Journal of Algebraic Combinaotrics, 7 (1998), 197–220
c 1998 Kluwer Academic Publishers. Manufactured in The Netherlands.
°
On Distance-Regular Graphs with Height Two, II
MASATO TOMIYAMA*
tomiyama@cc.osaka-kyoiku.ac.jp
Division of Mathematical Science, Osaka Kyoiku University, Asahigaoka 4-698-1, Kashiwara, Osaka 582, Japan
Received November 21, 1995; Revised December 4, 1996
Abstract. Let Γ be a distance-regular graph with diameter d ≥ 3 and height h = 2, where h = max{i : pdd,i 6= 0}.
Suppose that for every α in Γ and every β in Γd (α), the induced subgraph on Γd (α) ∩ Γ2 (β) is isomorphic to a
complete multipartite graph Kt×2 with t ≥ 2. Then d = 4 and Γ is isomorphic to the Johnson graph J(10, 4).
Keywords: distance-regular graph, height, Johnson graph, complete multipartite graph
1.
Introduction
Let Γ be a connected finite undirected graph without loops or multiple edges. We identify
Γ with the set of vertices. For vertices u and v, let ∂(u, v) denote the distance between u
and v, i.e., the length of a shortest path connecting u and v in Γ. Let d = d(Γ) denote the
diameter of Γ, i.e., the maximal distance of two vertices in Γ. We set
Γi (u) = {v ∈ Γ : ∂(u, v) = i}
(0 ≤ i ≤ d).
Γ is said to be distance-regular if the cardinality of the set Γi (x) ∩ Γj (y) depends only on
i, j and the distance between x and y. In this case we write
pli,j = pli,j (Γ) = |Γi (x) ∩ Γj (y)| (0 ≤ i, j, l ≤ d),
where l = ∂(x, y). Let
ki = ki (Γ) = p0i,i = |Γi (u)| (0 ≤ i ≤ d).
In particular k = k1 is the valency of Γ. Let
ci = ci (Γ) = pi1,i−1 , ai = ai (Γ) = pi1,i , bi = bi (Γ) = pi1,i+1
(0 ≤ i ≤ d).
They are called the intersection numbers of Γ, and


