Chapter 25 Balanced Conditions, II

Monday, March 29, 1993

Proof (Proof of Theorem 24.1 continued).

\((ib)\to(ic)\) Obvious.

\((ic)\to(ia)\) Without loss of generality, we may assume \(D\geq 3\), else trivial.

HS MEMO

The case \(D = 2\) should be treated somewhere, but the assumption \(D\geq 3\) is not used.

Fix \(w\in X\), and write \(E^*_i \equiv E^*_i(w)\), \(A^*_i\equiv A^*_i(w)\), \(A^*\equiv A^*_1\), and \(A_i\), \(i\)-th distance matrix. Set \[E \equiv E_1 = |X|^{-1}\sum_{i=0}^D \theta^*_i A_i.\] Since \((\rho, H)\) is nondegenerate, \[\theta^*_0 \neq \theta^*_h \; \text{for all }h\in \{1,2,\ldots, D\}\] See Lemma 23.1 \((iv)\).

Claim 1. Pick \(h\) \((1\leq h\leq D)\), and \(x,y\) with \((x,y)\in R_h\). Then \[\sum_{z\in X, (x,z)\in R_1, (y,z)\in R_2}\rho(z) - \sum_{z'\in X, (x,z')\in R_2, (y,z')\in R_1}\rho(z') = r^h_{12}(\rho(x)-\rho(y)),\] where \[r^h_{12} = p^h_{12}\frac{\theta_1^* - \theta^*_2}{\theta^*_0-\theta^*_h}.\]

Proof of Claim 1. By our assumption, \[\sum_{z\in X, (x,z)\in R_1, (y,z)\in R_2}\rho(z) - \sum_{z'\in X, (x,z')\in R_2, (y,z')\in R_1}\rho(z') = \alpha(\rho(x)-\rho(y)).\] Hence, \[\begin{align} |X|^{-1}p^h_{12}(\theta^*_1-\theta^*_2) & = \left\langle \sum_{z\in X, (x,z)\in R_1, (y,z)\in R_2}\rho(z) - \sum_{z'\in X, (x,z')\in R_2, (y,z')\in R_1}\rho(z'), \rho(x)\right\rangle \\ & = \alpha\langle \rho(x)-\rho(y), \rho(x)\rangle\\ & = \alpha |X|^{-1}(\theta_0^*-\theta^*_h). \end{align}\] We have \[\alpha = p^h_{12}\frac{\theta_1^* - \theta^*_2}{\theta^*_0-\theta^*_h}.\]

Claim 2. \({\displaystyle A_1A^*A_2 - A_2A^*A_1 = \sum_{h=1}^D r^h_{12}(A^*A_h - A_hA^*).}\)

Proof of Claim 2. The \(xy\) entry of the \(\mathrm{LHS} - \mathrm{RHS}\) is \[|X|\left\langle \sum_{z\in X, (x,z)\in R_1, (y,z)\in R_2}\rho(z) - \sum_{z'\in X, (x,z')\in R_2, (y,z')\in R_1}\rho(z') - r^h_{12}(\rho(x)-\rho(y)),\rho(w)\right\rangle,\] where \((x,y)\in R_h\), \(h = 1, 2, \ldots, D\), and the \(xy\) entry of the \(\mathrm{LHS} - \mathrm{RHS}\) is \(0\) if \(x = y\).

But the vector on the left in the above inner product is \(0\) by Claim 1, so the inner product is \(0\).

Thus, the \(xy\) entry of the \(\mathrm{LHS} - \mathrm{RHS}\) is always \(0\), and we have Claim 2.

Claim 3. \(A^*A_3 - A_3A^* \in \mathrm{Span}(AA^*A_2 - A_2A^*A, A^*A_2 - A_2A^*, A^*A-AA^*).\)

Proof of Claim 3. Since \(p^h_{12} = 0\), if \(h>3\), and \(p^h_{12} \neq 0\), if \(h=3\), we have \(r^h_{12} = 0\) if \(h > 0\), and \(r^h_{12} \neq 0\), if \(h = 3\). Note that \(\theta^*_1\neq \theta^*_2\).

Now we are done by Claim 2.

