A characterization of $Q$-polynomial distance-regular graphs using the intersection numbers

TL;DR

This paper characterizes $Q$-polynomial distance-regular graphs with diameter at least 3 by using intersection numbers to identify a positive semidefinite matrix with integer entries, where the matrix's determinant zero condition characterizes the $Q$-polynomial property.

Contribution

It introduces a new characterization of $Q$-polynomial distance-regular graphs via a specific positive semidefinite matrix derived from intersection numbers.

Findings

Matrix $G$ is positive semidefinite with integer entries.

Determinant of $G$ is zero if and only if the graph is $Q$-polynomial.

Provides a new algebraic criterion for $Q$-polynomial property.

Abstract

We consider a primitive distance-regular graph $\Gamma$ with diameter at least $3$. We use the intersection numbers of $\Gamma$ to find a positive semidefinite matrix $G$ with integer entries. We show that $G$ has determinant zero if and only if $\Gamma$ is $Q$-polynomial.

Tables

term	coefficient
$p_{ij}^0$	$k{k}_2$
$p_{ij}^1$	$k_2{a}_1{a}_2-k{b}_1^2$
$p_{ij}^2$	$k_2\left(c_2\left(b_1-1\right)-a_2\left(a_1+1\right)+b_2\left(c_3-1\right)\right)$
$p_{ij}^3$	$-k_3{c}_3^2$

$p_{1,h-1}^h=c_h$,		$p_{1h}^h=a_h$,		$p_{1,h+1}^h=b_h$,
$p_{h,h-1}^1=\frac{k_h{c}_h}{k}$,		$p_{hh}^1=\frac{k_h{a}_h}{k}$,		$p_{h,h+1}^1=\frac{k_h{b}_h}{k}$.

Equations

$$p_{ij}^h:=|{\Gamma }_i(x)\cap {\Gamma }_j(y)|$$

$$(A_i{)}_{xy}=\{\begin{array}{ll}1& \text{if }\partial (x,y)=i,\\ 0& \text{if }\partial (x,y)\ne i,\end{array}x,y\in X.$$

$$p_{0j}^h={\delta }_{hj}$$

$$p_{ij}^0={\delta }_{ij}{k}_i$$

$$p_{ij}^h=p_{ji}^h$$

$$k_h{p}_{ij}^h=k_i{p}_{hj}^i=k_j{p}_{ih}^j.$$

$$c_{i+1}{p}_{i+1,j}^h+a_i{p}_{ij}^h+b_{i-1}{p}_{i-1,j}^h=c_h{p}_{ij}^{h-1}+a_h{p}_{ij}^h+b_h{p}_{ij}^{h+1}$$

$$\lambda v_i=c_{i+1}{v}_{i+1}+a_i{v}_i+b_{i-1}{v}_{i-1}(1\le i\le D),$$

$$B=\left[\begin{array}{ccccc}{a}_0& b_0& & & 0\\ c_1& a_1& b_1& & \\ & c_2& a_2& \ddots & \\ & & \ddots & \ddots & b_{D-1}\\ 0& & & c_D& a_D\end{array}\right].$$

$$\lambda u_i=c_i{u}_{i-1}+a_i{u}_i+b_i{u}_{i+1}(1\le i\le D-1).$$

$$A_j=\sum _{i=0}^D{v}_j({\theta }_i)E_i$$

$$E_j=|X{|}^{-1}{m}_j\sum _{i=0}^D{u}_i({\theta }_j)A_i$$

$$E_j=\sum _{i=0}^D{S}_{ij}{A}_i$$

$$A_j=\sum _{i=0}^D(S^{-1}{)}_{ij}{E}_i$$

$$S_{ij}=|X{|}^{-1}{m}_j{u}_i({\theta }_j),(S^{-1}{)}_{ij}=v_j({\theta }_i).$$

$$\begin{array}{r}{E}_i\circ E_j=|X{|}^{-1}\sum _{h=0}^D{q}_{ij}^h{E}_h(0\le i,j\le D).\end{array}$$

$$(E_i^{\ast }{)}_{yy}=\{\begin{array}{ll}1& \text{if }\partial (x,y)=i,\\ 0& \text{if }\partial (x,y)\ne i,\end{array}y\in X.$$