 ∗ c1 c2 · · · ci · · · cd−1 cd 
ι(Γ) = a0 a1 a2 · · · ai · · · ad−1 ad


b0 b1 b2 · · · bi · · · bd−1 ∗
is called the intersection array of Γ.
* Supported in part by a grant from the Japan Society for Promotion of Sciences. Paper prepared while author
at Graduate School of Mathematics, Kyushu University
198
TOMIYAMA
The following are basic properties of intersection numbers, which we use implicitly in
this paper.
(1) ci + ai + bi = k (0 ≤ i ≤ d),
(2) 1 = c1 ≤ c2 ≤ c3 ≤ · · · ≤ cd−1 ≤ cd ≤ k,
(3) k = b0 > b1 ≥ b2 ≥ · · · ≥ bd−2 ≥ bd−1 ≥ 1,
(4) pli,j = plj,i (0 ≤ i, j, l ≤ d),
(5) pli,j = 0 if l > i + j or l < |i − j|,
(6) pli,j 6= 0 if l = i + j or l = |i − j|,
(7) pli,0 + pli,1 + · · · + pli,d = ki (0 ≤ i, l ≤ d),
(8) kl pli,j = ki pil,j = kj pjl,i (0 ≤ i, j, l ≤ d),
(9) ki bi = ki+1 ci+1 (0 ≤ i ≤ d − 1),
(10) ci ≤ bj if i + j ≤ d.
A graph is said to be strongly regular if it is distance-regular with diameter 2.
A graph is called a clique when any two of its vertices are adjacent. A coclique is a graph
in which no two vertices are adjacent.
Information about the general theory of distance-regular graphs is given in [1], [2] and
[3].
For some positive integers n and e with n ≥ 2e, let X be a finite set of cardinality n and
V = {T ⊂ X : |T | = e}. The Johnson graph J(n, e) is a graph whose vertex set is V
and two vertices x and y are adjacent if and only if |x ∩ y| = e − 1. It is well known that
J(n, e) is a distance-regular graph with diameter e.
The complete multipartite graph Km1 ,m2 ,···,mt is a graph whose vertex set is partitioned
into t parts M1 , M2 , · · · , Mt , where |Mi | = mi (1 ≤ i ≤ t), and two vertices are adjacent
if and only if they belong to different parts. We write Kt×m if m1 = m2 = · · · = mt = m.
In this paper we identify a subset A of Γ with the induced subgraph on A and define the
following terminology.
A subgraph A of Γ is called µ-closed if for every pair of vertices x and y in A with
∂(x, y) = 2 in Γ, Γ1 (x) ∩ Γ1 (y) ⊆ A, and λ-closed if for all adjacent vertices x and y in
A, Γ1 (x) ∩ Γ1 (y) ⊆ A.
Let h = max{i : pdd,i 6= 0} be the height of Γ. By definition, it is easy to see that h ≤ d.
Most known distance-regular graphs satisfy h = d. The Johnson graphs J(n, d) (n < 3d)
are examples which satisfy h < d. A distance-regular graph Γ is of height 0 if and only
if Γ is an antipodal 2-cover, and is of height 1 if and only if Γd (α) is a nontrivial clique
for every α in Γ. So if the height of Γ is 1, Γ is the distance-2 graph of a generalized odd
graph. (See Proposition 4.2.10 of [3] and Theorem III.4.2 of [1].) The next question is
what kind of distance-regular graphs are of height 2. This question is not easy in general,
but it is very interesting, because there are several intersection arrays for which no graphs
are known. (See Chapter 14 of [3].) Also, K. Nomura conjectured that there is no bipartite
distance-regular graph with diameter d ≥ 4 and height h = 2. (Conjecture 1.2 of [5].)
In [13] the author showed the following partial result on distance-regular graphs of height
2.
ON DISTANCE-REGULAR GRAPHS WITH HEIGHT TWO, II
199
Theorem 1.1 Let Γ be a distance-regular graph with diameter d ≥ 3 and height h = 2.
Suppose that for every α in Γ and every β in Γd (α), Γd (α) ∩ Γ2 (β) is isomorphic to a
clique. Then d = 3 and Γ is isomorphic to the Johnson graph J(8, 3).
In this theorem, we give a condition that Γd (α)∩Γ2 (β) is isomorphic to a clique for every
α, β ∈ Γ with ∂(α, β) = d. A clique is one of the natural graphs, because each vertex is
adjacent to all the other vertices. Instead of a clique, we consider a graph in which each
vertex is adjacent to all the other vertices but except one. That is a complete multipartite
graph Kt×2 . In this paper, we show the following theorem.
Theorem 1.2 Let Γ be a distance-regular graph with diameter d ≥ 3 and height h = 2.
Suppose that for every α in Γ and every β in Γd (α), Γd (α) ∩ Γ2 (β) is isomorphic to a
complete multipartite graph Kt×2 with t ≥ 2. Then d = 4 and Γ is isomorphic to the
Johnson graph J(10, 4).
Let Γ be a distance-regular graph with height 2 and suppose Γd (α) is connected for every
α ∈ Γ. The diameter of Γd (α) may be greater than 2. We consider the case the diameter of
Γd (α) is 2. For example, if Γ is isomorphic to the Hamming graphs H(2, q) (q ≥ 3), the
Johnson graphs J(n, 2) (n ≥ 6) or J(2d + 2, d) (d ≥ 2), then Γd (α) becomes strongly
regular. The known examples in which the diameter d ≥ 3 and the diameter of Γd (α) is
2 are J(2d + 2, d). For a given graph ∆ whose diameter is 2, is it possible to classify the
distance-regular graphs Γ whose antipodal structures Γd (α) are ∆? It is known that there
are finitely many distance-regular graphs in which Γd (α) ' ∆ for every α ∈ Γ. (See [6]
and [7].)
Let ∆ be a graph with diameter 2. Suppose Γd (α) ' ∆ for every α ∈ Γ. Then the height
of Γ becomes 2. In this situation, it is easy to see that ∆ is distance-degree regular, i.e.,
|∆1 (β)| = pdd,1 , |∆2 (β)| = pdd,2 do not depend on the choice of β in ∆. So we have the
following corollary of Theorem 1.2.
Corollary 1.3 Let Γ be a distance-regular graph with diameter d ≥ 3, and ∆ a distancedegree regular graph with diameter 2 such that ∆2 (β) is isomorphic to Kt×2 with t ≥ 2
for every β in ∆. Suppose Γd (α) is isomorphic to ∆ for every α in Γ. Then d = 4, Γ is
isomorphic to J(10, 4) and ∆ is isomorphic to J(6, 2).
We note that there are many distance-degree regular graphs ∆ such that d(∆) = 2 and
∆2 (β) ' Kt×2 for every β ∈ ∆. The complements of strongly regular graphs with a1 = 1
are in this class. It is not hard to construct graphs in this class which are not strongly regular.
For example, let a graph Λ be in this class, then we can construct a new graph ∆ in this class
from Λ. The construction can be done as follows. Take a positive integer s and consider
the s copies of each vertex in Λ. Let K u = {u1 , u2 , · · · , us } (u ∈ Λ). ∆ is a graph whose
vertex set is ∪u∈Λ K u and two distinct vertices ui ∈ K u and vj ∈ K v are adjacent if and
only if u = v, or u and v are adjacent in Λ, or u and v are not adjacent in Λ and i 6= j.
(In general, let Λ be a graph with diameter 2 and ∆ the graph constructed from Λ as above.
Then Λ2 (α) ' ∆2 (αi ) for every α ∈ Λ and αi ∈ ∆.)
200
TOMIYAMA
Though there are infinitly many graphs in this class, only J(6, 2) can be the antipodal
structrue of a distance-regular graph, and only J(10, 4) has the antipodal structure in this
class.
2.
Preliminaries
We shall introduce the intersection diagrams of rank l which we use as our main tool.
Let u, v ∈ Γ with ∂(u, v) = l. Set
Dji = Dji (u, v) = Γi (u) ∩ Γj (v)
(0 ≤ i, j ≤ d).
It is easy to see the following.
(1) Dji = φ if l > i + j or l < |i − j|,
i
6= φ if 0 ≤ i ≤ l,
(2) Dl−i
i
6= φ if 0 ≤ i ≤ d − l,
Dl+i
(3) There is no edge between Dji and Dgf if |i − f | > 1 or |j − g| > 1.
An intersection diagram of rank l with respect to (u, v) is the collection {Dji }i,j with
lines between Dji ’s and Dgf ’s. We draw a line
Dji ———Dgf
if there is the possibility of existence of edges between Dji and Dgf , and we erase the line
when we know there is no edge between Dji and Dgf . We also erase Dji when we know
Dji = φ. We say rank l diagram instead of intersection diagram of rank l.
For subsets A and B of Γ, let e(A, B) denote the number of edges between A and B, and
∂(A, B) = min{∂(x, y) : x ∈ A, y ∈ B}. Let e(γ, B) = e({γ}, B). We write α ∼ β,
when β ∈ Γ1 (α), and α 6∼ β, otherwise.
The following are straightforward and useful for determining the form of the intersection
diagrams.
For each γ ∈ Dji , we have the following.
i−1
i−1
) + e(γ, Dji−1 ) + e(γ, Dj−1
),
(4) ci = e(γ, Dj+1
i+1
i−1
i
cj = e(γ, Dj−1 ) + e(γ, Dj−1 ) + e(γ, Dj−1
),
i
i
i
(5) ai = e(γ, Dj+1 ) + e(γ, Dj ) + e(γ, Dj−1 ),
aj = e(γ, Dji+1 ) + e(γ, Dji ) + e(γ, Dji−1 ),
i+1
i+1
) + e(γ, Dji+1 ) + e(γ, Dj−1
),
(6) bi = e(γ, Dj+1
i+1
i−1
i
bj = e(γ, Dj+1
) + e(γ, Dj+1
) + e(γ, Dj+1
).
For the properties and applications of intersection diagrams, see for example [5], [6], [8],
[9] and [13].
Let α, β ∈ Γ with ∂(α, β) = d. We mainly use intersection diagrams of rank d with
respect to (α, β). So in this paper rank d diagram means rank d diagram with respect to
(α, β).
Let Γ be a distance-regular graph with height h = 2. We determine the shapes of some
diagrams of Γ and prove some lemmas.
ON DISTANCE-REGULAR GRAPHS WITH HEIGHT TWO, II
201
First we consider the rank d diagram. Suppose there is a vertex x ∈ Dji , for some i, j with
i + j ≥ d + 3. Then we can take y ∈ Γd (α) ∩ Γd−i (x). Since β, y ∈ Γd (α) and h = 2,
∂(β, y) ≤ 2. On the other hand, ∂(β, y) ≥ ∂(β, x) − ∂(x, y) ≥ 3, which is impossible.
So we get Dji = φ for i, j with i + j ≥ d + 3. Therefore the rank d diagram becomes as in
Figure 1.
Dd2
Dd1
Dd0
3
···
Dd−1
@
@
···
@
@
2
Dd−1
@
@
@
@
1
2
Dd−1
Dd−2
D2d
D4d−2
D3d−1
@
@
@
@
@
@
D1d
D2d−1
D3d−2
@
@
@
@
@
@
@
@
···
D0d
D3d−3
D2d−2
D1d−1
||
||
{α}
{β}
Figure 1.
j
Take any γ ∈ Dd2 , then Γj (α) ∩ Γj−2 (γ) ⊆ Dd−j+2
for 2 ≤ j ≤ d. As p2j,j−2 6= 0, we
get
j
6= φ and pdj,d−j+2 6= 0 for 2 ≤ j ≤ d.
Dd−j+2
Since kj pjd,d−j+2 = kd pdj,d−j+2 6= 0, we have
pjd,d−j+2 6= 0 for 2 ≤ j ≤ d.
Next take any u, v ∈ Γ with ∂(u, v) = i for 2 ≤ i ≤ d, and consider the rank i diagram
with respect to (u, v). Suppose there is x ∈ Dgf for some f + g ≥ 2d − i + 3. Take
y ∈ Γd (u) ∩ Γd−f (x). Then y ∈ Djd , where j = ∂(v, y) ≥ ∂(v, x) − ∂(x, y) ≥ d − i + 3.
d
, since pid,d−i 6= 0. So ∂(y, w) ≥ 3. This contradicts y, w ∈ Γd (u) and
We take w ∈ Dd−i
h = 2. So we have Dgf = φ for f + g ≥ 2d − i + 3. Hence the rank i diagram with respect
to (u, v) becomes as in Figure 2. As pid−i+2,d 6= 0, take z ∈ Ddd−i+2 . Then
d−i+2+j
for 0 ≤ j ≤ i − 2.
Γj (z) ∩ Γd−i+2+j (u) ⊆ Dd−j
Since pd−i+2
j,d−i+2+j 6= 0, we have
d−i+2+j
6= φ and pid−i+2+j,d−j 6= 0 for 0 ≤ j ≤ i − 2.
Dd−j
202
TOMIYAMA
d−i+3
d−1
d
···
Dd−i+2
Dd−1
Dd−i+3
Ddd−i+2
@
@
@
@
@
@
@
@
d−i+2
d−1
d−i+1
d
···
Dd−i+1
Dd−i+2
Dd−1
Dd
@
@
@
@
@
@
@
@
@
@
d−i+1
d−i+2
d−2
d−1
d
···
Dd−i
Dd−1
Dd−2
Dd−i+2
Dd−i+1
Ddd−i
@
@
@
@
@
@
@
@
@
@
d−i
d−i+1
d−2
d−1
···
Dd−1
Dd−2
Dd−i
Dd−i+1
@
@
@
@
@
@
@
@
@
@
d−i−1
d−i
d−i+1
d−3
d−2
d−1
···
Dd−1
Dd−2
Dd−3
Dd−i+1
Dd−i
Dd−i−1
@
@
@
@
@
.. @ ..
.. @ ..
.. @
.. @ ..
.. @ ..
..
.
.
.
.
.
.
.
.
.
.
1
Di+1
@
@
Di0
@
@
Di2
@
@
@
@
@
@
@
@
3
i−1
i
···
Di−1
D
D3
D1i+1
2
@
@
@
@
@
@
@
@
2
i−1
1
·
·
·
D
Di
D1i
D2
i−1
@
@
@
@
@
@
@
@
@
@
1
2
i−2
i−1
·
·
·
Di−1
Di−2
D0i
D2
D1
||
||
{u}
{v}
Figure 2.
Lemma 2.1 The following hold.
i
(1) For 0 ≤ i ≤ d − 1, we have pi+1
d,d−i−1 ≤ pd,d−i , and equality holds if and only if
bi = cd−i .
i
(2) For 2 ≤ i ≤ d, we have pd−i+2
d,i−2 ≤ pd,d−i+2 .
i
(3) For 2 ≤ i ≤ d − 1, we have pd,d−i+2 ≤ pi+1
d,d−i+1 and bd−i+1 ≤ ci+1 . Moreover
i+1
d−i+1
i
, Did−i+1 ) +
pd,d−i+2 = pd,d−i+1 if and only if bd−i+1 = ci+1 if and only if e(Di+1
d−i+1
d−i
e(Di+1 , Di ) = 0 in the rank d diagram.
(4) For 0 ≤ j ≤ i − 2 ≤ d − 2, take u, v ∈ Γ with ∂(u, v) = i. Then there is
z ∈ Γd−i+2+j (u) such that Γd (u) ∩ Γd−i (v) ⊆ Γd (u) ∩ Γi−j (z).
Proof: (1) By Lemma 4.1.7 of [3]
pid,d−i =
bi bi+1 bi+2 · · · bd−1
bi i+1
=
p
.
c1 c2 · · · cd−i−1 cd−i
cd−i d,d−i−1
ON DISTANCE-REGULAR GRAPHS WITH HEIGHT TWO, II
203
Since bi ≥ cd−i , we get (1).
Let u, v ∈ Γ such that ∂(u, v) = i, and consider the rank i diagram with respect to (u, v).
(2) For any x ∈ Ddd−i+2 ,
d
.
Γd (u) ∩ Γi−2 (x) ⊆ Dd−i+2
i
Since ∂(u, x) = d − i + 2, we get pd−i+2
d,i−2 ≤ pd,d−i+2 .
i+1
d
, ∂(z, y) ≥ ∂(v, z) − ∂(v, y) = d − i + 1. Since
(3) Take y ∈ D1 . For each z ∈ Dd−i+2
i+1
pd,j = 0 if j ≥ d − i + 2, we have ∂(z, y) = d − i + 1. So
d
⊆ Γd (u) ∩ Γd−i+1 (y),
Dd−i+2
and we have pid,d−i+2 ≤ pi+1
d,d−i+1 .
d−i+1
, we get
Next we use the rank d diagram. For any w ∈ Di+1
bd−i+1 = e(w, Did−i+2 ),
ci+1 = e(w, Did−i+2 ) + e(w, Did−i+1 ) + e(w, Did−i ).
So we have bd−i+1 ≤ ci+1 , and equality holds if and only if
d−i+1
d−i+1
, Did−i+1 ) + e(Di+1
, Did−i ) = 0
e(Di+1
Since
bi ki pid,d−i+2 = bi kd pdd−i+2,i ,
i+1
d
bi ki pd,d−i+1 = ci+1 ki+1 pi+1
d,d−i+1 = ci+1 kd pd−i+1,i+1 ,
d
d
pid,d−i+2 = pi+1
d,d−i+1 if and only if bi pd−i+2,i = ci+1 pd−i+1,i+1 .
In the rank d diagram,
d−i+1
d−i+1
d−i+1
, Did−i+2 ) + e(Di+1
, Did−i+1 ) + e(Di+1
, Did−i ) = ci+1 pdd−i+1,i+1 .
e(Di+1
d−i+1
, Did−i+2 ) = bi pdd−i+2,i .
e(Di+1
d−i+1
d−i+1
, Did−i+1 )+e(Di+1
, Did−i ) = 0
So bi pdd−i+2,i = ci+1 pdd−i+1,i+1 if and only if e(Di+1
if and only if bd−i+1 = ci+1 .
(4) Since pid−i+2+j,d−j 6= 0, take z ∈ Γd−i+2+j (u) ∩ Γd−j (v). In the rank d − i + 2 + j
i
d
. So Γd (u)∩Γd−i (v) ⊆ Di−j
= Γd (u)∩Γi−j (z).
diagram with respect to (u, z), v ∈ Dd−j
Lemma 2.2 Suppose p2d,d = 1. Then for every pair of adjacent vertices u and v, Γd (u) ∩
Γd (v) is a clique.
Proof: Suppose there are distinct vertices x, y ∈ Γd (u) ∩ Γd (v) such that x 6∼ y.
Then ∂(x, y) = 2 as h = 2. We have p2d,d ≥ 2 because u, v ∈ Γd (x) ∩ Γd (y).
204
TOMIYAMA
Lemma 2.3 Suppose bd−1 = 1. Then for every α ∈ Γ, Γd (α) is µ-closed and λ-closed.
Moreover Γd (α) becomes strongly regular with the intersection array


