Chapter 29 Tridiagonal Pair \(A, A^*\)
Wednesday, April 7, 1993
Introduction to Theorem 29.1
Let \(\Gamma = (X,E)\) be distance-regular with diameter \(D\geq 3\).
Assume \(\Gamma\) is \(Q\)-polynomial with respect to \(E_1\).
Fix a vertex \(x\in X\). Write \(E^*_i \equiv E^*_i(x)\), \(A^*_i \equiv A^*_i(x)\), \(A^* = A^*_1\).
We know for \(h, i, j\) \((0\leq h, i, j\leq D)\), \[\begin{align} E^*_iA_hE^*_j = O & \leftrightarrow p^h_{ij} = 0\\ E_iA_h^*E_j = O & \leftrightarrow q^h_{ij} = 0. \end{align}\] Also, for \(h, i, j\) \((0\leq h, i, j\leq D)\), \[\begin{align} h < |i-j| & \rightarrow p^h_{ij} = 0, \; q^h_{ij} = 0\\ h = |i-j| & \rightarrow p^h_{ij} \neq 0, \; q^h_{ij} \neq 0. \end{align}\]
Some \(A_h\) (resp. \(A^*_h\)) is a polynomial of degree exactly \(h\) in \(A\) (resp. \(A^*\)), it follows, for \(h, i, j\) \((0\leq h, i, j\leq D)\), \[E^*_iA^hE^*_j, \; E_i{A^*}^hE_j \quad \begin{cases} = 0 & \text{if } h < |i-j|,\\ \neq 0 & \text{if } h = |i-j|.\end{cases}\]
We saw that there exist \(\beta, \gamma, \delta\in \mathbb{R}\) such that \[0 = [A, A^2A^*-\beta AA^*A + A^*A^2 - \gamma(AA^*+A^*A) - \delta A^*].\] In fact, there exist \(\beta, \gamma^*, \delta^*\in \mathbb{R}\) such that \[0 = [A^*, {A^*}^2A-\beta A^*AA^* + A{A^*}^2 - \gamma^*(A^*A+AA^*) - \delta^* A]\] as well as we will now show.
Let \(K\) denote any field. Let \(V\) denote any vector space over \(K\) of finite positive dimension. Let \(\mathrm{End}_K(V)\) denote the \(K\)-algebra of all \(K\)-linear transformations \(V\to V\).
Theorem 29.1 Given semi-simple elements \(A, A^*\in \mathrm{End}_K(V)\), suppose \[\begin{align} E_i(A^*)^hE_j & \begin{cases} = 0 & \text{if } h < |i-j|,\\ \neq 0 & \text{if } h = |i-j|.\end{cases} \quad (0\leq h,i,j\leq D) \tag{29.1}\\ E^*_iA^hE^*_j & \begin{cases} = 0 & \text{if } h < |i-j|,\\ \neq 0 & \text{if } h = |i-j|.\end{cases} \quad (0\leq h,i,j\leq R) \tag{29.2} \end{align}\] for some ordering \(E_0, E_1, \ldots, E_D\) of the primitive idempotents for \(A\), and some ordering \(E^*_0, E^*_1, \ldots, E^*_R\) of primitive idempotents for \(A^*\). Then
\[\begin{align} 0 & = [A, A^2A^*-\beta AA^*A + A^*A^2 - \gamma(AA^*+A^*A) - \delta A^*] \tag{29.3}\\ & = A^3A^*-A^*A^3 - (\beta+1)(A^2A^*A-AA^*A^2)\\ & \quad -\gamma(A^2A^*-A^*A^2)-\delta(AA^*-A^*A) \tag{29.4}\\ 0 & = [A^*, {A^*}^2A-\beta A^*AA^* + A{A^*}^2 - \gamma^*(A^*A+AA^*) - \delta^* A] \tag{29.5}\\ & = {A^*}^3A - A{A^*}^3 - (\beta+1)({A^*}^2AA^*-A^*A{A^*}^2)\\ & \quad -\gamma^*({A^*}^2A-A{A^*}^2)-\delta^*(A^*A-AA^*). \tag{29.6} \end{align}\]
\[\begin{align} \beta & = \frac{\theta_i-\theta_{i+1}+\theta_{i+2}-\theta_{i+3}}{\theta_{i+1}-\theta_{i+2}} \quad (0\leq i\leq D-3) \tag{29.7}\\ & = \frac{\theta^*_i-\theta^*_{i+1}+\theta^*_{i+2}-\theta^*_{i+3}}{\theta^*_{i+1}-\theta^*_{i+2}} \quad (0\leq i\leq D-3) \tag{29.8}\\ \gamma & = \theta_i - \beta \theta_{i+1} + \theta_{i+2} \quad (0\leq i\leq D-2) \tag{29.9}\\ \gamma^* & = \theta^*_i - \beta \theta^*_{i+1} + \theta^*_{i+2} \quad (0\leq i\leq D-2) \tag{29.10}\\ \delta & = \theta^2_i - \beta \theta_i\theta_{i+1} + \theta^2_{i+1} - \gamma(\theta_i + \theta_{i+1}) \quad (0\leq i\leq D-1) \tag{29.11}\\ \delta^* & = {\theta^*}^2_i - \beta {\theta^*}_i{\theta^*}_{i+1} + {\theta^*}^2_{i+1} - \gamma^*({\theta^*}_i + {\theta^*}_{i+1}) \quad (0\leq i\leq D-1) \tag{29.12} \end{align}\] In particular, \(\beta, \gamma, \gamma^*, \delta, \delta^*\) are uniquely determined by \(A\), \(A^*\) and the above ordering of their primitive idempotents, whenever \(D\geq 3\).
