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\begin{document}
\title[Inner product for vector-valued Macdonald polynomials]{A positive-definite inner product for vector-valued
Macdonald polynomials}
\author{Charles F. Dunkl}
\address{Department. of Mathematics, University of Virginia  \\
Charlottes\-ville
VA 22904-4137, US}
\email[C.D.]{ cfd5z@virginia.edu}
\urladdr{http://people.virginia.edu/\symbol{126}cfd5z/home.html}
\date{}
\subjclass[2010]{Primary 20C08, 33D52; Secondary 05E10, 33D80}
\keywords{Hecke algebra, nonsymmetric Macdonald
polynomials}
\begin{abstract}In a previous paper J.-G. Luque and the author (Sem.\ Loth.\ Combin.\ 2011) developed the theory of nonsymmetric Macdonald polynomials
taking values in an irreducible module of the Hecke algebra of the symmetric group $\mathcal{S}_{N}$. The polynomials are parametrized by $\left (q ,t\right )$ and are simultaneous eigenfunctions of a commuting set of Cherednik
operators, which were studied by Baker and Forrester (IMRN 1997). In the Dunkl--Luque paper there is a construction of a pairing between $\left (q^{ -1} ,t^{ -1}\right )$-polynomials and $\left (q ,t\right )$-polynomials, and for which the Macdonald polynomials form a biorthogonal
set. The present work is a sequel with the purpose of constructing a symmetric bilinear form for which the Macdonald polynomials form an orthogonal basis
and of determining the region of $\left (q ,t\right )$-values for which the form is positive-definite. Irreducible representations
of the Hecke algebra are characterized by partitions of $N$. The positivity region depends only on the maximum hook-length of the Ferrers diagram of the partition. 
\end{abstract}
\maketitle

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\markright{\its S\'eminaire Lotharingien de
Combinatoire \bfs 80 \rms (2019), Article~B80a\hfill}
\def\thepage{}

\section{Introduction}
The theory of nonsymmetric Jack polynomials was generalized by Griffeth \cite{G2010}
to polynomials on the complex reflection groups of type $G \left (n ,p ,N\right )$ taking values in irreducible modules of the groups. This theory simplifies somewhat for the group $G \left (1 ,1 ,N\right )$, the symmetric group of $N$ objects, where any irreducible module is spanned by standard Young tableaux all of the same shape, corresponding to a partition
of $N$. Luque and the author \cite{DL2012} developed
an analogous theory for vector-valued Macdonald polynomials taking values in irreducible modules of the Hecke algebra of a symmetric group. The structure
has parameters $\left (q ,t\right )$ and depends on a commuting set of Cherednik operators whose simultaneous
eigenfunctions are the aforementioned Macdonald polynomials. The paper showed how to construct the polynomials by means of a Yang--Baxter graph (see \cite{L2001}). Also a bilinear form was defined which paired polynomials
for the parameters $\left (q^{ -1} ,t^{ -1}\right )$ with those parametrized by $\left (q ,t\right )$ and resulted in biorthogonality relations for the Macdonald polynomials.
The present paper is a sequel whose aim is to define a symmetric bilinear form for which these polynomials are mutually orthogonal. Some other natural conditions
are imposed on the form to force uniqueness. The form is positive-definite for a $\left (q ,t\right )$-region determined by the specific module. 

For
purposes of illustration the form is first defined for the scalar case, and leads to expressions only slightly different from the well-known hook-product
formulas. For the trivial representation of the Hecke algebra, corresponding to the one-part partition, the vector-valued polynomials specialize to the scalar polynomials. Section~\ref{VVMD} contains a short outline of representation theory of the Hecke algebra, the Yang--Baxter graph
of vector-valued Macdonald polynomials and the process leading to the definition of the symmetric bilinear form, followed by the characterization of $\left (q ,t\right )$-values yielding positivity of the form. The details of the construction
of the polynomials and related operators along with the proofs of their properties are found in \cite{DL2012}.

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\markboth{\SMALL CHARLES F. DUNKL}{\SMALL INNER PRODUCT FOR
VECTOR-VALUED MACDONALD POLYNOMIALS}


\subsection{Notation}
Let $\mathbb{N}_{0}:=\left \{0 ,1 ,2 ,3 ,\ldots \right \}$. The elements of $\mathbb{N}_{0}^{N}$ are called \textit{compositions,} and for \textit{}$\alpha  =\left (\alpha _{1} ,\ldots  ,\alpha _{N}\right ) \in \mathbb{N}_{0}^{N}$ let $\left \vert \alpha \right \vert :=\sum _{i =1}^{N}\alpha _{i}$. Let $\mathbb{N}_{0}^{N , +}$ denote the set of partitions $\left \{\lambda  \in \mathbb{N}_{0}^{N} :\lambda _{1} \geq \lambda _{2} \geq \ldots  \geq \lambda _{N}\right \}$, and let $\alpha ^{ +}$ denote the nonincreasing rearrangement of $\alpha $; for example, if $\alpha  =\left (1 ,2 ,1 ,4\right )$, then $\alpha ^{ +} =\left (4 ,2 ,1 ,1\right )$. There are two partial orders on compositions used in this work: for $\alpha  ,\beta  \in \mathbb{N}_{0}^{N}$ the relation $\alpha  \succ \beta $ means $\alpha  \neq \beta $ and $\sum _{i =1}^{j}\left (\alpha _{i} -\beta _{i}\right ) \geq 0$ for $1 \leq j \leq N$ (the dominance order), and $\alpha  \rhd \beta $ means $\left \vert \alpha \right \vert  =\left \vert \beta \right \vert $ and $\alpha ^{ +} \succ \beta ^{ +}$, or $\alpha ^{ +} =\beta ^{ +}$ and $\alpha  \succ \beta $. The rank function for $\alpha  \in \mathbb{N}_{0}^{N}$ is\begin{equation}r_{\alpha } \left (i\right ):=\# \left \{j :\alpha _{j} >\alpha _{i}\right \} +\# \left \{j :1 \leq j \leq i ,\alpha _{j} =\alpha _{i}\right \} ,~1 \leq i \leq N .
\end{equation} 
We have $\alpha  =\alpha ^{ +}$ if and only if $r_{\alpha } \left (i\right ) =i$ for all $i$. 

The symmetric group $\mathcal{S}_{N}$ is generated by the adjacent transpositions $s_{i}:=\left (i ,i +1\right )$ for $1 \leq i <N$, where $s_{i}$ acts on an $N$-tuple $a =\left (a_{1} ,\ldots  ,a_{N}\right )$ by $a .s_{i} =\left (\ldots  ,a_{i +1} ,a_{i} ,\ldots \right )$, interchanging entries $\# i$ and $\# \left (i +1\right )$. For a composition $\alpha  \in \mathbb{N}_{0}^{N}$ the inversion number is $\mathrm{i} \mathrm{n} \mathrm{v} \left (\alpha \right ):=\# \left \{\left (i ,j\right ) :1 \leq i <j \leq N ,\alpha _{i} <\alpha _{j}\right \}$. If $\alpha _{i} <\alpha _{i +1}$ then $\mathrm{i} \mathrm{n} \mathrm{v} \left (\alpha  .s_{i}\right ) =\mathrm{i} \mathrm{n} \mathrm{v} \left (\alpha \right ) -1$.

The space of polynomials is $\mathcal{P}:=\mathbb{K} \left [x_{1} ,x_{2} ,\ldots  ,x_{N}\right ]$, where $\mathbb{K}:=\mathbb{Q} \left (q ,t\right )$ and $q ,t$ are transcendental or generic, that is, complex numbers satisfying $q \neq 1 $ and $q^{a} t^{b} \neq 1$ for $a ,b \in \mathbb{Z}$ and $ -N \leq b \leq N$. For $\alpha  \in \mathbb{N}_{0}^{N}$ we write $x^{\alpha }$ for the \textit{monomial} $\prod \limits _{i =1}^{N}x_{i}^{\alpha _{i}}$. The space of homogeneous polynomials of degree $n$ is defined as $\mathcal{P}_{n} : =\mathrm{s} \mathrm{p} \mathrm{a} \mathrm{n}_{\mathbb{K}} \left \{x^{\alpha } :\left \vert \alpha \right \vert  =n\right \}$ for $n =0 ,1 ,2 ,\ldots $. The group $\mathcal{S}_{N}$ acts on polynomials by permutation of coordinates, $p \left (x\right ) \rightarrow \left (p s_{i}\right ) \left (x\right ):=p \left (x .s_{i}\right )$. 

The Hecke algebra $\mathcal{H}_{N} \left (t\right )$ is the associative algebra generated by\break 
$\left \{T_{1} ,T_{2} ,\ldots  ,T_{N -1}\right \}$ subject to the relations
\begin{align}\left (T_{i} +1\right ) \left (T_{i} -t\right ) &=0 , \label{Hrel} \\
T_{i} T_{i +1} T_{i} &=T_{i +1} T_{i} T_{i +1} ,\quad 1 \leq i \leq N -2 , \nonumber  \\
T_{i} T_{j} &=T_{j} T_{i} ,\quad 1 \leq i <j -1 \leq N -2. \nonumber \end{align}
The quadratic relation implies $T_{i}^{ -1} =\frac{1}{t} \left (T_{i} +1 -t\right ) \in \mathcal{H}_{N} \left (t\right )$ . For generic $t$ there is a linear isomorphism $\mathbb{K} \mathcal{S}_{N} \rightarrow \mathcal{H}_{N} \left (t\right )$ generated by $s_{i} \rightarrow T_{i}$. 

For $p \in \mathcal{P}$ and $1 \leq i <N$ define
\begin{equation}p \left (x\right ) T_{i}:=\left (1 -t\right ) x_{i +1} \frac{p \left (x\right ) -p \left (x .s_{i}\right )}{x_{i} -x_{i +1}} +t p \left (x .s_{i}\right ) . \label{pxTi}
\end{equation}It can be shown straightforwardly that these operators satisfy the defining relations of $\mathcal{H}_{N} \left (t\right )$. Also $p s_{i} =p$ (symmetry in $\left (x_{i} ,x_{i +1}\right )$) if and only if $p T_{i} =t p$ (because $p T_{i} -t p =\frac{t x_{i} -x_{i +1}}{x_{i} -x_{i +1}} \left (p -p s_{i}\right )$), and $p T_{i} = -p$ if and only if $p \left (x\right ) =\left (t x_{i} -x_{i +1}\right ) p_{0} \left (x\right )$ where $p_{0} \in \mathcal{P}$ and $p_{0} s_{i} =p_{0}$. Also $x_{i} T_{i} =x_{i +1}$ and $1 T_{i} =t$. 

