Linear Maps That Act Tridiagonally with Respect to Eigenbases of the Equitable Generators of U_q(sl₂)

Abstract

Let $\mathcal{F}$ denote an algebraically closed field; let q be a nonzero scalar in $\mathcal{F}$ such that q is not a root of unity; let d be a nonnegative integer; and let X, Y, Z be the equitable generators of ${\mathit{U}}_{\mathit{q}}\left({\mathit{sl}}_2\right)$ over $\mathcal{F}$. Let V denote a finite-dimensional irreducible ${\mathit{U}}_{\mathit{q}}\left({\mathit{sl}}_2\right)$-module with dimension $d+1$, and let R denote the set of all linear maps from V to itself that act tridiagonally on the standard ordering of the eigenbases for each of X, Y, and Z. We show that R has dimension at most seven. Indeed, we show that the actions of 1, X, Y, Z, $XY$, $YZ$, and $ZX$ on V give a basis for R when $d\ge 3$.

1. Introduction

We characterize the linear operators that act tridiagonally with respect to appropriately ordered eigenbases for all three equitable generators of ${\mathit{U}}_{\mathit{q}}\left({\mathit{sl}}_2\right)$ acting on its finite-dimensional irreducible modules. To state the main result, we first recall the equitable presentation of ${\mathit{U}}_{\mathit{q}}\left({\mathit{sl}}_2\right)$. Throughout this paper, let $\mathcal{F}$ denote an algebraically closed field, and let q be a nonzero scalar in $\mathcal{F}$ such that q is not a root of unity.

Lemma 1.

[Theorem 2.1] [1] The algebra ${\mathit{U}}_{\mathit{q}}\left({\mathit{sl}}_2\right)$ is isomorphic to the unital associative $\mathcal{F}$-algebra with generators $X^{\pm 1}$, Y, Z and the following relations:

$$X{X}^{-1}=X^{-1}X=1,$$

$$\frac{qXY-q^{-1}YX}{q-q^{-1}}=1,\frac{qYZ-q^{-1}ZY}{q-q^{-1}}=1,\frac{qZX-q^{-1}XZ}{q-q^{-1}}=1.$$

We call $X^{\pm 1}$, Y, Z the equitable generators for the quantum algebra ${\mathit{U}}_{\mathit{q}}\left({\mathit{sl}}_2\right)$.

The equitable presentation of this algebra was introduced in [1], where its relationship to the usual presentation in terms of the Chevalley generators [2] is discussed. The equitable presentation has been studied in connection with tridiagonal pairs [3,4], Leonard pairs [5], the q-tetrahedron algebra [6,7,8,9], bidiagonal pairs [10], Q-polynomial distance-regular graphs [11,12,13], in Poisson algebras [14], and the universal Askey–Wilson algebra [15].

Other relevant references include [16,17,18,19,20,21].

Definition 1.

[Definition 5.2] [1] Let $n_X$, $n_Y$, $n_Z$ denote the following elements of ${\mathit{U}}_{\mathit{q}}\left({\mathit{sl}}_2\right)$:

$$n_X=\frac{q(1-YZ)}{q-q^{-1}}=\frac{q^{-1}(1-ZY)}{q-q^{-1}},$$

$$n_Y=\frac{q(1-ZX)}{q-q^{-1}}=\frac{q^{-1}(1-XZ)}{q-q^{-1}},$$

$$n_Z=\frac{q(1-XY)}{q-q^{-1}}=\frac{q^{-1}(1-YX)}{q-q^{-1}}.$$

Definition 2.

Let V denote a vector space over $\mathcal{F}$ with dimension $d+1$. By a decomposition of V, we mean a sequence ${\left\{V_i\right\}}_{i=0}^d$ consisting of one-dimensional subspaces of V such that:

$$V=V_0+V_1+\dots +V_d,directsum.$$

Definition 3.

Let ${\left\{V_i\right\}}_{i=0}^d$ be a decomposition of V. For notational convenience, define $V_{-1}=0$ and $V_{d+1}=0$. For $A\in End\left(V\right)$, we say A raises ${\left\{V_i\right\}}_{i=0}^d$ whenever $A{V}_i=V_{i+1}$ for $0\le i\le d$. We say A lowers ${\left\{V_i\right\}}_{i=0}^d$ whenever $A{V}_i=V_{i-1}$ for $0\le i\le d$. An ordered pair A, B of elements in $End\left(V\right)$ is called LRwhenever there exists a decomposition of V that is lowered by A and raised by B. A three-tuple A, B, C of elements in $End\left(V\right)$ is called an LR triple whenever any two of A, B, C form an LR pair on V.

Definition 4.

Let $0\ne q\in \mathcal{F}$, $q^2\ne 1$; an LR pair A, B on V is said to be the q-Weyl type whenever:

$$\frac{qAB-q^{-1}BA}{q-q^{-1}}=I.$$

An LR triple A, B, C on V is said to be the q-Weyl type whenever the LR pairs A, B, B, C, and C, A all are the q-Weyl type.

Let A, B, C be an LR triple q-Weyl type on V. In [22], Nomura describes a family of linear maps that acts tridiagonally with respect to each of the $(A,B)$, $(B,C)$, and $(C,A)$ decompositions for V.

The point of view of our work is quite different. To state the main result of this paper, we use the following definition.

Definition 5.

A square matrix is said to be tridiagonal whenever each nonzero entry lies on either the diagonal, the subdiagonal, or the superdiagonal. A square matrix is said to be lower bidiagonal whenever each nonzero entry lies on either the diagonal or the subdiagonal; a square matrix is said to be upper bidiagonal whenever each nonzero entry lies on either the diagonal or the superdiagonal.

Our main result is the following; an ${\mathit{sl}}_2$ analogue appears in [23].

Theorem 1.

Let V be a finite-dimensional $U_q\left({\mathit{sl}}_2\right)$-module. Fix a linear map $\mathsf{\Psi }:V\to V$. Then, the following are equivalent.

(i) 

Ψ acts on V as a linear combination of one, X, Y, Z, $XY$, $YZ$, and $ZX$.

(ii) 

All three of the matrices representing Ψ with respect to standard X-, Y-, and Z-eigenbases are tridiagonal.

(iii) 

Any two of the matrices representing Ψ with respect to standard X-, Y-, and Z-eigenbases are tridiagonal.

Moreover, one, X, Y, Z, $XY$, $YZ$, and $ZX$ are linearly independent when $dimV\ge 3$.

2. Standard Eigenbases for ${\mathit{U}}_{\mathit{q}}\left({\mathit{sl}}_{\mathbf{2}}\right)$-Modules

In this section, we recall the finite-dimensional ${\mathit{U}}_{\mathit{q}}\left({\mathit{sl}}_2\right)$-modules and some distinguished bases.

Lemma 2.

