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\centerline {\bol SOME CLASSICAL EXPANSIONS FOR}
\centerline {\bol KNOP-SAHI AND MACDONALD POLYNOMIALS}
\sa
\centerline {\bol Jennifer  Morse}
\centerline {Department of Mathematics}
\centerline {University of California, San Diego}
\centerline {La Jolla, California 92093-0112}
\sap
{
\ninerm
\smallskip\midinsert\narrower\noindent\baselineskip12pt
\noindent
{\bols ABSTRACT}:\ssp
In recent simultaneous work, Knop and Sahi introduced
a non-homogeneous non-symmetric polynomial 
$G_\aaa(x;q,t)$ whose highest homogeneous component 
gives the non-symmetric Macdonald polynomial $E_\aaa(x;q,t)$. 
Macdonald shows that for any composition $\aaa$ 
that rearranges to a partition $\la$, an appropriate
Hecke algebra symmetrization of $E_\aaa$ yields the 
Macdonald polynomial $P_\la(x;q,t)$.
In the original papers all these polynomials 
are only shown to exist.  No explicit expressions are 
given relating them to the more classical bases.
Our basic discovery here is that $G_\aaa(x;q,t)$ appears to have
surprisingly elegant expansions in terms of the polynomials
$Z_\aaa(x_1,\ldots,x_n;q)=\prod_{i=1}^n(x_i;q)_{\aaa_i}$. 
In this paper we present 
the first results obtained in the 
problem of determining the connection coefficients 
relating these bases.  In particular
we give a solution to the problem of two variables. Our proofs rely 
on the theory of basic hypergeometric series and reveal 
a deep connection between this classical subject and the 
theory of Macdonald polynomials.
\endinsert}
\sap
\noindent{\bol Introduction}


The Macdonald basis $\{P_\la(x;q,t)\}_\la$ has recently become an intensive subject of 
study as a result of the many difficult conjectures that surround it. Its 
importance in the development of symmetric function theory is now widely
recognized. In addition to specializing to several fundamental bases,
(such as the Schur, the Hall-Littlewood, the Zonal, the Jack) its has
been conjectured to occur in a natural way [1] in representation theory 
and in some problems of particle mechanics [8]. One of the difficulties 
encountered in its study is the absence of explicit formulas expressing
$P_\la(x;q,t)$ in terms of more familiar bases. In fact, the connection 
coefficients relating a rescaled version of $\{P_\la(x;q,t)\}_\la$ to 
the modified Schur basis  $\{S_\la[X(1-t);q,t]\}_\la$ have only recently 
been shown to be polynomial functions of $q,t$ ([2],[4],[5],[7]). 
Macdonald in [12] shows that each $P_\la(x;q,t)$ decomposes into a sum of
homogeneous non-symmetric polynomials $E_\aaa(x;q,t)$ indexed by compositions.
More precisely, if $\aaa$ is any composition that
rearranges to $\la$ then
$$
P_\la \ses \sum_{\sig\in S_n}  t^{-length(\sig)}\ssp T_\sig\ssp E_\aaa
$$
where $T_\sig$ is an appropriately defined Hecke algebra operator.
Since the $E_\aaa$ are triangularly related to the monomial basis
$x^\aaa=x_1^{\aaa_1}x_2^{\aaa_2}\cdots x_n^{\aaa_n}$, they form themselves
a basis for the polynomials in $\xon$. In view of the fundamental nature
of the polynomials $P_\la(x;q,t)$ and their central place in symmetric
function theory, it is reasonable to assume that the $E_\aaa$ should
also play a central role in the study of polynomials in several variables.
It is shown in [12] that in fact the $E_\aaa$ themselves are but a special 
case of families of orthogonal polynomials associated to root
systems; the $E_\aaa$ being associated the root system $A_n$.
As for the $P_\la$, the $E_\aaa$ have only been shown to exist. They have
also been characterized as eigenfunctions of certain operators
and may be computed explicitly only through algorithms derived
from the recursions they satisfy. 
\sa

A breakthrough in the study of Macdonald polynomials 
is the simultaneous discovery
by Knop [5] and Sahi [14] of two 
closely related families $\{R_\la(x;q,t)\}_\la$
and $\{G_\aaa(x;q,t)\}_\aaa$  respectively indexed by partitions and 
compositions,
whose highest homogeneous components yield $\{P_\la(x;q,t)\}_\la$ 
and $\{E_\aaa(x;q,t)\}_\aaa$
respectively. Knop and Sahi show that $R_\la$
may be also be obtained as the Hecke algebra symmetrization of $G_\aaa$ for any 
composition $\aaa$ that rearranges to $\la$. What is remarkable about these 
new polynomials is that they may be chararacterized by very elementary 
vanishing properties which not only yield simple algorithms for their construction
but allow a quick and simple derivation of several heretofore difficult and
apparently deep properties of the Macdonald polynomials. The work of
Knop and Sahi 
brings to evidence that the $G_\aaa$ may be used as natural building
blocks in the construction of Macdonald polynomials, yet in a very concrete
sense, the $G_\la(x;q,t)$  are considerably easier to study than $E_\aaa$
and $P_\la$. This given we have set ourselves the task of finding some
classical basis in terms of which the $G_\la(x;q,t)$ may be given explicit, 
closed form expansions. Our discovery is that a most natural candidate 
to this effect appears to be the basis
$$
Z_\aaa(\xon;q)=\prod_{i=1}^n(1-x_i)(1-qx_i)\cdots (1-q^{\aaa_i-1}x_i)\ess .
\eqno {\rm I}.1
$$
In fact, as we shall see, the various properties of the polynomials $G_\aaa(x;q,t)$ established by
Knop and Sahi,
when expressed in terms of the connection coefficients relating the bases $G_\aaa(x;q,t)$ and 
$\{Z_\aaa(\xon;q)\}_\aaa$,  encode some of the less elementary identities [3]  of the theory of basic hypergeometric
series.
To see how all this comes about and to state our results we need to review the definitions and some of
the characterizing properties of the Knop-Sahi polynomials.
\sas
\def \oaa {\overline{\aaa}}
\def \obb {\overline{\bbb}}
\def \odd {\overline{\delta}}
\def \ogg {\overline{\gamma}}

We recall that by a composition we mean a vector $\aaa=(\aaa_1,\aaa_2,\ldots,\aaa_n)$
with non negative integral components. For convenience we set 
$$
|\aaa|\ses \aaa_1+\aaa_2+\cdots+\aaa_n\ess .
$$ 
We shall sometime express that $|\aaa|=m$ by saying that {\ita $\aaa$ is a composition of $m$}. 


We shall also denote by $\aaa^*$ the partition obtained by rearranging the components (parts)
of $\aaa$ in weakly decreasing order. If $\aaa$ has distinct parts then each $\aaa_i$ occupies
a well defined position $k_i=k_i(\aaa)$ in $\aaa^*$. By this we mean that $\aaa_i=\aaa_{k_i}^*$.
We can extend $k_i(\aaa)$ to the case when $\aaa$ has equal parts, breaking ties by
considering equal parts as decreasing from left to right. In other words if we label the parts of $\aaa$
from by decreasing size and from left  to right then $k_i(\aaa)$ gives the label of $\aaa_i$.
Here and after we refer to $k_i(\aaa)$ as the {\ita position} of $\aaa_i$ in $\aaa^*$.
Since Knop and Sahi use 
slightly different notations we will not be able to make our notation
consistent to both papers. We shall try as much as possible to adhere 
to Sahi's notation here.
This means that we will have to translate to Sahi notation some of those  ingredients and results 
that are in Knop's and not in Sahi's. This given, we recall that Sahi associates to each composition
$\aaa$ a vector of monomials $\oaa$ by setting
$$
\oaa_i\ses q^{-\aaa_i}t^{-n+k_i(\aaa)}
\eqno {\rm I}.2
$$
In [14] Sahi shows that if $\aaa$ is a composition of $m$  then 
in the linear span of the monomials $\{ x^\bbb\}_{|\bbb|\leq m}$
there exists a unique polynomial $G_\aaa(x;q,t)$ which satisfies the following two conditions
$$
\eqalign{
&a)\ess\ess G_\aaa(\obb;q,t)\ses 0\ess\ess\ess \hbox{for all $|\bbb|\leq|\aaa|$ and $\bbb\neq \aaa$\ssp ,}\cr
&b)\ess \ess  G_\aaa(\oaa;q,t)\ses 1\ssp .\cr}
\eqno {\rm I}.3
$$
Moreover it is shown that the coefficient of $x^\aaa$ in $G_\aaa$ doesnot vanish. We shall
express this by writing
$$
c)\ess\ess G_\aaa(x;q,t) \ssp |_{x^\aaa}\neq 0 
\eqno {\rm I}.4
$$
The uniqueness part of the Knop-Sahi result is relatively easy to show,
yet uniqueness permits
the immediate derivation of a number of surprising identities and recursions. Some of these
are given by Sahi in [14] and others only in  Knop [5]. We shall state them here in our present notation.
For the sake of completeness, we shall give 
some of the proofs as we need them in later sections.
To begin with we have the following immediate recursion.
\sas

\noindent{\bol Property I.1}\qquad
{\ita If $r=min\{\aaa_i\ssp :\ssp i=1..n\ssp \}> 0$ then setting $\gamma_i=\aaa_i-r$ we have
$$
G_\aaa(x;q,t)  \doteq \prod_{i=1}^n\ssp (x_i;q)_r \ess G_{\gamma}(q^r  x;q,t) 
\eqno {\rm I}.5
$$
where the symbol ``$\doteq$'' is to represent equality up to a scalar factor}. 
\sas

\noindent{\bol Property I.2}\qquad
{\ita If $\aaa_n>0$ then}
$$
G_\aaa(x;q,t)\ess \doteq \ess 
(1-x_n)\ssp G_{(\aaa_n-1,\aaa_1,\aaa_2,\ldots ,\aaa_{n-1})}(qx_n,x_1,x_2,\ldots,x_{n-1};q,t)\ess .
\eqno {\rm I}.6
$$
\sas

For $1\leq i\leq n-1$ let $s_i=(i,i+1)$ denote the transposition that interchanges $x_i$ and $x_{i+1}$
and set [9]
$$
T_{s_i}\ses s_i\sps {(1-t) \over x_i-x_{i+1}}\ssp x_i \ssp (1-s_i)\ess .
\eqno {\rm I}.7
$$
It is well known that the operators $T_{s_i}$ generate a faithful representation of the
Hecke algebra of $S_n$ in the space of polynomials in $\xon$. Indeed, it is easily verified
(nowadays using symbolic manipulation software) that we have
$$
\eqalign{ 
&a)\ess\ess T_{s_i} T_{s_j}\ses T_{s_j} T_{s_i}
\ess\ess\ess\ess\ess\ess\ess\ess {\ \over \ } 
\hbox{ for $\ess |i-j|>1$}\ess ,\cr
&b)\ess\ess T_{s_i} T_{s_{i+1}}T_{s_i} \ses T_{s_{i+1}}T_{s_i}T_{s_{i+1}}\ess\ess\ess\ess 
\hbox{ for $\ess i=1\dots n-1$}
,\cr
&c)\ess\ess
t\ssp T_{s_i}^{-1}\ses
 s_i\sps  {(1-t) \over x_i-x_{i+1}}\ssp  (1-s_i) \ssp x_{i+1}  
\ess .\cr
}
\eqno {\rm I}.8
$$
This permits us to extend the definition of $T$ to all permutations $\sig\in S_n$ by
setting for any reduced expression $\sig=s_{i_1}s_{i_2}\cdots s_{i_k}$
$$
T_\sig\ses T_{s_{i_1}}T_{s_{i_2}}\cdots T_{s_{i_k}}\ess .
\eqno {\rm I}.9
$$
\def \sdes {\ess \doteq \ess} 
\sas

\noindent{\bol Property I.3}
$$
\cases{
(a)\ess\ess\ess G_\aaa(x;q,t)\ses  T_{s_i}\ssp G_\aaa(x;q,t) \ses 
 s_i\ssp G_\aaa(x;q,t) 
& if $\aaa_i=\aaa_{i+1}$\cr
\cr
(b)\ess\ess\ess
G_{s_i\aaa}(x;q,t)\doteq (1-{\oaa_{i+1}\over\oaa_i})\ssp T_{s_i}G_\aaa(x;q,t) + (t-1)\ssp G_\aaa(x;q,t)
& if $\aaa_i\neq \aaa_{i+1}$\ess .
}
$$

\sas

It should be apparent that these three properties may be combined into a recursive algorithm
for computing the polynomials $G_\aaa$ starting from the initial condition
$$
G_{(o,o,\ldots ,o)}(x;q,t)\ses 1\ess .
\eqno {\rm I}.10
$$
This permits the rapid computation of extensive tables.
Now the particular nature of the right hand side of I.5 suggested that 
the basis defined in I.1 may turn out to be a natural tool for the study of these
polynomials. Our preliminary results  confirm this possibility.
The basic identity in the case of two variables  may be stated as follows:
\sa

\noindent{\bol Theorem I.1}\qquad
{\ita Setting   $x_1=x$ and $x_2=y$ we have for $m\geq 1$:}
$$
G_{(m,0)}(x,y;q,t)= {(-1)^m  q^{{m+1\choose 2}} 
\over (t;q)_{m+1}}
\sum_{0\leq j+k\leq m} 
{t^{k+j}  q^{j(k+1)} 
(t;q)_{m-k}\ssp 
(t;q)_{m+1-j}\ssp 
(x;q)_k \ssp
(y;q)_j
\over
(q;q)_k \ssp
(q;q)_j \ssp
(q;q)_{m-k-j} 
}
\eqno {\rm I}.11
$$

\noindent
Note that  Property I.1, in the two variable case, may be rewritten as
$$
G_{(a,b)}(x,y;q,t)\ses 
\cases
{ 
(x;q)_a(y;q)_a \ssp G_{(0,b-a)}(q^a x,q^a y;q,t) & if $a\leq b$ \cr
\cr
(x;q)_b(y;q)_b \ssp  G_{(a-b,0)}(q^b x,q^b y;q,t) & if $b< a$ \cr
}
\eqno {\rm I}.12
$$
In the same notation,  Property I.2 applied to the composition $\aaa=(0,b)$ gives: 
$$
G_{(0,b)}(x,y;q,t)\ses (1-y)G_{(b-1,0)}(qy,x)
\ess\ess \hbox{for $\ess b>0$}
\eqno {\rm I}.13
$$
Thus we see that the appropriate combination of I.11, I.12 and I.13 yields
entirely explicit formulas for the coefficients connecting the bases
$\{G_{(a,b)}(x,y;q,t)\}_{a,b\geq 0}$ and $\{(x;q)_a (y;q)_b\}_{a,b\geq 0}$.
\sa 

Finally, let us denote by $\phi$ the linear operator that sends a polynomial 
$Q=Q(\xon)$  into the polynomial
$$
\phi Q\ses Q(q x_n,x_1,x_2,\ldots ,x_{n-1})\ess .
\eqno {\rm I}.14
$$
This given, following Knop we set for $1\leq i\leq n-1$ 
$$
\Xi_i\ses {1\over x_i}\sps {1\over x_i}\ssp  T_{s_i}T_{s_{i+1}}\cdots T_{s_{n-1}}\ssp
(x_n-1)\ssp\phi\ssp  T_{s_1}T_{s_2}\cdots T_{s_{i-1}}
\eqno {\rm I}.15
$$
Translating theorem 3.6 of [6] to the present notation we can state that
\sas


\noindent{\bol Property I.4}
$$
\Xi_i\ssp G_\aaa\ses {1\over \oaa_i} \ssp  G_\aaa 
\bigsp
\hbox{ for $i=1,\ldots ,n-1$}
\eqno {\rm I}.16
$$
This remarkable result not only shows that the operators $\Xi_i$ form
a commuting family, but also constitute an independent characterization of
the Knop-Sahi polynomials.
\sas

This paper is divided into three sections. In the first section we rederive
the identities  I.5, I.6 , I.10 and I.16.  We also include a simple proof 
of the existence and uniqueness part of the Knop-Sahi result.  In the second
section, we use the vanishing property
of $G_{(m,o)}$ to give a short proof of Theorem I.1.  
Our formula can be used to give explicit expansions for the two variable case
of the  Macdonald polynomials $E_\aaa$ and $P_\la$. 
In section two we show that formula 4.9 [11], which gives $P_{(m)}$, can be 
derived from Theorem I.1.  

