Finite-Dimensional Irreducible Modules of the Racah Algebra at Characteristic
Zero

Finite-Dimensional Irreducible Modules
of the Racah Algebra at Characteristic Zero

Hau-Wen HUANG † and Sarah BOCKTING-CONRAD ‡

H.-W. Huang and S. Bockting-Conrad

† Department of Mathematics, National Central University, Chung-Li 32001,
Taiwan<EMAIL_ADDRESS>
‡ Department of Mathematical Sciences, DePaul University, Chicago, Illinois,
USA<EMAIL_ADDRESS>
Received November 12, 2019, in final form March 16, 2020; Published online
March 24, 2020

Assume that $\mathbb{F}$ is an algebraically closed field with characteristic
zero. The Racah algebra $\Re$ is the unital associative $\mathbb{F}$-algebra
defined by generators and relations in the following way. The generators are
$A$, $B$, $C$, $D$ and the relations assert that $[A,B]=[B,C]=[C,A]=2D$ and
that each of $[A,D]+AC-BA$, $[B,D]+BA-CB$, $[C,D]+CB-AC$ is central in $\Re$.
In this paper we discuss the finite-dimensional irreducible $\Re$-modules in
detail and classify them up to isomorphism. To do this, we apply an infinite-
dimensional $\Re$-module and its universal property. We additionally give the
necessary and sufficient conditions for $A$, $B$, $C$ to be diagonalizable on
finite-dimensional irreducible $\Re$-modules.

Racah algebra; quadratic algebra; irreducible modules; tridiagonal pairs;
universal property

81R10; 16S37

##  1 Introduction

Throughout this paper, we adopt the following conventions. Let $\mathbb{F}$
denote an algebraically closed field and let $\operatorname{char}\mathbb{F}$
denote the characteristic of $\mathbb{F}$. Let $\mathbb{Z}$ denote the set of
integers and let $\mathbb{N}$ denote the set of nonnegative integers. The
bracket $[\,,\,]$ stands for the commutator.

In this paper we consider the Racah algebra $\Re$ over $\mathbb{F}$ defined by
generators and relations as follows. The generators are $A$, $B$, $C$, $D$ and
the relations assert that

$\displaystyle[A,B]=[B,C]=[C,A]=2D$

and that each of

$\displaystyle[A,D]+AC-BA,\qquad[B,D]+BA-CB,\qquad[C,D]+CB-AC$

is central in $\Re$. The Racah algebra $\Re$ is a universal analog of the
original Racah algebras which first appeared in [17]. In that paper, the
original Racah algebras were used to establish a link between representation
theory and the quantum mechanical coupling of three angular momenta. Since
that time, the connections between the Racah algebras and many other areas
have been explored. We mention a few of them here. Their connections with the
additive double-affine Hecke algebra of type
$\big{(}C_{1}^{\vee},C_{1}\big{)}$, the Bannai–Ito algebra, and the Lie
algebras $\mathfrak{su}(2)$, $\mathfrak{su}(1,1)$ were investigated in [6, 9,
10, 14]. Their realizations via the Racah polynomials, the intermediate
Casimir operators, and the superintegrable models in two dimensions were
presented in [5, 7, 8, 9, 11]. For information concerning the higher rank
Racah algebras, see [3, 4].

We now mention an error in the literature on Racah algebras. In [2], the
authors considered the finite-dimensional irreducible modules of the original
Racah algebras when $\operatorname{char}\mathbb{F}=0$. In [2, Lemma 5.6], it
was claimed that the defining generators can be diagonalized on any finite-
dimensional irreducible module of the Racah algebras. This result was then
used in their classification of finite-dimensional irreducible modules of the
Racah algebras in [2, Section 6]. It turns out that [2, Lemma 5.6] is
conditional. We give the following example to help illustrate the issue
arising in [2].

######  Example 1.1.

It is routine to verify that there exists a five-dimensional $\Re$-module $V$
that has an $\mathbb{F}$-basis $\\{v_{i}\\}_{i=0}^{4}$ with respect to which
the matrices representing $A$, $B$, $C$, $D$ are $\frac{1}{4}$ times

$\displaystyle\begin{pmatrix}15&0&0&0&0\\\ 4&3&0&0&0\\\ 0&4&-1&0&0\\\
0&0&4&3&0\\\ 0&0&0&4&15\end{pmatrix},\qquad\begin{pmatrix}15&-36&0&0&0\\\
0&3&-6&0&0\\\ 0&0&-1&-6&0\\\ 0&0&0&3&-36\\\ 0&0&0&0&15\end{pmatrix},$
$\displaystyle\begin{pmatrix}-9&36&0&0&0\\\ -4&15&6&0&0\\\ 0&-4&23&6&0\\\
0&0&-4&15&36\\\ 0&0&0&-4&-9\end{pmatrix},\qquad\begin{pmatrix}18&-54&0&0&0\\\
6&-15&-3&0&0\\\ 0&2&0&3&0\\\ 0&0&-2&15&54\\\ 0&0&0&-6&-18\end{pmatrix},$

respectively. It follows that

$\displaystyle[A,D]+AC-BA=[B,D]+BA-CB=[C,D]+CB-AC=0$

on the $\Re$-module $V$. For each of $A$, $B$, $C$, it is straightforward to
verify that its minimal polynomial on $V$ is

$\displaystyle\left(x+\frac{1}{4}\right)\left(x-\frac{3}{4}\right)^{2}\left(x-\frac{15}{4}\right)^{2}.$

Therefore none of $A$, $B$, $C$ is diagonalizable on $V$. We now show that $V$
is in fact irreducible. Let $W$ denote a nonzero $\Re$-submodule of $V$. We
will show that $W=V$. Observe that the element $B$ has exactly three
eigenvalues on $V$, namely $\frac{15}{4}$, $\frac{3}{4}$, $-\frac{1}{4}$. A
direct calculation yields that the corresponding eigenspaces are each of
dimension $1$ and are spanned by

$\displaystyle v_{0},\qquad 3v_{0}+v_{1},\qquad 27v_{0}+12v_{1}+8v_{2},$ (1.1)

respectively. Since $W$ is nonzero, at least one of $\frac{15}{4}$,
$\frac{3}{4}$, $-\frac{1}{4}$ is an eigenvalue of $B$ on $W$. Therefore $W$
contains at least one of the elements listed in (1.1). Observe that the
$\Re$-module $V$ is generated by $v_{0}$. Thus, if $v_{0}\in W$ then $W=V$. If
$3v_{0}+v_{1}\in W$ then

$\displaystyle v_{0}=\frac{2}{9}\left(A-2D-\frac{9}{4}\right)(3v_{0}+v_{1})\in
W$

and hence $W=V$. If $27v_{0}+12v_{1}+8v_{2}\in W$ then

$\displaystyle
3v_{0}+v_{1}=-\frac{1}{18}\left(A+2D-\frac{11}{4}\right)(27v_{0}+12v_{1}+8v_{2})\in
W$

and hence $W=V$. Therefore $W=V$. Since the $\Re$-module $V$ is irreducible,
we now have a counterexample to [2, Lemma 5.6].

In light of the above example, we see that the finite-dimensional irreducible
$\Re$-modules are not yet completely classified. The goal of this paper is to
provide such a classification. The idea of our classification comes from [13].
We mention that a similar issue arises in the case of the Bannai–Ito algebra
$\mathfrak{BI}$ [12] which is addressed by the first author in [16]. The
result [15, Theorem 5.4] reveals that the Racah algebra $\Re$ is isomorphic to
an $\mathbb{F}$-subalgebra of $\mathfrak{BI}$. As an application of [16] and
this result, the lattices of $\Re$-submodules of finite-dimensional
irreducible $\mathfrak{BI}$-modules are classified in [14].

The outline of this paper is as follows. In Section 2 we state our
classification of finite-dimensional irreducible $\Re$-modules in Theorem 2.5.
In Section 3 we display an infinite-dimensional $\Re$-module and describe its
universal property. In Section 4 we give necessary and sufficient conditions
for the irreducibility of finite-dimensional $\Re$-modules. In Section 5 we
study the isomorphism classes of finite-dimensional irreducible $\Re$-modules.
In Section 6 we give our proof of Theorem 2.5.

