Chapter 34 Modules of Endpoint One
Friday, April 23, 1993
Let \(\Gamma = (X,E)\) be distance-regular of diameter \(D\geq 3\).
Assume \(\Gamma\) is \(Q\)-polynomial with respect to \(E_0, E_1, \ldots, E_D\). Write \[\tilde{A}_i = A_0 + A_1 + \cdots + A_i \quad i\in \{0, 1\ldots, D\}.\] Fix a vertex \(x\in X\), write \(E^*_i \equiv E^*_i(x)\), \(M^* \equiv M^*(x)\), \(T\equiv T(x)\).
Pick \(0\neq v\in (E^*_1V)_{new}\). Set \(v^* = |X|E_1v\). We will show that \[Tv = Mv + M^*v^*.\] We need a preliminary lemma.
Lemma 34.1 With the atove notation, we have the following.
\((E^*_{D+1} = A_{D+1} = O)\).
Proof.
\[\begin{align} E^*_hv^* & = |X|E^*_hE_1v\\ & = E^*_h\left(\sum_{i=0}^D\theta^*A_i\right)v\\ & = E^*_h\left(\sum_{i=0}^D\theta^*_i(\tilde{A}_i-\tilde{A}_{i-1})\right)v\\ & = E^*_h\left(\sum_{i=0}^{D-1}(\theta^*_i-\theta^*_{i+1})\tilde{A}_i\right)v + E^*_h\theta^*_D\tilde{A}_D v\\ & = E^*_h\left(\sum_{i=0}^{D-1}(\theta^*_i-\theta^*_{i+1})(E^*_{i+1}A_iv - E^*_iA_{i+1}v)\right)\\ & = (\theta^*_{h-1}-\theta^*_h)E^*_hA_{h-1}v - (\theta^*_h-\theta^*_{h+1})E^*_hA_{h+1}v. \end{align}\]
\[0^-, 0^+, 1^-, 1^+, 2^-, 2^+, \ldots .\] \(0^-\): Trivial.
HS MEMO
\[\begin{align} \mathrm{LHS} & = (\theta^*_0 - \theta^*_1)E^*_0A_1v \\ & = (\theta^*_{-1}-\theta^*_1)E^*_0A_{-1}v - E^*_hv^* \quad \text{(by $(ii)$)}\\ & = -E^*_0v^*\\ & = \mathrm{RHS}. \end{align}\]
\(i^+\): using \((i)\) and \(i^-\). \[\begin{align} \mathrm{LHS} & = (\theta^*_i - \theta^*_{i+1})E^*_{i+1}A_iv \\ & = (\theta^*_{i}-\theta^*_{i+1})E^*_{i}A_{i+1}v + (\theta^*_i-\theta^*_{i+1})\tilde{A}_iv \quad \text{(by $(i)$)}\\ & = \left(\sum_{h=0}^{i-1}(\theta^*_h-\theta^*_{i})A_h\right)v - \left(\sum_{h=0}^i E^*_h\right)v^* + (\theta^*_i-\theta^*_{i+1})\left(\sum_{h=0}^i A_h\right)v \quad \text{(by $i^-$)}\\ & = \left(\sum_{h=0}^i (\theta^*_h-\theta^*_{i+1})A_h\right)v - \left(\sum_{h=0}^i E^*_h\right)v^*. \end{align}\]
\(i^-\): using \((ii)\) and \((i-1)^+\). \[\begin{align} \mathrm{LHS} & = (\theta^*_i - \theta^*_{i+1})E^*_{i}A_{i+1}v \\ & = (\theta^*_{i-1}-\theta^*_{i})E^*_{i}A_{i-1}v - E^*_iv^* \quad \text{(by $(ii)$)}\\ & = \left(\sum_{h=0}^{i-1}(\theta^*_h-\theta^*_{i})A_h\right)v - \left(\sum_{h=0}^{i-1} E^*_h\right)v^* - E^*_iv^*\\ & = \left(\sum_{h=0}^{i-1} (\theta^*_h-\theta^*_{i})A_h\right)v - \left(\sum_{h=0}^i E^*_h\right)v^*. \end{align}\]
HS MEMO
\[\begin{align} Mv + M^*v^* & \subseteq \mathrm{Span}\{\tilde{A}_hv, E^*_hv^*\mid 0\leq h\leq D\}\\ & \subseteq \mathrm{Span}\{E^*_hA_{h-1}v, E^*_{h-1}A_hv\mid 1\leq h\leq D\} \end{align}\] by \((i)\) and \((ii)\).
