Double Lowering Operators on Polynomials

Double Lowering Operators on Polynomials

Paul TERWILLIGER

P. Terwilliger

Department of Mathematics, University of Wisconsin, Madison, WI 53706-1388,
USA<EMAIL_ADDRESS>
Received September 15, 2020, in final form January 19, 2021; Published online
January 28, 2021

Recently Sarah Bockting-Conrad introduced the double lowering operator $\psi$
for a tridiagonal pair. Motivated by $\psi$ we consider the following problem
about polynomials. Let $\mathbb{F}$ denote an algebraically closed field. Let
$x$ denote an indeterminate, and let $\mathbb{F}[x]$ denote the algebra
consisting of the polynomials in $x$ that have all coefficients in
$\mathbb{F}$. Let $N$ denote a positive integer or $\infty$. Let
$\\{a_{i}\\}_{i=0}^{N-1}$, $\\{b_{i}\\}_{i=0}^{N-1}$ denote scalars in
$\mathbb{F}$ such that $\sum_{h=0}^{i-1}a_{h}\not=\sum_{h=0}^{i-1}b_{h}$ for
$1\leq i\leq N$. For $0\leq i\leq N$ define polynomials
$\tau_{i},\eta_{i}\in\mathbb{F}[x]$ by $\tau_{i}=\prod_{h=0}^{i-1}(x-a_{h})$
and $\eta_{i}=\prod_{h=0}^{i-1}(x-b_{h})$. Let $V$ denote the subspace of
$\mathbb{F}[x]$ spanned by $\\{x^{i}\\}_{i=0}^{N}$. An element
$\psi\in\operatorname{End}(V)$ is called double lowering whenever
$\psi\tau_{i}\in\mathbb{F}\tau_{i-1}$ and
$\psi\eta_{i}\in\mathbb{F}\eta_{i-1}$ for $0\leq i\leq N$, where $\tau_{-1}=0$
and $\eta_{-1}=0$. We give necessary and sufficient conditions on
$\\{a_{i}\\}_{i=0}^{N-1}$, $\\{b_{i}\\}_{i=0}^{N-1}$ for there to exist a
nonzero double lowering map. There are four families of solutions, which we
describe in detail.

tridiagonal pair; $q$-exponential function; basic hypergeometric series;
$q$-binomial theorem

33D15; 15A21

##  1 Introduction

This paper is mainly about polynomials and special functions, but in order to
motivate things we first discuss a topic in linear algebra. The topic has to
do with tridiagonal pairs [17] and their associated double lowering operator
[7, 8, 9, 10]. A reader unfamiliar with tridiagonal pairs can safely skip to
Section 2. Let $V$ denote a vector space with finite positive dimension. A
tridiagonal pair on $V$ is an ordered pair of linear maps $A\colon V\to V$ and
$A^{*}\colon V\to V$ that satisfy the following conditions:

  1. (i)

each of $A$, $A^{*}$ is diagonalizable;

  2. (ii)

there exists an ordering $\\{V_{i}\\}_{i=0}^{d}$ of the eigenspaces of $A$
such that

$\displaystyle A^{*}V_{i}\subseteq V_{i-1}+V_{i}+V_{i+1},\qquad 0\leq i\leq
d,$

where $V_{-1}=0$ and $V_{d+1}=0$;

  3. (iii)

there exists an ordering $\\{V^{*}_{i}\\}_{i=0}^{\delta}$ of the eigenspaces
of $A^{*}$ such that

$\displaystyle AV^{*}_{i}\subseteq V^{*}_{i-1}+V^{*}_{i}+V^{*}_{i+1},\qquad
0\leq i\leq\delta,$

where $V^{*}_{-1}=0$ and $V^{*}_{\delta+1}=0$;

  4. (iv)

there does not exist a subspace $W\subseteq V$ such that $AW\subseteq W$ and
$A^{*}W\subseteq W$ and $W\not=0$ and $W\not=V$.

The tridiagonal pair concept originated in the theory of $Q$-polynomial
distance-regular graphs [4, 11] where it is used to describe how the adjacency
matrix is related to each dual adjacency matrix [17, Example 1.4], [29,
Section 3]. Since that origin, the tridiagonal pair concept has found
applications to all sorts of topics in special functions (orthogonal
polynomials of the Askey-scheme [3, 25, 27, 32, 33], the Askey–Wilson algebra
[14, 35, 40, 42]), Lie theory (the ${\mathfrak{sl}}_{2}$ loop algebra [19],
the tetrahedron algebra [15, 21]), statistical mechanics (the Onsager algebra
[12, 16] and $q$-Onsager algebra [5, 6, 23, 31, 37, 38]), and quantum groups
(the equitable presentation [1, 24, 36], the quantum affine
${\mathfrak{sl}}_{2}$ algebra [2, 18, 20, 22], $L$-operators [26, 39]). For
more information about the above topics, see [28, 34] and the references
therein.

Let $A$, $A^{*}$ denote a tridiagonal pair on $V$, as in the above definition.
By [17, Lemma 4.5] the integers $d$ and $\delta$ from (ii), (iii) are equal;
this common value is called the diameter of the pair. For $0\leq i\leq d$ let
$\theta_{i}$ (resp. $\theta^{*}_{i}$) denote the eigenvalue of $A$ (resp.
$A^{*}$) for the eigenspace $V_{i}$ (resp. $V^{*}_{i}$). By [17, Theorem 11.1]
the scalars

$\displaystyle\frac{\theta_{i-2}-\theta_{i+1}}{\theta_{i-1}-\theta_{i}},\qquad\frac{\theta^{*}_{i-2}-\theta^{*}_{i+1}}{\theta^{*}_{i-1}-\theta^{*}_{i}}$

are equal and independent of $i$ for $2\leq i\leq d-1$. For this constraint
the solutions can be given in closed form [17, Theorem 11.2]. The “most
general” solution is called $q$-Racah, and will be described shortly.

By construction the vector space $V$ has a direct sum decomposition into the
eigenspaces $\\{V_{i}\\}_{i=0}^{d}$ of $A$ and the eigenspaces
$\\{V^{*}_{i}\\}_{i=0}^{d}$ of $A^{*}$. The vector space $V$ has two other
direct sum decompositions of interest, called the first split decomposition
$\\{U_{i}\\}_{i=0}^{d}$ and second split decomposition
$\big{\\{}U^{\Downarrow}_{i}\big{\\}}_{i=0}^{d}$. By [17, Theorem 4.6] the
first split decomposition satisfies

$\displaystyle U_{0}+U_{1}+\cdots+U_{i}=V^{*}_{0}+V^{*}_{1}+\cdots+V^{*}_{i},$
$\displaystyle U_{i}+U_{i+1}+\cdots+U_{d}=V_{i}+V_{i+1}+\cdots+V_{d}$

for $0\leq i\leq d$. By [17, Theorem 4.6] the second split decomposition
satisfies

$\displaystyle
U^{\Downarrow}_{0}+U^{\Downarrow}_{1}+\cdots+U^{\Downarrow}_{i}=V^{*}_{0}+V^{*}_{1}+\cdots+V^{*}_{i},$
$\displaystyle
U^{\Downarrow}_{i}+U^{\Downarrow}_{i+1}+\cdots+U^{\Downarrow}_{d}=V_{0}+V_{1}+\cdots+V_{d-i}$

for $0\leq i\leq d$. By [17, Theorem 4.6],

$\displaystyle(A-\theta_{i}I)U_{i}\subseteq U_{i+1},\qquad$
$\displaystyle(A^{*}-\theta^{*}_{i}I)U_{i}\subseteq U_{i-1},$
$\displaystyle(A-\theta_{d-i}I)U^{\Downarrow}_{i}\subseteq
U^{\Downarrow}_{i+1},\qquad$
$\displaystyle(A^{*}-\theta^{*}_{i}I)U^{\Downarrow}_{i}\subseteq
U^{\Downarrow}_{i-1}$

for $0\leq i\leq d$, where $U_{-1}=0$, $U_{d+1}=0$ and
$U^{\Downarrow}_{-1}=0$, $U^{\Downarrow}_{d+1}=0$.

In [7, Sections 11 and 15] Sarah Bockting-Conrad introduces a linear map
$\Psi\colon V\to V$ such that

$\displaystyle\Psi U_{i}\subseteq U_{i-1},\qquad\Psi
U^{\Downarrow}_{i}\subseteq U^{\Downarrow}_{i-1},\qquad 0\leq i\leq d.$

This map is called the double lowering operator or Bockting operator. In [7,
Sections 9 and 15] Bockting-Conrad introduces an invertible linear map
$\Delta\colon V\to V$ that commutes with $\Psi$ and sends $U_{i}$ onto
$U^{\Downarrow}_{i}$ for $0\leq i\leq d$. The maps $\Psi$ and $\Delta$ are
related in the following way. For $0\leq i\leq d$ define two polynomials

$\displaystyle\tau_{i}=(x-\theta_{0})(x-\theta_{1})\cdots(x-\theta_{i-1}),$
(1.1)
$\displaystyle\eta_{i}=(x-\theta_{d})(x-\theta_{d-1})\cdots(x-\theta_{d-i+1})$
(1.2)

in a variable $x$. Define the scalars

$\displaystyle\vartheta_{i}=\sum_{h=0}^{i-1}\frac{\theta_{h}-\theta_{d-h}}{\theta_{0}-\theta_{d}},\qquad
1\leq i\leq d.$

By [7, Theorem 17.1],

$\displaystyle\Delta=\sum_{i=0}^{d}\frac{\eta_{i}(\theta_{0})}{\vartheta_{1}\vartheta_{2}\cdots\vartheta_{i}}\Psi^{i},\qquad\Delta^{-1}=\sum_{i=0}^{d}\frac{\tau_{i}(\theta_{d})}{\vartheta_{1}\vartheta_{2}\cdots\vartheta_{i}}\Psi^{i}$
(1.3)

provided that each of $\vartheta_{1},\vartheta_{2},\ldots,\vartheta_{d}$ is
nonzero.

Shortly we will describe $\Psi$ and $\Delta$ in more detail, but first we
restrict to the $q$-Racah case. In this case there exist nonzero scalars $a$,
$b$, $q$ such that $q^{4}\not=1$ and

$\displaystyle\theta_{i}=aq^{d-2i}+a^{-1}q^{2i-d},\qquad\theta^{*}_{i}=bq^{d-2i}+b^{-1}q^{2i-d}$

for $0\leq i\leq d$. Define a linear map $K\colon V\to V$ such that for $0\leq
i\leq d$, $U_{i}$ is an eigenspace of $K$ with eigenvalue $q^{d-2i}$. Define a
linear map $B\colon V\to V$ such that for $0\leq i\leq d$,
$U^{\Downarrow}_{i}$ is an eigenspace of $B$ with eigenvalue $q^{d-2i}$. For
notational convenience define
$\psi=\big{(}q-q^{-1}\big{)}\big{(}q^{d}-q^{-d}\big{)}\Psi$. By [9, Lemma 5.3]
and [8, Lemma 5.4],

$\displaystyle B\Delta=\Delta K,\qquad K\psi=q^{2}\psi K,\qquad
B\psi=q^{2}\psi B.$

By [8, Theorem 9.8] the map $\psi$ is equal to each of the following:

$\displaystyle\frac{I-BK^{-1}}{q\big{(}aI-a^{-1}BK^{-1}\big{)}},\qquad$
$\displaystyle\frac{I-KB^{-1}}{q\big{(}a^{-1}I-aKB^{-1}\big{)}},$
$\displaystyle\frac{q\big{(}I-K^{-1}B\big{)}}{aI-a^{-1}K^{-1}B},\qquad$
$\displaystyle\frac{q\big{(}I-B^{-1}K\big{)}}{a^{-1}I-aB^{-1}K}.$

This result is used in [8, Theorem 9.9] to obtain

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

By (1.3) and [9, Theorem 7.2],

$\displaystyle\Delta=\exp_{q}\left(\frac{a}{q-q^{-1}}\psi\right)\exp_{q^{-1}}\left(-\frac{a^{-1}}{q-q^{-1}}\psi\right).$

Motivated by this factorization, in [9, Sections 6,7] Bockting-Conrad
introduces an invertible linear map $M\colon V\to V$ such that

$\displaystyle
K\exp_{q}\left(\frac{a^{-1}}{q-q^{-1}}\psi\right)=\exp_{q}\left(\frac{a^{-1}}{q-q^{-1}}\psi\right)M,$
$\displaystyle
B\exp_{q}\left(\frac{a}{q-q^{-1}}\psi\right)=\exp_{q}\left(\frac{a}{q-q^{-1}}\psi\right)M.$

By [9, Section 6],

$\displaystyle M=\frac{aK-a^{-1}B}{a-a^{-1}},\qquad M\psi=q^{2}\psi M.$

By [9, Lemma 6.2], $M$ is equal to each of

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

By [9, Lemma 6.7],

$\displaystyle\frac{qM^{-1}K-q^{-1}KM^{-1}}{q-q^{-1}}=I,\qquad\frac{qM^{-1}B-q^{-1}BM^{-1}}{q-q^{-1}}=I.$

We just listed many results about $\psi$, $\Delta$, $K$, $B$, $M$. In the
present paper, we interpret these results in terms of polynomials. The
polynomials in question are essentially (1.1), (1.2) although we adopt a more
general point of view. In the next section we will describe a problem about
polynomials, and for the rest of the paper we will describe the solution. In
this description we will encounter analogs of the above results. We hope that
the above results are illuminated by our description.

##  2 Definitions and first steps

We now begin our formal argument. The following assumptions and notational
conventions apply throughout the paper. Recall the natural numbers
$\mathbb{N}=\\{0,1,2,\ldots\\}$ and integers $\mathbb{Z}=\\{0,\pm 1,\pm
2,\ldots\\}$. Let $\mathbb{F}$ denote an algebraically closed field. Every
vector space discussed in this paper is over $\mathbb{F}$. Every algebra
discussed in this paper is associative, over $\mathbb{F}$, and has a
multiplicative identity. Let $x$ denote an indeterminate. Let $\mathbb{F}[x]$
denote the algebra consisting of the polynomials in $x$ that have all
coefficients in $\mathbb{F}$. Throughout the paper we use the following
convention: the symbols $N$, $n$ refer to an integer or $\infty$; the symbols
$i$, $j$, $k$ refer to an integer.

We now describe a problem about polynomials. Let $N$ denote a positive integer
or $\infty$. We consider an ordered pair of sequences

$\displaystyle\\{a_{i}\\}_{i=0}^{N-1},\qquad\\{b_{i}\\}_{i=0}^{N-1}$ (2.1)

such that $a_{i},b_{i}\in\mathbb{F}$ for $0\leq i\leq N-1$. To avoid
degenerate situations, we assume that

$\displaystyle
a_{0}+a_{1}+\cdots+a_{i-1}\not=b_{0}+b_{1}+\cdots+b_{i-1},\qquad 1\leq i\leq
N.$ (2.2)

We call the ordered pair (2.1) the data. For $0\leq i\leq N$ define
polynomials $\tau_{i}$, $\eta_{i}\in\mathbb{F}[x]$ by

$\displaystyle\tau_{i}=(x-a_{0})(x-a_{1})\cdots(x-a_{i-1}),$ (2.3)
$\displaystyle\eta_{i}=(x-b_{0})(x-b_{1})\cdots(x-b_{i-1}).$ (2.4)

The polynomials $\tau_{i}$, $\eta_{i}$ are monic of degree $i$. For notational
convenience define $\tau_{-1}=0$ and $\eta_{-1}=0$. Let $V$ denote the
subspace of $\mathbb{F}[x]$ spanned by $\\{x^{i}\\}_{i=0}^{N}$. Each of
$\\{\tau_{i}\\}_{i=0}^{N}$, $\\{\eta_{i}\\}_{i=0}^{N}$ is a basis for $V$. Let
$\operatorname{End}(V)$ denote the algebra consisting of the
$\mathbb{F}$-linear maps from $V$ to $V$. Define
$\Delta\in\operatorname{End}(V)$ such that

$\displaystyle\Delta\tau_{i}=\eta_{i},\qquad 0\leq i\leq N.$ (2.5)

Note that $\Delta$ is invertible.

######  Definition 2.1.

An element $\psi\in\operatorname{End}(V)$ is called double lowering (with
respect to the given data) whenever both

$\displaystyle\psi\tau_{i}\in\mathbb{F}\tau_{i-1},\qquad\psi\eta_{i}\in\mathbb{F}\eta_{i-1}$

for $0\leq i\leq N$.

######  Definition 2.2.

Define

$\displaystyle\mathcal{L}=\\{\psi\in\operatorname{End}(V)\,|\,\psi\;{\mbox{\rm
is double lowering}}\\}.$

Note that $\mathcal{L}$ is a subspace of the vector space
$\operatorname{End}(V)$. We call $\mathcal{L}$ the double lowering space for
the given data.

######  Definition 2.3.

The data (2.1) is called double lowering whenever the double lowering space
$\mathcal{L}\not=0$.

######  Problem 2.4.

Find necessary and sufficient conditions for the data (2.1) to be double
lowering. In this case describe $\mathcal{L}$ and $\Delta$.

The above problem is solved in the present paper. The necessary and sufficient
conditions are given in Theorem 12.1. By that theorem, there are four cases.
For the first three cases, $\mathcal{L}$ and $\Delta$ are described in Section
6. For the fourth case, $\mathcal{L}$ and $\Delta$ are described in Section
13. We would like to acknowledge that the above problem was previously solved
by R. Vidunas under the assumption that $a_{i}=\theta_{i}$ and
$b_{i}=\theta_{N-i}$ for $0\leq i\leq N-1$, with $\\{\theta_{i}\\}_{i=0}^{N}$
mutually distinct [41].

We have some remarks. The polynomials (2.3), (2.4) satisfy

$\displaystyle\tau_{0}=1,\qquad\tau_{1}=x-a_{0},\qquad\eta_{0}=1,\qquad\eta_{1}=x-b_{0}.$
(2.6)

Moreover

$\displaystyle(x-a_{i})\tau_{i}=\tau_{i+1},\qquad(x-b_{i})\eta_{i}=\eta_{i+1},\qquad
0\leq i\leq N-1.$ (2.7)

######  Lemma 2.5.

Assume that $\psi\in\operatorname{End}(V)$ is double lowering. Then $\psi
1=0$. Moreover

$\displaystyle\psi\tau_{1}=\psi x=\psi\eta_{1},$ (2.8)

and this common value is contained in $\mathbb{F}$.

