Strongly Closed Subgraphs in a Distance-Regular Graph with c 2 > 1

Digital Object Identifier (DOI) 10.1007/s00373-008-0814-8Graphs and Combinatorics (2008) 24:537–550

Graphs andCombinatorics© Springer-Verlag 2008

Strongly Closed Subgraphs in a Distance-Regular Graphwith c2 > 1

Akira HirakiDivision of Mathematical Sciences, Osaka Kyoiku University, Osaka 582-8582, Japan.

Abstract. Let Γ be a distance-regular graph of diameter d ≥ 3 with c2 > 1. Let m be aninteger with 1 ≤ m ≤ d − 1. We consider the following conditions:

(SC)m : For any pair of vertices at distance m there exists a strongly closed subgraph ofdiameter m containing them.

(B B)m : Let (x, y, z) be a triple of vertices with ∂Γ (x, y) = 1 and ∂Γ (x, z) = ∂Γ (y, z) = m.Then B(x, z) = B(y, z).

(C A)m : Let (x, y, z) be a triple of vertices with ∂Γ (x, y) = 2, ∂Γ (x, z) = ∂Γ (y, z) = mand |C(z, x) ∩ C(z, y)| ≥ 2. Then C(x, z) ∪ A(x, z) = C(y, z) ∪ A(y, z).

Suppose that the condition (SC)m holds. Then it has been known that the condition(B B)i holds for all i with 1 ≤ i ≤ m. Similarly we can show that the condition (C A)i holdsfor all i with 1 ≤ i ≤ m. In this paper we prove that if the conditions (B B)i and (C A)i holdfor all i with 1 ≤ i ≤ m, then the condition (SC)m holds. Applying this result we give asufficient condition for the existence of a dual polar graph as a strongly closed subgraph in Γ.

Key words. distance-regular graph, strongly closed subgraph.

1. Introduction

The reader is referred to the next section or [1,2] for the definitions. Known distance-regular graphs have many subgraphs of high regularity. For example the Odd graphs,the doubled Odd graphs, the doubled Grassmann graphs, the Hamming graphs andthe dual polar graphs satisfy the following condition: “ For any pair of vertices thereexists a strongly closed subgraph containing them whose diameter is the distancebetween them. ” Here we do not assume that a strongly closed subgraph is distance-regular.

Our problem is to classify a distance-regular graph satisfying this condition. Alot of partial answers have been obtained. For example see [9,11,12], [14] and [15,16].We remark that a strongly closed subgraph in this paper is called a weak-geodeticallyclosed subgraph in [15,16].

Another problem is what kinds of conditions are sufficient for there exists astrongly closed subgraph. Several sufficient conditions have been obtained. For ex-ample see [2, § 4.3], [4–7] and [15]. In this paper we give another sufficient condition.

538 A. Hiraki

Let Γ be a distance-regular graph of diameter d ≥ 3 and valency k ≥ 3. Let mbe an integer with 1 ≤ m ≤ d − 1. In this paper we consider the following threeconditions:

(SC)m : For any pair of vertices at distance m there exists a strongly closed sub-graph of diameter m containing them.

(B B)m : Let (x, y, z) be a triple of vertices with ∂Γ (x, y) = 1 and ∂Γ (x, z) =∂Γ (y, z) = m. Then B(x, z) = B(y, z).

(C A)m : Let (x, y, z)be a triple of vertices with ∂Γ (x, y) = 2, ∂Γ (x, z)= ∂Γ (y, z) =m and |C(z, x)∩C(z, y)| ≥ 2. Then C(x, z)∪ A(x, z) = C(y, z)∪ A(y, z).

It is clear that B(x, z) = B(y, z) if and only if C(x, z) ∪ A(x, z) = C(y, z) ∪ A(y, z).The condition (B B)m is the same as the condition (SS)m in [5]. It had been shownthat the condition (SC)m implies the condition (B B)m (see [5, Proposition 2.1] orLemma 7).

In [8, Theorem 1] we proved that if the condition (SC)m holds, then the condition(SC)i holds for all i with 1 ≤ i ≤ m. It follows that if the condition (SC)m holds,then the condition (B B)i holds for all i with 1 ≤ i ≤ m. Suppose the condition(B B)m does not hold. Then there exists a triple (x, y, z) of vertices with ∂Γ (x, y) =1, ∂Γ (x, z) = ∂Γ (y, z) = m such that B(x, z) = B(y, z). Take w ∈ B(x, z)\B(y, z).Then ∂Γ (x, y) = ∂Γ (z, w) = 1, ∂Γ (x, w) = m + 1 and ∂Γ (x, z) = ∂Γ (y, z) =∂Γ (y, w) = m. Such a quadruple (x, y, z, w) of vertices is called a parallelogram oflength m + 1 in [16]. Conversely if there exists a parallelogram (x, y, z, w) of lengthm + 1, then (x, y, z) is triple of vertices with ∂Γ (x, y) = 1, ∂Γ (x, z) = ∂Γ (y, z) = msuch that B(x, z) = B(y, z). Hence the condition (B B)m holds if and only if thereis no parallelogram of length m + 1. For the case a1 = 0, C.-W. Weng [16, Theorem1] proved that if there is no parallelogram of length i with i ≤ m + 1, then thecondition (SC)i holds for all i with 1 ≤ i ≤ m. We consider the general case andgive a sufficient condition for that the condition (SC)m holds.

We consider a distance-regular graph Γ with a1 = a2 = 0 and c2 > 1. Then Γ

contains no triple (x, y, z) of vertices with ∂Γ (x, y) = 1, ∂Γ (x, z) = ∂Γ (y, z) = iwith i = 1, 2 and hence no parallelogram of length 2 and 3 (i.e., the condition(B B)1 and (B B)2 hold). So it seems that the condition (B B)i for all i with i ≤ mis not enough to show the condition (SC)m in general case. In this case a stronglyclosed subgraph of diameter 2 is the complete bipartite graph Kc2,c2 and hence thecondition (SC)2 holds if and only if the condition (C A)2 holds (see Lemma 6 ).The condition (C A)m is a generalization of this condition. We can show that thecondition (SC)m implies that the condition (C A)m (see Lemma 7). Therefore if thecondition (SC)m holds, then the conditions (B B)i and (C A)i hold for all i with1 ≤ i ≤ m. In this paper we prove that these conditions are necessary and sufficientfor the condition (SC)m . The following is our main result.