Section three includes the exploration of an
alternate path for proving Theorem I.1 
based on the characterization of the $G_\aaa$ as eigenfunctions of the operators $\Xi_i$.
In following this path we were able to determine explicit expressions for
the entries of the matrix expressing the action of $\Xi_1$ on the 
basis $\{(x;q)_a (y;q)_b\}_{a,b\geq 0}$.   

It is remarkable that consistency of these expressions, Theorem I.1 and Property I.4,  
as well as the computations carried out in section two, 
rest on
some of the deeper identities in the theory of basic hypergeometric series.
In particular, a crucial role is played by the summation formula for the well-poised 
$\ _6\Phi_5$. 
\sa


\noindent{\bol 1. The basic identities}

Since we are using a slightly different notation than in the original papers,
for convenience of the reader we shall rederive here the results of 
Knop and Sahi expressed by Properties I.1-I.4. 
Here as in [14] and [5],[6] all these properties
are derived from the vanishing conditions I.3 a) using the uniqueness portion
of the Knop-Sahi
 existence theorem. In each case we show that ``dot''-equality holds
by showing that the right-hand side has the same vanishing properties as the
the left-hand side. In order not to be unduly repetitious, in each of the 
following two proofs, $\aaa$ and $\bbb$ will be a generic pair of compositions 
satisfying
$$
\bbb\neq \aaa
\ess\ess\ess
{\rm and}
\ess\ess\ess
|\bbb|\leq |\aaa|
$$
\sas

\noindent{\bol Proof of Property I.1}\qquad
Set $ r'\ses min \{\bbb_i\ssp :\ssp 1\leq i\leq n\ssp \}\ess .  $
Assume first that $r'< r$ and let $j$ be the rightmost index such that $\bbb_j=r'$. 
Then $k_j(\bbb)=n$ and the definition in I.2 gives that $\obb_j= q^{-r'}$.
Consequently
$$
(\obb_j;q)_r\ses 0
$$
and this forces the vanishing of one of the factors in right-hand side of I.5. 

Assume next that $r'\geq r$ and set $\ddd=(\ddd_1,\ldots ,\ddd_n)$ with $\ddd_i=\bbb_i-r$.
Note that since the positions of $\bbb_i$ in $\bbb^*$ and $\ddd_i$ in $\ddd^*$ are the
same  we must have
$$
\obb_i= q^{-\bbb_i}t^{-n+k_i(\bbb)} \ses 
 q^{-r}q^{-\ddd_i}t^{-n+k_i(\ddd)} \ses  q^{-r}\ssp \odd_i\ess . 
$$   
This given, the definition of $G_\ggg$ and the fact that $\ddd\neq \ggg$ if and only if $\bbb\neq \aaa$
yield
$$
G_\ggg(q^r\obb)\ses 0\ess .
$$
The proof is completed by noting that the right-hand side of I.5 does not vanish
at $x=\oaa$.
\sa
\def \BR {{\bf R}}
\def \BD {{\bf \Delta}_q}

For convenience let $\BR$ and $\BD$ be the two ``affine'' rotation  operators
defined by setting for any composition $\ggg=(\ggg_1,\ggg_2,\ldots,\ggg_n)$
$$
\BR  \ggg = (\ggg_n-1,\ggg_1,\ldots  , \ggg_{n-1})
\ess\ess\ess
{\rm and}
\ess\ess\ess
\BD  \ggg= (q\ssp \ggg_n\ssp ,\ssp \ggg_1,\ldots  , \ggg_{n-1})\ess .
$$
Note that if $\ggg_n>0$ then $\BR \ggg$ is also a composition. Moreover it is
easy to see that the position of $\ggg_n$ in $\ggg^*$ is exactly the same as the position
of $\ggg_n-1$ in $(\BR \ggg)^*$. In other words we have $k_n(\ggg)=k_1(\BR \ggg)$ and thus 
$$
\overline {(\BR \ggg)}_1\ses q^{-\ggg_n +1}t^{-n+k_n(\ggg)}\ses {\overline \ggg}_n q\ess .
$$
Since we trivially have that $k_i(\BR \ggg)=k_{i-1}( \ggg)$ for $i=2,\ldots, n$ 
we immediately deduce that
$$
\overline {\BR \ggg}\ses \BD \ogg \ess .
\eqno 1.1
$$


\vbox{ 
\noindent{\bol Proof of Property I.2}\qquad
If $\bbb_n=0$ then $\obb_n=q^{-o}t^{-n+n}=1$ this gives $(1-\obb_n)=0$ which causes the vanishing of
the first factor in the right-hand side of I.6.

On the other hand if $\bbb_n>0$ then 1.1 gives that
$$
G_{\BR\aaa}(\BD\obb) \ses  G_{\BR\aaa}(\overline{\BR\bbb})\ses 0\ess , 
$$
(since $\BR\bbb\neq\BR\aaa$ if and only if $\bbb\neq \aaa$). This causes the vanishing
of the second factor in the right-hand side of I.6.

The proof is completed by noting that the right-hand side of I.6 does not vanish
at $x=\oaa$.
\sa
}

\def \sdes {\ess \doteq \ess} 
\noindent{\bol Proof of Property I.3} \qquad
We start by proving formula b). So suppose that $\aaa_i\neq \aaa_{i+1}$ and set $m=\aaa_{i+1}/\aaa_i$.
Substituting I.7 in the right-hand side of b) we derive that 
$$
\eqalign{
RHS &= \ssp (1-m)\ssp 
\Bigl\{
s_i G_\aaa  + (1-t) {x_i\over x_i-x_{i+1}} (1-s_i)G_\aaa
\Bigr\}
\sps
(t-1) G_\aaa\ses\cr
&=\ssp 
(1-m)\ssp 
\Bigl\{
1 - (1-t) {x_i\over x_i-x_{i+1}}
\Bigr\}
s_i G_\aaa \sps (1-t) \Bigl\{(1-m){x_i\over x_i-x_{i+1}} -1 \Bigr\} G_\aaa\ess .
\cr 
}
$$
Thus formula b) is equivalent to the identity
$$
G_{s_i  \aaa}(x;q,t)\sdes 
(1-m)\ssp 
\Bigl\{ {t x_i- x_{i+1} \over x_i-x_{i+1}} \Bigr\} s_i G_\aaa
\sps
(1-t) \Bigl\{ { x_{i+1}- m x_i \over x_i-x_{i+1}} 
\Bigr\} G_\aaa  \ess .
\eqno 1.2
$$
For convenience let $R_1$ and $R_2$ respectively denote the two summands on the right-hand side of 1.2.
To prove Property I.3  we must test the vanishing of $R_1+R_2$ for all $\bbb$ satisfying
$|\bbb|\leq|\aaa|$ and $\bbb\neq s_i\aaa$. 
Remarkably, it develops that $R_1$ and $R_2$ individually  vanish for all such $\bbb$'s!
We shall work with each separately.
\sas
\item{  $R_1:$}
\itemitem{(1)} If $ \bbb_i\neq\bbb_{i+1} $ 
then  the relations $k_i(\bbb)=k_{i+1}(s_i\bbb)$,  $k_{i+1}(\bbb)=k_{i}(s_i\bbb)$
and $k_j(\bbb)=k_j(s_i\bbb)$ for $j\neq i,i+1$ 
give that $s_i\obb=\overline{s_i\bbb}$. Thus, since $s_i\bbb\neq \aaa$ and $|s_i\bbb|=|\bbb|\leq | \aaa$
we get that
$$
s_iG_\aaa(\obb)\ses G_\aaa(s_i\obb)\ses G_\aaa(\overline{s_i\bbb}) \ses 0\ess .
$$ 
\itemitem{(2)} If $ \bbb_i=\bbb_{i+1} $ then $k_{i+1}(\bbb)=k_i(\bbb)+1$ gives 
$\obb_{i+1}=q^{\bbb_i}t^{-n+k_i(\bbb)+12}=t \obb_i$ and in this case it is the factor
$tx_i-x_{i+1}$ that forces the vanishing of $R_1$.
\sas
\vfill\supereject
\item{  $R_2:$}
\itemitem{(1)} If $\bbb\neq \aaa$ then the factor $G_\aaa$ itself vanishes for $x=\obb$.
Note that, since $\aaa_i\neq \aaa_{i+1}$ forces $s_i\aaa\neq \aaa$,
this factor will vanish  even for $\bbb=s_i\aaa$. 

\itemitem{(2)} If $\bbb= \aaa$ then the vanishing of $R_2$ is simply due to the factor
$x_{i+1}-m x_i$ which for $x=\oaa$  reduces to $\oaa_i-{\oaa_{i+1}\over \oaa_i} \oaa_i$.
\sa

To complete the proof of (b) we need only check that $R_1+R_2$ does not vanish for $x=s_i\aaa$.
However, since we noted that $R_2$ vanishes there, we need only show that $R_1$ doesn't vanish.
But this is immediate since $(1-m)(t\oaa_{i+1}-\oaa_i)\neq 0$ and $s_iG_\aaa(s_i\oaa)=G_\aaa(\oaa)=1$.
\sa

To prove (a) we start by noting that for any polynomial $Q$ we have
$$
\eqalign{
T_{s_i}Q\ses s_iQ \sps{(1-t)\ssp x_i\over x_i-x_{i+1}}\ssp (Q-s_iQ)
& \ses \bigl(1- {(1-t)\ssp x_i\over x_i-x_{i+1}}\bigr)\ssp s_i Q  \sps {(1-t)\ssp x_i\over x_i-x_{i+1}}\ssp Q  \cr
& \ses  {tx_i -x_{i+1}\over x_i-x_{i+1}}\ssp Q(s_ix)   \sps (1-t){x_i\over x_i-x_{i+1}}\ssp Q(x)   \cr}
\eqno 1.3
$$
Using the second form of $T_{s_i} Q$ we immediately see that $T_{s_i} Q=Q$ implies $s_i Q =Q$.
Thus we need only show the first of the equalities in (a). To this end note that 
for $\bbb\neq \aaa$ and $|\bbb|\leq |\aaa|$ the third equality in 1.3 for $Q=G_\aaa$ and $x=\obb$
yields that
$$
(T_si G_\aaa)(\obb)\ses {t\obb_i -\obb_{i+1}\over \obb_i-\obb_{i+1}}\ssp G_\aaa(s_i \obb)
\eqno 1.4
$$
now if $\bbb_i\neq \bbb_{i+1}$ then $s_i\obb=\overline{s_i\bbb}$ and $G_\aaa(s_i \obb)$ must vanish since
$\aaa_i= \aaa_{i+1}$ gives $s_i\bbb\neq \aaa$. On the other hand if  $\bbb_i=\bbb_{i+1}$
then $\obb_{i+1}=t \obb_i$ (as we have seen). This forces the vanishing of the first factor in 1.3.
Thus $T_si G_\aaa$ has the same vanishing properties as $G_\aaa$. 
This given, we  only need to compute its value at $x=\oaa$. But here, we have $t\oaa_i=\oaa_{i+1}$ 
and $G_\aaa(\oaa)=1$ which combined with the third equality in 1.3 yields that
$$
(T_{s_i} G_\aaa)(\oaa) \ses 
{t\oaa_i -\oaa_{i+1}\over \oaa_i-\oaa_{i+1}}\ssp Q(s_i\oaa)   \sps (1-t)\ssp {\oaa_i\over \oaa_i-\oaa_{i+1}}\ssp Q(\oaa)
\ses 1
$$ 
This completes the proof of Property I.3.
\sa

Before we can proceed to the proof of Property I.4 we need some preliminary observations.
We shall begin by rederiving Knop's beautiful result that all the operators $\Xi_i$ send
polynomials into polynomials. To this end, we note that $T_{s_i}$ and $T_{s_i}^{-1}$ may be 
given the alternate forms:
$$
\eqalign{
&a)\ess\ess T_{s_i}  \ses t\ssp  s_i\sps  {(1-t) \over x_i-x_{i+1}} (1-s_i)\ssp  x_i\ess ,\cr
&b)\ess\ess T_{s_i}^{-1}  =\ess  s_i\sps {(1-t)\over t} {x_{i+1}\over  x_i-x_{i+1}} (1-s_i)\ess .\cr
} 
\eqno 1.5
$$
This given, using 1.5 b) and then 1.5 a) we derive that for any polynomial $Q$ we have
$$
\eqalign{
{t \over x_{i+1}}T_{s_i}^{-1} Q &\ses t\ssp  s_i\ssp {1\over x_i} Q \sps {(1-t)\over x_i-x_{i+1}} (1-s_i)\ssp  Q\cr
&\ses \Bigl(t \ssp s_i \sps {(1-t) \over x_i-x_{i+1}} (1-s_i) x_i\Bigr) {1\over x_i}  Q\ssp \cr
&\ses  T_{s_i} {1\over x_i} \ssp Q  \ess .\cr
}
$$
Recalling the definition of in I.15 and using this relation we obtain for $i<n\ess $:

$$
\eqalign{
t \ssp \Xi_{i+1} T_{s_i}^{-1}\ssp Q
&\ses  {t \over x_{i+1}} T_{s_i}^{-1}\ssp Q \sps {t\over x_{i+1}} T_{s_i}^{-1}T_{s_i} 
T_{s_{i+1}}\cdots T_{s_{n-1}}(x_n-1)\phi T_{s_1}\cdots T_{s_i}  
T_{s_i}^{-1}  Q\cr 
&\ses  T_{s_i} \ssp {Q\over x_i} \sps T_{s_i} {1\over x_{i}}
T_{s_{i}}\cdots T_{s_{n-1}}(x_n-1)\phi T_{s_1}\cdots T_{s_{i-1}}  
\ssp   Q \cr
&\ses T_{s_i} \Xi_i\ssp Q \cr
}
$$
\def \To {T_{s_1}}
\def \Tn {T_{s_{n-1}}}
\def \Ti {T_{s_i}}
\def \Tim {T_{s_{i-1}}}
\def \Tip {T_{s_{i+1}}}


\noindent
Equivalently, for any polynomial $Q$ we have
$$
\Xi_i Q \ses t\ess T_{s_i}^{-1} \ssp \Xi_{i+1}\ssp  T_{s_i}^{-1}\ssp Q \ess .
\eqno 1.6
$$
Iterating this relation we finally obtain that
$$
\Xi_i Q \ses t^{n-i}\ess T_{s_i}^{-1}\cdots T_{s_{n-1}}^{-1} \Xi_n\ssp T_{s_{n-1}}^{-1}\cdots T_{s_i}^{-1}
\eqno 1.7
$$
This shows that if $\Xi_n$ sends polynomials into polynomials the same will be true
for all the other $\Xi_i$. To prove the result for $\Xi_n$ we follow Knop and write
$$
\Xi_n\ses \phi \ssp \To \cdots \Tn   \sps 
{1\over x_n} \bigl(I - \phi\ssp  \To\cdots \Tn  \bigr )\ess .
$$
Now the first term in this decomposition is clearly a polynomial operator. It develops
that the same is true for the second term for the simple reason that it is a sum
of divided difference operators. To see this we note that we may write
$$
\phi \ses \tau_n \ssp s_{n-1}\cdots s_2s_1
$$
where $\tau_n$ is the operator which replaces $\ssp x_n\ssp $ by $\ssp q  x_n\ssp $. 
This gives that
$$
\eqalign{
Q - \phi\ssp  \To\cdots \Tn  Q
&\ses Q- \phi\ssp s_1 \cdots s_{n-1} Q\sps 
\phi\ssp (s_1 \cdots s_{n-1}\sms \To \cdots \Tn ) \ssp Q
\cr
&\ses Q-\tau_n Q\sps  
\sum_{i=1}^{n-1} \phi\ssp s_1\cdots s_{i-1}\ssp\bigl( s_i -\Ti \bigr)\ssp \Tip\cdots \Tn\ssp Q
\cr
&\ses Q-\tau_n Q\sps  
(t-1)\sum_{i=1}^{n-1} \tau_n\ssp s_{n-1}\cdots s_{i}\ssp {x_i\over x_i-x_{i+1}}(1-s_i)
\ssp \Tip\cdots \Tn\ssp Q
\cr
&\ses Q-\tau_n Q\sps  
(t-1)\sum_{i=1}^{n-1} \tau_n\ssp  {x_n\over x_n-x_{i}}\bigl(1-(i,n)\bigr)
\ssp s_{n-1}\cdots s_{i}\ssp \Tip\cdots \Tn\ssp Q\ess ,
}
$$
where $(i,n)$ denotes the transposition which interchanges $x_i$ with $x_n$. 