Claim 4. There exist \(\beta, \gamma, \delta\in \mathbb{R}\) such that \[\begin{align} 0 & = [A, A^2A^*-\beta AA^*A + A^*A^2 - \gamma(AA^*+A^*A) - \delta A^*]\\ & = A^3A^* - A^*A^3 - (\beta+1)(A^2A^*A-AA^*A^2)-\gamma(A^2A^*-A^*A^2)-\delta(AA^*-A^*A). \end{align}\]

Proof of Claim 4. There exists \(f_i\in \mathbb{R}[\lambda]\), \(\deg f_i = i\) such that \(A_i = f_i(A_1)\).

Writing \(A_2\), \(A_3\) as polynomials in \(A\) in Claim 3 and simplifying, we find \[A^3A^*-A^*A^3 \in \mathrm{Span}(A^2A^*A-AA^*A^2, A^2A^*-A^*A^2, AA^*-A^*A).\]

HS MEMO

Let \(A_3 = \beta_3A^3 + \beta_2 A^2 + \beta_1 A + \beta_0 I\) with \(\beta_3\neq 0\), and \(A_2 = \gamma_2 A^2 + \gamma_1 A + \gamma_0 I\), with \(\gamma_2\neq 0\). Then \[\begin{align} A^*A_3-A_3A^* & = A^*(\beta_3 A^3 + \beta_2 A^2 + \beta_1 A + \beta_0 I) - (\beta_3 A^3 + \beta_2 A^2 + \beta_1 A + \beta_0 I)A^*.\\ A^3A^*-A^*A^3 & \in \mathrm{Span}(A^*A_3 - A_3A^*, A^2A^* - A^*A^2, AA^*-A^*A)\\ & \subseteq \mathrm{Span}(AA^*A_2 - A_2A^*A, A^*A_2-A_2A^*, A^2A^*-A^*A^2, AA^*-A^*A)\\ A^*A_2 - A_2A^* & = A^*(\gamma_2 A^2 + \gamma_1 A + \gamma_0 I) - (\gamma_2 A^2 + \gamma_1 A + \gamma_0 I)A^*\\ AA^*A_2 - A_2A^*A & = AA^*(\gamma_2 A^2 + \gamma_1 A + \gamma_0 I) - (\gamma_2 A^2 + \gamma_1 A + \gamma_0 I)A^*A\\ A^*A_2 - A_2A^* & \in \mathrm{Span}(A^2A^*-A^*A^2, AA^*-AA^*)\\ AA^*A_2 - A_2A^*A & \in \mathrm{Span}(A^2A^*A-AA^*A^2, AA^*-AA^*)\\ A^3A^*-A^*A^3 & \in \mathrm{Span}(A^2A^*A-AA^*A^2, A^2A^*-A^*A^2, AA^*-A^*A). \end{align}\]

Hence, we can find \(\delta, \gamma, \delta\) satisfying \[0 = A^3A^*-A^*A^3 - (\beta+1)(A^2A^*A-AA^*A^2)-\gamma(A^2A^*-A^*A^2)-\delta(AA^*-A^*A).\] On the other hand, \[\begin{align} & [A, A^2A^*-\beta AA^*A+A^*A^2-\gamma(AA^*+A^*A)-\delta A^*]\\ & \quad = A^3A^*-A^2A^*A-\beta A^2A^*A + \beta AA^*A^2 + AA^*A^2 - A^*A^3 \\ & \qquad\quad - \gamma A^2A^* - \gamma AA^*A + \gamma AA^*A + \gamma A^*A^2 - \delta AA^* + \delta A^*A\\ & \quad = A^3A^* - A^*A^3 - (\beta+1)(A^2A^*A-AA^*A^2)-\gamma(A^2A^*-A^*A^2)-\delta (AA^*-A^*A). \end{align}\] Thus we have \((i)\) and \((ii)\).

Define a diagram \(D_E\) on nodes \(0, 1, \ldots, D\).

Connect distinct nodes \(,\) by undirected arc if \(q^1_{ij}\neq 0\). (Note \(q^1_{ij} = q^1_{ji}\)).

Since \(q^1_{0j} = \delta_{1j}\), the \(0\)-node is adjacent to the \(1\)-node and no other node.

\(Y\) is \(Q\)-polynomial with respect to \(E\) if and only if \(D_E\) is a path.

Claim 5. \(D_E\) is connected.