$$\begin{array}{r}(A_i^{\ast }{)}_{yy}=|X|(E_i{)}_{xy}y\in X.\end{array}$$

$$A_j^{\ast }=m_j\sum _{i=0}^D{u}_i({\theta }_j)E_i^{\ast }$$

$$E_j^{\ast }=|X{|}^{-1}\sum _{i=0}^D{v}_j({\theta }_i)A_i^{\ast }$$

$$A_j^{\ast }=|X|\sum _{i=0}^D{S}_{ij}{E}_i^{\ast }$$

$$E_j^{\ast }=|X{|}^{-1}\sum _{i=0}^D(S^{-1}{)}_{ij}{A}_i^{\ast }$$

$$(S^{alt}{)}_{ij}=S_{ij}-S_{0j}$$

$$(S^{-1}{)}_{ij}^{alt}=(S^{-1}{)}_{ij}$$

$$A_2{A}_j^{\ast }A-A{A}_j^{\ast }{A}_2$$

$$=|X|(A_2{E}_0^{\ast }A-A{E}_0^{\ast }{A}_2)S_{0j}+|X|\sum _{i=1}^D(A_2{E}_i^{\ast }A-A{E}_i^{\ast }{A}_2)S_{ij}$$

$$=|X|(A_2(I-(E_1^{\ast }+\cdots +E_D^{\ast }))A-A(I-(E_1^{\ast }+\cdots +E_D^{\ast }))A_2)S_{0j}$$

$$+|X|\sum _{i=1}^D(A_2{E}_i^{\ast }A-A{E}_i^{\ast }{A}_2)S_{ij}$$

$$=|X|\sum _{i=1}^D(A_2{E}_i^{\ast }A-A{E}_i^{\ast }{A}_2)(S_{ij}-S_{0j})$$

$$=|X|\sum _{i=1}^D(A_2{E}_i^{\ast }A-A{E}_i^{\ast }{A}_2)(S^{alt}{)}_{ij}.$$

$$A_2{E}_j^{\ast }A-A{E}_j^{\ast }{A}_2$$

$$=|X{|}^{-1}(A_2{A}_0^{\ast }A-A{A}_0^{\ast }{A}_2)(S^{-1}{)}_{0j}$$

$$+|X{|}^{-1}\sum _{i=1}^D(A_2{A}_i^{\ast }A-A{A}_i^{\ast }{A}_2)(S^{-1}{)}_{ij}$$

$$=|X{|}^{-1}\sum _{i=1}^D(A_2{A}_i^{\ast }A-A{A}_i^{\ast }{A}_2)(S^{-1}{)}_{ij}$$

$$=|X{|}^{-1}\sum _{i=1}^D(A_2{A}_i^{\ast }A-A{A}_i^{\ast }{A}_2)(S^{-1}{)}_{ij}^{alt}.$$

$$S^{\prime }=\left[\begin{array}{cccc}{S}^{alt}& & & 0\\ & S^{alt}& & \\ & & S^{alt}& \\ 0& & & S^{alt}\end{array}\right],$$

$$\{A_2{E}_i^{\ast }A-A{E}_i^{\ast }{A}_2{\}}_{i=1}^D,\{A_3{E}_i^{\ast }-E_i^{\ast }{A}_3{\}}_{i=1}^D,\{A_2{E}_i^{\ast }-E_i^{\ast }{A}_2{\}}_{i=1}^D,\{A{E}_i^{\ast }-E_i^{\ast }A{\}}_{i=1}^D$$

$$\{A_2{A}_i^{\ast }A-A{A}_i^{\ast }{A}_2{\}}_{i=1}^D,\{A_3{A}_i^{\ast }-A_i^{\ast }{A}_3{\}}_{i=1}^D,\{A_2{A}_i^{\ast }-A_i^{\ast }{A}_2{\}}_{i=1}^D,\{A{A}_i^{\ast }-A_i^{\ast }A{\}}_{i=1}^D.$$

$$⟨A_i{E}_j^{\ast }{A}_h,A_r{E}_s^{\ast }{A}_t⟩=\sum _{\ell =0}^D{k}_{\ell }{p}_{ir}^{\ell }{p}_{js}^{\ell }{p}_{ht}^{\ell }.$$

Full text

1 Introduction

Let $\Gamma$ denote a distance-regular graph with diameter $D\geq 3$. In the literature there are a number of characterizations for the $Q$-polynomial condition on $\Gamma$. There is the balanced set characterization [9, Theorem 1.1], [10, Theorem 3.3]. There is a characterization involving the dual distance matrices [10, Theorem 3.3]. There is a characterization involving the intersection numbers $a_{i}$ [8, Theorem 3.8]; cf. [3, Theorem 5.1]. There is a characterization involving a tail in a representation diagram [5, Theorem 1.1]. There is a characterization involving a pair of primitive idempotents [6, Theorem 1.1]; cf. [7, Theorem 1.1].