In summary we have
$$
\Xi_n\ses \phi \ssp \To \cdots \Tn \sps \DDD_n\sps 
q \ssp (t-1)\sum_{i=1}^{n-1} \tau_n\ssp  D_{(i,n)} 
\ssp s_{n-1}\cdots s_{i}\ssp \Tip\cdots \Tn \ess ,
$$
where $\DD_n$ denotes the $n^{th}$ $q$-derivative operator and $D_{i,j}$ is the
divided difference operator acting on the pair $\ssp (x_i,x_n)\ssp $.
\sa

Our next observations reveal a remarkable property of the operator $(x_n-1)\phi$. To this end
note that the inverse of the operator $\BR$ defined in the introduction 
is obtained by setting
$$
\BR^{-1}\ssp (\ggg_1,\ggg_2,\ldots ,\ggg_n)\ses (\ggg_2,\ldots ,\ggg_n  ,\ggg_1+1)\ess .
$$
With this notation, we may rewrite Property I.2 by stating that 
$$
(x_n-1)\ssp \phi\ssp  G_\ggg(x;q,t) \ess  \doteq\ess  G_{\BR^{-1}\ggg}(x;q,t)   
\eqno 1.8
$$

\def \CG {{\cal G}}
\def \CR {{\cal R}}

\noindent
Now let $\CG_m$ denote the linear span of the collection of polynomials $\{\ssp G_\ggg \ssp\}_{|\ggg|=m}$.
Since the collection $\{G_\aaa\}_\aaa$ is a polynomial basis, we may view 1.8 as defining a linear operator
$\CR^{-1}$ which sends $\CG_m$ into the subspace $\CR^{-1}\CG_m$ of $\CG_{m+1}$ spanned by the
collection $\{ G_{\BR^{-1}\ggg}\}_{|\ggg|=m}$. Keeping all this in mind we are in a position to give
\sas

\noindent
{\bol The Proof of Property I.4}\qquad
Let $\aaa$ be a given composition. The definition in I.15 may now be written as 
$$
\Xi_i\ssp G_\aaa\ses 
{1\over x_i} G_\aaa \sps {1\over x_i} \ssp \Ti \cdots \Tn   \CR^{-1} \ssp \To\cdots T_{s_{i-1}}\ssp G_\aaa
\ess .
\eqno 1.9
$$
Since Property I.3 implies in particular that each $\Ti$ leaves $\CG_m$ invariant we see that
the polynomial
$$
\Ti \cdots \Tn   \CR^{-1} \ssp \To\cdots T_{s_{i-1}}\ssp G_\aaa
$$
will lie in the space $\CG_{|\aaa|+1}$. In particular it follows that the second term
in 1.9 will  necessarily vanish for all $|\bbb|\leq |\aaa|$. This immediately gives that
the right hand side of 1.9 vanishes for all 
$$
|\bbb|\leq |\aaa|\ess ,\ess 
\bbb\neq \aaa\ess .
$$
Moreover, evaluating both sides of 1.9 at $\oaa$ gives
$$
\Xi_i\ssp G_\aaa (\oaa;q,t)\ses {1\over \oaa_i}\ess .
$$  
This establishes I.16 and completes our proof.
\sap

\def \CB {{\cal B}}
\vskip -.3truein

Our treatment here would be completely self contained where it
not for the fact that we have made repeated  use of the uniqueness part of 
the Knop-Sahi existence theorem. For sake of completeness, we shall
terminate this section with a simple proof of this result.
To this end, for a given integers $n,m> 0$ let us 
denote by $\CB_m(n)$ the collection of all $n$-component compositions
of a number $\leq m$. In symbols
$$
\CB_m(n)\ses \{\ssp \aaa \ssp :\ssp |\aaa |\leq m\ssp \}\ess .
$$
For $n$ and $m$ being fixed, let
$$
\aaa^{(1)},\aaa^{(2)},\ldots ,\aaa^{(N)},
\eqno 1.10
$$ 
denote the elements of $\CB_m(n)$ in some fixed total order.
This given, the existence of the Sahi polynomial $G_\aaa$ for $|\aaa|=m$ 
depends on being able to construct coefficients $c_j(q,t)$ such that
$$
G_\aaa(\oaa^{(i)})\ses \sum_{j=1}^N\ssp c_j(q,t)\ssp \bigl[\oaa^{(i)}\bigr]^{\aaa^{(j)}}\ses 
\cases {
    0& if $\aaa^{(i)}\neq \aaa$\ess , \cr
&\cr
  1& if $\aaa^{(i)}= \aaa$ \ess .\cr
} 
$$
We can thus see that existence and uniqueness is assured at once for all compositions of $m$
by proving that the matrix
$$
\|\ssp \bigl[\oaa^{(i)}\bigr]^{\aaa^{(j)}}\ssp \|_{i,j=1..N}
\eqno 1.11
$$
has non vanishing determinant. Now it develops that this is but a very special case of a result which
may be stated as follows.
\sas

\noindent{\bol Proposition 1.1}\qquad
{\ita Let $\aaa^{(i)},\delta^{(i)}$ for $i=1..N$ be $n$-vectors with non-negative integral components.
then the polynomial 
$$
P(q,t)\ses det\ssp \|\ssp  q^{(\aaa^{(i)},\aaa^{(j)})}\ssp t^{(\aaa^{(i)},\delta^{(j)})}\ssp \|_{i,j=1..N}
\eqno 1.12
$$
cannot vanish identically.}

\noindent{\bol Proof}\qquad
We shall follow closely the argument used by Macdonald in the proof a similar result (see p. 334 of [11]).
We note first that we may write
$$
P(q,t)\ses \sum_{\sig\in S_N}\ess sign(\sig)\ess \ssp q^{\ess \sum_{i=1}^N\ssp (\aaa^{(i)},\aaa^{(\sig_i)})}\ess 
 t^{\ess \sum_{i=1}^N\ssp (\aaa^{(i)},\delta^{(\sig_i)})}\ess .
$$
Now, the simple inequality $ab\leq (a^2+b^2)/2$ valid  for any two numbers $a,b\geq 0$ immediately implies
that
$$
\sum_{i=1}^N\ssp \bigl(\aaa^{(i)},\aaa^{(\sig_i)}\bigr)\ssp \leq \ssp 
{1\over 2}\sum_{i=1}^N\ssp \bigl(\aaa^{(i)},\aaa^{(i)}\bigr)\sps 
{1\over 2}\sum_{i=1}^N\ssp \bigl(\aaa^{(\sig_i)},\aaa^{(\sig_i)}\bigr)
\ses
\sum_{i=1}^N\ssp \bigl(\aaa^{(i)},\aaa^{(i)}\bigr)\ess .
\eqno 1.13
$$
However, since $ab=(a^2+b^2)/2$ only if $a=b$ we see that equality can hold true in 1.13
only if $\aaa^{(\sig_i)}=\aaa^{(i)}$ for $i=1,..N$. This means that 
the term of highest $q$-degree in $P(q,t)$ can only come from the identity permutation.
Since its coefficient is 
$$
t^{\ess \sum_{i=1}^N\ssp (\aaa^{(i)},\delta^{(i)})}\not\equiv 0 
$$
the same must hold true for $P(q,t)$ itself.
\sas

To derive the non vanishing of the determinant of our matrix we only need to observe that
for any  $n$ component compositions $\aaa$ and $\bbb$ we have
$$
\obb^\aaa\ses q^{\ssp -  \sum_{i=1}^n \aaa_i\bbb_i}\ess t^{\ssp -\sum_{i=1}^n \aaa_i(n-k_i) }
\ses q^{\ssp -(\aaa,\bbb)}\ess  t^{\ssp - (\aaa,\delta(\bbb))}
$$
where we are letting $\delta(\bbb)$ denote the vector with components
$\delta_i(\bbb) =n-k_i(\bbb)$ (for $i=1..n$).
Thus our determinant is simply given by $P(1/q,1/t)$
when the $\aaa^{(i)}$'s are as given in 1.10 and $\ddd^{(i)}=\delta (\aaa^{(i)})$.

This completes our proof of existence and uniqueness of the Sahi polynomials.
\sa

\noindent{\bol Remark 1.1}\qquad
We should note that the non vanishing of our determinant, also shows that a polynomial
in $x_1,x_2,\ldots, x_n$ which is of degree $\leq m$ is completely determined by its values
at the points $\aaa^{(1)},\ldots ,\aaa^{(N)}$. It is then not difficult to derive from this fact that
the Knop-Sahi
polynomials do form a basis for the polynomials in $x_1,x_2,\ldots, x_n$.
\sap
\def \Ga {G_{(m,0)}(x,y;q,t)}
\def \Ex {K_m (x,y;q,t)}
 
\noindent{\bol 2. The two variable case and basic hypergeometric series.}
\sas
The theory of hypergeometric series comes into play in our first proof of
Theorem I.1.

\noindent
{\bol Proof of Theorem I.1}\qquad
For convenience let $\Ex$ denote the 
expression on the right-hand side of I.11. That is,
let us set for a moment
$$
\Ex={(-1)^m\ssp q^{m+1\choose 2}\over (t;q)_{m+1}}
\sum_{0\leq j+k\leq m} 
{t^{k+j} 
 q^{j(k+1)}
(t;q)_{m-k}\ssp 
(t;q)_{m+1-j}\ssp
(x;q)_k \ssp   
(y;q)_j    
\over 
(q;q)_k \ssp
(q;q)_j \ssp
(q;q)_{m-k-j} 
}\ess .
\eqno 2.1
$$
We recall that $\Ga$ may be characterized as the unique polynomial of degree $\leq m$ which satisfies I.3 and I.4  
for $\aaa=(m,0)$. Since $\Ex$ is clearly a polynomial of degree $\leq m$, to show I.11
we need only verify that $\Ex$ satisfies these two conditions. We start by verifying  I.3 a).
\sas

Transformations will be applied to 2.1 which reduce $\Ex$ to an expression 
that clearly vanishes for $(x,y)=\obb$ when $\beta=(a,b)$ with $a< b$.  Similar
transformations will make obvious the vanishing when $a\geq b$. 
\sa

Note that when $a<b$ we have
$$
\obb=({1\over q^a},{1\over q^b\,t})\ssp .
$$
Thus it will first be shown that 
$$
K_m(1/q^a,1/q^bt;q,t) \ses 0
\ess\ess\ess
\hbox{
when $\ess\ess  a<b\ess$  and $\ess a+b\leq m$.
}
\eqno 2.2
$$
\sas

The property of q-shifted factorials [3]
$$
(a;q)_{n-k}={(a;q)_{n}\over (q^{1-n}/a;q)_k}({-q\over a})^k\,
                q^{{k \choose 2}-nk}
\eqno 2.3
$$
will be used to modify several terms in 2.1. Namely we have
$$
\eqalign
{
&a)\ess\ess  (t;q)_{m-k}={(t;q)_{m}\over (q^{1-m}/t;q)_k}({-q\over t})^k\,
                q^{{k \choose 2}-mk} \cr
&b)\ess\ess(t;q)_{m+1-j}={(t;q)_{m+1}\over (q^{-m}/t;q)_j}
        ({-q\over t})^j\,q^{{j \choose 2}-(m+1)j}\cr
&c)\ess\ess(q;q)_{m-k-j}={(q;q)_{m-k}\over (q^{-m+k};q)_j}\,(-1)^j
                \,q^{{j \choose 2}-(m-k)j}\cr
&d)\ess\ess(q;q)_{m-k}={(q;q)_{m}\over (q^{-m};q)_k}\,(-1)^k
                \,q^{{k \choose 2}-mk}\cr
}
\eqno 2.4
$$
Substituting a) b) and c) (combined with d)) in 2.1 transforms it to
$$
\Ex= {(-1)^{m}q^{{m+1 \choose 2}}
(t;q)_{m} \over (q;q)_m}
\sum_{k\geq 0}^m
{q^{k}\ssp(x;q)_k \ssp(q^{-m};q)_k
        \over (q;q)_k\ssp (q^{1-m}/t;q)_{k}}
\sum_{j\geq 0}^{m-k}\,
{q^j(y;q)_j\ssp 
(q^{-m+k};q)_j \over(q;q)_j
\ssp (q^{-m}/t;q)_j}\ssp.
\eqno 2.5
$$
The sum over $j$ can be viewed as one going from $0$ to $\infty$
by virtue of $\,(q^{-m+k};q)_j$ vanishing  for all $j>m-k$.  This allows the
use of a $_2\phi_1$ summation identity, (II.7 in [3]),
$$ 
_2\phi_1(a,q^{-n};c;q,q)= {(c/a;q)_n\over (c;q)_n}\ssp  a^{n} 
\eqno 2.6
$$ 
with $a=y,\,q^{-n}= q^{-(m-k)}\,$ and $\,c=q^{-m}/t$.
The dependence on $j$ is nicely eliminated and the double sum becomes 
$${(-1)^{m}q^{{m+1 \choose 2}}
(t;q)_{m} \over (q;q)_m}
\sum_{k\geq 0}^m
{q^{k}\ssp (x;q)_k\ssp (q^{-m};q)_k
\over (q;q)_k\ssp(q^{1-m}/t;q)_{k}}
{(q^{-m}/ty;q)_{m-k}\ssp y^{m-k} \over(q^{-m}/t;q)_{m-k}}\ess .
$$
Evaluation at $x=1/q^a$ and $y=1/q^bt$ yields
$$
{(-1)^{m}q^{{m+1 \choose 2}}
(t;q)_{m} \over (q;q)_m}
\sum_{k\geq 0}^m
{q^{k}\ssp (q^{-a};q)_k\ssp (q^{-m};q)_k
\over (q;q)_k\ssp(q^{1-m}/t;q)_{k}}
{(q^{-m+b};q)_{m-k}\ssp (q^{-b}/t)^{m-k} \over(q^{-m}/t;q)_{m-k}}
\eqno 2.7
$$
Each term in this sum vanishes individually! This can be seen by assuming
the contradiction.  Suppose that the $k^{th}$ term in the
sum does not vanish.  This implies that 
$$
i)\ess\ess (q^{-a};q)_{k} \neq 0 
\ess\ess\ess 
{\rm AND}
\ess\ess\ess 
ii)\ess\ess  
(q^{-m+b};q)_{m-k}
 \neq 0\ess .
$$ 
However, $i)\Rightarrow a\geq k$ and  $ii) \Rightarrow k>b$. But this contradicts our
initial stipulation that $a<b$. Therefore 2.5 clearly vanishes as indicated,
which implies 2.1 as desired.
\sa