\section{\label{SMD}Scalar nonsymmetric Macdonald polynomials}
For $f \in \mathcal{P}$ define shift, Cherednik and Dunkl operators by (see \cite{BF1997},
and also \cite{DL2012})
\begin{align}f w \left (x\right ) & :=f \left (q x_{N} ,x_{1} ,x_{2} ,\ldots  ,x_{N -1}\right ) , \nonumber  \\
f \xi _{i} & :=t^{i -1} f T_{i -1}^{ -1} T_{i -2}^{ -1} \cdots  T_{1}^{ -1} w T_{N -1} T_{N -2} \cdots  T_{i} , \label{defxi0} \\
f \mathcal{D}_{N} & :=\left (f -f \xi _{N}\right )/x_{N} ,~f \mathcal{D}_{i}:=\frac{1}{t} f T_{i} \mathcal{D}_{i +1} T_{i} . \nonumber \end{align}
Note that $\xi _{i} =\frac{1}{t} T_{i} \xi _{i +1} T_{i}$. It is a nontrivial result that $D_{i}$ maps $\mathcal{P}_{n}$ to $\mathcal{P}_{n -1}$. The operators $\xi _{i}$ commute with each other and there is a basis of simultaneous eigenfunctions, the nonsymmetric Macdonald polynomials $M_{\alpha }$, labeled by $\alpha  \in \mathbb{N}_{0}^{N}$ with $ \rhd $-leading term $q^{a}t^{b}x^{\alpha }$ with $\alpha  ,\beta  \in \mathbb{N}_{0}$ such that
\begin{align}M_{\alpha } \left (x\right ) &  =^{}q^{a}t^{b}  x^{\alpha } +\sum _{\alpha  \rhd \beta }A_{\alpha  \beta } \left (q ,t\right ) x^{\beta } \\
M_{\alpha } \xi _{i} &  =q^{\alpha _{i}} t^{N -r_{\alpha } \left (i\right )} M_{\alpha } ,1 \leq i \leq N ; \nonumber \end{align}where the coefficients $A_{\alpha  \beta } \left (q ,t\right )$ are rational functions of $\left (q ,t\right )$ whose denominators are of the form $\left (1 -q^{a} t^{b}\right )$. The \textit{spectral vector} is $\zeta _{\alpha } \left (i\right ) =q^{\alpha _{i}} t^{N -r_{\alpha } \left (i\right )}$, $1 \leq i \leq N$. There is a simple relation between $M_{\alpha }$ and $M_{\alpha  .s_{i}}$ when $\alpha _{i} <\alpha _{i +1}$ and $\rho  =\zeta _{\alpha } \left (i +1\right )/\zeta _{\alpha } \left (i\right ) =q^{\alpha _{i +1} -\alpha _{i}} t^{r_{\alpha } \left (i\right ) -r_{\alpha } \left (i +1\right )}$, namely
\begin{align}M_{\alpha } T_{i} &  =M_{\alpha  s_{i}} -\frac{1 -t}{1 -\rho } M_{\alpha } , \label{MT1} \\
M_{\alpha  .s_{i}} T_{i} &  =\frac{\left (1 -\rho  t\right ) \left (t -\rho \right )}{\left (1 -\rho \right )^{2}} M_{\alpha } +\frac{\rho  \left (1 -t\right )}{\left (1 -\rho \right )} M_{\alpha  .s_{i}} , \label{MT2}\end{align}and $\zeta _{\alpha  .s_{i}} =\zeta _{\alpha } .s_{i}$. If $\alpha _{i} =\alpha _{i +1}$ then
\begin{equation}M_{\alpha } T_{i} =t M_{\alpha } . \label{MT3}
\end{equation}

The other step needed to construct any $M_{\alpha }$ starting from $1$ is the affine step
\begin{align}M_{\alpha  \Phi } &  =x_{N} \left (M_{\alpha } w\right ) , \label{MT4} \\
\alpha  \Phi  & :=\left (\alpha _{2} ,\alpha _{3} ,\ldots  ,\alpha _{N} ,\alpha _{1} +1\right ) , \nonumber  \\
\zeta _{\alpha  \Phi } &  =\left (\zeta _{\alpha } \left (2\right ) ,\ldots  ,\zeta _{\alpha } \left (N\right ) ,q \zeta _{\alpha } \left (1\right )\right ). \nonumber \end{align}

Formulas \eqref{MT1} and \eqref{MT4}
can be interpreted as edges of a Yang--Baxter graph for generating the polynomials (see \cite[Sec.~9]{L2001}). This graph has the root $\left (\mathbf{0} ,\left [t^{N -i}\right ]_{i =1}^{N} ,1\right )$ and nodes $\left (\alpha  ,\zeta _{\alpha } ,M_{\alpha }\right )$. There are steps $\left (\alpha  ,\zeta _{\alpha } ,M_{\alpha }\right )\overset{s_{i}}{ \rightarrow }\left (\alpha  .s_{i} ,\zeta _{\alpha } .s_{i} ,M_{\alpha  .s_{i}}\right )$ for $\alpha _{i +1} >\alpha _{i}$ given by 
$$M_{\alpha  s_{i}} =M_{\alpha } T_{i} +\frac{1 -t}{1 -\zeta _{\alpha } \left (i +1\right )/\zeta _{\alpha } \left (i\right )} M_{\alpha },$$ 
and affine steps $\left (\alpha  ,\zeta _{\alpha } ,M_{\alpha }\right )\overset{\Phi }{ \rightarrow }\left (\alpha  \Phi  ,\zeta _{\alpha  \Phi } ,M_{\alpha  \Phi }\right )$ (given by \eqref{MT4}). 

There is a short proof using Macdonald polynomials
that $\mathcal{D}_{N}$ maps $\mathcal{P}_{N}$ to $\mathcal{P}_{N -1}$: when $\alpha _{N} =0$ then $r_{\alpha } \left (N\right ) =N$, $\zeta _{\alpha } \left (N\right ) =1$ and $M_{\alpha } \xi _{N} =M_{\alpha }$, thus $M_{\alpha } \mathcal{D}_{N} =0$; if $\alpha _{N} \geq 1$ then the raising (affine) formula is $M_{\alpha } \left (x\right ) =x_{N} M_{\beta } w \left (x\right )$ where $\beta  =\left (\alpha _{N} -1 ,\alpha _{1} ,\alpha _{2} ,\ldots  ,\alpha _{N -1}\right )$, thus 
$$M_{\alpha } \left (1 -\xi _{N}\right ) =\left (1 -\zeta _{\alpha } \left (N\right )\right ) M_{\alpha },$$ 
which is divisible by $x_{N}$. 

Our logical outline is to first state a number of hypotheses to be satisfied by the inner product, then deduce
consequences leading to a formula which is used as a definition. To finish one has to show that the hypotheses are satisfied. The presentation is fairly
sketchy for the scalar case which is mostly intended as illustration. The material for vector-valued Macdonald polynomials is more detailed. 

\medskip
The \textbf{hypotheses} (BF1) for the symmetric bilinear form $ \langle  \cdot  , \cdot  \rangle $ on $\mathcal{P}$, with $w^{ \ast }:=T_{N -1}^{ -1} \cdots  T_{1}^{ -1} w T_{N -1} \cdots  T_{1}$, are (for $f ,g \in \mathcal{P}$):
\begin{align} \langle 1 ,1 \rangle  &  =1 , \label{BF1.1} \\
 \langle f T_{i} ,g \rangle  &  = \langle f ,g T_{i} \rangle  ,~1 \leq i <N, \label{BF1.2} \\
 \langle f \xi _{N} ,g \rangle  &  = \langle f ,g \xi _{N} \rangle  , \label{BF1.3} \\
 \langle f \mathcal{D}_{N} ,g \rangle  &  =\left (1 -q\right ) \langle f ,x_{N} \left (g w^{ \ast } w\right ) \rangle  . \label{BF1.4}\end{align}From the definition of $w^{ \ast } =T_{N -1}^{ -1} \cdots  T_{1}^{ -1} \xi _{1}$ it follows that
\begin{align} \langle f ,g w^{ \ast } \rangle  &  = \langle f ,g T_{N -1}^{ -1} \cdots  T_{1}^{ -1} \xi _{1} \rangle  = \langle f \xi _{1} ,g T_{N -1}^{ -1} \cdots  T_{1}^{ -1} \rangle  \\
 &  = \langle f \xi _{1} T_{1}^{ -1} \cdots  T_{N -1}^{ -1} ,g \rangle  = \langle f w ,g \rangle \text{.} \nonumber \end{align}(It is a trivial exercise to show $ \langle f T_{i}^{ -1} ,g \rangle  = \langle f ,g T_{i}^{ -1} \rangle $.) Here $w^{ \ast }$ is taken as a symbolic name without claiming that it is the adjoint. Since it is possible that there is a subspace $\mathcal{N}$ of $\mathcal{P}$ such that $\left \langle f ,h\right \rangle  =0$ for all $f \in \mathcal{P}$ and $h \in \mathcal{N}$, the adjoint of an operator is only defined modulo $\mathcal{N}$.   It follows from \eqref{BF1.2}, \eqref{BF1.3}
and $\xi _{i} =\frac{1}{t} T_{i} \xi _{i +1} T_{i}$ that $ \langle f \xi _{i} ,g \rangle  = \langle f ,g \xi _{i} \rangle $ for all $f ,g \in \mathcal{P}$ and all $i$. This implies the mutual orthogonality of $\left \{M_{\alpha } :\alpha  \in \mathbb{N}_{0}^{N}\right \}$ because the spectral vector $\zeta _{\alpha }$ determines $\alpha $. Implicitly $t \in \mathbb{R}$ since the eigenvalues of~$T_{i}$ are $t , -1$. If $\deg f \neq \deg g$ then $ \langle f ,g \rangle  =0$ because the Macdonald polynomials form a homogeneous basis. For convenience denote $ \langle f ,f \rangle  =\left \Vert f\right \Vert ^{2}$ (no claim is being made about positivity). 


\begin{definition}
For $z \in \mathbb{K}$ let\begin{equation}u \left (z\right ):=\frac{\left (t -z\right ) \left (1 -z t\right )}{\left (1 -z\right )^{2}}\text{.}
\end{equation} 
\end{definition}

Note that $u \left (z^{ -1}\right ) =u \left (z\right )$. 


\begin{proposition}
Suppose {\em(BF1)} holds, $a_{i} <\alpha _{i +1}$ and 
$$\rho  = 
q^{\alpha _{i +1} -\alpha _{i}} t^{r_{\alpha } \left (i\right ) -r_{\alpha } \left (i +1\right )}.$$ 
Then\begin{equation*}\left \Vert M_{\alpha  .s_{i}}\right \Vert ^{2} =u \left (\rho \right ) \left \Vert M_{\alpha }\right \Vert ^{2}.
\end{equation*} 
\end{proposition}


\begin{proof}
From equation \eqref{MT1} we infer $ \langle M_{\alpha } T_{i} ,M_{\alpha  .s_{i}} \rangle  =\left \Vert M_{\alpha  .s_{i}}\right \Vert ^{2}$ (by hypothesis $ \langle M_{\alpha } ,M_{\alpha  .s_{i}} \rangle  =0$), and by equation \eqref{MT2} we have 
\begin{align*} 
\langle M_{\alpha } T_{i} ,M_{\alpha  .s_{i}} \rangle  &= \langle M_{\alpha } ,M_{\alpha  .s_{i}} T_{i} \rangle \\
&=\frac{\left (1 -\rho  t\right ) \left (t -\rho \right )}{\left (1 -\rho \right )^{2}} \left \Vert M_{\alpha }\right \Vert ^{2} =u \left (\rho \right ) \left \Vert M_{\alpha }\right \Vert ^{2}.
\qedhere
\end{align*}
\end{proof}


\begin{definition}
For $\alpha  \in \mathbb{N}_{0}^{N}$ let\begin{equation}\mathcal{E} \left (\alpha \right ):=\prod _{1 \leq i <j \leq N ,\alpha _{i} <\alpha _{j}}u \left (q^{\alpha _{j} -\alpha _{i}} t^{r_{\alpha } \left (i\right ) -r_{\alpha } \left (j\right )}\right )\text{.}
\end{equation} 
\end{definition}


\begin{proposition}
\label{e-norm1}Suppose {\em(BF1)} holds and $\alpha  \in \mathbb{N}_{0}^{N}$. Then $\left \Vert M_{\alpha ^{ +}}\right \Vert ^{2} =\mathcal{E} \left (\alpha \right ) \left \Vert M_{\alpha }\right \Vert ^{2}$. 
\end{proposition}


