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%January\or February\or March\or April\or May\or June\or
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%\headline={\small Explicit formulas for  q,t -Kotska $\ess\ess\ess\ess\ess$ \hfill$\ess\ess\ess$\today$\ess\ess\ess\ess\ess\ess$ \folio } \footline={\hfil}

\headline={\tensc
   \ifodd\pageno
	\ifnum\pageno>1
	\hfill EXPLICIT FORMULAS FOR $q,t$-KOSTKA COEFFICIENTS \hfill
	\else
	\hfill
	\fi
   \else
	\hfill GARSIA, HAIMAN, AND TESLER \hfill
   \fi
}
\footline={\hss\twlrm\folio\hss}



\def\FOOTNOTE#1#2{{\ninepoint\parindent=1em\footnote{#1}{#2}}}
%\let\FOOTNOTE\footnote


% headings without page break -- kludge
\def\heading#1{\noindent{#1}\par\nobreak}


%\sap
\centerline {\bol  Explicit Plethystic Formulas 
}
\centerline {\bol for   }
\centerline {\bol Macdonald q,t-Kostka Coefficients }
\sa
\centerline {\ita
	A.\ M.\ Garsia${}^{\dag}$\FOOTNOTE{}
			{$\dag$\ess Work carried out under NSF grant support.},
	M.\ Haiman${}^{\dag}$,
	and G.\ Tesler${}^{\dag}$}
\sa


{\par\ninepoint\rm
\narrower
\noindent{\bol Abstract. }
For a partition $\mu=(\mu_1>\mu_2>\cdots >\mu_k>0)$ set 
$ B_\mu(q,t)=\sum_{i=1}^k\, t^{ i-1 }\,(1+ \cdots + q^{\mu_i-1}$).
In [8] Garsia-Tesler proved that if $\gamma$ is a partition of $k$ and $\la=(n-k,\gamma)$ is a partition
of $n$, then there is a unique symmetric polynomial $k_\gamma(x;q,t)$ of degree $\leq k$ 
with the property that $\TK_{\la\mu}(q,t)=k_\gamma[B_\mu(q,t);q,t]$ holds true 
for all  partitions $\mu$. It was shown there that these polynomials
have Schur function expansions of the form    
$k_\gamma(x;q,t)= \sum_{|\rho|\leq |\gamma|}S_\la(x)\, \,  k_{\rho,\gamma}(q,t)$
where the $ k_{\rho,\gamma}(q,t)$
are polynomials in $q,t,1/q,1/t$ with integer coefficients. 
This result yielded the first proof of the Macdonald polynomiality conjecture.
It also was used in a proof  [7]  of the positivity conjecture 
for the $\TK_{\la\mu}(q,t)$ for any $\la$ of the form $\la=(r,2,1^m)$ and
arbitrary $\mu$. In this paper we show that
the  polynomials $k_\gamma(x;q,t)$ may be given a very simple explicit expression
in terms of the operator $\nabla$ studied in [2]. In particular we also
obtain a new proof of the polynomiality of the coefficients $\TK_{\la\mu}(q,t)$.
Further byproducts  of these developments are a new explicit formula  for 
the polynomial $\TH_\mu[X;q,t]=\sum_\la\, S_\la[X]\TK_{\la\mu}(q,t)$
and a new derivation of the symmetric function results of Sahi [16] and Knop [11], [12]. 

}% end narrower

\sa

\heading{\bol Introduction}

To state our results we need to review some notation and recall some basic facts.
We work with the algebra $\La$ of symmetric functions in a formal infinite alphabet $X=x_1,x_2,\ldots,$ with coefficients 
in the field of rational functions ${\bf Q}(q,t)$. We also denote by $\La_{Z[q,t\, ]}$ the algebra of symmetric functions 
in $X $
with coefficients  in $Z[q,t\, ]$. We write $\La^{=d}$ for the space of symmetric functions homogeneous of degree
$d$. The spaces $\La^{\leq d}$ and $\La^{>d}$ are analogously defined.  We shall make extensive use here of 
``{\ita plethystic\/}''
notation. This is a notational device which simplifies manipulation of symmetric function identities. 
It can be easly defined and programmed in
$\scriptstyle MATHEMATICA$ or $\scriptstyle MAPLE$
if we view symmetric functions as formal power series in the power symmetric functions $p_k$.
To begin with, if  $E=E[t_1,t_2,t_3,\ldots ]$ is a formal Laurent series in the variables
$t_1,t_2,t_3,\ldots $ (which may include the parameters $q,t$) we set
$$
p_k[E]\ses E[t_1^k,t_2^k,t_3^k,\ldots ]\ess .
$$
More generally, if a certain symmetric function $F$ is expressed as the  formal power series
$$
F\ses Q[p_1,p_2,p_3,\ldots ]
$$
then we simply let 
$$
F[E]\ses Q[p_1,p_2,p_3,\ldots ]\Big|_{p_k \RA E[t_1^k,t_2^k,t_3^k,\ldots ]}\ess .
\eqno  {\rm I}.1
$$
and refer to it as ``{\ita plethystic substitution\/}''  of $E$ into the symmetric function $F$.  

We make the convention that inside the plethystic brackets ``$[\,\,]$'', $X$ and $X_n$ respectively stand for 
$x_1+x_2+x_3+\cdots$ and  $x_1+x_2 +\cdots +x_n$. In particular, one sees immediately from this definition that
if $f(x_1,x_2 ,\ldots,x_n)$ is a symmetric function then $f[X_n]=f(x_1,x_2 ,\ldots,x_n)$.
We shall also make use of the symbol $\OM(x)$ to represent the symmetric function
$$
\OM(x)\ses \prod_{i\geq 1}  {1\over 1-x_i }\ess .
$$
It is easily seen that in terms of it 
the Cauchy, Hall-Littlewood  and Macdonald kernels may be respectively be given the compact forms
$$
\OM[X_nY_m]\ess \scs\ess  \ess \OM[X_nY_m(1-t)]
\ess\ess\ess \hbox{and}\ess\ess\ess  \OM[X_nY_m{\textstyle{ 1-t\over 1-q}}]\ess.
$$
Indeed, since we may write
$$
\OM \ses   \exp\Big(\sum_{k\geq 1}
{p_k  \over k}\Big)\ess ,
$$
we see that the definition in I.2 gives
$$
\OM[X_nY_m]=\prod_{i=1}^n\prod_{j=1} ^m {1\over 1-x_iy_j}
\ess\ess\scs\ess\ess\ess 
\OM[X_nY_m(1-t)]=\prod_{i=1}^n\prod_{j=1} ^m {1-t \, x_iy_j\over 1-x_iy_j}
$$
and
$$
\OM[X_nY_m{\textstyle{ 1-t\over 1-q}}]\ses \prod_{i=1}^n\prod_{j=1} ^m \prod_{k=0}^\infty {1-t \,q^k x_iy_j\over 1-q^kx_iy_j}
\ess .
$$
In using plethystic notation we are forced to distinguish between two different
minus signs. Indeed note that the definition in I.1 yields that we have 
$$
p_k[-X_n] \ses p_k[-x_1-x_2-\cdots -x_n]\ses -x_1^k-x_2^k-\cdots -x_n^k\ses  -p_k[X_n]\ess .
$$
On the other hand, on using the ordinary meaning of the minus sign, we would obtain
$$
p_k[ X_n]\ssp \big|_{x_i\RA -x_i}\ses (-1)^k\ssp p_k[X_n]\ess .
$$
Since both operations will necessarily occur in our formulas, we shall adopt the convention that
when a certain variable has to be replaced by its negative, in the ordinary sense,
then that variable will be prepended by  a superscripted minus sign. For example, 
note that the $\om$ involution, which is customarily defined as the map
which interchanges the elementary and homogeneous bases, may also be defined
by setting
$$
\om\, p_k\ses (-1)^{k-1}\, p_k\ess .
$$
However, note that by the above conventions we obtain that
$$
p_k[-^-X_n]\ses (-1)^{k-1}\, p_k[X_n]\ess .
$$
In particular, for any symmetric polynomial $P$ of degree $\leq n$,
we may write
$$
\om\, P[X_n]\ses P[-^-X_n]\ess .
\eqno  {\rm I}.2
$$  
Sometimes it will be convenient to use the symbol ``$\eee$'' to represent ${}^-1$. The idea is that 
we should treat $\ssp \eee\ssp $ as any of the other variables in carrying out plethystic operations
and only at the end replace $\ssp \eee\ssp $ by $-1$ in the ordinary sense.



A partition $\mu$ will be represented and identified with its Ferrers diagram.
We shall use the French convention here and, 
given that the parts of $\mu$ are $\mu_1\geq \mu_2\geq \cdots\geq \mu_k>0$, we let  
the corresponding Ferrers diagram have  $\mu_i$ lattice cells in the $i^{th}$ row
(counting from the bottom up). It will be convenient to let $|\mu|$ and $l(\mu)$
denote respectively the sum of the parts and the number of nonzero parts of $\mu$. In this case
$|\mu|=\mu_1+\mu_2+\cdots +\mu_k$ and $l(\mu)=k$.
As customary the symbol ``$\mu \part n$'' will be
used to indicate that $|\mu|=n\ssp $. Following Macdonald, the {\ita arm, leg, coarm\/} and {\ita coleg\/} 
of a lattice square $s$ are the parameters $a_\mu(s),l_\mu(s),a_\mu'(s)$ and $l_\mu'(s)$ giving the number of cells of $\mu$ that are
respectively {\ita strictly} {\ninerm EAST, NORTH, WEST} and
{\ninerm SOUTH} of $s$ in $\mu$.

This given, here and after, for a partition $\mu=(\mu_1,\mu_2,\ldots ,\mu_k)$ we set
$$
n(\mu)=\sum_{i=1}^k (i-1)\mu_i\ses \sum_{s\in \mu}\, l'_\mu(s)\ses \sum_{s\in \mu}\, l_\mu(s)\ess .
$$
If $s$ is a cell of $\mu$ we shall refer to the monomial 
$w(s)=q^{a_\mu'(s)} t^{l_\mu'  (s)} $ as the {\ita weight} of $s$. The sum
of the weights of the cells of $\mu$ will be denoted by $B_\mu(q,t)$
and will be called the {\ita biexponent generator 
of }$\mu$. Note that we have
\def \II {{\rm I}}
$$
B_\mu(q,t)\ses \sum_{s\in \mu}\ssp q^{a_\mu'(s)} t^{l_\mu'  (s)}\ses
\sum_{i\geq 1}  t^{i-1}\ssp {1-q	^{\mu_i}\over 1-q\ess\ess}\ess .
\eqno {\rm I}.3
$$
If $\gamma\part k$ and $n-k \geq \max( \gamma) $, the partition of $n$ obtained by prepending
a part $n-k$ to $\gamma$ will be denoted by $(n-k,\gamma )$.
It will also be convenient to set
$$
T_\mu\,=\, t^{n(\mu)}q^{n(\mu')}\,=\, \prod_{s\in\mu} q^{a_\mu'(s)} t^{l_\mu'  (s)}
\ess\ess \hbox{and}\ess D_\mu \,=\, (1-t)(1-q)B_\mu(q,t)-1\,.\;\;
\eqno \II.4
$$
\sas

We shall work here with the symmetric polynomial $\TH_\mu[X;q,t]$  with Schur function expansion
$$
\TH_\mu[X;q,t]\ses \sum_{\la}\, S_\la[X]\, \TK_{\la\mu}(q,t)\ess ,
\eqno \II.5
$$
where the coefficients $\TK_{\la\mu}(q,t)$ are obtained from the Macdonald $q,t$-Kostka coefficients
by setting
$$
\TK_{\la\mu}(q,t)\ses t^{n(\mu)}\, K_{\la\mu}(q,1/t)\ess .
\eqno \II.6
$$ 
As we shall  see, most of the properties of $\TH_\mu[X;,q,t]$  we will need here can be routinely derived from the 
corresponding properties of the Macdonald's integral form $J_\mu[X;q,t]\, $ 
$(\dag)$ \FOOTNOTE{}{$(\dag)$\ess [15] Ch.\ VI, (8.3)},
via the identity 
$$
\TH_\mu[X;q,t]=t^{n(\mu)}J_\mu[{\textstyle{X\over 1-1/t}}\,; q,1/t\,]\ess .
\eqno \II.7
$$
This polynomial occurs naturally in our previous work, where it is conjectured to give
a representation theoretical interpretation to the coefficients  $\TK_{\la\mu}(q,t)$.
Another important ingredient in the present developments is the linear operator $\nabla$  
defined, in term of the basis $\{\TH_\mu[X;q,t]\}_\mu\, $, by setting
$$
\nabla\, \TH_\mu[X;q,t]\ses T_\mu\, \TH_\mu[X;q,t]\ess .
\eqno \II.8
$$
This operator also plays a crucial role in the developments relating Macdonald polynomials
to symmetric group representation theory [1], [3], [4], [5], [6]  and to geometry [9]. Computer experimentation
with $\nabla$ revealed that it has some truly remarkable properties. The reader is referred to
[2] for a collection of results and conjectures  about $\nabla$ that have emerged in  the few years
since its discovery.
\sas

It was shown in [8] that for any given $\gamma\part k$, there is a unique
symmetric polynomial $\bk_\gamma(x;q,t)$ of degree $\ssp \leq k\ssp $ yielding
$$
{\tilde K}_{(n-k,\gamma ),\mu}(q,t)\;=\;  \bk_\gamma[\ssp B_\mu(q,t)\ssp;q,t ]
\quad\quad (\ess \forall\ess\ess\mu\part n \geq k+ \max( \gamma)\ess )
\ess .
\eqno \II.9
$$
Although a formula for  $\bk_\gamma(x;q,t)$ could be extracted from the original proof of
this results (see [8] Th. 4.1), it was of such complexity that it yielded very little information
about the true nature of this polynomial. All that could be derived there is that $\bk_\gamma(x;q,t)$
has a Schur function expansion
of the form
$$
\bk_\gamma(x;q,t)\ses \sum_{|\rho|\leq k}\ssp S_\rho\ssp \bk_{\rho\gamma}(q,t)
\eqno \II.10
$$
with each $\bk_{\rho\gamma}(q,t)$  a Laurent polynomial in $q,t$ with integer 
coefficients. This result was sufficient to prove the integral polynomiality 
of the $K_{\la\mu}(q,t)$. Moreover, a relatively small number of these polynomials
already permitted the computation of extensive tables of the
polynomials $\TH_\mu[X;q,t]$. 
\sa

The remarkable development here is that, in terms of $\nabla$, the polynomial  
$\bk_\gamma(x;q,t)$ may be given a surprisingly simple expression.
\sas

%%%%\vbox{
\heading{\bol Theorem I.1}
\sas

{\ita For each $\gamma\part k$ let   
$$
\bk_\gamma  '(x;q,t)\ses \nabla^{-1}\ssp S_\gamma\big[
{\textstyle{1-\ ^-X\over (1-t)(1-q)}}-1\big]\ess . \ess\ess\ess (\dag)
\eqno \II.11
$$
\FOOTNOTE{}{$(\dag)\ess $
Here and in the following plethysms are to be carried out before operator actions.}
Then 
$$
{\tilde K}_{(n-k,\gamma ),\mu}(q,t)\ses
	 \bk_\gamma'[\ssp D_\mu(q,t)\ssp;q,t ]
\quad\quad (\; \forall\ess\mu\part n \geq k+ \max( \gamma)\ess )
\; .
\eqno \II.12
$$
In particular the symmetric polynomial uniquely characterized by I.9 and I.10 is given by 
the formula
$$
\bk_\gamma[X]\ses \bk_\gamma'[(1-t)(1-q)X-1]
\eqno \II.13
$$
}
%%%%}

 Let us recall that
the Hall scalar product for symmetric functions is defined by setting for the power basis
$\{p_\rho\}_\rho$ 
$$
\LL p_{\rho^{(1)}}\scs p_{\rho^{(2)}} \RR\ses
\cases{
 z_\rho & if $\rho^{(1)}=\rho^{(2)}=\rho$ \cr\cr
0 & otherwise \cr
}
$$
where for a partition $\rho=1^{\aaa_1},2^{\aaa_2},3^{\aaa_3},\cdots$ we set as customary
$$
z_\rho=1^{\aaa_1}2^{\aaa_2}3^{\aaa_3}\cdots \aaa_1!\aaa_2!\aaa_3!\cdots\ess .
$$
We shall also need here the scalar product $\LL\scs \RR_*$ defined by setting 
$$
\LL p_{\rho^{(1)}}\scs p_{\rho^{(2)}} \RR_* \,=\,
\cases{
(-1)^{ |\rho|-l(\rho) } \, z_\rho\,  p_\rho[{(1-t)(1-q)}]
		 & if $\rho^{(1)}=\rho^{(2)}=\rho\, $, \cr\cr
	0 & otherwise. \cr
}
\hskip-2pt
\eqno \II.14
$$
It will be convenient, here and in the following, to set for every $F[X ]\in \LA\,$,
$$
F^*[X ]\ses F\big[{\textstyle{X\over(1-t)(1-q) }} ]\ess . 
$$
Our main object here is the following very general result which has a variety of important
consequences including our formula I.11:

\sas
\heading{\bol Theorem I.2}
\sas

{\ita  
For each symmetric polynomial $f$  set
$$
\BPI_f  [X;q,t]\ses \nabla^{-1}\, f[X-^-1]
\eqno \II.15
$$
Then for all $\mu$ we have 
$$
{\bf \PI}_f'[D_\mu;q,t]\ses  \LL f\scs \TH_\mu[X+1]\RR_* 
\eqno \II.16
$$
Alternatively, if $f$ is homogeneous of degree $k$ and we also set
$$
{\bf \PI}_f[X;q,t]\ses \nabla^{-1}\, f[{\textstyle{1-^-X\over (1-t)(1-q)}}]\ess ,
\eqno \II.17
$$
then for all $\mu\part n\geq k$ we have
$$
\eqalign{
&a)\ess\ess\LL e_{n-k}^* \, f\sscs \TH_\mu\, \RR_* \ses {\bf \PI}_f' [D_\mu;q,t]\cr
&b)\ess\ess\LL h_{n-k} \, f\sscs \TH_\mu\, \RR \,\ses {\bf \PI}_f [D_\mu;q,t]\ess .\cr
}
\eqno \II.18
$$
}

\sas

We can define a skew version $\TH_{\mu/\nu}$ of the  symmetric polynomial $\TH_\mu$ yielding
the addition formula
$$
\TH_\mu[X+Y;q,t]\ses \sum_{\nu\con \mu}\, \TH_\nu[X ;q,t]\,  \TH_{\mu/\nu}[Y;q,t]\ess .
\eqno \II.19
$$
This can be derived from the analogous result for the Macdonald 
polynomial $Q_\la[X;q,t\, ]$ (see Ch.\ VI (7.9)). 
Now it develops that the identity in I.16 (with $\BPI_f$ given by I.15) is equivalent to
the following truly remarkable formula yielding the polynomial $ \TH_\mu$.
\sas

%%%%\vbox{
\heading{\bol Theorem I.3}
$$
\TH_\mu[X+1 \, ;q,t\, ]\ses  \Om  \big[{\textstyle{X \over M}}\big ]\, \nabla^{-1} \om \, 
\Om[{\textstyle{X \, D_\mu \over M}}\big]
\eqno \II.20 
$$
{\ita with}
$$
M\ses (1-t)(1-q)
\eqno \II.21 
$$
Another corollary of Theorem I.2 may be stated as follows.
\sas
%%%%}
%%%%\vbox{
\heading{\bol Theorem I.4}

{\ita For a partition $\mu$ set
$$
\delta_\mu[X;q,t\, ]\ses {\nabla^{-1}\TH_\mu[X-^-1] \over \th_\mu(q,t) \, \th_\mu'(q,t) } 
\eqno \II.22 
$$
with
$$
\th_\mu(q,t)=\prod_{s\in \mu}(q^{a_\mu(s)}-t^{l_\mu(s)+1})
\ess \scs \ess\ess
\th_\mu'(q,t)=\prod_{s\in \mu}(t^{l_\mu (s)}-q^{a_\mu (s)+1})\,\,  .
\eqno \II.23 
$$
Then
$$
\delta_\mu[D_\la\, ;q,t\, ]\ses
\cases{
\TH_{\la/\mu}[1;q,t\, ] & if $\mu\con \la$\cr\cr
0 & otherwise.
\cr
}
\eqno \II.24 
$$
}
%%%%} 

We shall see that the identity in  I.24 constitutes a new derivation and sharpening of the 
symmetric functions results of Sahi and Knop. 
\sas

In summary, the apparently simple identity in I.16 has astonishing consequences.
Several important results in the Theory of Macdonald polynomials may be derived from it.
Namely, 

\item {(1)} We recover the plethystic formulas  for the Macdonald coefficients
$K_{\la\mu}(q,t)$,  in a simpler and more effective form than in [7] and  [8];

\item{(2)} We obtain  a new and simple proof of the theorem
[7], [8], [10], [11], [12], [13], [16] that
the $K_{\la\mu}(q,t)$ are polynomials with integer coefficients.

