in Ω by Lemma 2.4(ii). Hence z is adjacent to c2−1 vertices in Γ1(y)\Ω.

Next, observe that each vertex in Γ₁(y) \ Ω is adjacent to at most c₂− 1 vertices in Γ1(x) \ Ω. Indeed pick w ∈ Γ1(y) \ Ω. Then by Lemma 2.3(iii),

δ(x, w) = 2.

Since y ∈ Ω ∩ Γ₁(x), w is adjacent to at most c₂− 1 vertices in Γ₁(x) \ Ω.

Now by counting edges between Γ1(x) \ Ω and Γ1(y) \ Ω, we have

¯¯

¯Γ1(x) \ Ω

¯¯

¯(c2− 1) ≤

¯¯

¯Γ1(y) \ Ω

¯¯

¯(c2− 1), (5.2)

and (5.1) follows since c2 > 1. This proves Lemma 5.1.

Lemma 5.2. Let Γ = (X, R) be a distance-regular graph with diameter D ≥ 2, and assume the intersection number c2 > 1. Let Ω denote a weak-geodetically closed subgraph of Γ. Then Ω is regular.

Proof. Suppose Ω is not regular. Since Ω is connected, there exist adjacent vertices x, y ∈ Ω such that

¯¯

¯Ω ∩ Γ1(x)

¯¯

¯ <

¯¯

¯Ω ∩ Γ1(y)

¯¯

¯, (5.3)

contradicting Lemma 5.1.

Corollary 5.3. Let Γ = (X, R) be a distance-regular graph with diameter D ≥ 2, and assume the intersection number c2 > 1. Let Ω denote a weak- geodetically closed subgraph of Γ. Then Ω is distance-regular with intersection array

{c1, c2, · · · , cd; b0− bd, b1− bd, · · · , bd−1− bd}, where d = d(Ω).

Proof. This is immediate from Lemma 5.2, Theorem 4.6 and (1.9).

Corollary 5.4. Let Γ = (X, R) be a distance-regular graph with diameter D ≥ 2, and assume the intersection number c2 > 1. Let Ω, Ω⁰ denote two weak- geodetically closed subgraphs such that Ω⁰ ⊆ Ω. Then the following (i)-(ii) are equivalent.

(i) Ω⁰ = Ω.

(ii) diam(Ω⁰) = diam(Ω).

Proof. (i)−→(ii). Clear.

(ii)−→(i). Ω, Ω⁰ are distance-regular with the same intersection array by Corol- lary 5.3. Now we have

¯¯

¯Ω

¯¯

¯ =

¯¯

¯Ω⁰

¯¯

¯, so Ω = Ω⁰.

Definition 5.5. Let Γ = (X, R) be a graph. For any vertex x ∈ X, and any subset C ⊆ X, define

[x, C] := {v ∈ X|there exists y ∈ C, such that δ(x, v) + δ(v, y) = δ(x, y)}.

The following proposition gives us a description of a weak-geodetically closed subgraph of a distance-regular graph with c₂ > 1.

Proposition 5.6. Let Γ = (X, R) be a distance-regular graph with diameter D ≥ 2, and assume the intersection number c2 > 1. Pick any subgraph Ω of Γ, and fix an integer d (0 ≤ d ≤ D). Then the following (i)-(ii) are equivalent.

(i) Ω is weak-geodetically closed with diameter d.

(ii) There exists a vertex x ∈ Ω that satisfies the following (iia)-(iid).

(iia) Ω is weak-geodetically closed with respect to x.

(iib)

¯¯

¯Ω ∩ Γ1(x)

¯¯

¯ = cd+ a_d.

(iic) Ω = [x, C] for some C ⊆ Γd(x).

(iid) For all v ∈ Ω and for all z ∈ X, if z is adjacent to two distinct vertices in C(v, x), then z ∈ Ω.

Proof. (i)−→(ii). Let x denote any vertex in Ω. (iia) is immediate from Definition 3.1. (iib) is immediate from Corollary 5.3 and (1.9). Suppose (iic) fails. Then there exists a vertex w ∈ Ω such that δ(x, w) < d and B(w, x)∩Ω = ∅.