\begin{proof}
Arguing by induction on $\mathrm{i} \mathrm{n} \mathrm{v} \left (\alpha \right )$ it suffices to show that $\alpha _{i} <\alpha _{i +1}$ implies $\mathcal{E} \left (\alpha \right )/\mathcal{E} \left (\alpha  .s_{i}\right ) =u \left (q^{\alpha _{i +1} -\alpha _{i}} t^{r_{\alpha } \left (i\right ) -r_{\alpha } \left (i +1\right )}\right )$. The factors corresponding to pairs $\left (l ,j\right )$ with $l ,j \neq i ,i +1$ are the same in the products, and the pairs with just one of $i ,i +1$ are interchanged in $\mathcal{E} \left (\alpha \right ) ,\mathcal{E} \left (\alpha  .s_{i}\right )$. There is only one factor in $\mathcal{E} \left (\alpha \right )$ that is not in $\mathcal{E} \left (\alpha  .s_{i}\right )$, namely $u \left (q^{\alpha _{i +1} -\alpha _{i}} t^{r_{\alpha } \left (i\right ) -r_{\alpha } \left (i +1\right )}\right )$ coming from $\left (i ,i +1\right )$. Thus 
\begin{equation*}
\mathcal{E} \left (\alpha \right ) \left \Vert M_{\alpha }\right \Vert ^{2} =\mathcal{E} \left (\alpha  .s_{i}\right ) \left \Vert M_{\alpha  .s_{i}}\right \Vert ^{2}.
\qedhere
\end{equation*}
\end{proof}


\begin{lemma}
Suppose $\alpha  \in \mathbb{N}_{0}^{N}$. Then $M_{\alpha  \Phi } \mathcal{D}_{N} =\left (1 -q \zeta _{\alpha } \left (1\right )\right ) M_{\alpha } w$. 
\end{lemma}


\begin{proof}
By definition we have 
\begin{align*}
M_{\alpha  \Phi } \mathcal{D}_{N} &=\left (1/x_{N}\right ) M_{\alpha  \Phi } \left (1 -\xi _{N}\right ) \\
&=\left (1/x_{N}\right ) \left (1 -\zeta _{\alpha  \Phi } \left (N\right )\right ) M_{\alpha  \Phi } =\left (1 -q \zeta _{\alpha } \left (1\right )\right ) M_{\alpha } w.
\qedhere
\end{align*}
\end{proof}


\begin{remark}
It is incompatible with \eqref{BF1.1}, \eqref{BF1.2} and \eqref{BF1.3}
to require either $ \langle f \mathcal{D}_{N} ,g \rangle  =c \langle f ,x_{N} g \rangle $ with some constant $c$, or $ \langle x_{N} f ,x_{N} g \rangle  = \langle f ,g \rangle $. Let $f =M_{\alpha  \Phi }$ and $g =M_{\beta } w$ with $\left \vert \alpha \right \vert  =\left \vert \beta \right \vert $; then $ \langle f ,x_{N} g \rangle  = \langle M_{\alpha  \Phi } ,M_{\beta  \Phi } \rangle $ while 
$$
 \langle f \mathcal{D}_{N} ,g \rangle  =\left (1 -q \zeta _{\alpha } \left (1\right )\right ) \langle M_{\alpha } w ,M_{\beta } w \rangle  
 =\left (1 -q \zeta _{\alpha } \left (1\right )\right ) \langle M_{\alpha } ,M_{\beta } w w^{ \ast } \rangle .
$$ 
If $\alpha  \neq \beta $ then $ \langle f ,x_{N} g \rangle  =0$ but in general $ \langle M_{\alpha } ,M_{\beta } w w^{ \ast } \rangle  \neq 0$; for example $\alpha  =\left (1 ,0 ,0 ,0\right )$ and $\beta  =\left (0 ,1 ,0 ,0\right )$. For the second part let $f =M_{\alpha } w$ so that $ \langle x_{N} f ,x_{N} g \rangle  = \langle M_{\alpha  \Phi } ,M_{\beta  \Phi } \rangle $, while $ \langle f ,g \rangle  = \langle M_{\alpha } ,M_{\beta } w w^{ \ast } \rangle $. 
\end{remark}


\begin{proposition}
\label{phinorm1}Suppose {\em(BF1)} holds and $\alpha  \in \mathbb{N}_{0}^{N}$. Then $\left \Vert M_{\alpha  \Phi }\right \Vert ^{2} =\frac{1 -q \zeta _{\alpha } \left (1\right )}{1 -q} \left \Vert M_{\alpha }\right \Vert ^{2}$. 
\end{proposition}


\begin{proof}
Let $g \in \mathcal{P}$ with $\deg g =\left \vert \alpha \right \vert $. Then by the previous lemma \begin{equation*} \langle M_{\alpha  \Phi } \mathcal{D}_{N} ,g \rangle  =\left (1 -q \zeta _{\alpha } \left (1\right )\right ) \langle M_{\alpha } w ,g \rangle  =\left (1 -q \zeta _{\alpha } \left (1\right )\right ) \langle M_{\alpha } ,g w^{ \ast } \rangle \text{.}
\end{equation*} Specialize to $g w^{ \ast } =M_{\alpha }$ to obtain\begin{equation*} \langle M_{\alpha  \Phi } \mathcal{D}_{N} ,M_{\alpha } \left (w^{ \ast }\right )^{ -1} \rangle  =\left (1 -q \zeta _{\alpha } \left (1\right )\right ) \langle M_{\alpha } ,M_{\alpha } \rangle \text{.}
\end{equation*} By \eqref{BF1.4} we have 
\begin{align*} 
\langle M_{\alpha  \Phi } \mathcal{D}_{N} ,M_{\alpha } \left (w^{ \ast }\right )^{ -1} \rangle  &=\left (1 -q\right ) \langle M_{\alpha  \Phi } ,x_{N} \left (M_{\alpha } \left (w^{ \ast }\right )^{ -1}w^{ \ast } w\right )  \rangle \\
& =\left (1 -q\right ) \langle M_{\alpha  \Phi } ,M_{\alpha  \Phi } \rangle .
\end{align*}
This completes the proof. 
\end{proof}

Next we use (BF1) to derive a formula for $\left \Vert M_{\lambda }\right \Vert ^{2}$ for any $\lambda  \in \mathbb{N}_{0}^{N , +}$. Suppose $\lambda _{m} \geq 1$ and $\lambda _{i} =0$ for $i >m$. Let  
\begin{align*}\alpha  &=\left (\lambda _{1} ,\ldots  ,\lambda _{m -1} ,0 ,\ldots  0 ,\lambda _{m}\right ) , \\
\beta  &=(\lambda _{m} -1 ,\lambda _{1} ,\ldots  ,\lambda _{m -1} ,0 ,\ldots ) , %\nonumber 
\end{align*} 
so that $\alpha  =\beta  \Phi $, and $\gamma  =\beta ^{ +} =\left (\lambda _{1} ,\ldots  ,\lambda _{m -1} ,\lambda _{m} -1 ,0 ,\ldots \right )$. Then $\left \Vert M_{\lambda }\right \Vert ^{2} =\mathcal{E} \left (\alpha \right ) \left \Vert M_{\alpha }\right \Vert ^{2}$, $\left \Vert M_{\alpha }\right \Vert ^{2} =\frac{1 -q \zeta _{\beta } \left (1\right )}{1 -q} \left \Vert M_{\beta }\right \Vert ^{2}$ and
 $\left \Vert M_{\gamma }\right \Vert ^{2} =\break\mathcal{E} \left (\beta \right ) \left \Vert M_{\beta }\right \Vert ^{2}$. The rank vectors for $\alpha  ,\beta $ are $\left (\ldots  ,m +1 ,\ldots  ,N ,m\right )$ and
 $\left (m ,1 ,2 ,\ldots  ,m -1 ,m +1 \ldots \right )$ respectively. Then
\begin{align}\mathcal{E} \left (\alpha \right ) &  =\prod \limits _{i =m +1}^{N}u \left (q^{\lambda _{m}} t^{i -m}\right ) =t^{N -m} \frac{\left (1 -q^{\lambda _{m}}\right ) \left (1 -q^{\lambda _{m}} t^{N -m +1}\right )}{\left (1 -q^{\lambda _{m}} t\right ) \left (1 -q^{\lambda _{m}} t^{N -m}\right )}\text{,} \\
\zeta _{\beta } \left (1\right ) &  =q^{\lambda _{m} -1} t^{N -m}\text{,} \nonumber  \\
\mathcal{E} \left (\beta \right ) &  =\prod \limits _{i =1}^{m -1}u \left (q^{\lambda _{i} -\lambda _{m} +1} t^{m -i}\right ) \nonumber \end{align}and\begin{equation*}\left \Vert M_{\lambda }\right \Vert ^{2} =\frac{1 -q^{\lambda _{m}} t^{N -m}}{1 -q} \frac{\mathcal{E} \left (\alpha \right )}{\mathcal{E} \left (\beta \right )} \left \Vert M_{\gamma }\right \Vert ^{2}\text{.}
\end{equation*} (The product $\mathcal{E} \left (\alpha \right )$ telescopes. If $m =N$ then $\mathcal{E} \left (\alpha \right ) =1.$) This is the key ingredient for an inductive argument. Denote the transpose of (the Ferrers diagram) $\lambda  \in \mathbb{N}_{0}^{N , +}$ by $\lambda ^{ \prime }$, so that $\mathrm{a} \mathrm{r} \mathrm{m} \left (\lambda  ;i ,j\right ) =\lambda _{i} -j$ and $\mathrm{l} \mathrm{e} \mathrm{g} \left (\lambda  ;i ,j\right ) =\lambda _{j}^{ \prime } -i$, and define the hook product\begin{equation}h_{q ,t} \left (\lambda  ;z\right ):=\prod _{\left (i ,j\right ) \in \lambda }\left (1 -z q^{\mathrm{a} \mathrm{r} \mathrm{m} \left (i ,j\right )} t^{\mathrm{l} \mathrm{e} \mathrm{g} \left (i ,j\right )}\right )\text{.}
\end{equation} The changes in the hook product going from $\lambda $ to $\gamma $ come from the hooks at $\left \{\left (i ,\lambda _{m}\right ) :1 \leq i \leq m -1\right \}$ and $\left \{\left (m ,j\right ) :1 \leq j \leq \lambda _{m}\right \}$. Thus\begin{equation}\frac{h_{q ,t} \left (\lambda  ;z\right )}{h_{q ,t} \left (\gamma  ;z\right )} =\prod \limits _{i =1}^{m -1}\frac{1 -z q^{\lambda _{i} -\lambda _{m}} t^{m -i}}{1 -z q^{\lambda _{i} -\lambda _{m}} t^{m -i -1}} \left (1 -z q^{\lambda _{m} -1}\right )\text{,}
\end{equation} because $\prod \limits _{j =1}^{\lambda _{m} -1}\frac{1 -z q^{\lambda _{m} -j}}{1 -z q^{\lambda _{m} -j -1}} \left (1 -z\right ) =1 -z q^{\lambda _{m} -1}$ by telescoping (this telescoping property is unique to the scalar case and the norm formulas for the vector-valued case look quite
different). Furthermore\begin{equation*}\frac{h_{q ,t} \left (\lambda  ;t z\right )}{h_{q ,t} \left (\gamma  ;t z\right )} \frac{h_{q ,t} \left (\gamma  ;z\right )}{h_{q ,t} \left (\lambda  ;z\right )} =t^{1 -m} \prod \limits _{i =1}^{m -1}u \left (z q^{\lambda _{i} -\lambda _{m}} t^{m -i}\right ) \frac{1 -z t q^{\lambda _{m} -1}}{1 -z q^{\lambda _{m} -1}}\text{.}
\end{equation*} Set $z =q$ to obtain\begin{equation*}\mathcal{E} \left (\beta \right ) =t^{m -1} \genfrac{(}{)}{}{}{1 -q^{\lambda _{m}}}{1 -t q^{\lambda _{m}}} \frac{h_{q ,t} \left (\lambda  ;t q\right )}{h_{q ,t} \left (\gamma  ;t q\right )} \frac{h_{q ,t} \left (\gamma  ;q\right )}{h_{q ,t} \left (\lambda  ;q\right )}
\end{equation*} and
\begin{align}\frac{\left \Vert M_{\lambda }\right \Vert ^{2}}{\left \Vert M_{\gamma }\right \Vert ^{2}} &  =t^{1 -m} \genfrac{(}{)}{}{}{1 -q^{\lambda _{m}} t^{N -m}}{1 -q} \genfrac{(}{)}{}{}{1 -q^{\lambda _{m}} t}{1 -q^{\lambda _{m}}} \\
 & \kern1cm \times \frac{h_{q ,t} \left (\lambda  ;q\right )}{h_{q ,t} \left (\gamma  ;q\right )} \frac{h_{q ,t} \left (\gamma  ;t q\right )}{h_{q ,t} \left (\lambda  ;t q\right )} t^{N -m} \frac{\left (1 -q^{\lambda _{m}}\right ) \left (1 -q^{\lambda _{m}} t^{N -m +1}\right )}{\left (1 -q^{\lambda _{m}} t\right ) \left (1 -q^{\lambda _{m}} t^{N -m}\right )} \nonumber  \\
 &  =t^{N -2 m +1} \genfrac{(}{)}{}{}{1 -q^{\lambda _{m}} t^{N -m +1}}{1 -q} \frac{h_{q ,t} \left (\lambda  ;q\right )}{h_{q ,t} \left (\gamma  ;q\right )} \frac{h_{q ,t} \left (\gamma  ;t q\right )}{h_{q ,t} \left (\lambda  ;t q\right )}\text{.} \nonumber \end{align}\_

