The shape of a tridiagonal pair\footnote{ {Keywords}. Leonard pair, tridiagonal pair, Askey scheme, $q$-Racah polynomial, subconstituent algebra. \hfil\break \noindent {2000 Mathematics Subject Classification}. Primary 17B37; Secondary 05E35, 15A21, 33C45, 33D45. }
Tatsuro Ito and Paul Terwilliger
Let $\fld $ denote an algebraically closed field with characteristic 0. Let $V$ denote a vector space over $\fld$ with finite positive dimension and let $A,A^*$ denote a tridiagonal pair on $V$. We make an assumption about this pair. Let $q$ denote a nonzero scalar in $\K$ which is not a root of unity. We assume $A$ and $A^*$ satisfy the $q$-Serre relations \begin{eqnarray*} A^3A^* - \lbrack 3 \rbrack A^2A^*A + \lbrack 3 \rbrack AA^*A^2 - A^*A^3&=&0, \\ A^{*3}A - \lbrack 3 \rbrack A^{*2}AA^* + \lbrack 3 \rbrack A^*AA^{*2} - AA^{*3}&=&0, \end{eqnarray*} where $\lbrack 3\rbrack=(q^3-q^{-3})/(q-q^{-1})$. Let $(\rho_0, \rho_1,\ldots,\rho_d)$ denote the shape vector for $A,A^*$. We show the entries in this shape vector are bounded above by binomial coefficients as follows: \begin{eqnarray*} \rho_i \;\leq \;\biggl( {{d}\atop {i}} \biggr) \qquad \qquad (0 \leq i \leq d). \end{eqnarray*} We obtain this result by displaying a spanning set for $V$.
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\title{ \bf The shape of    
 a tridiagonal pair\footnote{
{\bf Keywords}. Leonard pair, tridiagonal pair, Askey scheme,
 $q$-Racah polynomial, subconstituent algebra.
\hfil\break
\noindent
{\bf 2000 Mathematics Subject Classification}.
Primary 17B37; Secondary  05E35, 15A21, 33C45, 33D45.
}}
\author{Tatsuro Ito and Paul Terwilliger  
}
\date{}
%to get date printout, comment out the  line  \date{}
%if you dont want date printout, include \date{}
\maketitle
\begin{abstract} Let $\fld $ denote an algebraically closed field
with characteristic 0. 
Let $V$ denote a vector space
over  $\fld$ with finite positive dimension
and let $A,A^*$ denote a tridiagonal pair on $V$.
We make an assumption about this pair.
Let $q$ denote a nonzero scalar in $\K$ which is
not a root of unity. 
We assume $A$ and $A^*$ satisfy the $q$-Serre relations
\begin{eqnarray*}
A^3A^* - \lbrack 3 \rbrack A^2A^*A
+ \lbrack 3 \rbrack AA^*A^2 - A^*A^3&=&0, 
\\
A^{*3}A - \lbrack 3 \rbrack A^{*2}AA^*
+ \lbrack 3 \rbrack A^*AA^{*2} - AA^{*3}&=&0, 
\end{eqnarray*}
where $\lbrack 3\rbrack=(q^3-q^{-3})/(q-q^{-1})$.
Let $(\rho_0, \rho_1,\ldots,\rho_d)$ denote the shape vector
for $A,A^*$. We show the entries in this shape vector are bounded
above by binomial coefficients as follows:
\begin{eqnarray*}
\rho_i \;\leq \;\biggl( {{d}\atop {i}} \biggr) \qquad \qquad (0 \leq i \leq d).
\end{eqnarray*}
We obtain this result by displaying a spanning set for $V$.
\end{abstract}


%\noindent
%{\bf Keywords:} Tridiagonal system, Leonard pair, Leonard system,
%$q$-Racah polynomial,  
% Askey scheme, 
%subconstituent algebra,
%Terwilliger algebra.





\section{Introduction}


\noindent 
Throughout this paper,
 $\K$ will denote a field and  $V$ will denote
a vector space over $\K$ with finite positive dimension.
%A map 
%$A:V\rightarrow V$
%is said to be a 
%{\it linear transformation}
%whenever it is $\K$-linear.
 Let $A:V\rightarrow V$ denote a linear transformation and
 let $W$ denote a subspace of $V$. We call $W$ an {\it eigenspace} of $A$ whenever $W\not=0$ and there exists $\theta \in \K$ such that 
\begin{eqnarray*}
W=\lbrace v \in V \;\vert \;Av = \theta v\rbrace.
%\label{eq:defeigspace}
\end{eqnarray*}
%The scalar $\theta$ is called
%the {\it eigenvalue} of $A$ corresponding to $W$.
We say $A$ is {\it diagonalizable} whenever
$V$ is spanned by the eigenspaces of $A$.

\medskip
\noindent We now recall the notion of a {\it tridiagonal pair}.

\begin{definition}  
\cite{TD00}
\label{def:tdp}
\rm
By a {\it tridiagonal pair}  on $V$,
we mean an ordered pair $A,A^*$ where
$A:V\rightarrow V$ and 
$A^*:V\rightarrow V$ 
 are linear transformations which satisfy
the following four conditions.
\begin{enumerate}
\item Each of $A,A^*$ is diagonalizable.
\item There exists an ordering $V_0, V_1,\ldots, V_d$ of the  
eigenspaces of $A$ such that 
\begin{equation}
A^* V_i \subseteq V_{i-1} + V_i+ V_{i+1} \qquad \qquad (0 \leq i \leq d),
\label{eq:t1}
\end{equation}
where $V_{-1} = 0$, $V_{d+1}= 0$.
\item There exists an ordering $V^*_0, V^*_1,\ldots, V^*_\delta$ of
the  
eigenspaces of $A^*$ such that 
\begin{equation}
A V^*_i \subseteq V^*_{i-1} + V^*_i+ V^*_{i+1} \qquad \qquad (0 \leq i \leq \delta),
\label{eq:t2}
\end{equation}
where $V^*_{-1} = 0$, $V^*_{\delta+1}= 0$.
\item There does not exist a subspace $W$ of $V$ such  that $AW\subseteq W$,
$A^*W\subseteq W$, $W\not=0$, $W\not=V$.
\end{enumerate}
\end{definition}

\begin{note} \rm
According to a common 
notational convention, $A^*$ denotes the conjugate-transpose
of $A$.
 We are not using this convention.
In a tridiagonal pair $A,A^*$ the linear transformations 
$A$ and $A^*$ are arbitrary subject to (i)--(iv) above.
\end{note}

\noindent
In order to motivate our results we recall a few facts about 
tridiagonal pairs.
Let $A,A^*$ denote a tridiagonal pair on $V$
and let the integers $d, \delta$ be as in
Definition \ref{def:tdp}(ii), (iii) respectively.  By 
 \cite[Lemma 4.5]{TD00}
 we have
$d=\delta$; we call this common value the {\it diameter}
of $A,A^*$.  
An ordering of the eigenspaces of $A$ (resp. $A^*$)
is called
{\it standard} whenever it satisfies
%Let  $V_0, V_1,\ldots, V_d$ denote an ordering
%of the eigenspaces
%of $A$. We say this ordering is {\it standard}
%whenever it satisfies
(\ref{eq:t1}) (resp. 
(\ref{eq:t2})). 
We comment on the uniqueness of the standard ordering.
%Similarly an ordering
%$V^*_0, V^*_1,\ldots, V^*_d$ of the eigenspaces
%of $A^*$ is called {\it standard}
%whenever it satisfies
%(\ref{eq:t2}).
%We comment on the uniqueness of the standard order.
Let 
 $V_0, V_1,\ldots, V_d$ denote a standard ordering
of the eigenspaces of $A$.
Then the ordering 
 $V_d, V_{d-1},\ldots, V_0$ 
%(resp. $V^*_d, V^*_{d-1},\ldots, V^*_0$) 
is standard and no other ordering is standard.
A similar result holds for the 
 eigenspaces of $A^*$.
Let
 $V_0, V_1,\ldots, V_d$
(resp.  $V^*_0, V^*_1,\ldots, V^*_d$)
denote a standard ordering
of the eigenspaces of $A$ (resp. $A^*$).
By \cite[Corollary 5.7]{TD00},  for $0 \leq i \leq d$ 
the spaces $V_i$,  $V^*_i$ have the same dimension;
we denote this common dimension by $\rho_i$.
By the construction $\rho_i\not=0$.
 By  \cite[Corollary 5.7]{TD00} 
and \cite[Corollary 6.6]{TD00},
the sequence $\rho_0, \rho_1, \ldots, \rho_d$
is symmetric and unimodal; that is
$ \rho_i =\rho_{d-i}$ for $0 \leq i \leq d$
and 
$\rho_{i-1} \leq \rho_{i}$ for  $1 \leq i \leq d/2$.
We refer to the sequence
 $(\rho_0, \rho_1, \ldots, \rho_d)$ as the {\it shape vector} of $A,A^*$.

\medskip
\noindent
The following special case has received a lot of attention.
By a {\it Leonard pair} we mean a tridiagonal pair
which has shape vector $(1,1,\ldots, 1)$.
There is a natural correspondence between
the Leonard pairs and a family of orthogonal polynomials
consisting of
the $q$-Racah polynomials
 \cite{AWil}
and some related polynomials
of the Askey-scheme 
\cite{KoeSwa}, 
\cite{TLT:array},
\cite{TLTqrac}.
There is a classification of
Leonard pairs in
\cite{LS99}. This classification amounts to
a linear algebraic version of a theorem of
D. Leonard
\cite{BanIto}, \cite{Leon}
concerning the $q$-Racah polynomials.
See 
\cite{TersubI},
\cite{qSerre},
\cite{LS24},
\cite{conform},
\cite{lsint},
\cite{Terint},
\cite{TLT:split},
\cite{TLT:array},
\cite{TLTqrac}
for more information about Leonard pairs.

\medskip
\noindent In this paper our focus is on general tridiagonal pairs.
We will discuss the following conjecture.

\begin{conjecture} (Ito, Tanabe, Terwilliger \cite{TD00})
 \label{con:bc}
 Let  $(\rho_0, \rho_1, \ldots, \rho_d)$ denote the shape vector
of a tridiagonal pair.
Then
\begin{eqnarray*}
\rho_i \;\leq \;\biggl( {{d}\atop {i}} \biggr) \qquad \qquad (0 \leq i \leq d).
%\label{eq:upperbnd}
\end{eqnarray*}
\end{conjecture} 

\noindent In the present paper we will prove Conjecture
 \ref{con:bc} for a certain type of tridiagonal pair. We will describe this
type shortly. For now we review some more facts about general 
tridiagonal pairs.

\medskip
\noindent
Let $A,A^*$ denote a tridiagonal pair on $V$.
By
\cite[Theorem 10.1]{TD00}
there exists a sequence
$\beta, \gamma, \gamma^*, \varrho,
 \varrho^*$ of scalars taken
from $\K$ such that 
both
\begin{eqnarray}
\lbrack A,A^2A^*-\beta AA^*A + 
A^*A^2 -\gamma (AA^*+A^*A)-\varrho A^*\rbrack &=&0, 
\label{eq:m1n}
\\
\lbrack A^*,A^{*2}A-\beta A^*AA^* + AA^{*2} -\gamma^* (A^*A+AA^*)-
\varrho^* A\rbrack&=&0,
\label{eq:m2n}
\end{eqnarray}
where $\lbrack r,s \rbrack $ means $\,rs-sr$.
The sequence is unique if the diameter is at least $3$.
The equations 
(\ref{eq:m1n}),
(\ref{eq:m2n}) 
are known as the {\it tridiagonal relations}
\cite{TD00}, \cite{qSerre}.
An ordering of the eigenvalues of $A$ (resp. $A^*$)
will be called {\it standard}
whenever the corresponding ordering of the eigenspaces of $A$
(resp. $A^*$) is standard.
Let $\theta_0,\theta_1,\ldots, \theta_d$ (resp. 
 $\theta^*_0,\theta^*_1,\ldots, \theta^*_d$)
 denote a standard ordering
of the eigenvalues of $A$ (resp. $A^*$). By
\cite[Theorem 11.1]{TD00} we have
\begin{eqnarray}
\theta^2_{i-1}
-
\beta \theta_{i-1}\theta_i
+
\theta^2_i
-\gamma (\theta_{i-1}+\theta_i)
&=&\varrho 
\qquad \qquad 
(1 \leq i \leq d),
\label{eq:eigrec}
\\
\theta^{*2}_{i-1}
-
\beta \theta^*_{i-1}\theta^*_i
+
\theta^{*2}_i
-\gamma^* (\theta^*_{i-1}+\theta^*_i)
&=&\varrho^*
\qquad \qquad 
(1 \leq i \leq d).
\label{eq:eigrec2}
\end{eqnarray}
\noindent 
 By \cite[Theorem 4.6]{TD00}
there exists a unique sequence $U_0, U_1, \ldots, U_d$ consisting of
subspaces of $V$ such that
\begin{eqnarray}
&&\qquad V = U_0+ U_1 + \cdots + U_d \qquad \qquad (\hbox{direct sum}),
\label{eq:vdec}
\\
&&(A-\theta_iI)U_i\subseteq U_{i+1} \qquad \quad (0 \leq i <d),  
\qquad (A-\theta_dI)U_d=0,   
\label{eq:Aonui}
\\
&& (A^*-\theta^*_iI)U_i\subseteq U_{i-1} \qquad \quad (0 < i \leq d), 
\qquad (A^*-\theta^*_0I)U_0=0.    
\label{eq:Asonui}
\end{eqnarray}
%For notational convenience, we define $U_i=0$ for $i<0 $ or $i>d$.
By  \cite[Corollary 5.7]{TD00}   
the space $U_i$ has dimension $\rho_i$ for $0 \leq i \leq d$,
where $(\rho_0, \rho_1, \ldots, \rho_d)$ is the shape vector
for $A,A^*$.
We call the sequence $U_0,U_1,\ldots, U_d$
the {\it split decomposition} for $A,A^*$ (with respect to
the orderings $\theta_0,\theta_1,\ldots, \theta_d$ and
 $\theta^*_0,\theta^*_1,\ldots, \theta^*_d$).
For $0 \leq i \leq d$ let 
$F_i:V\rightarrow V$ denote the linear transformation which satisfies both
\begin{eqnarray}
&&\qquad (F_i-I)U_i=0,
\label{eq:fi1}
\\
&&F_iU_{j}=0 \quad \hbox{if}\quad j\not=i, \qquad (0 \leq j \leq d).
\label{eq:fi2}
\end{eqnarray}
In other words $F_i$  is the projection map from $V$ onto $U_i$.
We observe
\begin{eqnarray*}
&&F_iF_j = \delta_{ij}F_i \qquad \qquad (0 \leq i,j\leq d),
\label{eq:fifiidem}
\\
&&\qquad F_0+F_1+\cdots +F_d = I,
\label{eq:fisumtoident}
\\
&&F_iV = U_{i} \qquad \qquad (0 \leq i \leq d).
\label{eq:fivjustvi}
\end{eqnarray*}
In view of 
(\ref{eq:Aonui}), 
(\ref{eq:Asonui})
we define
\begin{eqnarray}
R&=&A-\sum_{h=0}^d \theta_hF_h,
\label{eq:defR}
\\
L&=&A^*-\sum_{h=0}^d \theta^*_hF_h
\label{eq:defL}
\end{eqnarray}
and observe
\begin{eqnarray}
&& RU_i \subseteq U_{i+1} \qquad \qquad (0 \leq i <d), \qquad RU_d=0,
\label{eq:raction}
\\
&&LU_i \subseteq U_{i-1} \qquad \qquad  (0 < i \leq d), \qquad LU_0=0.
\label{eq:laction}
\end{eqnarray}
Combining 
(\ref{eq:raction}),
(\ref{eq:laction}) with
(\ref{eq:vdec}) we find
\begin{eqnarray}
\label{eq:nil}
R^{d+1} = 0, \qquad \qquad 
L^{d+1} = 0.
\end{eqnarray}
We call $R$ (resp. $L$)
the {\it raising map} (resp. {\it lowering map}) for
$A,A^*$ 
with respect to $U_0, U_1, \ldots, U_d$.

\medskip
\noindent We now describe the type of 
tridiagonal pair for
 which we will prove
Conjecture \ref{con:bc}.
 

\begin{definition} 
\label{def:1p1}
\rm
For the rest of this paper, we assume $\K$ is algebraically closed
with characteristic 0.
We fix a nonzero scalar $q\in \fld$
which is not a root of unity.
We let $A,A^*$ denote a  tridiagonal pair on $V$
 which satisfies
(\ref{eq:m1n}),
(\ref{eq:m2n})
where
\begin{eqnarray}
\beta = q^2 + q^{-2},
\qquad \qquad  
\gamma = \gamma^* = \varrho= \varrho^* = 0. 
\label{eq:simpleassume}
\end{eqnarray}


\end{definition}
\begin{note}
\rm
The scalar $q$ which appears in \cite{TD00}   
is the same as $q^2$ in the present paper. We make this adjustment
for notational convenience.
\end{note}


\medskip
\noindent Before proceeding we discuss the significance of
assumption (\ref{eq:simpleassume}) from several points of view.
Let $A,A^*$ denote the tridiagonal pair
in Definition
\ref{def:1p1}.
Evaluating
 (\ref{eq:m1n}),
(\ref{eq:m2n})
using
(\ref{eq:simpleassume})
we obtain
\begin{eqnarray}
\label{eq:qs1}
 A^3A^* - \lbrack 3 \rbrack A^2A^*A
+ \lbrack 3 \rbrack AA^*A^2 - A^*A^3&=&0,
 \\
 A^{*3}A - \lbrack 3 \rbrack A^{*2}AA^*
+ \lbrack 3 \rbrack A^*AA^{*2} - AA^{*3}&=&0.
\label{eq:qs2} 
\end{eqnarray}
We are using the notation
\begin{eqnarray}
 \lbrack n\rbrack=\frac{q^n-q^{-n}}{q-q^{-1}}\qquad \qquad n=0,1,\ldots
\label{eq:qi}
\end{eqnarray}
The equations 
(\ref{eq:qs1}),
(\ref{eq:qs2}) are known as the {\it $q$-Serre relations}, and
are among the defining relations for the quantum affine algebra
$U_q(\widehat{ sl_2})$. See 
\cite{cp} and \cite{lu} for more information on $U_q(\widehat{ sl_2})$. 

\medskip
\noindent 
Assumption 
 (\ref{eq:simpleassume}) has the following significance 
for the eigenvalues.
Let $A,A^*$ denote the tridiagonal pair 
 in
Definition 
\ref{def:1p1}.
Let $\theta_0, \theta_1,\ldots, \theta_d$
(resp.
 $\theta^*_0, \theta^*_1,\ldots, \theta^*_d$)
denote a standard ordering of the eigenvalues of $A$ (resp. $A^*$).
Evaluating
(\ref{eq:eigrec})
using 
$\beta = q^2+q^{-2}$, 
$\gamma=0$, 
$\varrho=0$
we find either $\theta_i=q^2\theta_{i-1}$ for
$1 \leq i \leq d$
or $\theta_i=q^{-2}\theta_{i-1}$ for
$1 \leq i \leq d$. Replacing 
$\theta_0, \theta_1,\ldots, \theta_d$
by 
$\theta_d, \theta_{d-1},\ldots, \theta_0$ if necessary
we may assume 
 $\theta_i=q^2\theta_{i-1}$ for
$1 \leq i \leq d$. In this case
there exists
$a \in \fld$ such that
\begin{eqnarray}
\label{eq:eig}
\theta_i = a q^{2i-d} \qquad  (0 \leq i \leq d).
\end{eqnarray}
Similarly
replacing 
$\theta^*_0, \theta^*_1,\ldots, \theta^*_d$
by 
$\theta^*_d, \theta^*_{d-1},\ldots, \theta^*_0$ if necessary,
 there exists
$a^* \in\K$ such that
\begin{eqnarray}
\label{eq:eigs}
\theta^*_i = a^* q^{d-2i} \qquad \qquad (0 \leq i \leq d).
\end{eqnarray}

\noindent Assumption 
(\ref{eq:simpleassume}) has the following significance
for the raising and lowering maps.
Let $A,A^*$ denote the tridiagonal pair
in Definition 
\ref{def:1p1}. Let
$U_0, U_1, \ldots, U_d$ denote the split
decomposition
with respect to
(\ref{eq:eig}), 
(\ref{eq:eigs})
and let $R$ (resp. $L$)
denote the corresponding raising (resp. lowering)
map.
By \cite[Theorem 12.2]{TD00} 
the maps $R, L$ satisfy the $q$-Serre relations
\begin{eqnarray}
R^3L - \lbrack 3 \rbrack R^2LR+ \lbrack 3 \rbrack RLR^2 - LR^3&=&0, 
\label{eq:ms1}
\\
L^3R - \lbrack 3 \rbrack L^2RL+ \lbrack 3 \rbrack LRL^2 - RL^3&=&0.
\label{eq:ms2}
\end{eqnarray}
We emphasize that each of the pairs $R,L$ and $A,A^*$ satisfy the
$q$-Serre relations.  
Each of $R, L$
 is nilpotent on $V$ and each of $A, A^*$
is diagonalizable on $V$.

\medskip
\noindent
The  following is our main result.
\begin{theorem} 
\label{thm:bc}
Let $A,A^*$ denote the tridiagonal pair in
Definition 
\ref{def:1p1} and let 
$(\rho_0,\rho_1,\ldots, \rho_d)$ denote the 
corresponding shape vector. Then
\begin{eqnarray*}
\rho_i \;\leq \;\biggl( {{d}\atop {i}} \biggr) 
\qquad \qquad (0 \leq i \leq d).
%\label{eq:upperbnd}
\end{eqnarray*}
\end{theorem} 

\medskip
\noindent 
In order to prove
Theorem \ref{thm:bc}, for $0\leq i \leq d$
we display a spanning set for $U_i$ consisting
of 
$({{d}\atop{i}})$ vectors. We proceed as follows.

\begin{theorem} 
\label{con:Vspanguess}
Let $A,A^*$ denote the tridiagonal pair in 
Definition 
\ref{def:1p1}.
Let $U_0, U_1, \ldots, U_d$ denote the split
decomposition with respect to
(\ref{eq:eig}), 
(\ref{eq:eigs})
and let $R, L$
denote the corresponding raising and lowering
maps.
Let $v$ denote a nonzero vector in $U_0$.
Then $V$ is spanned by
the vectors of the form
\beast
L^{i_1}R^{i_2}L^{i_3}R^{i_4} \cdots R^{i_n}v,
\eeast
where 
$i_1, i_2, \ldots, i_n$ ranges over all sequences
such that $n$ is a nonnegative even integer, and 
 $i_1, i_2, \ldots i_n$ are integers
satisfying $0\leq i_1<i_2<\cdots <i_n\leq d$.
\end{theorem}

\medskip
\noindent We prove Theorem
\ref{con:Vspanguess}  at the end of Section 2.
In order to illustrate 
Theorem
\ref{con:Vspanguess} 
we now restate it
in concrete terms for $d=3$.


\begin{corollary} With reference to
Theorem
\ref{con:Vspanguess}, for
$d=3$
the space
$V$ is spanned by the vectors 
\beast
v,\quad  Rv, \quad R^2v, \quad  R^3v, 
\quad LR^2v, \quad LR^3v,
\quad L^2R^3v, \quad  
RL^2R^3v.
\eeast
\end{corollary}


\medskip
\noindent 
Combining Theorem
\ref{con:Vspanguess}
with 
(\ref{eq:raction}),
(\ref{eq:laction}) 
we routinely obtain the following theorem.



\begin{theorem} 
\label{thm:us}
Let $A,A^*$ denote the tridiagonal pair in 
Definition 
\ref{def:1p1}.
Let $U_0, U_1, \ldots, U_d$ denote the split
decomposition with respect to
(\ref{eq:eig}), 
(\ref{eq:eigs})
and let $R, L$
denote the corresponding raising and lowering
maps.
Let $v$ denote a nonzero vector in $U_0$.
Then for $0 \leq i \leq d$ the space $U_i$ is spanned by
the vectors of the form
\beast
L^{i_1}R^{i_2}L^{i_3}R^{i_4} \cdots R^{i_n}v,
\eeast
where 
$i_1, i_2, \ldots, i_n$ ranges over all sequences
such that (i) $n$ is a nonnegative even integer;
(ii)  $i_1, i_2, \ldots i_n$ are integers
satisfying $0\leq i_1<i_2<\cdots <i_n\leq d$ and $i=\sum_{h=1}^n i_h(-1)^h$.
\end{theorem}


\begin{corollary}
With reference to
Theorem
\ref{thm:us},
for $d=3$ 
the following (i)--(iv) hold.
\begin{enumerate}
\item
The space $U_0$ is spanned by
\beast
v.
\eeast
\item The space $U_1$ is spanned by
\beast
Rv, \quad LR^2v,\quad L^2R^3v.
\eeast
\item The space $U_2$ is spanned by
\beast
\quad R^2v, 
 \quad LR^3v,
 \quad  
RL^2R^3v.
\eeast
\item The space $U_3$ is spanned by
\beast
R^3v.
\eeast
\end{enumerate}
\end{corollary}

\section{The algebra $\mathcal A$}

\noindent Our next goal is to prove 
Theorem
\ref{con:Vspanguess}.
In order to do this we 
undertake a careful investigation of the 
$q$-Serre relations.
We begin with a definition.

\begin{definition}
\label{def:a}
\rm
We let $\mathcal A$ denote the associative $\fld$-algebra with identity
generated by symbols $x,y$ subject to the $q$-Serre relations
\begin{eqnarray*}
x^3y-\lbrack 3 \rbrack x^2yx+\lbrack 3 \rbrack xyx^2-yx^3&=&0,
\\
y^3x-\lbrack 3 \rbrack y^2xy+\lbrack 3 \rbrack yxy^2-xy^3&=&0.
\end{eqnarray*}
\end{definition}

\begin{remark}
\rm
The algebra $\mathcal A$ in Definition 
\ref{def:a}
is often called the
{\it positive part of 
$U_q({\widehat {sl_2}})$}.
\end{remark}

\begin{definition}
\label{def:s}
\rm
Referring to Definition \ref{def:a},
we let $\sigma_0:{\mathcal A}\rightarrow {\mathcal A}$ denote the $\fld$-algebra isomorphism
which sends $x$ to $y$ and $y$ to $x$.
We let $\sigma_1:{\mathcal A}\rightarrow{\mathcal A}$ denote the $\fld$-algebra antiisomorphism
which stabilizes each of 
$x,y$. The action of $\sigma_1$ inverts the word order.
We observe $\sigma_0, \sigma_1$  commute.
We define $\sigma=\sigma_0\sigma_1$ and
observe each of $\sigma_0, \sigma_1,\sigma$ is an involution.
\end{definition} 

\begin{example}
\label{ex:sig01}
Applying each of $\sigma_0, \sigma_1, \sigma$ to
$xyx^2y^2$ we get
\begin{eqnarray*}
yxy^2x^2,
\qquad  y^2x^2yx,
 \qquad x^2y^2xy
\end{eqnarray*}
 respectively.
\end{example}

\medskip
\noindent
We now state a lemma which we will find useful.
We first recall some notation.
Let $n$ denote a nonnegative integer.
With reference to 
(\ref{eq:qi})
we define
\begin{eqnarray*}
\lbrack n\rbrack !=
\lbrack n\rbrack 
\lbrack n-1\rbrack 
\cdots
\lbrack 2 \rbrack
\lbrack 1\rbrack.
\end{eqnarray*}
We interpret $\lbrack 0\rbrack !=1$.
We also define
\begin{eqnarray*}
\biggl\lbrack
{{n}\atop {i}}\biggr\rbrack
=\frac{\lbrack n\rbrack!}{\lbrack i\rbrack ! \lbrack n-i \rbrack !}
\qquad (0 \leq i\leq n).
\end{eqnarray*}

\begin{lemma}
\cite[Proposition 7.1.5]{lu}
\label{lem:ho}
The following relations
hold in $\mathcal A$.
Let $r$ denote a positive integer.
%and abbreviate $n=2r+1$. 
Then
\begin{eqnarray}
\sum_{i=0}^{2r+1}
(-1)^i
\biggl\lbrack
{{2r+1}\atop {i}}\biggr\rbrack
x^iy^rx^{2r+1-i} &=&0,
\label{eq:hoq}
\\
\sum_{i=0}^{2r+1}
(-1)^i
\biggl\lbrack
{{2r+1}\atop {i}}\biggr\rbrack
y^ix^ry^{2r+1-i} &=&0.
\label{eq:hoq2}
\end{eqnarray}
\end{lemma}

\begin{remark} \rm
The relations
(\ref{eq:hoq}), (\ref{eq:hoq2})
are often called the {\it higher order $q$-Serre relations} \cite[p.57]{lu}.
\end{remark}



\medskip
\noindent We now give the higher order $q$-Serre relations
in a modified form. We will use the following notation.
For all integers $n,k$ with $k$ nonnegative we define
\begin{eqnarray*}
%\label{eq:bnk}
\lbrack n\rbrack_k=
\lbrack n\rbrack
\lbrack n-1\rbrack
\cdots
\lbrack n-k+1\rbrack.
\end{eqnarray*}
We interpret $\lbrack n\rbrack_0=1$.
%We remark 
%$\lbrack n\rbrack_0=1$.