\item{(3)} We recover the vanishing theorem of Knop [11], [12]  and Sahi [16]
in a strong ``extended'' vanishing form, with an exact formula for their  vanishing 
polynomials and a natural interpretation for their values at the points where they do not
vanish. 

\item{(4)} Finally we shall see that the curious and remarkable Koornwinder-Macdonald reciprocity formula 
[15] (VI (6.6)) is but  a simple specialization of I.20.
\sas

\noindent
As we shall see the derivation of all these results is not difficult and uses no machinery other than 
well-known symmetric function theory. It does however depend on the discovery of certain 
plethystic operator identities that do provide a powerful insight into Macdonald Theory.
\sas 

This paper is divided into 4 sections. In the first section we introduce our basic tools
which consist of plethystic forms of  familiar symmetric function operations
and certain new plethystic operators which naturally
emerge in computations involving the polynomials $\TH_\mu$.
The identities we prove  there should have independent interest and have
been shown to have further important applications (see [2]). In Section 2 we prove Theorems I.1 -- I.4.
In Section 3 we give our applications including our derivation of the Sahi-Knop symmetric function results
and the reciprocity formula. Our developments rely on a number of identities for the polynomials
$\TH_\mu[X;q,t]$  that may be derived from corresponding identities for the Macdonald polynomials 
$P_\la[x;q,t]$. The derivations that are less accessible will be carried out in Section 4,
the others will be referred to the appropriate sources.
\sap

\heading{\bol 1. The basic tools}
\sas

We  shall start by reviewing a few facts about Schur functions we will need 
in our presentation. Recall that the Littlewood-Richardson coefficients $c_{\mu\nu}^\la$
occur in the expansion
$$
S_\mu \,   S_\nu \ses \sum_\la c_{\mu\nu}^\la\,  S_\la \ess ,
\eqno 1.1
$$
and in the addition formula
$$
S_\la[X+Y]\ses \sum_\mu\sum_\nu c_{\mu\nu}^\la\,S_{\mu}[X] \,  S_{\nu}[Y]\ess . 
\eqno 1.2
$$
The same coefficients are used to define the ``{\ita skew\/}'' Schur function $S_{\la/\mu}$  by setting
$$
\del_{S_\mu}\, S_\la\ses S_{\la/\mu}\ses \sum_\nu c_{\mu\nu}^\la\,  S_\nu \ess .
\eqno 1.3
$$
In the present context we shall interpret 1.1 and 1.3 as expressing the action, on the Schur basis, of the 
two operators ``$\uS_\mu$'' and ``$\del_{S_\mu}$'' respectively representing ``{\ita multiplication}''
and ``{\ita skewing}'' by $S_\mu$.
Note that since the orthogonality of Schur functions with respect to the  Hall scalar product
gives 
$$
\LL S_\mu\, S_\nu\scs S_\la\RR \ses  c_{\mu\nu}^\la \ses \LL S_\nu\scs S_{\la/\mu} \RR\ess ,
\eqno 1.4
$$
we see that 1.4 may be viewed as expressing that  $\del_{S_\mu}$ is the Hall scalar product adjoint of 
$\uS_\mu$. 
\sas

In the same vein we can define  two more general ``{\ita multiplication}'' and
 ``{\ita translation}''
 operators ``$\PAU_Y$'' and ``$\TAU_Y$'' by setting for any 
given  ``{\ita alphabet}'' $Y$
$\;\;(\dag)$
\FOOTNOTE{}{$(\dag)$\ess
We use the word ``alphabet'' here in a very general manner, since $Y$ itself
may be any algebraic expression that can be plethystically substituted into a 
symmetric function. For example see formulas 1.6 a) and b)  below.  },
and any symmetric function $Q[X]\in \LA$
$$
\eqalign{
a)\ess\ess\ess \TAU_Y\,\, Q[X] &\ses Q[X+Y]\cr
b)\ess\ess\ess \PAU_Y\,\, Q[X] &\ses \OM[XY]Q[X]\ess .\cr
}
\eqno 1.5
$$
These operators have the following useful ``Schur function''expansions:
\sas

\heading{\bol Theorem 1.1}
$$
\eqalign{
a)\ess\ess\ess \TAU_Y     &\ses \sum_\mu\,\, S_\mu[Y]\, \del_{S_\mu}\ess .\cr
b)\ess\ess\ess \PAU_Y  &\ses \sum_\mu\,\, S_\mu[Y]\,  {\uS_\mu}\ess .\cr
}
\eqno 1.6
$$
{\ita 
In particular we see that when $Y$ consists of a single variable $u$, we have}
$$
\TAU_u\, \ses \sum_{m\geq 0}u^m \,\, \del_{S_m}\ess .
\eqno 1.7
$$
\heading{\bol Proof}

Note that in view of 1.3, formula 1.2 may be written in the form
$$
S_\la[X+Y]\ses \sum_\nu\, S_\nu[Y]\, S_{\la/\nu}[X]\ess .
$$
In other words we have 
$$
\TAU_Y\, S_\la[X]\ses \sum_\nu\, S_\nu[Y]\, \del_{S_\nu}\, S_{\la}[X]\ess .
$$
This proves 1.6 a) when $\TAU_Y$ acts on the Schur basis. Thus the 
result must hold true for all symmetric functions.
To prove 1.6 b) we simply observe that from the Cauchy identity 
we derive that for $P[X]\in \Lambda$  
$$
\PAU_Y\, P[X] =  \OM[XY]\, P[X] =  \sum_\rho S_\rho[Y]\, S_\rho[X]\, P[X]
 =  \Big(\sum_\rho S_\rho[Y]\, \uS_\rho[X]\Big)\, P[X]\ess .
$$
Finally, we see that 1.6 a) reduces to 1.7 when  $Y=\{u\}$, because 
$S_\rho[u]$ fails to vanish identically only when 
$\rho=\{m\}$. This completes our proof.
\sas 
Our developments crucially depend on the operators  $D_k$
and $D_k^*$ defined by setting for every $F\in \La$:
$$\def\dsquad{\hskip6pt}
\dsmatrix{
a)& D_k\, F[X] \hfill& =\;  F\big[X\ssps {\TS {M \over z}}\,  \big]\, \OM[-z\, X\, ]\ssp \big|_{z^k}\hfill
\cr
b)&D_k^*\, F[X] \hfill& =\; F\big[X\ssms {\TS {\TM \over z}}\,  \big]\, \OM[z\, X\, ]\ssp \big|_{z^k}\hfill
}
\ess\ess  (\dag) \ess\ess
\hbox{for $\, -\infty<k<+\infty $}\, ,
\eqno 1.8
$$
\FOOTNOTE{}{$(\dag)\ess\ess$
 The symbol $\,\, $``$\,\, \big|_{z^k}$'' denotes  taking the coefficient
 of $z^k$ in the preceding expression.}%
where for convenience here and after we let
$$
M\ses (1-t)(1-q)\ess\ess\scs \ess\ess\ess \TM\ses (1-1/t)(1-1/q) .
\eqno 1.9
$$

\noindent
We should note that an expression such as ``$F\big[X\ssps {\TS {M \over z}}\,  \big]$'' is easily implemented on the computer
once $F$ is expanded in the power basis. In fact if $F=Q[p_1,p_2,p_3,\ldots ]$ then
$$
F\big[X\ssps {\TS {M \over z}}\,  \big]\ses Q[p_1,p_2,p_3,\ldots ]\ssp \big|_{p_k\RA p_k+{(1-t^k)(1-q^k)\over z^k}}\ess .
$$
It is also easily seen that the generating functions  of $ D_k\, $ and $ D_k^*$ have the following simple expressions 
in terms of the multiplication and translation  operators:
$$\dsmatrix{
a)\ess D(z) = \sum_{-\infty}^\infty\,\, z^k \, D_k= \PAU_{-z}\, \TAU_{M/z}
\ess\scs \ess
	\hfill\cr
b)\ess D^*(z) = \sum_{-\infty}^\infty\,\, z^k \, D_k^*= \PAU_{z}\, \TAU_{-\TM/z}\ess.
	\hfill\cr
}
\eqno 1.10
$$
\sas

The importance of these operators in the study of the polynomials $\TH_\mu[X;q,t]$ derives from the following
basic result.
\sas
%%%%\vbox{

\heading{\bol Theorem 1.2}

{\ita For $\mu\part n$ we have
$$
\eqalign{
&a)\ess  D_0\, \TH_\mu[X;q,t]  \, = -D_\mu(q,t)\,\, \TH_\mu[X;q,t]\ess ,
\ess\cr
\ess\ess
&b)\ess  D_0^*\, \TH_\mu[X;q,t]  \, = -D_\mu(1/q,1/t)\,\, \TH_\mu[X;q,t]\ess .
\cr
}
\eqno 1.11
$$
In particular $\TH_\mu[X;q,t]$ is uniquely characterized
by either one of a) or b) above and the normalization}
$$
 \TH_\mu[X;q,t] \,\, \big |_{S_{n}} \ses 1\ess .\ess \ess  (\dag)
\eqno 1.12
$$
\FOOTNOTE{}{$(\dag)\ess\ess $ The symbol ``$\ess \big |_{S_{n}}$'' represents taking the coefficient of
the Schur function $S_n[X]$ in the Schur function expansion of the preceding expression.}
%%%%}

\sas

The proof of this will be found in Section 4.



\sa
There are a number of  identities, involving various combinations of these operators, 
which we will need in our developments. 
Since they are of independent interest,  
we will give them as a series of propositions.
\sas

For  $F[X;q,t]\in \Lambda$ let us set
$$
\DA\,F[X;q,t\,]\ses \om\, F[X;1/q,1/t\,] \ses F[-^-X;1/q,1/t\,]\ess .
\eqno 1.13
$$
It is easily seen that the operator ``$\DA$'' is an involution. It also has 
the following useful properties:
\sas
 
\heading{\bol Proposition 1.1}

{\ita Using $\eee={}^-1$ we have}
$$
\eqalign{
&a)\ess \ess \ess \DA\, \TAU_1\,  \DA  \ses \TAU_{\eee }^{-1}
\cr
&b)\ess \ess \ess \DA\, \nabla\,  \DA  \ses \nabla ^{-1}
\cr
& c)\ess \ess \ess \DA \, D_k\,  \DA  \ses (-1)^k\, D_k^*\ess .
\cr}
\eqno 1.14
$$
{\bol Proof}

For any $P[X;q,t]\in \LA$ we have 
$$\eqalign{
\DA \TAU_1\DA P[X;q,t]&\ses \DA\TAU_1P[-^-X;1/q,1/t]
	\cr&\ses
	 \DA P[-^-(X+1);1/q,1/t]\ses P[X-\eee;q,t]\ess .
}
$$
This proves 1.14 a).
Next, we shall show in Section 4 that we have
$$
T_\mu\, \om  \TH_\mu[X;1/q,1/t\, ]\ses \TH_\mu[X;q,t\, ]\ess .
\eqno 1.15
$$
Now this may be rewritten as
$$
\DA\, \TH_\mu\ses {1\over T_\mu}\, \TH_\mu \ess .
\eqno 1.16
$$
Thus from the definition in I.8 we derive that
$$
\DA \, \nabla \, \DA\, \TH_\mu\ses \DA \, \nabla \,{1\over T_\mu}\, \TH_\mu
\ses
\DA \,  \,{T_\mu\over T_\mu}\, \TH_\mu\ses {1\over T_\mu}\, \TH_\mu\ses \nabla^{-1}\, \TH_\mu\ess .
$$
This proves 1.14 b) since the $\TH_\mu\, 's$ are a basis for $\LA$.  To prove 1.14 c) we note that 
for any $F[X;q,t]\in \LA$ the definitions in 1.6 a) and 1.13 give 
$$
\eqalign{
\DA\, D_k\, \DA \, F[X;q,t\, ]&\ses \DA\, D_k\, F[-^-X;1/q,1/t\, ]\cr
&\ses \DA\,F[-^-(X+{\textstyle{M/z}});1/q,1/t\, ]\OM[-zX]\, \big|_{z^k}
\cr
&\ses \DA\,F[-^- X-{\textstyle{M/ \ ^-z}}\, ;1/q,1/t\, ]\, \OM[-zX]\, \big|_{z^k}
\cr
&\ses  F[  X-{\textstyle{\TM/ \ ^-z}}\, ; q, t\, ]\, \OM[\ ^-zX]\, \big|_{z^k}
\cr
&\ses (-1)^k\,  F[  X-{\textstyle{\TM/  z}}\, ;q,t\, ]\, \OM[ zX]\, \big|_{z^k}
\bigsp  {\bf Q.E.D.}
\cr
}
$$
\sas

\heading{\bol Remark 1.1}

The identities in 1.14  can be used to systematically  derive results for the $D_k^*\, 's$ 
from corresponding results for the $D_k\, 's$. For instance note that to prove Theorem 1.2
we need only establish 1.11 a). Indeed 1.14 c), 1.11 a) and 1.16 give
$$
D_0^*\, \TH_\mu\,=\, \DA\, D_0\,\DA \,  \TH_\mu\,=\, \DA\,  D_0\, {1\over T_\mu}\, \TH_\mu
\,=\, \DA\,{-D_\mu(q,t)\over T_\mu}\,  \TH_\mu\,=\, -D_\mu(1/q,1/t)\, \TH_\mu\, .
$$

\sas


Let us now set
$$
\TOM[X]\;=\; \om\OM[X]\;=\; \OM[-^-X]\;=\;\prod_i (1+x_i)\;=\; \exp\Big[\sum_{k\geq 1}\, {(-1)^{k-1}p_k\over k}\Big]\; .
\eqno 1.17
$$ 
This given, we have the following basic expansions.
\sas

\heading{\bol Theorem 1.3}
 
$$
\eqalign{
a)\ess\ess & \TOM\big[{\textstyle{X\,  Y\over (1-q)(1-t) }}\big]\ses 
\sum_\rho {p_\rho[X]\, p_{\rho}  [Y]\over (-1)^{ |\rho|-l(\rho) }\,z_\rho\,  p_\rho[(1-t)(1-q)]}\ssp ,\cr
b)\ess\ess& \TOM\big[{\textstyle{X\,  Y\over (1-q)(1-t) }}\big]
\ses \sum_\la S_\la  \big[{\textstyle {X\over (1-q)(1-t)}} \big] \, S_{\la'}[Y] \ses 
\sum_\la S_\la^*[X]\, S_{\la'}  [Y]\ssp ,\cr
c)\ess\ess & \TOM\big[{\textstyle{X\,  Y\over (1-q)(1-t) }}\big]
\ses \sum_\mu {\TH_\mu[X;q,t\, ]\, \TH_\mu[Y;q,t\, ] \over \th_\mu(q,t)\, \th_\mu' (q,t)}\ssp .\cr
} 
\eqno 1.18
$$
{\bol Proof}

The identity in 1.18 a) is an immediate consequence of the definition in 1.17. Note that if we
make the plethystic substitution $X\RA X/M $ in  the classical expansion
$(\dag)$
\FOOTNOTE{}{$(\dag)\;\;$ $\chi^\la_\rho$ denotes the irreducible $S_n$ character indexed by $\la$ at permutations of cycle
structure
$\rho$.}
$$
p_\rho[X]\ses \sum_\la\, \chi^\la_\rho \, S_\la[X]\ess .
$$
and substitute the result in 1.18 a) we obtain
$$
\TOM\big[{\textstyle{X\,  Y\over (1-q)(1-t) }}\big]\ses 
\sum_\rho { p_{\rho}  [Y]\over (-1)^{ |\rho|-l(\rho) }\,z_\rho\, }\ssp
\sum_\la\, \chi^\la_\rho \, S_\la[{\textstyle{X\over (1-q)(1-t) }}]\ess . 
$$  
and 1.18 b) follows by interchanging the order of summation and using the identity
$$
S_{\la'}[Y]\ses \sum_\rho\, \chi^\la_\rho\,  { (-1)^{ |\rho|-l(\rho) } p_{\rho}  [Y]\over\,z_\rho\, }\ess .
$$
Formula 1.18 c) is another way of stating the ``Cauchy'' formula for Macdonald polynomials. The details of this derivation
can be found in Section 4. 
\sas

%%%%\vbox{
\heading{\bol Corollary 1.4}

{\ita The following three pairs are dual bases with respect to the $*$-scalar product:} 

$$
\eqalign{
& a)\ess\ess\ess \Big\{p_\rho[X]\Big\}_\rho\ess\ess \&\ess\ess \Big\{{ (-1)^{ |\rho|-l(\rho) }p_{\rho}  [X]/z_\rho\, }\Big\}_\rho
\cr
& b)\ess\ess\ess \Big\{ S_\la^*[X]\Big\}_\la\ess\ess \&\ess\ess \Big\{ S_{\la'}[X] \Big\}_\la
\cr
& c)\ess\ess\ess \Big\{ \TH_\mu[X;q,t\,]\Big\}_\mu\ess\ess \&\ess\ess \Big\{ \TH_\mu[X;q,t\,]/\th_\mu\, \th_{\mu'}\Big\}_\mu
\cr}
\eqno 1.19
$$
%%%%}
\noindent
{\bol Proof}

The definition in I.14 asserts that the pair of bases in 1.19 a) are $*$-dual. We thus derive from 1.18 a) that
$\TOM\big[{XY\over (1-t)(1-q)}\big]$ is the reproducing kernel of the 
$*$-scalar product. That is to say, for all $F[X]\in\LA$ we have
$$
F[Y]\ses \LL F[X]\scs \TOM\big[{\textstyle{XY\over (1-t)(1-q)}}\big]\RR_*  \ess . 
\eqno 1.20
$$
Using 1.18 b) and c) 1.20 yields the two expansions
$$
F[Y]\ses \sum_\la\, \LL F[X]\scs S_\la^*[X]\RR_*\, S_{\la'}[Y]
\eqno 1.21
$$ 
and
$$
F[Y]\ses \sum_\mu\, \LL F[X]\scs \TH_\mu [X;q,t\,]\RR_*\, {\TH_\mu [Y;q,t\,]\over \th_\mu\, \th_{\mu'}}
\eqno 1.22
$$ 
which are equivalent to the $*$-duality of the pairs in 1.19 b) and c).