[Lemma 4.2] [1] For each nonnegative integer d and for $ϵ\in \{1,-1\}$, there is an irreducible finite-dimensional ${\mathit{U}}_{\mathit{q}}\left({\mathit{sl}}_2\right)$-module $V_{d,ϵ}$ with basis $\{u_0,u_1,\dots ,u_d\}$ and action:

$$\begin{array}{ccc}\hfill (ϵX-q^{d-2i}I)u_i& =& 0(0\le i\le d),\hfill \\ \hfill (ϵY-q^{2i-d}I)u_i& =& (q^{-d}-q^{2i+2-d})u_{i+1}(0\le i\le d-1),\hfill \\ \hfill (ϵY-q^{d}I)u_d& =& 0,\hfill \\ \hfill (ϵZ-q^{-d}I)u_0& =& 0,\hfill \\ \hfill (ϵZ-q^{2i-d}I)u_i& =& (q^d-q^{2i-2-d})u_{i-1}(1\le i\le d).\hfill \end{array}$$

The basis $\{u_0,u_1,\dots ,u_d\}$ is called the standard X-eigenbasis of $V_{d,ϵ}$.

Since the module $V_{d,-1}$ can be treated similarly to $V_{d,1}$, we treat only the module $V_{d,1}$, and throughout this paper, we write $V_d$ to mean $V_{d,1}$. For any vector space V, $\mathrm{End}\left(V\right)$ is the $\mathcal{F}$-algebra of all $\mathcal{F}$-linear transformations from V to itself.

Corollary 1.

With reference to Lemma 2, the actions of X, Y, Z, on $V_d$ are each multiplicity free with eigenvalues ${\left\{q^{d-2i}\right\}}_{i=0}^d$. Moreover, each of X, Y, Z is invertible on $V_d$.

Definition 6.

Let $V_d$ be a finite-dimensional irreducible ${\mathit{U}}_{\mathit{q}}\left({\mathit{sl}}_2\right)$-module; for $\tau \in \{X,Y,Z\}$, define the decomposition $\left[\tau \right]$ of $V_d$ as follows. For $0\le i\le d$ the ith component of $\left[\tau \right]$ is the eigenspace for τ with eigenvalue $q^{d-2i}$.

Lemma 3.

With reference to Lemma 2, for $0\le i\le d$, let $U_i=span\left\{u_i\right\}$, then ${\left\{U_i\right\}}_{i=0}^d$ is the $\left[X\right]$ decomposition of $V_d$.

Proof. 

This is clear from Definition 6. □

Lemma 4.

[24] With reference to Definition 1, let $V_d$ be a finite-dimensional irreducible ${\mathit{U}}_{\mathit{q}}\left({\mathit{sl}}_2\right)$-module; the following hold:

(i) 

$n_X^d{V}_d$ is the eigenspace for Y $(resp.Z)$ on $V_d$ with eigenvalue $q^{-d}$$(resp.q^d)$.

(ii) 

$n_Y^d{V}_d$ is the eigenspace for Z $(resp.X)$ on $V_d$ with eigenvalue $q^{-d}$$(resp.q^d)$.

(iii) 

$n_Z^d{V}_d$ is the eigenspace for X $(resp.Y)$ on $V_d$ with eigenvalue $q^{-d}$$(resp.q^d)$.

Definition 7.

[24] Let $V_d$ be a finite-dimensional irreducible ${\mathit{U}}_{\mathit{q}}\left({\mathit{sl}}_2\right)$-module; for $\tau \in \{X,Y,Z\}$, a basis ${\left\{v_i\right\}}_{i=0}^d$ for V is said to be ${\left[\tau \right]}_{row}$ whenever the following hold:

(i) 

For $0\le i\le d$, the vector $v_i$ is contained in the component i of the decomposition $\left[\tau \right]$;

(ii) 

${\Sigma }_{i=0}^d{v}_i\in n_{\tau }^d{V}_d$.

Lemma 5.

With reference to Lemma 2, the basis $\mathbf{u}=\{u_0,u_1,\dots ,u_d\}$ is the ${\left[X\right]}_{row}$ basis for $V_d$.

Proof. 

Note that by Lemma 2,

$$\begin{array}{ccc}\hfill Y\left({\Sigma }_{i=0}^d{u}_i\right)& =& Y(u_0+u_1+u_2+\dots +u_{d-1}+u_d)\hfill \\ & =& Y{u}_0+Y{u}_1+Y{u}_2+\dots +Y{u}_{d-1}+Y{u}_d\hfill \\ & =& q^{-d}{u}_0+(q^{-d}-q^{2-d})u_1+q^{2-d}{u}_1+(q^{-d}-q^{4-d})u_2\hfill \\ & +& q^{4-d}{u}_2+(q^{-d}-q^{6-d})u_3+\cdots +q^{d-2}{u}_{d-1}+(q^{-d}-q^d)u_d+q^d{u}_d\hfill \\ & =& q^{-d}(u_0+u_1+u_2+\dots +u_{d-1}+u_d)\hfill \\ & =& q^{-d}\left({\Sigma }_{i=0}^d{u}_i\right),\hfill \end{array}$$

and:

$$\begin{array}{ccc}\hfill Z\left({\Sigma }_{i=0}^d{u}_i\right)& =& Z(u_0+u_1+u_2+\dots +u_{d-1}+u_d)\hfill \\ & =& Z{u}_0+Z{u}_1+Z{u}_2+\dots +Z{u}_{d-1}+Z{u}_d\hfill \\ & =& q^{-d}{u}_0+q^{2-d}{u}_1+(q^d-q^{-d})u_0+q^{4-d}{u}_2+(q^d-q^{2-d})u_1\hfill \\ & +& q^{6-d}{u}_3+(q^d-q^{4-d})u_2+\cdots +q^{d-2}{u}_{d-1}+(q^d-q^{d-4})u_{d-2}\hfill \\ & +& q^d{u}_d+(q^d-q^{d-2})u_{d-1}\hfill \\ & =& q^d(u_0+u_1+u_2+\dots +u_{d-1}+u_d)\hfill \\ & =& q^d\left({\Sigma }_{i=0}^d{u}_i\right).\hfill \end{array}$$

Hence, by Lemma 4, ${\Sigma }_{i=0}^d{u}_i\in n_x^d{V}_d$. Moreover, note that from Lemma 3:

$$V_d=U_1+U_2+\dots +U_d$$

and $u_i\in U_i$. Now, the result holds by Definition 7. □

Let ${\left\{a_i\right\}}_{i=0}^d$ and ${\left\{b_i\right\}}_{i=0}^d$ be two bases of the vector space V. By the transition matrix from ${\left\{a_i\right\}}_{i=0}^d$ to ${\left\{b_i\right\}}_{i=0}^d$, we mean the matrix $S\in Ma{t}_{d+1}\left(\mathcal{F}\right)$ such that $b_j={\Sigma }_{i=0}^d{S}_{ij}{a}_i$ for $0\le j\le d$.

For all integers k and for all nonnegative integers n, m, write:

$$\left[k\right]=\frac{q^k-q^{-k}}{q-q^{-1}},\left[n\right]!=\left[1\right]\left[2\right]\cdots \left[n\right],\left[\genfrac{}{}{0pt}{}{n}{m}\right]=\left\{\begin{array}{cc}{\displaystyle \frac{\left[n\right]!}{[n-m]!\left[m\right]!}}\hfill & \mathrm{if}n\ge m,\hfill \\ 0\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

Definition 8.

Let P and Q denote $(d+1)\times (d+1)$ matrices with entries $P_{ij}$ and $Q_{ij}$, respectively, where:

$$P_{ij}={(-1)}^j{q}^{(j-d)(i-1)}\left[\genfrac{}{}{0pt}{}{i}{d-j}\right](0\le i,j\le d),$$

$$Q_{ij}={(-1)}^j{q}^{j(d-i-1)}\left[\genfrac{}{}{0pt}{}{d-i}{j}\right](0\le i,j\le d).$$

Theorem 2.