%oldver
%\begin{lemma}
%(\cite{})
%Let $r$ denote a positive integer.
%Let $i,j$ denote integers such that
%$i\geq r+1$ and $j\geq r$.
%%Then
%\begin{eqnarray*}
%\label{eq:ho}
%x^iy^rx^j=
%\sum_{\eta=0}^r
%\frac{
%\lbrack i\rbrack_{r+1}
%\lbrack j\rbrack_{r}
%}
%{
%\lbrack i-\eta\rbrack
%\lbrack \eta \rbrack_{r+1}
%\lbrack i+j-\eta \rbrack_{r}
%}
%x^{\eta}y^rx^{i+j-\eta}
%+
%\sum_{\zeta=0}^{r-1}
%\frac{
%\lbrack i\rbrack_{r+1}
%\lbrack j\rbrack_{r}
%}
%{
%\lbrack j-\zeta\rbrack
%\lbrack i+j-\zeta\rbrack_{r+1}
%\lbrack \zeta\rbrack_{r}
%}
%x^{i+j-\zeta}y^rx^{\zeta}
%\end{eqnarray*}
%\end{lemma}

%newver
\begin{lemma}
\label{lem:homod}
The following relations hold in $\mathcal A$.
Let $r$ denote a positive integer
and let $n$ denote an integer at least $2r+1$.
Then for $0 \leq i\leq n$ we have
\begin{eqnarray}
\label{eq:ho}
&&x^iy^rx^{n-i}=
\sum_{\xi=0}^r
\frac{
\lbrack i\rbrack_{\xi}
\lbrack r-i\rbrack_{r-\xi}
\lbrack n-i\rbrack_{r}
}
{
\lbrack \xi \rbrack_{\xi}
\lbrack r-\xi \rbrack_{r-\xi}
\lbrack n-\xi \rbrack_{r}
}
x^{\xi}y^rx^{n-\xi}
\\
&&\qquad \qquad \qquad +
\sum_{\zeta=0}^{r-1}
\frac{
\lbrack i\rbrack_{r+1}
\lbrack i-n+r-1\rbrack_{r-\zeta-1}
\lbrack n-i\rbrack_{\zeta}
}
{
\lbrack n-\zeta \rbrack_{r+1}
\lbrack r-\zeta-1 \rbrack_{r-\zeta-1}
\lbrack \zeta \rbrack_{\zeta}
}
x^{n-\zeta}y^rx^{\zeta}.
\label{eq:zeta}
\end{eqnarray}
\end{lemma}
\noindent
%{\bf Proof:}
%We
%abbreviate $w_i=x^iy^rx^{n-i}$
%for $0 \leq i \leq n$. We obtain a system of $n-2r$ linear equations relating
%$w_0,w_1,\ldots,w_n$ as follows.
%For $0 \leq j\leq n-2r-1$ we multiply each term
%in 
%(\ref{eq:hoq}) on the left by $x^j$ and on the right by
%$x^{n-2r-1-j}$. The result is a linear equation
%relating
%$w_j,
%w_{j+1},\ldots,
%w_{j+2r+1}$.
%This yields a system of $n-2r$ linear equations
%involving
%$w_0,w_1,\ldots,w_n$.
%We now solve this system 
%to obtain each of
%$w_0,w_1,\ldots,w_n$
%in terms of $w_0,w_1,\ldots,w_r$ and
% $w_{n-r+1},w_{n-r+2},\ldots,w_n$.
%Let $C$ denote the coefficient matrix for the 
%system.
%Observe $C$ is $(n-2r)\times (n+1)$.
%For $0 \leq i \leq n-2r-1$
%and $0 \leq j \leq 2r+1$
%the entry $C_{i,i+j}=
%(-1)^j\biggl\lbrack
%{{2r+1}\atop {j}}\biggr
%\rbrack
%$. The remaining entries of $C$ are zero.
%The matrix $C$ is in row echelon form so it has rank $n-2r$.
%We define an
% $(n+1)\times (n+1)$ matrix $D$ as follows.
%For $0 \leq i\leq n$ and $ 0 \leq \xi \leq r$ the entry
%$D_{i\xi}$ is equal to the coefficient of $x^\xi y^r x^{n-\xi}$
%in 
%(\ref{eq:ho}).
%For $0 \leq i \leq n$ and 
% $0 \leq \zeta  \leq r-1$ the entry 
% $D_{i,n-\zeta}$ is 
%equal to the coefficient of $x^{n-\zeta}y^r x^\zeta $ in
%(\ref{eq:zeta}).
%All remaining entries of $D$ are zero.
%It suffices to show
%the row space of $D-I$ is contained in the row space
%of $C$. 
%For the  matrix
%$D$ 
%each of the columns $r+1, r+2, \ldots, n-r$ is zero.
%For the matrix $D-I$
%each of the rows $0,1,\ldots, r,n-r+1,n-r+2,\ldots, n$ is 0.
%From this we find $D$ has rank $2r+1$.
%Also $D^2=D$ so $(D-I)D=0$.
%Apparently the row space of $D-I$ is orthogonal
%to 
%the column 
%space of $D$ (with respect to the dot product.)
%We define an $(n+1)\times (2r+1)$ matrix $S$ as follows.
%For $0 \leq i \leq n $ and $-r \leq j \leq r$
%the entry $S_{ij}= q^{2ij}$.
%The matrix  $S$ has rank 
%$2r+1$ since it
%is essentially Vandermonde.
%Observe $CS=0$.
%Therefore the row space of $C$ is  orthogonal to
%the column space of $S$.
%By this and since the ranks of $C, S$ sum to $n+1$
%we find
%the row space of $C$ is equal to
%the orthogonal complement of the column space of $S$.
%Observe 
%the column space of $D$ is contained in the
%column space of $S$.
%By this and since 
%$D, S$ have the same
%rank we find
%the column space of $D$ is equal to
%the column space of
%$S$.
%By these remarks
%the row space of $D-I$ is contained
%in the  row space of $C$.
%The result follows.
%\hfill $\Box $ \\
{\bf Proof:}
We
abbreviate $w_i=x^iy^rx^{n-i}$
for $0 \leq i \leq n$. We obtain a system of $n-2r$ linear equations relating
$w_0,w_1,\ldots,w_n$ as follows.
For $0 \leq j\leq n-2r-1$ we multiply each term
in 
(\ref{eq:hoq}) on the left by $x^j$ and on the right by
$x^{n-2r-1-j}$. The result is a linear equation
relating
$w_j,
w_{j+1},\ldots,
w_{j+2r+1}$.
This yields a system of $n-2r$ linear equations
involving
$w_0,w_1,\ldots,w_n$.
We now solve this system 
to obtain each of
$w_0,w_1,\ldots,w_n$
in terms of $w_0,w_1,\ldots,w_r$ and
 $w_{n-r+1},w_{n-r+2},\ldots,w_n$.
We define a $(n+1)\times (2r+1)$ matrix $S$ as follows.
For
$0 \leq i \leq n$ and
$-r \leq j \leq r$ the entry
$S_{ij}=q^{2ij}$.
For $0 \leq i\leq n$ let $S_i$ denote row $i$ of $S$.
Observe that for $0 \leq j\leq n-2r-1$ the vectors
$S_j, S_{j+1}, \ldots, S_{j+2r+1}$ satisfy the same 
equation as the one for
$w_j,
w_{j+1},\ldots,
w_{j+2r+1}$ which we mentioned above.
Observe $S_0, S_1, \ldots, S_{2r}$
are linearly independent since
 $S$ is
essentially Vandermonde.
Therefore $S_0, S_1, \ldots, S_{2r}$ 
is a basis for $\K^{2r+1}$.
Consider the linear transformation $\varepsilon $
from $\K^{2r+1}$ to
$\mbox{Span}(
w_0, w_1, \ldots, w_n)$ which 
sends $S_i$ to $w_i$ for $0 \leq i \leq 2r$.
From our above comments $\varepsilon $ sends 
$S_i$ to $w_i$ for $0 \leq i \leq n$.
For $0 \leq i\leq n$ we have
\begin{eqnarray}
\label{eq:hoproof}
&&S_i=
\sum_{\xi=0}^r
\frac{
\lbrack i\rbrack_{\xi}
\lbrack r-i\rbrack_{r-\xi}
\lbrack n-i\rbrack_{r}
}
{
\lbrack \xi \rbrack_{\xi}
\lbrack r-\xi \rbrack_{r-\xi}
\lbrack n-\xi \rbrack_{r}
}
S_{\xi}
\\
&&\qquad \qquad \qquad +
\sum_{\zeta=0}^{r-1}
\frac{
\lbrack i\rbrack_{r+1}
\lbrack i-n+r-1\rbrack_{r-\zeta-1}
\lbrack n-i\rbrack_{\zeta}
}
{
\lbrack n-\zeta \rbrack_{r+1}
\lbrack r-\zeta-1 \rbrack_{r-\zeta-1}
\lbrack \zeta \rbrack_{\zeta}
}
S_{n-\zeta}.
\label{eq:zetaproof}
\end{eqnarray}
This can be verified using Cramer's rule.
Applying $\varepsilon$ to each term in 
(\ref{eq:hoproof}), 
(\ref{eq:zetaproof}) we obtain
(\ref{eq:ho}), 
(\ref{eq:zeta}).
\hfill $\Box $ \\


\begin{remark}
\label{rem:m}
 \rm
Applying each of $\sigma_0, \sigma_1,\sigma$
to the relations in Lemma
\ref{lem:homod}, we obtain additional relations involving $x, y$ which hold in 
$\mathcal A$.
\end{remark}


\noindent The following fact is an immediate consequence
of Lemma
\ref{lem:homod}
and Remark
\ref{rem:m}.


\begin{corollary}
\label{cor:dep}
The following linear dependencies hold in $\mathcal A$.
Let $r$ denote a positive integer and let $n$ denote an integer at least
$2r+1$. 
\begin{enumerate}
\item For $r+1\leq i \leq n-r$ the element $x^iy^rx^{n-i}$ is contained in
\begin{eqnarray*}
\mbox{Span}\lbrace x^{\xi}y^rx^{n-\xi} \;|\;0\leq \xi\leq r \rbrace+
\mbox{Span}\lbrace x^{n-\zeta}y^rx^{\zeta} \;|\;0\leq \zeta\leq r-1 \rbrace.
\end{eqnarray*}
\item For $r\leq i \leq n-r-1$ the element $x^iy^rx^{n-i}$ is contained in
\begin{eqnarray*}
\mbox{Span}\lbrace x^{\xi}y^rx^{n-\xi} \;|\;0\leq \xi\leq r-1 \rbrace+
\mbox{Span}\lbrace x^{n-\zeta}y^rx^{\zeta} \;|\;0\leq \zeta\leq r \rbrace.
\end{eqnarray*}
\item For $r+1\leq i \leq n-r$ the element $y^ix^ry^{n-i}$ is contained in
\begin{eqnarray*}
\mbox{Span}\lbrace y^{\xi}x^ry^{n-\xi} \;|\;0\leq \xi\leq r \rbrace+
\mbox{Span}\lbrace y^{n-\zeta}x^ry^{\zeta} \;|\;0\leq \zeta\leq r-1 \rbrace.
\end{eqnarray*}
\item For $r\leq i \leq n-r-1$ the element $y^ix^ry^{n-i}$ is contained in
\begin{eqnarray*}
\mbox{Span}\lbrace y^{\xi}x^ry^{n-\xi} \;|\;0\leq \xi\leq r-1 \rbrace+
\mbox{Span}\lbrace y^{n-\zeta}x^ry^{\zeta} \;|\;0\leq \zeta\leq r \rbrace.
\end{eqnarray*}
\end{enumerate}
\end{corollary}


\begin{definition} 
\label{def:word}
\rm
Let $n$ denote a nonnegative integer. By
a {\it word of length n} in $\mathcal A$, we mean an expression
of the form
\begin{equation}
a_1a_2\cdots a_n,
\label{eq:word}
\end{equation}
where $a_i=x$ or $a_i=y$ for $1 \leq i \leq n$.
We 
interpret the word of length 0 as the 
 identity element in $\mathcal A$.
We say this word is {\it trivial}.
By the {\it height} of the word
(\ref{eq:word}) we mean the integer
\begin{eqnarray*}
|\lbrace i\;|\;1 \leq i \leq  n, \quad a_i=x\rbrace | \;-\;
|\lbrace i\;|\;1 \leq i \leq  n, \quad a_i=y\rbrace |.
%\label{eq:leveldef}
\end{eqnarray*}

\begin{definition}\rm
A word is said to be {\it balanced}
whenever it has height 0. We observe that a balanced
word has even length.
\end{definition}

\begin{example} 
We list the balanced words of length $4$.
\begin{eqnarray*}
x^2y^2,
\qquad xyxy,
\qquad xy^2x,
\qquad y^2x^2,
\qquad yxyx,
\qquad yx^2y. 
\end{eqnarray*}
\end{example}
\end{definition}

\begin{definition} \rm
Let 
$a_1a_2\cdots a_n$ 
denote a word 
in $\mathcal A$.
By the {\it height vector} of 
 $a_1a_2\cdots a_n$ 
 we mean the sequence
$(h_0,h_1,\ldots, h_n)$,
where $h_i$ denotes the height of the word
$a_{i+1}a_{i+2}\cdots a_n$ for $0 \leq i \leq n$.
We observe that the height vector of
 $a_1a_2\cdots a_n$ 
is the unique sequence  $(h_0,h_1,\ldots, h_n)$ 
 such that
(i)
$h_n=0$;
(ii)
for $1\leq i \leq n$, $h_{i-1}-h_{i}=1$
if $a_i=x$ and  $h_{i-1}-h_{i}=-1$
if $a_i=y$.
\end{definition}

\begin{example}
The word $x^2yx^3y^2x$ has height vector
\begin{eqnarray*}
(3,2,1,2,1,0,-1,0,1,0).
\end{eqnarray*}
\end{example}


\begin{definition}
\rm
Let $a_1a_2\cdots a_n$ denote a word in $\mathcal A$
and let $(h_0, h_1,\ldots, h_n)$ denote the corresponding
height vector.
We say $a_1a_2\cdots a_n$ 
is
 {\it height-symmetric}
whenever $h_i=h_{n-i}$ for $0 \leq i \leq n$.
We observe that $a_1a_2\cdots a_n$ is height-symmetric
if and only if $a_i\not=a_{n-i+1}$ for
$1\leq i \leq n$.
We remark that a height-symmetric word is balanced and invariant 
under $\sigma$.
\end{definition}

\begin{example}
We list the height-symmetric words of length 4.
\begin{eqnarray*}
x^2y^2,
\qquad xyxy,
\qquad y^2x^2,
\qquad yxyx.
\end{eqnarray*}
\end{example}

\begin{definition}\rm
\rm
Let $a_1a_2\cdots a_n$ denote a word in $\mathcal A$
and let $(h_0, h_1,\ldots, h_n)$ denote the corresponding
height vector.
We say $a_1a_2\cdots a_n$ 
is
{\it nil}
whenever at least one of 
$h_0, h_1,\ldots, h_n$ is negative.
\end{definition}


\begin{example}
We list the nil balanced words of length 4.
\begin{eqnarray*}
x^2y^2, \qquad 
xyxy, 
\qquad xy^2x,
\qquad yx^2y. 
\end{eqnarray*}
\end{example}

\medskip
\noindent We mention a few subspaces of $\mathcal A$.

\begin{definition} \rm
We let $\mathcal B$ denote the linear subspace of $\mathcal A$
spanned by the balanced words. We observe $\mathcal B$ is a $\K$-subalgebra of
$\mathcal A$ which is invariant under each of
$\sigma_0, \sigma_1, \sigma$. We let 
${\mathcal B}^{sym}$
denote the linear subspace of $\mathcal B$ spanned by the height-symmetric
words. 
We let
%${\mathcal A}^{nil}$
${\mathcal B}^{nil}$
denote the linear subspace of 
%$\mathcal A$
%(resp. $\mathcal B$)
 $\mathcal B$
%spanned by the nil words (resp. nil balanced words.)
 spanned by the nil balanced words.
We observe 
%${\mathcal A}^{nil}$
%is a left ideal in $\mathcal A$.
%Moreover
 ${\mathcal B}^{nil}$
is a two sided ideal of $\mathcal B$ which is invariant under $\sigma$.
\end{definition}


%\begin{lemma}
%For $F \in {\mathcal F}$
% the following (i), (ii) are equivalent:
%\begin{enumerate}
%\item
%$F-F^{\sigma} \in {\mathcal F}^{nil}$.
%\item
%$F \in {\mathcal F}^{sym}+{\mathcal F}^{nil}$.
%\end{enumerate}
%\end{lemma}
%\noindent {\bf Proof:}
%(i)$\rightarrow $ (ii). Observe
%\begin{eqnarray*}
%F=\frac{F+F^\sigma}{2}+
%\frac{F-F^\sigma}{2}.
%\end{eqnarray*}
%Recall $\sigma$ is an involution so
%$F+F^\sigma \in {\mathcal F}^{sym}$. By assumption
%$F-F^\sigma \in {\mathcal F}^{nil}$. The result follows.
%\\
%\noindent (ii)$\rightarrow $(i).
%There exists $F_0\in {\mathcal F}^{sym}$
%and there exists $F_1\in {\mathcal F}^{nil}$
%such that
%$F=F_0+F_1$. Observe 
%$F_0^{\sigma}=F_0$ so $F-F^\sigma=F_1-F_1^{\sigma}$.
%By assumption $F_1 \in {\mathcal F}^{nil}$.
%Observe $\sigma$ leaves ${\mathcal F}^{nil}$ invariant so
%$F_1^{\sigma}\in {\mathcal F}^{nil}$.
%By these comments 
%$F-F^{\sigma} \in {\mathcal F}^{nil}$.
%\hfill $\Box $ \\

\begin{theorem}
\label{thm:m}
Let $i,j,m,n$ denote nonnegative integers such that $i+n=j+m$. Define
$b=y^ix^my^nx^j$ and observe $b$ is a balanced word. Then
\begin{eqnarray}
b  \in 
{\mathcal B}^{sym}+
{\mathcal B}^{nil}.
\label{eq:sn}
\end{eqnarray}
\end{theorem}
\noindent {\bf Proof:}
%We assume $i\geq m$; otherwise $b \in {\mathcal B}^{nil}$.
%Combining $i\geq m$ with $i+n=j+m$ we find
%$n\leq j$.
We proceed by induction on $\mbox{min}(m,n)$. First assume
 $\mbox{min}(m,n)=0$.
Then $b$ is height-symmetric.  Therefore
$b \in {\mathcal B}^{sym}$ so
(\ref{eq:sn}) holds.
Next assume
 $\mbox{min}(m,n)>0$.
Replacing $b$ by $b^\sigma$ if necessary we may assume
$m\geq n$.
Assume for the moment that $m=n$. Then 
$b$ is height-symmetric.
Therefore
$b \in {\mathcal B}^{sym}$ so
(\ref{eq:sn}) holds.
Next assume  $m>n$. 
We may assume $n\leq j$; otherwise $b \in 
{\mathcal B}^{nil}$.
By Corollary
\ref{cor:dep}(i),
$b$ is contained in 
\begin{eqnarray}
\mbox{Span}\lbrace y^{i}x^{\xi}y^nx^{i+n-\xi} \;|\;0\leq \xi\leq n \rbrace+
\mbox{Span}\lbrace y^{i}x^{i+n-\zeta}y^{n}x^{\zeta} \;|\;0\leq \zeta\leq n-1 
\rbrace.
\label{eq:sp}
\end{eqnarray}
By the induction hypothesis, for $0 \leq \xi \leq n-1$ the word
$y^{i}x^{\xi}y^nx^{i+n-\xi}$
is contained in
${\mathcal B}^{sym}+
{\mathcal B}^{nil}$.
For $\xi=n$ the word
$y^{i}x^{\xi}y^nx^{i+n-\xi}$ is height-symmetric
and therefore contained in
${\mathcal B}^{sym}$.
For $0\leq \zeta\leq n-1 $ the word
$y^{i}x^{i+n-\zeta}y^{n}x^{\zeta}$
is contained in 
${\mathcal B}^{nil}$.
Apparently the space displayed in
(\ref{eq:sp}) is contained in
${\mathcal B}^{sym}+{\mathcal B}^{nil}$.
It follows
$b \in {\mathcal B}^{sym}+{\mathcal B}^{nil}$ as desired.
\hfill $\Box $ \\


\begin{lemma}
\label{label:sym}
For  
$b \in {\mathcal B}^{sym}+{\mathcal B}^{nil}$ we have
$b-b^{\sigma} \in {\mathcal B}^{nil}$.
\end{lemma}
\noindent {\bf Proof:}
There exists $b_0\in {\mathcal B}^{sym}$
and there exists $b_1\in {\mathcal B}^{nil}$
such that
$b=b_0+b_1$. Observe 
$b_0^{\sigma}=b_0$ so $b-b^\sigma=b_1-b_1^{\sigma}$.
By assumption $b_1 \in {\mathcal B}^{nil}$.
Recall $\sigma$ leaves ${\mathcal B}^{nil}$ invariant so
$b_1^{\sigma}\in {\mathcal B}^{nil}$.
By these comments 
$b-b^{\sigma} \in {\mathcal B}^{nil}$.
\hfill $\Box $ \\

\begin{corollary}
\label{thm:mnew}
Let $i,j,m,n$ denote nonnegative integers such that $i+n=j+m$. Define
$b=y^ix^my^nx^j$ and recall $b$ is a balanced word. Then
\begin{eqnarray*}
b - b^{\sigma} \in {\mathcal B}^{nil}.
\end{eqnarray*}
\end{corollary}
\noindent {\bf Proof:}
Immediate from Theorem
\ref{thm:m}
and 
Lemma
\ref{label:sym}.
\hfill $\Box $ \\

%\medskip
%\noindent We state a special  case of Corollary
%\ref{thm:mnew}. We will use the following notation.
%
%
%\begin{definition}
%\rm
%For each nonnegative integer  $i$ we define
%\begin{eqnarray}
%\label{eq:del}
%\Delta_i =y^ix^i.
%\end{eqnarray}
%\end{definition}

\begin{corollary}
\label{cor:del}
For nonnegative integers $i,j$ we have
\begin{eqnarray*}
\lbrack y^ix^i, y^jx^j \rbrack \in {\mathcal B}^{nil}.
\end{eqnarray*}
\end{corollary}
\noindent {\bf Proof:}
Apply Corollary
\ref{thm:mnew} with $m=i$ and $n=j$.
\hfill $\Box $ \\

\noindent In order to state the next theorem we make a
definition.

\begin{definition}
\rm
Let $a_1a_2\cdots a_n$ 
denote a word in $\mathcal A$.
%By the {\it signature}
%of  $a_1a_2\cdots a_n$ 
Observe that there exists a unique sequence
 $(i_1, i_2, \ldots, i_r)$ consisting of positive
integers such that
 $a_1a_2\cdots a_n$ is one of
$x^{i_1}
y^{i_2}
x^{i_3}
\cdots
y^{i_r}$
or
$
x^{i_1}
y^{i_2}
x^{i_3}
\cdots
x^{i_r}$
or
$y^{i_1}
x^{i_2}
y^{i_3}
\cdots
x^{i_r}$
or
$
y^{i_1}
x^{i_2}
y^{i_3}
\cdots
y^{i_r}$.
We call the sequence
 $(i_1, i_2, \ldots, i_r)$ the {\it signature} of
$a_1a_2\cdots a_n$.
\end{definition}

\begin{example}
Each  of the words $yx^2y^2x$,
$xy^2x^2y$ has signature
$(1,2,2,1)$.
\end{example}

\begin{definition}
\label{def:irr}
\rm
Let $a_1a_2\cdots a_n$ denote a word in $\mathcal A$
and let $(i_1, i_2,\ldots, i_r)$
denote the corresponding signature.
We say  $a_1a_2\cdots a_n$
is {\it reducible} whenever 
there exists an integer $s$
$(2\leq s\leq r-1)$ such that
$i_{s-1}\geq i_s<i_{s+1}$.
We say a word is {\it irreducible} whenever it is
not reducible.
\end{definition}

\begin{example}
A word of length less than 4 is irreducible.
The only reducible words of length 4 are
 $xyx^2 $ and $yxy^2$. 
\end{example}

\medskip
\noindent In the following lemma we give a necessary and sufficient
condition for a given nontrivial word to be irreducible.

\begin{lemma}
\label{lem:red}
Let $a_1a_2\cdots a_n$ denote a  nontrivial word in $\mathcal A$
and let $(i_1, i_2,\ldots, i_r)$
denote the corresponding signature. 
Then the following (i), (ii)  are equivalent.
\begin{enumerate}
\item
 The word $a_1a_2\cdots a_n$
is irreducible.
\item  There exists an
integer $t$ $(1\leq t \leq r)$
such that 
\begin{eqnarray*}
i_1<i_2 < \cdots <i_{t-1}< i_t\geq i_{t+1} \geq i_{t+2}\geq \cdots \geq i_{r-1}\geq i_r.
\end{eqnarray*}
\end{enumerate}
\end{lemma}
\noindent
{\bf Proof:} 
Routine using
Definition
\ref{def:irr}.
\hfill $\Box $ \\

\medskip
\noindent 
Consider the algebra $\mathcal A$ 
as a vector space over $\K$. It turns out that
the set of irreducible words
in $\mathcal A$ forms a basis for $\mathcal A$. 
However, in order to prove
Theorem
 \ref{con:Vspanguess}
 all we need is that this set spans
 $\mathcal A$.
We will prove this much for now and use the result to obtain Theorem
 \ref{con:Vspanguess}.
For the sake of completeness, in Section 3 we will prove
that 
 the set of irreducible words
in $\mathcal A$ is a basis for $\mathcal A$.

\begin{theorem}
\label{thm:sp}
The irreducible words in $\mathcal A$ form a spanning set for $\mathcal A$.
\end{theorem}
{\bf Proof:} 
Let
$\lambda =(\lambda_1, \lambda_2, \ldots, \lambda_r)$
denote a finite sequence of positive integers.
 We say this sequence
is {\it  nonincreasing} whenever
$\lambda_{i-1}\geq \lambda_i$
for $2 \leq i \leq r$.
Let $\Psi$ denote the set consisting of the
 nonincreasing finite sequences of positive integers.
There exists a certain linear order on $\Psi$
called the 
reverse lexicographical order. This is defined as follows.
Let $\lambda =(\lambda_1, \lambda_2, \ldots, \lambda_r)$
and $\mu =(\mu_1, \mu_2, \ldots, \mu_s)$
denote elements in $\Psi$. Then
$\lambda$ is less than  $\mu$ in the reverse lexicographical order
 whenever (i) $r<s$; or
(ii) $r=s$ and there exists an integer $k$ $(1 \leq k \leq r)$
such that
$\lambda_i=\mu_i$ $(k+1\leq i \leq r)$, $\lambda_k<\mu_k$.
Let $\lambda =(\lambda_1, \lambda_2, \ldots, \lambda_r)$
denote a finite sequence of positive integers.
We let ${\overline {\lambda}}$ denote the rearrangement of
$\lambda $ into a nonincreasing sequence.
In other words
${\overline {\lambda}}= 
(\mu_1, \mu_2,\ldots, \mu_r)$ where
$\lbrace \mu_1, \mu_2,\ldots, \mu_r\rbrace =
\lbrace \lambda_1, \lambda_2, \ldots, \lambda_r\rbrace $
as  multisets and
the sequence 
$(\mu_1, \mu_2,\ldots, \mu_r)$ is nonincreasing. 
We call ${\overline {\lambda}}$
the {\it rearrangement} of $\lambda$. 
Let $\lambda =(\lambda_1, \lambda_2, \ldots, \lambda_r)$
denote a finite sequence of positive integers.
By an {\it inversion} in $\lambda $ we mean an ordered
pair $(i,j)$ of integers such that
 $1 \leq i<j\leq r$ and 
$\lambda_i<\lambda_j$. We observe $\lambda = \overline{\lambda }$
if and only if $\lambda $ has no inversions.
We assume the present theorem is false and obtain a contradiction.
By a {\it counterexample} we mean a word in $\mathcal A$
which is not contained in the span of the irreducible words.
By assumption there exists a counterexample.
Let $w$ denote a counterexample
 and let
$\lambda =(\lambda_1, \lambda_2, \ldots, \lambda_r)$ denote the corresponding 
signature.
Without loss, we assume
that among all the counterexamples, the rearrangement 
$\overline \lambda$ is minimal
with respect to the reverse lexicographical order.
Moreover, without loss we may assume
that among all the 
 counterexamples 
for which the
 rearranged signature is equal to $\overline \lambda$,
the signature
$\lambda$ has a minimal number of inversions. 
Since $w$ is a counterexample it is reducible.
Therefore there exists an integer $s$ $(2 \leq s \leq r-1)$ such 
that $\lambda_{s-1}\geq \lambda_s<\lambda_{s+1}$.
By the construction 
there exist words $w_1, w_2$ in $\mathcal A$
such that
 $w=w_1x^{\lambda_{s-1}}y^{\lambda_s}x^{\lambda_{s+1}}w_2$
or
 $w=w_1y^{\lambda_{s-1}}x^{\lambda_s}y^{\lambda_{s+1}}w_2$.
We treat the first case; the second case is treated in a similar
manner. 
By Corollary
\ref{cor:dep}(ii) (with $i=\lambda_{s-1}$, $r=\lambda_s$, 
$n=\lambda_{s-1}+\lambda_{s+1}$)
we find
$w$ is contained in 
\begin{eqnarray}
\label{eq:w1}
&&\mbox{Span}\lbrace w_1x^{\xi}y^{\lambda_s}x^{\lambda_{s-1}+\lambda_{s+1}-\xi}w_2 \;|\;0\leq \xi\leq \lambda_s-1 \rbrace\\
&&\qquad \qquad \qquad +
\quad 
\mbox{Span}\lbrace w_1x^{\lambda_{s-1}+\lambda_{s+1}-\zeta}y^{\lambda_s}x^{\zeta}w_2 \;|\;0\leq \zeta\leq \lambda_s \rbrace.
\label{x}
\end{eqnarray}
By our minimality assumptions no word which appears in 
(\ref{eq:w1}) is a counterexample
and no word which appears in
(\ref{x}) is a counterexample.
Therefore the sum displayed in
(\ref{eq:w1}),
(\ref{x}) is contained in the span of the irreducible words.
The word $w$ is contained in this space so $w$ is in the span of the
irreducible words. This contradicts our assumption that $w$ is a 
counterexample.
The result follows.
\hfill $\Box $ \\

\medskip
\noindent
We are now ready to prove 
Theorem \ref{con:Vspanguess}.