In this paper we obtain the following characterization of the $Q$-polynomial property. Assume $\Gamma$ is primitive. We use the intersection numbers of $\Gamma$ to find a positive semidefinite matrix $G$ with integer entries. We show that $G$ has determinant zero if and only if $\Gamma$ is $Q$-polynomial. Our main result is Theorem 18.

2 Preliminaries

Let $X$ denote a nonempty finite set. Let ${{\rm{Mat}}}_{X}(\mathbb{C})$ denote the $\mathbb{C}$-algebra consisting of the matrices whose rows and columns are indexed by $X$ and whose entries are in $\mathbb{C}$. For $B\in{{\rm{Mat}}_{X}(\mathbb{C})}$ let $\overline{B}$ and $B^{t}$ denote the complex conjugate and the transpose of $B$, respectively. Let $V=\mathbb{C}^{X}$ denote the vector space over $\mathbb{C}$ consisting of column vectors with coordinates indexed by $X$ and entries in $\mathbb{C}$. Observe that ${{\rm{Mat}}}_{X}(\mathbb{C})$ acts on $V$ by left multiplication. We endow $V$ with the Hermitean inner product $(\;,\;)$ such that $(u,v)=u^{t}\overline{v}$ for all $u,v\in{V}$. The inner product $(\;,\;)$ is positive definite. For $B\in{{\rm{Mat}}_{X}(\mathbb{C})}$ we obtain $(u,Bv)=(\overline{B}^{t}u,v)$ for all $u,v\in{V}$. We endow ${\rm{Mat}}_{X}(\mathbb{C})$ with the Hermitean inner product $\langle\;,\;\rangle$ such that $\langle R,S\rangle=tr(R^{t}\overline{S})$ for all $R,S\in{{\rm{Mat}}_{X}(\mathbb{C})}$. The inner product $\langle\;,\;\rangle$ is positive definite.

Let $\Gamma=(X,R)$ denote a finite, undirected, connected graph, without loops or multiple edges, with vertex set $X$ and edge set $R$. Let $\partial$ denote the shortest path-length distance function for $\Gamma$. Define the diameter $D:=\text{max}\{\partial(x,y)|x,y\in{X}\}$. For a vertex $x\in{X}$ and an integer $i\geq 0$ define $\Gamma_{i}(x)=\{y\in{X}|\partial(x,y)=i\}$. For notational convenience abbreviate $\Gamma(x)=\Gamma_{1}(x)$. For an integer $k\geq 0$, we call the graph $\Gamma$ regular with valency $k$ whenever $|\Gamma(x)|=k$ for all $x\in{X}$. We say that $\Gamma$ is distance-regular whenever for all integers $h,i,j$ $(0\leq h,i,j\leq D)$ and for all $x,y\in{X}$ with $\partial(x,y)=h$, the number

[TABLE]

is independent of $x$ and $y$. The integers $p^{h}_{ij}$ are called the intersection numbers of $\Gamma$. From now on we assume $\Gamma$ is distance-regular with diameter $D\geq 3$. We abbreviate $c_{i}:=p^{i}_{1,i-1}$ $(1\leq i\leq D)$, $a_{i}:=p^{i}_{1i}$ $(0\leq i\leq D)$, $b_{i}:=p^{i}_{1,i+1}$ $(0\leq i\leq D-1)$, $k_{i}:=p^{0}_{ii}$ $(0\leq i\leq D)$, and $c_{0}=0$, $b_{D}=0$. Observe that $\Gamma$ is regular with valency $k=b_{0}$ and $c_{i}+a_{i}+b_{i}=k$ $(0\leq i\leq D)$. By [2, p. 127] the following holds for $0\leq h,i,j\leq D$: (i) $p^{h}_{ij}=0$ if one of $h,i,j$ is greater than the sum of the other two; and (ii) $p^{h}_{ij}\neq 0$ if one of $h,i,j$ equals the sum of the other two. For $0\leq i\leq D$, let $A_{i}$ denote the matrix in ${\rm{Mat}}_{X}(\mathbb{C})$ with $(x,y)$-entry

[TABLE]

We call $A_{i}$ the $i$-th distance matrix of $\Gamma$. We call $A=A_{1}$ the adjacency matrix of $\Gamma$. Observe that $A_{i}$ is real and symmetric for $0\leq i\leq D$. Note that $A_{0}=I$, where $I$ is the identity matrix. Observe that $\sum_{i=0}^{D}A_{i}=J$, where $J$ is the all-ones matrix in ${\rm{Mat}}_{X}(\mathbb{C})$. Observe that for $0\leq i,j\leq D$, $A_{i}A_{j}=\sum_{h=0}^{D}p^{h}_{ij}A_{h}$. For integers $h,i,j$ $(0\leq h,i,j\leq D)$ we have

[TABLE]

For $0\leq i,j\leq D$ we have $A(A_{i}A_{j})=(AA_{i})A_{j}$. This gives a recursion

[TABLE]

that can be used to compute the intersection numbers.