[Theorem 16.3, Lemma 16.6] [24] With reference to Definition 8, let $V_d$ be a finite-dimensional irreducible ${\mathit{U}}_{\mathit{q}}\left({\mathit{sl}}_2\right)$-module, and let ${\left[X\right]}_{row}$, ${\left[Y\right]}_{row}$, ${\left[Z\right]}_{row}$ be bases for $V_d$, then:

(i) 

The matrix P is a transition matrix from ${\left[X\right]}_{row}$ to ${\left[Y\right]}_{row}$.

(ii) 

The matrix Q is a transition matrix from ${\left[X\right]}_{row}$ to ${\left[Z\right]}_{row}$.

Lemma 6.

With reference to Lemma 2 and Definition 8,

(i) 

Let $v_j={\sum }_{i=d-j}^d{P}_{ij}{u}_i$$(0\le j\le d)$, then $\mathbf{v}=\{v_0,v_1,\dots ,v_d\}$ is the ${\left[Y\right]}_{row}$ basis for $V_d$.

(ii) 

Let $w_j={\sum }_{i=0}^{d-j}{Q}_{ij}{u}_i$$(0\le j\le d)$, then $\mathbf{w}=\{w_0,w_1,\dots ,w_d\}$ is the ${\left[Z\right]}_{row}$ basis for $V_d$.

Proof. 

By Lemma 5, the basis $\mathbf{u}=\{u_0,u_1,\dots ,u_d\}$ is the ${\left[X\right]}_{row}$ basis for $V_d$. Hence, the results hold by Theorem 2. □

Theorem 3.

[Theorem 10.12] [24] With reference to Lemma 2, let $\mathbf{v}=\{v_0,v_1,\dots ,v_d\}$, and let $\mathbf{w}=\{w_0,w_1,\dots ,w_d\}$ be as in Lemma 6, then:

$$\begin{array}{cccc}\hfill (Y-q^{d-2i}I)v_i& =& 0\hfill & (0\le i\le d),\hfill \\ \hfill (Z-q^{2i-d}I)v_i& =& (q^{-d}-q^{2i+2-d})v_{i+1}\hfill & (0\le i\le d-1),\hfill \\ \hfill (Z-q^{d}I)v_d& =& 0,\hfill \\ \hfill (X-q^{-d}I)v_0& =& 0,\hfill & \\ \hfill (X-q^{2i-d}I)v_i& =& (q^d-q^{2i-2-d})v_{i-1}\hfill & (1\le i\le d).\hfill \\ \hfill (Z-q^{d-2i}I)w_i& =& 0\hfill & (0\le i\le d),\hfill \\ \hfill (X-q^{2i-d}I)w_i& =& (q^{-d}-q^{2i+2-d})w_{i+1}\hfill & (0\le i\le d-1),\hfill \\ \hfill (X-q^{d}I)w_d& =& 0,\hfill \\ \hfill (Y-q^{-d}I)w_0& =& 0,\hfill & \\ \hfill (Y-q^{2i-d}I)w_i& =& (q^d-q^{2i-2-d})w_{i-1}\hfill & (1\le i\le d).\hfill \end{array}$$

In view of Theorem 3, the bases ${\left[Y\right]}_{row}$ and ${\left[Z\right]}_{row}$ are called the standard Y- and Z-eigenbases of $V_d$, respectively.

Let ${\left[T\right]}_{\mathcal{B}}$ denote the matrix representing a linear operator T with respect to an ordered basis $\mathcal{B}$. We say that T acts upper bidiagonally, lower bidiagonally, or tridiagonally on $\mathcal{B}$ when the matrix ${\left[T\right]}_{\mathcal{B}}$ has the stated shape.

Lemma 7.

With reference to Lemma 2 and Theorem 3,

$${\left[X\right]}_{\mathbf{u}}={\left[Y\right]}_{\mathbf{v}}={\left[Z\right]}_{\mathbf{w}},$$

$${\left[X\right]}_{\mathbf{v}}={\left[Y\right]}_{\mathbf{w}}={\left[Z\right]}_{\mathbf{u}},$$

$${\left[X\right]}_{\mathbf{w}}={\left[Y\right]}_{\mathbf{u}}={\left[Z\right]}_{\mathbf{v}}.$$

Proof. 

This is clear from Theorem 3. □

Lemma 8.

With reference to Theorem 3, for all $s\in {\mathit{U}}_{\mathit{q}}\left({\mathit{sl}}_2\right)$,

$${\left[s\right]}_{\mathbf{u}}P=P{\left[s\right]}_{\mathbf{v}},{\left[s\right]}_{\mathbf{v}}P=P{\left[s\right]}_{\mathbf{w}},{\left[s\right]}_{\mathbf{w}}P=P{\left[s\right]}_{\mathbf{u}}.$$

Proof. 

By Theorem 2 and elementary linear algebra, for $s\in \{X,Y,Z\}$, ${\left[s\right]}_{\mathbf{u}}P=P{\left[s\right]}_{\mathbf{v}}$. Now, from Lemma 7, ${\left[Z\right]}_{\mathbf{u}}P=P{\left[Z\right]}_{\mathbf{v}}$ gives ${\left[X\right]}_{\mathbf{v}}P=P{\left[X\right]}_{\mathbf{w}}$ and ${\left[Y\right]}_{\mathbf{u}}P=P{\left[Y\right]}_{\mathbf{v}}$ gives ${\left[X\right]}_{\mathbf{w}}P=P{\left[X\right]}_{\mathbf{u}}$. Similarly, we can prove the result for Y and Z. Since these formulas hold on generators, they must hold for all $s\in {\mathit{U}}_{\mathit{q}}\left({\mathit{sl}}_2\right)$. □

Lemma 9.

[Lemmas 16.5 and 16.6] [24] $P^3=q^{d(d-1)}I$, $PQ={(-1)}^{d}I$.

3. Linear Combinations of $\mathbf{1},\mathbf{X},\mathbf{Y},\mathbf{Z},\mathbf{Xy},\mathbf{Yz},\mathbf{Zx}$

In this section, we define the linear transformation A and describe the action of A on the bases $\mathbf{u}$, $\mathbf{v}$, and $\mathbf{w}$ given in the previous section, which we will use later to prove Theorem 1 and some special cases of this theorem.

Lemma 10.

With reference to Lemma 2,

$$\begin{array}{cccc}\hfill XY{u}_i& =& u_i+(q^{-2(i+1)}-1)u_{i+1}\hfill & (0\le i\le d-1),\hfill \\ \hfill XY{u}_d& =& u_d,\hfill \\ \hfill ZX{u}_i& =& u_i+q^{2(d-i)}(1-q^{2(i-d-1)})u_{i-1}\hfill & (1\le i\le d),\hfill \\ \hfill ZX{u}_0& =& u_0,\hfill \\ \hfill YZ{u}_0& =& q^{-2d}{u}_0+q^{-2d}(1-q^2)u_1,\hfill \\ \hfill YZ{u}_d& =& q^{2d-2}(1-q^{-2})u_{d-1}+(q^{2d}+(1-q^{2d})(1-q^{-2})u_d,\hfill \\ \hfill YZ{u}_i& =& q^{2(i-1)}(1-q^{2(i-1-d)})u_{i-1}\hfill \\ & +& (q^{2(2i-d)}+(1-q^{2i})(1-q^{2(i-d-1)}))u_i\hfill \\ & +& q^{2(i-d)}(1-q^{2(i+1)})u_{i+1}\hfill & (1\le i\le d-1).\hfill \end{array}$$

Proof. 

