Journal of Algebraic Combinatorics, 7 (1998), 165–180
c 1998 Kluwer Academic Publishers. Manufactured in The Netherlands.
°
Imprimitive Q-polynomial Association Schemes*
hsuzuki@icu.ac.jp
HIROSHI SUZUKI
Department of Mathematics, International Christian University,
10-2, Osawa 3-chome, Mitaka-shi Tokyo 181, Japan
Received April 25, 1996; Revised January 13, 1997
Abstract. It is well known that imprimitive P -polynomial association schemes X = (X, {Ri }0≤i≤d ) with
k1 > 2 are either bipartite or antipodal, i.e., intersection numbers satisfy either ai = 0 for all i, or bi = cd−i for
all i 6= [d/2]. In this paper, we show that imprimitive Q-polynomial association schemes X = (X, {Ri }0≤i≤d )
with d > 6 and k1∗ > 2 are either dual bipartite or dual antipodal, i.e., dual intersection numbers satisfy either
a∗i = 0 for all i, or b∗i = c∗d−i for all i 6= [d/2].
Keywords: Q-polynomial, association scheme, imprimitivity, Krein parameters, distance-regular graph
1.
Introduction
A d-class symmetric association scheme is a pair X = (X, {Ri }0≤i≤d ), where X is a finite
set, each Ri is a nonempty subset of X × X for i = 0, 1, . . . , d satisfying the following.
(i) R0 = {(x, x)|x ∈ X}.
(ii) {Ri }0≤i≤d is a partition of X × X, i.e.,
X × X = R0 ∪ R1 ∪ · · · ∪ Rd , Ri ∩ Rj = ∅ if i 6= j.
(iii) t Ri = Ri for i = 0, 1, . . . , d, where t Ri = {(y, x)|(x, y) ∈ Ri }.
(iv) There exist integers phi,j such that for all x, y ∈ X with (x, y) ∈ Rh ,
phi,j = |{z ∈ X|(x, z) ∈ Ri , (z, y) ∈ Rj }|.
We refer to X as the vertex set of X , and to the integers phi,j as the intersection numbers
of X .
Let X = (X, {Ri }0≤i≤d ) be a symmetric association scheme. Let MatX (R) denote
the algebra of matrices over the reals R with rows and columns indexed by X. The i-th
adjacency matrix Ai ∈ MatX (R) of X is defined by
½
1 if (x, y) ∈ Ri
(x, y ∈ X).
(Ai )xy =
0 otherwise
From (i) − (iv) above, it is easy to see the following.
* This research was partially supported by the Grant-in-Aid for Scientific Research (No.06640075,
No. 09640062), the Ministry of Education, Science and Culture, Japan.
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SUZUKI
(i)0 A0 = I.
(ii)0 A0 + A1 + · · · + Ad = J, where J is the all-1s matrix, and Ai ◦ Aj = δi,j Ai for
0 ≤ i, j ≤ d, where ◦ denotes the entry-wise matrix product.
(iii)0
t
Ai = Ai for 0 ≤ i ≤ d.
(iv)0 Ai Aj =
d
X
phi,j Ah for 0 ≤ i, j ≤ d.
h=0
By the Bose-Mesner algebra of X we mean the subalgebra M of MatX (R) generated by
the adjacency matrices A0 , A1 , . . . , Ad . Observe by (iv)0 above that the adjacency matrices
form a basis for M. Moreover, M consists of symmetric matrices and it is closed under ◦.
In particular, M is commutative in both multiplications.
Since the algebra M consists of commutative symmetric matrices, there is a second basis
E0 , E1 , . . . , Ed satisfying the following.
(i)00 E0 =
1
J.
|X|
(ii)00 E0 + E1 + · · · + Ed = I, and Ei Ej = δi,j Ei for 0 ≤ i, j ≤ d.
(iii)00
t
Ei = Ei for 0 ≤ i ≤ d.
(iv)00 Ei ◦ Ej =
d
1 X h
h
qi,j Eh , (0 ≤ i, j ≤ d) for some real numbers qi,j
.
|X|
h=0
E0 , E1 , . . . , Ed are the primitive idempotents of the Bose-Mesner algebra. The parameters
h
are called Krein parameters.
qi,j
h
are zero if one of the indices h, i, j is out of range
Conventionally, we assume phi,j and qi,j
{0, 1, . . . , d} otherwise mentioned clearly.
A symmetric association scheme X = (X, {Ri }0≤i≤d ) with respect to the ordering
R0 , R1 , . . . , Rd of the relations is called a P -polynomial association scheme if the following
conditions are satisfied.
(P 1) phi,j = 0 if one of h, i, j is greater than the sum of the other two.
(P 2) phi,j 6= 0 if one of h, i, j is equal to the sum of the other two for 0 ≤ h, i, j ≤ d.
In this case we write ci = pii−1,1 , ai = pii,1 , bi = pii+1,1 and ki = p0i,i for i = 0, 1, . . . , d.
A symmetric association scheme X = (X, {Ri }0≤i≤d ) with respect to the ordering
E0 , E1 , . . . , Ed of the primitive idempotents of the Bose-Mesner algebra is called a Qpolynomial association scheme if the following conditions are satisfied.
h
= 0 if one of h, i, j is greater than the sum of the other two.
(Q1) qi,j
h
6= 0 if one of h, i, j is equal to the sum of the other two for 0 ≤ h, i, j ≤ d.
(Q2) qi,j
IMPRIMITIVE
Q-POLYNOMIAL SCHEMES
167
i
i
i
0
In this case we write c∗i = qi−1,1
, a∗i = qi,1
, b∗i = qi+1,1
and ki∗ = qi,i
for i = 0, 1, . . . , d.
