A characterization of Q-polynomial distance-regular graphs

Discrete Mathematics 308 (2008) 3090 – 3096 www.elsevier.com/locate/disc

A characterization of Q-polynomial distance-regular graphs Arlene A. Pascasio Department of Mathematics, De La Salle University, Manila, Philippines Received 30 August 2004; accepted 9 August 2007 Available online 24 September 2007

Abstract Let  denote a distance-regular graph with diameter D  3. Let  denote a nontrivial eigenvalue of  and let ∗0 , ∗1 , . . . , ∗D denote the corresponding dual eigenvalue sequence. In this paper we prove that  is Q-polynomial with respect to  if and only if the following (i)–(iii) hold: (i) There exist , ∗ ∈ C such that

∗ = ∗i−1 − ∗i + ∗i+1

(1  i  D − 1).

(1)

(ii) There exist , , ∗ ∈ C such that the intersection numbers ai satisfy ai (∗i − ∗i−1 )(∗i − ∗i+1 ) = ∗i 2 + ∗i + ∗ for 0  i  D, where ∗−1 and ∗D+1 are the scalars which satisfy Eq. (1) for i = 0, i = D, respectively. (iii) ∗i = ∗0 for 1  i  D. © 2007 Elsevier B.V. All rights reserved. Keywords: Distance-regular graph; Q-polynomial; Association scheme

1. Introduction Let  denote a distance-regular graph with diameter D 3 (see Section 2 for formal definitions). Let  denote a nontrivial eigenvalue of , and let ∗0 , ∗1 , . . . , ∗D denote the corresponding dual eigenvalue sequence. In this paper we give a simple condition involving ∗0 , ∗1 , . . . , ∗D and the intersection numbers of  that is satisfied if and only if  is Q-polynomial with respect to . In order to motivate our result we cite the following theorem of Brouwer, Cohen and Neumaier. Theorem 1.1 (Brouwer et al. [2, Theorem 8.2.1]). Let  denote a bipartite distance-regular graph with diameter D 3. Let  denote a nontrivial eigenvalue of  and let ∗0 , ∗1 , . . . , ∗D denote the corresponding dual eigenvalue sequence. Then  is Q-polynomial with respect to  if and only if the following (i), (ii) hold:

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Arlene A. Pascasio / Discrete Mathematics 308 (2008) 3090 – 3096

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(i) There exist , ∗ ∈ C such that ∗ = ∗i−1 − ∗i + ∗i+1

(1i D − 1).

(ii) ∗0 , ∗1 , . . . , ∗D are mutually distinct. In this paper we obtain an extension of the above theorem to arbitrary distance-regular graphs with diameter D 3. Our result is as follows. Theorem 1.2. Let  denote a distance-regular graph with diameter D 3. Let  denote a nontrivial eigenvalue of  and let ∗0 , ∗1 , . . . , ∗D denote the corresponding dual eigenvalue sequence. Then  is Q-polynomial with respect to  if and only if the following (i)–(iii) hold: (i) There exist , ∗ ∈ C such that ∗ = ∗i−1 − ∗i + ∗i+1

(1i D − 1).

(2)

(ii) There exist , , ∗ ∈ C such that the intersection numbers ai satisfy ai (∗i − ∗i−1 )(∗i − ∗i+1 ) = ∗i 2 + ∗i + ∗ for 0 i D, where ∗−1 and ∗D+1 are the scalars which satisfy Eq. (2) for i = 0, i = D, respectively. (iii) ∗i  = ∗0 for 1 i D. 2. Preliminaries In this section, we review some definitions and basic concepts. For more background information, the reader may refer to the books of Bannai and Ito [1], Brouwer et al. [2], or Godsil [3]. Let C denote the field of complex numbers. Let X denote a nonempty finite set, and let Mat X (C) denote the C-algebra of matrices with rows and columns indexed by X and whose entries are in C. Let  = (X, R) denote a finite, undirected, connected graph without loops or multiple edges, with vertex set X, edge set R, path-length distance function j and diameter D := max{j(x, y)|x, y ∈ X}. We say  is distance-regular whenever for all integers h, i, j (0 h, i, j D) and for all x, y ∈ X with j(x, y) = h, the number h pij := |{z ∈ X | j(x, z) = i, j(y, z) = j }| h are called the intersection numbers for . We abbreviate a := p i (0 i D), is independent of x and y. The integers pij i 1i i i bi := p1i+1 (0 i D − 1) and ci := p1i−1 (1i D). Set c0 := 0, bD := 0 and k := b0 . Observe that

ci + ai + bi = k

(0 i D).