Proof of Claim 5. Suppose there exists \(\Delta \subseteq \{0,1,\ldots, D\}\) such that \(i,j\) not connected for every \(i\in \Delta\) and \(j\in \{0,1,\ldots, D\}\setminus \Delta\).

Set \[f = \sum_{i\in \Delta}E_i.\] Observe \[\begin{align} fA^* & = \sum_{i\in \Delta} E_i A^* \left(\sum_{j=0}^D E_j\right)\\ & = \sum_{i\in \Delta, j\in \Delta}E_iA^*E_j \quad \text{(since $E_iA^*E_j=O$ if $q^1_{ij}=0$)}\\ & = fA^*f. \end{align}\] Also, \(A^*f = fA^*f\).

Hence, \(f\) commutes with \(A^*\).

But \(f\) is an element of the Bose-Mesner algebra \[f = \sum_{i=0}^D \alpha_i A_i \quad \text{for some $\alpha_0, \ldots, \alpha_D\in \mathbb{C}$}.\] We have \[0 = fA^*-A^*f = \sum_{i=1}^D \alpha_i(A_iA^*- A^*A_i).\] But \(\{A_hA^* - A^*A_h \mid 1\leq h\leq D\}\) are linearly independent. (The column \(w\) of \(A_hA^*-A^*A_h\) is \(\theta^*_h - \theta^*_0\) times the column \(w\) of \(A_h\).)

Hence, \(\alpha_1 = \cdots = \alpha_D = 0\), and \(f = \alpha_0 I\). Since \(f^2 = f\), \(\alpha_0\) or \(1\).

If \(\alpha_0 = 0\), \(f=O\) and \(\Delta = \emptyset\).

If \(\alpha_0 = 1\), \(f=I\) and \(\Delta = \{0, 1, \ldots, D\}\).

This proves Claim 5.

HS MEMO

Claim 5 proves the following in general.

Let \(Y = (X, \{R_i\}_{0\leq i\leq D})\) be a symmetric association scheme. Fix a vertex \(x\in X\), and let \[E = \frac{1}{|X|}\sum_{j=0}^D \theta^*_j A_j \quad (\theta^*_j = q_1(j) \; \text{ if $E = E_1$})\] be a primitive idempotent and \(E^*_j\equiv E^*_j(x)\). \[A^* = \sum_{j=0}^D \theta_j^*E^*_j.\] If \(\theta_0 = \theta^*_h\), \(h=1, \ldots, D\), then the following hold.

\((i)\) \(\{A_hA^* - A^*A_h \mid 1\leq h\leq D\}\) are linearly independent.

\((ii)\) The diagram \(D_E\) on nodes \(0, 1, \ldots, D\) defined by

\[i\sim j \Leftrightarrow E(E_i\circ E_j)\neq O\] is connected.

\((iii)\) \(C_M(A^*) = \{L\in M\mid LA^* = A^*L\} = \mathrm{Span}(I).\)

Proof.

\((i)\) The column \(x\) of \(A_hA^* - A^*(A_h)\) is \(\theta^*_0-\theta^*_h\) times the column \(x\) of \(A_h\).

\((iii)\) \({\displaystyle 0 = \left[\sum_{h=0}^D\alpha_hA_h, A^*\right] = \sum_{h=1}^D\alpha_h(A_hA^*-A^*A_h)}\). Hence, \(\alpha_0 = \cdots =\alpha_D = 0\).

\((ii)\) \(\Delta\) is a connected component. Let \(f = \sum_{i\in \Delta}E_i\), then \(f\in C_M(A^*)\).

Let \(Y = (X, \{R_i\}_{0\leq i\leq 2})\) be a symmetric association scheme with \(D = 2\). Let \[E = \frac{1}{|X|}\sum_{j=0}^2\theta^*_j A_j\] be a primitive idempotent.

Suppose \(\theta^*_0\neq \theta_1^*, \theta^*_2\). Then \(Y\) is \(Q\)-polynomial with respect to \(E\).

Proof. By the previous lemma, \(D_E\) is connected.

Note. It seems \(\theta^*_1 \neq \theta^*_2\) is necessary. Clarify the condition \(\theta^*_1 = \theta^*_2\).

Terwilliger claims that \(\theta^*_1 = \theta^*_2\) does not occur under the assumption \((ic)\). (March 7, 1995)