###### Proof.

Apply $\psi$ to each side of the equations in (2.6), and use Definition 2.1. ∎

For $0\leq n\leq N$ define

$\displaystyle V_{n}=\operatorname{Span}\big{\\{}x^{i}\big{\\}}_{i=0}^{n}.$
(2.9)

We have $V_{N}=V$. We have $\dim(V_{i})=i+1$ for $0\leq i\leq N$, and
$V_{i-1}\subseteq V_{i}$ for $1\leq i\leq N$. For notational convenience
define $V_{-1}=0$.

######  Lemma 2.6.

For $0\leq n\leq N$, each of

$\displaystyle\\{\tau_{i}\\}_{i=0}^{n},\qquad\\{\eta_{i}\\}_{i=0}^{n}$

is a basis for $V_{n}$.

###### Proof.

Each of $\tau_{i}$, $\eta_{i}$ has degree $i$ for $0\leq i\leq N$. ∎

######  Lemma 2.7.

For $\psi\in\mathcal{L}$ and $0\leq i\leq N$,

$\displaystyle\psi V_{i}\subseteq V_{i-1}.$

Moreover $\psi^{i+1}V_{i}=0$.

###### Proof.

By Definition 2.1 and Lemma 2.6. ∎

For $T\in\operatorname{End}(V)$, $T$ is called nilpotent whenever there exists
a positive integer $j$ such that $T^{j}=0$. The map $T$ is called locally
nilpotent whenever for all $v\in V$ there exists a positive integer $j$ such
that $T^{j}v=0$. If $T$ is nilpotent then $T$ is locally nilpotent. For
$N\not=\infty$, if $T$ is locally nilpotent then $T$ is nilpotent.

######  Lemma 2.8.

Each element of $\mathcal{L}$ is locally nilpotent. If $N\not=\infty$ then
each element of $\mathcal{L}$ is nilpotent.

###### Proof.

By Lemma 2.7 and the comments below it. ∎

######  Lemma 2.9.

For $1\leq i\leq N$,

  1. $(i)$

in $\tau_{i}$ the coefficient of $x^{i-1}$ is

$\displaystyle-a_{0}-a_{1}-\cdots-a_{i-1};$

  2. $(ii)$

in $\eta_{i}$ the coefficient of $x^{i-1}$ is

$\displaystyle-b_{0}-b_{1}-\cdots-b_{i-1}.$

###### Proof.

Use (2.3), (2.4). ∎

######  Lemma 2.10.

For $1\leq i\leq N$,

  1. $(i)$

$\eta_{i}-\tau_{i}\in V_{i-1}$;

  2. $(ii)$

in $\eta_{i}-\tau_{i}$ the coefficient of $x^{i-1}$ is

$\displaystyle a_{0}+a_{1}+\cdots+a_{i-1}-b_{0}-b_{1}-\cdots-b_{i-1}.$

###### Proof.

(i) Each of $\tau_{i}$, $\eta_{i}$ is monic with degree $i$. (ii) Use Lemma
2.9. ∎

######  Definition 2.11.

Define a map $A\colon V_{N-1}\to V_{N}$, $v\mapsto xv$. Note that $A$ is
$\mathbb{F}$-linear.

######  Lemma 2.12.

We have

$\displaystyle AV_{n-1}\subseteq V_{n},\qquad 1\leq n\leq N.$

###### Proof.

By (2.9). ∎

######  Lemma 2.13.

For $0\leq i\leq N-1$,

$\displaystyle A\tau_{i}=a_{i}\tau_{i}+\tau_{i+1},\qquad
A\eta_{i}=b_{i}\eta_{i}+\eta_{i+1}.$

###### Proof.

By (2.7) and Definition 2.11. ∎

We mention an elementary result for later use.

######  Lemma 2.14.

Assume that $T\in\operatorname{End}(V)$ is locally nilpotent. Then $I-T$ is
invertible, and $(I-T)^{-1}=\sum_{i=0}^{N}T^{i}$.

##  3 Adjusting the data

In this section we describe how the double lowering space is affected when we
adjust the data in an affine way.

Let ${\rm GL}_{2}(\mathbb{F})$ denote the group of invertible $2\times 2$
matrices that have all entries in $\mathbb{F}$.

######  Definition 3.1.

Let $G$ denote the subgroup of ${\rm GL}_{2}(\mathbb{F})$ consisting of the
matrices

$\displaystyle\left(\begin{matrix}1&t\\\ 0&s\end{matrix}\right),\qquad
0\not=s\in\mathbb{F},\qquad t\in\mathbb{F}.$

The above matrix is denoted $g(s,t)$.

######  Lemma 3.2.

With reference to Definition 3.1,

  1. $(i)$

$g(s,t)g(S,T)=g(sS,T+tS)$;

  2. $(ii)$

the inverse of $g(s,t)$ is $g\big{(}s^{-1},-s^{-1}t\big{)}$.

###### Proof.

Routine matrix multiplication. ∎

For an algebra $\mathcal{A}$, an automorphism of $\mathcal{A}$ is an algebra
isomorphism $\mathcal{A}\to\mathcal{A}$.

######  Lemma 3.3.

The group $G$ acts on the algebra ${\mathbb{F}}[x]$ as a group of
automorphisms in the following way: each element $g(s,t)\in G$ sends $x\mapsto
sx+t$.

###### Proof.

This is routinely checked using Lemma 3.2. ∎

Recall from Definition 2.2 the double lowering space $\mathcal{L}$ for the
data (2.1). Pick $0\not=s\in\mathbb{F}$ and $t\in\mathbb{F}$. Let
$\mathcal{L}^{\prime}$ denote the double lowering space for the data

$\displaystyle\\{sa_{i}+t\\}_{i=0}^{N-1},\qquad\\{sb_{i}+t\\}_{i=0}^{N-1}.$

######  Proposition 3.4.

The following $(i)$–$(iii)$ hold for the above scalars $s$, $t$ and
$g=g(s,t)$:

  1. $(i)$

there exists an $\mathbb{F}$-linear map $\mathcal{L}\to\mathcal{L}^{\prime}$,
$\psi\mapsto g^{-1}\psi g$;

  2. $(ii)$

there exists an $\mathbb{F}$-linear map $\mathcal{L}^{\prime}\to\mathcal{L}$,
$\zeta\mapsto g\zeta g^{-1}$;

  3. $(iii)$

the maps in $(i)$, $(ii)$ above are inverses, and hence bijections.

###### Proof.

(i) For $\psi\in\mathcal{L}$ we show that $g^{-1}\psi
g\in\mathcal{L}^{\prime}$. For $\alpha\in\mathbb{F}$ define
$\alpha^{\prime}=s\alpha+t$. For $0\leq i\leq N$ define

$\displaystyle\tau^{\prime}_{i}=(x-a^{\prime}_{0})(x-a^{\prime}_{1})\cdots(x-a^{\prime}_{i-1}),$
$\displaystyle\eta^{\prime}_{i}=(x-b^{\prime}_{0})(x-b^{\prime}_{1})\cdots(x-b^{\prime}_{i-1}).$

For notational convenience define $\tau^{\prime}_{-1}=0$ and
$\eta^{\prime}_{-1}=0$. Pick an integer $i$, $0\leq i\leq N$. One checks that
$g$ sends

$\displaystyle\tau^{\prime}_{i}\mapsto
s^{i}\tau_{i},\qquad\eta^{\prime}_{i}\mapsto s^{i}\eta_{i}.$

By Definition 2.1, $\psi$ sends $\tau_{i}$ (resp. $\eta_{i}$) to a scalar
multiple of $\tau_{i-1}$ (resp. $\eta_{i-1}$). By these comments $g^{-1}\psi
g$ sends $\tau^{\prime}_{i}$ (resp. $\eta^{\prime}_{i}$) to a scalar multiple
of $\tau^{\prime}_{i-1}$ (resp. $\eta^{\prime}_{i-1}$). Therefore $g^{-1}\psi
g\in\mathcal{L}^{\prime}$. We have shown that there exists a map
$\mathcal{L}\to\mathcal{L}^{\prime}$, $\psi\mapsto g^{-1}\psi g$. By
construction this map is $\mathbb{F}$-linear.

(ii) Similar to the proof of (i) above. (iii) By construction. ∎

######  Corollary 3.5.

Pick $0\not=s\in\mathbb{F}$ and $t\in\mathbb{F}$. Then the data (2.1) is
double lowering if and only if the data

$\displaystyle\\{sa_{i}+t\\}_{i=0}^{N-1},\qquad\\{sb_{i}+t\\}_{i=0}^{N-1}$

is double lowering.

###### Proof.

By Definition 2.3 and Proposition 3.4. ∎

We have a comment.

######  Lemma 3.6.

Referring to the data (2.1), for distinct
$a^{\prime}_{0},b^{\prime}_{0}\in\mathbb{F}$ define

$\displaystyle s=\frac{a^{\prime}_{0}-b^{\prime}_{0}}{a_{0}-b_{0}},\qquad
t=\frac{a_{0}b^{\prime}_{0}-a^{\prime}_{0}b_{0}}{a_{0}-b_{0}}.$

Then $s\not=0$ and

$\displaystyle sa_{0}+t=a^{\prime}_{0},\qquad sb_{0}+t=b^{\prime}_{0}.$

###### Proof.

Routine. ∎

Corollary 3.5 and Lemma 3.6 show that for double lowering data (2.1), the
scalars $a_{0}$ and $b_{0}$ are “free”, with $a_{0}\not=b_{0}$ being the only
constraint.

##  4 The parameters $\boldsymbol{\vartheta_{i}}$

We continue to discuss the double lowering space $\mathcal{L}$ for the data
(2.1). In this section we use $\mathcal{L}$ to define some scalars
$\\{\vartheta_{i}\\}_{i=0}^{N}$ that will play a role in our theory.

######  Definition 4.1.

For $0\leq i\leq N$ define

$\displaystyle\vartheta_{i}=\frac{a_{0}+a_{1}+\cdots+a_{i-1}-b_{0}-b_{1}-\cdots-
b_{i-1}}{a_{0}-b_{0}}.$

Referring to Definition 4.1, we have $\vartheta_{0}=0$ and $\vartheta_{1}=1$.
By (2.2) we have

$\displaystyle\vartheta_{i}\not=0,\qquad 1\leq i\leq N.$

######  Proposition 4.2.

The following $(i)$–$(iii)$ hold for $\psi\in\mathcal{L}$ and $1\leq i\leq N$:

  1. $(i)$

for the polynomial $\psi x^{i}$ the coefficient of $x^{i-1}$ is
$\vartheta_{i}\psi x$;

  2. $(ii)$

$\psi\tau_{i}=(\vartheta_{i}\psi x)\tau_{i-1}$;

  3. $(iii)$

$\psi\eta_{i}=(\vartheta_{i}\psi x)\eta_{i-1}$.

###### Proof.

We use induction on $i$. First assume that $i=1$, and recall
$\vartheta_{1}=1$. Assertion (i) is vacuously true. Assertions (ii), (iii)
hold by (2.8) and since $\tau_{0}=1=\eta_{0}$. Next assume that $i\geq 2$. By
Lemma 2.7, $\psi x^{i}\in V_{i-1}$. Let the scalar $\alpha_{i}$ be the
coefficient of $x^{i-1}$ in $\psi x^{i}$. Since $\psi\in\mathcal{L}$,
$\psi\tau_{i}\in\mathbb{F}\tau_{i-1}$. Since each of $\tau_{i}$, $\tau_{i-1}$
is monic,

$\displaystyle\psi\tau_{i}=\alpha_{i}\tau_{i-1}.$ (4.1)

Similarly,

$\displaystyle\psi\eta_{i}=\alpha_{i}\eta_{i-1}.$ (4.2)

We show that $\alpha_{i}=\vartheta_{i}\psi x$. Using (4.1), (4.2) we see that

$\displaystyle(a_{0}-b_{0})\vartheta_{i-1}x^{i-2}$ (4.3)

times

$\displaystyle\alpha_{i}-\vartheta_{i}\psi x$ (4.4)

is equal to

$\displaystyle\psi(\eta_{i}-\tau_{i})-(a_{0}-b_{0})\vartheta_{i}\psi x^{i-1}$
(4.5)

minus $\alpha_{i}$ times

$\displaystyle\eta_{i-1}-\tau_{i-1}-(a_{0}-b_{0})\vartheta_{i-1}x^{i-2}$ (4.6)

plus $(a_{0}-b_{0})\vartheta_{i}$ times

$\displaystyle\psi x^{i-1}-(\psi x)\vartheta_{i-1}x^{i-2}.$ (4.7)

By Lemma 2.10 and Definition 4.1,

$\displaystyle\eta_{i}-\tau_{i}-(a_{0}-b_{0})\vartheta_{i}x^{i-1}\in V_{i-2}.$
(4.8)

In this inclusion, we apply $\psi$ to each side and use Lemma 2.7 to find that
(4.5) is contained in $V_{i-3}$. In (4.8) we replace $i$ by $i-1$, to find
that (4.6) is contained in $V_{i-3}$. By induction (4.7) is contained in
$V_{i-3}$. By these comments the polynomial (4.3) times the scalar (4.4) is
contained in $V_{i-3}$. Consider the factors in the polynomial (4.3). We have
$a_{0}-b_{0}\not=0$ and $\vartheta_{i-1}\not=0$ and $x^{i-2}\not\in V_{i-3}$.
So the polynomial (4.3) is not contained in $V_{i-3}$. Consequently the scalar
(4.4) is zero, so $\alpha_{i}=\vartheta_{i}\psi x$. The result follows from
this and (4.1), (4.2). ∎

######  Corollary 4.3.

The map $\mathcal{L}\to\mathbb{F}$, $\psi\mapsto\psi x$ is injective.

###### Proof.

For $\psi\in\mathcal{L}$ such that $\psi x=0$, we show that $\psi=0$. The
vector space $V$ has a basis $\\{\tau_{i}\\}_{i=0}^{N}$. By Definition 2.1,
$\psi\tau_{0}=0$. By Proposition 4.2(ii), $\psi\tau_{i}=0$ for $1\leq i\leq
N$. By these comments $\psi=0$. ∎

######  Corollary 4.4.

Assume that $\mathcal{L}\not=0$. Then the map in Corollary 4.3 is a bijection.
Moreover $\mathcal{L}$ has dimension $1$.

###### Proof.

By Corollary 4.3. ∎

######  Definition 4.5.

An element $\psi\in\mathcal{L}$ is called normalized whenever $\psi x=1$.

######  Lemma 4.6.

The double lowering space $\mathcal{L}$ contains a normalized element if and
only if $\mathcal{L}\not=0$; in this case the normalized element is unique.

###### Proof.

By Corollary 4.4 and Definition 4.5. ∎

######  Lemma 4.7.

Assume that $\mathcal{L}\not=0$. Then the vector space $\mathcal{L}$ has a
basis consisting of its normalized element.

###### Proof.

The vector space $\mathcal{L}$ has dimension 1, and its normalized element is
nonzero. ∎

######  Lemma 4.8.

For $\psi\in\operatorname{End}(V)$ the following are equivalent:

  1. $(i)$

$\psi\in\mathcal{L}$ and $\psi$ is normalized;

  2. $(ii)$

for $0\leq i\leq N$ both

$\displaystyle\psi\tau_{i}=\vartheta_{i}\tau_{i-1},\qquad\psi\eta_{i}=\vartheta_{i}\eta_{i-1}.$
(4.9)

###### Proof.

(i) $\Rightarrow$ (ii) Set $\psi x=1$ in Proposition 4.2(ii), (iii). (ii)
$\Rightarrow$ (i) We have $\psi\in\mathcal{L}$ by Definition 2.2 and (4.9). To
see that $\psi$ is normalized, set $i=1$ in (4.9) and use Lemma 2.5 to obtain
$\psi x=1$. ∎

##  5 Describing $\boldsymbol{\mathcal{L}}$ using $\boldsymbol{\Delta}$

We continue to discuss the double lowering space $\mathcal{L}$ for the data
(2.1). In this section we describe $\mathcal{L}$ using the map $\Delta$ from
(2.5).

######  Proposition 5.1.

For $\psi\in\operatorname{End}(V)$ the following $(i)$–$(iii)$ are equivalent:

  1. $(i)$

$\Delta\psi=\psi\Delta$ and

$\displaystyle\psi\tau_{i}\in\mathbb{F}\tau_{i-1},\qquad 0\leq i\leq N;$ (5.1)

  2. $(ii)$

$\Delta\psi=\psi\Delta$ and

$\displaystyle\psi\eta_{i}\in\mathbb{F}\eta_{i-1},\qquad 0\leq i\leq N;$ (5.2)

  3. $(iii)$

$\psi\in\mathcal{L}$.

