Chapter 40 Structure of 1-Thin DRG

Friday, May 7, 1993

Lemma 40.1 With the above notation, let \(W\) denota a thin irreducible \(T\)-module of endpoint \(0\) or \(1\). Pick \(0\neq v\in E^*_1V\). Then the following hold.

\((i)\) Eigenvalue for \(\tilde{J}\) is \(0\) if \(r(W)=1\), and \(k\) if \(r(W) = 0\).

\((ii)\) Eigenvalue for \(E^*_1\) is \(1\) if \(r(W)=1\), and \(1\) if \(r(W) = 0\).

\((iii)\) Eigenvalue for \(\tilde{A}\) is \(a_0(W)\) if \(r(W)=1\), and \(a_1\) if \(r(W) = 0\).

\((iv)\) Eigenvalue for \(A^+\) is \(a^+(W) = \frac{\gamma_1}{c_2}-1-\Psi\) if \(r(W)=1\), and \(\frac{a_2}{c_2}-\Psi\) if \(r(W) = 0\).

\((v)\) Eigenvalue for \(A^-\) is \(a^-(W) = a_0(W)\left(\frac{\gamma_1}{c_2}-1-2\Psi\right)-\frac{a_2}{c_2}\) if \(r(W)=1\),

where \[\gamma_0 = 1+a_0(W), \text{ and } \gamma_1 = \frac{c_2b_2\gamma_0}{b_1+\gamma_0(a_1+2-c_2)-\gamma^2_0}\] as in Theorem 14.2. (The eigenvalue for \(A^-\) on \(v\) will be discussed later in this lecture.)

Proof.

\((i)-(iii)\) Clear.

\((iv)\) We have

\[\begin{align} A^+ & = R^{-1}E^*_2A_2E^*_1 - \Psi E^*_1,\\ A_2 & = \frac{A^2-a_1A - kI}{c_2},\\ E^*_2A_2E^*_1 & = E^*_2\left(\frac{A^2-a_1A - kI}{c_2}\right)E^*_1\\ & = \frac{1}{c_2}(RF + FR - a_1R)E^*_1. \end{align}\]

If \(r(W) = 1\), \[\begin{align} A^+v & = \frac{1}{c_2}(R^{-1}RFv + R^{-1}FRv - a_1R^{-1}Rv) - \Psi v\\ & = \frac{1}{c_2}(R^{-1}Ra_0(W)v + R^{-1}a_1(W)Rv - a_1R^{-1}Rv) - \Psi v\\ & = \frac{1}{c_2}\left(a_0(W) + a_1(W) - a_1) - \Psi\right)v. \end{align}\] But, \[a_1(W) = \gamma_1 - \gamma_0 + a_1 + 1 - c_2, \quad \gamma_0 = a_0(W) + 1\] by Theorem 16.1.

So, \[\begin{align} A^+v & = \left(\frac{1}{c_2}(a_0(W) + \gamma_1 - \gamma_0 + a_1 + 1 - c_2 - a_1) - \Psi)\right)v\\ & = \left(\frac{\gamma_1}{c_2}-1-\Psi\right)v. \end{align}\]

If \(r(W) = 0\), \[\begin{align} A^+v & = \frac{1}{c_2}(R^{-1}RFv + R^{-1}FRv - a_1R^{-1}Rv) - \Psi v\\ & = \frac{1}{c_2}(R^{-1}Ra_1v + R^{-1}a_2Rv - a_1R^{-1}Rv) - \Psi v\\ & = \left(\frac{a_2}{c_2}-\Psi\right)v. \end{align}\]

\((v)\) Immediate from \((iv)\), and

\[A^- = \tilde{A}A^+ - \left(\frac{a_2}{c_2} - \Psi\right)E^*_1 + \Psi(\tilde{J}-\tilde{A}-E^*_1).\]

HS MEMO

If \(r(W) =1\), \[\begin{align} A^-v & = \left(a_0(W)\left(\frac{\gamma_1}{c_2}-1-\Psi\right) - \left(\frac{c_2}{a_2}-\Psi\right) + \Psi(-a_0(W)-1)\right) v\\ & = \left(a_0(W)\left(\frac{\gamma_1}{c_2}-1-2\Psi\right) - \frac{c_2}{a_2}\right)v. \end{align}\] If \(r(W) = 0\), \[\begin{align} A^-v & = \left(a_1\left(\frac{a_2}{c_2}-\Psi\right) - \left(\frac{a_2}{c_2}-\Psi\right) + \Psi(k-a_1-1)\right) v\\ & = \left((a_1-1)\frac{a_2}{c_2}+ (k-2a_1)\Psi\right)v. \end{align}\]

This completes the proof.