This contradicts Corollary 5.3. Hence we have (iic). To prove (iid), suppose z is adjacent to distinct vertices w, w⁰ ∈ C(v, x). Then w, w⁰ ∈ Ω by Lemma 2.3(ii), so z ∈ Ω by Lemma 2.4(ii).

(ii)−→(i). First, we prove Ω is weak-geodetically closed. To do this, by parts (ii), (iii) of Theorem 4.6, it suffices to show that Ω is regular. We will show each vertex in Ω has valency cd+ ad. Note that by Lemma 2.6(i), for all w ∈ C,

¯¯

¯Ω ∩ Γ1(w)

¯¯

¯ = cd+ ad. (5.4)

Claim. For all integers j (1 ≤ j ≤ d), and for all pairs of adjacent vertices u, v ∈ Ω such that u ∈ Γj−1(x) and v ∈ Γj(x), we have

¯¯

¯Ω ∩ Γ1(u)

¯¯

¯ ≥

¯¯

¯Ω ∩ Γ1(v)

¯¯

¯. (5.5)

Proof of Claim. Since Γ is regular, to prove (5.5), it suffices to prove

¯¯

¯Γ1(u) \ Ω

¯¯

¯ ≤

¯¯

¯Γ1(v) \ Ω

¯¯

¯. (5.6)

We will count the edges between Γ1(u) \ Ω and Γ1(v) \ Ω in two ways to establish (5.6). On the one hand, we prove that each vertex in Γ₁(u) \ Ω is adjacent to exactly c2− 1 vertices in Γ1(v) \ Ω. Pick z ∈ Γ1(u) \ Ω. Note that z ∈ Γj(x) by Lemma 2.3(iii). We now show δ(z, v) = 2. Obviously δ(z, v) ≤ 2, since z, u, v is a path. Observe z 6= v by construction. Also z, v are not adjacent, otherwise z ∈ A(v, x) ⊆ Ω by Lemma 2.3(ii), a contradiction. Hence δ(z, v) = 2. Next we show

Ω ∩ C(z, v) = {u}. (5.7)

To see this, pick w ∈ Ω ∩ C(z, v) and suppose w 6= u. Note that w 6∈ C(v, x), otherwise z is adjacent to u, w ∈ C(v, x), putting z ∈ Ω by (iid), a contradiction.

Note that w 6∈ A(v, x), otherwise w ∈ A(v, x) and z ∈ A(w, x), putting z ∈ Ω by Lemma 2.3(ii), a contradiction. Hence w ∈ B(v, x). Now z ∈ C(w, x), putting z ∈ Ω by Lemma 2.3(ii), a contradiction. Hence w = u and we have (5.7). Now observe that by (5.7),

¯¯

¯Γ1(z) ∩ (Γ₁(v) \ Ω)

¯¯

¯ =

¯¯

¯C(z, v) \ {u}

¯¯

¯

= c2− 1.

Hence z is adjacent to exactly c₂− 1 vertices in Γ₁(v) \ Ω.

On the other hand, we show that each vertex in Γ1(v) \ Ω is adjacent to at most c2− 1 vertices in Γ1(u) \ Ω. Pick a vertex z ∈ Γ1(v) \ Ω. Observe δ(x, z) = j + 1 by Lemma 2.3(iii). Observe δ(u, z) = 2. Now we have the desired property, since z is adjacent to c2 vertices in Γ1(u) and v ∈ Ω is one of them.

Using above the two ways to count the edges between vertices in Γ1(u) \ Ω and vertices in Γ₁(v) \ Ω, we have

¯¯

¯Γ1(u) \ Ω

¯¯

¯(c2− 1) ≤

¯¯

¯Γ1(v) \ Ω

¯¯

¯(c2− 1), and (5.6) follows since c2 > 1. This proves the claim.

To show Ω is regular, fix any geodesic path x = x₀, x₁, · · · , x_d, where x_d ∈ C, and set

t_l:=

¯¯

¯Ω ∩ Γ1(x_l)

¯¯

¯ (0 ≤ l ≤ d).