Theorem 1. Let Γ be a distance-regular graph of diameter d ≥ 3 and c2 > 1. Let mbe an integer with 1 ≤ m ≤ d − 1. Then the following conditions are equivalent.

(i) The condition (SC)m holds.(ii) The conditions (B B)i and (C A)i hold for all i with 1 ≤ i ≤ m.

Strongly Closed Subgraphs in a Distance-Regular graph with c2 > 1 539

Theorem 1 is a generalization of the result of [16]. In [10] we gave a sufficientcondition for that there exists a bipartite geodetically closed subgraph of diameterm. Theorem 1 is also a generalization of this result for the case a1 = · · · = am = 0.

In [5] we introduced the condition (C R)m and showed that it is a necessary andsufficient for the condition (SC)m . Theorem 1 is also a refinement of this result forthe case c2 > 1.

We will be able to obtain lots of applications of Theorem 1. In this paper weprove the following result as an application of Theorem 1.

Proposition 2. Let Γ be a distance-regular graph of diameter d ≥ 5. Let q and m beintegers with q > 1 and 4 ≤ m ≤ d − 1. Suppose for any three distinct vertices thenumber of their common neighbors is 0, 1 or q + 1. Then the following conditions areequivalent.

(i) ci = 1 + q + · · · + qi−1 and ai = 0 for all i with 1 ≤ i ≤ m.

(ii) The condition (SC)m holds. Moreover q is a prime power and any strongly closedsubgraph of diameter m is the dual polar graph on [Dm(q)].

This paper is organized as follows. In Section 2, we recall some definitions andbasic terminologies for distance-regular graphs and strongly closed subgraphs. Wecollect several known results for strongly closed subgraphs and give some conse-quences. In section 3, we collect several basic results for a distance-regular graphsatisfying the conditions (B B)i and (C A)i . We prove Theorem 1 and Proposition 2in Section 4 and Section 5, respectively.

2. Preliminaries

First we recall our notation and terminology. Let Γ = (V Γ, EΓ ) be a connectedgraph with usual distance ∂Γ and diameter d = d(Γ ). For a vertex u in Γ we denoteby Γ j (u) the set of vertices which are at distance j from u. For two vertices x and yin Γ with ∂Γ (x, y) = j, let

C(x, y) :=Γ j−1(x)∩Γ1(y), A(x, y) :=Γ j (x)∩Γ1(y), B(x, y) :=Γ j+1(x)∩Γ1(y).

Definition 3. Let i be an integer with 0 ≤ i ≤ d.

(i) We say ci (Γ )-exists if ci (Γ ) = |C(x, y)| is a constant whenever ∂Γ (x, y) = i.(ii) We say ai (Γ )-exists if ai (Γ ) = |A(x, y)| is a constant whenever ∂Γ (x, y) = i.

(iii) We say bi (Γ )-exists if bi (Γ ) = |B(x, y)| is a constant whenever ∂Γ (x, y) = i.

A connected graph Γ of diameter d is said to be distance-regular if ci (Γ )-existsand bi (Γ )-exists for all i = 0, 1, . . . , d. Then Γ is a regular graph of valency k =k(Γ ) = b0(Γ ) and ai (Γ )-exists with ai (Γ ) = k(Γ ) − ci (Γ ) − bi (Γ ) for all i =0, 1, . . . , d. We remark that c0(Γ ) = a0(Γ ) = bd(Γ ) = 0 and c1(Γ ) = 1. Theconstants ci (Γ ), ai (Γ ) and bi (Γ ) ( i = 0, 1 . . . , d) are called the intersection numbersof Γ.

540 A. Hiraki

For more background information about distance-regular graphs we refer thereader to [1,2].

Let Γ be a distance-regular graph of diameter d = d(Γ ) ≥ 2 and valencyk = k(Γ ) ≥ 3. We denote by ci , ai and bi the intersection numbers ci (Γ ), ai (Γ )

and bi (Γ ) of Γ.

A quadruple (x, y, z, w)of vertices is called a quadrangle if ∂Γ (x, y) = ∂Γ (y, z) =∂Γ (z, w) = ∂Γ (w, x) = 1 and ∂Γ (x, z) = ∂Γ (y, w) = 2.

Let ∆ be a subset of vertices in Γ. We identify ∆ with the induced subgraph onit. Let x ∈ ∆. ∆ is called strongly closed (resp. geodetically closed) with respect tox if C(x, y) ∪ A(x, y) ⊆ ∆ (resp. C(x, y) ⊆ ∆) for any y ∈ ∆. ∆ is called stronglyclosed (resp. geodetically closed) if it is strongly closed (resp. geodetically closed) withrespect to x for any x ∈ ∆. It is clear that a strongly closed subgraph is geodeticallyclosed.

Let ∆ be a geodetically closed subgraph of Γ with diameter m := d(∆). For anyvertices x and y in ∆, a shortest path connecting x and y in Γ is contained in ∆. Sothe distance ∂∆ in ∆ is coincide with the distance ∂Γ in Γ. It follows that ci (∆)-existswith ci (∆) = ci for all i = 0, 1, . . . , m. Moreover if ∆ is strongly closed, then ai (∆)-exists with ai (∆) = ai for all i = 0, 1, . . . , m. Hence if a strongly closed subgraph∆ is regular of valency k(∆), then bi (∆)-exists with bi (∆) = k(∆) − ci − ai for alli = 0, 1, . . . , m and hence it is distance-regular. However we remark that there existexamples of non-regular strongly closed subgraphs in a distance-regular graph (see[9,11,14]).