If X = (X, {Ri }0≤i≤d ) is a P -polynomial association scheme with respect to the ordering
R0 , R1 , . . . , Rd , then the graph Γ = (X, R1 ) with vertex set X, edge set defined by R1
becomes a distance-regular graph. In this case,
Ri = {(x, y) ∈ X × X|∂(x, y) = i},
where ∂(x, y) denotes the distance between x and y. Conversely, every distance-regular
graph is obtained in this way.
Q-polynomial association schemes appear in design theory in connection with tight conditions, but it is not much studied compared with P -polynomial association schemes, though
there are extensive studies of P - and Q-polynomial association schemes.
A symmetric association scheme X = (X, {Ri }0≤i≤d ) is said to be imprimitive if it
satisfies one of the following equivalent conditions.
(A) By a suitable rearrangement of indices 1, 2, . . . , d, there exists an index s (0 < s < d)
such that Ai Aj is a linear combination of A0 , A1 , . . . , As for all i, j (0 ≤ i, j ≤ s).
(E) By a suitable rearrangement of indices 1, 2, . . . , d, there exists an index t (0 < t < d)
such that Ei ◦ Ej is a linear combination of E0 , E1 , . . . , Et for all i, j (0 ≤ i, j ≤ t).
The imprimitivity of association schemes including the equivalence of the above definitions were first studied in [3]. We also refer the readers to sections 2.4, 2.9 and 3.6 in [1]
and sections 2.4, 4.1 and 4.2 in [2].
The following is well known. See the references above.
Theorem 1 Let X = (X, {Ri }0≤i≤d ) be an imprimitive P -polynomial association scheme
with respect to the ordering R0 , R1 , . . . , Rd of the relations. If k1 > 2, then one of the
following holds.
(i) ai = 0 for all i = 0, 1, . . . , d.
(ii) bi = cd−i for all i = 0, 1, . . . , d except possibly for i = [d/2].
If the condition (i) is satisfied, the scheme is called bipartite, and if the condition (ii) is
satisfied, it is called antipodal, by adopting the terminologies of the distance-regular graph
associated with the P -polynomial structure.
The following is our main result in this paper.
Theorem 2 Let X = (X, {Ri }0≤i≤d ) be an imprimitive Q-polynomial association scheme
with respect to the ordering E0 , E1 , . . . , Ed of the primitive idempotents. If d > 6 and
k1∗ > 2, then one of the following holds.
(i) a∗i = 0 for all i = 0, 1, . . . , d.
(ii) b∗i = c∗d−i for all i = 0, 1, . . . , d except possibly for i = [d/2].
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SUZUKI
If the condition (i) is satisfied, the scheme is called dual bipartite, and if the condition
(ii) is satisfied, it is called dual antipodal. It is known that if k1∗ = 2, then X is an ordinary
polygon.
The proof of Theorem 1 is relatively easy and uses the inequalities based on the combinatorial structure of distance-regular graphs. We substitute that part by matrix identities to
prove Theorem 2. These identities were used in Dickie’s paper [5], which is a part of [4,
Chapter 4].
2.
P -polynomial C-algebra
We begin with a definition of P -polynomial C-algebra.
Let d be a positive integer and let ci+1 , ai , bi−1 (i = 0, 1, . . . , d) be real numbers satisfying
the following.
(i) a0 = b−1 = cd+1 = 0 and c1 = 1.
(ii) ci + ai + bi = b0 = cd + ad for i = 1, . . . , d − 1.
(iii) bi ci+1 > 0 for i = 0, 1, . . . , d − 1.
A P -polynomial C-algebra is an algebra over the reals R with basis x0 , x1 , . . . , xd , which
satisfies the following.
x0 x0 = x0 ,
x1 xi = bi−1 xi−1 + ai xi + ci+1 xi+1 , (0 ≤ i ≤ d),
(1)
where x−1 and xd+1 are indeterminates. Then xi can be written as a polynomial of x1 of
degree i and x0 = 1, the unit element in this algebra. Define constants phi,j by the following.
xi xj =
d
X
phi,j xh , 0 ≤ i, j ≤ d.
(2)
h=0
Since the algebra becomes commutative, phi,j = phj,i . Let ki = p0i,i , n = k0 + k1 + · · · + kd ,
and ne0 = x0 + x1 + · · · + xd . Then it is easy to check by (i) and (ii) that k1 = b0 and
that x1 (ne0 ) = k1 (ne1 ).
The algebra M =< x0 , x1 , . . . , xd > defined above becomes a C-algebra in the sense
defined in [1, Section 2.5]. See also [1, Section 3.6] and (2) in the following lemma. In
particular, M has another basis {e0 , e1 , . . . , ed } consisting of primitive idempotents and
the dual algebra M∗ defined by xi ◦ xj = δi,j xi becomes a C-algebra with respect to the
basis ne0 = x0 + x1 + · · · + xd , ne1 , . . . , ned . Let
ei ◦ ej =
d
1X h
qi,j eh .
n
h=0
As the intersection numbers and the Krein parameters, by convention we assume the pah
of C-algebras are zero if one of the indices h, i, j is out of range
rameters phi,j and qi,j
{0, 1, . . . , d}.
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Q-POLYNOMIAL SCHEMES
169
Lemma 1 Let M =< x0 , x1 , . . . , xd > be a P -polynomial C-algebra. Let ki = p0i,i .
Then the following hold.
(1) phi+1,j ci+1 = phi,j−1 bj−1 + phi,j (aj − ai ) + phi,j+1 cj+1 − phi−1,j bi−1 .
(2) p0i,j = δi,j ki , kh phi,j = ki pij,h and ki > 0 for i = 0, 1, . . . , d. In particular, phi,j = 0 if
and only if pih,j = 0.
(3) phi,j = 0 if one of h, i, j is greater than the sum of the other two.
(4) phi,j 6= 0 if one of h, i, j is equal to the sum of the other two for 0 ≤ h, i, j ≤ d.
i+h
(5) pi+h
i,h+1 ch+1 = pi,h (ai + · · · + ai+h − a1 − · · · − ah ).
