Some q-Exponential Formulas Involving the Double Lowering Operator ψ for a Tridiagonal Pair (Research)

Abstract

Let $\mathbb {K}$ denote an algebraically closed field and let V  denote a vector space over $\mathbb {K}$ with finite positive dimension. Let A, A ^∗ denote a tridiagonal pair on V . We assume that A, A ^∗ belongs to a family of tridiagonal pairs said to have q-Racah type. Let $\{U_i\}_{i=0}^d$ and $\{U_i^\Downarrow \}_{i=0}^{d}$ denote the first and second split decompositions of V . In an earlier paper we introduced a double lowering operator ψ : V → V  with the notable feature that both ψU _i ⊆ U _i−1 and $\psi U_i^\Downarrow \subseteq U_{i-1}^\Downarrow$ for 0 ≤ i ≤ d, where U ₋₁ = 0 and $U_{-1}^\Downarrow =0$. In the same paper, we showed that there exists a unique linear transformation Δ : V → V  such that $\Delta (U_i)\subseteq U_i^{\Downarrow }$ and ( Δ − I)U _i ⊆ U ₀ + U ₁ + ⋯ + U _i−1 for 0 ≤ i ≤ d. In the present paper, we show that Δ can be expressed as a product of two linear transformations; one is a q-exponential in ψ and the other is a q ⁻¹-exponential in ψ. We view Δ as a transition matrix from the first split decomposition of V  to the second. Consequently, we view the q ⁻¹-exponential in ψ as a transition matrix from the first split decomposition to a decomposition of V  which we interpret as a kind of halfway point. This halfway point turns out to be the eigenspace decomposition of a certain linear transformation $\mathcal {M}$. We discuss the eigenspace decomposition of $\mathcal {M}$ and give the actions of various operators on this decomposition.

1 Introduction

Throughout this paper, $\mathbb {K}$ denotes an algebraically closed field. We begin by recalling the notion of a tridiagonal pair. We will use the following terms. Let V  denote a vector space over $\mathbb {K}$ with finite positive dimension. For a linear transformation A : V → V  and a subspace W ⊆ V , we say that W is an eigenspace of A whenever W ≠ 0 and there exists $\theta \in \mathbb {K}$ such that W = {v ∈ V |Av = θv}. In this case, θ is called the eigenvalue of A associated with W. We say that A is diagonalizable whenever V  is spanned by the eigenspaces of A.

Definition 1 ([9, Definition 1.1])

Let V  denote a vector space over $\mathbb {K}$ with finite positive dimension. By a tridiagonal pair (or TD pair) on V  we mean an ordered pair of linear transformations A : V → V  and A ^∗ : V → V  that satisfy the following four conditions.

• (i) Each of A, A ^∗ is diagonalizable.

• (ii) There exists an ordering $\{V_i\}_{i=0}^d$ of the eigenspaces of A such that

  $$\displaystyle \begin{aligned} A^{\ast} V_i \subseteq V_{i-1} + V_i+ V_{i+1} \qquad \qquad (0 \leq i \leq d), \end{aligned} $$

  (1)

  where V ₋₁ = 0 and V _d+1 = 0.

• (iii) There exists an ordering $\{V^{\ast }_i\}_{i=0}^{\delta }$ of the eigenspaces of A ^∗ such that

  $$\displaystyle \begin{aligned} A V^{\ast}_i \subseteq V^{\ast}_{i-1} + V^{\ast}_i+ V^{\ast}_{i+1} \qquad \qquad (0 \leq i \leq \delta), \end{aligned} $$

  (2)

  where $V^{\ast }_{-1} = 0$ and $V^{\ast }_{\delta +1}= 0$.

• (iv) There does not exist a subspace W of V  such that AW ⊆ W, A ^∗W ⊆ W, W ≠ 0, W ≠ V .

We say the pair A, A ^∗ is over $\mathbb {K}$.

Note 1

According to a common notational convention A ^∗ denotes the conjugate-transpose of A. We are not using this convention. In a TD pair A, A ^∗ the linear transformations A and A ^∗ are arbitrary subject to (i)–(iv) above.

Referring to the TD pair in Definition 1, by [9, Lemma 4.5] the scalars d and δ are equal. We call this common value the diameter of A, A ^∗. To avoid trivialities, throughout this paper we assume that the diameter is at least one.

TD pairs first arose in the study of Q-polynomial distance-regular graphs and provided a way to study the irreducible modules of the Terwilliger algebra associated with such a graph. Since their introduction, TD pairs have been found to appear naturally in a variety of other contexts including representation theory [1, 7, 10,11,12, 14, 15, 25], orthogonal polynomials [23, 24], partially ordered sets [22], statistical mechanical models [3, 6, 19], and other areas of physics [16, 18]. As a result, TD pairs have become an area of interest in their own right. Among the above papers on representation theory, there are several works that connect TD pairs to quantum groups [1, 5, 7, 11, 12]. These papers consider certain special classes of TD pairs. We call particular attention to [5], in which the present author describes a new relationship between TD pairs in the q-Racah class and quantum groups. The present paper builds off of this work.

In the present paper, we give a new relationship between the maps Δ, ψ : V → V  introduced in [4], as well as describe a new decomposition of the underlying vector space that, in some sense, lies between the first and second split decompositions associated with a TD pair. In order to motivate our results, we now recall some basic facts concerning TD pairs. For the rest of this section, let A, A ^∗ denote a TD pair on V , as in Definition 1. Fix an ordering $\{V_i\}_{i=0}^d$ (resp. $\{V_i^{\ast }\}_{i=0}^d$) of the eigenspaces of A (resp. A ^∗) which satisfies (1) (resp. (2)). For 0 ≤ i ≤ d let θ _i (resp. $\theta _i^{\ast }$) denote the eigenvalue of A (resp. A ^∗) corresponding to V _i (resp. $V_i^{\ast }$). By [9, Theorem 11.1] the ratios

$$\displaystyle \begin{aligned} \frac{\theta_{i-2}-\theta_{i+1}}{\theta_{i-1}-\theta_i},\qquad \qquad \frac{\theta_{i-2}^{\ast}-\theta_{i+1}^{\ast}}{\theta_{i-1}^{\ast}-\theta_i^{\ast}} \end{aligned}$$

are equal and independent of i for 2 ≤ i ≤ d − 1. This gives two recurrence relations, whose solutions can be written in closed form. There are several cases [9, Theorem 11.2]. The most general case is called the q-Racah case [12, Section 1]. We will discuss this case shortly.

We now recall the split decompositions of V  [9]. For 0 ≤ i ≤ d define

$$\displaystyle \begin{aligned} \begin{array}{rcl} U_i&\displaystyle = (V^{\ast}_0+V^{\ast}_1+\cdots + V^{\ast}_i)\cap (V_i+V_{i+1}+\cdots + V_d),\\ U_i^{\Downarrow} &\displaystyle = (V^{\ast}_0+V^{\ast}_1+\cdots + V^{\ast}_i)\cap (V_0+V_1+\cdots + V_{d-i}). \end{array} \end{aligned} $$

By [9, Theorem 4.6], both the sums $V=\sum _{i=0}^d U_i$ and $V=\sum _{i=0}^d U_i^{\Downarrow }$ are direct. We call $\{U_i\}_{i=0}^d$ (resp. $\{U_i^{\Downarrow }\}_{i=0}^d$) the first split decomposition (resp. second split decomposition) of V . In [9], the authors showed that A, A ^∗ act on the first and second split decomposition in a particularly attractive way. This will be described in more detail in Sect. 3.

We now describe the q-Racah case. We say that the TD pair A, A ^∗ has q-Racah type whenever there exist nonzero scalars $q,a,b\in \mathbb {K}$ such that q ⁴ ≠ 1 and

$$\displaystyle \begin{aligned} \theta_i=aq^{d-2i}+a^{-1}q^{2i-d},\qquad \qquad \theta_i^{\ast}=bq^{d-2i}+b^{-1}q^{2i-d} \end{aligned}$$

for 0 ≤ i ≤ d. For the rest of this section assume that A, A ^∗ has q-Racah type.

We recall the maps K and B [13, Section 1.1]. Let K : V → V  denote the linear transformation such that for 0 ≤ i ≤ d, U _i is an eigenspace of K with eigenvalue q ^d−2i. Let B : V → V  denote the linear transformation such that for 0 ≤ i ≤ d, $U_i^{\Downarrow }$ is an eigenspace of B with eigenvalue q ^d−2i. The relationship between K and B is discussed in considerable detail in [5].

We now bring in the linear transformation Ψ : V → V  [4, Lemma 11.1]. As in [5], we work with the normalization ψ = (q − q ⁻¹)(q ^d − q ^−d) Ψ. A key feature of ψ is that by [4, Lemma 11.2, Corollary 15.3],

$$\displaystyle \begin{aligned} \psi U_i\subseteq U_{i-1},\qquad \qquad \psi U_i^\Downarrow\subseteq U_{i-1}^\Downarrow \end{aligned}$$

for 1 ≤ i ≤ d and both ψU ₀ = 0 and $\psi U_0^\Downarrow =0$. In [5], it is shown how ψ is related to several maps, including the maps K, B, as well as the map Δ which we now recall. By [4, Lemma 9.5], there exists a unique linear transformation Δ : V → V  such that

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle \Delta U_i\subseteq U_i^{\Downarrow} &\displaystyle \ (0\leq i \leq d),\\ &\displaystyle (\Delta -I)U_i\subseteq U_0+U_1+\cdots +U_{i-1} &\displaystyle \ (0\leq i \leq d). \end{array} \end{aligned} $$

In [4, Theorem 17.1], the present author showed that both

$$\displaystyle \begin{aligned} \Delta = \sum_{i=0}^d \left(\prod_{j=1}^i \frac{aq^{j-1}-a^{-1}q^{1-j}}{q^j-q^{-j}}\right)\psi^i, \ \Delta^{-1} =\sum_{i=0}^d \left(\prod_{j=1}^i \frac{a^{-1}q^{j-1}-aq^{1-j}}{q^j-q^{-j}}\right)\psi^i. \end{aligned}$$