Performroutine calculations using the action of X, Y, and Z on the basis $\mathbf{u}$ in Lemma 2. □

Lemma 11.

With reference to Lemma 2 and Theorem 3,

$${\left[XY\right]}_{\mathbf{u}}={\left[YZ\right]}_{\mathbf{v}}={\left[ZX\right]}_{\mathbf{w}},$$

$${\left[XY\right]}_{\mathbf{v}}={\left[YZ\right]}_{\mathbf{w}}={\left[ZX\right]}_{\mathbf{u}},$$

$${\left[XY\right]}_{\mathbf{w}}={\left[YZ\right]}_{\mathbf{u}}={\left[ZX\right]}_{\mathbf{v}}.$$

Proof. 

By elementary linear algebra and Lemma 7,

$${\left[XY\right]}_{\mathbf{u}}={\left[X\right]}_{\mathbf{u}}{\left[Y\right]}_{\mathbf{u}}={\left[Y\right]}_{\mathbf{v}}{\left[Z\right]}_{\mathbf{v}}={\left[YZ\right]}_{\mathbf{v}},$$

$${\left[XY\right]}_{\mathbf{u}}={\left[X\right]}_{\mathbf{u}}{\left[Y\right]}_{\mathbf{u}}={\left[Z\right]}_{\mathbf{w}}{\left[X\right]}_{\mathbf{w}}={\left[ZX\right]}_{\mathbf{w}}.$$

Similarly, we can prove the other results. □

Definition 9.

Let d be a nonnegative integer, and consider $V_d$. Let $A\in \mathrm{End}\left(V_d\right)$ denote any linear combination of $\{1$, X, Y, Z, $XY$, $YZ$, $ZX\}$. Write:

$$A=a_{{}_I}1+a_{{}_x}X+a_{{}_y}Y+a_{{}_z}Z+a_{{}_{xy}}XY+a_{{}_{yz}}YZ+a_{{}_{zx}}ZX.$$

Lemma 12.

With reference to Lemma 2 and Definition 9, the action of A on the basis $\mathbf{u}$ is given by:

$$\begin{array}{ccc}\hfill A{u}_0& =& {\alpha }_0{u}_0+{\beta }_1{u}_1,\hfill \\ \hfill A{u}_i& =& {\gamma }_{i-1}{u}_{i-1}+{\alpha }_i{u}_i+{\beta }_{i+1}{u}_{i+1}(1\le i\le d-1),\hfill \\ \hfill A{u}_d& =& {\gamma }_{d-1}{u}_{d-1}+{\alpha }_d{u}_d,\hfill \end{array}$$

where:

$$\begin{array}{ccc}\hfill {\alpha }_i& =& a_{{}_I}+q^{d-2i}{a}_{{}_x}+q^{2i-d}(a_{{}_y}+a_{{}_z})+a_{{}_{xy}}+a_{{}_{zx}}\hfill \\ & & \phantom{X}+(q^{2(2i-d)}+(1-q^{2(i-d-1)})(1-q^{2i}))a_{{}_{yz}}(0\le i\le d),\hfill \\ \hfill {\gamma }_i& =& (1-q^{2(i-d)})(q^d{a}_{{}_z}+q^{2i}{a}_{{}_{yz}}+q^{2(d-i-1)}{a}_{{}_{zx}})(1\le i\le d),\hfill \\ \hfill {\beta }_i& =& (1-q^{2i})(q^{-d}{a}_{{}_y}+q^{-2i}{a}_{{}_{xy}}+q^{2(i-d-1)}{a}_{{}_{yz}})(0\le i\le d-1).\hfill \end{array}$$

Proof. 

Use the actions of X, Y, and Z from Lemma 2 and the actions of $XY$, $YZ$, and $ZX$ from Lemma 10 on the basis $\mathbf{u}$ to get the result. □

Lemma 13.

With reference to Theorem 3 and Definition 9, the action of A on the basis $\mathbf{v}$ is given by:

$$\begin{array}{ccc}\hfill A{v}_0& =& {\alpha }_0^{\prime }{v}_0+{\beta }_1^{\prime }{v}_1,\hfill \\ \hfill A{v}_i& =& {\gamma }_{i-1}^{\prime }{v}_{i-1}+{\alpha }_i^{\prime }{v}_i+{\beta }_{i+1}^{\prime }{v}_{i+1}(1\le i\le d-1),\hfill \\ \hfill A{v}_d& =& {\gamma }_{d-1}^{\prime }{v}_{d-1}+{\alpha }_d^{\prime }{v}_d,\hfill \end{array}$$

where:

$$\begin{array}{ccc}\hfill {\alpha }_i^{\prime }& =& a_{{}_I}+q^{d-2i}{a}_{{}_y}+q^{2i-d}(a_{{}_z}+a_{{}_x})+a_{{}_{yz}}+a_{{}_{xy}}\hfill \\ & & \phantom{X}+(q^{2(2i-d)}+(1-q^{2(i-d-1)})(1-q^{2i}))a_{{}_{zx}}(0\le i\le d),\hfill \\ \hfill {\gamma }_i^{\prime }& =& (1-q^{2(i-d)})(q^d{a}_{{}_x}+q^{2i}{a}_{{}_{zx}}+q^{2(d-i-1)}{a}_{{}_{xy}})(1\le i\le d),\hfill \\ \hfill {\beta }_i^{\prime }& =& (1-q^{2i})(q^{-d}{a}_{{}_z}+q^{-2i}{a}_{{}_{yz}}+q^{2(i-d-1)}{a}_{{}_{zx}})(0\le i\le d-1).\hfill \end{array}$$

Proof. 

The actions of X, Y, and Z on the basis $\mathbf{v}$ are given in Theorem 3, and by Lemma 11, the actions of $XY$, $YZ$, and $ZX$ on the basis $\mathbf{v}$ have the same coefficients of the actions of $ZX$, $XY$, and $YZ$ on the basis $\mathbf{u}$, respectively, which appear in Lemma 10. Now, expand the actions of these components on the basis $\mathbf{v}$ to get the result. □

Lemma 14.

With reference to Theorem 3 and Definition 9, the action of A on the basis $\mathbf{w}$ is given by:

$$\begin{array}{ccc}\hfill A{w}_0& =& {\alpha }_0^{*}{w}_0+{\beta }_1^{*}{w}_1,\hfill \\ \hfill A{w}_i& =& {\gamma }_{i-1}^{*}{w}_{i-1}+{\alpha }_i^{*}{w}_i+{\beta }_{i+1}^{*}{w}_{i+1}(1\le i\le d-1),\hfill \\ \hfill A{w}_d& =& {\gamma }_{d-1}^{*}{w}_{d-1}+{\alpha }_d^{*}{w}_d,\hfill \end{array}$$

where:

$$\begin{array}{ccc}\hfill {\alpha }_i^{*}& =& a_{{}_I}+q^{d-2i}{a}_{{}_z}+q^{2i-d}(a_{{}_x}+a_{{}_y})+a_{{}_{zx}}+a_{{}_{yz}}\hfill \\ & & \phantom{X}+(q^{2(2i-d)}+(1-q^{2(i-d-1)})(1-q^{2i}))a_{{}_{xy}}(0\le i\le d),\hfill \\ \hfill {\gamma }_i^{*}& =& (1-q^{2(i-d)})(q^d{a}_{{}_y}+q^{2i}{a}_{{}_{xy}}+q^{2(d-i-1)}{a}_{{}_{yz}})(1\le i\le d),\hfill \\ \hfill {\beta }_i^{*}& =& (1-q^{2i})(q^{-d}{a}_{{}_x}+q^{-2i}{a}_{{}_{zx}}+q^{2(i-d-1)}{a}_{{}_{xy}})(0\le i\le d-1).\hfill \end{array}$$

Proof. 