\sa


Note next that the operators $\TAU$ and  $\PAU$ commute in the following manner:
\sas

\heading{\bol Proposition 1.2}

{\ita For any two alphabets $Z$ and $Y$ we have
}
$$
\TAU_Y\,\, \PAU_Z\ses  \OM[ZY]\, \PAU_Z\,\,\TAU_Y\ess . 
\eqno 1.23
$$
{\bol Proof}

For $Q\in \LA$ we obtain
$$
\eqalign{
\TAU_Y\,\, \PAU_Z\, Q[X]&\ses \TAU_Y\,\OM[XZ]\, Q[X]\ses \OM\big[(X+Y) Z\big]\, Q[X+Y] 
\cr
&\ses\OM\big[Y Z\big]\, \OM\big[XZ\big]\, Q[X+Y]\cr
& \ses  
\OM\big[Y Z\big]\,\PAU_Z\, \TAU_Y\, Q[X]
\bigsp\bigsp\bigsp {\bol Q.E.D. }
\cr
}
$$
\sas

%%%%\vbox{
\heading{\bol Proposition 1.3}

$$
\eqalign{
a)\ess\ess  D_k\, \del_{S_m}\ssms \del_{S_m}\, D_k &\ses D_{k-1}\,\del_{S_{m-1}} \ess ,
 \cr
b)\ess\ess  D_k^*\, \del_{S_m}\ssms \del_{S_m}\, D_k^* &\ses - D_{k-1}^*\,\del_{S_{m-1}}\ess . 
  \cr}
\eqno 1.24
$$
{\ita In particular we also have}
$$
\eqalign{
a)\ess\ess  D_k\, \del_{S_1}\ssms \del_{S_1}\, D_k &\ses D_{k-1}  \ess ,
 \cr
b)\ess\ess  D_k^*\, \del_{S_1}\ssms \del_{S_1}\,  D_k^* &\ses - D_{k-1}^* \ess . 
  \cr}
\eqno 1.25
$$
%%%%}
\heading{\bol Proof}

We may view the identity in 1.7 as expressing that the operator $\TAU_u$ 
is the generating function of the operators $\del_{S_m}$. Note then
that we may write
$$
\del_{S_m}\, D_k\ses  \TAU_u\, D(z)\, \big|_{u^mz^k}\ess .
$$
This given, using 1.10 a) and 1.23 we get
$$
\eqalign{
 \TAU_u\, D(z)
&\ses 
 \TAU_u\, \PAU_{-z}\,\TAU_{M/z}\,\ses \OM[-zu]\, \PAU_{-z}\, \TAU_u\,\TAU_{M/z}\, 
\cr
&\ses 
 \OM[-zu]\, \PAU_{-z}\,\TAU_{M/z}\,   \TAU_u 
\cr
&\ses
(1-u\, z)\, \, D(z)\,   \TAU_u\ess , 
\cr}
\eqno 1.26
$$ 
and 1.24 a) is obtained by equating coefficients of
$u^m z^k$  on both sides.
We also clearly see that equating coefficients of $u  z^k$
yields the special case in 1.25 a). This given, 1.24 b) and 1.25 b)
may be obtained by means of 1.14 c). 
\sas

\heading{\bol Remark 1.2}

Since the operator $\del_{S_1}$ will occur in many of our identities,  it will be convenient
to simply denote it by $\del_1$. Note also that in this particular case, $\del_1$
reduces to differentiation with respect to the power function $p_1$. More precisely,
if $F=Q(p_1,p_2,p_3,\ldots )$ 
is a symmetric function expressed in the power basis, then 
$$  
\del_1\, F\ses \del_{p_1}\, Q(p_1,p_2,p_3,\ldots )\ssp .
$$
Note also that iterations of the identities in 1.25  yield  
$$
\eqalign{D_{-k}&\ses\sum_{i=0}^k {k\choose i}(-1)^i\,\,  \del_1^i\, D_0\, \del_1^{k-i}\ess ,\cr
D_{-k}^*&\ses\sum_{i=0}^k {k\choose i}(-1)^{k-i}\,\,  \del_1^i\, D_0^*\, \del_1^{k-i}\ess .
\cr
}
\bigsp (\ess \forall\ess\ess k\geq 1\ess)
\eqno 1.27
$$
\sas

%%%%\vbox{
The relations in 1.25 and 1.27 have the following degree-raising counterparts: 
\sa


\heading{\bol Proposition 1.4}

{\ita For all $k\in(-\infty,+\infty)$:
$$
\eqalign{
a)\ess\ess  D_k\,\, \ue_1\ssms \ue_1\, D_k &\ses M\,  D_{k+1} 
 \cr
b)\ess\ess  D_k^*\,\, \ue_1\ssms \ue_1\, D_k^* &\ses - \TM\,  D_{k+1}^* \ess ,
  \cr}
\ess .
\eqno 1.28
$$
 and by iteration we deduce that we must have  }
$$
\eqalign{
a)\ess\ess D_{ k}&\ses{1\over M^k }\sum_{i=0}^k {k\choose i}(-1)^i\,\,  \ue_1^i\, D_0\,\, \ue_1 ^{k-i}\cr
b)\ess\ess D_{ k}^*&\ses{1\over \TM^k }\sum_{i=0}^k {k\choose i}(-1)^{k-i}\,\,  \ue_1 ^i\, D_0^*\,\, \ue_1 ^{ i}
\cr
}
\bigsp (\ess \forall\ess\ess k\geq 1\ess)
\eqno 1.29
$$
%%%%}
\heading{\bol Proof}

Note that the definition in 1.8 a) gives that for any $F\in \La$ we have
$$
\eqalign{
D_k\,\,\, \ue_1\,  F[X]&\ses \big( e_1\sps  {\textstyle {M\over z}}\big) \, F[X+{\textstyle {M\over z}}]\,
\Om[-zX]\, \big|_{z^k}\cr
&\ses   \ue_1\, D_k\, F[X]\sps  M  \, F[X+{\textstyle {M\over z}}]\,
\Om[-zX]\, \big|_{z^{k+1}}\cr
&\ses  \ue_1\, D_k\, F[X]\sps M\, D_{k+1}\, F[X]  
\cr}
$$
This given, 1.28 b) follows from 1.14 c).
\sas

The operators $D_k$ and $D_k^*$ are tied to $\nabla$ via the following basic relations
\sas

\heading{\bol Proposition 1.5}
$$
\dsmatrix{
a)\; D_0\, \del_1\,-\, \del_1D_0\;=\; M \, \nabla^{-1}\del_1 \, \nabla  
	\;,\hfill&
a^*)\; D_0^*\, \del_1\,-\, \del_1D_0^* \;=\; \TM \, \nabla\, \del_1 \, \nabla^{-1}
	\;,\hfill\cr\noalign{\smallskip}
b)\; D_0\, \ue_1\,-\, \ue_1D_0\;=\; -M \, \nabla\, \ue_1 \, \nabla^{-1}  
	\;,\hfill&
b^*)\; D_0^*\, \ue_1\,-\, \ue_1D_0^* \;=\; -\TM \, \nabla^{-1}\ue_1 \, \nabla
	\;.\hfill\cr  
}
\eqno 1.30
$$ 
\heading{\bol Proof}

It follows from the Macdonald Pieri rules (see [4] Proposition 1.3) that there are certain
coefficients
$c_{\mu\nu}(q,t)$ and $d_{\mu\nu}(q,t)$ giving 
$$
a)\ess \del_1\TH_\mu\ses \sum_{\nu\RA \mu}\, c_{\mu\nu}(q,t)\, \TH_\nu\ess\scs \ess \ess
b)\ess \ue_1\TH_\nu\ses \sum_{\mu\leftarrow\nu }\, d_{\mu\nu}(q,t)\, \TH_\mu
\eqno 1.31
$$
where the symbol ``$\nu \RA \mu$'' means that $\nu$ is obtained by removing a corner of $\mu\,\,$. 
Combining 1.31 a) with   1.11 a) gives
$$
\eqalign{
D_0\, \del_1\, \TH_\mu&\ses  \sum_{\nu\RA\mu}\,c_{\mu\nu}(q,t)\, \big(-D_\nu(q,t)\big)\, \TH_\nu\ess ,
\cr
\del_1\,D_0\, \TH_\mu&\ses  \sum_{\nu\RA\mu}\,c_{\mu\nu}(q,t)\, \big(-D_\mu(q,t)\big)\, \TH_\nu\ess .
\cr
}
$$
Subtracting and using I.4 then gives 
$$
\big(D_0\, \del_1\ssms \del_1\, D_0\big)\, \TH_\mu
\ses
M\,  \sum_{\nu\RA\mu}\,c_{\mu\nu}(q,t)\, \big(B_\mu(q,t)-B_\nu(q,t)\big)\, \TH_\nu\ess .
\eqno 1.32
$$
On the other hand, from the definition I.8 we get that
$$
M\, \nabla^{-1}\,\del_1\,\,  \nabla\, \TH_\mu\ses
M\, \sum_{\nu\RA\mu}\,c_{\mu\nu}(q,t)\, \big(T_\mu/T_\nu\big)\, \TH_\nu\ess .
$$
Comparing this with 1.32 we see that 1.31 a)  will hold true  
if and only if 
$$
B_\mu(q,t)-B_\nu(q,t)\ses T_\mu/T_\nu\ess .
\eqno 1.33
$$
But this is a simple consequence of the fact that the monomial $T_\mu/T_\nu$ is 
precisely the weight of the cell we must add to $\nu$ to get $\mu$.

Similarly, from 1.31 b) we derive that
$$
\eqalign{
\big(D_0\,\,  \ue_1\ssms \ue_1  D_0\big)\, \TH_\nu
&\ses
M\,  \sum_{\mu\leftarrow  \nu}\,d_{\mu\nu}(q,t)\, \big(-B_\mu(q,t)+B_\nu(q,t)\big)\, \TH_\mu
\cr
&\ses - M\,  \sum_{\mu\leftarrow  \nu}\,d_{\mu\nu}(q,t)\,\,\,  T_\mu/T_\nu\,\, \TH_\mu\cr
&\ses - M\,  \nabla \, \ue_1 \nabla^{-1}\, \TH_\mu\ess .
\cr
}
\eqno 1.34
$$
This proves 1.30 b). The remaining relations may now be derived from 1.14 c).
This completes our proof.
\sa

\sas

\heading{\bol Proposition 1.6}
$$
a)\ess \PAU_{ 1/M}\, D_k\,  \PAU_{-1/M}\,=\,   D_k-D_{k+1} 
\;\scs \ess
b)\ess \PAU_{ -1/\TM}\, D_k^*\,\,  \PAU_{1/\TM}\,=\,   D_k^*-D_{k+1}^* 
	\;
\eqno 1.35
$$
{\bol Proof}

From 1.10 a) and 1.23 we get that
$$
\eqalign{
\PAU_{ 1/M}\, D(z)\, \PAU_{-1/M}
&\ses \PAU_{ 1/M}\, \PAU_{-z}\,\TAU_{M/z} \PAU_{-1/M}
\cr
&\ses
\PAU_{ 1/M}\, \PAU_{-z}\,\OM[-1/z]\, \PAU_{-1/M}\, \TAU_{M/z} 
\cr
&\ses
\PAU_{-z}\, (  1-1/z ) \, \TAU_{M/z}\ses (1-1/z)\, D(z)\ess ,  
\cr}
$$
and 1.35 a) follows by equating coefficients of $z^k$ on both sides. Similarly, 
from 1.10 b) we get
$$
\eqalign{
\PAU_{ -1/\TM}\, D^*(z)\, \PAU_{1/\TM}&\ses \PAU_{ -1/\TM}\, \PAU_{z}\,\TAU_{-\TM/z} \PAU_{ 1/\TM}\cr
&\ses
\PAU_{ -1/\TM}\, \PAU_{z}\,\OM[-1/z]\, \PAU_{ 1/\TM}\, \TAU_{-\TM/z} 
\cr
&\ses (1-1/z)\, D^*(z)\ess ,
\cr
}
$$
and 1.35 b) follows again by equating coefficients of $z^k$.
\sas

\heading{\bol Proposition 1.7}
%
{\ita Again with $\eee={}^-1$ we have  }
$$
\eqalign{
a)\ess\TAU_1\,\,  D_k\,\,  \TAU_1^{-1}\, \;=\; D_k\,-\, D_{k-1} 
	\, ,\ess\ess\ess&
a^*)\ess\TAU_\eee^{-1}\,\,  D_k^*\,\,  \TAU_\eee \,  \;=\; D_k^*\,+\, D_{k-1}^*
	\, ,
\cr
b) \ess\TAU_\eee\,\,  D_k\,\,  \TAU_\eee^{-1}\, \;=\; D_k\,+\, D_{k-1} 
	\, ,\ess\ess\ess&
b^*)\ess\TAU_1^{-1}\,\,  D_k^*\,\,  \TAU_1 \, \;=\; D_k^*\,-\, D_{k-1}^*
	\, .
\cr
}
\eqno 1.36
$$
{\bol Proof}

Equating coefficients of $z^k$  in 1.26 we get
$$
\TAU_u\, D_k\ses (D_k-u\, D_{k-1})\, \TAU_u\ess.
$$
Now $u=1$ gives 1.36 a) and $u=\eee$ gives 1.36 b).
This given, 1.36 a*) and b*) follow by applications of 1.14 a) and  c).
\sas

To carry out our proofs we need a few properties of the $*$-scalar product and its relations  
to our operators. We shall start with its relation to the ordinary Hall scalar product:
\sas

\heading{\bol Proposition 1.8}

{\ita For all symmetric functions $P$ and $Q$ we have
$$
\LL P\sscs Q\RR_* \ses \LL \phi\, \om P\sscs Q\RR 
\ses \LL \om\, \phi\, P\sscs Q\RR 
\eqno 1.37 
$$
where $\phi$ is the operator defined by the plethysm
$$
\phi \, P[X]\ses P\big[MX\big]\ses P\big[{\textstyle{  (1-t)(1-q)M}}\big]
\eqno 1.38 
$$
} 
\heading{\bol Proof}

Note first that since by I.2 we have
$$
 \om\, \phi^{-1}\, P[X]\ses P\big[{\textstyle{-^-X\over M}}\big]\ses P\big[{\textstyle -^-\big({X\over M}}\big)\big]
\ses \phi^{-1}\, \om\, P[X]\ess ,
\eqno 1.39 
$$
we see that the two operators $\om$ and $\phi$ do commute with each other, and therefore 
the last equality in 1.37 must hold true.

To prove the first equality, we set $P=\phi^{-1} \, \om \, p_{\rho^{(1)}}$
and $Q=p_{\rho^{(2)}}$ and observe that the definition in I.14 gives that
for $\rho^{(1)}=\rho^{(2)}=\rho$, we have
$$
\LLL \phi^{-1} \, \om \, p_{\rho^{(1)}} \scs p_{\rho^{(2)}}\RRR_{*}  \ses 
((-1)^{|\rho|-l(\rho)})^2z_\rho  \ssp 
{\prod_i{(1-q^{\rho_i})(1-t^{\rho_i})}\over p_{\rho }[(1-t)(1-q)]}
\ses z_\rho\ess .
$$
Since for $\rho^{(1)}\neq \rho^{(2)}$ we get
$$
\LLL \phi^{-1} \, \om \, p_{\rho^{(1)}} \scs p_{\rho^{(2)}}\RRR_{*} \ses 0\ses 
\LLL   p_{\rho^{(1)}} \scs p_{\rho^{(2)}}\RRR\ess ,
$$
it follows that the identity 
$$
\LL \phi^{-1} \om\, P\sscs Q\RR_* \ses \LL   P\sscs Q\RR 
\eqno 1.40
$$
must hold true for all pairs of symmetric functions $P$ and $Q\, .$
However, this is just another way of stating 1.37.
\sas

\heading{\bol Proposition 1.9}

{\ita The operators $D_0,D_0^*$ and $\nabla$  are all self-adjoint
with respect to the $*$-scalar product. Moreover, for any pair of symmetric
functions $P$ and $Q$ we have}
$$
\LL   e_1^*\, P\sscs Q\RR_*\ses  \ssp \LL P\sscs \del_1\, Q\RR_*  \ess .
\eqno 1.41 
$$ 
\heading{\bol Proof}

The identity in 1.18 c) and the definition I.8 give  that
$$
\nabla^x\, \TOM[{\textstyle{XY \over (1-t)(1-q)}}]
\ses \sum_\mu\ssp { T_\mu\, \TH_\mu(x;q,t)\, \TH_\mu(y;q,t)\over \th_\mu(q,t)\th_\mu  '(q,t)}
\ses \nabla^y \, \TOM[{\textstyle{XY \over (1-t)(1-q)}}]
\ess .
\eqno 1.42  
$$
where $\nabla^x$ and $\nabla^y$ denote $\nabla$  acting on symmetric function in the
alphabets $X$ and $Y$ respectively. However, since $\TOM[{\textstyle{XY \over (1-t)(1-q)}}]$
is the reproducing kernel of the $*$-scalar product, the relation in 1.42 is equivalent to
the identity
$$
\LL   \nabla  P\sscs Q\RR_*\ses    \LL P\sscs \nabla\,  Q\RR_* \ess . 
\eqno 1.43 
$$ 
Entirely analogous arguments based on 1.11 a) and b) yield  the identities 
$$
\LL   D_0 \, P\sscs Q\RR_*=    \LL P\sscs D_0\,  Q\RR_*   
\scs \ess\ess
\LL   D_0^* \, P\sscs Q\RR_*=   \LL P\sscs D_0^*\,  Q\RR_*  \ess .
$$  
 Finally, recalling that $\del_1$ is the Hall scalar product 
adjoint of multiplication by $h_1$ (or $e_1$), we see that 
1.37 gives
$$
\LL e_1^*\, P \,,\, Q\RR_*
		=
    \LL\phi\om  (e_1^* P) \,,\, Q\RR
		\,=\,
    \LL e_1\, \phi\om P\,,\, Q\RR
		\,=\,  
   \LL \phi\om  P\,,\, \del_1 Q\RR
		\,=\,
    \LL    P\,,\, \del_1 Q\RR_*\, .
$$
\hfill {\bol Q.E.D.}
\sas
\sas

\heading{\bol Proposition 1.10}

{\ita For $k\geq 1$, the operators $D_k$ and $D_k^*$ are $*$-adjoint to $(-1)^kD_{-k}$ and $(-qt)^kD_{-k}^*$
respectively.
}