On the other hand, \[E^*hA_{h-1}v, E^*_{h-1}A_hv\in Mv + M^*v^* \quad i\in \{1, 2, \ldots, D\}\] by \((iii)\) and \((iv)\).
Lemma 34.2 With the notation of Lemma 34.1, assume \(0\neq v\in (E^*_1V)_{new}\) is an eigenvector for \(\tilde{A}:=E^*_1AE^*_1\). Then
Proof.
\[v^* \in Mv \subseteq Tv.\] Hence, \(M^*v^* \subseteq Tv\).
\(\subseteq\): It suffices to show that \(Mv + M^*v^*\) is a \(T\)-module (since it clearly contains \(v\)).
Show:
Proof of \((a)\). By the transpose of \((i)\) in Lemma 33.2, \[M^*ME^*_1 = ME^*_1 + M^*E_0E^*_1 + M^*E_1E^*_1.\] Since \(v\in E^*_1V\), \(E^*_1v = v\) and \[M^*Mv = Mv + M^*E_0v + M^*E_1v.\] But also \(E_0v = 0\) since \(v\) is orthogonal to the trivial \(T\)-module. Since \(E_1v = |X|^{-1}v^*\), \[M^*Mv = Mv + M^*v^*\] as desired.
HS MEMO
\[\begin{align} MM^*v & = MM^*E_1v^*\\ & = M^*E_1v^* + ME^*_0E_1v^* + ME^*_1E_1v^*\\ & = M^*v^* + ME^*_0v^* + ME^*_1v^*. \end{align}\] \(E^*_0v^*\in Tv\) and \(E^*_0Tv = 0\) as \(v\in (E^*_1V)_{new}\). So, \(E^*_0v^* = 0\). \[\begin{align} E^*_1v^* & = |X|E^*_1E_1v \\ & = |X|E^*_1E_1E^*_1v\\ & = ((\theta^*_0-\theta^*_2)E^*_1 + (\theta^*_1-\theta^*_2)E^*_1AE^*_1 + \theta^*_2|X|E^*_1E_0E^*_1)v\\ & = (\theta^*_0-\theta^*_2)v + (\theta^*_1-\theta^*_2)E^*_1AE^*_1v + \theta^*_2|X|E^*_1E_0v\\ & \in \mathrm{Span}\{v\}, \end{align}\] as \(E_0v = 0\), and \(v\) is an eigenvector of \(E^*_1AE^*_1\).
\(\star\) \(v\in (E^*_1V)_{new}\). If \(v\) is an eigenvector of \(E^*_1AE^*_1\), \[E^*_1v^* \in \mathrm{Span}\{v\}.\]
\[\begin{align} Tv & = Mv + M^*v^*\\ & = \mathrm{Span}\{E^*_iA_{i-1}v, E^*_{i-1}A_iv\mid 1\leq i\leq D\}\\ & = \mathrm{Span}\{v^+_i, v^-_{i-1}\mid 1\leq i\leq D\}\\ & = \mathrm{Span}\{v^+_1, v^+_2, \ldots, v^*_D, v_0^-, \ldots, v^-_{D-1}\} \end{align}\] by Lemma 34.1 \((v)\).
But \(v^-_0 = E^*_0A_1v = 0\) since \(v\in (E^*_1V)_{new}\), and \(v^-_1\in \mathrm{Span}\{v^+_1\}\).
Indeed, \[v^-_1 = E^*_1A_2v = (-1-a_0(Tv))v^+_1.\] where \(a_0(Tv)\) is the eigenvalue of \(v\) associated with \(\tilde{A}\).