The primary goal of this paper is to provide factorizations of these power series in ψ and to investigate the consequences of these factorizations. We accomplish this goal using a linear transformation $\mathcal {M}:V\to V$ given by

$$\displaystyle \begin{aligned} \mathcal{M} =\frac{aK-a^{-1}B}{a-a^{-1}}. \end{aligned}$$

By construction, $\mathcal {M}^\Downarrow =\mathcal {M}$. One can quickly check that $\mathcal {M}$ is invertible. We show that the map $\mathcal {M}$ is equal to each of

$$\displaystyle \begin{aligned} (I-a^{-1}q\psi)^{-1}K, \quad K(I-a^{-1}q^{-1}\psi)^{-1}, \quad (I-aq\psi)^{-1}B, \quad B(I-aq^{-1}\psi)^{-1}. \end{aligned}$$

We give a number of different relations involving the maps $\mathcal {M},K,B,\psi$, the most significant of which are the following:

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle K \exp_q\left(\frac{a^{-1}}{q-q^{-1}}\psi\right)=\exp_q\left(\frac{a^{-1}}{q-q^{-1}}\psi\right) \mathcal{M},\\ &\displaystyle B \exp_q\left(\frac{a}{q-q^{-1}}\psi\right)=\exp_q\left(\frac{a}{q-q^{-1}}\psi\right) \mathcal{M}. \end{array} \end{aligned} $$

Using these equations, we obtain our main result which is that both

$$\displaystyle \begin{aligned} \begin{array}{rcl} \Delta&\displaystyle =\exp_{q}\left(\frac{a}{q-q^{-1}}\psi\right)\exp_{q^{-1}}\left(-\frac{a^{-1}}{q-q^{-1}}\psi\right),\\ \Delta^{-1}&\displaystyle =\exp_{q}\left(\frac{a^{-1}}{q-q^{-1}}\psi\right)\exp_{q^{-1}}\left(-\frac{a}{q-q^{-1}}\psi\right). \end{array} \end{aligned} $$

Due to its important role in the factorization of Δ, we explore the map $\mathcal {M}$ further. We show that $\mathcal {M}$ is diagonalizable with eigenvalues q ^d, q ^d−2, q ^d−4, …, q ^−d. For 0 ≤ i ≤ d, let W _i denote the eigenspace of $\mathcal {M}$ corresponding to the eigenvalue q ^d−2i. We show that for 0 ≤ i ≤ d,

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle U_i=\exp_q\left(\frac{a^{-1}}{q-q^{-1}}\psi\right) W_i,\qquad \qquad &\displaystyle U_i^\Downarrow=\exp_q\left(\frac{a}{q-q^{-1}}\psi\right) W_i,\\ &\displaystyle W_i=\exp_{q^{-1}}\left(-\frac{a^{-1}}{q-q^{-1}}\psi\right) U_i, \qquad \qquad &\displaystyle W_i=\exp_{q^{-1}}\left(-\frac{a}{q-q^{-1}}\psi\right) U_i^\Downarrow. \end{array} \end{aligned} $$

In light of this result, we interpret the decomposition $\{W_i\}_{i=0}^d$ as a sort of halfway point between the first and second split decompositions. We explore this decomposition further and give the actions of ψ, K, B,  Δ, A, A ^∗ on $\{W_i\}_{i=0}^d$. We then give the actions of $\mathcal {M}^{\pm 1}$ on $\{U_i\}_{i=0}^d,\{U_i^\Downarrow \}_{i=0}^d, \{V_i\}_{i=0}^d, \{V_i^{\ast }\}_{i=0}^d$. We conclude the paper with a discussion of the special case when A, A ^∗ is a Leonard pair.

The present paper is organized as follows. In Sect. 2 we discuss some preliminary facts concerning TD pairs and TD systems. In Sect. 3 we discuss the split decompositions of V  as well as the maps K and B. In Sect. 4 we discuss the map ψ. In Sect. 5 we recall the map Δ and give Δ as a power series in ψ. In Sect. 6 we introduce the map $\mathcal {M}$ and describe its relationship with A, K, B, ψ. In Sect. 7 we express Δ as a product of two linear transformations; one is a q-exponential in ψ and the other is a q ⁻¹-exponential in ψ. In Sect. 8 we describe the eigenvalues and eigenspaces of $\mathcal {M}$ and discuss how the eigenspace decomposition of $\mathcal {M}$ is related to the first and second split decompositions. In Sect. 9 we discuss the actions of ψ, K, B,  Δ, A, A ^∗ on the eigenspace decomposition of $\mathcal {M}$. In Sect. 10 we describe the action of $\mathcal {M}$ on the first and second split decompositions of V , as well as on the eigenspace decompositions of A, A ^∗. In Sect. 11 we consider the case when A, A ^∗ is a Leonard pair.

2 Preliminaries

When working with a tridiagonal pair, it is useful to consider a closely related object called a tridiagonal system. In order to define this object, we first recall some facts from elementary linear algebra [9, Section 2].

We use the following conventions. When we discuss an algebra, we mean a unital associative algebra. When we discuss a subalgebra, we assume that it has the same unit as the parent algebra.

Let V  denote a vector space over $\mathbb {K}$ with finite positive dimension. By a decomposition of V, we mean a sequence of nonzero subspaces whose direct sum is V . Let End(V ) denote the $\mathbb {K}$-algebra consisting of all linear transformations from V  to V . Let A denote a diagonalizable element in End(V ). Let $\{V_i\}_{i=0}^d$ denote an ordering of the eigenspaces of A. For 0 ≤ i ≤ d let θ _i be the eigenvalue of A corresponding to V _i. Define E _i ∈End(V ) by (E _i − I)V _i = 0 and E _iV _j = 0 if j ≠ i (0 ≤ j ≤ d). In other words, E _i is the projection map from V  onto V _i. We refer to E _i as the primitive idempotent of A associated with θ _i. By elementary linear algebra, (i) AE _i = E _iA = θ _iE _i (0 ≤ i ≤ d); (ii) E _iE _j = δ _ijE _i (0 ≤ i, j ≤ d); (iii) V _i = E _iV  (0 ≤ i ≤ d); (iv) $I=\sum _{i=0}^d E_i$. Moreover

$$\displaystyle \begin{aligned} E_i = \prod_{\genfrac{}{}{0 pt}{}{0 \leq j \leq d}{j\neq i}} \frac{A-\theta_j I}{\theta_i-\theta_j}\qquad \qquad (0 \leq i \leq d).{} \end{aligned}$$

Let M denote the subalgebra of End(V ) generated by A. Note that each of $\{ A^i\}_{i=0}^d$, $\{E_i\}_{i=0}^d$ is a basis for the $\mathbb {K}$-vector space M.

Let A, A ^∗ denote a TD pair on V . An ordering of the eigenspaces of A (resp. A ^∗) is said to be standard whenever it satisfies (1) (resp. (2)). Let $\{V_i\}_{i=0}^d$ denote a standard ordering of the eigenspaces of A. By [9, Lemma 2.4], the ordering $\{V_{d-i}\}_{i=0}^d$ is standard and no further ordering of the eigenspaces of A is standard. A similar result holds for the eigenspaces of A ^∗. An ordering of the primitive idempotents of A (resp. A ^∗) is said to be standard whenever the corresponding ordering of the eigenspaces of A (resp. A ^∗) is standard.

Definition 2 ([17, Definition 2.1])

Let V  denote a vector space over $\mathbb {K}$ with finite positive dimension. By a tridiagonal system (or TD system) on V, we mean a sequence

$$\displaystyle \begin{aligned} \Phi = (A; \{ E_i\}_{i=0}^d; A^{\ast};\{ E_i^{\ast}\}_{i=0}^d) \end{aligned}$$

that satisfies (i)–(iii) below.

• (i) A, A ^∗ is a tridiagonal pair on V .

• (ii) $\{ E_i\}_{i=0}^d$ is a standard ordering of the primitive idempotents of A.

• (iii) $\{ E_i^{\ast }\}_{i=0}^d$ is a standard ordering of the primitive idempotents of A ^∗.

We call d the diameter of Φ, and say Φ is over $\mathbb {K}$. For notational convenience, set E ₋₁ = 0, E _d+1 = 0, $E^{\ast }_{-1}=0$, $E^{\ast }_{d+1}=0$.

In Definition 2 we do not assume that the primitive idempotents $\{ E_i\}_{i=0}^d,\{ E_i^{\ast }\}_{i=0}^d$ all have rank 1. A TD system for which each of these primitive idempotents has rank 1 is called a Leonard system [20]. The Leonard systems are classified up to isomorphism [20, Theorem 1.9].

For the rest of this paper, fix a TD system Φ on V  as in Definition 2. Our TD system Φ can be modified in a number of ways to get a new TD system [9, Section 3]. For example, the sequence

$$\displaystyle \begin{aligned} \Phi^{\Downarrow} = (A;\{ E_{d-i}\}_{i=0}^d;A^{\ast};\{ E_i^{\ast}\}_{i=0}^d) \end{aligned}$$

is a TD system on V . Following [9, Section 3], we call Φ^⇓ the second inversion of Φ. When discussing Φ^⇓, we use the following notational convention. For any object f associated with Φ, let f ^⇓ denote the corresponding object associated with Φ^⇓.

Definition 3

For 0 ≤ i ≤ d let θ _i (resp. $\theta _i^{\ast }$) denote the eigenvalue of A (resp. A ^∗) associated with E _i (resp. $E_i^{\ast }$). We refer to $\{\theta _i\}_{i=0}^d$ (resp. $\{\theta _i^{\ast }\}_{i=0}^d$) as the eigenvalue sequence (resp. dual eigenvalue sequence) of Φ.

By construction $\{\theta _i\}_{i=0}^d$ are mutually distinct and $\{\theta _i^{\ast }\}_{i=0}^d$ are mutually distinct. By [9, Theorem 11.1], the scalars

$$\displaystyle \begin{aligned} \frac{\theta_{i-2}-\theta_{i+1}}{\theta_{i-1}-\theta_i},\qquad \qquad \frac{\theta_{i-2}^{\ast}-\theta_{i+1}^{\ast}}{\theta_{i-1}^{\ast}-\theta_i^{\ast}}{} \end{aligned}$$

are equal and independent of i for 2 ≤ i ≤ d − 1. For this restriction, the solutions have been found in closed form [9, Theorem 11.2]. The most general solution is called q-Racah [12, Section 1]. This solution is described as follows.