1
c2 
 ∗
0
a1
ad − c2 .
ι(Γd (α)) =


ad ad − a1 − 1
∗
Proof: Take any α ∈ Γ, and any β, γ ∈ Γd (α). Consider the rank d diagram. Since
bd−1 = 1, we know e(D1d−1 , D2d ) + e(D1d−1 , D1d ) = 0.
If ∂(β, γ) = 1, then γ ∈ D1d and we get Γ1 (β) ∩ Γ1 (γ) ⊆ D1d ⊆ Γd (α). So we have
Γd (α) is λ-closed. If ∂(β, γ) = 2, then γ ∈ D2d and Γ1 (β) ∩ Γ1 (γ) ⊆ D1d ⊆ Γd (α). Hence
Γd (α) is µ-closed. Therefore Γd (α) becomes strongly regular with c2 (Γd (α)) = c2 and
a1 (Γd (α)) = a1 .
Lemma 2.4 For some i with 1 ≤ i ≤ d − 1 and every u, v ∈ Γ with ∂(u, v) = i, suppose
Γd (u) ∩ Γd−i (v) is a clique. Then the following hold.
j
j+1
, Dd−j+1
) = 0 for i ≤ j ≤ d − 1.
(1) In the rank d diagram, e(Dd−j
(2) Γd (α) is µ-closed and d(Γd (α)) = 2 for every α ∈ Γ.
j
j+1
, Dd−j+1
) 6= 0 for some j with i ≤ j ≤ d − 1. Then there is an
Proof: Suppose e(Dd−j
j
j+1
. Take z ∈ Γd (α)∩Γd−j−1 (y). Then z ∈ D2d .
edge x ∼ y such that x ∈ Dd−j , y ∈ Dd−j+1
So z 6∼ β. Take w ∈ Γi (α) ∩ Γj−i (x). Then z, β ∈ Γd (α) ∩ Γd−i (w). This contradicts
Γd (α) ∩ Γd−i (w) being a clique. So we have (1). (2) follows from e(D1d−1 , D2d ) = 0.
Lemma 2.3 Suppose d ≥ 3. And suppose Γd (x) ∩ Γd−l+2 (y) is a clique for some l with
3 ≤ l ≤ d and every x, y ∈ Γ with ∂(x, y) = l. Then Γd (u) ∩ Γ1 (v) is a clique for every
u, v ∈ Γ with ∂(u, v) = d − 1.
Proof: By Lemma 2.1(4) with i = d − 1 and j = l − 3, for every u, v ∈ Γ with
∂(u, v) = d − 1, there is z ∈ Γl (u) such that Γd (u) ∩ Γ1 (v) ⊆ Γd (u) ∩ Γd−l+2 (z). Since
Γd (u) ∩ Γd−l+2 (z) is a clique, we get the assertion.
Lemma 2.6 Suppose d ≥ 3. Then we have c3 > 1.
Proof: Suppose c3 = 1. We use the rank d diagram. Then by Lemma 2.1(3) with
i = 2, bd−1 = c3 = c2 = 1 and e(D3d−1 , D2d−1 ) + e(D3d−1 , D2d−2 ) = 0. By Lemma 2.3,
e(D1d−1 , D2d ) + e(D1d−1 , D1d ) = 0 and Γd (α) is µ-closed. The rank d diagram becomes as
in Figure 3.
1
and let ∆ = Γd (γ).
Take γ ∈ Dd−1
Claim.
∆ ⊆ D3d−1 ∪ D2d−1 ∪ D2d ∪ D1d , ∆ ∩ D3d−1 6= φ and |∆ ∩ D1d | = 1.
Take x ∈ ∆. We have ∂(x, α) ≥ ∂(x, γ) − ∂(γ, α) = d − 1. So we get ∆ ⊆
d−1
= Γd (γ) ∩ Γ3 (β) 6= φ.
D3d−1 ∪ D2d−1 ∪ D1d−1 ∪ D2d ∪ D1d . Since pd−1
d,3 6= 0, ∆ ∩ D3
ON DISTANCE-REGULAR GRAPHS WITH HEIGHT TWO, II
3
···
Dd−1
@
@
···
Dd2
Dd1
Dd0
205
@
@
2
Dd−1
@
@
1
2
Dd−1
Dd−2
D2d
D4d−2
D3d−1
@
@
@
@
d−2
d−1
D1d
D3
D2
@
@
@
@
@
@
@
@
···
D0d
D3d−3
D2d−2
D1d−1
||
||
{α}
{β}
Figure 3.
Take y ∈ ∆ ∩ D3d−1 . Suppose there is some vertex z ∈ ∆ ∩ D1d−1 . Then ∂(y, z) ≥ 3,
which contradicts y, z ∈ Γd (γ) and h = 2. So we have ∆ ∩ D1d−1 = φ. |∆ ∩ D1d | =
|Γd (γ) ∩ Γ1 (β)| = bd−1 = 1. Hence we have the claim.
Let {u} = ∆∩D1d . Since |∆1 (u)| = ad > e(u, D1d ∪D2d ), there exists v ∈ D2d−1 ∩∆1 (u).
Then ∂(v, y) = 2 as h = 2. By Lemma 2.3, ∆ is µ-closed. By Claim, Γ1 (v)∩Γ1 (y) ⊆ D2d .
Hence {u}∪(Γ1 (v) ∩ Γ1 (y)) ⊆ Γd (α) ∩ Γ1 (v). This contradicts bd−1 = 1.
Lemma 2.7 Suppose d ≥ 3 and Γd (x) ∩ Γd−1 (y) is a clique for every x, y ∈ Γ with
∂(x, y) = 3. Then for every u ∈ Γd−1 (x) ∩ Γd (y) and every v ∈ Γd (x) ∩ Γd−1 (y), we
have u ∼ v. Moreover we have bd−1 = p3d,d−1 .
Proof: Use the rank 3 diagram with respect to (x, y).
d
d
. By way of contradiction, we assume u 6∼ v. Since Dd−1
Then u ∈ Ddd−1 and v ∈ Dd−1
is a clique, we get ∂(u, v) = 2.
As p2d,d 6= 0, we take z ∈ Γd (u) ∩ Γd (v). Since x, z ∈ Γd (v) and h = 2, ∂(x, z) ≤ 2.
Similarly ∂(y, z) ≤ 2. So z ∈ D21 ∪ D22 ∪ D12 . We may assume ∂(x, z) = 2.
d
is a clique, Γ1 (v) ⊇ Γd (x) ∩ Γ1 (u). By Lemma 2.5 with l = 3 and 2.4(2)
Since Dd−1
with i = d − 1, Γd (z) is µ-closed. Since u, v ∈ Γd (z) with ∂(u, v) = 2, we get
Γd (z) ⊇ (Γ1 (u) ∩ Γ1 (v)) ∪ {v} ∪ {u}
⊇ (Γd (x) ∩ Γ1 (u)) ∪ {v}.
Hence
Γd (x) ∩ Γd (z) ⊇ (Γd (x) ∩ Γ1 (u)) ∪ {v}.
Claim 1.
bd−1 = c3 .
206
TOMIYAMA
d
Suppose there is γ ∈ Dd−1
such that γ 6∈ Γd (x) ∩ Γd (z). Then ∂(z, γ) = d − 1 because
d
Dd−1 is a clique and ∂(z, v) = d. So
Γd (z) ∩ Γ1 (γ) ⊇ ((Γd (x) ∩ Γ1 (u)) ∪ {v}) ∩ Γ1 (γ)
= (Γd (x) ∩ Γ1 (u)) ∪ {v}
d
, i.e.,
In this case bd−1 ≥ bd−1 + 1, which is impossible. Hence Γd (x) ∩ Γd (z) ⊇ Dd−1
2
3
2
3
pd,d ≥ pd,d−1 . By Lemma 2.1(3) with i = 2, we have pd,d = pd,d−1 and bd−1 = c3 .
Claim 2.
For any α, δ ∈ Γ such that ∂(α, δ) = 3, Γ2 (α) ∩ Γ1 (δ) is a clique.
For any α, δ ∈ Γ such that ∂(α, δ) = 3, take β ∈ Γd (α) ∩ Γd−1 (δ) and consider the
3
. By Claim 1 and Lemma 2.1(3) with i = 2, we get
rank d diagram. Then δ ∈ Dd−1
3
2
3
2
e(Dd−1 , Dd−1 ) + e(Dd−1 , Dd−2 ) = 0. So Γ2 (α) ∩ Γ1 (δ) = Γd (β) ∩ Γ1 (δ). By Lemma
2.5 with l = 3, Γd (β) ∩ Γ1 (δ) is a clique. So we get Γ2 (α) ∩ Γ1 (δ) is a clique.
By Claim 2 and Lemma 5.5.2 of [3], we get c3 = c2 = 1. This contradicts Lemma
d
2.6. So we have u ∼ v. Therefore bd−1 = |Γd (x) ∩ Γ1 (u)| = |Dd−1
| = p3d,d−1 .
Lemma 2.8 Suppose d ≥ 3 and d(Γd (α)) = 2 for every α ∈ Γ. Then, in the rank d
i+1
i
i
, Dd−i
) 6= 0 for 1 ≤ i ≤ d − 1. In particular, Dd−i+1
6= φ for
diagram, e(Dd−i+1
1 ≤ i ≤ d.
i+1
i
, Dd−i
) = 0 for some i with 2 ≤ i ≤ d − 1. Take any x ∈ D1d .
Proof: Suppose e(Dd−i+1
i
Then Γi (α) ∩ Γd−i (x) ⊆ Dd−i
. Since both sizes are pdi,d−i , we have
i
.
Γi (α) ∩ Γd−i (x) = Dd−i
i
,
Hence for any y ∈ Dd−i
Γd (α) ∩ Γd−i (y) ⊇ D1d ∪ {β}.
Since d(Γd (α)) = 2, for every z ∈ D2d there is w ∈ D1d such that z ∼ w. So ∂(y, z) ≤
d − i + 1, and Γd (α) ∩ Γd−i+2 (y) = φ. This contradicts pid,d−i+2 6= 0. Hence we have
i+1
i
2
, Dd−i
) 6= 0 for 2 ≤ i ≤ d − 1. By symmetry we have e(Dd1 , Dd−1
) 6= 0.