##  2 Statement of results

In this section we more formally introduce the Racah algebra $\Re$ and state
the main result of the paper which gives a classification of the finite-
dimensional irreducible modules of the Racah algebra $\Re$. This main result
will be proved later in Section 6.

######  Definition 2.1 ([1, Definition 3.1]).

The Racah algebra $\Re$ is the unital associative $\mathbb{F}$-algebra defined
by generators and relations in the following way. The generators are $A$, $B$,
$C$, $D$. The relations state that

$\displaystyle[A,B]=[B,C]=[C,A]=2D$ (2.1)

and that each of

$\displaystyle[A,D]+AC-BA,\qquad[B,D]+BA-CB,\qquad[C,D]+CB-AC$

is central in $\Re$.

It follows from the above definition that the element $A+B+C$ is also central
in $\Re$. For notational convenience, we let

$\displaystyle\alpha=[A,D]+AC-BA,$ (2.2) $\displaystyle\beta=[B,D]+BA-CB,$
(2.3) $\displaystyle\gamma=[C,D]+CB-AC,$ (2.4) $\displaystyle\delta=A+B+C.$
(2.5)

######  Lemma 2.2.

  1. $(i)$

The Racah algebra $\Re$ is generated by the elements $A$, $B$, $C$.

  2. $(ii)$

The Racah algebra $\Re$ is generated by the elements $A$, $B$, $\delta$.

###### Proof.

(i) By (2.1) the element $D$ can be expressed in terms of $A$, $B$. Hence (i)
follows from Definition 2.1. (ii) By (2.5) the element $C$ can be expressed in
terms of $A$, $B$, $\delta$. Hence (ii) follows from (i). ∎

######  Lemma 2.3.

The $\mathbb{F}$-algebra $\Re$ has a presentation given by generators $A$,
$B$, $\alpha$, $\beta$, $\delta$ and relations

$\displaystyle A^{2}B-2ABA+BA^{2}-2AB-2BA=2A^{2}-2A\delta+2\alpha,$ (2.6)
$\displaystyle AB^{2}-2BAB+B^{2}A-2AB-2BA=2B^{2}-2B\delta-2\beta,$ (2.7)
$\displaystyle\alpha A=A\alpha,\qquad\beta A=A\beta,\qquad\delta A=A\delta,$
$\displaystyle\alpha B=B\alpha,\qquad\beta B=B\beta,\qquad\delta
B=B\delta,\qquad\alpha\delta=\delta\alpha,\qquad\beta\delta=\delta\beta.$

###### Proof.

We know from Lemma 2.2(ii) that $A$, $B$, $\alpha$, $\beta$, $\delta$ generate
$\Re$. Observe that $C=\delta-A-B$ by (2.5) and $D=\frac{1}{2}[A,B]$ by (2.1).
The result can now be obtained by either using these two facts to eliminate
$C$, $D$ from the presentation of $\Re$ given in Definition 2.1 or by using
$D=\frac{1}{2}[A,B]$ to eliminate $D$ from the presentation of $\Re$ given in
[1, Proposition 3.4]. ∎

In the following proposition, we assert the existence of certain finite-
dimensional $\Re$-modules and describe the actions of the generators of $\Re$
on these modules. A reader familiar with the theory of tridiagonal pairs will
immediately recognize the form of the matrices representing $A$ and $B$ as
precisely those given in Terwilliger’s 2001 seminal work on tridiagonal pairs
[18, Theorem 3.2].

######  Proposition 2.4.

For any $a,b,c\in\mathbb{F}$ and any $d\in\mathbb{N}$ there exists a
$(d+1)$-dimensional $\Re$-module $R_{d}(a,b,c)$ satisfying each of the
following conditions:

  1. $(i)$

There exists an $\mathbb{F}$-basis $\\{v_{i}\\}_{i=0}^{d}$ for $R_{d}(a,b,c)$
with respect to which the matrices representing $A$ and $B$ are

$\displaystyle\begin{pmatrix}\theta_{0}&&&&{\bf 0}\\\ 1&\theta_{1}\\\
&1&\theta_{2}\\\ &&\ddots&\ddots\\\ {\bf
0}&&&1&\theta_{d}\end{pmatrix},\qquad\begin{pmatrix}\theta_{0}^{*}&\varphi_{1}&&&{\bf
0}\\\ &\theta_{1}^{*}&\varphi_{2}\\\ &&\theta_{2}^{*}&\ddots\\\
&&&\ddots&\varphi_{d}\\\ {\bf 0}&&&&\theta_{d}^{*}\end{pmatrix},$

respectively, where

$\displaystyle\theta_{i}=\big{(}a+\textstyle\frac{d}{2}-i\big{)}\big{(}a+\textstyle\frac{d}{2}-i+1\big{)},\qquad
0\leq i\leq d,$
$\displaystyle\theta^{*}_{i}=\big{(}b+\textstyle\frac{d}{2}-i\big{)}\big{(}b+\textstyle\frac{d}{2}-i+1\big{)},\qquad
0\leq i\leq d,$
$\displaystyle\varphi_{i}=i(i-d-1)\big{(}a+b+c+\textstyle\frac{d}{2}-i+2\big{)}\big{(}a+b-c+\textstyle\frac{d}{2}-i+1\big{)},\qquad
1\leq i\leq d.$

  2. $(ii)$

The elements $\alpha$, $\beta$, $\delta$ act on $R_{d}(a,b,c)$ as scalar
multiplication by

$\displaystyle(c-b)(c+b+1)\big{(}a-\textstyle\frac{d}{2}\big{)}\big{(}a+\textstyle\frac{d}{2}+1\big{)},$
$\displaystyle(a-c)(a+c+1)\big{(}b-\textstyle\frac{d}{2}\big{)}\big{(}b+\textstyle\frac{d}{2}+1\big{)},$
$\displaystyle\textstyle\frac{d}{2}\big{(}\textstyle\frac{d}{2}+1\big{)}+a(a+1)+b(b+1)+c(c+1),$

respectively.

###### Proof.

Using Lemma 2.3, this result can be verified through routine, though tedious,
computations. ∎

In order to state our main result more succinctly, we will use the following
conventions and definitions. Let $d\in\mathbb{N}$ and let
$\mathbf{P}=\mathbf{P}_{d}$ denote the set of all $(a,b,c)\in\mathbb{F}^{3}$
that satisfy

$\displaystyle
a+b+c+1,-a+b+c,a-b+c,a+b-c\not\in\big{\\{}\tfrac{d}{2}-i\,\big{|}\,i=1,2,\ldots,d\big{\\}}.$

We define an action of the abelian group $\\{\pm 1\\}^{3}$ on $\mathbf{P}$ by

$\displaystyle(a,b,c)^{(-1,1,1)}=(-a-1,b,c),\qquad(a,b,c)^{(1,-1,1)}=(a,-b-1,c),$
$\displaystyle(a,b,c)^{(1,1,-1)}=(a,b,-c-1)$

for all $(a,b,c)\in\mathbf{P}$. We let $\mathbf{P}/\\{\pm 1\\}^{3}$ denote the
set of the $\\{\pm 1\\}^{3}$-orbits of $\mathbf{P}$. For
$(a,b,c)\in\mathbf{P}$, let $[a,b,c]$ denote the $\\{\pm 1\\}^{3}$-orbit of
$\mathbf{P}$ that contains $(a,b,c)$. We are now ready to state the
classification of finite-dimensional irreducible $\Re$-modules.

######  Theorem 2.5.

Assume that $\mathbb{F}$ is algebraically closed with
$\operatorname{char}\mathbb{F}=0$. Let $d$ denote a nonnegative integer. Let
$\mathbf{M}$ denote the set of all isomorphism classes of irreducible
$\Re$-modules that have dimension $d+1$. Then there exists a bijection
$\mathcal{R}\colon\mathbf{P}/\\{\pm 1\\}^{3}\to\mathbf{M}$ given by

$\displaystyle[a,b,c]\mapsto\hbox{the isomorphism class of $R_{d}(a,b,c)$}$

for all $[a,b,c]\in\mathbf{P}/\\{\pm 1\\}^{3}$.

We will give a proof of Theorem 2.5 in Section 6.