The following lemma is a well known fact for a strongly closed subgraph.

Lemma 4. Let ∆ ⊆ V Γ, x, y ∈ ∆ and z ∈ V Γ. Suppose ∆ is strongly closed withrespect to x and ∂Γ (x, z) + ∂Γ (z, y) ≤ ∂Γ (x, y) + 1. Then z ∈ ∆.

Proof. We prove the assertion by induction on ∂Γ (z, y). Suppose ∂Γ (z, y) = 1.

Then we have z ∈ C(x, y) ∪ A(x, y) ⊆ ∆. Suppose ∂Γ (z, y) ≥ 2. Take z′ ∈ C(y, z).Then ∂Γ (x, z′) + ∂Γ (z′, y) ≤ ∂Γ (x, z) + ∂Γ (z, y) ≤ ∂Γ (x, y) + 1 and thus z′ ∈ ∆

by the inductive hypothesis. Since ∂Γ (x, z) ≤ ∂Γ (x, y) + 1 − ∂Γ (z, y) = ∂Γ (x, y) −∂Γ (z′, y) ≤ ∂Γ (x, z′), z ∈ C(x, z′) ∪ A(x, z′) ⊆ ∆. The assertion is proved.

Since a strongly closed subgraph of diameter 1 is a clique of size a1 + 2, it isstraightforward to see the following fact.

Lemma 5. The following conditions are equivalent.

(i) The condition (B B)1 holds.(ii) Γ contains no induced subgraph K2,1,1.

(iii) Each edge lies on a clique of size a1 + 2.

(iv) The condition (SC)1 holds.

Suppose Γ is a distance-regular graph of diameter d ≥ 3 such that a1 = a2 = 0and c2 = 2. Then each pair of vertices at distance 2 and their two common neighborsinduce a quadrangle which is strongly closed in Γ. Hence the condition (SC)2 alwaysholds. For the case c2 > 2 we have the following fact.


Lemma 6. Suppose a1 = a2 = 0 and c2 > 2. Then the following conditions areequivalent.

(i) The condition (C A)2 holds.(ii) For any three distinct vertices the number of their common neighbors is 0, 1 or

c2.

(iii) The condition (SC)2 holds. In particular, any strongly closed subgraph of diam-eter 2 is the complete bipartite graph Kc2,c2 .

Proof. (i) ⇒ (ii): Suppose three distinct vertices x, y and z have at least two commonneighbors. Then they are at distance 2 each other since a1 = 0. Thus we haveC(x, z) = C(y, z). It follows that x, y and z have exactly c2 common neighbors.(ii) ⇒ (iii): Let (u1, u2) be a pair of vertices at distance 2. Let w1 = w2 ∈ C(u1, u2)

and Λ = C(u1, u2)∪C(w1, w2). Let w3 ∈ C(u1, u2)\w1, w2. Then w1, w2, w3 haveat least two common neighbors, namely u1, u2. Hence exactly c2 common neighbors.This implies that w3 is adjacent to each vertex in C(w1, w2). Similarly any vertexu3 in C(w1, w2) is adjacent to each vertex in C(u1, u2). Then Λ is the completelybipartite graph Kc2,c2 which is strongly closed in Γ.

(iii) ⇒ (i): Let (x, y, z) be a triple of vertices with ∂Γ (x, y) = ∂Γ (x, z) = ∂Γ (y, z) =2 such that there exist distinct vertices u and v in C(z, x) ∩ C(z, y). Let ∆ be astrongly closed subgraph of diameter 2 containing x and z. Then u, v ∈ C(z, x) ⊆∆, ∂Γ (u, v) = 2 and y ∈ C(u, v) ⊆ ∆. Hence C(x, z) = Γ1(z) ∩ ∆ = C(y, z). Thelemma is proved.

Let m be an integer with 1 ≤ m ≤ d − 1. In [5, Proposition 2.1] we showedthat if the condition (SC)m holds, then the condition (B B)m holds (the condition(B B)m is the same as the condition (SS)m in [5]). Here we reprove this result. Usingthe same proof we can show that if the condition (SC)m holds, then the condition(C A)m holds.

Lemma 7. If the condition (SC)m holds, then the conditions (B B)m and (C A)m hold.

Proof. Let (x, y, z) be a triple of vertices with ∂Γ (x, z) = ∂Γ (y, z) = m. Let ∆ bea strongly closed subgraph of diameter m containing x and z. We prove y ∈ ∆ ineach case. Then

C(x, z) ∪ A(x, z) = Γ1(z) ∩ ∆ = C(y, z) ∪ A(y, z).

If ∂Γ (x, y) = 1, then y ∈ A(z, x) ⊆ ∆. The condition (B B)m holds.If ∂Γ (x, y) = 2 and there exist two distinct vertices w and w′ in C(z, x) ∩ C(z, y),

then w,w′ ∈ C(z, x) ⊆ ∆ and thus y ∈ Γ1(w)∩Γ1(w′) ⊆ ∆. The conditions (C A)m

holds.

In [8, Theorem 1] we have proved that if the condition (SC)m holds, then thecondition (SC)i holds for all i with 1 ≤ i ≤ m. Hence we have the following corollaryas a direct consequence.

542 A. Hiraki

Corollary 8. Suppose the condition (SC)m holds. Then the conditions (SC)i , (B B)i

and (C A)i hold for all i with 1 ≤ i ≤ m.

This corollary shows that the statement (i) implies the statement (ii) in Theo-rem 1.

The rest of this section we recall several basic results for a distance-regular graphsatisfying the condition (B B)m .