\heading{\bol Proof}

We need only show this for one of the pairs since the other pair can be dealt with in exactly the same way.
Now, the statement that $D_k^*$ and $(-qt)^kD_{-k}^*$ are $*$-adjoint is equivalent to the identity
$$
{}^xD_k^*\, \, \TOM\Big[{X\, Y\over M}\Big ] \ses (-qt)^k\, \ ^yD_{-k}^*\, \, \TOM\Big[{X\, Y\over M}\Big ]
\eqno 1.44 
$$
where ``$\ ^xD_k^*$'' and ``$ \ ^yD_{-k}^*$'' represent these operators acting on the $X$ and $Y$ alphabets
respectively. However, 1.8 b) gives
%%%%\supereject
$$
\eqalign{
{}^xD_k^*\, \, \TOM\Big[{X\, Y\over M}\Big ] 
&=\; \TOM\Big[{(X-{\textstyle{\TM\over z}})\, Y\over M}\Big ]
 \OM\Big[ {zX  }\Big ]\ssp \Big|_{z^k}
\;=\; 
\TOM\Big[{X\, Y\over M}\Big ]\TOM\Big[{ -{\textstyle{ \TM\over z}} \, Y\over M}\Big ] \OM\Big[ {zX  }\Big ]\ssp
\Big|_{z^k}
\cr & \hskip-.25in =\;
\TOM\Big[{X\, Y\over M}\Big ]
\TOM\Big[  -{Y\over z\,  t\, q  }   \Big ] \OM\Big[ {zX  }\Big ]\ssp \Big|_{z^k}
\;=\;
\TOM\Big[{X\, Y\over M}\Big ]
\OM\Big[ {Y\over \ ^- z\,  t\, q  }   \Big ] \OM\Big[ {zX  }\Big ]\ssp \Big|_{z^k}
\cr
}
$$
and similarly
$$
\eqalign{
{}^yD_{-k}^*\, \, \TOM\Big[{X\, Y\over M}\Big ] &= \TOM\Big[{X\, (Y-{\textstyle{\TM\over z}}) \over M}\Big ]
  \OM\Big[ {zY  }\Big ]\, \Big|_{z^{-k}}
= 
\TOM\Big[{X\, Y\over M}\Big ]\TOM\Big[{ -{\textstyle{ \TM\over z}}  X\over M}\Big ] \OM\Big[ {zY  }\Big ]
	\,\Big|_{z^{-k}}
\cr & \hskip-.5in =
\TOM\Big[{X\, Y\over M}\Big ] 
\TOM\Big[  -{X\over  z\, t\, q }   \Big ] \OM\Big[ {zY  }\Big ]\ssp \Big|_{z^{-k}}=
\TOM\Big[{X\, Y\over M}\Big ] 
\OM\Big[ {X\over \ ^- z\, t\, q }   \Big ] \OM\Big[ {zY  }\Big ]\ssp \Big|_{z^{-k}}\ess .
\cr
}
$$
Then 1.44 follows since for any two formal power series $\Phi(z)$ and $\Psi(z)$ we have
$$
\Phi\Big({1\over\ ^-z\,q \, t }\Big)\, \Psi\big( z\big)  \Big|_{z^k}\ses 
\Big(-{1\over q\, t }\Big)^k\,\, \Phi\big( z\big)\Psi\Big({1\over\ ^-z\,q \, t }\Big)\,   \Big|_{z^{-k}}
\ess .
$$

The expansion in 1.7
has the following surprising corollary.
\sas

\heading{\bol Proposition 1.11}

{\ita If $P$ and $Q$ are homogeneous polynomials of degrees $k$ and $n-k$
respectively we have}
$$
\eqalign{
&a)\ess\ess\ess \LL h_{n-k}\, P\sscs Q\RR  \ses \LL P\sscs \TAU_1\, Q \RR
\cr
&b)\ess\ess\ess \LL e_{n-k}^*\, P\sscs Q\RR_* \,\,  = \,\, \LL P\sscs \TAU_1\, Q \RR_*
\cr
}
\eqno 1.45 
$$
\heading{\bol Proof}

From 1.7 with $u=1$ and the Hall adjointness of $\uS_m$ and $\del_{S_m}$
we get 
$$
\LL P\sscs \TAU_1\, Q \RR\ses
\sum_{m\geq 0}\, \LL\, P\sscs \del_{S_m} Q\RR
\ses
\sum_{m\geq 0}\LL h_m \, P\sscs Q\RR\ess .
$$
However this reduces to 1.45 a) since $\LL h_m \, P\sscs Q\RR\neq 0 $ only when
${\rm deg} ( h_m \, P)={\rm deg} (Q)$,
and that is when $m=n-k$. 

To prove 1.45 b) note that 1.37, 1.45 a)  and 1.37 give 
$$
\eqalign{
\LL P\sscs \TAU_1\, Q \RR_*&\ses \LL \phi\,\om  P\sscs \TAU_1\, Q \RR \cr
&\ses\LL h_{n-k}\, \phi\,\om \, P\sscs   Q \RR
\cr
&\ses\LL \phi\, \om  (e_{n-k}^*\,   P)\sscs   Q \RR
\cr
&\ses\LL  e_{n-k}^*\,     P \sscs   Q \RR_*\ess .
\cr} 
$$
This completes our proof.
\sas

The last item we need to deal with here is the definition of the ``{\ita skewed}''
version of the polynomials $\TH_\mu(x;q,t)$. To this end we need the following auxiliary result: 

\sas

%%%%\vbox{
\heading{\bol Proposition 1.12}

{\ita There are rational functions $d_{\mu\nu}^\la(q,t)$ such that}
$$
\TH_\mu \, \TH_\nu \ses \sum_{\la\, \supseteq\, \mu \, , \nu}
\ssp d_{\mu\nu}^\la(q,t)\ssp \TH_\la \ess .
\eqno 1.46
$$
%%%%}
\heading{\bol Proof}

The $*$-duality of the bases $\{\TH_\la\}_\la$ and $\{\TH_\la /\th_\la\th_\la'\}_\la$
gives that these coefficients are given by the formula
$$
d_{\mu\nu}^\la(q,t)\ses \LL\TH_\mu \, \TH_\nu \sscs \TH_\la /\th_\la\th_\la'\RR_*\ess ,
\eqno 1.47
$$
from which the rationality easily follows. The fact that the sum in 1.46
runs only over pairs partitions $\la$ which contain both $\mu$ and $\nu$
is an immediate consequence of the Macdonald Pieri formulas (see [15] Ch VI (7.1') and (7.4)).
\sas

We have the following immediate consequence of 1.46.
\sas

\heading{\bol Theorem 1.3}

{\ita For any two alphabets $X$ and $Y$ we have 
$$
\TH_\la[X+Y;q,t]\ses 
\sum_{\mu,\, \nu\, \subseteq \, \la}\, \TH_\mu[X;q,t]\, \TH_\nu[Y;q,t]\, c_{\mu\, , \nu}^\la(q,t)
\eqno 1.48
$$
with
}
$$
c_{\mu\, , \nu}^\la \ses {
d_{\mu\, , \nu}^\la \, \th_\la \th_\la' 
\over 
\th_\mu \th_\mu' \ssp \th _\nu \th_\nu ' 
}\ess .
\eqno 1.49
$$
{\bol Proof}

Note that if $Z$ is an additional auxiliary alphabet, 
and we make the replacements $X\ra X+Y$, $Y\RA Z$
in 1.18 c), we obtain 
$$
\TOM\big[{\textstyle{X\, Z\over M}}\big]\, \TOM\big[{\textstyle{Y\, Z\over M}}\big]
\ses
\TOM\big[{\textstyle{(X\ssps Y)\, Z\over M}}\big]
\ses \sum_\la\, {\TH_\la[X+Y;q,t]\, \TH_\la[Z;q,t] \over  \th_\la \th_\la' }\ess .
\eqno 1.50
$$
On the other hand again from 1.18 c)  we get
$$
\TOM\big[{\textstyle{X\, Z\over M}}\big]\, \TOM\big[{\textstyle{Y\, Z\over M}}\big]
\ses \sum_{\mu\, , \nu} \,
{\TH_\mu[X ]\, \TH_\nu[Y]  \over  \th_\mu \th_\mu'\, \th_\nu \th_\nu'  }\ess \TH_\mu[Z;q,t]\, \TH_\nu[Z;q,t]\ess .
\eqno 1.51
$$ 
Combining  1.50 and 1.51 and using 1.46, we finally obtain that
$$
\eqalign{
\sum_\la\, {\TH_\la[X+Y;q,t]\, \TH_\la[Z;q,t] \over  \th_\la \th_\la' }
&=
\sum_{\mu\, , \nu}\,
{ \TH_\mu[X ]\, \TH_\nu[Y]  \over  \th_\mu \th_\mu'\, \th_\nu \th_\nu'  }
\sum_{\la\, \supseteq\, \mu \, , \nu}
\ssp d_{\mu\nu}^\la(q,t)\ssp \TH_\la[Z;q,t]
\cr
&\hskip-10pt =
\sum_{\la}\, \TH_\la[Z;q,t]\ssp \sum_{\mu\, , \nu\subseteq\, \la}\,
\ssp d_{\mu\nu}^\la(q,t)\ssp
{ \TH_\mu[X ]\, \TH_\nu[Y]  \over  \th_\mu \th_\mu'\, \th_\nu \th_\nu'  }
\ssp 
\cr}
$$ 
and  1.48 (with 1.49) follows by equating coefficients of $ \TH_\la[Z;q,t]$.
\sas

In analogy with the Schur function case (as well as definition 7.5, p.~344 of [15])
we shall here and after set, for any alphabet $Y$,
$$
\TH_{\la/\mu}[Y;q,t\, ]\ses  \sum_{\nu\, \con\,  \la}\, c_{\mu\, \nu}^\la(q,t) \, \TH_\nu [Y;q,t\, ]\ess .
\eqno 1.52
$$
This permits us to write the addition formula 1.31 in the form
$$
\TH_\la[X+Y;q,t]\ses 
\sum_{\mu\, \subseteq \, \la}\, \TH_\mu[X;q,t]\, \TH_{\la/\mu}[Y;q,t\, ]\ess .
\eqno 1.53
$$
\vskip -.1in
\heading{\bol Remark 1.1}

An easy calculation yields that $\TH_{11/1}=(1+t)S_1$ and $\TH_{21/2}={t^2-q\over t-q}S_1$.
This given, word of caution should be added here concerning the subscript $\la/\mu$ appearing
in the left-hand side of 1.52. We have used this notation mainly as 
a reminder that  $\TH_{\la/\mu}$ is defined by 1.52 only for $\mu\con \la$. This
should not be taken to mean that this polynomial depends  only on the diagram of the skew
partition $\la/\mu$. The best way to interpret the meaning of 
our definition is that $\TH_{\la/\mu}$ is simply an abbreviation for the
right-hand side of 1.52 when $\mu\con \la$ and is equal to $0$
when $\mu\not\con \la$.
\sas

%%%%\vbox{
\heading{\bol Remark 1.2}

Note that since the definitions in 1.46 and 1.49 give
$$
 \Big\langle \,\,{\TH_\mu\over \th_\mu\th_\mu'}\, \, \TH_\nu\sscs \TH_\la\Big\rangle_*\ses 
{\th_\la\th_\la' \over \th_\mu\th_\mu'}\ess d_{\mu\, ,\nu}^\la
\ses  c_{\mu\, ,\nu}^\la\,\,\th_\nu\th_\nu'\ses \LL    \TH_\nu\sscs \TH_{\la/\mu}\RR_*\ess ,
$$
we see that the linear extension of the map
$$
\TH_\mu\,\, \RA\,\, \TH_{\la/\mu}
\eqno 1.54
$$
may be viewed as the $*$-scalar product adjoint of multiplication by
$\TH_\mu/\th_\mu\th_\mu'$.
%%%%}
%%%%\vskip -.3truein



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%\headline={\small (Preliminary Version)\hfill\small Explicit formulas for  q,t -Kotska $\ess\ess\ess\ess\ess$ 
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\sa 


\heading{\bol 2. Proofs of the main results.}

Our arguments here hinge on the following fundamental fact:
\sas

\heading{\bol Theorem 2.1}

{\ita Every symmetric polynomial $P$, homogeneous of degree $k\geq 1$, may be written in the form 
$$
P\ses D_1\, A\sps \ue_1\, B
\eqno 2.1
$$
with $A,B $ homogeneous symmetric polynomials of degree $k-1$. Moreover, if $P\in \LA_{Z[q,t]}$ 
then 2.1 holds true with $A=R[X;q,t\, ]/M^{k-1}$ and  $B=S[X;q,t\, ]/M^{k-1} ,$ with $R$ and $S$
polynomials in $\LA_{Z[q,t]}$. Of course,
the same result holds true with $D_1$ replaced by $D_1^*$ in 2.1.
}

\noindent
{\bol Proof}

It is sufficient to work with $D_1$ since the result for $D_1^*$ immediately follows by an application of 1.14~c).  For convenience, we shall write
$$
U[X;q,t\, ]\ssp \equiv _{\,\, \ue_1}\ssp V[X;q,t\, ]
\eqno 2.2
$$
to indicate that $\,\, U[X;q,t\, ]-V[X;q,t\, ]=e_1\,S[X;q,t\, ] $ 
with $S[X;q,t\, ]\in \LA_{Z[q,t\, ]}$. This given, we shall show that
for every elementary basis element $e_\aaa=e_{\aaa_1}e_{\aaa_2}\cdots e_{\aaa_m}$ 
with $\aaa\part k$, we have an identity of the form
$$
M^{k-1}\,  e_\aaa=M^{k-1}\,  e_{\aaa_1}e_{\aaa_2}\cdots e_{\aaa_m}
\ssp \equiv _{\,\, \ue_1}\ssp D_1\, R[X;q,t\, ]
\eqno 2.3
$$
(with $R[X;q,t\, ]\in \LA_{Z[q,t\, ]}$\ssp) .

We shall prove 2.3 by a process which was first used in [7]. The idea is to
proceed by an induction which ``descends`` on the largest part of $\aaa$.
To begin with note that we have
$$
\eqalign{
D_1\, e_1^{k-1}&\ses \big(e_1\big[X+{\textstyle{M\over z}}\big]\big)^{k-1}\OM[-zX]\, \Big|_z\cr
& \ses
\big(e_1[X]+{\textstyle{M\over z}}\big)^{k-1}\OM[-zX]\, \Big|_z\cr
&\ess \equiv _{\,\, \ue_1} {M^{k-1}\over z^{k-1}}\,\, \OM[-zX]\, \Big|_z\ses (-1)^k\, M^{k-1} e_k[X]\ess .
\cr
}
$$
In other words 
$$
M^{k-1}\,  e_k[X] \ess \equiv _{\,\, \ue_1} \ess  (-1)^k   \ssp D_1\, e_1^{k-1}\ess .
\eqno 2.4
$$
This proves 2.3 when the largest part of $\aaa$ is as large as possible. Let us then assume that we  
have 
$$
M^{k-1}\, e_{\bbb_1}e_{\bbb_2}\cdots e_{\bbb_m}\;\equiv _{\,\, \ue_1}\; D_1\, R_\bbb[X;q,t\,]
\ess\ess (\hbox{with $R_\bbb[X;q,t]\in \LA_{Z[q,t ]}$})
\eqno 2.5
$$
when $\,\, \bbb_1>a$. Our goal is to use this to prove 2.3 for
$$
\aaa=(a\geq \aaa_2\geq \cdots \geq \aaa_k\geq 0)\ess .
$$
To this end we note that 
$$
\eqalign{
D_1\, e_1^{a-1}\, e_{\aaa_2}\cdots e_{\aaa_k} 
&=\;
\big(e_1+{\textstyle{M\over z}}\big)^{a-1}\, 
\left(\,  \prod_{i=2}^k\Big(\sum_{r_i=0}^{\aaa_i}e_{\aaa_i-r_i}[X]\, 
{\textstyle{ e_{r_i}[M]\over z^{r_i}}}\Big)\right)\, \OM[-zX]\, \Big|_{z}
\cr
&\equiv _{\,\, \ue_1}\ssp
 {M^{a-1}\over z^{a }} 
\left(\,  \prod_{i=2}^k\Big(\sum_{r_i=0}^{\aaa_i}e_{\aaa_i-r_i}[X]\, 
{\textstyle{ e_{r_i}[M]\over z^{r_i}}}\Big)\right)\, \OM[-zX]\, \Big|_{z^0}\ess .
\cr
}
$$
Multiplying both sides by $M^{k-a}$, expanding the product and arranging the resulting terms according to increasing
powers of $z$ gives
$$
\dsmatrix{
M^{k-a}\, D_1\, e_1^{a-1}\, e_{\aaa_2}\cdots e_{\aaa_k}\ssp
	\hfill\cr\hskip .3in
	%&
\equiv _{\,\, \ue_1}
M^{k-1}\bigg({e_{\aaa_2}\cdots e_{\aaa_k}\over z^a}+ 
\sum_{b>a}
\ssp \sum_{\bbb_2\geq \cdots \geq\bbb_k}c_{b, \bbb_2,  \cdots ,   \bbb_k   }[M]
\ssp {e_{\bbb_2}\cdots e_{\bbb_k}\over z^b}\bigg)\OM[-zX]\, \Big|_{z^0}
	\hfill\cr\hskip.3in
	%&
\equiv _{\,\, \ue_1}
M^{k-1}
\bigg(
(-1)^a\, e_a e_{\aaa_2}\cdots e_{\aaa_k} + 
\sum_{b>a}(-1)^b
 \;\sum_{\hbox to0pt{\hss$\scriptstyle\bbb_2\geq \cdots \geq\bbb_k$\hss} }
	c_{b, \bbb_2,  \cdots ,   \bbb_k   }[M]
\ssp  e_{b}\, e_{\bbb_2}\cdots e_{\bbb_k} \bigg)  
	\hfill\cr
}$$
with $c_{b, \bbb_2,  \cdots ,   \bbb_k   }[M]$ an elementary basis element plethystically 
evaluated at $M$. In other words we obtain that
$$\eqalign{
M^{k-1} e_a e_{\aaa_2}\cdots e_{\aaa_k}
	&\equiv _{\,\, \ue_1}
 { (-1)^a M^{k-a}}
 D_1\, e_1^{a-1}\, e_{\aaa_2}\cdots e_{\aaa_k}
	\cr&\hskip.4in\sms 
\sum_{b>a}(-1)^{b-a}
 \hskip- .16in \sum_{\bbb_2\geq \cdots \geq\bbb_k }c_{b, \bbb_2,  \cdots ,   \bbb_k   }[M]
\ssp M^{k-1} e_{b}e_{\bbb_2}\cdots e_{\bbb_k}
}
$$
and the induction hypothesis in 2.5  yields 2.3 as desired.
\sa

We are now in a position to give our 

\sas

\heading{\bol Proof of Theorem I.2}

We shall begin by showing I.16. To this end it will be convenient to write our operator $\BPI$ in the form (see
I.15)
$$
\BPI\ses \nabla^{-1}\TAU_\eee^{-1}\bigsp\bigsp (\hbox{with $\eee=^-1$}).
\eqno 2.6
$$
This given, we are to show that for any homogeneous polynomial $f$, of degree $d(f)$,  we have
$$
\LL f\scs \TH_\mu[X+1;q,t\,]\RR_*\ses\BPI_f\big[D_\mu\big]\bigsp\bigsp (\hbox{for all $\mu$}).
\eqno 2.7
$$

We shall proceed by induction on $d(f)$. Since $\TH_\emptyset=1$ we have
$$
\nabla\, 1\ses \nabla^{-1}\, 1\ses 1\ess .
\eqno 2.8
$$
Thus 2.6 gives $\BPI_f\equiv 1$ for $f\equiv 1$. On the other hand, the expansion in I.5 
 and the normalization in 1.12 yield that
$$
\LL 1\scs \TH_\mu[X+1;q,t\,]\RR_*\ses\TH_\mu[1;q,t\,]\ses \TK_{n,\mu}(q,t)\ses 
\TH_\mu \, \big|_{S_n}\ses 1\ess .
$$
Thus 2.7 is trivially true when $f$ is a constant, and we can start our induction at $d(f)=0$.
Let us then assume that 2.7 is true for all $\mu\part n$ and for $d(f)<k$.
Now, since both sides of 2.7 are linear in $f$, we can use Theorem 2.1 and complete the induction 
argument by a direct verification of $2.7$ when $f=D_1\, A$ and $f=\ue_1\, B$. 
\sas

\itemitem {\ita Case 1)} {\ita  $f=D_1\, A$ with $A$ homogeneous of degree $k-1$} 

