Chapter 9 Thin \(T\)-Module, I

Monday, February 8, 1993

Let \(\Gamma = (X, E)\) be any graph.

\(M\): Bose-Mesner algebra over \(K/\mathbb{C}\) generated by the adjacency matrix \(A\).

\(\quad M = \mathrm{Span}(E_0, \ldots, E_R)\).

\(M\) acts on the standard module \(V = \mathbb{C}^{|X|}\).

Fix \(x\in X\), let \(D \equiv D(x)\) be the \(x\)-diameter, and \(k = k(x)\) be the valency of \(x\).

Definition 9.1 Pick \(x\in X\) and write \(E_i^* \equiv E_i^*(x)\) and \(T \equiv T(x)\).

Let \(W\) be an irreducible thin \(T\)-module with endpoint \(r\), diameter \(d\).

Let \(a_i = a_i(W)\in \mathbb{C}\) satisfying \[E_{r+i}^*A{E^*_{r+i}|}_{E_{r+i}^*W} = a_i1|_{E_{r+i}^*} \quad (0\leq i\leq d).\] Let \(x_i = x_i(W)\in \mathbb{C}\) satisfying \[E_{r+i-1}^*A{E^*_{r+i}AE^*_{r+i-1}|}_{E_{r+i-1}^*W} = x_i1|_{E_{r+i-1}^*} \quad (0\leq i\leq d).\]

Lemma 9.1 With above notation, the following hold.

\((i)\) \(a_i\in \mathbb{R} \quad (0\leq i\leq d)\).

\((ii)\) \(x_i\in \mathbb{R}^{>0} \quad (0\leq i\leq d)\).

\((iii)\) Pick \(0\neq w_0\in E^*_rW\). Set \(w_i = E^*_{r+i}A^iw_0\) for all \(i\). Then

\(\quad (iiia)\) \(w_0, w_1, \ldots, w_d\) is a basis for \(W\), \(w_{-1} = w_{d+1} = 0\).
\(\quad (iiib)\) \(Aw_i = w_{i+1} + a_iw_{i} + x_iw_{i-1} \quad (0\leq i\leq d)\).

\((iv)\) Define \(p_0, p_1, \ldots, p_{d+1}\in \mathbb{R}[\lambda]\) by

\[p_0 = 1, \quad \lambda p_i = p_{i+1} + a_i p_i + x_i p_{i-1} \quad (0\leq i\leq d),\quad p_{-1} = 0.\]

\(\quad (iva)\) \(p_i(A)w_0 = w_i, \quad (0\leq i\leq d+1)\).
\(\quad (ivb)\) \(p_{d+1}\) is the minimal polynomial of \(A|_W\).

Proof. \((i)\) \(a_i\) is an eigenvalue of a real symmetric matrix \(E_{r+i}^*AE^*_{r+i}\).

\((ii)\) \(x_i\) is an eigenvalue of a real symmetrix matrix \(B^\top B\), where \[B = E^*_{r+i}AE^*_{r+i-1}.\] Hence, \(x_i\in \mathbb{R}\).

Since \(B^\top B\) is positive semidefinite, \[x_i \geq 0.\] Pf. If \(B^\top Bv = \sigma v\) for some \(\sigma \in \mathbb{R}\), \(v\in \mathbb{R}^m \setminus \{0\}\), then \[0\leq \|Bv\|^2 = v^\top B^\top Bv = \sigma v^\top v = \sigma \|v\|^2, \quad \|v\|^2 >0.\] Hence, \(\sigma \geq 0\).

Moreover, \(x_i\neq 0\) by Lemma 4.1 \((iv)\).

\((iiia)\) Observe \[w_i = E^*_{r+i}AE^*_{r+i-1}w_{i-1} \quad (1\leq i \leq d).\] So \(w_i \neq 0 \quad (0\leq i \leq d)\) by Lemma 4.1 \((iv)\).

Hence, \[W = \mathrm{Span}(w_0, \ldots, w_d)\] by Lemma 4.1. \((iii)\).

\((iiib)\) We have that \[\begin{align} Aw_i & = E^*_{r+i+1}Aw_i + E_{r+i}^*Aw_i + E^*_{r+i-1}Aw_i\\ & = w_{i+1} + E^*_{r+i}AE^*_{r+i}w_i + E^*_{r+i-1}AE^*_{r+i}AE^*_{r+i-1}w_{i-1}\\ & = w_{i+1} + a_iw_{i} + x_iw_{i-1}. \end{align}\]

\((iva)\) Clear for \(i=0\). Assume it is valid for \(0, \ldots, i\). \[p_{i+1}(A)w_0 = (A-a_iI)w_i - x_iw_{i-1} = w_{i+1}.\]

\((ivb)\) By definition, \[p_{d+1}(A)w_0 = 0.\] Moreover, \(p_{d+1}(A)W = 0\) because of the following.