Definition 4

Let Φ denote a TD system on V  as in Definition 2. We say that Φ has q-Racah type whenever there exist nonzero scalars $q,a,b\in \mathbb {K}$ such that such that q ⁴ ≠ 1 and

$$\displaystyle \begin{aligned} \theta_i=aq^{d-2i}+a^{-1}q^{2i-d},\qquad \qquad \theta_i^{\ast}=bq^{d-2i}+b^{-1}q^{2i-d} {} \end{aligned} $$

(3)

for 0 ≤ i ≤ d.

Note 2

Referring to Definition 4, the scalars q, a, b are not uniquely defined by Φ. If q, a, b is one solution, then their inverses give another solution.

For the rest of the paper, we make the following assumption.

Assumption 1

We assume that our TD system Φ has q-Racah type. We fix q, a, b as in Definition 4.

Lemma 1 ([5, Lemma 2.4])

With reference to Assumption 1 , the following hold.

• (i) Neither of a ², b ² is among q ^2d−2, q ^2d−4, …, q ^2−2d.

• (ii) q ²ⁱ ≠ 1 for 1 ≤ i ≤ d.

Proof

The result follows from the comment below Definition 3. □

3 The First and Second Split Decomposition of V 

Recall the TD system Φ from Assumption 1. In this section we consider two decompositions of V  associated with Φ, called the first and second split decomposition.

For 0 ≤ i ≤ d define

$$\displaystyle \begin{aligned} U_i = (E^{\ast}_0V+E^{\ast}_1V+\cdots + E^{\ast}_iV)\cap (E_iV+E_{i+1}V+\cdots + E_dV). \end{aligned}$$

For notational convenience, define U ₋₁ = 0 and U _d+1 = 0. Note that for 0 ≤ i ≤ d,

$$\displaystyle \begin{aligned} U_i^{\Downarrow}=(E_0^{\ast}V+E_1^{\ast}V +\cdots + E_i^{\ast} V)\cap (E_0V+E_1V+\cdots + E_{d-i} V). \end{aligned}$$

By [9, Theorem 4.6], the sequence $\{U_i\}_{i=0}^d$ (resp. $\{U_i^\Downarrow \}_{i=0}^d$) is a decomposition of V . Following [9], we refer to $\{U_i\}_{i=0}^d$ (resp. $\{U_i^\Downarrow \}_{i=0}^d$) as the first split decomposition (resp. second split decomposition) of V  with respect to Φ. By [9, Corollary 5.7], for 0 ≤ i ≤ d the dimensions of E _iV , $E_i^{\ast }V$, U _i, $U_i^\Downarrow$ coincide; we denote the common dimension by ρ _i. By [9, Theorem 4.6],

$$\displaystyle \begin{aligned} \begin{array}{rcl} E_iV+E_{i+1}V+\cdots+E_dV=U_{i}+U_{i+1}+\cdots+U_d,{} \end{array} \end{aligned} $$

(4)

$$\displaystyle \begin{aligned} \begin{array}{rcl} E_0V+E_1V+\cdots+E_iV=U_{d-i}^\Downarrow+U_{d-i+1}^\Downarrow+\cdots+U_d^\Downarrow,{} \end{array} \end{aligned} $$

(5)

$$\displaystyle \begin{aligned} \begin{array}{rcl} E_0^{\ast}V+E_1^{\ast}V+\cdots E_i^{\ast}V=U_0+U_1+\cdots +U_i=U_0^\Downarrow+U_1^\Downarrow+\cdots +U_i^\Downarrow. {} \end{array} \end{aligned} $$

(6)

By [9, Theorem 4.6], A and A ^∗ act on the first split decomposition in the following way:

$$\displaystyle \begin{aligned} \begin{array}{rcl} (A-\theta_i I)U_i\subseteq U_{i+1} \qquad &\displaystyle (0\leq i \leq d-1 ), \qquad &\displaystyle (A-\theta_d I)U_d=0,\\ (A^{\ast}-\theta_i^{\ast} I)U_i\subseteq U_{i-1} \qquad &\displaystyle (1\leq i\leq d ), \qquad &\displaystyle (A^{\ast}-\theta_0^{\ast} I)U_0=0. \end{array} \end{aligned} $$

By [9, Theorem 4.6], A and A ^∗ act on the second split decomposition in the following way:

$$\displaystyle \begin{aligned} \begin{array}{rcl} (A-\theta_{d-i} I)U_i^{\Downarrow}\subseteq U_{i+1}^{\Downarrow}\qquad &\displaystyle (0\leq i\leq d-1 ), \qquad &\displaystyle (A-\theta_0 I)U_d^{\Downarrow}=0,\\ (A^{\ast}-\theta_i^{\ast} I)U_i^{\Downarrow}\subseteq U_{i-1}^{\Downarrow}\qquad &\displaystyle (1\leq i\leq d ), \qquad &\displaystyle (A^{\ast}-\theta_0^{\ast} I)U_0^{\Downarrow}=0. \end{array} \end{aligned} $$

Definition 5 ([5, Definitions 3.1 and 3.2])

Define K, B ∈End(V ) such that for 0 ≤ i ≤ d, U _i (resp. $U_i^\Downarrow$) is the eigenspace of K (resp. B) with eigenvalue q ^d−2i. In other words,

$$\displaystyle \begin{aligned} (K-q^{d-2i}I)U_i=0, \qquad \qquad (B-q^{d-2i}I)U_i^\Downarrow=0 \qquad \qquad (0\leq i\leq d).{} \end{aligned} $$

(7)

Observe that B = K ^⇓.

By construction each of K, B is invertible and diagonalizable on V .

We now describe how K and B act on the eigenspaces of the other one.

Lemma 2 ([5, Lemma 3.3])

For 0 ≤ i ≤ d,

$$\displaystyle \begin{aligned} \begin{array}{rcl} (B-q^{d-2i}I)U_i&\displaystyle \subseteq U_0+U_1+\cdots +U_{i-1},{} \end{array} \end{aligned} $$

(8)

$$\displaystyle \begin{aligned} \begin{array}{rcl} (K-q^{d-2i}I)U_i^\Downarrow &\displaystyle \subseteq U_0^\Downarrow+U_1^\Downarrow+\cdots +U_{i-1}^\Downarrow.{} \end{array} \end{aligned} $$

(9)

Next we describe how A, K, B are related.

Lemma 3 ([13, Section 1.1])

Both

$$\displaystyle \begin{aligned} \frac{qKA-q^{-1}AK}{q-q^{-1}}=aK^2+a^{-1}I,\qquad \qquad \frac{qBA-q^{-1}AB}{q-q^{-1}}=a^{-1}B^2+aI. {} \end{aligned} $$

(10)

Lemma 4 ([5, Theorem 9.9])

We have

$$\displaystyle \begin{aligned} aK^2-\frac{a^{-1}q-aq^{-1}}{q-q^{-1}}\ KB-\frac{aq-a^{-1}q^{-1}}{q-q^{-1}}\ BK+a^{-1}B^2=0.{} \end{aligned} $$

(11)

4 The Linear Transformation ψ

We continue to discuss the situation of Assumption 1. In [4, Section 11] we introduced an element Ψ ∈ End(V ). In [5] we used the normalization ψ = (q − q ⁻¹)(q ^d − q ^−d) Ψ. In [5, Theorem 9.8], we showed that ψ is equal to some rational expressions involving K, B. We now recall this result. We start with a comment.

Lemma 5 ([5, Lemma 9.7])

Each of the following is invertible:

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle aI-a^{-1}BK^{-1}, \qquad \qquad &\displaystyle a^{-1}I-aKB^{-1},{} \end{array} \end{aligned} $$

(12)

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle aI-a^{-1}K^{-1}B, \qquad \qquad &\displaystyle a^{-1}I-aB^{-1}K. \end{array} \end{aligned} $$

(13)

Lemma 6 ([5, Theorem 9.8])

The following four expressions coincide:

$$\displaystyle \begin{aligned} \begin{array}{rcl} \frac{I-BK^{-1}}{q(aI-a^{-1}BK^{-1})}, \qquad \qquad &\displaystyle &\displaystyle \frac{I-KB^{-1}}{q(a^{-1}I-aKB^{-1})},{} \end{array} \end{aligned} $$

(14)

$$\displaystyle \begin{aligned} \begin{array}{rcl} \frac{q(I-K^{-1}B)}{aI-a^{-1}K^{-1}B}, \qquad \qquad &\displaystyle &\displaystyle \frac{q(I-B^{-1}K)}{a^{-1}I-aB^{-1}K}.{} \end{array} \end{aligned} $$

(15)

In (14), (15) the denominators are invertible by Lemma 5.

Definition 6

Define ψ ∈End(V ) to be the common value of the four expressions in Lemma 6.

We now recall some facts concerning ψ.

Lemma 7 ([5, Lemma 5.4])

Both

$$\displaystyle \begin{aligned} K\psi =q^2\psi K, \qquad \qquad B\psi =q^2\psi B.{} \end{aligned} $$

(16)

Lemma 8 ([4, Lemma 11.2, Corollary 15.3])

We have

$$\displaystyle \begin{aligned} \psi U_i\subseteq U_{i-1},\qquad \qquad \psi U_i^\Downarrow\subseteq U_{i-1}^\Downarrow\qquad \qquad (1\leq i\leq d) \end{aligned} $$

(17)

and also ψU ₀ = 0 and $\psi U_0^\Downarrow =0$. Moreover ψ ^d+1 = 0.

In Lemma 6 we obtained ψ as a rational expression in BK ⁻¹ or K ⁻¹B. Next we solve for BK ⁻¹ and K ⁻¹B as a rational function in ψ. In order to state the answer, we will need the following result.