###### Proof.

(i) $\Rightarrow$ (ii) We show (5.2). Using (2.5),

$\displaystyle\psi\eta_{i}=\psi\Delta\tau_{i}=\Delta\psi\tau_{i}\in\mathbb{F}\Delta\tau_{i-1}=\mathbb{F}\eta_{i-1}.$

(ii) $\Rightarrow$ (i) Similar to the proof of (i) $\Rightarrow$ (ii). (i),
(ii) $\Rightarrow$ (iii) By Definitions 2.1, 2.2. (iii) $\Rightarrow$ (i) We
have (5.1) by Definition 2.2. We show $\Delta\psi=\psi\Delta$. By Lemma 4.7 we
may assume that $\psi$ is normalized. The vector space $V$ has a basis
$\\{\tau_{i}\\}_{i=0}^{N}$. By Lemma 4.8 we have

$\displaystyle\psi\tau_{i}=\vartheta_{i}\tau_{i-1},\qquad\psi\eta_{i}=\vartheta_{i}\eta_{i-1},\qquad
0\leq i\leq N.$

So for $0\leq i\leq N$,

$\displaystyle\Delta\psi\tau_{i}=\vartheta_{i}\Delta\tau_{i-1}=\vartheta_{i}\eta_{i-1}=\psi\eta_{i}=\psi\Delta\tau_{i}.$

By these comments $\Delta\psi=\psi\Delta$. ∎

We introduce some notation. For $0\leq i\leq j\leq N$ define

$\displaystyle\left[\begin{matrix}j\\\
i\end{matrix}\right]_{\vartheta}=\frac{\vartheta_{j}\vartheta_{j-1}\cdots\vartheta_{j-i+1}}{\vartheta_{1}\vartheta_{2}\cdots\vartheta_{i}}.$
(5.3)

Note that

$\displaystyle\left[\begin{matrix}j\\\
i\end{matrix}\right]_{\vartheta}=\left[\begin{matrix}j\\\
j-i\end{matrix}\right]_{\vartheta},\qquad 0\leq i\leq j\leq N.$

######  Proposition 5.2.

The following $(i)$–$(iii)$ are equivalent:

  1. $(i)$

$\mathcal{L}\not=0$;

  2. $(ii)$

for $0\leq j\leq N$,

$\displaystyle\eta_{j}=\sum_{i=0}^{j}\eta_{j-i}(a_{0})\left[\begin{matrix}j\\\
i\end{matrix}\right]_{\vartheta}\tau_{i};$ (5.4)

  3. $(iii)$

for $0\leq j\leq N$,

$\displaystyle\tau_{j}=\sum_{i=0}^{j}\tau_{j-i}(b_{0})\left[\begin{matrix}j\\\
i\end{matrix}\right]_{\vartheta}\eta_{i}.$

Assume that $(i)$–$(iii)$ hold, and let $\psi\in\mathcal{L}$ be normalized.
Then

$\displaystyle\Delta=\sum_{i=0}^{N}\frac{\eta_{i}(a_{0})}{\vartheta_{1}\vartheta_{2}\cdots\vartheta_{i}}\psi^{i},$
(5.5)
$\displaystyle\Delta^{-1}=\sum_{i=0}^{N}\frac{\tau_{i}(b_{0})}{\vartheta_{1}\vartheta_{2}\cdots\vartheta_{i}}\psi^{i}.$
(5.6)

###### Proof.

(i) $\Rightarrow$ (ii) Let $j$ be given. There exist scalars
$\\{\alpha_{i}\\}_{i=0}^{j}$ in $\mathbb{F}$ such that

$\displaystyle\eta_{j}=\sum_{i=0}^{j}\alpha_{i}\tau_{i}.$ (5.7)

For $0\leq i\leq j$ we show

$\displaystyle\alpha_{i}=\eta_{j-i}(a_{0})\left[\begin{matrix}j\\\
i\end{matrix}\right]_{\vartheta}.$ (5.8)

Let $\psi\in\mathcal{L}$ be normalized. In (5.7) we apply $\psi^{i}$ to each
side, and evaluate the result using Lemma 4.8. We then set $x=a_{0}$, and use
the fact that $\tau_{0}=1$ and $\tau_{k}(a_{0})=0$ for $1\leq k\leq N$. By
these comments

$\displaystyle\eta_{j-i}(a_{0})\vartheta_{j}\vartheta_{j-1}\cdots\vartheta_{j-i+1}=\alpha_{i}\vartheta_{1}\vartheta_{2}\cdots\vartheta_{i}.$

From this equation we obtain (5.8). By (5.7), (5.8) we obtain (5.4), so (ii)
holds.

(ii) $\Rightarrow$ (i) Define $\psi\in\operatorname{End}(V)$ such that
$\psi\tau_{i}=\vartheta_{i}\tau_{i-1}$ for $0\leq i\leq N$. We have
$\psi\not=0$ since $N\geq 1$ and $\vartheta_{1}=1$, $\tau_{0}=1$. We show
$\psi\in\mathcal{L}$. To do this, it is convenient to first show that $\psi$
satisfies (5.5). To obtain (5.5), for $0\leq j\leq N$ we apply each side of
(5.5) to $\tau_{j}$. Concerning the left-hand side of (5.5), we have
$\Delta\tau_{j}=\eta_{j}$ by (2.5). Concerning the right-hand side of (5.5),

$\displaystyle\sum_{i=0}^{j}\frac{\eta_{i}(a_{0})}{\vartheta_{1}\vartheta_{2}\cdots\vartheta_{i}}\psi^{i}\tau_{j}$
$\displaystyle=\sum_{i=0}^{j}\frac{\eta_{i}(a_{0})\vartheta_{j}\vartheta_{j-1}\cdots\vartheta_{j-i+1}}{\vartheta_{1}\vartheta_{2}\cdots\vartheta_{i}}\tau_{j-i}=\sum_{i=0}^{j}\eta_{i}(a_{0})\left[\begin{matrix}j\\\
i\end{matrix}\right]_{\vartheta}\tau_{j-i}$
$\displaystyle=\sum_{i=0}^{j}\eta_{j-i}(a_{0})\left[\begin{matrix}j\\\
i\end{matrix}\right]_{\vartheta}\tau_{i}=\eta_{j}.$

We have shown that each side of (5.5) sends $\tau_{j}\mapsto\eta_{j}$ for
$0\leq j\leq N$. Therefore (5.5) holds. By (5.5), $\Delta$ is a polynomial in
$\psi$. Consequently $\psi\Delta=\Delta\psi$. The map $\psi$ satisfies
Proposition 5.1(i), so $\psi\in\mathcal{L}$ by Proposition 5.1(i), (iii).
Therefore $\mathcal{L}\not=0$.

(i) $\Leftrightarrow$ (iii) Interchange the roles of
$\\{a_{i}\\}_{i=0}^{N-1}$, $\\{b_{i}\\}_{i=0}^{N-1}$ in the proof of (i)
$\Leftrightarrow$ (ii).

Now assume that (i)–(iii) hold. We saw in the proof of (ii) $\Rightarrow$ (i)
that (5.5) holds. Interchanging the roles of $\\{a_{i}\\}_{i=0}^{N-1}$,
$\\{b_{i}\\}_{i=0}^{N-1}$ in that proof, we see that (5.6) holds. ∎

Later in the paper, we will obtain necessary and sufficient conditions for the
data (2.1) to satisfy conditions (i)–(iii) in Proposition 5.2; our result is
Theorem 12.1. In order to motivate this result, we look at some examples of
double lowering data. This will be done in the next section.

##  6 First examples of double lowering data

We continue to discuss the double lowering space $\mathcal{L}$ for the data
(2.1). In this section we give three assumptions under which this data is
double lowering. Under each assumption we describe the polynomials
$\\{\tau_{i}\\}_{i=0}^{N-1}$, $\\{\eta_{i}\\}_{i=0}^{N-1}$ from (2.3), (2.4),
the parameters $\\{\vartheta_{i}\\}_{i=0}^{N}$ from Definition 4.1, and the
map $\Delta$ from (2.5).

As a warmup, we examine the condition (5.4) for some small values of $j$.

######  Lemma 6.1.

The following $(i)$–$(iv)$ hold.

  1. $(i)$

$\eta_{0}=\tau_{0}$.

  2. $(ii)$

$\eta_{1}=\eta_{1}(a_{0})\tau_{0}+\tau_{1}$.

  3. $(iii)$

For $N\geq 2$,

$\displaystyle\eta_{2}=\eta_{2}(a_{0})\tau_{0}+\vartheta_{2}\eta_{1}(a_{0})\tau_{1}+\tau_{2}.$

  4. $(iv)$

For $N\geq 3$,

$\displaystyle\eta_{3}=\eta_{3}(a_{0})\tau_{0}+\vartheta_{3}\eta_{2}(a_{0})\tau_{1}+\vartheta_{3}\eta_{1}(a_{0})\tau_{2}+\tau_{3}+(x-a_{0})\varepsilon,$

where

$\displaystyle\varepsilon=(b_{0}-a_{1})(b_{2}-a_{1})-(a_{0}-b_{1})(a_{2}-b_{1}).$

###### Proof.

To verify these equations, evaluate the terms using (2.3), (2.4) and
Definition 4.1. ∎

######  Lemma 6.2.

Assume that $N\leq 2$. Then $\mathcal{L}\not=0$.

###### Proof.

By Proposition 5.2(i), (ii) and Lemma 6.1. ∎

######  Lemma 6.3.

Assume that $N=3$. Then $\mathcal{L}\not=0$ if and only if

$\displaystyle(a_{0}-b_{1})(a_{2}-b_{1})=(b_{0}-a_{1})(b_{2}-a_{1}).$

###### Proof.

By Proposition 5.2(i),(ii) and Lemma 6.1. ∎

######  Lemma 6.4.

Assume that $a_{i-1}=b_{i}$ for $1\leq i\leq N-1$. Then the following (i)–(v)
hold:

  1. $(i)$

$\mathcal{L}\not=0$;

  2. $(ii)$

$\eta_{i}=(x-b_{0})\tau_{i-1}$ for $1\leq i\leq N$;

  3. $(iii)$

$\eta_{i}(a_{0})=0$ for $2\leq i\leq N$;

  4. $(iv)$

$\vartheta_{i}=\frac{a_{i-1}-b_{0}}{a_{0}-b_{0}}$ for $1\leq i\leq N$;

  5. $(v)$

$\Delta=I+(a_{0}-b_{0})\psi$, where $\psi\in\mathcal{L}$ is normalized.

###### Proof.

(ii) By (2.3), (2.4). (iii) By (ii) and since $\tau_{j}(a_{0})=0$ for $1\leq
j\leq N$. (iv) Use Definition 4.1. (i) Apply Proposition 5.2(i), (ii). (v) By
(5.5) and (iii) above. ∎

######  Lemma 6.5.

Assume that $a_{i}=b_{i-1}$ for $1\leq i\leq N-1$. Then the following
$(i)$–$(v)$ hold:

  1. $(i)$

$\mathcal{L}\not=0$;

  2. $(ii)$

$\tau_{i}=(x-a_{0})\eta_{i-1}$ for $1\leq i\leq N$;

  3. $(iii)$

$\tau_{i}(b_{0})=0$ for $2\leq i\leq N$;

  4. $(iv)$

$\displaystyle{\vartheta_{i}=\frac{a_{0}-b_{i-1}}{a_{0}-b_{0}}}$ for $1\leq
i\leq N$;

  5. $(v)$

$\Delta^{-1}=I+(b_{0}-a_{0})\psi$, where $\psi\in\mathcal{L}$ is normalized.

###### Proof.

Interchange the roles of $\\{a_{i}\\}_{i=0}^{N-1}$, $\\{b_{i}\\}_{i=0}^{N-1}$
in Lemma 6.4. ∎

For the rest of this section, assume that $N\geq 2$. Also for the rest of this
section, fix $\theta\in\mathbb{F}$ and assume

$\displaystyle a_{0}\not=\theta,\qquad b_{0}\not=\theta;$ (6.1) $\displaystyle
a_{i}=\theta,\qquad b_{i}=\theta,\qquad 1\leq i\leq N-2;$ (6.2)
$\displaystyle\frac{\theta-a_{N-1}}{\theta-b_{0}}=\frac{\theta-
b_{N-1}}{\theta-a_{0}}\qquad\text{if}\ N\not=\infty.$ (6.3)

Using Definition 4.1,

$\displaystyle\vartheta_{i}=1,\qquad 1\leq i\leq N-1$

and for $N\not=\infty$,

$\displaystyle a_{N-1}=b_{0}+\vartheta_{N}(\theta-b_{0}),\qquad
b_{N-1}=a_{0}+\vartheta_{N}(\theta-a_{0}).$

Using (5.3),

$\displaystyle\left[\begin{matrix}j\\\
i\end{matrix}\right]_{\vartheta}=1,\qquad 0\leq i\leq j\leq N-1$

and for $N\not=\infty$,

$\displaystyle\left[\begin{matrix}N\\\
i\end{matrix}\right]_{\vartheta}=\vartheta_{N},\qquad 1\leq i\leq N-1.$

For $0\leq i\leq N$ the polynomials $\tau_{i}$, $\eta_{i}$ are described in
the table below:

$i$ | $\tau_{i}$ | $\eta_{i}$
---|---|---
$0$ | $1$ | $1$
$1\leq i\leq N-1$ | $(x-a_{0})(x-\theta)^{i-1}$ | $(x-b_{0})(x-\theta)^{i-1}$
$N$ | $(x-a_{0})(x-\theta)^{N-2}(x-a_{N-1})$ | $(x-b_{0})(x-\theta)^{N-2}(x-b_{N-1})$

For $1\leq i\leq N$ the values of $\eta_{i}-\tau_{i}$ and $\eta_{i}(a_{0})$
are described in the table below:

$i$ | $\eta_{i}-\tau_{i}$ | $\eta_{i}(a_{0})$
---|---|---
$1\leq i\leq N-1$ | $(a_{0}-b_{0})(x-\theta)^{i-1}$ | $(a_{0}-b_{0})(a_{0}-\theta)^{i-1}$
$N$ | $\vartheta_{N}(a_{0}-b_{0})(x-\theta)^{N-1}$ | $\vartheta_{N}(a_{0}-b_{0})(a_{0}-\theta)^{N-1}$

######  Lemma 6.6.

Under assumptions (6.1)–(6.3) the following $(i)$–$(iii)$ hold:

  1. $(i)$

$\mathcal{L}\not=0$;

  2. $(ii)$

$\Delta=\frac{I+(\theta-b_{0})\psi}{I+(\theta-a_{0})\psi}$;

  3. $(iii)$

$\Delta^{-1}=\frac{I+(\theta-a_{0})\psi}{I+(\theta-b_{0})\psi}$.

In the above lines $\psi\in\mathcal{L}$ is normalized.

###### Proof.

(i) We invoke Proposition 5.2(i), (ii). For $0\leq j\leq N$ we verify (5.4).
We may assume that $2\leq j\leq N$; otherwise we are done by Lemma 6.1. For
$N\not=\infty$ we separate the cases $2\leq j\leq N-1$ and $j=N$. It suffices
to show that

$\displaystyle\eta_{j}=\tau_{j}+\eta_{j}(a_{0})+\sum_{i=1}^{j-1}\eta_{j-i}(a_{0})\tau_{i},\qquad
2\leq j\leq N-1,$ (6.4)
$\displaystyle\eta_{N}=\tau_{N}+\eta_{N}(a_{0})+\vartheta_{N}\sum_{i=1}^{N-1}\eta_{N-i}(a_{0})\tau_{i},\qquad\text{if}\
N\not=\infty.$ (6.5)

For $2\leq j\leq N$ the values of $\eta_{j}-\tau_{j}$ and $\eta_{j}(a_{0})$
are given in the table above the lemma statement. Also for $2\leq j\leq N$,

$\displaystyle\sum_{i=1}^{j-1}\eta_{j-i}(a_{0})\tau_{i}$
$\displaystyle=(a_{0}-b_{0})(x-a_{0})\sum_{i=1}^{j-1}(a_{0}-\theta)^{j-i-1}(x-\theta)^{i-1}$
$\displaystyle=(a_{0}-b_{0})(x-a_{0})(a_{0}-\theta)^{j-2}\sum_{k=0}^{j-2}\left(\frac{x-\theta}{a_{0}-\theta}\right)^{k}$
$\displaystyle=(a_{0}-b_{0})(x-\theta)^{j-1}-(a_{0}-b_{0})(a_{0}-\theta)^{j-1}.$

Using the above comments we routinely verify (6.4), (6.5).

(ii) We will verify the equation by showing that the two sides agree on
$V_{j}$ for $2\leq j\leq N$. Using (5.5) and $\psi^{j+1}V_{j}=0$ we see that
on $V_{j}$,

$\displaystyle\Delta-I$
$\displaystyle=\sum_{i=1}^{j}\frac{\eta_{i}(a_{0})}{\vartheta_{1}\vartheta_{2}\cdots\vartheta_{i}}\psi^{i}=\frac{\eta_{j}(a_{0})}{\vartheta_{j}}\psi^{j}+\sum_{i=1}^{j-1}\eta_{i}(a_{0})\psi^{i}=(a_{0}-b_{0})\psi\sum_{k=0}^{j-1}(a_{0}-\theta)^{k}\psi^{k}$
$\displaystyle=(a_{0}-b_{0})\psi\frac{I-(a_{0}-\theta)^{j}\psi^{j}}{I-(a_{0}-\theta)\psi}=\frac{(a_{0}-b_{0})\psi}{I-(a_{0}-\theta)\psi}.$

The result follows. (iii) By (ii) above. ∎

We just gave some examples of double lowering data. There is another example
that is somewhat more involved; it will be described later in the paper.