e(Dd−i+1
3.
Some lemmas
In the rest of this paper, we assume the hypothesis of Theorem 1.2. In the rank d diagram,
we have D2d ' Kt×2 .
Let κ1 = pdd,1 = ad and κ2 = pdd,2 = 2t. Then kd = 1 + κ1 + κ2 . Take any γ ∈ D2d , then
we have e(γ, D2d ) = κ2 − 2, e(γ, D1d ) = κ1 − κ2 + 2 and e(γ, D1d−1 ) = c2 − (κ1 − κ2 + 2)
by our assumption.
Lemma 3.1 For every α ∈ Γ and every β, γ ∈ Γd (α) with ∂(β, γ) = 2, we have
|{δ ∈ Γd (α) : ∂(δ, β) = ∂(δ, γ) = 2}| = 1.
ON DISTANCE-REGULAR GRAPHS WITH HEIGHT TWO, II
207
Proof: Consider the rank d diagram. Then γ ∈ D2d . As D2d ' Kt×2 , we get |{δ ∈
Γd (α) : ∂(δ, β) = ∂(δ, γ) = 2}| = |{δ ∈ D2d : δ 6∼ γ}| = 1.
Lemma 3.2 We have bd−1 ≥ 2.
To prove Lemma 3.2, we need the following lemma.
Lemma 3.3 Let ∆ be a strongly regular graph with p22,2 (∆) = 1. Then we have c2 (∆) ≥ 2
and k1 (∆) ≥ k2 (∆). Moreover one of the following holds.
(1) ∆ is isomorphic to the complete multipartite graph K3,3 .
(2) ∆ is isomorphic to the Hamming graph H(2, 3).
(3) ∆1 (α) is connected for every α ∈ ∆, i.e., ∆ is locally connected.
Proof: Let κ1 = k1 (∆), κ2 = k2 (∆) and λ = a1 (∆). As p22,2 (∆) = 1, we have
a2 (∆) = κ2 − 2 and c2 (∆) = κ1 − κ2 + 2.
If c2 (∆) = 1, then κ1 + 1 = κ2 = κ1 b1 (∆), which is a contradiction. So c2 (∆) ≥ 2
and κ1 ≥ κ2 .
Take any α ∈ ∆ and x, y ∈ ∆1 (α) with x 6∼ y. Let X = ∆1 (α) ∩ ∆1 (x), Y = ∆1 (α) ∩
∆1 (y) and {z} = ∆2 (x) ∩ ∆2 (y). Then by p22,2 (∆) = 1, ∆1 (α) ⊆ {x, y, z} ∪ X ∪ Y .
So we have κ1 ≤ 2λ + 3.
Suppose κ1 = 2λ+3. Then X ∩Y = φ and {z} ⊆ ∆1 (α). Since {x} = ∆2 (y)∩∆2 (z),
X ⊆ ∆1 (z). Similarly Y ⊆ ∆1 (z). So we have X ∪ Y ⊆ ∆1 (α) ∩ ∆1 (z) and 2λ ≤ λ.
Hence λ = 0 and κ1 = 3. So we have ∆ ' K3,3
Suppose κ1 = 2λ + 2. Then since
λ + 1 = κ1 − λ − 1 = b1 (∆) =
we get
κ2 c2 (∆)
κ2 (2(λ + 1) − κ2 + 2)
=
κ1
2(λ + 1)
(κ2 − (λ + 2))2 + (λ2 − 2) = 0.
Hence λ = 1, κ2 = 4 and we easily get ∆ ' H(2, 3).
Suppose κ1 ≤ 2λ + 1, then X ∩ Y 6= φ. So ∆ is locally connected.
Proof of Lemma 3.2: Suppose bd−1 = 1. Take x ∈ Γ. Then by Lemma 2.3, Γd (x)
becomes strongly regular. Let ∆ = Γd (x), then by Lemma 3.1, ∆ has the property that
p22,2 (∆) = 1. Hence by Lemma 3.3, ∆ ' K3,3 or ∆ ' H(2, 3) or ∆ is locally connected.
A. Hiraki and H. Suzuki showed that there is no distance-regular graph Γ with diameter
d ≥ 3 such that Γd (α) ' Kt×s (t ≥ 2, s ≥ 2). (See [13].) So the case ∆ ' K3,3 does not
occur.
Case 1.
∆ ' H(2, 3).
Since a1 = 1 and c2 = 2, we have k2 = k(k − 2)/2. Since kp1d,d = kd pd1,d = 36,
k2 p2d,d = kd pd2,d = 36 and 4 = ad < k, we get k = 6 and k2 = 12. Therefore we get
cd = 2. This is a contradiction because c2 ≥ 2 implies cd > c2 (see Theorem 5.4.1 of [3]).
Case 2.
∆ is locally connected.
208
TOMIYAMA
Take y ∈ Γ with ∂(x, y) = i (i = 1, 2). Suppose Γd (x) ∩ Γd (y) contains at least two
vertices, then it contains an edge u ∼ v since it is µ-closed. So ∆1 (u) ⊆ Γd (x) ∩ Γd (y)
because Γd (x) ∩ Γd (y) is λ-closed. Since c2 > 1 and Γd (x) ∩ Γd (y) is λ-closed and
µ- closed, ∆ ⊆ Γd (x) ∩ Γd (y). Hence pid,d = kd and pid,d−i = 0, which is impossible.
Therefore we get p1d,d = p2d,d = 1 and k = kd κ1 , k2 = kd κ2 . By Lemma 3.3, k ≥ k2 .
This also contradicts Lemma 5.1.2 of [3].
Lemma 3.4 Suppose d ≥ 4. For 3 ≤ i ≤ d − 1, take every u, v ∈ Γ with ∂(u, v) = i.
Then Γd (u) ∩ Γd−i (v) and Γd (u) ∩ Γd−i+2 (v) are cliques of size at least 2.
i
3
Proof: By Lemmas 2.1 and 3.2, pid,d−i ≥ pd−1
d,1 = bd−1 ≥ 2 and pd,d−i+2 ≥ pd,d−1 ≥
pd−1
d,1 ≥ 2.
For some u, v ∈ Γ with ∂(u, v) = i, suppose there are x, y ∈ Γd (u) ∩ Γd−i (v) such that
d
with ∂(x, y) = 2.
x 6∼ y. Use the rank i diagram with respect to (u, v). Then x, y ∈ Dd−i
i
d
Since pd,d−i+2 ≥ 2, there are z, w ∈ Dd−i+2 . Clearly ∂(z, x) = ∂(z, y) = ∂(w, x) =
∂(w, y) = 2. So in Γd (u) there are at least two vertices at distance 2 from x and y, which
contradicts Lemma 3.1. So Γd (u) ∩ Γd−i (v) is a clique.
Similarly Γd (u) ∩ Γd−i+2 (v) is a clique.
Lemma 3.5 Suppose d ≥ 4. Then we have the following.
i+1
i
, Dd−i+1
) = 0 for 1 ≤ i ≤ d − 1.
(1) In the rank d diagram, e(Dd−i
(2) Γd (α) is µ-closed and d(Γd (α)) = 2 for every α ∈ Γ.
i
6= φ for 1 ≤ i ≤ d.
(3) Dd−i+1
i+1
i
, Dd−i+1
) = 0 for 3 ≤ i ≤ d − 1. Suppose
Proof: (1) From Lemmas 3.4 and 2.4, e(Dd−i
2
3
2
3
such that x ∼ y. By Lemma
e(Dd−2 , Dd−1 ) 6= 0. Then there are x ∈ Dd−2 , y ∈ Dd−1
d
3.4, we take z, w ∈ Γd (α) ∩ Γd−3 (y) ⊆ D2 . By our assumption, (i.e., that D2d ' Kt×2 ,)
for any γ ∈ Γd (α), ∂(γ, β) ≤ 1 or ∂(γ, z) ≤ 1 or ∂(γ, w) ≤ 1. Therefore we get
2
3
Γd (α) ∩ Γd (x) = φ. This is impossible because p2d,d 6= 0. So we get e(Dd−2
, Dd−1
) = 0.
1
2
By symmetry, e(Dd−1 , Dd ) = 0.
(2)(3) It is clear from Lemmas 2.4(2), 2.8 and 3.4.
Lemma 3.6 Suppose d ≥ 4. Then we have bd−1 ≤
κ2
2 .
Proof: Consider the rank d diagram. For any γ ∈ D3d−1 ,
Γd (α) ∩ Γ1 (γ) ⊆ D2d .
By Lemma 3.4 with i = d − 1, Γd (α) ∩ Γ1 (γ) is a clique. On the other hand, D2d ' Kt×2
by our assumption. Since the maximal cliques of Kt×2 are size t, we have
κ2
.
bd−1 ≤ t =
2
ON DISTANCE-REGULAR GRAPHS WITH HEIGHT TWO, II
209
The case d ≥ 5
4.
In this section we assume d ≥ 5 and show this case does not occur.
Proposition 4.1 Let Γ be a distance-regular graph with diameter d and height h = 2.
Suppose that for every α in Γ and every β in Γd (α), Γd (α) ∩ Γ2 (β) is isomorphic to Kt×2
with t ≥ 2. Then d ≤ 4.
By Lemma 3.5, the rank d diagram becomes as in Figure 4.