Let \(W_1, W_2, W_3, W_4\) denote \(4\) possible isomorphism classes of \(T\)-modules of endpoint \(1\). Then \(a_0(W_1), a_0(W_2), a_0(W_3), a_0(W_4)\) are roots of a fourth degree polynomial whose coefficients are determined from intersection numbers of \(\Gamma\).

So, \(a_0(W_1), a_0(W_2), a_0(W_3), a_0(W_4)\) are determined by intersection numbers.

Let \(\widetilde{m_i}\) denote the multiplicity of \(W_i\) \((1\leq i\leq 4\)), which is equal to the multiplicity of \(a_0(W)\) as eigenvalue \(1\) of \(\tilde{A}|_{(E^*_1V)_{new}}\).

Lemma 40.2 With the above notation, we have the following.

\((i)\) \(\tilde{m}_1\), \(\tilde{m}_2\), \(\tilde{m}_3\), \(\tilde{m}_4\) are determined from intersection numbers and \(\Psi\).

\((ii)\) \(\tilde{m}_i\) is independent of vertex \(x\). \((1\leq i\leq 4)\).

\((iii)\) \(\ell:= \dim E^*_1TE^*_1\) is independent of \(x\).

Proof.

\((i)\) Let \(e_i\in E^*_1TE^*_1\) \((1\leq i\leq 4)\) denote the orthogonal projection on to the maximal eigenspace of \((E^*_1V)_{new}\) corresponding to \(\lambda_i\). (\(e=0\) if and only if \(\lambda_i\) does not appear.) Set

\[e_0 = \frac{1}{k}\tilde{J}.\] Then eigenvalues for each \(e_1, e_1, e_3, e_4\) are as follows. \[\begin{array}{|c|ccccc|} \hline \text{} & e_0 & e_1 & e_2 & e_3 & e_4\\ \hline \tilde{J} & k & 0 & 0 & 0 & 0\\ E^*_1 & 1 & 1 & 1 & 1 & 1 \\ \tilde{A} & a_1 & a_0(W_1) & a_0(W_2) & a_0(W_3) & a_0(W_4)\\ A^+ & \frac{a_2}{c_2}-\Psi & a^+(W_1) & a^+(W_2) & a^+(W_3) & a^+(W_4)\\ A^- & \star & a^-(W_1) & a^-(W_2) & a^-(W_3) & a^-(W_4)\\ \hline \end{array}\]

Observe that \(e^2_i = e_i\), \(\mathrm{trace} \:e_i = \mathrm{rank}\:e_i = \tilde{m}_i\) \((1\leq i\leq 4)\), and \(\mathrm{trace}\: e_0 = \mathrm{rank}\:e_0 = 1\).

By taking the trace of \(\tilde{J}, E^*_1, \tilde{A}, A^+, A^-\), we have \[\begin{align} k & = k,\\ k & = 1 + \tilde{m}_1 + \tilde{m}_2 + \tilde{m}_3 + \tilde{m}_4,\\ 0 & = a_1 + a_0(W_1)\tilde{m}_1 + a_0(W_2)\tilde{m}_2 + a_0(W_3)\tilde{m}_3 + a_0(W_4)\tilde{m}_4,\\ 0 & = \left(\frac{a_2}{c_2}-\Psi\right) + a^+(W_1)\tilde{m}_1 + a^+(W_2)\tilde{m}_2 + a^+(W_3)\tilde{m}_3 + a^+(W_4)\tilde{m}_4,\\ 0 & = \left(\star\right) + a^-(W_1)\tilde{m}_1 + a^-(W_2)\tilde{m}_2 + a^-(W_3)\tilde{m}_3 + a^-(W_4)\tilde{m}_4. \end{align}\]

The coefficient matrix for \(\tilde{m}_1,\tilde{m}_2, \tilde{m}_3,\tilde{m}_4\) is nonsingular (this is what you need to check and show).

HS MEMO

Complutation is not completed.

\((ii)\) \(\Psi\) is independent of base vertex \(x\).

\((iii)\) We have

\[\begin{align} \dim E^*_1TE^*_1 & = |\{i\mid 1\leq i\leq 4, \; e_i\neq 0\}| + 1\\ & = |\{i\mid 1\leq i\leq 4, \; \tilde{m}_i\neq 0\}| + 1. \end{align}\]

This completes the proof of the lemma.

Let \(\Gamma = (X, E)\) be thin distance regular of diameter \(D\geq 5\), and \(Q\)-polynomial with respect to \(E_0, E_1, \ldots, E_D\).

Fix vertices \(x,y\in X\) with \(\partial(x,y) = 1\), \[E^*_{ij} \equiv E^*_i(x)E^*_j(y), \quad \delta_{ij} = E^*_{ij}\delta.\] We saw \[T(x)\hat{y} = T(y)\hat{x}.\] Hence, \[H := T(x)\hat{y} = T(y)\hat{x}\] is a \(T(x,y)\) module. \(T(x,y) \subseteq \mathrm{Mat}_X(\mathbb{C})\) is generated by \(M\), \(M^*(x)\), \(M^*(y)\).