Lemma 9 ([5, Lemma 9.2])

Each of the following is invertible:

$$\displaystyle \begin{aligned} I-aq\psi,\qquad I-a^{-1}q\psi, \qquad I-aq^{-1}\psi,\qquad I-a^{-1}q^{-1}\psi.{} \end{aligned} $$

(18)

Their inverses are as follows:

$$\displaystyle \begin{aligned} \begin{array}{rcl} (I-aq\psi)^{-1}&\displaystyle =\sum_{i=0}^da^iq^i\psi^i,\qquad (I-a^{-1}q\psi)^{-1}&\displaystyle =\sum_{i=0}^d a^{-i}q^i\psi^i,{} \end{array} \end{aligned} $$

(19)

$$\displaystyle \begin{aligned} \begin{array}{rcl} (I-aq^{-1}\psi)^{-1}&\displaystyle =\sum_{i=0}^d a^iq^{-i}\psi^i,\qquad (I-a^{-1}q^{-1}\psi)^{-1}&\displaystyle =\sum_{i=0}^d a^{-i}q^{-i}\psi^i.{} \end{array} \end{aligned} $$

(20)

The next result is an immediate consequence of Lemma 6, Definition 6, and Lemma 9.

Theorem 1 ([5, Theorem 9.4])

The following hold:

$$\displaystyle \begin{aligned} \begin{array}{rcl} BK^{-1}&\displaystyle =\displaystyle\frac{I-aq\psi}{I-a^{-1}q\psi}, \qquad \qquad &\displaystyle KB^{-1}=\frac{I-a^{-1}q\psi}{I-aq\psi},{} \end{array} \end{aligned} $$

(21)

$$\displaystyle \begin{aligned} \begin{array}{rcl} K^{-1}B&\displaystyle =\displaystyle\frac{I-aq^{-1}\psi}{I-a^{-1}q^{-1}\psi}, \qquad \qquad &\displaystyle B^{-1}K=\frac{I-a^{-1}q^{-1}\psi}{I-aq^{-1}\psi}.{} \end{array} \end{aligned} $$

(22)

In (21), (22) the denominators are invertible by Lemma 9.

Lemma 10 ([5, Equation (22)])

We have

$$\displaystyle \begin{aligned} \frac{\psi A-A\psi}{q-q^{-1}}=\left(I-aq\psi\right)K-\left(I-a^{-1}q^{-1}\psi\right)K^{-1}. \end{aligned} $$

(23)

Proof

This result is a reformulation of [5, Equation (22)] using [5, Equation (14)]. □

5 The Linear Transformation Δ

We continue to discuss the situation of Assumption 1. In [4, Section 9] we introduced an invertible element Δ ∈ End(V ). In [4] we showed that Δ, ψ commute and in fact both Δ, Δ⁻¹ are power series in ψ. These power series will be the central focus of this paper. We will show that each of those power series factors as a product of two power series, each of which is a quantum exponential in ψ.

Lemma 11 ([4, Lemma 9.5])

There exists a unique Δ ∈End(V ) such that

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle \Delta U_i\subseteq U_i^{\Downarrow}\qquad &\displaystyle \ (0\leq i \leq d),{} \end{array} \end{aligned} $$

(24)

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle (\Delta -I)U_i\subseteq U_0+U_1+\cdots +U_{i-1} \qquad &\displaystyle \ (0\leq i \leq d).{} \end{array} \end{aligned} $$

(25)

Lemma 12 ([4, Lemmas 9.3 and 9.6])

The map Δ is invertible. Moreover Δ ⁻¹ = Δ ^⇓ and

$$\displaystyle \begin{aligned} (\Delta^{-1} -I)U_i\subseteq U_0+U_1+\cdots +U_{i-1} \qquad \qquad (0\leq i\leq d). {} \end{aligned} $$

(26)

Lemma 13

The map Δ − I is nilpotent. Moreover ΔK = B Δ.

Proof

The first assertion follows from (25). The last assertion follows from (24) and Definition 5. □

The map Δ is characterized as follows.

Lemma 14 ([4, Lemma 9.8])

The map Δ is the unique element of End(V ) such that

$$\displaystyle \begin{aligned} \begin{array}{rcl} (\Delta -I)E_i^{*} V\subseteq E_0^{*}V+E_1^{*}V+\cdots +E_{i-1}^{*}V \quad &\displaystyle \ (0\leq i \leq d),\\{} \end{array} \end{aligned} $$

(27)

$$\displaystyle \begin{aligned} \begin{array}{rcl} \Delta (E_iV+E_{i+1}V+\cdots+E_d V)= E_0 V+E_{1}V+\cdots+E_{d-i} V\quad &\displaystyle (0\leq i \leq d). \\{} \end{array} \end{aligned} $$

(28)

Theorem 2 ([4, Theorem 17.1])

Both

$$\displaystyle \begin{aligned} \begin{array}{rcl} \Delta &\displaystyle =&\displaystyle \sum_{i=0}^d \left(\prod_{j=1}^i \frac{aq^{j-1}-a^{-1}q^{1-j}}{q^j-q^{-j}}\right)\psi^i, {} \end{array} \end{aligned} $$

(29)

$$\displaystyle \begin{aligned} \begin{array}{rcl} \Delta^{-1} &\displaystyle =&\displaystyle \sum_{i=0}^d \left(\prod_{j=1}^i \frac{a^{-1}q^{j-1}-aq^{1-j}}{q^j-q^{-j}}\right)\psi^i. {} \end{array} \end{aligned} $$

(30)

In (29) and (30), the elements Δ, Δ⁻¹ are expressed as a power series in ψ. In the present paper, we factor these power series and interpret the results. This interpretation will involve a linear transformation $\mathcal {M}$. We introduce $\mathcal {M}$ in the next section.

6 The Linear Transformation $\mathcal {M}$

We continue to discuss the situation of Assumption 1. In this section we introduce an element $\mathcal {M} \in \mathrm {End}(V)$. We explain how $\mathcal {M}$ is related to K, B, ψ, A.

Definition 7

Define $\mathcal {M} \in \mathrm {End}(V)$ by

$$\displaystyle \begin{aligned} \mathcal{M} =\frac{aK-a^{-1}B}{a-a^{-1}}.{} \end{aligned} $$

(31)

By construction, $\mathcal {M}^\Downarrow =\mathcal {M}$. Evaluating (31) using Lemma 5, we see that $\mathcal {M}$ is invertible.

Lemma 15

The map $\mathcal {M}$ is equal to each of:

$$\displaystyle \begin{aligned} (I-a^{-1}q\psi)^{-1}K, \qquad K(I-a^{-1}q^{-1}\psi)^{-1}, \qquad (I-aq\psi)^{-1}B, \qquad B(I-aq^{-1}\psi)^{-1}. \end{aligned}$$

Proof

We first show that $\mathcal {M}=(I-a^{-1}q\psi )^{-1}K$. By Definition 7,

$$\displaystyle \begin{aligned} (a-a^{-1})\mathcal{M} K^{-1}=aI-a^{-1}BK^{-1}. \end{aligned}$$

The result follows from this fact along with the equation on the left in (21).

The remaining assertions follow from Theorem 1. □

Lemma 15 can be reformulated as follows.

Lemma 16

We have

$$\displaystyle \begin{aligned} \begin{array}{rcl} K&\displaystyle =\left(I-a^{-1}q\psi\right)\mathcal{M}, \qquad \qquad &\displaystyle K=\mathcal{M} \left(I-a^{-1}q^{-1}\psi\right), {} \end{array} \end{aligned} $$

(32)

$$\displaystyle \begin{aligned} \begin{array}{rcl} B&\displaystyle =\left(I-aq\psi\right)\mathcal{M},\qquad \qquad &\displaystyle B=\mathcal{M}\left(I-aq^{-1}\psi\right). {} \end{array} \end{aligned} $$

(33)

For later use, we give several descriptions of $\mathcal {M}^{\pm 1}$.

Lemma 17

The map $\mathcal {M}^{-1}$ is equal to each of:

$$\displaystyle \begin{aligned} K^{-1}(I-a^{-1}q\psi), \quad (I-a^{-1}q^{-1}\psi)K^{-1}, \quad B^{-1}(I-aq\psi), \quad (I-aq^{-1}\psi)B^{-1}. \end{aligned}$$

Proof

Immediate from Lemma 15. □

Lemma 18

The map $\mathcal {M}$ is equal to each of:

$$\displaystyle \begin{aligned} K\sum_{n=0}^d a^{-n}q^{-n}\psi^n,\qquad \sum_{n=0}^d a^{-n}q^{n}\psi^nK,\qquad B\sum_{n=0}^d a^{n}q^{-n}\psi^n,\qquad \sum_{n=0}^d a^{n}q^{n}\psi^n B \end{aligned} $$

(34)

Proof

Use Lemmas 9 and 15. □

We now give some attractive equations that show how $\mathcal {M}$ is related to ψ, K, B, A.

Lemma 19

We have

$$\displaystyle \begin{aligned} \mathcal{M} \psi=q^2\psi \mathcal{M} . \end{aligned} $$

(35)

Proof

Use Lemma 7 and Definition 7. □

Lemma 20

We have

$$\displaystyle \begin{aligned} \frac{q\mathcal{M}^{-1}K-q^{-1}K\mathcal{M}^{-1}}{q-q^{-1}}=I,\qquad \qquad \frac{q\mathcal{M}^{-1}B-q^{-1}B\mathcal{M}^{-1}}{q-q^{-1}}=I.{} \end{aligned} $$

(36)

Proof

Use Lemma 17. □

Lemma 21

We have

$$\displaystyle \begin{aligned} \frac{qA\mathcal{M}^{-1}-q^{-1}\mathcal{M}^{-1}A}{q-q^{-1}}=(a+a^{-1})I-(q+q^{-1})\psi.{} \end{aligned} $$

(37)

Proof

Use Lemmas 3, 7, 10, and 17. □

Lemma 22

We have

$$\displaystyle \begin{aligned} \mathcal{M}^{-2}A-(q^2+q^{-2})\mathcal{M}^{-1}A\mathcal{M}^{-1}+A\mathcal{M}^{-2}=-(q-q^{-1})^2(a+a^{-1})\mathcal{M}^{-1}. \end{aligned} $$

(38)

Proof

Use Lemmas 19 and 21. □

7 A Factorization of Δ

We continue to discuss the situation of Assumption 1. We now bring in the q-exponential function [8]. In [4, Theorem 17.1] we expressed Δ as a power series in ψ. In this section we strengthen this result in the following way. We express Δ as a product of two linear transformations; one is a q-exponential in ψ and the other is a q ⁻¹-exponential in ψ.