Proof.
Since \(A\) is semisimple with exactly \(D+1\) distinct eigenvalues, the minimal polynomial of \(A\) has degree \(D+1\).
Since \(R \geq D+1\), \[A^R \in \mathrm{Span}\{A^j\mid 0\leq j\leq D\}.\] Multiplying each term on the left by \(E^*_R\) and on the right by \(E^*_0\), we find \[\begin{equation} E^*_RA^RE_0^* \in \mathrm{Span}\{E^*_RA^jE^*_0\mid 0\leq j\leq D\}. \tag{29.13} \end{equation}\] But by (29.2), the left side of (29.13) is nonzero and the right side of (29.13) is \(0\), a contradiction.
Hence \(D\geq R\).
Recalling the definitions, we have \[\begin{align} A & = \sum_{i=0}^D \theta_i E_i,\\ A^* & = \sum_{i=0}^D \theta_i^* E_i^*,\\ AE_i & = E_iA = \theta_iE_i \quad (0\leq i\leq D),\\ A^*E_i^* & = E_i^*A^* = \theta_i^*E_i^* \quad (0\leq i\leq D). \end{align}\]
Claim 1. For all integers \(i,j,k,\ell\) \((0\leq i, j, k, \ell\leq D)\) such that \(j+k \leq i-\ell\), \[\begin{equation} E^*_iA^jA^*A^kE^*_\ell = \begin{cases} \theta^*_{\ell+k}E^*_iA^{j+k}E^*_\ell & \text{if } j+k = i-l,\\ O & \text{if } j+k < i-\ell.\end{cases} \tag{29.14} \end{equation}\]
Proof of Claim 1. The product (29.14) eqia;s \[E^*_iA^j\left(\sum_{h=0}^D \theta_h^* E_h^*\right)A^kE^*_\ell = \sum_{h=0}^D \theta^*_h E^*_iA^jE^*_hA^kE^*_\ell.\] Now pick any \(h\) \((0\leq h\leq D)\), where \[E^*_iA^jE^*_hA^kE^*_\ell \neq O.\] Then by (29.2), \(j\geq |i-h|\), otherwise \[E^*_iA^jE^*_h = O\] and by (29.1), \(k\geq |h-\ell|\) otherwise \[E^*_hA^kE^*_\ell = O.\] Hence, \[j+k \geq |i-h|+|h-\ell| \geq |i-\ell| \geq i-\ell.\] Now if \(j+k < i-\ell\), we see there is no such \(h\), so (29.14) holds.
(Pf. Suppose \(i=j+k+\ell\) with \(0\leq i, j, k,\ell, h\leq D\).
Then \(i \geq j, k, \ell\). Since \(k = |h-\ell|\), if \(h\neq \ell+k\), \(h = \ell-k\) and \(j - i-h\), \(\ell-h+i-h = i-\ell\) implies \(h=\ell\), \(k = 0\) and \(h = \ell + k\).)
This proves Claim 1.
Let \(M\) denote the subalgebra of \(\mathrm{End}_K(V)\) generated by \(A\). Observe that \(M\) has a basis \(E_0, \ldots, E_D\) as a vector space over \(K\). Set \[L:= \mathrm{Span}\{mA^*m-nA^*m \mid m, n\in M\}.\]
Claim 2. \(\dim L \leq D\).
Proof of Claim 2. Since \(E_0, \ldots, E_D\) span \(M\), \[\begin{align} L & = \mathrm{Span}\{E_iA^*E_j - E_jA^*E_i \mid 0\leq i < j \leq D\}\\ & = \mathrm{Span}\{E_{j-1}A^*E_j - E_jA^*E_{j-1} \mid 1\leq j \leq D\} \end{align}\] by (29.1).
