Chapter 39 \(A^+\) and \(A^-\)

Wednesday, May 5, 1993

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

Fix a vertex \(x\in X\), write \(E^*_i\equiv E^*_i(x)\), \(R\equiv R(x)\), \(T\equiv T(x)\).

Pick \(y\in X\) with \(\partial(x,y) = 1\). Write \(E^*_{i,j} \equiv E^*(x)E^*(y)\), \(\delta_{ij} = E^*_{ij}\delta\), and \(\tilde{A} = E^*_1AE^*_1\).

Recall that \(\delta^+_{11}\in E^*_{11}V\) and \[R^{-1}E^*_2A_2E^*_1\hat{y} = \delta^+_{11} + \Psi(x,y)\hat{y}.\] We saw \(\Psi(x,y) = \Psi(y,x)\). We shall show below that \(\Psi(x,y)\) is independent of edge \(xy\).

Lemma 39.1 With the above notation, set \(\Psi:=\Psi(x,y)\). Then the following hold.

\((i)\) \(\delta^-_{11} = \tilde{A}\delta^+_{11} - \left( \frac{a_2}{c_2} - \Psi\right)\hat{y} + \Psi \delta_{12}\in E^*_{11}V\).

\((ii)\) \(\delta^-_{11}(x,y) = \delta^{-1}_{11}(y,x)\).

Proof.

\((i)\) \(\delta^-_{12}\in E^*_{12}V\), \(\delta^-_{11}\in E^*_{11}V\) and \(\delta^-_{10}\in E^*_{10}V\), and

\[\begin{equation} \tilde{A}\delta^+_{11} = \delta^-_{12} + \delta^-_{11} + \delta^-_{10}, \tag{39.1} \end{equation}\] \[\begin{equation} \delta^-_{12} = E^*_{12}AE^*_{11}\delta^+_{11} = -\Psi(x,y)\delta_{12} \tag{39.2}, \end{equation}\] by Lemma 38.2 \((v)\).

Also, \(\delta^-_{10} = \sigma\hat{y}\) for some \(\sigma\in \mathbb{C}\), where \[\begin{equation} \sigma = \langle \tilde{A}\delta^+_{11}, \hat{y}\rangle = \langle \delta^+_{11}, \tilde{A}\hat{y}\rangle = \langle \delta^+_{11}, \delta_{11}\rangle = \frac{a_2}{c_2}-\Psi. \tag{39.3} \end{equation}\]

Solving for \(\delta^-_{11}\) in (39.1), using (39.2) and (39.3), we have \[\begin{align} \delta^-_{11} & = \tilde{A}\delta^+_{11} - \delta^-_{12} - \delta^-_{10}\\ & = A\delta^+_{11} + \Psi \delta_{12} - \left(\frac{a_2}{c_2} - \Psi\right)\hat{y}. \end{align}\]

\((ii)\) Since

\[\delta^-_{11} = E^*_{11}AE^*_{11}\delta^+_{11},\] we have \(\delta^+_{11}(x,y) = \delta^+_{11}(y,x)\).

Lemma 39.2 With the above noation, \(\Psi = \Psi(u,v)\) is independent of \(u, v\), where \(u,v\in X\), with \(\partial(u,v)=1\).

Proof. Let \(x,y\) be as above (\(x\sim y\)), and pick \(z\in X\) such that \(\partial(x,z) = 1\), but \(z\neq y\). Then it suffices to show: \[\Psi(x,y) = \Psi(x,z).\]

Case: \(\partial(y,z) = 2\).

Set \(\Delta: = \tilde{A}R^{-1}E^*_2A_2E^*_1\).

Observe: \(\Delta \in E^*_1TE^*_1\) and \(E^*_1TE^*_1\) is symmetrix by Lemma 33.4.

Hence, \(\Delta_{yz} = \Delta_{zy}\).

Since \(\Delta\in \mathrm{Mat}_X(\mathbb{R})\), \[\langle \Delta\hat{y}, \hat{z}\rangle = \langle \Delta \hat{z}, \hat{y}\rangle.\] But, \[\begin{align} \langle \Delta\hat{y}, \hat{z}\rangle & = \langle \tilde{A}\delta^+_{11} + \Psi(x,y)\hat{y}, \hat{z}\rangle \\ & = \langle \tilde{A}\delta^+_{11}, \hat{z}\rangle\\ & = \Bigl\langle \delta^-_{11} + \left(\frac{a_2}{c_2}-\Psi\right)\hat{y} - \Psi(x,y)\delta_{12}, \hat{z}\Bigr\rangle\\ & = -\Psi(x,y). \end{align}\] Note that \(\partial(x,y) = 2\) by Lemma 39.1 \((i)\).