Define the generalized $q ,t$ factorial for $\lambda  \in \mathbb{N}_{0}^{N , +}$ by $\left (z ;q ,t\right )_{\lambda } =\prod \limits _{i =1}^{N}\left (z t^{1 -i} ;q\right )_{\lambda _{i}}$ , where $\left (z ;q\right )_{n}:=\prod \limits _{i =1}^{N}\left (1 -z q^{i -1}\right )$. 


\begin{theorem}
Suppose {\em(BF1)} holds and $\lambda  \in \mathbb{N}_{0}^{N , +}$. Then
\begin{align}\left \Vert M_{\lambda }\right \Vert ^{2} &  =t^{k \left (\lambda \right )} \frac{h_{q ,t} \left (\lambda  ;q\right )}{h_{q ,t} \left (\lambda  ;q t\right )} \frac{\left (q t^{N -1} ;q\right )_{\lambda }}{\left (1 -q\right )^{\left \vert \lambda \right \vert }} , \label{normdef1} \\
k \left (\lambda \right ) & :=\sum _{i =1}^{N}\left (N -2 i +1\right ) \lambda _{i} . \nonumber \end{align}
\end{theorem}


\begin{proof}
The formula gives the trivial result $\left \Vert 1\right \Vert ^{2} =1$, where $M_{\mathbf{0}} =1$. One needs only check  
$$\frac {\left (q t^{N -1} ;q ,t\right )_{\lambda }} {\left (q t^{N -1} ;q ,t\right )_{\gamma }} = 
\frac {\left (q t^{N -m} ;q\right )_{\lambda _{m}}} {\left (q t^{N -m} ;q\right )_{\lambda _{m} -1} =1 -q^{\lambda _{m}} t^{N -m}}
$$ and $k \left (\lambda \right ) -k \left (\gamma \right ) =N -2 m +1$. 
\end{proof}

Note that $k \left (\lambda \right ) =\sum _{i =1}^{\left \lfloor N/2\right \rfloor }\left (\lambda _{i} -\lambda _{N +1 -i}\right ) \left (N -2 i +1\right ) \geq 0$. We now use formula \eqref{normdef1}, together with $ \langle M_{\alpha } ,M_{\beta } \rangle  =0$ for $\alpha  \neq \beta $ and $\left \Vert M_{\alpha }\right \Vert ^{2} =\mathcal{E} \left (\alpha \right )^{ -1} \left \Vert M_{\alpha ^{ +}}\right \Vert ^{2}$, as definition of the form. It is straightforward to check properties \eqref{BF1.1}, \eqref{BF1.2}
and \eqref{BF1.3}. For \eqref{BF1.4} we need to show $\left \Vert M_{\alpha  \Phi }\right \Vert ^{2} =\frac{1 -q \zeta _{\alpha } \left (1\right )}{1 -q} \left \Vert M_{\alpha }\right \Vert ^{2}$ (detailed argument in Section~\ref{VVMD}) and the formula $ \langle f \mathcal{D}_{N} ,g \rangle  =\left (1 -q\right ) \langle f ,x_{N} \left (g w^{ \ast } w\right ) \rangle $. It suffices to prove this for $f =M_{\gamma }$ and $g w^{ \ast } =M_{\beta }$ with $\left \vert \gamma \right \vert  =\left \vert \beta \right \vert  +1$; indeed $ \langle M_{\gamma } \mathcal{D}_{N} ,M_{\beta } \left (w^{ \ast }\right )^{ -1} \rangle  = \langle M_{\gamma } \mathcal{D}_{N} w^{ -1} ,M_{\beta } \rangle $ and $ \langle M_{\gamma } ,x_{N} M_{\beta } w \rangle  = \langle M_{\gamma } ,M_{\beta  \Phi } \rangle $. If $\gamma  =\alpha  \Phi $ for some $\alpha $ then both terms vanish for $\alpha  \neq \beta $, otherwise the equation $\left \Vert M_{\alpha  \Phi }\right \Vert ^{2} =\frac{1 -q \zeta _{\alpha } \left (1\right )}{1 -q} \left \Vert M_{\alpha }\right \Vert ^{2}$ holds. If $\gamma _{N} =0$ then $M_{\gamma } \mathcal{D}_{N} =0$ and $ \langle M_{\gamma } ,M_{\beta  \Phi } \rangle  =0$ (since $\gamma  \neq \beta  \Phi $). 

The last of our concerns here is to determine the $\left (q ,t\right )$ region of positivity of $ \langle  \cdot  , \cdot  \rangle $. Inspection of the norm formula shows that there
is an even number of factors of the form $1 -q^{a} t^{b}$ where $a \geq 1$ and $0 \leq b \leq N$. There are two possibilities: either each such factor is positive or each is negative. Always assume $q ,t >0$ and $q \neq 1$. If each is positive then $0 <q <1$ and $q^{a} t^{b} \leq q t^{b}$. If $0 <t \leq 1$ then $q t^{b} \leq q <1$, or if $t >1$ then $q t^{b} \leq q t^{N} <1$, that is $q <t^{ -N}$. If each factor is negative then $q >1$: if $t \geq 1$ then $q^{a} t^{b} \geq q >1$, or if $0 <t \leq 1$ then $q^{a} t^{b} \geq q t^{b} \geq q t^{N} >1$, that is, $q >t^{ -N}$. 


\begin{proposition}
The inner product $ \langle  \cdot  , \cdot  \rangle $ is positive-definite, that is, $ \langle M_{\alpha } ,M_{\alpha } \rangle  >0$ for all $\alpha  \in \mathbb{N}_{0}^{N}$ provided $q ,t >0$, and $q <\min \left (1 ,t^{ -N}\right )$ or $q >\max \left (1 ,t^{ -N}\right )$. 
\end{proposition}

Figure \ref{posit10} is an illustration of the positivity region
with $N =4$ using logarithmic coordinates.

      
\begin{figure}\centering 
\setlength\fboxrule{0in}\setlength\fboxsep{0.2in}\fcolorbox[HTML]{000000}{FFFFFF}{\includegraphics[ width=3.1514in, height=3.1514in,]{posreg1}
}
\caption{\label{posit10}Logarithmic coordinates, N = 4}\end{figure}


\section{\label{VVMD}Vector-valued Macdonald polynomials.}
These are polynomials whose values lie in an irreducible $\mathcal{H}_{N} \left (t\right )$-module. The generating relations for the Hecke algebra $\mathcal{H}_{N} \left (t\right )$ are stated in \eqref{Hrel}. For the purpose of constructing a positive symmetric bilinear form we make the
restriction $t >0$. Also throughout $q ,t \neq 0 ,1$. 

\subsection{Representations of the Hecke algebra}
The irreducible modules of $\mathcal{H}_{N} \left (t\right )$ correspond to partitions of $N$ and are constructed in terms of Young tableaux (see \cite{DJ1986}).


Let $\tau $ be a partition of $N\text{,}$ that is, $\tau  \in \mathbb{N}_{0}^{N , +}$ and $\left \vert \tau \right \vert  =N$. Thus $\tau  =\left (\tau _{1} ,\tau _{2} ,\ldots \right )$ and often the trailing zero entries are dropped when writing $\tau $. The length of $\tau $ is $\ell  \left (\tau \right ) =\max \left \{i :\tau _{i} >0\right \}$. There is a Ferrers diagram of shape $\tau $ (given the same label), with boxes at points $\left (i ,j\right )$ with $1 \leq i \leq \ell  \left (\tau \right )$ and $1 \leq j \leq \tau _{i}$. A \textit{tableau} of shape $\tau $ is a filling of the boxes with numbers, and a \textit{reverse standard Young tableau} (RSYT) is a filling with
the numbers $\left \{1 ,2 ,\ldots  ,N\right \}$ so that the entries decrease in each row and each column.
Denote the set of RSYT's of shape $\tau $ by $\mathcal{Y} \left (\tau \right )$ and let $V_{\tau } =\mathrm{s} \mathrm{p} \mathrm{a} \mathrm{n}_{\mathbb{K}} \left \{S :S \in \mathcal{Y} \left (\tau \right )\right \}$ with orthogonal basis $\mathcal{Y} \left (\tau \right )$ (recall $\mathbb{K} =\mathbb{Q} \left (q ,t\right )$). Set $n_{\tau }:=\dim  V_{\tau } =\# \mathcal{Y} \left (\tau \right )$. The formula for the dimension is a hook-length product. For $1 \leq i \leq N$ and $S \in \mathcal{Y} \left (\tau \right )$ the entry $i$ is at coordinates $\left (\ensuremath{\operatorname*{row}} \left (i ,S\right ) ,\ensuremath{\operatorname*{col}} \left (i ,S\right )\right )$ and the \textit{content} of the entry is $c \left (i ,S\right ):=\ensuremath{\operatorname*{col}} \left (i ,S\right ) -\ensuremath{\operatorname*{row}} \left (i ,S\right )$. Each $S \in \mathcal{Y} \left (\tau \right )$ is uniquely determined by its \textit{content vector} $\left [c \left (i ,S\right )\right ]_{i =1}^{N}$. For example let $\tau  =\left (4 ,3\right )$ and $S =$ $\begin{array}{cccc}4 & 3 & 1 & \, \\
7 & 6 & 5 & 2\end{array}$. Then the content vector is $\left [1 ,3 ,0 , -1 ,2 ,1 ,0\right ]$. There is a representation of $\mathcal{H}_{N} \left (t\right )$ on $V_{\tau }$, also denoted by $\tau $ (slight abuse of notation). The description will be given in terms of the actions of $\left \{T_{i}\right \}$ on the basis elements. 