##  3 An infinite-dimensional $\boldsymbol{\Re}$-module and its universal
property

In this section we introduce an infinite-dimensional $\Re$-module and its
universal property in order to prove Theorem 2.5. For convenience the
following conventions are used throughout the rest of this paper. We let $a$,
$b$, $c$, $\nu$ denote any scalars in $\mathbb{F}$. We define the following
families of parameters associated with $a$, $b$, $c$, $\nu$:

$\displaystyle\theta_{i}=\big{(}a+\textstyle\frac{\nu}{2}-i\big{)}\big{(}a+\textstyle\frac{\nu}{2}-i+1\big{)}\qquad\hbox{for
all $i\in\mathbb{Z}$},$ (3.1)
$\displaystyle\theta^{*}_{i}=\big{(}b+\textstyle\frac{\nu}{2}-i\big{)}\big{(}b+\textstyle\frac{\nu}{2}-i+1\big{)}\qquad\hbox{for
all $i\in\mathbb{Z}$},$ (3.2)
$\displaystyle\phi_{i}=i(i-\nu-1)\big{(}a-b+c-\textstyle\frac{\nu}{2}+i\big{)}\big{(}a-b-c-\textstyle\frac{\nu}{2}+i-1\big{)}\qquad\hbox{for
all $i\in\mathbb{Z}$},$ (3.3)
$\displaystyle\varphi_{i}=i(i-\nu-1)\big{(}a+b+c+\textstyle\frac{\nu}{2}-i+2\big{)}\big{(}a+b-c+\textstyle\frac{\nu}{2}-i+1\big{)}\qquad\hbox{for
all $i\in\mathbb{Z}$},$ (3.4)
$\displaystyle\zeta=(c-b)(c+b+1)\big{(}a-\textstyle\frac{\nu}{2}\big{)}\big{(}a+\textstyle\frac{\nu}{2}+1\big{)},$
(3.5)
$\displaystyle\zeta^{*}=(a-c)(a+c+1)\big{(}b-\textstyle\frac{\nu}{2}\big{)}\big{(}b+\textstyle\frac{\nu}{2}+1\big{)},$
(3.6)
$\displaystyle\eta=\textstyle\frac{\nu}{2}\big{(}\textstyle\frac{\nu}{2}+1\big{)}+a(a+1)+b(b+1)+c(c+1).$
(3.7)

######  Proposition 3.1.

There exists an $\Re$-module $M_{\nu}(a,b,c)$ satisfying each of the following
conditions:

  1. (i)

There exists an $\mathbb{F}$-basis $\\{m_{i}\\}_{i=0}^{\infty}$ for
$M_{\nu}(a,b,c)$ with respect to which the matrices representing $A$ and $B$
are

$\displaystyle\begin{pmatrix}\theta_{0}&&&&&{\bf 0}\\\ 1&\theta_{1}\\\
&1&\theta_{2}\\\ &&\cdot&\cdot\\\ &&&\cdot&\cdot\\\ {\bf
0}&&&&\cdot&\cdot\end{pmatrix},\qquad\begin{pmatrix}\theta_{0}^{*}&\varphi_{1}&&&&{\bf
0}\\\ &\theta_{1}^{*}&\varphi_{2}\\\ &&\theta_{2}^{*}&\cdot\\\
&&&\cdot&\cdot\\\ &&&&\cdot&\cdot\\\ {\bf 0}&&&&&\cdot\end{pmatrix},$

respectively.

  2. (ii)

The elements $\alpha$, $\beta$, $\delta$ act on $M_{\nu}(a,b,c)$ as scalar
multiplication by $\zeta$, $\zeta^{*}$, $\eta$, respectively.

###### Proof.

Using Lemma 2.3, this result can be verified through routine computations. ∎

Throughout the rest of this paper we will let $\\{m_{i}\\}_{i=0}^{\infty}$
denote the $\mathbb{F}$-basis for $M_{\nu}(a,b,c)$ from Proposition 3.1(i).
The following result is an immediate consequence of Proposition 3.1(i).

######  Lemma 3.2.

$m_{j+1}=\prod\limits_{h=i}^{j}(A-\theta_{h})m_{i}$ for any $i,j\in\mathbb{N}$
with $i\leq j$.

Shortly we will describe the $\Re$-module $M_{\nu}(a,b,c)$ in an alternate
way. To aid us in this endeavor, we first recall a Poincaré–Birkhoff–Witt
basis for $\Re$.

######  Lemma 3.3 ([1, Theorem 5.1]).

The elements

$\displaystyle A^{i}D^{j}B^{k}\alpha^{r}\delta^{s}\beta^{t}\qquad\hbox{for all
$i,j,k,r,s,t\in\mathbb{N}$}$

form an $\mathbb{F}$-basis of $\Re$.

Let $I_{\nu}(a,b,c)$ denote the left ideal of $\Re$ generated by the elements

$\displaystyle B-\theta_{0}^{*},$ (3.8)
$\displaystyle(B-\theta_{1}^{*})(A-\theta_{0})-\varphi_{1},$ (3.9)
$\displaystyle\alpha-\zeta,\qquad\beta-\zeta^{*},\qquad\delta-\eta.$ (3.10)

We now consider certain cosets of $\Re/I_{\nu}(a,b,c)$.

######  Lemma 3.4.

For each $n\in\mathbb{N}$, each of the following holds:

  1. $(i)$

$BA^{n}+I_{\nu}(a,b,c)$ is an $\mathbb{F}$-linear combination of
$A^{i}+I_{\nu}(a,b,c)$ for all $0\leq i\leq n$,

  2. $(ii)$

$DA^{n}+I_{\nu}(a,b,c)$ is an $\mathbb{F}$-linear combination of
$A^{i}+I_{\nu}(a,b,c)$ for all $0\leq i\leq n+1$,

  3. $(iii)$

$D^{n}+I_{\nu}(a,b,c)$ is an $\mathbb{F}$-linear combination of
$A^{i}+I_{\nu}(a,b,c)$ for all $0\leq i\leq n$.

###### Proof.

(i) We proceed by induction on $n$. Since $I_{\nu}(a,b,c)$ contains the
element (3.8), the statement holds for $n=0$. Since $I_{\nu}(a,b,c)$ contains
both of the elements (3.8) and (3.9), the statement holds for $n=1$. Now
suppose $n\geq 2$. Right multiplying each side of (2.6) by $A^{n-2}$ yields
that

$\displaystyle
A^{2}BA^{n-2}-2ABA^{n-1}+BA^{n}-2ABA^{n-2}-2BA^{n-1}=2A^{n}-2A^{n-1}\delta+2A^{n-2}\alpha.$

Since $I_{\nu}(a,b,c)$ contains each of the elements listed in (3.10), it
follows that $BA^{n}$ is congruent to

$\displaystyle 2ABA^{n-1}+2ABA^{n-2}-A^{2}BA^{n-2}+2BA^{n-1}+2A^{n}-2\eta
A^{n-1}+2\zeta A^{n-2}$ (3.11)

modulo $I_{\nu}(a,b,c)$. By the inductive hypothesis, the element (3.11) is
congruent to an $\mathbb{F}$-linear combination of $A^{i}$, for all $0\leq
i\leq n$, modulo $I_{\nu}(a,b,c)$. Therefore (i) follows.

(ii) Observe that $DA^{n}=\frac{1}{2}\big{(}ABA^{n}-BA^{n+1}\big{)}$ by (2.1).
In light of this fact, the result now follows from Lemma 3.4(i).

(iii) We proceed by induction on $n$. The statement holds trivially for $n=0$.
Now suppose that $n\geq 1$. By the inductive hypothesis, $D^{n}=DD^{n-1}$ is
congruent to an $\mathbb{F}$-linear combination of

$\displaystyle DA^{i},\qquad 0\leq i\leq n-1,$ (3.12)

modulo $I_{\nu}(a,b,c)$. By Lemma 3.4(ii) each of the elements listed in
(3.12) is an $\mathbb{F}$-linear combination of $A^{k}$, for all $0\leq k\leq
n$, modulo $I_{\nu}(a,b,c)$. Therefore the result follows. ∎

######  Lemma 3.5.

The $\mathbb{F}$-vector space $\Re/I_{\nu}(a,b,c)$ is spanned by

$\displaystyle A^{i}+I_{\nu}(a,b,c)\qquad\hbox{for all $i\in\mathbb{N}$}.$

###### Proof.

By Lemma 3.3, the $\mathbb{F}$-vector space $\Re/I_{\nu}(a,b,c)$ is spanned by

$\displaystyle
A^{i}D^{j}B^{k}\alpha^{r}\delta^{s}\beta^{t}+I_{\nu}(a,b,c)\qquad\hbox{for all
$i,j,k,r,s,t\in\mathbb{N}$}.$ (3.13)

Since $I_{\nu}(a,b,c)$ contains the elements listed in (3.8) and (3.10), each
of the elements listed in (3.13) can be expressed as an $\mathbb{F}$-linear
combination of

$\displaystyle A^{i}D^{j}+I_{\nu}(a,b,c)\qquad\hbox{for all
$i,j\in\mathbb{N}$}.$

The result now follows from these facts along with Lemma 3.4(iii). ∎

We are now ready to give our second description of $M_{\nu}(a,b,c)$.