In particular, \(L\) has a spanning set of order \(D\).
So, Claim 2 holds.
Claim 3. \(\{A^iA^* - A^*A^i\mid 1\leq i\leq D\}\) is a basis for \(L\).
Proof of Claim 3. Since \[A^iA^*-A^*A^i = A^iA^*I - IA^*A^i\] is contained in \(L\) \((1\leq i\leq D)\), and since \(\dim L \leq D\), it suffices to show the given elements are linearly independent.
Suppose they are dependent. Then there exists an integer \(i\) \((1\leq i\leq D)\) such that \[\begin{equation} A^iA^*-A^*A^i \in \mathrm{Span}(A^jA^*-A^*A^j\mid 1\leq j < i). \tag{29.15} \end{equation}\] Multiplying each term in (29.15) on the left by \(E^*_i\), and on the left by \(E^*_0\), and simplifying using \[E^*_i(A^\ell A^*-A^*A^\ell)E^*_0 = (\theta^*_0-\theta^*_i)E^*_iA^\ell E^*_0,\] we find \[\begin{equation} E^*_iA^\ell E^*_0 \in \mathrm{Span}(E^*_iA^jE^*_0\mid 1\leq j < i). \tag{29.16} \end{equation}\] But the left side of (29.16) is nonzero.
A contradiction.
Since \(A^2A^*A-AA^*A^2\) is contained in \(L\), we find by Claim 2, \[\begin{equation} A^2A^*A - AA^*A^2 = \sum_{i=1}^D\alpha_i(A^iA^*-A^*A^i) \tag{29.17} \end{equation}\] for some \(\alpha_0, \ldots, \alpha_D\in K\).
Claim 4. \(\alpha_i = 0 \quad (3<i\leq D)\).
Proof of Claim 4. Suppose not, and set \[t = \max\{i\mid 3<i\leq D, \; \alpha_i\neq 0\}.\] Then by (29.17), and Claim 1, \[\begin{align} 0 & = E^*_t\left(A^2A^*A - AA^*A^2 - \sum_{i=1}^D\alpha_i(A^iA^*-A^*A^i)\right)E^*_0\\ & = \alpha_t(\theta^*_t-\theta^*_0)E^*_tA^tE^*_0\\ & \neq O. \end{align}\]
(Since \(\alpha_i=0\) if \(i>t\), \[\begin{align} E^*_tA^2A^*AE^*_0 & = E^*_tAA^*A^2E^*_0 = O \quad (\text{as } 2+1 < t-0)\\ E^*_tA^iA^*E^*_0 & = E^*_tA^*A^iE^*_0 = O\\ E^*_tA^tA^*E^*_0 & = \theta^*_0 E^*_tA^tE^*_0,\\ E^*_tA^*A^tE^*_0 & = \theta^*_tE^*_tA^*A^tE^*_0.)\\ \end{align}\]
A contradiction. This proves Claim 4.
Claim 5. Suppose \(D\geq 3\). Then \[\begin{equation} \alpha_3 = \frac{\theta^*_{i+1}-\theta^*_{i+2}}{\theta^*_i - \theta^*_{i+3}} \quad \text{for all }i, \; (0\leq i\leq D-3). \tag{29.18} \end{equation}\] In particular, \(\alpha\neq 0\).
Proof of Claim 5. Fix and integer \(i\) \((0\leq i\leq D-3\)). Then by (29.14) and (29.17), \[\begin{align} O & = E^*_{i+3}\left(A^2A^*A - AA^*A^2 - \sum_{j=1}^3\alpha_j(A^iA^*-A^*A^i)\right)E^*_i\\ & = (\theta^*_{i+1}-\theta^*_{i+2}-\alpha_3(\theta^*_i-\theta^*_{i+3}))E^*_{i+3}A^3E^*_i. \end{align}\] But \(E^*_{i+3}A^3E^*_i\neq O\) by (29.2), so (29.18) holds.
This proves Claim 5.
Claim 6. Lines (29.3), (29.4), (29.8) hold.
Proof of Claim 6. First suppose \(D\geq 3\). Then by (29.17), Claims 4, and 5, \[\begin{equation} A^2A^*A-AA^*A^2 = \alpha_3(A^3A^*-A^*A^3)+\alpha_2(A^2A^*-A^*A^2) + \alpha_1(AA^*-A^*A), \tag{29.19} \end{equation}\] where \(\alpha_3\neq 0\). Hence \[A^3A^*-A^*A^3-\frac{1}{\alpha_3}(A^2A^*A-AA^*A^2)+\frac{\alpha_2}{\alpha_3}(A^2A^*-A^*A^2)+ \frac{\alpha_1}{\alpha_3}(AA^*-A^*A)=O.\] Now (29.4) is immediate, where \[\begin{align} \beta & = \frac{1}{\alpha_3}-1, \tag{29.20}\\ \gamma & = -\frac{\alpha_2}{\alpha_3}, \tag{29.21}\\ \delta & = -\frac{\alpha_1}{\alpha_3}. \tag{29.22} \end{align}\] The line (29.3) follows from the definition of \([\text{ }, \text{ }]\).
