Chapter 27 \(P\)-and \(Q\)-Polynomial Schemes
Friday, April 2, 1993
Theorem 27.1 Let \(\Gamma = (X,E)\) be a distance-regular graph of diameter \(D\geq 3\).
Let \(\theta\) denote an eigenvalue of \(\Gamma\) with associated primitive idempotent \[E = \frac{1}{|X|}\sum_{i=0}^D \theta^*_iA_i.\] Then the following are equivalent.
\[\begin{align} & c_i\left(\theta^*_2 - \theta^*_i - \frac{(\theta^*_1-\theta^*_{i-1})^2}{\theta^*_0-\theta^*_i}\right) + b_{i-1}\left(\theta^*_2 - \theta^*_{i-1} - \frac{(\theta^*_1-\theta^*_{i})^2}{\theta^*_0-\theta^*_{i-1}}\right)\\ & = (k-\theta)(\theta^*_1+\theta^*_2-\theta^*_{i-1}-\theta^*_i)-(\theta+1)(\theta^*_0-\theta^*_2) \tag{27.1} \end{align}\]
HS MEMO
Note (27.1) is trivial for \(i = 1, 2\).
\(i=1\): \[\begin{align} \mathrm{LHS} & = \left(\theta^*_2-\theta^*_1 - \frac{(\theta^*_1-\theta^*_0)^2}{\theta^*_0-\theta^*_1}\right)+ k(\theta^*_2-\theta^*_0)\\ & = \theta^*_2-\theta^*_1 - \theta^*_0 + \theta^*_1 + k(\theta_2^*-\theta^*_0)\\ & = (k+1)(\theta^*_2-\theta^*_0)\\ \mathrm{RHS} & = (k-\theta)(\theta^*_1+\theta^*_2-\theta^*_0-\theta^*_1)-(\theta+1)(\theta^*_0-\theta^*_2)\\ & = (k+1)(\theta^*_2-\theta^*_0). \end{align}\]
\(i=2\): \[\begin{align} \mathrm{LHS} & = b_1\left(\theta^*_2-\theta^*_1 - \frac{(\theta^*_1-\theta^*_0)^2}{\theta^*_0-\theta^*_1}\right)\\ & = b_1\frac{(\theta^*_2-\theta^*_1)(\theta^*_0-\theta^*_1-\theta^*_2+\theta^*_1)}{\theta^*_0-\theta^*_1}\\ & = b_1\frac{(\theta^*_2-\theta^*_1)(\theta^*_0-\theta^*_2)}{\theta^*_0-\theta^*_1}\\ \mathrm{RHS} & = -(\theta+1)(\theta^*_0-\theta^*_2). \end{align}\] Hence, \[\begin{align} \mathrm{LHS}=\mathrm{RHS} & \leftrightarrow b_1\frac{\theta^*_2-\theta^*_1}{\theta^*_0-\theta^*_1} + (\theta + 1) = 0\\ & \leftrightarrow b_1(\theta^*_2-\theta^*_1)+(\theta+1)(\theta^*_0-\theta^*_1) = 0. \end{align}\] On the other hand, \[\begin{align} b_1\theta^*_2 + a_1\theta^*_1 + c_1\theta^*_0 & = \theta \theta^*_1\\ b_1\theta^*_1 + a_1\theta^*_1 + c_1\theta^*_1 & = k \theta^*_1, \end{align}\] as \(\theta \theta^*_0 = k\theta^*_1\). We have \[b_1(\theta^*_2 - \theta^*_1) + (\theta^*_0-\theta^*_1) = \theta(\theta^*_1-\theta^*_0).\]
Proof. Immediate from the proof of Theorem 2.1 in ‘A new inequality for distance-regular graphs’ (Terwilliger 1995) and Theorem 24.1.