######  Theorem 3.6.

There exists a unique $\Re$-module homomorphism

$\displaystyle\Phi\colon\ \Re/I_{\nu}(a,b,c)\to M_{\nu}(a,b,c)$

that sends $1+I_{\nu}(a,b,c)$ to $m_{0}$. Moreover, $\Phi$ is an isomorphism.

###### Proof.

Consider the $\Re$-module homomorphism $\Psi\colon\Re\to M_{\nu}(a,b,c)$ that
sends $1$ to $m_{0}$. By Proposition 3.1(i), the elements (3.8) and (3.9) are
in the kernel of $\Psi$. By Proposition 3.1(ii), the elements listed in (3.10)
are also in the kernel of $\Psi$. Hence $I_{\nu}(a,b,c)$ is contained in the
kernel of $\Psi$. It follows that $\Psi$ induces an $\Re$-module homomorphism
$\Phi\colon\Re/I_{\nu}(a,b,c)\to M_{\nu}(a,b,c)$ that maps $1+I_{\nu}(a,b,c)$
to $m_{0}$. Observe that $\Phi$ is the unique $\Re$-module homomorphism with
the desired property since $\Re/I_{\nu}(a,b,c)$ is generated by
$1+I_{\nu}(a,b,c)$ as an $\Re$-module.

By Lemma 3.2 the homomorphism $\Phi$ sends

$\displaystyle\prod_{h=1}^{i-1}(A-\theta_{h})+I_{\nu}(a,b,c)$ (3.14)

to $m_{i}$ for all $i\in\mathbb{N}$. Since $\\{m_{i}\\}_{i=0}^{\infty}$ are
linearly independent, the cosets (3.14) are linearly independent. Combining
this with Lemma 3.5, we see that the cosets (3.14) are an $\mathbb{F}$-basis
for $\Re/I_{\nu}(a,b,c)$. Therefore $\Phi$ is an isomorphism. ∎

As a consequence of Theorem 3.6, the $\Re$-module $M_{\nu}(a,b,c)$ satisfies
the following universal property.

######  Proposition 3.7.

If $V$ is an $\Re$-module which has a vector $v\in V$ satisfying

$\displaystyle Bv=\theta_{0}^{*}v,$
$\displaystyle(B-\theta_{1}^{*})(A-\theta_{0})v=\varphi_{1}v,$
$\displaystyle\alpha v=\zeta v,\qquad\beta v=\zeta^{*}v,\qquad\delta v=\eta
v,$

then there exists a unique $\Re$-module homomorphism $M_{\nu}(a,b,c)\to V$
that sends $m_{0}$ to $v$.

For the rest of the present paper, we will consider the case $\nu=d$. Define
$N_{d}(a,b,c)$ to be the $A$-cyclic $\mathbb{F}$-subspace of $M_{d}(a,b,c)$
generated by the element $m_{d+1}$.

######  Lemma 3.8.

$N_{d}(a,b,c)$ is an $\Re$-submodule of $M_{d}(a,b,c)$ with the
$\mathbb{F}$-basis $\\{m_{i}\\}_{i=d+1}^{\infty}$.

###### Proof.

Recall from Lemma 3.2 that

$\displaystyle(A-\theta_{i})m_{i}=m_{i+1}\qquad\hbox{for all $i\geq d+1$}.$

It follows from this fact that $\\{m_{i}\\}_{i=d+1}^{\infty}$ is an
$\mathbb{F}$-basis for $N_{d}(a,b,c)$.

We now show that $N_{d}(a,b,c)$ is an $\Re$-submodule of $M_{d}(a,b,c)$. By
Proposition 3.1(i),

$\displaystyle(B-\theta_{i}^{*})m_{i}=\varphi_{i}m_{i-1}\qquad\hbox{for all
$i\geq d+1$}.$

By (3.4), the scalar $\varphi_{d+1}=0$ when $\nu=d$. Hence $N_{d}(a,b,c)$ is
$B$-invariant. By Proposition 3.1(ii), the element $\delta$ acts on
$N_{d}(a,b,c)$ as scalar multiplication by $\eta$. It now follows from Lemma
2.2(ii) that $N_{d}(a,b,c)$ is an $\Re$-submodule of $M_{d}(a,b,c)$. ∎

Recall the $\Re$-module $R_{d}(a,b,c)$ from Proposition 2.4. In the sequel we
display how the $\Re$-module $R_{d}(a,b,c)$ is connected to $M_{\nu}(a,b,c)$.
For convenience we let $\\{v_{i}\\}_{i=0}^{d}$ denote the $\mathbb{F}$-basis
for $R_{d}(a,b,c)$ from Proposition 2.4(i) in the rest of this paper.

######  Lemma 3.9.

There exists a unique $\Re$-module isomorphism

$\displaystyle M_{d}(a,b,c)/N_{d}(a,b,c)\to R_{d}(a,b,c)$

that sends $m_{i}+N_{d}(a,b,c)$ to $v_{i}$ for all $0\leq i\leq d$.

###### Proof.

By Lemma 3.8, $M_{d}(a,b,c)/N_{d}(a,b,c)$ is a $(d+1)$-dimensional
$\Re$-module with the $\mathbb{F}$-basis

$\displaystyle\\{m_{i}+N_{d}(a,b,c)\\}_{i=0}^{d}.$ (3.15)

Observe that the matrices representing $A$ and $B$ with respect to the
$\mathbb{F}$-basis $\\{v_{i}\\}_{i=0}^{d}$ for $R_{d}(a,b,c)$ are identical
with the matrices representing $A$ and $B$ with respect to the
$\mathbb{F}$-basis (3.15) for $M_{d}(a,b,c)/N_{d}(a,b,c)$ by Propositions
2.4(i) and 3.1(i). By Propositions 2.4(ii) and 3.1(ii), the actions of
$\delta$ on $R_{d}(a,b,c)$ and $M_{d}(a,b,c)/N_{d}(a,b,c)$ are scalar
multiplication by the same scalar $\eta$. In light of these comments, the
result now follows from Lemma 2.2(ii). ∎

######  Proposition 3.10.

Suppose that $V$ is an $\Re$-module which has a vector $v\in V$ satisfying

$\displaystyle\prod_{i=0}^{d}(A-\theta_{i})v=0.$ (3.16)

If there is an $\Re$-module homomorphism $M_{d}(a,b,c)\to V$ that sends
$m_{0}$ to $v$, then there exists an $\Re$-module homomorphism
$R_{d}(a,b,c)\to V$ that sends $v_{0}$ to $v$.

###### Proof.

Let $\varrho$ denote the $\Re$-module homomorphism $M_{d}(a,b,c)\to V$ that
sends $m_{0}$ to $v$. By Lemma 3.2, we have

$\displaystyle m_{d+1}=\prod_{i=0}^{d}(A-\theta_{i})m_{0}.$

Combining this with (3.16), we see that $m_{d+1}$ is in the kernel of
$\varrho$. Therefore $N_{d}(a,b,c)$ is contained in the kernel of $\varrho$.
By Lemma 3.8, there exists an $\Re$-module homomorphism
$M_{d}(a,b,c)/N_{d}(a,b,c)\to V$ that sends $m_{0}+N_{d}(a,b,c)$ to $v$. The
result follows from this fact along with Lemma 3.9. ∎

##  4 Conditions for the irreducibility of $\boldsymbol{R_{d}(a,b,c)}$

In this section, we derive the necessary and sufficient conditions for
$R_{d}(a,b,c)$ to be irreducible in terms of the parameters $a$, $b$, $c$,
$d$. Throughout this section, we let

$\displaystyle w_{i}=\prod_{h=0}^{i-1}(A-\theta_{d-h})v_{0},\qquad 0\leq i\leq
d.$ (4.1)

######  Lemma 4.1.

If the $\Re$-module $R_{d}(a,b,c)$ is irreducible, then each of the following
holds:

  1. $(i)$

$\operatorname{char}\mathbb{F}=0$ or $\operatorname{char}\mathbb{F}>d$,

  2. $(ii)$

$a+b+c+1,a+b-c\not\in\big{\\{}{\frac{d}{2}-i}\,\big{|}\,i=1,2,\ldots,d\big{\\}}$.