##  7 Extending the data

Throughout this section, we assume that $N$ is an integer at least 2. Recall
the data $\\{a_{i}\\}_{i=0}^{N-1}$, $\\{b_{i}\\}_{i=0}^{N-1}$ from (2.1), and
assume that this data is double lowering. Let $a_{N},b_{N}\in\mathbb{F}$
satisfy

$\displaystyle a_{0}+a_{1}+\cdots+a_{N}\not=b_{0}+b_{1}+\cdots+b_{N},$

giving data

$\displaystyle\\{a_{i}\\}_{i=0}^{N},\qquad\\{b_{i}\\}_{i=0}^{N}.$ (7.1)

In this section we obtain necessary and sufficient conditions on $a_{N}$,
$b_{N}$ for the data (7.1) to be double lowering. By (2.7),

$\displaystyle\tau_{N+1}=(x-a_{N})\tau_{N},\qquad\eta_{N+1}=(x-b_{N})\eta_{N}.$

######  Lemma 7.1.

The following $(i)$–$(iii)$ are equivalent:

  1. $(i)$

the data (7.1) is double lowering;

  2. $(ii)$

we have

$\displaystyle\eta_{N+1}=\sum_{i=0}^{N+1}\eta_{N-i+1}(a_{0})\left[\begin{matrix}N+1\\\
i\end{matrix}\right]_{\vartheta}\tau_{i};$ (7.2)

  3. $(iii)$

we have

$\displaystyle\tau_{N+1}=\sum_{i=0}^{N+1}\tau_{N-i+1}(b_{0})\left[\begin{matrix}N+1\\\
i\end{matrix}\right]_{\vartheta}\eta_{i}.$

###### Proof.

By Proposition 5.2. ∎

######  Lemma 7.2.

We have

$\displaystyle\eta_{N+1}=\sum_{i=0}^{N}\eta_{N-i}(a_{0})\left[\begin{matrix}N\\\
i\end{matrix}\right]_{\vartheta}(a_{i}-b_{N})\tau_{i}+\sum_{i=1}^{N+1}\eta_{N-i+1}(a_{0})\left[\begin{matrix}N\\\
i-1\end{matrix}\right]_{\vartheta}\tau_{i},$ (7.3)
$\displaystyle\tau_{N+1}=\sum_{i=0}^{N}\tau_{N-i}(b_{0})\left[\begin{matrix}N\\\
i\end{matrix}\right]_{\vartheta}(b_{i}-a_{N})\eta_{i}+\sum_{i=1}^{N+1}\tau_{N-i+1}(b_{0})\left[\begin{matrix}N\\\
i-1\end{matrix}\right]_{\vartheta}\eta_{i}.$ (7.4)

###### Proof.

We show (7.3). Using Proposition 5.2,

$\displaystyle\eta_{N+1}$
$\displaystyle=(x-b_{N})\eta_{N}=(x-b_{N})\sum_{i=0}^{N}\eta_{N-i}(a_{0})\left[\begin{matrix}N\\\
i\end{matrix}\right]_{\vartheta}\tau_{i}$
$\displaystyle=\sum_{i=0}^{N}\eta_{N-i}(a_{0})\left[\begin{matrix}N\\\
i\end{matrix}\right]_{\vartheta}\bigl{(}(a_{i}-b_{N})\tau_{i}+\tau_{i+1}\bigr{)}$
$\displaystyle=\sum_{i=0}^{N}\eta_{N-i}(a_{0})\left[\begin{matrix}N\\\
i\end{matrix}\right]_{\vartheta}(a_{i}-b_{N})\tau_{i}+\sum_{i=1}^{N+1}\eta_{N-i+1}(a_{0})\left[\begin{matrix}N\\\
i-1\end{matrix}\right]_{\vartheta}\tau_{i}.$

Line (7.4) is similarly obtained. ∎

######  Proposition 7.3.

The following $(i)$–$(iii)$ are equivalent:

  1. $(i)$

the data (7.1) is double lowering;

  2. $(ii)$

for $0\leq i\leq N-1$ such that $\eta_{i}(a_{0})\not=0$,

$\displaystyle(a_{0}+\cdots+a_{i}-b_{0}-\cdots-b_{i})(a_{N-i}-b_{N})$
$\displaystyle\qquad{}=(a_{0}-b_{i})(a_{N-i}+\cdots+a_{N}-b_{N-i}-\cdots-
b_{N});$

  3. $(iii)$

for $0\leq i\leq N-1$ such that $\tau_{i}(b_{0})\not=0$,

$\displaystyle(b_{0}+\cdots+b_{i}-a_{0}-\cdots-a_{i})(b_{N-i}-a_{N})$
$\displaystyle\qquad{}=(b_{0}-a_{i})(b_{N-i}+\cdots+b_{N}-a_{N-i}-\cdots-
a_{N}).$

###### Proof.

(i) $\Leftrightarrow$ $(ii)$ We invoke Lemma 7.1(i), (ii). Subtract (7.2) from
(7.3) to obtain an equation $0=\sum_{i=1}^{N}d_{i}\tau_{i}$ where

$\displaystyle d_{i}=\eta_{N-i}(a_{0})\left[\begin{matrix}N\\\
i\end{matrix}\right]_{\vartheta}(a_{i}-b_{N})+\eta_{N-i+1}(a_{0})\left[\begin{matrix}N\\\
i-1\end{matrix}\right]_{\vartheta}-\eta_{N-i+1}(a_{0})\left[\begin{matrix}N+1\\\
i\end{matrix}\right]_{\vartheta}$ (7.5)

for $1\leq i\leq N$. Note that (7.2) holds iff $0=\sum_{i=1}^{N}d_{i}\tau_{i}$
iff $d_{i}=0$ for $1\leq i\leq N$. For $1\leq i\leq N$ we simplify (7.5) using

$\displaystyle\eta_{N-i+1}(a_{0})=\eta_{N-i}(a_{0})(a_{0}-b_{N-i})$

and

$\displaystyle\left[\begin{matrix}N\\\
i-1\end{matrix}\right]_{\vartheta}=\left[\begin{matrix}N\\\
i\end{matrix}\right]_{\vartheta}\frac{\vartheta_{i}}{\vartheta_{N-i+1}},\qquad\left[\begin{matrix}N+1\\\
i\end{matrix}\right]_{\vartheta}=\left[\begin{matrix}N\\\
i\end{matrix}\right]_{\vartheta}\frac{\vartheta_{N+1}}{\vartheta_{N-i+1}}.$

We find that $d_{i}$ is equal to

$\displaystyle\frac{\eta_{N-i}(a_{0})}{\vartheta_{N-i+1}}\left[\begin{matrix}N\\\
i\end{matrix}\right]_{\vartheta}$

times

$\displaystyle(a_{i}-b_{N})\vartheta_{N-i+1}+(a_{0}-b_{N-i})(\vartheta_{i}-\vartheta_{N+1})$

for $1\leq i\leq N$. Therefore, (7.2) holds if and only if

$\displaystyle\eta_{N-i}(a_{0})=0\qquad\text{or}\qquad(a_{i}-b_{N})\vartheta_{N-i+1}=(a_{0}-b_{N-i})(\vartheta_{N+1}-\vartheta_{i})$

for $1\leq i\leq N$. Replacing $i$ by $N-i$, we see that (7.2) holds if and
only if

$\displaystyle\eta_{i}(a_{0})=0\qquad\text{or}\qquad(a_{N-i}-b_{N})\vartheta_{i+1}=(a_{0}-b_{i})(\vartheta_{N+1}-\vartheta_{N-i})$

for $0\leq i\leq N-1$. The result follows in view of Definition 4.1.

(i) $\Leftrightarrow$ (iii) Similar to the proof of (i) $\Leftrightarrow$
(ii). ∎

Our next general goal is to solve the equations in Proposition 7.3(ii), (iii).
The main solution will involve a type of sequence, said to be recurrent.

##  8 Recurrent sequences

Throughout this section let $n$ denote an integer at least 2, or $\infty$. let
$\\{a_{i}\\}_{i=0}^{n}$ denote scalars in $\mathbb{F}$.

######  Definition 8.1 (see [30, Definition 8.2]).

Let $\beta$, $\gamma$, $\varrho$ denote scalars in $\mathbb{F}$.

  1. (i)

The sequence $\\{a_{i}\\}_{i=0}^{n}$ is said to be
$(\beta,\gamma,\varrho)$-recurrent whenever

$a^{2}_{i-1}-\beta a_{i-1}a_{i}+a^{2}_{i}-\gamma(a_{i-1}+a_{i})=\varrho$ (8.1)

for $1\leq i\leq n$.

  2. (ii)

The sequence $\\{a_{i}\\}_{i=0}^{n}$ is said to be $(\beta,\gamma)$-recurrent
whenever

$a_{i-1}-\beta a_{i}+a_{i+1}=\gamma$ (8.2)

for $1\leq i\leq n-1$.

  3. (iii)

The sequence $\\{a_{i}\\}_{i=0}^{n}$ is said to be $\beta$-recurrent whenever

$a_{i-2}-(\beta+1)a_{i-1}+(\beta+1)a_{i}-a_{i+1}$ (8.3)

is zero for $2\leq i\leq n-1$.

  4. (iv)

The sequence $\\{a_{i}\\}_{i=0}^{n}$ is said to be recurrent whenever there
exists $\beta\in\mathbb{F}$ such that $\\{a_{i}\\}_{i=0}^{n}$ is
$\beta$-recurrent.

######  Lemma 8.2.

The following are equivalent:

  1. $(i)$

the sequence $\\{a_{i}\\}_{i=0}^{n}$ is recurrent;

  2. $(ii)$

there exists $\beta\in\mathbb{F}$ such that $\\{a_{i}\\}_{i=0}^{n}$ is
$\beta$-recurrent.

###### Proof.

By Definition 8.1. ∎

######  Lemma 8.3.

For $\beta\in\mathbb{F}$ the following are equivalent:

  1. $(i)$

the sequence $\\{a_{i}\\}_{i=0}^{n}$ is $\beta$-recurrent;

  2. $(ii)$

there exists $\gamma\in\mathbb{F}$ such that $\\{a_{i}\\}_{i=0}^{n}$ is
$(\beta,\gamma)$-recurrent.

###### Proof.

(i) $\Rightarrow$ (ii) For $2\leq i\leq n-1$, the expression (8.3) is zero by
assumption, so

$\displaystyle a_{i-2}-\beta a_{i-1}+a_{i}=a_{i-1}-\beta a_{i}+a_{i+1}.$

The left-hand side of (8.2) is independent of $i$, and the result follows.

(ii) $\Rightarrow$ (i) For $2\leq i\leq n-1$, subtract the equation (8.2) at
$i$ from the corresponding equation obtained by replacing $i$ by $i-1$, to
find (8.3) is zero. ∎

######  Lemma 8.4.

The following $(i)$, $(ii)$ hold for all $\beta,\gamma\in\mathbb{F}$.

  1. $(i)$

Suppose $\\{a_{i}\\}_{i=0}^{n}$ is $(\beta,\gamma)$-recurrent. Then there
exists $\varrho\in\mathbb{F}$ such that $\\{a_{i}\\}_{i=0}^{n}$ is
$(\beta,\gamma,\varrho)$-recurrent.

  2. $(ii)$

Suppose $\\{a_{i}\\}_{i=0}^{n}$ is $(\beta,\gamma,\varrho)$-recurrent, and
that $a_{i-1}\not=a_{i+1}$ for $1\leq i\leq n-1$. Then $\\{a_{i}\\}_{i=0}^{n}$
is $(\beta,\gamma)$-recurrent.

###### Proof.

Let $p_{i}$ denote the expression on the left in (8.1), and observe

$\displaystyle p_{i}-p_{i+1}=(a_{i-1}-a_{i+1})(a_{i-1}-\beta
a_{i}+a_{i+1}-\gamma)$

for $1\leq i\leq n-1$. Assertions (i), (ii) are both routine consequences of
this. ∎

######  Definition 8.5.

Assume that $\\{a_{i}\\}_{i=0}^{n}$ is recurrent. By a parameter triple for
$\\{a_{i}\\}_{i=0}^{n}$ we mean a 3-tuple $\beta$, $\gamma$, $\varrho$ of
scalars in $\mathbb{F}$ such that $\\{a_{i}\\}_{i=0}^{n}$ is
$(\beta,\gamma)$-recurrent and $(\beta,\gamma,\varrho)$-recurrent.

Note that a recurrent sequence has at least one parameter triple.

##  9 Recurrent sequences in closed form

In this section, we describe the recurrent sequences in closed form. Let $n$
denote an integer at least 2, or $\infty$.

######  Lemma 9.1 (see [30, Lemma 9.2]).

The recurrent sequences $\\{a_{i}\\}_{i=0}^{n}$ are described in the table
below:

case | $a_{i}$ | comments
---|---|---
I | $\alpha_{1}+\alpha_{2}q^{i}+\alpha_{3}q^{-i}$ | $q\not\in\\{0,1,-1\\}$
II | $\alpha_{1}+\alpha_{2}i+\alpha_{3}\binom{i}{2}$ |
III | $\alpha_{1}+\alpha_{2}(-1)^{i}+\alpha_{3}i(-1)^{i}$ | ${\rm char}(\mathbb{F})\not=2$

In the above table $q$, $\alpha_{1}$, $\alpha_{2}$, $\alpha_{3}$ are scalars
in $\mathbb{F}$.

######  Lemma 9.2.

The following scalars $\beta$, $\gamma$, $\varrho$ give a parameter triple for
the recurrent sequence $\\{a_{i}\\}_{i=0}^{n}$ in Lemma 9.1.

Case I:

$\displaystyle\beta=q+q^{-1},\qquad\gamma=-\alpha_{1}(q-1)^{2}q^{-1},\qquad\varrho=\alpha_{1}^{2}(q-1)^{2}q^{-1}-\alpha_{2}\alpha_{3}\big{(}q-q^{-1}\big{)}^{2}.$

Case II:

$\displaystyle\beta=2,\qquad\gamma=\alpha_{3},\qquad\varrho=\alpha_{2}^{2}-\alpha_{2}\alpha_{3}-2\alpha_{1}\alpha_{3}.$

Case III:

$\displaystyle\beta=-2,\qquad\gamma=4\alpha_{1},\qquad\varrho=\alpha_{3}^{2}-4\alpha_{1}^{2}.$

###### Proof.

This is routinely checked using Definition 8.5. ∎

######  Lemma 9.3.

Referring to Lemma 9.1, for $0\leq i\leq n+1$ the sum
$a_{0}+a_{1}+\cdots+a_{i-1}$ is given in the table below:

case | $a_{0}+a_{1}+\cdots+a_{i-1}$
---|---
I | $\alpha_{1}i+\alpha_{2}\frac{1-q^{i}}{1-q}+\alpha_{3}\frac{1-q^{-i}}{1-q^{-1}}$
II | $\alpha_{1}i+\alpha_{2}\binom{i}{2}+\alpha_{3}\binom{i}{3}$
III, $i$ even | $\alpha_{1}i-\alpha_{3}i/2$
III, $i$ odd | $\alpha_{1}i+\alpha_{2}+\alpha_{3}(i-1)/2$

###### Proof.

Use induction on $i$. ∎

######  Note 9.4.

Referring to Case III of the above table, the subcases $i$ even and $i$ odd
can be handled in the following uniform way. For $0\leq i\leq n+1$,

$\displaystyle
a_{0}+a_{1}+\cdots+a_{i-1}=\frac{2\alpha_{2}-\alpha_{3}+4\alpha_{1}i+(\alpha_{3}-2\alpha_{2})(-1)^{i}-2\alpha_{3}i(-1)^{i}}{4}.$

##  10 Twin recurrent sequences

Let $n$ denote an integer at least 2, or $\infty$. Let
$\\{a_{i}\\}_{i=0}^{n}$, $\\{b_{i}\\}_{i=0}^{n}$ denote scalars in
$\mathbb{F}$.

######  Definition 10.1.

Assume that the sequences $\\{a_{i}\\}_{i=0}^{n}$, $\\{b_{i}\\}_{i=0}^{n}$ are
recurrent. These sequences are called twins whenever they have a parameter
triple in common.

######  Lemma 10.2.

Assume that the sequences $\\{a_{i}\\}_{i=0}^{n}$, $\\{b_{i}\\}_{i=0}^{n}$ are
recurrent. These sequences are twins if and only if they are related in the
following way:

Case I:

$\displaystyle a_{i}=\alpha_{1}+\alpha_{2}q^{i}+\alpha_{3}q^{-i},\qquad
b_{i}=\alpha_{1}+\alpha^{\prime}_{2}q^{i}+\alpha^{\prime}_{3}q^{-i},\qquad\alpha_{2}^{\prime}\alpha_{3}^{\prime}=\alpha_{2}\alpha_{3}.$

Case II:

$\displaystyle a_{i}=\alpha_{1}+\alpha_{2}i+\alpha_{3}\binom{i}{2},\qquad
b_{i}=\alpha^{\prime}_{1}+\alpha^{\prime}_{2}i+\alpha_{3}\binom{i}{2},$
$\displaystyle(\alpha_{2}-\alpha^{\prime}_{2})(\alpha_{2}+\alpha^{\prime}_{2}-\alpha_{3})=2(\alpha_{1}-\alpha^{\prime}_{1})\alpha_{3}.$

Case III:

$\displaystyle a_{i}=\alpha_{1}+\alpha_{2}(-1)^{i}+\alpha_{3}i(-1)^{i},\qquad
b_{i}=\alpha_{1}+\alpha^{\prime}_{2}(-1)^{i}+\alpha^{\prime}_{3}i(-1)^{i},$
$\displaystyle\alpha^{\prime}_{3}=\alpha_{3}\qquad\text{or}\qquad\alpha^{\prime}_{3}=-\alpha_{3}.$

###### Proof.

First assume that the sequences $\\{a_{i}\\}_{i=0}^{n}$,
$\\{b_{i}\\}_{i=0}^{n}$ are related in the specified way. Then these sequences
share the parameter triple $\beta$, $\gamma$, $\varrho$ from Lemma 9.2.
Therefore these sequences are twins. Next assume that the sequences
$\\{a_{i}\\}_{i=0}^{n}$, $\\{b_{i}\\}_{i=0}^{n}$ are twins. It follows from
Definition 8.1 and Lemma 9.1 that they are related in the specified way. ∎

##  11 A characterization of twin recurrent sequences

In this section we explain what twin recurrent sequences have to do with the
equations in Proposition 7.3. Let $n$ denote an integer at least 2, or
$\infty$. Let $\\{a_{i}\\}_{i=0}^{n}$, $\\{b_{i}\\}_{i=0}^{n}$ denote scalars
in $\mathbb{F}$.

######  Definition 11.1.

For $0\leq i\leq j\leq n$ let $E(i,j)$ denote the equation

$\displaystyle(a_{0}+\cdots+a_{i}-b_{0}-\cdots-
b_{i})(a_{j-i}-b_{j})=(a_{0}-b_{i})(a_{j-i}+\cdots+a_{j}-b_{j-i}-\cdots-
b_{j}).$

######  Lemma 11.2.

The equations $E(0,j)$ and $E(j,j)$ hold for $0\leq j\leq n$.

###### Proof.

This is routinely checked. ∎

######  Proposition 11.3.

Assume that the sequences $\\{a_{i}\\}_{i=0}^{n}$, $\\{b_{i}\\}_{i=0}^{n}$ are
recurrent and twins. Then $E(i,j)$ holds for $0\leq i\leq j\leq n$.

###### Proof.

This is routinely verified for each Case I–III in Lemma 10.2. To carry out the
verification, use the formulas in Lemma 9.3. ∎

In the next two lemmas, we give some additional solutions for the equations
$E(i,j)$ in Definition 11.1.

######  Lemma 11.4.

Assume that $a_{i}=b_{i-1}$ for $1\leq i\leq n$. Then $E(i,j)$ holds for
$0\leq i\leq j\leq n$.

###### Proof.

For $0\leq i\leq j\leq n$, each side of $E(i,j)$ is equal to
$(a_{0}-b_{i})(a_{j-i}-b_{j})$. ∎

######  Lemma 11.5.

Pick $\theta\in\mathbb{F}$ and assume

$\displaystyle a_{i}=\theta,\qquad b_{i}=\theta,\qquad 1\leq i\leq n-1,$
$\displaystyle(\theta-a_{0})(\theta-a_{n})=(\theta-b_{0})(\theta-
b_{n})\qquad\text{if}\ n\not=\infty.$ (11.1)

Then $E(i,j)$ holds for $0\leq i\leq j\leq n$.

###### Proof.

By Lemma 11.2 it suffices to verify $E(i,j)$ for $1\leq i<j\leq n$. Let $i,j$
be given. For $j<n$, each side of $E(i,j)$ is zero. For $n\not=\infty$ and
$j=n$, the equation $E(i,j)$ becomes

$\displaystyle(a_{0}-b_{0})(\theta-b_{n})=(a_{0}-\theta)(a_{n}-b_{n})$

which is a reformulation of (11.1). ∎

######  Proposition 11.6.

Assume that $a_{1}\not=b_{0}$ and $a_{1}\not=b_{1}$. Further assume that
$E(i,j)$ holds for $1\leq i\leq 2$ and $i+1\leq j\leq n$. Then the sequences
$\\{a_{i}\\}_{i=0}^{n}$, $\\{b_{i}\\}_{i=0}^{n}$ are recurrent and twins.