Note. Suppose \((i)-(iii)\) hold. In particular, \(\theta^*_0, \theta^*_1, \ldots, \theta^*_D\) are distinct. Then, \[c_i + a_i + b_i = k\quad (0\leq i\leq D).\] \[c_i\theta^*_{i-1} + a_i\theta^*_i + b_i\theta^*_{i+1} = \theta \theta^*_j \quad (0\leq i\leq D).\] \[\frac{\theta^*_i-\theta^*_{i+1}+\theta^*_{i+2}-\theta^*_{i-3}}{\theta^*_{i+1}-\theta^*_{i+2}}\quad \text{is independent of $i$}\quad (0\leq i\leq D-3).\] \[\begin{align} & c_i\left(\theta^*_2 - \theta^*_i - \frac{(\theta^*_1-\theta^*_{i-1})^2}{\theta^*_0-\theta^*_i}\right) + b_{i-1}\left(\theta^*_2 - \theta^*_{i-1} - \frac{(\theta^*_1-\theta^*_{i})^2}{\theta^*_0-\theta^*_{i-1}}\right)\\ & = (k-\theta)(\theta^*_1+\theta^*_2-\theta^*_{i-1}-\theta^*_i)-(\theta+1)(\theta^*_0-\theta^*_2). \end{align}\]
Furthermore, we can solve for \(c_1, \ldots, c_D\), \(a_1, \ldots, a_D\), \(b_0, b_1, \ldots, b_{D-1}\) in terms of five free parameters.
In general, we can take the five parameters to be \[D, q, s^*, r_1, r_2\] and get \[\begin{align} b_i & = \frac{h(1-q^{i-D})(1-s^*q^{i+1})(1-r_1q^{i+1})(1-r_2q^{i+1})}{(1-s^*q^{2i+1})(1-s^*q^{2i+2})} \quad (0\leq i\leq D),\\ c_i & = \frac{h(1-q^{i})(1-s^*q^{D+i+1})(r_1-s^*q^{i})(r_2-s^*q^{i})}{s^*q^D(1-s^*q^{2i})(1-s^*q^{2i+1})} \quad (0\leq i\leq D),\\ a_i & = b_0 - c_i - b_i \quad (0\leq i\leq D), \end{align}\] where \(h\) variable is chosen so that \(c_1 = 1\).
(We must also consider limiting cases \(h\to 0\), \(s^*\to 0\), \(q^*\to \pm 1\).)
See Theorem 2.1 in “The subconstituent algebra of an association scheme, I, II, III, (Terwilliger 1992), (Terwilliger 1993a), (Terwilliger 1993b).
Definition 27.1 Let \(\Gamma = (X,E)\) be a distance-regular graph of diameter \(D\geq 3\). Choose \(q \in \mathbb{R}\setminus \{0,-1\}\), set \[\begin{bmatrix}{i}\\{1}\end{bmatrix} = 1 + q + \cdots + q^{i-1} = \begin{cases} \frac{q^i-1}{q-1} & q\neq 1\\ i & q = 1.\end{cases}\]
Definition 27.2 \(\Gamma\) has classical parameters if \[\begin{align} c_i & = \begin{bmatrix}{i}\\{1}\end{bmatrix}\left(1+ \alpha \begin{bmatrix}{i-1}\\{1}\end{bmatrix}\right) \tag{27.2}\\ b_i & = \left(\begin{bmatrix}{D}\\{1}\end{bmatrix}-\begin{bmatrix}{i}\\{1}\end{bmatrix}\right)\left(\sigma- \alpha \begin{bmatrix}{i}\\{1}\end{bmatrix}\right) \tag{27.3} \end{align}\] for some \(\sigma, \alpha\in \mathbb{R}\).
(This happens for essentially all known families of distance-regular graphs with unbounded diameter, and is essentially equivalent to \(s^*=0\).)
Lemma 27.1 With above notation, suppose (27.2), (27.3) hold. Then,
\[\frac{\theta^*_i}{\theta^*_0} = 1 + \left(\frac{\theta}{k}-1\right)\begin{bmatrix}{i}\\{1}\end{bmatrix} q^{1-i} \quad (0\leq i\leq D).\] In particular, \(\theta^*_i \neq \theta^*_0\) for all \(i\in \{1, 2 \ldots, D\}\).
