Chapter 19 Commutative Association Schemes
Friday, March 5, 1993
Lemma 19.1 Let \(Y = (X, \{R_i\}_{0\leq i\leq D})\) be a commutative scheme with Bose-Mesner algebra \(M\).
Then there exists a basis \(E_0, E_1, \ldots, E_D\) for \(M\) such that
Proof. \(M\) acts on Hermitean space \(V = \mathbb{C}^n\) \((n = |X|)\).
If \(W\) is an \(M\)-module, so is \(W^\bot\).
Each irreducible \(M\)-module is \(1\) dimensional by commutativity of \(M\). So \(V\) is orthognal direct sum of \(1\)-dimensional \(M\)-modules.
Let \(v_1, \ldots, v_n\) be an orthonormal basis for \(V\) consisiting of eigenvectors for all \(m\in M\).
Set \(P\in \mathrm{Mat}_X(\mathbb{C})\) so that the \(i\)-th column of \(P\) is equal to \(v_i\). So, \[\bar{P}^\top P = I = P\bar{P}^\top = \bar{P}P^\top,\] and \(P\) is unitary.
Also, for all \(m\in M\), \[\begin{align} P^{-1}mP & = \text{diagonal}\\ & = \mathrm{diag}(\theta_1(m), \ldots, \theta_n(m)). \end{align}\] for some functions \[\theta_i: M \longrightarrow \mathbb{C}.\] Observe: each \(\theta = \theta_i\) is a character of \(M\), i.e., \[\theta: M\longrightarrow \mathbb{C}\] is a \(\mathbb{C}\)-algebra homomorphism.
Observe: the \(\theta_1, \ldots, \theta_n\) are not all distinct.
Let \(\sigma_0, \ldots, \sigma_r\) denote distinct elements of \[\theta_1, \ldots, \theta_n.\] Say \(\sigma_i\) appears \(m_i\) times. Without loss of generality, we may assume that \[P^{-1}mP = \begin{pmatrix}\sigma_0(m)I_{m_0} & O & O & O \\ O & \sigma_1(m)I_{m_1} & O & O\\ O & O & \ddots & O \\ O & O & O & \sigma_r(m)I_{m_r} \end{pmatrix}.\] Set \[E_i = P\begin{pmatrix} O & O & O\\ O & I_{m_i} & O\\ O & O & O \end{pmatrix}P^{-1},\] where \(I_{m_i}\) is in the \(i\)-th block.
Then, \[E_iE_j = \delta_{ij}E_i \quad (0\leq i,j\leq r),\] \[E_0 + E_1 + \cdots + E_r = I.\] Hence for all \(m\in M\), \[m = \sum_{i=0}^r \sigma_i(m)E_i \in \mathrm{Span}(E_0, \ldots, E_r).\] So, \[M \subseteq \mathrm{Span}(E_0, \ldots, E_r).\] Since \(E_0,\ldots, E_r\) are linearly independent, \(r\geq D\).
Show \(E_i\in M\).
Claim 1. For all distinct \(i, j\) \(\quad (0\leq i, j\leq D)\), there exists \(m\in M\) such that \(\sigma_i(m)\neq 0\), \(\sigma_j(m)=0\).
Pf of Claim 1. \(\sigma_i\neq \sigma_j\) implies that there exists \(m'\in M\) such that \(\sigma_i(m')\neq \sigma_j(m')\).
Set \(m = m'-\sigma_j(m')I\). Then, \[\begin{align} \sigma_j(m) & = \sigma_j(m') - \sigma_j(m') = 0,\\ \sigma_i(m) & = \sigma_i(m') - \sigma_j(m') \neq 0. \end{align}\]
Claim 2. \(E_i\in M\) \(\quad (0\leq i \leq D)\).