Similarly, \[\langle\Delta\hat{z}, \hat{y}\rangle = -\Psi(x,z).\] Hence, \(\Psi(x,y) = \Psi(x,z)\).

Case: \(\partial(y,z) = 1\).

By Lemma 38.1 \((ii)\), there exists \(w\in X\) such that \[\partial(x,z) = 1, \; \partial(w,y)=2, \; \partial(w,z)=2.\]

Now, \[\Psi(x,y) = \Psi(x,w) = \Psi(x,z)\] from the first case.

Lemma 39.3 With the above notation, the following hold.

\((i)\) \(A^+: = R^{-1}E^*_2A_2E^*_1 - \Psi E^*_1\), and

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

are both generalized adjacency matrices for the subgraph induced on the first subconstituent with respect to \(x\).

Moreover, \(A^+\), \(A^-\) have \(0\) diagonal.

Proof. Pick vertices \(y,z\in X\) such that \(\partial(x,y) = \partial(x,z) = 1\).

Show that \(A^+_{yz}\), \(A^-_{yz}\) are both \(0\) if \(\partial(y,z) = 0\) or \(2\).

Since \(A^+_{yz} = R^{-1}E^*_2A_2E^*_1\hat{y} - \Psi E^*_1\hat{y} = \delta^+_{11}\), \[A^+_{yz} = \langle A^+\hat{y}, \hat{z}\rangle = \langle \delta^+_{11},\hat{z}\rangle = 0,\] if \(\partial(y,z) = 0\) or \(2\).

Since \[\begin{align} A^-\hat{y} & = \tilde{A}A^+\hat{y} - \left(\frac{a_2}{c_2}-\Psi\right)E^*_1\hat{y} + \Psi(\tilde{J} - \tilde{A} - E^*_1)\hat{y}\\ & = \tilde{A}\delta^+_{11} - \left(\frac{a_2}{c_2}-\Psi\right)\hat{y} + \Psi\delta_{12}\\ & = \delta^-_{11}, \end{align}\] \[A^-_{yz} = \langle A^-\hat{y}, \hat{z}\rangle = \langle \delta^-_{11}, \hat{z}\rangle = 0,\] if \(\partial(y,z) = 0\) or \(2\).

Since \(E^*_1TE^*_1 = \mathrm{Span}(\tilde{J},E^*_1, \tilde{A}, \tilde{A}^2, \ldots)\) by Lemma 33.4.

\(A^+, A^-\) are both generalized matrices for the adjacency subgraph induced on the first subconstituent concerning \(x\).

Similarly, \[E^*_1TE^*_1 \ni \tilde{J}, E^*_1, \tilde{A}, A^+, A^-,\] and \(\dim E^*_1TE^*_1 \leq 5\).

Fact: With the above assumption, \[E^*_1TE^*_1 = \mathrm{Span}(\tilde{J}, E^*_1, \tilde{A}, A^+, A^-)\] (may not be independent).

Lemma 39.4 If \(\partial(x,y) = 1\), then \[T(y)\hat{y} = T(x)\hat{y}.\]

Proof. \[\begin{align} T(x)\hat{x} & = T(x)E^*_1\hat{y}\\ & = M(E^*_0+E^*_1)T(x)E^*_1\hat{y} \quad (\text{as $\Gamma$ is thin})\\ & = M\hat{x} + ME^*_1TE^*_1\hat{y}\\ & = M\hat{x} + M\mathrm{Span}(\tilde{J}, E^*_1, \tilde{A}, A^+, A^-)\hat{y}\\ & = M\hat{x} + M\mathrm{Span}(\delta_{12}+\delta_{11}+\delta_{10}, \delta_{10}, \delta_{11}, \delta^+_{11}, \delta^-_{11})\\ & = M\mathrm{Span}(\delta_{01}, \delta_{10}, \delta_{11}, \delta^+_{11}, \delta^-_{11}). \end{align}\]

But the identity of these conditions does not change if we interchange \(x\) and \(y\).

Hence, \[T(y)\hat{y} = T(x)\hat{y}.\] This proves the lemma.