Proof.
\[c_i\theta^*_{i-1} + a_i\theta^*_i + b_i\theta^*_{i+1} = \theta \theta^*_i \quad (0\leq i\leq D),\] where \(a_i = k-c_i - b_i\) \(\quad (0\leq i \leq D)\).
(equivalently: check \[\begin{equation} c_i(\theta^*_{i-1}-\theta^*_i) + b_i(\theta^*_i -\theta^*_{i+1}) = (\theta-k) \theta^*_i \quad (0\leq i\leq D),\tag{27.4} \end{equation}\] where \(c_i, b_i, \theta^*_i, \theta\) are as given.)
HS MEMO
\[\theta = \frac{b_1}{q}-1, \; \frac{\theta^*_i}{\theta^*_0} = 1 + \left(\frac{\theta}{k}-1\right)\begin{bmatrix}{i}\\{1}\end{bmatrix} q^{1-i}, \; b_0 = \begin{bmatrix}{D}\\{1}\end{bmatrix}\sigma.\] \(i=0\). \[\frac{\theta^*_i}{\theta^*_0} = \frac{\theta}{k}, \quad - k\left(1-\frac{\theta^*_1}{\theta^*_0}\right)= -k\left(1-\frac{\theta}{k}\right) = \theta -k.\] \[\frac{\theta^*_{i-1}-\theta^*_i}{\theta^*_0} = \left(\frac{\theta}{k}-1\right)\left(\begin{bmatrix}{i-1}\\{1}\end{bmatrix}q^{2-i}-\begin{bmatrix}{i}\\{1}\end{bmatrix}q^{1-i}\right)=-\left(\frac{\theta}{k}-1\right)q^{1-i}.\] \[\theta-k=\left(\begin{bmatrix}{D}\\{1}\end{bmatrix}-1\right)(\sigma-\alpha)/q-1 - \begin{bmatrix}{D}\\{1}\end{bmatrix}\sigma = \begin{bmatrix}{D-1}\\{1}\end{bmatrix}(\sigma-\alpha)-1-\begin{bmatrix}{D}\\{1}\end{bmatrix}\sigma.\] \[\begin{align} & (c_i(\theta^*_{i-1}-\theta^*_i) + b_i(\theta^*_i -\theta^*_{i+1}) - (\theta-k) \theta^*_i)/\theta^*_0\\ & \quad = -\begin{bmatrix}{i}\\{1}\end{bmatrix}\left(1+\alpha \begin{bmatrix}{i-1}\\{1}\end{bmatrix}\right)\left(\frac{\theta}{k}-1\right)q^{1-i} + \left(\begin{bmatrix}{D}\\{1}\end{bmatrix}-\begin{bmatrix}{i}\\{1}\end{bmatrix}\right)\left(\sigma- \alpha \begin{bmatrix}{i}\\{1}\end{bmatrix}\right)\left(\frac{\theta}{k}-1\right)q^{-i}\\ & \qquad\qquad -(\theta-k)\left(1+\left(\frac{\theta}{k}-1\right)\begin{bmatrix}{i}\\{1}\end{bmatrix}q^{1-i}\right)\\ & \quad = \left(\frac{\theta}{k}-1\right)\left(-\begin{bmatrix}{i}\\{1}\end{bmatrix}\left(1+ \alpha\begin{bmatrix}{i-1}\\{1}\end{bmatrix}\right)q^{1-i}+\begin{bmatrix}{D-i}\\{1}\end{bmatrix}\left(\sigma-\alpha \begin{bmatrix}{i}\\{1}\end{bmatrix}\right)\right.\\ & \qquad\qquad\left.