Pf of Claim 2. Fix a vertex \(x\in X\). For all \(j\neq i\), there exists \(m_j\in M\) such that \[\sigma_i(m_j)\neq 0, \quad \sigma_j(m_j) = 0, \quad i\neq j.\] Observe \[s = \sigma_i\left(\prod_{\ell\neq i}m_\ell\right) \neq 0.\] Set \[m^* = \left(\prod_{\ell\neq i}m_\ell\right) s^{-1}.\] Observe \[\sigma_i(m^*) =1, \quad \sigma_j(m^*) = 0, \quad \text{for all }j\neq i \quad (0\leq j\leq D).\] So \[P^{-1}m^*P = \begin{pmatrix} O & O & O\\ O & I_{m_i} & O\\ O & O & O \end{pmatrix}.\] We have \[E_i = m^*\in M.\] Now \(r = D\), \(M = \mathrm{Span}(E_0, \ldots, E_D)\) and \(E_0, \ldots, E_D\) is a basis for \(M\).
Observe \[P^{-1}E_iP = \begin{pmatrix} O & O & O\\ O & I_{m_i} & O\\ O & O & O \end{pmatrix}\] implies \[P^{-1}\overline{E_i}^\top P = \bar{P}^\top \overline{E_i}^\top \overline{P^{-1}}^\top = \begin{pmatrix} O & O & O\\ O & I_{m_i} & O\\ O & O & O \end{pmatrix}^\top = P^{-1}E_i P.\] Hence, \[\overline{E_i}^\top = E_i.\] \(E_0^\top, \ldots, E_D^\top\) are nonzero matrices satisfying \[E_i^\top E_j^\top = \delta_{ij}E_i^\top,\] \[E_0^\top + E_1^\top + \cdots + E_D^\top = I.\] Each \(E_i^\top\) is a linear combination of \(E_0, \ldots, E_D\) with coefficientss that are \(0\) or \(1\), and for no two \(E_i\)’s are coefficients of any \(E_j\) both \(1\)’s.
So, \(E_0^\top, \ldots, E_D^\top\) is a permutation of \(E_0, \ldots, E_D\).
Observe \(J = A_0 + \cdots + A_D\in M\).
The matrix \(|X|^{-1}J\) is an idempotent of rank \(1\).
So, without loss of generality we may assume that \[E_0 = \frac{1}{|X|}J.\] We have the assertions.
Define entry-wise product \(\circ\) on \(\mathrm{Mat}_X(\mathbb{C})\). \[A_i \circ A_j = \delta_{ij}A_i.\] So, \(M\) is closed under \(\circ\). \[E_i \circ E_j = \frac{1}{|X|}\sum_{h=0}^D q^h_{ij}E_h.\] The numbers \(q^h_{ij}\) is called Krein parameters of \(Y\).
Claim. \(q^h_{ij}\in \mathbb{R}\).
Pf. \[\begin{align} \frac{1}{|X|}\sum_{h=0}^D \overline{q^h_{ij}}E_h & = \frac{1}{|X|}\sum_{h=0}^D \overline{q^h_{ij}}\overline{E_h}^\top \\ & = (\overline{E_i\circ E_j})^\top \\ & = E_i\circ E_j \\ & = \frac{1}{|X|}\sum_{h=0}^D q^h_{ij}E_h. \end{align}\] Hence, \(q^h_{ij} = \overline{q^h_{ij}}\).