The actions of X, Y, and Z on the basis $\mathbf{w}$ are given in Theorem 3, and by Lemma 11, the actions of $XY$, $YZ$, and $ZX$ on the basis $\mathbf{w}$ have the same coefficients of the actions of $YZ$, $ZX$, and $XY$ on the basis $\mathbf{u}$, respectively, which appear in Lemma 10. Now, expand the actions of these components on the basis $\mathbf{w}$ to get the result. □

Corollary 2.

With reference to Definition 9, the matrices representing A with respect to the ${\left[X\right]}_{row}$, ${\left[Y\right]}_{row}$, and ${\left[Z\right]}_{row}$ bases for $V_d$ are tridiagonal.

Proof. 

This is clear from Lemmas 12–14. □

Note that Lemmas 12–14 give that (i) implies (ii) in Theorem 1.

Routine calculations using Lemmas 12–14 and taking the advantage of the symmetry of the actions of X, Y, Z on the bases $\mathbf{u}$, $\mathbf{v}$, $\mathbf{w}$ give expressions for $a_{{}_I}$, $a_{{}_x}$, $a_{{}_y}$, $a_{{}_z}$, $a_{{}_{xy}}$, $a_{{}_{yz}}$, $a_{{}_{zx}}$, which appear in the following corollaries.

Corollary 3.

With reference to Lemma 12, assume $d\ge 2$. Then:

$$\begin{array}{ccc}\hfill a_{{}_I}& =& \left(q^{-4}-q^{3-2d}(q-q^{-1})+\frac{\left[4\right]}{\left[2\right]}\right)a_{{}_{yz}}+\frac{\left[2\right]{\beta }_1-q{\gamma }_{d-1}-{\beta }_2/\left[2\right]}{(q-q^{-1})}\hfill \\ & & \phantom{X}-\frac{q^{-d}{\alpha }_0-q^d{\alpha }_d}{q^d-q^{-d}}-\frac{q^{2d-2}{\beta }_1+q^{2-2d}{\gamma }_{d-1}}{{(q-q^{-1})}^2}+\frac{q^{2d-1}{\beta }_2+q^{1-2d}{\gamma }_{d-2}}{\left[2\right]{(q-q^{-1})}^2},\hfill \\ \hfill a_{{}_x}& =& -(q^{d-4}+q^{2-d})a_{{}_{yz}}\hfill \\ & & \phantom{X}+\frac{{\alpha }_0-{\alpha }_d}{(q^d-q^{-d})}+\frac{q^{d-2}{\beta }_1+q^{2-d}{\gamma }_{d-1}}{{(q-q^{-1})}^2}+\frac{q^{d-1}{\beta }_2+q^{1-d}{\gamma }_{d-2}}{\left[2\right]{(q-q^{-1})}^2},\hfill \\ \hfill a_{{}_y}& =& -q^{1-d}(q+q^{-1})a_{{}_{yz}}+\frac{q^{d-2}\left[2\right]{\beta }_1-q^{d-1}{\beta }_2}{\left[2\right]{(q-q^{-1})}^2},\hfill \\ \hfill a_{{}_z}& =& -q^{d-3}(q+q^{-1})a_{{}_{yz}}+\frac{q^{2-d}\left[2\right]{\gamma }_{d-1}-q^{1-d}{\gamma }_{d-2}}{\left[2\right]{(q-q^{-1})}^2},\hfill \\ \hfill a_{{}_{xy}}& =& q^{4-2d}{a}_{{}_{yz}}+\frac{q^2\left[2\right]{\beta }_1-q{\beta }_2}{\left[2\right]{(q-q^{-1})}^2},\hfill \\ \hfill a_{{}_{zx}}& =& q^{2d-4}{a}_{{}_{yz}}+\frac{\left[2\right]{\gamma }_{d-1}-q{\gamma }_{d-2}}{\left[2\right]{(q-q^{-1})}^2}.\hfill \end{array}$$

If $d\ge 3$, then:

$$a_{{}_{yz}}=\frac{-q^{2d-4}\left({\beta }_1-{\beta }_2+{\beta }_3/\left[3\right]\right)}{(q^2-q^{-2}){(q-q^{-1})}^2}=\frac{q^{6-2d}\left({\gamma }_{d-1}-{\gamma }_{d-2}+{\gamma }_{d-3}/\left[3\right]\right)}{(q^2-q^{-2}){(q-q^{-1})}^2}.$$

Corollary 4.

With reference to Lemma 13, assume $d\ge 2$. Then:

$$\begin{array}{ccc}\hfill a_{{}_I}& =& \left(q^{-4}-q^{3-2d}(q-q^{-1})+\frac{\left[4\right]}{\left[2\right]}\right)a_{{}_{zx}}+\frac{\left[2\right]{\beta }_1^{\prime }-q{\gamma }_{d-1}^{\prime }-{\beta }_2^{\prime }/\left[2\right]}{(q-q^{-1})}\hfill \\ & & \phantom{X}-\frac{q^{-d}{\alpha }_0^{\prime }-q^d{\alpha }_d^{\prime }}{q^d-q^{-d}}-\frac{q^{2d-2}{\beta }_1^{\prime }+q^{2-2d}{\gamma }_{d-1}^{\prime }}{{(q-q^{-1})}^2}+\frac{q^{2d-1}{\beta }_2^{\prime }+q^{1-2d}{\gamma }_{d-2}^{\prime }}{\left[2\right]{(q-q^{-1})}^2},\hfill \\ \hfill a_{{}_y}& =& -(q^{d-4}+q^{2-d})a_{{}_{zx}}\hfill \\ & & \phantom{X}+\frac{{\alpha }_0^{\prime }-{\alpha }_d^{\prime }}{(q^d-q^{-d})}+\frac{q^{d-2}{\beta }_1^{\prime }+q^{2-d}{\gamma }_{d-1}^{\prime }}{{(q-q^{-1})}^2}+\frac{q^{d-1}{\beta }_2^{\prime }+q^{1-d}{\gamma }_{d-2}^{\prime }}{\left[2\right]{(q-q^{-1})}^2},\hfill \\ \hfill a_{{}_z}& =& -q^{1-d}(q+q^{-1})a_{{}_{zx}}+\frac{q^{d-2}\left[2\right]{\beta }_1^{\prime }-q^{d-1}{\beta }_2^{\prime }}{\left[2\right]{(q-q^{-1})}^2},\hfill \\ \hfill a_{{}_x}& =& -q^{d-3}(q+q^{-1})a_{{}_{zx}}+\frac{q^{2-d}\left[2\right]{\gamma }_{d-1}^{\prime }-q^{1-d}{\gamma }_{d-2}^{\prime }}{\left[2\right]{(q-q^{-1})}^2},\hfill \\ \hfill a_{{}_{yz}}& =& q^{4-2d}{a}_{{}_{zx}}+\frac{q^2\left[2\right]{\beta }_1^{\prime }-q{\beta }_2^{\prime }}{\left[2\right]{(q-q^{-1})}^2},\hfill \\ \hfill a_{{}_{xy}}& =& q^{2d-4}{a}_{{}_{zx}}+\frac{\left[2\right]{\gamma }_{d-1}^{\prime }-q{\gamma }_{d-2}^{\prime }}{\left[2\right]{(q-q^{-1})}^2}.\hfill \end{array}$$