(3)

We say  is bipartite whenever each of a0 , a1 , . . . , aD is zero. From now on we assume  = (X, R) is a distance-regular graph with diameter D 3. We recall the Bose–Mesner algebra of . For each integer i (0 i D), let Ai denote the matrix in Mat X (C) with x, y entry  1 if j(x, y) = i (Ai )xy = (x, y ∈ X). 0 if j(x, y)  = i The matrices A0 , A1 , . . . , AD are called the distance matrices of . We set A := A1 and call this the adjacency matrix of t = A (0 i D), A A = D p h A (0 i, j D), where A = J, A = A (0 i D), A . Observe A0 = I, D i i i i i j i=0 h=0 ij h i I denotes the identity matrix and J denotes the all ones matrix. It follows from these that the matrices A0 , A1 , . . . , AD form a basis for a commutative semi-simple C-algebra M. Moreover A generates M. We call M the Bose–Mesner algebra of .  t By [1, pp. 59, 64] M has a second basis E0 , E1 , . . . , ED such that E0 =|X|−1 J, D i=0 Ei =I, Ei =Ei (0 i D), Ei = Ei (0 i D), Ei Ej = ij Ei (0 i, j D). The E0 , E1 , . . . , ED are called the primitive idempotents of , and E0 is called the trivial idempotent.

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We recall the eigenvalues and  dual eigenvalues of . Since E0 , E1 , . . . , ED form a basis for M, there exist 0 , 1 , . . . , D ∈ C such that A = D i=0 i Ei . By [1, p. 197] the scalars 0 , 1 , . . . , D are in R. Observe the scalars 0 , 1 , . . . , D are mutually distinct since A generates M. Moreover AE i = Ei A = i Ei (0 i D). We refer to i as the eigenvalue of  associated with Ei . We call 0 the trivial eigenvalue of . Let  denote an eigenvalue of  and let E denote the associated primitive idempotent. Since A0 , A1 , . . . , AD form a basis for M there exist ∗0 , ∗1 , . . . , ∗D ∈ C  ∗ ∗ ∗ ∗ such that E = |X|−1 D i=0 i Ai . We call 0 , 1 , . . . , D the dual eigenvalue sequence associated with E (or ). It is ∗ known 0  = 0 [1, p. 94]. We recall the Q-polynomial property. Let ◦ denote entrywise multiplication in Mat X (C). Observe Ai ◦ Aj = ij Ai (0 i, j D) so M is closed under ◦. Thus there exist scalars qijh ∈ C (0 h, i, j D) such that Ei ◦ Ej =  h h |X|−1 D h=0 qij Eh (0 i, j D). We call the qij the Krein parameters of . We say  is Q-polynomial (with respect to the given ordering E0 , E1 , . . . , ED of the primitive idempotents) whenever for all integers h, i, j (0 h, i, j D) qijh = 0

if one of h, i, j is greater than the sum of the other two, and

qijh  = 0

if one of h, i, j is equal to the sum of the other two.

Let  denote a nontrivial eigenvalue of  and let E denote the corresponding primitive idempotent. We say  is Q-polynomial with respect to E whenever there exists a Q-polynomial ordering E0 , E1 , . . . , ED of the primitive idempotents such that E = E1 . We end this section with some results concerning dual eigenvalue sequences which will be useful to prove our main theorem. Lemma 2.1 (Brouwer et al. [2, Theorem 8.1.1(ii)]). Let  denote a distance-regular graph with diameter D 3. Let  denote a nontrivial eigenvalue of  and let ∗0 , ∗1 , . . . , ∗D denote the corresponding dual eigenvalue sequence. Assume  is Q-polynomial with respect to . Then ∗0 , ∗1 , . . . , ∗D are mutually distinct. Lemma 2.2 (Brouwer et al. [2, p. 131]). Let  denote a distance-regular graph with diameter D 3. Let  denote an eigenvalue of  and let ∗0 , ∗1 , . . . , ∗D denote the corresponding dual eigenvalue sequence. Then the intersection numbers of  satisfy ci ∗i−1 + ai ∗i + bi ∗i+1 = ∗i

(0 i D).