Proof: (1) Compute the coefficient of xh in the expression of (x1 xi )xj = (x1 xj )xi by
applying (1) and then (2), and we obtain the formula.
(2) First we prove that ci+1 p0i+1,j+1 = δi,j bj p0i,j for 0 ≤ i ≤ j ≤ d − 1 by induction
on i. If i = 0, then this is obvious. Compute the coefficient of x0 in the expression of
(x1 xi )xj = xi (x1 xj ) in two ways. By induction hypothesis p0l,m = 0 for l < i + 1, m,
we have ci+1 p0i+1,j+1 = bj p0i,j . Since p0i,j = δi,j p0i,i , we have the assertion. Hence we
have p0i,j = δi,j ki and ki bi = ki+1 ci+1 . By our assumption bi ci+1 > 0, we have ki > 0 as
k0 = 1.
Next compute the coefficient of x0 in the expression of (xi xj )xh = (xj xh )xi in two
ways using the formula p0i,j = δi,j ki just shown above, and we obtain the second formula
kh phi,j = ki pij,h .
(3) By (2), we may assume that h > i + j. Since xi is expressed as a polynomial of x1
of degree i, we have the assertion.
i+j
(4) By (2), we may assume that h = i + j. Then by (1), pi+j
i,j ci = pi−1,j+1 cj+1 . Hence
we have the assertion by induction on i.
(5) This follows by induction on h using (1).
By definition, it is easy to see that the Bose-Mesner algebra M of a P -polynomial association scheme becomes a P -polynomial C-algebra with respect to the basis A0 , A1 , . . . , Ad .
Moreover, if we take ◦ product, the dual Bose-Mesner algebra M∗ of Q-polynomial association scheme becomes a P -polynomial C-algebra with respect to the basis |X|E0 , |X|E1 ,
. . . , |X|Ed .
In both of these cases, the structure constants and Krein parameters are nonnegative, i.e.,
h
≥ 0. The latter inequality is called the Krein condition.
phi,j ≥ 0 and qi,j
Lemma 2 Let M =< x0 , x1 , . . . , xd > be a P -polynomial C-algebra. Suppose the
structure constants phi,j are all nonnegative. Then the following hold.
(1) If phi+1,j−1 = phi+1,j = phi+1,j+1 = 0 for 0 ≤ i < d, then phi,j = phi+2,j = 0.
(2) If phl,j−l+i = phl,j−l+i+1 = · · · = phl,j+l−i = 0 for i ≤ l and 0 ≤ i < d, then
phi,j = ph2l−i,j = 0.
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(3) For all i, j with 0 ≤ i, h, i + h ≤ d, ai = ai+1 = · · · = ai+h = 0 implies a1 = · · · =
ah = 0.
(4) For all h and i with 0 ≤ h, i, i + h ≤ d, the following hold.
(i) If phi,i+h−1 = 0, then ai ≤ ai+h . Moreover if ai = ai+h , then phi+1,i+h = 0.
(ii) If phi+1,i+h = 0, then ai ≥ ai+h . Moreover if ai = ai+h , then phi,i+h−1 = 0.
(iii) If phi,i+h−1 = phi+1,i+h = 0, then ai = ai+h .
(5) For all h and i with 0 ≤ i ≤ h ≤ d, the following hold.
(i) If phi,h−i+1 = 0, then ai ≤ ah−i . Moreover if ai = ah−i , then phi+1,h−i = 0.
(ii) If phi+1,h−i = 0, then ai ≥ ah−i . Moreover if ai = ah−i , then phi,h−i+1 = 0.
(iii) If phi,h−i+1 = phi+1,h−i = 0, then ai = ah−i .
Proof: (1) Replacing i by i + 1, by Lemma 1 (1) we have
phi,j bi + phi+2,j ci+2 = phi+1,j−1 bj−1 + phi+1,j (aj − ai+1 ) + phi+1,j+1 cj+1 .
Since i < d by our assumption, bi > 0 and we have the assertion. Note that bi = pi1,i+1
with i < d is nonzero by the definition of P -polynomial C-algebra and it is nonnegative
by our assumption.
(2) We prove the assertion by induction on m = l − i. If l = i, there is nothing to prove.
Suppose the assertion holds for m = l − i − 1 ≥ 0. Then
phi+1,j−1 = phi+1,j = phi+1,j+1 = ph2l−i−1,j−1 = ph2l−i−1,j = ph2l−i−1,j+1 = 0.
By (1), we have phi,j = ph2l−i,j = 0.
(3) This follows from Lemma 1 (4), (5) and the nonnegativity of the aj ’s.
(4) Since phi−1,i+h = phi,i+h+1 = 0 by Lemma 1 (3), it follows from Lemma 1 (1) by
setting j = i + h that
phi+1,i+h ci+1 + phi,i+h ai = phi,i+h−1 bi+h−1 + phi,i+h ai+h .
Since phi,i+h 6= 0, we have the assertion.
(5) This is similar to (4). Consider the following.
phi+1,h−i ci+1 + phi,h−i ai = phi,h−i+1 ch−i+1 + phi,h−i ah−i .
Lemma 3 Let M =< x0 , x1 , . . . , xd > be a P -polynomial C-algebra such that the
structure constants phi,j are all nonnegative. Suppose for a positive integer α, pα
i,jα 6= 0
only if i ≡ 0 (mod α). Then pα
l,m 6= 0 only if l ≡ m or − m (mod α).