\medskip
\noindent{\bf Proof of Theorem 
\ref{con:Vspanguess}:}
By 
(\ref{eq:ms1}),
(\ref{eq:ms2}) there exists an $\mathcal A$-module structure
 on $V$
such that  $x.v=Rv$ and $y.v=Lv$ for all $v \in V$.
Let $b$ denote a balanced word in $\mathcal A$.
Using (\ref{eq:raction})
 and 
(\ref{eq:laction}) we find
each of $U_0,U_1,\ldots, U_d$ is invariant under $b$.
In particular $U_0$ is invariant under $b$.
Let $i$ denote a nonnegative integer.
The word $y^ix^i$ is balanced so it leaves $U_0$ invariant.
From 
(\ref{eq:nil})
we find $x^{d+1}$ vanishes on $V$.
In particular $x^{d+1}$ vanishes on $U_0$.
Therefore $y^ix^i$ vanishes on $U_0$ for $i>d$.
We claim the elements $y^ix^i$ $(0 \leq i \leq d)$ mutually commute on
$U_0$.
To see this, observe by
(\ref{eq:raction}),
(\ref{eq:laction}) that
each nil word of $\mathcal A$ vanishes
on $U_0$. In particular each nil word 
in
$\mathcal B$ vanishes on
$U_0$ so ${\mathcal B}^{nil}$ vanishes on $U_0$. 
By this and Corollary
\ref{cor:del}
we find $\lbrack y^ix^i,y^jx^j\rbrack$ vanishes on $U_0$
 for $0 \leq i,j\leq d$.
We have now shown the elements $y^ix^i$ $(0 \leq i \leq d)$ mutually commute
on $U_0$.
Since $\fld$ is algebraically closed there exists a nonzero 
$v \in U_0$ which is a common eigenvector of 
the $y^ix^i$ $(0 \leq i \leq d)$. 
Let $W$ denote the subspace of $V$ spanned by the
vectors
of the form
\begin{eqnarray}
\label{eq:zsp}
y^{i_1}x^{i_2}y^{i_3}x^{i_4}\cdots x^{i_n}.v,
\end{eqnarray}
where 
$i_1, i_2, \ldots, i_n$ ranges over all sequences
such that $n$ is a nonnegative even integer, and 
 $i_1, i_2, \ldots i_n$ are integers
satisfying $0\leq i_1<i_2<\cdots <i_n\leq d$.
We show $W=V$.
Observe $v \in W$ so $W\not=0$.
By Definition
\ref{def:tdp}(iv)
the space $V$ is irreducible as a module for
$A,A^*$. In order to show $W=V$ we show
$W$ is invariant under each of $A, A^*$.
Using
Lemma
\ref{lem:red}
and
 Theorem
\ref{thm:sp}
 we routinely find
${\mathcal A}.v=W$. From this we find
$W$ is invariant under each of $x,y$.
It follows $W$ is invariant under each of $R,L$.
Using
(\ref{eq:raction}),
(\ref{eq:laction})
we find the vector 
(\ref{eq:zsp})
is contained in $U_i$
where
$i=\sum_{h=1}^n i_h(-1)^h$.
By this and
(\ref{eq:fi1}),
(\ref{eq:fi2})
the vector 
(\ref{eq:zsp})
is an eigenvector for each of $F_0, F_1, \ldots, F_d$.
We now see $W$ is invariant under each of
$F_0, F_1, \ldots, F_d$.
By these comments and 
(\ref{eq:defR}), (\ref{eq:defL})
we find $W$ is invariant under each of $A, A^* $.
We conclude $W=V$ and
the result follows.
\hfill $\Box $ \\

\medskip
\noindent {\bf Proof of Theorem
\ref{thm:us}:} Combine Theorem
\ref{con:Vspanguess} with 
(\ref{eq:raction}) and
(\ref{eq:laction}).
\hfill $\Box $ \\

\medskip
\noindent {\bf Proof of Theorem 
\ref{thm:bc}:}
For $0 \leq i \leq d$, the spanning set for $U_i$ given in
Theorem
\ref{thm:us} 
has cardinality
$\bigl( {{d}\atop {i}} \bigr)$.
\hfill $\Box $ \\

\section{Comments and suggestions for further research}

\noindent
In this section we give some comments and suggestions for further research.
 We begin with a comment.

\begin{lemma}
\label{lem:fc}
The quotient algebra ${\mathcal B}/{\mathcal B}^{nil}$ is generated by
the elements
\begin{eqnarray*}
y^ix^i+{\mathcal B}^{nil} \qquad (0\leq i < \infty).
\end{eqnarray*}
Moreover 
 ${\mathcal B}/{\mathcal B}^{nil}$ is commutative.
\end{lemma}
{\bf Proof:} 
The first assertion is immediate from Theorem
\ref{thm:sp}.
The last assertion follows from this and Corollary
\ref{cor:del}.
\hfill $\Box $ \\

\medskip
\noindent Just before 
Theorem
\ref{thm:sp}
 we asserted that 
the set of irreducible words in $\mathcal A$ is a basis
for $\mathcal A$. We will now prove this assertion.


\begin{theorem}
The set of irreducible words in $\mathcal A$ is a basis
for $\mathcal A$.
\end{theorem}
{\bf Proof:}
Let $X$ denote the set of irreducible words in $\mathcal A$.
Then  $X$ spans $\mathcal A$
by Theorem
\ref{thm:sp}.
For $0 \leq n < \infty $ let 
$X_n$ denote the set of irreducible words in $\mathcal A$ which have
length $n$. 
Of course $X=\cup_{n=0}^\infty X_n$.
For $0 \leq n < \infty $ let 
${\mathcal A}_n$
denote the subspace of $\mathcal A$ spanned by the
words of length $n$.
Since the $q$-Serre relations
 are homogeneous we have
\begin{eqnarray}
\label{eq:dir}
{\mathcal A}
=\sum_{n=0}^\infty {\mathcal A}_n \qquad \qquad (\mbox{direct sum}).
\end{eqnarray}
By the construction $X_n \subseteq {\mathcal A}_n$ for 
$0 \leq n< \infty$.
By these comments we find $X_n$ spans ${\mathcal A}_n$ for
$0 \leq n < \infty $. 
We show
\begin{eqnarray}
|X_n|=\mbox{dim}({\mathcal A}_n)
\qquad \qquad 
 (0 \leq n < \infty).
\label{eq:req}
\end{eqnarray}
Let $v$ denote an indeterminate. In what follows, we consider
formal power series in $v$ which have coefficients in $\C$. 
Line 
 (\ref{eq:req}) will follow once
we  show
\begin{eqnarray}
\sum_{n=0}^\infty |X_n|v^n=\sum_{n=0}^\infty \mbox{dim}({\mathcal A}_n)v^n.
\label{eq:req2}
\end{eqnarray}
%The following result is 
%well known 
%in the theory of quantum groups \cite{matsuo}:
We claim 
\begin{eqnarray}
\label{eq:dimgen}
\sum_{n=0}^\infty \mbox{dim}({\mathcal A}_n)v^n=
\prod_{m=1}^\infty (1-v^{2m})^{-1}(1-v^{2m-1})^{-2}.
\end{eqnarray}
To see
(\ref{eq:dimgen}), recall $\mathcal A$ is the positive
part of 
$U_q(\widehat{sl_2})$ and hence isomorphic to the 
Verma module for 
$\widehat{sl_2}$ \cite[p. 123]{waki}
as a graded vector space.  
Apparently the left-hand side of
(\ref{eq:dimgen}) is equal to the formal character of the Verma module.
This character is inverse to the principally specialized Weyl
denominator and is therefore equal to the right-hand side of
(\ref{eq:dimgen}) \cite[p. 181]{kac}. We now have 
(\ref{eq:dimgen}).
For $0 \leq n < \infty$,
by a {\it partition of $n$} we mean a 
 sequence of positive integers
$\lambda=(\lambda_1, \lambda_2,\ldots, \lambda_r)$ 
such that $\lambda_{i-1}\geq \lambda_i$ for $2 \leq i \leq r$
and $\sum_{i=1}^r \lambda_i=n$.
We call
$\lambda_1, \lambda_2, \ldots, \lambda_r$ the {\it parts} of $\lambda$.
We sometimes write $n=|\lambda |$.
Let $p_n$ denote the number of partitions of $n$.
Let $p'_n$ denote the number of partitions of $n$ whose parts are
mutually distinct.
The following two generating functions are well known:
\begin{eqnarray*}
\sum_{n=0}^\infty p_n v^n &=&\prod_{m=1}^\infty (1-v^m)^{-1},
\\
\sum_{n=0}^\infty p'_n v^n &=&\prod_{m=1}^\infty (1-v^{2m-1})^{-1}.
\end{eqnarray*}
See for example
\cite[Theorem 1.1]{Andrews}.
 For $0 \leq n < \infty $ let $Y_n$ denote the set of ordered pairs
$(\lambda, \mu )$
 such that (i) $\lambda$ is a partition whose parts are mutually
distinct; (ii) $\mu $ is a partition; (iii) $|\lambda |+|\mu|=n$.
From the construction
\begin{eqnarray}
\sum_{n=0}^\infty |Y_n|v^n &=&
\biggl(\sum_{n=0}^\infty p'_n v^n\biggr)
\biggl(\sum_{n=0}^\infty p_n v^n\biggr)
\nonumber
\\
 &=&\prod_{m=1}^\infty 
(1-v^{2m})^{-1}
(1-v^{2m-1})^{-2}.
\label{eq:pr}
\end{eqnarray}
We show
\begin{eqnarray}
|X_n|=|Y_n| \qquad \qquad 
(0 \leq n < \infty).
\label{eq:yx}
\end{eqnarray}
To obtain 
(\ref{eq:yx}),
for $0 \leq n < \infty$
 we 
display a bijection $\varepsilon :Y_n \rightarrow X_n$. 
Let $(\lambda, \mu)$ denote an element of $Y_n$ and write
$\lambda 
=(\lambda_1, \lambda_2, \ldots, \lambda_r)$,
$\mu =(\mu_1, \mu_2, \ldots, \mu_s)$.
The image of $(\lambda, \mu)$ under $\varepsilon$ is
 the following word in $\mathcal A$:
\begin{eqnarray}
\label{eq:finword}
\cdots x^{\lambda_4}
y^{\lambda_3}
x^{\lambda_2}
y^{\lambda_1}
x^{\mu_1}
y^{\mu_2}
x^{\mu_3}
y^{\mu_4}
\cdots.
\end{eqnarray}
By the construction
the word
(\ref{eq:finword}) is irreducible with length $n$.
Therefore the word (\ref{eq:finword}) is contained in $X_n$.
Using Lemma
\ref{lem:red}
we  find $\varepsilon :
Y_n \rightarrow X_n$ is a  bijection.
We now have
(\ref{eq:yx}).
Combining
(\ref{eq:dimgen}),
(\ref{eq:pr}),
(\ref{eq:yx})
we obtain
(\ref{eq:req2}) and
(\ref{eq:req}) 
follows.
We conclude
$X_n$ is a basis of ${\mathcal A}_n$ for $0\leq n< \infty$.
By this and
(\ref{eq:dir})
we find $X$ is a basis for $\mathcal A$.
\hfill $\Box $ \\

\begin{remark}
\rm
The following is a specialization of the Jacobi triple 
product identity \cite[p. 185]{kac}:
%By \cite[p. 105]{Sloane} we have
\begin{eqnarray}
\sum_{n=-\infty}^\infty (-v)^{n^2}=
\prod_{m=1}^\infty 
(1-v^{2m})
(1-v^{2m-1})^{2}.
\label{eq:thf}
\end{eqnarray}
If we set 
 $v=e^{\pi i z}$ for $i, z \in \C$ with $i^2=-1$ and 
 $\mbox{Im}(z)>0$, then
 either side of 
(\ref{eq:thf})
is equal to the theta function $\theta_4(z)$.
See 
\cite[pp. 103--105]{Sloane} for more detail.   
\end{remark}


\medskip
\noindent We give a suggestion for future research.

\begin{problem}
\label{prob:basis}
\rm
Let $\beta, \gamma, \gamma^*, \varrho, \varrho^*$ denote
a sequence of scalars taken from $\K$.
Let $\mathcal T$ denote the associative $\K$-algebra with identity generated
by symbols $x,y$ subject to the tridiagonal relations
\begin{eqnarray*}
\lbrack x,x^2y-\beta xyx + 
yx^2 -\gamma (xy+yx)-\varrho y\rbrack &=&0, 
\\
\lbrack y,y^{2}x-\beta yxy + xy^{2} -\gamma^* (yx+xy)-
\varrho^* x\rbrack&=&0.
\end{eqnarray*}
Find analogs of the higher order $q$-Serre relations which hold in
$\mathcal T$. Find a  basis for the $\K$-vector space $\mathcal T$.
\end{problem}

\begin{problem}
\rm
Let the algebra $\mathcal T$ be as in Problem
\ref{prob:basis}. An element of $\mathcal T$
is called {\it central} whenever it commutes with every
element of $\mathcal T$. By definition 
the {\it center} of $\mathcal T$
is the $\K$-subalgebra of $\mathcal T$ consisting of
the central elements of $\mathcal T$. Describe the center
of $\mathcal T$. Find a generating set for this center.
\end{problem}





%\begin{conjecture}
%$0 \leq i \leq d$ let $\gamma_i$ denote the eigenvalue of the
%$y^ix^i$-action on $U_0$. We conjecture that the isomorphism class of
%the tridiagonal pair $A,A^*$ from Definition
%\ref{def:1p1}
%is determined by  
%$\gamma_0, \gamma_1,\ldots, \gamma_d$, together with the scalars
%$a, a^*$ from 
%
%\end{conjecture}





\section{Acknowledgment}  The authors thank 
Georgia Benkart,
Vyjayanthi Chari, 
Arjeh Cohen,
Atsushi Matsuo, Anne Schilling, and
Kenichiro Tanabe
for helpful discussions on the subject of this paper.
The authors thank Michio Jimbo for 
providing a proof of 
(\ref{eq:dimgen}).


\begin{thebibliography}{10}

\bibitem{Andrews}
G.~Andrews.
\newblock {\em The Theory of Partitions}.
\newblock Cambridge University Press, Cambridge, 1984.


\bibitem{AWil}
R.~Askey and J.A.~Wilson.
\newblock A set of orthogonal polynomials that 
generalize the {R}acah coefficients or $6-j$ symbols. 
\newblock {\em SIAM J. Math. Anal.}, 10:1008--1016, 1979. 

\bibitem{BanIto}
E.~Bannai and T.~Ito.
\newblock {\em Algebraic Combinatorics {I}: Association Schemes}.
\newblock Benjamin/Cummings, London, 1984.


\bibitem{cp}
V.~Chari and A.~Pressley.
\newblock Quantum affine algebras.
\newblock {\em Comm. Math. Phys.}, 142:261--283, 1991.

\bibitem{Sloane}
J.~H.~Conway and N.~J.~A.~Sloane.
\newblock {\em Sphere packings, Lattices and Groups}.
\newblock Springer-Verlag, New York, 1988.


\bibitem{TD00}
T.~Ito, K.~Tanabe, and P.~Terwilliger.
\newblock Some algebra related to ${P}$- and ${Q}$-polynomial association
  schemes. In
\newblock {\em Codes and Association Schemes (Piscataway NJ, 1999)}, Amer.
Math. Soc., Providence, RI, 56:167--192, 2001.

\bibitem{kac}
V.~Kac.
\newblock {\em Infinite dimensional Lie algebras}.
\newblock {Cambridge U. Press, Cambridge}, 1995.

\bibitem{KoeSwa}
R.~Koekoek and R.~Swarttouw.
\newblock {\em The Askey-scheme of hypergeometric orthogonal polyomials and its
  $q$-analog}, volume 98-17 of {\em Reports of the faculty of Technical
  Mathematics and Informatics}.
\newblock Delft, The Netherlands, 1998.
Available at
 \newblock{ \tt http://aw.twi.tudelft.nl/{\~{}}koekoek/research.html}

\bibitem{Leon}
D.~Leonard.
\newblock Orthogonal polynomials, duality, and association
schemes.
\newblock {\em SIAM J. Math. Anal.}, 13(4):656--663, 1982.

\bibitem{lu}
G.~Lusztig.
\newblock {\em Introduction to quantum groups.}
\newblock {Progress in Mathematics 110.}
\newblock {Birkhauser, Boston, 1993.}

%\bibitem{matsuo}
%Atsushi~Matsuo.
%\newblock Personal communication.


\bibitem{TersubI}
P.~Terwilliger.
\newblock The subconstituent algebra of an association scheme. 
\newblock {\em J. Algebraic Combin.}, 1(4):363--388, 1992.

\bibitem{LS99}
P.~Terwilliger.
\newblock Two linear transformations each tridiagonal with respect to an
  eigenbasis of the other.
\newblock {\em Linear Algebra Appl.},  330:149--203, 2001. 

\bibitem{qSerre}
P.~Terwilliger.
\newblock Two relations that generalize the $q$-Serre relations and the 
Dolan-Grady relations. 
\newblock {\em {P}hysics and {C}ombinatorics 1999 (Nagoya),
377--398.} World Sci. Publishing,
River Edge, NJ, 2001. 

\bibitem{LS24}
P.~Terwilliger.
\newblock  Leonard pairs from 24 points of view.
\newblock {\em Rocky mountain J.  Math.}, 32(2):827--888, 2002. 

\bibitem{conform}
P.~Terwilliger.
\newblock Two linear transformations each tridiagonal with respect to an
  eigenbasis of the other; the $TD$-$D$ canonical form
  and the $LB$-$UB$ canonical form.
\newblock {\em J. Algebra}, submitted.


\bibitem{lsint}
P.~Terwilliger.
\newblock Introduction to Leonard pairs.
(OPSFA Rome 2001),
\newblock {\em J. Comput. Appl. Math.}, 153(2):463--475, 2003.

\bibitem{Terint}
P.~Terwilliger.
\newblock Introduction to {L}eonard pairs and 
	     {L}eonard systems.
  \newblock {\em S\=urikaisekikenky\=usho K\=oky\=uroku},
  \newblock 
   (1109):67--79, 1999.
    \newblock Algebraic combinatorics  (Kyoto, 1999).


\bibitem{TLT:split}
P.~Terwilliger.
\newblock Two linear transformations each tridiagonal with respect to an
  eigenbasis of the other; comments on the split decomposition.
\newblock {\em Indag. Math.}, submitted.


\bibitem{TLT:array}
P.~Terwilliger.
\newblock Two linear transformations each tridiagonal with respect to an
  eigenbasis of the other; comments on the parameter array.
\newblock {\em Geometric and Algebraic Combinatorics 2,
(Oisterwijk, The Netherlands 2002)},
submitted.

\bibitem{TLTqrac}
P.~Terwilliger.
\newblock Leonard pairs and the $q$-Racah polynomials.
\newblock {\em Linear Algebra Appl.}, submitted. 

\bibitem{waki}
Minoru~Wakimoto.
\newblock {\em Infinite-dimensional Lie algebras}.
\newblock Translations of Mathematical Monographs 195, 
Amer. Math.
Soc. Providence, RI, 2001.

\end{thebibliography}




\medskip
\noindent
Tatsuro Ito \hfil\break
Department of Computational Science\hfil\break
Faculty of Science \hfil\break
Kanazawa University \hfil\break
Kakuma-machi, Kanazawa 920--1192, Japan \hfil\break
E-mail: ito@kappa.s.kanazawa-u.ac.jp \hfil\break

\medskip
\noindent
Paul Terwilliger \hfil\break
Department of Mathematics\hfil\break
University of Wisconsin \hfil\break
480 Lincoln drive  \hfil\break
Madison, Wisconsin, 53706, USA \hfil\break
E-mail: terwilli@math.wisc.edu \hfil\break




\end{document}








\section{The map $K$}


\begin{definition}
\label{def:K}
Let us define
\begin{eqnarray}
K= \sum_{i=0}^d q^{2i-d} F_i,
\label{eq:kdef}
\end{eqnarray}
where the $F_i$ are from
(\ref{eq:fi1}), 
(\ref{eq:fi2}). From those lines we observe 
\begin{eqnarray}
K^{-1}= \sum_{i=0}^d q^{d-2i} F_i.
\label{eq:kinv}
\end{eqnarray}
\end{definition}

\begin{lemma} We have
\begin{eqnarray}
A &=& aK + R,
\label{eq:aandk}
\\
A^* &=& a^*K^{-1} +L, 
\label{eq:asandk}
\end{eqnarray}
where $a$ and $a^*$ are from
(\ref{eq:thetaform}), 
(\ref{eq:thetasform}), respectively. 
We also have
\begin{eqnarray}
K R = q^2 R K,
\qquad  \qquad 
K L = q^{-2} L K.
\label{eq:krl}
\end{eqnarray}
\end{lemma}
\begin{proof}
Lines
(\ref{eq:aandk}), 
(\ref{eq:asandk})
are immediate from
 (\ref{eq:defR}), 
(\ref{eq:defL}).
Line 
(\ref{eq:krl}) follows from
(\ref{eq:randf}),
(\ref{eq:landf}).
\end{proof}

\noindent The relations 
(\ref{eq:krl}) have the following interpretation. 

\begin{lemma}   
\label{eq:krlag}
Suppose we are given  
any linear transformation  $X:V\rightarrow V$.
Then 
$K X = q^2 X K $ if and only if
\begin{eqnarray}
X U_i \subseteq U_{i+1} \qquad \qquad (0 \leq i \leq d-1),  \qquad XU_d=0. 
\end{eqnarray}
Similarily
$ K X = q^{-2} X K$
 if and only if
\begin{eqnarray}
X U_i \subseteq U_{i-1} \qquad \qquad (1 \leq i \leq d),  \qquad XU_0=0. 
\end{eqnarray}
\end{lemma}



\section {The maps $l$ and $r$}

\begin{lemma}
\label{lem:uitiddle}
Let $\alpha_0, \alpha_1, \ldots, \alpha_d$ denote any sequence of
scalars taken from $\fld$ such that $\alpha_i + \alpha_{d-i} = 0 $
for $0 \leq i \leq d$. Then there exists a linear transformation
$l : V \rightarrow V$ satisfying 
\begin{eqnarray}
&&lU_i \subseteq U_{i-1} \qquad \qquad  (0 < i \leq d), \qquad lU_0=0,
\label{eq:slaction}
\end{eqnarray}
and 
\begin{eqnarray}
lR - Rl = \sum_{i=0}^d \alpha_i F_i.
\label{eq:slaction2}
\end{eqnarray}
The map $l$  is uniquely determined by 
$\alpha_0, \alpha_1, \ldots, \alpha_d$.
\end{lemma}

\begin{proof} For all integers $i$ $(0 \leq i \leq d/2)$, we define
a subspace 
\begin{eqnarray}
{\tilde U}_i = \lbrace u \in U_i \,\vert \,R^{d-2i+1}u = 0\rbrace .
\label{eq:defutilde}
\end{eqnarray}
Using Lemma 5.5 in \cite{} we find that for $0 \leq j \leq d$,
\begin{eqnarray}
U_j = R^j {\tilde U}_0 +  
 R^{j-1} {\tilde U}_1 +  \cdots + R^{j-i}{\tilde U}_i \qquad \qquad 
 (\hbox{direct sum}),
\label{eq:tildesum}
\end{eqnarray}
where $i = \hbox{min} \lbrace j,d-j \rbrace $.  Combinining this with
(\ref{eq:vdec}), we find
\begin{eqnarray}
V = \sum R^{j-i} {\tilde U}_i \qquad \qquad 
 (\hbox{direct sum}),
\label{eq:vtildesum}
\end{eqnarray}
where the sum is over all ordered
pairs of integer $i,j$ such that
$0 \leq i \leq d/2$ and $ i \leq j \leq d-i$. To define $l$,
it suffices to define $l$ on each summand in  
(\ref{eq:vtildesum}).
 Pick any integer $i$ $(0 \leq i \leq d/2)$.
 We define $l$ on  ${\tilde U}_i$ to be 0. Now pick any integer
 $j $ $(i<j\leq d-i)$. To define $l$ on $R^{j-i}{\tilde U}_i$, we first
 observe by Lemma 5.5 in \cite {} that the map
\begin{eqnarray}
{{ R^{j-i-1} {\tilde U}_i \quad \rightarrow \quad R^{j-i}{\tilde U}_i
}\atop {
\quad v \quad  \rightarrow \quad R v}}
\label{eq:lsettup}
\end{eqnarray}
is a bijection. We define $l$ on  
  $R^{j-i}{\tilde U}_i$ to be the inverse of this map,
 times the  scalar
\begin{eqnarray}
\alpha_i + \alpha_{i+1} + \cdots + \alpha_{j-1}.
\label{eq:scalarrinv}
\end{eqnarray}
 We have now defined $l$  on   each summand
in 
(\ref{eq:vtildesum}), 
 and hence on $V$.
The map $l$ satisfies 
(\ref{eq:slaction}) by
(\ref{eq:raction}) and the construction.
To obtain
(\ref{eq:slaction2}),
we observe that for
$0 \leq j\leq d$,
\begin{eqnarray}
lR - Rl -\alpha_j I
\label{eq:gettingl}
\end{eqnarray}
 vanishes on each summand on the  
right in 
(\ref{eq:tildesum}), and hence on  
all of $U_j$. 
Combining this with 
(\ref{eq:vdec}),
(\ref{eq:fi1}), 
(\ref{eq:fi2}),  we obtain
(\ref{eq:slaction2}). 
Our  uniqueness assertion is routinely verified.

\end{proof}





\begin{lemma}
\label{lem:uitiddle2}
Let $\alpha_0, \alpha_1, \ldots, \alpha_d$ denote any sequence of
scalars taken from $\fld$ such that $\alpha_i + \alpha_{d-i} = 0 $
for $0 \leq i \leq d$. Then there exists a linear transformation
$r : V \rightarrow V$ satisfying 
\begin{eqnarray}
&&rU_i \subseteq U_{i+1} \qquad \qquad  (0 \leq i < d), \qquad rU_d=0,
\label{eq:sraction}
\end{eqnarray}
and 
\begin{eqnarray}
rL - Lr = \sum_{i=0}^d \alpha_i F_i.
\label{eq:sraction2}
\end{eqnarray}
The map $r$  is uniquely determined by 
$\alpha_0, \alpha_1, \ldots, \alpha_d$.
\end{lemma}

\begin{proof} Similar to the proof of
Lemma \ref{lem:uitiddle2}.