(4)

3. Proof of main theorem This section contains the proof of Theorem 1.2. We begin by proving the necessity of the conditions (i)–(iii). With the assumptions of Theorem 1.2, observe that condition (i) follows by Leonard’s Theorem [2, Theorem 8.1.2]. Condition (ii) follows from [7, Theorem 5.3(iv)], and condition (iii) follows from Lemma 2.1. The remainder of this paper establishes the sufficiency of conditions (i)–(iii) in Theorem 1.2. We adopt the following notation and assumptions. Assumption 3.1. Let  denote a distance-regular graph with diameter D 3. Let  denote a nontrivial eigenvalue of  and let ∗0 , ∗1 , . . . , ∗D denote the corresponding dual eigenvalue sequence. Assume that the following (i)–(iii) hold. (i) There exist , ∗ ∈ C such that ∗ = ∗i−1 − ∗i + ∗i+1

(1i D − 1).

(5)

(ii) There exist , , ∗ ∈ C such that ai (∗i − ∗i−1 )(∗i − ∗i+1 ) = ∗i 2 + ∗i + ∗

(0 i D),

where ∗−1 and ∗D+1 are the scalars which satisfy Eq. (5) for i = 0, i = D, respectively. (iii) ∗0  = ∗i for 1i D.

(6)

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Lemma 3.2. With reference to Assumption 3.1, for 1 i D − 1 we have (∗0 − ∗i )(∗2 − ∗i ) = (∗1 − ∗i−1 )(∗1 − ∗i+1 ).

(7)

Proof. Observe there exists 0  = q ∈ C such that q + q −1 = . First assume q  = 1, q  = −1. Then by (5) there exist r, s ∗ , t ∗ ∈ C such that ∗i = r + s ∗ q i + t ∗ q −i

(0 i D).

(8)

Evaluating each side of (7) using (8) we find they agree. Next assume q = 1. Then by (5) there exist r, s ∗ , t ∗ ∈ C such that ∗i = r + s ∗ i + t ∗ i 2

(0 i D).

(9)

Evaluating each side of (7) using (9) we find they agree. Next assume q = −1. Then by (5) there exist r, s ∗ , t ∗ ∈ C such that ∗i = r + s ∗ (−1)i + it ∗ (−1)i

(0 i D).

(10)

Evaluating each side of (7) using (10) we find they agree. Lemma 3.3. With reference to Assumption 3.1, the following hold. (i) ∗1  = ∗2 . (ii) ∗1  = ∗D . (iii) ∗2  = ∗3 . Proof. (i) Suppose ∗1 = ∗2 . Evaluating (5) at i = 1 and at i = 2 gives ∗ = ∗0 + ∗1 (1 − ) and ∗ = ∗3 + ∗1 (1 − ). It follows ∗0 = ∗3 , a contradiction to Assumption 3.1(iii). Thus ∗1  = ∗2 as desired. (ii) Suppose ∗1 = ∗D .

(11)

Setting i = D − 1 in (7) we get (∗0 − ∗D−1 )(∗2 − ∗D−1 ) = 0. But ∗0  = ∗D−1 by Assumption 3.1(iii) so ∗2 = ∗D−1 .

(12)

Setting i = 1 in (5) we get ∗ = ∗0 − ∗1 + ∗2 .

(13)

By Assumption 3.1(ii) ∗D+1 satisfies ∗ = ∗D−1 − ∗D + ∗D+1 .

(14)

Eliminating ∗D−1 , ∗D in (14) using (11), (12) and subtracting the resulting equation from (13) we get ∗0 = ∗D+1 .

(15)

Now evaluating (6) at i = 1 and i = D we get a1 (∗1 − ∗0 )(∗1 − ∗2 ) = ∗1 2 + ∗1 + ∗ ,

(16)

aD (∗D − ∗D−1 )(∗D − ∗D+1 ) = ∗D 2 + ∗D + ∗ .

(17)

Simplifying (17) using (11), (12), (15) and subtracting the resulting equation from (16) we get a1 = aD ,

(18)

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Arlene A. Pascasio / Discrete Mathematics 308 (2008) 3090 – 3096

in view of (i) and Assumption 3.1(iii). Evaluating (4) at i = 1 and simplifying the result using c1 = 1 and b1 = k − a1 − 1 we get a1 (∗1 − ∗2 ) + k∗2 − ∗1 = ∗2 − ∗0 .