\noindent 
We start by noting that we have (using  Propositions 1.10 \& 1.9, 1.36 a) for $k=0$,  and 1.11 a)):
$$
\eqalign{
\LL f\scs \TH_\mu[X+1;q,t\,]\RR_*&\ses\LL D_1 A\scs \TAU_1\TH_\mu\RR_*\ses -\,\LL  A\scs D_{-1} \TAU_1\TH_\mu\RR_*\cr
&\ses    \LL  A\scs  \TAU_1D_0\TH_\mu\RR_*\sms \LL  A\scs  D_0\TAU_1\TH_\mu\RR_*\cr
&\ses  -D_\mu(q,t) \LL  A\scs  \TAU_1 \TH_\mu\RR_*\sms \LL D_0 A\scs  \TAU_1\TH_\mu\RR_*
\cr
}
$$
Since by assumption $ A$ is homogeneous of degree $k-1$ and $D_0$ preserves degree, we can use the induction
hypothesis on $A$ and $D_0A$ and finally obtain that
$$
\LL f\scs \TH_\mu[X+1;q,t\,]\RR_*\ses -D_\mu(q,t)\,{\bf \BPI}_{A}[D_\mu(q,t)]
\sms {\bf \BPI}_{D_0A}[D_\mu(q,t)]
 \ess .
$$ 
In conclusion, the validity of 2.7 in this case will be established if we can show that
 we have 
$$
{\bf \BPI}_{D_1A}[D_\mu(q,t)]\,=\, -D_\mu(q,t)\,{\bf \BPI}_{A}[D_\mu(q,t)]
\,-\, {\bf \BPI}_{D_0A}[D_\mu(q,t)]\, ,\quad  (\hbox{for all $\mu\,$})
$$
or, equivalently, that 
$$
{\bf \BPI}_{D_1A}[X]\ses -e_1[X]\,{\bf \BPI}_{A}[X]
\sms {\bf \BPI}_{D_0A}[X]\ess .
$$
Recalling the definition of $\BPI$ in   2.6, we are brought to verify  the operator identity
$$
\nabla^{-1}\TAU_\eee^{-1}D_1\ses -\ue_1\nabla^{-1}\TAU_\eee^{-1}\sms \nabla^{-1} \TAU_\eee^{-1}D_0\ess .
\eqno 2.10
$$ 
To prove this, note that equating the left hand side of  1.28 a) (with $k=0$) with the left hand side 
of 1.30 b) we derive that
$$
D_1\ses -\nabla\, \ue_1 \nabla ^{-1}\ess .
\eqno 2.11
$$
On the other hand, 1.36 b) gives
$$
\TAU_\eee D_1 \TAU_\eee^{-1}\ses D_0+D_1\ess . 
\eqno 2.12
$$ 
Combining these two identities yields
$$
-\,\,  \TAU_\eee\nabla\, \ue_1 \nabla ^{-1}\TAU_\eee^{-1}
\ses \TAU_\eee D_1\TAU_\eee^{-1} 
\ses D_0+D_1 
$$
which is easily seen to be just another way of writing 2.10.  This completes the proof of the first case.
\sa

\itemitem {\ita Case 2)} {\ita  $f=e_1 B$ with $B$ homogeneous of degree $k-1$}
\sas

\noindent
We start by noting that $\del_1$ and $\TAU_1$ are commuting operators. This is easily verified
by having them alternately act on any power basis element. This given, since $e_1 =Me_1^*$,
the identity in 1.31 a) gives
$$
\eqalign{
\LL  e_1 B\scs \TH_\mu[X+1;q,t\,]\RR_*&\ses M\, \LL  B\scs\del_1 \TAU_1\TH_\mu\RR_*
\cr
&\ses M\, \LL  B\scs\TAU_1 \del_1 \TH_\mu\RR_*
\cr
&\ses M \sum_{\nu\RA \mu}c_{\mu\nu}(q,t)\, \LL  B\scs \TH_\nu[X+1;q,t\, ]\RR_*\ess . 
\cr
}\eqno 2.13
$$
Now it was shown in [8] (Theorem 2.2) that the following identities hold true for every partition $\mu$:
$$
\sum_{\nu\RA\mu}c_{\mu\nu}(q,t)(T_{\mu/\nu})^r\ses  
\cases{
{t\, q\over M}\, h_{r+1}\big[D_\mu(q,t)/t q\big]  & if $r>0$\ssp ,
\cr\cr
B_\mu(q,t) & if $r=0$\ssp .
\cr
}
\eqno 2.14
$$
This given, we see that 2.13 for $k=1$ and $B=1$ reduces to
$$
\LL  e_1 \scs \TH_\mu[X+1;q,t\,]\RR_* \ses M\, \sum_{\nu\RA\mu}c_{\mu\nu}(q,t)\ses M B_\mu(q,t)\ses D_\mu(q,t)+1 
$$
Thus the validity of 2.7 for the case $f=e_1\cdot 1$ requires that 
$$
\nabla^{-1}\TAU_{\eee}^{-1}e_1 \ses  e_1+1\ess .
\eqno 2.15
$$
To verify this, we note that definition in 1.8 a), 2.8  and  2.11 give
$$
-e_1=\OM[-zX]\big|_{z}\ses D_1\cdot 1 \ses D_1 \nabla\, 1\ses -\nabla e_1\cdot 1\ses -\nabla e_1\ess .
$$
In other words 
$$
\nabla^{-1}e_1\ses e_1\ess .
$$
We have then
$$
\nabla^{-1}\TAU_\eee^{-1}e_1\ses \nabla^{-1}(e_1-e_1[\eee])\ses \nabla^{-1}(e_1+1)\ses e_1+1\ssp ,
$$
as desired. 

This establishes Case 2) for $k=1$. Let us now deal with the case when $B$ is of degree $k-1>0$. 
To this end, we start by using the induction hypothesis in 2.13 and get
$$
\LL  e_1 B\scs \TH_\mu[X+1;q,t\, ]\RR_* \ses M \sum_{\nu\RA \mu}c_{\mu\nu}(q,t)\, 
\BPI_B\big[ D_\nu(q,t)\big]\ess . 
\eqno 2.16
$$
Since $D_\nu=D_\mu-M T_{\mu/\nu}$, by I.4 and 1.33 we may write
$$
\BPI_B\big[ D_\nu(q,t)\big]\ses \sum_{r= 0}^{k-1}\BPI_B\big[ D_\mu-M/z\big]\big|_{z^{-r}} (T_{\mu/\nu})^r\ess .
$$ 
Substituting this back into 2.16 gives
$$
\eqalign{
\LL  e_1 B\,, \TH_\mu[X+1;q,t ]\RR_*
	&= M \sum_{\nu\RA \mu}c_{\mu\nu}(q,t)\, 
	 \sum_{r= 0}^{k-1}\BPI_B\big[ D_\mu-M/z\big]\big|_{z^{-r}}
			 (T_{\mu/\nu})^r\cr
	&= M 
	 \sum_{r= 0}^{k-1}\BPI_B\big[ D_\mu-M/z\big]\big|_{z^{-r}}
		 \sum_{\nu\RA \mu}c_{\mu\nu}(q,t)\, 
			(T_{\mu/\nu})^r\, . 
\cr
}\hskip -.3in
\eqno 2.17
$$ 
We now use 2.14 and get
$$
\eqalign{
\LL  e_1 B \,,\, \TH_\mu[X+1;q,t\, ]\RR_*
	\hskip-1in\cr
&=\, M\, \BPI_B\big[ D_\mu\big] B_\mu(q,t)  
\cr
&
\hskip.5in
\sps M 
\sum_{r= 1}^{k-1} \BPI_B\big[ D_\mu-M/z\big]\big|_{z^{-r}}\times{t\, q\over M}\, h_{r+1}\big[D_\mu(q,t)/t q\big] 
\cr
&=\, M\, \BPI_B\big[ D_\mu\big] B_\mu(q,t)\sms \BPI_B\big[ D_\mu\big] D_\mu(q,t)
\cr
&
\hskip .5in
\sps 
\sum_{r= 0}^{k-1}
   \BPI_B\big[ D_\mu-M/z\big]\big|_{z^{-r}}\times {t\, q}\,\OM\big[zD_\mu(q,t)/t q\big]\big|_{z^{r+1}} 
\cr
&=\,  \BPI_B\big[ D_\mu\big] +
\sum_{r= 0}^{k-1} \BPI_B\big[ D_\mu-M/z\big]\big|_{z^{-r}}\times{t\, q}\,\OM\big[zD_\mu(q,t)/t q\big]\big|_{z^{r+1}}\ssp .
\cr}
$$
In other words we have
$$
\LL  e_1 B\,,\, \TH_\mu[X+1;q,t]\RR_*
	 =\,
\BPI_B\big[ D_\mu\big] \,+\, \BPI_B\big[ D_\mu-M/tq z\big]
			      \OM\big[zD_\mu(q,t)\big]\Big|_{z} \, .
\;
\eqno 2.18
$$
This equality results from the fact that for any two formal power series 
$\Phi(z),\Psi(z)$, we have
$$
\sum_{r\geq 0}\Phi(1/z)\Big|_{z^{-r}}\times q\, t\, \Psi(z/q\, t)\big|_{z^{r+1}}\,=\, q\, t\, \Phi(1/z)\Psi(z/q\, t)\big|_{z}
\,=\, \Phi(1/tqz)\Psi(z)\big|_{z}\, .
$$
Recalling the definition of $D_1^*$ given in 1.8, we see that 2.18 is none other than
$$
\LL  e_1 B\scs \TH_\mu[X+1;q,t\, ]\RR_*\ses\BPI_B\big[ D_\mu\big] \sps (D_1^*\BPI_B)\big[ D_\mu\big]\ess .
$$
Thus the validity of 2.7 for $f=e_1B$ is reduced to showing that we have
$$
\BPI_{e_1B}\ses \BPI_B  \sps  D_1^*\BPI_B \ess .
\eqno 2.19
$$
For this to hold for all $k$ we must have
$$
\nabla^{-1}\TAU_\eee^{-1}\ue_1\ses \nabla^{-1}\TAU_\eee^{-1}\sps D_1^*\, \nabla^{-1}\TAU_\eee^{-1}\ess ,
$$
or better, multiplying on the right by $\TAU_\eee\nabla$,
$$
\nabla^{-1}\TAU_\eee^{-1}\ue_1\,\TAU_\eee\nabla\ses {1}\sps  D_1^*\ess .
\eqno 2.20
$$
Now to show this, note that for any polynomial $P[X]$ we have 
$$
\TAU_\eee^{-1}\ue_1\,\TAU_\eee \, P[X]\ses 
\TAU_\eee^{-1} \ue_1 \,  P[X+\eee\,]\ses
(\ue_1-e_1[\eee])P[X]\ses (\ue_1+1)P[X]\ess .
$$
Thus 2.20 is equivalent to 
$$
\nabla^{-1}(\ue_1+1)\nabla\ses {  1}\sps  D_1^*\ess .
$$
Namely
$$
\nabla^{-1} \ue_1 \nabla\ses \,   D_1^*\ess .
\eqno 2.21
$$
But this is none other than what we obtain by equating the right hand side of 1.28 b) to the right hand
side of 1.30 b*). This completes the proof of I.16. Note then that 
when $d(f)=k\leq n$ we can use 1.45 b) and obtain that
$$
\LLL f\scs \TH_\mu[X+1;q,t\, ]\RR_*\ses \LL e_{n-k}^*f\scs \TH_\mu\RR_*
\eqno 2.22
$$
which yields I.18 a). As for I.18 b), we note that 2.22 for $f\, \RA\,  f[{-^-X\over M}]=\om\phi^{-1} f $ gives
$$
\eqalign{
\LLL \om \phi^{-1}f\scs \TH_\mu[X+1;q,t\, ]\RR_* &\ses
 \LL (\om\phi^{-1} h_{n-k} )\, \om\phi^{-1} f\scs \TH_\mu\RR_*
	\cr&\ses
 \LL \om\phi^{-1} (h_{n-k}   f)\scs \TH_\mu\RR_*
	\ses
 \LL  h_{n-k}\,    f \scs \TH_\mu\RR \ess ,
}
$$
and I.18 a) gives I.18 b) with
$$
{\bf \PI}_f[X;q,t\,]\ses \BPI_{\om\phi^{-1} f}[X;q,t\,]\ses \nabla^{-1}\TAU_\eee^{-1} f\big[{\textstyle {-^-X\over
M}}\big]
\ses\nabla^{-1}f\big[{\textstyle{1-^-X\over (1-t)(1-q) }}\big]
$$
completing the proof of Theorem I.2.
\sa

\heading{\bol Proof of Theorem I.3}

For convenience  let us set
$$
E_\mu[X;q,t\, ]\ses \OM[{\textstyle{X\over M}}]   \,\nabla^{-1}\,\om\,
\OM[{\textstyle{D_\mu\over M}}X]\ess . 
\eqno 2.23
$$ 
This given, the equality
$$
E_\mu[X;q,t\, ]=\TH_\mu[X+1;q,t\,] 
\eqno 2.24
$$ 
is a consequence 
of the following  remarkable sequence of equivalent expressions:
$$\hskip-2pt
\def\dsquad{\hskip 6pt}
\dsmatrix{
\hfill
	\BPI_f  [D_\mu\, ;q,t]
&=&
	\LL \nabla^{-1}\, f\, [X-^-1;q,t]\sscs \OM[D_\mu X]\RR 
\hfill&(1)\cr
%
\hfill
	\hbox{\rm (1.40)}\RA\;
&=&
	\LL \nabla^{-1}\, f\, [X-^-1;q,t]\sscs \om\, \OM[{\textstyle{D_\mu \over M}}X]\RR_*
\hfill&(2)\cr
%
\hfill
	\hbox{\rm(Prop.\ 1.9)}\RA\;
&=&
	\LL  f\, [X-^-1;q,t]\sscs \nabla^{-1}\,\om\, \OM[{\textstyle{D_\mu \over M}}X]\RR_* 
\hfill&(3)\cr
%
\hfill
	\hbox{\rm (def.\  1.5a) \&  1.37)}\RA\; 
&=&
	 \LL \TAU_{-^-1} f\, [X;q,t]\sscs \phi\, \om\, \nabla^{-1}\,\om\,
	\OM[{\textstyle{D_\mu \over M}}X]\RR 
\hfill&(4)\cr
%
\hfill
	\hbox{\rm (def.\ 1.5$\,$b)$\,$)}\RA\; 
&=&
	\LL  f\, [X;q,t]\sscs \PAU_{-^-1}\, \phi\, \om\, \nabla^{-1}\,\om\,
	\OM[{\textstyle{D_\mu\over M}}X]\RR 
\hfill&(5)\cr
%
\hfill
	\hbox{\rm (I.2 \& def.\ 1.38)}\RA\; 
&=&
	\Big\langle  f\, [X;q,t]\sscs \phi\, \om\,\Big(  \OM[{\textstyle{X\over M}}]   \,\nabla^{-1}\,\om\,
	\OM[{\textstyle{D_\mu\over M}}X]\Big)\Big\rangle
\hfill&(6)\cr
%
\hfill
	\hbox{\rm (1.37)}\RA\; 
&=&
	 \LL  f\, [X;q,t]\sscs    \OM[{\textstyle{X\over M}}]   \,\nabla^{-1}\,\om\,
	\OM[{\textstyle{D_\mu\over M}}X] \RR_*   \ess .
\hfill&(7)
}
\eqno 2.25
$$
In other words, we must have
$$
\BPI_f [D_\mu\, ;q,t]\ses \LL  f\,  \sscs  E_\mu[X;q,t]\RR_*\ess .
$$
Combining this with 2.7 we obtain that
$$
 \LL  f \sscs \,  \TH_\mu[X+1]\, \RR_*\ses \LL  f \sscs  E_\mu[X;q,t]\RR_* 
\eqno 2.26
$$
must hold true for any symmetric polynomial $f$, forcing the equality in 2.24.
This completes our proof of Theorem I.3.
\sas

\heading{\bol Remark 2.1}

We should note that,
conversely, starting from  2.24 and carrying out the steps in 2.25 in 
the reverse order $(7)\RA(6) \RA\cdots\RA (2)\RA(1)$ shows that I.20  and  I.16 are simply equivalent statements.
\sas

\sas

\heading{\bol Remark 2.2}

We should also mention that the identity in I.16 contains the remarkable fact that
for any homogeneous polynomial $P$  of degree $d(P)$ we have
$$
\BPI_P[D_\mu(q,t)\,]\ses 0\bigsp\ess\ess\hbox{for all $|\mu|<d(P)$}
\eqno 2.27
$$
Indeed, according to I.16,  the right hand side of this equality should be given by 
$\LL P,\TH_\mu[X+1;q,t]\RR_*$,
but this vanishes simply because $\TH_\mu[X+1;q,t\, ]$ has degree less than $d(P)$. 
\sa

This brings us to the
\sas

\heading{\bol Proof of Theorem I.4}

Applying I.16  with the replacements $\mu\RA\la$ and  $f\RA\TH_\mu$, and using the expansion in 1.53 gives  
$$
\eqalign{
\BPI_{\TH_\mu}[D_\la(q,t)]&\ses \LL\TH_\mu,\TH_\la[X+1]\RR_*\cr
&\ses \sum_{\rho\con \la}\,\LL \TH_\mu\scs \TH_\rho\RR_*\, \TH_{\la/\rho}[1;q,t\, ]\cr
&\ses \th_\mu\th_\mu'\, \TH_{\la/\mu}[1;q,t\, ]
\cr
}
$$ 
the last equality resulting from the $*$--duality of the pair of bases in 1.19 c).
This proves the first
part of I.24 when $\mu\con \la$.  The second part follows from the observations in Remark 1.1.
In particular this shows that the polynomial $\delta_\mu[X;q,t\, ]$ has much stronger
vanishing properties than the more general polynomials $\BPI_P[X;q,t\, ]$.
\sas