Lemma 9. Let x, y, w be vertices with ∂Γ (x, w) = m+1, ∂Γ (x, y) = 1 and ∂Γ (y, w) =m. Suppose the condition (B B)m holds. Then C(x, w) ∩ A(y, w) = ∅ and A(y, w) ⊆A(x, w). In particular, [ A(x, w)\ A(y, w) ] ⊆ B(y, w) and [ C(x, w)\C(y, w) ] ⊆B(y, w).

Proof. If there exists z ∈ C(x, w)∩ A(y, w), then ∂Γ (x, z) = ∂Γ (y, z) = m and w ∈B(x, z)\B(y, z). This contradicts the condition (B B)m . Hence C(x, w)∩A(y, w) = ∅and A(y, w) ⊆ A(x, w). The second assertion follows from the first one.

Lemma 10. Suppose the condition (B B)h holds for all h with 1 ≤ h ≤ m. Then thefollowing hold.

(i) For any integers i, j with i + j ≤ m + 1, there is no quadruple (x, y, z, w)

of vertices such that ∂Γ (x, y) = 1, ∂Γ (x, z) = ∂Γ (y, z) = i, ∂Γ (z, w) =j, ∂Γ (y, w) = i + j − 1 and ∂Γ (x, w) = i + j.

(ii) For any pair (u, v) of vertices at distance j + 1 with j ≤ m, there are no edgesin C(u, v).

Proof. (i) We prove the assertion by induction on j. Suppose there exists a quadruple(x, y, z, w) of vertices as in the statement and derive a contradiction. If j = 1, thenw ∈ B(x, z)\ B(y, z) which contradicts the condition (B B)i . If j ≥ 2, then takez′ ∈ C(w, z). We have z′ ∈ B(x, z) = B(y, z) by the condition (B B)i . The quadruple(x, y, z′, w) of vertices contradicts the inductive hypothesis.(ii) Suppose there exists an edge (y, z) in C(u, v). Then the quadruple (v, y, z, u) ofvertices contradicts (i).

3. Some Basic Results

Let m and d be integers with 2 ≤ m ≤ d − 1. Throughout this section Γ denotesa distance-regular graph of diameter d and c2 > 1 which satisfies the conditions(B B)h and (C A)h for all h with 1 ≤ h ≤ m.

Lemma 11. Let j be an integer with 1 ≤ j ≤ m. Let u be a vertex in Γ. Then there isno quadrangle (x, y, z, w) such that x, y, z ∈ Γ j (u) and w ∈ Γ j+1(u).

Proof. We prove the assertion by induction on j. Suppose there exists a quadrangle(x, y, z, w) of vertices as in the statement and derive a contradiction. Suppose j =


1. Then (x, z, y, u) forms K2,1,1 which contradicts the condition (B B)1. Supposej ≥ 2. Let u′ ∈ C(x, u) ⊆ C(w, u). If ∂Γ (u′, z) = j + 1, then ∂Γ (u′, y) = j and(u, u′, y) contradicts the condition (B B) j as z ∈ B(u′, y)\B(u, y). If ∂Γ (u′, z) = j,then (u, u′, z) contradicts the condition (B B) j as w ∈ B(u, z)\ B(u′, z). Assume∂Γ (u′, z) = j − 1. If ∂Γ (u′, y) = j − 1, then u′ and (x, y, z, w) contradicts theinductive hypothesis. If ∂Γ (u′, y) = j, then (y, w, u′) contradicts the condition(C A) j since x, z ⊆ C(u′, y) ∩ C(u′, w) and u ∈ B(w, u′)\ B(y, u′). The desiredresult is proved.

Lemma 12. Let j be an integer with 1 ≤ j ≤ m. Let (u, w, z) be a triple of verticessuch that ∂Γ (u, w) = j + 1, ∂Γ (w, z) = 1 and ∂Γ (u, z) = j. Then the following hold.

(i) Let x and x ′ be distinct vertices in C(u, z). Then C(x, w) ∩ C(x ′, w) = z.(ii) Let y ∈ A(u, z). Then ∂Γ (y, w) = 2 and [ C(y, w)\z ] ⊆ Γ j+1(u).

(iii) Let y and y′ be distinct vertices in A(u, z). Then C(y, w) ∩ C(y′, w) = z.(iv) Let p and p′ be distinct vertices in C(u, z)∪A(u, z). Then C(p, w)∩C(p′, w) =

z.

Proof. (i) Since x, x ′ ⊆ C(u, z),we have j ≥ 2.We prove the assertion by inductionon j. Suppose there exists z′ ∈ [ C(x, w)∩C(x ′, w) ]\z to derive a contradiction. Ifj = 2, then (x ′, w, x) contradicts the condition (C A)2 as z, z′ ⊆ C(x, x ′)∩C(x, w)

and u ∈ B(w, x)\B(x ′, x). Assume j ≥ 3. Let v ∈ C(x, u). Then ∂Γ (v,w) = j and∂Γ (v, z) = ∂Γ (v, z′) = j − 1. If x ′ ∈ C(v, z), then we have a contradiction by theinductive hypothesis. If x ′ ∈ C(v, z), then x ′ ∈ [ C(u, z)\C(v, z) ] ⊆ B(v, z) byLemma 9. Then (x ′, w, v) contradicts the condition (C A) j since z, z′ ⊆ C(v, x ′)∩C(v,w) and u ∈ B(w, v)\B(x ′, v). The desired result is proved.(ii) We have ∂Γ (y, w) = 2 by Lemma 10 (ii). Let z′ ∈ C(y, w)\z. Then (z, y, z′, w)

is a quadrangle. Hence z′ ∈ Γ j+1(u) by Lemma 11.(iii) Suppose there exists w′ ∈ [ C(y, w) ∩ C(y′, w) ]\z. Then w′ ∈ Γ j+1(u) by (ii).Hence the quadrangle (y, z, y′, w′) contradicts Lemma 11.(iv) This follows by (i)–(iii). The lemma is proved.

As a direct consequence of the above lemma we have the following.