Let $M$ denote the subalgebra of ${\rm{Mat}}_{X}(\mathbb{C})$ generated by $A$. By [2, p. 127] the matrices $A_{0},A_{1},...,A_{D}$ form a basis for $M$. We call $M$ the Bose-Mesner algebra of $\Gamma$. By [2, p. 45], $M$ has a basis $E_{0},E_{1},...,E_{D}$ such that (i) $E_{0}=|X|^{-1}J$; (ii) $\sum_{i=0}^{D}E_{i}=I$; (iii) $E_{i}^{t}=E_{i}$ $(0\leq i\leq D)$; (iv) $\overline{E_{i}}=E_{i}$ $(0\leq i\leq D)$; (v) $E_{i}E_{j}=\delta_{ij}E_{i}$ $(0\leq i,j\leq D)$. The matrices $E_{0},E_{1},...,E_{D}$ are called the primitive idempotents of $\Gamma$, and $E_{0}$ is called the trivial idempotent. For $0\leq i\leq D$ let $m_{i}$ denote the rank of $E_{i}$. Let $\lambda$ denote an indeterminate. Define polynomials $\{v_{i}\}_{i=0}^{D+1}$ in $\mathbb{C}[\lambda]$ by $v_{0}=1$, $v_{1}=\lambda$, and

[TABLE]

where $c_{D+1}:=1$. By [2, p. 128] and [11, Lemma 3.8], the following hold: (i) $deg$ $v_{i}=i$ $(0\leq i\leq D+1)$; (ii) the coefficient of $\lambda^{i}$ in $v_{i}$ is $(c_{1}c_{2}\dots c_{i})^{-1}$ $(0\leq i\leq D+1)$; (iii) $v_{i}(A)=A_{i}$ $(0\leq i\leq D)$; (iv) $v_{D+1}(A)=0$; (v) the distinct eigenvalues of $\Gamma$ are precisely the zeros of $v_{D+1}$; call these $\theta_{0},\theta_{1},...,\theta_{D}$. Define a matrix $B\in{{\rm{Mat}}_{D+1}(\mathbb{C})}$ as follows:

[TABLE]

We call $B$ the intersection matrix of $\Gamma$. Note that $A$ has the same minimal polynomial as $B$. Moreover the minimal polynomial of $B$ is the characteristic polynomial of $B$. For an eigenvalue $\theta$ of $B$ we have $vB=\theta v$ where $v$ is a row vector $v=(v_{0}(\theta),v_{1}(\theta),...,v_{D}(\theta))$. Define polynomials $\{u_{i}\}_{i=0}^{D}$ in $\mathbb{C}[\lambda]$ by $u_{0}=1$, $u_{1}=\lambda/k$, and

[TABLE]

Observe that $u_{i}=v_{i}/k_{i}$ $(0\leq i\leq D)$. For an eigenvalue $\theta$ of $B$ we have $Bu=\theta u$ where $u$ is a column vector $u=(u_{0}(\theta),u_{1}(\theta),...,u_{D}(\theta))^{t}$. By [2, p. 131, 132],

[TABLE]

Since $E_{i}E_{j}=\delta_{ij}E_{i}$ and by (6), (7) we have $A_{j}E_{i}=E_{i}A_{j}$ $(0\leq i,j\leq D)$.

For $1\leq i\leq D$ let $\Gamma_{i}$ denote the graph with vertex set $X$ where vertices are adjacent in $\Gamma_{i}$ whenever they are at distance $i$ in $\Gamma$. We observe that $\Gamma_{1}=\Gamma$. The graph $\Gamma$ is said to be primitive whenever $\Gamma_{i}$ is connected for $1\leq i\leq D$.

Lemma 1**.**

$\rm{(See~{}\cite[cite]{[\@@bibref{}{BCN}{}{}, Proposition~{}4.4.7]}.)}$* Assume $\Gamma$ is primitive. Then $u_{i}(\theta_{j})\neq 1$ for $1\leq i,j\leq D$.*

We now define a matrix $S\in{{\rm{Mat}}_{D+1}(\mathbb{C})}$.

Definition 2**.**

Let $S\in{{\rm{Mat}}_{D+1}(\mathbb{C})}$ denote the transition matrix from the basis $\{A_{i}\}_{i=0}^{D}$ of $M$ to the basis $\{E_{i}\}_{i=0}^{D}$ of $M$. Thus

[TABLE]

Lemma 3**.**

The entries of $S$ and $S^{-1}$ are given below. For $0\leq i,j\leq D$,

[TABLE]

Proof.