We terminate this section by establishing the main result of the paper:
\sas

\heading{\bol Proof of Theorem I.1}

Our point of departure is the identity
$$
\TK_{(n-k,\ggg),\mu}(q,t)\ses \LL S_{(n-k,\ggg)'}^* \scs \TH_\mu\RR_*
\eqno 2.28
$$
which is an immediate consequence of the expansion in I.5 and the $*$-duality of the pair of bases in 1.19 b).
This given, our argument is based on the fact that $D_k^*|_{q=t=\infty}$ is the ``creation'' operator
for Schur functions. More precisely, if we set for any $P\in \LA$
$$
\BH_k\, P[X]\ses P[X-{\textstyle{1\over z}}]\, \OM[zX]\big|_{z^k}
\eqno 2.29
$$  
then for $\la=(\la_1\geq \la_2\geq\cdots \geq\la_k>0)$ we have
$$
S_\la[X]\ses \BH_{\la_1}\BH_{\la_2}\cdots \BH_{\la_k}\, {  1}\ess .
\eqno 2.30
$$
This is a classical result whose proof may be found in Section 4.  This given we see that
when $\la$ is of the form $(n-k,\ggg)$ with $\ggg\part k$, we can write
$$
\eqalign{
S_{(n-k,\ggg)}[X]&\ses \BH_{n-k}\, S_\ggg[X]\ses S_\ggg[X-{\textstyle{1\over z}}]\, \OM[zX]\big|_{z^{n-k}}\cr
&\ses\sum_{i=0}^{|\ggg|}S_\ggg[X-{\textstyle{1\over z}}]\, \big|_{z^{-i}}\,\,\OM[zX]\big|_{z^{n-k+i}}\cr
&\ses \sum_{i=0}^{|\ggg|}\, h_{n-k+i}[X]\, S_\ggg[X-{\textstyle{1\over z}}]\, \big|_{z^{-i}}\ess .
\cr
}
$$
In particular, we derive that
$$
S_{(n-k,\ggg)'}^*[X]\ses  \sum_{i=0}^{|\ggg|}\, e_{n-k+i}^*[X]\, 
S_\ggg[{\textstyle{X\over M}}-{\textstyle{1\over z}}]\, \big|_{z^{-i}} \ess .
$$
Substituting this in 2.28 gives
$$
\TK_{(n-k,\ggg),\mu}(q,t)\ses \sum_{i=0}^{|\ggg|}\,
\LL e_{n-k+i}^*[X]\, S_\ggg[{\textstyle{X\over M}}-{\textstyle{1\over z}}] 
 \scs \TH_\mu\RR_*\, \Big|_{z^{-i}}\ess . 
\eqno 2.31
$$
From I.18 a) we then get that
$$
\LL e_{n-k+i}^*[X]\, S_\ggg[{\textstyle{X\over M}}-{\textstyle{1\over z}}]  \scs \TH_\mu\RR_*\, \Big|_{z^{-i}}\ses
\BPI_{P_i}[D_\mu(q,t)\,]
$$
with
$$
P_i\ses  S_\ggg[{\textstyle{X\over M}}-{\textstyle{1\over z}}]\, \Big|_{z^{-i}}\ess .
$$
Thus
$$
\TK_{(n-k,\ggg),\mu}(q,t)\ses \sum_{i=0}^{|\ggg|}\,\BPI_{P_i}[D_\mu(q,t)\,]\ess .
\eqno 2.32
$$
Now the definition in 2.6 gives 
$$
\BPI_{P_i}\ses \nabla^{-1}S_\ggg[{\textstyle{X-^-1\over M}}-{\textstyle{1\over z}}]\, \Big|_{z^{-i}}\ess .
$$
Substituting this in 2.32, we finally get
$$
\eqalign{
\TK_{(n-k,\ggg),\mu}(q,t)&\ses \sum_{i=0}^{|\ggg|}\,
\nabla^{-1}S_\ggg[{\textstyle{X-^-1\over M}}-{\textstyle{1\over z}}]\, \Big|_{z^{-i}}\ess\bigg|_{X\RA D_\mu(q,t)}
\cr
&\ses \nabla^{-1}S_\ggg[{\textstyle{X-^-1\over M}}-1]\ess\bigg|_{X\RA D_\mu(q,t)}\ess .
\cr
} 
$$
This completes our proof of Theorem I.1.
\sas

\heading{\bol Remark 2.3}

It follows from Theorem I.3 that the expression 
$$
E_\mu[X;q,t\, ]\ses \OM[{\textstyle{X\over M}}]   \,\nabla^{-1}\,\om\,
\OM[{\textstyle{D_\mu\over M}}X]\ess 
\eqno 2.33
$$ 
defines a polynomial of degree $|\mu|$. However, there is a more 
illuminating way to see this. To begin with, note that 
from 1.18 c) and the definition in I.8, we get that

$$
\nabla^{-1}\om\OM[{\textstyle{D_\mu\over M}}X] \ses 
\sum_\aaa\, {\TH_\aaa[X;q,t\, ]\over T_\aaa\,}\, { \TH_\aaa[D_\mu]\over \th_\aaa\th_\aaa ' }\ess .
\eqno 2.34
$$ 
On the other hand, since we have
$$
\OM[{\textstyle{X\over M}}] \ses   \tilde \OM[{\textstyle{-^-X\over M}}] \ess ,
$$
again from 1.18 c) we obtain that$$
\OM[{\textstyle{X\over M}}] \ses
\sum_\bbb\, \TH_\bbb[X;q,t\, ]\, { \TH_\bbb[-^-1]\over \th_\bbb\th_\bbb' }\ess .
\eqno 2.35
$$
Multiplying 2.35 and 2.34, the definition in 2.33 gives
$$
E_\mu[X;q,t\, ]
\ses 
\sum_\aaa\,\sum_\bbb\, \TH_\aaa[X  ]\,\TH_\bbb[X ]\,
{ \TH_\aaa[D_\mu]\over T_\aaa\,\th_\aaa\th_\aaa ' }\, { \TH_\bbb[-^-1]\over \th_\bbb\th_\bbb' }\ess .
\eqno 2.36
$$
We can thus apply 1.46 and 1.49, and then 1.48 to obtain
$$
\eqalign{
E_\mu[X;q,t\, ]
&\ses 
\sum_\aaa \sum_\bbb\, \Big(\sum_{\la\supseteq \, \aaa , \bbb}\, d_{\aa,\bbb}^\la\, \TH_\la[X]\Big)
{ \TH_\aaa[D_\mu]\over T_\aaa\,\th_\aaa\th_\aaa ' }\, { \TH_\bbb[-^-1]\over \th_\bbb\th_\bbb' }
\cr
&\ses 
\sum_{\la}\,\TH_\la[X]\, \sum_{\aaa , \bbb\con \la}\,  
\, d_{\aa,\bbb}^\la\,{ \TH_\aaa[D_\mu]\over T_\aaa\,\th_\aaa\th_\aaa ' }\, { \TH_\bbb[-^-1]\over \th_\bbb\th_\bbb'
} 
\cr
&\ses 
\sum_{\la}\,{\TH_\la[X]\over\th_\la\th_\la ' }\,
\Big( \sum_{\aaa , \bbb\con \la}\,  
\, {c_{\aa,\bbb}^\la\over T_\aaa\,}\,{ \TH_\aaa[D_\mu]  }\, { \TH_\bbb[-^-1]  }\Big)
\cr
&\ses 
\sum_{\la}\,{\TH_\la[X]\over\th_\la\th_\la ' }\, 
\Big( \sum_{\aaa , \bbb\con \la}\,  
\, {c_{\aa,\bbb}^\la\over T_\aaa\,}\,{ \TH_\aaa[Y]  }\, { \TH_\bbb[-^-1]  }\Big)\ssp \Big|_{Y\RA D_\mu}
\cr
&\ses 
\sum_{\la}\, \TH_\la[X] \ess
{\nabla_y^{-1}\,\TH_\la[Y-^-1]\over\th_\la\th_\la ' }\ssp \Big|_{Y\RA D_\mu}
\ses \sum_{\la}\, \TH_\la[X] \ess \delta_\la\big [D_\mu \big]\ess ,
\cr
} 
$$
which shows that the polynomiality of $E_\mu[X;q,t\,]$ is a direct consequence of the
vanishing properties of $\delta_\la$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\hsize 6truein
%\def\del {\partial}
%\def \con {\subseteq} 
%\def \scong {\ess \cong \ess }
%\def \brkpg {\supereject}
%\def\lll{\lambda}
%\def\La {\Lambda}
%\def\sig{\sigma}
%\def\dia{\diamondsuit}
%\def \ggg {\gamma}
%\def \GG {\Gamma}
%\def\PI{\Pi}
%\def \-> {\rightarrow}
%\def\LL{\big\langle}
%\def \ra {\rightarrow}
%\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\TOM {\tilde {\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\ttt{\theta}
%\def\ggg {\gamma}
%\def\aa {\alpha}
%\def\bb {\beta}
%\def\dd{\delta}
%\def\gg {\gamma}
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%\def \ses {\enskip = \enskip}
%\def \sps {\enskip + \enskip}
%\def \ssms {\, -\,\,}
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%\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}}
%\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}
%
%\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 \BPI {{\bf \PI'}}
%\def \DA {{\downarrow}}
%\def \TM {{\tilde M}}
%\def \om {\omega}
%\def \BH {{\bf H}}
%\def \TK {{\tilde K}}
%\def \Om {\Omega}
%\def \bk {{\bf k}}
%\def \TH {{\tilde H}}
%\def \th {{\tilde h}}
%\def \xon {x_1,\ldots ,x_n}
%\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 \TAU  {{\cal T}}
%\def \PAU  {{\cal P}}
%\def \uS  {{\underline S}}
%\def \ue  {{\underline e}}
%\def \LA {\Lambda}
%\def \TS {\textstyle}
%\def \om {\omega}
%\def \TK {{\tilde K}}
%\def \TH {{\tilde H}}
%\def \xon {x_1,\ldots ,x_n}
%\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 \da {\downarrow}
%\def \sda {\hskip -.03truein\downarrow}
%\def \dan {\hskip -.03truein\downarrow_n}
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%\def \sua {\hskip -.03truein\uparrow}
%\def \os {\overline {s}}
%\def \ox {\overline {x}}
%\def \uan {\hskip -.03truein\uparrow^n}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\heading{\bol Remark 2.4}

Soon after the original conjecture of formula 2.24, we discovered the following extremely simple ``{\ita proof\/}''. 
First rewrite the formula as
$$
\TH_\mu[X;q,t\,]\ses \TAU_{-1}\PAU_{1/M} \nabla^{-1}\om\OM\big[{\textstyle{X D_\mu\over M}}\big]
\ess .
\eqno 2.37
$$ 
Now $\TH_\mu$ is uniquely characterized up to a scalar factor as the eigenfunction of $D_0$ with eigenvalue $D_\mu$,
so we must verify that the right hand side of 2.37
has the same property.  Now we have
$$
\dsmatrix{
D_0 &=& \TAU_1^{-1}(D_0-D_{-1})\TAU_1	\hfill
    &\quad\quad&(\hbox {by 1.36 a)})    \hfill \cr
    &=& -\ssp\TAU_1^{-1} \PAU_{1/M}D_{-1}\PAU_{-1/M}\TAU_1 \hfill
    &          &(\hbox {by 1.35 a)})	\hfill \cr
    &=&-M\ssp\TAU_1^{-1} \PAU_{1/M}
       \nabla^{-1}\del_1\nabla
       \PAU_{-1/M}\TAU_1		\hfill
    &          &(\hbox {by 1.24 a) \& 1.30 a)}) \hfill
}
$$
and this immediately yields 
$$
\eqalign{
D_0\ssp  \TAU_{-1}\PAU_{1/M} \nabla^{-1}\om\OM\big[{\textstyle{X D_\mu\over M}}\big]
	\hskip-.75in\cr
&\ses
-M\ssp\TAU_1^{-1} \PAU_{1/M}
\nabla^{-1}\del_1\nabla
\PAU_{-1/M}\TAU_1\ess \TAU_{-1}\PAU_{1/M} \nabla^{-1}\om\OM\big[{\textstyle{X D_\mu\over M}}\big]
\cr
&\ses
-M\ssp\TAU_1^{-1} \PAU_{1/M}
\nabla^{-1}\del_1\om\OM\big[{\textstyle{X D_\mu\over M}}\big]
\cr
{(\dag)}
&\ses
-M\ssp\TAU_1^{-1} \PAU_{1/M}
\nabla^{-1}{D_\mu\over M}\,\,\om\OM\big[{\textstyle{X D_\mu\over M}}\big]
\cr
&\ses
-D_\mu\,\,\TAU_1^{-1} \PAU_{1/M}
\nabla^{-1}\,\om\OM\big[{\textstyle{X D_\mu\over M}}\big]
\cr}
$$
\FOOTNOTE{}{$(\dag)\ess$ This is because for any $A$ we have
$\del_1\om \,{\rm exp}(p_1[XA])=A\,\om \,{\rm exp}(p_1[XA])\ssp . $}%
as desired. The missing scalar factor is easily shown to be $1$ by setting $X=0$ in (2.23). 
For this proof we don't need the deeper identity 2.14; we only need that 
$\del_1\TH_\mu=\sum_{\nu\RA\mu}c_{\mu\nu}\TH_\nu$ without explicit knowledge of the coefficients
$c_{\mu\nu}$, (that is only formula (6.7) p.~332 of [15]) .  

On further reflection, however, it is clear that something must be  wrong with this argument.
Indeed, if $A$ is any quantity whatsoever, it appears to show that the expression 
$$
\TAU_1^{-1}\PAU_{1/M}\nabla^{-1}\om \OM\big[{\textstyle{X A\over M}}\big]
$$
is an eigenfunction of $D_0$ with eigenvalue $A$, a highly unlikely possibility!

The problem is that the right hand side of 2.33 is, {\ita a priori}, a formal series, containing terms
of unbounded degree. One cannot apply the operator $\TAU^{-1}$ to such a series, just as one cannot
substitute $x-1$ for $x$ in a formal power series in one variable.  

It is possible to evade this difficulty to a certain extent by proving instead that $E_\mu[X;q,t\, ]$
is an eigenfunction of the operator $\TAU_1  D_0\TAU_1^{-1}$. This  makes sense because the latter operator is equal to 
$D_0-D_{-1}$ which {\bol can} be applied to a formal series. The problem this causes is that if we admit formal
series as eigenfunctions, the inhomogeneous operator $D_0-D_{-1}$ no longer has a ``discrete spectrum'': it has in fact
infinitely many independent 
eigenfunctions with any given eigenvalue $A$. All we can say is that $\TH_\mu[X+1;q,t\, ]$ is its unique
``{\ita polynomial\/}'' eigenvector
with eigenvalue $-D_\mu$. Absent a separate 
$(\dag\dag)$
\FOOTNOTE{}{$(\dag\dag)$\ess One that does not use Theorem I.2 .}
demonstration that $E_\mu[X;q,t\, ]$ is a polynomial,
this particular would-be ''{\ita proof\/}'' is incomplete.
\sa

%%%%\vbox{
\heading{\bol 3. Some applications}

Formula I.11 yields yet one more path for establishing the integrality of the Macdonald
$q,t$-Kostka coefficients. To see this we need a few preliminary observations. To begin with 
it follows from the Macdonald ``duality'' result ([15] (5.1) p.~327) that we have
$$
\TK_{\la'\mu}(q,t)\ses t^{n(\mu)}q^{n(\mu')}\TK_{\la\mu}(1/q,1/t)\ess .
\eqno 3.1
$$
In particular we see that if $\TK_{\la\mu}(q,t)$ is a polynomial then it must be 
of degrees $\leq n(\mu)$ in $t$ and  $\leq n(\mu')$ in $q$. Thus the definition
$\TK_{\la\mu}(q,t)= t^{n(\mu)}K_{\la\mu}(q,1/t)$ guarantees that
%%%%}
$$
a)\ess\ess \TK_{\la\mu}(q,t)\in {\bf Z}[q,t]\ess\ess\ess \Longleftrightarrow\ess\ess\ess
b)\ess\ess K_{\la\mu}(q,t)\in {\bf Z}[q,t]\ess .
\eqno 3.2
$$
There are a number of algorithms for constructing the $K_{\la\mu}(q,t)$ that stem from
the various identities established in Macdonald's original papers [14], [15]. All of these algorithms
introduce denominators of one kind or another. The simplest and most remarkable of these
algorithms is one discovered by Vinet-Lapointe [13]. They observed that the Macdonald
``integral form'' 
$$
J_\mu[X_n;q,t]=\sum_{\la\part n}\, S_\la[X_n(1-t)] K_{\la\mu}(q,t)
\eqno 3.3
$$
may be constructed ``one column at a time'' by applications of successive  specializations of 
the Macdonald operator $D_n(X;q,t)$. More precisely they set
$$
{\bf B_n}^{(r)}\ses D(-1/qt^{n-r-1};q,t)\, \ue_r
$$
and note that if $\nu$ is any partition with no more than $r\leq n$ parts, and $\mu$ is obtained by adding a column of
length $r$ to $\nu$, then 
$$
J_\mu[X;q,t]\ses {1 \over \prod_{i=r+1}^n(1-q^{-1}t^{r-i+1}) }\ess {\bf B}_n^{(r)}\, J_\nu[X;q,t\, ]\ess . 
\eqno 3.4
$$
Recalling that we have set
$$
\TH_\mu[X;q,t]\ses t^{n(\mu)}J_\mu[{\textstyle{X\over 1-1/t}};q,1/t\,]\ess ,
\eqno 3.5
$$
we can easily see that when  $\TH_\mu[X;q,t]$ is constructed by combining 3.5 with the recurrence in 3.4,  the 
$\TK_{\la\mu}(q,t)\, 's$ will necessarily  come out as polynomials
in ${\bf Z}[q,t]$ divided by factors of the form
$$
1-t^r\ess\scs\ess\ess\ess 
q-t^r\ess\scs\ess\ess\ess 
t^rq^s\ess .
$$
Now another consequence of the Macdonald duality theorem is that we have
$$
\TK_{\la\mu}(q,t)\ses \TK_{\la\mu'}(t,q)\ess .
\eqno 3.6
$$ 
This shows that $\TK_{\la\mu}(q,t)$ itself may also be given an expression consisting of a polynomial
in ${\bf Z}[q,t]$ divided by factors of the form
$$
1-q^r\ess\scs\ess\ess\ess 
t-q^r\ess\scs\ess\ess\ess 
t^rq^s\ess .
$$
Comparing these two sets of factors we see that each of these two different expressions for
$\TK_{\la\mu}(q,t)$ must in the end simplify to the point that the only remaining factors are of the form
$$
1-t\ess\scs\ess\ess\ess 
1-q\ess\scs\ess\ess\ess 
t-q\ess\scs\ess\ess\ess 
t^rq^s\ess .
$$
Now specializations at $t=1$ or at $q=1$ have been given by Macdonald
(see ex.~7, p.~364 of [15])
yielding, for instance, that
$$
\TH_\mu[X;q,1]\ses \prod_{i=1}^{l(\mu)}h_{\mu_i}\big[{\textstyle {X\over 1-q}\big](1-q)(1-q^2)\cdots (1-q^{\mu_i})}
$$
from which we can easily derive that $\TK_{\la\mu}(q,1)\in {\bf Z}[q,t]$. This excludes at once both 
$1-q$  and $1-t$ as possible denominator factors. Similarly, it is also shown in [15] ((8.12) p.~354) that
$K_{\la\mu}(0,t)$ is none other than the ``Kostka-Foulkes'' coefficient. This, together with 3.6, eliminates at once   
both factors $t^r$ and $q^s\, ,$ leaving only powers of
$$
t-q
\eqno 3.7
$$
as possible denominators!

In conclusion, to complete the proof of 3.2 a) and b), we only need a result expressing $\TK_{\la\mu}(q,t)$ 
as a rational function with
denominator factors coprime with
$t-q$. Our formula I.11 provides precisely such an expression. In fact, the two sources of denominators in I.11 are
the application of $\nabla^{-1}$ and the  plethystic substitution of $1/(1-t)(1-q)$. However, it is easily seen 
from the definition in I.1 that the latter 
only introduces denominator factors of the form $(1-t^i)(1-q^i)$, and this is sufficient for our purposes
here.  As for  $\nabla^{-1}$, we can use the identities we have already collected in this section and derive that the
only   denominator factors it can possibly introduce are powers of $t$, $q$ and $M$. 

To see this,
let us assume, by induction, that 
for all $g\in \LA_{{\bf Z}[q,t]}$, which are homogeneous of degree $k-1$,
we have  $\nabla^{-1}\, g\in \LA_{{\bf Z}[q,t,1/q,1/t,1/M]}$. 
By Theorem 2.1, we can complete the induction by proving that we also
have $\nabla^{-1}\, f\in \LA_{{\bf Z}[q,t,1/q,1/t,1/M]}$ when 
$f=D_1 g\ess $
or $f=\ue_1 g\ssp $, with $g=S[X;q,t]/M^{k-1}$ and $S$ homogeneous of degree $k-1$.
 Now in the first case  the identity in 2.11 gives
$$
  \nabla^{-1}\, f\ses  \nabla^{-1}\,D_1\, S[X;q,t]/ M^{k-1}\ses 
 -\ue_1\, \nabla^{-1} S[X;q,t]/ M^{k-1}\ess ,
$$
and in the second case, we can apply 2.21 and derive that 
$$
  \nabla^{-1}\, f\ses  \nabla^{-1}\,\ue_1\, S[X;q,t]/ M^{k-1}\ses 
D_1^*\, \nabla^{-1} S[X;q,t]/ M^{k-1}\ess .
$$
Since $\ue_1$ introduces no denominators and $D_1^*$ (see 1.8 b),  at the worst, introduces
powers of $qt$ in the denominator, the induction hypothesis yields that in both cases  we must  
have  $f\in \LA_{{\bf Z}[q,t,1/q,1/t,1/M]}$ as desired.
This completes our proof that $\TK_{\la\mu}(q,t)\in {\bf Z}[q,t]$.