To see this, observe \[\begin{align} 0 & = \tilde{J}v\\ & = E^*_1\left(\sum_{i=0}^DA_i\right)E^*_1v\\ & = E^*_1\left(\sum_{i=0}^2A_i\right)E^*_1v\\ & = v + a_0(Tv)v + v^-_1. \end{align}\] Therefore, \[Tv = \mathrm{Span}\{v^+_1, v^+_2, \ldots, v^+_D, v_2^-, \ldots, v^-_{D-1}\}.\]
\[E^*_1Tv = E^*_1W_1 + E^*_1W_2\] has dimension \(1\) by \((iii)\). Assume \(v\in E^*_1W_1\). Then \(Tv \subseteq W_1\), a contradiction.
Lemma 34.3 With the notation of Lemma 34.1, assume \(0\neq v\in (E^*_1V)_{new}\) is an eigenvector for \(\tilde{A}:=E^*_1AE^*_1\).
\[W = Tv'\] for some \(0\neq v'\in (E^*_1V)_{new}\) that is an eigenvector of \(\tilde{A}\).
Then \(Tv\), \(Tv'\) are isomorphic \(T\)-module if and only if \(a_0(Tv) = a_0(Tv')\).
\[\tilde{J}, E^*_1, \tilde{A}, \tilde{A}^2, \ldots, \tilde{A}^{\ell-1},\] where \(\ell\) is the number of mutually nonisomorphic \(T\)-modules with endpoint \(1\).
Proof.
HS MEMO
Originally, the statement was \(Tv\) is thin if and only if \(M^*v^* = Mv\). This is not the case in general. Suppose \(\Gamma\) is thin. Let \(W\) be an irreducible \(T\)-module of endpoint \(1\). Then, that \(W \cap E^*_1V \ni v \neq 0\) implies \(v^*\in W\cap E_1V\) gives one to one and \(k\leq m\).
However, by ‘Distance-Regular Graphs’ (A. E. Brouwer 1989),
\(J(v,d)\): \(v\geq 2d\) \[\begin{align} b_j = (d-j)(v-d-j), & \quad c_j = j^2,\\ \theta_j = (d-j)(v-d-j)-j, & \quad m_j = \binom{v}{j}-\binom{v}{j-1}. \end{align}\] In particular, \[k = b_0 = d(v-d) > m_1 = v-1 \quad \text{ if }d\geq 2,\] and \(J(v,d)\) is thin.
So \(|X|E_1v = v^*\) may be \(0\) sometimes. But as \(Tv\) is dual thin of diameter at least \(D-2\). The dual endpint \(r^*\leq 2\), so in that case, \(E_2v \neq 0\). Hence, if \(D\geq 3\), \(E_2v\neq 0\) always.
HS MEMO
Now assume \(M^*v^*\subseteq Mv = Tv\). Then \[Mv = \{f(A)v\mid f(\lambda)\in \mathbb{C}[\lambda]\}.\] So, \[E_iTv = E_iMv \in \mathrm{Span}(E_iv).\] Hence, \(Tv\) is dual thin.
Now we can construct a basis, \(0\neq w^*_0\in E_{r^*}W\), where \(r^*\) is the dual endpoint, and \[w^*_0, w^*_1, \ldots, w^*_d\in W = Tv,\] where \(w^*_i = E^*_{r^*+i}{A^*_1}^iw^*_0\). \[A^*_1w^*_i = w^*_{i+1}+ a_i^*w^*_i + x^*_iw^*_{i-1},\] and \(w^*_i = p^*_i(A^*)w^*_0\). \[E^*_{r^*+i}A^*_1E_{r^*+i}|_{E_{r^*+i}W} = a^*_i\cdot 1|_{E_{r^*+i}W},\] \[E^*_{r^*+i-1}A^*_1E_{r^*+i}A^*E_{r^*+i-1}|_{E_{r^*+i-1}W} = x^*_i\cdot 1|_{E_{r^*+i-1}W}.\] See Lemma 9.1, and Lemma 22.2.
From above, \(Tv = M^*w^*_0\). So, \[E^*_iTv = E^*_iM^*w^*_0 \in \mathrm{Span}\{E^*_iw^*_0\}.\] Thus, \(Tv\) is thin.
Need to write down the dual at least for Lemma 9.1, Corollary 9.1.
Since \(W\) is irreducible, \(Tv' = W\).
