Chapter 35 \(\dim E\)1*\(TE\)1* \(\leq 5\)
Monday, April 26, 1993
Theorem 35.1 Let \(\Gamma = (X, E)\) be distance regular of diameter \(D\geq 3\). Assume \(\Gamma\) is \(Q\)-polynomial with respect to \(E_0, E_1, \ldots, E_D\). Fix a vertex \(x\in X\), and write \(E^*_i\equiv E^*_i(x)\), \(T\equiv T(x)\).
\[\dim E^*_1TE^*_1 \leq 5.\]
Proof.
Claim 1. \(E^*_1ME^*_1 = \mathrm{Span}(\tilde{J}, E^*_1, \tilde{A})\).
Proof of Claim 1. \[E^*_1ME^*_1 = \mathrm{Span}\{E^*_1, E^*_1AE^*_1, E^*_1A_2E^*_1, E^*_1A_3E^*_1, \ldots\}.\] But \(E^*_1A_hE^*_h = O\) if \(h >2\) (by Lemma 16.1). So, \[E^*_1ME^*_1 = \mathrm{Span}\{E^*_1, E^*_1AE^*_1, E^*_1A_2E^*_1\}.\] Also, \[\begin{align} \tilde{J} & = E^*_1JE^*_1 \\ & = E^*_1\left(\sum_{h=0}^DA_h\right)E^*_1\\ & = E^*_1 + E^*_1AE^*_1 + E^*_1A_2E^*_1. \end{align}\] So, \[E^*_1ME^*_1 = \mathrm{Span}\{E^*_1, E^*_1AE^*_1, \tilde{J}\}.\] We are done, since \(\tilde{A} = E^*_1AE^*_1\).
Claim 2. \(E^*_1MM^*ME^*_1 = \mathrm{Span}(\tilde{J}, E^*_1, \tilde{A}, \tilde{A}^2)\).
Proof of Claim 2. \(\supseteq\): Clear.
\(\subseteq\): In Lemma 33.4 \((i)\), we say \[E^*_1T = E^*_1E_0M^* + E^*_1M + E^*_1E_1M^* + E^*E_1E^*_1M + \cdots.\] In fact, the proof of that lemma gives a sequence; \[\begin{align} E^*_1MM^* & = E^*_1E_0M^* + E^*_1M + E^*_1E_1M^*,\\ E^*_1MM^*M & = E^*_1E_0M^* + E^*_1M + E^*_1E_1M^* + E^*E_1E^*_1M, \tag{35.1}\\ E^*_1MM^*MM^* & = E^*_1E_0M^* + E^*_1M + E^*_1E_1M^* + E^*E_1E^*_1M + E^*E_1E^*_1MM^*,\\ & \qquad \vdots \end{align}\]
Multiply (35.1) through on the right by \(E^*_1\) to get \[E^*_1MM^*ME^*_1 = E^*_1ME^*_1 + E^*_1E_1E^*_1ME^*_1 = \mathrm{Span}\{\tilde{J}, E^*_1, \tilde{A}, \tilde{A}^2\},\] since \(\tilde{J}^2, \tilde{A}\tilde{J} = \tilde{J}\tilde{A}\in \mathrm{Span}\{\tilde{J}\}\).
This proves Claim 2.
Now, let \(W\) denote any irreducible \(T\)-module with endpoint \(1\), and pick \(0\neq v\in E^*_1W\). Set \[v^+_i = E^*_iA_{i-1}E^*_1v, \quad v^-_i = E^*_iA_{i+1}E^*_1v, \; i\in \{1, \ldots, D\}.\] We know by Lemma 34.2 \((ii)\) that \(W\) is thin if and only if \(v^+_i, v^-_i\) are linearly dependent for all \(i\in \{2, \ldots, D-1\}\).
In general, \[\Phi_i = \det \begin{pmatrix}\|v^+_i\|^2 & \langle v^+_i, v^-_i\rangle\\ \langle v^+_i,v^-_i\rangle & \|v^-_i\|^2 \end{pmatrix} \geq 0\] with equality if and only if \(v^+_i, v^-_i\) are linearly dependent, (because \(\Phi_i\) is the determinant of a Gram matrix).
Let \(i\) be an integer in \(\{2, \ldots, D-1\}\).
Claim 3. There exists \(p^{++}\in \mathrm{C}[\lambda]\), \(\deg p^{++}\leq 2\) (that depends only on the intersection numbers) such that \[\|v^+_i\|^2 = \|v\|^2 p^{++}(a_0(W)).\]
Proof of Claim 3. \[\|v^+_i\|^2 = \bar{v}^\top E^*_1A_{i-1}E^*_iE^*_iA_{i-1}E^*_1v = \bar{v}^\top E^*_1A_{i-1}E^*_iA_{i-1}E^*_1v.\] But, \[E^*_1A_{i-1}E^*_iA_{i-1}E^*_1 \in E^*_1MM^*ME^*_1 = \mathrm{Span}(\tilde{J}, E^*_1, \tilde{A}, \tilde{A}^2)\] by Claim 2.
