\newdimen\mylength %\mylength=1pt %\def\cornd{6} \mylength=.3pt % unit coordinate \def\cornd{18} % side length of corner squares in above units \def\corndl{\cornd\mylength} % place #3 at coordinates #1,#2 % #1,#2 \def\myput(#1,#2)#3{\raise#2\mylength\hbox to 0pt{\hskip #1\mylength #3\hss}} % \hseg(#1,#2;#3) draws a line from (#1,#2) to (#1+#3,#2) \def\hseg(#1,#2;#3){\myput(#1,#2){\vrule width#3\mylength depth.4pt}} % \vseg(#1,#2;#3) draws a line from (#1,#2) to (#1,#2+#3) \def\vseg(#1,#2;#3){\myput(#1,#2){\vrule height#3\mylength width.4pt depth.4pt}} % \corn(#1,#2) draws a corner cell with given upper right coordinate \def\corn(#1,#2){\myput(#1,#2){\vbox to0pt{\hbox to0pt{\hss% \vbox{\hrule width \corndl\hbox to\corndl{% \vrule height \corndl \hfil \vrule height \corndl}% \hrule width \corndl}% \kern-.4pt}\vss}}} % \labUR(#1,#2;#3), \labLL(#1,#2;#3) % put label #3 beyond upper right (\labUR) or lower left (\labLL) vertex of a % \corn(#1,#2) (but don't draw the \corn) \def\labUR(#1,#2;#3){\myput(#1,#2){\hskip 2pt$#3$}} \def\labLL(#1,#2;#3){\myput(#1,#2){\myput(-\cornd,-\cornd){% \vbox to0pt{\hbox to0pt{\hss $#3$\hskip 1pt}\vss}}}} % \drawL(x1,x2,y1,y2;A,B) draws % A % y2 +-------+ % O O| % B | % | % y1 + % x1 x2 % where O is corner square, A, B are labels \newcount\drawcalc \def\drawL(#1,#2,#3,#4;#5,#6){% % \vseg(x2,y1;y2-y1)\corn(x2,y2)\labUR(x2,y2;A)% \drawcalc=-#3 \advance\drawcalc#4% \vseg(#2,#3;\drawcalc)\corn(#2,#4)\labUR(#2,#4;#5)% % % \hseg(x1,y2;x2-x1)\corn(x1,y2)\labLL(x1,y2;B)% \drawcalc=-#1 \advance\drawcalc#2% \hseg(#1,#4;\drawcalc)\corn(#1,#4)\labLL(#1,#4;#6)% } \overfullrule=0pt \baselineskip 14pt \settabs 8 \columns \+ \hfill&\hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & & \hfill \hfill\cr %\voffset=1truein tells how low in the page should we start! %\hoffset=.5truein %tells how far to right should we start! \parindent=.5truein \hfuzz=3.44182pt \hsize 6truein %\font\normal=Palatino at 10pt \font\ita=cmsl10 \font\bols=cmbx9 %\font\bol=PalatinoB at 10pt %\font\title=PalatinoB scaled\magstep2 %\font\ninerm=Palatino at 9pt \font\small=cmr6 \font\title=cmbx10 scaled\magstep2 \font\normal=cmr10 \font\ninerm=cmr9 \font\ita=cmti10 \font\bol=cmbx10 \def\del {\partial} \def \con {\subseteq} \def \scong {\ess \cong \ess } \def \brkpg {\supereject} \def\lll{\lambda} \def\sig{\sigma} \def\dia{\diamondsuit} \def \ggg {\gamma} \def\PI{\Pi} \def \-> {\rightarrow} \def\LL{\big\langle} \def\RR {\big\rangle} \def\splangle{\langle\ssp } \def\sprangle {\ssp \rangle} \def \UU {{\scriptscriptstyle U} } \def \HHH {{\bf H}} \def\OO {\Omega} \def\DD {\Delta} \def\OM {\Omega} \def\om {\omega} \def\la {\lambda} \def \RA {\rightarrow} \def \LA {\lefttarrow} \def\xon {x_1,x_2,\ldots ,x_n} \def\GA {\Gamma} \def \sas {\vskip .06truein} \def\sa{{\vskip .125truein}} \def\msas{{\vskip -.01in}} \def\sap{{\vskip .25truein}} \def\sapp {{\vskip .5truein}} \def\sss {\sigma} \def \eee {\epsilon} \def\DDD {\Delta} \def\aaa {\alpha} \def\bbb {\beta} \def\ddd{\delta} \def\ggg {\gamma} \def\aa {\alpha} \def\bb {\beta} \def\dd{\delta} \def\gg {\gamma} \def\con {\subseteq} \def \ses {\enskip = \enskip} \def \s=s {\enskip = \enskip} \def \sps {\enskip + \enskip} \def \sms {\enskip -\enskip} \def \lag {\langle\ssp } \def \shhh {{\scriptstyle \cup\hskip -.18em\cup}} \def \rag {\ssp \rangle} \def \scs {\ssp , \ssp} \def \ess {\enskip} \def \ssp {\hskip .25em} \def \bigsp {\hskip .5truein} \def \part {\vdash} \def \scos {\ssp :\ssp} \def \DD {\Delta} \def \X {{X}} \def \Y {{Y}} \def \Sl {S_\lambda} \def \oom {{\omega}} \def \II {{\rm I}} \def \bk {{\bf k}} \normal \def\uul{\underline} \vsize=8truein %\normalbaselineskip=12pt 18pt for space and a half, %\normalbaselines24pt for double space \sap \def\today{\ifcase\month\or January\or February\or March\or April\or may\or June\or July\or August\or September\or October\or November\or December\fi \space\number\day, \number\year} \font\ita=cmsl10 \def \LRA {{\ssp \longrightarrow \ssp}} \def \RA {{ \rightarrow }} \def \LA {{ \leftarrow }} \def \BH {{\bf H}} \def \CC {{\cal C}} \def \CR {{\cal R}} \def \BV {{\bf V}} \def \BR {{\bf R}} \def \BQ {{\bf Q}} \def \BM {{\bf M}} \def \om {\omega} \def \BH {{\bf H}} \def \TK {{\tilde K}} \def \TH {{\tilde H}} \def \th {{\tilde h}} \def \yon {y_1, \ldots ,y_n} \def \LLL {\langle} \def \RRR {\rangle} \def \BH {{\bf H}} \def \BR {{\bf R}} \def \BQ {{\bf Q}} \def \om {\omega} \def \TK {{\tilde K}} \def \TH {{\tilde H}} \def \yon {y_1, \ldots ,y_n} \def \LLL {\langle} \def \RRR {\rangle} \def \xom {x_1,\ldots ,x_m} \def \sgmu {\ ^{sg}\BH_\mu} \def \bhmu {\BH_\mu} \def \pmu {\rho_\mu} \def \FP {{\cal FP}[X,Y]} \def \gamu {{\bf \Gamma}_\mu} \def \mgamu {m({\bf \Gamma}_\mu)} \def \RImu {{\cal RI}_\mu} \def \xom {x_1,\ldots ,x_m} \def \sgmu {\ ^{sg}\BH_\mu} \def \bhmu {\BH_\mu} \def \pmu {\rho_\mu} \def \FP {{\cal FP}[X,Y]} \def \gamu {{\bf \Gamma}_\mu} \def \mgamu {m({\bf \Gamma}_\mu)} \def \RImu {{\cal RI}_\mu} \def \Ex {K_m(x,y;q,t)} \def\multi#1{\vbox{\baselineskip=0pt\halign{\hfil$\scriptstyle\vphantom{(_)}##$\hfil\cr#1\crcr}}} \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 $i 0$ 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 $am-k$. 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 $aa\geq b\ess$ 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 . $$ But then $i)\ess \Rightarrow \ssp ab$ and $a=b$. \noindent Let $a>b$. Then $$ \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 & $\ess\ess\ess ab$\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\, {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 $$ M_{(a,b;k,l)}= \cases { \cases{t\ssp q^a & if $a\geq b$ \cr\cr q^a & if $a< b$ \cr} & for $\ess k=a$ and $\ess l=b$\cr \cr \cr {(t-1)\,q^{l^2-la-lb+ab-l+a+b}\, (q;q)_{k-b}\,(q;q)_{k-a} \over (q;q)_{k+l-a-b}\,(q;q)_{a-l}\,(q;q)_{b-l}} & for $k>a$ and $\ess l\leq b\ess $ when $\ess k>l$ \cr \cr {(1-t)\,q^{k^2+k-ka-kb+ab}\, (q;q)_{l-b-1}\,(q;q)_{l-a-1} \over (q;q)_{k+l-a-b}\,(q;q)_{a-k-1}\,(q;q)_{b-k-1}} & for $l>a$ and $\ess k< b\ess $ when $\ess kl \cr k>a \ssp \&\ssp l\leq b \cr a+b\leq l+k\leq m}} m_{k,l}^{(1)}\,C_m^{(k,l)} + \sum_{\multi{ ka \ssp \&\ssp k< b \cr a+b\leq l+k\leq m}} m_{k,l}^{(2)}\,C_m^{(k,l)} \eqno 3.4 $$ 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} $$ \Xi_1\;z_k\,w_l = \cases { t\,q^k\,z_k\,w_l +(t-1)\sum_{\multi{l\leq a\leq k-1 \cr l\leq b\leq k+l-a }} {q^{l^2-la-bl+ab-l+a+b} (q;q)_{k-b}\,(q;q)_{k-a} \over (q;q)_{k+l-a-b}\,(q;q)_{a-l}\,(q;q)_{b-l}}\, z_a\,w_b & if $k\geq l$ \cr \cr q^k\,z_k\,w_l +(1-t)\sum_{\multi{k+1\leq a\leq l-1\cr k+1\leq b\leq l+k-a}} {q^{k^2+k-ak-kb+ab} (q;q)_{l-b-1}\,(q;q)_{l-a-1} \over (q;q)_{k+l-a-b}\,(q;q)_{a-k-1}\,(q;q)_{b-k-1}}\, z_a\,w_b & if $kb-l\cr k>0\ssp \& \ssp l\geq 0\cr l\leq k \cr k-l+a+b\leq m}} m_{(k+a,b-l)}^{(1)}\, C_m^{(k+a,b-l)}\,+ \sum_{\multi{ a+k>b-l\cr k>0\ssp \& \ssp l>0\cr l\leq k\ssp \cr k-l+a+b\leq m}} m_{(b-l,k+a)}^{(2)}\,C_m^{(b-l,k+a)} \ssp. $$ 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. 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