Observe

t0 = ad+ cd (5.8)

by assumption (iib),

t_d = a_d+ c_d (5.9)

by (5.4), and

tl−1 ≥ tl (1 ≤ l ≤ d) (5.10)

by the claim. It follows from (5.8)-(5.10) that

tl = ad+ cd (0 ≤ l ≤ d).

By Definition 5.5, Ω is the union of geodesic paths of the above type, and we conclude every vertex in Ω has valency ad + cd. Now Ω is weak-geodetically closed by Theorem 4.6. It remains to show Ω has diameter d. This holds, since Ω is distance-regular by Corollary 5.3 and diamx(Ω) = d by (iic). This proves Proposition 5.6.

If we assume d = 2 and a₂ 6= 0 in Proposition 5.6, we get the following improve- ment.

Proposition 5.7. Let Γ = (X, R) be a distance-regular graph with diameter D ≥ 2, and assume the intersection numbers c₂ > 1, a₂ 6= 0. Then for any subgraph Ω of Γ, the following (i)-(ii) are equivalent.

(i) Ω is weak-geodetically closed with diameter 2.

(ii) There exists a vertex x ∈ Ω that satisfies the following (iia)-(iic).

(iia) Ω is weak-geodetically closed with respect to x.

(iib)

¯¯

¯Ω ∩ Γ1(x)

¯¯

¯ = a2+ c2. (iic) diam_x(Ω) = 2.

Proof. (i)−→(ii). (iia)-(iic) are immediate from Proposition 5.6 (iia)-(iic) (with d = 2).

(ii)−→(i). First, we prove Ω is weak-geodetically closed. To do this, in view of parts (ii), (iii) of Theorem 4.6, it suffices to show Ω is regular. We show each vertex in Ω has valency c2 + a2. Observe by Lemma 2.6(i) that for all

y ∈ Ω ∩ Γ2(x), ¯

¯¯Ω ∩ Γ1(y)

¯¯

¯ = c2+ a₂. (5.11)

Observe by Lemma 5.1 that for all z ∈ Ω ∩ Γ1(x),

¯¯

¯Ω ∩ Γ1(z)

¯¯

¯ ≤ c2+ a2. (5.12)

It remains to show equality holds in (5.12) for all z ∈ Ω ∩ Γ1(x). We suppose this is not the case and get a contradiction. Pick z⁰ ∈ Ω∩Γ1(x) such that

¯¯

¯Ω∩Γ1(z⁰)

¯¯

¯ is minimal and assume ¯

¯¯Ω ∩ Γ1(z⁰)

¯¯

¯ < c2+ a₂. (5.13)

Claim 1. There exists a vertex y ∈ Ω ∩ Γ2(x) that is not adjacent to z⁰.

Proof of Claim 1. If this fails, then z⁰ is adjacent to each vertex in Ω ∩ Γ2(x).

Hence by Lemma 2.3(ii) and the construction, for all z ∈ Ω ∩ Γ1(x),

¯¯

¯Ω ∩ B(z, x)

¯¯

¯ =

¯¯

¯Ω ∩ Γ1(z)

¯¯

¯ − c1− a1

≥

¯¯

¯Ω ∩ Γ1(z⁰)

¯¯

¯ − c1− a₁

=

¯¯

¯Ω ∩ Γ2(x)

¯¯

¯.

Then every vertex in Ω ∩ Γ1(x) is adjacent to every vertex in Ω ∩ Γ2(x). But this is inconsistent with (iib) and a2 6= 0. Hence we have Claim 1.

We fix y, z⁰ for the rest of this proof. Observe, by (iib), Lemma 2.6(iv), C(x, y) ∪ A(x, y) = Ω ∩ Γ1(x).

Now set

γ := 1 c2

X

z∈C(x,y)

¯¯

¯Ω ∩ Γ1(z)

¯¯

¯, (5.14)

and observe γ is the average valency (in Ω) of a vertex in C(x, y). Similarly, set λ := 1

a₂ X

z∈A(x,y)

¯¯

¯Ω ∩ Γ1(z)

¯¯

¯, (5.15)

and observe λ is the average valency (in Ω) of a vertex in A(x, y).

