Chapter 26 Representation Diagrams
Wednesday, March 31, 1993
Proof (Proof of Theorem 24.1 continued). Assume \(Y = (X, \{R_i\}_{0\leq i\leq D})\) is \(P\)-polynomial. Let \(E\) be a primitive idempotent of \(Y\) such that the corresponding representation \((\rho, H)\) is nondegenerate.
Show for all \(x, y\in X\), \[\sum_{z\in X, (x,z)\in R_1, (y,z)\in R_2}\rho(z) - \sum_{z'\in X, (x,z')\in R_2, (y,z')\in R_1}\rho(z') \in \mathrm{Span}(\rho(x)-\rho(y))\] implies that \(Y\) is \(Q\)-polynomial with respect to \(E\).
Define a diagram \(D_E\) on nodes \(0, 1, \ldots, D\), for \(i\neq j\), \[i \frown j \leftrightarrow q^1_{ij}\neq 0\] by setting \(E = E_1\).
We showed that \(0 \frown j \leftrightarrow j = 1\) \((1\leq j\leq D)\) and \(D_E\) is connected.
Now it is sufficient to show the following.
Claim 6. Let \(i\) be a node in \(D_E\). Then \(i\) is adjacent to at most \(2\) arcs.
Proof of Claim 6. Suppose the node \(j\) is adjacent to \(i\) in \(D_E\). By Claim 4, \[\begin{align} 0 & = E_i(A^3A^* - A^*A^3 - (\beta+1)(A^2A^*A-AA^*A^2) - \gamma(A^2A^*-A^*A) - \delta(AA^*-A^*A))E_j\\ & = E_iA^*E_j(\theta^3_i-\theta^3_j-(\beta+1)(\theta^2\theta_j - \theta_i\theta_j^2)-\gamma(\theta_i^2-\theta_j^2)-\delta(\theta_i-\theta_j))\\ & = E_iA^*A_j(\theta_i-\theta_j)p(\theta_i, \theta_j), \end{align}\] where \[p(s,t) = s^2 - \beta st + t^2 - \gamma(s+t) - \delta.\]
HS MEMO
\[\begin{align} & (\theta_i-\theta_j)(\theta_i^2 - \beta \theta_i\theta_j + \theta_j^2 - \gamma(\theta_i + \theta_j) - \delta)\\ & = \quad \theta^3_i-\theta^3_j-(\beta+1)(\theta^2\theta_j - \theta_i\theta_j^2)-\gamma(\theta_i^2-\theta_j^2)-\delta(\theta_i-\theta_j) \end{align}\]
Since \(i\) is adjacent to \(j\), \(q^1_{ij}\neq 0\) and \[E_iA^*E_j\neq O\] by Lemma 20.3 \((ii)\). Since \(Y\) is \(P\)-polynomial, \[\theta_i \neq \theta_j \quad \text{ if }\; i\neq j.\] Hence \(p(\theta_i,\theta_j) = 0\). But \(p\) is quadratic in \(t\). So \(p(\theta_i,t) = 0\) has at most two solutions for \(\theta_j\).
Now \(D_E\) is a pth, and \(\Gamma\) is \(Q\)-polynomial with respect to \(E\).
This proves Theorem 24.1.
Corollary 26.1 Assume \(Y = (X, \{R_i\}_{0\leq i\leq D})\) is \(P\)-polynomial, and \(Q\)-polynomial with respect to a primitive idempotent \[E = \frac{1}{|X|}\sum_{i=0}^D \theta^*_i A_i.\] Then, \[\beta = \frac{\theta^*_i - \theta^*_{i+1} + \theta^*_{i+2}-\theta^*_{i+3}}{\theta^*_{i+1}-\theta^*_{i+2}}\] is independent of \(i\) \((0\leq i\leq D-3)\).
Proof. Fix \(i\). Without loss of generality, \(D\geq 3\), else vacuous.
Pick \(x,y\in X\) with \((x,y)\in R_3\).