\sas

\heading{\bol Remark 3.1}


We should mention that a more refined argument (see [2]) proves that 
$\nabla$ itself is ``integral`` and $\nabla^{-1}$ is ``Laurent''. More precisely we have 
$$
\nabla  \LA_{{\bf Z}[q,t]}\con \LA_{{\bf Z}[q,t]}
\ess\ess\ess 
\hbox{and}\ess\ess\ess\ess
\nabla^{-1}\LA_{{\bf Z}[q,t]}\con \LA_{{\bf Z}[q,t,{1\over t},{1\over q}]}\ess ,
$$
and this is best possible.
\sas

The next application is our derivation of the symmetric function results of 
Sahi [16] and Knop~[11], [12]. Since these two works are very closely related
we shall deal only with Sahi's case here.  
\sas

The results we are concerned with here may be stated as follows:
\sas

\heading{\bol Theorem 3.1 \rm (Sahi)}

{\ita For any $\mu \part n$  
 there is a unique polynomial $R_\mu[X;q,t\, ]\in \LA_{{\bf Z}[q,t]}$
with the vanishing properties
$$
R_\mu\big[\,{\textstyle {\sum_{i=1}^n}t^{-n+i}q^{-\nu_i}};q,t\, ]\ses 0
\ess\ess\ess\ess 
\hbox{for all  $|\nu|\leq |\mu|\ssp \&\ssp   \nu\neq \mu  $ }
\eqno 3.8
$$
and the normalization
$$
R_\mu\big[\,{\textstyle {\sum_{i=1}^n}t^{-n+i}q^{-\mu_i}};q,t\, ]\ses 1\ssp .
\eqno 3.9
$$
This polynomial can also be characterized, up to a scalar factor,  by the difference equation 
$$
{\tilde D}_1\, R_\mu[X;q,t\, ]\ses \Big( {\textstyle{{\, \, \, 1-t^n\over 1-t}\, -\,  \sum_{i=1}^n}t^{n-i}q^{\mu_i}}\Big)
R_\mu[X;q,t\, ]\, ,
\eqno 3.10
$$
where $\tilde D_1$ is the non-homogeneous difference operator
$$
{\tilde D}_1\ses \sum_{i=1}^n 
\Big( 
{\prod_{\multi{j=1 \cr j\neq i\cr }}^n {\textstyle tx_i-x_j\over x_i-x_j}}
\Big)
\big( 1- {\textstyle {1\over
x_i}}\big)\big( 1- T_q^{(i)}\big)
\eqno 3.11
$$
and $T_q^{(i)}$ is the operator that changes $x_i$ into $qx_i$
in a polynomial in%
	\break
$x_1,x_2,\ldots,x_n$.
}
\sas

Our results not only explicitly identify $R_\mu[X;q,t\, ]$ as an image of the Macdonald polynomial $P_\mu$, 
but also determine the
values taken by the left hand side of 3.8 for all the other choices of $\mu$.  Moreover we can show 
that the difference equation in 3.10 is itself the appropriately ``shifted'' image of the Macdonald difference operator. 
\sa

To be precise we have: 

\heading{\bol Theorem 3.2}

{\ita The polynomial $R_\mu[X;q,t\, ]$ may be obtained by deforming the polynomial
defined in I.22 according to the following equation:
$$
R_\mu[X;q,t\, ]\ses \delta_\mu\big[ t^n(1-1/t)X-t^n; q^{-1},t\,\big]\ess .
\eqno 3.12 
$$
In particular we must also have

$$
R_\mu\Big[{\textstyle{\sum_{i=1}^n t^{-n+i}q^{-\nu_i}}} \,\,  ;q \,,t\Big]\ses
\cases { \TH_{\nu/\mu}[1;q^{-1},t\, ] & if $\mu\con \nu\,$,\cr\cr
0 & otherwise.
\cr
} 
\eqno 3.13 
$$
Moreover, the difference equation in 3.10 may be obtained by applying the corresponding  deformation
to the  difference equation
$$
D_0^*\, \delta_\mu[X ;q,t\, ]\sms {\textstyle{M\over tq}}\del_1\,\,\delta_\mu[X ;q,t\, ] \ses -D_\mu(1/q,1/t)\, \delta_\mu[X
;q,t\, ]
\eqno 3.14 
$$ 
which characterizes our polynomial $\delta_\mu[X;q,t\, ]$.
}
\sas

\heading{\bol Proof}

For convenience, let us set as Sahi does 
$$
\overline \mu(q,t)\ses  \sum_{i=1}^n t^{-n+i}q^{-\mu_i}\ess .
$$
The definition in I.3 can then be rewritten as
$$
MB_\mu(q,t)\ses 1-t^n-t^{n-1}(1-t)\,  {\overline \mu}(q^{-1},t)\ess .
$$ 
Thus from I.4 we obtain that 
$$
D_\nu(q^{-1},t)\ses t^n(1-1/t)\,\, {\overline \nu}(q ,t)- t^n\ess .
\eqno 3.15
$$ 
Making the replacements $q\rightarrow q^{-1}$ and  $\la\RA\nu$ in  I.24, we get
$$
\delta_\mu\Big[t^n(1-1/t)\,\, {\overline \nu}(q ,t)- t^n\,\, ;q^{-1},t\Big]\ses
\cases { \TH_{\nu/\mu}[1;q^{-1},t\, ] & if $\mu\con \nu\,$,\cr\cr
0 & otherwise.
\cr
} 
\eqno 3.16
$$
In particular, we obtain that the polynomial $\delta_\mu\Big[t^n(1-1/t)\,\, X- t^n\,\, ;q^{-1},t\Big]$ 
satisfies the conditions in 3.8 and 3.9 that characterize $R_\mu[X;q,t\, ]$. This proves the identity in 3.12
and thus 3.13 follows from 3.16 . 

To prove 3.14 we start by noting that 1.36 b) gives 
$$
\TAU_\eee^{-1} D_0^*\ses D_0^*\, \TAU_\eee^{-1}\sps D_{-1}^* \TAU_\eee^{-1}\ess .
$$
Thus the Macdonald equation 
$$
D_0^*\, \TH_\mu[X;q,t\, ] \ses -D_\mu({\textstyle{1 \over q},{1\over t}} )\,\TH_\mu[X;q,t\, ] 
$$
may be converted to 
$$
 D_0^*\, \TAU_\eee^{-1}\, \TH_\mu[X;q,t\, ]  \sps D_{-1}^* \TAU_\eee^{-1}\, \, \TH_\mu[X;q,t\, ] 
\ses -D_\mu({\textstyle{1 \over q},{1\over t}} )\, \TAU_\eee^{-1}\TH_\mu[X;q,t\, ] \ess .
$$
Applying $\nabla^{-1}$ to both sides and using the commutativity of $D_0^*$ and $\nabla$, we can write 
$$\displaylines{
 D_0^*\,\nabla^{-1} \TAU_\eee^{-1}\, \TH_\mu [X;q,t\, ]+ \nabla^{-1} D_{-1}^*\nabla \nabla^{-1}  \TAU_\eee^{-1}\, \,
\TH_\mu[X;q,t\, ] 
	\hskip.5in\cr\hskip2.5in
= -D_\mu({\textstyle{1 \over q},{1\over t}})\,\nabla^{-1} \TAU_\eee^{-1}\TH_\mu[X;q,t\, ] \ess .
	\cr}
$$
Thus the definition in I.22 gives
$$
D_0^*\delta_\mu [X;q,t\, ]\sps \nabla^{-1} D_{-1}^*\nabla\,\delta_\mu [X;q,t\, ]\ses
-D_\mu({\textstyle{1 \over q},{1\over t}})\, \delta_\mu [X;q,t\, ]\ess . 
\eqno 3.17
$$ 
On the other hand, 1.24 b) and 1.30 a*) give 
$$
-D_{-1}^*\ses D_0^*\del_1 -\del_1 D_0^*\ses {\textstyle{M\over qt}}\, \nabla \del_1\nabla^{-1}
$$
reducing 3.17 to
$$
D_0^*\delta_\mu [X;q,t\, ]\sms {\textstyle{M\over qt}} \del_1\,\delta_\mu [X;q,t\, ]\ses
-D_\mu({\textstyle{1 \over q},{1\over t}})\, \delta_\mu [X;q,t\, ] 
\eqno 3.18
$$ 
as desired.
\sas

We are left to   show that 3.10 is just another way of writing 3.18. 
To this end, we recall the definition in 1.8 b), and write 3.18 as
$$
\delta_\mu\big[X-{\textstyle{M\over  z qt}}\, ;q,t\, ]\OM[zX]\, \big|_{z^0}\sms 
{\textstyle{M\over   qt}}\, \del_1\, \delta_\mu[X;q,t\, ]\ses -D_\mu({\textstyle{1 \over q},{1\over t}})
\delta_\mu[X;q,t\, ]\ess .
$$
Making the replacements $q\RA1/q\scs X\RA t^n(1-1/t)X-t^n$ gives
$$
\eqalign{
\delta_\mu\big[t^n&(1-1/t)X-t^n\,-\, {\textstyle{(1-1/t)(1-q)\over  z }}\, ;q^{-1},t\, ] \OM[z(t^n(1-1/t)X-t^n)]\,
\big|_{z^0}\cr &  \sms 
{\textstyle{(1-1/t)(1-q)}}\, (\del_1\, \delta_\mu)[t^n(1-1/t)X-t^n;q^{-1},t\, ]\cr
&\hskip 1.2in
=\;-D_\mu({\textstyle{  q},{1\over t}})
\delta_\mu[t^n(1-1/t)X-t^n;q^{-1},t\, ]\ess .
\cr
}
\eqno 3.19
$$
Since
$$
(\del_1\, \delta_\mu)[t^n(1-1/t)X-t^n;q^{-1},t ] \,=\,
{1\over t^n(1-1/t)}\,  \del_1  \Big(\delta_\mu [t^n(1-1/t)X-t^n;q^{-1},t ]\Big)
$$
we can use  3.12 and rewrite  3.19 as
$$
\eqalign{
R_\mu\big[ X \,-\, &{\textstyle{  1-q \over t^n z }}\, ;q ,t\, ]\ssp  \OM\big[z t^n(1-1/t)X\big](1-t^nz)]\,
\big|_{z^0}\cr &\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess \ess\ess\ess\ess\ess\ess\ess\ess \sms 
{\textstyle{  1-q \over t^n}}\,  \del_1\, R_\mu [ X ;q ,t\, ]\ses \cr
&\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess
\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess
-D_\mu({\textstyle{  q},{1\over t}})\,  
R_\mu[X;q  ,t\, ]\, .
\cr
}
$$
Since we can make the replacement $zt^n\RA z$ before taking the coefficient of $z^0$, this
equation is equivalent to
$$
\eqalign{
R_\mu\big[ X \,-\, &{\textstyle{ 1-q\over   z }}\, ;q ,t\, ]\ssp  \OM\big[z  (1-1/t)X\big](1- z)]\,
\big|_{z^0}\cr &\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\sms 
{\textstyle{ 1-q\over t^n}}\,  \del_1\, R_\mu [ X ;q ,t\, ]\ses \cr
&\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess
\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess
-D_\mu({\textstyle{  q},{1\over t}})\,  
R_\mu[X;q  ,t\, ]\, .
\cr
}
\hskip -.5in
\eqno 3.20
$$
Simple manipulations yield that
$$
-t^n\, D_\mu(q,1/t)\, \ses 1-(1-t)\sum_{i=1}^n\, t^{n-i}q^{\mu_i}\ess .
$$
This given, multiplying  3.20 by $\, t^n/(1-t)$, and adding ${t^n\over t-1}R_\mu$
to both sides we finally obtain that
$$
\matrix{
{\textstyle{ t^n\over  t-1 }}\, R_\mu[X;q,t\, ]
	\hfill\cr\hskip.5in
	\;+\;
{\textstyle{ t^n\over  1-t }}\, 
R_\mu\big[ X - {\textstyle{ 1-q\over   z }}\, ;q ,t\, \big]\ssp  \OM\big[z  (1-1/t)X\big](1- z)]\,
\big|_{z^0}
	\hfill
	\cr\hskip 1in
\sms{\textstyle{  1-q \over 1-t}}\,  \del_1\, R_\mu [ X ;q ,t\,]
	\ses
\big({\textstyle{{\,\,\,1-t^n\over 1-t}-\sum_{i=1}^n\, t^{n-i}q^{\mu_i}}}\big)\ess R_\mu[X;q  ,t\, ]
\cr
}
\hskip-.4in
\eqno 3.21
$$
%$$
%\eqalign{
%{\textstyle{ t^n\over  t-1 }}\, R_\mu[X;q,t\, ]+{\textstyle{ t^n\over  1-t }}\, 
%R_\mu\big[ X - &{\textstyle{ 1-q\over   z }}\, ;q ,t\, \big]\ssp  \OM\big[z  (1-1/t)X\big](1- z)]\,
%\big|_{z^0}\cr
%&\ess\ess\ess\ess\ess\ess\ess\ess\bigsp 
%\sms{\textstyle{  1-q \over 1-t}}\,  \del_1\, R_\mu [ X ;q ,t\,
%]\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess\ess 3.21\cr &\ess\ess\ess\ess\ess\ess\bigsp\bigsp
%\ses\big({\textstyle{{\,\,\,1-t^n\over 1-t}-\sum_{i=1}^n\, t^{n-i}q^{\mu_i}}}\big)\ess R_\mu[X;q  ,t\, ]
%\cr
%}
%$$
Now we shall show in Section 4 that
the Sahi operator $\tilde D_1$  may also be given the plethystic form
$$
{\tilde D_1}P[X]= {\textstyle{ t^n\over t-1}}\,P[X]+{\textstyle{ t^n\over  1-t }}\,
P[X-{\textstyle{ 1-q\over   z }}\big]\ssp  \OM\big[z  (1-1/t)X\big](1- z)\,
\big|_{z^0}-{\textstyle{  1-q \over 1-t}}\,  \del_1\,P[X]
	\, .
\eqno 3.22
$$ 
Thus 3.20 reduces to 3.10 as desired, and our proof is complete.
\sa

Our final application is the Macdonald-Koornwinder reciprocity formula
([15] eq.~(6.6) p.~332).
Simple manipulations allow us to state this identity in the following plethystic form.
\sas

\heading{\bol Theorem 3.3}

{\ita For all pairs of partitions $\la\scs \mu$ we have}
$$
\TH_\mu[1+u\, D_\la(q,t);q,t ]\,  \OM[uB_\mu (q,t)]  \,=\,
\TH_\la[1+u\, D_\mu(q,t);q,t ]\, \OM[uB_\la (q,t)] \, .
\eqno 3.23
$$
{\bol Proof}

Since $\OM[XY]$ is the reproducing kernel for the Hall scalar product, we have
$$
\TH_\mu[1+u\, D_\la;q,t\, ]\ses \LL \TH_\mu[X+1;q,t\, ]\scs \OM[uXD_\la ]\RR\ess .
$$
Thus, expressing $\TH_\mu[X+1;q,t\, ]$ by means of our  formula I.20, we obtain 
the following remarkable sequence of equalities.
$$ 
\eqalign{
\TH_\mu[1+u\, D_\la;q,t\, ]\, \OM[uB_\mu ]
	\hskip -.5in\cr
&\ses 
 \OM[uB_\mu  ]\,  
\Big\langle  \PAU_{1/M}\nabla^{-1}\om  \OM\big[{\textstyle{XD_\mu\over M}}\big]\scs \OM[uXD_\la ]\Big\rangle 
\cr
&\ses 
\OM[uB_\mu  ]\,  
\Big\langle  \nabla^{-1}\om  \OM\big[{\textstyle{XD_\mu\over M}}\big]\scs \TAU_{1/M} \OM[uXD_\la ]\Big\rangle 
\cr
&\ses
\OM[uB_\mu  ]\,\OM\big[{\textstyle{uD_\la\over M}}\big]\,
\Big\langle  \nabla^{-1}  \OM\big[{\textstyle{XD_\mu\over M}}\big]\scs \OM[uXD_\la
]\Big\rangle 
\cr
&\ses
\OM[uB_\mu  ]\,\OM[uB_\la  ]\,\OM\big[-{\textstyle{u\over M}}\big]\,
\Big\langle  \nabla^{-1}\om  \OM\big[{\textstyle{XD_\mu\over M}}\big]\scs \OM[uXD_\la
]\Big\rangle 
\cr
{(\rm by\ess  1.40)}&\ses
\OM[uB_\mu  ]\,\OM[uB_\la  ]\,\OM\big[-{\textstyle{u\over M}}\big]\,
\Big\langle  \nabla^{-1}\om  \OM\big[{\textstyle{XD_\mu\over M}}\big]\scs \om \OM\big[{\textstyle{XD_\la\over M}}\big]
\Big\rangle_*
\cr
}
$$
and this proves 3.23 since the last expression is symmetric in $\mu$ and $\la$ by virtue of the
$*$-self-adjointness of the operator $\nabla$.
\sas

As a corollary we immediately obtain our version of the Macdonald specialization ([15] (6.17) p.~338).
\sas

\heading{\bol Theorem 3.4}
$$
\TH_\mu[1-u;q,t\, ]\ses \prod_{s\in\mu}\big(1-u \, t^{l_\mu'(s)}q^{a_\mu'(s)}\big)
\eqno 3.24
$$ 
{\bol Proof}

Simply set $\la=\emptyset$ in 3.23.