\end{proof}




\begin{corollary}
\label{cor:fixingrl}
There exists a unique linear transformation $l:V \rightarrow V$
such that both
\begin{eqnarray}
 K l = q^{-2} l K, \qquad \qquad 
lR-Rl = {{K^{-1}-K}\over {q-q^{-1}}}. 
\label{eq:getl}
\end{eqnarray}
Moreover, there exists
 a unique linear transformation $r:V \rightarrow V$
such that both
\begin{eqnarray}
&&K r = q^2 r K, \qquad \qquad  
rL-Lr = {{K-K^{-1}}\over {q-q^{-1}}}.
\label{eq:getr}
\end{eqnarray}

\end{corollary}

\begin{proof}  To get $l$, 
apply Lemma 
\ref{lem:uitiddle} with 
$\alpha_i = \lbrack d-2i \rbrack $ for $0 \leq i \leq d$, and
apply the second assertion of Lemma 
\ref{eq:krlag}. To get $r$, 
apply Lemma 
\ref{lem:uitiddle2} with 
$\alpha_i = \lbrack 2i-d \rbrack $ for $0 \leq i \leq d$, and
apply the first assertion of Lemma 
\ref{eq:krlag}. 


\end{proof}

\section{ Two $U_q(sl_2)$ actions}

\noindent In this section, we present two actions of
the quantum algebra
$U_q(sl_2)$ on $V$. The definition of 
$U_q(sl_2)$ appears in Appendix 1.  


\begin{theorem}
\label{thm:uqsl2} 
Let the maps $r, l$ be as in Corollary
\ref{cor:fixingrl}.
Then the space $V$ carries the structure of
a $U_q(sl_2)$ module with action $e \leftrightarrow l$,
 $f \leftrightarrow R$, $k \leftrightarrow K^{-1}$.
 Moreover 
 $V$ carries the structure of
a $U_q(sl_2)$ module with action $e \leftrightarrow L$,
 $f \leftrightarrow r$ , 
  $k \leftrightarrow K^{-1}$.
\end{theorem}

\begin{proof} The first structure follows from
(\ref{eq:getl}) and the equation on the left in
(\ref{eq:krl}). The second structure follows from
(\ref{eq:getr}) and the equation on the right in
(\ref{eq:krl}).

\end{proof}

\noindent In Appendix 1, we defined the notion of
a $U_q(sl_2)$ module of type 1.

\begin{lemma}
\label{lem:firststr}
Refering to either one of the 
$U_q(sl_2)$ module structures on $V$ mentioned
in 
Theorem \ref{thm:uqsl2},
all irreducible 
$U_q(sl_2)$-submodules are of type 1,
with highest weight $q^{d-2i}$ for some integer
$i$ $(0 \leq i \leq d/2)$.
For
each integer $i$ $(0 \leq i \leq d/2)$, 
the irreducible 
$U_q(sl_2)$-module of type 1 with highest weight
$q^{d-2i}$ appears in $V$ with multiplicity $\rho_i-\rho_{i-1}$,
where $\rho_{-1}=0$.

\end{lemma}  

\begin{proof} This follows from the construction of $l$ and $r$. 

\end{proof}


\noindent Later in the paper we show the maps $R,r,L,l$ satisfy
some additional relations. To get these, we need more information
about how $R$ and $L$ interact. This is the subject of the 
next section.

\section{Linear dependencies on words}

\begin{definition} Let $n$ denote a nonnegative integer. By
a word of length n in $R$ and $L$, we mean any  product of the form
\begin{equation}
X_1X_2\cdots X_n,
\label{eq:word}
\end{equation}
where $X_i=R$ or $X_i=L$ for $1 \leq i \leq n$.
If $n=0$, we
 interpret 
(\ref{eq:word}) as 
 the identity map on $V$.
By the level of the word
(\ref{eq:word}) we mean the integer
\begin{equation}
|\lbrace i\;|\;1 \leq i \leq  n, \quad X_i=R\rbrace | \;-\;
|\lbrace i\;|\;1 \leq i \leq  n, \quad X_i=L\rbrace |.
\label{eq:leveldef}
\end{equation}
\end{definition}

\noindent
Let $t$ denote an integer, and 
let $W$ denote a word of level $t$ in $R$ and $L$. We observe
$WU_i \subseteq U_{i+t} $ for $0 \leq i \leq d$. In particular,
if $t<0$, then $WU_0=0$.
This has the following consequence.

\begin{lemma}
\label{lem:1p5}
Let $n$ denote a nonnegative integer, and let $X_1X_2\cdots X_n$
denote a word of length $n$ in $R$ and $L$. Suppose the level
of $X_iX_{i+1}\cdots X_n$ is negative for at least one $i$ $(1 \leq i \leq n)$.
Then 
 $X_1X_2\cdots X_n$ vanishes on $U_0$.
\end{lemma}
For the rest of this section, we consider the action
of certain words in $R$ and
$L$ on the subspace $U_0$. We find a number of linear dependencies
among these words. We begin by considering some words of length 4.

\begin{lemma} The following expression vanishes on $U_0$.
\begin{equation}
RLR^2 - R^2LR - \lbrack 3 \rbrack^{-1} LR^3.
\label{eq:length3}
\end{equation}
\end{lemma}

\begin{proof} Divide      
(\ref{eq:ms1}) by $\lbrack 3 \rbrack$, apply the result to $U_0$, 
 and recall $LU_0=0$.

\end{proof}

\noindent We now consider words in $R$ and $L$ that
have length 5 and level 3. 


\begin{lemma} The following  vanish on $U_0$.
\begin{eqnarray}
&&R^3LR - \lbrack 3 \rbrack R^2LR^2+ \lbrack 3 \rbrack RLR^3 - LR^4, 
\label{eq:1R}
\\
&& - \lbrack 3 \rbrack R^3LR+ \lbrack 3 \rbrack R^2LR^2 - RLR^3.
\label{eq:R1}
\end{eqnarray}
\end{lemma}

\begin{proof} To get (\ref{eq:1R}), multiply (\ref{eq:ms1})
on the right by $R$.
 To get (\ref{eq:R1}), multiply (\ref{eq:ms1})
on the left by $R$, and evaluate the result using Lemma
\ref{lem:1p5}.

\end{proof}

\noindent We now use the above lemma to obtain the action on $U_0$ of
$ RLR^3$ and $R^2LR^2$, in terms of the remaining words of length 5, level 3. 

\begin{corollary} The following vanish on $U_0$.
\begin{eqnarray}
&& RLR^3 - 
{{\lbrack 2 \rbrack}\over { \lbrack 4 \rbrack}} LR^4
- R^3LR,
\label{eq:rlrrr}
\\
&& R^2LR^2- 
{{\lbrack 2 \rbrack}\over {\lbrack 3 \rbrack \lbrack 4 \rbrack }} LR^4
- {{\lbrack 2 \rbrack^2}\over {\lbrack 3 \rbrack}} R^3LR.
\label{eq:rrlrr}
\end{eqnarray}
\end{corollary}

\begin{proof} Set 
(\ref{eq:1R}), 
(\ref{eq:R1})  equal to zero, and solve for $RLR^3$,  $R^2LR^2$.

\end{proof}

\noindent We now consider words in $R$ and $L$ that have length 6 and level 0.

\begin{lemma} The following vanish on $U_0$.
\begin{eqnarray}
&&- \lbrack 3 \rbrack L^2R^2LR+ \lbrack 3 \rbrack L^2RLR^2 - L^3R^3, 
\label{eq:LL1}
\\
&&
 - \lbrack 3 \rbrack LRL^2R^2+ \lbrack 3 \rbrack L^2RLR^2 - L^3R^3.
\label{eq:2RR}
\end{eqnarray}
\end{lemma}

\begin{proof} To get 
(\ref{eq:LL1}), multiply (\ref{eq:ms1}) on the left by $L^2$, and
apply Lemma \ref{lem:1p5}.
To get 
(\ref{eq:2RR}), multiply (\ref{eq:ms2}) on the right by $R^2$,
apply Lemma \ref{lem:1p5}.
\end{proof}

\begin{corollary}
The maps $LR, L^2R^2$ commute on $U_0$.

\end{corollary} 

\begin{proof} Subtracting (\ref{eq:LL1}) from (\ref{eq:2RR}),
and dividing the result by $\lbrack 3 \rbrack$, we find 
$L^2R^2LR -  LRL^2R^2 $ vanishes on $U_0$.

\end{proof}

\noindent We now consider words in $R$ and $L$ that have length 7 and
level 1.

\begin{lemma} The following vanish on $U_0$.
\begin{eqnarray}
&& - \lbrack 3 \rbrack RL^2R^2LR+ \lbrack 3 \rbrack RL^2RLR^2 - RL^3R^3, 
\label{eq:RL^21}
\\
&&- \lbrack 3 \rbrack LRLR^2LR+ \lbrack 3 \rbrack LRLRLR^2 - LRL^2R^3, 
\label{eq:LRL1}
\\
&& - \lbrack 3 \rbrack L^2R^3LR+ \lbrack 3 \rbrack L^2R^2LR^2 - L^2RLR^3, 
\label{eq:LLR1}
\\
&&L^2R^3LR - \lbrack 3 \rbrack L^2R^2LR^2+ 
\lbrack 3 \rbrack L^2RLR^3 - L^3R^4, 
\label{eq:LL1R}
\\
&& - \lbrack 3 \rbrack LR^2LRLR+ \lbrack 3 \rbrack LRLR^2LR - L^2R^3LR, 
\label{eq:L1LR}
\\
 &&RL^3R^3 - \lbrack 3 \rbrack LRL^2R^3+ \lbrack 3 \rbrack L^2RLR^3 - L^3R^4.
\label{eq:2RRR}
 \\
 && - \lbrack 3 \rbrack RLRL^2R^2+ \lbrack 3 \rbrack RL^2RLR^2 - RL^3R^3.
\label{eq:R2RR}
\end{eqnarray}
\end{lemma}

\begin{proof} To get 
(\ref{eq:RL^21})--(\ref{eq:L1LR}),  
let $\Delta $ denote the expression in (\ref{eq:ms1}),
and 
evaluate $RL^2 \Delta$, $LRL \Delta$, $L^2R \Delta $, $L^2 \Delta R $, $L\Delta LR$,
respectively using Lemma \ref{lem:1p5}. To obtain
(\ref{eq:2RRR})--(\ref{eq:R2RR}), let $\Psi$ denote the expression in
(\ref{eq:ms2}), and evaluate  $\Psi R^3$, $ R\Psi R^2$ using Lemma \ref{lem:1p5}.

\end{proof} 

\noindent We now use the above lemma to 
obtain the action on $U_0$ 
of each of 
\begin{equation}
L^2R^2LR^2,  \;
L^2RLR^3, \;
LRLR^2LR,\;
LRL^2R^3,\;
LRLRLR^2,\;
RL^2R^2LR,\;
RLRL^2R^2,
\label{eq:indvars}
\end{equation}
in terms of the remaining words of length 7 and level 1.


\begin{lemma} The following vanish on $U_0$.
\begin{eqnarray}
&&L^2R^2LR^2- {{\lbrack 2\rbrack^2}\over {\lbrack 3 \rbrack}} L^2R^3LR
-{{\lbrack 2 \rbrack}\over{\lbrack 3 \rbrack \lbrack 4 \rbrack}} L^3R^4,
\\
&& L^2RLR^3 -  L^2R^3LR -{{\lbrack 2 \rbrack}\over {\lbrack 4 \rbrack  }} L^3R^4
\\
&&
LRLR^2LR -  LR^2LRLR  -{{1}\over {\lbrack 3 \rbrack}} L^2R^3LR  
\\
&&
LRL^2R^3 - 
{{1}\over {\lbrack 3 \rbrack}} RL^3R^3 - L^2R^3LR 
-{{\lbrack 2 \rbrack }\over{\lbrack 3 \rbrack \lbrack 4 \rbrack}} L^3R^4,
\label{eq:lrllrrrv}
\\
&&
LRLRLR^2 - 
{{1}\over {\lbrack 3 \rbrack^2}} RL^3R^3 -  LR^2LRLR -{{2}\over {\lbrack 3 \rbrack }} L^2R^3LR
-{{\lbrack 2 \rbrack }\over{\lbrack 3 \rbrack^2 \lbrack 4 \rbrack}} L^3R^4,
\\
&&
RL^2R^2LR - 
RL^2RLR^2 +{{1}\over {\lbrack 3 \rbrack}}RL^3R^3 
\\
&&
RLRL^2R^2  - 
RL^2RLR^2 +{{1}\over {\lbrack 3 \rbrack}}RL^3R^3 
\end{eqnarray}
\end{lemma}

\begin{proof} Set the expressions in  
(\ref{eq:RL^21})--(\ref{eq:R2RR}) equal to zero, and solve
the resulting linear system for the expressions in  
(\ref{eq:indvars}).

\end{proof}

\noindent We now consider words in $R$ and $L$ that have
length 7 and level 3. 

\begin{lemma} the following vanish on $U_0$.
\begin{eqnarray}
&& - \lbrack 3 \rbrack R^2LR^2LR+ \lbrack 3 \rbrack R^2LRLR^2 -
R^2L^2R^3, 
\label{eq:RRL1}
\\
&&- \lbrack 3 \rbrack RLR^3LR+ \lbrack 3 \rbrack RLR^2LR^2 - RLRLR^3, 
\label{eq:RLR1}
\\
&& - \lbrack 3 \rbrack LR^4LR+ \lbrack 3 \rbrack LR^3LR^2 - LR^2LR^3, 
\label{eq:LRR1}
\\
&&LR^4LR - \lbrack 3 \rbrack LR^3LR^2+ \lbrack 3 \rbrack LR^2LR^3 - LRLR^4, 
\label{eq:LR1R}
\\
&&RLR^3LR - \lbrack 3 \rbrack RLR^2LR^2+ \lbrack 3 \rbrack RLRLR^3 - RL^2R^4, 
\label{eq:RL1R}
\\
&&LR^3LR^2 - \lbrack 3 \rbrack LR^2LR^3+ \lbrack 3 \rbrack LRLR^4 - L^2R^5, 
\label{eq:L1RR}
\\
&&R^4L^2R - \lbrack 3 \rbrack R^3LRLR+ \lbrack 3 \rbrack R^2LR^2LR - RLR^3LR, 
\label{eq:R1LR}
\\
&&R^3LRLR - \lbrack 3 \rbrack R^2LR^2LR+ \lbrack 3 \rbrack RLR^3LR - LR^4LR, 
\label{eq:1RLR}
\\
&&R^3L^2R^2 - \lbrack 3 \rbrack R^2LRLR^2+ \lbrack 3 \rbrack RLR^2LR^2 -
LR^3LR^2, 
\label{eq:1LRR}
\end{eqnarray}
\end{lemma}

\begin{proof} Let $\Delta$ denote the expression
in (\ref{eq:ms1}), and evaluate 
$R^2L\Delta $,
$RLR\Delta $, 
$LR^2\Delta $, 
$LR\Delta R$, 
$RL\Delta R$, 
$L\Delta R^2$, 
$R\Delta LR$, 
$\Delta RLR$, 
$\Delta LR^2$, using Lemma \ref{lem:1p5}.

\end{proof}

\noindent We now use the previous lemma to obtain the action on $U_0$ of each
of 
\begin{eqnarray}
&& R^2LR^2LR, \;
R^2L^2R^3,\;
R^2LRLR^2,\;
RLR^3LR,\;
RLR^2LR^2, 
\label{eq:depvar73}
\\
&&
RLRLR^3,\;
LR^3LR^2,\;
LR^2LR^3,\;
LRLR^4,
\label{eq:depvar73p}
\end{eqnarray}
in terms of terms of the remaining  words of length 7 and level 3. 


\begin{corollary}
The following vanish on $U_0$.
\begin{eqnarray}
&&
R^2LR^2LR
-{{\lbrack 2 \rbrack^2}\over{\lbrack 3 \rbrack}} R^3LRLR -
{{\lbrack 2 \rbrack}\over {\lbrack 3 \rbrack \lbrack 4 \rbrack }}LR^4LR, 
\\
&&
R^2L^2R^3-
R^3L^2R^2-{{\lbrack 2 \rbrack}\over {\lbrack 4 \rbrack }} RL^2R^4 +{{\lbrack 2 \rbrack }\over {\lbrack 4 \rbrack \lbrack 5 \rbrack}} L^2R^5,
\label{eq:rrllrrrv}
\\
&&
R^2LRLR^2-
{{1}\over {\lbrack 3 \rbrack}}R^3L^2R^2
\\
&& \quad 
-{{\lbrack 2 \rbrack^2}\over {\lbrack 3 \rbrack }}R^3LRLR- 
{{\lbrack 2 \rbrack}\over {\lbrack 3 \rbrack \lbrack 4 \rbrack}}RL^2R^4
-
{{\lbrack 2 \rbrack}\over {\lbrack 3 \rbrack \lbrack 4 \rbrack}} LR^4LR
+
{{\lbrack 2 \rbrack}\over
{\lbrack 3 \rbrack \lbrack 4 \rbrack \lbrack 5\rbrack}} 
L^2R^5,
\\
&&
RLR^3LR-
R^3LRLR 
-
{{\lbrack 2 \rbrack}\over {\lbrack 4 \rbrack}} LR^4LR,
\\
&&
RLR^2LR^2-
{{\lbrack 2 \rbrack^2}\over {\lbrack 3 \rbrack }}R^3LRLR- 
{{\lbrack 2 \rbrack}\over {\lbrack 3 \rbrack \lbrack 4 \rbrack}} RL^2R^4
-
{{\lbrack 2 \rbrack^3}\over {\lbrack 3 \rbrack \lbrack 4 \rbrack}}LR^4LR,
\\
&&
RLRLR^3-
R^3LRLR- 
{{\lbrack 2 \rbrack}\over {\lbrack 4 \rbrack}} RL^2R^4
-
{{\lbrack 2 \rbrack}\over {\lbrack 4 \rbrack}}LR^4LR,
\\
&&
LR^3LR^2-
{{\lbrack 2 \rbrack\lbrack 3 \rbrack}\over { \lbrack 4 \rbrack}} LR^4LR
-
{{\lbrack 2 \rbrack}\over
{\lbrack 4 \rbrack \lbrack 5\rbrack}} 
L^2R^5,
\\
&&
LR^2LR^3-
{{\lbrack 2 \rbrack\lbrack 3 \rbrack}\over { \lbrack 4 \rbrack}} LR^4LR
-
{{\lbrack 2 \rbrack \lbrack 3 \rbrack}\over
{\lbrack 4 \rbrack \lbrack 5\rbrack}} 
L^2R^5,
\\
&&
LRLR^4-
LR^4LR -
{{\lbrack 3 \rbrack}\over
{ \lbrack 5\rbrack}} 
L^2R^5.
\end{eqnarray}
\end{corollary}

\begin{proof} In the previous lemma, set each of 
(\ref{eq:RRL1})--(\ref{eq:1LRR})  equal to zero, and
solve the resulting linear system for
the expressions in (\ref{eq:depvar73}), 
(\ref{eq:depvar73p}).

\end{proof}

\noindent We now consider words
in $R$ and $L$ that have length  8 and level 0.

\begin{lemma}
\label{lem:80}
The following vanish on $U_0$. 
\begin{eqnarray}
&&
 - \lbrack 3 \rbrack LRL^2R^2LR+ \lbrack 3 \rbrack LRL^2RLR^2
- LRL^3R^3, 
\label{eq:LRLL1}
\\
&&
 - \lbrack 3 \rbrack L^2RLR^2LR+ \lbrack 3 \rbrack L^2RLRLR^2 
- L^2RL^2R^3, 
\label{eq:LLRL1}
\\
&&
 - \lbrack 3 \rbrack L^3R^3LR+ \lbrack 3 \rbrack L^3R^2LR^2 - L^3RLR^3, 
\label{eq:LLLR1}
\\
&&
L^3R^3LR - \lbrack 3 \rbrack L^3R^2LR^2+ \lbrack 3 \rbrack L^3RLR^3 - L^4R^4,
\label{eq:LLL1R}
\\
&&
 - \lbrack 3 \rbrack L^2R^2LRLR+ \lbrack 3 \rbrack L^2RLR^2LR
- L^3R^3LR, 
\label{eq:LL1LR}
\\
&&
 - \lbrack 3 \rbrack LRL^3R^3+ \lbrack 3 \rbrack L^2RL^2R^3 
- L^3RLR^3,
\label{eq:2LRRR}
\\
&&
 - \lbrack 3 \rbrack LRL^2RLR^2+ \lbrack 3 \rbrack L^2RLRLR^2
 - L^3R^2LR^2,
\label{eq:2RLRR}
\\
&&
 - \lbrack 3 \rbrack LRL^2R^2LR+ \lbrack 3 \rbrack L^2RLR^2LR 
- L^3R^3LR,
\label{eq:2RRLR}
\\
&&
 LRL^3R^3 - \lbrack 3 \rbrack L^2RL^2R^3+ \lbrack 3 \rbrack L^3RLR^3 - L^4R^4,
\label{eq:L2RRR}
\\
&&
  - \lbrack 3 \rbrack LRLRL^2R^2+ \lbrack 3 \rbrack LRL^2RLR^2
 - LRL^3R^3.
\label{eq:LR2RR}
\end{eqnarray}
\end{lemma}

\begin{proof} 
to get
(\ref{eq:LRLL1})--(\ref{eq:LL1LR}), 
let $\Delta $ denote the expression in (\ref{eq:ms1}),
and 
evaluate
$LRL^2\Delta$, $L^2RL \Delta$, $L^3R \Delta $, $L^3 \Delta R $,
$L^2\Delta LR$,
respectively using Lemma \ref{lem:1p5}. To obtain
(\ref{eq:2LRRR})--(\ref{eq:LR2RR}), let $\Psi$ denote the expression in
(\ref{eq:ms2}), and evaluate  $\Psi LR^3$, $ \Psi RLR^2$, $\Psi R^2LR$,
$L\Psi R^3$, $LR \Psi R^2$, respectively  using Lemma \ref{lem:1p5}.

\end{proof} 




\noindent  We now use  the previous lemma to obtain
 the action on $U_0$ of each of  
\begin{eqnarray}
&&L^2R^2LRLR, \;
L^2RLR^2LR,\;  
L^3R^3LR,\; 
LRLRL^2R^2,\; 
LRL^2RLR^2,
\label{eq:word81}
\\
&&
L^3R^2LR^2,\; 
LRL^3R^3,\; 
L^2RL^2R^3,\; 
L^2RLRLR^2, 
\label{eq:word82}
\end{eqnarray}
in terms of the remaining words of length 8 and level 0.


\begin{corollary}
\label{cor:80}
The following vanish on $U_0$.
\begin{eqnarray}
&&
L^2R^2LRLR - LRL^2R^2LR,
\\
&&L^2RLR^2LR - LRL^2R^2LR - 
{{1}\over {\lbrack 3 \rbrack }}L^3RLR^3  
+ {{\lbrack 2 \rbrack}\over {\lbrack 3 \rbrack \lbrack 4 \rbrack }}
L^4R^4,
\\
&&L^3R^3LR 
-L^3RLR^3  
+ {{\lbrack 2 \rbrack}\over { \lbrack 4 \rbrack }}
L^4R^4,
\label{eq:LLLRRRLR}
\\
&&
LRLRL^2R^2 - LRL^2R^2LR,
\\
&&LRL^2RLR^2 - LRL^2R^2LR - 
{{1}\over {\lbrack 3 \rbrack }}L^3RLR^3  
+ {{\lbrack 2 \rbrack}\over {\lbrack 3 \rbrack \lbrack 4 \rbrack }}
L^4R^4,
\\
&&
L^3R^2LR^2 
-{{\lbrack 2 \rbrack^2}\over {\lbrack 3 \rbrack }}L^3RLR^3  
+ {{\lbrack 2 \rbrack}\over {\lbrack 4 \rbrack }}
L^4R^4,
\\
&&LRL^3R^3 
-L^3RLR^3  
+ {{\lbrack 2 \rbrack}\over { \lbrack 4 \rbrack }}
L^4R^4,
\label{eq:LRLLLRRR}
\\
&&
L^2RL^2R^3- 
{{\lbrack 2 \rbrack^2}\over {\lbrack 3 \rbrack }}L^3RLR^3  
+ {{\lbrack 2 \rbrack}\over {\lbrack 4 \rbrack }}
L^4R^4,
\\
&&
L^2RLRLR^2 - LRL^2R^2LR- 
{{2\lbrack 3 \rbrack +1}\over {\lbrack 3 \rbrack^2 }}L^3RLR^3  
+ {{2\lbrack 2 \rbrack}\over {\lbrack 3 \rbrack \lbrack 4 \rbrack }}
L^4R^4
\end{eqnarray}
\end{corollary}

\begin{proof} In the previous lemma, set the expressions equal to
zero and solve for  the expressions in 
(\ref{eq:word81}),
(\ref{eq:word82}).

\end{proof}

\begin{corollary}
The maps $LR$, $L^3R^3$ commute on $U_0$.

\end{corollary}

\begin{proof} Compare
(\ref{eq:LLLRRRLR}) and
(\ref{eq:LRLLLRRR}).

\end{proof}



\begin{lemma}
The maps $L^2R^2$, $L^3R^3$ commute on $U_0$.

\end{lemma}

\begin{proof} (sketch): Work with words of 
length 10 and level 0, as in  Lemma  
\ref{lem:80} and Corollary
\ref{cor:80}. There are 42 equations, so I wont write them
all down. Project: find a simpler way to do this. 

\end{proof}

\noindent We have now shown that $LR$, $L^2R^2$, $L^3R^3$ mutually
commute on $U_0$. This is good enough for our purposes if $d=3$,
but in general we will need the following.

\begin{conjecture}
\label{conj:lrcom}
The maps $LR, L^2R^2, \ldots, L^dR^d$ mutually
commute on $U_0$.

\end{conjecture}


\section{The zigzag basis}

\begin{definition}
\label{def:lam}
Assume either $d=3$, or that    
 Conjecture \ref{conj:lrcom} holds. Recall the 
field $\fld$  is algebraically closed, so there exists a nonzero
vector $v \in U_0$ that is a common eigenvector for
$LR, L^2R^2,\ldots, L^dR^d$.  For $0 \leq i \leq d$, We let $ 
\sigma_i$ denote the eigenvalue of $L^iR^i$ associated with $v$.
We observe $\sigma_0=1$.
For notational convenience, we abbreviate $\sigma = \sigma_1$,
$\lambda = \sigma_2$, $\varepsilon=\sigma_3$.

\end{definition}

\begin{lemma}
\label{lem:epsnot0} 
 Assume either $d=3$, or that    
 Conjecture \ref{conj:lrcom} holds.
Then refering to Definition
\ref{def:lam}, the scalar
$\sigma_d \not=0$. In particular,
if $d=3$ then 
$\varepsilon\not=0$.

\end{lemma}

\begin{proof} 
We saw in \cite{} that the restriction of $R^d $ to 
$U_0$ is a bijection onto $U_d$, and that the restriction of
$L^d$ to $U_d$ is a bijection onto $U_0$. It follows that the
restriction of $L^dR^d$ to $U_0$ is invertible. By definition
$\sigma_d$
is an eigenvalue for this map, so  
$\sigma_d\not=0$.

\end{proof}




\begin{lemma}
\label{lem:rlaction}
For $d=3$, we have
\begin{eqnarray}
Lv &=&0,
\label{eq:Lv}
\\
LRv &=& \sigma v,
\label{eq:LRv}
\\
L^2R^2v &=& \lambda v, 
\label{eq:LLRRv}
\\
L^3R^3v &=& \varepsilon v, 
\label{eq:LLLRRRv}
\\
R^4v &=& 0, 
\label{eq:RRRRv}
\\
RLR^2v &=& \sigma R^2v+ \lbrack 3 \rbrack^{-1} LR^3v, 
\label{eq:RLRRv}
\\
RLR^3v &=& \sigma R^3v, 
\label{eq:RLRRRv}
\\
R^2L^2R^3v &=& \lambda R^3v, 
\label{eq:RRLLRRRv}
\\
LRL^2R^3v &=& \varepsilon \lbrack 3 \rbrack^{-1} Rv + \sigma L^2R^3v.
\label{eq:LRLLRRRv}
\end{eqnarray}
\end{lemma}

\begin{proof}  Lines (\ref{eq:Lv})--(\ref{eq:RRRRv}) are routine.
To obtain 
(\ref{eq:RLRRv}), apply 
 (\ref{eq:length3}) to $v$, and use 
(\ref{eq:LRv}).
To obtain (\ref{eq:RLRRRv}), 
apply (\ref{eq:rlrrr}) to $v$, and use 
(\ref{eq:LRv}),
(\ref{eq:RRRRv}).
To obtain 
(\ref{eq:RRLLRRRv}), apply
(\ref{eq:rrllrrrv}) to $v$, 
and use
(\ref{eq:LLRRv}), (\ref{eq:RRRRv}).
To obtain 
(\ref{eq:LRLLRRRv}), apply
(\ref{eq:lrllrrrv}) to $v$, and use (\ref{eq:LRv}), (\ref{eq:LLRRv}),
(\ref{eq:RRRRv}).