-\left(\begin{bmatrix}{D}\\{1}\end{bmatrix}\sigma + \left(\begin{bmatrix}{D-1}\\{1}\end{bmatrix}(\sigma-\alpha)-1-\begin{bmatrix}{D}\\{1}\end{bmatrix}\sigma\right)\begin{bmatrix}{i}\\{1}\end{bmatrix}q^{1-i}\right)\right)\\ & \quad = \left(\frac{\theta}{k}-1\right)\left(-\begin{bmatrix}{i}\\{1}\end{bmatrix}q^{1-i}-\alpha\left(\begin{bmatrix}{i}\\{1}\end{bmatrix}\begin{bmatrix}{i-1}\\{1}\end{bmatrix}q^{1-i}+\begin{bmatrix}{D-i}\\{1}\end{bmatrix}\begin{bmatrix}{i}\\{1}\end{bmatrix}-\begin{bmatrix}{D-1}\\{1}\end{bmatrix}\begin{bmatrix}{i}\\{1}\end{bmatrix}q^{1-i}\right)\right.\\ & \qquad\qquad \left.+\sigma\left(\begin{bmatrix}{D-i}\\{1}\end{bmatrix}-\begin{bmatrix}{D}\\{1}\end{bmatrix}-\begin{bmatrix}{D-1}\\{1}\end{bmatrix}\begin{bmatrix}{i}\\{1}\end{bmatrix}q^{1-i}+\begin{bmatrix}{D}\\{1}\end{bmatrix}\begin{bmatrix}{i}\\{1}\end{bmatrix}q^{1-i}\right)+\begin{bmatrix}{i}\\{1}\end{bmatrix}q^{1-i}\right) \end{align}\]
Check \(\theta\neq k\). Suppose \(\theta = k\). Then \[\frac{b_1}{q}-1 = k, \; \text{ and }\; q>0.\] By (27.2), (27.3), \[\begin{align} qc_i - b_i - q(qc_{i-1}-b_{i-1}) & = (k-\theta)q \quad (1\leq i\leq D)\\ & = 0. \end{align}\]
HS MEMO
With the notation of Lemma 27.1, we have the above equality in general. \[\begin{align} & qc_i-b_i - q(qc_{i-1}-b_{i-1})\\ & \quad = q\begin{bmatrix}{i}\\{1}\end{bmatrix}\left(1+\alpha \begin{bmatrix}{i-1}\\{1}\end{bmatrix}\right)-\left(\begin{bmatrix}{D}\\{1}\end{bmatrix}-\begin{bmatrix}{i}\\{1}\end{bmatrix}\right)\left(\sigma-\alpha\begin{bmatrix}{i}\\{1}\end{bmatrix}\right)\\ & \qquad\qquad -q\left(q\begin{bmatrix}{i-1}\\{1}\end{bmatrix}\left(1+\alpha\begin{bmatrix}{i-2}\\{1}\end{bmatrix}\right)-\left(\begin{bmatrix}{D}\\{1}\end{bmatrix}-\begin{bmatrix}{i-1}\\{1}\end{bmatrix}\right)\left(\sigma-\alpha\begin{bmatrix}{i-1}\\{1}\end{bmatrix}\right)\right)\\ & \quad = \left(q\begin{bmatrix}{i}\\{1}\end{bmatrix}-q^2\begin{bmatrix}{i-1}\\{1}\end{bmatrix}\right) \\ & \qquad\qquad + \alpha\left(q\begin{bmatrix}{i}\\{1}\end{bmatrix}\begin{bmatrix}{i-1}\\{1}\end{bmatrix}+\begin{bmatrix}{D}\\{1}\end{bmatrix}\begin{bmatrix}{i}\\{1}\end{bmatrix}-\begin{bmatrix}{i}\\{1}\end{bmatrix}\begin{bmatrix}{i}\\{1}\end{bmatrix}\right.\\ & \qquad\qquad \left.