Dd2
Dd1
Dd0
3
···
Dd−1
@
@
···
@
@
2
Dd−1
@
@
@
@
1
2
Dd−1
Dd−2
D2d
D4d−2
D3d−1
@
@
@
@
@
@
D1d
D2d−1
D3d−2
@
@
@
@
@
@
@
@
···
D0d
D3d−3
D2d−2
D1d−1
||
||
{α}
{β}
Figure 4.
Lemma 4.2 We have the following.
(1) b2 ≥ cd−1 .
(2) bd−2 = c3 .
Proof: First we show b2 ≥ cd−1 and bd−2 ≥ c3 . From Lemma 2.1(3), we have bi ≤
cd−i+2 for i = 2, d − 2. If bi = cd−i+2 , then we get
bi = cd−i+2 ≥ cd−i+1 .
So we may assume bi < cd−i+2 . Then, in the rank d diagram, there is an edge x ∼ y such
that x ∈ Did−i+2 and y ∈ Did−i+1 .
Claim 1.
e(y, Did−i ) = 0 for i = 2, d − 2.
Suppose there is z ∈ Did−i such that y ∼ z. If i = 2, then β, x ∈ Γd (α) ∩ Γ2 (z) with
β 6∼ x. This contradicts Lemma 3.4 with i = d − 2. If i = d − 2, then from Lemma 3.4
with i = 4, we can take u, v ∈ Γd (α) ∩ Γd−4 (x). Then u, v ∈ D2d and u, v ∈ Γd−2 (z).
Take any γ ∈ Γd (α). If γ ∈ D1d , then ∂(z, γ) ≤ d − 1. If γ ∈ D2d , then by our assumption,
γ ∼ u of γ ∼ v. So we have ∂(z, γ) ≤ d − 1. Hence we have ∂(z, γ) ≤ d − 1 for any
γ ∈ Γd (α). ∂(z, γ) ≤ d − 1. So Γd (α) ∩ Γd (z) = φ. This contradicts p2d,d 6= 0. So we
have the claim.
210
TOMIYAMA
By Claim 1, for i = 2 and d − 2, we get
d−i
d−i+1
d−i
) ≤ e(y, Di+1
) + e(y, Di+1
) = bi .
cd−i+1 = e(y, Di+1
So we have (1) and c3 ≤ bd−2 .
We know bd−3 ≥ c3 . If bd−3 = c3 , then we get
bd−2 ≤ bd−3 = c3 .
So we may assume bd−3 > c3 , then there is an edge x ∼ y such that x ∈ D3d−3 and
y ∈ D3d−2 .
Claim 2.
e(y, D3d−1 ) = 0.
Suppose there is z ∈ D3d−1 such that y ∼ z. Then by Lemma 3.2, take u, v ∈ Γd (α) ∩
Γ1 (z) ⊆ D2d . By our assumption, for any γ ∈ Γd (α), ∂(x, γ) ≤ 4. So Γd (α) ∩ Γ5 (x) = φ.
This contradicts pd−3
d,5 6= 0. Hence we have the claim.
By Claim 2, we have
bd−2 = e(y, D2d−1 ) ≤ e(y, D2d−1 ) + e(y, D2d−2 ) = c3 .
Lemma 4.3 In the rank d diagram, we have the following.
3
, Γd (x) ⊆ D3d−3 ∪ D3d−2 ∪ D2d−2 ∪ D3d−1 ∪ D2d−1 ∪ D1d−1 .
(1) For every x ∈ Dd−1
2
, Γd (y) ⊆ D4d−2 ∪ D3d−2 ∪ D3d−1 ∪ D2d−1 ∪ D2d . In particular
(2) For every y ∈ Dd−2
Γd (y) ∩ (D2d−1 ∪ D2d ) is a clique, Γd (y) ∩ D2d 6= φ and |Γd (y) ∩ D4d−2 | ≥ 2.
Proof: (1) Take any z ∈ Γd (x). Since ∂(α, x) = 3 and p3i,d = 0 if i ≤ d − 4 or i = d, we
know d − 3 ≤ ∂(α, z) ≤ d − 1. Similarly 1 ≤ ∂(β, z) ≤ 3. So we get (1).
(2) Take any w ∈ Γd (y). Since ∂(α, y) = 2 and ∂(β, y) = d − 2, ∂(α, w) ≥ d − 2 and
∂(β, w) ≥ 2. So we get
Γd (y) ⊆ D4d−2 ∪ D3d−2 ∪ D2d−2 ∪ D3d−1 ∪ D2d−1 ∪ D2d .
Claim.
Γd (y) ∩ D2d 6= φ, Γd (y) ∩ D2d−2 = φ and Γd (y) ∩ (D2d−1 ∪ D2d ) is a clique.
Take u ∈ Γd (y) ∩ Γd (α), then u ∈ Γd (y) ∩ D2d . So Γd (y) ∩ D2d 6= φ. Since Γd (y) ∩
(D2d−2 ∪ D2d−1 ∪ D2d ) = Γd (y) ∩ Γ2 (β) is a clique by Lemma 3.4 with i = d − 2, we get
the claim.
By Lemma 3.4 with i = d − 2 we get
|Γd (y) ∩ D4d−2 | = |Γd (y) ∩ Γ4 (β)| ≥ 2.
2
3
Lemma 4.4 In the rank d diagram, we have ∂(Dd−2
, Dd−1
) ≥ 3.
ON DISTANCE-REGULAR GRAPHS WITH HEIGHT TWO, II
211
3
2
Proof: By way of contradiction, assume that there are x ∈ Dd−1
and y ∈ Dd−2
such that
d−1
∂(x, y) = 2. Then we can take u ∈ Γd (x) ∩ Γd (y). By Lemma 4.3, u ∈ D2 ∪ D3d−1 ∪
D3d−2 .
Case 1.
u ∈ D2d−1 .
Since Γd (y) ∩ (D2d−1 ∪ D2d ) is a clique and Γd (y) ∩ D2d 6= φ, there is v ∈ D2d such that
u ∼ v. By Lemma 3.4 with i = 3, we take an edge γ ∼ δ in Γd (α) ∩ Γd−3 (x). Then γ,
δ ∈ D2d . By our assumption, v ∼ γ or v ∼ δ. Therefore we get ∂(u, x) ≤ d − 1, which
contradicts u ∈ Γd (x).
Case 2.
u ∈ D3d−1 .
We can take v ∈ Γd (α) ∩ Γ1 (u). Then v ∈ D2d . Similar to Case 1, this case does not
occur.
Case 3.
u ∈ D3d−2 .
Since Γd (y) ∩ D2d 6= φ and |Γd (y) ∩ D4d−2 | ≥ 2, take z ∈ Γd (y) ∩ D2d and ε, ζ ∈
Γd (y) ∩ D4d−2 . Then ∂(z, ε) = ∂(z, ζ) = ∂(z, u) = 2 and z, ε, ζ, u ∈ Γd (y). Hence
by Lemma 3.1 u ∼ ε or u ∼ ζ. We may assume u ∼ ε. Then we take an edge η ∼ θ in
Γd (β) ∩ Γd−4 (ε) and ξ ∈ Γd (β) ∩ Γ1 (x). Then ξ, η, θ ∈ Dd2 , so ξ ∼ η or ξ ∼ θ and
∂(x, u) ≤ d − 1. We get a contradiction.
Hence we have the assertion.
2
Proof of Proposition 4.1: Suppose bd−2 > c2 , then there is an x ∼ y such that x ∈ Dd−2
2
and y ∈ Dd−1 . By Lemma 4.2(1),
3
3
) + e(y, Dd−2
)
b2 = e(y, Dd−1
3
2
3
) ≥ e(y, Dd−2
) + e(y, x).
≥ cd−1 = e(y, Dd−2 ) + e(y, Dd−2
3
such that y ∼ z, which contradicts Lemma 4.4. Therefore we get
So there is z ∈ Dd−1
bd−2 = c2 . By Lemma 4.2(2), we get
c3 = bd−2 = c2 .
c3 = c2 implies c3 = 1. (See Theorem 5.4.1 of [3].) This contradicts Lemma 2.6.
5.
The case d = 3
In this section we assume d = 3 and prove the following proposition.
Proposition 5.1 Let Γ be a distance-regular graph with height h = 2. For every α in Γ
and every β in Γd (α), Γd (α) ∩ Γ2 (β) is isomorphic to Kt×2 with t ≥ 2. Then d 6= 3.
Lemma 5.2 For every u, v ∈ Γ with ∂(u, v) = 2, Γ3 (u) ∩ Γ3 (v) is a clique.
Proof: Suppose there are x, y ∈ Γ3 (u) ∩ Γ3 (v) such that x 6∼ y. Then ∂(x, y) = 2. Since
b2 ≥ 2, there are z, w ∈ Γ3 (u) ∩ Γ1 (v). Then x, y, z, w ∈ Γ3 (u) and ∂(z, x) = ∂(z, y) =
∂(w, x) = ∂(w, y) = 2. This contradicts Lemma 3.1.
212
TOMIYAMA
Lemma 5.3 We have the following.
(1) p23,3 = 1, k2 = κ2 (1 + κ1 + κ2 ) and c3 = κ2 b2 .
(2) For every edge u ∼ v, Γ3 (u) ∩ Γ3 (v) is a clique.
Proof: (1) By way of contradiction, we assume p23,3 ≥ 2.
Similar to Lemma 5.2, we have Γ3 (u) ∩ Γ1 (v) is a clique for every u, v ∈ Γ with
∂(u, v) = 2.
Consider the rank 3 diagram. By Lemma 2.4 with i = 2, e(D21 , D32 ) = e(D12 , D23 ) = 0
and Γ3 (α) is µ-closed for every α ∈ Γ.