\end{proof}
 
\begin{lemma} 
\label{lem:uispan}
For $d=3$,  the following (i)--(iv) hold.
\begin{enumerate}
\item The subspace $U_0$ is spanned by
\begin{equation}
v.
\label{eq:u0span}
\end{equation}
\item The subspace $U_1$ is spanned by
\begin{equation}
Rv, \quad LR^2v, \quad L^2R^3v.
\label{eq:u1span}
\end{equation}
\item The subspace $U_2$ is spanned by
\begin{equation}
R^2v,  \quad RL^2R^3v, \quad LR^3v
\label{eq:u2span}
\end{equation}
\item The subspace $U_3$ is spanned by
\begin{equation}
R^3v.
\label{eq:u3span}
\end{equation}
\end{enumerate}
Moreover, $V$ is spanned by the eight vectors 
(\ref{eq:u0span})--(\ref{eq:u3span}).

\end{lemma} 

\begin{proof} By definition $v$ is contained in $U_0$.
Applying (\ref{eq:raction}), 
(\ref{eq:laction}),  
we routinely find the vectors (\ref{eq:u1span}) are contained in
$U_1$, the vectors  
 (\ref{eq:u2span}) are contained in
$U_2$,  and the vector 
 (\ref{eq:u3span}) is contained in
$U_3$. 
In view of
(\ref{eq:vdec}), to finish the proof it suffices   to show
 the eight vectors 
(\ref{eq:u0span})--(\ref{eq:u3span}) span $V$. Let $W$ denote
the subspace of $V$ spanned by these eight vectors. We show
$W=V$. By the definition of a TD system, there is no subspace
of $V$  that is invariant under both $A$ and $A^*$, other than
$0$ and $V$. Therefore, to show $W=V$,
we show $AW\subseteq W$ and $A^*W\subseteq W$.
By 
(\ref{eq:fi1}), 
(\ref{eq:fi2}), we find that for 
$0 \leq i \leq 3$, the eight vectors 
(\ref{eq:u0span})--(\ref{eq:u3span}) are eigenvectors
for $F_i$. It follows $F_iW\subseteq W$.
Using Lemma 
\ref{lem:rlaction}, we routinely  find $RW\subseteq W$ and $LW\subseteq W$. 
Now $AW\subseteq W$ by 
(\ref{eq:defR}) and 
 $A^*W\subseteq W$ by 
(\ref{eq:defL}). Observe $W$ is nonzero, since it contains
the nonzero vector $v$, so $W=V$.

\end{proof}

\begin{lemma}
\label{lem:shapeV}
For $d=3$, the shape of $\Phi$ is one of
$1111$, $1221$, $1331$.

\end{lemma}

\begin{proof}
Recall the sequence
 $\rho_0, \rho_1, \rho_2, \rho_3$ is unimodal and symmetric.
By Lemma   
\ref{lem:uispan}, we have $\rho_0=1$ and $ \rho_2\leq 3$.
The result follows.

\end{proof}

\noindent It turns out that each of the three cases 
mentioned in Lemma 
\ref{lem:shapeV} can occur. Of the three,
the last case is the most general,
and in a sense the first two  cases are limiting cases of this one.
Therefore, for the time being and to keep things simple, 
we will focus on the last case.  

\begin{definition}
\label{def:generic}
For $d=3$, we will say the TD system $\Phi$  is generic 
whenever it has shape
  $1331$. 
In this case, the vectors
\begin{equation}
v, \quad Rv, \quad LR^2v, \quad L^2R^3v, 
\quad R^2v,  \quad RL^2R^3v, \quad LR^3v, \quad
R^3v
\label{eq:firstbase}
\end{equation}
form a basis for $V$, by Lemma 
\ref{lem:uispan}. We refer to this as the zigzag basis.

\end{definition}


\begin{lemma}
\label{lem:matsforf}
Assume $d=3$, and that $\Phi$ is generic. With respect to the 
zigzag basis 
(\ref{eq:firstbase}), the matrices representing 
$F_0, F_1, F_2, F_3$ are as follows.

\begin{eqnarray}
F_0: &&\quad \hbox{diag}(1,0,0,0,0,0,0,0),
\label{eq:f0mat}
\\
F_1:  &&\quad \hbox{diag}(0,1,1,1,0,0,0,0),
\label{eq:f1mat}
\\
F_2: &&\quad \hbox{diag}(0,0,0,0,1,1,1,0),
\label{eq:f2mat}
\\
F_3: &&\quad \hbox{diag}(0,0,0,0,0,0,0,1).
\label{eq:f3mat}
\end{eqnarray}
\end{lemma}

\begin{proof} Immediate from 
(\ref{eq:fi1}), 
(\ref{eq:fi2}),  and  Lemma
\ref{lem:uispan}.

\end{proof}


\begin{lemma}
\label{kmatt}
Assume $d=3$, and that $\Phi$ is generic. Then with
respect to the zigzag basis
(\ref{eq:firstbase}), the matrix representing $K$ is
\begin{eqnarray}
K: &&\quad \hbox{diag}(q^{-3},q^{-1},q^{-1},q^{-1},q,q,q,q^3).
\label{eq:Kmat}
\end{eqnarray}
\end{lemma}

\begin{proof} Immediate from 
(\ref{eq:kdef}) and 
Lemma \ref{lem:matsforf}.

\end{proof}


\begin{lemma}
\label{lem:matsforRL}
Assume $d=3$, and that $\Phi$ is generic. With respect to the 
zigzag basis 
(\ref{eq:firstbase}), the matrix representing $R$ is
\begin{eqnarray}
R: \qquad \left(
\begin{array}{ c c c c c c c c }
0 & 0  &  0    & 0 & 0 & 0 & 0 & 0  \\
1 & 0  &  0   &  0  & 0 & 0 & 0 & 0 \\
0 & 0  &  0    & 0 & 0 & 0 & 0 & 0  \\
0 & 0  &  0    & 0 & 0 & 0 & 0 & 0  \\
0 & 1  &  \sigma  & 0 & 0 & 0 & 0 & 0  \\
0 & 0  &  0    & 1 & 0 & 0 & 0 & 0  \\
0 & 0  &  \lbrack 3 \rbrack^{-1}  & 0 & 0 & 0 & 0 & 0  \\
0 & 0  &  0    & 0 & 1 & \lambda & \sigma & 0  \\
\end{array}
\right)  , 
\label{eq:rmat}
\end{eqnarray}
and the matrix representing $L$ is
\begin{eqnarray}
L: \qquad \left(
\begin{array}{ c c c c c c c c }
0 & \sigma  &  \lambda    & \varepsilon & 0 & 0 & 0 & 0  \\
0 & 0  &  0   &  0  & 0 &  
  \lbrack 3 \rbrack^{-1} \varepsilon& 0&0 \\
0 & 0  &  0    & 0 & 1 & 0 & 0 & 0  \\
0 & 0  &  0    & 0 & 0 &  \sigma & 1& 0  \\
0 & 0  &  0    & 0 & 0 & 0 & 0 & 0  \\
0 & 0  &  0    & 0 & 0 & 0 & 0 & 0  \\
0 & 0  &  0    & 0 & 0 & 0 & 0 & 1  \\
0 & 0  &  0    & 0 & 0 & 0 & 0 & 0  \\
\end{array}
\right)  . 
\label{eq:lmat}
\end{eqnarray}

\end{lemma}

\begin{proof} This is a routine consequence of 
Lemma 
\ref{lem:rlaction}.

\end{proof}


\begin{lemma} 
\label{lem:uqaction}
Assume $d=3$, and that $\Phi$ is generic. Then
with respect to the zigzag basis 
(\ref{eq:firstbase}), the matrix representing $r$ is
\begin{eqnarray}
r: \qquad \left(
\begin{array}{ c c c c c c c c }
0 & 0  &  0    & 0 & 0 & 0 & 0 & 0  \\
0 & 0  &  0   &  0  & 0 & 0 & 0 & 0 \\
0 & 0  &  0    & 0 & 0 & 0 & 0 & 0  \\
\lbrack 3 \rbrack \varepsilon^{-1} & 0  &  0    & 0 & 0 & 0 & 0 & 0  \\
0 & 0  &  1  & 0 & 0 & 0 & 0 & 0  \\
0 &  \lbrack 3 \rbrack \varepsilon^{-1} &0 &   0 & 0 & 0 & 0 & 0  \\
0 & 0 & \lbrack 3 \rbrack \lambda \varepsilon^{-1}  &  \lbrack 2 \rbrack^{2}  &  0 & 0 & 0 & 0  \\
0 & 0  &  0    & 0 & \lbrack 3 \rbrack \lambda \varepsilon^{-1} & 
\lbrack 2 \rbrack^2\sigma &
\lbrack 3 \rbrack  
& 0  \\
\end{array}
\right)  , 
\label{eq:smallrmat}
\end{eqnarray}
and the matrix representing $l$ is
\begin{eqnarray}
l: \qquad \left(
\begin{array}{ c c c c c c c c }
0 & \lbrack 3 \rbrack  &  \lbrack 2 \rbrack^2\sigma    & \lbrack 3 \rbrack \lambda & 0 & 0 & 0 & 0  \\
0 & 0  &  0   &  0   &  
  \lbrack 2 \rbrack^{2} &  \lbrack 3\rbrack \lambda &0 & 0\\
0 & 0  &  0    & 0 & 0 & 0 &\lbrack 3 \rbrack  & 0  \\
0 & 0  &  0    & 0 & 0 &  1 & 0&  0  \\
0 & 0  &  0    & 0 & 0 & 0 & 0 & \lbrack 3 \rbrack  \\
0 & 0  &  0    & 0 & 0 & 0 & 0 & 0  \\
0 & 0  &  0    & 0 & 0 & 0 & 0 & 0  \\
0 & 0  &  0    & 0 & 0 & 0 & 0 & 0  \\
\end{array}
\right)  . 
\label{eq:smalllmat}
\end{eqnarray}

\end{lemma}


\begin{proof} One readily checks
the matrices 
(\ref{eq:lmat}),
(\ref{eq:Kmat}), 
(\ref{eq:smallrmat}) satisfy the relations
(\ref{eq:getr}). It follows   
the matrix (\ref{eq:lmat})  represents 
$r$ by Corollary 
\ref{cor:fixingrl}.
 Similarily, one readily checks
the matrices 
(\ref{eq:rmat}),
(\ref{eq:Kmat}), 
(\ref{eq:smalllmat}) satisfy the relations
(\ref{eq:getl}).  
 It follows   
the matrix (\ref{eq:rmat})  represents 
$l$ by Corollary 
\ref{cor:fixingrl}.

\end{proof}



\noindent Referring to the generic case with $d=3$, the reader might
wonder if 
 there any restrictions on $\sigma $, $\lambda$, $\varepsilon$.
As we will see, the sequence $\sigma, \lambda, \varepsilon$
cannot take certain singular values as a consequence of irreducibility.
However, there are no equations relating 
 $\sigma, \lambda, \varepsilon$.
Indeed we have the following.

\begin{lemma}
\label{lem:serrecheck}
Let $\sigma$, $\lambda$, and $\varepsilon$ denote arbitrary scalars
in $\fld $. Let $q$ denote any  nonzero scalar in $\fld$ that is not
a root of unity. Then the matrices
(\ref{eq:rmat}), 
(\ref{eq:lmat}) 
satisfy the Serre relations
(\ref{eq:ms1}), 
(\ref{eq:ms2}), where $\lbrack 3 \rbrack $ is as in
(\ref{eq:brackndef}).

\end{lemma}

\begin{proof} Routine verification.

\end{proof}


\section{A 
$U_q(\widehat{ sl_2})$ action}
\noindent 
In this section, we   
construct an action of
the quantum affine algebra 
$U_q(\widehat{ sl_2})$ on $V$ that is closely related
to our maps $R,L,r,l$.
The definition of 
$U_q(\widehat{ sl_2})$ is given in Appendix 2.


%\noindent  We need the following technical result. 
%
%\begin{lemma}
%\label{lem:commm}
%Assume $d=3$, and that $\Phi$ is generic.
%\begin{enumerate}
%\item If a linear transformation 
% $X:V \rightarrow V$ satisfies
%both 
%\begin{eqnarray}
%&&XL=LX, \qquad \qquad 
%K X = q^2 X K,
%\end{eqnarray}
%then $X=0$.  
%\item
%If a  
% linear transformation on $Y:V\rightarrow V$ satisfies both 
%\begin{eqnarray}
%&&YR=RY,
%\qquad \qquad   K Y = q^{-2} Y K,
%\end{eqnarray}
%then $Y=0$. 
%\end{enumerate}
%\end{lemma}
%
%\begin{proof} Routine linear algebra.
%\end{proof}

%\begin{lemma} 
%\label{lem:uqaction}
%Assume $d=3$, and that $\Phi$ is generic. Then there
%exists a unique linear transformation $r:V \rightarrow V$
%such that both
%\begin{eqnarray}
%&&K r = q^2 r K, \qquad \qquad  
%rL-Lr = {{K-K^{-1}}\over {q-q^{-1}}}.
%\label{eq:getr}
%\end{eqnarray}
%With respect to the zigzag basis 
%(\ref{eq:firstbase}), the matrix representing $r$ is
%\begin{eqnarray}
%r: \qquad \left(
%\begin{array}{ c c c c c c c c }
%0 & 0  &  0    & 0 & 0 & 0 & 0 & 0  \\
%0 & 0  &  0   &  0  & 0 & 0 & 0 & 0 \\
%0 & 0  &  0    & 0 & 0 & 0 & 0 & 0  \\
%\lbrack 3 \rbrack \varepsilon^{-1} & 0  &  0    & 0 & 0 & 0 & 0 & 0  \\
%0 & 0  &  1  & 0 & 0 & 0 & 0 & 0  \\
%0 &  \lbrack 3 \rbrack \varepsilon^{-1} &0 &   0 & 0 & 0 & 0 & 0  \\
%0 & 0 & \lbrack 3 \rbrack \lambda \varepsilon^{-1} 
%&  \lbrack 2 \rbrack^{2}  &  0 & 0 & 0 & 0  \\
%0 & 0  &  0    & 0 & \lbrack 3 \rbrack \lambda \varepsilon^{-1} & 
%\lbrack 2 \rbrack^2\sigma &
%\lbrack 3 \rbrack  
%& 0  \\
%\end{array}
%\right)  . 
%\label{eq:smallrmat}
%\end{eqnarray}
%Similarily, 
%there exists a unique linear transformation $l:V \rightarrow V$
%such that both
%\begin{eqnarray}
% K l = q^{-2} l K, \qquad \qquad 
%lR-Rl = {{K^{-1}-K}\over {q-q^{-1}}}. 
%\label{eq:getl}
%\end{eqnarray}
%With respect to the zigzag basis 
%(\ref{eq:firstbase}), the matrix representing $l$ is
%\begin{eqnarray}
%l: \qquad \left(
%\begin{array}{ c c c c c c c c }
%0 & \lbrack 3 \rbrack  &  \lbrack 2 \rbrack^2\sigma  
%& \lbrack 3 \rbrack \lambda & 0 & 0 & 0 & 0  \\
%0 & 0  &  0   &  0   &  
%  \lbrack 2 \rbrack^{2} &  \lbrack 3\rbrack \lambda &0 & 0\\
%0 & 0  &  0    & 0 & 0 & 0 &\lbrack 3 \rbrack  & 0  \\
%0 & 0  &  0    & 0 & 0 &  1 & 0&  0  \\
%0 & 0  &  0    & 0 & 0 & 0 & 0 & \lbrack 3 \rbrack  \\
%0 & 0  &  0    & 0 & 0 & 0 & 0 & 0  \\
%0 & 0  &  0    & 0 & 0 & 0 & 0 & 0  \\
%0 & 0  &  0    & 0 & 0 & 0 & 0 & 0  \\
%\end{array}
%\right)  . 
%\label{eq:smalllmat}
%\end{eqnarray}
%
%\end{lemma}
%
%
%\begin{proof} Concerning the existance of $r$, one readily checks
%the matrices 
%(\ref{eq:lmat}),
%(\ref{eq:Kmat}), 
%(\ref{eq:smallrmat}) satisfy the relations
%(\ref{eq:getr}). The uniqueness of $r$ follows from
%Lemma
%\ref{lem:commm}. 
% Concerning the existance of $l$, one readily checks
%the matrices 
%(\ref{eq:rmat}),
%(\ref{eq:Kmat}), 
%(\ref{eq:smalllmat}) satisfy the relations
%(\ref{eq:getl}). The uniqueness of $l$ follows from
%Lemma
%\ref{lem:commm} .
%\end{proof}
%

\begin{theorem}
Assume $d=3$, and that $\Phi$ is generic. Then
the maps $r$ and $l$  from
Corollary
\ref{cor:fixingrl}
satisfy
\begin{eqnarray}
&&lL = Ll , \qquad \qquad rR=Rr,
\label{eq:lll}
\\
&&
0= r^3l - \lbrack 3 \rbrack r^2lr+ \lbrack 3 \rbrack rlr^2 - lr^3, 
\label{eq:smallserre1}
\\
&&0= rl^3 - \lbrack 3 \rbrack lrl^2+ \lbrack 3 \rbrack l^2rl - l^3r.
\label{eq:smallserre2}
\end{eqnarray}

\end{theorem}

\begin{proof} One readily verifies
the matrices
(\ref{eq:rmat}),
(\ref{eq:lmat}),
(\ref{eq:smallrmat}),
(\ref{eq:smalllmat})
 satisfy 
(\ref{eq:lll})--(\ref{eq:smallserre2}).
\end{proof}

\begin{theorem}
\label{thm:extend}
Assume $d=3$, and that  $\Phi$ is generic.
Then $V$ carries the structure of a
$U_q(\widehat{ sl_2})$ module
with 
action
$K\leftrightarrow K_0$,
 $K^{-1}\leftrightarrow K_1$,
$L\leftrightarrow e^{-}_0$,
$R\leftrightarrow e^{-}_1$,
 $l\leftrightarrow e^{+}_1$,
$r\leftrightarrow e^{+}_0$.
Moreover, $V$ is irreducible as a
$U_q(\widehat{ sl_2})$ module.

\end{theorem}

\begin{proof} The relations have all been verified.
$V$ is irreducible as an  
$U_q(\widehat{ sl_2})$-module in view of
(\ref{eq:aandk}),
(\ref{eq:asandk}), and since there is no subspace of $V$
that is invariant under  both $A$ and $A^*$, other than $0$
and $V$.
\end{proof}



\section{The weights}


\noindent We now introduce some parameters that will
help us describe our TD pair.

\begin{definition} 
Assume either $d=3$, or that    
 Conjecture \ref{conj:lrcom} holds.
We let $w_1, w_2, \ldots, w_d$   denote the
roots of the polynomial
\begin{eqnarray}
\sum_{i=0}^d {{\sigma_i (-1)^i x^{d-i}}\over 
{\lbrack 1\rbrack^2\lbrack 2 \rbrack^2\cdots \lbrack i \rbrack^2 }},
\end{eqnarray}
where the $\sigma_i$ are from Definition
\ref{def:lam}.  We refer to these roots as the weights of $\Phi$.
These weights are only defined up to permutation, and are not
assumed to 
be distinct.
\end{definition}

\begin{example}
\label{def:weights}
For $d=3$, the weights $w_1, w_2, w_3$ are the roots
of the polynomial 
\begin{equation}
x^3  - \sigma x^2 + {{\lambda }\over {\lbrack 2 \rbrack^2 }} x
- {{\varepsilon }\over {\lbrack 2 \rbrack^2 \lbrack 3 \rbrack^2}},
\label{eq:weights}
\end{equation}
where $\sigma$, $\lambda$, $\varepsilon$ are from
Definition \ref{def:lam}.
We observe 
\begin{eqnarray}
\sigma &=& w_1 + w_2 + w_3, \label{eq:wsum}
\\
\lambda &=& \lbrack 2 \rbrack^2(  w_1 w_2+ w_1 w_3 + w_2 w_3),
\label{eq:w2sum}
\\
\varepsilon &=& 
\lbrack 2 \rbrack^2 \lbrack 3 \rbrack^2  w_1 w_2 w_3.
\label{eq:wprod}
\end{eqnarray}
\end{example}



\begin{lemma}
Assume $d=3$, or that    
 Conjecture \ref{conj:lrcom} holds.
Then the weights $w_1, w_2,  \ldots, w_d$ are nonzero.
\end{lemma}


\begin{proof} The product $w_1w_2\cdots w_d$ is a nonzero
scalar multiple of 
 $\sigma_d $, and 
 $\sigma_d 
 \not=0$  by Lemma
\ref{lem:epsnot0}.

\end{proof}

\noindent We now use the weights to describe the shape of
$\Phi$.

\begin{definition}
By a $q$-string, we mean a 
 set of scalars in $\fld$ 
of the form 
$w, wq^2, wq^4,\ldots, wq^{2n}$, where $n$ is a nonnegative
integer. We observe
the elements of a $q$-string are distinct.
By the length the a $q$-string, we mean the number of scalars
in that string, which we observe equals $n+1$.

\end{definition}

\begin{lemma}
\label{eq:wrest}
 Assume $d=3$. Then the following (i)--(iii) hold.
\begin{enumerate}
\item $\Phi$ has shape 1111 if and only if
the set $w_1, w_2, w_3$ is a $q$-string.
\item
$\Phi$ has shape 1221 if and only if
the longest $q$-string contained in 
$w_1, w_2, w_3$ has length 2.
\item 
$\Phi$ has shape 1331 if and only if
the longest $q$-string contained in 
$w_1, w_2, w_3$ has length 1.
\end{enumerate}
\end{lemma}

\noindent To guess at what the pattern is for general $d$,
we use the following concept.  

\begin{definition}(Chari and Pressley, p275)
Two $q$-strings $S_1$ and $S_2$ are said to be in general position
if either
\begin{enumerate}
\item $S_1\cup S_2$ is not a $q$-string, or 
\item $S_1\subseteq S_2$ or $S_2\subseteq S_1$.
\end{enumerate}
\end{definition}

\begin{lemma}(Chari and Pressley, p276)
Any finite multi-set of scalars in $\fld$ can be written
uniquely as  a union of $q$-strings, any two of which are in
general position.
\end{lemma}

\begin{conjecture} 
\label{con:shape}
Assume either $d=3$, or that    
 Conjecture \ref{conj:lrcom} holds.
Write the weights $w_1, w_2, \ldots, w_d$
as a union of $q$-strings, 
any two of which are in
general position. Let $n$ denote the number of 
$q$-strings in this decomposition, and let   
$d_1, d_2, \ldots, d_n$ denote the lengths of these strings.
Then the polynomial
\begin{eqnarray}
&& \qquad \sum_{i=0}^d \rho_i t^i =
\\
&&(1+t+t^2+\cdots + t^{d_1})
(1+t+t^2+\cdots + t^{d_2})
\cdots
(1+t+t^2+\cdots + t^{d_n}).
\end{eqnarray}
\end{conjecture}

%\begin{definition}
%Assume 
% Conjecture \ref{conj:lrcom} 
%and Conjecure \label{con:shape}
%holds. Expanding on 
%Lemma \ref{eq:wrest}(iii), we say $\Phi$ is
%generic whenever the set $w_1, w_2, \ldots, w_d$ contains
%no $q$-strings of length $\geq 2$. 
%\end{definition}
%%%%%%%%
\section{Some bilinear forms}

\begin{lemma}
\label{lem:form1}
Assume $d=3$, and that $\Phi$ is generic.
Then there exists a nonzero symmetric bilinear form
$\langle \,,\,\rangle : V \times V \rightarrow \fld$ such
that 
\begin{eqnarray}
\langle Ru, v \rangle &=& 
\langle u, Lv \rangle 
\label{eq:effectonr}
\end{eqnarray}
for all $u,v \in V$.
The form is nondegenerate, and unique up to multiplication by
a nonzero scalar.
With respect to
the zigzag basis 
(\ref{eq:firstbase})
 the matrix representing 
$\langle \,,\,\rangle $
is block diagonal 
\begin{eqnarray}
 \alpha \hbox{diag} (M_0, M_1, M_2, M_3), 
\label{eq:mdef}
\end{eqnarray}
where $\alpha $ is a nonzero scalar, and 
\begin{eqnarray}
M_0 &=& 
 \;\left(
\begin{array}{ c  }
1 \\
\end{array}
\right)  , 
\label{eq:m0}
\\
M_1 &=& 
 \;\left(
\begin{array}{ c c c  }
\sigma & \lambda &
\varepsilon    \\
\lambda  & \sigma \lambda + {{\varepsilon}\over {\lbrack 3 \rbrack}} & 
{{\sigma \varepsilon \lbrack 2 \rbrack^2}\over {\lbrack 3 \rbrack }}\\
\varepsilon & 
{{\sigma \varepsilon \lbrack 2 \rbrack^2}\over {\lbrack 3 \rbrack }}
&
\lambda \varepsilon \\
\end{array}
\right)  , 
\label{eq:m1ag}
\\
M_2 &=& 
 \;\left(
\begin{array}{ c c c  }
\lambda & 
{{\sigma \varepsilon \lbrack 2 \rbrack^2}\over {\lbrack 3 \rbrack }}
&\varepsilon
\\
{{\sigma \varepsilon \lbrack 2 \rbrack^2}\over {\lbrack 3 \rbrack }}
&
\sigma \lambda \varepsilon + {{\varepsilon^2}\over {\lbrack 3 \rbrack}} & 
 \lambda \varepsilon
\\
\varepsilon
& \lambda \varepsilon & \sigma \varepsilon
\end{array}
\right)  , 
\label{eq:m2ag}
\\
M_3 &=& 
 \;\left(
\begin{array}{ c   }
\varepsilon
 \\
\end{array}
\right)  . 
\label{eq:m3}
\end{eqnarray}
\end{lemma}

\begin{proof} Concerning the existance of 
$\langle\,,\,\rangle $, one readily verifies
the transpose of the matrix  
(\ref{eq:rmat}) 
times the matrix 
(\ref{eq:mdef}),
equals the matrix 
(\ref{eq:mdef})
times the matrix
(\ref{eq:lmat}). 
If one writes the determinants of $M_0, M_1, M_2, M_3$ in
terms of the weights $w_1, w_2, w_3$,  one readily finds
these determinants are nonzero. It follows 
$\langle \,,\,\rangle $ is nondegenerate.
Using 
(\ref{eq:effectonr}), one routinely
obtains the uniqueness of 
$\langle\,,\,\rangle $.
\end{proof}

\begin{lemma} 
\label{lem:whatabtrl}
Assume $d=3$, and that $\Phi$ is generic. Let 
the  bilinear form
$\langle \,,\,\rangle $ be as in 
Lemma 
\ref{lem:form1}. Then the map $ K$ from
(\ref{eq:kdef})
and the maps $r,l$ from
Lemma \ref{lem:uqaction}
satisfy
\begin{eqnarray}
\langle Ku,v \rangle &=& \langle u, Kv \rangle
\label{eq:anglek}
\\
\langle ru,v \rangle &=& \langle u, lv \rangle,
\label{eq:angler}
\end{eqnarray}
for all $u,v \in V$.
\end{lemma}

\begin{proof} To obtain
(\ref{eq:anglek}), observe the matrix
(\ref{eq:Kmat})
is symmetric, and commutes with the matrix
(\ref{eq:mdef}). To obtain
(\ref{eq:angler}),
observe the transpose of the matrix
(\ref{eq:smallrmat})
times 
the matrix 
(\ref{eq:mdef})
equals 
the matrix 
(\ref{eq:mdef})
times
the matrix
(\ref{eq:smalllmat}).

