Chapter 6 \(T\)-Modules of \(H(D,2)\), II

Monday, February 1, 1993

Proof (Proof of Theorem 5.1 Continued).

\((iii)\) Let \(r = r'\),

\(w_0,\ldots, w_d\): a basis for \(W\) with \(w_i\in E^*_iW\), and

\(w_0', \ldots, w_d'\): a basis for \(W'\) with \(w_i'\in E^*_iW'\).

Then \(d = D-2r = D-2r' = d'\), and \[\sigma: W \to W' \quad (w_i\mapsto w_i')\] is an isomorsphism of \(T\)-modules by \((i)\).

If \(r\neq r'\), then \[d = D-2r \neq D-2r' = d',\] hence, \(\dim W \neq \dim W'\).

\((iv)\) Let \(W_i\) be an irreducible \(T\)-module with endpoint \(i\). Then

\[\dim E_r^*V = \binom{D}{r} = \sum_{i=0}^r \mathrm{mult}(W_i).\] Hence, we have that \[\mathrm{mult}(W_r) = \binom{D}{r} - \binom{D}{r-1}\] by induction on \(r\).

Theorem 6.1 Let \(\Gamma = H(D,2)\) with \(D\geq 1\). Fix a vertex \(x\in X\) and write \[E^*_i \equiv E^*_i(x), \quad T = T(x), \text{ and } A^* \equiv \sum_{i=0}^D(D-2i)E^*_i.\] Let \(W\) be an irreducible \(T\)-module with endpoint \(r\) with \(0\leq r\leq D/2\). Then,

\((i)\) \(W\) has a basis

\[w_0^*, w_1^*, \ldots, w_d^* \quad(d = D-2r), \; \text{ such that }\; w_i^*\in E_{i+r}W\; (0\leq i \leq d)\] with respect to which the matrix corresponding to \(A^*\) is \[\begin{pmatrix} 0 & d & 0 & & & & \\ 1 & 0 & d-1 & & & & \\ 0 & 2 & 0 & & & & \\ & & \ddots & \ddots & \ddots & & \\ & & & & 0 & 2 & 0 \\ & & & & d-1 & 0 & 1\\ & & & & 0 & d & 0 \end{pmatrix}.\] In particular,

\((ii)\) \(E_iA^*E_j = 0\) if \(|i-j|\neq 1\) for \(0 \leq i, j\leq D\).

Proof. We use the notation, \[[\alpha, \beta] = \alpha\beta - \beta\alpha \; (=-[\beta, \alpha]).\]

Recall that

\((a)\) \([L, R] = A^*\),

\((b)\) \([A^*, L] = wL\),

\((c)\) \([A^*, R] = -2R\),

and \(A = L + R\).

Write \((a) - (c)\) in terms of \(A\) and \(A^*\), we have, \[[A, A^*] = [L, A^*]+ [R, A^*] = 2(R-L).\] \[\begin{cases} R + L & = A\\ R-L & = [A,A^*]/2. \end{cases}.\] Hence, \[\begin{align} R & = \frac{1}{4}(2A + [A, A^*]) \quad \text{ and }\\ L & = \frac{1}{4}(2A - [A, A^*]). \end{align}\]

Now \((a)\), \((b)\) become \[\begin{align} A^2A^* - 2AA^*A + A^*A^2 - 4A^* & = 0 \tag{6.1}\\ {A^*}^2A - 2A^*AA^* + A{A^*}^2 - 4A & = 0 \tag{6.2} \end{align}\] Pf. By \((b)\), \[\begin{align} 2A - AA^* + A^*A & = 4L\\ & = 2[A^*, L]\\ & = A^* \frac{2A - [A,A^*]}{2} - \frac{2A - [A, A^*]}{2}A^*\\ & = A^*A-AA^* + \frac12(-A^*AA^* + {A^*}^2A + A{A^*}^2 - A^*AA^*) \end{align}\] So we have (6.2) \[ {A^*}^2A - 2A^*AA^* + A{A^*}^2 - 4A = 0. \]

By \((a)\), \[\begin{align} -16A^* & = [2A + [A, A^*], 2A - [A, A^*]]\\ & = (2A + [A, A^*])(2A - [A, A^*]) - (2A - [A,A^*])(2A + [A, A^*])\\ & = [4A^2 - 2A[A, A^*] + [A, A^*](2A) - [A,A^*]^2\\ & \quad - 4A^2 - 2A[A, A^*] + [A, A^*](2A) + [A, A^*]^2\\ & = -4A^2A^* + 4AA^*A + 4AA^*A - 4A^*A^2. \end{align}\] So, \[A^2A^* - 2AA^*A + A^*A^2 - 4A^* = 0.\]

Claim: \(E_i^*A^*E_j = 0\) if \(|i-j| \neq 1\) for \(0\leq i, j\leq D\).