###### Proof.

Using $E(1,2)$,

$\displaystyle(a_{0}-b_{1})(a_{2}-b_{1})=(b_{0}-a_{1})(b_{2}-a_{1}).$ (11.2)

Since $a_{1}\not=b_{1}$, there exists a unique pair $\beta$, $\gamma$ of
scalars in $\mathbb{F}$ such that

$\displaystyle a_{0}-\beta a_{1}+a_{2}=\gamma,\qquad b_{0}-\beta
b_{1}+b_{2}=\gamma.$

Using these equations we eliminate $a_{2}$, $b_{2}$ in (11.2):

$\displaystyle(a_{0}-b_{1})(\gamma+\beta
a_{1}-a_{0}-b_{1})=(b_{0}-a_{1})(\gamma+\beta b_{1}-b_{0}-a_{1}).$

In this equation we rearrange terms to get

$\displaystyle a^{2}_{0}-\beta
a_{0}a_{1}+a^{2}_{1}-\gamma(a_{0}+a_{1})=b^{2}_{0}-\beta
b_{0}b_{1}+b^{2}_{1}-\gamma(b_{0}+b_{1}).$

Let $\varrho$ denote this common value. We show that each of
$\\{a_{i}\\}_{i=0}^{n}$, $\\{b_{i}\\}_{i=0}^{n}$ is $(\beta,\gamma)$-recurrent
and $(\beta,\gamma,\varrho)$-recurrent. To do this, we show that for $2\leq
j\leq n$, each of $\\{a_{i}\\}_{i=0}^{j}$, $\\{b_{i}\\}_{i=0}^{j}$ is
$(\beta,\gamma)$-recurrent and $(\beta,\gamma,\varrho)$-recurrent. We will use
induction on $j$. First assume that $j=2$. By construction
$\\{a_{i}\\}_{i=0}^{2}$ is $(\beta,\gamma)$-recurrent. By construction and
Lemma 8.4(i), the sequence $\\{a_{i}\\}_{i=0}^{2}$ is
$(\beta,\gamma,\varrho)$-recurrent. Similarly $\\{b_{i}\\}_{i=0}^{2}$ is
$(\beta,\gamma)$-recurrent and $(\beta,\gamma,\varrho)$-recurrent. We are done
for $j=2$. Next assume that $j\geq 3$. By $E(1,j)$,

$\displaystyle(a_{0}+a_{1}-b_{0}-b_{1})(a_{j-1}-b_{j})=(a_{0}-b_{1})(a_{j-1}+a_{j}-b_{j-1}-b_{j}).$
(11.3)

By $E(2,j)$,

$\displaystyle(a_{0}+a_{1}+a_{2}-b_{0}-b_{1}-b_{2})(a_{j-2}-b_{j})$
$\displaystyle\qquad{}=(a_{0}-b_{2})(a_{j-2}+a_{j-1}+a_{j}-b_{j-2}-b_{j-1}-b_{j}).$
(11.4)

The equations (11.3), (11.4) give a linear system in the unknowns $a_{j}$,
$b_{j}$. For this system the coefficient matrix is

$\displaystyle C=\left(\begin{matrix}a_{0}-b_{1}&a_{1}-b_{0}\\\
a_{0}-b_{2}&a_{1}+a_{2}-b_{0}-b_{1}\end{matrix}\right).$

We have

$\displaystyle\det(C)=(a_{0}-b_{1})(a_{1}+a_{2}-b_{0}-b_{1})-(a_{0}-b_{2})(a_{1}-b_{0}).$

In the above equation we simplify the right-hand side using (11.2), to obtain

$\displaystyle\det(C)=(a_{1}-b_{0})(a_{1}-b_{1}).$

Therefore $\det(C)\not=0$, so the system (11.3), (11.4) has a unique solution
for $a_{j}$, $b_{j}$. We now describe the solution. By induction the sequences
$\\{a_{i}\\}_{i=0}^{j-1}$, $\\{b_{i}\\}_{i=0}^{j-1}$ are
$(\beta,\gamma)$-recurrent and $(\beta,\gamma,\varrho)$-recurrent. Define
$a^{\prime}_{j}$, $b^{\prime}_{j}$ such that

$\displaystyle a_{j-2}-\beta a_{j-1}+a^{\prime}_{j}=\gamma,\qquad
b_{j-2}-\beta b_{j-1}+b^{\prime}_{j}=\gamma.$

Consider the two sequences

$\displaystyle a_{0},a_{1},\ldots,a_{j-1},a^{\prime}_{j};$ (11.5)
$\displaystyle b_{0},b_{1},\ldots,b_{j-1},b^{\prime}_{j}.$ (11.6)

By construction, each of (11.5), (11.6) is $(\beta,\gamma)$-recurrent. By
construction and Lemma 8.4(i), each of (11.5), (11.6) is
$(\beta,\gamma,\varrho)$-recurrent. We show that $a_{j}=a^{\prime}_{j}$ and
$b_{j}=b^{\prime}_{j}$. The sequences (11.5), (11.6) are recurrent and twins,
so by Proposition 11.3 they satisfy $E(1,j)$ and $E(2,j)$. Therefore, the
equations (11.3), (11.4) still hold if we replace $a_{j}$, $b_{j}$ by
$a^{\prime}_{j}$, $b^{\prime}_{j}$. We mentioned earlier that the system
(11.3), (11.4) has a unique solution for $a_{j}$, $b_{j}$. By these comments
$a_{j}=a^{\prime}_{j}$ and $b_{j}=b^{\prime}_{j}$. Consequently each of
$\\{a_{i}\\}_{i=0}^{j}$, $\\{b_{i}\\}_{i=0}^{j}$ is $(\beta,\gamma)$-recurrent
and $(\beta,\gamma,\varrho)$-recurrent. The above argument shows that each of
$\\{a_{i}\\}_{i=0}^{n}$, $\\{b_{i}\\}_{i=0}^{n}$ is $(\beta,\gamma)$-recurrent
and $(\beta,\gamma,\varrho)$-recurrent. ∎

######  Lemma 11.7.

Assume that $a_{i-1}=b_{i}$ for $1\leq i\leq n$. Then the equation $E(1,j)$
holds for $1\leq j\leq n$. However, in general it is not the case that
$E(i,j)$ holds for $0\leq i\leq j\leq n$.

###### Proof.

For $0\leq i\leq j\leq n$ the equation $E(i,j)$ becomes

$\displaystyle(a_{i}-b_{0})(a_{j-i}-b_{j})=(a_{0}-b_{i})(a_{j}-b_{j-i}).$
(11.7)

If $i=1$ then each side of (11.7) is zero, so $E(1,j)$ holds. Assume that
$n=3$ and

$\displaystyle a_{0}=0=b_{1},\qquad a_{1}=1=b_{2},\qquad a_{2}=0=b_{3},\qquad
a_{3}=1,\qquad b_{0}=0.$

Then (11.7) fails for $i=2$ and $j=3$. ∎

##  12 The classification of the double lowering data

Recall the data $\\{a_{i}\\}_{i=0}^{N-1}$, $\\{b_{i}\\}_{i=0}^{N-1}$ from
(2.1). In this section we obtain necessary and sufficient conditions for this
data to be double lowering. In view of Lemma 6.2 we assume $N\geq 3$.

######  Theorem 12.1.

Let $N$ denote an integer at least $3$, or $\infty$. Let

$\displaystyle\\{a_{i}\\}_{i=0}^{N-1},\qquad\\{b_{i}\\}_{i=0}^{N-1}$ (12.1)

denote scalars in $\mathbb{F}$ such that

$\displaystyle
a_{0}+a_{1}+\cdots+a_{i-1}\not=b_{0}+b_{1}+\cdots+b_{i-1},\qquad 1\leq i\leq
N.$ (12.2)

Then the data (12.1) is double lowering if and only if at least one of the
following $(i)$–$(iv)$ holds:

  1. $(i)$

$a_{i-1}=b_{i}$ for $1\leq i\leq N-1$;

  2. $(ii)$

$a_{i}=b_{i-1}$ for $1\leq i\leq N-1$;

  3. $(iii)$

there exists $\theta\in\mathbb{F}$ such that

$\displaystyle a_{0}\not=\theta,\qquad b_{0}\not=\theta,\qquad
a_{i}=\theta,\qquad b_{i}=\theta,\qquad 1\leq i\leq N-2,$
$\displaystyle\frac{\theta-a_{N-1}}{\theta-b_{0}}=\frac{\theta-
b_{N-1}}{\theta-a_{0}}\qquad\text{if}\ N\not=\infty.$

  4. $(iv)$

the sequences (12.1) are recurrent and twins.

###### Proof.

First assume that at least one of (i)–(iii) holds. Then (12.1) is double
lowering, by Lemmas 6.4, 6.5, 6.6. Next assume that (iv) holds and
$N\not=\infty$. By Proposition 11.3 (with $n=N-1$) the equations $E(i,j)$ hold
for $0\leq i\leq j\leq N-1$. We delete $a_{N-1}$, $b_{N-1}$ from (12.1) and
consider the data

$\displaystyle\\{a_{i}\\}_{i=0}^{N-2},\qquad\\{b_{i}\\}_{i=0}^{N-2}.$ (12.3)

By Lemma 6.2 and induction on $N$, we may assume that the sequences (12.3) are
double lowering. By Proposition 7.3(i), (ii) (with $N$ replaced by $N-1$) we
find that (12.1) is double lowering. Next assume that (iv) holds and
$N=\infty$. Then for all integers $j\geq 2$ the sequences
$\\{a_{i}\\}_{i=0}^{j}$, $\\{b_{i}\\}_{i=0}^{j}$ are recurrent and twins.
Consequently the data $\\{a_{i}\\}_{i=0}^{j}$, $\\{b_{i}\\}_{i=0}^{j}$ is
double lowering. Therefore the data $\\{a_{i}\\}_{i=0}^{\infty}$,
$\\{b_{i}\\}_{i=0}^{\infty}$ is double lowering. We are done in one direction.

We now reverse the direction. Next assume that (12.1) is double lowering. We
break the argument into cases.

Case $a_{0}=b_{1}$. We show that (i) holds. We have $\eta_{i}(a_{0})\not=0$
for $0\leq i\leq 1$. By assumption the data (12.1) is double lowering, so the
data $\\{a_{i}\\}_{i=0}^{j}$, $\\{b_{i}\\}_{i=0}^{j}$ is double lowering for
$1\leq j\leq N-1$. Applying Proposition 7.3(i), (ii) repeatedly (with $N$
replaced by $2,3,\ldots,N-1$) we find that $E(1,j)$ holds for $2\leq j\leq
N-1$. Using these equations and (12.2) we routinely obtain $a_{j-1}=b_{j}$ for
$2\leq j\leq N-1$. This and $a_{0}=b_{1}$ implies (i).

Case $a_{1}=b_{0}$. Interchanging the roles of $\\{a_{i}\\}_{i=0}^{N-1}$,
$\\{b_{i}\\}_{i=0}^{N-1}$ in the previous case, we find that (ii) holds.

Case $a_{0}\not=b_{1}$, $a_{1}\not=b_{0}$, $a_{1}=b_{1}$. We show that (iii)
holds. Define $\theta=a_{1}=b_{1}$, and note that $a_{0}\not=\theta$,
$b_{0}\not=\theta$. We have $\eta_{i}(a_{0})\not=0$ for $0\leq i\leq 2$. By
assumption the data (12.1) is double lowering, so the data
$\\{a_{i}\\}_{i=0}^{j}$, $\\{b_{i}\\}_{i=0}^{j}$ is double lowering for $1\leq
j\leq N-1$. Applying Proposition 7.3(i), (ii) repeatedly (with $N$ replaced by
$2,3,\ldots,N-1$) we find that $E(i,j)$ holds for $1\leq i\leq 2$ and $i+1\leq
j\leq N-1$. Next we show that $a_{k}=b_{k}=\theta$ for $2\leq k\leq N-2$. We
will use induction on $k$. Assume $N\geq 4$; otherwise there is nothing to
prove. Using $E(1,2)$, $E(1,3)$ and $a_{0}\not=b_{1}$ we obtain

$\displaystyle a_{2}=b_{2}\zeta+\theta(1-\zeta),$ (12.4) $\displaystyle
a_{3}=b_{3}\zeta+b_{2}\big{(}1-\zeta^{2}\big{)}-\theta\zeta(1-\zeta),$ (12.5)

where

$\displaystyle\zeta=\frac{b_{0}-\theta}{a_{0}-\theta}.$ (12.6)

In the equation $E(2,3)$ we eliminate $a_{2}$, $a_{3}$ using (12.4), (12.5).
We evaluate the result using (12.2) (with $i=3$), to obtain $b_{2}=\theta$. In
(12.4) we set $b_{2}=\theta$ to obtain $a_{2}=\theta$. Next assume that $3\leq
k\leq N-2$. By induction each of $a_{2},a_{3},\ldots,a_{k-1}$,
$b_{2},b_{3},\ldots,b_{k-1}$ is equal to $\theta$. Using this we evaluate
$E(1,k)$, $E(1,k+1)$ to obtain

$\displaystyle a_{k}=b_{k}\zeta+\theta(1-\zeta),$ (12.7) $\displaystyle
a_{k+1}=b_{k+1}\zeta+b_{k}\big{(}1-\zeta^{2}\big{)}-\theta\zeta(1-\zeta).$
(12.8)

Using (12.7), (12.8) we evaluate $E(2,k+1)$ to obtain $b_{k}=\theta$. In
(12.7) we set $b_{k}=\theta$ to obtain $a_{k}=\theta$. We have shown that
$a_{k}=b_{k}=\theta$ for $2\leq k\leq N-2$. Now for $N\not=\infty$ we use
$E(1,N-1)$ to obtain

$\displaystyle a_{N-1}=b_{N-1}\zeta+\theta(1-\zeta).$

Evaluating this using (12.6) we obtain

$\displaystyle\frac{\theta-a_{N-1}}{\theta-b_{0}}=\frac{\theta-
b_{N-1}}{\theta-a_{0}}.$

We have shown that (iii) holds.

Case $a_{0}\not=b_{1}$, $a_{1}\not=b_{0}$, $a_{1}\not=b_{1}$. We show that
(iv) holds. Using Proposition 7.3(i), (ii) as in the previous case, we find
that $E(i,j)$ holds for $1\leq i\leq 2$ and $i+1\leq j\leq N-1$. Now by
Proposition 11.6 (with $n=N-1$), the sequences (12.1) are recurrent and twins.
We have shown that (iv) holds. ∎

##  13 $\boldsymbol{\mathcal{L}}$ and $\boldsymbol{\Delta}$ for twin recurrent
data

Our goal for the rest of the paper is to give a comprehensive description of
$\mathcal{L}$ and $\Delta$ for twin recurrent data. We will focus on Case I in
Lemma 10.2, or more precisely, an adjusted version of this case as described
in Section 3.

For the rest of this paper we assume that $N$ is an integer at least 2, or
$\infty$. Recall the double lowering space $\mathcal{L}$ for the data (2.1).
For the rest of this paper, fix nonzero $a,b,q\in\mathbb{F}$ and assume

$\displaystyle a_{i}=aq^{i}+a^{-1}q^{-i},\qquad b_{i}=bq^{i}+b^{-1}q^{-i}$
(13.1)

for $0\leq i\leq N-1$. By (2.2) we have $a\not=b$ and

$\displaystyle q^{i}\not=1,\qquad abq^{i-1}\not=1,\qquad 1\leq i\leq N.$

######  Note 13.1.

The data $\\{a_{i}\\}_{i=0}^{N-1}$, $\\{b_{i}\\}_{i=0}^{N-1}$ is unchanged if
we replace

$\displaystyle q\mapsto q^{-1},\qquad a\mapsto a^{-1},\qquad b\mapsto b^{-1}.$

######  Lemma 13.2.

The sequences (2.1) are recurrent and twins.

###### Proof.

By Lemmas 9.1 and 10.2. ∎

######  Corollary 13.3.

The data (2.1) is double lowering.

###### Proof.

By Lemma 13.2 along with Lemma 6.2 and Theorem 12.1(iv). ∎

Our next general goal is to describe the polynomials
$\\{\tau_{i}\\}_{i=0}^{N-1}$, $\\{\eta_{i}\\}_{i=0}^{N-1}$ from (2.3), (2.4),
the parameters $\\{\vartheta_{i}\\}_{i=0}^{N}$ from Definition 4.1, and the
map $\Delta$ from (2.5).

We mention some formulas for later use.

######  Lemma 13.4.

For $1\leq i\leq N-1$,

$\displaystyle qa_{i}-a_{i-1}=\big{(}q-q^{-1}\big{)}aq^{i},\qquad$
$\displaystyle qb_{i}-b_{i-1}=\big{(}q-q^{-1}\big{)}bq^{i},$ $\displaystyle
a_{i}-a_{i-1}=(q-1)\big{(}aq^{i-1}-a^{-1}q^{-i}\big{)},\qquad$ $\displaystyle
b_{i}-b_{i-1}=(q-1)\big{(}bq^{i-1}-b^{-1}q^{-i}\big{)},$ $\displaystyle
a_{i}-qa_{i-1}=\big{(}1-q^{2}\big{)}a^{-1}q^{-i},\qquad$ $\displaystyle
b_{i}-qb_{i-1}=\big{(}1-q^{2}\big{)}b^{-1}q^{-i}.$

###### Proof.

Use (13.1). ∎

######  Lemma 13.5.

For $0\leq i\leq N$,

$\displaystyle
a_{0}+a_{1}+\cdots+a_{i-1}=\frac{1-q^{i}}{1-q}\big{(}a+a^{-1}q^{1-i}\big{)},$
$\displaystyle
b_{0}+b_{1}+\cdots+b_{i-1}=\frac{1-q^{i}}{1-q}\big{(}b+b^{-1}q^{1-i}\big{)}.$

###### Proof.

Use (13.1). ∎

Next we describe $\\{\vartheta_{i}\\}_{i=0}^{N}$. We give two versions.

######  Lemma 13.6.

For $0\leq i\leq N$,

  1. $(i)$

$\displaystyle{\vartheta_{i}=\frac{1-q^{i}}{1-q}\frac{1-abq^{i-1}}{1-ab}q^{1-i}}$;

  2. $(ii)$

$\displaystyle{\vartheta_{i}=\frac{1-q^{-i}}{1-q^{-1}}\frac{1-a^{-1}b^{-1}q^{1-i}}{1-a^{-1}b^{-1}}q^{i-1}}$.