-q^2\begin{bmatrix}{i-1}\\{1}\end{bmatrix}\begin{bmatrix}{i-2}\\{1}\end{bmatrix}-q\begin{bmatrix}{D}\\{1}\end{bmatrix}\begin{bmatrix}{i-1}\\{1}\end{bmatrix}+q\begin{bmatrix}{i-1}\\{1}\end{bmatrix}\begin{bmatrix}{i-1}\\{1}\end{bmatrix}\right)\\ & \qquad\qquad + \sigma\left(-\begin{bmatrix}{D}\\{1}\end{bmatrix}+\begin{bmatrix}{i}\\{1}\end{bmatrix} + q\begin{bmatrix}{i-1}\\{1}\end{bmatrix}\right)\\ & \quad = q+\alpha\left(-\begin{bmatrix}{i}\\{1}\end{bmatrix}+\begin{bmatrix}{D}\\{1}\end{bmatrix}+q\begin{bmatrix}{i-1}\\{1}\end{bmatrix}\right)+\sigma(q^D-1+1)\\ & \quad = q\left(1 + \begin{bmatrix}{D-1}\\{1}\end{bmatrix}\alpha + q^{D-1}\sigma\right)\\ & \quad = q\left(\begin{bmatrix}{D}\\{1}\end{bmatrix}\sigma - \begin{bmatrix}{D-1}\\{1}\end{bmatrix}\sigma + \begin{bmatrix}{D-1}\\{1}\end{bmatrix}\alpha + 1\right)\\ & \quad = q\left(k-\frac{\begin{bmatrix}{D}\\{1}\end{bmatrix}-1}{q}(\sigma-\alpha)+1\right)\\ & \quad = q(k-\theta). \end{align}\]
Hence, \[\begin{align} qc_i - b_i & = q(qc_{i-1}-b_{i-1})\quad (1\leq i\leq D)\\ & = q^i(qc_0 - b_0)\\ & = -q^ik. \end{align}\] If \(i = D\), \(qc_D = -q^Dk\), \(c_D = -q^{D-1}k < 0\), a contradiction.
HS MEMO
\(\theta^*_0\neq \theta^*_h\) for all \(h\in \{1,2,\ldots, D\}\) and \[c_3\left(\theta^*_2 - \theta^*_3 - \frac{(\theta^*_1-\theta^*_{2})^2}{\theta^*_0-\theta^*_3}\right) - b_{2} \frac{(\theta^*_1-\theta^*_{3})^2}{\theta^*_0-\theta^*_{2}} = (k-\theta)(\theta^*_1-\theta^*_3)-(\theta+1)(\theta^*_0-\theta^*_2).\] Pf. \[\begin{align} \frac{\mathrm{LHS}}{\theta^*_0} & = \begin{bmatrix}{3}\\{1}\end{bmatrix}\left(1+\alpha \begin{bmatrix}{2}\\{1}\end{bmatrix}\right)\left(1-\frac{\theta}{k}\right)\left(q^{-2} - \frac{q^{-2}}{\begin{bmatrix}{3}\\{1}\end{bmatrix}q^{-2}}\right)\\ & \qquad -\left(\begin{bmatrix}{D}\\{1}\end{bmatrix}-\begin{bmatrix}{2}\\{1}\end{bmatrix}\right)\left(\sigma-\alpha \begin{bmatrix}{2}\\{1}\end{bmatrix}\right)\left(1-\frac{\theta}{k}\right)\frac{\left(\begin{bmatrix}{3}\\{1}\end{bmatrix}q^{1-3}-1\right)^2}{\begin{bmatrix}{2}\\{1}\end{bmatrix}q^{-1}}\\ & = \left(1-\frac{\theta}{k}\right)\left(\left(1+\alpha\begin{bmatrix}{2}\\{1}\end{bmatrix}\right)q^{-2}\begin{bmatrix}{2}\\{1}\end{bmatrix}-\left(\begin{bmatrix}{D}\\{1}\end{bmatrix}-\begin{bmatrix}{2}\\{1}\end{bmatrix}\right)\left(\sigma-\alpha\begin{bmatrix}{2}\\{1}\end{bmatrix}\right)\begin{bmatrix}{2}\\{1}\end{bmatrix}q^{-3}\right)\\ & = \left(1-\frac{\theta}{k}\right)\left(q^{-2}\begin{bmatrix}{2}\\{1}\end{bmatrix}+\alpha\left(q^{-2}\begin{bmatrix}{2}\\{1}\end{bmatrix}\begin{bmatrix}{2}\\{1}\end{bmatrix}+q^{-1}\begin{bmatrix}{2}\\{1}\end{bmatrix}\begin{bmatrix}{2}\\{1}\end{bmatrix}\begin{bmatrix}{D-2}\\{1}\end{bmatrix}\right)\right.