Recall \(\sigma s = s\sigma\) for all \(s\in T\). \[\mathrm{Span}\{\sigma v\} = \sigma E^*_1Tv = E^*_1\sigma Tv = E^*_1Tv' = \mathrm{Span}\{v'\}.\] Hence, \[a_0(Tv)\sigma v = \sigma(a_0(Tv)v) = \sigma \tilde{A}v = \tilde{A}\sigma v = a_0(Tv')\sigma v.\] Since \(\sigma v \neq 0\), \(a_0(Tv) = a_0(Tv')\).
Now suppose \(a_0(Tv)=a_0(Tv')\). Show \[\sigma: Tv \to Tv' \quad (sv \mapsto sv') \quad (s\in T)\] is an isomorphism of \(T\)-modules.
Pick \(s\in T\). Require \(sv = 0\) if and only if \(sv' = 0\).
Without loss of generality , \(s\in TE^*_1\), since \(v, v'\in E^*_1V\).
Now \(0 = sv\) if and only if \[0 = \|sv\|^2 = \bar{v}^\top \bar{s}^\top sv.\] But, \(\bar{s}^\top s\in E^*_1TE^*_1\).
Hence, by Lemma 33.4 \((iii)\), \[\bar{s}^\top s = \alpha \tilde{J} + p(\tilde{A})\] for some \(\alpha\in \mathbb{C}\) and \(p(\lambda)\in \mathbb{C}[\lambda]\).
Thus, using the fact that \(\tilde{J}v = 0\), \[0 = \|sv\|^2 = \bar{v}^\top (\alpha\tilde{J}+p(\tilde{A}))v = \|v\|^2 p(a_0(Tv))\] if and only if \(0 = p(a_0(Tv))\).
Replacing \(v\) by \(v'\), we have \[\begin{align} 0 = sv' & \leftrightarrow 0 = p(a_0(Tv'))\\ & \leftrightarrow 0 = p(a_0(Tv))\\ & \leftrightarrow 0 = sv \end{align}\] as desired.
\[\begin{align} \ell & = \text{the number of mutually nonisomorphic $T$-modules with endpoint $1$}\\ & = \text{the number of distinct eigenvalues of $\tilde{A}:(E^*_1V)_{new}\to (E^*_1V)_{new}$}\\ & = \text{the degree of minimal polynomial of $\tilde{A}:(E^*_1V)_{new}\to (E^*_1V)_{new}$.} \end{align}\]
Claim 1. \(\tilde{J}. E^*_1, \tilde{A}, \ldots, \tilde{A}^{\ell-1}\) are linearly independent.
Proof of Claim 1. Suppose not. Then \[\alpha \tilde{J} + p(\tilde{A}) = O\] for some \(\alpha\in \mathrm{C}\) and \(p(\lambda)\in \mathbb{C}[\lambda]\) with \(\deg p \leq \ell -1\).
But \(\tilde{J}|_{(E^*_1V)_{new}} = O\) impiles \(p(\tilde{A})|_{(E^*_1V)_{new}}=O\).
Since \[\deg p < \text{the degree of minimal polynomial of $\tilde{A}|_{(E^*_1V)_{new}}$},\] we find \(p\) is identically \(0\).
Then \(\alpha\) is identically \(0\) also.
Claim 2. \(\tilde{J}. E^*_1, \tilde{A}, \ldots, \tilde{A}^{\ell-1}\) span \(E^*_iTE^*_i\).
Proof of Claim 2. It needs to show \[\begin{equation} \tilde{J}. E^*_1, \tilde{A}, \ldots, \tilde{A}^{\ell} \; \text{are linearly dependent}.\tag{34.1} \end{equation}\] Let \(m\) denote the minimal polynomial of \(\tilde{A}|_{(E^*_1V)_{new}}\). So, \[m(\tilde{A}|_{(E^*_1V)_{new}}) = 0.\]
Observe that \[E^*_1V = (E^*_1V)_{new} + \mathrm{Span}\{A\hat{x}\}.\] (direct sum of \(E^*_1TE^*_1\)-modules.) \[m(\tilde{A})A\hat{x} = f\cdot A\hat{x}\quad \text{for some }f\in \mathbb{C}.\]
On the other hand, \[\tilde{J}A\hat{x} = kA\hat{x} \quad (k: \text{valency of }\Gamma).\] Therefore, \[m(\tilde{A}) - \frac{f}{k}\tilde{J} = O,\] and (34.1) holds.