\begin{definition}
\label{deftauTi}The representation $\tau $ of $\mathcal{H}_{N} \left (t\right )$ is defined by the action of the generators specified as follows: for $1 \leq i <N$ and $S \in \mathcal{Y} \left (\tau \right )$,
\begin{enumerate}
\item if\/ $\ensuremath{\operatorname*{row}} \left (i ,S\right ) =\ensuremath{\operatorname*{row}} \left (i +1 ,S\right )$ (implying $\ensuremath{\operatorname*{col}} \left (i ,S\right ) =\ensuremath{\operatorname*{col}} \left (i +1 ,S\right ) +1$ and $c \left (i ,S\right ) -c \left (i +1 ,S\right ) =1$) then\begin{equation*}S \tau  \left (T_{i}\right ) =t S ;
\end{equation*}  
\item if\/ $\ensuremath{\operatorname*{col}} \left (i ,S\right ) =\ensuremath{\operatorname*{col}} \left (i +1 ,S\right )$ (implying $\ensuremath{\operatorname*{row}} \left (i ,S\right ) =\ensuremath{\operatorname*{row}} \left (i +1 ,S\right ) +1$ and $c \left (i ,S\right ) -c \left (i +1 ,S\right ) = -1$) then\begin{equation*}S \tau  \left (T_{i}\right ) = -S ;
\end{equation*} 
\item if\/ $\ensuremath{\operatorname*{row}} \left (i ,S\right ) <\ensuremath{\operatorname*{row}} \left (i +1 ,S\right )$ and $\ensuremath{\operatorname*{col}} \left (i ,S\right ) >\ensuremath{\operatorname*{col}} \left (i +1 ,S\right )$ then $c \left (i ,S\right ) -c \left (i +1 ,S\right ) \geq 2$; the tableau $S^{\left (i\right )}$ obtained from $S$ by exchanging $i$ and $i +1$, is an element of $\mathcal{Y} \left (\tau \right )$ and\begin{equation*}S \tau  \left (T_{i}\right ) =S^{\left (i\right )} +\dfrac{t -1}{1 -t^{c \left (i +1 ,S\right ) -c \left (i ,S\right )}} S ;
\end{equation*} 
\item if\/ $c \left (i ,S\right ) -c \left (i +1 ,S\right ) \leq  -2$, thus $\ensuremath{\operatorname*{row}} \left (i ,S\right ) >\ensuremath{\operatorname*{row}} \left (i +1 ,S\right )$ and $\ensuremath{\operatorname*{col}} \left (i ,S\right ) <\ensuremath{\operatorname*{col}} \left (i +1 ,S\right )$, then with $b =c \left (i ,S\right ) -c \left (i +1 ,S\right )$,\begin{equation*}S \tau  \left (T_{i}\right ) =\frac{t \left (t^{b +1} -1\right ) \left (t^{b -1} -1\right )}{\left (t^{b} -1\right )^{2}} S^{\left (i\right )} +\frac{t^{b} \left (t -1\right )}{t^{b} -1} S\text{.}
\end{equation*} 
\end{enumerate}
\end{definition}

The formulas in (4) are consequences of those in (3) by interchanging
$S$ and $S^{\left (i\right )}$ and applying the relations $\left (\tau  \left (T_{i}\right ) +I\right ) \left (\tau  \left (T_{i}\right ) -t I\right ) =0$ (where $I$ denotes the identity operator on $V_{\tau }$). There is a partial order on $\mathcal{Y} \left (\tau \right )$ related to the inversion number, namely
\begin{equation}\mathrm{i} \mathrm{n} \mathrm{v} \left (S\right ):=\# \left \{\left (i ,j\right ) :1 \leq i <j \leq N ,c \left (i ,S\right ) \geq c \left (j ,S\right ) +2\right \}\text{,}
\end{equation} so $\mathrm{i} \mathrm{n} \mathrm{v} \left (S^{\left (i\right )}\right ) =\mathrm{i} \mathrm{n} \mathrm{v} \left (S\right ) -1$ in (3) above. The {\it inv}-maximal element $S_{0}$ of $\mathcal{Y} \left (\tau \right )$ has the numbers $N ,N -1 ,\ldots  ,1$ entered column-by-column, and the {\it inv}-minimal element $S_{1}$ of $\mathcal{Y} \left (\tau \right )$ has the numbers $N ,N -1 ,\ldots  ,1$ entered row-by-row. The set $\mathcal{Y} \left (\tau \right )$ can be given the structure of a Yang--Baxter graph, with root $S_{0}$, sink $S_{1}$ with arrows labeled by $T_{i}$ joining $S$ to $S^{\left (i\right )}$ as in (3). Some properties can be proved by induction on the inversion number. Recall $u \left (z\right ) =\frac{\left (t -z\right ) \left (1 -t z\right )}{\left (1 -z\right )^{2}} =u \left (z^{ -1}\right )$. 


\begin{definition}
The bilinear symmetric form $ \langle  \cdot  , \cdot  \rangle _{0}$ on $V_{\tau }$ is defined to be the linear extension of \begin{equation} \langle S ,S^{ \prime } \rangle _{0} =\delta _{S ,S^{ \prime }} \underset{c \left (j ,S\right ) -c \left (i ,S\right ) \geq 2}{\prod _{i <j}}u \left (t^{c \left (i ,S\right ) -c \left (j ,S\right )}\right )\text{.}
\end{equation} 
\end{definition}


\begin{proposition}
\label{Sform}Suppose $f ,g \in V_{\tau }$. Then $ \langle f \tau  \left (T_{i}\right ) ,g \rangle _{0} = \langle f ,g \tau  \left (T_{i}\right ) \rangle _{0}$ for $1 \leq i <N$. If $c \left (i ,S\right ) -c \left (i +1 ,S\right ) \geq 2$ for some $i ,S$ then $ \langle S^{\left (i\right )} ,S^{\left (i\right )} \rangle _{0} =u \left (t^{c \left (i ,S\right ) -c \left (i +1 ,S\right )}\right )  \langle S ,S \rangle _{0}$. 
\end{proposition}


\begin{proof}
If $\ensuremath{\operatorname*{row}} \left (i ,S\right ) =\ensuremath{\operatorname*{row}} \left (i +1 ,S\right )$ or $\ensuremath{\operatorname*{col}} \left (i ,S\right ) =\ensuremath{\operatorname*{col}} \left (i +1 ,S\right )$ then 
$$ \langle S \tau  \left (T_{i}\right ) ,S \rangle _{0} =t  \langle S ,S \rangle _{0} = \langle S ,S \tau  \left (T_{i}\right ) \rangle _{0}$$ 
or 
$$ \langle S \tau  \left (T_{i}\right ) ,S \rangle _{0} = - \langle S ,S \rangle _{0} = \langle S ,S \tau  \left (T_{i}\right ) \rangle _{0}$$ 
respectively. If $c \left (i ,S\right ) -c \left (i +1 ,S\right ) \geq 2$ and $b = 
c \left (i +1 ,S\right ) -c \left (i ,S\right )$ then $ \langle S^{\left (i\right )} ,S^{\left (i\right )} \rangle _{0}/ \langle S ,S \rangle _{0} =u \left (t^{ -b}\right )$ (in the product the only difference is the term $\left (i ,i +1\right )$, appearing in $ \langle S^{\left (i\right )} ,S^{\left (i\right )} \rangle _{0}$). Then
\begin{align*} \langle S \tau  \left (T_{i}\right ) ,S^{\left (i\right )} \rangle _{0} &  = \langle S^{\left (i\right )} ,S^{\left (i\right )} \rangle _{0} +\dfrac{t -1}{1 -t^{b}}  \langle S^{\left (i\right )} ,S \rangle _{0} = \langle S^{\left (i\right )} ,S^{\left (i\right )} \rangle _{0}\text{,} \\
 \langle S^{\left (i\right )} \tau  \left (T_{i}\right ) ,S \rangle _{0} &  =\frac{t \left (t^{b +1} -1\right ) \left (t^{b -1} -1\right )}{\left (t^{b} -1\right )^{2}}  \langle S ,S \rangle _{0} +\frac{t^{b} \left (t -1\right )}{t^{b} -1}  \langle S^{\left (i\right )} ,S \rangle _{0} \\
 &  =\frac{t \left (t^{b +1} -1\right ) \left (t^{b -1} -1\right )}{\left (t^{b} -1\right )^{2}}  \langle S ,S \rangle _{0} =u \left (t^{b}\right )  \langle S ,S \rangle _{0}\text{,}\end{align*}thus $ \langle S \tau  \left (T_{i}\right ) ,S^{\left (i\right )} \rangle _{0} = \langle S^{\left (i\right )} \tau  \left (T_{i}\right ) ,S \rangle _{0}$. These statements imply that $ \langle f \tau  \left (T_{i}\right ) ,g \rangle _{0} = \langle f ,g \tau  \left (T_{i}\right ) \rangle _{0}$ for $f ,g \in V_{\tau }$. 
\end{proof}

Furthermore if $t >0$ then $ \langle S ,S \rangle _{0} \geq 0$; each term is of the form $\frac{\left (t -t^{m}\right ) \left (1 -t^{m +1}\right )}{\left (1 -t^{m}\right )^{2}}$ with $m \geq 2$; either all parts are positive or all are negative depending on $0 <t <1$ or $t >1$ respectively (the limit as $t \rightarrow 1$ is $\frac{m^{2} -1}{m^{2}} >0$). Denote $ \langle f ,f \rangle _{0} =\left \Vert f\right \Vert _{0}^{2}$ for $f \in V_{\tau }$. 

There is a commutative set of Jucys--Murphy elements in $\mathcal{H}_{N} \left (t\right )$ which are diagonalized with respect to the basis $\mathcal{Y} \left (\tau \right )$. 


\begin{definition}
Set $\phi _{N}:=1$ and $\phi _{i}:=\frac{1}{t} T_{i} \phi _{i +1} T_{i}$ for $1 \leq i <N$. 
\end{definition}


\begin{proposition}
\label{eigenphi}Suppose $1 \leq i \leq N$ and $S \in \mathcal{Y} \left (\tau \right )$. Then $S \tau  \left (\phi _{i}\right ) =t^{c \left (i ,S\right )} S$. 
\end{proposition}


\begin{proof}
Arguing inductively suppose that $S \tau  \left (\phi _{i +1}\right ) =t^{c \left (i +1 ,S\right )} S$ for all $S \in \mathcal{Y} \left (\tau \right )$; this is trivially true for $i =N -1$ since $c \left (N ,S\right ) =0$ and $\phi _{N} =1$. If $\ensuremath{\operatorname*{row}} \left (i ,S\right ) =\ensuremath{\operatorname*{row}} \left (i +1 ,S\right )$ then $S \tau  \left (\phi _{i}\right ) =\frac{1}{t} S \tau  \left (T_{i}\right ) \tau  \left (\phi _{i +1}\right ) \tau  \left (T_{i}\right ) =t^{c \left (i +1 ,S\right ) +1} S$ and $c \left (i ,S\right ) =c \left (i +1 ,S\right ) +1$. If $\ensuremath{\operatorname*{col}} \left (i ,S\right ) =\ensuremath{\operatorname*{col}} \left (i +1 ,S\right )$ then $S \tau  \left (\phi _{i}\right ) =\frac{1}{t} S \tau  \left (T_{i}\right ) \tau  \left (\phi _{i +1}\right ) \tau  \left (T_{i}\right ) =\frac{1}{t} t^{c \left (i +1 ,S\right )} S$ and $c \left (i ,S\right ) = \break
c \left (i +1 ,S\right ) -1$ (since $S \tau  \left (T_{i}\right ) = -S$). Suppose $c \left (i ,S\right ) -c \left (i +1 ,S\right ) \geq 2$. Then the matrices $\mathcal{T} ,\Phi $ of $\tau  \left (T_{i}\right ) ,\tau  \left (\phi _{i +1}\right )$ respectively with respect to the basis $\left [S ,S^{\left (i\right )}\right ]$ are\begin{equation*}\mathcal{T} =\left [\begin{array}{cc} -\frac{1 -t}{1 -\rho } & 1 \\
\frac{\left (1 -\rho  t\right ) \left (t -\rho \right )}{\left (1 -\rho \right )^{2}} & \frac{\rho  \left (1 -t\right )}{1 -\rho }\end{array}\right ] ,~\Phi  =\left [\begin{array}{cc}t^{c \left (i +1 ,S\right )} & 0 \\
0 & t^{c \left (i ,S\right )}\end{array}\right ]\text{,}
\end{equation*} where $\rho  =t^{c \left (i +1 ,S\right ) -c \left (i ,S\right )}$. A simple calculation shows \begin{equation*}\frac{1}{t} \mathcal{T} \Phi  \mathcal{T} =\left [\begin{array}{cc}t^{c \left (i ,S\right )} & 0 \\
0 & t^{c \left (i +1 ,S\right )}\end{array}\right ] .
\qedhere
\end{equation*} 
\end{proof}