Proof: It suffices to consider pα
l,m with 0 < m − l < α by Lemma 1 (3). We may assume
that (2i − 1)α < l + m < 2iα or 2iα < l + m < (2i + 1)α. In the first case, there
exists 0 ≤ β ≤ [α/2] − 1 such that m = iα − β or iα + β as l < m. Similarly, in the
IMPRIMITIVE
Q-POLYNOMIAL SCHEMES
171
latter case, there exists 0 ≤ β ≤ [α/2] − 1 such that l = iα − β or iα + β. Define γ by
the following: l = (i − 1)α + β + γ in the first case and m = (i + 1)α − β − γ in the
latter. Since 0 < m − l < α and m + l is in the corresponding range, in each case we have
1 ≤ γ ≤ α − 1 and that 2β + γ < α. Thus there are four cases.
(i) l = (i − 1)α + β + γ and m = iα − β.
(ii) l = (i − 1)α + β + γ and m = iα + β.
(iii) l = iα − β and m = (i + 1)α − β − γ.
(iv) l = iα + β and m = (i + 1)α − β − γ.
α
α
We apply Lemma 2 (2). Since pα
(i−1)α+γ,iα = · · · = p(i−1)α+2β+γ,iα = 0, pl,m = 0 in
α
α
α
the first two cases. Since piα,(i+1)α−2β−γ = · · · = piα,(i+1)α−γ = 0, pl,m = 0 in the last
two cases.
The following is Proposition 6.2 in [1] but the description of it involves an error. Hence
we restate the corrected version below. Note that we do not know if bt = ct+1 when
α = 2t + 1.
Proposition 1 Let M =< xi | 0 ≤ i ≤ d > be a P -polynomial C-algebra with respect to
h
≥ 0 for all h, i, j. Let < xβ | β ∈ T >
the basis x0 , x1 , . . . , xd . Assume phi,j ≥ 0 and qi,j
be a proper C-subalgebra of M. Then
2d
d 2d + 1
,
}.
T = {0, α, 2α, 3α, . . .} for some α ∈ {2, d, ,
s 2s + 1 2s + 1
Let the following be the array of defining parameters,

  
 ci   ∗ 1 c2 · · · cd−1 cd 
ai
0 a1 a2 · · · ad−1 ad .
=

  
bi
b0 b1 b2 · · · bd−1 ∗
Then M has a C-subalgebra < xβ | β ∈ T > with (i) α = 2, (ii) α = d, (iii)
2d
α ∈ { ds , 2d+1
2s+1 , 2s+1 } respectively if and only if the following hold.
(i) a2 = a4 = · · · = 0 and a1 = a3 = · · ·.
(ii) bi = cd−i for all i except possibly for i = [d/2].
(iii) The parameters ch , ah , bh satisfy the following for 0 ≤ h ≤ d − 1. bi = cα−i =
bjα+i = c(j+1)α−i for all 1 ≤ i ≤ α − 1 and 1 ≤ j except for i = [α/2], ai = aα−i =
ajα+i = a(j+1)α−i for all 0 ≤ i ≤ α and 1 ≤ j except for i = [α/2], [(α + 1)/2] with
odd α. Moreover,

if α = ds
 (b0 , 0)
(cd , ad ) = (c(α−1)/2 , a(α−1)/2 + b(α−1)/2 ) if α = 2d+1
2s+1

2d
(cα/2 + bα/2 , aα/2 )
if α = 2s+1
.
Note that (i) and (ii) are special cases of (iii) for α = 2 and α = d, respectively.
172
3.
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Vanishing Conditions of Krein Parameters
h
of symmetric association schemes
Only a few restrictions of the Krein parameters qi,j
are known except those derived algebraically using Lemma 1. We first list them in the
following.
Proposition 2 Let X = (X, {Ri }0≤i≤d ) be a symmetric association scheme. Let E0 , E1 ,
h
be the Krein parameters. Then the following
. . . , Ed be primitive idempotents and let qi,j
hold.
h
≥ 0 for all 0 ≤ h, i, j ≤ d.
(1) qi,j
(2) For 0 ≤ h, i, j ≤ d, we have
h
=0⇔
qi,j
X
(Eh )ux (Ei )uy (Ej )uz = 0 for all x, y, z ∈ X.
u∈X
Proposition 2 (1) is known as Krein condition and (2) is in [3]. See also [1, Theorem
2.3.8, Proposition 2.8.3].
Lemma 4 Let X = (X, {Ri }0≤i≤d ) be a symmetric association scheme. Let
h
be the Krein parameters. Suppose
E0 , E1 , . . . , Ed be primitive idempotents and let qi,j
i
i
{i | qj,k ql,m 6= 0} ⊂ {h}. Then for all integers 0 ≤ h, i, j, k, l, m ≤ d and all vertices
a, a0 , b, b0 , the following hold.
(1)
X
(Ej )ea (Ek )ea0 (El )eb (Em )eb =
e∈X
(2)
X
h
X
ql,m
(Ej )ea (Ek )ea0 (Eh )eb .
|X|
e∈X
X
(Ej )ea (Ek )ea0 (El )eb (Em )eb0 =
(Ej )ea (Ek )ea0 (Eh )ee0 (El )e0 b (Em )e0 b0 .
e,e0 ∈X
e∈X
Proof: (1) By Proposition 2 (2), we have
X
(Ej )ea (Ek )ea0 (El )eb (Em )eb
e∈X
=
X
(Ej )ea (Ek )ea0 (El ◦ Em )eb
e∈X
=
d
1 X i X
q
(Ej )ea (Ek )ea0 (Ei )eb
|X| i=0 l,m
e∈X
=
h
ql,m
X
|X|
e∈X
(Ej )ea (Ek )ea0 (Eh )eb .
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Q-POLYNOMIAL SCHEMES
(2) Since I = E0 + E1 + · · · + Ed , similarly we have
X
(Ej )ea (Ek )ea0 (El )eb (Em )eb0
e∈X
=
X
(Ej )ea (Ek )ea0 (I)ee0 (El )e0 b (Em )e0 b0
e,e0 ∈X
=
d
X
X
(Ej )ea (Ek )ea0 (Ei )ee0 (El )e0 b (Em )e0 b0
i=0 e,e0 ∈X
=
d X
X
i=0
=
Ã
e0 ∈X
d X
X
!