Lemma 13. Let j be an integer with 1 ≤ j ≤ m. Let (u, w, z) be a triple of verticessuch that ∂Γ (u, w) = j + 1, ∂Γ (w, z) = 1 and ∂Γ (u, z) = j. Then

⋃

x∈C(u,z)

[ C(x, w)\z ] ⊆ [ C(u, w)\z ] (1)

and ⋃

y∈A(u,z)

[ C(y, w)\z ] ⊆ [ A(u, w)\ A(z, w) ], (2)

where the left-hand sides of (1) and (2) are disjoint. In particular,

c j (c2 − 1) + c1 ≤ c j+1 (3)

544 A. Hiraki

anda j (c2 − 1) + a1 ≤ a j+1 (4)

hold.

The inequalities (3) and (4) had been proved in [16, Proposition 3.2] under theassumption that there exists a strongly closed subgraph of diameter m. We areinterested in the case that the equalities hold in (3) and (4). Here we considerthe refinement of this result. The similar result had been shown in [12] under theassumption that the condition (SC)m holds.

Lemma 14. Let i be an integer with 1 ≤ i ≤ m. Let (u, v, w, z) be a quadruple ofvertices such that ∂Γ (u, w) = i +1, ∂Γ (u, v) = ∂Γ (w, z) = 1, ∂Γ (v,w) = ∂Γ (u, z) =i and ∂Γ (v, z) = i − 1. Then

⋃

x∈[ C(u,z)\C(v,z) ][ C(x, w)\z ] ⊆ [ C(u, w)\C(v,w) ] (5)

and ⋃

y∈[ A(u,z)\A(v,z) ][ C(y, w)\z ] ⊆ [ A(u, w)\ A(v,w) ], (6)

where the left-hand sides of (5) and (6) are disjoint. In particular,

(ci − ci−1)(c2 − 1) ≤ ci+1 − ci (7)

and(ai − ai−1)(c2 − 1) ≤ ai+1 − ai (8)

hold.

Proof. Let x ∈ C(u, z)\C(v, z). Then x ∈ B(v, z) by Lemma 9. Let z′ ∈ C(x, w)\z. If z′ ∈ C(v,w), then (x, w, v) contradicts the condition (C A)i since z, z′ ⊆C(v, x) ∩ C(v,w) and u ∈ B(w, v)\ B(x, v). Hence z′ ∈ [ C(u, w)\C(v,w) ] andthus (5) holds. Let y ∈ A(u, z)\ A(v, z). Then y ∈ B(v, z) by Lemma 9. Let z′′ ∈C(y, z)\z. Then z′′ ∈ B(u, y) = B(v, y) by Lemma 12 (ii) and the condition (B B)i .

It follows that z′′ ∈ [ A(u, w)\ A(v,w) ] and thus (6) holds. The left-hand sides of(5) and (6) are disjoint by Lemma 12 (i),(iii). The lemma is proved.

As a direct consequence of these lemmas we have the following result.

Corollary 15. Let j be an integer with 1 ≤ j ≤ m. Then the following conditions areequivalent:

(i) The equality holds in (3). (resp. in (4).)(ii) The equality holds in (1) (resp. in (2)) for any triple (u, w, z) of vertices as in

Lemma 13.(iii) The equality holds in (1) (resp. in (2)) for some triple (u, w, z) of vertices as in

Lemma 13.


(iv) The equality holds in (7) (resp. in (8)) for all i with 1 ≤ i ≤ j.(v) The equality holds in (5) (resp. in (6)) for all i with 1 ≤ i ≤ j and for any

quadruple (u, v, w, z) of vertices as in Lemma 14.(vi) For all i with 1 ≤ i ≤ j the equality holds in (5) (resp. in (6)) for some quadruple

(u, v, w, z) of vertices as in Lemma 14.

Proof. The equations (3) and (4) can be obtained by adding the equations (7) and(8) for i = 1, 2, . . . , j, respectively. The assertion follows from Lemma 13 andLemma 14.

We remark that if Γ is either a Hamming graph or a dual polar graph, then theconditions (SC)h, (B B)h and (C A)h hold for all h with 1 ≤ h ≤ d − 1 (see [12]).Also the equalities hold in (3), (4), (7) and (8) for all i and j with 1 ≤ i, j ≤ d − 1.

To close this section we prove the following lemma.

Lemma 16. Let i be an integer with 1 ≤ i ≤ m. Let (x, y, z) be a triple of verticessuch that ∂Γ (x, z) = i, ∂Γ (x, y) = 1, and ∂Γ (y, z) = i − 1. Then the following hold.

(i) Let p, p′ ∈ B(y, z)\B(x, z) with ∂Γ (p, p′) = 2. Then C(p, p′)∩Γi+1(y) = ∅.

(ii) Let p, p′ ∈ B(y, z)\B(x, z). Then B(p, y) = B(p′, y).

Proof. (i) Suppose there exists w ∈ C(p, p′)∩Γi+1(y) and derive a contradiction. Ifp ∈ A(x, z), then w ∈ B(y, p) = B(x, p) by the condition (B B)i . Apply Lemma 12(ii) to (x, w, p). Then p′ ∈ [ C(z, w)\p ] ⊆ Γi+1(x) which is a contradiction. Ifp ∈ C(x, z), then apply Lemma 14 to (y, x, w, p). Then p′ ∈ [ C(z, w)\p ] ⊆[ C(y, w)\C(x, w) ] ⊆ B(x, w) by Lemma 9 which is a contradiction.(ii) If ∂Γ (p, p′) ≤ 1, then the assertion is clear by the condition (B B)i . Assume∂Γ (p, p′) = 2. Suppose there exists u ∈ B(w, y)\B(p, y) and derive a contradiction.Apply Lemma 14 to a quadruple (u, y, p′, z) of vertices. If p ∈ C(u, z), then p ∈C(u, z)\C(y, z) and thus

∅ = [ C(p, p′)\z ] ⊆ [ C(u, p′)\C(y, p′) ] ⊆ B(y, p′) ⊆ Γi+1(y)

by Lemma 9. This contradicts (i). If p ∈ A(u, z), then p ∈ A(u, z)\ A(y, z) andhence

∅ = [ C(p, p′)\z ] ⊆ [ A(u, p′)\ A(y, p′) ] ⊆ B(y, p′) ⊆ Γi+1(y)

by Lemma 9. This contradicts (i). The lemma is proved.