If $d\ge 3$, then:

$$a_{{}_{zx}}=\frac{-q^{2d-4}\left({\beta }_1^{\prime }-{\beta }_2^{\prime }+{\beta }_3^{\prime }/\left[3\right]\right)}{(q^2-q^{-2}){(q-q^{-1})}^2}=\frac{q^{6-2d}\left({\gamma }_{d-1}^{\prime }-{\gamma }_{d-2}^{\prime }+{\gamma }_{d-3}^{\prime }/\left[3\right]\right)}{(q^2-q^{-2}){(q-q^{-1})}^2}.$$

Corollary 5.

With reference to Lemma 14, assume $d\ge 2$. Then:

$$\begin{array}{ccc}\hfill a_{{}_I}& =& \left(q^{-4}-q^{3-2d}(q-q^{-1})+\frac{\left[4\right]}{\left[2\right]}\right)a_{{}_{xy}}+\frac{\left[2\right]{\beta }_1^{*}-q{\gamma }_{d-1}^{*}-{\beta }_2^{*}/\left[2\right]}{(q-q^{-1})}\hfill \\ & & \phantom{X}-\frac{q^{-d}{\alpha }_0^{*}-q^d{\alpha }_d^{*}}{q^d-q^{-d}}-\frac{q^{2d-2}{\beta }_1^{*}+q^{2-2d}{\gamma }_{d-1}^{*}}{{(q-q^{-1})}^2}+\frac{q^{2d-1}{\beta }_2^{*}+q^{1-2d}{\gamma }_{d-2}^{*}}{\left[2\right]{(q-q^{-1})}^2},\hfill \\ \hfill a_{{}_z}& =& -(q^{d-4}+q^{2-d})a_{{}_{xy}}\hfill \\ & & \phantom{X}+\frac{{\alpha }_0^{*}-{\alpha }_d^{*}}{(q^d-q^{-d})}+\frac{q^{d-2}{\beta }_1^{*}+q^{2-d}{\gamma }_{d-1}^{*}}{{(q-q^{-1})}^2}+\frac{q^{d-1}{\beta }_2^{*}+q^{1-d}{\gamma }_{d-2}^{*}}{\left[2\right]{(q-q^{-1})}^2},\hfill \\ \hfill a_{{}_x}& =& -q^{1-d}(q+q^{-1})a_{{}_{xy}}+\frac{q^{d-2}\left[2\right]{\beta }_1^{*}-q^{d-1}{\beta }_2^{*}}{\left[2\right]{(q-q^{-1})}^2},\hfill \\ \hfill a_{{}_y}& =& -q^{d-3}(q+q^{-1})a_{{}_{xy}}+\frac{q^{2-d}\left[2\right]{\gamma }_{d-1}^{*}-q^{1-d}{\gamma }_{d-2}^{*}}{\left[2\right]{(q-q^{-1})}^2},\hfill \\ \hfill a_{{}_{zx}}& =& q^{4-2d}{a}_{{}_{xy}}+\frac{q^2\left[2\right]{\beta }_1^{*}-q{\beta }_2^{*}}{\left[2\right]{(q-q^{-1})}^2},\hfill \\ \hfill a_{{}_{yz}}& =& q^{2d-4}{a}_{{}_{xy}}+\frac{\left[2\right]{\gamma }_{d-1}^{*}-q{\gamma }_{d-2}^{*}}{\left[2\right]{(q-q^{-1})}^2}.\hfill \end{array}$$

If $d\ge 3$, then:

$$a_{{}_{xy}}=\frac{-q^{2d-4}\left({\beta }_1^{*}-{\beta }_2^{*}+{\beta }_3^{*}/\left[3\right]\right)}{(q^2-q^{-2}){(q-q^{-1})}^2}=\frac{q^{6-2d}\left({\gamma }_{d-1}^{*}-{\gamma }_{d-2}^{*}+{\gamma }_{d-3}^{*}/\left[3\right]\right)}{(q^2-q^{-2}){(q-q^{-1})}^2}.$$

4. Tridiagonal Operators

In this section, we prove that $\left(\mathrm{iii}\right)$ implies $\left(\mathrm{i}\right)$ in Theorem 1.

Lemma 15.

Assume $d\ge 2$. Fix $\mathsf{\Psi }\in \mathrm{End}\left(V_d\right)$. Let $\mathbf{u}=\{u_0,u_1,\dots ,u_d\}$ and $\mathbf{v}=\{v_0,v_1,\dots ,v_d\}$ be ${\left[X\right]}_{row}$ and ${\left[Y\right]}_{row}$ bases for $V_d$, respectively. Assume that Ψ acts tridiagonally on $\mathbf{u}$ and $\mathbf{v}$, so for some ${\alpha }_i$, ${\beta }_i$, ${\gamma }_i$, ${\alpha }_i^{\prime }$, ${\beta }_i^{\prime }$, ${\gamma }_i^{\prime }\in \mathcal{F}$:

$$\begin{array}{cccccc}\hfill \mathsf{\Psi }{u}_0& =& {\alpha }_0{u}_0+{\beta }_1{u}_1,\hfill & \hfill \mathsf{\Psi }{v}_0& =& {\alpha }_0^{\prime }{v}_0+{\beta }_1^{\prime }{v}_1,\hfill \\ \hfill \mathsf{\Psi }{u}_i& =& {\gamma }_{i-1}{u}_{i-1}+{\alpha }_i{u}_i+{\beta }_{i+1}{u}_{i+1},\hfill & \hfill \mathsf{\Psi }{v}_i& =& {\gamma }_{i-1}^{\prime }{v}_{i-1}+{\alpha }_i^{\prime }{v}_i+{\beta }_{i+1}^{\prime }{v}_{i+1}\hfill \\ & & & & & (1\le i\le d-1),\hfill \\ \hfill \mathsf{\Psi }{u}_d& =& {\gamma }_{d-1}{u}_{d-1}+{\alpha }_d{u}_d,\hfill & \hfill \mathsf{\Psi }{v}_d& =& {\gamma }_{d-1}^{\prime }{v}_{d-1}+{\alpha }_d^{\prime }{v}_d.\hfill \end{array}$$

(i) 

${\alpha }_i$, ${\beta }_i$, ${\gamma }_i$, ${\alpha }_i^{\prime }$, ${\beta }_i^{\prime }$, ${\gamma }_i^{\prime }$, and $a_{{}_I}$, $a_{{}_x}$, $a_{{}_y}$, $a_{{}_z}$, $a_{{}_{xy}}$, $a_{{}_{yz}}$, $a_{{}_{zx}}$ are related by Lemmas 12 and 13.