\end{proof}

\noindent Here is second bilinear form.

\begin{lemma}
\label{lem:form2}
Assume $d=3$, and that $\Phi$ is generic.
Then there exists a nonzero symmetric bilinear form
$( \,,\,) : V \times V \rightarrow \fld$ such
that 
\begin{eqnarray}
(Ru, v ) &=& 
(u, Rv ), 
\label{eq:f2effectonr}
\\
(Lu, v )&=& 
( u, Lv ) 
\label{eq:f2effectonl}
\end{eqnarray}
for all $u,v \in V$.
The form is nondegenerate, and unique up to multiplication by
a nonzero scalar.
With respect to
the zigzag basis 
(\ref{eq:firstbase}),
 the matrix representing 
$(\,,\,) $ has the form 
\begin{eqnarray}
 \;\gamma \left(
\begin{array}{ c c c c }
0 & 0 & 0 & 1
\\
0  & 0 &  P & 0 
\\
0 &  P^t & 0 & 0 
\\
1 & 0 & 0 &0\\
\end{array}
\right)  , 
\label{eq:shapep}
\end{eqnarray}
where $\gamma $ is a nonzero scalar, and 
where 
\begin{eqnarray}
P &=& 
 \;\left(
\begin{array}{ c c c  }
1 & \lambda &
\sigma     \\
{{\sigma \lbrack 2 \rbrack^2}\over {\lbrack 3 \rbrack }}
&
 \sigma \lambda + {{\varepsilon}\over {\lbrack 3 \rbrack}} & 
\lambda \\
\lambda 
&
{{\sigma \varepsilon \lbrack 2 \rbrack^2}\over {\lbrack 3 \rbrack }}
&
\varepsilon \\
\end{array}
\right)  . 
\label{eq:pp}
\end{eqnarray}
\end{lemma}

\begin{proof}
Concerning the  existance of
$( \,,\,)$ 
 observe 
the transpose of the matrix
(\ref{eq:rmat}) 
times
the matrix (\ref{eq:shapep}) equals
the matrix (\ref{eq:shapep}) 
times (\ref{eq:rmat}) , 
and also
the transpose of the matrix
(\ref{eq:lmat}) 
times
the matrix (\ref{eq:shapep}) equals
the matrix (\ref{eq:shapep}) 
times (\ref{eq:lmat}) . 
If one writes the determinant of $P$ in terms of
the weights $w_1, w_2, w_3$, one readily finds it is nonzero.
It follows 
 the form 
$( \,,\,)$ 
 in nondegenerate.
Using 
(\ref{eq:f2effectonr}), (\ref{eq:f2effectonl}),
one readily obtains the uniqueness of
$( \,,\,)$.

\end{proof}

\begin{lemma}
Assume $d=3$, and that $\Phi$ is generic.
Let the  bilinear form
$( \,,\,)$
be as in Lemma
\ref{lem:form2}. Then
 the map $ K$ from
(\ref{eq:kdef})
and the maps $r,l$ from
Lemma \ref{lem:uqaction}
satisfy
\begin{eqnarray}
\langle Ku,v \rangle &=& \langle u, K^{-1}v \rangle,
\label{eq:angleks}
\\
\langle ru,v \rangle &=& \langle u, rv \rangle,
\label{eq:anglers}
\\
\langle lu,v \rangle &=& \langle u, lv \rangle
\label{eq:anglels}
\end{eqnarray}
for all $u,v \in V$.
\end{lemma}

\begin{proof} To obtain
(\ref{eq:angleks}), observe the  transpose of
the matrix (\ref{eq:Kmat})
times the matrix 
(\ref{eq:shapep})
equals the matrix 
(\ref{eq:shapep}) times the inverse of
the matrix (\ref{eq:Kmat}).
 To obtain
(\ref{eq:anglers}),
observe the transpose of the matrix
(\ref{eq:smallrmat})
times 
the matrix 
(\ref{eq:shapep})
equals 
the matrix 
(\ref{eq:shapep})
times
the matrix
(\ref{eq:smallrmat}).
 To obtain
(\ref{eq:anglels}),
observe the transpose of the matrix
(\ref{eq:smalllmat})
times 
the matrix 
(\ref{eq:shapep})
equals 
the matrix 
(\ref{eq:shapep})
times
the matrix
(\ref{eq:smalllmat}).

\end{proof}

\noindent The following seems tricky; Im still not sure if
it is true or not.
\begin{conjecture}
Assume $d=3$, and that $\Phi$ is generic. Then
the following are equivalent.
\begin{enumerate}
\item
There exists a nondegenerate symmetric bilinear form 
$<,>: V \times V \rightarrow \fld $ 
such that
\begin{eqnarray}
< Au,v> &=& <u,Av>,
\\
< A^*u,v> &=& <u,A^*v>
\end{eqnarray}
for all $u,v \in V$.
\item The weights $w_1, w_2, w_3$ are distinct.
\end{enumerate}
\end{conjecture}



\section{The elements $q^{-1}RL-qLR$ and $qRL-q^{-1}LR$}

In this section, we 
consider the following two maps:
\begin{eqnarray}
 q^{-1} RL - q LR, \qquad \qquad qRL-q^{-1} LR.
\label{eq:ddef}
\end{eqnarray}
\noindent 
We show the eigenvalues of these maps are closely related to the
weights. In what follows, we will focus on $q^{-1}RL-qLR$.
To get the corresponding results for $qRL-q^{-1}LR$, just
replace $q$ by $q^{-1}$ everywhere.


%\noindent It is well known (at least by me) that 
%TD pairs come up in  combinatorics, in the theory of association
%schemes. For the TD pairs that arise  in this manner,
%the products $E^*_iAE^*_i$ are
%always diagonalizable, because they can be represented by
%symmetric matrices with real entries. In this case,
%$C$ is diagonalizable in view of 
%(\ref{eq:eae}).  
%
%\noindent Let us now consider when $C$ is diagonalizable.


\begin{lemma}
\label{lem:dblocks}
Assume $d=3$, and that $\Phi$ is generic. With respect to the 
zig-zag basis 
(\ref{eq:firstbase}), the matrix representing $q^{-1}RL-qLR$ is block
diagonal
\begin{eqnarray}
  \hbox{diag} (C_0, C_1, C_2, C_3), 
\end{eqnarray}
where
\begin{eqnarray}
C_0 &=& 
 \;\left(
\begin{array}{ c  }
-q \sigma  \\
\end{array}
\right)  , 
\label{eq:d0}
\\
C_1 &=& 
 \;\left(
\begin{array}{ c c c  }
{{\sigma}\over {q}} & {{\lambda}\over {q}} &
{{\varepsilon \lbrack 2 \rbrack }\over { q^2 \lbrack 3 \rbrack}}   \\
-q & -q\sigma & 0  \\
0 & -{{q}\over {\lbrack 3 \rbrack}}& -q \sigma \\
\end{array}
\right)  , 
\label{eq:d1}
\\
C_2 &=& 
 \;\left(
\begin{array}{ c c c  }
{{\sigma}\over{q}}  & 
{{ \varepsilon  }\over { q \lbrack 3 \rbrack}} &0   \\
0
&  
{{\sigma}\over {q}}
&
{{1}\over {q}} 
\\
-{{q^2\lbrack 2 \rbrack }\over {\lbrack 3 \rbrack}} & -q\lambda &  -q \sigma  \\
\end{array}
\right)  , 
\label{eq:d2}
\\
C_3 &=& 
 \;\left(
\begin{array}{ c   }
q^{-1} \sigma
 \\
\end{array}
\right)  . 
\label{eq:d3}
\end{eqnarray}
\end{lemma}

\begin{proof} Routine computation using
(\ref{eq:rmat}), 
(\ref{eq:lmat}), 
(\ref{eq:ddef}).

\end{proof}

\begin{lemma} Assume $d=3$.
Then the matrices $C_0, C_1, C_2, C_3$
in Lemma 
\ref{lem:dblocks} satisfy the following (i)--(iv).
\begin{enumerate}
\item The eigenvalue of $C_0$ is
\begin{eqnarray}
- q w_1 - q w_2 - q w_3.
\label{eq:d0eig}
\end{eqnarray}
\item The eigenvalues of $C_1$ are 
\begin{eqnarray}
 q^{-1} w_1 - q w_2 - q w_3, \qquad 
- q w_1 + q^{-1} w_2 - q w_3, \qquad 
- q w_1 - q w_2 + q^{-1} w_3.
\label{eq:d1eig}
\end{eqnarray}
\item The eigenvalues of $C_2$ are 
\begin{eqnarray}
 q^{-1} w_1 + q^{-1} w_2 - q w_3, \qquad 
 q^{-1} w_1 - q w_2 + q^{-1} w_3, \qquad 
- q w_1 + q^{-1} w_2 + q^{-1} w_3.
\label{eq:d2eig}
\end{eqnarray}
\item The eigenvalue of $C_3$ is 
\begin{eqnarray}
 q^{-1} w_1 + q^{-1} w_2 + q^{-1} w_3. 
\label{eq:d3eig}
\end{eqnarray}
\end{enumerate}
\end{lemma}

\begin{proof}  
In 
(\ref{eq:d0})--(\ref{eq:d3}), 
eliminate $\sigma, \lambda, \varepsilon $ using
(\ref{eq:wsum})--(\ref{eq:wprod}),
and compute the eigenvalues of the resulting matrices.

\end{proof}



%We observe  the expressions in
%(\ref{eq:d0eig})--(\ref{eq:d3eig}) equal the expressions
%$\lambda_x$ given in Appendix 2.

%\begin{definition}
%\label{def:ceig}
%For all subsets $x \subseteq \lbrace 1,2,\ldots d\rbrace$,
%we define 
%\begin{eqnarray}
%\lambda_x = q^{-1}\sum_{i \in x} w_i - q\sum_{j\in y} w_j,
%\label{eq:ceig}
%\end{eqnarray}
%where $y$ denotes the complement of $x$ in 
%$ \lbrace 1,2,\ldots, d\rbrace $.
%
%\end{definition}
%%%%%%%%%%

%\begin{conjecture} Assume
% Conjecture \ref{conj:lrcom} holds. Further assume $\Phi$ is
% generic, and that $w_1, w_2, \ldots, w_d$ are distinct.
%Then the eigenvalues of $C$ are the $\lambda_x$, where $x$
%runs over all subsets of 
%$ \lbrace 1,2,\ldots, d\rbrace $.
%\end{conjecture}

\noindent Given the remarkable 
pattern among the eigenvalues of
(\ref{eq:ddef}), 
 it is natural
to consider the corresponding eigenvectors. Unfortunately,
it turns out these maps are  
 not always diagonalizable, even in the generic case
with $d=3$.
We now investigate how this can occur.

\begin{lemma}
\label{lem:minchar}
Assume $d=3$, and that $\Phi$ is generic. Then refering to 
Lemma 
\ref{lem:dblocks},
the minimal and characteristic polynomial of $C_1$  coincide.
Moreover,  
the minimal and characteristic polynomial of $C_2$  coincide.
\end{lemma}

\begin{proof} 
The $31$ entry of $C_1^2$ is 
$q^2\lbrack 3 \rbrack^{-1} $, and hence nonzero. The $31$ entry of
both $C_1$ and $I$ are zero.  It follows 
$C^2_1$ is not dependent on $I$, $C_1$, so the minimal polynomial
of $C_1$ has degree 3. 
Similarily, the $21$ entry of $C_2^2$ is 
$-q \lbrack 2 \rbrack \lbrack 3 \rbrack^{-1}$,
and hence nonzero. The $21$ entry of $C_2$ and $I$ are zero.
It follows $C_2^2$ is not linearly dependent on $C_2$ and $I$,
so the minimal polynomial of $C_2$ has degree 3.
The  result follows.

\end{proof}

\begin{corollary}
\label{cor:whenddiag} Assume $d=3$, and that $\Phi$ is generic. 
Then the following are equivalent:
\begin{enumerate}
\item $q^{-1}RL-qLR$ is diagonalizable.
\item The eigenvalues of $C_1$ are distinct, and the eigenvalues of
$C_2$ are distinct.
\item The weights $w_1, w_2, w_3$ are distinct. 
\end{enumerate}
\end{corollary}

\begin{proof} $(i)\leftrightarrow (ii)$ Immediate from Lemma
\ref{lem:minchar}

\noindent 
 $(ii)\leftrightarrow (iii)$ From 
(\ref{eq:d1eig}) and  
(\ref{eq:d2eig}),
we see the
eigenvalues of  $C_1$   are distinct if and
only if
the eigenvalues of  $C_2$   are distinct if and
only if 
$w_1, w_2, w_3$ are distinct.

\end{proof}

\noindent We finish this section with a few comments.
\begin{lemma}
\label{lem:candform1}
Assume $d=3$, and that $\Phi$ is
generic.
Write $C^\pm:=q^{\pm}RL-q^{\mp}LR$.  
\begin{enumerate}
\item The 
 bilinear form
$\langle \,,\,\rangle $
from  
(\ref{eq:effectonr}) satisfies
\begin{eqnarray}
\langle C^\pm u, v \rangle = \langle u,C^\pm v \rangle
\label{eq:candform1}
\end{eqnarray}
for all $u,v \in V$.
\item The  bilinear form 
$( \,,\,) $
from  
Lemma \ref{lem:form2} satisfies
\begin{eqnarray}
( C^\pm u, v ) = (u,C^\mp v ) 
\label{eq:candform2}
\end{eqnarray}
for all $u,v \in V$.
\end{enumerate}
\end{lemma}

\begin{proof} Routine.

\end{proof}

%%pt jan31

\begin{lemma}
\label{lem:cCcommute}
Assume $d=3$, and that $\Phi$
is generic. Then the maps
\begin{eqnarray}
q^{-1}RL - qLR, \qquad \qquad qrl-q^{-1}lr
\end{eqnarray}
commute. Moroever, the maps
\begin{eqnarray}
qRL - q^{-1}LR, \qquad \qquad q^{-1}rl-qlr
\end{eqnarray}
commute. 
\end{lemma}

\begin{proof} Follows from the relations
(\ref{eq:getr}),
(\ref{eq:getl}),
(\ref{eq:lll}).

\end{proof}





\section{On finding a nice basis}

\noindent Assume $d=3$, and  that $\Phi$ is generic. We already
have one basis for $V$,  namely the zig-zag basis. However,
we would like to find a better one. Here are several ways
to proceed:
\begin{enumerate}
\item Find a basis for $V$  consisting of eigenvectors for
$q^{-1}RL-qLR$. (When $w_1, w_2, w_3$ are distinct). 
\item Find a basis for $V$  consisting of eigenvectors for
$qRL-q^{-1}LR$. (When $w_1, w_2, w_3$ are distinct). 
\item Find a basis for $V$ with respect to which the matrix
representing
$q^{-1}RL-qLR$ is lower triangular, and the matrix representing
$qRL-q^{-1}LR$ is upper triangular.
\item Find a basis for $V$ with respect to which the matrix
representing
$q^{-1}RL-qLR$ is upper triangular, and the matrix representing
$qRL-q^{-1}LR$ is lower triangular.
\end{enumerate}
There is no substantial difference between projects (i), (ii),
since $q$ is only defined up to inverse. Similarily, there is
no substantial difference between (iii), (iv). So we will focus
on (i) and (iii).

\section{An eigenbasis for $q^{-1}RL-qLR$}

\noindent In this section, we  assume $d=3$, and that $\Phi$ is
generic. We further
assume the weights $w_1, w_2, w_3$ are distinct,  
so
that the map $q^{-1}RL-qLR$
is diagonalizable. 
For each subset 
$x \subseteq  \lbrace 1,2,3 \rbrace $,
we define a vector $v_x$, and show this is an eigenvector
for $q^{-1}RL-qLR$.
We compute the action of $R, L, r, l$ on the basis $v_x$.

\begin{definition}
\label{def:basis2}
Assume $d=3$, and that $\Phi$ is generic.
Further assume the weights  $w_1, w_2, w_3$ from 
Definition \ref{def:weights} are distinct. For each subset
$x \subseteq  \lbrace 1,2,3 \rbrace $, we define a vector $v_x$
as follows.
\begin{eqnarray}
&&v_{\emptyset} = v,
\label{eq:vempty}
\\
&&v_1 = {{w_1^2 Rv}\over{(w_1-w_2)(w_1-w_3)}}  \;-\;
{{q w_1 LR^2v}\over {\lbrack 2 \rbrack (w_1-w_2)(w_1-w_3)}}  
\nonumber\\
&&  \quad \qquad \qquad   +\;
{{q^2 L^2R^3v}\over 
{\lbrack 2 \rbrack^2\lbrack 3 \rbrack (w_1-w_2)(w_1-w_3)}},  
\\
&&v_2 = {{w_2^2 Rv}\over{(w_2-w_1)(w_2-w_3)}}  \;-\;
{{q w_2 LR^2v}\over {\lbrack 2 \rbrack (w_2-w_1)(w_2-w_3)}}  
\nonumber\\
&&  \quad \qquad \qquad   +\;
{{q^2 L^2R^3v}\over 
{\lbrack 2 \rbrack^2\lbrack 3 \rbrack (w_2-w_1)(w_2-w_3)}},  
\\
&&v_3 = {{w_3^2 Rv}\over{(w_3-w_1)(w_3-w_2)}}  \;-\;
{{q w_3 LR^2v}\over {\lbrack 2 \rbrack (w_3-w_1)(w_3-w_2)}}  
\nonumber\\
&&  \quad \qquad \qquad   +\;
{{q^2 L^2R^3v}\over 
{\lbrack 2 \rbrack^2\lbrack 3 \rbrack (w_3-w_1)(w_3-w_2)}},  
\\
&&v_{12} = {{w_1 w_2 R^2v}\over{\lbrack 2 \rbrack (w_1-w_3)(w_2-w_3)}} \;+\;
  {{q^2w_3LR^3v}\over{\lbrack 2 \rbrack \lbrack 3 \rbrack (w_1-w_3)(w_2-w_3)}}
  \\
 && \qquad \qquad  \quad-  
 \;{{q RL^2R^3v}\over
 {\lbrack 2 \rbrack^2 \lbrack 3 \rbrack (w_1-w_3)(w_2-w_3)}},
\\
&&v_{13} = {{w_1 w_3 R^2v}\over{\lbrack 2 \rbrack (w_1-w_2)(w_3-w_2)}} \;+\;
  {{q^2w_2LR^3v}\over{\lbrack 2 \rbrack \lbrack 3 \rbrack (w_1-w_2)(w_3-w_2)}}
  \\
 && \qquad \qquad  \quad-  
 \;{{q RL^2R^3v}\over
 {\lbrack 2 \rbrack^2 \lbrack 3 \rbrack (w_1-w_2)(w_3-w_2)}},
\\
\\
&&v_{23} = {{w_2 w_3 R^2v}\over{\lbrack 2 \rbrack (w_2-w_1)(w_3-w_1)}} \;+\;
  {{q^2w_1LR^3v}\over{\lbrack 2 \rbrack \lbrack 3 \rbrack (w_2-w_1)(w_3-w_1)}}
  \\
 && \qquad \qquad  \quad-  
 \;{{q RL^2R^3v}\over
 {\lbrack 2 \rbrack^2 \lbrack 3 \rbrack (w_2-w_1)(w_3-w_1)}},
\\
&&v_{123} = {{R^3v}\over{\lbrack 2 \rbrack \lbrack 3 \rbrack}}.
\label{eq:v123s}
\end{eqnarray}
(The vector $v$ is from Definition 
\ref{def:lam}).
\end{definition}

\begin{lemma}
\label{lem:stbasis}
Assume $d=3$, and that $\Phi$ is generic.
Assume further the weights  $w_1, w_2, w_3$ from 
Definition \ref{def:weights} are distinct. 
Then with reference to Definition
\ref{def:basis2}, we have the following (i)--(iv). 
\begin{enumerate}
\item The subspace $U_0$ has basis 
\begin{equation}
v_\emptyset.
\label{eq:u0span2}
\end{equation}
\item The subspace $U_1$ has basis 
\begin{equation}
v_1, \quad v_2, \quad v_3.
\label{eq:u1span2}
\end{equation}
\item The subspace $U_2$ has basis 
\begin{equation}
v_{12}, \quad v_{13}, \quad v_{23}.
\label{eq:u2span2}
\end{equation}
\item The subspace $U_3$ has basis
\begin{equation}
v_{123}.
\label{eq:u3span2}
\end{equation}
\end{enumerate}
Moreover, $V$ has a basis consisting of the eight vectors 
(\ref{eq:u0span2})--(\ref{eq:u3span2}).
We refer to this as the standard
eigenbasis for $q^{-1}RL-qLR$.
\end{lemma} 

\begin{proof}
Assertions (i) and (iv) are clear.
To obtain (ii), one readily  verifies 
the coefficient matrix giving 
(\ref{eq:u1span2}) in terms of 
(\ref{eq:u1span}) is invertible. 
To obtain (iii), one readily  verifies 
the coefficient matrix giving 
(\ref{eq:u2span2}) in terms of 
(\ref{eq:u2span}) is invertible. 

\end{proof}

\begin{lemma}
\label{lem:newvsold}
 Assume $d=3$, and that $\Phi$ is generic.
Further assume the weights  $w_1, w_2, w_3$ from 
Definition \ref{def:weights} are distinct.
Then with reference to Definition
\ref{def:basis2}, we have:
\begin{eqnarray}
&& v = v_\emptyset,
\\
&&Rv = v_1 + v_2 + v_3,
\\
&&
LR^2v = {{\lbrack 2 \rbrack (w_2+w_3) v_1 }\over {q}}  \;+\;
 {{\lbrack 2 \rbrack (w_1+w_3) v_2 }\over {q}} 
\; +\;{{\lbrack 2 \rbrack (w_1+w_2) v_3 }\over {q}},
\\
&&L^2R^3v = 
 {{\lbrack 2 \rbrack^2\lbrack 3 \rbrack w_2w_3 v_1  }\over {q^2}}
\;+\;
 {{\lbrack 2 \rbrack^2\lbrack 3 \rbrack w_1w_3 v_2  }\over {q^2}}
\;+\; {{\lbrack 2 \rbrack^2\lbrack 3 \rbrack w_1w_2 v_3  }\over {q^2}},
\\
&&R^2v = \lbrack 2 \rbrack v_{12} + \lbrack 2 \rbrack  v_{13}
+ \lbrack 2 \rbrack v_{23},
\\
&&RL^2R^3 = 
 {{\lbrack 2 \rbrack^2\lbrack 3 \rbrack (w_1+w_2)w_3 v_{12}  }\over {q}}
\;+\;
 {{\lbrack 2 \rbrack^2\lbrack 3 \rbrack (w_1+w_3)w_2 v_{13}  }\over {q}}
\\
&& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 
 +\; {{\lbrack 2 \rbrack^2\lbrack 3 \rbrack (w_2+w_3)w_1 v_{23}  }\over {q}},
\\
&&
LR^3v = {{\lbrack 2 \rbrack \lbrack 3 \rbrack w_3 v_{12} }\over {q^2}}  \;+\;
 {{\lbrack 2 \rbrack \lbrack 3 \rbrack w_2 v_{13} }\over {q^2}} 
 \;+\;{{\lbrack 2 \rbrack \lbrack 3 \rbrack  w_1 v_{23} }\over {q^2}},
\\
&&R^3v = \lbrack 2 \rbrack \lbrack 3 \rbrack v_{123}.
\end{eqnarray}
\end{lemma}

\begin{proof} Solve the equations 
in Definition  \ref{def:basis2} using elementary linear algebra.  

\end{proof}

\begin{lemma}
\label{lem:rlnew}
 Assume $d=3$, and that $\Phi$ is generic.
Further assume the weights  $w_1, w_2, w_3$ from 
Definition \ref{def:weights} are distinct.
Then $R$ and $L$ act on the standard eigenbasis as follows.
\begin{eqnarray}
&&Rv_{\emptyset} = v_1 + v_2 + v_3,
\\
&& Rv_1 =  {{qw_2-q^{-1}w_1}\over {w_2-w_1}} v_{12} 
\;+\;  {{qw_3-q^{-1}w_1}\over {w_3-w_1}} v_{13},
\\
&& Rv_2 =  {{qw_1-q^{-1}w_2}\over {w_1-w_2}} v_{12} 
\;+\;  {{qw_3-q^{-1}w_2}\over {w_3-w_2}} v_{23},
\\
&& Rv_3 =  {{qw_1-q^{-1}w_3}\over {w_1-w_3}} v_{13} 
\;+\;  {{qw_2-q^{-1}w_3}\over {w_2-w_3}} v_{23},
\\
&&
Rv_{12} = {{(qw_3-q^{-1}w_1)(qw_3-q^{-1}w_2)}\over {(w_3-w_1)(w_3-w_2)}}
v_{123},
\\
&&
Rv_{13} = {{(qw_2-q^{-1}w_1)(qw_2-q^{-1}w_3)}\over {(w_2-w_1)(w_2-w_3)}}
v_{123},
\\
&&
Rv_{23} = {{(qw_1-q^{-1}w_2)(qw_1-q^{-1}w_3)}\over {(w_1-w_2)(w_1-w_3)}}
v_{123},
\\
&&
Rv_{123} = 0,
\end{eqnarray}
\begin{eqnarray}
&&Lv_{\emptyset} = 0,
\\
&&
Lv_1 = {{q^2w_1(q^{-1}w_1-qw_2)(q^{-1}w_1-qw_3)}\over {(w_1-w_2)(w_1-w_3)}}
v_{\emptyset},
\\
&&
Lv_2 = {{q^2w_2(q^{-1}w_2-qw_1)(q^{-1}w_2-qw_3)}\over {(w_2-w_1)(w_2-w_3)}}
v_{\emptyset},
\\
&&
Lv_3 = {{q^2w_3(q^{-1}w_3-qw_1)(q^{-1}w_3-qw_2)}\over {(w_3-w_1)(w_3-w_2)}}
v_{\emptyset},
\\
&& Lv_{12} =  {{w_2(q^{-1}w_2-qw_3)}\over {w_2-w_3}} v_1 
\;+\;  {{w_1(q^{-1}w_1-qw_3)}\over {w_1-w_3}} v_2,
\\
&& Lv_{13} =  {{w_3(q^{-1}w_3-qw_2)}\over {w_3-w_2}} v_1 
\;+\;  {{w_1(q^{-1}w_1-qw_2)}\over {w_1-w_2}} v_3,
\\
&& Lv_{23} =  {{w_3(q^{-1}w_3-qw_1)}\over {w_3-w_1}} v_2 
\;+\;  {{w_2(q^{-1}w_2-qw_1)}\over {w_2-w_1}} v_3,
\\
&&Lv_{123} = {{w_3v_{12} + w_2v_{13} + w_1v_{23}}\over {q^2}}.
\end{eqnarray}
\end{lemma}

\begin{proof}  Elementary linear algebra.

\end{proof} 

\begin{lemma}
\label{lem:rlsmallnew}
 Assume $d=3$, and that $\Phi$ is generic.
Further assume the weights  $w_1, w_2, w_3$ from 
Definition \ref{def:weights} are distinct.
Then the maps $r$ and $l$ from
Lemma \ref{lem:uqaction}
act on the standard eigenbasis as follows.
\begin{eqnarray}
&&rv_{\emptyset} = {{v_1}\over {q^2w_1}} + {{v_2}\over{q^2w_2}}
+ {{v_3}\over{q^2w_3}},
\\
&& rv_1 =  {{qw_2-q^{-1}w_1}\over {w_2(w_2-w_1)}} v_{12} 
\;+\;  {{qw_3-q^{-1}w_1}\over {w_3(w_3-w_1)}} v_{13},
\\
&& rv_2 =  {{qw_1-q^{-1}w_2}\over {w_1(w_1-w_2)}} v_{12} 
\;+\;  {{qw_3-q^{-1}w_2}\over {w_3(w_3-w_2)}} v_{23},
\\
&& rv_3 =  {{qw_1-q^{-1}w_3}\over {w_1(w_1-w_3)}} v_{13} 
\;+\;  {{qw_2-q^{-1}w_3}\over {w_2(w_2-w_3)}} v_{23},
\\
&&
rv_{12} = {{q^2(qw_3-q^{-1}w_1)(qw_3-q^{-1}w_2)}\over {w_3(w_3-w_1)(w_3-w_2)}}
v_{123},
\\
&&
rv_{13} = {{q^2(qw_2-q^{-1}w_1)(qw_2-q^{-1}w_3)}\over {w_2(w_2-w_1)(w_2-w_3)}}
v_{123},
\\
&&
rv_{23} = {{q^2(qw_1-q^{-1}w_2)(qw_1-q^{-1}w_3)}\over {w_1(w_1-w_2)(w_1-w_3)}}
v_{123},
\\
&&
rv_{123} = 0,
\end{eqnarray}
\begin{eqnarray}
&&lv_{\emptyset} = 0,
\\
&&
lv_1 = {{(q^{-1}w_1-qw_2)(q^{-1}w_1-qw_3)}\over {(w_1-w_2)(w_1-w_3)}}
v_{\emptyset},
\\
&&
lv_2 = {{(q^{-1}w_2-qw_1)(q^{-1}w_2-qw_3)}\over {(w_2-w_1)(w_2-w_3)}}
v_{\emptyset},
\\
&&
lv_3 = {{(q^{-1}w_3-qw_1)(q^{-1}w_3-qw_2)}\over {(w_3-w_1)(w_3-w_2)}}
v_{\emptyset},
\\
&& lv_{12} =  {{(q^{-1}w_2-qw_3)}\over {w_2-w_3}} v_1 
\;+\;  {{(q^{-1}w_1-qw_3)}\over {w_1-w_3}} v_2,
\\
&& lv_{13} =  {{(q^{-1}w_3-qw_2)}\over {w_3-w_2}} v_1 
\;+\;  {{(q^{-1}w_1-qw_2)}\over {w_1-w_2}} v_3,
\\
&& lv_{23} =  {{(q^{-1}w_3-qw_1)}\over {w_3-w_1}} v_2 
\;+\;  {{(q^{-1}w_2-qw_1)}\over {w_2-w_1}} v_3,
\\
&&lv_{123} = v_{12} + v_{13} + v_{23}.
\end{eqnarray}
\end{lemma}

\begin{proof}  Elementary linear algebra.