For every \(w\in W\), write \[\begin{align} w & = \sum_{i=0}^d \alpha_i w_i \\ & = \sum_{i=0}^d \alpha_i p_i(A)w_0 && \text{for some }\alpha_i\in\mathbb{C}\\ & = p(A)w_0 && \text{for some }p\in \mathbb{C}[\lambda]. \end{align}\] Hence, \[\begin{align} p_{d+1}(A)w & = p_{d+1}(A)p(A)w_0\\ & = p(A)p_{d+1}(A)w_0\\ & = 0. \end{align}\]

Note that \(p_{d+1}\) is the minimal polynomial.

Pf. Suppose \(q(A)W = 0\) for some \(0\neq q\in \mathbb{C}[\lambda]\) with \(\deg q < \deg p_{d+1} = d+1\). Then, \[q = \sum_{i=0}^d\beta_ip_i \quad \text{for some }\beta_i\in \mathbb{C}.\] We have, \[0 = q(A)w_0 = \sum_{i=0}^d \beta_iw_i.\] Hence \(\beta_0 = \cdots = \beta_d = 0\) by \((iiia)\). Thus \(q = 0\), and a contradiction.

Corollary 9.1 Let \(\Gamma\), \(W\), \(r\), \(d\) be as above. Then

\((i)\) \(W\) is dual thin, that is,

\[\dim E_iW \leq 1 \quad (1\leq i \leq d).\]

\((ii)\) \(d = |\{i \mid E_iW \neq 0\}| - 1.\)

Proof. \((i)\) Set as in Lemma 9.1, \[w_i = p_i(A)w_0\in E^*_{r+i}W.\] Then \(w_0, w_1, \ldots, w_d\) is a basis for \(W\). We have \[W = Mw_0.\] So, \[E_iW = E_iMw_0 = \mathrm{Span}(E_iw_0).\] Thus, \[\dim E_iW = \begin{cases}1 & \text{if } E_iw_0\neq 0,\\ 0 & \text{if }E_iw_0 = 0.\end{cases}\] In particular, \[\dim E^*_iW \leq 1.\]

\((ii)\) Immediate as \[\dim W = d+1.\] This proves the lemma.

Lemma 9.2 Given an irreducible \(T(x)\)-module \(W\) with endpoint \(r = r(W)\), diameter \(d = d(W)\). Write \[x_i = x_i(W) \; (0\leq i\leq d), \quad w_i = p_i(A)w_0\in E^*_{r+i}W \; (0\leq i\leq d), \quad 0\neq w_0 \in E^*_rW.\] Then, \[\frac{\|w_i\|^2}{\|w_0\|^2} = x_1x_2\cdots x_i \quad (1\leq i\leq d).\]

Proof. It suffices to show that \[\|w_i\|^2 = x_i\|w_{i-1}\|^2 \quad (1\leq i\leq d).\] Recall by Lemma 9.1 \((iiib)\) that \[Aw_j = w_{j+1} + a_jw_j + x_jw_{j-1} \quad (0\leq j\leq d), \quad w_{-1} = w_{d+1} = 0.\] Now observe, \[\begin{align} \langle w_{i-1}, Aw_i\rangle & = \langle w_{i-1}, w_{i+1}+ a_iw_i + x_iw_{i-1}\rangle\\ & = \overline{x_i}\|w_{i-1}\|^2\\ & = x_i\|w_{i-1}\|^2. \end{align}\] by Lemma 9.1 \((ii)\). Also, \[\begin{align} \langle w_{i-1}, Aw_i\rangle & = \langle Aw_{i-1}, w_i\rangle \quad (\text{since}\; \bar{A}^\top = A)\\ & = \langle w_i + a_{i-1}w_{i-1} + x_{i-1}w_{i-2}, w_i\rangle\\ & = \|w_i\|^2. \end{align}\] This proves the lemma.

Definition 9.2 Let \(W\) be an irreducible thin \(T(x)\) module with endpoint \(r\), \(E^*_i \equiv E_i^*(x)\).

The measure \(m = m_W\) is the function \[m: \mathbb{R} \to \mathbb{R}\] such that \[ m(\theta) = \begin{cases}\frac{\|E_iw\|^2}{\|w\|^2} & \text{where } 0\neq w \in E^*_rW\\ & \text{ if $\theta = \theta_i$ is an eigenvalue for $\Gamma$,}\\ 0 & \text{if $\theta$ is not an eigenvalue for $\Gamma$.} \end{cases}\]