###### Proof.

Suppose that there is an integer $i$, with $1\leq i\leq d$, such that
$\varphi_{i}=0$. By Proposition 2.4, the $\mathbb{F}$-subspace $W$ of
$R_{d}(a,b,c)$ spanned by $\\{v_{h}\\}_{h=i}^{d}$ is invariant under $A$, $B$,
$\delta$. It follows from Lemma 2.2(ii) that $W$ is an $\Re$-submodule of
$R_{d}(a,b,c)$, a contradiction to the irreducibility of $R_{d}(a,b,c)$.
Therefore $\varphi_{i}\not=0$ for all $1\leq i\leq d$, which is equivalent to
(i) and (ii) by (3.4). ∎

######  Lemma 4.2.

The elements $\\{w_{i}\\}_{i=0}^{d}$ form an $\mathbb{F}$-basis for
$R_{d}(a,b,c)$.

###### Proof.

It follows from Proposition 2.4(i) that

$\displaystyle v_{i}=\prod_{h=0}^{i-1}(A-\theta_{h})v_{0},\qquad 0\leq i\leq
d.$

Comparing this with (4.1), the result now follows. ∎

######  Proposition 4.3.

The $\Re$-module $R_{d}(a,b,c)$ is isomorphic to the $\Re$-module
$R_{d}(-a-1,b,c)$. Moreover, the matrices representing $A$ and $B$ with
respect to the $\mathbb{F}$-basis $\\{w_{i}\\}_{i=0}^{d}$ for $R_{d}(a,b,c)$
are

$\displaystyle\begin{pmatrix}\theta_{d}&&&&{\bf 0}\\\ 1&\theta_{d-1}\\\
&1&\theta_{d-2}\\\ &&\ddots&\ddots\\\ {\bf
0}&&&1&\theta_{0}\end{pmatrix},\qquad\begin{pmatrix}\theta_{0}^{*}&\phi_{1}&&&{\bf
0}\\\ &\theta_{1}^{*}&\phi_{2}\\\ &&\theta_{2}^{*}&\ddots\\\
&&&\ddots&\phi_{d}\\\ {\bf 0}&&&&\theta_{d}^{*}\end{pmatrix},$ (4.2)

respectively.

###### Proof.

By Proposition 2.4(i), there exists an $\mathbb{F}$-basis
$\\{u_{i}\\}_{i=0}^{d}$ for $R_{d}(-a-1,b,c)$ with respect to which the
matrices representing $A$ and $B$ are equal to the matrices displayed in
(4.2). By Lemma 4.2, it suffices to show that there is an $\Re$-module
homomorphism $R_{d}(a,b,c)\to R_{d}(-a-1,\allowbreak b,c)$ that sends $w_{i}$
to $u_{i}$ for all $0\leq i\leq d$.

Observe that $Bu_{0}=\theta_{0}^{*}u_{0}$ and a direct calculation yields that

$\displaystyle(B-\theta_{1}^{*})(A-\theta_{0})u_{0}=\varphi_{1}u_{0}.$

By Proposition 2.4(ii), the elements $\alpha$, $\beta$, $\delta$ act on
$R_{d}(-a-1,b,c)$ as scalar multiplication by $\zeta$, $\zeta^{*}$, $\eta$,
respectively. According to Proposition 3.7, there exists a unique $\Re$-module
homomorphism $M_{d}(a,b,c)\to R_{d}(-a-1,b,c)$ that sends $m_{0}$ to $u_{0}$.
By inspecting the matrix representing $A$ given in (4.2) we see that

$\displaystyle\prod_{i=0}^{d}(A-\theta_{i})u_{0}=0.$

Hence there exists a $\Re$-module homomorphism

$\displaystyle R_{d}(a,b,c)\to R_{d}(-a-1,b,c)$

that maps $v_{0}$ to $u_{0}$ by Proposition 3.10. It now follows from (4.1)
that this homomorphism sends $w_{i}$ to $u_{i}$ for all $0\leq i\leq d$. The
result follows. ∎

######  Lemma 4.4.

If the $\Re$-module $R_{d}(a,b,c)$ is irreducible, then each of the following
holds:

  1. (i)

$\operatorname{char}\mathbb{F}=0$ or $\operatorname{char}\mathbb{F}>d$,

  2. (ii)

$a+b+c+1,-a+b+c,a-b+c,a+b-c\not\in\big{\\{}\frac{d}{2}-i\,\big{|}\,i=1,2,\ldots,d\big{\\}}$.

###### Proof.

By Proposition 4.3, the $\Re$-module $R_{d}(a,b,c)$ is isomorphic to
$R_{d}(-a-1,b,c)$. Hence the result follows by applying Lemma 4.1 to both
$R_{d}(a,b,c)$ and $R_{d}(-a-1,b,c)$. ∎

Shortly we will show that the converse of Lemma 4.4 is also true. To aid us in
doing so, we establish the following notation. We define

$\displaystyle R=\prod_{h=1}^{d}(B-\theta^{*}_{h}),$ $\displaystyle
S_{i}=\prod_{h=1}^{d-i}(A-\theta_{d-h+1}),\qquad 0\leq i\leq d.$

It follows from Proposition 2.4(i) that $Rv$ is a scalar multiple of $v_{0}$
for all $v\in R_{d}(a,b,c)$. Thus, for any integers $i$, $j$ with $0\leq
i,j\leq d$, there exists a unique $L_{ij}\in\mathbb{F}$ such that

$\displaystyle RS_{i}v_{j}=L_{ij}v_{0}.$ (4.3)

By examining Proposition 2.4(i) further, we see that

$\displaystyle L_{ij}=0,\qquad 0\leq i<j\leq d,$ (4.4) $\displaystyle
L_{ij}=(\theta_{i}-\theta_{j-1})L_{i,j-1}+L_{i-1,j-1},\qquad 1\leq j\leq i\leq
d.$ (4.5)

It follows from Proposition 4.3 that

$\displaystyle
L_{i0}=\prod_{h=1}^{i}(\theta_{0}^{*}-\theta_{d-h+1}^{*})\prod_{h=1}^{d-i}\phi_{h},\qquad
0\leq i\leq d.$ (4.6)

Solving the recurrence relation (4.5) with the initial conditions (4.4) and
(4.6) yields that

$\displaystyle L_{ij}={d-i+j\choose j}{i\choose j}{d\choose
j}^{-1}\prod_{h=1}^{i-j}(\theta_{0}^{*}-\theta_{d-h+1}^{*})\prod_{h=1}^{d-i}\phi_{h}\prod_{h=1}^{j}\varphi_{h},\qquad
0\leq j\leq i\leq d.$ (4.7)

######  Theorem 4.5.

The $\Re$-module $R_{d}(a,b,c)$ is irreducible if and only if both of the
following conditions hold:

  1. $(i)$

$\operatorname{char}\mathbb{F}=0$ or $\operatorname{char}\mathbb{F}>d$,

  2. $(ii)$

$a+b+c+1,-a+b+c,a-b+c,a+b-c\not\in\big{\\{}\frac{d}{2}-i\,\big{|}\,i=1,2,\ldots,d\big{\\}}$.

###### Proof.

$(\Rightarrow)$ This is immediate from Lemma 4.4.