Take x, y ∈ D23 such that ∂(x, y) = 2. Since p23,3 ≥ 2, we take γ ∈ Γ3 (x) ∩ Γ3 (y) with
γ 6= α. Then γ ∈ D31 or γ ∈ D21 because Γ3 (x) ∩ Γ3 (y) is a clique by Lemma 5.2.
Case 1.
γ ∈ D31 .
Since x, y ∈ Γ3 (γ), D23 ' Kt×2 and Γ3 (γ) is µ-closed, we get D23 ⊆ Γ3 (γ). Since
p33,2 = |Γ3 (γ) ∩ Γ2 (β)| = |Γ3 (γ) ∩ (D21 ∪ D22 ∪ D23 )|
≥ |Γ3 (γ) ∩ D23 | = |D23 | = p33,2 ,
we have Γ3 (γ) ∩ (D22 ∪ D21 ) = φ.
Take z ∈ Γ3 (γ) ∩ Γ2 (α). Then z ∈ D12 because Γ3 (γ) ∩ D22 = φ. Since ∂(x, z) = 2
and Γ3 (γ) is µ-closed,
Γ1 (x) ∩ Γ1 (z) ⊆ D13 ∩ Γ1 (z) ⊆ Γ3 (α) ∩ Γ1 (z) \ {β}.
So
c2 = |Γ1 (x) ∩ Γ1 (z)| ≤ |Γ3 (α) ∩ Γ1 (z) \ {β}| = b2 − 1.
If there is δ ∈ D32 such that γ ∼ δ, then we take ε ∈ D23 such that δ ∼ ε. So ∂(γ, D23 ) = 2,
which contradicts Γ3 (γ) ⊇ D23 . Hence e(γ, D32 ) = 0. So we can take ζ ∈ D22 such that
γ ∼ ζ. Since Γ3 (γ) ⊇ D23 , we get e(ζ, D23 ) = 0 and b2 ≤ c2 . This contradicts c2 ≤ b2 − 1.
Therefore this case does not occur.
Case 2.
γ ∈ D21 .
Similar to Case 1, we get D23 ⊆ Γ3 (γ).
Clearly we have Γ3 (γ) ∩ Γ3 (β) ⊆ D32 . Since Γ3 (γ) ∩ Γ3 (β) is a clique and D32 ' Kt×2 ,
we take w ∈ D32 such that w 6∈ Γ3 (γ). Then ∂(w, γ) = 2 because e(D21 , D32 ) = 0. By
p23,3 ≥ 2, there are η, θ in Γ3 (γ) ∩ Γ3 (β). Then η, θ ∈ D32 . Since D32 ' Kt×2 , we have
w ∼ η or w ∼ θ. We may assume w ∼ η. Then
Γ3 (α) ∩ Γ1 (w) ⊆ D23 ∩ Γ1 (w) ⊆ Γ3 (γ) ∩ Γ1 (w).
So
b2 + 1 = |(Γ3 (α) ∩ Γ1 (w)) ∪ {η}| ≤ |Γ3 (γ) ∩ Γ1 (w)| = b2 .
This is a contradiction. Hence this case is also impossible.
Therefore we get p23,3 = 1. So we have k2 = k2 p23,3 = k3 p33,2 = κ2 (1 + κ1 + κ2 ).
c3 = κ2 b2 follows from k2 b2 = k3 c3 .
(2) It is clear from (1) and Lemma 2.2.
ON DISTANCE-REGULAR GRAPHS WITH HEIGHT TWO, II
213
Lemma 5.4 For every u, v ∈ Γ with ∂(u, v) = 2, Γ3 (u) ∩ Γ1 (v) ' Ks×2 for some s ≤ t.
Proof: By Lemma 2.1(4) with d = 3, i = 2, j = 0 and our assumption Γ3 (u) ∩ Γ1 (v) is
a subgraph of Kt×2 . So we may assume b2 < κ2 .
Claim 1.
In the rank 3 diagram, e(γ, D23 ) ≤ 1 for every γ ∈ D12 .
3
If e(γ, D2 ) ≥ 2 for some γ ∈ D12 , then Γ3 (α) ∩ Γ3 (γ) = φ because D23 ' Kt×2 . This
contradicts p23,3 6= 0. So we get the claim.
Suppose e(γ, D23 ) = 1 for every α, β ∈ Γ with ∂(α, β) = 3 and γ ∈ D12 . Then since α
and β are arbitary, we get Γ3 (α) ∩ Γ1 (γ) ' Ks×2 for every α, γ with ∂(α, γ) = 2. So by
way of contradiction, we assume e(γ, D23 ) = 0 for some α, β ∈ Γ with ∂(α, β) = 3 and
γ ∈ D12 . Then by Claim 1,
p33,2 (c2 − (κ1 − κ2 + 2)) = e(D12 , D23 ) < p32,1 .
So
κ2 (c2 − (κ1 − κ2 + 2)) < c3 = κ2 b2 .
So we have
c2 − (κ1 − κ2 + 2) ≤ b2 − 1.
Claim 2.
c2 = (κ1 − κ2 + 2) + (b2 − 1).
In the rank 3 diagram, we can take x ∈ D32 , y ∈ D23 such that ∂(x, y) = 2 because
b2 < κ2 . Let {z} = Γ3 (x) ∩ Γ3 (y), then z ∈ D21 ∪ D22 ∪ D12 . So we may assume
∂(α, z) = 2. We know Γ3 (y) ⊆ {α} ∪ D21 ∪ D22 ∪ D12 because Γ3 (β) ∩ Γ3 (y) = {α}.
Since z ∈ Γ3 (y) ∩ Γ2 (α) = Γ3 (y) ∩ (D22 ∪ D12 ) ' Kt×2 , e(x, Γ3 (y) ∩ D22 ) ≤ 1. Hence
e(x, D21 ) ≥ e(x, Γ3 (y) ∩ D21 ) ≥ b2 − 1. Therefore we get c2 − (κ1 − κ2 + 2) = b2 − 1.
Claim 3.
For every α ∈ Γ and β, γ ∈ Γ3 (α) with ∂(β, γ) = 2, b2 − 1 vertices of
Γ1 (β) ∩ Γ1 (γ) are at distance 2 from α and κ1 − κ2 + 2 vertices of Γ1 (β) ∩ Γ1 (γ) are at
distance 3 from α
Consider the rank 3 diagram. Then γ ∈ D23 . Since e(γ, D12 ) = b2 − 1, by Claim 2, and
e(γ, D13 ) = κ1 − κ2 + 2, we get the claim.
Claim 4.
There are α, x ∈ Γ with ∂(α, x) = 2 such that Γ3 (α) ∩ Γ1 (x) is not a clique.
Take α ∈ Γ and β, γ ∈ Γ3 (α) with ∂(β, γ) = 2. By Claim 3 and b2 ≥ 2, we can take
x ∈ Γ2 (α) such that x ∼ γ, x ∼ β.
Claim 5.
b2 ≤ t.
In the rank 3 diagram, take x ∈ D32 , y ∈ D23 such that ∂(x, y) = 2. Let {z} =
Γ3 (x) ∩ Γ3 (y). We may assume ∂(α, z) = 2. Let
A = {u ∈ Γ3 (α) ∩ Γ1 (x) : u ∼ y},
B = {u ∈ Γ3 (β) ∩ Γ1 (y) : u ∼ x}.
Since Γ3 (α) ∩ Γ1 (x) ⊆ D23 ' Kt×2 and y ∈ D23 , we have |A| = b2 or b2 − 1. Similarly
|B| = b2 or b2 − 1. As {y} = Γ3 (α) ∩ Γ3 (z), no vertex in Γ3 (α) ∩ Γ1 (x) is at distance 3
from z. So all vertices in A are at distance 2 from z. By Claim 3 with α = z, β = x and
γ = y, all vertices in B are at distance 3 from z. Hence
Γ3 (β) ∩ Γ3 (z) ⊇ B ∪ {x}.
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TOMIYAMA
By Lemma 5.3, we get ∂(β, z) = 1 and B ∪ {x} is a clique of size b2 or b2 + 1. Since
B ∪ {x} ⊆ D32 ' Kt×2 and the maximal cliques of Kt×2 are size t, we have b2 ≤ t.
By Claim 4, we take α, x ∈ Γ with ∂(α, x) = 2 such that Γ3 (α) ∩ Γ1 (x) is not a
clique. Let {β} = Γ3 (α) ∩ Γ3 (x) and consider the rank 3 diagram, then x ∈ D32 . We can
take y ∈ D23 such that Γ3 (α) ∩ Γ1 (x) ⊆ Γ1 (y) because Γ3 (α) ∩ Γ1 (x) ⊆ D23 ' Kt×2 ,
Γ3 (α) ∩ Γ1 (x) is not a clique and b2 ≤ t.
Let A, B be as above. Then A = Γ3 (α) ∩ Γ1 (x). Let {z} = Γ3 (x) ∩ Γ3 (y). Then
z ∈ D21 ∪ D22 ∪ D12 .
If z ∈ D21 , then by an arguement similar to that in the proof of Claim 5, A ∪ {y} ⊆
Γ3 (α) ∩ Γ3 (z). This is impossible because A is not a clique and Γ3 (α) ∩ Γ3 (z) is a clique.
If z ∈ D22 ∪ D12 , then similarly all vertices in A are at distance 2 from z. Since |A| = b2 ,
this contradicts Claim 3.
Hence we have the assertion.
Proof of Proposition 5.1: Consider the rank 3 diagram. If D13 ia a clique, then {β} ∪ D13
is a clique of size a3 + 1. This contradicts Γ3 (α) being diameter 2. So D13 is not a clique.
Take x ∈ D31 . Since Γ3 (x) ∩ Γ3 (α) is a clique, Γ3 (x) ∩ Γ3 (α) ⊆ D13 ∪ {β}. D13 is not a