\subsection{Polynomials and operators}
Let $\mathcal{P}_{\tau }:=\mathcal{P} \otimes V_{\tau }$. The equation $\deg \left (f\right ) =n$ means $f \in \mathcal{P}_{n} \otimes V_{\tau }$. The action of $\mathcal{H}_{N} \left (t\right )$ and the operators are defined as follows: with $p \in \mathcal{P} ,S \in \mathcal{Y} \left (\tau \right )$ and $1 \leq i <N$,
\begin{align}\left (p \left (x\right ) \otimes S\right ) \boldsymbol{T}_{i} & :=\left (1 -t\right ) x_{i +1} \frac{p \left (x\right ) -p \left (x .s_{i}\right )}{x_{i} -x_{i +1}} \otimes S +p \left (x .s_{i}\right ) \otimes S \tau  \left (T_{i}\right )\text{,} \\
\omega  & :=T_{1} T_{2} \cdots  T_{N -1}\text{,} \\
\left (p \left (x\right ) \otimes S\right ) \boldsymbol{w} & :=p \left (q x_{N} ,x_{1} ,\ldots  ,x_{N -1}\right ) \otimes S \tau  \left (\omega \right ) \text{,}\\
\mathbf{\xi }_{i} & :=t^{i -N} \boldsymbol{T}_{i -1}^{ -1} \cdots  \boldsymbol{T}_{1}^{ -1} \boldsymbol{w} \boldsymbol{T}_{N -1} \cdots  \boldsymbol{T}_{i} \text{,}\\
\mathcal{D}_{N} & :=\left (1 -\mathbf{\xi }_{N}\right )/x_{N} ,~\mathcal{D}_{i}:=\frac{1}{t} \boldsymbol{T}_{i} \mathcal{D}_{i +1} \boldsymbol{T}_{i}\text{.}\end{align}By the braid relations we have
\begin{align*}T_{i +1} \omega  &  =T_{1} \cdots  T_{i +1} T_{i} T_{i +1} T_{i +2} \cdots  T_{N -1} \\
 &  =T_{1} \cdots  T_{i} T_{i +1} T_{i} T_{i +2} \cdots  T_{N -1} =\omega  T_{i}\text{,}\end{align*}for $1 \leq i <N -1$. It follows that $\boldsymbol{T}_{i +1} \boldsymbol{w} =\boldsymbol{w} \boldsymbol{T}_{i}$ acting on $\mathcal{P}_{\tau }$. The operators $\left \{\mathbf{\xi }_{i}\right \}$ mutually commute and the simultaneous polynomial eigenfunctions
are the vector-valued (nonsymmetric) Macdonald polynomials. The factor $t^{i -N}$ in $\mathbf{\xi }_{i}$ appears to differ from the scalar case, but if $\tau  =\left (N\right )$, the trivial representation, then $S \tau  \left (T_{i}\right ) =t S$ (the unique RSYT of shape $\left (N\right )$) and $S \tau  \left (\omega \right ) =t^{N -1} S$, and thus $\mathbf{\xi }_{i}$ coincides with \eqref{defxi0}. The operator $\mathbf{\xi }_{i}$ acting on constants coincides with $I \otimes \tau  \left (\phi _{i}\right )$:
\begin{align*}\left (1 \otimes S\right ) \mathbf{\xi }_{i} &  =t^{i -N} \otimes S \tau  \left (T_{i -1}^{ -1} \cdots  T_{1}^{ -1} T_{1} T_{2} \cdots  T_{N -1} T_{N -1} \cdots  T_{i}\right ) \\
 &  =t^{i -N} \otimes S \tau  \left (T_{i} \cdots  T_{N -1} T_{N -1} \cdots  T_{i}\right ) \\
 &  =1 \otimes S \tau  \left (\phi _{i}\right ) =t^{c \left (i ,S\right )} \left (1 \otimes S\right )\text{.}\end{align*}For each $\alpha  \in \mathbb{N}_{0}^{N}$ and $S \in \mathcal{Y} \left (\tau \right )$ there is an $\left \{\mathbf{\xi }_{i}\right \}$ eigenfunction\begin{equation}M_{\alpha  ,S} \left (x\right ) =\eta  \left (\alpha  ,S\right ) x^{\alpha } \otimes S \tau  \left (R_{\alpha }\right ) +\sum _{\alpha  \rhd \beta }x^{\beta } \otimes B_{\alpha  ,\beta  ,S} \left (q ,t\right )\text{,}
\end{equation} where $\eta  \left (\alpha  ,S\right ) =q^{a} t^{b}$ with $a ,b \in \mathbb{N}_{0}$, $R_{\alpha } \in \mathcal{H}_{N}\left (t\right ) $, $B_{\alpha  ,\beta  ,S}\left (q ,t\right ) \in V_{\tau }\text{. }$Furthermore $R_{\alpha }$ is an analog of $r_{\alpha }$ (see \cite[p.~9]{DL2012});
if $\alpha  \in \mathbb{N}_{0}^{N , +}$ then $R_{\alpha } =I$, and if $\alpha _{i} <\alpha _{i +1}$ then $R_{\alpha  .s_{i}} =R_{\alpha } T_{i}$ (there is a definition of $R_{\alpha }$ below). Furthermore
\begin{align}M_{\alpha  ,S} \mathbf{\xi }_{i} &  =\zeta _{\alpha  ,S} \left (i\right ) M_{\alpha  ,S} ,1 \leq i \leq N\text{,} \\
\zeta _{\alpha  ,S} \left (i\right ) &  =q^{\alpha _{i}} t^{c \left (r_{\alpha } \left (i\right ) ,S\right )}\text{.} \nonumber \end{align}These polynomials are produced with the Yang--Baxter graph. The typical node (labeled by $\left (\alpha  ,S\right )$) is\begin{equation*}\left (\alpha  ,S ,\zeta _{\alpha  ,S} ,R_{\alpha } ,M_{\alpha  ,S}\right )
\end{equation*} and the root is $\left (\mathbf{0} ,S_{0} ,\left [t^{c \left (i ,S_{0}\right )}\right ]_{i =1}^{N} ,I ,1 \otimes S_{0}\right )$. 

There are steps: 


\begin{itemize}
\item if $\alpha _{i} <\alpha _{i +1}$ there is a step labeled $s_{i}$
\begin{align}\left (\alpha  ,S ,\zeta _{\alpha  ,S} ,R_{\alpha } ,M_{\alpha  ,S}\right ) &  \rightarrow \left (\alpha  .s_{i} ,S ,\zeta _{\alpha  .s_{i} ,S} ,R_{\alpha  .s_{i}} ,M_{\alpha  .s_{i} ,S}\right ) , \nonumber  \\
M_{\alpha  .s_{i} ,S} &  =M_{\alpha  ,S} \boldsymbol{T}_{i} +\frac{t -1}{\zeta _{\alpha  ,S} \left (i +1\right )/\zeta _{\alpha  ,S} \left (i\right ) -1} M_{\alpha  ,S } , \label{MT&gt;i} \\
R_{\alpha  .s_{i}} &  =R_{\alpha } T_{i} ,~\eta  \left (\alpha  .s_{i} ,S\right ) =\eta  \left (\alpha  ,S\right )  \nonumber \end{align}(note that $\left (x_{i}^{\alpha _{i}} x_{i +1}^{\alpha _{i +1}} \otimes S R_{\alpha }\right ) \boldsymbol{T}_{i} =x_{i}^{\alpha _{i +1}} x_{i +1}^{\alpha _{i}} \otimes S \tau  \left (R_{\alpha } T_{i}\right ) +\cdots $); 

\item if $\alpha _{i} =\alpha _{i +1}$, $j =r_{\alpha } \left (i\right )$ (thus $j +1 =r_{\alpha } \left (i +1\right )$, and $R_{\alpha } T_{i} =T_{j} R_{\alpha }$; see \cite[Lemma~2.14]{DL2012})
and $c \left (j ,S\right ) -c \left (j +1 ,S\right ) \geq 2$ there is a step
\begin{align}\left (\alpha  ,S ,\zeta _{\alpha  ,S} ,R_{\alpha } ,M_{\alpha  ,S}\right ) &  \rightarrow \left (\alpha  ,S^{\left (j\right )} ,\left (\zeta _{\alpha  ,S}\right ) .s_{i} ,R_{\alpha } ,M_{\alpha  ,S^{\left (j\right )}}\right ) , \nonumber  \\
M_{\alpha  ,S^{\left (j\right )}} &  =M_{\alpha  ,S} \boldsymbol{T}_{i} +\frac{t -1}{\zeta _{\alpha  ,S} \left (i +1\right )/\zeta _{\alpha  ,S} \left (i\right ) -1} M_{\alpha  ,S} , \label{MT=i} \\
\frac{\zeta _{\alpha  ,S} \left (i +1\right )}{\zeta _{\alpha  ,S} \left (i\right )} &  =t^{c \left (j +1 ,S\right ) -c \left (j ,S\right )} ,\eta  \left (\alpha  ,S^{\left (j\right )}\right ) =\eta  \left (\alpha  ,S\right ) ; \nonumber \end{align}\end{itemize}


For these formulas to be valid it is required that the
denominators $\zeta _{\alpha  ,S} \left (i +1\right )/\zeta _{\alpha  ,S} \left (i\right ) -1 \neq 0$, that is, $q^{\alpha _{i +1} -\alpha _{i}}t^{c\left (r_{\alpha }\left (i +1\right ) ,S\right ) -c\left (r_{\alpha }\left (i\right ) ,S\right )}$$ \neq 1$. From the bound $\left \vert c \left (j ,S\right ) -c \left (j^{ \prime } ,S\right )\right \vert  \leq \tau _{1} +\ell  \left (\tau \right ) -2$ we obtain the necessary condition $q^{q} t^{b} \neq 1$ for $a \geq 0$ and $\left \vert b\right \vert  \leq \tau _{1} +\ell  \left (\tau \right ) -2$. These conditions are satisfied in the region of positivity described in Proposition~\ref{regpos2}.


The other possibilities for the action of $\boldsymbol{T}_{i}$ are: 


\begin{itemize}
\item if $\alpha _{i} >\alpha _{i +1}$ set $\rho :=\zeta _{\alpha  ,S} \left (i\right )/\zeta _{\alpha  ,S} \left (i +1\right )$ then
\begin{equation}M_{\alpha  ,S} \boldsymbol{T}_{i} =\frac{\left (1 -t \rho \right ) \left (t -\rho \right )}{\left (1 -\rho \right )^{2}} M_{\alpha  .s_{i} ,S} +\frac{\rho  \left (1 -t\right )}{\left (1 -\rho \right )} M_{\alpha  ,S} ; \label{MT<i}
\end{equation}

\item if $\alpha _{i} =\alpha _{i +1}$ and $j =r_{\alpha } \left (i\right ) ,~c \left (j ,S\right ) -c \left (j +1 ,S\right ) \leq 2 ,~\rho  =t^{c \left (j ,S\right ) -c \left (j +1 ,S\right )}$ then
\begin{equation}M_{\alpha  ,S} \boldsymbol{T}_{i} =\frac{\left (1 -t \rho \right ) \left (t -\rho \right )}{\left (1 -\rho \right )^{2}} M_{\alpha  ,S^{\left (j\right )}} +\frac{\rho  \left (1 -t\right )}{\left (1 -\rho \right )} M_{\alpha  ,S} ; \label{MT=2}
\end{equation}

\item if $\alpha _{i} =\alpha _{i +1}$ and $j =r_{\alpha } \left (i\right ) ,\ensuremath{\operatorname*{row}} \left (j ,S\right ) =\ensuremath{\operatorname*{row}} \left (j +1 ,S\right )$ then $M_{\alpha  ,S} \boldsymbol{T}_{i} =t M_{\alpha  ,S}$; 

\item if $\alpha _{i} =\alpha _{i +1}$ and $j =r_{\alpha } \left (i\right ) ,\ensuremath{\operatorname*{col}} \left (j ,S\right ) =\ensuremath{\operatorname*{col}} \left (j +1 ,S\right )$ then $M_{\alpha  ,S} \boldsymbol{T}_{i} = -M_{\alpha  ,S}$ . \end{itemize}