(ii) 

Ψ acts on $V_d$ as $A=a_{{}_I}1+a_{{}_x}X+a_{{}_y}Y+a_{{}_z}Z+a_{{}_{xy}}XY+a_{{}_{yz}}YZ+a_{{}_{zx}}ZX$.

(iii) 

Ψ acts tridiagonally on any ${\left[Z\right]}_{row}$ basis.

Proof. 

By Lemma 8, ${\left[\mathsf{\Psi }\right]}_{\mathbf{u}}P-P{\left[\mathsf{\Psi }\right]}_{\mathbf{v}}=0$. Take ${\beta }_0={\beta }_0^{\prime }={\gamma }_d={\gamma }_d^{\prime }=0$. For $0\le i,j\le d$, matrix multiplication gives the $(i,j)$-entry:

$$P_{ij}({\alpha }_i-{\alpha }_j^{\prime })+P_{i-1j}{\beta }_i+P_{i+1j}{\gamma }_i-P_{ij-1}{\gamma }_{j-1}^{\prime }-P_{ij+1}{\beta }_{j+1}^{\prime }=0.$$

(1)

It is routine to verify that the parameters given in terms of $a_{{}_I}$, $a_{{}_x}$, $a_{{}_y}$, $a_{{}_z}$, $a_{{}_{xy}}$, $a_{{}_{yz}}$, and $a_{{}_{zx}}$ by Lemmas 12 and 13 are a solution. we now show that there is a unique solution in the parameters ${\alpha }_0$, ${\alpha }_d$, ${\beta }_1$, ${\beta }_2$, ${\gamma }_{d-1}$, ${\gamma }_{d-2}$, and ${\gamma }_{d-3}$.

Respectively taking i to be $d-j-1$, $d-j$, and $d-j+1$ in (1) gives:

$$\begin{array}{ccc}\hfill {\beta }_k^{\prime }& =& -q^{2(k-d)}{\gamma }_{d-k}(1\le k\le d),\hfill \\ \hfill {\alpha }_k^{\prime }& =& {\alpha }_{d-k}+q^{k-d}[d-k+1]{\gamma }_{d-k}+q^{d-k-1}[d-k]{\beta }_{k+1}^{\prime }(0\le k\le d),\hfill \\ \hfill {\gamma }_k^{\prime }& =& q^{d-k-1}[d-k]({\alpha }_{k+1}^{\prime }-{\alpha }_{d-k})-q^{2(d-k-1)}{\beta }_{d-k}-\frac{[d-k][d-k+1]}{\left[2\right]}{\gamma }_{d-k}\hfill \\ & -& \frac{q^{2(d-k-1)}[d-k-1][d-k]}{\left[2\right]}{\beta }_{k+2}^{\prime }(0\le k\le d-1).\hfill \end{array}$$

Substituting the above into (1) at $j=d$, $d-1$, and $d-2$ gives respectively:

$$\begin{array}{ccc}\hfill {\alpha }_k+{\beta }_k+{\gamma }_k& =& q^{1-k}\left[k\right]({\alpha }_1+{\beta }_1+{\gamma }_1)\hfill \\ & +& (1-q^{1-k}\left[k\right])({\alpha }_0+{\gamma }_0)(2\le k\le d),\hfill \end{array}$$

(2)

$$\begin{array}{ccc}& & q\left[k\right]{\alpha }_k+q^2[k-1]{\beta }_k+[k+1]{\gamma }_k=(q^k-q\left[k\right]-\frac{(q^3-q\left[2\right])[k-1]\left[k\right]}{q^{k-1}\left[2\right]}){\gamma }_0\hfill \\ & & +(1-q^{2-k}[k-1])\left[k\right](q{\alpha }_1+\left[2\right]{\gamma }_1)\hfill \\ & & +\frac{[k-1]\left[k\right]}{q^{k-2}\left[2\right]}(q\left[2\right]{\alpha }_2+q^2{\beta }_2+\left[3\right]{\gamma }_2)(3\le k\le d),\hfill \end{array}$$

(3)

$$\begin{array}{ccc}& & q^2[k-1]\left[k\right]{\alpha }_k+q^2[k-2][k-1]{\beta }_k+\left[k\right][k+1]{\gamma }_k\hfill \\ & & =-(q^{4-k}(q-\left[2\right])[k-2][k-1]\left[k\right]+\left[2\right]\left[k\right](q[k-1]-q^{k-1})){\gamma }_1\hfill \\ & & +(1-q^{3-k}[k-2][k-1]\left[k\right](q^2{\alpha }_2+\left[3\right]{\gamma }_2)\hfill \\ & & +\frac{[k-2][k-1]\left[k\right]}{q^{k-3}}(q^2{\alpha }_3+q^4\frac{{\beta }_3}{\left[3\right]}+\frac{\left[4\right]{\gamma }_3}{\left[2\right]})(4\le k\le d).\hfill \end{array}$$

(4)

For $4\le k\le d$, the system of equations in ${\alpha }_k$, ${\beta }_k$, ${\gamma }_k$ has a coefficient matrix with determinant $(q^2+1)q^{3(1-k)}\ne 0$, so these parameters are uniquely determined by those with a lower index. Use (2) at $k=2,3$ and (3) at $k=3$ to solve for ${\alpha }_2$, ${\alpha }_3$, and ${\beta }_3$. Since ${\gamma }_d=0$, (2), (3), and (4) at $k=d$ can be solved for ${\alpha }_1$, ${\gamma }_3$, and ${\beta }_d$ in term of ${\alpha }_0$, ${\alpha }_d$, ${\beta }_1$, ${\beta }_2$, ${\gamma }_0$, ${\gamma }_1$, and ${\gamma }_2$. We may use (2) at $k=d-1,d-2$, and $d-3$ to replace ${\gamma }_0$, ${\gamma }_1$, and ${\gamma }_2$ with ${\gamma }_{d-1}$, ${\gamma }_{d-2}$, and ${\gamma }_{d-3}$ to get a unique solution in terms of ${\alpha }_0$, ${\alpha }_d$, ${\beta }_1$, ${\beta }_2$, ${\gamma }_{d-1}$, ${\gamma }_{d-2}$, and ${\gamma }_{d-3}$. Now, $\left(\mathrm{i}\right)$ follows.

For $\left(\mathrm{ii}\right)$, expand the actions of X, Y, and Z on the basis $\mathbf{u}$ to verify that the action of A agrees with the action of $\mathsf{\Psi }$ on the basis $\mathbf{u}$. Now, $\left(\mathrm{iii}\right)$ follows from $\left(\mathrm{ii}\right)$ since the operators in the sum act tridiagonally on any ${\left[Z\right]}_{row}$ basis by Theorem 3. □

We can use the same argument to prove the result in Lemma 15 when $\mathsf{\Psi }$ acts tridiagonally on the bases $\mathbf{v}$ and $\mathbf{w}$ or when $\mathsf{\Psi }$ acts tridiagonally on the bases $\mathbf{w}$ and $\mathbf{u}$.

5. Main Results

In this section, we prove the main result of this paper, and then, we give some special cases.

Theorem 4.