Immediate from Definition 2 and (6), (7). ∎

We recall the $Q$-polynomial property. Let $\circ$ denote the entry-wise multiplication in ${\rm{Mat}}_{X}(\mathbb{C})$. Note that $A_{i}\circ A_{j}=\delta_{ij}A_{i}$ for $0\leq i,j\leq D$. So $M$ is closed under $\circ$. By [11, p. 377], there exist scalars $q^{h}_{ij}\in{\mathbb{C}}$ such that

[TABLE]

We call the $q^{h}_{ij}$ the Krein parameters of $\Gamma$. By [2, p. 48, 49], these parameters are real and nonnegative for $0\leq h,i,j\leq D$. The graph $\Gamma$ is said to be $Q$-polynomial with respect to the ordering $E_{0},E_{1},...,E_{D}$ whenever the following hold for $0\leq h,i,j\leq D$: (i) $q^{h}_{ij}=0$ if one of $h,i,j$ is greater than the sum of the other two; and (ii) $q^{h}_{ij}\neq 0$ if one of $h,i,j$ equals the sum of the other two. Let $E$ denote a primitive idempotent of $\Gamma$. We say that $\Gamma$ is $Q$-polynomial with respect to $E$ whenever there exists a $Q$-polynomial ordering $E_{0},E_{1},...,E_{D}$ of the primitive idempotents such that $E=E_{1}$.

We recall the dual Bose-Mesner algebra of $\Gamma$. Fix a vertex $x\in{X}$. For $0\leq i\leq D$ let $E_{i}^{*}=E_{i}^{*}(x)$ denote the diagonal matrix in ${\rm{Mat}}_{X}(\mathbb{C})$ with $(y,y)$-entry

[TABLE]

We call $E_{i}^{*}$ the $i$-th dual idempotent of $\Gamma$ with respect to $x$. Observe that (i) $\sum_{i=0}^{D}E_{i}^{*}=I$; (ii) $E_{i}^{*t}=E_{i}^{*}$ $(0\leq i\leq D)$; (iii) $\overline{E_{i}^{*}}=E_{i}^{*}$ $(0\leq i\leq D)$; (iv) $E_{i}^{*}E_{j}^{*}=\delta_{ij}E_{i}^{*}$ $(0\leq i,j\leq D)$. By construction $E_{0}^{*},E_{1}^{*},...,E_{D}^{*}$ are linearly independent. Let $M^{*}=M^{*}(x)$ denote the subalgebra of ${\rm{Mat}}_{X}(\mathbb{C})$ with basis $E_{0}^{*},E_{1}^{*},...,E_{D}^{*}$. We call $M^{*}$ the dual Bose-Mesner algebra of $\Gamma$ with respect to $x$.

We now recall the dual distance matrices of $\Gamma$. For $0\leq i\leq D$ let $A_{i}^{*}=A_{i}^{*}(x)$ denote the diagonal matrix in ${\rm{Mat}}_{X}(\mathbb{C})$ with $(y,y)$-entry

[TABLE]

We call $A_{i}^{*}$ the dual distance matrix of $\Gamma$ with respect to $x$ and $E_{i}$. By [11, p. 379], the matrices $A_{0}^{*},A_{1}^{*},...,A_{D}^{*}$ form a basis for $M^{*}$. Observe that (i) $A_{0}^{*}=I$; (ii) $\sum_{i=0}^{D}A_{i}^{*}=|X|E_{0}^{*}$; (iii) $A_{i}^{*t}=A_{i}^{*}$ $(0\leq i\leq D)$; (iv) $\overline{A_{i}^{*}}=A_{i}^{*}$ $(0\leq i\leq D)$; (v) $A_{i}^{*}A_{j}^{*}=\sum_{h=0}^{D}q^{h}_{ij}A_{h}^{*}$ $(0\leq i,j\leq D)$. From (6), (7) we have

[TABLE]

Lemma 4**.**

The matrix $|X|S$ is the transition matrix from the basis $\{E^{*}_{i}\}_{i=0}^{D}$ of $M^{*}$ to the basis $\{A^{*}_{i}\}_{i=0}^{D}$ of $M^{*}$. Thus

[TABLE]

Proof.

Immediate from Lemma 3 and (10), (11). ∎

3 The matrices $S^{alt},(S^{-1})^{alt},S^{\prime}$

Recall the matrix $S$ from Definition 2. We now modify the matrices $S,S^{-1}$ to obtain $D\times D$ matrices $S^{alt},(S^{-1})^{alt}$ as follows:

[TABLE]

Lemma 5**.**

The following (i)–(iv) hold.