The degree-raising operation, namely, the affine step, takes $\alpha $ to $\alpha  \Phi :=\left (\alpha _{2} ,\alpha _{3} ,\ldots  ,\alpha _{N} ,\alpha _{1} +1\right )$:
\begin{align}\left (\alpha  ,S ,\zeta _{\alpha  ,S} ,R_{\alpha } ,M_{\alpha  ,S}\right ) &\rightarrow \left (\alpha \Phi  ,S ,\zeta _{\alpha \Phi  ,S} ,R_{\alpha \Phi } ,M_{\alpha \Phi  ,S}\right ) , \nonumber  \\
M_{\alpha \Phi  ,S} &=x_{N}\left (M_{\alpha  ,S}\boldsymbol{w}\right ) , \\
\alpha \Phi  &=\left (\alpha _{2} ,\alpha _{3} ,\ldots  ,\alpha _{N} ,\alpha _{1} +1\right ) , \nonumber  \\
\zeta _{\alpha \Phi  ,S} &=\left (\zeta _{\alpha  ,S}\left (2\right ) ,\ldots  ,\zeta _{\alpha  ,S}\left (N\right ) ,q\zeta _{\alpha  ,S}\left (1\right )\right ) . \nonumber \end{align}

The inversion number $\mathrm{i} \mathrm{n} \mathrm{v} \left (\alpha \right )$ of $\alpha  \in \mathbb{N}_{0}^{N}$ is the length of the shortest product $g =s_{i_{1}} s_{i_{2}} \cdots  s_{i_{m}}$ such that $\alpha  .g =\alpha ^{ +}$. From this and the Yang--Baxter graph we deduce that the series of steps $s_{i_{1}} ,s_{i_{2}} ,\cdots  ,s_{i_{m}}$ leads from $M_{\alpha  ,S}$ to $M_{\alpha ^{ +} ,S}$ and $R_{\alpha } T_{i_{i}} T_{i_{2}} \cdots  T_{i_{m}} =R_{\alpha ^{ +}} =I$. 


\begin{definition}
Suppose $\alpha  \in \mathbb{N}_{0}^{N}$. Then $R_{\alpha }:=\left (T_{i_{i}} T_{i_{2}} \cdots  T_{i_{m}}\right )^{ -1}$ where $\alpha  .s_{i_{1}} s_{i_{2}} \cdots  s_{i_{m}} =\alpha ^{ +}$ and $m =\mathrm{i} \mathrm{n} \mathrm{v} \left (\alpha \right )$. 
\end{definition}

There may be different products $\alpha  .s_{j_{1}} s_{j_{2}} \cdots  s_{i j} =\alpha ^{ +}$ of length $\mathrm{i} \mathrm{n} \mathrm{v} \left (\alpha \right )$ but they all give the same value of $R_{\alpha }$ by the braid relations. It is shown in \cite[p.~10, Eq.~(2.15)]{DL2012}
that $R_{\alpha } \omega  =t^{N -m} \phi _{m} R_{\alpha  \Phi }$ with $m =r_{\alpha } \left (1\right )$. 

\subsection{The bilinear symmetric form}
We will define a symmetric bilinear form $ \langle  \cdot  , \cdot  \rangle $ on $\mathcal{P}_{\tau }$ satisfying certain postulates, using the same logical outline as in Section~\ref{SMD}; first we derive
consequences from these, then state the definition and show that the desired properties apply. 

The \textbf{hypotheses} (BF2) for the
symmetric bilinear form $ \langle  \cdot  , \cdot  \rangle $ on $\mathcal{P}_{\tau }$, with $\boldsymbol{w}^{ \ast }:=\boldsymbol{T}_{N -1}^{ -1} \cdots  \boldsymbol{T}_{1}^{ -1} \boldsymbol{w} \boldsymbol{T}_{N -1} \cdots  \boldsymbol{T}_{1}$, are (for $f ,g \in \mathcal{P}_{\tau } ,~S ,S^{ \prime } \in \mathcal{Y} \left (\tau \right ) ,~1 \leq i <N$):
\begin{subequations}
\begin{align} \langle 1 \otimes S ,1 \otimes S^{ \prime } \rangle  &  = \langle S ,S^{ \prime } \rangle _{0} , \label{BF2.1} \\
 \langle f \boldsymbol{T}_{i} ,g \rangle  &  = \langle f ,g \boldsymbol{T}_{i} \rangle  , \label{BF2.2} \\
 \langle f \mathbf{\xi }_{N} ,g \rangle  &  = \langle f ,g \mathbf{\xi }_{N} \rangle  , \label{BF2.3} \\
 \langle f \mathcal{D}_{N} ,g \rangle  &  =\left (1 -q\right ) \langle f ,x_{N} \left (g \boldsymbol{w}^{ \ast } \boldsymbol{w}\right ) \rangle.  \label{BF2.4}\end{align}
\end{subequations}
Properties \eqref{BF2.2} and \eqref{BF2.3} imply
$ \langle f \mathbf{\xi }_{i} ,g \rangle  = \langle f ,g \mathbf{\xi }_{i} \rangle $ for each $i$ and thus the $M_{\alpha  ,S}$'s are mutually orthogonal. As in the scalar case $ \langle f \boldsymbol{w} ,g \rangle  = \langle f ,g \boldsymbol{w}^{ \ast } \rangle $. As before denote $ \langle f ,f \rangle  =\left \Vert f\right \Vert ^{2}$. First we will show that these hypotheses determine the form uniquely when $q ,t \neq 0 ,1$ without recourse to the Macdonald polynomials. We use the commutation relationships $\left (x_{i +1} f\right ) \boldsymbol{T}_{i} =x_{i} \left (f \boldsymbol{T}_{i}\right ) +\left (t -1\right ) x_{i +1} f$ and $\left (x_{j} f\right ) \boldsymbol{T}_{i} =x_{j} \left (f \boldsymbol{T}_{i}\right )$ for $f \in \mathcal{P}_{\tau }$ and $j \neq i ,i +1$ (a simple direct computation).
\begin{proposition}
\label{fDxg}Suppose {\em(BF2)} holds. Then for $1 \leq i \leq j \leq N$ and $q ,t \neq 0 ,1$ there are operators $A_{i ,j} ,B_{i ,j}$,on $\mathcal{P}_{\tau }$ preserving degree of homogeneity such that $A_{i ,i}$ and $B_{i ,i}$ are invertible and for $f ,g \in \mathcal{P}_{\tau }$
\begin{align*} \langle f \mathcal{D}_{i} ,g \rangle  &  =\sum _{j =i}^{N} \langle f ,x_{j} \left (g A_{i ,j}\right ) \rangle  \text{,}\\
 \langle f ,x_{i} g \rangle  &  =\sum _{j =i}^{N} \langle f \mathcal{D}_{j} ,g B_{i ,j} \rangle \text{.}\end{align*}
\end{proposition}


\begin{proof}
Suppose $i =N$. Then $A_{N ,N} =\left (1 -q\right ) \boldsymbol{w}^{ \ast } \boldsymbol{w}$ and $B_{N ,N} =A_{N ,N}^{ -1}$ by \eqref{BF2.4}. Arguing by induction suppose the statement is true for $k +1 \leq i \leq N$. Then for any $f ,g \in \mathcal{P}_{\tau }$
\begin{align*} \langle f \mathcal{D}_{k} ,g \rangle  &  =\frac{1}{t} \langle f \boldsymbol{T}_{k} \mathcal{D}_{k +1} \boldsymbol{T}_{k} ,g \rangle  =\frac{1}{t} \langle f \boldsymbol{T}_{k} \mathcal{D}_{k +1} ,g \boldsymbol{T}_{k} \rangle  \\
 &  =\frac{1}{t} \sum _{j =k +1}^{N} \langle f \boldsymbol{T}_{k} ,x_{j} \left (g \boldsymbol{T}_{k} A_{k +1 ,j}\right ) \rangle  \\
 &  =\frac{1}{t} \sum _{j =k +1}^{N} \langle f ,\left \{x_{j} \left (g \boldsymbol{T}_{k} A_{k +1 ,j}\right )\right \} \boldsymbol{T}_{k} \rangle \text{.}\end{align*}Hence
\begin{align*}\left \{x_{k +1} \left (g \boldsymbol{T}_{k} A_{k +1 ,k +1}\right )\right \} \boldsymbol{T}_{k} &  =x_{k} \left (g \boldsymbol{T}_{k} A_{k +1 ,k +1} \boldsymbol{T}_{k}\right ) \\
&\kern1cm +\left (t -1\right ) x_{k +1} \left (g \boldsymbol{T}_{k} A_{k +1 ,k +1}\right )\text{,} \\
\left \{x_{j} \left (g \boldsymbol{T}_{k} A_{k +1 ,j}\right )\right \} \boldsymbol{T}_{k} &  =x_{j} \left (g \boldsymbol{T}_{k} A_{k +1 ,j} \boldsymbol{T}_{k}\right )\text{.}\end{align*}Thus set $A_{k ,k}:=\frac{1}{t} \boldsymbol{T}_{k} A_{k +1 ,k +1} \boldsymbol{T}_{k}$, $A_{k ,k +1}:=\frac{t -1}{t} \boldsymbol{T}_{k} A_{k +1 ,k +1}$ and $A_{k ,j}:=\frac{1}{t} \boldsymbol{T}_{k} A_{k +1 ,j} \boldsymbol{T}_{k}$ for $j >k +1$. Next \begin{equation*} \langle f ,x_{k} \left (g A_{k ,k}\right ) \rangle  = \langle f \mathcal{D}_{k} ,g \rangle  -\sum _{j =k +1}^{N} \langle f ,x_{j} \left (g A_{k ,j}\right ) \rangle \text{.}
\end{equation*} Replace $g$ by $g A_{k ,k}^{ -1}$ and use the inductive hypothesis to get
\begin{align*} \langle f ,x_{k} g \rangle  &  = \langle f \mathcal{D}_{k} ,g A_{k ,k}^{ -1} \rangle  +\sum _{m =k +1}^{N} \langle f \mathcal{D}_{m} ,g B_{k ,m} \rangle \text{,} \\
B_{k ,m} & := -\sum _{j =k +1}^{m}A_{k ,k}^{ -1} A_{k ,j} B_{j ,m} ,B_{k ,k}:=A_{k ,k}^{ -1}\text{.}\end{align*}This completes the induction. 
\end{proof}


\begin{corollary}
The symmetric bilinear form is uniquely determined by the hypotheses {\em(BF2)}. If $\deg \left (f\right ) <\deg \left (g\right )$ then  $\left \langle f ,g\right \rangle  =0$.


\begin{proof}
If $\deg \left (g\right ) =n \geq 1$ then $g$ can be expressed as a sum $g =\sum _{i =1}^{N}x_{i}g_{i}$ with $\deg \left (g_{i}\right ) =n -1$ for $i$ such that $g_{i} \neq 0$. This shows that if $f =1 \otimes S$ and $\deg \left (g\right ) \geq 1$ then $\left \langle f ,g\right \rangle  =0$ because $f\mathcal{D}_{i} =0$ for all $i$. Arguing inductively suppose the stated orthogonality property holds for all $h\text{ with }\text{}\deg \left (h\right ) \leq k$  and let $\deg \left (f\right ) =k +1 ,\deg \left (g\right ) >k$. Then $\left \langle f ,x_{i}g\right \rangle  =\sum _{j =i}^{N}$$\left \langle f\mathcal{D}_{j} ,gB_{i ,j}\right \rangle  =0$ because $\deg \left (f\mathcal{D}_{j}\right ) =k <\deg \left (gB_{i ,j}\right )$. Thus the orthogonality property holds for $k +1.$ The form is uniquely defined for $\mathcal{P}_{0} \otimes V_{\tau }$ and a similar inductive argument shows that $\left \langle f ,g\right \rangle $  is uniquely determined when $\deg \left (f\right ) =\deg \left (g\right ) >0$.
\end{proof}
\end{corollary}

 However the result does not prove existence. A closer look at the formulas shows that $\left (1 -q\right )^{\left \vert \alpha \right \vert } \langle x^{\alpha } \otimes S ,x^{\beta } \otimes S^{ \prime } \rangle $ is a Laurent polynomial in $q ,t$ (a sum of $q^{a} t^{b}$ with $a ,b \in \mathbb{Z}$) for any $\alpha  ,\beta  \in \mathbb{N}_{0}^{N}$, $S ,S^{ \prime } \in \mathcal{Y} \left (\tau \right )$. 

