Proof.

1. (i): Immediate from (1.7).

2. (ii): Immediate from (i).

3. (iii): By Definition 1.4, statement (iii) is immediate from (i) and (ii).

Proof. Let V denote a nonzero $\triangle _q$ -module of dimension less than . Then V contains an irreducible $\triangle _q$ -submodule W. Since , the $\triangle _q$ -module W has a marginal weight by Theorem 2.1. The lemma is now immediate from Lemma 2.2(iii).

Proof.

1. (i): Immediate from [Reference Huang21, Lemma 6.2].

2. (ii): By Lemma 2.6(i), a routine induction on i yields (ii).

Proof. Let W denote the subspace of V spanned by $\{A^i v\}_{i\in \mathbb N}$ . Then W is A-invariant and the vectors $\{A^i v\}_{i=0}^{\dim W-1}$ are a basis for W. By Lemma 2.6(i), W is B-invariant. By Schur’s lemma, the central elements $\alpha ,\beta ,\gamma $ act on V, as well as W, as scalar multiplication. Hence W is a $\triangle _q$ -submodule of V by Proposition 1.2. Since $v\in W$ and $v\not =0$ , it follows that $W=V$ . The lemma follows.

Proof. By Theorem 1.5, condition (i) implies (3.1), (3.3), (3.4) and the following equations:

(3.5) $$ \begin{align} (B-\theta_1^*)(A-\theta_0) v&=\varphi_1 v, \end{align} $$

(3.6) $$ \begin{align} \alpha v&=\omega v, \end{align} $$

where $\omega $ is the scalar (1.12). By [Reference Huang16, Proposition 3.2], equations (3.1) and (3.3)–(3.6) imply condition (i). Observe that (3.2) is identical to (3.5) when (3.1) holds. Applying both sides of (1.5) to v and evaluating the resulting equation using (3.1), (3.2), and (3.4), we obtain equation (3.6). Therefore (3.1)–(3.4) hold if and only if (3.1) and (3.3)–(3.6) hold. The proposition follows.

Proof. Statement (i) is exactly Definition 3.2(i).

( $\Rightarrow $ ): Statement (ii) is immediate from Proposition 3.1 and Definition 3.2.

( $\Leftarrow $ ): Since (ii) holds, it follows from Proposition 3.1 that equations 3.1–3.4 hold. By (i), the scalar $\theta _1^*$ is equal to $\mu q^2+\mu ^{-1}q^{-2}$ . Since $v\not =0$ , comparing (3.2) with (3.8) yields Definition 3.2(ii). For similar reasons, Definition 3.2(iii) and (iv) follow. Therefore $(a,b,c,\lambda )$ is feasible for $(\mu ,\varphi ,\omega ^*,\omega ^\varepsilon )$ .

Proof. By Lemma 2.5, there exists a marginal weight vector v of V with weight $\mu $ . Hence (3.7) follows. By Definition 2.4 and since $q^2\not =1$ , there is a scalar $\varphi \in \mathbb F$ such that (3.8) holds. By Schur’s lemma, there are $\omega ^*,\omega ^\varepsilon \in \mathbb F$ such that (3.9) and (3.10) hold.

Proof.

1. (i): By Theorem 1.5(i), it is routine to verify statement (i).

2. (ii): By Theorem 1.5(i), a straightforward calculation shows that

   $$ \begin{align*} (B-b^{-1}\lambda q^2-b\lambda^{-1}q^{-2}) Am_0= \varphi m_0 + (q^2-q^{-2})(b\lambda^{-1}-b^{-1}\lambda) m_1 \end{align*} $$

for some scalar $\varphi \in \mathbb F$ . Statement (ii) follows from the above equation.

Proof. It is routine to verify the lemma by using Proposition 1.2.

Proof. If $n=1$ , then the lemma is immediate from (ii). Now suppose that $n\geq 2$ . Since $q^4\not =1$ , the roots $1$ , $q^2$ , $q^{-2}$ of the characteristic equation for the linear homogeneous recurrence (i) are mutually distinct. Thus there are $c_0,c_1,c_2\in \mathbb F$ such that

$$ \begin{align*}h_i=c_0 + q^{2i} c_1 + q^{-2i} c_2 \qquad (0\leq i\leq n+1). \end{align*} $$

By (ii), the following linear equations hold:

$$ \begin{align*} \left\{ \begin{array}{ll} c_0 + q^{2j}c_1 + q^{-2j} c_2 =0, \\ c_0 + q^{2k}c_1 + q^{-2k} c_2 =0, \\ c_0 + q^{2\ell}c_1 + q^{-2\ell} c_2 =0. \end{array} \right. \end{align*} $$

The determinant of the coefficient matrix for the above linear equations is equal to

$$ \begin{align*}(q^{j-k}-q^{k-j})(q^{k-\ell}-q^{\ell-k})(q^{\ell-j}-q^{j-\ell}). \end{align*} $$

Since each of $k-j$ , $\ell -k$ , $\ell -j$ is a positive integer less than , none of $q^{2(k-j)}$ , $q^{2(\ell -k)}$ , $q^{2(\ell -j)}$ is equal to one. Therefore the determinant is nonzero. Since the coefficient matrix is invertible, each of $c_0, c_1, c_2$ is zero. The lemma follows.

Proof of Theorem 1.6

By Lemma 2.3, there exists a marginal weight $\mu $ of the $\triangle _q$ -module V. By Lemma 3.4, there are a nonzero vector v of V and three scalars $\varphi ,\omega ^*,\omega ^\varepsilon \in \mathbb F$ satisfying equations (3.7)–(3.10).

By Lemma 2.3, there exists a marginal weight $\kappa $ of the $\triangle _q$ -module $V^\vee $ . Let n denote the dimension of V minus one. Set

(4.1) $$ \begin{align} (a,b,\lambda)=(\kappa q^n,\mu q^n,q^n) \end{align} $$

and let c be the scalar in $\mathbb F^\times $ satisfying

(4.2) $$ \begin{align} c+c^{-1}=&\left\{ \begin{array}{ll} \displaystyle \frac{\omega^\varepsilon-(a+a^{-1})(b+b^{-1})}{\lambda q+\lambda^{-1}q^{-1}} \qquad &\text{if } n=0, \\ \displaystyle \frac{\varphi+(a+a^{-1})(bq-b^{-1}q^{-1})}{\lambda-\lambda^{-1}} \qquad &\text{if }n\geq 1. \end{array} \right. \end{align} $$

Since , it follows that $\lambda ^2=1$ if and only if $n=0$ . Since $q^4\not =1$ , it follows that $\lambda ^2 \not =-q^{-2}$ when $n=0$ . Hence the denominators in the right-hand side of (4.2) are nonzero. Since $\mathbb F$ is algebraically closed, the existence of c follows.