(19)

Evaluating (4) at i = D and simplifying the result using bD = 0 and cD = k − aD we get aD (∗D − ∗D−1 ) + k∗D−1 − ∗D = 0.

(20)

Simplifying (20) using (11), (12), (18) and subtracting the resulting equation from (19) we get ∗0 = ∗2 , a contradiction. Thus ∗1  = ∗D , as desired. (iii) We assume ∗2 = ∗3 and get a contradiction. First assume D = 3. Evaluating (4) at i = 3 and simplifying the result using c3 = k − a3 we get (k − )∗2 = 0.

(21)

Evaluating (4) at i = 2 and simplifying the result using b2 = k − a2 − c2 we get c2 ∗1 + (k − c2 − )∗2 = 0.

(22)

Combining (21), (22) we find c2 (∗1 − ∗2 ) = 0. But c2  = 0 by construction and ∗1 − ∗2  = 0 by (i) above, for a contradiction. Next assume D 4. Evaluating (7) at i = 3 we get (∗1 − ∗2 )(∗1 − ∗4 ) = 0. It follows from (i) that ∗1 = ∗4 . This gives a contradiction in case D = 4, in view of (ii). In case D 5, we proceed by setting i = 4 in (7). Simplifying this equation using ∗1 = ∗4 , ∗2 = ∗3 and (i) we get ∗0 = ∗5 , a contradiction to Assumption 3.1(iii).  Lemma 3.4. With reference to Assumption 3.1, the following hold. ∗ = =

∗1 (∗1 + ∗3 ) − ∗2 (∗0 + ∗2 ) . ∗1 − ∗2

(23)

∗0 − ∗1 + ∗2 − ∗3 . ∗1 − ∗2

(24)

Proof. Setting i = 1 and i = 2 in (5) gives a system of linear equations in ∗ and . Solving this system for ∗ ,  gives the desired result.  Corollary 3.5. With reference to Assumption 3.1, ∗D+1 is given by ∗D+1 =

(∗1 − ∗2 )(∗1 + ∗2 − ∗D−1 − ∗D ) − ∗0 (∗2 − ∗D ) + ∗3 (∗1 − ∗D ) . ∗1 − ∗2

Proof. Set i = D in (5). In the resulting equation eliminate ∗ and  using (23) and (24).

(25) 

Lemma 3.6. With reference to Assumption 3.1, the following equations hold for 0 i D.   ∗0 (∗i 2 + ∗i + ∗ ) = (∗i − ∗i+1 ) k(∗1 ∗i − ∗0 ∗i−1 ) + bi ∗0 (∗i−1 − ∗i+1 ) ,   ∗0 (∗i 2 + ∗i + ∗ ) = (∗i − ∗i−1 ) k(∗1 ∗i − ∗0 ∗i+1 ) − ci ∗0 (∗i−1 − ∗i+1 ) .

(26) (27)

Proof. We mentioned in the introduction that ∗0  = 0. Setting i = 0 in (4) we find =

k∗1 . ∗0

(28)

In (4), we eliminate  using (28) and we eliminate ci using (3) to obtain ai ∗0 (∗i − ∗i−1 ) = k(∗1 ∗i − ∗0 ∗i−1 ) + bi ∗0 (∗i−1 − ∗i+1 ). Eliminating ai (∗i − ∗i−1 ) in (6) using (29) we get line (26). Line (27) is obtained similarly.

(29) 

Arlene A. Pascasio / Discrete Mathematics 308 (2008) 3090 – 3096

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Lemma 3.7. With reference to Assumption 3.1, ∗ ,  and  satisfy the following. ∗ (∗0 − ∗D )(∗1 − ∗D ) = k∗1 ∗D+1 (∗0 ∗D−1 − ∗1 ∗D ) + k∗0 ∗1 ∗D (∗1 − ∗D−1 ) − ∗0 ∗D (∗0 − ∗D )(∗0 + (k − 1)∗2 ),

(30)

∗0 (∗0 − ∗D )(∗1 − ∗D ) = k(∗0 ∗D−1 − ∗1 ∗D )(∗0 (∗D − ∗D+1 ) − ∗1 ∗D+1 )

+ k∗0 ∗1 (∗D−1 ∗D − ∗0 ∗1 ) + ∗0 (∗0 2 − ∗D 2 )(∗0 + (k − 1)∗2 ),

∗0 (∗0 − ∗D )(∗1 − ∗D ) = (∗0 − ∗D )(k(∗1 2 − ∗0 ∗2 ) − ∗0 (∗0 − ∗2 )) − k(∗D − ∗D+1 )(∗0 ∗D−1 − ∗1 ∗D ).