The line (29.8) is immediate from (29.18) and (29.20).
Now suppose \(D < 3\). Then the line (29.8) is vacuously true, so consider (29.4).
Let \(\alpha_3\) denote any nonzoro element of \(K\).
Then \(A^2A^*-A^*A^2, AA^*-A^*A\) certainly span \(L\) by Claim 3.
So, (29.19) holds for appropriate \(\alpha_1\) and \(\alpha_2\in K\).
Now, (29.4) holds, where \(\beta\), \(\gamma\), \(\delta\) are given by (29.20), (29.21), (29.22).
Claim 7. Lines (29.7), (29.9), (29.11) hold.
Proof of Claim 7. Pick an integer \(i\) \((0\leq i\leq D-1)\).
By (29.4), we have \[\begin{align} O & = E_i(A^3A^*-A^*A^3-(\beta+1)(A^2A^*A-AA^*A^2)-\gamma(A^2A^*-A^*A^2)-\delta(AA^*-A^*A))E_{i+1}\\ & = E_iA^*E_{i+1}(\theta^3_i-\theta^3_{i+1}-(\beta+1)(\theta^*_i\theta_{i+1}-\theta_i\theta^2_{i+1})-\gamma(\theta^2_i-\theta^2_{i+1})-\delta(\theta_i-\theta_{i+1}))\\ & = E_iA^*E_{i+1}(\theta_i-\theta_{i+1})(\theta^2_i+\theta_i\theta_{i+1}+\theta_{i+1}^2-(\beta+1)\theta_i\theta_{i+1}-\gamma(\theta_i+\theta_{i+1})-\delta)\\ & = E_iA^*E_{i+1}(\theta_i-\theta_{i+1})(\theta^2_i-\beta\theta_i\theta_{i+1}+\theta_{i+1}^2-\gamma(\theta_i+\theta_{i+1})-\delta). \end{align}\]
But \(E_iA^*E_{i+1} \neq O\) by (29.1), and of course, \(\theta_i\neq \theta_{i+1}\), so \[0 = \theta^2_i-\beta\theta_i\theta_{i+1}+\theta_{i+1}^2-\gamma(\theta_i+\theta_{i+1})-\delta.\] This proves (29.11).
To obtain (29.9), pick any integer \(i\) \((0\leq i\leq D-2)\). Then by (29.11), \[\begin{align} 0 & = \theta^2_i-\beta\theta_i\theta_{i+1}+\theta_{i+1}^2-\gamma(\theta_i+\theta_{i+1})-\delta\\ & \qquad -(\theta^2_{i+1}-\beta\theta_{i+1}\theta_{i+2}+\theta_{i+2}^2-\gamma(\theta_{i+1}+\theta_{i+2})-\delta)\\ & = \theta_i^2 - \beta\theta_i\theta_{i+1}-\gamma\theta_i + \beta\theta_{i+1}\theta_{i+2}-{\theta_{i+2}}^2 + \gamma\theta_{i+2}\\ & = (\theta_i-\theta_{i+2})(\theta_i - \beta\theta_{i+1} + \theta_{i+2} - \gamma). \end{align}\] So \(0 = \theta_i - \beta\theta_{i+1} + \theta_{i+2} - \gamma\).
This gives (29.9).
To see (29.7), pick an integer \(i\) \((0\leq i\leq D-3)\).
Then by (29.9), \[\begin{align} 0 & = (\theta_i - \beta\theta_{i+1} + \theta_{i+2} - \gamma) - (\theta_{i+1} - \beta\theta_{i+2} + \theta_{i+3} - \gamma) \\ & = \theta_i-(\beta+1)\theta_{i+1} + (\beta+1)\theta_{i+2} - \theta_{i+3}. \end{align}\] We have \[\beta = \frac{\theta_i-\theta_{i+3}}{\theta_{i+1}-\theta_{i+2}} - 1 = \frac{\theta_i-\theta_{i+1}+\theta_{i+2}-\theta_{i+3}}{\theta_{i+1}-\theta_{i+2}},\] as desired.
This proves Claim 7.
We have now proved (29.3), (29.4), (29.7), (29.8), (29.9), (29.11).
Interchanging the roles of \(A\) and \(A^*\), we obtain (29.5), (29.6), (29.10), (29.12).