Note that when $\bbb=(a,b)$ with $a\geq b$ we have
$$
\obb\ses ({1\over q^at},{1\over q^b})
$$
So we must show next that  
$$
K_m({1\over q^at},{1\over q^b};q,t)\ses 0\ess\ess\ess
\hbox{
when $\ess\ess  m>a\geq b\ess$  and $\ess a+b\leq m$.
}
\eqno 2.8
$$
In this case we use $a)$, $ b) $ of  2.4 and the combination of
$$
\eqalign{
&d)\ess\ess (q;q)_{m-j-k}={(q;q)_{m-j}\over (q^{-m+j};q)_k}\,(-1)^k
                \,q^{{k \choose 2}-(m-j)k}\ess ,\cr
&e)\ess\ess
(q;q)_{m-j}={(q;q)_{m}\over (q^{-m};q)_j}\,(-1)^j
                \,q^{{j \choose 2}-mj}\ess . \cr
}
$$ 
With these modifications formula 2.1 becomes
$$ 
\Ex\ses
{(-1)^{m}\, q^{{m+1 \choose 2}}\,
(t;q)_{m}\,
\over (q;q)_m}
\sum_{j\geq 0}^m
{q^{j}\, (y;q)_j\,(q^{-m};q)_j
\over (q;q)_j
\,(q^{-m}/t;q)_j}
\sum_{k\geq 0}^{m-j}
{q^{k}\,(x;q)_k\,
(q^{-m+j};q)_k \over
(q;q)_k\,(q^{1-m}/t;q)_{k}}\ssp.
$$
Letting $a=x$, $\,q^{-n}= q^{-(m-j)}$ and $c=q^{-m}/t$,
2.6 can again be applied to eliminate the dependence on $k$ giving 
$$
\Ex\ses{(-1)^{m}\, q^{{m+1 \choose 2}}\,
(t;q)_{m} \over (q;q)_m}\,
\sum_{j\geq 0}^m
{q^{j}\,(y;q)_j\,(q^{-m};q)_j
\over (q;q)_j\,(q^{-m}/t;q)_{j}}
{(q^{1-m}/tx;q)_{m-j}\,x^{m-j} \over(q^{1-m}/t;q)_{m-j}}\ssp .
\eqno 2.9
$$
Evaluation at $x=1/q^at$ and $y=1/q^b$ then yields
$$
{(-1)^{m}\, q^{{m+1 \choose 2}}\,
(t;q)_{m} \over (q;q)_m}\,
\sum_{j\geq 0}^m
{q^{j}\,(q^{-b};q)_j\,(q^{-m};q)_j
\over (q;q)_j\,(q^{-m}/t;q)_{j}}
{(q^{1-m+a};q)_{m-j}\,(q^{-a}/t)^{m-j} \over(q^{1-m}/t;q)_{m-j}}\ssp .
$$
Proceding as before, let us assume that the $j^{th}$ term in the sum does not vanish.
We must then have
$$
i)\ess\ess
(q^{a+1-m};q)_{m-j}\neq 0
\ess\ess
{\rm and}
\ess\ess
ii)\ess\ess
(q^{-b};q)_j \neq 0\ess .
$$
But then $i)\ess \Rightarrow \ssp a<j$ and $ii)  \Rightarrow b\geq j$
which contradict our initial assumption that $a\geq b$. This proves 2.8 as desired.
We have thus verified that $K_m$ vanishes for all $\obb$ when
$\beta=(a,b) \neq (m,0)$ and $|\beta|\leq |\alpha|$.
\sas

We must next check condition I.3.b. We must then evaluate $K_m$ at $\oaa$ when $\aaa=(m,0)$.
For this purpose we can use formula 2.9 which was shown to be equivalent to 2.1.
Now, $\oaa=(m,0)$ implies
$$
\oaa=({1\over q^m\,t},1)\ess .
$$
Substituting $x={1\over q^m}$ and $y=1$ in 2.9 produces
$$
{(-1)^{m}\, q^{{m+1 \choose 2}}\,
(t;q)_{m} \over (q;q)_{m}}
\sum_{j\geq 0}^m
{q^{j}\,(1;q)_j\,(q^{-m};q)_j \,(q;q)_{m-j}\,(q^{-m}/t)^{m-j}
\over (q;q)_j\,(q^{-m}/t;q)_{j}(q^{1-m}/t;q)_{m-j}}\ssp.
$$
Note that the occurence of $(1;q)_j$ in the numerator forces all the terms
to vanish except the $j=0$ term.  Formula 2.9 reduces to  
$$
{(-1)^{m}\, q^{{m+1 \choose 2}}\,
(t;q)_{m}\,(q;q)_{m}\,(q^{-m}/t)^m
\over (q;q)_m\,
(q^{1-m}/t;q)_{m}}\ssp.
$$
\noindent 
The q-shifted factorial property, (I.8 in [3])
$$
(a\,q^{-n};q)_n=({q\over a};q)_n\,(-{a\over q})^n\,q^{-{n\choose 2}}
\ssp,
\eqno 2.10
$$
converts the term $(q^{1-m}/t;q)_{m}$ to $(t;q)_m\,(-{1\over t})^m\,
q^{-{m\choose 2}}$ 
beautifully cancelling all the terms proving that the expression 2.9 evaluates to 1 as desired!
This completes our proof of Theorem I.1.
\sas

\noindent{\bol Remark 2.1}\qquad
We should also mention that in the Sahi paper the condition
$G_\aaa(x;q,t)\ssp |_{x^\aaa}=1$ replaces I.3 b). To work under this alternate 
characterization, we need only check that the polynomials we are constructing satisfy
I.4. This presents no additional difficulty. For instance,
in the case of $G_{(m,0)}$, we only need to show that the coefficient of $x^m$ in $E_m$ does not vanish.
Now, the occurrence of $x^m$ in $(x;q)_k$ where $0\leq k\leq m$ is only possible when $k=m\ssp .$
However, the restriction that $j+k\leq m$ forces $j$ to be zero, thus the coefficient
of $x^m\,y^0$ can be computed exactly to be
$$
{(-1)^m\ssp 
q^{{m+1\choose 2}+{m\choose 2}}\ssp
t^{m} 
\over (q;q)_m \ssp}
\eqno 2.11
$$
\noindent
which is clearly non-zero!
\sap
Macdonald gives an explicit formula for the polynomial $P_\la$ when $\la$ is a one part partition.
Namely, (see eq. 4.9 p. 323 of [11]) he shows that
$$
P_{(m)}(x,y;q,t)\ses {(q;q)_m\over (t;q)_m}\sum_{|\mu|=m}
{(t;q)_{\mu}\over (q;q)_{\mu}}\ess m_{\mu}\ess .
\eqno 2.12
$$
Macdonald in [11] shows that the polynomial $P_\la$ (up to a scalar factor) can be obtained  by a Hecke
algebra symmetrization of his polynomial $E_\aaa$ whenever $\aaa$ is a composition
that rearranges to $\la$. 
Knop [5] and 
Sahi [14] show that $E_\aaa$ may be recovered from the top
of homogeneous component of their polynomials. Denoting the top component of $G_\aaa$ by
${G}^{top}$, we deduce that, whenever $\aaa$ rearranges to $\la$, we must have
$$
P_{\lambda}(x;q,t)\ssp \doteq  \ssp 
\sum_{\omega\in S_n}t^{-l(\omega)}\,T_{\omega}\,
{G}^{top}_{\aaa}(x;q,t)\ess .
\eqno 2.13
$$
Here again, we have used the ``$\doteq$'' sign since due to different normalizations used in this
paper from those adopted in [12], [14] and [5], equality holds true up to a scalar factor.
This given, we may use our formula 2.1 to obtain explicit expressions for the Macdonald polynomials. 
We shall  carry this out here in the two variable case when  $\aaa=(m,0)$. 
\sa

Taking the top component of the right hand side of 2.1 we get
$$
G^{top}_{(m,0)}(x,y;q,t)
\ses  
{(-1)^m\ssp q^{{m+1\choose 2}+{m \choose 2}} t^m\over (t;q)_{m+1}}\ess 
\sum_{j=0}^m\ess 
{q^j\ssp 
(t;q)_{j}\ssp  
(t;q)_{m+1-j}\ssp
\over     
(q;q)_{m-j} \ssp
(q;q)_j
}\ssp x^{m-j}\ssp y^j\ess .
$$
Two applications of formula 2.3 transforms this to
$$
G^{top}_{(m,0)}(x,y;q,t)
\ses  
{(-1)^m\ssp q^{{m+1\choose 2}+{m\choose 2}}\; t^m\over (q;q)_{m}}
\sum_{j=0}^m
{q^j\,  
(t;q)_{j}\ssp
(q^{-m};q)_j
\over     
t^{j}\,
(q;q)_j
\ssp (q^{-m}/t;q)_j
}\;
x^{m-j}\ssp y^j\ess .
\eqno 2.14
$$ 
\noindent
The two variable case of the operator $T_{s_i}$, defined in I.7, may be written as
$$
T_s\ses s\sps { (1-t) \over x-y}\ssp x \ssp (1-s)
$$
where $s$ denotes the transposition which exchanges $x$ and $y$. Setting
$$
S  \ses  T_{id}+{1\over t}\,T_{s} 
\ses 1+{1\over t}\,s+{(1-t)\over t}{x\over (x-y)}\,(1-s)
$$
2.13 implies that
$$
P_{(m)}\ssp \doteq \ssp S\ess {G}^{top}_{(m,0)}(x,y;q,t)\ess .
$$ 
It is interesting to see how this result can be directly derived from our
explicit formulas. This is the contents of our next result.
\sas

\noindent{\bol Theorem 2.1}
$$
P_{(m)}
\ses { (-1)^m\ssp (q;q)_m  \over  q^{{m+1\choose 2}+{m\choose 2}}\; t^{m-1}}
\ess {   (1- t q^m)\over (1-t^2 q^m) }\ess
S\ssp {G}^{top}_{(m,0)}(x,y;q,t)\ess . 
\eqno 2.15
$$
To verify  this identity we need to prove three auxiliary lemmas.

For convenience let us set
$$
\overline{G}^{top}_{(m,0)}(x,y;q,t)\ses { (-1)^m\ssp (q;q)_{m} \over  q^{{m+1\choose 2}+{m\choose 2}}\; t^{m-1}}
\ess {   (1-t q^m)\over (1-t^2 q^m) }\ess
{G}^{top}_{(m,0)}(x,y;q,t)
\eqno 2.16
$$
and note that from 2.14 and 2.16 we derive that
$$
\overline{G}^{top}_{(m,0)}(x,y;q,t)\ses \sum_{j=0}^m\ssp c_j\ssp x^{m-j} y^j
$$
with
$$ 
{c}_j \ses
{(1-tq^m)\over (1-t^2q^m)}
\ssp
{q^j\,
(t;q)_{j}\ssp
(q^{-m};q)_j
\over     
t^{j-1}\,
(q;q)_j
\ssp (q^{-m}/t;q)_j
}
\ess .
\eqno 2.17
$$

\noindent{\bol Lemma 2.1}
\sa
$$
S\;\overline{G}^{top}_{\aaa}(x,y;q,t) \ses
\sum_{a=0}^m\left(\,
({1\over t}-1)\, \sum_{j=0}^{a-1}\, 
(\bar{c}_j- {c}_{m-j})
\;+\;{ {c}_a\over t}\;+\; {c}_{m-a}\right)
\, x^{a}\,y^{m-a}
\eqno 2.18
$$
\sa
\noindent{\bol Proof}\qquad 
First examine the action of $S$ on an arbitrary monomial,
$x^a\,y^b$.  There are three separate cases to consider, $a<b$, $a>b$  and $a=b$.

\noindent Let $a>b$. Then  
$$
\eqalign
{
S\,x^a\,y^b & =
\Bigl(1+{1\over t}\,s+{(1-t)\over t}{x\over (x-y)}\,(1-s)
\Bigr)\,x^a\,y^b = 
{(x^{a+1}\,y^b-x^b\,y^{a+1})
+ t\,(x^{b+1}\,y^a-x^a\,y^{b+1})
\over t\,(x-y) }\cr
\cr
& = 
{1\over t}\,\sum_{r=0}^{a-b}
x^{a-r}\,y^{r+b}
\,-\,\sum_{r=1}^{a-b-1} x^{a-r}\,y^{r+b}\ses ({1\over t}-1)\ssp \sum_{r=1}^{a-b-1} x^{a-r}\,y^{r+b}
\sps { x^ay^b+x^by^a \over t}
}
$$

\noindent The two remaining cases are determined in a similar
fashion to obtain:
$$
S\,x^a\,y^b
\ses 
\cases
{
(1-{1\over t})\ssp f(b,a)\sps   \, x^ay^b\,+\, x^by^a  & $\ess\ess\ess a<b$\cr
\cr
({1\over t}-1)\ssp f(a,b)\ess +   {1\over t}\, x^ay^b+ {1\over t}\, x^by^a     & $\ess\ess\ess a>b$\cr
\cr
(1+{1\over t})\,x^a\,y^a & $\ess\ess\ess a=b$
}
$$
\noindent Where for convenience we have set
$$
f(a,b)\ses x^{a-1} y^{b+1}\sps \cdots \sps x^{b+1} y^{a-1}  \ess\ess\ess\ess {\rm for}\ess\ess\ess a>b \ess .
\eqno 2.19
$$
Thus, applying $S$ to 2.17 we get
$$
\eqalign{
S\ssp \overline{G}^{top}_{(m,0)}  = \sum_{\ess 0\ssp \leq\ssp  j\ssp < \ssp m/2} 
c_j\ssp\Bigl[  (\textstyle {1\over t}-1)\ssp& f(m-j,j)\ess + 
 \textstyle {1\over t}\, x^{m-j} y^j+ \textstyle {1\over t}\, x^j y^{m-j} \ssp \Bigr]\sps\cr
\sum_{\ess m/2\ssp <\ssp  j\ssp \leq \ssp m}&
c_j\ssp\Bigl[ (1-\textstyle {1\over t})\ssp f(j,m-j)\sps   x^{m-j} y^j+x^j y^{m-j} \ssp \Bigr]\sps\cr
&\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess
\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess
c_{m/2}\ssp (1+\textstyle {1\over t})\ssp  x^{m/2}y^{m/2}\ess .\cr
}
$$
with the convention to set $c_{m/2}=0$ when $m$ is not even.