Claim 2. λ < a2+ c2.

Proof of Claim 2. This is immediate from (5.12), (5.13), (5.15), and the obser- vation z⁰ ∈ A(x, y).

Now set

∆ := {w ∈ Ω|δ(x, w) = 2, δ(y, w) ≥ 2}.

Claim 3. ¯¯

¯∆

¯¯

¯c2+ a₂c₂+ c₂ = c₂(γ − a₁− 1) + a₂(λ − a₁ − 1). (5.16) Proof of Claim 3. Let e denote the number of edges connecting vertices in Ω ∩ Γ1(x) to vertices in Ω ∩ Γ2(x). We count e in two ways. On the one hand,

e =

¯¯

¯Ω ∩ Γ2(x)

¯¯

¯c2

=

¯¯

¯∆ ∪ A(y, x) ∪ {y}

¯¯

¯c2

= (

¯¯

¯∆

¯¯

¯ + a2+ 1)c₂. (5.17)

On the other hand, e =

¯¯

¯C(x, y)

¯¯

¯(γ − a1− 1) +

¯¯

¯A(x, y)

¯¯

¯(λ − a1− 1)

= c₂(γ − a₁− 1) + a₂(λ − a₁− 1). (5.18) Line (5.16) is immediate from (5.17), (5.18), and Claim 3 is proved.

Claim 4.

¯¯

¯∆

¯¯

¯c2+ a2c2+ c2 ≥ c2(γ − a1− 1) + a2(a2+ c2− a1− 1). (5.19) Proof of Claim 4. Let f denote the number of edges connecting vertices in Ω ∩ Γ1(y) to vertices in Ω ∩ Γ2(y). Again, we count f in two ways. On the one hand,

f ≤

¯¯

¯Ω ∩ Γ2(y)

¯¯

¯c2

≤

¯¯

¯∆ ∪ A(x, y) ∪ {x}

¯¯

¯c2

=

¯¯

¯∆

¯¯

¯c2+ a2c2+ c2, (5.20)

and on the other hand, using (5.11), f ≥

¯¯

¯C(y, x)

¯¯

¯(γ − a1− 1) +

¯¯

¯A(y, x)

¯¯

¯(a2+ c2− a1− 1)

= c2(γ − a1− 1) + a2(a2+ c2− a1− 1). (5.21) (5.19) is immediate from (5.20), (5.21), and Claim 4 is proved.

Now subtracting (5.16) from (5.19), we find 0 ≥ a2(a2+ c2− λ).

But this is impossible since a2 > 0 by assumption, and a2+ c2− λ > 0 by Claim 2. Hence equality holds in (5.12) for all z ∈ Ω ∩ Γ₁(x). Now Ω is regular by (iib), (5.11), so Ω is weak-geodetically closed by Theorem 4.6. It remains to show Ω has diameter 2. This holds, since Ω is distance-regular by Corollary 5.3, and since diamx(Ω) = 2 by (iic). This proves Proposition 5.7.

6. Distance-regular graphs with many weak-geodetically closed sub- graphs.

In this section, we obtain our first major result, Theorem 6.4. To describe it, we need a few definitions.

Definition 6.1. Let Γ = (X, R) be a graph with diameter D, and let i denote an integer (0 ≤ i ≤ D). Then Γ is said to be i − bounded, if for all integers j (0 ≤ j ≤ i), and for all x, y ∈ X such that δ(x, y) = j, x, y are contained in a common weak-geodetically closed subgraph of diameter j.

Lemma 6.2. Let Γ = (X, R) be a graph with diameter D ≥ 1. Then the following (i)-(iii) hold.

(i) Γ is 0-bounded.

(ii) For each integer i (1 ≤ i ≤ D), if Γ is i-bounded then Γ is (i − 1)-bounded.

(iii) Suppose Γ is (D − 1)-bounded. Then Γ is D-bounded.

Proof. (i)-(iii) are clear from Definition 6.1.

In Theorem 6.4, we obtain a simple criterion for a distance-regular graph Γ to be i-bounded. We will use the following notation.