4. Strongly Closed Subgraphs

Let m and d be integers with 2 ≤ m ≤ d − 1. Throughout this section Γ denotesa distance-regular graph of diameter d and c2 > 1 which satisfies the conditions(B B)h and (C A)h for all h with 1 ≤ h ≤ m. Then c j < c j+1 and a j ≤ a j+1 for all jwith 1 ≤ j ≤ m by Lemma 13. For any vertices x and y we denote by

Π(x, y) = w ∈ V Γ | ∂Γ (x, w) + ∂Γ (w, y) = ∂Γ (x, y)the set of vertices lying on a shortest path connecting x and y.

546 A. Hiraki

Definition 17. Let (u, v) be a pair of vertices at distance m. Let

Ω := w ∈ Γm(u) | C(w, u) ∪ A(w, u) = C(v, u) ∪ A(v, u)and

∆ = ∆(u, v) :=⋃

w∈Ω

Π(u, w).

For any x ∈ ∆, we denote by κ(x) := |Γ1(x) ∩ ∆| that is the valency of x in ∆.

Fix a pair (u, v) of vertices at distance m and let ∆ = ∆(u, v) be as in Defini-tion 17. We prove that ∆ is a strongly closed subgraph of diameter m.

We remark that C(x, z) ∪ A(x, z) = C(y, z) ∪ A(y, z) if and only if B(x, z) =B(y, z).

Lemma 18. Let ∆ = ∆(u, v) be as in Definition 17. Then the following hold.

(i) Let p ∈ ∆ with ∂Γ (u, p) ≤ m − 1. Then B(u, p) ∩ ∆ = ∅.

(ii) Γi (u) ∩ ∆ = ∅ if and only if 0 ≤ i ≤ m.

(iii) Let w ∈ Ω. Then Γ1(w) ∩ ∆ = C(u, w) ∪ A(u, w) and κ(w) = cm + am .

(iv) ∆ is strongly closed with respect to u.

(v) Let w ∈ Ω. Then Γ1(u) ∩ ∆ = C(w, u) ∪ A(w, u) and κ(u) = cm + am .

Proof. (i),(ii) These are clear from the definition of ∆.

(iii) We have B(u, w)∩∆ = ∅ and C(u, w) ⊆ Π(u, w) ⊆ ∆. Take any w′ ∈ A(u, w).

Then B(w′, u) = B(w, u) = B(v, u) by the condition (B B)m and w ∈ Ω. Hencew′ ∈ Ω ⊆ ∆.

(iv) Let x ∈ ∆ and j := ∂Γ (u, x). We prove C(u, x) ∪ A(u, x) ⊆ ∆ by induction onm − j. If ∂Γ (u, x) = m, then the claim follows by (iii). Suppose ∂Γ (u, x) ≤ m − 1.

Then there exists w ∈ Ω such that x ∈ Π(u, w). Let z ∈ C(w, x). We have C(u, x) ⊆Π(u, w) ⊆ ∆ and z ∈ Π(u, w) ⊆ ∆. Let y ∈ A(u, x). Then we have ∂Γ (z, y) = 2and ∅ = [ C(y, z)\x ] ⊆ Γ j+1(u) by Lemma 12 (ii). Let z′ ∈ C(y, z)\x. Thenz′ ∈ A(u, z) ⊆ ∆ and y ∈ C(u, z′) ⊆ ∆ by the inductive hypothesis.(v) We have C(w, u)∪ A(w, u) ⊆ Γ1(u)∩∆ by (iv) and Lemma 4. Let p ∈ Γ1(u)∩∆.

Then there exists w′ ∈ Ω such that p ∈ Π(u, w′). Thus

p ∈ C(w′, u) ⊆ C(w′, u) ∪ A(w′, u) = C(w, u) ∪ A(w, u)

since w,w′ ∈ Ω. Therefore Γ1(u) ∩ ∆ = C(w, u) ∪ A(w, u). The lemma is proved.

Lemma 19. Let x ∈ ∆. Then the following hold.

(i) Let p ∈ ∆ with ∂Γ (x, p) ≤ m − 1. Then B(x, p) ∩ ∆ = ∅.

(ii) Γi (x) ∩ ∆ = ∅ if and only if 0 ≤ i ≤ m.

(iii) Let z ∈ Γm(x) ∩ ∆. Then Γ1(z) ∩ ∆ = C(x, z) ∩ A(x, z).(iv) Let w ∈ ∆ with ∂Γ (x, u) + ∂Γ (u, w) ≥ m. Then κ(w) = cm + am .


(v) ∆ is strongly closed with respect to x .

(vi) Let z ∈ Γm(x) ∩ ∆. Then Γ1(x) ∩ ∆ = C(z, x) ∩ A(z, x) and κ(x) = cm + am .

Proof. We prove the statements by induction on j := ∂Γ (u, x). Lemma 18 showsthat the statements are true for the case j = 0. We assume j ≥ 1 and the statementsare true for any x ′ ∈ ∆ with ∂Γ (u, x ′) ≤ j − 1. Let y ∈ C(u, x). Then y ∈ ∆ with∂Γ (u, y) = j − 1.