Splitting the two sums and making the change of variables $j\rightarrow m-j$ in the
first portion of the  second sum, we can regroup the resulting terms and obtain
$$
\eqalign{
S\ssp \overline{G}^{top}_{(m,0)}  & = \hskip -.1truein\sum_{\ess 0\ssp \leq\ssp  j\ssp < \ssp m/2} 
 (c_j-c_{m-j})\ssp  ({\textstyle {1\over t}}-1)\ssp f(m-j,j)\sps\ess \cr
&\ess\ess\ess 
\sum_{\ess 0\ssp \leq\ssp  j\ssp < \ssp m/2} 
({\textstyle {1\over t}}\, c_j   + c_{m-j})
\ssp    x^{m-j} y^j   \sps \hskip -.1truein
\sum_{\ess m/2\ssp <\ssp  j\ssp \leq \ssp m} 
(c_j+ {\textstyle {1\over t}}\, c_{m-j})\, x^j y^{m-j} \ess\ess\ess\ess\ess\ess\ess\ess \ess\ess {\rm 2.20}\cr
&\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess
\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess
\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess
\ess\ess\ess 
\sps c_{m/2}\ssp (1+\textstyle {1\over t})\ssp  x^{m/2}y^{m/2}\ess .\cr
}
$$
Note now that 2.19 for $a=m-j$ and $b=j$ gives
$$
f(m-j,j)= \ssp x^{m-j-1}y^{j+1}+ \cdots +  x^{j+1}y^{m-j-1}
\ses \sum_{a=0}^m\ssp x^{m-a}y^a \ess \chi \bigl(j+1\leq a\leq m-j-1\bigr)\ess .
$$
Substituting this in the first sum on the right-hand side of 2.20 and, for a moment,
calling the result $S_1$ we get
$$
\eqalign{
S_1 &\ses \sum_{a=0}^m\ssp x^{m-a}y^a \sum_{0\leq j< \, m/2}\ssp (c_j-c_{m-j})(\, {\textstyle {1\over t}}-1)
\ess \chi \bigl(j+1\leq a\leq m-j-1\bigr)\cr
&
\ses \sum_{0\leq a\leq \, m/2}\ssp x^{m-a}y^a \sum_{0\leq j\leq a-1}\ssp (c_j-c_{m-j})(\, {\textstyle {1\over t}}-1)
\cr
&
\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess
\sps\sum_{m/2\, <a\leq m}\ssp x^{m-a}y^a \sum_{0\leq j\leq m- a-1}\ssp (c_j-c_{m-j})(\, {\textstyle {1\over t}}-1)
\ess .}
$$
Substituting this in 2.20 and grouping terms we may write
$$
\eqalign{
S\, \overline G_\aaa^{top} 
& = \sum_{0\leq a\leq \, m/2}\ssp x^{m-a}y^a 
\Bigl(\sum_{0\leq j\leq a-1}\ssp (c_j-c_{m-j})(\, {\textstyle {1\over t}}-1)\sps 
{\textstyle {1\over t}}\, c_a + c_{m-a}\Bigr)\cr 
&\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess
\sps \sum_{m/2\, <a\leq\,  m}\ssp x^{m-a}y^a 
\Bigl(\sum_{0\leq j\leq m-a-1}\ssp (c_j-c_{m-j})(\, {\textstyle {1\over t}}-1)\sps 
 c_a +{\textstyle {1\over t}}\, c_{m-a}\Bigr)\ess ,\cr
}
$$
and this gives 2.18 since for $a>{m\over 2}$ we do have
$$
\sum_{0\leq j\leq\,  m-a-1}\ssp (c_j-c_{m-j})(\, {\textstyle {1\over t}}-1)\ess +\ssp
 c_a +{\textstyle {1\over t}}\, c_{m-a}
=
\sum_{0\leq j\leq a-1}\ssp (c_j-c_{m-j})(\, {\textstyle {1\over t}}-1)\ess +\ssp 
{\textstyle {1\over t}}\, c_a + c_{m-a}
$$
This completes the proof of lemma 2.1.
\sa
\noindent{\bol Lemma 2.2}
$$ {c}_j- {c}_{m-j}={(1-tq^m)\,(t;q)_j\,(q^{-m};q)_j\,(1-q^{-m+2j})
\over t^{j}\, (1-t^2q^m)\,(q;q)_j\,(q^{-m}/t;q)_{j+1}}
\eqno 2.21
$$
\sa
\noindent{\bol Proof}\qquad 
The definition in 2.17 directly implies
$$ {c}_j- {c}_{m-j}\ses {(1-tq^m)\over (1-t^2q^m)}
\left(
{q^j\, (t;q)_j\,(q^{-m};q)_j
\over t^{j-1}\,  (q;q)_j\,(q^{-m}/t;q)_{j}}
\;-\;
{q^{m-j}\, (t;q)_{m-j}\,(q^{-m};q)_{m-j}
\over t^{m-j-1}\,  (q;q)_{m-j}\,(q^{-m}/t;q)_{m-j}}
\right)\ssp.
\eqno 2.22
$$

\noindent Application of the q-shifted factorial property 2.3 and
$$ 
(aq^{-n};q)_{n-k}\ses {(q/a;q)_n\over (q/a;q)_k}
\left({a\over q}\right)^{n-k}q^{{k\choose 2}-{n\choose 2}}
\eqno 2.23
$$
allow the expression in 2.22 to take the form
$$
\eqalign
{
{c}_j &-  {c}_{m-j}\ses \cr\cr
& =  { 1-tq^m  \over  1-t^2q^m }
\left(
{
q^j\,(t;q)_j\,(q^{-m};q)_j
\over t^{j-1}\, (q;q)_j\,(q^{-m}/t;q)_{j}
} 
\sms {q^{-{m\choose 2}}\,(t;q)_m\,(q;q)_m\,(qt;q)_{j}\,(q^{-m};q)_{j}\,
\over 
t^{m+j-1}\,(q;q)_{m}\, (q^{-m}/t;q)_{m} (q;q)_{j}\,(q^{-m+1}/t;q)_{j}}
\right)
\cr 
\cr
& ={(1-tq^m q^j)\,\ssp (t;q)_j\,(q^{-m};q)_j
\over(1-t^2q^m )\,  t^{j-1}\,(q;q)_j\,(q^{-m}/t;q)_{j+1}}
\left((1-q^{-m+j}/t)\;-\;
{(t;q)_m\,(1-tq^j)\,(1-q^{-m}/t)\,(-1)^m
\over q^{{m\choose 2}+j}\,t^{m}\, (q^{-m}/t;q)_{m}\,
(1-t)}\right)\ssp .
}
$$
Using the  transformations
$$
(t;q)_m\,(1-q^{-m}/t)\ses{-q^{-m}\,(t;q)_{m+1}\over t}
$$
$$
(q^{-m}/t;q)_m\,(1-t)\ses(-1)^m\,{(t;q)_{m+1}\over t^m\,q^{m+1\choose 2}}
$$
we obtain
$$
\eqalign
{
 {c}_j- {c}_{m-j}
& \ses
{q^j\,(1-tq^m)\,(t;q)_j\,(q^{-m};q)_j
\over t^{j-1}\,(1-t^2q^m)\,(q;q)_j\,(q^{-m}/t;q)_{j+1}}
\left((1-q^{-m+j}/t)\;+\; {q^{-j}\,(1-tq^j)\over t}
\right)\cr
\cr
& \ses
{(1-tq^m)\,(t;q)_j\,(q^{-m};q)_j\,(1-q^{-m+2j})
\over t^{j}\,(1-t^2q^m)\,(q;q)_j\,(q^{-m}/t;q)_{j+1}}\ssp.
}
$$
This completes the proof of Lemma 2.2. 
\sa
Now, in the two variable case, the Macdonald polynomial given in 2.12  
reduces to 
$$ 
\eqalign
{
P_{(m)}(x,y;q,t)
& \ses
{(q;q)_{m}
\over (t;q)_m}
\sum_{a=0}^m
{(t;q)_{m-a}\,(t;q)_{a}
\over (q;q)_{m-a}\,(q;q)_{a}}\;x^{m-a}y^{a}\cr
\cr
&\ses
\sum_{a=0}^m\,
{q^a\,(t;q)_a\,(q^{-m};q)_a\over t^a\,(q^{1-m}/t;q)_a
\,(q;q)_a}\;x^{m-a}\,y^a
\ssp .
}
\eqno 2.24
$$
Denoting the coefficient of $x^{m-a}\,y^a$ by $m_a$, that is
$$
m_a\ses
{q^a\,(t;q)_a\,(q^{-m};q)_a\over t^a\,(q^{1-m}/t;q)_a
\,(q;q)_a}\ssp ,
\eqno 2.25
$$
we see that 
Lemma 2.1 immediately allows the identity in 2.15 to be expressed in the
following manner:
$$
\sum_{a=0}^m\;m_a\;x^{m-a}\,y^a
\ses
\sum_{a=0}^m\left(\,
({1\over t}-1)\, \sum_{j=0}^{a-1}\, 
( {c}_j- {c}_{m-j})
\;+\;{ {c}_a\over t}\;+\; {c}_{m-a}\right)
\; x^{m-a}\,y^{a}
\ssp .
$$
Thus to prove Theorem 2.1 we need only verify that
$$
m_a \;-\;{ {c}_a\over t}\;-\; {c}_{m-a}
\ses
({1\over t}-1)\, \sum_{j=0}^{a-1}\, 
( {c}_j- {c}_{m-j})
\ssp.
\eqno 2.26
$$

\sa
\noindent{\bol Lemma 2.3} \qquad
{Setting  $R_a= m_{a}\, -\, {c}_a/t\,-\, {c}_{m-a}$\ita we have } 
$$
R_a \ses 
{(1-tq^m)\over
(1-t^2q^m)}\ssp 
{(t;q)_a (q^{-m};q)_a\over t^a (q^{-m}/t;q)_a (q;q)_{a-1}}
$$
\sa
\noindent{\bol Proof}\qquad
Definitions 2.17 and 2.25 directly give
$$ 
\eqalign{
R_a &\ses
{q^a\,(q^{-m};q)_a\,(t;q)_a\over
t^a\,(q^{1-m}/t;q)_a\,(q;q)_a}\cr
&\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess
\sms {q^a\,(1-tq^m)\,(q^{-m};q)_a\,(t;q)_a\over
t^a\,(q^{-m}/t;q)_a\,(q;q)_a\,(1-t^2q^m)}
\cr
&\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess
\sms {t\,q^{m-a}\,(1-tq^m)\,(q^{-m};q)_{m-a}\,(t;q)_{m-a}\over
t^{m-a}\,(q^{-m}/t;q)_{m-a}\,(q;q)_{m-a}\,(1-t^2q^m)}
\ess .
\cr
}
\eqno 2.27
$$
Transforming the third term by means of 
formula 2.3 and 2.23, we get
$$
{t\,q^{m-a}\,(1-tq^m)\,(q^{-m};q)_{m-a}\,(t;q)_a\over
t^{m-a}\,(q^{-m}/t;q)_{m-a}\,(q;q)_{m-a}\,(1-t^2q^m)}
\ses 
{(q^{-m};q)_a\,(tq;q)_a (1-tq^m)\,(t;q)_m\,\over
t^{m+a}\,(q^{-m};q)_m\,(q^{1-m}/t;q)_a\,(q;q)_a\,(1-t^2q^m)}
\ess  .
$$
This allows the expression in the right-hand side of 2.27 to be reduced to the form:
$$
\eqalign{
{(1-tq^m)\over (1-t^2q^m)}\ess & {\,(t;q)_a\,(q^{-m};q)_a\over
 t^a\,(q^{-m}/t;q)_a\,(q;q)_{a-1}}\ess \times \ess\cr \cr
&\left(
{(1-t^2q^m)\,(1-q^{-m}/t)\,q^a\over (1-tq^m)\,(1-q^{-m+a}/t)\,(1-q^a)}
-{q^a\over (1-q^a)}-{t\,(1-tq^a)\,(1-q^{-m}/t)\over q^{-m}\,
(1-tq^m)\,(1-q^a) \,(1-q^{-m+a}/t)}\right)
\ssp.
\cr
}
$$
\noindent The expression in the parentheses simplifies to $1$
proving Lemma 2.3. 
\sa

\noindent{\bol Proof of Theorem 2.1}\qquad
Combining Lemma 2.2 and Lemma 2.3 reduces 2.26 to the form
$$
{(1-tq^m)\,(t;q)_a\,(q^{-m};q)_a\over
(1-t^2q^m)\,t^a\,(q;q)_{a-1}\, (q^{-m}/t;q)_a}
\ses
{(1/t-1)\,(1-tq^m)\over (1-t^2q^m)}
\sum_{j=0}^{a-1}\,{(t;q)_j\,(q^{-m};q)_j\,
(1-q^{-m+2j})\over
t^{j}\,(q;q)_j\,(q^{-m}/t;q)_{j+1}}
$$
Thus we are left to verify that
$$
{(t;q)_a\,(q^{-m};q)_a\over
t^a\,(q;q)_{a-1}\, (q^{-m}/t;q)_a}
\ses
{(1-t)\over t}\ssp 
\sum_{j=0}^{a-1}\,{(t;q)_j\,(q^{-m};q)_j\,
(1-q^{-m+2j})\over
t^{j}\,(q;q)_j\,(q^{-m}/t;q)_{j+1}}\ess .
$$
However this is immediate since for all $j>0$ we have
$$
{(1-t)\over t}\ssp 
{(t;q)_j\,(q^{-m};q)_j\,
(1-q^{-m+2j})\over
t^{j}\,(q;q)_j\,(q^{-m}/t;q)_{j+1}}
\ses
{(t;q)_{j+1}\,(q^{-m};q)_{j+1}\over
t^{j+1}\,(q;q)_j\, (q^{-m}/t;q)_{j+1}}\sms 
{(t;q)_j\,(q^{-m};q)_j\over
t^j\,(q;q)_{j-1}\, (q^{-m}/t;q)_j}\ess ,
$$
and moreover this relation remains true even for $j=0$ provided we set $(q;q)_{-1}=\infty\ssp .$
This completes our proof.
\sap


\noindent{\bol 3. The matrix of the Knop operator $\Xi_1$ and a $\ _6\Phi_5$ summation formula.}
\sas

Our original proof of Theorem I.1 was based on the characterization of 
the polynomials $G_\aaa$ as eigenfunctions of the Knop-operators $\Xi_i$.  
This approach required the construction of explicit formulas for the entries of the
matrix expressing the action of $\Xi_1$ on the basis $\{(x;q)_k (y;q)_l\}_{k,l}$.
The computations and identities that result  from this proof turn out to
be quite interesting in themselves. In particular they reveal an intimate connection 
between the Knop-Sahi 
polynomials and some of the deeper identities of the theory of
basic hypergeometric series. In this section we shall give an outline of this alternate proof
focussing on the salient features and omitting some of the more laborious details.
The reader is referred to [13] for the complete treatment.
\sas