We are now going to show that $(a,b,c,\lambda )$ is feasible for $(\mu ,\varphi ,\omega ^*,\omega ^\varepsilon )$ . Apparently, Definition 3.2(i) is immediate from (4.1). Due to (4.2), we divide the argument into two cases: $n=0$ and $n\geq 1$ .

$(n=0)$ : In this case, v is a basis for V and $(a,b,\lambda )=(\kappa ,\mu ,1)$ . Definition 3.2(iv) is immediate from (4.2). Since a is a weight of $V^\vee $ , it follows from Lemma 4.1 that

(4.3) $$ \begin{align} Av=(a+a^{-1}) v \end{align} $$

on V. Evaluating the left-hand side of (3.8) by using (3.7) and (4.3) yields

$$ \begin{align*}\varphi=-(a+a^{-1})(b q-b^{-1}q^{-1}). \end{align*} $$

Definition 3.2(ii) follows. Applying both sides of (1.6) to v, we evaluate the resulting equation by using (3.7) and (4.3). It follows that $\omega ^*$ is equal to

$$ \begin{align*}(a+a^{-1}) \cdot \frac{\omega^\varepsilon-(a+a^{-1})(b+b^{-1})}{q+q^{-1}} + (q+q^{-1})(b+b^{-1}). \end{align*} $$

Definition 3.2(iii) follows by substituting Definition 3.2(iv) into the above scalar. Therefore $(a,b,c,\lambda )$ is feasible for $(\mu ,\varphi ,\omega ^*,\omega ^\varepsilon )$ when $n=0$ .

$(n\geq 1)$ : Definition 3.2(ii) is immediate from (4.2). Set

(4.4) $$ \begin{align} v_0=v, \qquad v_i= & (A-\theta_{i-1}) v_{i-1} \qquad (1\leq i\leq n). \end{align} $$

By Lemma 2.7, the vectors $\{v_i\}_{i=0}^n$ are a basis for V. By Lemma 2.5, there exists a marginal weight vector w of the $\triangle _q$ -module $V^\vee $ with weight $\kappa $ . It follows from Lemma 2.6(ii) that $\prod _{i=0}^n (A-\theta _i)$ vanishes at $\{B^i w\}_{i=0}^n$ on the $\triangle _q$ -module V. By Lemma 2.7, the vectors $\{B^i w\}_{i=0}^n$ are also a basis for V. Hence $ \prod _{i=0}^n (A-\theta _i) $ vanishes on V. Combined with (4.4), this implies that

(4.5) $$ \begin{align} (A-\theta_n)v_n&=0. \end{align} $$

Let $\varphi ^{(i)}_j\in \mathbb F$ for all integers $i,j$ with $0\leq i,j\leq n$ such that

(4.6) $$ \begin{align} (B-\theta_i^*)v_i= & \sum_{j=0}^n \varphi^{(i)}_j v_j. \end{align} $$

By Lemma 2.6(i), the coefficients

(4.7) $$ \begin{align} \varphi^{(i)}_j= & 0 \qquad (0\leq i\leq j\leq n). \end{align} $$

Applying (3.7) and (3.8) yields that

$$ \begin{align*}\varphi^{(1)}_0=(q-q^{-1})\varphi-\theta_0(\theta_0^*-\theta_1^*). \end{align*} $$

Evaluating the right-hand side of the above equation by using Definition 3.2(ii) yields that

(4.8) $$ \begin{align} \varphi^{(1)}_0= & \varphi_1. \end{align} $$

For any integer i with $1\leq i\leq n$ , we apply both sides of (1.6) to $v_{i-1}$ and evaluate the coefficient of $v_i$ in the resulting equation using (3.7) and (4.4)–(4.7). It follows that

(4.9) $$ \begin{align} \varphi^{(i+1)}_i -(q^2+q^{-2}) \varphi^{(i)}_{i-1} +\varphi^{(i-1)}_{i-2} +c_i +(q-q^{-1})^2\omega^\varepsilon =0 \qquad (1\leq i\leq n). \end{align} $$

Here, $\varphi ^{(0)}_{-1}=\varphi ^{(n+1)}_n=0$ and

$$ \begin{align*}c_i=(\theta_i+\theta_{i-1})(\theta_i^*+\theta_{i-1}^*)-(q^2+q^{-2})(\theta_i\theta_i^*+\theta_{i-1}\theta_{i-1}^*) \qquad (1\leq i\leq n). \end{align*} $$

Applying (4.9) yields that

$$ \begin{align*} \varphi^{(i+2)}_{i+1} -(q^2+1+q^{-2}) (\varphi^{(i+1)}_i-\varphi^{(i)}_{i-1}) -\varphi^{(i-1)}_{i-2} =c_i-c_{i+1} \qquad (1\leq i\leq n-1). \end{align*} $$

On the other hand, a direct calculation shows that $\{\varphi _i\}_{i=0}^{n+1}$ also satisfy the same recurrence. Hence the scalars

$$ \begin{align*}h_i=\varphi^{(i)}_{i-1}-\varphi_i \qquad (0\leq i\leq n+1) \end{align*} $$

satisfy Lemma 4.2(i). By (4.8) and since $\varphi _0=\varphi _{n+1}=\varphi ^{(0)}_{-1}=\varphi ^{(n+1)}_n=0$ , it follows that $h_0=h_1=h_{n+1}=0$ . Applying Lemma 4.2 yields

$$ \begin{align*} \varphi^{(i)}_{i-1}=\varphi_i \qquad (0\leq i\leq n+1). \end{align*} $$

It is now routine to verify Definition 3.2(iv) by substituting the above equations into (4.9).

For any integer i with $0\leq i\leq n$ , we apply both sides of (1.6) to $v_i$ and evaluate the coefficient of $v_i$ in the resulting equation. It follows that

(4.10) $$ \begin{align} (q-q^{-1})(q^2-q^{-2})\omega^*= \varphi^{(i+2)}_i -(q^2+q^{-2}) \varphi^{(i+1)}_{i-1} + \varphi^{(i)}_{i-2} + c_i \qquad (0\leq i\leq n). \end{align} $$