(31)

(32)

Proof. Setting i = 0 in (6) and using the fact that a0 = 0 we get ∗0 2 + ∗0 + ∗ = 0.

(33)

Setting i = 1 in (27) and using the fact that c1 = 1 we get ∗0 (∗1 2 + ∗1 + ∗ ) = (∗1 − ∗0 )(k(∗1 2 − ∗0 ∗2 ) − ∗0 (∗0 − ∗2 )).

(34)

Setting i = D in (26) and using the fact that bD = 0 we get ∗0 (∗D 2 + ∗D + ∗ ) = k(∗D − ∗D+1 )(∗1 ∗D − ∗0 ∗D−1 ).

(35)

Solving for ,  and ∗ using Eqs. (33)–(35) we get lines (30)–(32) as desired.



Theorem 3.8. With reference to Assumption 3.1,  is Q-polynomial with respect to . Proof. We apply [6, Theorem 3.3]. We show that   (∗1 − ∗2 )2 (∗1 − ∗3 )2 ∗ ∗ c3 + b −  +  = ( − k)(∗1 − ∗3 ) + ( + 1)(∗0 − ∗2 ). 2 2 3 ∗0 − ∗3 ∗0 − ∗2

(36)

To show (36) we first show b2

(∗1 − ∗3 )2 (∗1 − ∗3 )(∗2 2 + ∗2 + ∗ ) k∗1 ∗ = + ∗ (1 − ∗3 ). ∗0 − ∗2 (∗0 − ∗2 )(∗2 − ∗3 ) 0

(37) ∗ −∗

To get (37), set i = 2 in (26) and multiply the resulting equation by ∗ (∗ −1 ∗ )(3 ∗ −∗ ) . Next we show 0 0 2 2 3   (∗1 − ∗2 )2 −(∗1 − ∗2 )(∗3 2 + ∗3 + ∗ ) ∗ ∗ c3 ∗ ∗ −  2 + 3 = 0 − 3 (∗0 − ∗3 )(∗2 − ∗3 ) +

k ∗ ∗ ( ( − ∗2 ) − ∗0 (∗2 − ∗3 )). ∗0 1 1

(38)

To get (38), we first set i = 3 in (5) and eliminate ∗ and  in the resulting equation using (23), (24). This gives ∗4 = ∗1 −

(∗0 − ∗3 )(∗2 − ∗3 ) . (∗1 − ∗2 )

(39)

Now set i = 3 in (27), eliminate ∗4 using (39), and multiply the resulting equation by (∗1 − ∗2 )/∗0 (∗0 − ∗3 )(∗2 − ∗3 ); the result is line (38) as desired. We can now easily verify (36). Evaluating the left-hand side of (36) using (37), (38) and simplifying the result using (25), (30)–(32) and the fact that  = k∗1 (∗0 )−1 we get the right-hand side of (36) as desired. By [6, Theorem 3.3], we conclude that  is Q-polynomial with respect to . 

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Acknowledgments This work was supported by the De La Salle University Research Faculty grant. The author would like to thank Professor Paul Terwilliger for proposing the problem and for his many helpful suggestions. References [1] [2] [3] [6] [7]

E. Bannai, T. Ito, Algebraic Combinatorics I: Association Schemes, Benjamin/Cummings, London, 1984. A.E. Brouwer, A.M. Cohen, A. Neumaier, Distance-Regular Graphs, Springer, Berlin, 1989. C.D. Godsil, Algebraic Combinatorics, Chapman & Hall, New York, 1993. P. Terwilliger, A new inequality for distance-regular graphs, Discrete Math. 137 (1995) 319–332. P. Terwilliger, R. Vidunas, Leonard pairs and the Askey–Wilson relations, J. Algebra Appl. 3 (4) (2004) 411–426.