Observe \(A_0, \ldots, A_D\), \(E_0, \ldots, E_D\) are bases for \(M\). Hence, there exist \(p_i(j)\), \(q_i(j)\in \mathbb{C}\) such that \[\begin{align} A_i & = \sum_{j = 0}^D p_i(j)E_j\\ E_i & = \frac{1}{|X|}\sum_{j=0}^D q_i(j)A_j. \end{align}\] Taking transpose and conjugate we find, \[\begin{align} \overline{p_i(j)} & = p_i(j) = p_{i'}(\hat{j}) && (0\leq i,j\leq D)\\ \overline{q_i(j)} & = q_i(j) = q_{\hat{i}}({j}') && (0\leq i,j\leq D). \end{align}\]
Fix a vertex \(x\in X\). Define \[E^*_i \equiv E^*_i(x) \in \mathrm{Mat}_X(\mathbb{C})\] to be a diagonal matrix such that \[(E^*_i)_{xy} = \begin{cases} 1 & \text{if } (x,y)\in R_i\\ 0 & \text{if } (x,y)\not\in R_i \end{cases} \quad (0\leq i\leq D, y\in X.)\] Then, \[E^*_iE^*_j = \delta_{ij}E^*_i,\] \[E^*_0 + \cdots + E^*_D = I,\] \[(E^*_i)^\top = \overline{E^*_i} = E^*_i.\]
Definition 19.1 Dual Bose-Mesner algebra : \(M^* \equiv M^*(x)\) with respect to \(x\) is \[\mathrm{Span}(E^*_0, \ldots, E^*_D).\]
Define dual associate matrices \(A_0^*, \ldots, A^*_D\). Indeed \(A^*_i \equiv A^*_i(x)\in \mathrm{Mat}_X(\mathbb{C})\) is a diagonal matrix with \[(A_i^*)_{yy} = |X|(E_i)_{xy}\quad (y\in X).\] \(A^*_i\) is a diagonal matrix having the row \(x\) of \(E_i^*\) on the diagonal.
Observe \[\begin{align} A^*_i & = \sum_{j=0}^Dq_i(j)E^*_j \quad \left(E_i = \frac{1}{|X|}\sum_{j=0}^D q_i(j)A_j\right)\\ E^*_i & = \frac{1}{|X|}\sum_{j=0}^D p_i(j)A^*_j \quad \left(A_i = \sum_{j=0}^D p_i(j)E_j\right). \end{align}\] So, \(A^*_0, \ldots, A^*_D\) form a basis for \(M^*\).
Also, \[A^*_iE^*_j = q_i(j)E^*_j.\] \[\left(A^*_iE^*_j = \sum_{h=0}^D q_i(h)E^*_hE^*_j = q_i(j)E^*_j.\right)\] So, \(q_i(j)\) are dual eigenvalues of \(A^*_i\).
Observe, \[A^*_0 = I, \quad A^*_0 + \cdots + A^*_D = |X|E^*_0, \quad \overline{A^*_i} = A^*_{\hat{i}},\] \[A^*_iA^*_j = \sum_{h=0}^D q^h_{ij}A^*_h \quad (0\leq i,j\leq D).\]
HS MEMO
Proof. \[(A^*_0)_{yy} = |X|(E_0)_{xy} = (J)_{xy} = 1.\] \[A^*_0 + \cdots + A^*_D = \sum_{i=0}^D\sum_{j=0}^D q_i(j)E^*_j = |X|E^*_0.\] Note that \[I = E_0 + \cdots + E_D = \frac{1}{|X|}\sum_{i=0}^D\sum_{j=0}^D q_i(j)A_j.\] \[\sum_{i=0}^D q_i(j) = \delta_{j0}|X|.\] \[\overline{A^*_i} = \sum_{j=0}^D\overline{q_i(j)}E^*_j = \sum_{j=0}^D q_{\hat{i}}(j)E^*_j = A^*_{\hat{i}}.\] \[\begin{align} (A^*_iA^*_j)_{yy} & = |X|^2 (E_i)_{xy}(E_j)_{xy}\\ & = |X|^2(E_i\circ E_j)_{xy}\\ & = |X|\sum_{h=0}^D q^h_{ij}(E_h)_{xy}\\ & = \sum_{h=0}^D q^h_{ij}(A^*_h)_{yy}. \end{align}\]
The following statements will be proved after a couple of lemmas in the next lecture.
Lemma. Let \(Y = (X, \{R_i\}_{0\leq i\leq D})\) be a commutative scheme. Fix a vertex \(x\in X\), and set \(E^*\equiv E^*_i(x)\) and \(A^*_i \equiv A^*(x)\). Then the following hold.