Definition 6.3. Let Γ = (X, R) be a graph with diameter D ≥ 2. Pick an integer i (2 ≤ i ≤ D). By a parallelogram of length i in Γ, we mean a 4-tuple xyzw of vertices of X such that

δ(x, y) = δ(z, w) = 1, δ(x, w) = i,

δ(x, z) = δ(y, z) = δ(y, w) = i − 1.

We now state the first main theorem of our paper.

Theorem 6.4. Let Γ = (X, R) be a distance-regular graph with diameter D ≥ 2, and assume the intersection numbers c₂ > 1, a₁ 6= 0. Pick an integer i (1 ≤ i < D). Then the following (i)-(ii) are equivalent.

(i) Γ is i-bounded.

(ii) Γ contains no parallelogram of length ≤ i + 1.

The following Lemma proves Theorem 6.4(i)−→(ii).

Lemma 6.5. Let Γ = (X, R) be a distance-regular graph with diameter D ≥ 2.

Pick an integer i (1 ≤ i < D). Suppose Γ is i-bounded. Then Γ contains no parallelogram of length ≤ i + 1.

Proof. Suppose Γ contains a parallelogram xyzw of some length j ≤ i+1. Then δ(y, z) = j − 1 ≤ i. By Definition 6.1, there exists a weak-geodetically closed subgraph Ω of Γ that has diameter j − 1 and contains y, z. Observe x ∈ A(y, z) and w ∈ A(z, y), so x, w ∈ Ω by Lemma 2.3(ii). But δ(x, w) = j, contradicting our assumption that Ω has diameter j − 1. This proves the lemma.

We prove Theorem 6.4 (ii)−→(i) by induction on i. We deal with the case i = 1 in Lemma 6.6, the case i = 2 in Proposition 6.7, prove some general results in Proposition 6.8-Lemma 6.13, and then proceed to the case i ≥ 3 at the end of this section.

Lemma 6.6. Let Γ = (X, R) be a distance-regular graph with diameter D ≥ 2.

Suppose that Γ contains no parallelogram of length 2. Then Γ is 1-bounded.

Proof. Pick x, y ∈ X with δ(x, y) = 1, and set Ω := A(x, y) ∪ {x, y}. Ω is a clique of size a1+ 2 since Γ contains no parallelograms of length 2; in particular, Ω has diameter 1. Also Ω is weak-geodetically closed by Theorem 4.6, since Ω is regular and equality holds in (4.12). This proves Lemma 6.6

Our proof of the case i = 2 in Theorem 6.4 (ii)−→(i) is different from our proof for the case i ≥ 3, also, we can prove it under the assumption a2 6= 0 instead of a₁ 6= 0(One can easily show if Γ contains no parallelogram of length 2 then a1 6= 0 implies a2 6= 0). Hence we prove it separately.

Proposition 6.7. Let Γ = (X, R) be a distance-regular graph with diameter D ≥ 2. Assume that the intersection numbers c2 > 1, a2 6= 0. Suppose that Γ contains no parallelogram of length ≤ 3. Then Γ is 2-bounded.

Proof. Pick any vertices x, y ∈ X with δ(x, y) = 2. Let C be the connected component of Γ₂(x) containing y. Set Ω := [x, C] as in Definition 5.5. We prove

Ω is weak-geodetically closed of diameter 2. To do this, we show Ω satisfies (iia)-(iic) of Proposition 5.7.

Claim 1. Ω is weak-geodetically closed with respect to x. In particular, (iia) of Proposition 5.7 is satisfied.