Pf. We have, \[\begin{align} 0 & = E_i(A^2A^* - 2AA^*A + A^*A^2 - 4A^*)E_j\\ & = E_iA^*E_j(\theta_i^2 - 2\theta_i\theta_j + \theta_j^2 - 4)\\ & \quad (AE_j = \theta_jE_j, \; E_iA = (AE_j)^\top = (\theta_iE_i)^\top = \theta_iE_i)\\ & = E_iA^*E_j(\theta_i - \theta_j -2)(\theta_i - \theta_j + 2)\\ & = E_iA^*E_j(D-2i - (D-2j)-2)(D-2i - (D-2j) + 2)\\ & \quad (\theta_k = D-2k)\\ & = E_iA^*E_j \cdot 4(i-j+1)(i-j-1) \end{align}\] and \(i-j+1 \neq 0\), \(i-j-1\neq 0\). Hence, \(E_i^*A^*E_j = 0\).

Now define “dual raising matrix”, \[R^* = \sum_{i=0}^D E_{i+1}A^*E_i.\] So, \[R^*E_iV \subseteq E_{i+1}V, \quad (0\leq i\leq D, \; E_{D+1}V = 0).\] Define “dual lowering matrix” \[L^* = \sum_{i=0}^D E_{i-1}A^*E_i.\] Then \[L^*E_iV \subseteq E_{i-1}V \quad (0\leq i\leq D, \; E_{-1}V = 0).\] Observe that \[A^* = \left(\sum_{i=0}^DE_i\right)A^*\left(\sum_{j=0}^DE_j\right) = L^* + R^*\] by Claim 1.

Claim 2. We have

\((a)\) \([L^*, R^*] = A\),

\((b)\) \([A, L^*] = 2L^*\),

\((c)\) \([A, R^*] = -2R^*\).

Pf. \((b)\) \[\begin{align} AL^* - L^*A & = \sum_{i=0}^D(AE_{i-1}A^*E_i - E_{i-1}A^*E_iA)\\ & = \sum_{i=0}^D E_{i-1}A^*E_i (\theta_{i-1} - \theta_i)\\ & \quad (\theta_k = D-2k, \quad \theta_{i-1}- \theta_i = 2I - 2(i-1) = 2\\ & = 2L^*. \end{align}\]

\((c)\) Similar.

HS MEMO

\[\begin{align} AR^* - R^*A & = \sum_{i=0}^D (AE_{i+1}A^*E_i - E_{i+1}A^*E_iA)\\ & = \sum_{i=0}^D E_{i+1}A^*E_i (\theta_{i+1} - \theta_i)\\ & = -2R^*. \end{align}\]

\((a)\) We have, by \((b)\), \((c)\) \[\begin{equation} [A, A^*] = [A, L^*] + [A, R^*] = 2(L^* - R^*). \end{equation}\] Since \(A^* = L^* + R^*\), \[R^* = \frac{2A^* + [A^*, A]}{4}, \quad L^* = \frac{2A^* - [A^* - A]}{4}.\] Now \((a)\) is seen to be equivalent to (6.2) upon evaluation. This proves Claim 2.

HS MEMO

\[\begin{align} [L^*,R^*] & = \frac{1}{16}((2A^*-[A^*,A])(2A^*+[A^*,A]) - (2A^*+[A^*,A])(2A^*- [A,A^*]))\\ & = \frac{1}{16}(4{A^*}^2 + 2A^*[A^*,A] - [A^*,A]2A^* - [A^*,A]^2 - 4{A^*}^2\\ & \qquad + 2A^*[A^*,A] - [A^*,A]2A^* + [A^*,A]^2)\\ & = \frac{1}{4}({A^*}^2A - 2A^*AA^* + A{A^*}^2)\\ & = A, \end{align}\] by (6.2).