X
(Ej )ea (Ek )ea0 (Ei )ee0
e∈X
Ã
(Ej )ea (Ek )ea0
X
(El )e0 b (Em )e0 b0
!
(Ei )ee0 (El )e0 b (Em )e0 b0
.
e0 ∈X
i=0 e∈X
Comparing the last two expressions using Proposition 2 (2) we have the right hand side of
(2) by our assumption.
Proposition 3 Let X = (X, {Ri }0≤i≤d ) be a Q-polynomial association scheme with
respect to the ordering E0 , E1 , . . . , Ed of primitive idempotents. Suppose that
l
l
qi−j,h+j
6= 0} ⊂ {h + i − j}.
{l | qj,h+i
h+j
h+i
= 0 implies that qj,h+j
= 0.
Then for h ≥ 0, i ≥ j ≥ 1 with h + i + j ≤ d, qi,h+j
h+i
= 0, by Proposition 2,
Proof: Since qi,h+j
i
X
qj,i−j
(Eh+i )ux (Ei )uy (Eh+j )uz
|X|
u∈X
Ã
!
i
X qj,i−j
(Ei )uy (Eh+i )ux (Eh+j )uz
=
|X|
u∈X
X
=
((Ej ◦ Ei−j )Ei )uy (Eh+i )ux (Eh+j )uz
0 =
u∈X
=
X X
(Ej )uv (Ei−j )uv (Ei )vy (Eh+i )ux (Eh+j )uz
u∈X v∈X
=
X
v∈X
Ã
(Ei )vy
X
!
(Ej )uv (Eh+i )ux (Ei−j )uv (Eh+j )uz
.
u∈X
l
l
qi−j,h+j
6= 0} ⊂ {h + i − j}, by Lemma 4 (2),
Since {l | qj,h+i
X X X
(Ei )vy (Ej )uv (Eh+i )ux (Eh+i−j )uw (Ei−j )wv (Eh+j )wz .
=
u∈X v∈X w∈X
174
SUZUKI
Since this holds for arbitrary x, y, z, we have
X
(Eh+i+j )xy (Eh+j )yz (Ej )xz ×
0 =
x,y,z∈X
X
(Ei )vy (Ej )uv (Eh+i )ux (Eh+i−j )uw (Ei−j )wv (Eh+j )wz
u,v,w∈X
=
X
(Eh+j )yz (Ei )vy (Ei−j )wv (Eh+j )wz ×
y,z,v,w∈X
X
(Eh+i+j )xy (Ej )xz (Eh+i )ux (Ej )uv (Eh+i−j )uw .
x,u∈X
l
l
qj,h+i−j
6= 0} ⊂ {h + i}, by Lemma 4 (2) we have
Since {l | qh+i+j,j
X
(Eh+j )yz (Ei )vy (Ei−j )wv (Eh+j )wz ×
=
y,z,v,w∈X
X
(Eh+i+j )xy (Ej )xz (Ej )xv (Eh+i−j )xw
x∈X
=
X
(Eh+j )wz (Ej )xz (Eh+i−j )xw ×
x,z,w∈X
X
(Eh+i+j )yx (Eh+j )yz (Ei )yv (Ej )xv (Ei−j )wv .
y,v∈X
l
l
qj,i−j
6= 0} ⊂ {i}, by Lemma 4 (2) we have
Since {l | qh+i+j,h+j
X
(Eh+j )wz (Ej )xz (Eh+i−j )xw ×
=
x,z,w∈X
X
(Eh+i+j )yx (Eh+j )yz (Ej )xy (Ei−j )wy
y∈X
=
X
(Eh+j )wz (Ej )xz (Eh+i−j )xw ×
x,z,w∈X
X
(Eh+j )yz (Ei−j )yw (Ej )yx (Eh+i+j )yx .
y∈X
l
l
qj,h+i+j
6= 0} ⊂ {h + i}, by Lemma 4 (1) we have
Since {l | qh+j,i−j
=
h+i
qj,h+i+j
X
(Eh+j )wz (Ej )xz (Eh+i−j )xw ×
|X|
x,z,w∈X
X
(Eh+j )yz (Ei−j )yw (Eh+i )yx
y∈X
=
h+i
X
qj,h+i+j
(Eh+j )yz ×
|X|
y,z∈X
X
(Eh+i )xy (Ej )xz (Eh+i−j )xw (Ei−j )wy (Eh+j )wz .
x,w∈X
IMPRIMITIVE
175
Q-POLYNOMIAL SCHEMES
l
l
Since {l | qh+i,j
qi−j,h+j
6= 0} ⊂ {h + i − j}, by Lemma 4 (2), we have
=
=
=
=
=
=
h+i
X
qj,h+i+j
|X|
y,z∈X
h+i
X
qj,h+i+j
|X|
X
(Eh+i )xy (Ej )xz (Ei−j )xy (Eh+j )xz
x∈X
(Ej )xz (Eh+j )xz
x,z∈X
h+i
X
qj,h+i+j
|X|
(Eh+j )yz
(Ei−j )xy (Eh+i )xy (Eh+j )yz
y∈X
(Ej )xz (Eh+j )xz
x,z∈X
X
((Ei−j ) ◦ (Eh+i ))xy (Eh+j )yz
y∈X
h+j
h+i
X
qj,h+i+j
qi−j,h+i
|X|2
X
(Ej )xz (Eh+j )xz (Eh+j )xz
x,z∈X
h+j
h+i
X
qj,h+i+j
qi−j,h+i
Ã
X
!
((Ej ) ◦ (Eh+j ))xz (Eh+j )zx
x∈X z∈X
h+j
h+j
h+i
X
qi−j,h+i
qj,h+j
qj,h+i+j
(Eh+j )xx .