1. (i)

$S^{alt}$* is the transition matrix from $\{A_{2}E_{i}^{*}A-AE_{i}^{*}A_{2}\}_{i=1}^{D}$ to $\{A_{2}A_{i}^{*}A-AA_{i}^{*}A_{2}\}_{i=1}^{D}$.* 2. (ii)

$S^{alt}$* is the transition matrix from $\{A_{3}E_{i}^{*}-E_{i}^{*}A_{3}\}_{i=1}^{D}$ to $\{A_{3}A_{i}^{*}-A_{i}^{*}A_{3}\}_{i=1}^{D}$.* 3. (iii)

$S^{alt}$* is the transition matrix from $\{A_{2}E_{i}^{*}-E_{i}^{*}A_{2}\}_{i=1}^{D}$ to $\{A_{2}A_{i}^{*}-A_{i}^{*}A_{2}\}_{i=1}^{D}$.* 4. (iv)

$S^{alt}$* is the transition matrix from $\{AE_{i}^{*}-E_{i}^{*}A\}_{i=1}^{D}$ to $\{AA_{i}^{*}-A_{i}^{*}A\}_{i=1}^{D}$.* 5. (v)

$(S^{-1})^{alt}$* and $S^{alt}$ are inverses.*

Proof.

$(i),(v)$ For $1\leq j\leq D$ we write $A_{2}A_{j}^{*}A-AA_{j}^{*}A_{2}$ in terms of $\{A_{2}E_{i}^{*}A-AE_{i}^{*}A_{2}\}_{i=1}^{D}$. By Lemma 4 and (12) and since $\sum_{i=0}^{D}E_{i}^{*}=I$, we have

[TABLE]

Next, for $1\leq j\leq D$ we write $A_{2}E_{j}^{*}A-AE_{j}^{*}A_{2}$ in terms of $\{A_{2}A_{i}^{*}A-AA_{i}^{*}A_{2}\}_{i=1}^{D}$. By Lemma 4 and (13) and since $A_{0}^{*}=I$, we find

[TABLE]

The result follows.

$(ii)-(iv)$ Similar to the proof of $(i)$. ∎

Define a matrix

[TABLE]

where $S^{alt}$ is from (12). Observe that $S^{\prime}$ is $4D\times 4D$.

Lemma 6**.**

det(S^{\prime})=\big{(}det(S^{alt})\big{)}^{4}. Moreover $S^{\prime}$ is invertible.

Proof.

Immediate from the construction of $S^{\prime}$. ∎

Corollary 7**.**

The matrix $S^{\prime}$ is the transition matrix from

[TABLE]

to

[TABLE]

Proof.

Immediate from Lemma 5. ∎

4 The bilinear form $\langle\;,\;\rangle$

Recall the positive definite Hermitean bilinear form $\langle\;,\;\rangle$ on ${\rm{Mat}}_{X}(\mathbb{C})$.

Lemma 8**.**

$\rm{(See~{}\cite[cite]{[\@@bibref{}{T_subconstituentI}{}{}, Lemma~{} 3.2]}.)}$* For $0\leq h,i,j,r,s,t\leq D$,*

1. (i)

$\langle E_{i}^{*}A_{j}E_{h}^{*},E_{r}^{*}A_{s}E_{t}^{*}\rangle=\delta_{ir}\delta_{js}\delta_{ht}k_{h}p^{h}_{ij}$, 2. (ii)

$\langle E_{i}A_{j}^{*}E_{h},E_{r}A_{s}^{*}E_{t}\rangle=\delta_{ir}\delta_{js}\delta_{ht}m_{h}q^{h}_{ij}$.

Corollary 9**.**

$\rm{(See~{}\cite[cite]{[\@@bibref{}{T_subconstituentI}{}{}, Lemma~{} 3.2]}.)}$* For $0\leq h,i,j\leq D$,*

1. (i)

$E_{i}^{*}A_{j}E_{h}^{*}=0$* if and only if $p^{h}_{ij}=0$,* 2. (ii)

$E_{i}A_{j}^{*}E_{h}=0$* if and only if $q^{h}_{ij}=0$.*

Lemma 10**.**

For $0\leq h,i,j,r,s,t\leq D$ we have

[TABLE]

Proof.

Since $A_{i}A_{j}=\sum_{h=0}^{D}p^{h}_{ij}A_{h}$ and $E_{i}E_{j}=\delta_{ij}E_{i}$ $(0\leq h,i,j\leq D)$ and by Lemma 8 and (4), we obtain

[TABLE]

Definition 11**.**

Let $G$ denote the matrix of inner products for

[TABLE]

where $1\leq i\leq D$. Thus the matrix $G$ is $4D\times 4D$.