Our point of departure is the following simple observation.
\sas

\vbox{
\noindent{\bol Proposition 3.1}\qquad
{\ita If a polynomial $P(x,y;q,t)$ is of degree $m$ and satisfies $\Xi_1\ssp P(x,y;q,t)= q^m t \ssp P(x,y;q,t)$
then it is necessarily a multiple of $G_{(m,0)}(x,y;q,t)$.}
\sas
}

\def \Ex {K_m(x,y;q,t)}
\noindent{\bol Proof}\qquad
Formula I.2 gives that $\oaa_1=q^m t$ 
if and only if $m=\aaa_1\geq \aaa_2$. Thus 
from Property I.4 we derive that $\Xi_1\ssp G_\aaa = q^m t \ssp G_\aaa$ if and only if 
$\aaa=(m,i) $ for some $0\leq i\leq m$. Thus the elements,
$
\{G_{(m,0)},G_{(m,1)},\ldots,G_{(m,m)}\},
$
form a basis for the $q^mt$-eigenspace of $\Xi_1$. In particular, this gives that our
polynomial $P(x,y;q,t)$ must have the expansion
$$
P(x,y;q,t)\ses \sum_{i=0}^m d_i\, G_{(m,i)}(x,y;q,t)\ess .
\eqno 3.1
$$
Note now that the term $x^my^m$ must occur in $G_{(m,m)}$ and at the same time it cannot occur
anywhere else in 3.1 since all the other polynomials (including $P$) have degree strictly less than $2m$.
This forces $d_m=0$. Similar reasoning recursively applied yields $d_i=0$ for all $i\geq 1$.
Thus  $P(x,y;q,t)\ses d_0\ssp G_{(m,0)}(x,y;q,t)$ as asserted.
\sas

Note that since the polynomial $ K_m(x,y;q,t)$ given in 2.1 is clearly of degree $m$ 
and, as we have seen in section 1, it statisfies the normalization $K_m((\overline{m,0});q,t)=1$,
Proposition 3.1 reduces the proof of Theorem I.1 to showing that
$$
\ssp\Xi_1K_m(x,y;q,t) =q^mt\;K_m(x,y;q,t)\ess .
\eqno 3.2
$$
\sas

\noindent
To this end let us set 
$$
\Ex=\sum_{k,l}z_k\ssp w_l\,C_m^{(k,l)}\ess\ess {\rm and} \ssp \ess
\Xi_1\ssp z_k\ssp w_l= \sum_{a,b}\ssp z_a\ssp w_b \ssp M_{(a,b;k,l)}
$$
where for simplicity we let $z_k=(x;q)_k$ and $w_l=(y;q)_l\ess $. Substituting in 3.2 and
equating coefficients of $z_a\ssp w_b$ we derive that 3.2 holds if and only if we have 
$$
\sum_{k,l}M_{(a,b;k,l)}\,C_m^{(k,l)}
\ses q^mt\;C_m^{(a,b)}\ess .
\eqno 3.3
$$
\sas
Now it develops that the coefficients $M_{(a,b;k,l)}$ may be given the following explicit expressions:

\vbox{
\noindent{\bol Theorem 3.1}\qquad
{\ita For $a+b\leq k+l\leq m$ we have 
$$
M_{(a,b;k,l)}=
\cases
   {
\cases{t\ssp q^a & if $a\geq b$ \cr\cr q^a & if  $a< b$ \cr}
& for $\ess k=a$ and $\ess l=b$\cr
\cr
        \cr
    {(t-1)\,q^{l^2-la-lb+ab-l+a+b}\,
    (q;q)_{k-b}\,(q;q)_{k-a} \over
    (q;q)_{k+l-a-b}\,(q;q)_{a-l}\,(q;q)_{b-l}} 
& for $k>a$ and $\ess l\leq b\ess $ when $\ess k>l$  \cr
        \cr
        {(1-t)\,q^{k^2+k-ka-kb+ab}\,
        (q;q)_{l-b-1}\,(q;q)_{l-a-1} \over 
        (q;q)_{k+l-a-b}\,(q;q)_{a-k-1}\,(q;q)_{b-k-1}}
& for $l>a$ and $\ess k< b\ess $ when $\ess k<l$  \cr
    } 
$$
and $\ess M_{(a,b;k,l)}=0$ otherwise.}
}
\sa

In view of 3.3 and the preceeding observations, 
the proof of Theorem I.1 by this approach reduces to the following identity:
\sas

\noindent{\bol Proposition 3.2}
$$
q^m t \ssp C_m^{(a,b)}
\ses C_m^{(a,b)} \times \left\{\matrix{ q^{a}\,t\ssp  & \hbox{if $a\geq b$} \cr q^a\ssp & \hbox{ if $a<b$} \cr}\right\}
 +
        \sum_{\multi{
                k>l \cr
                k>a \ssp \&\ssp l\leq b \cr
                a+b\leq l+k\leq m}}
            m_{k,l}^{(1)}\,C_m^{(k,l)}
      + \sum_{\multi{   
                k<l \cr
                l>a \ssp \&\ssp k< b \cr
                a+b\leq l+k\leq m}} 
        m_{k,l}^{(2)}\,C_m^{(k,l)}
\eqno 3.4
$$
where 
$$
\eqalign{
& a)\ess\ess\ess  m_{k,l}^{(1)} =  {(t-1)\,(q;q)_{k-b}\,(q;q)_{k-a}\,
                q^{l^2-al-bl-l+ab+a+b} \over
                (q;q)_{b-l}\,(q;q)_{a-l}\,(q;q)_{l+k-a-b}}\ess ,\cr
\cr
& b)\ess\ess\ess m_{k,l}^{(2)} =  {(1-t)\,(q;q)_{l-a-1}\,(q;q)_{l-b-1}\,
                q^{k^2+k-ak+ab-bk} \over
                (q;q)_{b-k-1}\,(q;q)_{a-k-1}\,(q;q)_{l+k-a-b}}
\ess .
}
\eqno 3.5
$$
\sas

The expressions for the entries $M_{(a,b;k.l)}$ of the matrix of the operator $\Xi_1$ 
given by Theorem 3.1 are an immediate consequences of the following
\sas

\noindent{\bol Proposition 3.3}
$$
\Xi_1\;z_k\,w_l =
\cases
{
t\,q^k\,z_k\,w_l
+(t-1)\sum_{\multi{l\leq a\leq k-1 \cr 
                l\leq b\leq k+l-a }}
{q^{l^2-la-bl+ab-l+a+b}
(q;q)_{k-b}\,(q;q)_{k-a} \over
(q;q)_{k+l-a-b}\,(q;q)_{a-l}\,(q;q)_{b-l}}\,
z_a\,w_b & if $k\geq l$ \cr
\cr
q^k\,z_k\,w_l
+(1-t)\sum_{\multi{k+1\leq a\leq l-1\cr k+1\leq b\leq l+k-a}}
{q^{k^2+k-ak-kb+ab}
(q;q)_{l-b-1}\,(q;q)_{l-a-1} \over 
(q;q)_{k+l-a-b}\,(q;q)_{a-k-1}\,(q;q)_{b-k-1}}\,
z_a\,w_b & if $k<l$ \cr
}
\eqno 3.6
$$
The proof of this proposition depends primarily on the following
cute identity.
\sa
\noindent{\bol Theorem 3.2}
$$
{(x;q)_n-(y;q)_n\over
x-y}=\sum_{0\leq a+b \leq n-1}-q^{ab+a+b}\,
{(q;q)_{n-b-1}\,(q;q)_{n-a-1}\over
(q;q)_{n-a-b-1}\,(q;q)_{a}\,(q;q)_b}\,
(x;q)_a\,(y;q)_b 
$$

\noindent{\bol Proof}\qquad 
Express both sides in powers of $x$ and $y$.
$$
\eqalign{
\sum_{0<k\leq n}{(q^{-n};q)_k\,q^{n\,k}\over (q;q)_k}\,\left[
        {x^k-y^k\over x-y} \right]
 = &
\sum_{0\leq a+b \leq n-1}
\Big({-q^{ab+a+b}\,
(q;q)_{n-b-1}\,(q;q)_{n-a-1}\over
(q;q)_{n-a-b-1}\,(q;q)_{a}\,(q;q)_b}\ess \times \cr
&\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess
 \times
\sum_{i=0}^a{(q^{-a};q)_i\,(x\,q^{a})^i\over
                (q;q)_i}
\sum_{j=0}^b{(q^{-b};q)_j\,(y\,q^{b})^j\over
                (q;q)_j}\Big)
}
$$
Expand the left hand side.
$$
\eqalign{
\sum_{k<n}{(q^{-n};q)_k\,q^{n\,k}\over (q;q)_k}\,
        \sum_{0\leq l\leq k-1} x^l\,y^{k-1-l} 
\ses & 
\sum_{0\leq a+b\leq n-1}
\Big({-q^{ab+a+b}\,
(q;q)_{n-b-1}\,(q;q)_{n-a-1}\over
(q;q)_{n-a-b-1}\,(q;q)_{a}\,(q;q)_b}\ess \times \cr
&\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess \times
\sum_{i=0}^a \sum_{j=0}^b
{q^{a\,i+b\,j}\,
(q^{-a};q)_i\,
(q^{-b};q)_j
\,x^i\,y^j\,
\over 
(q;q)_i\, (q;q)_j}\Big)
}
$$
Take the coefficient of $x^ry^s$ on both sides, where
$r,s \geq 0$. 
$$
{(q^{-n};q)_{1+r+s}\,q^{n\,(r+s+1)}\over (q;q)_{r+s+1}}
=
\sum_{\multi{0\leq a+b \leq n-1\cr
        r\leq a\cr
        s\leq b}}
-q^{ab+a+b+a\,r+b\,s}\, {(q;q)_{n-b-1}\,(q;q)_{n-a-1}
(q^{-a};q)_r\,(q^{-b};q)_s
\over
(q;q)_{n-a-b-1}\,(q;q)_{a}\,(q;q)_b\,(q;q)_r\,(q;q)_s}
$$
Transforming the factors
$(q;q)_{n-a-1}$ and $(q;q)_{n-a-b-1}$ using 2.3,  and $(q^{-a};q)_r$ and $(q^{-b};q)_s$ using 2.10,
reduces this equality to a form
in which the factors depending on $n$ can be removed from the
summand, and the dependence on $a$ and $b$ will appear only in the 
indices of the q-shifted factorials.
$$
{ q^{n(r+s+1)}
(q;q)_r\,(q;q)_s\,(q^{-n};q)_{1+r+s}\,
\over (-1)^{r+s}\,
q^{{r\choose 2}+{s\choose 2}}\, (q;q)_{n-1}\,(q;q)_{r+s+1}}
=
\sum_{r\leq a \leq n-1\atop
        s\leq b\leq n-1-a}
{-q^{a+b}\,
(q^{-n+1};q)_{a+b}\,
\over
(q^{-n+1};q)_a\,(q^{-n+1};q)_b\,
(q;q)_{a-r}\,(q;q)_{b-s}}
$$
We now make the change of variables  
$a\rightarrow a+r$ and $b\rightarrow b+s$, obtaining
$$
{ q^{n(r+s+1)}
(q;q)_r\,(q;q)_s\,(q^{-n};q)_{1+r+s}\,
\over 
(-1)^{r+s}\,
q^{{r\choose 2}+{s\choose 2}}\,
 (q;q)_{n-1}\,(q;q)_{r+s+1}}
=
\sum_{r\leq a+r \leq n-1\atop
        s\leq b+s\leq n-1-a-r}
{-q^{a+r+s+b}\,
(q^{-n+1};q)_{a+b+r+s}\,
\over
(q^{-n+1};q)_{r+a}\,(q^{-n+1};q)_{s+b}\,
(q;q)_{a}\,(q;q)_{b}}
$$
The q-shifted factorial property,
$$
(a;q)_{k+n}=(a;q)_k\,(aq^k;q)_n
\eqno 3.7
$$ 
modifies the summands to consist of factors indexed by
only one of the variables. This allows
one variable to be considered fixed while summing over the other.
$$
\eqalign
{&
{q^{n(r+s+1)}
\,(q^{-n+1};q)_r\,(q^{-n+1};q)_s\,(q;q)_r\,(q;q)_s\,(q^{-n};q)_{1+r+s}\,
\over 
(-1)^{r+s+1}\,
q^{r+s+{r\choose 2}+{s\choose 2}}\,(q;q)_{n-1}\,(q;q)_{r+s+1}\,
(q^{-n+1};q)_{r+s}}
\ses 
\cr
\cr
& \ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess \ses   
\sum_{a=0}^{n-1-r} 
q^{a}\, {(q^{-n+1+r+s};q)_{a}\,
\over
(q^{-n+1+r};q)_a\,(q;q)_{a}}
\sum_{b=0}^{n-1-a-s-r}
q^{b}\, {(q^{-n+1+r+s+a};q)_b
\over (q^{-n+1+s};q)_b\,(q;q)_{b}}
}
\eqno 3.8
$$

This places us in a position to use the summation identity
$$
_2\phi_1(0,q^{-n};c;q,q) \ses {(-1)^n\,c^n\,q^{n\choose 2} \over (c;q)_n}\ess ,
\eqno 3.9
$$
which is the particular case of 2.6 obtained by making the replacement 
$$
{(c/a;q)_n\over (c;q)_n}\,a^n
\ses (-1)^n\,c^n\,q^{n\choose 2}\,
{(a\,q^{-n+1}/c;q)_n\over (c;q)_n}\ssp,
$$
an letting $a\rightarrow 0$.
\sas

Applying 3.9  with $n=n-1-r-s-a$ and $c=q^{-n+1+s}$, the double sum in 3.8 reduces to a 
single sum and 3.8 becomes:
$$
\eqalign{
&{(-1)^{r+s+1}\,q^{n(r+s+1)}
\,(q^{-n+1};q)_r\,(q^{-n+1};q)_s\,(q;q)_r\,(q;q)_s\,(q^{-n};q)_{1+r+s}\,
\over q^{r+s+{r\choose 2}+{s\choose 2}}\,(q;q)_{n-1}\,(q;q)_{r+s+1}\,
(q^{-n+1};q)_{r+s}} \ses\cr\cr
&
=
\sum_{a=0}^{n-1-r}
q^{a}\, {(q^{-n+1+r+s};q)_{a}\,
\over
(q^{-n+1+r};q)_a\,(q;q)_{a}}
{(-1)^{n-1-r-s-a}\,(q^{-n+1+s})^{n-1-r-s-a}\,q^{n-1-r-s-a\choose 2}
\over (q^{-n+1+s};q)_{n-1-r-s-a}}
}
\eqno 3.10
$$

\vbox{
\noindent
Applying 2.3 to $(q^{-n+1+s};q)_{n-1-r-s-a}$ transforms 3.10 to
$$
{(-1)^{n}\,q^{n(r+s+1)}
\,(q^{-n+1+s};q)_{n-1-r-s}\,
(q^{-n+1};q)_r\,(q^{-n+1};q)_s\,(q;q)_r\,(q;q)_s\,(q^{-n};q)_{1+r+s}\,
\over q^{r+s+{r\choose 2}+{s\choose 2}}\,(q;q)_{n-1}\,(q;q)_{r+s+1}\,
(q^{-n+1};q)_{r+s}}
$$
$$
\ses
\sum_{1\leq a \leq n-1-r}
q^{a}\, {(q^{-n+1+r+s};q)_{a}\,(q^{1+r};q)_a
\over
(q^{-n+1+r};q)_a\,(q;q)_{a}}\ssp.
\eqno 3.11
$$
}
Let $c=q^{1+r-n}$, $n=n-1-r-s$ and $a=q^{1+r}$ to permit the application of
identity 2.6.  The sum is thus eliminated and 3.11 reduces to 
$$
{(-1)^{n} \,q^{n^2+n\,r-r-r^2-{n\choose 2}}
\,(q^{-n+1+s};q)_{n-1-r-s}\,
(q^{-n+1};q)_r\,(q^{-n+1};q)_s\,(q;q)_r\,(q;q)_s\,(q^{-n};q)_{1+r+s}\,
\over (q;q)_{n-1}\,(q;q)_{r+s+1}\,(q^{-n+1};q)_{r+s}}
$$
$$
=
{(q^{-n};q)_{n-1-r-s}\,
(q^{1+r})^{n-1-r-s}
\over
(q^{-n+1+r};q)_{n-1-r-s} 
}
$$
This equality can be easily be seen to hold, proving the theorem. 
\sa
\noindent {\bol Proof of Proposition 3.3}\qquad
Note that we may express the 
action of $\Xi_1$ in terms of $T_{s_1}$ and $\phi$ in the
form
$$
\Xi_1\ses t\,T_{s_1}^{-1}\,( \phi\,T_{s_1}+1/y-1/y\,\phi\,T_{s_1})\,T_{s_1}^{-1}
\ssp.
$$
This can be seen by combining 1.6 with $i=1$ with the definition I.15 for $n=i=2$
and $x_2=y$. 