clique, so we take y ∈ D13 such that ∂(x, y) = 2. Let {z} = Γ3 (x) ∩ Γ3 (y), then we know
z ∈ D22 .
Claim. Γ3 (z) ⊆ D31 ∪ D21 ∪ D22 ∪ D12 ∪ D13 .
Since p23,3 = 1, {x} = Γ3 (z) ∩ Γ3 (β) and {y} = Γ3 (z) ∩ Γ3 (α), We get the claim.
By Lemma 5.4, Γ3 (z) ∩ Γ1 (α) ' Γ3 (z) ∩ Γ1 (β) ' Ks×2 . So we can take u ∈
Γ3 (z) ∩ Γ1 (α) such that x 6∼ u and v ∈ Γ3 (z) ∩ Γ1 (β) such that y 6∼ v. Then u ∈ D21 and
v ∈ D12 . So ∂(u, x) = ∂(u, y) = ∂(v, x) = ∂(v, y) = 2, which contradicts Lemma 3.1.
Therefore we have the assertion.
6.
Proof of Theorem 1.2
In this section we prove our main theorem. By Proposition 4.1 and 5.1, we may assume
d = 4.
By Lemma 3.5(1), the rank 4 diagram becomes as in Figure 5.
Lemma 6.1 Take every x, y ∈ Γ with ∂(x, y) = 3. For every u ∈ Γ3 (x) ∩ Γ4 (y) and every
v ∈ Γ4 (x) ∩ Γ3 (y), we have u ∼ v. Moreover we have b3 = p34,3 .
Proof: It is clear from Lemma 3.4 with i = 3 and Lemma 2.7.
Lemma 6.2 Take every α, β ∈ Γ with ∂(α, β) = 4. For every x ∈ Γ2 (α) ∩ Γ4 (β) and
every y ∈ Γ4 (α) ∩ Γ2 (β), we have ∂(x, y) = 2. Moreover we have p24,2 = κ2 .
215
ON DISTANCE-REGULAR GRAPHS WITH HEIGHT TWO, II
D42
@
@
D32
D41
@
@
D31
D40
D33
D24
@
@
D23
@
@
D22
@
@
D14
@
@
D13
@
@
D04
||
||
{α}
{β}
Figure 5.
Proof: Use the rank 4 diagram, then x ∈ D42 and y ∈ D24 . Suppose ∂(x, y) 6= 2. Since
b3 ≥ 2, we get ∂(x, y) = 3. Consider Γ4 (x), then we know
Γ4 (x) ⊆ D22 ∪ D23 ∪ D13 ∪ D14 ∪ {β}.
Similarly, we get
Γ4 (y) ⊆ D22 ∪ D32 ∪ D31 ∪ D41 ∪ {α}.
Suppose Γ4 (x) ∩ Γ1 (y) ⊆ Γ4 (x) ∩ D14 . Then
p24,4 = |Γ4 (α) ∩ Γ4 (x)| = |Γ4 (x) ∩ (D14 ∪ {β})|
≥ |Γ4 (x) ∩ Γ1 (y)| + |{β}|
= b3 + 1 > p34,3 .
This contradicts Lemma 2.1(3) with i = 2. So there is z ∈ D23 such that z ∈ Γ4 (x)∩Γ1 (y).
Suppose there is no vertex u in Γ4 (x) ∩ D13 such that ∂(y, u) = 3. Then
Γ4 (x) ∩ Γ3 (y) ⊆ (D22 ∪ D23 ) ∩ Γ4 (x).
From Lemma 6.1, |Γ4 (x) ∩ Γ3 (y)| = p34,3 = b3 ≥ 2. So there are at least two vertices
γ, δ ∈ Γ4 (x) ∩ Γ3 (y). Then ∂(γ, β) = ∂(δ, β) = 2 and ∂(γ, z) = ∂(δ, z) = 2. This
contradicts Lemma 3.1. So there is u ∈ Γ4 (x) ∩ D13 such that ∂(u, y) = 3. Similarly there
is v ∈ Γ4 (y) ∩ D31 such that ∂(v, x) = 3. Then u 6∼ v.
Now consider the rank 3 diagram with respect to (x, y). Then u ∈ D34 , v ∈ D43 with
u 6∼ v. This contradicts Lemma 6.1.
Finally,
p24,2 = |Γ4 (α) ∩ Γ2 (x)| = |D24 | = κ2 .
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TOMIYAMA
Lemma 6.3 We have the following.
(1) p24,4 = 1 and k2 = κ2 (1 + κ1 + κ2 ).
(2) c2 = κ2 − 2.
(3) For every edge u ∼ v, Γ4 (u) ∩ Γ4 (v) is a clique.
Proof: (1) In the rank 4 diagram take γ ∈ D22 . We can take x ∈ Γ4 (β) ∩ Γ4 (γ) and
y ∈ Γ4 (α) ∩ Γ4 (γ). Then x ∈ D42 and y ∈ D24 . By Lemma 6.2, ∂(x, y) = 2. Hence there
is a unique vertex z ∈ Γ4 (γ) such that ∂(x, z) = ∂(y, z) = 2 by our assumption. Since
z ∈ Γ4 (γ) ∩ Γ2 (y) and α ∈ Γ2 (γ) ∩ Γ4 (y), we get ∂(α, z) = 2 from Lemma 6.2. Similarly
we get ∂(β, z) = 2. We have z ∈ D22 . Suppose p24,4 ≥ 2, then we take w ∈ Γ4 (α) ∩ Γ4 (γ)
with y 6= w. Then ∂(x, z) = ∂(y, x) = ∂(y, z) = ∂(w, x) = ∂(w, z) = 2, which
contradicts Lemma 3.1. So we have p24,4 = 1. Since k2 = k2 p24,4 = k4 p44,2 , we have
k2 = κ2 (1 + κ1 + κ2 ).
(2) Since Γ4 (γ) is µ-closed and e(D22 , D33 ) = 0,
Γ1 (y) ∩ Γ1 (z) ⊆ Γ4 (γ) ∩ D23 ⊆ Γ4 (γ) ∩ (D22 ∪ D23 ∪ D24 ) ⊆ Γ4 (γ) ∩ Γ2 (x) ' Kt×2 .
So we have c2 = κ2 − 2.
(3) It is clear from Lemma 2.2.
Lemma 6.4 In the rank 4 diagram, for every x ∈ D42 ∪ D33 and every u ∈ D23 , we have
∂(x, u) 6= 4.
Proof: Suppose ∂(x, u) = 4. If x ∈ D42 , then α ∈ Γ2 (x)∩Γ4 (β) and u ∈ Γ4 (x)∩Γ2 (β).
By Lemma 6.2, we have ∂(α, u) = 2. This contradicts α ∈ D23 . If x ∈ D33 , similarly by
Lemma 6.1, we have a contradiction.
Lemma 6.5 In the rank 4 diagram, for any u ∈ D23 , e(u, D22 ) 6= 0.
Proof: Take any u ∈ D23 .
Claim 1. Γ4 (u) ⊆ D41 ∪ D31 ∪ D32 ∪ D22 ∪ D23 .
Take any v ∈ Γ4 (u), then 1 ≤ ∂(α, v) ≤ 3, 2 ≤ ∂(β, v) ≤ 4. If v ∈ D42 ∪ D33 , then by
Lemma 6.4, ∂(v, u) 6= 4. Hence we get the claim.
Claim 2. Γ4 (u) ∩ D22 6= φ.
By Lemma 3.6, 6.1 and 6.2,
|(D22
|D23 ∩ Γ4 (u)| = |Γ3 (α) ∩ Γ4 (u)| = p34,3 = b3
∪ D23 ) ∩ Γ4 (u)| = |Γ2 (β) ∩ Γ4 (u)| = p22,4 = κ2 > b3 .
Hence we have Γ4 (u) ∩ D22 6= φ.
Take γ ∈ Γ4 (u) ∩ D22 and take x, y, z ∈ Γ4 (γ) as in the proof of Lemma 6.3. Then
y ∼ u ∼ z because y, u, z ∈ Γ4 (γ) ∩ Γ2 (x) ' Kt×2 and ∂(y, z) = 2. Therefore we get
e(u, D22 ) 6= 0.
ON DISTANCE-REGULAR GRAPHS WITH HEIGHT TWO, II
217
Lemma 6.6 Take α, γ ∈ Γ with ∂(α, γ) = 3. In rank 3 diagram with respect to (α, γ), for
every edge x ∼ y in D14 , Γ4 (α) ∩ Γ2 (x) ∩ Γ2 (y) ⊆ D34 .
Proof: Suppose there is β ∈ Γ4 (α) ∩ Γ2 (x) ∩ Γ2 (y) such that β ∈ D24 . Consider the
rank 4 diagram. Then γ ∈ D23 , x, y ∈ D24 , γ ∼ x, γ ∼ y and x ∼ y. By Lemma 6.5,
there is δ ∈ D22 such that δ ∼ γ. Then Γ4 (α) ∩ Γ4 (δ) = φ, which contradicts p24,4 6= 0.
Lemma 6.7 For every α ∈ Γ and every edge β ∼ γ in Γ4 (α), there is δ ∈ Γ3 (α) such that
δ ∼ β, δ ∼ γ.
Proof: Consider the rank 4 diagram. Then γ ∈ D14 .
Claim 1.
Γ4 (γ) ⊆ {α} ∪ D41 ∪ D31 ∪ D32 .
For every x ∈ Γ4 (γ), ∂(α, x) ≤ 2 and ∂(β, x) ≥ 3. If ∂(β, x) = 4, then α ∼ x by
Lemma 6.3(3). Hence we get the claim.
Take y ∈ Γ2 (α) ∩ Γ4 (γ). Then by claim, y ∈ D32 . Since γ ∈ Γ4 (y) ∩ Γ1 (β) and b3 ≥ 2,
we can take δ ∈ Γ4 (y)∩Γ1 (β) such that δ 6= γ. Then δ ∈ D13 because {γ} = Γ4 (α)∩Γ4 (y).
By Lemma 3.4 with i = 3, γ ∼ δ.
Lemma 6.8 For every α ∈ Γ, Γ4 (α) is a strongly regular graph with intersection array