Here, $\varphi ^{(0)}_{-2}=\varphi ^{(1)}_{-1}=\varphi ^{(n+1)}_{n-1}=\varphi ^{(n+2)}_n=0$ and

$$ \begin{align*} \begin{split} c_i &= (\theta_i-\theta_{i+1})\varphi_i +(\theta_i-\theta_{i-1})\varphi_{i+1} -(q-q^{-1})^2\theta_i^2\theta_i^* \\ & \qquad +\,(q^2-q^{-2})^2\theta_i^* +(q-q^{-1})^2\theta_i\omega^\varepsilon \end{split} \qquad (0\leq i\leq n), \end{align*} $$

where $\theta _{-1}=a q^{-n-2}+a^{-1} q^{n+2}$ . A direct calculation shows that $c_i$ is equal to $(q-q^{-1})(q^2-q^{-2})$ times the right-hand side of Definition 3.2(iii) for each $i=0,1,\ldots ,n$ . Combined with (4.10), this implies that the values of $ \varphi ^{(i+2)}_i -(q^2+q^{-2})\varphi ^{(i+1)}_{i-1} +\varphi ^{(i)}_{i-2} $ are identical for all $i=0,1,\ldots ,n$ . It follows that

$$ \begin{align*}\varphi^{(i+2)}_{i}-(q^2+1+q^{-2})(\varphi^{(i+1)}_{i-1}-\varphi^{(i)}_{i-2})-\varphi^{(i-1)}_{i-3}=0 \qquad (1\leq i\leq n). \end{align*} $$

Applying Lemma 4.2 yields

$$ \begin{align*}\varphi^{(i)}_{i-2}=0 \qquad (0\leq i\leq n+2). \end{align*} $$

Definition 3.2(iii) follows by substituting the above equations into (4.10). Therefore, $(a,b,c,\lambda )$ is feasible for $(\mu ,\varphi ,\omega ^*,\omega ^\varepsilon )$ when $n\geq 1$ .

Since $(a,b,c,\lambda )$ is feasible for $(\mu ,\varphi ,\omega ^*,\omega ^\varepsilon )$ , it follows from Theorem 3.3 that there exists a $\triangle _q$ -module homomorphism

(4.11) $$ \begin{align} M_\lambda(a,b,c)\to V \end{align} $$

that sends $m_0$ to v. By (4.3) and (4.5), the vector $m_{n+1}$ lies in the kernel of (4.11). Recall from Section 1 that the $\triangle _q$ -submodule $N_\lambda (a,b,c)$ of $M_\lambda (a,b,c)$ is generated by $m_{n+1}$ . Hence (4.11) induces the $\triangle _q$ -module homomorphism

(4.12) $$ \begin{align} V_n(a,b,c)\to V \end{align} $$

that sends $m_0+N_\lambda (a,b,c)$ to v. Since $V_n(a,b,c)$ and V have the same dimension $n+1$ and by the irreducibility of the $\triangle _q$ -module V, the map (4.12) is a $\triangle _q$ -module isomorphism. The result follows.

Proof. Let

(5.1) $$ \begin{align} (a,b)=(\kappa\lambda,\mu\lambda). \end{align} $$

Clearly, Definition 3.2(i) holds. Then $(a,b,c,\lambda )$ is feasible for $(\mu ,\varphi ,\omega ^*,\omega ^\varepsilon )$ if and only if Definition 3.2(ii)–(iv) hold.

( $\Rightarrow $ ): Suppose that Definition 3.2(ii)–(iv) hold and we show (i)–(iii). Condition (iii) is immediate from Definition 3.2(ii) if $\lambda ^2\not =1$ ; condition (iii) is immediate from Definition 3.2(iv) if $\lambda ^2=1$ .

Using Definition 3.2(ii) and (iv), it is routine to verify that

(5.2) $$ \begin{align} \frac{\omega^\varepsilon-q\varphi}{q+q^{-1}}&=\lambda^{-1}(c+c^{-1})+bq(a+a^{-1}), \end{align} $$

(5.3) $$ \begin{align} \frac{\omega^\varepsilon+q^{-1}\varphi}{q+q^{-1}}&=\lambda(c+c^{-1})+b^{-1}q^{-1}(a+a^{-1}). \end{align} $$

We multiply (5.2) and (5.3) by $\lambda $ and $\lambda ^{-1}$ , respectively. The difference between the resulting equations gives

$$ \begin{align*}\lambda\frac{\omega^\varepsilon-q\varphi}{q+q^{-1}}-\lambda^{-1}\frac{\omega^\varepsilon+q^{-1}\varphi}{q+q^{-1}} =(a+a^{-1})(b\lambda q-b^{-1} \lambda^{-1} q^{-1}). \end{align*} $$

Condition (ii) follows by substituting (5.1) into the above equation.

We multiply (5.2) and (5.3) by $a^{-1}\lambda $ and $a\lambda ^{-1}$ , respectively. The sum of the resulting equations gives

(5.4) $$ \begin{align} a^{-1}\lambda \frac{\omega^\varepsilon-q\varphi}{q+q^{-1}}+a\lambda^{-1} \frac{\omega^\varepsilon+q^{-1}\varphi}{q+q^{-1}} =& (a+a^{-1})(a^{-1}b\lambda q+ab^{-1}\lambda^{-1}q^{-1}+c+c^{-1}). \end{align} $$

Subtracting Definition 3.2(iii) from (5.4) yields

$$ \begin{align*}a^{-1}\lambda \frac{\omega^\varepsilon-q\varphi}{q+q^{-1}} +a\lambda^{-1} \frac{\omega^\varepsilon+q^{-1}\varphi}{q+q^{-1}} -\omega^* = (ab^{-1}-a^{-1}b)(a\lambda^{-1}q^{-1}-a^{-1}\lambda q). \end{align*} $$

Condition (i) follows by substituting (5.1) into the above equation. The “only if” part follows.

( $\Leftarrow $ ): Suppose that (i)–(iii) hold. We show Definition 3.2(ii)–(iv). Definition 3.2(ii) is immediate from (iii) if $\lambda ^2\not =1$ ; Definition 3.2(ii) is obtained from condition (ii) if $\lambda ^2=1$ .

Definition 3.2(iv) is immediate from (iii) if $\lambda ^2=1$ . Suppose that $\lambda ^2\not =1$ . Applying (ii) yields

$$ \begin{align*}\frac{\lambda-\lambda^{-1}}{q+q^{-1}}\omega^\varepsilon=(\kappa\lambda+\kappa^{-1}\lambda^{-1})(\mu\lambda^2 q-\mu^{-1}\lambda^{-2} q^{-1}) +\frac{\lambda q+\lambda^{-1} q^{-1}}{q+q^{-1}}\varphi. \end{align*} $$

To get Definition 3.2(iv), we evaluate the right-hand side of the above equation by using Definition 3.2(ii) and replacing $\kappa $ and $\mu $ with $a\lambda ^{-1}$ and $b\lambda ^{-1}$ .

Applying (i) yields

$$ \begin{align*}\omega^*=(\kappa^{-1}q-\kappa q^{-1}) \left( \kappa\mu^{-1}-\kappa^{-1}\mu-\frac{\varphi}{q+q^{-1}} \right) +\frac{\kappa+\kappa^{-1}}{q+q^{-1}}\omega^\varepsilon. \end{align*} $$

To get Definition 3.2(iii), we evaluate the right-hand side of the above equation by using Definition 3.2(ii) and (iv) and replacing $\kappa $ and $\mu $ with $a\lambda ^{-1}$ and $b\lambda ^{-1}$ . The “if” part follows.