Let \((\rho, H)\) be the representation for \(E\). \[\begin{equation} \sum_{z\in X, (x,z)\in R_1, (y,z)\in R_2}\rho(z) - \sum_{z'\in X, (x,z')\in R_2, (y,z')\in R_1}\rho(z') = \frac{p^3_{12}(\theta^*_1-\theta^*_2)}{\theta^*_0-\theta^*_3}(\rho(x)-\rho(y)), \tag{26.1} \end{equation}\] and \(p^3_{12} = c_3\).
Since \(p^3_{i,i+3} \neq 0\), there exists \(w\in X\) such that \((x,w)\in R_{i+3}\), \((y,w) \in R_i\).
Take inner product of (26.1) with \(\rho(w)\). We have \[\begin{align} P^3_{12}(x,y) & \subseteq P^{i+3}_{1,i+2}(x,w)\cap P^i_{2,i+2}(y,w)\\ P^3_{21}(x,y) & \subseteq P^{i+3}_{2,i+1}(x,w)\cap P^i_{2,i+1}(y,w). \end{align}\] Hence, \[\left\langle \sum_{z\in X, (x,z)\in R_1, (y,z)\in R_2}\rho(z) - \sum_{z'\in X, (x,z')\in R_2, (y,z')\in R_1}\rho(z'), \rho(w)\right\rangle = |X|^{-1}c_3(\theta^*_{i+2}-\theta^*_{i+1}),\] \[\left\langle \frac{c_3(\theta^*_1-\theta^*_2)}{\theta^*_0-\theta^*_3}(\rho(x)-\rho(y)), \rho(w)\right\rangle = \frac{c_3(\theta^*_1-\theta^*_2)}{\theta^*_0-\theta^*_3}|X|^{-1}(\theta^*_{i+3}-\theta^*_{i+1}).\] We have, \[\sigma = \frac{\theta^*_{i+1}-\theta^*_{i+2}}{\theta^*_i-\theta^*_{i+3}} = \frac{\theta^*_1-\theta^*_2}{\theta^*_0-\theta^*_3}.\]
HS MEMO
Note that since \(Y\) is \(P\) and \(Q\) with respect to \(A_1\) and \(E_1\), \(\theta^*_0, \theta^*_1, \ldots, \theta^*_D\), \(\theta_0, \theta_1, \ldots, \theta_D\) are all distinct.
So \[\beta = \frac{1}{\sigma}-1 = \frac{\theta^*_i - \theta^*_{i+1} + \theta^*_{i+2}-\theta^*_{i+3}}{\theta^*_{i+1}-\theta^*_{i+2}} = \frac{\theta^*_0 - \theta^*_{1} + \theta^*_{2}-\theta^*_{3}}{\theta^*_{1}-\theta^*_{2}}.\] We have the assertion.
Given the intersection number of a distance-regular graph \(\Gamma\). The following two lemmas give an efficient method to determine if \(\Gamma\) is \(Q\)-polynomial with respect to some primitive idempotent.
Lemma 26.1 Let \(\Gamma\) be a distance-regular graph of diameter \(D\geq 1\). Pick \(\theta, \theta^*_0, \theta^*_1, \ldots, \theta^*_D\in \mathbb{R}\) such that \(\theta^*_0 \neq 0\), and set \[E = \frac{1}{|X|}\sum_{i=0}^D\theta^*_iA_i.\]
\[\begin{pmatrix} a_0 & b_0 & 0 & \cdots & \cdots & 0\\ c_1 & a_1 & b_1 & 0 & \cdots & 0\\ 0 & c_2 & a_2 & b_2 & \ddots & \vdots\\ \cdots & \ddots & \ddots & \ddots & \ddots & 0 \\ 0 & \cdots & 0 & c_{D-1} & a_{D-1} & b_{D-1}\\ 0 & \cdots & \cdots & 0 & c_D & a_D \end{pmatrix} \begin{pmatrix} \theta^*_0\\\theta^*_1\\\vdots \\\vdots \\\vdots\\ \theta^*_D\end{pmatrix} = \theta\cdot \begin{pmatrix} \theta^*_0\\\theta^*_1\\\vdots\\\vdots\\ \vdots \\ \theta^*_D\end{pmatrix}, \] and \(\theta^*_0 = \mathrm{rank}E\).