For an integer n, define

$$\displaystyle \begin{aligned}{}[n]_q=\frac{q^n-q^{-n}}{q-q^{-1}} \end{aligned} $$

(39)

and for n ≥ 0, define

$$\displaystyle \begin{aligned}{}[n]_q^!=[n]_q[n-1]_q\cdots [1]_q. \end{aligned} $$

(40)

We interpret $[0]_q^!=1$.

We now recall the q-exponential function [8]. For a nilpotent T ∈End(V ),

$$\displaystyle \begin{aligned} \exp_q(T)=\sum_{n=0}^\infty \frac{q^{\binom{n}{2}}}{[n]_q^!} T^n.{} \end{aligned} $$

(41)

The map exp_q(T) is invertible. Its inverse is given by

$$\displaystyle \begin{aligned} \exp_{q^{-1}}(-T)=\sum_{n=0}^\infty \frac{(-1)^n q^{-\binom{n}{2}}}{[n]_q^!} T^n.{} \end{aligned} $$

(42)

Using (41) we obtain

$$\displaystyle \begin{aligned} (I-(q^2-1)T)\exp_q(q^2T)=\exp_q(T).{} \end{aligned} $$

(43)

For S ∈End(V ) such that ST = q ²TS, we have

$$\displaystyle \begin{aligned} S\exp_q(T)S^{-1}=\exp_q(STS^{-1})=\exp_q(q^2T). \end{aligned}$$

Consequently

$$\displaystyle \begin{aligned} S\exp_q(T)=\exp_q(q^2T)S.{} \end{aligned} $$

(44)

Combining (43) and (44),

$$\displaystyle \begin{aligned} (I-(q^2-1)T)S\exp_q(T)=\exp_q(T)S.{} \end{aligned} $$

(45)

We return our attention to $K,B,\psi ,\mathcal {M}$.

Proposition 1

Both

$$\displaystyle \begin{aligned} \begin{array}{rcl} K \exp_q\left(\frac{a^{-1}}{q-q^{-1}}\psi\right)&\displaystyle =&\displaystyle \exp_q\left(\frac{a^{-1}}{q-q^{-1}}\psi\right) \mathcal{M} ,{} \end{array} \end{aligned} $$

(46)

$$\displaystyle \begin{aligned} \begin{array}{rcl} B \exp_q\left(\frac{a}{q-q^{-1}}\psi\right)&\displaystyle =&\displaystyle \exp_q\left(\frac{a}{q-q^{-1}}\psi\right) \mathcal{M} .{} \end{array} \end{aligned} $$

(47)

Proof

Recall from Lemma 19 that $\mathcal {M} \psi =q^2\psi \mathcal {M}$. We first obtain (46). To do this, in (45) take $S=\mathcal {M}$ and $T=\frac {a^{-1}}{q-q^{-1}}\psi$. Evaluate the result using the equation $\mathcal {M}=(I-a^{-1}q\psi )^{-1}K$ from Lemma 15.

Next we obtain (47). To do this, in (45) take $S=\mathcal {M}$ and $T=\frac {a}{q-q^{-1}}\psi$. Evaluate the result using the equation $\mathcal {M}=(I-aq\psi )^{-1}B$ from Lemma 15. □

The following is our main result.

Theorem 3

Both

$$\displaystyle \begin{aligned} \begin{array}{rcl} \Delta&\displaystyle =&\displaystyle \exp_{q}\left(\frac{a}{q-q^{-1}}\psi\right)\exp_{q^{-1}}\left(-\frac{a^{-1}}{q-q^{-1}}\psi\right),{} \end{array} \end{aligned} $$

(48)

$$\displaystyle \begin{aligned} \begin{array}{rcl} \Delta^{-1}&\displaystyle =&\displaystyle \exp_{q}\left(\frac{a^{-1}}{q-q^{-1}}\psi\right)\exp_{q^{-1}}\left(-\frac{a}{q-q^{-1}}\psi\right).{} \end{array} \end{aligned} $$

(49)

Proof

We first show (48). Let $\tilde \Delta$ denote the expression on the right in (48). Combining (46) and (47), we see that $\tilde \Delta K=B\tilde \Delta$. Therefore $\tilde \Delta U_i=U_i^\Downarrow$ for 0 ≤ i ≤ d. Observe that $\tilde \Delta -I$ is a polynomial in ψ with zero constant term. By Lemma 8, $(\tilde \Delta -I)U_i\subseteq U_0+U_1+\cdots +U_{i-1}$ for 0 ≤ i ≤ d. By Lemma 11, $\tilde \Delta =\Delta$.

To obtain (49) from (48), use (42). □

Corollary 1

We have

$$\displaystyle \begin{aligned} \begin{array}{rcl} \exp_{q}\left(\frac{a}{q-q^{-1}}\psi\right)\exp_{q^{-1}}\left(-\frac{a^{-1}}{q-q^{-1}}\psi\right)&\displaystyle =&\displaystyle \sum_{i=0}^d \left(\prod_{j=1}^i \frac{aq^{j-1}-a^{-1}q^{1-j}}{q^j-q^{-j}}\right)\psi^i,\\ \exp_{q}\left(\frac{a^{-1}}{q-q^{-1}}\psi\right)\exp_{q^{-1}}\left(-\frac{a}{q-q^{-1}}\psi\right)&\displaystyle =&\displaystyle \sum_{i=0}^d \left(\prod_{j=1}^i \frac{a^{-1}q^{j-1}-aq^{1-j}}{q^j-q^{-j}}\right)\psi^i. \end{array} \end{aligned} $$

Proof

Combine Theorems 2 and 3. The equations can also be obtained directly by expanding their left-hand sides using (41) and (42), and evaluating the results using the q-binomial theorem [2, Theorem 10.2.1]. □

8 The Eigenvalues and Eigenspaces of $\mathcal {M}$

We continue to discuss the situation of Assumption 1. In Sect. 6 we introduced the linear transformation $\mathcal {M}$. Proposition 1 indicates the role of $\mathcal {M}$ in the factorization of Δ in Theorem 3. In this section we show that $\mathcal {M}$ is diagonalizable. We describe the eigenvalues and eigenspaces of $\mathcal {M}$. We also explain how the eigenspace decomposition for $\mathcal {M}$ is related to the first and second split decompositions.

Lemma 23

The map $\mathcal {M}$ is diagonalizable with eigenvalues q ^d, q ^d−2, q ^d−4, …, q ^−d.

Proof

Let $E=\exp _{q}\left (\frac {a^{-1}}{q-q^{-1}}\psi \right )$. By (46), $\mathcal {M}=E^{-1}KE$. By construction K is diagonalizable with eigenvalues q ^d, q ^d−2, q ^d−4, …, q ^−d. The result follows. □

Definition 8

For 0 ≤ i ≤ d, let W _i denote the eigenspace of $\mathcal {M}$ corresponding to the eigenvalue q ^d−2i. Note that $\{W_i\}_{i=0}^d$ is a decomposition of V , and that $W_i^\Downarrow =W_i$ for 0 ≤ i ≤ d. For notational convenience, let W ₋₁ = 0 and W _d+1 = 0.

Proposition 2

For 0 ≤ i ≤ d,

$$\displaystyle \begin{aligned} \begin{array}{rcl} U_i&\displaystyle =\exp_q\left(\frac{a^{-1}}{q-q^{-1}}\psi\right) W_i, \qquad &\displaystyle U_i^\Downarrow=\exp_q\left(\frac{a}{q-q^{-1}}\psi\right) W_i, {} \end{array} \end{aligned} $$

(50)

$$\displaystyle \begin{aligned} \begin{array}{rcl} W_i&\displaystyle =\exp_{q^{-1}}\left(-\frac{a^{-1}}{q-q^{-1}}\psi\right) U_i,\qquad &\displaystyle W_i=\exp_{q^{-1}}\left(-\frac{a}{q-q^{-1}}\psi\right) U_i^\Downarrow.{} \end{array} \end{aligned} $$

(51)

Proof

Define E as in the proof of Lemma 23. We show that U _i = EW _i. By (46), $KE=E\mathcal {M}$. Recall that U _i (resp. W _i) is the eigenspace of K (resp. $\mathcal {M}$) corresponding to the eigenvalue q ^d−2i. By these comments U _i = EW _i.

Define $F=\exp _{q}(\frac {a}{q-q^{-1}}\psi )$. We show $U_i^\Downarrow =FW_i$. By (47), $BF=F\mathcal {M}$. Recall that $U_i^\Downarrow$ (resp. W _i) is the eigenspace of B (resp. $\mathcal {M}$) corresponding to the eigenvalue q ^d−2i. By these comments $U_i^\Downarrow =FW_i$.

To obtain (51) from (50), use (42). □

Lemma 24

For 0 ≤ i ≤ d, the dimension of W _i is ρ _i.

Proof

This follows from Proposition 2 and the fact that $U_i,U_i^\Downarrow$ have dimension ρ _i. □

Recall from (6) that

$$\displaystyle \begin{aligned} \sum_{h=0}^i E_h^{\ast}V=\sum_{h=0}^i U_h=\sum_{h=0}^i U_h^{\Downarrow} {} \end{aligned} $$

(52)

for 0 ≤ i ≤ d.

Lemma 25

For 0 ≤ i ≤ d, the sum $\sum _{h=0}^iW_h$ is equal to the common value of (52).

Proof

Define $W=\sum _{h=0}^i W_h$ and let U denote the common value of (52). We show that W = U. By Lemma 8 and the equation on the left in (51), W ⊆ U. By Lemma 24, W and U have the same dimension. Thus W = U. □

9 The Actions of ψ, K, B,  Δ, A, A ^∗ on $\{W_i\}_{i=0}^d$

We continue to discuss the situation of Assumption 1. Recall the eigenspace decomposition $\{W_i\}_{i=0}^d$ for $\mathcal {M}$. In this section, we discuss the actions of ψ, K, B,  Δ, A, A ^∗ on $\{W_i\}_{i=0}^d$.