###### Proof.

(i) By Definition 4.1 and Lemma 13.5. (ii) By Note 13.1 and (i) above. ∎

We mention some formulas for later use.

######  Lemma 13.7.

For $0\leq i\leq N-1$,

$\displaystyle q\vartheta_{i+1}-\vartheta_{i}=\frac{q+ab-(q+1)abq^{i}}{1-ab},$
$\displaystyle\vartheta_{i+1}-\vartheta_{i}=\frac{q^{-i}-abq^{i}}{1-ab},$
$\displaystyle\vartheta_{i+1}-q\vartheta_{i}=\frac{(1+q)q^{-i}-q-ab}{1-ab}.$

###### Proof.

Use Lemma 13.6. ∎

We recall some notation. For an element $\alpha$ in any algebra, define

$\displaystyle(\alpha;q)_{i}=(1-\alpha)(1-\alpha q)\cdots\big{(}1-\alpha
q^{i-1}\big{)},\qquad i\in\mathbb{N}.$

We interpret $(\alpha;q)_{0}=1$. We remark that for $j\geq i\geq 0$,

$\displaystyle\big{(}q^{-j};q\big{)}_{i}(q;q)_{j-i}=(-1)^{i}(q;q)_{j}q^{\binom{i}{2}}q^{-ij}.$
(13.2)

######  Lemma 13.8.

For $0\leq i\leq N$,

$\displaystyle\vartheta_{1}\vartheta_{2}\cdots\vartheta_{i}=\frac{(q;q)_{i}(ab;q)_{i}q^{-\binom{i}{2}}}{(1-q)^{i}(1-ab)^{i}}.$
(13.3)

For $0\leq i\leq j\leq N$,

$\displaystyle\vartheta_{j}\vartheta_{j-1}\cdots\vartheta_{j-i+1}=\frac{\big{(}q^{-j};q\big{)}_{i}\big{(}a^{-1}b^{-1}q^{1-j};q\big{)}_{i}q^{i(j-i)}q^{\binom{i}{2}}}{\big{(}1-q^{-1}\big{)}^{i}\big{(}1-a^{-1}b^{-1}\big{)}^{i}}.$
(13.4)

###### Proof.

To obtain (13.3), use Lemma 13.6(i). To obtain (13.4), use Lemma 13.6(ii). ∎

######  Lemma 13.9.

For $0\leq i\leq j\leq N$,

$\displaystyle\left[\begin{matrix}j\\\
i\end{matrix}\right]_{\vartheta}=\frac{\big{(}q^{-j};q\big{)}_{i}\big{(}a^{-1}b^{-1}q^{1-j};q\big{)}_{i}q^{ij}a^{i}b^{i}}{(q;q)_{i}(ab;q)_{i}}.$

###### Proof.

Evaluate (5.3) using Lemma 13.8. ∎

We comment on the notation. Let $y$ denote an indeterminate. Let
${\mathbb{F}}\big{[}y,y^{-1}\big{]}$ denote the algebra consisting of the
Laurent polynomials in $y$ that have all coefficients in $\mathbb{F}$. This
algebra has an automorphism that sends $y\mapsto y^{-1}$. An element of
${\mathbb{F}}\big{[}y,y^{-1}\big{]}$ that is fixed by the automorphism is
called symmetric. The symmetric elements form a subalgebra of
${\mathbb{F}}\big{[}y,y^{-1}\big{]}$ called its symmetric part. There exists
an injective algebra homomorphism
$\iota\colon{\mathbb{F}}[x]\to{\mathbb{F}}\big{[}y,y^{-1}\big{]}$ that sends
$x\mapsto y+y^{-1}$. The image of ${\mathbb{F}}[x]$ under $\iota$ is the
symmetric part of ${\mathbb{F}}[y,y^{-1}]$. Via $\iota$ we identify
${\mathbb{F}}[x]$ with the symmetric part of ${\mathbb{F}}[y,y^{-1}]$.

######  Lemma 13.10.

For $0\leq i\leq N$,

  1. $(i)$

$\tau_{i}=(-1)^{i}a^{-i}q^{-\binom{i}{2}}(ay;q)_{i}\big{(}ay^{-1};q\big{)}_{i}$;

  2. $(ii)$

$\eta_{i}=(-1)^{i}b^{-i}q^{-\binom{i}{2}}(by;q)_{i}\big{(}by^{-1};q\big{)}_{i}$.

In the above lines $x=y+y^{-1}$.

###### Proof.

(i) We use (2.3) and (13.1). For $0\leq j\leq i-1$,

$\displaystyle
x-a_{j}=y+y^{-1}-aq^{j}-a^{-1}q^{-j}=-a^{-1}q^{-j}\big{(}1-ayq^{j}\big{)}\big{(}1-ay^{-1}q^{j}\big{)}.$

The result follows. (ii) Similar to the proof of (i) above. ∎

######  Lemma 13.11.

For $0\leq i\leq N$,

  1. $(i)$

$\tau_{i}(b_{0})=(-1)^{i}a^{-i}q^{-\binom{i}{2}}(ab;q)_{i}\big{(}ab^{-1};q\big{)}_{i}$;

  2. $(ii)$

$\eta_{i}(a_{0})=(-1)^{i}b^{-i}q^{-\binom{i}{2}}(ab;q)_{i}\big{(}a^{-1}b;q\big{)}_{i}$.

###### Proof.

(i) Set $y=b$ in Lemma 13.10(i), and use $b_{0}=b+b^{-1}$. (ii) Similar to the
proof of (i) above. ∎

Our data is double lowering, so $\mathcal{L}\not=0$. For the rest of the
paper, let $\psi\in\mathcal{L}$ be normalized. The maps $\Delta$, $\psi$ are
related by (5.5), (5.6). Our next goal is to interpret (5.5), (5.6) using the
$q$-exponential function. This function is defined as follows. For locally
nilpotent $T\in\operatorname{End}(V)$,

$\displaystyle\exp_{q}(T)=\sum_{i=0}^{N}\frac{q^{\binom{i}{2}}(1-q)^{i}T^{i}}{(q;q)_{i}}.$
(13.5)

The map $\exp_{q}(T)$ is invertible; its inverse is

$\displaystyle\exp_{q^{-1}}(-T)=\sum_{i=0}^{N}\frac{(-1)^{i}(1-q)^{i}T^{i}}{(q;q)_{i}}.$
(13.6)

######  Lemma 13.12.

For locally nilpotent $T\in\operatorname{End}(V)$,

$\displaystyle\bigl{(}1-(q-1)T\bigr{)}\exp_{q}(qT)=\exp_{q}(T).$

###### Proof.

To verify this equation, for $0\leq i\leq N$ compare the coefficient of
$T^{i}$ on each side. ∎

######  Proposition 13.13.

We have

$\displaystyle\Delta=\exp_{q}\big{(}a^{-1}\xi\psi\big{)}\exp_{q^{-1}}\big{(}{-}b^{-1}\xi\psi\big{)},$
(13.7)

where $\xi=1-ab$.

###### Proof.

For $0\leq j\leq N$ we compare the coefficient of $\psi^{j}$ on each side of
(13.7). For the left-hand side these coefficients are obtained from (5.5). We
require

$\displaystyle\frac{\eta_{j}(a_{0})}{\vartheta_{1}\vartheta_{2}\cdots\vartheta_{j}}=\sum_{i=0}^{j}\frac{q^{\binom{i}{2}}a^{-i}(1-q)^{i}\xi^{i}}{(q;q)_{i}}\frac{(-1)^{j-i}b^{i-j}(1-q)^{j-i}\xi^{j-i}}{(q;q)_{j-i}}.$
(13.8)

By (13.3) and the construction,

$\displaystyle\vartheta_{1}\vartheta_{2}\cdots\vartheta_{j}=(q;q)_{j}(ab;q)_{j}q^{-\binom{j}{2}}(1-q)^{-j}\xi^{-j}.$

By Lemma 13.11(ii),

$\displaystyle\eta_{j}(a_{0})=(-1)^{j}b^{-j}q^{-\binom{j}{2}}(ab;q)_{j}\big{(}a^{-1}b;q\big{)}_{j}.$

Using these comments and (13.2), the equation (13.8) becomes

$\displaystyle\big{(}zq^{-j};q\big{)}_{j}=\sum_{i=0}^{j}\frac{\big{(}q^{-j};q\big{)}_{i}z^{i}}{(q;q)_{i}},$
(13.9)

where $z=a^{-1}bq^{j}$. Basic hypergeometric series are discussed in [13, 25].
In (13.9) the sum on the right is the basic hypergeometric series

$\displaystyle{}_{1}\phi_{0}\left(\genfrac{}{}{0.0pt}{}{q^{-j}}{-}\,\bigg{|}\,q;z\right).$

This observation reveals that (13.9) is an instance of the $q$-binomial
theorem [13, Section 1.3]. The result follows. ∎

Proposition 13.13 gives a factorization of $\Delta$. We now investigate the
factors.

######  Lemma 13.14.

For $0\leq i\leq N$,

$\displaystyle\exp_{q^{-1}}\big{(}{-}a^{-1}\xi\psi\big{)}\eta_{i}=\exp_{q^{-1}}\big{(}{-}b^{-1}\xi\psi\big{)}\tau_{i},$
(13.10)
$\displaystyle\exp_{q}\big{(}b^{-1}\xi\psi\big{)}\eta_{i}=\exp_{q}\big{(}a^{-1}\xi\psi\big{)}\tau_{i},$
(13.11)

where $\xi=1-ab$.

###### Proof.

By (13.7) and the comments above Lemma 13.12,

$\displaystyle\exp_{q^{-1}}\big{(}{-}a^{-1}\xi\psi\big{)}\Delta=\exp_{q^{-1}}\big{(}{-}b^{-1}\xi\psi\big{)}.$
(13.12)

To obtain (13.10), apply each side of (13.12) to $\tau_{i}$ and evaluate the
result using (2.5). For the equation (13.7), the two factors on the right
commute; swapping these factors and proceeding as above,

$\displaystyle\exp_{q}\big{(}b^{-1}\xi\psi\big{)}\Delta=\exp_{q}\big{(}a^{-1}\xi\psi\big{)}.$
(13.13)

To obtain (13.11), apply each side of (13.13) to $\tau_{i}$ and evaluate the
result using (2.5). ∎

######  Definition 13.15.

For $0\leq i\leq N$ let $w_{i}$ (resp. $w^{\prime}_{i}$) denote the common
value of (13.10) (resp. (13.11)). For notational convenience define $w_{-1}=0$
and $w^{\prime}_{-1}=0$.

######  Lemma 13.16.

For $0\leq i\leq N$,

$\displaystyle\tau_{i}=\exp_{q}\big{(}b^{-1}\xi\psi\big{)}w_{i},\qquad$
$\displaystyle w_{i}=\exp_{q^{-1}}\big{(}{-}b^{-1}\xi\psi\big{)}\tau_{i},$
$\displaystyle\eta_{i}=\exp_{q}\big{(}a^{-1}\xi\psi\big{)}w_{i},\qquad$
$\displaystyle w_{i}=\exp_{q^{-1}}\big{(}{-}a^{-1}\xi\psi\big{)}\eta_{i}$

and

$\displaystyle\tau_{i}=\exp_{q^{-1}}\big{(}{-}a^{-1}\xi\psi\big{)}w^{\prime}_{i},\qquad$
$\displaystyle w^{\prime}_{i}=\exp_{q}\big{(}a^{-1}\xi\psi\big{)}\tau_{i},$
$\displaystyle\eta_{i}=\exp_{q^{-1}}\big{(}{-}b^{-1}\xi\psi\big{)}w^{\prime}_{i}\qquad$
$\displaystyle w^{\prime}_{i}=\exp_{q}\big{(}b^{-1}\xi\psi\big{)}\eta_{i}.$

In the above lines $\xi=1-ab$.

###### Proof.

By Definition 13.15 and the comments above Lemma 13.12. ∎

######  Note 13.17.

Referring to Definition 13.15, the polynomials
$\\{w^{\prime}_{i}\\}_{i=0}^{N}$ are obtained from the polynomials
$\\{w_{i}\\}_{i=0}^{N}$ by replacing $q\mapsto q^{-1}$, $a\mapsto a^{-1}$,
$b\mapsto b^{-1}$.

######  Example 13.18.

The following (i)–(iii) hold:

  1. (i)

$w_{0}=1$;

  2. (ii)

$w_{1}$ is equal to each of

$\displaystyle\tau_{1}-(1-ab)b^{-1}\tau_{0},\qquad\eta_{1}-(1-ab)a^{-1}\eta_{0},\qquad
x-a^{-1}-b^{-1};$

  3. (iii)

$w_{2}$ is equal to each of

$\displaystyle\tau_{2}-\big{(}q^{-1}+1\big{)}(1-abq)b^{-1}\tau_{1}+q^{-1}(1-ab)(1-abq)b^{-2}\tau_{0},$
$\displaystyle\eta_{2}-\big{(}q^{-1}+1\big{)}(1-abq)a^{-1}\eta_{1}+q^{-1}(1-ab)(1-abq)a^{-2}\eta_{0},$
$\displaystyle\big{(}x-a^{-1}-b^{-1}\big{)}\big{(}x-q^{-1}a^{-1}-q^{-1}b^{-1}\big{)}+\big{(}q^{-1}-1\big{)}\big{(}1-a^{-1}b^{-1}\big{)}.$

To get $w^{\prime}_{0}$, $w^{\prime}_{1}$, $w^{\prime}_{2}$ replace $q\mapsto
q^{-1}$, $a\mapsto a^{-1}$, $b\mapsto b^{-1}$ in (i)–(iii) above.

######  Lemma 13.19.

The following $(i)$–$(iii)$ hold:

  1. $(i)$

for $0\leq i\leq N$ the polynomials $w_{i}$, $w^{\prime}_{i}$ are monic with
degree $i$;

  2. $(ii)$

for $0\leq n\leq N$, each of $\\{w_{i}\\}_{i=0}^{n}$,
$\\{w^{\prime}_{i}\\}_{i=0}^{n}$ is a basis for the vector space $V_{n}$;

  3. $(iii)$

each of $\\{w_{i}\\}_{i=0}^{N}$, $\\{w^{\prime}_{i}\\}_{i=0}^{N}$ is a basis
for the vector space $V$.

###### Proof.

(i) By (4.9) and Definition 13.15. (ii), (iii) By (i) above. ∎

######  Lemma 13.20.

For $0\leq j\leq N$,

$\displaystyle w^{\prime}_{j}$
$\displaystyle=a^{-j}(ab;q)_{j}\sum_{i=0}^{j}\frac{\big{(}q^{-j};q\big{)}_{i}(ay;q)_{i}\big{(}ay^{-1};q\big{)}_{i}q^{i}}{(ab;q)_{i}(q;q)_{i}}$
$\displaystyle=a^{-j}(ab;q)_{j}\,{}_{3}\phi_{2}\left(\genfrac{}{}{0.0pt}{}{q^{-j},ay,ay^{-1}}{ab,0}\,\bigg{|}\,q;q\right),$

where $x=y+y^{-1}$. To get $w_{j}$ from $w^{\prime}_{j}$, replace $q\mapsto
q^{-1}$, $a\mapsto a^{-1}$, $b\mapsto b^{-1}$.

###### Proof.

In the equation $w^{\prime}_{j}=\exp_{q}\big{(}a^{-1}\xi\psi\big{)}\tau_{j}$,
expand the $q$-exponential using (13.5), and evaluate the result using the
equation on the left in (4.9). This yields
$w^{\prime}_{j}=\sum_{i=0}^{j}\alpha_{i}\tau_{i}$ where for $0\leq i\leq j$,

$\displaystyle\alpha_{i}=\frac{q^{\binom{j-i}{2}}(1-q)^{j-i}a^{i-j}\xi^{j-i}\vartheta_{j}\vartheta_{j-1}\cdots\vartheta_{i+1}}{(q;q)_{j-i}}.$

Evaluating this using

$\displaystyle\vartheta_{j}\vartheta_{j-1}\cdots\vartheta_{i+1}=\frac{\vartheta_{1}\vartheta_{2}\cdots\vartheta_{j}}{\vartheta_{1}\vartheta_{2}\cdots\vartheta_{i}}$

and (13.2), (13.3) we obtain

$\displaystyle\alpha_{i}=\frac{(-1)^{i}a^{i-j}(ab;q)_{j}\big{(}q^{-j};q\big{)}_{i}q^{\binom{i}{2}}q^{i}}{(ab;q)_{i}(q;q)_{i}}.$

The polynomial $\tau_{i}$ is given in Lemma 13.10(i). The result follows. ∎

######  Note 13.21.

The polynomials $\\{w_{i}\\}_{i=0}^{N}$ and $\\{w^{\prime}_{i}\\}_{i=0}^{N}$
are in the Al-Salam/Chihara family [25, Section 14.8] if $N=\infty$, and the
dual $q$-Krawtchouk family [25, Section 14.17] if $N\not=\infty$. The Al-
Salam/Chihara and dual $q$-Krawtchouk polynomials satisfy a 3-term recurrence;
the details will be given in Lemmas 13.67 and 13.68 below.

Going forward we focus on $\\{w_{i}\\}_{i=0}^{N}$; similar results hold for
$\\{w^{\prime}_{i}\\}_{i=0}^{N}$.

######  Lemma 13.22.

We have

$\displaystyle\psi w_{i}=\vartheta_{i}w_{i-1},\qquad 0\leq i\leq N.$

###### Proof.

By Definition 13.15 and since $\psi\tau_{i}=\vartheta_{i}\tau_{i-1}$. ∎

Our next general goal is to describe in more detail how the bases
$\\{\tau_{i}\\}_{i=0}^{N}$, $\\{\eta_{i}\\}_{i=0}^{N}$,
$\\{w_{i}\\}_{i=0}^{N}$ are related. To this end, we introduce some maps
$K,B,M\in\operatorname{End}(V)$.

######  Definition 13.23.

Define $K,B,M\in\operatorname{End}(V)$ such that for $0\leq i\leq N$,

$\displaystyle K\tau_{i}=q^{-i}\tau_{i},\qquad B\eta_{i}=q^{-i}\eta_{i},\qquad
Mw_{i}=q^{-i}w_{i}.$

Each of $K$, $B$, $M$ is invertible.

######  Lemma 13.24.

The following $(i)$–$(iii)$ hold:

  1. $(i)$

$K\psi=q\psi K$;

  2. $(ii)$

$B\psi=q\psi B$;

  3. $(iii)$

$M\psi=q\psi M$.