Proof of Claim 1. Fix z ∈ Ω. By Lemma 2.3(ii) and the construction, it suffices to show A(z, x) ⊆ Ω. This clearly holds if z = x or z ∈ Ω ∩ Γ2(x), so assume z ∈ Ω ∩ Γ1(x). Pick w ∈ A(z, x). By construction, there exists z⁰ ∈ C such that z ∈ C(z⁰, x). Observe that δ(w, z⁰) = 2; otherwise δ(w, z⁰) = 1 and xzwz⁰ is a parallelogram of length 2, a contradiction. Pick w⁰ ∈ C(w, z⁰) \ {z}. Observe that w⁰ ∈ C(z⁰, x)∪A(z⁰, x). Suppose w⁰ ∈ C(z⁰, x). Then δ(z, w⁰) = 2; otherwise δ(z, w⁰) = 1 and z⁰zw⁰x is a parallelogram of length 2, a contradiction. But now w⁰wxz is a parallelogram of length 2, a contradiction. Hence w⁰ ∈ A(z⁰, x), forcing w⁰ ∈ C by construction. Now w ∈ Ω by construction. This proves Claim 1.

Claim 2. For all adjacent vertices z, z⁰ ∈ C, B(x, z) = B(x, z⁰). In particular, B(x, w) = B(x, w⁰) for all w, w⁰ ∈ C.

Proof of Claim 2. Fix adjacent vertices z, z⁰ ∈ C. By symmetry, it suffices to prove B(x, z) ⊆ B(x, z⁰). Suppose there exists a vertex p ∈ B(x, z) \ B(x, z⁰). Of course δ(x, p) = 1, δ(z, p) = 3 by construction, so δ(z⁰, p) = 2 by the triangular inequality. Now the 4-tuple pxz⁰z is a parallelogram of length 3, a contradiction.

Hence B(x, z) = B(x, z⁰). Since C is connected, we have B(x, w) = B(x, w⁰) for all w, w⁰ ∈ C. This proves Claim 2.

Claim 3.

¯¯

¯Ω ∩ Γ1(x)

¯¯

¯ = c2+ a₂. In particular, (iib) of Proposition 5.7 is satisfied.

Proof of Claim 3. Pick z ∈ C. Then it suffices to show Ω ∩ Γ1(x) = C(x, z) ∪ A(x, z). By Claim 1, Ω is weak-geodetically closed with respect to x. Hence by Lemma 2.6(ii), C(x, z) ∪ A(x, z) ⊆ Ω ∩ Γ1(x). Since Γ1(x) = C(x, z) ∪ A(x, z) ∪ B(x, z), it remains to show Ω∩B(x, z) = ∅. Suppose there exists w ∈ Ω∩B(x, z).

By construction, there exists w⁰ ∈ C such that w ∈ C(x, w⁰). But w ∈ B(x, z) = B(x, w⁰) by Claim 2, a contradiction. Hence Ω ∩ B(x, z) = ∅, as desired. This proves Claim 3.

Note that (iic) of Proposition 5.7 is satisfied by the construction. Hence Ω is weak-geodetically closed with diameter 2 by Proposition 5.7. We now have Proposition 6.7.

Proposition 6.8. Let Γ = (X, R) be a distance-regular graph with diameter D ≥ 2. Assume the intersection numbers c₂ > 1, a₂ 6= 0. Suppose Γ contains no parallelogram of length 2 and suppose there exists a weak-geodetically closed subgraph Ω of diameter 2. Fix a vertex x ∈ Ω. Then Ω ∩ Γ₂(x) is connected.

Proof. Note that Ω is distance-regular by Corollary 5.3. Suppose that Ω∩Γ2(x) is not connected. Pick u, v ∈ Ω ∩ Γ2(x) such that there is no path in Ω ∩ Γ2(x) connecting u, v. Observe δ(u, v) = 2, since Ω has diameter 2. For each vertex z ∈ C(u, v), we must have z ∈ C(u, x), otherwise δ(x, z) = 2 and u, z, v is a path in Ω ∩ Γ₂(x). Hence we have C(u, v) ⊆ C(u, x). Now C(u, v) = C(u, x), since both sets have the same cardinality c2. Similarly, we have C(u, v) = C(v, x).

Pick w ∈ A(u, v). Observe δ(x, w) = 2, since w 6∈ C(u, v) = C(u, x). We do not have a path in Ω ∩ Γ2(x) connecting w, v, otherwise we can extend this path to a path in Ω ∩ Γ₂(x) connecting u, v. By the same argument as above, we have C(w, v) = C(w, x) = C(v, x). Now we have

C(u, v) = C(v, x)

= C(w, v).