3
|X|
x∈X
|X|2
Since Eh+j is a nonzero idempotent,
X
(Eh+j )xx = trace(Eh+j ) = rank(Eh+j ) 6= 0.
x∈X
h+j
h+j
h+i
6= 0 and qi−j,h+i
6= 0 by (Q2). Hence qj,h+j
= 0.
Moreover, qj,h+i+j
Corollary 1 Let X = (X, {Ri }0≤i≤d ) be a Q-polynomial association scheme with respect
to the ordering E0 , E1 , . . . , Ed of primitive idempotents.
(1) For h ≥ 0, i ≥ 1 with h + i + 1 ≤ d,
h+i
h+i
h+1
= q1,h+i
= 0 implies that q1,h+1
= 0.
qi,h+1
(2) For h ≥ 0, i ≥ 2 with h + i + 2 ≤ d,
h+i
h+i
h+i
h+2
= q2,h+i
= q2,h+i−1
= 0 implies that q2,h+2
= 0.
qi,h+2
h+i
= 0, by (Q2) we have the following.
Proof: (1) Since q1,h+i
l
l
qi−1,h+1
6= 0} ⊂ {h + i − 1}.
{l | q1,h+i
Hence we have the assertion from Proposition 3 by setting j = 1.
h+i
h+i
= q2,h+i−1
, by (Q2) we have the following.
(2) Since q2,h+i
l
l
qi−2,h+2
6= 0} ⊂ {h + i − 2}.
{l | q2,h+i
176
SUZUKI
Hence we have the assertion from Proposition 3 by setting j = 2.
By setting h = 0 in Corollary 1, we obtain the result of G. A. Dickie in [4, 5]. Hence
the proposition is a generalization of it. The following result for P -polynomial association
schemes is not used elsewhere in this paper but it is the dual of the result above, which can
be proved very similarly. The proof suggests a possible way to find vanishing conditions
and its proof in Q-polynomial association schemes.
Proposition 4 Let X = (X, {Ri }0≤i≤d ) be a P -polynomial association scheme with
respect to the ordering R0 , R1 , . . . , Rd of relations. Suppose that
{l | plj,h+i pli−j,h+j 6= 0} ⊂ {h + i − j}.
h+j
Then for h ≥ 0, i ≥ j ≥ 1, and h + i + j ≤ d, ph+i
i,h+j = 0 implies that pj,h+j = 0.
Proof: Suppose ph+j
j,h+j 6= 0. Then there are vertices α, β, γ ∈ X such that
(α, β), (α, γ) ∈ Rh+j and (β, γ) ∈ Rj .
Since ph+j
i−j,h+i 6= 0 by (P 2), there exists a vertex δ ∈ X such that (α, δ) ∈ Ri−j and
(δ, β) ∈ Rh+i . Consider two triples (δ, β, γ) and (δ, α, γ). Since {l | plh+i,j pli−j,h+j 6=
0} ⊂ {h + i − j}, (δ, γ) ∈ Rh+i−j . Since (β, δ) ∈ Rh+i and ph+i
h+i+j,j 6= 0 by (P 2), there
exists a vertex ² ∈ X such that (β, ²) ∈ Rh+i+j and (², δ) ∈ Rj . Consider two triples
(², β, α) and (², δ, α). Since {l | plh+i+j,h+j plj,i−j 6= 0} ⊂ {i}, we have (², α) ∈ Ri .
Next consider two triples (², β, γ) and (², δ, γ). Since {l | plh+i+j,j plj,h+i−j 6= 0} ⊂
{h + i}, we have (², γ) ∈ Rh+i . Finally consider a triple (², α, γ). Since
(², γ) ∈ Rh+i , (², α) ∈ Ri , and (α, γ) ∈ Rh+j ,
ph+i
i,h+j 6= 0, which is a contradiction.
4.
Proof of Main Theorem
In this section, we prove the following result. It is obvious that Theorem 2 is a direct
consequence of it.
Theorem 3 Let X = (X, {Ri }0≤i≤d ) be a Q-polynomial association scheme with respect
to the ordering E0 , E1 , . . . , Ed of the primitive idempotents. Suppose X is imprimitive. Or
more precisely, suppose the linear span of {Ei | i ∈ T } is closed under ◦ product for some
proper subset T of {0, 1, . . . , d} with T 6= {0}. In addition, assume that k1∗ > 2. Then
T = {0, α, 2α, 3α, . . .} for some α ≥ 2, and one of the following holds.
(i) α = 2 and a∗i = 0 for all i.
(ii) α = d and b∗i = c∗d−i for all i = 0, 1, . . . , d except possibly for i = [d/2].
IMPRIMITIVE
177
Q-POLYNOMIAL SCHEMES
(iii) d = 4, α = 3, and the parameters satisfy the following conditions.

 ∗ 
1 
1
c∗2 c∗3
 ci   ∗
a∗
0
0
a∗2 0 k ∗ − 1 .
=

 ∗i   ∗ ∗
k k − 1 1 b∗3
bi
∗
(iv) d = 6, α = 3, and the parameters satisfy the following conditions.

 ∗ 
1
c∗2
c∗3 1 c∗5 k ∗ 
 ci   ∗
a∗
0
0
a∗4 + a∗5 0 a∗4 a∗5 0
.
=

 ∗i   ∗ ∗
bi
1
b∗3 b∗4 1 ∗
k k −1
It is not difficult to see from Proposition 1 that if one of the conditions (i) − (iv) of the
theorem above holds, then the linear span of {Ei | i ∈ T } is closed under ◦ product, where
T = {0, α, 2α, 3α, . . .}. In particular, the association scheme X is imprimitive.
Throughout this section assume the following:
X = (X, {Ri }0≤i≤d ) is a Q-polynomial association scheme with respect to the ordering
E0 , E1 , . . . , Ed of the primitive idempotents such that the linear span of {Ei | i ∈ T } is
closed under ◦ product for some proper subset T of {0, 1, . . . , d} with T 6= {0}.