Lemma 26

For 0 ≤ i ≤ d,

$$\displaystyle \begin{aligned} \psi W_i\subseteq W_{i-1}. \end{aligned} $$

(53)

Proof

Use Lemma 19. □

Lemma 27

For 0 ≤ i ≤ d,

$$\displaystyle \begin{aligned} (K-q^{d-2i}I)W_i\subseteq W_{i-1},\qquad \qquad (B-q^{d-2i}I)W_i\subseteq W_{i-1}. \end{aligned} $$

(54)

Proof

Use Lemmas 16 and 26. □

Lemma 28

For 0 ≤ i ≤ d,

$$\displaystyle \begin{aligned} \begin{array}{rcl} (\Delta-I)W_i&\displaystyle \subseteq W_0+W_1+\cdots +W_{i-1},{} \end{array} \end{aligned} $$

(55)

$$\displaystyle \begin{aligned} \begin{array}{rcl} (\Delta^{-1}-I)W_i&\displaystyle \subseteq W_0+W_1+\cdots +W_{i-1}{}. \end{array} \end{aligned} $$

(56)

Proof

To show (55), use (25) and Lemma 25.

To show (56), use (26) and Lemma 25. □

Lemma 29

For 0 ≤ i ≤ d,

$$\displaystyle \begin{aligned} (A-(a+a^{-1})q^{d-2i}I)W_i\subseteq W_{i-1}+W_{i+1}. \end{aligned} $$

(57)

Proof

By Lemma 22, the expression

$$\displaystyle \begin{aligned} (\mathcal{M}^{-1}-q^{2i+2-d}I)(\mathcal{M}^{-1}-q^{2i-2-d}I)(A-(a+a^{-1})q^{d-2i}I) \end{aligned}$$

vanishes on W _i. Therefore $(\mathcal {M}^{-1}-q^{2i+2-d}I)(\mathcal {M}^{-1}-q^{2i-2-d}I)$ vanishes on (A − (a + a ⁻¹)q ^d−2iI)W _i. The result follows. □

Lemma 30

For 0 ≤ i ≤ d,

$$\displaystyle \begin{aligned} (A^{\ast}-\theta_i^{\ast}I)W_i\subseteq W_0+W_1+\cdots +W_{i-1}. \end{aligned} $$

(58)

Proof

Use $(A^{\ast }-\theta _i^{\ast }I)E_i^{\ast }V=0$ together with (25) and Lemma 25. □

10 The Actions of $\mathcal {M}^{\pm 1}$ on $\{U_i\}_{i=0}^d$,$\{U_i^\Downarrow \}_{i=0}^d$, $\{E_iV\}_{i=0}^d$, $\{E_i^{\ast }V\}_{i=0}^d$

We continue to discuss the situation of Assumption 1. In Sect. 8 we saw how various operators act on the decomposition $\{W_i\}_{i=0}^d$. In this section we investigate the action of $\mathcal {M}$ on the first and second split decompositions of V , as well as on the eigenspace decompositions of A, A ^∗.

Lemma 31

For 0 ≤ i ≤ d,

$$\displaystyle \begin{aligned} \begin{array}{rcl} (\mathcal{M} -q^{d-2i}I)U_i&\displaystyle \subseteq U_0+U_1+\cdots+U_{i-1},{} \end{array} \end{aligned} $$

(59)

$$\displaystyle \begin{aligned} \begin{array}{rcl} (\mathcal{M} -q^{d-2i}I)U_i^\Downarrow&\displaystyle \subseteq U_0^\Downarrow+U_1^\Downarrow+\cdots+U_{i-1}^\Downarrow.{} \end{array} \end{aligned} $$

(60)

Proof

To show (59), use Definition 5, Lemma 2, and Definition 7.

To show (60), use (59) applied to Φ^⇓, along with $\mathcal {M}^\Downarrow =\mathcal {M}$. □

Lemma 32

For 0 ≤ i ≤ d,

$$\displaystyle \begin{aligned} (\mathcal{M}^{-1}-q^{2i-d}I)U_i\subseteq U_{i-1},\qquad \qquad (\mathcal{M}^{-1}-q^{2i-d}I)U_i^\Downarrow \subseteq U_{i-1}^\Downarrow.{} \end{aligned} $$

(61)

Proof

We first show the equation on the left in (61). By Lemma 17,

$$\displaystyle \begin{aligned} \mathcal{M}^{-1}=(I-a^{-1}q^{-1}\psi)K^{-1}. \end{aligned} $$

(62)

From this and Definition 5, it follows that on U _i,

$$\displaystyle \begin{aligned} \mathcal{M}^{-1}-q^{2i-d}I=a^{-1}q^{2i-d-1}\psi. \end{aligned} $$

(63)

The result follows from this along with Lemma 8.

The proof of the equation on the right in (61) follows from the equation on the left in (61) applied to Φ^⇓, along with the fact that $\mathcal {M}^\Downarrow =\mathcal {M}$. □

Lemma 33

For 0 ≤ i ≤ d,

$$\displaystyle \begin{aligned} \mathcal{M}^{-1}E_iV\subseteq E_{i-1}V+E_iV+E_{i+1}V. \end{aligned} $$

(64)

Proof

We first show that $\mathcal {M}^{-1}E_iV\subseteq \sum _{h=0}^{i+1}E_hV$. Recall from (5) that $E_iV\subseteq \sum _{h=d-i}^d U_h^\Downarrow$. By this, Lemma 32, and (5), we obtain $\mathcal {M}^{-1}E_iV\subseteq \sum _{h=0}^{i+1} E_hV$.

We now show that $\mathcal {M}^{-1}E_iV\subseteq \sum _{h=i-1}^d E_hV$. Recall from (4) that $E_iV\subseteq \sum _{h=i}^d U_h$. By this, Lemma 32, and (4), we obtain $\mathcal {M}^{-1}E_iV\subseteq \sum _{h=i-1}^d E_hV$.

Thus $\mathcal {M}^{-1}E_iV$ is contained in the intersection of $\sum _{h=0}^{i+1} E_hV$ and $\sum _{h=i-1}^d E_hV$, which is E _i−1V + E _iV + E _i+1V . □

Lemma 34

For 0 ≤ i ≤ d,

$$\displaystyle \begin{aligned} \begin{array}{rcl} (\mathcal{M} -q^{d-2i}I)E_i^{\ast}V&\displaystyle \subseteq E_0^{\ast}V+E_1^{\ast}V+\cdots+E_{i-1}^{\ast}V,\\ (\mathcal{M}^{-1} -q^{2i-d}I)E_i^{\ast}V&\displaystyle \subseteq E_0^{\ast}V+E_1^{\ast}V+\cdots+E_{i-1}^{\ast}V. \end{array} \end{aligned} $$

Proof

Note that $E_i^{\ast }V\subseteq E_0^{\ast }V+E_1^{\ast }V+\cdots +E_i^{\ast }V= W_0+W_1+\cdots +W_i$ by Lemma 25. The result follows from this fact along with Definition 8. □

11 When Φ Is a Leonard System

We continue to discuss the situation of Assumption 1. For the rest of the paper we assume ρ _i = 1 for 0 ≤ i ≤ d. In this case Φ is called a Leonard system.

We use the following notational convention. Let $\{V_i\}_{i=0}^d$ denote a basis for V . The sequence of subspaces $\{\mathbb {K} v_i\}_{i=0}^d$ is a decomposition of V  said, to be induced by the basis $\{V_i\}_{i=0}^d$.

We display a basis $\{U_i\}_{i=0}^d$ (resp. $\{U_i^\Downarrow \}_{i=0}^d$) (resp. $\{W_i\}_{i=0}^d$) that induces the decomposition $\{U_i\}_{i=0}^d$ (resp. $\{U_i^\Downarrow \}_{i=0}^d$) (resp. $\{W_i\}_{i=0}^d$). We find the actions of ψ, K, B, Δ^±1, A on these bases. We also display the transition matrices between these bases.

For the rest of this section fix 0 ≠ u ₀ ∈ U ₀. Let M denote the subalgebra of End(V ) generated by A. By [21, Lemma 5.1], the map M → V , X↦Xu ₀ is an isomorphism of vector spaces. Consequently, the vectors $\{A^iu_0\}_{i=0}^d$ form a basis for V .

We now define a basis $\{U_i\}_{i=0}^d$ of V  that induces $\{U_i\}_{i=0}^d$. For 0 ≤ i ≤ d, define

$$\displaystyle \begin{aligned} u_i=\left(\prod_{j=0}^{i-1}\left(A-\theta_jI\right)\right) u_0.{} \end{aligned} $$

(65)

Observe that u _i ≠ 0. By [9, Theorem 4.6], u _i ∈ U _i. So u _i is a basis for U _i. Consequently, $\{U_i\}_{i=0}^d$ is a basis for V  that induces $\{U_i\}_{i=0}^d$.

Next we define a basis $\{U_i^\Downarrow \}_{i=0}^d$ of V  that induces $\{U_i^\Downarrow \}_{i=0}^d$. For 0 ≤ i ≤ d, define

$$\displaystyle \begin{aligned} u_i^\Downarrow=\left(\prod_{j=0}^{i-1}\left(A-\theta_{d-j}I\right)\right) u_0.{}\end{aligned} $$

(66)

Observe that $u_i^\Downarrow \neq 0$. By Lemma 11, $u_i^\Downarrow \in U_i^\Downarrow$. So $U_i^\Downarrow$ is a basis for $U_i^\Downarrow$. Consequently, $\{U_i^\Downarrow \}_{i=0}^d$ is a basis for V  that induces $\{U_i^\Downarrow \}_{i=0}^d$.