Pick distinct vertices z, z⁰ ∈ C(u, v) = C(w, v). If δ(z, z⁰) = 1 then the 4-tuple uzz⁰v is a parallelogram of length 2, a contradiction. If δ(z, z⁰) = 2 then the 4-tuple zuwz⁰ is a parallelogram of length 2, another contradiction. Hence we prove Ω ∩ Γ2(x) is connected.

Note. Proposition 6.8 tells us that in the case d = 2 of Proposition 5.6(iic), some C ⊆ Γd(x) is connected. We do not know if this is true in general situation d > 2.

Lemma 6.9. Let Γ = (X, R) be a distance-regular graph with diameter D ≥ 3. Suppose the intersection numbers c2 > 1, a2 6= 0. Pick an integer i (2 ≤ i < D), and suppose Γ contains no parallelogram of any length ≤ i + 1.

Let x be a vertex of Γ, and let Ω be a weak-geodetically closed subgraph of Γ with diameter 2. Suppose there exists a vertex u ∈ Ω ∩ Γ_i−1(x), and suppose Ω ∩ Γi+1(x) 6= ∅. Then for all t ∈ Ω, we have δ(x, t) = i − 1 + δ(u, t).

Proof. We prove this by induction on the integer i. The case i = 2 is immediate from Lemma 2.4(i), so suppose i > 2. Note that

Ω ⊆ Γ_i−1(x) ∪ Γ_i(x) ∪ Γ_i+1(x),

since diam(Ω) = 2, Ω ∩ Γi−1(x) 6= ∅, and Ω ∩ Γi+1(x) 6= ∅. We need to prove Ω ∩ Γ₁(u) ⊆ Γ_i(x) and Ω ∩ Γ₂(u) ⊆ Γ_i+1(x). It suffices to prove that Ω ∩ Γ2(u) ⊆ Γi+1(x), since Ω is distance-regular with diameter 2 and for each vertex w ∈ Ω ∩ Γ₁(u), w ∈ C(z, u) for some vertex z ∈ Ω ∩ Γ₂(u). Suppose that Ω ∩ Γ2(u) 6⊆ Γi+1(x). Since Ω ∩ Γ2(u) is connected by Proposition 6.8, and since Ω ∩ Γ₂(u) ∩ Γ_i+1(x) = Ω ∩ Γ_i+1(x) 6= ∅, there exist adjacent vertices v, v⁰ ∈ Γ2(u) ∩ Ω such that v ∈ Γi+1(x) and v⁰ ∈ Γi(x). Pick x⁰ ∈ C(x, u). Then δ(x⁰, u) = i − 2 and δ(x⁰, v) = i. By induction hypothesis, we have δ(x⁰, v⁰) = i.

Now the 4-tuple xx⁰v⁰v is a parallelogram of length i + 1, a contradiction. This proves the lemma.

Corollary 6.10. Let Γ = (X, R) be a distance-regular graph with diameter D ≥ 3. Assume the intersection numbers c2 > 1, a2 6= 0. Pick any integer i (2 ≤ i < D), and suppose Γ contains no parallelogram of any length ≤ i + 1. Let x be a vertex of Γ, and let Ω be a weak-geodetically closed subgraph of diameter 2. If there exist 2 distinct vertices u, v in Ω such that δ(x, u) = δ(x, v) = i − 1, then δ(x, t) ≤ i for all vertices t ∈ Ω.

Proof. Suppose this is false. Then Ω∩Γi+1(x) 6= ∅, so δ(x, v) = δ(x, u)+δ(u, v) by Lemma 6.9. Since δ(x, v) = δ(x, u) = i − 1, we have δ(u, v) = 0 and hence u = v, a contradiction. This proves the corollary.

Definition 6.11. Let Γ = (X, R) be a graph with diameter D ≥ 2. Pick an integer i (2 ≤ i ≤ D). By a kite of length i, we mean a 4-tuple xyzw of vertices of Γ such that {x, y, z} is a clique, and w is at distances

δ(w, x) = i, δ(w, y) = i − 1, δ(w, z) = i − 1.