Theorem 12**.**

The entries of $G$ are as follows: For $1\leq i,j\leq D$,

where $\phi/2$ is a weighted sum involving the following terms and coefficients:

Proof.

By Lemma 10 and using (1)–(5), we obtain

[TABLE]

Similarly, for $1\leq h\leq 3$,

[TABLE]

Similarly, for $1\leq h,\ell\leq 3$,

[TABLE]

The result follows. ∎

In Appendix 2, we give the matrix $G$ for $D=3$.

Definition 13**.**

For $1\leq i\leq D$ let $B_{i}$ denote the matrix of inner products for

[TABLE]

So the matrix $B_{i}$ is $4\times 4$.

Definition 14**.**

Let $\widetilde{G}$ denote the matrix of inner products for

[TABLE]

where $1\leq i\leq D$. Thus the matrix $\widetilde{G}$ is $4D\times 4D$.

Lemma 15**.**

The matrix $\widetilde{G}$ has the form

[TABLE]

where $B_{1},B_{2},...,B_{D}$ are from Definition 13.

Proof.

Recall that $A_{0}^{*},A_{1}^{*},...,A_{D}^{*}$ form a basis for $M^{*}$. Therefore the sum $MM^{*}M=\sum_{i=0}^{D}MA_{i}^{*}M$ is direct. The summands are mutually orthogonal by Lemma 8(ii). The result follows. ∎

Lemma 16**.**

$det(\widetilde{G})=\displaystyle\prod_{i=1}^{D}det(B_{i})$.

Proof.

Immediate from Lemma 15. ∎

5 The main result

In this section we obtain our main result, which is Theorem 18.

Lemma 17**.**

The following (i)–(iii) hold.

1. (i)

$\widetilde{G}=(S^{\prime})^{t}GS^{\prime}$. 2. (ii)

det(G)=\big{(}det(S^{\prime})\big{)}^{-2}det(\widetilde{G}). 3. (iii)

det(G)=\big{(}det(S^{alt})\big{)}^{-8}\displaystyle\prod_{i=1}^{D}det(B_{i}).

Proof.

$(i)$ Immediate from Definition 11, Definition 14, and Corollary 7.

$(ii)$ Follows from $(i)$.

$(iii)$ Follows from $(ii)$ and Lemmas 6, 16. ∎

Theorem 18**.**

Let $\Gamma$ denote a primitive distance-regular graph with diameter $D\geq 3$. Then $\Gamma$ is $Q$-polynomial if and only if $det(G)=0$.

Proof.

To prove the theorem in one direction, assume that $\Gamma$ is $Q$-polynomial with respect to the ordering $E_{0},E_{1},...,E_{D}$. Write $A^{*}=A^{*}_{1}$. By [10, Theorem 3.3] and Lemma 1, we obtain $A^{*}A_{3}-A_{3}A^{*}\in{Span\{AA^{*}A_{2}-A_{2}A^{*}A,A^{*}A_{2}-A_{2}A^{*},A^{*}A-AA^{*}\}}$. Thus $AA^{*}A_{2}-A_{2}A^{*}A,A^{*}A_{3}-A_{3}A^{*},A^{*}A_{2}-A_{2}A^{*},A^{*}A-AA^{*}$ are linearly dependent. Now the matrix $B_{1}$ from Definition 13 has determinant zero. Now $det(G)=0$ by Lemma 17(iii).

For the other direction, assume $det(G)=0$. By Lemma 17(iii) and since $S^{alt}$ is invertible, there exists an integer $t$ $(1\leq t\leq D)$ such that $det(B_{t})=0$. Now $AA_{t}^{*}A_{2}-A_{2}A_{t}^{*}A,A_{t}^{*}A_{3}-A_{3}A_{t}^{*},A_{t}^{*}A_{2}-A_{2}A_{t}^{*},A_{t}^{*}A-AA_{t}^{*}$ are linearly dependent. We will show that $A_{t}^{*}A_{3}-A_{3}A_{t}^{*}\in{Span\{AA_{t}^{*}A_{2}-A_{2}A_{t}^{*}A,A_{t}^{*}A_{2}-A_{2}A_{t}^{*},A_{t}^{*}A-AA_{t}^{*}\}}$. To do this we show that $AA_{t}^{*}A_{2}-A_{2}A_{t}^{*}A,A_{t}^{*}A_{2}-A_{2}A_{t}^{*},A_{t}^{*}A-AA_{t}^{*}$ are linearly independent. Suppose not. Then there exist scalars $a,b,c$, not all zero, such that