$(\Leftarrow)$ To see the irreducibility of $R_{d}(a,b,c)$, we assume that $W$
is a nonzero $\Re$-submodule of $R_{d}(a,b,c)$ and show that $W=R_{d}(a,b,c)$.
Pick a nonzero vector $w\in W$. Since $W$ is invariant under $A$ and $B$, it
follows that

$\displaystyle RS_{i}w\in W,\qquad 0\leq i\leq d.$ (4.8)

Since $\\{v_{i}\\}_{i=0}^{d}$ is an $\mathbb{F}$-basis for $R_{d}(a,b,c)$,
there exist $a_{j}\in\mathbb{F}$, for $0\leq j\leq d$, such that

$\displaystyle w=\sum_{j=0}^{d}a_{j}v_{j}.$

It now follows from (4.3) that

$\displaystyle RS_{i}w=\left(\sum_{j=0}^{d}L_{ij}a_{j}\right)v_{0},\qquad
0\leq i\leq d.$ (4.9)

Recall the parameters $\\{\phi_{i}\\}_{i\in\mathbb{Z}}$ and
$\\{\varphi_{i}\\}_{i\in\mathbb{Z}}$ from (3.3) and (3.4), respectively. It
follows from our assumptions (i) and (ii) that the scalars $\varphi_{i}\not=0$
and $\phi_{i}\not=0$ for all $1\leq i\leq d$. Let $L$ denote the
$(d+1)\times(d+1)$ matrix, indexed by $0,1,\ldots,d$, with $(i,j)$-entry given
by $L_{ij}$ for all $0\leq i,j\leq d$. By (4.4), the square matrix $L$ is
lower triangular. By (4.7), the diagonal entries of $L$ are

$\displaystyle
L_{ii}=\prod_{h=1}^{d-i}\phi_{h}\prod_{h=1}^{i}\varphi_{i},\qquad 0\leq i\leq
d,$

which we know to be nonzero. Therefore the matrix $L$ is nonsingular. Since
$w$ is nonzero at least one of $\\{a_{j}\\}_{j=0}^{d}$ is nonzero. Hence there
exists an integer $i$ with $0\leq i\leq d$ such that

$\sum_{j=0}^{d}L_{ij}a_{j}\not=0.$ (4.10)

Combining (4.10) with (4.8) and (4.9), we find that $v_{0}\in W$. Since the
$\Re$-module $R_{d}(a,b,c)$ is generated by $v_{0}$, it follows that
$W=R_{d}(a,b,c)$ and so $R_{d}(a,b,c)$ is irreducible. ∎

##  5 The isomorphism class of the $\boldsymbol{\Re}$-module
$\boldsymbol{R_{d}(a,b,c)}$

In Proposition 4.3, we showed that the $\Re$-module $R_{d}(a,b,c)$ is
isomorphic to the $\Re$-module $R_{d}(-a-1,b,c)$. In this section, we discuss
the isomorphism class of $R_{d}(a,b,c)$ in further detail.

######  Proposition 5.1.

The $\Re$-module $R_{d}(a,b,c)$ is isomorphic to the $\Re$-module
$R_{d}(a,b,-c-1)$.

###### Proof.

By Proposition 2.4(i), there are $\mathbb{F}$-bases for $R_{d}(a,b,c)$ and
$R_{d}(a,b,-c-1)$ with respect to which the matrices representing $A$ and $B$
are the same. By Proposition 2.4(ii), the actions of $\delta$ on
$R_{d}(a,b,c)$ and $R_{d}(a,b,-c-1)$ are both scalar multiplication by the
same scalar $\eta$. Hence $R_{d}(a,b,c)$ is isomorphic to $R_{d}(a,b,-c-1)$ by
Lemma 2.2(ii). ∎

######  Proposition 5.2.

If the $\Re$-module $R_{d}(a,b,c)$ is irreducible, then $R_{d}(a,b,c)$ is
isomorphic to the $\Re$-module $R_{d}(a,-b-1,c)$.

###### Proof.

By Proposition 2.4(i), there is an $\mathbb{F}$-basis $\\{u_{i}\\}_{i=0}^{d}$
for $R_{d}(a,-b-1,c)$ with respect to which the matrices representing $A$ and
$B$ are

$\displaystyle\begin{pmatrix}\theta_{0}&&&&{\bf 0}\\\ 1&\theta_{1}\\\
&1&\theta_{2}\\\ &&\ddots&\ddots\\\ {\bf
0}&&&1&\theta_{d}\end{pmatrix},\qquad\begin{pmatrix}\theta_{d}^{*}&\phi_{d}&&&{\bf
0}\\\ &\theta_{d-1}^{*}&\phi_{d-1}\\\ &&\theta_{d-2}^{*}&\ddots\\\
&&&\ddots&\phi_{1}\\\ {\bf 0}&&&&\theta_{0}^{*}\end{pmatrix},$ (5.1)

respectively. Since the $\Re$-module $R_{d}(a,b,c)$ is irreducible, it follows
from Theorem 4.5 that $\phi_{i}\not=0$ for all $1\leq i\leq d$. Thus we may
set

$\displaystyle
v=\sum_{i=0}^{d}\prod_{h=1}^{i}\frac{\theta_{0}^{*}-\theta_{d-h+1}^{*}}{\phi_{d-h+1}}u_{i}.$

A direct calculation yields that $Bv=\theta_{0}^{*}v$ and

$\displaystyle(B-\theta_{1}^{*})(A-\theta_{0})v=\varphi_{1}v.$

By Proposition 2.4(ii), the elements $\alpha$, $\beta$, $\delta$ act on
$R_{d}(a,-b-1,c)$ as scalar multiplication by $\zeta$, $\zeta^{*}$, $\eta$,
respectively. According to Proposition 3.7, there exists a unique $\Re$-module
homomorphism $M_{d}(a,b,c)\to R_{d}(a,-b-1,c)$ that maps $m_{0}$ to $v$. By
inspecting the matrix representing $A$ given in (5.1), we see that

$\displaystyle\prod_{i=0}^{d}(A-\theta_{i})v=0.$

Hence there exists an $\Re$-module homomorphism

$\displaystyle R_{d}(a,b,c)\to R_{d}(a,-b-1,c)$ (5.2)

that sends $v_{0}$ to $v$ by Proposition 3.10. Since the $\Re$-module
$R_{d}(a,b,c)$ is irreducible, the $\Re$-module $R_{d}(a,-b-1,c)$ is also
irreducible by Theorem 4.5. Therefore (5.2) is an isomorphism. ∎

We end this section with a simple combination of Propositions 4.3, 5.1, and
5.2.

######  Theorem 5.3.

If the $\Re$-module $R_{d}(a,b,c)$ is irreducible, then $R_{d}(a,b,c)$ is
isomorphic to each of the $\Re$-modules $R_{d}(-a-1,b,c)$, $R_{d}(a,-b-1,c)$
and $R_{d}(a,b,-c-1)$.

##  6 The proof of Theorem 2.5

Theorems 4.5 and 5.3 indicate that the map $\mathcal{R}$ in Theorem 2.5 is
well-defined. In this section, we shall show that $\mathcal{R}$ is a
bijection.

######  Lemma 6.1.

Assume that $\mathbb{F}$ is algebraically closed. If $V$ is a finite-
dimensional irreducible $\Re$-module, then each central element of $\Re$ acts
on $V$ as scalar multiplication.

###### Proof.

This result follows from applying Schur’s lemma to $\Re$. ∎

######  Lemma 6.2.

For any $i\in\mathbb{Z}$, each of the following hold:

  1. $(i)$

$\theta_{i+1}+\theta_{i-1}=2(\theta_{i}+1)$,

  2. $(ii)$

$\theta_{i+1}\theta_{i-1}=\theta_{i}(\theta_{i}-2)$.

###### Proof.

The result can be routinely verified using (3.1). ∎

######  Theorem 6.3.

Assume that $\mathbb{F}$ is algebraically closed with
$\operatorname{char}\mathbb{F}=0$. Let $d$ denote a nonnegative integer. If
$V$ is a $(d+1)$-dimensional irreducible $\Re$-module, then there exist
$a,b,c\in\mathbb{F}$ such that the $\Re$-module $R_{d}(a,b,c)$ is isomorphic
to $V$.

###### Proof.

Given any scalar $\kappa\in\mathbb{F}$, we define

$\displaystyle\vartheta_{i}(\kappa)=(\kappa-i)(\kappa-i+1)\qquad\hbox{for all
$i\in\mathbb{Z}$}.$

Since $\operatorname{char}\mathbb{F}=0$, for any distinct integers $i,j$, the
scalars $\vartheta_{i}(\kappa)$ and $\vartheta_{j}(\kappa)$ are equal if and
only if $i+j=2\kappa+1$. In particular
$\\{\vartheta_{i}(\kappa)\\}_{i=0}^{-\infty}$ contains infinitely many values.