###### Proof.

(i) The vectors $\\{\tau_{i}\\}_{i=0}^{N}$ form a basis for $V$. For $0\leq
i\leq N$,

$\displaystyle
K\psi\tau_{i}=\vartheta_{i}K\tau_{i-1}=\vartheta_{i}q^{1-i}\tau_{i-1},\qquad
q\psi K\tau_{i}=q^{1-i}\psi\tau_{i}=\vartheta_{i}q^{1-i}\tau_{i-1}.$

Therefore $K\psi=q\psi K$. (ii), (iii) Similar to the proof of (i) above. ∎

######  Lemma 13.25.

The following $(i)$–$(iii)$ hold:

  1. $(i)$

$B\Delta=\Delta K$;

  2. $(ii)$

$K\exp_{q}\big{(}b^{-1}\xi\psi\big{)}=\exp_{q}\big{(}b^{-1}\xi\psi\big{)}M$;

  3. $(iii)$

$B\exp_{q}\big{(}a^{-1}\xi\psi\big{)}=\exp_{q}\big{(}a^{-1}\xi\psi\big{)}M$.

###### Proof.

(i) For $0\leq i\leq N$, apply each side to $\tau_{i}$ and use (2.5) along
with Definition 13.23.

(ii), (iii) For $0\leq i\leq N$, apply each side to $w_{i}$ and use Lemma
13.16 along with Definition 13.23. ∎

######  Proposition 13.26.

The following $(i)$–$(iv)$ hold:

  1. $(i)$

$KM^{-1}=I+(q-1)\big{(}a-b^{-1}\big{)}\psi$;

  2. $(ii)$

$M^{-1}K=I+(q^{-1}-1)\big{(}b^{-1}-a\big{)}\psi$;

  3. $(iii)$

$BM^{-1}=I+(q-1)\big{(}b-a^{-1}\big{)}\psi$;

  4. $(iv)$

$M^{-1}B=I+\big{(}q^{-1}-1\big{)}\big{(}a^{-1}-b\big{)}\psi$.

###### Proof.

(i) The map $T=b^{-1}\xi\psi$ is locally nilpotent. We have $KTK^{-1}=qT$ by
Lemma 13.24(i), and $K\exp_{q}(T)=\exp_{q}(T)M$ by Lemma 13.25(ii). By these
comments and Lemma 13.12,

$\displaystyle\exp_{q}(T)$
$\displaystyle=\big{(}I-(q-1)T\big{)}\exp_{q}(qT)=\big{(}I-(q-1)T\big{)}\exp_{q}\big{(}KTK^{-1}\big{)}$
$\displaystyle=\big{(}I-(q-1)T\big{)}K\exp_{q}(T)K^{-1}=\bigl{(}I-(q-1)T\bigr{)}\exp_{q}(T)MK^{-1}$
$\displaystyle=\exp_{q}(T)\big{(}I-(q-1)T\big{)}MK^{-1}.$

By this and since $\exp_{q}(T)$ is invertible,

$\displaystyle I=\bigl{(}I-(q-1)T\bigr{)}MK^{-1}.$

The result follows from this and $\xi=1-ab$.

(ii) By (i) above and $M\psi=q\psi M$. (iii), (iv) Similar to the proof of
(i), (ii) above. ∎

######  Corollary 13.27.

We have

$\displaystyle\psi=\frac{bq}{1-ab}\frac{M^{-1}K-KM^{-1}}{(q-1)^{2}},$ (13.14)
$\displaystyle\psi=\frac{aq}{1-ab}\frac{M^{-1}B-BM^{-1}}{(q-1)^{2}}.$ (13.15)

###### Proof.

To get (13.14) use Proposition 13.26(i), (ii).

To get (13.15) use Proposition 13.26(iii), (iv). ∎

######  Proposition 13.28.

We have

$\displaystyle M=\frac{bK-aB}{b-a}.$

###### Proof.

Compute $b$ times Proposition 13.26(i) minus $a$ times Proposition 13.26(iii).
∎

######  Lemma 13.29.

Each of the following is invertible:

$\displaystyle aI-bB^{-1}K,\qquad bI-aK^{-1}B,\qquad
a^{-1}I-b^{-1}BK^{-1},\qquad b^{-1}I-a^{-1}KB^{-1}.$

###### Proof.

By Proposition 13.28 and since $M$ is invertible. ∎

Our next goal is to describe how $K$, $B$ are related. We will use the
following result.

######  Lemma 13.30.

We have

$\displaystyle\psi M=\frac{1}{1-q}\frac{ab}{1-ab}\frac{K-B}{a-b},$ (13.16)
$\displaystyle M\psi=\frac{q}{1-q}\frac{ab}{1-ab}\frac{K-B}{a-b}.$ (13.17)

###### Proof.

To get (13.16), subtract Proposition 13.26(iii) from Proposition 13.26(i). To
get (13.17) from (13.16), use $M\psi=q\psi M$. ∎

######  Proposition 13.31.

We have

$\displaystyle(bK-aB)(K-B)=q(K-B)(bK-aB).$

###### Proof.

We have $M\psi=q\psi M$ so $M(\psi M)=q(\psi M)M$. Evaluate this using
Proposition 13.28 and (13.16). ∎

We mention some reformulations of Proposition 13.31.

######  Corollary 13.32.

We have

$\displaystyle 0=aB^{2}-\frac{bq-a}{q-1}BK-\frac{aq-b}{q-1}KB+bK^{2}$

and

$\displaystyle\big{(}bI-
aK^{-1}B\big{)}\big{(}I-KB^{-1}\big{)}=q\big{(}I-K^{-1}B\big{)}\big{(}aI-
bKB^{-1}\big{)},$ $\displaystyle\big{(}aI-
bB^{-1}K\big{)}\big{(}I-KB^{-1}\big{)}=q\big{(}I-B^{-1}K\big{)}\big{(}aI-
bKB^{-1}\big{)},$ $\displaystyle\big{(}aI-
bB^{-1}K\big{)}\big{(}I-BK^{-1}\big{)}=q\big{(}I-B^{-1}K\big{)}\big{(}bI-
aBK^{-1}\big{)},$ $\displaystyle\big{(}bI-
aK^{-1}B\big{)}\big{(}I-BK^{-1}\big{)}=q\big{(}I-K^{-1}B\big{)}\big{(}bI-
aBK^{-1}\big{)}.$

######  Proposition 13.33.

We have

$\displaystyle
KB^{-1}=\frac{I+(q-1)\big{(}a-b^{-1}\big{)}\psi}{I+(q-1)\big{(}b-a^{-1}\big{)}\psi},$
(13.18) $\displaystyle
BK^{-1}=\frac{I+(q-1)\big{(}b-a^{-1}\big{)}\psi}{I+(q-1)\big{(}a-b^{-1}\big{)}\psi},$
(13.19) $\displaystyle
K^{-1}B=\frac{I+\big{(}q^{-1}-1\big{)}\big{(}a^{-1}-b\big{)}\psi}{I+\big{(}q^{-1}-1\big{)}\big{(}b^{-1}-a\big{)}\psi},$
(13.20) $\displaystyle
B^{-1}K=\frac{I+\big{(}q^{-1}-1\big{)}\big{(}b^{-1}-a\big{)}\psi}{I+\big{(}q^{-1}-1\big{)}\big{(}a^{-1}-b\big{)}\psi}.$
(13.21)

In the above fractions the denominator is invertible since $\psi$ is locally
nilpotent.

###### Proof.

To get (13.18), equate the two expressions for $M^{-1}$ obtained from
Proposition 13.26(i) and (iii). To get (13.19) from (13.18), compute the
inverse of each side. To get (13.20), equate the two expressions for $M^{-1}$
obtained from Proposition 13.26(ii) and (iv). To get (13.21) from (13.20),
compute the inverse of each side. ∎

######  Lemma 13.34.

The following mutually commute:

$\displaystyle\psi,\qquad KB^{-1},\qquad BK^{-1},\qquad K^{-1}B,\qquad
B^{-1}K.$

###### Proof.

By Proposition 13.33. ∎

######  Proposition 13.35.

We have

$\displaystyle\psi=\frac{1}{q-1}\frac{1}{1-ab}\frac{I-KB^{-1}}{b^{-1}I-a^{-1}KB^{-1}},$
$\displaystyle\psi=\frac{1}{q-1}\frac{1}{1-ab}\frac{I-BK^{-1}}{a^{-1}I-b^{-1}BK^{-1}},$
$\displaystyle\psi=\frac{1}{q^{-1}-1}\frac{1}{1-a^{-1}b^{-1}}\frac{I-K^{-1}B}{bI-
aK^{-1}B},$
$\displaystyle\psi=\frac{1}{q^{-1}-1}\frac{1}{1-a^{-1}b^{-1}}\frac{I-B^{-1}K}{aI-
bB^{-1}K}.$

In the above fractions the denominator is invertible by Lemma 13.29.

###### Proof.

In Proposition 13.33 solve for $\psi$. ∎

######  Proposition 13.36.

We have

$\displaystyle\frac{qM^{-1}K-KM^{-1}}{q-1}=I,$ (13.22)
$\displaystyle\frac{qM^{-1}B-BM^{-1}}{q-1}=I.$ (13.23)

###### Proof.

To get (13.22) use Proposition 13.26(i) and (ii). To get (13.23) use
Proposition 13.26(iii) and (iv). ∎

We have a comment.

######  Lemma 13.37.

The relations in Proposition 13.31 and Corollary 13.32 still hold if we
replace

$\displaystyle q\mapsto q^{-1},\qquad a\mapsto a^{-1},\qquad b\mapsto
b^{-1},\qquad K\mapsto K^{-1},\qquad B\mapsto B^{-1}.$

###### Proof.

Use Proposition 13.31 and Lemma 13.34. ∎

We recall some notation. Let ${\rm Mat}_{N+1}(\mathbb{F})$ denote the set of
$N+1$ by $N+1$ matrices that have all entries in $\mathbb{F}$. We index the
rows and columns by $0,1,\ldots,N$. Let $\\{v_{i}\\}_{i=0}^{N}$ denote a basis
for $V$. For $M\in{\rm Mat}_{N+1}(\mathbb{F})$ and
$T\in\operatorname{End}(V)$, we say that $M$ represents $T$ with respect to
$\\{v_{i}\\}_{i=0}^{N}$ whenever $Tv_{j}=\sum_{i=0}^{N}M_{i,j}v_{i}$ for
$0\leq j\leq N$.

Our next goal is to display the matrices that represent $\psi$, $K^{\pm 1}$,
$M^{\pm 1}$, $B^{\pm 1}$ with respect to the bases $\\{\tau_{i}\\}_{i=0}^{N}$,
$\\{w_{i}\\}_{i=0}^{N}$, $\\{\eta_{i}\\}_{i=0}^{N}$ of $V$.

######  Definition 13.38.

Let $\widehat{\psi}$ denote the matrix in ${\rm Mat}_{N+1}(\mathbb{F})$ that
has $(i-1,i)$-entry $\vartheta_{i}$ for $1\leq i\leq N$, and all other entries
0. Thus

$\displaystyle{\widehat{\psi}}=\left(\begin{matrix}0&\vartheta_{1}&&&&{\bf
0}\\\ &0&\vartheta_{2}&&&\\\ &&0&\cdot&&\\\ &&&\cdot&\cdot&\\\
&&&&\cdot&\vartheta_{N}\\\ {\bf 0}&&&&&0\end{matrix}\right).$

######  Lemma 13.39.

The matrix $\widehat{\psi}$ represents $\psi$ with respect to
$\\{\tau_{i}\\}_{i=0}^{N}$ and $\\{w_{i}\\}_{i=0}^{N}$ and
$\\{\eta_{i}\\}_{i=0}^{N}$.

###### Proof.

By (4.9) and Lemma 13.22. ∎

######  Lemma 13.40.

The matrix ${\rm diag}\big{(}1,q^{-1},q^{-2},\ldots,q^{-N}\big{)}$ represents
$K$ $($resp. $M)$ $($resp. $B)$ with respect to $\\{\tau_{i}\\}_{i=0}^{N}$
$($resp. $\\{w_{i}\\}_{i=0}^{N})$ $($resp. $\\{\eta_{i}\\}_{i=0}^{N})$.

###### Proof.

By Definition 13.23. ∎

######  Lemma 13.41.

We give the matrix in ${\rm Mat}_{N+1}(\mathbb{F})$ that represents $K$ with
respect to $\\{w_{i}\\}_{i=0}^{N}$. The $(i,i)$-entry is $q^{-i}$ for $0\leq
i\leq N$. The $(i-1,i)$-entry is
$\big{(}1-q^{-i}\big{)}\big{(}a-b^{-1}q^{1-i}\big{)}$ for $1\leq i\leq N$. All
other entries are $0$.

###### Proof.

Use $KM^{-1}=I+(q-1)\big{(}a-b^{-1}\big{)}\psi$ and Lemmas 13.6(ii), 13.39,
13.40. ∎

######  Lemma 13.42.

We give the matrix in ${\rm Mat}_{N+1}(\mathbb{F})$ that represents $B$ with
respect to $\\{w_{i}\\}_{i=0}^{N}$. The $(i,i)$-entry is $q^{-i}$ for $0\leq
i\leq N$. The $(i-1,i)$-entry is
$\big{(}1-q^{-i}\big{)}\big{(}b-a^{-1}q^{1-i}\big{)}$ for $1\leq i\leq N$. All
other entries are $0$.

###### Proof.

Use $BM^{-1}=I+(q-1)\big{(}b-a^{-1}\big{)}\psi$ and Lemmas 13.6(ii), 13.39,
13.40. ∎

######  Lemma 13.43.

We give the matrix in ${\rm Mat}_{N+1}(\mathbb{F})$ that represents $M^{-1}$
with respect to $\\{\tau_{i}\\}_{i=0}^{N}$. The $(i,i)$-entry is $q^{i}$ for
$0\leq i\leq N$. The $(i-1,i)$-entry is
$\big{(}q^{i}-1\big{)}\big{(}aq^{i-1}-b^{-1}\big{)}$ for $1\leq i\leq N$. All
other entries are $0$.

###### Proof.

Use $M^{-1}K=I+\big{(}q^{-1}-1\big{)}\big{(}b^{-1}-a\big{)}\psi$ and Lemmas
13.6(i), 13.39, 13.40. ∎

######  Lemma 13.44.

We give the matrix in ${\rm Mat}_{N+1}(\mathbb{F})$ that represents $M^{-1}$
with respect to $\\{\eta_{i}\\}_{i=0}^{N}$. The $(i,i)$-entry is $q^{i}$ for
$0\leq i\leq N$. The $(i-1,i)$-entry is
$\big{(}q^{i}-1\big{)}\big{(}bq^{i-1}-a^{-1}\big{)}$ for $1\leq i\leq N$. All
other entries are $0$.

###### Proof.

Use $M^{-1}B=I+\big{(}q^{-1}-1\big{)}\big{(}a^{-1}-b\big{)}\psi$ and Lemmas
13.6(i), 13.39, 13.40. ∎

We give a variation on Proposition 13.26.

######  Lemma 13.45.

We have

$\displaystyle
MK^{-1}=\sum_{i=0}^{N}(-1)^{i}(1-q)^{i}(1-ab)^{i}b^{-i}\psi^{i},$ (13.24)
$\displaystyle
K^{-1}M=\sum_{i=0}^{N}(-1)^{i}(1-q)^{i}q^{-i}(1-ab)^{i}b^{-i}\psi^{i},$
(13.25) $\displaystyle
MB^{-1}=\sum_{i=0}^{N}(-1)^{i}(1-q)^{i}(1-ab)^{i}a^{-i}\psi^{i},$ (13.26)
$\displaystyle
B^{-1}M=\sum_{i=0}^{N}(-1)^{i}(1-q)^{i}q^{-i}(1-ab)^{i}a^{-i}\psi^{i}.$
(13.27)

###### Proof.

For each equation in Proposition 13.26, take the inverse of each side and
evaluate the result using Lemma 2.14. ∎

######  Lemma 13.46.

We give the matrix in ${\rm Mat}_{N+1}(\mathbb{F})$ that represents $M$ with
respect to $\\{\tau_{i}\\}_{i=0}^{N}$. The $(i,j)$-entry is

$\displaystyle\frac{(-1)^{j-i}b^{i-j}(ab;q)_{j}(q;q)_{j}q^{\binom{i}{2}-\binom{j}{2}-j}}{(ab;q)_{i}(q;q)_{i}}$

for $0\leq i\leq j\leq N$. All other entries are $0$.

###### Proof.

Use (13.3), (13.24) and Lemmas 13.39, 13.40. ∎

######  Lemma 13.47.

We give the matrix in ${\rm Mat}_{N+1}(\mathbb{F})$ that represents $K^{-1}$
with respect to $\\{w_{i}\\}_{i=0}^{N}$. The $(i,j)$-entry is

$\displaystyle\frac{(-1)^{j-i}b^{i-j}(ab;q)_{j}(q;q)_{j}q^{\binom{i}{2}-\binom{j}{2}+i}}{(ab;q)_{i}(q;q)_{i}}$

for $0\leq i\leq j\leq N$. All other entries are $0$.

###### Proof.

Use (13.3), (13.25) and Lemmas 13.39, 13.40. ∎

######  Lemma 13.48.

We give the matrix in ${\rm Mat}_{N+1}(\mathbb{F})$ that represents $M$ with
respect to $\\{\eta_{i}\\}_{i=0}^{N}$. The $(i,j)$-entry is

$\displaystyle\frac{(-1)^{j-i}a^{i-j}(ab;q)_{j}(q;q)_{j}q^{\binom{i}{2}-\binom{j}{2}-j}}{(ab;q)_{i}(q;q)_{i}}$

for $0\leq i\leq j\leq N$. All other entries are $0$.

###### Proof.

Use (13.3), (13.26) and Lemmas 13.39, 13.40. ∎

######  Lemma 13.49.

We give the matrix in ${\rm Mat}_{N+1}(\mathbb{F})$ that represents $B^{-1}$
with respect to $\\{w_{i}\\}_{i=0}^{N}$. The $(i,j)$-entry is

$\displaystyle\frac{(-1)^{j-i}a^{i-j}(ab;q)_{j}(q;q)_{j}q^{\binom{i}{2}-\binom{j}{2}+i}}{(ab;q)_{i}(q;q)_{i}}$

for $0\leq i\leq j\leq N$. All other entries are $0$.

###### Proof.

Use (13.3), (13.27) and Lemmas 13.39, 13.40. ∎

We give a variation on Proposition 13.33.