Proof.

1. (i): Since $\mathbb F$ is algebraically closed, statement (i) is immediate from Theorem 5.1.

2. (ii): Since $(a,b,c,\lambda )$ is feasible for $(\mu ,\varphi ,\omega ^*,\omega ^\varepsilon )$ , we may substitute Definition 3.2(i)–(iv) into the polynomials given in Theorem 5.1(i)–(iii). By Theorem 5.1, one may factor those polynomials into linear factors to obtain all quadruples that are feasible for $(\mu ,\varphi ,\omega ^*,\omega ^\varepsilon )$ . By Table A1, these quadruples are all elements of the equivalence class of $(a,b,c,\lambda )$ under .

Proof.

1. (i): Since , the vector . Combined with Theorem 1.5(i), statement (i) follows.

2. (ii): Immediate from Theorem 1.5(ii).

3. (iii): Using Lemma 6.1(i) yields that for all . Since $W_\lambda ^\delta (a,b,c)$ is spanned by , statement (iii) follows.

Proof. ( $\Rightarrow $ ): Composing the $\triangle _q$ -module homomorphism $W_\lambda ^\delta (a,b,c)\to V$ with the canonical map $M_\lambda (a,b,c)\to W_\lambda ^\delta (a,b,c)$ yields condition (i). Condition (ii) is immediate from Lemma 6.1(iii).

( $\Leftarrow $ ): By (ii), the vector

lies in the kernel of the $\triangle _q$ -module homomorphism $M_\lambda (a,b,c)\to V$ described in (i). By Theorem 1.5(i), the above subtrahend is equal to . Since the $\triangle _q$ -submodule $O_\lambda ^{\delta }(a,b,c)$ of $M_\lambda (a,b,c)$ is generated by , this induces the desired $\triangle _q$ -module homomorphism. The proposition follows.

Proof. Let y be another indeterminate over $\mathbb F$ commuting with x. In the proof, we view as a polynomial in x with coefficients in the Laurent polynomial ring $\mathbb F[y^{\pm 1}]$ . Since , it follows from (6.1) that

Since are mutually distinct elements of $\mathbb F[y^{\pm 1}]$ and is a monic polynomial of degree , it follows that

The lemma follows.

Proof. Immediate from Lemma 6.3 and Theorem 6.4.

Proof of Theorem 1.9

Let $\mu $ denote a marginal weight of the $\triangle _q$ -module V. By Lemma 3.4, there are a nonzero vector v of V and three scalars $\varphi ,\omega ^*,\omega ^\varepsilon \in \mathbb F$ satisfying equations (3.7)–(3.10).

By Theorem 5.2(i), there exists a quadruple $(a,b,c,\lambda )\in {\mathbb F^\times }^4$ that is feasible for $(\mu ,\varphi ,\omega ^*,\omega ^\varepsilon )$ . By Theorem 3.3, there exists a unique $\triangle _q$ -module homomorphism

$$ \begin{align*}M_\lambda(a,b,c)\to V \end{align*} $$

that sends $m_0$ to v. By Lemma 6.5, the element is central in $\triangle _q$ . By Schur’s lemma, this central element acts on V as scalar multiplication by some $\delta \in \mathbb F$ . Consequently, by Proposition 6.2, there exists a unique $\triangle _q$ -module homomorphism

(6.2) $$ \begin{align} W_\lambda^\delta (a,b,c)\to V \end{align} $$

that sends $w_0$ to v. Since $W_\lambda ^\delta (a,b,c)$ and V have the same dimension and by the irreducibility of the $\triangle _q$ -module V, the map (6.2) is a $\triangle _q$ -module isomorphism. The result follows.

Proof. (ii) and (iii) $\Leftrightarrow $ (iv) and (v): Trivial.

(i) $\Rightarrow $ (ii) and (iii): By Theorem 5.2(ii), condition (i) implies that $(a,b,c,\lambda )$ and $(\bar {a},\bar {b},\bar {c},\bar {\lambda })$ are feasible for the same element of $\mathbb F^\times \times \mathbb F^3$ . Conditions (ii) and (iii) then follow from Theorem 3.3.

(ii) $\Rightarrow $ (i): By Lemma 3.5(i), the vector $m_0$ is a marginal weight vector of $M_\lambda (a,b,c)$ with weight $b\lambda ^{-1}$ . By Lemma 3.5(ii) and since $\bar {m}_0$ and $\bar {m}_1$ are linearly independent, condition (ii) implies that $\bar {b}\bar {\lambda }^{-1}=b\lambda ^{-1}$ . Applying Theorem 3.3 yields that $(a,b,c,\lambda )$ and $(\bar {a},\bar {b},\bar {c},\bar {\lambda })$ are feasible for the same element of $\mathbb F^\times \times \mathbb F^3$ . Condition (i) then follows from Theorem 5.2(ii).

(iii) $\Rightarrow $ (i): Similar to the proof of (ii) $\Rightarrow $ (i).

Proof. Composing the given $\triangle _q$ -module homomorphism with the canonical map $M_\lambda (a,b,c)\to W_\lambda ^\delta (a,b,c)$ yields the $\triangle _q$ -module homomorphism $ M_\lambda (a,b,c) \to W_{\bar {\lambda }}^{\bar {\delta }}(\bar {a},\bar {b},\bar {c}) $ that sends $m_0$ to $\bar {w}_0$ . By Lemma 3.5(i), the vector $m_0$ is a marginal weight vector of $M_\lambda (a,b,c)$ with weight $b\lambda ^{-1}$ . Since $\bar {w}_0$ and $\bar {w}_1$ are linearly independent and by Lemma 3.5(ii), the scalar $\bar {b}\bar {\lambda }^{-1}=b\lambda ^{-1}$ . Applying Theorem 3.3 yields that $(a,b,c,\lambda )$ and $(\bar {a},\bar {b},\bar {c},\bar {\lambda })$ are feasible for the same element of $\mathbb F^\times \times \mathbb F^3$ . The lemma then follows from Theorem 5.2(ii).

Proof. By Proposition 6.2, it suffices to show that equation (1.15) holds if and only if the following equation holds:

(7.1)

To see this, we utilize the following two equations: It follows from Lemma 6.1(iii) that

(7.2)

It follows from Lemma 6.3 that

(7.3)

(7.1) $\Rightarrow $ (1.15): We replace x by A and then apply both sides of (7.3) to $\bar {w}$ . Simplifying the resulting equation using (7.1) and (7.2), we obtain

Since $\bar {w}\not =0$ , equation (1.15) follows.