Note. A kite of length 2 is the same thing as a parallelogram of length 2.

Lemma 6.12. Let Γ = (X, R) be a graph with diameter D ≥ 2. Fix an integer i (2 ≤ i ≤ D). Suppose Γ contains no parallelogram of any length ≤ i.

Then Γ contains no kite of any length ≤ i.

Proof. Suppose Γ contains a kite of length ≤ i. Of all these kites, pick a kite xyzw with minimal length j. Observe j 6= 2, otherwise xyzw is a parallelogram of length 2. Now pick a ∈ C(w, z). Note that δ(a, z) = j − 2. Observe

δ(a, y) ≤ δ(a, z) + δ(z, y)

= j − 2 + 1

= j − 1, and

δ(a, y) ≥ δ(y, w) − δ(a, w)

= j − 1 − 1

= j − 2,

so δ(a, y) = j − 2 or δ(a, y) = j − 1. If δ(a, y) = j − 2, then the 4-tuple xyza is a kite of length j − 1, contradicting our construction, so δ(a, y) = j − 1. Now the 4-tuple wayx is a parallelogram of length j, a contradiction. This proves the lemma.

Lemma 6.13. Let Γ = (X, R) be a distance-regular graph with diameter D ≥ 3. Assume the intersection numbers c₂ > 1 and a₁ 6= 0. Pick an integer i

(2 ≤ i < D), and suppose Γ contains no parallelogram of any length ≤ i + 1.

Let x be a vertex of X, and let Ω be a weak-geodetically closed subgraph of diameter 2. Set j :=max{δ(x, w) |w ∈ Ω}, and assume j ≤ i. Then Ω ∩ Γj(x) is connected.

Proof. Note that Ω is distance-regular by Corollary 5.3. Since Γ contains no parallelogram of length 2, for any vertex u ∈ Ω, Ω ∩ Γ1(u) is a disjoined union of cliques of size a1+ 1. Let l denote the number of these cliques in Ω ∩ Γ1(u).

Suppose Ω∩Γj(x) is not connected. Then there exist two vertices t, s ∈ Ω∩Γj(x) such that there is no path in Ω ∩ Γj(x) connecting t and s. Note that δ(t, s) = 2, since diam(Ω) = 2. Consider the set

N := {t} ∪ (Γj(x) ∩ Ω ∩ Γ1(t)).

Claim 1.

¯¯

¯N

¯¯

¯ ≥ 1 + la1.

Proof of Claim 1. Γ contains no kite of length j by Lemma 6.12, , so

¯¯

¯K ∩ Γj−1(x)

¯¯

¯ ≤ 1 for all maximal cliques K ⊆ Ω ∩ Γ1(t). Hence

¯¯

¯K ∩ N

¯¯

¯ ≥ a1 for all maximal cliques K ⊆ Ω ∩ Γ1(t). Since there are l such cliques,

¯¯

¯N

¯¯

¯ ≥

¯¯

¯{x}

¯¯

¯ + la1

= 1 + la1, as desired.

Claim 2. δ(z, s) = 2 for any z ∈ N.

Proof of Claim 2. δ(z, s) ≤ 2, since diam(Ω) = 2. z 6= s by construction. Also δ(z, s) 6= 1, otherwise t, z, s is a path in Ω ∩ Γj(x), a contradiction. Hence δ(z, s) = 2.

Now consider the set

M := [

z∈N

C(s, z).

Claim 3.

¯¯

¯M

¯¯

¯ ≤ l.

Proof of Claim 3. Note that M ⊆ Ω ∩ Γ1(s) by construction and Lemma 2.3(ii), so there is no vertex in M with distance j + 1 to x. If the distance from x to a vertex in M is j, then we find a path in Ω ∩ Γj(x) connecting t and s, a contradiction. Thus

M ⊆ Ω ∩ Γ1(s) ∩ Γj−1(x).

Since δ(x, s) = j and since Γ contains no kite of length j by Lemma 6.12, each maximal clique in Ω ∩ Γ₁(s) contains at most one vertex in M. Hence |M | ≤ l.