(i) Let i := ∂Γ (x, p). Then ∂Γ (y, p) ∈ i −1, i, i +1. If ∂Γ (y, p) = i +1, then thereexists p′ ∈ [ C(y, p)\C(x, p) ] as ci < ci+1. Then p′ ∈ B(x, p) ∩ ∆ by Lemma 9and the inductive hypothesis. If ∂Γ (y, p) = i, then B(x, p) ∩ ∆ = B(y, p) ∩ ∆ = ∅by the condition (B B)i . If ∂Γ (y, p) = i − 1, then there exists p1 ∈ B(y, p) ∩ ∆ andp2 ∈ B(y, p1)∩∆. Let z ∈ C(p, p2)\p1. Then z ∈ C(y, p2) ⊆ ∆. If p1 ∈ B(x, p),

then p1 ∈ B(x, p) ∩ ∆. If p1 ∈ A(x, p), then p2 ∈ B(y, p1) = B(x, p1). ApplyLemma 12 (ii) to (x, p2, p1). Then z ∈ Γi+1(x) and hence z ∈ B(x, p) ∩ ∆. Ifp1 ∈ C(x, p), then p ∈ C(y, p1)\C(x, p1). Apply Lemma 14 to (y, x, p2, p1). Then

z ∈ [ C(p, p2)\p1 ] ⊆ [ C(y, p2)\C(x, p2) ] ⊆ Γi+1(x).

by Lemma 9. Hence z ∈ B(x, p) ∩ ∆. The assertion is proved.(ii) If there exists w ∈ Γm+1(x)∩∆, then w ∈ Γm(y)∩∆. It follows, by the inductivehypothesis, that x ∈ Γ1(y) ∩ ∆ = C(w, y) ∪ A(w, y) which is a contradiction. Thedesired result follows by (i).(iii) If ∂Γ (y, z) = m, then C(x, z) ∪ A(x, z) = C(y, z) ∪ A(y, z) = Γ1(z) ∩ ∆ bythe condition (B B)m . Suppose ∂Γ (y, z) = m − 1. There exists w ∈ B(y, z) ∩ ∆.

We have Γ1(z) ∩ ∆ ⊆ C(x, z) ∪ A(x, z) by (ii). If ∂Γ (y, u) + ∂Γ (u, z) ≥ m, thenκ(z) = cm +am by the inductive hypothesis, and thus Γ1(z)∩∆ = C(x, z)∪ A(x, z).We may assume that ∂Γ (y, z) = ∂Γ (y, u) + ∂Γ (u, z). Let p ∈ C(x, z) ∪ A(x, z).It is enough to show p ∈ ∆. Since ∆ is strongly closed with respect to y, we mayassume that p ∈ B(y, z) and ∂Γ (w, p) = 2. Suppose j = 1. Then y = u andw ∈ Ω. Hence Lemma 16 (ii) implies that C(p, u) ∪ A(p, u) = C(w, u) ∪ A(w, u)

and thus p ∈ Ω ⊆ ∆. Suppose j ≥ 2. Let z′ ∈ C(p, w)\z. Then Lemma 16 (i)implies z′ ∈ Γm+1(y) and thus z′ ∈ C(y, w) ∪ A(y, w) ⊆ ∆. Since ∆ is stronglyclosed with respect to u, we may assume that p ∈ B(u, z′) and thus z′ ∈ C(u, p).

Let u′ ∈ C(w, u) \ C(z, u). Then u′ ∈ ∆ and u′ ∈ B(z, u) by Lemma 9. Since∂Γ (w, p) = 2, ∂Γ (u, w) = ∂Γ (u, p) = m − j + 1 and z, z′ ⊆ C(u, w) ∩ C(u, p),

the condition (C A)m− j+1 implies

u′ ∈ C(w, u) ⊆ C(w, u) ∪ A(w, u) = C(p, u) ∪ A(p, u).

Hence p ∈ C(u′, z) ∪ A(u′, z) ⊆ ∆ by the inductive hypothesis.(iv) If ∂Γ (y, u) + ∂Γ (u, w) ≥ m, then the assertion follows by the inductive hypoth-esis. Suppose ∂Γ (y, u) + ∂Γ (u, w) ≤ m − 1. Then

m ≤ ∂Γ (x, u) + ∂Γ (u, w) = 1 + ∂Γ (y, u) + ∂Γ (u, w) ≤ 1 + (m − 1)

and thus ∂Γ (u, w) = m − j. Since cm + am = κ(u) > cm− j + am− j , there exists u1 ∈B(w, u) ∩ ∆. If j = 1, then w ∈ Γm(u1) ∩ ∆ and Γ1(w) ∩ ∆ = C(u1, w) ∪ A(u1, w)

by (iii). So κ(w) = cm +am . If j ≥ 2, then cm +am = κ(u1) > cm− j+1 +am− j+1 and

548 A. Hiraki

thus there exists u2 ∈ B(w, u1) ∩ ∆. Inductively we can take ui+1 ∈ B(w, ui ) ∩ ∆

for i = 1, . . . , j − 1. Then ∂Γ (u, u j ) = j and w ∈ Γm(u j ) ∩ ∆. It follows, by (iii),that κ(w) = cm + am . The assertion is proved.(v) Let z ∈ ∆and i := ∂Γ (x, z).We prove C(x, z)∪A(x, z) ⊆ ∆by induction on m−i.Suppose ∂Γ (x, z) = m. Then the claim follows by (iii). Suppose ∂Γ (x, z) ≤ m − 1.

Then ∂Γ (y, z) ∈ i − 1, i, i + 1. If ∂Γ (y, z) = i + 1, then C(x, z) ∪ A(x, z) ⊆C(y, z)∪A(y, z) ⊆ ∆. If ∂Γ (y, z) = i, then C(x, z)∪A(x, z) = C(y, z)∪A(y, z) ⊆ ∆

by the condition (B B)i . Assume ∂Γ (y, z) = i − 1. Let p ∈ C(x, z) ∪ A(x, z). It isenough to show p ∈ ∆. There exists w ∈ B(x, z)∩∆ by (i). Then ∂Γ (p, w) = 2 andthere exists z′ ∈ C(w, p)\z. If p ∈ A(x, z), then z′ ∈ Γi+1(x)by Lemma 12 (ii). Thusz′ ∈ A(x, w) ⊆ ∆ and p ∈ C(x, z′) ⊆ ∆. If p ∈ C(y, z), then p ∈ ∆ as ∆ is stronglyclosed with respect to y. If p ∈ C(x, z)\C(y, z), then z′ ∈ [ C(x, w)\C(y, w) ]by Lemma 14. Hence z′ ∈ C(x, w) ⊆ ∆ by the inductive hypothesis. Thereforep ∈ C(y, z′) ⊆ ∆.