Since $\mathbb{F}$ is algebraically closed, we may choose a scalar
$\kappa\in\mathbb{F}$ such that $\vartheta_{0}(\kappa)$ is an eigenvalue of
$A$ on $V$. Since $V$ is of dimension $d+1$, there are at most $d+1$ distinct
eigenvalues of $A$ on $V$. Thus, there exists an integer $j\leq 0$ such that
$\vartheta_{j}(\kappa)$ is an eigenvalue of $A$ but $\vartheta_{j-1}(\kappa)$
is not an eigenvalue of $A$ on $V$. Set

$\displaystyle a=\kappa-j-\tfrac{d}{2}.$

Similarly, there exists a scalar $\lambda\in\mathbb{F}$ and an integer $k\leq
0$ such that $\vartheta_{k}(\lambda)$ is an eigenvalue of $B$ but
$\vartheta_{k-1}(\lambda)$ is not an eigenvalue of $B$ in $V$. We set

$\displaystyle b=\lambda-k-\tfrac{d}{2}.$

Observe that under these settings, we have

$\displaystyle\theta_{i}=\vartheta_{i+j}(\kappa)\qquad\hbox{for all
$i\in\mathbb{Z}$},$ (6.1)
$\displaystyle\theta_{i}^{*}=\vartheta_{i+k}(\lambda)\qquad\hbox{for all
$i\in\mathbb{Z}$}.$ (6.2)

By Lemma 6.1, the element $\delta$ acts on $V$ as scalar multiplication. Since
$\mathbb{F}$ is algebraically closed, there exists a scalar $c\in\mathbb{F}$
such that the action of $\delta$ on $V$ is the scalar multiplication by

$\displaystyle\eta=\textstyle\frac{d}{2}\big{(}\frac{d}{2}+1\big{)}+a(a+1)+b(b+1)+c(c+1).$

To prove the theorem, it now suffices to show that there exists an
$\Re$-module isomorphism from $R_{d}(a,b,c)$ into $V$.

Given any $T\in\Re$ and $\theta\in\mathbb{F}$, we let

$\displaystyle V_{T}(\theta)=\\{v\in V\,|\,Tv=\theta v\\}.$

Pick any $v\in V_{A}(\theta_{0})$. Applying each side of (2.6) to $v$ and
using Lemma 6.2 to simplify the result, we obtain that

$\displaystyle(A-\theta_{-1})(A-\theta_{1})Bv=2(\theta_{0}(\theta_{0}-\eta)+\alpha)v.$
(6.3)

Left multiplying each side of (6.3) by $(A-\theta_{0})$, we obtain that

$\displaystyle(A-\theta_{-1})(A-\theta_{0})(A-\theta_{1})Bv=0.$

By (6.1), the scalar $\theta_{-1}$ is not an eigenvalue of $A$ in $V$. Hence

$\displaystyle(A-\theta_{0})(A-\theta_{1})Bv=0.$

In other words $(A-\theta_{1})Bv\in V_{A}(\theta_{0})$ and therefore
$V_{A}(\theta_{0})$ is invariant under $(A-\theta_{1})B$. Since $\mathbb{F}$
is algebraically closed, there exists an eigenvector $u$ of $(A-\theta_{1})B$
in $V_{A}(\theta_{0})$. Similarly, there exists an eigenvector $w$ of
$(B-\theta_{1}^{*})A$ in $V_{B}(\theta_{0}^{*})$. Define

$\displaystyle u_{i}=\prod_{h=0}^{i-1}(B-\theta_{h}^{*})u\qquad\hbox{for all
$i\in\mathbb{N}$},$ (6.4) $\displaystyle
w_{i}=\prod_{h=0}^{i-1}(A-\theta_{h})w\qquad\hbox{for all $i\in\mathbb{N}$}.$
(6.5)

We now proceed by induction to show that

$\displaystyle(A-\theta_{i})u_{i}\in\operatorname{span}_{\mathbb{F}}\\{u_{0},u_{1},\ldots,u_{i-1}\\}\qquad\hbox{for
all $i\in\mathbb{N}$}.$ (6.6)

Since $u$ is an eigenvector of $(A-\theta_{1})B$ in $V_{A}(\theta_{0})$, the
claim is true for $i=0,1$. Now suppose that $i\geq 2$. Applying each side of
(2.7) to $u_{i-2}$, we obtain that

$\displaystyle\big{(}AB^{2}-2BAB+B^{2}A-2AB-2BA-2B^{2}+2\eta
B\big{)}u_{i-2}=-2\beta u_{i-2}.$ (6.7)

By Lemma 6.1, the right-hand side of (6.7) is a scalar multiple of $u_{i-2}$.
Using the inductive hypothesis, (6.4), and Lemma 6.2(i), we find that the
left-hand side of (6.7) is equal to

$\displaystyle(A-\theta_{i})u_{i}$ (6.8)

plus an $\mathbb{F}$-linear combination of $u_{0},u_{1},\ldots,u_{i-1}$.
Combining the above results, the claim (6.6) follows.

Next, we show that $\\{u_{i}\\}_{i=0}^{d}$ is an $\mathbb{F}$-basis for $V$.
Suppose on the contrary that there is an integer $h$ with $0\leq h\leq d-1$
such that $u_{h+1}$ is an $\mathbb{F}$-linear combination of
$u_{0},u_{1},\ldots,u_{h}$. Let $W$ denote the $\mathbb{F}$-subspace of $V$
spanned by $u_{0},u_{1},\ldots,u_{h}$. Since $W$ is $B$-invariant by (6.4) and
$A$-invariant by (6.6), it follows that $W$ is an $\Re$-submodule of $V$ by
Lemma 2.2(ii). Since $V$ is irreducible, this forces that $W=V$. By
construction, $W$ is of dimension at most $d$, a contradiction. Therefore
$\\{u_{i}\\}_{i=0}^{d}$ is an $\mathbb{F}$-basis for $V$. By a similar
argument, it follows that

$\displaystyle(B-\theta_{i}^{*})w_{i}\in\operatorname{span}_{\mathbb{F}}\\{w_{0},w_{1},\ldots,w_{i-1}\\}\qquad\hbox{for
all $i\in\mathbb{N}$}$ (6.9)

and $\\{w_{i}\\}_{i=0}^{d}$ is an $\mathbb{F}$-basis for $V$.

By (6.6), the matrix representing $A$ with respect to the $\mathbb{F}$-basis
$\\{u_{i}\\}_{i=0}^{d}$ is upper triangular with diagonal entries
$\\{\theta_{i}\\}_{i=0}^{d}$. By the Cayley–Hamilton theorem, we have

$\displaystyle\prod\limits_{i=0}^{d}(A-\theta_{i})w_{0}=0.$ (6.10)

In other words $Aw_{d}=\theta_{d}w_{d}$ by (6.5). Hence the matrix
representing $A$ with respect to the $\mathbb{F}$-basis
$\\{w_{i}\\}_{i=0}^{d}$ is

$\displaystyle\begin{pmatrix}\theta_{0}&&&&{\bf 0}\\\ 1&\theta_{1}\\\
&1&\theta_{2}\\\ &&\ddots&\ddots\\\ {\bf 0}&&&1&\theta_{d}\end{pmatrix}.$

By (6.9), the matrix representing $B$ with respect to the $\mathbb{F}$-basis
$\\{w_{i}\\}_{i=0}^{d}$ is upper triangular with diagonal entries
$\\{\theta_{i}^{*}\\}_{i=0}^{d}$. We let
$\\{\varphi_{i}^{\prime}\\}_{i=1}^{d}$ denote its superdiagonal entries as
follows

$\displaystyle\begin{pmatrix}\theta_{0}^{*}&\varphi_{1}^{\prime}&&&{\bf*}\\\
&\theta_{1}^{*}&\varphi_{2}^{\prime}\\\ &&\theta_{2}^{*}&\ddots\\\
&&&\ddots&\varphi_{d}^{\prime}\\\ {\bf 0}&&&&\theta_{d}^{*}\end{pmatrix}.$
(6.11)

Applying each side of (2.6) to $w_{i-1}$, we obtain from the coefficients of
$w_{i}$ that

$\displaystyle\varphi_{i+1}^{\prime}-2\varphi_{i}^{\prime}+\varphi_{i-1}^{\prime}=(\theta_{i}-\theta_{i-1}+2)\theta_{i}^{*}-(\theta_{i}-\theta_{i-1}-2)\theta_{i-1}^{*}+2(\theta_{i}+\theta_{i-1}-\eta)$
(6.12)

for all $1\leq i\leq d$, where $\varphi_{0}^{\prime}$ and
$\varphi_{d+1}^{\prime}$ are interpreted as zero. It is straightforward to
verify that $\\{\varphi_{i}\\}_{i=1}^{d}$ also satisfy the recurrence relation
(6.12). Since $\operatorname{char}\mathbb{F}=0$, the corresponding homogeneous
recurrence relation

$\displaystyle\sigma_{i+1}-2\sigma_{i}+\sigma_{i-1}=0,\qquad 1\leq i\leq d,$

with the initial values $\sigma_{0}=0$ and $\sigma_{d+1}=0$, has the unique
solution $\sigma_{i}=0$ for all $0\leq i\leq d+1$. Therefore
$\varphi_{i}^{\prime}=\varphi_{i}$ for all $1\leq i\leq d$.