Let V be a finite-dimensional $U_q\left({\mathit{sl}}_2\right)$-module. Fix a linear map $\mathsf{\Psi }:V\to V$. Then, the following are equivalent.

(i) 

Ψ acts on V as a linear combination of one, X, Y, Z, $XY$, $YZ$, and $ZX$.

(ii) 

All three of the matrices representing Ψ with respect to standard X-, Y-, and Z-eigenbases are tridiagonal.

(iii) 

Any two of the matrices representing Ψ with respect to standard X-, Y-, and Z-eigenbases are tridiagonal.

Moreover, one, X, Y, Z, $XY$, $YZ$, and $ZX$ are linearly independent when $dimV\ge 3$.

Proof. 

The theorem holds for $d<2$ since (i), (ii), and (iii) do. Suppose $d\ge 2$. Corollary 2 and Lemma 15 show that (i) implies (ii). Clearly, (ii) implies (iii). Finally, (iii) implies (i) by Lemma 15. Thus, the result holds. □

Corollary 6.

With reference to Lemma 2 and Theorem 3, let $\mathsf{\Psi }\in \mathrm{End}\left(V_d\right)$. Then, the following (i)–(iii) are equivalent.

(i) 

Ψ acts on $V_d$ as a linear combination of I, Y, Z, and $YZ$.

(ii) 

All three of (a), (b), (c) below hold.

(iii) 

Any two of (a), (b), (c) below hold.

(a) 

Ψ acts tridiagonally on a ${\left[X\right]}_{row}$ basis.

(b) 

Ψ acts lower bidiagonally on a ${\left[Y\right]}_{row}$ basis.

(c) 

Ψ acts upper bidiagonally on a ${\left[Z\right]}_{row}$ basis.

Proof. 

The actions of Y, Z, and $YZ$ are described in Lemma 2, and Theorem 3 shows that (a), (b), and (c) hold when $\mathsf{\Psi }$ acts as a linear combination of I, Y, Z, and $YZ$. If all three of (a), (b), and (c) hold, then clearly so do any two of them. Finally, suppose any two of (a), (b), (c) hold, so one of (b) and (c) holds. Then, the matrices representing $\mathsf{\Psi }$ with respect to standard X-, Y-, and Z-bases are all tridiagonal, so Lemma 15 gives that $\mathsf{\Psi }$ acts on $V_d$ as $A=a_{{}_I}1+a_{{}_x}X+a_{{}_y}Y+a_{{}_z}Z+a_{{}_{xy}}XY+a_{{}_{yz}}YZ+a_{{}_{zx}}ZX$ with the coefficients as described in that lemma. Now, (b) implies that all ${\gamma }_i^{\prime }=0$, so $a_{{}_x}=a_{{}_{zx}}=a_{{}_{xy}}=0$ by Corollary 4. Similarly, (c) implies that all ${\beta }_i^{*}=0$, so $a_{{}_x}=a_{{}_{zx}}=a_{{}_{xy}}=0$ by Corollary 5. Thus, $\mathsf{\Psi }$ acts on $V_d$ as a linear combination of I, Y, Z, and $YZ$. □

Corollary 7.

With reference to Lemma 2 and Theorem 3, let $\mathsf{\Psi }\in \mathrm{End}\left(V_d\right)$. Then, the following (i)–(iii) are equivalent.

(i) 

Ψ acts on $V_d$ as a linear combination of I, Z, X, and $ZX$.

(ii) 

All three of (a), (b), (c) below hold.

(iii) 

Any two of (a), (b), (c) below hold.

(a) 

Ψ acts tridiagonally on a ${\left[Y\right]}_{row}$ basis.

(b) 

Ψ acts lower bidiagonally on a ${\left[Z\right]}_{row}$ basis.

(c) 

Ψ acts upper bidiagonally on a ${\left[X\right]}_{row}$ basis.

Proof. 

This is similar to the proof of Corollary 6. □

Corollary 8.

With reference to Lemma 2 and Theorem 3, let $\mathsf{\Psi }\in \mathrm{End}\left(V_d\right)$. Then, the following (i)–(iii) are equivalent.

(i) 

Ψ acts on $V_d$ as a linear combination of I, X, Y, and $XY$.

(ii) 

All three of (a), (b), (c) below hold.

(iii) 

Any two of (a), (b), (c) below hold.

(a) 

Ψ acts tridiagonally on a ${\left[Z\right]}_{row}$ basis.

(b) 

Ψ acts lower bidiagonally on a ${\left[X\right]}_{row}$ basis.

(c) 

Ψ acts upper bidiagonally on a ${\left[Y\right]}_{row}$ basis.

Proof. 

This is similar to the proof of Corollary 6. □

Corollary 9.

With reference to Lemma 2 and Theorem 3, let $\mathsf{\Psi }\in \mathrm{End}\left(V_d\right)$. Then, the following (i)–(iii) are equivalent.

(i) 

Ψ acts on $V_d$ as a linear combination of I, X.

(ii) 

All three of (a), (b), (c) below hold.

(iii) 

Any two of (a), (b), (c) below hold.

(a) 

Ψ acts diagonally on a ${\left[X\right]}_{row}$ basis.

(b) 

Ψ acts upper bidiagonally on a ${\left[Y\right]}_{row}$ basis.

(c) 

Ψ acts lower bidiagonally on a ${\left[Z\right]}_{row}$ basis.

Proof. 

We argue as in the proof of Corollary 6. The result is clear if (a) holds. Note that (b) implies that all ${\beta }_i^{\prime }=0$, so $a_{{}_z}=a_{{}_{zx}}=a_{{}_{yz}}=0$ by Corollary 4, and (c) implies that all ${\gamma }_i^{*}=0$, so $a_{{}_y}=a_{{}_{yz}}=a_{{}_{xy}}=0$ by Corollary 5. Thus, $\mathsf{\Psi }$ acts on $V_d$ as a linear combination of I, X. □

Corollary 10.

With reference to Lemma 2 and Theorem 3, let $\mathsf{\Psi }\in \mathrm{End}\left(V_d\right)$. Then, the following (i)–(iii) are equivalent.

(i) 

Ψ acts on $V_d$ as a linear combination of I, Y.

(ii) 

All three of (a), (b), (c) below hold.

(iii) 

Any two of (a), (b), (c) below hold.

(a) 

Ψ acts diagonally on a ${\left[Y\right]}_{row}$ basis.

(b) 

Ψ acts upper bidiagonally on a ${\left[Z\right]}_{row}$ basis.

(c) 

Ψ acts lower bidiagonally on a ${\left[X\right]}_{row}$ basis.

Proof. 

This is similar to the proof of Corollary 9. □

Corollary 11.

With reference to Lemma 2 and Theorem 3, let $\mathsf{\Psi }\in \mathrm{End}\left(V_d\right)$. Then, the following (i)–(iii) are equivalent.

(i) 

Ψ acts on $V_d$ as a linear combination of I, Z.

(ii) 

All three of (a), (b), (c) below hold.

(iii) 

Any two of (a), (b), (c) below hold.

(a) 

Ψ acts diagonally on a ${\left[Z\right]}_{row}$ basis.

(b) 

Ψ acts upper bidiagonally on a ${\left[X\right]}_{row}$ basis.

(c) 

Ψ acts lower bidiagonally on a ${\left[Y\right]}_{row}$ basis.

Proof. 

This is similar to the proof of Corollary 9. □