(1.15) $\Rightarrow $ (7.1): Simplifying (7.3) by using (1.15) yields that

We replace x by A and then apply both sides of the above equation to $\bar {w}$ . Since the right-hand side of the resulting equation is equal to the zero vector by (7.2), equation (7.1) follows.

Proof. (ii) and (iii) $\Leftrightarrow $ (iv) and (v): Trivial.

(ii) and (iii) $\Rightarrow $ (i): Immediate from Lemma 7.2 and Proposition 7.3.

(i) $\Rightarrow $ (ii): Since by Definition 1.10(ii), the map mentioned in Theorem 7.1(ii) exists. Composing this map with the canonical map $ M_{\bar {\lambda }}(\bar {a},\bar {b},\bar {c}) \to W_{\bar {\lambda }}^{\bar {\delta }}(\bar {a},\bar {b},\bar {c}) $ yields the $\triangle _q$ -module homomorphism $ M_\lambda (a,b,c) \to W_{\bar {\lambda }}^{\bar {\delta }}(\bar {a},\bar {b},\bar {c}) $ that maps $m_0$ to $\bar {w}_0$ . Since (1.15) holds by Definition 1.10(ii), condition (ii) is now immediate from Proposition 7.3.

(i) $\Rightarrow $ (iii): Similar to the proof of (i) $\Rightarrow $ (ii).

Proof. Suppose, to the contrary, that $\delta =0$ and $\lambda ^2=q^{2(i-1)}$ for some integer i with . Let W be the subspace of $W_\lambda ^\delta (a,b,c)$ spanned by . By Lemma 6.1(ii), the elements $\alpha ,\beta ,\gamma $ act on W as scalar multiplication. Since $\delta =0$ and $\varphi _i=0$ by (1.11), it follows from Lemma 6.1(i) that W is invariant under A and B. Hence W is a proper $\triangle _q$ -submodule of $W_\lambda ^\delta (a,b,c)$ by Proposition 1.2, a contradiction.

Proof of the implication (i) $\Rightarrow $ (ii) of Theorem 1.11

By Theorem 7.4, the $\triangle _q$ -module $W_\lambda ^{\delta }(a,b,c)$ is isomorphic to $W_{\bar {\lambda }}^{\bar {\delta }}(\bar {a},\bar {b},\bar {c})$ for any . Combining this with Lemma 8.1, the implication (i) $\Rightarrow $ (ii) follows.

Proof. By Lemma 6.1(iii), the characteristic polynomial of A on $W_\lambda ^\delta (a,b,c)$ is equal to the right-hand side of (8.6). According to Lemma 6.3, we have

Combining this with (8.5), the lemma follows.

Proof. Immediate from (8.7).

Proof. Applying Lemma 6.1(i), a direct calculation shows that $(A-\vartheta _i)e_i$ is equal to

A change of indices shows that the above vector is equal to the scalar multiple of $w_0$ by

which is equal to zero by (8.6). Therefore, $e_i$ is a weight vector of $W_\lambda ^\delta (a,b,c)^\vee $ with weight $\nu ^{-1} q^{2i}$ or $\nu q^{-2i}$ . By Lemma 6.1(i), the rank of $A-\vartheta _i$ on $W_\lambda ^\delta (a,b,c)$ is at least . The lemma now follows from the rank-nullity theorem.

Proof. (i) $\Leftrightarrow $ (ii): Immediate from Definition 1.4 and Lemma 8.5(i).

(ii) $\Leftrightarrow $ (iii): Immediate from Definition 2.4 and Lemma 8.5(i).

(ii) $\Leftrightarrow $ (iv): Applying Lemma 1.3(ii) to $W_\lambda ^\delta (a,b,c)^\vee $ yields that

(8.8) $$ \begin{align} (A-\vartheta_{i+1})(A-\vartheta_i)Be_i\in W_\lambda^\delta(a,b,c)^\vee(\nu^{-1}q^{2(i-1)}). \end{align} $$

By Lemma 8.5(i), the left-hand side of (8.8) is a scalar multiple of $e_{i-1}$ , where $e_{i-1}$ is interpreted as if $i=0$ . In view of Lemma 8.4, condition (ii) holds if and only if the coefficient of in the left-hand side of (8.8) is equal to zero. Applying Lemma 6.1(i), a direct calculation shows that this coefficient is equal to $b^{-1}\lambda ^{-1} q^{-1}(q-q^{-1})(q^2-q^{-2})$ times

$$ \begin{align*} (q^{1-i}-bc \nu^{-1} q^i) (q^{1-i}-bc^{-1} \nu^{-1} q^i) (q^{i-1}-a \lambda \nu q^{1-i}) (q^{i-1}-a^{-1} \lambda \nu q^{1-i}). \end{align*} $$

The equivalence of (ii) and (iv) follows.

Proof. Similar to the proof of Lemma 8.6.

Proof. Immediate from Lemmas 8.5–8.7.

Proof. ( $\Rightarrow $ ): Since the $\triangle _q$ -module $W_\lambda ^\delta (a,b,c)$ is reducible, there exists a proper irreducible $\triangle _q$ -submodule W of $W_\lambda ^\delta (a,b,c)$ . By Lemma 2.3, there exists a marginal weight $\mu $ of $W^\vee $ . It follows from Lemma 2.2(iii) that $\mu $ is also a marginal weight of $W_\lambda ^\delta (a,b,c)^\vee $ . By Lemma 8.3, there exists an integer i with such that $\mu =\nu q^{-2i}$ or $\mu =\nu ^{-1}q^{2i}$ . Combining this with Lemma 8.8, condition (i) follows.

Since $\mu $ is a weight of $W^\vee $ , it follows from Lemma 8.5 that $e_i\in W$ . Combined with (8.10), the vector . Since the $\triangle _q$ -module $W_\lambda ^\delta (a,b,c)$ is generated by $w_0$ , the vector $w_0\not \in W$ . Condition (ii) follows.

( $\Leftarrow $ ): Suppose, to the contrary, that the $\triangle _q$ -module $W_\lambda ^\delta (a,b,c)$ is irreducible. Applying Lemma 2.7 to $W_\lambda ^\delta (a,b,c)^\vee $ , the vectors form a basis for $W_\lambda ^\delta (a,b,c)$ . However, by condition (ii), the right-hand side of (8.10) is zero. This implies that are linearly dependent, a contradiction.

Proof. Fix an integer i with . Recall the coefficients from (8.9). Applying Lemma 6.1(i) yields that

(8.11)

(8.12)

By (8.7), we have

(8.13)

The coefficients are uniquely determined by the recurrence (8.12) and the initial conditions (8.11) and (8.13). To see (i)–(iv), it is routine but tedious to verify that the given formulas satisfy (8.11)–(8.13).