[TABLE]

Abbreviate $\theta^{*}_{i}=m_{t}u_{i}(\theta_{t})$ $(0\leq i\leq D)$. So $A^{*}_{t}=\sum_{i=0}^{D}\theta^{*}_{i}E^{*}_{i}$. By Lemma 1,

[TABLE]

For $1\leq h\leq 3$ pick $z\in X$ such that $\partial(x,z)=h$. Compute the $(x,z)$-entry in (17). For $h=3$ we get $ac_{3}(\theta^{*}_{1}-\theta^{*}_{2})=0$. For $h=2$ we get $aa_{2}(\theta^{*}_{1}-\theta^{*}_{2})+b(\theta^{*}_{0}-\theta^{*}_{2})=0$. For $h=1$ we get $ab_{1}(\theta^{*}_{1}-\theta^{*}_{2})+c(\theta^{*}_{0}-\theta^{*}_{1})=0$. Solving these equations we obtain $a(\theta^{*}_{1}-\theta^{*}_{2})=0$ and $b=0$, $c=0$. Recall that $a,b,c$ are not all zero, so $a\neq 0$ and $\theta^{*}_{1}=\theta^{*}_{2}$. Now (17) becomes $AA_{t}^{*}A_{2}-A_{2}A_{t}^{*}A=0$. Recall $c_{2}A_{2}=A^{2}-a_{1}A-kI$. We have $AA^{*}_{t}A^{2}+kA^{*}_{t}A=A^{2}A^{*}_{t}A+kAA^{*}_{t}$. Thus $[A,AA^{*}_{t}A+kA^{*}_{t}]=0$. For $0\leq i,j\leq D$ such that $i\neq j$ we have $E_{i}A^{*}_{t}E_{j}(\theta_{i}\theta_{j}+k)=0$. By Corollary 9, $E_{i}A^{*}_{t}E_{j}\neq 0$ if and only if $q^{t}_{ij}\neq 0$, and in this case $\theta_{i}\theta_{j}+k=0$. Since $q^{t}_{0t}=1$ and $\theta_{0}=k$, we have $k\theta_{t}+k=0$ and hence $\theta_{t}=-1$. Define a diagram with nodes $0,1,...,D$. There exists an arc between nodes $i,j$ if and only if $i\neq j$ and $q^{t}_{ij}\neq 0$. Observe that node [math] is connected to node $t$ and no other nodes. By [2, Proposition 2.11.1] and Lemma 1, the diagram is connected. Thus there exists a node $s$ with $s\neq 0$ and $s\neq t$ that is connected to node $t$ by an arc. In other words $q^{t}_{st}\neq 0$. So $\theta_{s}\theta_{t}+k=0$ and hence $\theta_{s}=k$, a contradiction. Therefore $AA_{t}^{*}A_{2}-A_{2}A_{t}^{*}A,A_{t}^{*}A_{2}-A_{2}A_{t}^{*},A_{t}^{*}A-AA_{t}^{*}$ are linearly independent. So $A_{t}^{*}A_{3}-A_{3}A_{t}^{*}\in{Span\{AA_{t}^{*}A_{2}-A_{2}A_{t}^{*}A,A_{t}^{*}A_{2}-A_{2}A_{t}^{*},A_{t}^{*}A-AA_{t}^{*}\}}$. Now by [10, Theorem 3.3] and (18), $\Gamma$ is a $Q$-polynomial with respect to $E=E_{t}$. ∎

6 Appendix 1

Recall the distance-regular graph $\Gamma$ with diameter $D$. Recall for $0\leq h\leq D$

We now give $p^{h}_{2j}$ for $h-2\leq j\leq h+2$.

[TABLE]

We now give $p^{h}_{3j}$ for $h-3\leq j\leq h+3$.

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[TABLE]

7 Appendix 2

Recall the matrix $G$ from Theorem 12. In this appendix we give $G$ for $D=3$.

Example 19**.**

Assume $D=3$. The rows and columns of $G$ are indexed by the following matrices, in the specified order:

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So the matrix $G$ is $12\times 12$. $G$ has the form

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where each block is a $3\times 3$ symmetric matrix as shown below.

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The matrix $\mathbb{W}$ is symmetric with entries

[TABLE]

From Appendix 1, we find

[TABLE]

8 Acknowledgement

The author would like to thank Professor Paul Terwilliger for many valuable ideas and suggestions. This paper was written while the author was an Honorary Fellow at the University of Wisconsin-Madison (January 2017- January 2018) supported by the Development and Promotion of Science and Technology Talents (DPST) Project, Thailand.