######  Lemma 13.50.

We have

$\displaystyle
KB^{-1}=I+\frac{b-a}{b}\sum_{i=1}^{N}(-1)^{i}(1-q)^{i}(1-ab)^{i}a^{-i}\psi^{i},$
(13.28) $\displaystyle
BK^{-1}=I+\frac{a-b}{a}\sum_{i=1}^{N}(-1)^{i}(1-q)^{i}(1-ab)^{i}b^{-i}\psi^{i},$
(13.29) $\displaystyle
K^{-1}B=I+\frac{a-b}{a}\sum_{i=1}^{N}(-1)^{i}(1-q)^{i}q^{-i}(1-ab)^{i}b^{-i}\psi^{i},$
(13.30) $\displaystyle
B^{-1}K=I+\frac{b-a}{b}\sum_{i=1}^{N}(-1)^{i}(1-q)^{i}q^{-i}(1-ab)^{i}a^{-i}\psi^{i}.$
(13.31)

###### Proof.

Evaluate each equation in Proposition 13.33 using Lemma 2.14. ∎

######  Lemma 13.51.

We give the matrix in ${\rm Mat}_{N+1}(\mathbb{F})$ that represents $K$ with
respect to $\\{\eta_{i}\\}_{i=0}^{N}$. The $(i,i)$-entry is $q^{-i}$ for
$0\leq i\leq N$. The $(i,j)$-entry is

$\displaystyle\frac{b-a}{b}\frac{(-1)^{j-i}a^{i-j}(ab;q)_{j}(q;q)_{j}q^{\binom{i}{2}-\binom{j}{2}-j}}{(ab;q)_{i}(q;q)_{i}}$

for $0\leq i<j\leq N$. All other entries are $0$.

###### Proof.

Use (13.3), (13.28) and Lemmas 13.39, 13.40. ∎

######  Lemma 13.52.

We give the matrix in ${\rm Mat}_{N+1}(\mathbb{F})$ that represents $B$ with
respect to $\\{\tau_{i}\\}_{i=0}^{N}$. The $(i,i)$-entry is $q^{-i}$ for
$0\leq i\leq N$. The $(i,j)$-entry is

$\displaystyle\frac{a-b}{a}\frac{(-1)^{j-i}b^{i-j}(ab;q)_{j}(q;q)_{j}q^{\binom{i}{2}-\binom{j}{2}-j}}{(ab;q)_{i}(q;q)_{i}}$

for $0\leq i<j\leq N$. All other entries are $0$.

###### Proof.

Use (13.3), (13.29) and Lemmas 13.39, 13.40. ∎

######  Lemma 13.53.

We give the matrix in ${\rm Mat}_{N+1}(\mathbb{F})$ that represents $K^{-1}$
with respect to $\\{\eta_{i}\\}_{i=0}^{N}$. The $(i,i)$-entry is $q^{i}$ for
$0\leq i\leq N$. The $(i,j)$-entry is

$\displaystyle\frac{a-b}{a}\frac{(-1)^{j-i}b^{i-j}(ab;q)_{j}(q;q)_{j}q^{\binom{i}{2}-\binom{j}{2}+i}}{(ab;q)_{i}(q;q)_{i}}$

for $0\leq i<j\leq N$. All other entries are $0$.

###### Proof.

Use (13.3), (13.30) and Lemmas 13.39, 13.40. ∎

######  Lemma 13.54.

We give the matrix in ${\rm Mat}_{N+1}(\mathbb{F})$ that represents $B^{-1}$
with respect to $\\{\tau_{i}\\}_{i=0}^{N}$. The $(i,i)$-entry is $q^{i}$ for
$0\leq i\leq N$. The $(i,j)$-entry is

$\displaystyle\frac{b-a}{b}\frac{(-1)^{j-i}a^{i-j}(ab;q)_{j}(q;q)_{j}q^{\binom{i}{2}-\binom{j}{2}+i}}{(ab;q)_{i}(q;q)_{i}}$

for $0\leq i<j\leq N$. All other entries are $0$.

###### Proof.

Use (13.3), (13.31) and Lemmas 13.39, 13.40. ∎

We recall some notation. Let $\\{u_{i}\\}_{i=0}^{N}$ and
$\\{v_{i}\\}_{i=0}^{N}$ denote bases for $V$. By the transition matrix from
$\\{u_{i}\\}_{i=0}^{N}$ to $\\{v_{i}\\}_{i=0}^{N}$ we mean the matrix
$T\in{\rm Mat}_{N+1}(\mathbb{F})$ such that $v_{j}=\sum_{i=0}^{N}T_{i,j}u_{i}$
for $0\leq j\leq N$.

Our next goal is to display the transition matrices between the bases
$\\{\tau_{i}\\}_{i=0}^{N}$, $\\{w_{i}\\}_{i=0}^{N}$,
$\\{\eta_{i}\\}_{i=0}^{N}$. Recall the notation $\xi=1-ab$.

Consider the following matrices:

$\displaystyle\exp_{q}\big{(}a^{-1}\xi\widehat{\psi}\big{)},\qquad\exp_{q}\big{(}b^{-1}\xi\widehat{\psi}\big{)}.$
(13.32)

Their inverses are

$\displaystyle\exp_{q^{-1}}\big{(}{-}a^{-1}\xi\widehat{\psi}\big{)},\qquad\exp_{q^{-1}}\big{(}{-}b^{-1}\xi\widehat{\psi}\big{)}.$
(13.33)

The matrices (13.32), (13.33) are upper triangular. We now give their entries.

######  Lemma 13.55.

For $0\not=z\in\mathbb{F}$ the matrix
$\exp_{q}\big{(}z\xi\widehat{\psi}\big{)}$ is upper triangular. Its
$(i,j)$-entry is

$\displaystyle\frac{(-1)^{i}z^{j-i}(ab;q)_{j}(q^{-j};q)_{i}q^{i+\binom{i}{2}}}{(ab;q)_{i}(q;q)_{i}}$

for $0\leq i\leq j\leq N$. The matrix
$\exp_{q^{-1}}\big{(}{-}z\xi\widehat{\psi}\big{)}$ is upper triangular. Its
$(i,j)$-entry is

$\displaystyle\frac{(-1)^{j}z^{j-i}(ab;q)_{j}(q^{-j};q)_{i}q^{ij-\binom{j}{2}}}{(ab;q)_{i}(q;q)_{i}}$

for $0\leq i\leq j\leq N$.

###### Proof.

Use (13.2), (13.3), (13.5), (13.6) and Definition 13.38. ∎

######  Lemma 13.56.

The transition matrices between the basis $\\{w_{i}\\}_{i=0}^{N}$ and the
bases $\\{\tau_{i}\\}_{i=0}^{N}$, $\\{\eta_{i}\\}_{i=0}^{N}$ are given in the
table below:

from | to | transition matrix
---|---|---
$\\{\tau_{i}\\}_{i=0}^{N}$ | $\\{w_{i}\\}_{i=0}^{N}$ | $\exp_{q^{-1}}\big{(}{-}b^{-1}\xi\widehat{\psi}\big{)}$
$\\{w_{i}\\}_{i=0}^{N}$ | $\\{\tau_{i}\\}_{i=0}^{N}$ | $\exp_{q}\big{(}b^{-1}\xi\widehat{\psi}\big{)}$
$\\{\eta_{i}\\}_{i=0}^{N}$ | $\\{w_{i}\\}_{i=0}^{N}$ | $\exp_{q^{-1}}\big{(}-a^{-1}\xi\widehat{\psi}\big{)}$
$\\{w_{i}\\}_{i=0}^{N}$ | $\\{\eta_{i}\\}_{i=0}^{N}$ | $\exp_{q}\big{(}a^{-1}\xi\widehat{\psi}\big{)}$

###### Proof.

By Lemmas 13.16, 13.39. ∎

Next we consider the product

$\displaystyle\exp_{q}\big{(}a^{-1}\xi\widehat{\psi}\big{)}\exp_{q^{-1}}\big{(}-b^{-1}\xi\widehat{\psi}\big{)}.$
(13.34)

The inverse of (13.34) is

$\displaystyle\exp_{q}\bigl{(}b^{-1}\xi\widehat{\psi}\bigr{)}\exp_{q^{-1}}\bigl{(}-a^{-1}\xi\widehat{\psi}\bigr{)}.$
(13.35)

The matrices (13.34), (13.35) are upper triangular. Shortly we will give their
entries.

######  Lemma 13.57.

The matrix (13.34) is the transition matrix from the basis
$\\{\tau_{i}\\}_{i=0}^{N}$ to the basis $\\{\eta_{i}\\}_{i=0}^{N}$. The matrix
(13.35) is the transition matrix from the basis $\\{\eta_{i}\\}_{i=0}^{N}$ to
the basis $\\{\tau_{i}\\}_{i=0}^{N}$.

###### Proof.

By Lemma 13.56. ∎

######  Lemma 13.58.

The matrix (13.34) represents $\Delta$ with respect to
$\\{\tau_{i}\\}_{i=0}^{N}$ and $\\{w_{i}\\}_{i=0}^{N}$ and
$\\{\eta_{i}\\}_{i=0}^{N}$. The matrix (13.35) represents $\Delta^{-1}$ with
respect to $\\{\tau_{i}\\}_{i=0}^{N}$ and $\\{w_{i}\\}_{i=0}^{N}$ and
$\\{\eta_{i}\\}_{i=0}^{N}$.

###### Proof.

By Proposition 13.13 and Lemma 13.39. ∎

######  Lemma 13.59.

The matrix (13.34) is upper triangular, with $(i,j)$-entry

$\displaystyle\eta_{j-i}(a_{0})\left[\begin{matrix}j\\\
i\end{matrix}\right]_{\vartheta}$ (13.36)

for $0\leq i\leq j\leq N$. The matrix (13.35) is upper triangular, with
$(i,j)$-entry

$\displaystyle\tau_{j-i}(b_{0})\left[\begin{matrix}j\\\
i\end{matrix}\right]_{\vartheta}$ (13.37)

for $0\leq i\leq j\leq N$. To express (13.36), (13.37) in terms of $a$, $b$,
$q$ use Lemmas 13.9, 13.11.

###### Proof.

By Lemma 13.57, the matrix (13.34) is the transition matrix from the basis
$\\{\tau_{i}\\}_{i=0}^{N}$ to the basis $\\{\eta_{i}\\}_{i=0}^{N}$. The
entries of this matrix are obtained from Proposition 5.2(ii). The entries of
the matrix (13.35) are similarly obtained. ∎

Our next goal is to show how the map $A$ from Definition 2.11 is related to
$\psi$, $K$, $B$, $M$.

######  Lemma 13.60.

On $V_{N-1}$,

$\displaystyle\frac{qKA-AK}{q-1}=a^{-1}K^{2}+aI,$ (13.38)
$\displaystyle\frac{qBA-AB}{q-1}=b^{-1}B^{2}+bI.$ (13.39)

###### Proof.

We first show (13.38). For $0\leq i\leq N-1$ apply each side of (13.38) to
$\tau_{i}$, and evaluate the result using Lemma 2.13 along with (13.1) and
Definition 13.23. We have

$\displaystyle
KA\tau_{i}=K(a_{i}\tau_{i}+\tau_{i+1})=q^{-i}a_{i}\tau_{i}+q^{-i-1}\tau_{i+1},$
$\displaystyle AK\tau_{i}=q^{-i}A\tau_{i}=q^{-i}(a_{i}\tau_{i}+\tau_{i+1}),$
$\displaystyle\big{(}a^{-1}K^{2}+aI\big{)}\tau_{i}=\big{(}a^{-1}q^{-2i}+a\big{)}\tau_{i}=q^{-i}a_{i}\tau_{i}.$

By these comments we obtain (13.38). Equation (13.39) is similarly obtained. ∎

######  Lemma 13.61.

On $V_{N-1}$,

$\displaystyle q\psi A-A\psi=\frac{(q+1)abM^{-1}-(q+ab)I}{ab-1}.$ (13.40)

###### Proof.

For $0\leq i\leq N-1$ apply each side of (13.40) to $\tau_{i}$. Using Lemma
2.13 and (4.9),

$\displaystyle\psi
A\tau_{i}=\psi(a_{i}\tau_{i}+\tau_{i+1})=a_{i}\vartheta_{i}\tau_{i-1}+\vartheta_{i+1}\tau_{i},$
$\displaystyle
A\psi\tau_{i}=\vartheta_{i}A\tau_{i-1}=\vartheta_{i}(a_{i-1}\tau_{i-1}+\tau_{i}).$

Using Lemma 13.6(i) and Lemma 13.43,

$\displaystyle
M^{-1}\tau_{i}=q^{i}\tau_{i}+(q-1)\big{(}a-b^{-1}\big{)}q^{i-1}\vartheta_{i}\tau_{i-1}.$

By the above comments and Lemmas 13.4, 13.7 we get the result. ∎

######  Lemma 13.62.

On $V_{N-1}$,

$\displaystyle\frac{qAM^{-1}-M^{-1}A}{q-1}=\big{(}a^{-1}+b^{-1}\big{)}I+\big{(}q-q^{-1}\big{)}\big{(}1-a^{-1}b^{-1}\big{)}\psi.$
(13.41)

###### Proof.

For $0\leq i\leq N-1$ apply each side of (13.41) to $\tau_{i}$. Evaluate the
result using (4.9) and Lemmas 2.13, 13.43 along with Lemmas 13.4, 13.7. ∎

######  Proposition 13.63.

On $V_{N-2}$,

$\displaystyle A^{2}\psi-\big{(}q+q^{-1}\big{)}A\psi A+\psi
A^{2}+\big{(}q-q^{-1}\big{)}^{2}\psi=\frac{(1-q)\big{(}1+q^{-1}ab\big{)}}{1-ab}A+\frac{\big{(}q-q^{-1}\big{)}(a+b)}{1-ab}I.$

###### Proof.

Let $X$ denote the expression on either side of (13.40). Compute

$\displaystyle\frac{qAX-XA}{q-1}$

and evaluate the result using Lemma 13.62. ∎

######  Proposition 13.64.

On $V_{N-1}$,

$\displaystyle\psi^{2}A-\big{(}q+q^{-1}\big{)}\psi
A\psi+A\psi^{2}=\frac{(1-q)\big{(}1+q^{-1}ab\big{)}}{1-ab}\psi.$

###### Proof.

Let $X$ denote the expression on either side of (13.40). Compute $qX\psi-\psi
X$ and evaluate the result using $qM^{-1}\psi=\psi M^{-1}$. ∎

######  Proposition 13.65.

On $V_{N-2}$,

$\displaystyle
A^{2}M^{-1}-\big{(}q+q^{-1}\big{)}AM^{-1}A+M^{-1}A^{2}+\big{(}q-q^{-1}\big{)}^{2}M^{-1}$
$\displaystyle\qquad{}=(q-1)\big{(}q-q^{-1}\big{)}\big{(}q^{-1}+a^{-1}b^{-1}\big{)}I-q^{-1}(q-1)^{2}\big{(}a^{-1}+b^{-1}\big{)}A.$

###### Proof.

Let $Y$ denote the expression on either side of (13.41). Compute $qYA-AY$ and
evaluate the result using Lemma 13.61. ∎

######  Proposition 13.66.

On $V_{N-1}$,

$\displaystyle
M^{-2}A-\big{(}q+q^{-1}\big{)}M^{-1}AM^{-1}+AM^{-2}=(q-1)\big{(}q^{-1}-1\big{)}\big{(}a^{-1}+b^{-1}\big{)}M^{-1}.$

###### Proof.

Let $Y$ denote the expression on either side of (13.41). Compute
$qM^{-1}Y-YM^{-1}$ and evaluate the result using $qM^{-1}\psi=\psi M^{-1}$. ∎

In Note 13.21 we mentioned a 3-term recurrence satisfied by the polynomials
$\\{w_{i}\\}_{i=0}^{N}$ and $\\{w^{\prime}_{i}\\}_{i=0}^{N}$. Our next goal is
to describe this recurrence.

######  Lemma 13.67.

We have

$\displaystyle
xw_{i}=w_{i+1}+q^{-i}\big{(}a^{-1}+b^{-1}\big{)}w_{i}+\big{(}1-q^{-i}\big{)}\big{(}1-q^{1-i}a^{-1}b^{-1}\big{)}w_{i-1}$

for $0\leq i\leq N-1$, where $w_{0}=1$ and $w_{-1}=0$.

###### Proof.

The result holds for $i=0$, since $w_{1}=x-a^{-1}-b^{-1}$ by Example
13.18(ii). Assume that $i\geq 1$. By Lemma 13.19(i) there exist scalars
$\\{\alpha_{k}\\}_{k=0}^{i+1}$ in $\mathbb{F}$ such that
$xw_{i}=\sum_{k=0}^{i+1}\alpha_{k}w_{k}$ and $\alpha_{k+1}=1$. To obtain
$\\{\alpha_{k}\\}_{k=0}^{i}$, apply each side of (13.41) to $w_{i}$ and
evaluate the result using Lemma 13.22 along with $M^{-1}w_{j}=q^{j}w_{j}$ for
$0\leq j\leq i$. After a brief calculation this yields
$\alpha_{i}=q^{-i}\big{(}a^{-1}+b^{-1}\big{)}$ and
$\alpha_{i-1}=\big{(}1-q^{-i}\big{)}\big{(}1-q^{1-i}a^{-1}b^{-1}\big{)}$ and
$\alpha_{k}=0$ for $0\leq k\leq i-2$. The result follows. ∎

######  Lemma 13.68.

We have

$\displaystyle
xw^{\prime}_{i}=w^{\prime}_{i+1}+q^{i}(a+b)w^{\prime}_{i}+\big{(}1-q^{i}\big{)}\big{(}1-q^{i-1}ab\big{)}w^{\prime}_{i-1}$

$0\leq i\leq N-1$, where $w^{\prime}_{0}=1$ and $w^{\prime}_{-1}=0$.

###### Proof.

In Proposition 13.67 replace $q\mapsto q^{-1}$, $a\mapsto a^{-1}$, $b\mapsto
b^{-1}$ and use Note 13.17. ∎

This completes our description of $\cal L$ and $\Delta$ for the twin recurrent
data from Case I of Lemma 10.2. In this description we encountered analogs of
the results from Section 1 about the double lowering operator of a tridiagonal
pair. This connection suggests that double lowering operators on polynomials
can be used to further develop the theory of tridiagonal pairs; we hope to
pursue this in the future.

### Acknowledgement

The author would like to thank Kazumasa Nomura for giving this paper a close
reading and offering many valuable comments.

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