Lemma 40.3 With the above notation, we have the following.

\((i)\) \(E^*_{i,i+1}H = \mathrm{Span}(\delta_{i,i+1}) \quad (0\leq i\leq D-1)\).

\((ii)\) \(E^*_{i+1,i}H = \mathrm{Span}(\delta_{i+1,i}) \quad (0\leq i\leq D-1)\).

\((iii)\) \(E^*_{i,i}H = \ell-2 \leq 3 \quad (1\leq i\leq D-1)\).

Proof.

\((i)\) \(\supseteq\): We have

\[\delta_{i,i+1} = E^*_iA_{i+1}\hat{y} \in T(x)\hat{y} = H.\]

\(\subseteq\): Pick \(h\in E^*_{i,i+1}H\). Then \(h = R^{i-1}v\), where \(v = (R^{-1})^{i-1}h\in E^*_1V\).

So, \(v\in \mathrm{Span}(\delta_{12}, \delta_{11}, \delta_{10}, \delta^+_{11}, \delta^-_{11})\).

HS MEMO

\[\begin{align} v &\in E^*_1V \cap T(x)\hat{y}\\ & = E^*_1T(x)E^*_1\hat{y}\\ & = \mathrm{Span}(\tilde{J}, E^*_1, \tilde{A}, A^+, A^-)\hat{y}\\ & = \mathrm{Span}(\delta_{10}+ \delta_{11}+ \delta_{12}, \delta_{10}, \delta_{11}, \delta^+_{11}, \delta^-_{11})\\ & = \mathrm{Span}(\delta_{10}, \delta_{11}, \delta_{12}, \delta^+_{11}, \delta^-_{11}). \end{align}\]

Hence, there exists \(\alpha \in \mathbb{C}\) such that \[v-\alpha \delta_{12} \in \mathrm{Span}(\delta_{10}, \delta_{11}, \delta^+_{11}, \delta^-_{11}) = E^*_{11}H + E^*_{10}H.\] So, \[v-\alpha(\delta_{12} + \delta_{11} + \delta_{10}) \in E^*_{11}H + E^*_{10}H.\] \[\begin{align} E^*_{ii}H + E^*_{i,i-1}H & \ni R^{i-1}(v-\alpha(\delta_{12}+\delta_{11}+\delta_{10}))\\ & = h - \alpha'(\delta_{i,i+1}+\delta_{ii}+\delta_{i,i-1}). \end{align}\] Hence, \[h-\alpha'\delta_{i,i+1}\in (E^*_{ii}H + E^*_{i,i-1}H)\cap E^*_{i,i+1}H.\] Thus, \[h = \alpha'\delta_{i,i+1} \in \mathrm{Span}(\delta_{i,i+1}).\]

\((ii)\) By symmetry, we have the assertion.

\((iii)\) \(E^*_i H = E^*_{i,i+1}H + E^*_{i,i}H + E^*_{i,i-1}H\), and \(\dim E^*_iH = \ell\), \(\dim E^*_{i,i+1}H =1\), and \(\dim E^*_{i,i-1}H = 1\).

Hence, \(\dim E^*_{i,i}H = \ell -2\).

HS MEMO

Since \(H = T(x)\hat{y} \subseteq T(x)E^*_{1}(x)V\), and \[(R^{-1})^{i-1}: E^*_iH \to E^*_1H\] is one-to-one and onto if \(i\leq D\).

Theorem 40.1 Let \(\Gamma = (X, E)\) be thin distance regular of diameter \(D\geq 5\), and \(Q\)-polynomial with respect to \(E_0, E_1, \ldots, E_D\).

Pick \(i\) \((2\leq i\leq D)\), and pick \(x, y, z\in X\) such that \(\partial(x,y) =1\), \(\partial(y,z) = i-1\), \(\partial(x,z) = i\).

Then, \[z_i = |\{w\mid w\in W, \partial(x,w) =1, \partial(y,w)=1, \partial(z,w)=i-1\}|\] is independent of \(x, y, z\).

Proof. Observe that \(z_i\) is the \(zx\) entry in \[\Delta = E^*_{i-1}(y)A_{i-1}E^*_1(y)AE^*_1(y)\] as \[\Delta \hat{x} = \sum_{z\in X, \partial(x,z)=i, \partial(y,z)=i-1}z_i(x,y,z)\hat{z}.\] Hence, \(z_i(x,y,z)\) is independent of \(z\).

So, \(z_i(x,y,z)\) is determined by intersection numbers and \(\Psi = \Psi(x,y)\), which is independent of \(x, y\) as well.