Proof of the implication (ii)  $\Rightarrow $  (i) of Theorem 1.11

Suppose, to the contrary, that Theorem 1.11(i) fails. By Theorem 8.9, there is an integer i with such that Theorem 8.9(i) and (ii) hold. By Lemma 8.8, there are four possible cases: (a) $\nu q^{-2i}\in \{a\lambda ^{-1}q^{-2},a^{-1}\lambda q^2\}$ ; (b) $\nu q^{-2i}\in \{a\lambda q^2,a^{-1}\lambda ^{-1} q^{-2}\}$ ; (c) $\nu q^{-2i}\in \{bcq^{-1},b^{-1}c^{-1}q\}$ ; and (d) $\nu q^{-2i}\in \{bc^{-1}q^{-1},b^{-1}cq\}$ .

(a): Since is the order of $q^2$ , the scalar . It follows from (8.5) that $\delta =0$ . By Lemma 8.2(i), condition (8.1) holds. Hence $\varphi _h\not =0$ for all by (1.11). Then by Proposition 8.10(i), which contradicts Theorem 8.9(ii).

(b): Since is the order of $q^2$ , it follows that . Using (8.5) yields . By Lemma 8.2(ii), condition (8.2) holds. Then by Proposition 8.10(ii), which contradicts Theorem 8.9(ii).

(c): Since is the order of $q^2$ , it follows that . Using (8.5) yields . By Lemma 8.2(iii), condition (8.3) holds. Then by Proposition 8.10(iii), which contradicts Theorem 8.9(ii).

(d): Since is the order of $q^2$ , it follows that . Using (8.5) yields . By Lemma 8.2(iv), condition (8.4) holds. Then by Proposition 8.10(iv), which contradicts Theorem 8.9(ii).

Proof. Immediate from Lemma 6.1(i).

Proof. Since $\varphi _i=0$ , the first assertion follows from Lemma 6.1(i) and the characteristic polynomial of B on W is

(9.2) $$ \begin{align} \prod_{h=i}^j(x-\theta_h^*). \end{align} $$

Recall from (9.1) that $w_{ij}$ is a linear combination of $\{w_h\}_{h=i}^j$ . Hence $w_{ij}\in W$ . It follows from Lemma 9.2 that $(B-\theta _j^*)w_{ij}=0$ . To see the second assertion, under assumption (i) or (ii), we need to show that (a) the $\theta _j^*$ -eigenspace of B on W is one-dimensional and (b) the vector $w_{ij}$ is nonzero.

Suppose that (i) holds. Then the rank of $B-\theta _j^*$ on W is equal to $j-i$ . By the rank-nullity theorem, condition (a) follows. Since the coefficient of $w_i$ in $w_{ij}$ is $\varphi _{i+1}\varphi _{i+2}\dots \varphi _j \not =0$ , condition (b) follows. Suppose that (ii) holds. Then $\theta _j^*$ is a simple root of (9.2). Hence condition (a) follows. Since the coefficient of $w_j$ in $w_{ij}$ is $ (\theta _j^*-\theta _i^*)(\theta _j^*-\theta _{i+1}^*)\dots (\theta _j^*-\theta _{j-1}^*) \not =0 $ , condition (b) follows. The lemma follows.

Proof. Since $\lambda ^2=q^{2(i-1)}$ and $(\bar {b},\bar {\lambda })=(b,\lambda ^{-1}q^{-2})$ , it follows that

(9.3) $$ \begin{align} \bar{b}\bar{\lambda}^{-1}=b \lambda^{-1} q^{2i}. \end{align} $$

Condition (i) follows from this and (1.10). Let $\mu $ denote the scalar (9.3). Then $(\mu +\mu ^{-1},\mu q^2+\mu ^{-1} q^{-2})=(\theta _i^*,\theta _{i+1}^*)$ . Since $\lambda ^2=q^{2(i-1)}$ , it follows from (1.11) that $\varphi _i=0$ . By Lemma 9.2, the vector $w_i$ is a weight vector of $W_\lambda ^{\delta }(a,b,c)$ with weight $\mu $ . Let

$$ \begin{align*}\varphi=\frac{\varphi_{i+1}+\theta_i(\theta_i^*-\theta_{i+1}^*)}{q-q^{-1}}. \end{align*} $$

Using Lemma 6.1(i), a routine calculation yields that $ (B-\theta _{i+1}^*)Aw_i=(q-q^{-1})\varphi \, w_i. $ Condition (ii) follows.

Under the hypothesis $\lambda ^2=q^{2(i-1)}$ , the quadruple $(\bar {a},\bar {b},\bar {c},\bar {\lambda })$ is feasible for $(\mu ,\varphi ,\omega ^*,\omega ^\varepsilon )$ , where $\omega ^*$ and $\omega ^{\varepsilon }$ are the scalars (1.13) and (1.14), respectively. By Theorem 3.3, there exists a $\triangle _q$ -module homomorphism $ M_{\bar {\lambda }}(\bar {a},\bar {b},\bar {c})\to W_\lambda ^{\delta }(a,b,c) $ that sends $\bar {m}_0$ to $w_i$ . Since , equation (1.15) holds. Condition (iii) now follows from Proposition 7.3.

Proof. Since $(\bar {b},\bar {\lambda })=(b^{-1},\lambda ^{-1}q^{-2})$ , it follows that

(9.4) $$ \begin{align} \bar{b}\bar{\lambda}^{-1}=b^{-1}\lambda q^2. \end{align} $$

Condition (i) follows. Let $\mu $ denote the scalar (9.4). Then . Observe that condition (a) of Definition 1.12(iii) is equivalent to

(9.5)

By Lemma 9.5, the vector is a basis for the weight space $W_\lambda ^\delta (a,b,c)(\mu )$ . Under condition (b) of Definition 1.12(iii), Lemma 9.4(iii) implies that

Since $q^2\not =1$ and is a basis for $W_\lambda ^\delta (a,b,c)(\mu )$ , there exists a scalar $\varphi \in \mathbb F$ such that Condition (ii) follows.

Applying Lemma 6.1(i), a direct calculation shows that

By Definition 3.2, it is straightforward to verify that $(\bar {a},\bar {b},\bar {c},\bar {\lambda })$ is feasible for $(\mu ,\varphi ,\omega ^*,\omega ^\varepsilon )$ , where $\omega ^*$ and $\omega ^{\varepsilon }$ are the scalars (1.13) and (1.14), respectively. By Theorem 3.3, there exists a $\triangle _q$ -module homomorphism $ M_{\bar {\lambda }}(\bar {a},\bar {b},\bar {c})\to W_\lambda ^{\delta }(a,b,c) $ that sends $\bar {m}_0$ to . Since , equation (1.15) holds. Condition (iii) now follows from Proposition 7.3.