Lemma 35

For 0 ≤ i ≤ d,

$$\displaystyle \begin{aligned} u_i^\Downarrow=\Delta u_i.\end{aligned} $$

(67)

Proof

By Lemma 11, $\Delta U_i=U_i^\Downarrow$. So there exists $0\neq \lambda \in \mathbb {K}$ such that $\Delta u_i=\lambda u_i^\Downarrow$. We show that λ = 1. By [4, Lemma 7.3] and (25), Δu _i − A ⁱu is a linear combination of $\{A^ju\}_{j=0}^{i-1}$. Also, $u_i^\Downarrow -A^iu$ is a linear combination of $\{A^ju\}_{j=0}^{i-1}$. The vectors $\{A^ju\}_{j=0}^{i-1}$ are linearly independent. By these comments λ = 1. □

We next define a basis $\{W_i\}_{i=0}^d$ of V  that induces $\{W_i\}_{i=0}^d$. For 0 ≤ i ≤ d, define

$$\displaystyle \begin{aligned} w_i=\exp_{q^{-1}}\left(-\frac{a^{-1}}{q-q^{-1}}\ \psi\right) u_i.{}\end{aligned} $$

(68)

Since $\{u_i \}_{i=0}^d$ is a basis of V  and $\exp _{q^{-1}}(-\frac {a^{-1}}{q-q^{-1}}\ \psi )$ is invertible, w _i is a basis for W _i. Consequently, $\{W_i\}_{i=0}^d$ is a basis for V  that induces $\{W_i\}_{i=0}^d$.

Lemma 36

For 0 ≤ i ≤ d,

$$\displaystyle \begin{aligned} \begin{array}{rcl} u_i&\displaystyle =\exp_q\left(\frac{a^{-1}}{q-q^{-1}}\psi\right) w_i, \qquad \qquad u_i^\Downarrow&\displaystyle =\exp_q\left(\frac{a}{q-q^{-1}}\psi\right) w_i, {} \end{array} \end{aligned} $$

(69)

$$\displaystyle \begin{aligned} \begin{array}{rcl} w_i&\displaystyle =\exp_{q^{-1}}\left(-\frac{a^{-1}}{q-q^{-1}}\psi\right) u_i, \qquad \qquad w_i&\displaystyle =\exp_{q^{-1}}\left(-\frac{a}{q-q^{-1}}\psi\right) u_i^\Downarrow.{} \end{array} \end{aligned} $$

(70)

Proof

Use (68) to obtain the equations on the left in (69),(70). To obtain the equations on the right in (69),(70), use Theorem 3, Lemma 35, and (68). □

We now describe the actions of $\psi ,K,B,\mathcal {M}, \Delta ,A$ on the bases $\{U_i\}_{i=0}^d$, $\{U_i^\Downarrow \}_{i=0}^d$, $\{W_i\}_{i=0}^d$. First we recall a notion from linear algebra. Let $\mathrm {Mat}_{d+1}(\mathbb {K})$ denote the $\mathbb {K}$-algebra of (d + 1) × (d + 1) matrices that have all entries in $\mathbb {K}$. We index the rows and columns by 0, 1, …, d. Let $\{V_i\}_{i=0}^d$ denote a basis of V . For T ∈End(V ) and $X\in \mathrm {Mat}_{d+1}(\mathbb {K})$, we say that X represents T with respect to $\{V_i\}_{i=0}^d$ whenever $Tv_j=\sum _{i=0}^d X_{ij}v_i$ for 0 ≤ j ≤ d.

By (65) and (66), the matrices that represent A with respect to $\{U_i\}_{i=0}^d$ and $\{U_i^\Downarrow \}_{i=0}^d$ are, respectively,

(71)

By construction, the matrix diag(q ^d, q ^d−2, …, q ^−d) represents K with respect to $\{U_i\}_{i=0}^d$, and B with respect to $\{U_i^\Downarrow \}_{i=0}^d$, and $\mathcal {M}$ with respect to $\{W_i\}_{i=0}^d$.

Definition 9

We define a matrix $\widehat {\psi }\in \mathrm {Mat}_{d+1}(\mathbb {K})$. For 1 ≤ i ≤ d, the (i − 1, i)-entry is (q ⁱ − q ⁻ⁱ)(q ^d−i+1 − q ^i−d−1). All other entries are 0.

Proposition 3

The matrix $\widehat \psi$ represents ψ with respect to each of the bases $\{U_i\}_{i=0}^d$, $\{U_i^\Downarrow \}_{i=0}^d$, $\{W_i\}_{i=0}^d$.

Proof

By [5, Line (23)], $\widehat \psi$ represents ψ with respect to $\{U_i\}_{i=0}^d$. The remaining assertions follow from Lemma 36. □

Next we give the matrices that represent $\mathcal {M}^{\pm 1}$ with respect to the bases $\{U_i\}_{i=0}^d$, $\{U_i^\Downarrow \}_{i=0}^d$.

Lemma 37

We give the matrix in $\mathrm {Mat}_{d+1}(\mathbb {K})$ that represents $\mathcal {M}$ with respect to $\{U_i\}_{i=0}^d$. This matrix is upper triangular. For 0 ≤ i ≤ j ≤ d, the (i, j)-entry is

$$\displaystyle \begin{aligned} a^{i-j}q^{d-j-i}\left(q-q^{-1}\right)^{2(j-i)} \frac{\ \ [j]_q^![d-i]_q^!\ \ }{[i]_q^![d-j]_q^!}. \end{aligned} $$

(72)

Proof

The matrix diag(q ^d, q ^d−2, …, q ^−d) represents K with respect to $\{U_i\}_{i=0}^d$. Use this fact along with Lemma 18 and Proposition 3. □

Lemma 38

We give the matrix in $\mathrm {Mat}_{d+1}(\mathbb {K})$ that represents $\mathcal {M}^{-1}$ with respect to $\{U_i\}_{i=0}^d$. For 0 ≤ i ≤ d, the (i, i)-entry is q ^2i−d. For 1 ≤ i ≤ d, the (i − 1, i)-entry is

$$\displaystyle \begin{aligned} -a^{-1}q^{2i-d-1}\left(q^i-q^{-i}\right)\left(q^{d-i+1}-q^{i-d-1}\right). \end{aligned}$$

All other entries are zero.

Proof

The matrix diag(q ^−d, q ^2−d, …, q ^d) represents K ⁻¹ with respect to $\{U_i\}_{i=0}^d$. Use this fact along with Lemma 17 and Proposition 3. □

Lemma 39

We give the matrix in $\mathrm {Mat}_{d+1}(\mathbb {K})$ that represents $\mathcal {M}$ with respect to $\{U_i^\Downarrow \}_{i=0}^d$. This matrix is upper triangular. For 0 ≤ i ≤ j ≤ d, the (i, j)-entry is

$$\displaystyle \begin{aligned} a^{j-i}q^{d-j-i}\left(q-q^{-1}\right)^{2(j-i)} \frac{\ \ [j]_q^![d-i]_q^!\ \ }{[i]_q^![d-j]_q^!}. \end{aligned} $$

(73)

Proof

The matrix diag(q ^d, q ^d−2, …, q ^−d) represents B with respect to $\{U_i^\Downarrow \}_{i=0}^d$. Use this fact along with Lemma 18 and Proposition 3. □

Lemma 40

We give the matrix in $\mathrm {Mat}_{d+1}(\mathbb {K})$ that represents $\mathcal {M}^{-1}$ with respect to $\{U_i^\Downarrow \}_{i=0}^d$. For 0 ≤ i ≤ d, the (i, i)-entry is q ^2i−d. For 1 ≤ i ≤ d, the (i − 1, i)-entry is

$$\displaystyle \begin{aligned} -aq^{2i-d-1}\left(q^i-q^{-i}\right)\left(q^{d-i+1}-q^{i-d-1}\right). \end{aligned}$$

All other entries are zero.

Proof

The matrix diag(q ^−d, q ^2−d, …, q ^d) represents B ⁻¹ with respect to $\{U_i^\Downarrow \}_{i=0}^d$. Use this fact along with Lemma 17 and Proposition 3. □

Next we give the matrices that represent K with respect to the bases $\{U_i^\Downarrow \}_{i=0}^d$, $\{W_i\}_{i=0}^d$.

Lemma 41

We give the matrix in $\mathrm {Mat}_{d+1}(\mathbb {K})$ that represents K with respect to $\{U_i^\Downarrow \}_{i=0}^d$. For 0 ≤ i ≤ d, the (i, i)-entry is q ^d−2i. For 0 ≤ i < j ≤ d, the (i, j)-entry is

$$\displaystyle \begin{aligned} \left(1-a^{-2}\right)a^{j-i}q^{d-j-i} \left(q-q^{-1}\right)^{2(j-i)} \frac{\ \ [j]_q^![d-i]_q^!\ \ }{[i]_q^![d-j]_q^!}. \end{aligned} $$

(74)

All other entries are zero.

Proof

Evaluating the equation on the right in (14) using the equation on the left in (12) we get

$$\displaystyle \begin{aligned} K=\left(a^{-2}I+(1-a^{-2})\sum_{n=0}^d a^nq^{n}\psi^n\right)B. {} \end{aligned} $$

(75)

The result follows from this along with Proposition 3 and the fact that the matrix diag(q ^d, q ^d−2, …, q ^−d) represents B with respect to $\{U_i^\Downarrow \}_{i=0}^d$. □

Lemma 42

We give the matrix in $\mathrm {Mat}_{d+1}(\mathbb {K})$ that represents K with respect to $\{W_i\}_{i=0}^d$. For 0 ≤ i ≤ d, the (i, i)-entry is q ^d−2i. For 1 ≤ i ≤ d, the (i − 1, i)-entry is

$$\displaystyle \begin{aligned} -a^{-1}q^{d-2i+1} (q^{i}-q^{-i})(q^{d-i+1}-q^{i-d-1}). \end{aligned}$$

All other entries are zero.

Proof

The matrix diag(q ^d, q ^d−2, …, q ^−d) represents $\mathcal {M}$ with respect to $\{W_i\}_{i=0}^d$. Use this fact along with Proposition 3 and the equation on the left in (32). □

Next we give the matrices that represent B with respect to the bases $\{U_i\}_{i=0}^d$, $\{W_i\}_{i=0}^d$.