(vi) We have C(z, x) ∪ A(z, x) ⊆ ∆ by (v) and Lemma 4. We count the number ofedge (p, p′) in ∆ such that (x, y, p, p′) is a quadrangle. Then we have

( κ(x) − a1 − 1 )(c2 − 1) = ( κ(y) − a1 − 1 )(c2 − 1).

Hence κ(x) = κ(y) = cm + am and C(z, x) ∪ A(z, x) = Γ1(x) ∩ ∆. The lemma isproved.

Proof of Theorem 1. (i) ⇒ (ii): This is proved in Corollary 8.(ii) ⇒ (i): Let (u, v) be a pair of vertices at distance m in Γ. Define ∆ = ∆(u, v)

as in Definition 17. Then Lemma 19 shows that ∆ is a strongly closed subgraph ofdiameter m containing u and v. The theorem is proved.

5. An Application of the Theorem

The reader is referred to [2, § 9.2, 9.4] for the definitions of the d-cube and the dualpolar graph. Let Γ be a distance-regular graph of diameter d ≥ 4 and m be aninteger with 1 ≤ m ≤ d − 1. If c j = j and a j = 0 for all j with 1 ≤ j ≤ m, thenfor any pair (x, y) of vertices at distance m, Π(x, y) is the m-cube which is stronglyclosed in Γ (see [2, § 1.13, § 9.2 ] or [10, Proposition 5.3]).

Let q and i be integers with q > 1 and i ≥ 1. We denote by[

i1

]= qi − 1

q − 1= 1 + q + · · · + qi−1

the Gaussian binomial coefficient with basis q.

Then it had been known that the following result (see [3] or [2, Theorem 9.4.5]).

Theorem 20. Let q > 1 be an integer. Let Γ be a bipartite distance-regular graph of

diameter d ≥ 4 with c j =[

j1

]for all j with 1 ≤ j ≤ d. If for any three distinct

vertices the number of their common neighbors is 0, 1 or q +1, then q is a prime powerand Γ is the dual polar graph on [Dd(q)].


In this section we prove Proposition 2 by using this theorem and applyingTheorem 1.

We first show the following lemma.

Lemma 21. Let Γ be a distance-regular graph of diameter d ≥ 4 with c2 > 1. Let jbe an integer with 2 ≤ j ≤ d − 2. Suppose the condition (SC) j holds. If a1 = · · · =a j+1 = 0 and c j+1 = c j (c2 − 1)+ 1. Then the conditions (C A) j+1 and (SC) j+1 hold.

Proof. The conditions (B B)i and (C A)i hold for all i with 1 ≤ i ≤ j by Corollary 8.The condition (B B) j+1 also holds as a j+1 = 0. We prove that the condition (C A) j+1holds. Then the desired result follows by Theorem 1. Let (w,w′, v) be a triple ofvertices with ∂Γ (w,w′) = 2 and ∂Γ (v,w) = ∂Γ (v,w′) = j + 1 such that there existdistinct vertices z and z′ in C(v,w) ∩ C(v,w′). Suppose there exists u ∈ B(w, v)\B(w′, v) and derive a contradiction. Then ∂Γ (u, w′) = j as a j+1 = 0. Since c j+1 =c j (c2 − 1) + 1, we have

⋃

x∈C(v,z)

[ C(x, w)\z ] = [ C(v,w)\z ] (9)

by Lemma 13 and Corollary 15. Hence there exists x ∈ C(v, z) such that z′ ∈C(x, w)\z. Next we apply Lemma 14 and Corollary 15 to a quadruple (u, v, z, x)

of vertices. Then⋃

y∈[ C(u,x)\C(v,x) ][ C(y, z)\x ] = [ C(u, z)\C(v, z) ]. (10)

Since w′ ∈ C(u, z)\C(v, z), there exists y ∈ C(u, x)\C(v, x) such that w′ ∈ C(y, z)\x. Then ∂Γ (u, w) = j +2, ∂Γ (u, y) = j −1 and thus ∂Γ (y, w) = 3. It follows that∂Γ (w,w′) = ∂Γ (x, w) = ∂Γ (x, w′) = 2 with z, z′ ⊆ C(x, w) ∩ C(x, w′) such thaty ∈ B(w, x)\ B(w′, x). This contradicts the condition (C A)2. The desired result isproved. Proof of Proposition 2. (i) ⇒ (ii): Lemma 5 and Lemma 6 show that the condition(SC)i holds for i = 1, 2. Let j be an integer with 2 ≤ j ≤ m − 1. Then

c j+1 =[

j + 11

]=

[j1

]q + 1 = c j (c2 − 1) + 1.

It follows, by Lemma 21 and induction on i , that the condition (SC)i holds for alli with 1 ≤ i ≤ m. The rest of the assertion follows by Theorem 20.(ii) ⇒ (i): Let Λ be a strongly closed subgraph of diameter m. Then

ci = ci (Λ) =[

i1

], ai = ai (Λ) = 0

for all i with 1 ≤ i ≤ d(Λ) = m. The proposition is proved. We will be able to obtain lots of applications of Theorem 1 (see [13]).

Acknowledgements. This work was supported by the Grant-in-Aid for Scientific Research,the Ministry of Education, Science and Culture, JAPAN.

550 A. Hiraki

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Received: July 3, 2007Final Version received: August 29, 2008

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Strongly Closed Subgraphs in a Distance-Regular Graph with c 2 > 1