\noindent Thus, application of $\Xi_1$ to $z_kw_l$ immediately 
produces 
$$
\Xi_1\,z_k\ssp w_l\ses t\,T_{s_1}^{-1}\,\phi \,
z_k\ssp w_l\,
+ t\,T_{s_1}^{-1}\, 1/y\,T_{s_1}^{-1}\,
z_k\ssp w_l\,
-t\,T_{s_1}^{-1}\,1/y\,\phi\,
z_k\ssp w_l\ssp.
$$
The definition of $\phi$ and 1.5 b) yields
$$
\Xi_1\,(x;q)_k\,(y;q)_l\ses 
{1\over x}(x;q)_k\,(y;q)_l\,+\,
{x-1\over x}{(t\,x-y)\over (x-y)}\,(qx;q)_k\,(y;q)_l
\,+\,
{(y-1) (1-t)\over (x-y)}\,(qy;q)_k\,(x;q)_l\ssp.
$$
Manipulation gives 
$$
\Xi_1\,(x;q)_k\,(y;q)_l
\ses 
{q^k}\,(x;q)_k\,(y;q)_l\,+\,
{(1-t)\over x-y}
[(x;q)_{k+1}\,(y;q)_l\,-\,
(y;q)_{k+1}\,(x;q)_l]
\ssp.
$$
This reduces to two cases;
$$
\Xi_1\,z_k\,w_l\ses
\cases
{
q^k\,z_k\,w_l +{(1-t)\over (x-y)}[(x\,q^l;q)_{k-l+1}-(y\,q^l;q)_{k-l+1}]
 z_l\,w_l & if $k\geq l$ \cr
\cr
q^k\,z_k\,w_l +{(1-t)\over (x-y)}[(y\,q^{k+1};q)_{l-k-1}-(x\,q^{k+1};q)_{l-k-1}]
 z_{k+1}\,w_{k+1} & if $k<l$\cr
}
$$
This given, two applications of Theorem 3.2 yield  that for $k\geq l$ we have
$$
\Xi_1\,z_kw_l\ses
q^k\,z_k\,w_l \,+\,(t-1)
\sum_{0\leq a\leq k-l\atop 0\leq b\leq k-l-a}
{q^{ab+a+l+b}(q;q)_{k-b-l}\,(q;q)_{k-l-a}\over
(q;q)_{k-l-a-b}\,(q;q)_k\,(q;q)_l}\,
z_{l+a}\,w_{l+b}\ssp,
$$
and for $k<l$,
$$
\Xi_1\,z_kw_l\ses
q^k\,z_k\,w_l\,-\,(t-1)
\sum_{0\leq a\leq l-k-2\atop 0\leq b\leq l-k-2-a}{q^{ab+b+k+a+1}
(q;q)_{l-k-b-2}\,(q;q)_{l-k-a-2}\over
(q;q)_{l-k-a-b-2}\,(q;q)_a\,(q;q)_b}\,
z_{k+1+a}\,w_{k+1+b} \ssp.
$$
A change of variables allows the action to be described
in terms of basis elements indexed only by $a$ and $b$.
For $k\geq l$, make the change of variables, $a\rightarrow a-l$ and
$b\rightarrow b-l$, and for $k<l$, let $a\rightarrow a-k-1$
and $b\rightarrow b-k-1$.
$$
\Xi_1\,z_kw_l\ses
\cases
{
q^k\,z_k\,w_l \,+\,(t-1)
\sum_{l\leq a\leq k\atop l\leq b\leq k+l-a}
{q^{ab-al-lb+l^2+a-l+b}\,
(q;q)_{k-b}\,(q;q)_{k-a}\over
(q;q)_{k+l-a-b}\,(q;q)_{a-l}\,(q;q)_{b-l}}\,
z_{a}\,w_{b} & for $k\geq l$\cr 
\cr
\cr
q^k\,z_k\,w_l\,-\,(t-1)
\sum_{k+1\leq a\leq l-1\atop
     k+1\leq b\leq l+k-a}
{q^{ab-ak-kb+k^2+k}\,
(q;q)_{l-b-1}\,(q;q)_{l-a-1}\over
(q;q)_{k+l-a-b}\,(q;q)_{a-k-1}\,(q;q)_{b-k-1}}\,
z_{a}\,w_{b} & for $k<l$\cr
}
$$
Subtraction of the $k=a$ term from the sum in the first case yields the equalities in 3.6 and completes
the proof of the proposition.
\sa
\noindent {\bol Proof of Proposition 3.2}\qquad
Only the case $a\geq b$ will be given here.  The case of $a<b$ can be seen in a similar manner.
The variable changes $k\rightarrow a+k\ess ,$$\ess l\rightarrow b-l$ in the first sum of 3.4,
and $k\rightarrow b-l\scs l\rightarrow a+k$ in the second sum reduces the problem to showing 
$$
(q^m\,t-q^{a}\,t)\,C_m^{(a,b)} =
 \sum_{\multi{ a+k>b-l\cr 
     k>0\ssp \& \ssp l\geq 0\cr
                l\leq k \cr
        k-l+a+b\leq m}}
         m_{(k+a,b-l)}^{(1)}\, C_m^{(k+a,b-l)}\,+
        \sum_{\multi{ a+k>b-l\cr
        k>0\ssp \& \ssp l>0\cr
        l\leq k\ssp \cr
        k-l+a+b\leq m}}
         m_{(b-l,k+a)}^{(2)}\,C_m^{(b-l,k+a)}
\ssp.
$$
Denote the right hand side by $R_m$ and   
substitute our explicit expressions for $C_{m}^{(k,l)}$, $m_{k,l}^{(1)}$, $m_{k,l}^{(2)}$
given by 2.1 and 3.5 a) and b).
Because we are only considering the case $a\geq b$ the condition
$a+k>b-l$ is redundant and can therefore be eliminated. This gives
$$ 
R_m=
 \sum_{ \multi{ 
        l\leq k\ssp \cr
        k>0\ssp \& \ssp l\geq 0\cr
        k-l+a+b\leq m}}\hskip -.08truein
{(-1)^{m}\,(t-1)\,
q^{{m+1\choose 2}+l^2-lk-lb+ba+bk+a+b}\,
(t;q)_{m-a-k}\,(t;q)_{m-b+l+1}\,(q;q)_{a-b+k}\,(q;q)_k
\over 
t^{l-a-k-b}\,
(t;q)_{m+1}\,
(q;q)_{a+k}\,(q;q)_{b-l}\,(q;q)_{m-a-k-b+l}\,(q;q)_{a-b+l}\,
(q;q)_l\,(q;q)_{k-l}}
$$
$$+ \sum_{\multi{ 
        l\leq k\ssp \cr
        k>0\ssp \& \ssp l>0\cr
        k-l+a+b\leq m}} \hskip -.08truein
{(-1)^{m}\,(1-t)
\,q^{{m+1\choose 2}+l^2-lk-lb+bk+ba+k+a}
\,(t;q)_{m-a-k+1}\,(t;q)_{m-b+l}
\,(q;q)_{a-b+k-1}
\,(q;q)_{k-1}
\over
t^{l-a-k-b}\,
(t;q)_{m+1}
(q;q)_{a+k}\,(q;q)_{b-l}\,(q;q)_{m-a-k-b+l}\,
(q;q)_{a-b+l-1}\,
(q;q)_{l-1}\,
(q;q)_{k-l}}
$$
\noindent
Transformation of terms using properties 2.3 and 3.7 
allows the summands to be expressed with 
q-shifted factorials that are either indexed by $k$ and independent of $l$
or indexed by $l$ with a possible dependence on $k$, but no further 
dependence on $l$. This  converts $R_m$to the form
$$ 
\eqalign{
&
\sum_{\multi{
        l\leq k\ssp \cr
        k>0\ssp \& \ssp l\geq 0\cr
        k-l+a+b\leq m}}
\bigg(
{(-1)^{m}\, t^{a+b}\, q^{{m+1\choose 2}+b(a+1)+a+k}
\over
t^{l}(t;q)_{m+1}\,(q;q)_{a}\,
(q;q)_{b}\,(q;q)_{m-a-b}}\biggr)
\ssp \times\cr
&
{q^l\, (t-1)(t;q)_{m-a}\,(t;q)_{m-b+1}
(q^{a+b-m};q)_k\,(q^{a-b+1};q)_k\,(tq^{m-b+1};q)_l\,
(q^{-b};q)_l\,(q^{-k};q)_l
\over
(q^{1-m+a}/t;q)_k\,(q^{a+1};q)_k\,
(q^{-a-b+m-k+1};q)_l\,(q^{a-b+1};q)_l\,(q;q)_l} \sps\cr
& \ess\ess\ess\ess\ess 
\sps
\sum_{\multi{
        l\leq k\ssp \cr
        k>0\ssp \& \ssp l> 0\cr
        k-l+a+b\leq m}}
\bigg(
{(-1)^{m}\, t^{a+b}\, q^{{m+1\choose 2}+b(a+1)+a+k}
\over
t^{l}(t;q)_{m+1}\,(q;q)_{a}\,
(q;q)_{b}\,(q;q)_{m-a-b}}\biggr)
\ssp \times\cr
&
{(1-t) (t;q)_{m-a+1}\,(t;q)_{m-b}\,(tq^{m-b};q)_{l}\,(q;q)_{k-1}
\,(q^{a-b};q)_{k}\,(q^{-b};q)_{l}\,(q^{a+b-m};q)_{k}\,(q^{-k},q)_l
\over 
(q;q)_{k}\,(q^{-m+a}/t;q)_{k}\,
(q^{a+1};q)_{k}\,(q;q)_{l-1}\,(q^{1-k-a-b+m};q)_{l}\,(q^{a-b};q)_{l}}
\Bigg) }
$$
\noindent
We can see now that 
$C_m^{(a,b)}={(-1)^{m}\,t^{a+b}\,
                q^{{m+1\choose 2}+b(a+1)}\,
                (t;q)_{m-b+1}\,(t;q)_{m-a}\over
                (t;q)_{m+1}\,(q;q)_{a}\,(q;q)_{b}\,(q;q)_{m-a-b}}$
can be factored out of $R_m$ reducing the problem to verifying the
equality:
$$
t\,q^m-t\,q^{a} 
\ses \sum_{\multi{
        l\leq k\ssp \cr
        k>0\ssp \& \ssp l\geq 0\cr
        k-l+a+b\leq m}}
{q^{k+l+a}\,(t-1)\,
(q^{a-b+1};q)_k\,(q^{a+b-m};q)_k\,(tq^{m-b+1};q)_l\,
(q^{-b};q)_l\,(q^{-k};q)_l\over t^l\,
(q^{1-m+a}/t;q)_k\,(q^{a+1};q)_k\,(q^{-a-b+m-k+1};q)_l\,
(q^{a-b+1};q)_l\,(q;q)_l}
$$ 
$$
\ssp +
 \sum_{\multi{
        l\leq k\ssp \cr
        k>0\ssp \& \ssp l>0\cr
        k-l+a+b\leq m}}
{q^{m}\,(1-t)\,
(q^{a+b-m};q)_k\,(q^{a-b+1};q)_{k-1}\,
(tq^{m-b+1};q)_{l-1}\,(q^{-b};q)_l\,(q^{-k+1};q)_{l-1}\over 
t^{l-1}\,
(q^{1-m+a}/t;q)_{k-1}\,(q^{a+1};q)_k\,(q^{-a-b+m-k+1};q)_l
\,(q^{a-b+1};q)_{l-1}\,(q;q)_{l-1}}
\ssp.
$$
Recall that we are working only with the case $a\geq b$.
This condition assures that $(q^{a-b+1};q)_k$ does not vanish, thus
the only zero in the  denominators of the summands may come from the factor
$(q^{-a-b+m-k+1};q)_l$. Observe now that since both sums are over $l\leq k$
we may apply the following transformation 
$$
{(q^{a+b-m};q)_{k}\over
(q^{-a-b+m-k+1};q)_{l}}=
(-1)^l\,q^{-l(1-a-b+m)+kl-{l\choose 2}}\,
(q^{a+b-m};q)_{k-l}\ssp.
\eqno 3.12
$$
which removes all denominators zeros and assures that the summands
will vanish only if there is at least one zero in the numerator.
Refer again to the previous equation.  Because this term 
vanishes for $k-l+a+b>m$, the sumands will vanish if
$k-l+a+b>m$. Thus the restriction $k-l+a+b\leq m$  
be eliminated from both sums and we are left with showing the equality 
$$
t\,q^m-t\,q^{a} 
\ses  
\sum_{\multi{1\leq k< \infty\cr
                0\leq l < \infty\cr
                l\leq k}}
{q^{k+l+a}\,(t-1)\,
(q^{a-b+1};q)_k\,(q^{a+b-m};q)_k\,(tq^{m-b+1};q)_l\,
(q^{-b};q)_l\,(q^{-k};q)_l\over t^l\,
(q^{1-m+a}/t;q)_k\,(q^{a+1};q)_k\,(q^{-a-b+m-k+1};q)_l\,
(q^{a-b+1};q)_l\,(q;q)_l}
$$
\hfill 3.13
$$
\sum_{\multi{1\leq k<\infty\cr
                1\leq l <\infty \cr
                l\leq k}}
{q^{m}\,(1-t)\,(q^{a+b-m};q)_k\,(q^{a-b+1};q)_{k-1}\,
(tq^{m-b+1};q)_{l-1}\,(q^{-b};q)_l\,(q^{-k+1};q)_{l-1}\over t^{l-1}\,
(q^{1-m+a}/t;q)_{k-1}\,(q^{a+1};q)_k\,(q^{-a-b+m-k+1};q)_l
\,(q^{a-b+1};q)_{l-1}\,(q;q)_{l-1}}\ssp.
$$
Adding a $(k,l)=(0,0)$ term to the first
sum converts it to a sum over $0\leq k\leq \infty$, and the change of variables,
$k\rightarrow k+1\scs l\rightarrow l+1$, in the second allows the sums to be combined. 
The right hand side of 3.13 can thus be transformed as follows:
$$  -q^{a}\,(t-1)\,
+ \sum_{\multi{0\leq k< \infty\cr
                0\leq l < \infty\cr
                l\leq k}}
{q^{k+l+a}\,(t-1)\,
(q^{a-b+1};q)_k\,(q^{a+b-m};q)_k\,(tq^{m-b+1};q)_l\,
(q^{-b};q)_l\,(q^{-k};q)_l\over t^l\,
(q^{1-m+a}/t;q)_k\,(q^{a+1};q)_k\,(q^{-a-b+m-k+1};q)_l\,
(q^{a-b+1};q)_l\,(q;q)_l}
$$
$$+ \sum_{\multi{0\leq k<\infty\cr
                0\leq l <\infty\cr
                l\leq k}}
{q^{m}\,(1-t)\,(q^{a+b-m};q)_{k+1}\,(q^{a-b+1};q)_{k}\,
(tq^{m-b+1};q)_{l}\,(q^{-b};q)_{l+1}\,(q^{-k};q)_{l}\over t^l\,
(q^{1-m+a}/t;q)_{k}\,(q^{a+1};q)_{k+1}\,(q^{-a-b+m-k};q)_{l+1}
\,(q^{a-b+1};q)_{l}\,(q;q)_{l}}
$$
\sa
$$ 
\eqalign{
\ses -q^{a}\,(t-1)\;\;+ & \sum_{\multi{0\leq k< \infty\cr
                0\leq l < \infty\cr
                l\leq k}}
\Bigg({(t-1)\,(q^{a-b+1};q)_k\,(q^{a+b-m};q)_k\,(tq^{m-b+1};q)_l\,
(q^{-b};q)_l\,(q^{-k};q)_l\over
t^l\,(q^{1-m+a}/t;q)_k\,(q^{a+1};q)_k\,
(q^{-a-b+m-k+1};q)_l\,(q^{a-b+1};q)_l\,(q;q)_l}\cr
& \ssp\ssp\ssp \times 
q^{k+b+a}\,{(1-q^{a+k+1-b+l})\over(1-q^{a+k+1})}
\Bigg)
}
$$
\sa
\noindent The proof of proposition 3.2 is now a matter of
validating the equality
\sas