Up to this point, we have shown that

$\displaystyle Bw_{0}=\theta_{0}^{*}w_{0},$ (6.13)
$\displaystyle(B-\theta_{1}^{*})(A-\theta_{0})w_{0}=\varphi_{1}w_{0},$ (6.14)
$\displaystyle\delta w_{0}=\eta w_{0}.$ (6.15)

Applying each side of (2.7) to $w_{0}$ and using (6.14) to simplify the
resulting equation, we find that

$\displaystyle\beta w_{0}=\zeta^{*}w_{0}.$ (6.16)

Using logic similar to what was used to show (6.14), we obtain
$(A-\theta_{1})(B-\theta_{0}^{*})u_{0}=\varphi_{1}u_{0}$. Now, by applying
each side of (2.6) to $u_{0}$ and using the above equation to simplify the
resulting equation, we find that $\alpha u_{0}=\zeta u_{0}$. It follows from
Lemma 6.1 that

$\displaystyle\alpha w_{0}=\zeta w_{0}.$ (6.17)

In view of (6.13)–(6.17), it follows from Proposition 3.7 that there exists a
unique $\Re$-module homomorphism $M_{d}(a,b,c)\to V$ that sends $m_{0}$ to
$w_{0}$. By Proposition 3.1(i), the entries above the superdiagonal in (6.11)
are zero. Combining the above $\Re$-module homomorphism $M_{d}(a,b,c)\to V$
with (6.10), there is an $\Re$-module homomorphism

$\displaystyle R_{d}(a,b,c)\to V$ (6.18)

that sends $v_{0}$ to $w_{0}$ by Proposition 3.10. Since the $\Re$-module $V$
is irreducible, the homomorphism (6.18) is onto. Since $R_{d}(a,b,c)$ and $V$
are both of dimension $d+1$, it follows that (6.18) is an isomorphism. The
result follows. ∎

######  Lemma 6.4.

The traces of $A$, $B$, $C$ on the $\Re$-module $R_{d}(a,b,c)$ are equal to
$d+1$ times

$\displaystyle a^{2}+a+\frac{d(d+2)}{12},\qquad
b^{2}+b+\frac{d(d+2)}{12},\qquad c^{2}+c+\frac{d(d+2)}{12},$

respectively.

###### Proof.

To compute the traces of $A$, $B$ on $R_{d}(a,b,c)$, use Proposition 2.4(i).
By (2.5), the trace of $C$ is equal to the trace of $\delta$ minus the trace
of $A+B$ on $R_{d}(a,b,c)$. Use the above facts along with Proposition 2.4(ii)
to compute the trace of $C$ on $R_{d}(a,b,c)$. ∎

The following is a quick consequence of Theorems 5.3, 6.3 and Lemma 6.4.

######  Corollary 6.5.

Assume that $\mathbb{F}$ is algebraically closed with
$\operatorname{char}\mathbb{F}=0$. Let $V$ denote a $(d+1)$-dimensional
irreducible $\Re$-module. Let $\operatorname{tr}A$, $\operatorname{tr}B$,
$\operatorname{tr}C$ denote the traces of $A$, $B$, $C$ on $V$, respectively.
Then the $\Re$-module $R_{d}(a,b,c)$ is isomorphic to $V$ if and only if $a$,
$b$, $c$ are the roots of

$\displaystyle x^{2}+x+\frac{d(d+2)}{12}=\frac{\operatorname{tr}A}{d+1},$
$\displaystyle x^{2}+x+\frac{d(d+2)}{12}=\frac{\operatorname{tr}B}{d+1},$
$\displaystyle x^{2}+x+\frac{d(d+2)}{12}=\frac{\operatorname{tr}C}{d+1},$

respectively.

We are now ready to prove Theorem 2.5.

###### Proof of Theorem 2.5.

By Theorems 4.5 and 5.3, the map $\mathcal{R}$ is well-defined. By Theorem
6.3, the map $\mathcal{R}$ is onto. Since any element of $\Re$ has the same
trace on the isomorphic finite-dimensional $\Re$-modules, it follows from
Lemma 6.4 that $\mathcal{R}$ is one-to-one. ∎

In Example 1.1 we showed a five-dimensional irreducible $\Re$-module on which
none of $A$, $B$, $C$ is diagonalizable. Note that the $\Re$-module is
isomorphic to $R_{4}\big{(}{-}\frac{1}{2},-\frac{1}{2},-\frac{1}{2}\big{)}$.
We finish this paper with the necessary and sufficient conditions for $A$,
$B$, $C$ to be diagonalizable on finite-dimensional irreducible $\Re$-modules.

######  Theorem 6.6.

Assume that $\mathbb{F}$ is algebraically closed with
$\operatorname{char}\mathbb{F}=0$. For any $a,b,c\in\mathbb{F}$ and
$d\in\mathbb{N}$ satisfying the conditions $(i)$ and $(ii)$ of Theorem 4.5,
the following statements are equivalent:

  1. $(i)$

$A$ (resp. $B$) (resp. $C$) is diagonalizable on $R_{d}(a,b,c)$,

  2. $(ii)$

$a$ (resp. $b$) (resp. $c$) is not in

$\displaystyle\big{\\{}\tfrac{i-d-1}{2}\,\big{|}\,i=1,2,\ldots,2d-1\big{\\}}.$
(6.19)

###### Proof.

With reference to Proposition 2.4(i), the minimal polynomial of $A$ in
$R_{d}(a,b,c)$ is

$\displaystyle\prod_{i=0}^{d}(x-\theta_{i}).$

Thus $A$ is diagonalizable on $R_{d}(a,b,c)$ if and only if the scalars
$\\{\theta_{i}\\}_{i=0}^{d}$ are mutually distinct. A direct calculation
yields that the latter holds if and only if $a$ is not in (6.19). By a similar
argument $B$ is diagonalizable on $R_{d}(a,b,c)$ if and only if $b$ is not in
(6.19).

To derive the condition for $C$ as diagonalizable on $R_{d}(a,b,c)$, we
consider the $(d+1)$-dimensional $\Re$-module $R_{d}(b,c,a)$. By Theorem 4.5
the $\Re$-module $R_{d}(b,c,a)$ is irreducible. By Definition 2.1 or [1,
Proposition 4.1] there exists a unique $\mathbb{F}$-algebra automorphism
$\varrho$ of $\Re$ that sends $A$, $B$, $C$, $D$ to $C$, $A$, $B$, $D$,
respectively. Let $R_{d}(b,c,a)^{\varrho}$ denote the $\Re$-module obtained by
pulling back the $\Re$-module $R_{d}(b,c,a)$ via $\varrho$. Observe that $C$
is diagonalizable on $R_{d}(b,c,a)^{\varrho}$ if and only if $c$ is not in
(6.19). Using Corollary 6.5 yields that $R_{d}(b,c,a)^{\varrho}$ is isomorphic
to the $\Re$-module $R_{d}(a,b,c)$. The result follows. ∎

### Acknowledgements

The research of the first author is supported by the Ministry of Science and
Technology of Taiwan under the project MOST 106-2628-M-008-001-MY4.

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