1
κ1 − κ2 + 2 


 ∗ 2
κ −κ −κ κ +κ2 −2κ2
κ2 − 2
0 1 1 1κ12 2
,
ι(Γ4 (α)) =


κ2 (κ1 −κ2 +2)

κ
∗
1
κ1
and p12,2 (Γ4 (α)) = b3 .
Proof: Let ∆ = Γ4 (α). It is clear that k(∆) = κ1 , k2 (∆) = κ2 , c2 (∆) = κ1 − κ2 + 2.
We only need to show |∆2 (x) ∩ ∆1 (y)| is constant for every edge x ∼ y in ∆. To show
this, we prove |∆2 (x) ∩ ∆2 (y)| is constant as |∆2 (x) ∩ ∆1 (y)| + |∆2 (x) ∩ ∆2 (y)| = κ2 .
By Lemma 6.7, there is γ ∈ Γ3 (α) such that γ ∼ x, γ ∼ y. Consider the rank 3 diagram
with respect to (α, γ). Then x, y ∈ D14 . By Lemma 6.1 and 6.6,
|∆2 (x) ∩ ∆2 (y)| = |D34 | = b3 .
So |∆2 (x) ∩ ∆2 (y)| is constant for every edge x ∼ y in ∆. >From k(∆)b1 (∆) =
2 +2)
.
k2 (∆)c2 (∆), we have b1 (∆) = κ2 (κ1κ−κ
1
Now we complete the proof of Theorem 1.2.
Proof of Theorem 1.2:
By Lemmas 6.3, 3.5(2) and 6.8, κ2 − 2 = c2 = c2 (∆) = κ1 − κ2 + 2. So
κ1 = 2κ2 − 4.
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TOMIYAMA
As κ2 = 2t, then the intersection array of ∆ = Γ4 (α) becomes


∗
1
2(t − 1) 

0
3t − 5 2(t − 1) .
ι(∆) =


4(t − 1)
t
∗
Claim 1. c2 = 2(t − 1), a4 = 4(t − 1), k4 = 6t − 3, k2 = 2t(6t − 3) and b3 = t.
c2 = c2 (∆), a4 = k(∆), k4 = 1 + k(∆) + k2 (∆), k2 = κ2 (1 + κ1 + κ2 ). Since
b3 = p12,2 (∆) = k2 (∆) − p12,1 (∆) = t, therefore we get the claim.
Claim 2. b2 = 4(t − 1), c4 = 8(t − 1), k = 12(t − 1) and c3 = t2 .
By Lemma 6.2, p24,2 = 2t. By Lemma 4.1.7 of [3], p24,2 = b2c2b3 . So we get b2 = 4(t − 1).
By Lemma 6.1, p34,1 = p34,3 . So p41,3 = p43,3 . Consider the rank 4 diagram, then
e(D33 , D24 ) = p43,3 b3 = c4 b3
and
e(D33 , D24 ) = p44,2 b2 = 2tb2 .
Hence we get c4 = 8(t − 1) and k = 12(t − 1). Since k2 b2 b3 = k4 c4 c3 , we get c3 = t2 .
Claim 3. a1 = 5t − 7, b1 = 7t − 6 and t = 3.
In the rank 4 diagram, take any γ ∈ D14 . Then
e(D13 , D14 ) = p43,1 (b3 − 1) = 8(t − 1)2
and
e(D13 , D14 ) = p44,1 (a1 − a1 (∆)) = 4(t − 1)(a1 − (3t − 5)).
So we get a1 = 5t − 7 and b1 = 7t − 6. Since kb1 = k2 c2 , we get t = 3.
By these claims, we know all the intersection numbers of Γ. By the uniqueness ([4] and
[12]), we get
Γ ' J(10, 4).
Therefore we have completed the proof of Theorem 1.2.
7.
Remarks
In Theorem 1.2 we assume t ≥ 2, i.e., Γd (α) ∩ Γ2 (β) is not a coclique of size 2. If the
assumption t ≥ 2 is removed, does any other graph but J(10, 4) appear? More generally,
for every α, β ∈ Γ with ∂(α, β) = d, which distance-regular graph satisfies d ≥ 3, h = 2
and Γd (α) ∩ Γ2 (β) is a coclique of size s? In this situation we do not have a complete
answer, but in the case s = 2, i.e., t = 1 in Theorem 1.2, there is only one graph.
Lemma 7.1 Let Γ be a distance-regular graph with d ≥ 3 and height h = 2. Suppose
Γd (α) ∩ Γ2 (β) is a coclique of size s for every α, β ∈ Γ with ∂(α, β) = d. Then Γd (α) is
a coclique of size s + 1 for every α ∈ Γ.
ON DISTANCE-REGULAR GRAPHS WITH HEIGHT TWO, II
219
Proof: Suppose ad = pdd,1 6= 0. Then in the rank d diagram, for every γ ∈ D2d and every
δ ∈ D1d , γ ∼ δ. So we know Γd (α) ' Kτ ×(s+1) with τ ≥ 2 for every α ∈ Γ. A. Hiraki
and H. Suzuki showed that there is no such graph. (See Appendix of [13].)
We may assume ad = 0. In this case we get Γd (α) is a coclique of size s + 1 for every
α ∈ Γ.
Lemma 7.2 Let Γ be a distance-regular graph with d ≥ 3 and height h = 2. Suppose
Γd (α) ∩ Γ2 (β) is a coclique of size 2 for every α, β ∈ Γ with ∂(α, β) = d. Then the
intersection array of Γ becomes


∗ 1 2 4
ι(Γ) = 0 0 0 0 .


4 3 2 ∗
Proof: By the previous lemma, Γd (α) is a coclique of size 3 for every α ∈ Γ. Since
2 < k < k2 and k2 divides kd pdd,2 = 6, we get k2 = 6. So k = 3 or 4 or 5. Since
kb1 = k2 c2 = 6c2 and c2 < k, k 6= 5. Since kb1 · · · bd−1 = kd cd cd−1 · · · c1 and cd = k,
3 divides b1 · · · bd−1 . So k 6= 3. Hence k = 4. Then from kb1 = k2 c2 , we have b1 = 3
and c2 = 2. Since c2 = 2, ci ≥ 3 (3 ≤ i ≤ d − 1). (See 5.4.1 of [3].) By using
k2 b2 · · · bd−1 = kd cd cd−1 · · · c3 , ci ≥ 3 (3 ≤ i ≤ d − 1) and bi ≤ 2 (2 ≤ i ≤ d − 1), we
have d = 3 and b2 = 2. Thus we have the assertion.
Remarks. 1. The array in Lemma 7.2 uniquely determines a graph. (See [11] or Theorem
7.5.3 of [3].)
2. Every bipartite distance-regular graph Γ with h = 2 satisfies Γd (α) ∩ Γ2 (β) is a coclique
for every α, β ∈ Γ with ∂(α, β) = d.
3. K. Nomura conjectured that there is no bipartite distance-regular graph Γ with diameter
d ≥ 4 and height h = 2. (Conjecture 1.2 of [5].) But the following counter example is
known for d = 4. A graph with the array


 ∗ 1 5 12 15 
0 0 0 0 0
ι(Γ) =


15 14 10 3 ∗
satisfies h = 2. (See Section 6 of [8].) In [10] H. Suzuki showed that this conjecture is true
if d is odd. If d is even, then each bipartite half of Γ has height 1.
References
1. E. Bannai and T. Ito, Algebraic Combinatorics I, Benjamin-Cummings, California, 1984.
2. N. L. Biggs, Algebraic Graph Theory, Cambridge University Press, Cambridge, 1974.
3. A. E. Brouwer, A. M. Cohen and A. Neumaier, Distance-Regular Graphs, Springer, Berlin-Heidelberg,
1989
220
TOMIYAMA
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