Chapter 38 An Injection from E₁₁^* to E₂₂^*

Monday, May 3, 1993

Lemma 38.1 Let \(\Gamma = (X, E)\) be a thin distance-regular graph of diameter \(D\geq 5\), and \(Q\)-polynomial with respect to \(E_0, E_1, \ldots, E_D\). Pick vertices \(x,y\in X\) such that \(\partial(x,y) = 1\), and write \(E^*_{ij}:=E^*_i(x)E^*_j(y)\) for \(i,j\in \{0, 1, \ldots, D\}\). Then the following hold.

\((i)\) \(E^*_{22}AE^*_{11}: E^*_{11}V \to E^*_{22}V\) is one-to-one.

\((ii)\) For every \(z\in X\) such that \(\partial(x,z) = \partial(y,z) = 1\), there is \(w\in X\) such that

\[\partial(w,x) = \partial(w,y) = 2, \; \partial(w,z)=1.\]

Proof.

\((i)\) Write \(E^*_i\equiv E^*_i(x)\), \(R \equiv R(x)\), \(F \equiv F(x)\), \(L \equiv L(x)\), and \(T \equiv T(x)\).

Suppose there exists \[\begin{equation} 0\neq v\in E^*_{11}V \; \text{ such that }\; E^*_{22}AE^*_{11}v = 0. \tag{38.1} \end{equation}\]

Claim 1. \(E^*_{34}A^2E^*_{12}AE^*_{11}v \neq 0\).

Proof of Claim 1. Recall by Lemma 32.1 \((ii)\), \((3 \leq 5-2 \leq D-2t)\), \[R(y)^3: E^*_1(y)V \to E^*_4(y)V\] is one-to-one.

Since \(v\in E^*_1(y)V\), we find \[\begin{align} 0 & \neq R^3(y)v\\ & = E^*_4(y)A^3E^*_1(y)v\\ & = E^*_4(y)A^2E^*_2(y)AE^*_{11}v\\ & = E^*_4(y)A^2\left(\sum_{h=0}^D E^*_{h,2}\right)AE^*_{11}v\\ & = E^*_4(y)A^2(E^*_{12} + E^*_{22})AE^*_{11}v\\ & = E^*_4(y)A^2E^*_{12}AE^*_{11}v \\ & = E^*_{34}(y)A^2E^*_{12}AE^*_{11}v, \end{align}\] by (38.1). This proves the claim.

By Theorem 30.1 \((i)\), \[\begin{equation} 0 = (g_3^-R^2F + RFR + g^+_3FR^2 - \gamma R^2)E^*_1. \tag{38.2} \end{equation}\]

HS MEMO

Theorem 30.1 \((i)\) states \[(g_i^-FL^2 + LFL + g^+_iL^2F - \gamma L^2)E^*_i = O \;\text{ for }i\in \{2, \ldots, D\}.\] For \(i = 3\), \[E^*_1(g^-_3 FL^2 + LFL + g^+_3 L^2F - \gamma L^2)E^*_3 = O.\] Taking the transpose, we have

\[E^*_3(g^{-}_3 R^2 F + RFR + g^{+}_3 FR^2 - \gamma R^2) E^*_1 = O.\] Hence, we have (38.2).

Multiplying each term on the left by \(E^*_4(y)\), on the right by \(E^*_1(y)\), we find \[\begin{align} O & = g^-_3E^*_{34}R^2FE^*_{11} + E^*_{34}RFRE^*_{11} + g^+_3E^*_{34}FR^2E^*_{11}-\gamma E^*_{34}R^2E^*_{11}\\ & = g^-_3E^*_{34}A^2E^*_{12}AE^*_{11} + E^*_{34}AE^*_{23}AE^*_{22}AE^*_{11} + g^+_3E^*_{34}AE^*_{33}AE^*_{22}AE^*_{11}. \tag{38.3} \end{align}\] Applying this to \(v\), we find by (38.1) that \[0 = g^-_3E^*_{34}A^2E^*_{12}AE^*_{11}v.\] So, \(g^-_3 = 0\) by Claim 1. But by Lemma 30.1, \[g^-_3 = \frac{\theta^*_1-\theta^*_0}{\theta^*_1-\theta^*_3} \neq 0,\] a contradiction.

Let \(\Gamma\), \(x, y\) be as in Lemma 38.1. We saw in Lemma 37.2, \[R^{-1}E^*_2A_2E^*_1\hat{y} = \delta^+_{10} + \delta^+_{11},\] where \[\delta^+_{10}\in E^*_{10}V = \mathrm{Span}(\hat{y}), \quad \delta^+_{11}\in E^*_{11}V.\]

Definition 38.1 Define \(\Psi = \Psi(x,y) \in \mathbb{C}\) by \(\delta^+_{10} = \Psi\hat{y}\).