\end{proof} 



\begin{lemma}
\label{lem:eigvect}
 Assume $d=3$, and that $\Phi$ is generic.
Further assume the weights  $w_1, w_2, w_3$ from 
Definition \ref{def:weights} are distinct.
Pick any subset 
 $x \subseteq  \lbrace 1,2,3 \rbrace $, and let
 $\overline x$ denote the complement of $x$ in 
 $\lbrace 1,2,3 \rbrace $.
Then
\begin{enumerate}
\item  $v_x$ is an eigenvector for $q^{-1}RL-qLR$, with eigenvalue
\begin{eqnarray}
q^{-1}\sum_{i \in x} w_i - q\sum_{j\in {\overline x}} w_j.
\label{eq:lamdeff}
\end{eqnarray}
\item  $v_x$ is an eigenvector for $qrl-q^{-1}lr$, with eigenvalue
\begin{eqnarray}
q\sum_{i \in x} w^{-1}_i - q^{-1}\sum_{j\in {\overline x}} w^{-1}_j.
\end{eqnarray}
\end{enumerate}
\end{lemma}

\begin{proof} Routine calculation using
(\ref{eq:ddef}) and Lemma 
\ref{lem:rlnew}.

\end{proof}
\noindent The following is now obvious. 

\begin{theorem} Assume $d=3$, and that $\Phi$ is generic.
Then the 
$U_q(\widehat{ sl_2})$ module structure on $V$
obtained in 
Theorem \ref{thm:extend}
is isomorphic to the $d=3$ case in 
Lemma 
\ref{lem:uqmoreaction} of 
Appendix 2.

\end{theorem}

\section{A basis with respect to which $q^{-1}RL-qLR$ is lower
triangular and $qRL-q^{-1}LR$ is upper triangular}

\noindent In this section, we assume $d=3$, and that $\Phi$
is generic. For each subset
$x\subseteq 
\lbrace 1,2,3\rbrace$, we define a vector $u_x$ in $V$.
We show the $u_x$ form a basis 
 for $V$ with respect to which
 $q^{-1}RL-qLR$ is lower
triangular and $qRL-q^{-1}LR$ is upper triangular.

\begin{lemma}
\label{lem:thewx}
Assume $d=3$, and that $\Phi$
is generic.
There exists vectors 
$u_1, u_2, u_3$ in $V$  such that
\begin{eqnarray}
Rv &=& u_1  + q^{-1}u_2 + q^{-2}u_3,
\label{eq:coeff1}
\\
q\lbrack 2 \rbrack^{-1} LR^2v &=& (w_2+w_3)u_1  + 
(qw_1+q^{-1}w_3)u_2 + (w_1+w_2)u_3,
\label{eq:coeff2}
\\
\lbrack 2 \rbrack^{-2}\lbrack 3 \rbrack^{-1} L^2R^3v &=&
q^{-2}w_2w_3 u_1 + q^{-1}w_1w_3u_2 + w_1w_2u_3.
\label{eq:coeff3}
\end{eqnarray}
Moreover,  there exists vectors 
$u_{12}, u_{13}, u_{23}$ in $V$ such that 
\begin{eqnarray}
\lbrack 2 \rbrack^{-1}R^2v &=& u_{12}  + q^{-1}u_{13} + q^{-2}u_{23},
\\
q\lbrack 2 \rbrack^{-2}\lbrack 3 \rbrack^{-1} RL^2R^3v &=&
w_3(w_1+w_2)u_{12}  + 
w_2(qw_1+q^{-1}w_3)u_{13} + w_1(w_2+w_3)u_{23},
\\
\lbrack 2 \rbrack^{-1}\lbrack 3 \rbrack^{-1} LR^3v &=&
q^{-2}w_3 u_{12} + q^{-1}w_2u_{13} + w_1u_{23}.
\end{eqnarray}

\end{lemma} 

\begin{proof}  To obtain the first assertion,
 we show the coefficient matrix for the system
(\ref{eq:coeff1})--(\ref{eq:coeff3}) is nonsingular.
One can routinely verify the determinant is
\begin{eqnarray}
(qw_1-q^{-1}w_2)(qw_1-q^{-1}w_3)(qw_2-q^{-1}w_3)q^{-3},
\label{eq:coeffdet}
\end{eqnarray}
and hence nonzero in view of
Lemma \ref{eq:wrest}(iii).
It follows the 
 the coefficient matrix for the system
(\ref{eq:coeff1})--(\ref{eq:coeff3}) is nonsingular,
and we have our first assertion. 
The second assertion is similar.

\end{proof}

\begin{definition}
\label{def:ubasis}
Assume $d=3$, and that $\Phi$
is generic.
We define $u_\emptyset = v$. We let $u_1, u_2, u_3$
and $u_{12}, u_{13}, u_{23}$ be as in Lemma 
\ref{lem:thewx}. We set 
$u_{123}=\lbrack 2 \rbrack^{-1}\lbrack 3 \rbrack^{-1}R^3v$.
\end{definition}

\begin{lemma}
\label{lem:ronu}
Assume $d=3$, and that $\Phi$
is generic.  
Then with reference to Definition
\ref{def:ubasis}, we have the following (i)--(iv). 
\begin{enumerate}
\item The subspace $U_0$ has basis 
\begin{equation}
u_\emptyset.
\label{eq:uu0span2}
\end{equation}
\item The subspace $U_1$ has basis 
\begin{equation}
u_1, \quad u_2, \quad u_3.
\label{eq:uu1span2}
\end{equation}
\item The subspace $U_2$ has basis 
\begin{equation}
u_{12}, \quad u_{13}, \quad u_{23}.
\label{eq:uu2span2}
\end{equation}
\item The subspace $U_3$ has basis
\begin{equation}
u_{123}.
\label{eq:uu3span2}
\end{equation}
\end{enumerate}
Moreover, $V$ has a basis consisting of the eight vectors 
(\ref{eq:uu0span2})--(\ref{eq:uu3span2}).
We refer to this as the $LU$-basis.
\end{lemma} 

\begin{proof} Routine.

\end{proof}

\begin{lemma}
\label{ronu}
Assume $d=3$, and that $\Phi$
is generic. Then 
the maps $R$ and $L$ act on the $LU$ basis as follows.

\begin{eqnarray}
Ru_\emptyset &=& u_1 + q^{-1}u_2 + q^{-2}u_3,
\\
R u_1 &=& qu_{12} + u_{13},
\\
Ru_2 &=& u_{12} + u_{23},
\\
Ru_3 &=& u_{13} + q^{-1}u_{23},
\\
Ru_{12} &=& q^2u_{123},
\\
Ru_{13} &=& qu_{123},
\\
Ru_{23} &=& u_{123},
\\
Ru_{123} &=& 0.
\end{eqnarray}
\begin{eqnarray}
L u_\emptyset &=& 0,
\\
L u_1 &=& w_1 u_\emptyset,
\\
L u_2 &=& qw_2 u_\emptyset,
\\
L u_3 &=& q^2w_3 u_\emptyset,
\\
L u_{12} &=& q^{-1}w_2 u_1 + w_1 u_2,
\\
L u_{13} &=& w_3 u_1 + w_1 u_3,
\\
L u_{23} &=& w_3 u_2 + qw_2 u_3,
\\
L u_{123} &=& q^{-2}w_3u_{12}+q^{-1}w_2u_{13} + w_1u_{23}.
\end{eqnarray}
\end{lemma}

\begin{proof} Routine linear algebra.

\end{proof}

%%pt jan31
\begin{lemma}
\label{lem:sractonu}
Assume $d=3$, and that $\Phi$
is generic. Then 
the maps $r$ and $l$ act
on the $LU$ basis as follows.

\begin{eqnarray}
ru_\emptyset &=& q^{-2}w^{-1}_1u_1 + q^{-1}w^{-1}_2u_2 + w^{-1}_3u_3,
\\
r u_1 &=& q^{-1}w^{-1}_2u_{12} + w^{-1}_3u_{13},
\\
ru_2 &=& w^{-1}_1u_{12} + w^{-1}_3u_{23},
\\
ru_3 &=& w^{-1}_1u_{13} + qw^{-1}_2u_{23},
\\
ru_{12} &=& w^{-1}_3u_{123},
\\
ru_{13} &=& qw^{-1}_2u_{123},
\\
ru_{23} &=& q^2w^{-1}_1u_{123},
\\
ru_{123} &=& 0.
\end{eqnarray}
\begin{eqnarray}
l u_\emptyset &=& 0,
\\
l u_1 &=& q^2 u_\emptyset,
\\
l u_2 &=& q u_\emptyset,
\\
l u_3 &=&  u_\emptyset,
\\
l u_{12} &=& q u_1 +  u_2,
\\
l u_{13} &=&  u_1 +  u_3,
\\
l u_{23} &=&  u_2 + q^{-1}u_3,
\\
l u_{123} &=& u_{12}+q^{-1}u_{13} + q^{-2}u_{23}.
\end{eqnarray}
\end{lemma}

\begin{proof} Routine linear algebra.

\end{proof}
%%jan31
\noindent We comment the LU basis  $u_x$ underlies 
the decompostion of $V$ into a $q$-tensor product of
two dimensional evaluation representations. See
Chari and  Pressley Ch4, p271.

\begin{lemma}
Assume $d=3$, and that $\Phi$
is generic. Then the map 
$C^-:=q^{-1}RL-qLR$  acts on the $LU$ basis as follows.
\begin{eqnarray}
C^- u_\emptyset &=& -q(w_1+w_2+w_3)u_\emptyset,
\\
C^- u_1 &=& 
(q^{-1}w_1-qw_2-qw_3)u_1 +
(q^{-2}-q^2)w_1 u_2 +
(q^{-3}-q)w_1u_3,
\\
C^- u_2 &=&  
 (-qw_1+q^{-1}w_2-qw_3)u_2 + 
(q^{-2}-q^2)w_2 u_3, 
\\
C^- u_3 &=& 
(-qw_1-qw_2+q^{-1}w_3)u_3,
\\
C^- u_{12} &=& (q^{-1}w_1+q^{-1}w_2-qw_3)u_{12}
+ (q^{-2}-q^2)w_2 u_{13} +
(q^{-1}-q^3)w_1u_{23}, 
\\
C^- u_{13} &=&
(q^{-1}w_1-qw_2+q^{-1}w_3)u_{13} +
(q^{-2}-q^2)w_1 u_{23}, 
\\
C^- u_{23} &=&
(-qw_1+q^{-1}w_2+q^{-1}w_3)u_{23},
\\
C^- u_{123} &=& q^{-1}(w_1+w_2+w_3)u_{123}.
\end{eqnarray}
The map 
$C^+:=qRL-q^{-1}LR$  acts on the $LU$ basis as follows.
\begin{eqnarray}
C^+ u_\emptyset &=& -q^{-1}(w_1+w_2+w_3)u_\emptyset,
\\
C^+ u_1 &=& 
(qw_1-q^{-1}w_2-q^{-1}w_3)u_1,
%(q^{2}-q^{-2})w_2u_2 + (q^{3}-q^{-1})w_3 u_3, 
\\
C^+ u_2 &=&
(q^{2}-q^{-2})w_2u_1 
+
(-q^{-1}w_1+qw_2-q^{-1}w_3)u_2, 
\\
%(q^{2}-q^{-2})w_3 u_3 ,
C^+ u_3 &=& 
 (q^{3}-q^{-1})w_3 u_1 +
(q^{2}-q^{-2})w_3u_2  
+
(-q^{-1}w_1-q^{-1}w_2+qw_3)u_3,
\\
C^+ u_{12} &=& 
(qw_1+qw_2-q^{-1}w_3)u_{12},
%(q^{2}-q^{-2})w_3u_{13} + (q-q^{-3})w_3 u_{23}, 
\\
C^+ u_{13} &=&
(q^{2}-q^{-2})w_3u_{12} 
+
(qw_1-q^{-1}w_2+qw_3)u_{13}, 
%(q^{2}-q^{-2})w_2 u_{23} ,
\\
C^+ u_{23} &=&
 (q-q^{-3})w_3 u_{12} +
(q^{2}-q^{-2})w_2 u_{13}+
(-q^{-1}w_1+qw_2+qw_3)u_{23},
\\
C^+ u_{123} &=& q(w_1+w_2+w_3)u_{123}.
\end{eqnarray}
The map 
$c^+:=qrl-q^{-1}lr$  acts on the $LU$ basis as follows.
\begin{eqnarray}
c^+ u_\emptyset &=& -q^{-1}(w^{-1}_1+w^{-1}_2+w^{-1}_3)u_\emptyset,
\\
c^+ u_1 &=& 
(qw^{-1}_1-q^{-1}w^{-1}_2-q^{-1}w^{-1}_3)u_1 +
(q^{2}-q^{-2})w^{-1}_2 u_2 +
(q^{3}-q^{-1})w^{-1}_3u_3,
\\
c^+ u_2 &=&  
 (-q^{-1}w^{-1}_1+qw^{-1}_2-q^{-1}w^{-1}_3)u_2 + 
(q^{2}-q^{-2})w^{-1}_3 u_3, 
\\
c^+ u_3 &=& 
(-q^{-1}w^{-1}_1-q^{-1}w^{-1}_2+qw^{-1}_3)u_3,
\\
c^+ u_{12} &=& (qw^{-1}_1+qw^{-1}_2-q^{-1}w^{-1}_3)u_{12}
+ (q^{2}-q^{-2})w^{-1}_3 u_{13} +
(q-q^{-3})w^{-1}_3u_{23}, 
\\
c^+ u_{13} &=&
(qw^{-1}_1-q^{-1}w^{-1}_2+qw^{-1}_3)u_{13} +
(q^{2}-q^{-2})w^{-1}_2 u_{23}, 
\\
c^+ u_{23} &=&
(-q^{-1}w^{-1}_1+qw^{-1}_2+qw^{-1}_3)u_{23},
\\
c^+ u_{123} &=& q(w^{-1}_1+w^{-1}_2+w^{-1}_3)u_{123}.
\end{eqnarray}
The map 
$c^-:=q^{-1}rl-qlr$  acts on the $LU$ basis as follows.
\begin{eqnarray}
&&c^- u_\emptyset = -q(w^{-1}_1+w^{-1}_2+w^{-1}_3)u_\emptyset,
\\
&&c^- u_1 = 
(q^{-1}w^{-1}_1-qw^{-1}_2-qw^{-1}_3)u_1,
\\
&&c^- u_2 =
(q^{-2}-q^{2})w^{-1}_1u_1 
+
(-qw^{-1}_1+q^{-1}w^{-1}_2-qw^{-1}_3)u_2, 
\\
&&c^- u_3 = 
 (q^{-3}-q)w^{-1}_1 u_1 +
(q^{-2}-q^{2})w^{-1}_2u_2  
+
(-qw^{-1}_1-qw^{-1}_2+q^{-1}w^{-1}_3)u_3,
\\
&&c^- u_{12} = 
(q^{-1}w^{-1}_1+q^{-1}w^{-1}_2-qw^{-1}_3)u_{12},
\\
&&c^- u_{13} =
(q^{-2}-q^{2})w^{-1}_2u_{12} 
+
(q^{-1}w^{-1}_1-qw^{-1}_2+q^{-1}w^{-1}_3)u_{13}, 
\\
&&c^- u_{23} =
 (q^{-1}-q^{3})w^{-1}_1 u_{12} +
(q^{-2}-q^{2})w^{-1}_1 u_{13}
\nonumber
\\
&&\qquad \qquad \qquad \qquad \qquad+ (-qw^{-1}_1+q^{-1}w^{-1}_2+q^{-1}w^{-1}_3)u_{23},
\\
&&c^- u_{123} = q^{-1}(w^{-1}_1+w^{-1}_2+w^{-1}_3)u_{123}.
\end{eqnarray}

\end{lemma}


\section{Replacing $q$ by $q^{-1}$}

\noindent Observe that in Definition 
\ref{def:1p1}, the scalar  $q$ is only defined up to reciprocal,
so everything we have done so far still
holds
if we  replace $q$ by $q^{-1}$. This replacement
has no effect on
$R,L, r, l$ and $w_1, w_2, w_3$,  but it does alter 
the standard eigenbasis
from Definition
\ref{def:basis2}.

\begin{definition}
\label{def:vxp}
Assume $d=3$, and that $\Phi$ is generic. Further assume the
weights $w_1, w_2, w_3$ are distinct.
For any subseteq $x \subseteq \lbrace 1,2,3\rbrace $, Let $v'_x$
denote the vector obtained from $v_x$ in
(\ref{eq:vempty})--(\ref{eq:v123s})
by replacing $q$ by $q^{-1}$.
We refer to the $v'_x$ as  standard eigebasis for $qRL-q^{-1}LR$. 
\end{definition}

\begin{lemma}
\label{lem:alt}
Assume $d=3$, and that $\Phi$ is generic. Further assume the
weights $w_1, w_2, w_3$ are distinct.
Then the standard eigenbasis  for $q^{-1}RL-qLR$ and
the standard eigenbasis 
 $qRL-q^{-1}LR$
  are related as follows.
\begin{eqnarray}
&&v'_{\emptyset} = v_\emptyset,
\\
&&v'_1 = {{(qw_1-q^{-1}w_2)(
 qw_1-q^{-1}w_3) v_1}
 \over 
{(w_1-w_2)
 (w_1-w_3)q^2}}
\;+\;
 {{(q-q^{-1})w_1
 (qw_1-q^{-1}w_3)v_2}\over{ 
q^2 
 (w_1-w_2)
 (w_1-w_3)}}
\\
&&\qquad \qquad 
\;+\;
 {{(q-q^{-1})w_1
(qw_1-q^{-1}w_2) v_3
}\over {q^2(w_1-w_3)(w_1-w_2)}},
\\
&&v'_2 =
 {{(q-q^{-1})w_2
 (qw_2-q^{-1}w_3)v_1}\over{ 
q^2 
 (w_2-w_1)
 (w_2-w_3)}}
\;+\;
{{(qw_2-q^{-1}w_1)(
 qw_2-q^{-1}w_3) v_2}
 \over 
{(w_2-w_1)
 (w_2-w_3)q^2}}
\\
&&\qquad \qquad 
\;+\;
 {{(q-q^{-1})w_2
(qw_2-q^{-1}w_1) v_3
}\over {q^2(w_2-w_3)(w_2-w_1)}},
\\
&&v'_3 =
%%
 {{(q-q^{-1})w_3
(qw_3-q^{-1}w_2) v_1
}\over {q^2(w_3-w_1)(w_3-w_2)}} \;+\;
 {{(q-q^{-1})w_3
 (qw_3-q^{-1}w_1)v_2}\over{ 
q^2 
 (w_3-w_2)
 (w_3-w_1)}}
\\
&&
\qquad \qquad \;+\;{{(qw_3-q^{-1}w_1)(
 qw_3-q^{-1}w_2) v_3}
 \over 
{(w_3-w_1)
 (w_3-w_2)q^2}}
\\
&&v'_{12} 
=
{{(qw_1-q^{-1}w_3)(qw_2-q^{-1}w_3)v_{12}}\over {(w_1-w_3)(w_2-w_3)q^2}}
\;+\;
{{(q-q^{-1})w_2(qw_1-q^{-1}w_3)v_{13}}\over{q^2(w_2-w_3)(w_1-w_3)}}
\\
&&
\qquad \qquad \;+\; 
{{(q-q^{-1})w_1(qw_2-q^{-1}w_3)v_{23}}\over {q^2(w_1-w_3)(w_2-w_3)}}
\\
&&v'_{13} 
=
{{(q-q^{-1})w_3(qw_1-q^{-1}w_2)v_{12}}\over{q^2(w_3-w_2)(w_1-w_2)}}
\;+\;
{{(qw_1-q^{-1}w_2)(qw_3-q^{-1}w_2)v_{13}}\over {(w_1-w_2)(w_3-w_2)q^2}}
\;+\;
\\
&&
\qquad \qquad \;+\; 
{{(q-q^{-1})w_1(qw_3-q^{-1}w_2)v_{23}}\over {q^2(w_1-w_2)(w_3-w_2)}}
\\
&&v'_{23} 
=
{{(q-q^{-1})w_3(qw_2-q^{-1}w_1)v_{12}}\over{q^2(w_3-w_1)(w_2-w_1)}}
\;+\;
{{(q-q^{-1})w_2(qw_3-q^{-1}w_1)v_{13}}\over {q^2(w_2-w_1)(w_3-w_1)}}
\\
&&\qquad \qquad \;+\; 
{{(qw_2-q^{-1}w_1)(qw_3-q^{-1}w_1)v_{23}}\over {(w_2-w_1)(w_3-w_1)q^2}}
\\
&&v'_{123}=  v_{123}.
\end{eqnarray}
Moreover, the above equations still hold if we interchange $v_x, v'_x$
for all $x$, and replace $q$ by $q^{-1}$.  
\end{lemma}

%%ptjan31
\begin{problem}
 Assume $d=3$, and that $\Phi$ is generic. Further assume the
weights $w_1, w_2, w_3$ are distinct.
Find the transition matrices between the $u_x$ basis,
and the $v_x$ basis;  also between  
 the $u_x$ basis,
and the $v'_x$ basis.
\end{problem}



\section{ More on bilinear forms}


\begin{lemma} 
 Assume $d=3$, and that $\Phi$ is generic. Further assume the
weights $w_1, w_2, w_3$ are distinct.
Let the bilinear form
 $\langle\;, \;\rangle $  be as in
Lemma \ref{lem:form1}.
Then the standard basis is orthogonal with respect to 
 $\langle\;, \;\rangle $.
Moreover, after setting $\alpha=1$ in Lemma 
\ref{lem:form1},
\begin{eqnarray}
\Vert v_\emptyset \Vert^2 &=&  1,
\\
\Vert v_1 \Vert^2 &=&  {{q^2w_1(q^{-1}w_1-qw_2)(q^{-1}w_1-qw_3)}\over
{(w_1-w_2)(w_1-w_3)}},
\\
\Vert v_2 \Vert^2 &=&  {{q^2w_2(q^{-1}w_2-qw_1)(q^{-1}w_2-qw_3)}\over
{(w_2-w_1)(w_2-w_3)}},
\\
\Vert v_3 \Vert^2 &=&  {{q^2w_3(q^{-1}w_3-qw_1)(q^{-1}w_3-qw_2)}\over
{(w_3-w_1)(w_3-w_2)}},
\\
\Vert v_{12} \Vert^2 &=&  {{q^2w_1w_2(q^{-1}w_1-qw_3)(q^{-1}w_2-qw_3)}\over
{(w_1-w_3)(w_2-w_3)}},
\\
\Vert v_{13} \Vert^2 &=&  {{q^2w_1w_3(q^{-1}w_1-qw_2)(q^{-1}w_3-qw_2)}\over
{(w_1-w_2)(w_3-w_2)}},
\\
\Vert v_{23} \Vert^2 &=&  {{q^2w_2w_3(q^{-1}w_2-qw_1)(q^{-1}w_3-qw_1)}\over
{(w_2-w_1)(w_3-w_1)}},
\\
\Vert v_{123} \Vert^2 &=&  w_1w_2w_3. 
\end{eqnarray}
\end{lemma}

\begin{proof} Routine linear algebra.

\end{proof}


\noindent We remark a similar result holds for the
alternative standard basis.

\begin{lemma}
\label{lem:form2ag}
 Assume $d=3$, and that $\Phi$ is generic.
Further assume the weights  $w_1, w_2, w_3$ from 
Definition \ref{def:weights} are distinct.
Let the bilinear form  $(\, ,\,)$  be as in Lemma
\ref{lem:form2}. Then for all subsets 
$x,y\subseteq \lbrace 1,2,3\rbrace $,
$(v_x,v'_y) = 0 $ unless $y$ is the complement of $x$ in
$\lbrace 1,2,3\rbrace $. Morever, after setting
$\gamma=\lbrack 2 \rbrack^{-1} \lbrack 3 \rbrack^{-1}$
in Lemma 
\ref{lem:form2},
\begin{eqnarray}
(v_{\emptyset}, v'_{123}) &=& 1,
\\
(v_1, v'_{23}) &=& 
  {{(q^{-1}w_1-qw_2)(q^{-1}w_1-qw_3)}\over {(w_1-w_2)(w_1-w_3)}},
\\
(v_2, v'_{13}) &=& 
  {{(q^{-1}w_2-qw_1)(q^{-1}w_2-qw_3)}\over {(w_2-w_1)(w_2-w_3)}},
\\
(v_3, v'_{12}) &=& 
  {{(q^{-1}w_3-qw_1)(q^{-1}w_3-qw_2)}\over {(w_3-w_1)(w_3-w_2)}},
\\
(v_{12}, v'_{3}) &=& 
 {{(qw_3-q^{-1}w_1)(qw_3-q^{-1}w_2)}\over {(w_3-w_1)(w_3-w_2)}},
\\
(v_{13}, v'_{2}) &=& 
{{(qw_2-q^{-1}w_1)(qw_2-q^{-1}w_3)}\over {(w_2-w_1)(w_2-w_3)}},
\\
(v_{23}, v'_{1}) &=& 
 {{(qw_1-q^{-1}w_2)(qw_1-q^{-1}w_3)}\over {(w_1-w_2)(w_1-w_3)}},
\\
(v_{123}, v'_{\emptyset}) &=& 1. 
\end{eqnarray}
\end{lemma}

\begin{proof} Routine linear algebra.

\end{proof}

%%ptjan31
\begin{problem}
Assume $d=3$, and that $\Phi$ is generic.
Find the matrix  of inner products for
the bilinear form
$\langle \,,\,\rangle $, with respect to 
the LU basis $u_x$ . Also, 
find the matrix  of inner products for
the bilinear form
$(\,,\,) $, with respect to 
the LU basis $u_x$ .
\end{problem}


\section{The action of $A$ and $A^*$}

%\noindent Throughout this section, we assume 
%  Conjecture \ref{conj:lrcom} holds. 


\noindent To describe the action of $A$, 
it is helpful to decompose it as follows. 

\begin{definition} We define
\begin{eqnarray}
A^+ &=& \sum_{i=0}^{d-1} E^*_{i+1}AE^*_i,
\\
A^0 &=& \sum_{i=0}^{d} E^*_iAE^*_i,
\\
A^- &=& \sum_{i=1}^{d} E^*_{i-1}AE^*_i.
\end{eqnarray}
We observe
\begin{eqnarray}
A = A^{+} + A^{0} + A^{-},
\label{eq:asplit}
\end{eqnarray}
and that
\begin{eqnarray}
&&A^+ E^*_iV \subseteq E^*_{i+1}V \qquad \qquad (0 \leq i < d),\qquad  
A^+ E^*_dV  = 0,
\\
&&A^0 E^*_iV \subseteq E^*_iV \qquad \qquad (0 \leq i \leq  d),  
\\
&&A^- E^*_iV \subseteq E^*_{i-1}V \qquad \qquad (0 < i \leq d),\qquad  
A^- E^*_0V  = 0.
\end{eqnarray}
\end{definition}

\noindent We now consider when is $A^0$  diagonalizable.
To do this, we introduce the map $S$.

\begin{lemma}
\label{lem:smat} Consider the map
\begin{equation}
S= \sum_{i=0}^d E^*_iF_i.
\label{eq:smat}
\end{equation}
Then $S^{-1} $ exists, and is given by
\begin{equation}
S^{-1}= \sum_{i=0}^d F_iE^*_i.
\label{eq:sinvmat}
\end{equation}
Moreover, for  $0 \leq i \leq d$, the restriction of $S$ to $U_i$ 
is a bijection onto $E^*_iV$.

\end{lemma}

\begin{proof} This is a rephrasing of Lemma 4.6(ii) in \cite{}.