Under the assumption above, M∗ =< |X|E0 , |X|E1 , . . . , |X|Ed > with ◦ product is a
h
. Hence we can apply Proposition
P -polynomial C-algebra with nonnegative phi,j and qi,j
1. In particular, we have the following two lemmas as direct consequences.
Lemma 5 (1) T = {0, α, 2α, 3α, . . .} for some α ≥ 2.
(2) Let β = [α/2]. Then d ≡ 0 or β
(mod α).
Lemma 6 Let β = [α/2]. Then the following hold.
α
6= 0 only if l ≡ m or − m
(1) ql,m
(mod α).
α
= 0, unless α = 2β + 1 with h = β + 1.
(2) qα−h+1,h
α
= 0, unless α = 2β + 1 with h ≡ β + 1
(3) qα+h−1,h
(mod α).
α
6= 0, unless α = 2β with h = β + 1.
(4) Suppose α > 2 and 2 ≤ h ≤ α. Then qα−h+2,h
h
≥ 0 for all h, i and j. Hence this is a direct consequence
Proof: (1) By Proposition 2, qi,j
of Lemma 3.
(2) By (1) we have that 2h − 1 ≡ 0 (mod α), if the value is not zero. Hence α is odd
and h = β + 1.
(3) This is similar to (2).
(4) By (1) we have that 2h − 2 ≡ 0 (mod α), if the value is not zero. Since 2 ≤ h ≤ α,
α is even and h = β + 1.
178
SUZUKI
Lemma 7 If α < d, then a∗i = 0 for all i = 0, 1, . . . , α except for i = β + 1 with
α = 2β + 1.
α
= 0, unless α = 2β + 1 with i = β + 1. Since
Proof: By Lemma 6 (2), qα−i+1,i
α
i
= 0 by Corollary 1 (1) as desired.
q1,α = 0 by our assumption, we have a∗i = q1,i
Lemma 8 The following hold.
(1) Suppose α = 2β. Then for each 0 ≤ h ≤ α and i ≥ 0 with 0 ≤ h + iα ≤ d,
a∗h = a∗h+iα .
(2) Suppose α = 2β + 1. Then for each 0 ≤ h ≤ α and i ≥ 0 with 0 ≤ h + iα ≤ d,
a∗h = a∗h+iα unless h = β, β + 1. Moreover, a∗(i−1)α+β ≤ a∗iα+β , and a∗(i−1)α+β+1 ≥
a∗iα+β+1 .
α
= 0, unless α = 2β + 1 with h ≡ β + 1
Proof: By Lemma 6 (3), we have that qα+h−1,h
(mod α).
α
= 0 for every h. Hence by Lemma 2 (4)(iii), we
(1) Suppose α = 2β. Then qα+h−1,h
∗
∗
have that ah = ah+iα , for each 0 ≤ h ≤ α and i ≥ 0 with 0 ≤ h + iα ≤ d.
α
(2) Suppose α = 2β + 1. Then qα+h−1,h
= 0, unless h ≡ β + 1 (mod α). Hence
∗
∗
by Lemma 2 (4), ah = ah+iα unless h = β, β + 1, for each 0 ≤ h ≤ α and i ≥ 0
with 0 ≤ h + iα ≤ d. Moreover, a∗(i−1)α+β ≤ a∗iα+β , and a∗(i−1)α+β+1 ≥ a∗iα+β+1 .
Lemma 9 If α = 2β with α < d, then a∗i = 0 for all i.
Proof: By Lemma 7, a∗i = 0 for all i = 0, 1, . . . , α. Hence we have the assertion by
Lemma 8 (1).
Lemma 10 Suppose α = 2β + 1. If a∗h = 0 for all h = 0, 1, . . . , α. Then a∗j 6= 0 only
when j = d and d ≡ β (mod α).
Proof: Choose an integer i so that iα + 1 ≤ j ≤ (i + 1)α. We prove by induction on i.
There is nothing to prove when i = 0.
By induction hypothesis and Lemma 8, we may assume that j = iα+β or j = iα+β +1.
Since a∗(i−1)α+β+1 ≥ a∗iα+β+1 , a∗iα+β+1 = 0 by induction hypothesis.
α
= 0 and a∗(i−1)α+β+1 = a∗iα+β+1
Suppose iα + β < d. Then q(i−1)α+β+2,iα+β+1
α
α
implies q(i−1)α+β+1,iα+β
= 0 by Lemma 2 (4)(ii). Since q(i−1)α+β,iα+β−1
= 0, we have
∗
∗
0 = a(i−1)α+β = aiα+β by Lemma 2 (4)(iii) as desired.
Lemma 11 Suppose α = 2β + 1 ≥ 5 with α < d. Then a∗i = 0 for all i < d. Moreover,
a∗d 6= 0 only if d ≡ β (mod α).
Proof: By Lemma 10 and Lemma 7, it suffices to show that a∗β+1 = 0.
Assume that α ≥ 7. Then we have
a∗β+2 = · · · = a∗α = a∗α+1 = a∗α+2
IMPRIMITIVE
179
Q-POLYNOMIAL SCHEMES
by Lemma 8 (2) as a∗1 = a∗2 = 0. Note that d ≥ α + 2 by Lemma 5 (2). Since d ≥ α + 2,
we have by Lemma 2 (3) that a∗β+1 = 0 as (α + 2) − (β + 2) = β + 1.
Suppose α = 5. Then we have
a∗1 = a∗2 = a∗4 = a∗5 = a∗6 = 0.
5
4
= 0 by Lemma 6. Since q2,5
= 0 and a∗2 = a∗6 , by Lemma 2 (4)(i), we have
Now q2,4
4
6
6
q3,6 = 0. Hence q4,3 = q1,6 = 0, and by Corollary 1 (1), a∗3 = 0 as d ≥ 7 by Lemma 5 (2).