\noindent
$$
1-tq^{m-a}\ses \sum_{\multi{0\leq k< \infty\cr
                0\leq l < \infty\cr
                l\leq k}}
\Bigg(
{q^{k+b}\,(1-t)\over
t^l}
\bigsp\bigsp\bigsp\bigsp\bigsp{\rm 3.14}
$$
$$
\bigsp\bigsp\times {(q^{a-b+1};q)_k\,(q^{a+b-m};q)_k\,
(tq^{m-b+1};q)_l\,(q^{-b};q)_l\,(q^{-k};q)_l\,
(1-q^{a+k+1-b+l})
\over
(q^{1-m+a}/t;q)_k
\,(q^{a+1};q)_k\,(q^{-a-b+m-k+1};q)_l\,(q^{a-b+1};q)_l\,(q;q)_l\,
(1-q^{a+k+1})}\Bigg)
\ssp.
$$
Let the right hand side be denoted, $F_m$. Note that the simple equality 
$$
{(1-q^{a+k+1-b+l})\over(1-q^{a+k+1})}= 
{(1-q^{a+k-b+1})\,(q^{a-b+k+2};q)_l\over(1-q^{a+k+1})\,
(q^{a-b+k+1};q)_l}
$$
allows $F_m$ to be expressed entirely in terms of q-shifted 
factorials and factors of $q$ and $t$, yielding

$$
F_m = \sum_{\multi{0\leq k< \infty\cr
                0\leq l < \infty\cr
                l\leq k}}
{q^{k+b}\,(1-t)\,(q^{a-b+1};q)_{k+1}\,(q^{a+b-m};q)_k\,
(tq^{m-b+1};q)_l\,(q^{-b};q)_l\,(q^{-k};q)_l
\,(q^{a-b+k+2};q)_l \over t^l\,(q^{1-m+a}/t;q)_k
\,(q^{a+1};q)_{k+1}\,(q^{-a-b+m-k+1};q)_l\,(q^{a-b+1};q)_l\,(q;q)_l
(q^{a-b+k+1};q)_l}
$$
Next, the change of variables  $j+l=k$ gives
$$
\eqalign{
F_m\ses& \sum_{\multi{0\leq j< \infty\cr
                0\leq l < \infty}}
\Bigg({q^{j+l+b}\,(1-t)\over t^l}\cr
& \times
{(q^{a-b+1};q)_{j+l+1}\,(q^{a+b-m};q)_{j+l}\,
(tq^{m-b+1};q)_l\,(q^{-b};q)_l\,(q^{-j-l};q)_l
\,(q^{a-b+j+l+2};q)_l \over (q^{1-m+a}/t;q)_{j+l}
\,(q^{a+1};q)_{j+l+1}\,(q^{-a-b+m-j-l+1};q)_l\,(q^{a-b+1};q)_l\,(q;q)_l
(q^{a-b+j+l+1};q)_l}
\Bigg)
}
$$
\sa
We now need to transform the summand into one that has only factors that
are indexed by $j$ and entirely independent of $l$, and factors  
indexed by $l$ that have no other dependence on $l$.   
To this end, we apply  property 3.7 to the factors
$(q^{a-b+1};q)_{j+1+l}$, $\,(q^{a+b-m};q)_{j+l}$,
$\,(q^{1-m+a}/t;q)_{j+l}$, and $\,(q^{a+1};q)_{j+1+l},\,$
obtaining
$$
\eqalign{
& F_m  \ses
 \sum_{\multi{0\leq j< \infty\cr
                0\leq l < \infty}}
\Bigg({q^{j+b}\,(1-t)\,(q^{a-b+1};q)_{j+1}\,(q^{a+b-m};q)_{j}\,
\over (q^{1-m+a}/t;q)_j\,(q^{a+1};q)_{j+1}}\cr
& \ssp\ssp \times
{q^{l(a+b-m)}\,(q^{a-b+2+j};q)_{l}\,(q^{a+b-m+j};q)_{l}\,
(tq^{m-b+1};q)_l\,(q^{-b};q)_l\,(q^{j+1};q)_l
\,(q^{a-b+j+2};q)_{2l}(q^{a-b+j+1};q)_l
\over t^l\,(q^{1-m+a+j}/t;q)_{l}\,(q^{a+j+2};q)_{l}
\,(q^{a-b+1};q)_l\,(q;q)_l\,(q^{a+b-m+j};q)_l\,
(q^{a-b+j+1};q)_{2l}\,(q^{a-b+j+2};q)_l}\Bigg)
}
$$
The additional equivalence 
$$
{(q^{a-b+j+2};q)_{2l}\over (q^{a-b+j+1};q)_{2l}} 
\ses{(q^{1/2(a-b+j+3)};q)_l\,( -q^{1/2(a-b+j+3)};q)_{l}\over
     (q^{1/2(a-b+j+1)};q)_l\,(-q^{1/2(a-b+j+1)};q)_{l}}
$$
puts the sum into a form in which all the q-shifted factorials 
are indexed as desired, yielding our final expression 
$$
\eqalign{
& F_m\ses
 \sum_{0\leq j< \infty}
\Bigg({q^{j+b}\,(1-t)\,(q^{a-b+1};q)_{j+1}\,(q^{a+b-m};q)_{j}\,
\over (q^{1-m+a}/t;q)_j\,(q^{a+1};q)_{j+1}}\cr
&\ssp \times \sum_{0\leq l< \infty}
\left({q^{a+b-m}\over t}\right)^l{(q^{-b};q)_l
(tq^{m-b+1};q)_l(q^{j+1};q)_l
(q^{a-b+1+j};q)_{l}(q^{1/2(a-b+j+3)};q)_{l}
(-q^{1/2(a-b+j+3)};q)_{l} \over
(q^{1-m+a+j}/t;q)_{l}(q^{a+j+2};q)_{l}
(q^{a-b+1};q)_l(q;q)_l(q^{1/2(a-b+j+1)};q)_l
(-q^{1/2(a-b+j+1)};q)_l}\Bigg)\ess .
}
$$
Our efforts are now rewarded by a rather pleasing discovery that
the inner sum may be evaluated by means of the basic summation formula
$$
_6\phi_5 \left[\matrix{ a,\, qa^{1/2},\, -qa^{1/2},\,b,\,c,\,q^{-n}\cr
                                a^{1/2},\,-a^{1/2},\,aq/b,\,aq/c,\,aq^{n+1}}
                  ;q,{aq^{n+1}\over bc}
            \right]=
        {(aq,aq/bc;q)_n\over (aq/b,aq/c;q)_n} \ssp.
$$
In fact, if we let $a=q^{a-b+j+1},\,b=t\,q^{m-b+1},\,$ and $\,c=q^{j+1}$
the $q$-shifted factorials appearing in the inner sum precisely fit the
pattern needed for an application of this remarkable identity.
This reduces $F_m$ to a single sum and permits the following
series of reductions:   
$$
\eqalign
{
F_m &\ses  \sum_{0\leq j< \infty}
\left({q^{j+b}\,(1-t)\,(q^{a-b+1};q)_{j+1}\,(q^{a+b-m};q)_{j}\,
\over (q^{1-m+a}/t;q)_j\,(q^{a+1};q)_{j+1}}\cdot
{(q^{a-b+2+j};q)_{b}\,(q^{-m+a}/t;q)_{b} \over
(q^{1-m+a+j}/t;q)_{b}\,(q^{a-b+1};q)_{b}}\right)\cr
&\ses {q^{b}\,(1-t)\,(q^{-m+a}/t;q)_{b}\over(q^{a-b+1};q)_{b}}
\sum_{0\leq j< \infty}
{q^{j}\,(q^{a-b+1};q)_{j+1}\,(q^{a+b-m};q)_{j}\,
(q^{a-b+2+j};q)_{b}\over (q^{1-m+a}/t;q)_j
\,(q^{a+1};q)_{j+1}\,(q^{1-m+a+j}/t;q)_{b}}\cr
&\ses {q^{b}\,(1-t)\,(q^{-m+a}/t;q)_{b}\over(q^{a-b+1};q)_{b}}
\sum_{0\leq j< \infty}
{q^{j}\,(q^{a-b+1};q)_{b+1+j}\,(q^{a+b-m};q)_{j}\over
(q^{1-m+a}/t;q)_{b+j}\,(q^{a+1};q)_{j+1}}\cr
&\ses{q^{b}\,(1-t)\,(q^{-m+a}/t;q)_{b}\,(q^{a-b+1};q)_{b+1}
\over(q^{a-b+1};q)_{b}\,(q^{1-m+a}/t;q)_{b} }
\sum_{0\leq j< \infty}
{q^{j}\,(q^{a-b+b+2};q)_{j}\,(q^{a+b-m};q)_{j}\over
(q^{1-m+a+b}/t;q)_{j}\,(q^{a+1};q)_{j+1}}\cr
&\ses {q^{b}\,(1-t)\,(q^{-m+a}/t;q)_{b}\,(q^{a-b+1};q)_{b+1}
\over(q^{a-b+1};q)_{b}\,(q^{1-m+a}/t;q)_{b}\,(1-q^{a+1})}
\sum_{0\leq j< \infty}
{q^{j}\,(q^{a+b-m};q)_{j}\over(q^{1-m+a+b}/t;q)_{j}}
}
$$
Now this last sum can be easily evaluated using identity 2.6 with
$a=q,\,q^{-n}= q^{-(m-a-b)},\,$ and $c=q^{-m+a+b+1}/t.$
$$
\eqalign
{
F_m &\ses {q^{b}\,(1-t)\,(q^{-m+a}/t;q)_{b}\,(q^{a-b+1};q)_{b+1}
\over(q^{a-b+1};q)_{b}\,(q^{1-m+a}/t;q)_{b}\,(1-q^{a+1})}\cdot
{q^{m-a-b}\,(q^{a+b-m}/t;q)_{m-a-b}\over(q^{-m+a+b+1}/t;q)_{m-a-b}}\cr
&\ses {q^{m-a}\,(1-t)\,(1-q^{-m+a}/t)\over (1-1/t)}\cr
&\ses 1-tq^{m-a}
}
$$
This establishes 3.14  and completes the proof of Proposition 3.2.
\sap
\centerline {\bol REFERENCES}
\sa
\parindent=.25truein


\item{[1]}
A. Garsia and M. Haiman, {\ita  Some bigraded $S_n$-modules and the Macdonald q,t-Kostka 
coefficients}, Electronic Journal of Algebraic Combinatorics,  
Foata Festschrift, Paper R24, (web site http://ejc.math.gatech.edu:8080/Journal/journalhome.html).
\sas


\item {[2]}
A. M. Garsia and G. Tessler,
{\ita Plethystic Formulas for the Macdonald $q,t$-Kostka coefficients},
Advances in Mathematics, (to appear).
\sas

\item {[3]}
G. Gasper and M. Rahman, {\ita Basic Hypergeometric Series}, Cambridge U. press,
Cambridge (1990).
\sas

\item {[4]}
A. Kirillov and M. Noumi,
{\ita Raising operators for Macdonald Polynomials},
(preprint).
\sas

\item {[5]}
F. Knop, {\ita  Integrality of Two Variable Kostka Functions},
(preprint).
\sas

\item {[6]}
F. Knop, {\ita Symmetric and non-symmetric Quantum Capelli Polynomials},
(preprint).
\sas

\item {[7]}
L. Lapointe and L. Vinet,
{\ita A short proof of the integrality of the Macdonald (q,t)-Kostka coefficients},
Centre de Recherches Mathematiques (preprint \#2360).

\item {[8]}
L. Lapointe and L. Vinet,
{\ita Exact operator solution of the Calogero-Sutherland model},
Centre de Recherches Mathematiques (preprint \#2368).

\item {[8]}
L. Lapointe and L. Vinet,
{\ita Exact operator solution of the Calogero-Sutherland model},
Centre de Recherches Mathematiques (preprint \#2368).

\item {[9]}
A. Lascoux and M. P. Sh\"utzenberger, {\ita Symmetry and Flag
manifolds}, Invariant Theory, Springer L.N. 996 (1983) 118-144.

\item {[10]}
I. G. Macdonald, {\ita  A new class of symmetric functions}, 
Actes du $20^e$ S\'eminaire Lotharingien, 
Publ. I.R.M.A. Strasbourg, (1988)
131-171.
\sas

\item {[11]}
I. G. Macdonald, {\ita Symmetric functions and Hall polynomials},
Second Edition, Clarendon Press, Oxford (1995).
\sas

\item {[12]}
I. G. Macdonald, {\ita Affine Hecke Algebras and orthogonal Polynomials},
S\'eminaire Bourbaki, (1995) \#  797.
\sas

\item {[13]}
J. Morse, {\ita Symmetric Functions and Basic Hypergeometric Series},
UCSD Doctoral Thesis.
\sas

\item {[14]}
S. Sahi, {\ita Interpolation and integrality for Macdonald's Polynomials},
(preprint)
\end 