\end{proof}

\begin{lemma} 
\label{lem:apiece}
Pick any integer $i$ $(0 \leq i \leq d)$. On the subspace
$U_i$,
\begin{enumerate}
\item $S^{-1}A^+S$ agrees with $R$,
\item $S^{-1}A^0S$ agrees with
\begin{eqnarray}
\theta_iI \;+\; {{RL}\over {\theta^*_i-\theta^*_{i-1}}}\;+\;
{{LR}\over {\theta^*_i-\theta^*_{i+1}}},
\label{eq:cpreq}
\end{eqnarray}
\item $S^{-1}A^-S$ agrees with
\begin{eqnarray}
{{L^2R}\over {(\theta^*_{i-1}-\theta^*_i)(\theta^*_{i-1}-\theta^*_{i+1})}}
\;-\;
{{LRL}\over {(\theta^*_{i-1}-\theta^*_i)^2}}
\;+\;
{{RL^2}\over {(\theta^*_{i}-\theta^*_{i-1})(\theta^*_{i}-\theta^*_{i-2})}}
\nonumber
\\
 \qquad \qquad 
\;-\;
{{(\theta_{i-1}-\theta_i)L}\over {(\theta^*_{i-1}-\theta^*_{i})}}.
&&
\label{eq:ccpreq}
\end{eqnarray}
\end{enumerate}
\end{lemma}


\begin{lemma}
\label{lem:commmm}
Referring to Lemma 
\ref{lem:apiece}, we have the following.
\begin{enumerate}
\item The expression 
(\ref{eq:cpreq})  equals $\theta^{*-1}_i$ times
\begin{eqnarray}
aa^*I - {{C}\over {q-q^{-1}}}.
\label{eq:cpfin}
\end{eqnarray}
\item The expression 
(\ref{eq:ccpreq}) equals 
 $\theta^{*-1}_{i-1}\theta^{*-1}_i$ times
\begin{eqnarray}
aa^*L \;+\; {{CL-LC}\over {(q-q^{-1})(q^2-q^{-2})}}.
\label{eq:ccpfin}
\end{eqnarray}
\end{enumerate}
We remark $a$ is from 
(\ref{eq:thetaform}), $a^*$ is from 
(\ref{eq:thetasform}), and 
$C$  is from
(\ref{eq:ddef}).
\end{lemma}

\begin{lemma} The map $C$ from 
(\ref{eq:ddef})
is diagonalizable if and only if $A^0$ is diagonalizable.

\end{lemma}

\begin{proof} Observe $A^0$ is diagonalizable if and only if
$S^{-1}A^0S$ is 
 diagonalizable. The result now follows from 
Lemma \ref{lem:apiece} (ii) and 
Lemma \ref{lem:commmm}(i).

\end{proof}

\noindent For the rest of this section, we restrict our
attention to the $d=3$ case.

\begin{lemma} Assume $d=3$, and that $\Phi $ is generic.
Further assume $w_1, w_2, w_3$ are distinct.
Then for $0 \leq i \leq d$, The  following vectors give a basis for
 $E^*_iV$.
\begin{eqnarray}
S v_x \qquad \qquad x \subseteq \lbrace 1,2,\ldots, d\rbrace,\qquad \vert x \vert = i.
\end{eqnarray}
Moroever, The following vectors give a basis for $V$.
\begin{eqnarray}
S v_x \qquad \qquad x \subseteq \lbrace 1,2,\ldots, d\rbrace.
\label{eq:svxbasis}
\end{eqnarray}
\end{lemma}

\begin{proof} Immediate from Lemma 
\ref{lem:smat}.
\end{proof}


\begin{lemma} Assume $d=3$, and that $\Phi $ is generic.
Further assume $w_1, w_2, w_3$ are distinct.
Then for all $x \subseteq \lbrace 1,2,\ldots, d \rbrace $,
\begin{eqnarray}
A^* Sv_x = a^*q^{d-2\vert x \vert} S v_x.
\end{eqnarray}
\end{lemma}



\begin{lemma} Assume $d=3$, and that $\Phi $ is generic.
Further assume $w_1, w_2, w_3$ are distinct.
Then for all $x \subseteq \lbrace 1,2,\ldots, d \rbrace $,
\begin{eqnarray}
A^0 Sv_x =  
 \biggl(aa^*-{{\lambda_x}\over {q-q^{-1}}}\biggr)
 {{q^{2\vert x \vert-d}}\over {a^*}} Sv_x,
\end{eqnarray}
where $\lambda_x$ denotes the expession in
(\ref{eq:lamdeff}).
\end{lemma}

\begin{lemma} 
\label{lem:aplusaction}
Assume $d=3$, and that $\Phi $ is generic.
Further assume $w_1, w_2, w_3$ are distinct.
Then the action of $A^+$ and $A^-$ on the basis 
(\ref{eq:svxbasis}) is as follows.
For all 
subsets $
 x \subseteq \lbrace 1,2,\ldots, d\rbrace $,
\begin{eqnarray}
&&A^+ S v_x = \sum_{i \in  y} S v_{x\cup i} 
\prod_{j \in x} {{qw_i-q^{-1}w_j}\over {w_i-w_j}},
\\
&&A^-Sv_x =  \sum_{i \in x} S v_{x\backslash i}
 \biggl(aa^*-{{w_i}\over {(q-q^{-1})^2}}\biggr) 
{{w_iq^{2\vert x \vert -d-1}}\over {a^{*2}}}
\prod_{j \in y} {{q^{-1}w_i-qw_j}\over {w_i-w_j}},
\end{eqnarray}
where $y$ denotes the complement of $x$ in   
$\lbrace 1,2,\ldots, d \rbrace $.
\end{lemma}

\begin{lemma} Assume $d=3$, and that $\Phi $ is generic.
Further assume $w_1, w_2, w_3$ are distinct.
Then 
\begin{eqnarray}
w_i \not =  aa^*(q-q^{-1})^2 \qquad \qquad (1 \leq i \leq d).
\end{eqnarray}
\end{lemma}


\section{ When $\Phi$ has shape $111\cdots 1$}


\begin{lemma} Assume $\Phi$ has shape $111\cdots 1$. Then
there exists a nonzero scalar $w \in \fld$
and there exists a basis $v_0, v_1, \ldots v_d$  for $V$ such that
\begin{eqnarray}
&&v_i \in U_i \qquad \qquad (0 \leq i \leq d),
\\
&&
Rv_i = \lbrack i+1 \rbrack v_{i+1} \qquad \qquad (0 \leq i \leq d-1),
\qquad  Rv_d = 0,
\\
&&
Lv_i = w \lbrack d-i+1 \rbrack v_{i-1} \qquad \qquad (1 \leq i \leq d),
\qquad  Lv_0 = 0,
\\
&&
rv_i = \lbrack i+1 \rbrack w^{-1} v_{i+1} \qquad \qquad (0 \leq i \leq d-1),
\qquad  rv_d = 0,
\\
&&
lv_i =  \lbrack d-i+1 \rbrack v_{i-1} \qquad \qquad (1 \leq i \leq d),
\qquad lv_0 = 0,
\\
&&
Kv_i = q^{2i-d}v_i \qquad \qquad (0 \leq i \leq d).
\end{eqnarray}
Moreover, the weights form a $q$-string $wq^{1-d}, wq^{3-d},\ldots, wq^{d-1}$.


\end{lemma}

\section{Appendix 1: The algebra 
$U_q({ sl_2})$ }

In this section, we recall the definition of
the quantum algebra
$U_q({ sl_2})$.

\noindent Throughout this section,  
 $\fld$ will  denote an algebraically closed field, and
 $q$ will denote a nonzero element in $\fld$ which is not
a root of unity.


\begin{definition}
\label{def:uqdef}
The quantum  algebra 
$U_q( sl_2)$ is the associative $\fld$-algebra 
with generators $e,f,k,k^{-1}$ 
and the following relations:
\begin{eqnarray}
kk^{-1} &=& 
k^{-1}k =  1,
\\
ke &=& q^2ek, \qquad \qquad 
kf = q^{-2}fk,  
\label{eq:kef}
\\
ef-fe &=& {{k-k^{-1}}\over {q-q^{-1}}}.
\label{eq:efk}
\end{eqnarray}

\end{definition}

We present a finite dimensional
$U_q( sl_2)$ module.

\begin{lemma}
\label{lem:uqmodule}
Let $d$ denote a nonnegative integer, and let $V$ denote a vector
space over $\fld$ with a basis $v_0, v_1, \ldots v_d$. Let $e,f,K$
denote the linear transformations from $V$ to $V$ that satisfy
\begin{eqnarray}
e v_i &=& [d-i+1] v_{i-1}       \qquad \qquad (0< i \leq d), 
\qquad ev_0=0,
\\
f v_i &=& [i+1] v_{i+1}       \qquad \qquad (0 \leq i < d), 
\qquad fv_d=0,
\\
k v_i &=& q^{d-2i}v_i \qquad \qquad (0 \leq i \leq d).
\end{eqnarray}
Then $e,f,k$ satisfy the relations of 
(\ref{eq:kef}),
(\ref{eq:efk}). In particular, $V$ has the structure of 
a $U_q(sl_2)$ module. This module is irreducible.
We refer to this module as type 1, with highest weight $q^{d}$.
\end{lemma}


We present a second finite dimensional
$U_q( sl_2)$ module.

\begin{lemma}
\label{lem:uqmodulemin}
Let $d$ denote a nonnegative integer, and let $V$ denote a vector
space over $\fld$ with a basis $v_0, v_1, \ldots v_d$. Let $e,f,K$
denote the linear transformations from $V$ to $V$ that satisfy
\begin{eqnarray}
e v_i &=& [d-i+1] v_{i-1}       \qquad \qquad (0< i \leq d), 
\qquad ev_0=0,
\\
f v_i &=& -[i+1] v_{i+1}       \qquad \qquad (0 \leq i < d), 
\qquad fv_d=0,
\\
k v_i &=& -q^{d-2i}v_i \qquad \qquad (0 \leq i \leq d).
\end{eqnarray}
Then $e,f,k$ satisfy the relations of 
(\ref{eq:kef}),
(\ref{eq:efk}). In particular, $V$ has the structure of 
a $U_q(sl_2)$ module. This module is irreducible.
We refer to this module as type -1, with highest weight $-q^{d}$.
\end{lemma}

\noindent Up to isomorphism,
any finite dimensional irreducible  $U_q(sl_2)$ module
is type $1$ or type $-1$.



\section{Appendix 2: The algebra 
$U_q(\widehat{ sl_2})$ }

In this section, we recall the definition of
the quantum affine algebra 
$U_q(\widehat{ sl_2})$, and give a finite dimensional
irreducible module.

\medskip
\noindent Throughout this section,  
 $\fld$ will  denote an algebraically closed field, and
 $q$ will denote a nonzero element in $\fld$ which is not
a root of unity.

\begin{definition} (Chari and Pressley, p262) 
The quantum affine algebra 
$U_q(\widehat{ sl_2})$ is the associative $\fld$-algebra 
with generators
$e^{\pm}_i$, $K_i^{{\pm}1}$, $i\in \lbrace 0,1\rbrace $
and the following relations:
\begin{eqnarray}
K_iK^{-1}_i &=& 
K^{-1}_iK_i =  1,
\\
K_0K_1&=& K_1K_0,
\\
K_ie^{\pm}_iK^{-1}_i &=& q^{{\pm}2}e^{\pm}_i,
\\
K_ie^{\pm}_jK^{-1}_i &=& q^{{\mp}2}e^{\pm}_j, \qquad i\not=j,
\\
\lbrack e^+_i, e^-_i\rbrack &=& {{K_i-K^{-1}_i}\over {q-q^{-1}}},
\\
\lbrack e^{\pm}_0, e^{\mp}_1\rbrack &=& 0,
\end{eqnarray}
\begin{eqnarray}
(e^{\pm}_i)^3e^{\pm}_j -  
\lbrack 3 \rbrack (e^{\pm}_i)^2e^{\pm}_j e^{\pm}_i 
+\lbrack 3 \rbrack e^{\pm}_ie^{\pm}_j (e^{\pm}_i)^2 - 
e^{\pm}_j (e^{\pm}_i)^3 =0, \qquad i\not=j.
\end{eqnarray}


\end{definition}

We present a finite dimensional
$U_q(\widehat{ sl_2})$ module.


\begin{definition}
\label{lem:fullactiond1}
Let $d$ denote a positive integer,
and let
 $w_1, w_2, \ldots, w_d$ denote nonzero scalars in
 $\fld$. Let $V$ denote a vector space over $\fld$,
 with a basis $u_x$ indexed by the  subsets $x$ of  
$\lbrace 1,2,\ldots, d\rbrace  $.
Let 
 $R, L, r, l,K$ denote the linear transformations from 
$V$ to $V$ that satisfy the following:
For all $x \subseteq \lbrace 1,2,\ldots, d \rbrace $,
\begin{eqnarray}
R u_x &=& \sum_{i \in {\overline x}} u_{x\cup i}
q^{\vert \lbrace 1,2,\ldots , i-1\rbrace \cap x \vert - 
\vert \lbrace 1,2,\ldots , i-1 \rbrace \cap {\overline x}\vert},
\\
L u_x &=& \sum_{i \in x} u_{x\backslash i} w_i 
q^{\vert \lbrace 1,2,\ldots , i-1\rbrace \cap {\overline x} \vert - 
\vert \lbrace 1,2,\ldots , i-1 \rbrace \cap  x\vert},
\\
r u_x &=& \sum_{i \in {\overline x}} u_{x\cup i} w^{-1}_i
q^{\vert \lbrace i+1,i+2,\ldots , d\rbrace \cap x \vert - 
\vert \lbrace i+1,i+2,\ldots , d \rbrace \cap {\overline x}\vert},
\\
l u_x &=& \sum_{i \in x} u_{x\backslash i}  
q^{\vert \lbrace i+1,i+2,\ldots , d\rbrace \cap {\overline x} \vert - 
\vert \lbrace i+1,i+2,\ldots , d \rbrace \cap  x\vert},
\\
K u_x &=& q^{2\vert x \vert -d} u_x.
\end{eqnarray}
In the above lines,  $\overline  x$ denotes the complement of $x$ in
 $\lbrace 1,2,\ldots d\rbrace$.

\end{definition}


\begin{lemma}
\label{lem:uqmoreaction}
The maps $R, L, r, l, K$ in 
 Definition
\ref{lem:fullactiond1} 
satisfy the following.
\begin{eqnarray}
&&K R = q^2 R K,
\qquad  \qquad 
K L = q^{-2} L K,
\label{eq:app1}
\\
&&
K r = q^2 r K, \qquad \qquad  
 K l = q^{-2} l K,  
\label{eq:appr}
\\
&&
rL-Lr = {{K-K^{-1}}\over {q-q^{-1}}},
\qquad \qquad 
lR-Rl = {{K^{-1}-K}\over {q-q^{-1}}},
\label{eq:app3}
\\
&&
lL = Ll , \qquad \qquad rR=Rr,
\label{eq:app4}
\\
&&
0= R^3L - \lbrack 3 \rbrack R^2LR+ \lbrack 3 \rbrack RLR^2 - LR^3, 
\label{eq:app5}
\\
&&
0= RL^3 - \lbrack 3 \rbrack LRL^2+ \lbrack 3 \rbrack L^2RL - L^3R.
\label{eq:app6}
\\
&&
0= r^3l - \lbrack 3 \rbrack r^2lr+ \lbrack 3 \rbrack rlr^2 - lr^3, 
\label{eq:app7}
\\
&&
0= rl^3 - \lbrack 3 \rbrack lrl^2+ \lbrack 3 \rbrack l^2rl - l^3r.
\label{eq:app8}
\end{eqnarray}
Moreover, $V$ carries the structure of a 
$U_q(\widehat{ sl_2})$ module
with 
action
$K\leftrightarrow K_0$,
 $K^{-1}\leftrightarrow K_1$,
$L\leftrightarrow e^{-}_0$,
$R\leftrightarrow e^{-}_1$,
 $l\leftrightarrow e^{+}_1$,
$r\leftrightarrow e^{+}_0$.
We call this the standard module. We call $w_1, w_2, \ldots, w_d$
the weights of the module. 
\end{lemma}

\begin{conjecture}
\label{conj:appirred}
The 
$U_q(\widehat{ sl_2})$ module in Lemma
\ref{lem:uqmoreaction} is irreducible if and only if 
$w_i\not=q^2w_j$ for $1 \leq i,j\leq d$.
\end{conjecture}  

\begin{lemma}
\label{lem:fullaction} 
Refering to the $U_q({\widehat {sl_2}})$-module in
Definition
\ref{lem:fullactiond1}, assume
the scalars $w_1, w_2, \ldots, w_d$ are distinct.
Then $V$ has a  
 basis $v_x$ indexed by the  subsets $x$ of  
$\lbrace 1,2,\ldots, d\rbrace  $ such that
for all $x \subseteq \lbrace 1,2,\ldots, d \rbrace $,
\begin{eqnarray}
R v_x &=& \sum_{i \in y} v_{x\cup i}
\prod_{j\in x} {{q w_i-q^{-1}w_j}\over {w_i-w_j}},
\\
L v_x &=& q^{d+1-2\vert x \vert}\sum_{i \in x} v_{x\backslash i}
w_i 
\prod_{j\in y} {{q^{-1} w_i-qw_j}\over {w_i-w_j}}
\\
r v_x &=& 
q^{2\vert x \vert -d+1}
\sum_{i \in y} v_{x\cup i}
w^{-1}_i\prod_{j\in x}
{{q w_i-q^{-1}w_j}\over {w_i-w_j}},
\\
l v_x &=& \sum_{i \in x} v_{x\backslash i}
\prod_{j\in y} {{q^{-1} w_i-qw_j}\over {w_i-w_j}},
\\
K v_x &=& q^{2\vert x \vert -d} v_x,
\end{eqnarray}
where $y$ denotes the complement of $x$ in   
$\lbrace 1,2,\ldots, d \rbrace $.
\end{lemma}



\begin{lemma}
\label{def:ceigap} With reference to
Lemma  \ref{lem:fullaction},
pick any subset  
 $x \subseteq \lbrace 1,2,\ldots d\rbrace$, and let
$\overline  x$ denote the complement of $x$ in
 $\lbrace 1,2,\ldots d\rbrace$.
Then
\begin{enumerate}
\item $v_x$ is an eigenvector for $ 
q^{-1}RL-qLR $, with
eigenvalue
\begin{eqnarray}
 q^{-1}\sum_{i \in x} w_i - q\sum_{j\in {\overline {x}}} w_j.
\end{eqnarray}
\item $v_x$ is an eigenvector for $ 
qrl-q^{-1}lr $, with
eigenvalue
\begin{eqnarray}
 q\sum_{i \in x} w^{-1}_i - q^{-1}\sum_{j\in {\overline{x}}} w^{-1}_j.
\end{eqnarray}
\end{enumerate}
\end{lemma}


\section{The bottom line}

\noindent Based on everything I have learned, the following
conjecture seems very  plausable. To describe it,  we use the following
notation. Following Chari and Pressley,
let us say a finite dimensional irreducible 
$U_q(\widehat{ sl_2})$-module  is {\it type (1,1)} whenever
$K_0K_1$
acts as the identity. In this case, we abbreviate
$K\leftrightarrow K_0$,
 $K^{-1}\leftrightarrow K_1$,
$L\leftrightarrow e^{-}_0$,
$R\leftrightarrow e^{-}_1$,
 $l\leftrightarrow e^{+}_1$,
$r\leftrightarrow e^{+}_0$, and observe
(\ref{eq:app1})--(\ref{eq:app8})
hold .


\begin{conjecture}
Let $A,A^*$ denote any TD pair on $V$ satisfying the
conditions of Definition 
\ref{def:1p1} and
Lemma \ref{lem:eigform}.
Then $V$ possesses an irreducible 
$U_q(\widehat{ sl_2})$-module structure of type (1,1) 
 such that
\begin{eqnarray}
A = aK + R, \qquad  \qquad A^*=a^*K^{-1} + L.
\end{eqnarray}
Moreover,  none of the weights equal $aa^*(q-q^{-1})^2$.
Conversely, let $V$ denote a finite dimensional 
irreducible 
$U_q(\widehat{ sl_2})$-module of type (1,1). Let
$a$ and $a^*$ denote nonzero scalars in $\fld$,
and assume none of the weights of $V$  equal 
$a a^* (q-q^{-1})^2$.
Set 
\begin{eqnarray}
A = aK + R,
\qquad \qquad 
A^* &=& a^*K^{-1} + L.
\end{eqnarray}
Then $A,A^*$ forms a TD pair on $V$ that  satisfies 
the conditions of
Defintion \ref{def:1p1}
and Lemma \ref{lem:eigform}. 
\end{conjecture}.




\section{Misc. remarks and questions}

\begin{conjecture} Assume $\fld$ is algebraically closed,
and let $q$ denote any scalar in $\fld$ that is not a root of
unity.
Let $V$ denote a vector space over
$\fld$ with finite positive dimension. Let $R$ and $L$
denote any nilpotent linear transformations from  $V$ to $V$,
and assume $R$ and $L$ satisfy 
the two Serre relations
(\ref{eq:ms1}),
(\ref{eq:ms2}).
Further assume there is no subspace $W \subseteq V$ such that
both $RW\subseteq W$, 
 $LW\subseteq W$,  other than $W=0$ and $W=V$.
Then we conjecture 
\begin{enumerate}
\item There exists a nonnegative integer $d$ and a  sequence  
$U_0, U_1, \ldots, U_d$ of nonzero 
subspaces of $V$ such that
\begin{eqnarray}
&&\qquad V = U_0+ U_1 + \cdots + U_d \qquad \qquad (\hbox{direct sum}),
\label{eq:vcdec}
\\
&&RU_i\subseteq U_{i+1} \qquad \quad (0 \leq i <d),  
\qquad RU_d=0,   
\label{eq:Ronci}
\\
&&LU_i\subseteq U_{i-1} \qquad \quad (0 < i \leq d),  
\qquad LU_0=0 .  
\label{eq:Lcui}
\end{eqnarray}
\item Assuming (ii) holds, we further conjecture
that for $0 \leq i\leq j\leq d$ and $i+j=d$, the restriction of
$R^{j-i}$ to $U_i$ is a bijection onto $U_j$, and the 
restriction of $L^{j-i}$ to $U_j$ is a bijection onto $U_i$. 
\end{enumerate}.
If (i), (ii) do not follow from the stated assumptions, find a counterexample.
\end{conjecture}

\noindent The reason the above conjecture is interesting is this:
Consider the associative algebra $T$ on two generators $R, L$ subject
to the quantum Serre relations
(\ref{eq:ms1}),
(\ref{eq:ms2}).
Let $V$ denote a finite dimensional irreducible $T$-module on which
both $R$ and $L$ act nilpotently. If the conjecture is true,
it appears very likely that the action of $T$ on $V$ can be exptended
to an action of 
$U_q(\widehat{ sl_2})$. This suggests there is very little difference
between $T$-modules and 
$U_q(\widehat{ sl_2})$-modules.



\bibliographystyle{plain}

\bibliography{master}

\medskip
\noindent
Georgia Benkart, Department of Mathematics,
University of Wisconsin, 480 Lincoln drive, 
Madison, Wisconsin, 53706, USA \hfil\break
E-mail: benkart@math.wisc.edu \hfil\break


\medskip
\noindent
Tatsuro Ito, Department of Computational Science, Faculty of Science,
Kanazawa University, Kakuma-machi, Kanazawa 920--1192, Japan \hfil\break
E-mail: ito@kappa.s.kanazawa-u.ac.jp \hfil\break


\medskip
\noindent 
Kenichiro Tanabe,
 Graduate School of Mathematics, 
          Kyushu University, 33
        Fukuoka 812-8581, Japan
\hfil \break
 E-mail : tanabe@math.kyushu-u.ac.jp
 \hfil\break

%\medskip
%\noindent 
%Kenichiro Tanabe, Department of Math, Kyushu University, Fukuoka 812,
%Japan \hfil \break
%E-mail:  tanabe@math.kyushu-u.ac.jp \hfil\break

\medskip
\noindent
Paul Terwilliger, Department of Mathematics,
University of Wisconsin, 480 Lincoln drive, 
Madison, Wisconsin, 53706, USA \hfil\break
E-mail: terwilli@math.wisc.edu \hfil\break




\end{document}

%%%%%%%%%%%%%%%%

% alt address for tanabe:
%         Combinatorial and Computational Mathematics Center
%          Pohang University of Science and Technology
%          San 31 Hyoja Dong, Nam-Gu, Pohang, 790-784, Korea
%
% e-mail : tanabe@com2mac.postech.ac.kr
%


--------------1D80B32B7240993B2E03FB0A--


math0304244.extracted_bib
\begin{thebibliography}{10}

\bibitem{Andrews}
G.~Andrews.
\newblock {\em The Theory of Partitions}.
\newblock Cambridge University Press, Cambridge, 1984.


\bibitem{AWil}
R.~Askey and J.A.~Wilson.
\newblock A set of orthogonal polynomials that 
generalize the {R}acah coefficients or $6-j$ symbols. 
\newblock {\em SIAM J. Math. Anal.}, 10:1008--1016, 1979. 

\bibitem{BanIto}
E.~Bannai and T.~Ito.
\newblock {\em Algebraic Combinatorics {I}: Association Schemes}.
\newblock Benjamin/Cummings, London, 1984.


\bibitem{cp}
V.~Chari and A.~Pressley.
\newblock Quantum affine algebras.
\newblock {\em Comm. Math. Phys.}, 142:261--283, 1991.

\bibitem{Sloane}
J.~H.~Conway and N.~J.~A.~Sloane.
\newblock {\em Sphere packings, Lattices and Groups}.
\newblock Springer-Verlag, New York, 1988.


\bibitem{TD00}
T.~Ito, K.~Tanabe, and P.~Terwilliger.
\newblock Some algebra related to ${P}$- and ${Q}$-polynomial association
  schemes. In
\newblock {\em Codes and Association Schemes (Piscataway NJ, 1999)}, Amer.
Math. Soc., Providence, RI, 56:167--192, 2001.

\bibitem{kac}
V.~Kac.
\newblock {\em Infinite dimensional Lie algebras}.
\newblock {Cambridge U. Press, Cambridge}, 1995.

\bibitem{KoeSwa}
R.~Koekoek and R.~Swarttouw.
\newblock {\em The Askey-scheme of hypergeometric orthogonal polyomials and its
  $q$-analog}, volume 98-17 of {\em Reports of the faculty of Technical
  Mathematics and Informatics}.
\newblock Delft, The Netherlands, 1998.
Available at
 \newblock{ \tt http://aw.twi.tudelft.nl/{\~{}}koekoek/research.html}

\bibitem{Leon}
D.~Leonard.
\newblock Orthogonal polynomials, duality, and association
schemes.
\newblock {\em SIAM J. Math. Anal.}, 13(4):656--663, 1982.

\bibitem{lu}
G.~Lusztig.
\newblock {\em Introduction to quantum groups.}
\newblock {Progress in Mathematics 110.}
\newblock {Birkhauser, Boston, 1993.}

%\bibitem{matsuo}
%Atsushi~Matsuo.
%\newblock Personal communication.


\bibitem{TersubI}
P.~Terwilliger.
\newblock The subconstituent algebra of an association scheme. 
\newblock {\em J. Algebraic Combin.}, 1(4):363--388, 1992.

\bibitem{LS99}
P.~Terwilliger.
\newblock Two linear transformations each tridiagonal with respect to an
  eigenbasis of the other.
\newblock {\em Linear Algebra Appl.},  330:149--203, 2001. 

\bibitem{qSerre}
P.~Terwilliger.
\newblock Two relations that generalize the $q$-Serre relations and the 
Dolan-Grady relations. 
\newblock {\em {P}hysics and {C}ombinatorics 1999 (Nagoya),
377--398.} World Sci. Publishing,
River Edge, NJ, 2001. 

\bibitem{LS24}
P.~Terwilliger.
\newblock  Leonard pairs from 24 points of view.
\newblock {\em Rocky mountain J.  Math.}, 32(2):827--888, 2002. 

\bibitem{conform}
P.~Terwilliger.
\newblock Two linear transformations each tridiagonal with respect to an
  eigenbasis of the other; the $TD$-$D$ canonical form
  and the $LB$-$UB$ canonical form.
\newblock {\em J. Algebra}, submitted.


\bibitem{lsint}
P.~Terwilliger.
\newblock Introduction to Leonard pairs.
(OPSFA Rome 2001),
\newblock {\em J. Comput. Appl. Math.}, 153(2):463--475, 2003.

\bibitem{Terint}
P.~Terwilliger.
\newblock Introduction to {L}eonard pairs and 
	     {L}eonard systems.
  \newblock {\em S\=urikaisekikenky\=usho K\=oky\=uroku},
  \newblock 
   (1109):67--79, 1999.
    \newblock Algebraic combinatorics  (Kyoto, 1999).


\bibitem{TLT:split}
P.~Terwilliger.
\newblock Two linear transformations each tridiagonal with respect to an
  eigenbasis of the other; comments on the split decomposition.
\newblock {\em Indag. Math.}, submitted.


\bibitem{TLT:array}
P.~Terwilliger.
\newblock Two linear transformations each tridiagonal with respect to an
  eigenbasis of the other; comments on the parameter array.
\newblock {\em Geometric and Algebraic Combinatorics 2,
(Oisterwijk, The Netherlands 2002)},
submitted.

\bibitem{TLTqrac}
P.~Terwilliger.
\newblock Leonard pairs and the $q$-Racah polynomials.
\newblock {\em Linear Algebra Appl.}, submitted. 

\bibitem{waki}
Minoru~Wakimoto.
\newblock {\em Infinite-dimensional Lie algebras}.
\newblock Translations of Mathematical Monographs 195, 
Amer. Math.
Soc. Providence, RI, 2001.

\end{thebibliography}