Lemma 12 Suppose α = 3 < d. Then one of the following holds.
(1) a∗i = 0 for all i < d, and a∗d 6= 0 only if d ≡ β
(mod 3).
(2) d = 6, a∗1 = a∗3 = a∗6 = 0, and a∗2 = a∗4 + a∗5 6= 0.
(3) d = 4, a∗1 = a∗3 = 0, and a∗2 6= 0.
6
= 0,
Proof: It is easy to see that a∗1 = a∗3i = 0 for every i. Suppose d ≥ 7. Since q3,4
∗
3
∗
∗
3
∗
a4 = 0 by Corollary 1 (1). Since q1,3 = 0 and a1 = a4 = 0, q2,4 = 0 and a2 = a∗5 by
4
= 0 with a∗4 = 0 implies a∗2 = 0 by Corollary 1
Lemma 2 (4) (i), (iii). Moreover, q3,2
∗
∗
∗
(1). Therefore we have a1 = a2 = a3 = 0. Hence we have (1) in this case.
If d ≤ 6, then d = 4 or 6 by Lemma 5 (2). If d = 4 or 6, then we have a∗1 = a∗3i = 0 for
5
= 0, a∗4 + a∗5 = a∗2 . Clearly if a∗2 = 0, we have case (1) by
every i. Moreover, since q3,3
Lemma 10. Hence we have one of the three cases above.
Lemma 13 Suppose d ≥ α + 2 with α > 2 and a∗i = 0 for all i = 1, 2, . . . , d − 1. Then
k1∗ = 2.
α
= 0, unless α = 2β and
Proof: Suppose k1∗ > 2. Observe by Lemma 6 that qα−h+2,h
α
α
h = β + 1 for 2 ≤ h ≤ α. Moreover q2,α = q2,α−1 = 0 by our assumption. Hence by
h
α
Corollary 1 (2), q2,h
= 0 when qα−h+2,h
= 0.
h
Suppose α = 2β + 1. Then q2,h = 0 for 2 ≤ h ≤ α. Hence
c∗h b∗h−1 + b∗h c∗h+1 − k1∗ = 0.
Now by induction we show that c∗h ≤ 1 when h is odd. The assertion is trivial if h = 1.
h
= 0, we have
Suppose c∗h−1 ≤ 1. Since q2,h
0 =
=
≥
≥
≥
c∗h b∗h−1 + b∗h c∗h+1 − k1∗
c∗h (k1∗ − c∗h−1 ) + b∗h c∗h+1 − k1∗
c∗h (k1∗ − 1) + b∗h + b∗h (c∗h+1 − 1) − k1∗
c∗h + b∗h − k1∗ + b∗h (c∗h+1 − 1)
b∗h (c∗h+1 − 1)
180
SUZUKI
Thus c∗h+1 − 1 ≤ 0. Since α is odd, c∗α−2 ≤ 1. Now by Proposition 1, b∗2 = c∗α−2 ≤ 1.
2
= 0, we have
Note that this holds for α = 5 as well because a∗2 = 0 in our case. Since q2,2
0 = c∗2 b∗1 + b∗2 c∗3 − k1∗ > c∗2 b∗1 − k1∗ ≥ (k1∗ − 1)2 − k1∗ .
This is impossible. Hence we have the assertion when α is odd.
Suppose α = 2β. Then the argument above shows that c∗h ≤ 1 when h is odd and
β
= 0. Suppose h is odd and h ≤ β. Then
h ≤ β + 1. Note that q2,β
0 = c∗h b∗h−1 + b∗h c∗h+1 − k1∗ > (k1∗ − 1)c∗h+1 − k1∗ .
Since k1∗ ≥ 3 as k1∗ is an integer, c∗h+1 < 3/2. Therefore, we have the following.
c∗h ≤ 1 if h is odd and h ≤ β + 1.
c∗h < 3/2 if h is even and h ≤ β + 1.
Suppose β is odd. Then b∗β−1 = c∗β+1 < 3/2 by Proposition 1. Since c∗β−1 < 3/2,
3/2 > b∗β−1 = k1∗ − c∗β−1 > k1∗ − 3/2. Thus k1∗ < 3. This contradicts our assumption.
Suppose β is even. Then c∗β−1 ≤ 1 and b∗β−1 = c∗β+1 ≤ 1. Thus we obtain that k1∗ ≤ 2.
This proves the assertion.
Proof of Theorem 3: Suppose α = 2. Then by Lemma 9, we have (i). Suppose α = d.
Then by Proposition 1, we have (ii). Suppose 2 < α < d. If α is even, then a∗i = 0 for
every i by Lemma 9. If α is odd, then by Lemma 11, Lemma 12 and Proposition 1, we have
a∗i = 0 for all i = 1, 2, . . . , d − 1 unless α = 3 and (iii) or (iv) holds. Now by Lemma
13, we cannot have other cases.
This completes the proof of Theorem 3.
Acknowledgments
The author is very grateful to Dr. Garth A. Dickie for sending him preprints. The author
learned a lot from these papers. He is also grateful to Dr. Masato Tomiyama for his careful
reading of the manuscript. Dr. Tomiyama pointed out a serious gap in the first manuscript.
The author also would like to thank the anonymous referee who greatly improved the clarity
of this paper by providing many pages of helpful comments.
References
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3. P. J. Cameron, J. M. Goethals and J. J. Seidel, “The Krein condition, spherical designs, Norton algebras and
permutation groups,” Indag. Math. 40 (1978), 196–206.
4. G. A. Dickie, “Q-polynomial structures for association schemes and distance-regular graphs,” Ph.D. Thesis,
University of Wisconsin, 1995.
5. G. A. Dickie, “A note on Q-polynomial association schemes,” preprint.