Now apply same argument as for (6.1), (6.2) of Theorem 5.1 and observe \(A^*\) has \(D+1\) distinct eigenvalues. So, \[A^* = \sum_{i=0}^D(D-2i)E^*_i\] generates \[M^* = \mathrm{Span}(E^*_0, \ldots, E^*_D).\] Hence, \(E_0, \ldots, E_D, \; A^*\) generates \(T\).

Take an irreducible \(T\)-module \(W\) with endpoint \(r\) with \(0\leq r \leq D/2\). Set \(t = \min\{i\mid E_iW\}\).

Pick \(0\neq w_0^*\in E_tW\). Set \[w_i^* = \frac{1}{i!}{R^*}^i w_0^* \in E_{t+i}W \quad \text{for all }i.\] Then, \[R^*w_i^* = (i+1)w_{i+1}^* \quad \text{for all }i.\] By \((a)\), we get by induction, \(L^*w_i^* = (D-2t-i+1)w^*_{i-1}\), \[\begin{align} L^*w_i^* & = \frac{1}{i}L^*R^*w_{i-1}^* \\ & = \frac{1}{i}(A + R^*L^*)w_{i-1}^* \\ & = \frac{1}{i}((D-2(t+i-1))w^*_{i-1} + (i-1)(D-2t-i+2)w_{i-1}^*)\\ & = (D-2t - i + 1)w_{i-1}^*. \end{align}\]

So \(\mathrm{Span}(w_0^*, w_1^*, \ldots )\) is \(L^*\), \(R^*\), \(A^*\)-invariant. Hence, \(W = \mathrm{Span}(w_0^*, w_1^*, \ldots, w_d^*)\), \(w_0^*, w_1^*, \ldots, w_d^* \neq 0\), \(w^*_i = 0\) for every \(i>d\) by dimension.

Thus \(d = D-2t\).

Pf. \[\begin{align} (D -2(t+d))w^*_d & = Aw_d^* \\ & = (L^*R^* - R^*L^*)w_d^*\\ & = -(D-2t - d + 1)R^*w_{d-1}^*\\ & = -(D-2t - d +1)dw^*_d. \end{align}\] Hence, \[0 = d^2 + (2t - D - 1 + 2)d - (D-2t) = (d-D+2t)(d+1)\] So \(d = D-2t\).

Definition 6.1 For any graph \(\Gamma = (X, E)\), pick a vertex \(x\in X\), and set \(E^*_i \equiv E^*_i(x)\) and \(T \equiv T(x)\).

\((i)\) An irreducible \(T\)-module \(W\) is thin if \(\dim E_i^*W \leq 1\) for every \(i\).

\((ii)\) \(\Gamma\) is thin with respet to \(x\), if every irreducible \(T(x)\)-module is thin,

\((iii)\) An irreducible \(T\)-module \(W\) is dual thin if \(\dim E_iW \leq 1\) for every \(i\).

\((iv)\) \(\Gamma\) is dual thin with respect to \(x\), if every irreducible \(T(x)\)-module is dual thin.

Observe: \(H(D,2)\) is thin, dual thin with respect to each \(x\in X\).

Definition 6.2 With above notation, write \(D \equiv D(x)\).

\((i)\) An ordering \(E_0, E_1, \ldots, E_R\) of primitive idempotents of \(\Gamma\) is restricted if \(E_0\) corresponds to the maximal eigenvalue.

Fix a restricted ordering,

\((ii)\) \(\Gamma\) is \(Q\)-polynomial with respect to \(x\), above ordering if there exists \(A^* \equiv A^*(x)\) such that

\(\quad (a)\) \(E_0^*V, \ldots, E_D^*V\) are the maximal eigenspaces for \(A^*\).

\(\quad (b)\) \(E_iA^*E_j = 0\) if \(|i-j| > 1\) for \(0\leq i,j\leq R\).

Observe \(H(D,2)\) is \(Q\)-polynomial with respect to the natural ordering of the idempotents and every vetex.

Program. Study graphs that are thin and \(Q\)-polynomial with respect to each vertex.

(In fact, thin with respect to \(x\) implies dual thin with respect to \(x\).)

Get a situation like \(H(D,2)\), where \(T\) is generated by \(A\), \(A^*\). Except \(\mathrm{sl}_2(\mathbb{C})\) is replaced by a quantum Lie algebra.