Proof. Immediate from Theorem 7.4 and Lemmas 9.6 and 9.7.

Proof. According to Table A1, the scalar $b\lambda ^{-1}=\bar {b}\bar {\lambda }^{-1}$ . Together with (1.10), this implies the lemma.

Proof. Since $\varphi _i=0$ , it follows from (1.11) that

$$ \begin{align*}q^{2(i-1)}\in\{\lambda^2, a^{-1}b^{-1}c^{-1}\lambda q^{-1},a^{-1}b^{-1}c\lambda q^{-1}\}. \end{align*} $$

We set $\bar {\delta }=\delta $ and select $(\bar {a},\bar {b},\bar {c},\bar {\lambda })$ from the $2$ nd, $4$ th, and $7$ th rows of Table A1 as follows:

$$ \begin{align*} (\bar{a},\bar{b},\bar{c},\bar{\lambda}) =\left\{ \begin{array}{ll} (a,b,c,\lambda) \qquad &\text{if }q^{2(i-1)}=\lambda^2, \\ (\frac{a}{\sqrt{abc\lambda q}}, \frac{b}{\sqrt{abc\lambda q}}, \frac{c}{\sqrt{abc\lambda q}}, \frac{\lambda}{\sqrt{abc\lambda q}}) \qquad &\text{if }q^{2(i-1)}=a^{-1}b^{-1}c^{-1}\lambda q^{-1}, \\ (\frac{ac}{\sqrt{abc\lambda q}}, \frac{bc}{\sqrt{abc\lambda q}}, \frac{1}{\sqrt{abc\lambda q}}, \frac{c\lambda}{\sqrt{abc\lambda q}}) \qquad &\text{if }q^{2(i-1)}=a^{-1}b^{-1}c\lambda q^{-1}. \end{array} \right. \end{align*} $$

By construction, condition (i) holds. Since $\bar {a}\bar {\lambda }^{-1}=a\lambda ^{-1}$ , condition (ii) holds by (1.9). Since by Definition 1.10(i), condition (iii) is immediate from Lemma 10.1. It is straightforward to verify condition (iv) using (1.11).

Since $(\bar {a}\bar {\lambda }^{-1},\bar {\delta })=(a\lambda ^{-1},\delta )$ , equation (1.15) holds. Hence by Definition 1.10(ii). By Theorem 7.4, there exists a $\triangle _q$ -module isomorphism

$$ \begin{align*}f:W_{\bar{\lambda}}^{\bar{\delta}}(\bar{a},\bar{b},\bar{c})\to W_\lambda^\delta(a,b,c) \end{align*} $$

that maps $\bar {w}_0$ to $w_0$ . Applying (ii) and Lemma 6.1(i) yields that $f(\bar {w}_h)=w_h$ for all . Condition (v) follows.

Proof. Immediate from Theorem 1.11 and Proposition 9.8.

Proof. If $i=0$ , then the lemma follows by selecting $(\bar {a},\bar {b},\bar {c},\bar {\lambda },\bar {\delta })=(a,b,c,\lambda ,\delta )$ . Suppose that $i\geq 1$ . By Lemma 10.2, we may assume that $\lambda ^2=q^{2(i-1)}$ instead of $\varphi _i=0$ . The lemma is now immediate from Lemma 9.6.

Proof. The proof proceeds by induction on $j-i$ . By Lemma 10.4, there exists a quintuple $(\hat {a},\hat {b},\hat {c},\hat {\lambda },\hat {\delta })\in {\mathbb F^\times }^4\times \mathbb F$ such that $(a,b,c,\lambda ,\delta )\simeq (\hat {a},\hat {b},\hat {c},\hat {\lambda },\hat {\delta })$ and the following conditions hold:

1. (1) $\hat {\theta }_h^*=\theta _{i+h}^*$ for all $h\in \mathbb N$ .

2. (2) There is a $\triangle _q$ -module homomorphism $f_1:W_{\hat {\lambda }}^{\hat {\delta }}(\hat {a},\hat {b},\hat {c})\to W_\lambda ^\delta (a,b,c)$ that maps $\hat {w}_0$ to $w_i$ .

Since and by Lemma 10.3, the quintuple . Since the $\triangle _q$ -module $W_\lambda ^\delta (a,b,c)$ is irreducible by Theorem 1.11, it follows that $f_1$ is a $\triangle _q$ -module isomorphism. If $i=j$ , then the lemma follows by selecting $k=i$ and $(\bar {a},\bar {b},\bar {c},\bar {\lambda },\bar {\delta })=(\hat {a},\hat {b},\hat {c},\hat {\lambda },\hat {\delta })$ .

Suppose that $j> i$ . Since $\hat {\varphi }_0=0$ , there is an integer $i'$ with $0\leq i'\leq j-i$ such that $\hat {\varphi }_{i'}=0$ and $\hat {\varphi }_{i'+1}\hat {\varphi }_{i'+2}\dots \hat {\varphi }_{j-i}\not =0$ . Suppose that $i'\geq 1$ . By the induction hypothesis, there are an integer $k'$ with $i'\leq k'\leq j-i$ and a quintuple such that $(\hat {a},\hat {b},\hat {c},\hat {\lambda },\hat {\delta })\simeq (\bar {a},\bar {b},\bar {c},\bar {\lambda },\bar {\delta })$ and the following conditions hold:

1. (1’) $\bar {\varphi }_1\bar {\varphi }_2\dots \bar {\varphi }_{j-i-k'}\not =0$ .

2. (2’) $\bar {\theta }_h^*=\hat {\theta }_{k'+h}^*$ for all $h\in \mathbb N$ .

3. (3’) There is a $\triangle _q$ -module isomorphism $f_2:W_{\bar {\lambda }}^{\bar {\delta }}(\bar {a},\bar {b},\bar {c})\to W_{\hat {\lambda }}^{\hat {\delta }}(\hat {a},\hat {b},\hat {c})$ that maps $\bar {w}_{0,j-i-k'}$ to $\hat {w}_{i',j-i}$ .

If $i'=0$ , then the above statement remains true by selecting $k'=0$ and $(\bar {a},\bar {b},\bar {c},\bar {\lambda },\bar {\delta })=(\hat {a},\hat {b},\hat {c},\hat {\lambda },\hat {\delta })$ . By the transitivity of $\simeq $ , the quintuple $(a,b,c,\lambda ,\delta )\simeq (\bar {a},\bar {b},\bar {c},\bar {\lambda },\bar {\delta })$ . Set

$$ \begin{align*}k=i+k'. \end{align*} $$

Then condition (i) is exactly condition (1’). Condition (ii) is immediate from (1) and (2’).