We will show that \(\Psi(x,y)\) is independent of \(x, y\).

Observe \(R^{-1}, A_i, E^*_i\in \mathrm{Mat}_X(\mathbb{Q})\). So \(\Psi\in \mathbb{Q}\).

Firstly, show \[\Psi(x,y) = \Psi(y,x).\]

Lemma 38.2 With the notation of Lemma 38.1, the following hold.

\((i)\) \(E^*_{22}AE^*_{11}\delta^+_{11} = \delta_{22}\).

\((ii)\) \(E^*_{21}AE^*_{11}\delta^+_{11} = -\Psi(x,y)\delta_{21}\).

\((iii)\) \(\langle \delta^+_{11}, \delta_{11}\rangle = \frac{a_2}{c_2} - \Psi(x,y)\).

\((iv)\) \(\Psi(x,y) = \Psi(y,x)\).

\((v)\) \(E^*_{12}AE^*_{11}\delta^+_{11} = -\Psi(x,y)\delta_{12}\).

Proof. Write \(\Psi \equiv \Psi(x,y)\), \(R \equiv R(x)\), \(E^*_i \equiv E^*_i(x)\), etc.

\((i)\) We have

\[\begin{align} R(\delta^+_{11} + \Psi \hat{y}) & = R(\delta^+_{11} + \delta^+_{10})\\ & = R(R^{-1}(E^*_2A_2E^*_1))\hat{y}\\ & = E^*_2A_2E^*_1\hat{y}\\ & = \delta_{22}. \end{align}\] So, \[\begin{align} \delta_{22} & = R(\delta^+_{11} + \Psi \hat{y})\\ & = E^*_2AE^*_1(\delta^+_{11} + \Psi \hat{y})\\ & = E^*_{22}AE^*_{11}\delta^+_{11} + \Psi E^*_{22}A E^*_{10}\hat{y}. \end{align}\] The second term is zero.

\((ii)\) We have

\[\begin{align} 0 & = E^*_{21}\delta_{22}\\ & = E^*_{21}R(\delta^+_{11} + \Psi \hat{y})\\ & = E^*_{21}AE^*_{11}\delta^+_{11} + \Psi E^*_{21}AE^*_{10}\hat{y}\\ & = E^*_{21}AE^*_{11} + \Psi \delta_{21}. \end{align}\]

\((iii)\) We have

\[\begin{align} p^{1}_{22} & = \|\delta_{22}\|^2 \\ & = \langle \delta_{22}, \delta_{21}+\delta_{22}+\delta_{23}\rangle\\ & = \langle R(\delta^+_{11} + \Psi \hat{y}), \delta_{21}+\delta_{22}+\delta_{23}\rangle\\ & = \langle \delta^+_{11} + \Psi \hat{y}, L(\delta_{21}+\delta_{22}+\delta_{23})\rangle\\ & = b_1\langle \delta^+_{11} + \Psi \hat{y}, \delta_{10}+\delta_{11}+\delta_{12}\rangle\\ & = b_1(\langle \delta^+_{11}, \delta_{11}\rangle + \Psi). \end{align}\] So, \[\langle \delta^+_{11}, \delta_{11}\rangle = b_1^{-1}p^{1}_{22}- \Psi = \frac{a_2}{c_2}-\Psi.\]

HS MEMO

\[b^{-1}_1p^1_{22} = b^{-1}_1\frac{k_1}{k_1}p^1_{22} = b^{-1}_1\frac{1}{k_1}k_2p^{2}_{12} = b^{-1}_1\frac{b_1}{c_2}a_2 = \frac{a_2}{c_2}.\]

\((iv)\) Interchanging roles of \(x, y\) above, we find there exists \(\delta^{+'}_{11}\in E^*_{11}V\) such that

\[R(y)^{-1}E^*_2(y)A_2E^*_1(y)\hat{x} = \delta^{+'}_{11} + \Psi(y,x)\hat{y}.\]

Then, \[E^*_{22}AE^*_{11}(\delta^{+'}_{11}) = \delta_{22}.\] So, \[E^*_{22}AE^*_{11}(\delta^{+}_{11}-\delta^{+'}_{11}) = 0.\] Hence, \(\delta^+_{11} = \delta^{+'}_{11}\) since \[E^*_{22}AE^*_{11}: E^*_{11}V \to E^*_{22}V\] is one-to-one.

Now, \[\frac{a_2}{c_2}-\Psi(x,y) = \langle \delta^+_{11}, \delta_{11}\rangle = \langle \delta^{+'}_{11}, \delta_{11}\rangle = \frac{a_2}{c_2}-\Psi(y,x).\] Thus, \[\Psi(x,y) = \Psi(y,x).\]

\((v)\) Immediate from \((ii)\), \((iv)\).