\sa

\heading{\bol 4. Auxiliary identities}

We shall begin by converting some of the basic difference operators to plethystic form.
\sas

\heading{\bol Theorem 4.1}

{\ita  For any $P\in \LA$ set
$$
\BH_mP]X]\ses P\big[X-{\textstyle {1\over z}}\big]\OM[zX]\, \big|_{z^m}
\ess .
\eqno 4.1
$$
Then for $\la=(\la_1\geq \la_2\geq \cdots \geq \la_n\geq 0)$ we have
}
$$
S_\la [X]\ses \BH_{\la_1}\BH_{\la_2}\cdots \BH_{\la_n}\cdot 1 \ess .
\eqno 4.2
$$
{\bol Proof}

The bideterminatal formula for Schur functions
$(\dag)$\FOOTNOTE{}{$(\dag)$\ess (3.1) p.~40 of [15]}
may be written in the form
$$
S_{\la_1,\ldots ,\la_n}[X_n]\ses \sum_{\sig\in S_{[1,n]}}\sig 
\Big( {x_{1}^{n-1+\la_1}x_{2}^{n-2+\la_2}\cdots x_{n}^{n-n+\la_n} \over 
\prod_{1\leq i<j \leq n}(x_{i}-x_{j})}\Big).
\eqno 4.3
$$
Now, by means of the left coset decomposition 
$$\sum_{\sig\in S_{[1,n]}}\sig \ses\sum_{i=1}^n\,  (i,1)\, \sum_{\aaa\in S_{[2,n]}}\aaa \ess ,
$$
we can readily transforms 4.3 into the recursion  
$$
S_{\la_1,\ldots ,\la_n}[X_n]\ses \sum_{i=1}^n\, {x_i^{\la_1+n-1}\over \prod_{\multi{j=1\cr j\neq i}}^n(x_i-x_j) }
\,\, S_{\la_2,\ldots ,\la_n}[X_n-x_i\,]\ess .
\eqno 4.4
$$
Let us then set for $P\in \LA$
$$
\BH_m^{(n)}P[X_n  ]\ses \sum_{i=1}^n\,A_i(x)\, x_i^m\,\,
\,\,  P[X_n-x_i\, ]\ess ,
\eqno 4.5
$$
where for convenience we let
$$
A_i(x)\ses {x_i^{n-1}\over \prod_{\multi{j=1\cr j\neq i}}^n(x_i-x_j) } \ess .
$$
This given, to prove 4.2 we only need to show that $\BH_m^{(n)}$ also has  the plethystic form
$$
\BH_m^{(n)}\, P[X_n]\ses P\big[X_n-{\textstyle {1\over z}}\big]\OM[zX_n]\, \big|_{z^m}\ess .
$$
To this end we note that we can write, for an arbitrary alphabet $Y$ 
$$
\eqalign{
{\BH_m^{(n)}\,\OM[X_nY]\over \OM[X_nY]}&\ses\sum_{i=1}^n\, A_i(x)\, x_i^m\,\OM\big[-x_iY\big] \cr
&\ses\sum_{i=1}^n\, A_i(x)\, x_i^m\, \sum_{k\geq 0}x_i^k\, h_k[-Y]  \cr 
&\ses
\sum_{k\geq 0} \, h_k[-Y]
\sum_{i=1}^n\, A_i(x)\, x_i^{m+k}\, .\cr
}
\eqno 4.6
$$
Now from the partial fraction expansion
$$
\OM[zX_n]\ses \prod_{i=1}^n{1\over 1-zx_i}\ses \sum_{i=1}^n\, A_i(x)\, {1\over 1-zx_i}\ess ,
$$
we derive that for all $m+k\geq 0$, we have
$$
\sum_{i=1}^n\, A_i(x)\, x_i^{m+k}\ses 
\OM[zX_n]\, \big|_{z^{m+k}}\ess .
$$
Substituting this in 4.6 gives 
$$
\eqalign{
{\BH_m^{(n)}\,\OM[X_nY]\over \OM[X_nY]}
&\ses\sum_{k\geq 0} \, h_k[-Y]\, \OM[zX_n]\, \big|_{z^{m+k}}
\cr
&\ses\sum_{k\geq 0} \, \OM\big[-{\textstyle{1\over z}}Y\big]\, \big|_{z^{-k }}\, \OM[zX_n]\, \big|_{z^{m+k}}
\cr
&\ses \OM\big[-{\textstyle{1\over z}}Y\big] \, \OM[zX_n]\, \big|_{z^{m }}\ess . 
\cr
}
$$
That is 
$$
\BH_m^{(n)}\,\OM[X_nY]\ses \OM\big[(X_n-{\textstyle{1\over z}})Y\big] \, \OM[zX_n]\, \big|_{z^{m }}\ess .
$$
Equating coefficients of $S_\la[Y]$ on both sides of this equation yields that 4.5 is true for the Schur functions
and therefore must be true for all $P\in \LA$ as desired, completing our proof
\sas

We shall prove next a similar result for the Macdonald and Sahi operators.
\sas

\heading{\bol Theorem 4.2}

{\ita For any $P\in \LA$ set
$$
D_1^{(n)}P[X_n]\ses \sum_{i=1}^n \,\,  A_i(x;t)\,T_q^{(i)} 
\eqno 4.7
$$
and
$$
{\tilde D}_1^{(n)}P[X_n]\ses \sum_{i=1}^n \, A_i(x;t)\big({\textstyle 1-{1\over x_i}}\big)(1-T_q^{(i)})
\eqno 4.8
$$
where 
$$
A_i(x;t)\ses \prod_{\multi{j=1\cr j\neq i}}^n\, {tx_i-x_j\over x_1-x_j }\ess .
\eqno 4.9
$$
Then for all $P\in \LA$ we have
$$
D_1^{(n)}P[X_n]= {\textstyle{ 1\over  1-t }}\,P[X_n]-{\textstyle{ t^n\over  1-t }}\,
P[X_n-{\textstyle{ 1-q\over   z }}\big]\ssp  \OM\big[z  (1-1/t)X_n\big]\,
\big|_{z^0}\ess
,
\eqno 4.10
$$
and}
$$\eqalign{
{\tilde D_1}^{(n)}P[X_n]
	&=
	 {\textstyle{ t^n\over  t-1 }}\,P[X_n]
		+
	{\textstyle{ t^n\over  1-t }}\,
	P[X_n-{\textstyle{ 1-q\over   z }}\big]\ssp
		  \OM\big[z  (1-1/t)X_n\big](1- z)]\,\big|_{z^0}
	\cr&\hskip .5in
		-
	{\textstyle{  1-q \over 1-t}}\,  \del_1\,P[X_n]\ess
.
}
	\hskip -.2em
\eqno 4.11
$$ 
{\bol Proof}

The crucial ingredient here is the partial fraction expansion
$$
\OM\big[({\textstyle{1-{1/t}}})zX_n\big] \ses \prod_{i=1}^n\, {1-zx_i/t\over1-zx_i }
\ses   {1\over t^n} \sps  {t-1\over t^n}\, \sum_{i=1}^n\, {A_i(x;t)\over 1-z x_i}\ess ,
\eqno 4.12
$$
which gives
$$
\sum_{i=1}^n\,A_i(x;t)\, x_i^m\ses {t^n\over t-1}\,\OM\big[({\textstyle{1-{1/t}}})zX_n\big]\, \Big|_{z^m}
\bigsp (\hbox{for all $m\geq 1$}) \ssp .
\eqno 4.13
$$
We should also note that setting $z=0$ in 4.12 yields 
$$
\sum_{i=1}^n\,A_i(x;t)\ses {t^n-1 \over t-1}\ess . 
\eqno 4.14
$$
This given we have
$$
\eqalign{
{{\tilde D_1}^{(n)}\OM[X_nY]\over \OM[X_nY]}
&\ses 
\sum_{i=1}^n\,A_i(x;t)({\textstyle1-{1\over x_i}})\big(1-\OM[(q-1)x_iY]\big)
\cr
&\ses 
\sms\sum_{i=1}^n\,A_i(x;t)({\textstyle1-{1\over x_i}})\sum_{m\geq 1}\, h_m[(q-1)Y]\, x_i^m
\cr
&\ses 
\sms\sum_{m\geq 1}\, h_m[(q-1)Y]\sum_{i=1}^n\,A_i(x;t)({\textstyle1-{1\over x_i}})  \, x_i^m\ess  .
\cr}
\eqno 4.15
$$
Now using 4.13 we get that
$$
\eqalign{
\sum_{m\geq 1}\, h_m[(q-1)Y]\sum_{i=1}^n\,A_i(x;t)   \, x_i^m
	\hskip-1.2in\cr
&= {\textstyle{t^n\over t-1}} \sum_{m\geq 1}\, h_m[(q-1)Y]\,\OM\big[({\textstyle{1-{1/t}}})zX_n\, \big]\, \Big|_{z^m}
\cr
&= {\textstyle{t^n\over t-1}} \sum_{m\geq 0}\, h_m[(q-1)Y]\,\OM\big[({\textstyle{1-{1/t}}})zX_n\, \big]\, \Big|_{z^m}
- {\textstyle{t^n\over t-1}}
\cr
&= {\textstyle{t^n\over t-1}} \sum_{m\geq 0}\, \OM[(q-1)Y/z]\,\Big|_{z^{-m}}
\OM\big[({\textstyle{1-{1/t}}})zX_n\, \big]\, \Big|_{z^m}
- {\textstyle{t^n\over t-1}}
\cr
&= {\textstyle{t^n\over t-1}}  \OM[(q-1)Y/z]\, 
\OM\big[({\textstyle{1-{1/t}}})zX_n\, \big]\, \Big|_{z^0}
-{\textstyle{t^n\over t-1}}
%\ess\ess\ess\ess\ess\ess\ess\ess 4.16
\cr}
	\hskip -1em
\eqno 4.16
$$
Similarly, using 4.13 and  4.14  we get
$$
\eqalign{
\sum_{m\geq 1}\, h_m[(q-1)Y]\sum_{i=1}^n\,A_i(x;t)   \, x_i^{m-1}
	\hskip-1.9in\cr
&= {\textstyle{t^n\over t-1}} \sum_{m\geq 2} h_m[(q-1)Y] \OM\big[({\textstyle{1-{1/t}}})zX_n\, \big]\,  \Big|_{z^{m-1}}
\hskip -.04in+ {\textstyle{(q-1)( t^n-1)\over  t-1}}e_1[Y]
\cr
&= {\textstyle{t^n\over t-1}} \sum_{m\geq 1} h_m[(q-1)Y] \OM\big[({\textstyle{1-{1/t}}})zX_n\, \big]\,  \Big|_{z^{m-1}}
\hskip -.04in- {\textstyle{ q-1  \over t-1}}e_1[Y]
\cr
&= {\textstyle{t^n\over t-1}}  \OM [(q-1)Y/z\, ] \OM\big[({\textstyle{1-{1/t}}})zX_n\, \big]\,  \Big|_{z^{-1} }
\hskip -.04in- {\textstyle{ q-1  \over t-1}}e_1[Y]
\cr
}
\eqno 4.17
$$
Substituting 4.16 and 4.17 into 4.15 gives
$$
{{\tilde D_1}^{(n)}\OM[X_nY]\over \OM[X_nY]}
	\,=\,
{\textstyle{t^n\over t-1}}+
{\textstyle{t^n\over 1-t}}  \OM [(q-1)Y/z\, ] \OM\big[({\textstyle{1-{1/t}}})zX_n\, \big]\, (1-z) \Big|_{z^0 }
-\ssp
{\textstyle{ q-1  \over t-1}}e_1[Y]\, .
$$
In other words, we must have
$$\eqalign{
{\tilde D_1}^{(n)}\OM[X_nY]&=
	 {\textstyle{t^n\over t-1}}\OM[X_nY]
		+
	{\textstyle{t^n\over 1-t}}\,\OM \big[{\textstyle{(X-{q-1\over z}}})Y\, ]
	\OM\big[({\textstyle{1-{1/t}}})zX_n\, \big]\, (1-z)\Big|_{z^0 }
	\cr & \hskip .8in
		 -
	{\textstyle{ q-1  \over t-1}}\del_1\OM[X_nY]\, ,
}
\eqno 4.18
$$
since
$$
e_1[Y]\OM[X_nY]\ses \del_1\OM[X_nY]\ess .
$$
Equating  coefficients of $S_\la[Y]$ in 4.18 proves 4.11 for the Schur function basis 
and therefore establishes the validity of 4.11 for all symmetric polynomials.

To prove 4.10, note first that it follows immediately from the definitions in 4.7 and 4.8
that the Macdonald and Sahi operators are related by the identity 
$$
 {}^{top}{\tilde D}_1^{(n)}\ses \sum_{i=1}^n A_i\ssp - D_1^{(n)}\ess . 
$$
Where the symbol ``${}^{top}{\tilde D}_1^{(n)}$'' is to represent the 
highest homogeneous component of ${\tilde D}_1^{(n)}$. Using 4.14 this can be written
as 
$$
D_1^{(n)}\ses {\textstyle{\,\,\, 1- t^n\over  1-t }}\ssp - \ ^{top}{\tilde D}_1^{(n)}\ess . 
\eqno 4.19
$$ 
Now from 4.11 we derive that
$$
 {}^{top}{\tilde D}_1^{(n)}P[X_n]\ses
{\textstyle{ t^n\over  t-1 }}\,P[X_n]+{\textstyle{ t^n\over  1-t }}\,
P[X_n-{\textstyle{ 1-q\over   z }}\big]\ssp  \OM\big[z  (1-1/t)X_n\big]\, \big|_{z^0}\ess.
\eqno 4.20 
$$
and 4.10 follows by combining 4.19 with 4.20. This completes our proof.
\sas

We can now complete our
\sas

\heading{\bol Proof of Theorem 1.2}

We start by recalling the Macdonald identity (see [15] (4.15) p.~324)
$$
D_1^{(n)}P_\mu[X_n;q,t\,]\ses \Big(\sum_{i=1}^n t^{n-i}q^{\mu_i}\Big)\,P_\mu[X_n;q,t\,]\ess . 
\eqno 4.21 
$$
Now the ``integral form'' $J_\mu[X;q,t\, ]$ defined in p.~352 of [15], can be written as
$$
J_\mu[X;q,t\, ]\ses h_\mu(q,t)P_\mu[X_n;q,t\,]\ses h_\mu'(q,t) Q_\mu(X;q,t)\ess ,
\eqno 4.22
$$
with
$$
h_\mu(q,t)= \prod_{s\in \mu}\big(1-q^{a_\mu(s)}t^{l_\mu(s)+1}\big)
\ess\ess \hbox{and}\ess\ess 
h_\mu'(q,t)= \prod_{s\in \mu}\big(1-t^{l_\mu(s)}q^{a_\mu(s)+1}\big)\, .
\eqno 4.23
$$
Combining, 4.22 with 4.21 and 4.10 we derive that
$$\eqalign{
 {\textstyle{ 1\over  1-t }}\,J_\mu[X_n;q,t\,]
		-
  {\textstyle{ t^n\over  1-t }}\,
	  J_\mu[X_n-{\textstyle{ 1-q\over   z }};q,t\,\big]\ssp
	  \OM\big[z  (1-1/t)X_n\big]\,\big|_{z^0}
	\hskip-1.3in\cr
	&=
 \Big(\sum_{i=1}^n t^{n-i}q^{\mu_i}\Big)\,J_\mu[X_n;q,t\,]\ssp .
}
\eqno 4.24
$$
Making the replacements $t\RA 1/t$ and $X_n\RA X_n/(1-1/t)$ and multiplying both sides by $t^{n-1}$
we can write this in the form
$$
 \eqalign{
{\textstyle{ t^n\over  t-1 }}\,J_\mu[{\textstyle{X_n  \over  1-1/t }};q,1/t\,] -{\textstyle{ 1\over  t-1 }} \,
J_\mu[{\textstyle{ X_n +(1-t)(1-q)/tz\over   1-1/t }};&q,1/t\,\big]\ssp   \OM\big[-tz X_n\big]\,
\big|_{z^0}\cr
&\hskip-.5in= \Big(\sum_{i=1}^n t^{i-1}q^{\mu_i}\Big)\,J_\mu[{\textstyle{X_n \over 1-1/t }};q,1/t\,]\ess .
\cr
}
$$
Multiplying by $1-t$ and making the replacement $tz\RA z$, before taking the coefficient of $z^0$,
from I.7 we get that
$$\eqalign{
-t^n\TH_\mu[X_n;q,t\,]\sps \TH_\mu[X_n+{\textstyle{M  \over  z }};q,t\, ]\OM\big[-z X_n\big]\,
\big|_{z^0}
	\hskip-1.5in\cr &=
 \Big((1-t)\sum_{i=1}^n t^{i-1}q^{\mu_i}\Big)\,\TH_\mu[X_n;q,t\,]\ess.
}
\eqno 4.25
$$
Now simple manipulations give the identity
$$
 (1-t)\sum_{i=1}^n t^{i-1}q^{\mu_i} \ses -t^n-D_\mu(q,t)\ess .
$$
Substituting this in 4.25 finally yields  
$$
 \TH_\mu[X_n+{\textstyle{M  \over  z }};q,t\, ]\OM\big[-z X_n\big]\,
\big|_{z^0}\ses-D_\mu(q,t)\,\TH_\mu[X_n;q,t\,]\ess,
$$
which is 1.11 a). We have seen (Remark 1.1) that 1.11 a) implies 1.1 b). Thus the only thing that
remains is to verify the normalization in 1.12. However, this follows from the identity
$$
K_{(n),\mu}(q,t)\ses t^{n(\mu)}
$$ 
which is proved in ex.~2, p.~362 of [15]. 
\sas

\heading{\bol Proof of 1.15}

It is shown in [15] ((5.13) (iv) p.~324) that
$$
P_\mu[X ;q,t\, ]\ses P_\mu[X ;1/q,1/t\, ] \ess .
$$
This given, from 4.22 and 4.23 we get that
$$
\eqalign{
J_\mu[X ,q,t]&\ses h_\mu(q,t)\, P_\mu[X ;1/q,1/t\, ]\cr
&\ses t^{n(\mu)}q^{n(\mu')}(-t)^{|\mu|}h_\mu(1/q,1/t)\,P_\mu[X ;1/q,1/t\, ]\cr 
&\ses  t^{n(\mu)}q^{n(\mu')}(-t)^{|\mu|}J_\mu[X ,1/q,1/t] \ess .
\cr
}
$$
Thus making the replacements $t\RA1/t\scs X \RA X /(1-1/t)$ and using I.7 we obtain
$$
\eqalign{
\TH_\mu[X_n,q,t]&\ses 
q^{n(\mu')}(-t)^{-|\mu|}J_\mu[{\textstyle{X \over 1-1/t}},1/q,t\, ]\cr
&\ses 
q^{n(\mu')}(-1)^{|\mu|}J_\mu[{\textstyle{-X \,\,\over 1-t}},1/q,t\, ]\cr
&\ses 
q^{n(\mu')}t^{n(\mu)}(-1)^{|\mu|}\TH_\mu[{\textstyle{-X }},1/q,1/t\, ]\ess ,
\cr
}
$$
and I.2 gives 1.15.
\sas

We terminate with the  
\sas

\heading{\bol Proof of formula 1.18 c)}

The starting point is the Macdonald ``Cauchy'' formula ((4.13) p.~324 of [15])
$$
\OM\big[XY{\textstyle{1-t\over 1-q}}\big]\ses \sum_{\mu}\, P_\mu[X;q,t]Q_\mu[Y;q,t]\ess .
$$
Using 4.22 we can rewrite this as
$$
\OM\big[XY{\textstyle{1-t\over 1-q}}\big]\ses \sum_{\mu}\, {J_\mu[X;q,t]J_\mu[Y;q,t] \over h_\mu(q,t)h_\mu'(q,t)}\ess .
$$
Making the replacements $t\RA1/t$ and then $X\RA X/(1-1/t)\scs Y\RA Y/(1-1/t)$
we get (recalling I.23)
$$
\eqalign{
\OM\big[{\textstyle{-t\, XY\over(1-t)( 1-q)}}\big]
&\ses 
\sum_{\mu}\, {J_\mu[{\textstyle{X\over 1-1/t }};q,1/t]J_\mu[{\textstyle{Y\over 1-1/t }};q,1/t] \over
\th_\mu(q,t)\th_\mu'(q,t)}(-t)^{|\mu|}t^{2n(\mu)}
\cr
&\ses 
\sum_{\mu}\, {\TH_\mu[{\textstyle{X }};q, t\,]\TH_\mu[{\textstyle{Y }};q, t\, ] \over
\th_\mu(q,t)\th_\mu'(q,t)}(-t)^{|\mu|} 
\cr
&\ses 
\sum_{\mu}\, {\TH_\mu[{\textstyle{ ^-tX }};q, t\,]\TH_\mu[{\textstyle{Y }};q, t\, ] \over
\th_\mu(q,t)\th_\mu'(q,t)} \ess ,
\cr
}
$$
and 1.18 c) follows by making the replacement $ ^-tX\RA X$ and using  1.17.
\sa

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\sas 

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\end