It follows from Lemma 9.2 that $(B-\bar {\theta }_{j-k}^*)\bar {w}_{0,j-k}=0$ . Since $\bar {\theta }_{j-k}^*=\theta _j^*$ by (1) and (2’), applying $f_1\circ f_2$ to the above equation yields

$$ \begin{align*}(B-\theta_j^*)(f_1\circ f_2)(\bar{w}_{0,j-k})=0. \end{align*} $$

Applying Lemma 9.5 to (1’) yields that $\bar {w}_{0,j-k}$ is nonzero. Since $f_1$ and $f_2$ are two $\triangle _q$ -module isomorphisms, the vector $(f_1\circ f_2)(\bar {w}_{0,j-k})$ is nonzero. By (3’), the vector $(f_1\circ f_2)(\bar {w}_{0,j-k})=f_1(\hat {w}_{i',j-i})$ . Applying Lemma 6.1(i) to (2) yields that $f_1(\hat {w}_{i',j-i})$ is a linear combination of $\{w_h\}_{h=i}^{j}$ . Since $\varphi _i=0$ and $\varphi _{i+1}\varphi _{i+2}\dots \varphi _j\not =0$ , it follows from Lemma 9.5 that there exists $\varepsilon \in \mathbb F^\times $ such that

$$ \begin{align*}f_1(f_2(\bar{w}_{0,j-k}))=\varepsilon w_{ij}. \end{align*} $$

The $\triangle _q$ -module isomorphism $\varepsilon ^{-1}\cdot (f_1\circ f_2)$ satisfies condition (iii). The lemma follows.

Proof. (ii) $\Rightarrow $ (i): Immediate from Lemma 9.3.

(i) $\Rightarrow $ (ii): Since $\varphi _1\varphi _2\dots \varphi _{i-1}\not =0$ , it follows from Lemma 9.5 that $w_{0,i-1}$ is nonzero. Together with the assumption $\varphi _i\not =0$ , the implication (i) $\Rightarrow $ (ii) follows from (1.11) and Lemma 9.3.

Proof. If $i=0$ , then the lemma follows by selecting $(\bar {a},\bar {b},\bar {c},\bar {\lambda },\bar {\delta })=(a,b,c,\lambda ,\delta )$ . Suppose that $i\geq 1$ . It follows from Lemma 10.6 that $q^{2(i-1)}\in \{b^{-2}q^{-2}, ab^{-1}c\lambda q^{-1}, ab^{-1}c^{-1}\lambda q^{-1}\}$ . We select $(\hat {a},\hat {b},\hat {c},\hat {\lambda })$ from the $10$ th, $13$ th, and $18$ th rows of Table A1 as follows:

$$ \begin{align*} (\hat{a},\hat{b},\hat{c},\hat{\lambda}) =\left\{ \begin{array}{ll} (c, \lambda^{-1}q^{-1}, a, b^{-1}q^{-1}) \qquad &\text{if }q^{2(i-1)}=b^{-2}q^{-2}, \\ (\frac{c}{\sqrt{abc\lambda q}}, \frac{abc}{\sqrt{abc\lambda q}}, \frac{a}{\sqrt{abc\lambda q}}, \frac{ac\lambda}{\sqrt{abc\lambda q}}) \qquad &\text{if }q^{2(i-1)}=ab^{-1}c\lambda q^{-1}, \\ (\frac{1}{\sqrt{abc\lambda q}}, \frac{ab}{\sqrt{abc\lambda q}}, \frac{ac}{\sqrt{abc\lambda q}}, \frac{a\lambda}{\sqrt{abc\lambda q}}) \qquad &\text{if }q^{2(i-1)}=ab^{-1}c^{-1}\lambda q^{-1}. \end{array} \right. \end{align*} $$

By construction, $\hat {\lambda }^2=q^{2(i-1)}$ . Note that by Definition 1.10(i). Set

Then by Definition 1.10(ii). By Theorem 7.4, there exists a $\triangle _q$ -module isomorphism

$$ \begin{align*}f_1:W_{\hat{\lambda}}^{\hat{\delta}}(\hat{a},\hat{b},\hat{c})\to W_\lambda^\delta(a,b,c) \end{align*} $$

that maps $\hat {w}_0$ to $w_0$ . Applying Lemma 9.6, there exists a quintuple $(\bar {a},\bar {b},\bar {c},\bar {\lambda },\bar {\delta })\in {\mathbb F^\times }^4\times \mathbb F$ such that $(\hat {a},\hat {b},\hat {c},\hat {\lambda },\hat {\delta })\sim (\bar {a},\bar {b},\bar {c},\bar {\lambda },\bar {\delta })$ and there is a $\triangle _q$ -module homomorphism

$$ \begin{align*}f_2:W_{\bar{\lambda}}^{\bar{\delta}}(\bar{a},\bar{b},\bar{c})\to W_{\hat{\lambda}}^{\hat{\delta}}(\hat{a},\hat{b},\hat{c}) \end{align*} $$

that maps $\bar {w}_0$ to $\hat {w}_i$ . Since $(\hat {a},\hat {b},\hat {c},\hat {\lambda },\hat {\delta })\sim (\bar {a},\bar {b},\bar {c},\bar {\lambda },\bar {\delta })$ and $(a,b,c,\lambda ,\delta )\sim (\hat {a},\hat {b},\hat {c},\hat {\lambda },\hat {\delta })$ by Definition 1.12(i), it follows from Definition 1.13 that $ (a,b,c,\lambda ,\delta )\simeq (\bar {a},\bar {b},\bar {c},\bar {\lambda },\bar {\delta })$ .

Since $\hat {\lambda }^2=q^{2(i-1)}$ , it follows from (1.11) that $\hat {\varphi }_i=0$ . Hence $ (B-\hat {\theta }_i^*)\hat {w}_i=0$ by Lemma 9.2. Since $\theta _i^*=\hat {\theta }_i^*$ by Lemma 10.1, applying $f_1$ to the above equation yields

$$ \begin{align*}(B-\theta_i^*)f_1(\hat{w}_i)=0. \end{align*} $$

Since $f_1$ is a $\triangle _q$ -module isomorphism, the vector $f_1(\hat {w}_i)$ is nonzero. Since $f_1(\hat {w}_0)=w_0$ and by Lemma 6.1(i), the vector $f_1(\hat {w}_i)$ is a linear combination of $\{w_h\}_{h=0}^i$ . Since $\varphi _1\varphi _2\dots \varphi _i\not =0$ , it follows from Lemma 9.5 that there exists $\varepsilon \in \mathbb F^\times $ such that

$$ \begin{align*}(f_1\circ f_2)(\bar{w}_0)=f_1(\hat{w}_i)=\varepsilon w_{0i}. \end{align*} $$

The $\triangle _q$ -module homomorphism $\varepsilon ^{-1}\cdot (f_1\circ f_2)$ maps $\bar {w}_0$ to $w_{0i}$ . The lemma follows.