So, there exists \(\alpha \in \mathbb{C}\), and \(p^{++}\in \mathbb{C}[\lambda]\) with \(\deg p^{++}\leq 2\) such that \[E^*_1A_{i-1}E^*_iA_{i-1}E^*_1 = \alpha \tilde{J} + p^{++}(\tilde{A}), \quad (\tilde{A}^0 = E^*_1).\] Now, \[\|v^+_i\|^2 = \bar{v}^\top (\alpha \tilde{J} + p^{++}(\tilde{A}))v = \|v\|^2 p^{++}(a_0(W)),\] since \(\tilde{J}v = 0\), and \(\tilde{A}v = a_0(W)v\).
This proves Claim 3.
Similarly, there exist \(p^{--}, p^{+-}\in \mathbb{C}[\lambda]\) with \(\deg p^{--}, \deg p^{+-}\leq 2\) such that \[\|v^-_i\|^2 = \|v\|^2p^{--}p(a_0(W)), \; \langle v^+_i, v^-_i\rangle = \|v\|^2 p^{+-}(a_0(W)).\]
Claim 4. \(E^*_1A_{i-1}E^*_iA_{i+1}E^*_1 = (\tilde{J}-\tilde{A}-E^*_1)p^2_{i-1,i+1}\). In particular, \[p^{+-}(\lambda) = -p^2_{i-1,i+1}(\lambda +1).\]
Proof of Claim 4. Pick vertices \(y,z\in X\) such that \(\partial(x,y) = \partial(x,z) = 1\).
\[\begin{align} (\mathrm{LHS})_{yz} & = \sum_{w\in X}(E^*_1A_{i-1}E^*_i)_{yw}(E^*_iA_{i+1}E^*_1)_{wz}\\ & = \sum_{w\in X, \partial(y,w)=i-1,\partial(x,w)=i, \partial(w,z)=i+1}1\\ & = \begin{cases} 0 & \text{if } \partial(y,z) =0,\\ 0 & \text{if } \partial(y,z) = 1, \\ p^2_{i-1,i+1} & \text{if }\partial(y,z)=2, \end{cases}\\ & = \mathrm{RHS}_{yz}. \end{align}\] Note that \(E^*_1A_2E^*_1 = \tilde{J}- \tilde{A} - E^*_1\).
Now, \[\begin{align} \langle v^+_i, v^-_i\rangle & = \bar{v}^\top E^*_1A_{i-1}E^*_iA_{i+1}E^*_1v\\ & = p^2_{i-1,i+1}(\bar{v}^\top (\tilde{J}-\tilde{A}-E^*_1)v)\\ & = -(a_0(W)+1)p^2_{i-1,i+1}\|v\|^2. \end{align}\]
Claim 5. \(\deg p^{++} = \deg p^{--} = 2\). (only need for some \(i\))
Proof of Claim 5. We need to calculate \(p^{++}\), \(p^{--}\).
HS MEMO
Pick vertices \(y,z\in X\) such that \(\partial(x,y) = \partial(x,z) =1\). Then \[(E^*_1A_{i-1}E^*_iA_{i-1}E^*_1)_{yz} = |\Gamma_{i-1}(y)\cap \Gamma_i(x)\cap \Gamma_{i-1}(z)|,\] which is equal to \(p^{1}_{i-1,i}\) if \(\partial(y,z)=0\). \[(E^*_1A_{i+1}E^*_iA_{i+1}E^*_1)_{yz} = |\Gamma_{i+1}(y)\cap \Gamma_i(x)\cap \Gamma_{i+1}(z)|,\] which is equal to \(p^{1}_{i+1,i}\) if \(\partial(y,z)=0\).
Conclusion. \[\begin{align} \Phi_i & = \det \begin{pmatrix}\|v^+_i\|^2 & \langle v^+_i, v^-_i\rangle\\ \langle v^+_i,v^-_i\rangle & \|v^-_i\|^2 \end{pmatrix} \geq 0\\ & = \|v\|^4(p^{++}(\lambda)p^{--}(\lambda) - (p^2_{i-1,i+1})^2(\lambda+1)^2\\ & \geq 0, \end{align}\] where \(\lambda = a_0(W)\).
\(W\) is thin if and only if \(\Phi_i(\lambda) = 0\) for all \(i\in \{2, \ldots, D-1\}\).
Each \(\Phi_i\) is degree \(4\) solutions for \(\lambda\). Since \(\lambda\) determines the isomorphism class of \(W\) by Lemma 34.3 (iii), there are at most \(4\) different thin irreducible modules \(W\) of endpoint \(1\) up to isomorphism.
Note. In fact \(\Phi_i(\lambda)\) is independent of \(i\) up to scalar multiple for \(i\in \{2,\ldots, D-1\}\).
If \(\Gamma\) has classical parameters \((q,D, \alpha, \beta)\), the roots are; \[\beta-\alpha-1, -1, -q-1, dq\frac{q^{D-1}-1}{q-1}-1.\]