\[\frac{\theta^*_1}{\theta^*_0}, \ldots, \frac{\theta^*_D}{\theta^*_0}\] can be computed from \(\theta\) using \[\frac{\theta^*_i}{\theta^*_0} = \frac{p_i(\theta)}{kb_1\cdots b_{i-1}} \quad (1\leq i\leq D),\] where \(p_0 = 1\), \(p_1(\lambda) = \lambda\), and \[\lambda p_i(\lambda) = p_{i+1}(\lambda) + a_ip_i(\lambda) + b_{i-1}c_ip_{i-1}(\lambda)\quad (0\leq i\leq D).\]
Proof.
\[\begin{align} (ia) & \leftrightarrow (A- \theta I)E = 0 \text{ and } E^2 = E\\ & \leftrightarrow 0 = \sum_{i=0}^D(A-\theta I)\theta^*_i A_i \text{ and $\mathrm{rank}E = \mathrm{trace}E = \theta^*_0$}\\ & \qquad = \sum_{i=0}^D\theta^*_i(c_{i+1}A_{i+1}+ a_iA_i + b_{i-1}A_{i-1}-\theta A_i)\\ & \qquad = \sum_{j=0}^D A_j(c_j\theta^*_{j-1}+a_j\theta^*_j+b_j\theta^*_{j+1}-\theta \theta^*_j)\\ & \leftrightarrow c_j \theta^*_{j-1} + a_j\theta^*_j + b_j\theta^*_{j+1} = \theta \theta^*_j \; (0\leq j \leq D) \text{ and }\mathrm{rank}E = \theta^*_0\\ & \leftrightarrow (ib). \end{align}\]
HS MEMO
The first \(\leftrightarrow\). \(\rightarrow\) is clear.
\(\leftarrow\): By the first condition, \(AE = \theta E\). So \(E\) is a scalar multiple of the primitive idempotent corresponding to \(\theta\). Hence, \(\mathrm{rank}E = \mathrm{trace}E\) implies \(E\) is the primitive idempotent.
\(i = 0\) is trivial.
\(i=1\): Set \(j = 0\) above \(c_0 = 0, a_0 = 0, b_0 = k\). We have \[k\theta^*_1 = \theta \theta^*_0.\] So \[\frac{\theta^*_1}{\theta^*_0} = \frac{\theta}{k} = \frac{p_1(\theta)}{k}.\]
\(i\geq 2\): Set \(j=i-1\) above. We have \[c_{i-2}\theta^*_{i-2} + a_{i-1}\theta^*_{i-1} + b_{i-1}\theta^*_i = \theta \theta^*_{i-1}.\] So, \[\begin{align} \frac{\theta^*_i}{\theta^*_0} & = \frac{\theta\theta^*_{i-1}-a_{i-1}\theta^*_{i-1}-c_{i-1}\theta^*_{i-2}}{b_{i-1}\theta^*_0}\\ & = \left((\theta-a_{i-1})\frac{\theta^*_{i-1}}{\theta^*_0}-c_{i-1}\frac{\theta^*_{i-2}}{\theta^*_0}\right)\frac{1}{b_{i-1}}\\ & = \left((\theta-a_{i-1})\frac{p_{i-1}(\theta)}{kb_1\cdots b_{i-2}}-c_{i-1}\frac{p_{i-2}(\theta)}{kb_1\cdots b_{i-3}}\right)\frac{1}{b_{i-1}}\\ & = \frac{p_i(\theta)}{kb_1\cdots b_{i-2}b_{i-1}}, \end{align}\] as desired.