Lemma 43

We give the matrix in $\mathrm {Mat}_{d+1}(\mathbb {K})$ that represents B with respect to $\{U_i\}_{i=0}^d$. For 0 ≤ i ≤ d, the (i, i)-entry is q ^d−2i. For 0 ≤ i < j ≤ d, the (i, j)-entry is

$$\displaystyle \begin{aligned} \left(1-a^{2}\right)a^{i-j}q^{d-j-i}\left(q-q^{-1}\right)^{2(j-i)} \frac{\ \ [j]_q^![d-i]_q^!\ \ }{[i]_q^![d-j]_q^!}. \end{aligned} $$

(76)

All other entries are zero.

Proof

Evaluating the equation on the left in (14) using the equation on the right in (12) we get

$$\displaystyle \begin{aligned} B=\left(a^{2}I+(1-a^{2})\sum_{n=0}^d a^{-n}q^{n}\psi^n\right)K. {} \end{aligned} $$

(77)

The result follows from this along with Proposition 3 and the fact that the matrix diag(q ^d, q ^d−2, …, q ^−d) represents K with respect to $\{U_i\}_{i=0}^d$. □

Lemma 44

We give the matrix in $\mathrm {Mat}_{d+1}(\mathbb {K})$ that represents B with respect to $\{W_i\}_{i=0}^d$. For 0 ≤ i ≤ d, the (i, i)-entry is q ^d−2i. For 1 ≤ i ≤ d, the (i − 1, i)-entry is

$$\displaystyle \begin{aligned} -aq^{d-2i+1} (q^{i}-q^{-i})(q^{d-i+1}-q^{i-d-1}). \end{aligned}$$

All other entries are zero.

Proof

The matrix diag(q ^d, q ^d−2, …, q ^−d) represents $\mathcal {M}$ with respect to $\{W_i\}_{i=0}^d$. Use this fact along with Proposition 3 and the equation on the left in (33). □

Next we consider the matrices

$$\displaystyle \begin{aligned} \exp_q\left(\frac{a}{q-q^{-1}}\widehat{\psi}\right),\qquad \qquad \exp_q\left(\frac{a^{-1}}{q-q^{-1}}\widehat{\psi}\right).{} \end{aligned} $$

(78)

Their inverses are

$$\displaystyle \begin{aligned} \exp_{q^{-1}}\left(-\frac{a}{q-q^{-1}}\widehat{\psi}\right),\qquad \qquad \exp_{q^{-1}}\left(-\frac{a^{-1}}{q-q^{-1}}\widehat{\psi}\right){} \end{aligned} $$

(79)

respectively. The matrices in (78), (79) are upper triangular. We now consider the entries of (78), (79).

Lemma 45

For $0\neq x\in \mathbb {K}$, the matrix $\exp _q(x\widehat {\psi })$ is upper triangular. For 0 ≤ i ≤ j ≤ d, the (i, j)-entry is

$$\displaystyle \begin{aligned} x^{j-i}q^{\binom{j-i}{2}}\left(q-q^{-1}\right)^{2(j-i)}\cdot \frac{ [j]_q^![d-i]_q^! }{[i]_q^![j-i]_q^![d-j]_q^!}. \end{aligned} $$

(80)

The matrix $exp_{q^{-1}}(x\widehat {\psi })$ is upper triangular. For 0 ≤ i ≤ j ≤ d, the (i, j)-entry is

$$\displaystyle \begin{aligned} x^{j-i}q^{-\binom{j-i}{2}}\left(q-q^{-1}\right)^{2(j-i)}\cdot \frac{ [j]_q^![d-i]_q^! }{[i]_q^![j-i]_q^![d-j]_q^!}. \end{aligned} $$

(81)

Lemma 46

The transition matrices between the basis $\{W_i\}_{i=0}^d$ and the bases $\{U_i\}_{i=0}^d$, $\{U_i^\Downarrow \}_{i=0}^d$ are given in the table below.

From	To	Transition matrix
$\{U_i\}_{i=0}^d$	$\{W_i\}_{i=0}^d$	$\exp _{q^{-1}}\left (-\frac {a^{-1}}{q-q^{-1}}\widehat \psi \right )$
$\{W_i\}_{i=0}^d$	$\{u_i\}_{i=0}^d$	$\exp _q\left (\frac {a^{-1}}{q-q^{-1}}\widehat \psi \right )$
$\{U_i^\Downarrow \}_{i=0}^d$	$\{W_i\}_{i=0}^d$	$\exp _{q^{-1}}\left (-\frac {a}{q-q^{-1}}\widehat \psi \right )$
$\{W_i\}_{i=0}^d$	$\{U_i^\Downarrow \}_{i=0}^d$	$\exp _q\left (\frac {a}{q-q^{-1}}\widehat \psi \right )$

Proof

Use Lemma 36 and Proposition 3. □

We next consider the product

$$\displaystyle \begin{aligned} \exp_q\left(\frac{a}{q-q^{-1}}\widehat\psi\right)\exp_{q^{-1}}\left(-\frac{a^{-1}}{q-q^{-1}}\widehat\psi\right).{} \end{aligned} $$

(82)

The inverse of (82) is

$$\displaystyle \begin{aligned} \exp_q\left(\frac{a^{-1}}{q-q^{-1}}\widehat\psi\right)\exp_{q^{-1}}\left(-\frac{a}{q-q^{-1}}\widehat\psi\right). {} \end{aligned} $$

(83)

The matrices in (82), (83) are upper triangular.

Lemma 47

The transition matrices between the bases $\{U_i\}_{i=0}^d$, $\{U_i^\Downarrow \}_{i=0}^d$ are given in the table below.

From	To	Transition matrix
$\{U_i\}_{i=0}^d$	$\{u^\Downarrow _i\}_{i=0}^d$	$\exp _q\left (\frac {a}{q-q^{-1}}\widehat \psi \right )\exp _{q^{-1}}\left (-\frac {a^{-1}}{q-q^{-1}}\widehat \psi \right )$
$\{u^\Downarrow _i\}_{i=0}^d$	$\{U_i\}_{i=0}^d$	$\exp _q\left (\frac {a^{-1}}{q-q^{-1}}\widehat \psi \right )\exp _{q^{-1}}\left (-\frac {a}{q-q^{-1}}\widehat \psi \right )$

Proof

Use Lemma 46. □

Lemma 48

With respect to each of the bases $\{u_i\}_{i=0}^d, \{u_i^\Downarrow \}_{i=0}^d, \{w_i\}_{i=0}^d$ , the matrices that represent Δ and Δ ⁻¹ are $\exp _q\left (\frac {a}{q-q^{-1}}\widehat \psi \right )\exp _{q^{-1}}\left (-\frac {a^{-1}}{q-q^{-1}}\widehat \psi \right )$ and

$\exp _q\left (\frac {a^{-1}}{q-q^{-1}}\widehat \psi \right )\exp _{q^{-1}}\left (-\frac {a}{q-q^{-1}}\widehat \psi \right )$ respectively.

Proof

Use Theorem 3 and Proposition 3. □

We give the entries of the matrices representing Δ, Δ⁻¹ in the following lemma.

Lemma 49

The matrix in (82) is upper triangular. For 0 ≤ i ≤ j ≤ d, the (i, j)-entry of (82) is

$$\displaystyle \begin{aligned} \frac{\left(q-q^{-1}\right)^{j-i} [j]_q^![d-i]_q^! }{[i]_q^![j-i]_q^![d-j]_q^! } \ \prod_{n=1}^{j-i} \left(aq^{n-1}-a^{-1}q^{1-n}\right). \end{aligned} $$

(84)

The matrix in (83) is upper triangular. For 0 ≤ i ≤ j ≤ d, the (i, j)-entry of (83) is

$$\displaystyle \begin{aligned} \frac{\left(q-q^{-1}\right)^{j-i} [j]_q^![d-i]_q^! }{[i]_q^![j-i]_q^![d-j]_q^! } \ \prod_{n=1}^{j-i} \left(a^{-1}q^{n-1}-aq^{1-n}\right). \end{aligned} $$

(85)

Proof

Use Corollary 1 and Proposition 3. □

We finish the paper by giving the matrix that represents A with respect to $\{W_i\}_{i=0}^d$.

Lemma 50

We give the matrix in $\mathrm {Mat}_{d+1}(\mathbb {K})$ that represents A with respect to $\{W_i\}_{i=0}^d$. For 1 ≤ i ≤ d, the (i, i − 1)-entry is 1. For 0 ≤ i ≤ d, the (i, i)-entry is (a + a ⁻¹)q ^d−2i. For 1 ≤ i ≤ d, the (i − 1, i)-entry is

$$\displaystyle \begin{aligned} -q^{d-2i+1}(q^i-q^{-i})(q^{d-i+1}-q^{i-d-1}). \end{aligned}$$

All other entries are zero.

Proof

Let $\mathcal {A}$ denote the matrix that represents A with respect to $\{W_i\}_{i=0}^d$. By Lemma 29, $\mathcal {A}$ is tridiagonal with (i, i)-entry given by (a + a ⁻¹)q ^d−2i for 0 ≤ i ≤ d.

We now show that the subdiagonal entries of $\mathcal {A}$ are all 1. Let $\mathcal {A}^{\prime }$ denote the matrix that represents A with respect to $\{U_i\}_{i=0}^d$. Recall that this matrix is displayed on the left in (71). Observe that $\mathcal {A}$ is equal to $\exp _{q^{-1}}(-\frac {a^{-1}}{q-q^{-1}}\widehat \psi ) \mathcal {A}' \exp _{q}(\frac {a^{-1}}{q-q^{-1}}\widehat \psi )$. It follows from this fact that the subdiagonal entries of $\mathcal {A}$ are all 1.

We next obtain the superdiagonal entries of $\mathcal {A}$. Let 0 ≤ i ≤ d. Apply both sides of (37) to w _i. Evaluate the result using Proposition 3 and the fact that the w _i is an eigenvector for $\mathcal {M}$ with eigenvalue q ^2i−d. Analyze the result in light of the above comments concerning the entries of $\mathcal {A}$ to obtain the desired result. □