\\ & \qquad \left.-q^{-1}\begin{bmatrix}{2}\\{1}\end{bmatrix}\begin{bmatrix}{D-2}\\{1}\end{bmatrix}\sigma\right)\\ \frac{\mathrm{RHS}}{\theta^*_0} & = \left(\begin{bmatrix}{D}\\{1}\end{bmatrix}\sigma-\begin{bmatrix}{D-1}\\{1}\end{bmatrix}(\sigma-\alpha)+1\right)\left(1-\frac{\theta}{k}\right)\left(\begin{bmatrix}{3}\\{1}\end{bmatrix}q^{-2}-1\right)\\ & \qquad -\begin{bmatrix}{D-1}\\{1}\end{bmatrix}(\sigma-\alpha)\left(1-\frac{\theta}{k}\right)\begin{bmatrix}{2}\\{1}\end{bmatrix}q^{-1}\\ & = \left(1-\frac{\theta}{k}\right)\left(q^{-2}\begin{bmatrix}{2}\\{1}\end{bmatrix}+\begin{bmatrix}{2}\\{1}\end{bmatrix}q^{-1}\sigma\left(q^{D-2}-\begin{bmatrix}{D-1}\\{1}\end{bmatrix}\right)\right.\\ & \qquad \left. +\begin{bmatrix}{2}\\{1}\end{bmatrix}q^{-2}\alpha\left(\begin{bmatrix}{D-1}\\{1}\end{bmatrix}+q\begin{bmatrix}{D-1}\\{1}\end{bmatrix}\right)\right)\\ & = \left(1-\frac{\theta}{k}\right)\left(q^{-2}\begin{bmatrix}{2}\\{1}\end{bmatrix}-\sigma q^{-1}\begin{bmatrix}{2}\\{1}\end{bmatrix}\begin{bmatrix}{D-2}\\{1}\end{bmatrix}+\alpha q^{-2}\begin{bmatrix}{2}\\{1}\end{bmatrix}\begin{bmatrix}{2}\\{1}\end{bmatrix}\begin{bmatrix}{D-1}\\{1}\end{bmatrix}\right) \end{align}\]
Example 27.1 \(Q\)-polynomial distance-regular graphs with classical parameters.
has classical parameters: \((q,\alpha, \sigma) = (1, 0, 1)\).
\(c_i = i^2\), \(b_i = (D-i)(N-D-i)\) has classical parameters \((q,\alpha, \sigma) = (1, 1, N-D)\).
\[c_i = \left(\frac{q^i-1}{q-1}\right)^2 = \begin{bmatrix}{i}\\{1}\end{bmatrix}^2, \quad b_i = \frac{q(q^D-q^i)(q^{N-D}-q^i)}{(q-1)^2}\] has classical parameters \[(q,\alpha, \sigma) = \left(q, q, \left(\frac{q^{N-D+1}-1}{q-1}\right)-1\right) = \left(q, q, \begin{bmatrix}{N-D+1}\\{1}\end{bmatrix}-1\right).\]
HS MEMO
\[\begin{align} b_i & = \left(\begin{bmatrix}{D}\\{1}\end{bmatrix}-\begin{bmatrix}{i}\\{1}\end{bmatrix}\right)\left(\begin{bmatrix}{N-D+1}\\{1}\end{bmatrix}-1-q\begin{bmatrix}{i}\\{1}\end{bmatrix}\right)\\ & = \left(\begin{bmatrix}{D}\\{1}\end{bmatrix}-\begin{bmatrix}{i}\\{1}\end{bmatrix}\right)\left(\begin{bmatrix}{N-D+1}\\{1}\end{bmatrix}-\begin{bmatrix}{i+1}\\{1}\end{bmatrix}\right)\\ & = \frac{q(q^D-q^i)(q^{N-D}-q^i)}{(q-1)^2}. \end{align}\]