Recall $u \left (z\right ):=\frac{\left (1 -t z\right ) \left (t -z\right )}{\left (1 -z\right )^{2}}$. 


\begin{lemma}
Suppose {\em(BF2)} holds and suppose $\left (\alpha  ,S\right )$ satisfies $\alpha _{i} <\alpha _{i +1}$. Then with $\rho  =\zeta _{\alpha  ,S} \left (i +1\right )/\zeta _{\alpha  ,S} \left (i\right )$ we have \begin{equation*}\left \Vert M_{\alpha  .s_{i} ,S}\right \Vert ^{2} =u \left (\rho \right ) \left \Vert M_{\alpha  ,S}\right \Vert ^{2}\text{.}
\end{equation*} 
\end{lemma}


\begin{proof}
From \eqref{MT&gt;i} and \eqref{MT<i} we have
\begin{align*}M_{\alpha  ,S} \boldsymbol{T}_{i} &  = -\frac{1 -t}{1 -\rho } M_{\alpha  ,S} +M_{\alpha  .s_{i} ,S}\text{,} \\
M_{\alpha  .s_{i} ,S} \boldsymbol{T}_{i} &  =\frac{\left (1 -t \rho \right ) \left (t -\rho \right )}{\left (1 -\rho \right )^{2}} M_{\alpha  ,S} +\frac{\rho  \left (1 -t\right )}{\left (1 -\rho \right )} M_{\alpha  .s_{i} ,S}\text{.}\end{align*}Take the inner product of the first equation with $M_{\alpha  .s_{i} ,S}$  and use $ \langle M_{\alpha  ,S} ,M_{\alpha  .s_{i} ,S} \rangle  =0$, then take the inner product of the second equation with $M_{\alpha  ,S}$ and again use $ \langle M_{\alpha  ,S} ,M_{\alpha  .s_{i} ,S} \rangle  =0$ to obtain
\begin{align*} \langle M_{\alpha  ,S} \boldsymbol{T}_{i} ,M_{\alpha  .s_{i} ,S} \rangle  &  =\left \Vert M_{\alpha  .s_{i} ,S}\right \Vert ^{2}\text{,} \\
 \langle M_{\alpha  ,S} ,M_{\alpha  .s_{i} ,S} \boldsymbol{T}_{i} \rangle  &  =\frac{\left (1 -t \rho \right ) \left (t -\rho \right )}{\left (1 -\rho \right )^{2}} \left \Vert M_{\alpha  ,S}\right \Vert ^{2}\text{.}\end{align*}The hypothesis $ \langle M_{\alpha  ,S} \boldsymbol{T}_{i} ,M_{\alpha  .s_{i} ,S} \rangle  = \langle M_{\alpha  ,S} ,M_{\alpha  .s_{i} ,S} \boldsymbol{T}_{i} \rangle $ completes the proof. 
\end{proof}


\begin{lemma}
Suppose {\em(BF2)} holds and suppose $\left (\alpha  ,S\right )$ satisfies $\alpha _{i} =\alpha _{i +1} $, $j =r_{\alpha } \left (i\right ) $, $c \left (j ,S\right ) -c \left (j +1 ,S\right ) \geq 2$. Then with 
$$\rho  = 
\zeta _{\alpha  ,S} \left (i +1\right )/\zeta _{\alpha  ,S} \left (i\right ) =t^{c \left (j +1 ,S\right ) -c \left (j ,S\right )}$$
we have
\begin{equation*}\left \Vert M_{\alpha  ,S^{\left (j\right )}}\right \Vert ^{2} =u \left (\rho \right ) \left \Vert M_{\alpha  ,S}\right \Vert ^{2} =\frac{\left \Vert S^{\left (j\right )}\right \Vert _{0}^{2}}{\left \Vert S\right \Vert _{0}^{2}} \left \Vert M_{\alpha  ,S}\right \Vert ^{2}\text{.}
\end{equation*} 
\end{lemma}


\begin{proof}
Using the same argument as in the previous lemma on formulas \eqref{MT=i} and \eqref{MT=2}, one
shows $\left \Vert M_{\alpha  ,S^{\left (j\right )}}\right \Vert ^{2} =u \left (\rho \right ) \left \Vert M_{\alpha  ,S}\right \Vert ^{2}$. Proposition~\ref{Sform} asserted that $u \left (\rho \right ) =\left \Vert S^{\left (j\right )}\right \Vert _{0}^{2}/\left \Vert S\right \Vert _{0}^{2}$ . 
\end{proof}


\begin{definition}
For $\alpha  \in \mathbb{N}_{0}^{N} ,S \in \mathcal{Y} \left (\tau \right )$ let\begin{equation}\mathcal{E} \left (\alpha  ,S\right ):=\underset{\alpha _{i} <\alpha _{j}}{\prod _{1 \leq i <j \leq N}}u \left (q^{\alpha _{j} -\alpha _{i}} t^{c \left (r_{\alpha } \left (j\right ) ,S\right ) -c \left (r_{\alpha } \left (i\right ) ,S\right )}\right )\text{.}
\end{equation} 
\end{definition}

There are $\mathrm{i} \mathrm{n} \mathrm{v} \left (\alpha \right )$ terms in $\mathcal{E} \left (\alpha  ,S\right )$. 


\begin{lemma}
\label{Mdiff2}Suppose $\alpha  \in \mathbb{N}_{0}^{N} ,S \in \mathcal{Y} \left (\tau \right )$. Then 
$$M_{\alpha  \Phi  ,S} \mathcal{D}_{N} = 
\left (1 -q \zeta _{\alpha  ,S} \left (1\right )\right ) M_{\alpha  ,S} \boldsymbol{w}.$$ 
\end{lemma}


\begin{proof}
By definition we have
\begin{align*}M_{\alpha  \Phi  ,S} \mathcal{D}_{N} &  =\left (1/x_{N}\right ) M_{\alpha  \Phi  ,S} \left (I -\mathbf{\xi }_{N}\right ) =\left (1/x_{N}\right ) \left (1 -\zeta _{\alpha  \Phi  ,S} \left (N\right )\right ) M_{\alpha  \Phi  ,S} \\
 &  =\left (1 -q \zeta _{\alpha  ,S} \left (1\right )\right ) M_{\alpha  ,S} \boldsymbol{w}\text{.}\qedhere\end{align*}
\end{proof}

The following is proved exactly like Propositions~\ref{e-norm1}
and~\ref{phinorm1}. 


\begin{proposition}
\label{ephinorm}Suppose {\em(BF2)} holds and $\alpha  \in \mathbb{N}_{0}^{N} $, $S \in \mathcal{Y} \left (\tau \right )$. Then
\begin{align*}\left \Vert M_{\alpha ^{ +} ,S}\right \Vert ^{2} &  =\mathcal{E} \left (\alpha  ,S\right ) \left \Vert M_{\alpha  ,S}\right \Vert ^{2}\text{,} \\
\left \Vert M_{\alpha  ,\Phi  ,S}\right \Vert ^{2} &  =\frac{1 -q^{\alpha _{1} +1} t^{c \left (r_{\alpha } \left (1\right ) ,S\right )}}{1 -q} \left \Vert M_{\alpha  ,S}\right \Vert ^{2}\text{.}\end{align*}
\end{proposition}

The intention here is to find the explicit formula for $\left \Vert M_{\alpha  ,S}\right \Vert ^{2}$ implied by (BF2) and then prove that, as a definition, it satisfies (BF2). We use the same inductive scheme as in Section~\ref{SMD}.


Suppose (BF2) holds and $\lambda  \in \mathbb{N}_{0}^{N , +} ,S \in \mathcal{Y} \left (\tau \right )$ and $\lambda _{m} >0 =\lambda _{m +1}$. Then set
\begin{align}\alpha  & :=\left (\lambda _{1} ,\ldots  ,\lambda _{m -1} ,0 ,\ldots  0 ,\lambda _{m}\right ) , \label{abgamma} \\
r_{\alpha } &  =\left (1 ,\ldots  ,m -1 ,m +1 ,\ldots  ,N ,m\right ) , \nonumber  \\
\beta  & :=\left (\lambda _{m} -1 ,\lambda _{1} ,\ldots  ,\lambda _{m -1} ,0 ,\ldots \right ) , \nonumber  \\
r_{\beta } &  =\left (m ,1 ,\ldots  ,m -1 ,m +1 ,\ldots  ,N\right ) , \nonumber  \\
\gamma  & :=\left (\lambda _{1} ,\ldots  ,\lambda _{m -1} ,\lambda _{m} -1 ,0 ,\ldots \right ) =\beta ^{ +} . \nonumber \end{align}Thus $\left \Vert M_{\lambda  ,S}\right \Vert ^{2} =\mathcal{E} \left (\alpha  ,S\right ) \left \Vert M_{\alpha  ,S}\right \Vert ^{2}$ and $\left \Vert M_{\beta  ,S}\right \Vert ^{2} =\mathcal{E} \left (\beta  ,S\right )^{ -1} \left \Vert M_{\gamma  ,S}\right \Vert ^{2}$; by Proposition~\ref{ephinorm} we have $\left \Vert M_{\alpha  ,S}\right \Vert ^{2} =\frac{1 -q^{\lambda _{m}} t^{c \left (m ,S\right )}}{1 -q} \left \Vert M_{\beta  ,S}\right \Vert ^{2}$. Also\break 
 $\alpha  .\left (s_{N -1} s_{N -2} \cdots  s_{m}\right ) =\lambda $ and $\beta  .\left (s_{1} s_{2} \cdots  s_{m -1}\right ) =\gamma $ thus 
 $R_{\alpha } =\break T_{m}^{ -1} \cdots  T_{N -1}^{ -1}$ and $R_{\beta } =T_{m -1}^{ -1} \cdots  T_{1}^{ -1}$. The leading term of $M_{\beta  ,S}$ is $\eta  \left (\beta  ,S\right ) x^{\beta } \otimes S \tau  \left (R_{\beta }\right )$, so the leading term of $M_{\gamma  ,S}$ is $\eta  \left (\beta  ,S\right ) x^{\gamma } \otimes S$ (and $\eta  \left (\gamma  ,S\right ) =\eta  \left (\beta  ,S\right )$). 

Apply $\mathbf{w}$ to $M_{\beta  ,S}$. Then we have
\begin{align}x_{N} \left (\left (x^{\beta }\right ) w\right ) S \tau  \left (R_{\beta } \omega \right ) &  =q^{\beta _{1}} x^{\alpha } \otimes S \tau  \left (\left (T_{m -1}^{ -1} \cdots  T_{1}^{ -1}\right ) T_{1} \cdots  T_{N -1}\right ) \\
 &  =q^{\beta _{1}} x^{\alpha } \otimes S \tau  \left (T_{m} \cdots  T_{N -1}\right )\text{,} \nonumber \end{align}and
\begin{align}S \tau  \left (T_{m} \cdots  T_{N -1}\right ) &  =S \tau  \left (\left (T_{m} \cdots  T_{N -1}\right ) \left (T_{N -1} \cdots  T_{m}\right ) R_{\alpha }\right ) \\
 &  =t^{N -m} S \tau  \left (\phi _{m} R_{\alpha }\right ) =t^{N -m +c \left (m ,S\right )} S \tau  \left (R_{\alpha }\right )\text{.} \nonumber \end{align}Thus
\begin{align}\eta  \left (\alpha  ,S\right ) &  =q^{\lambda _{m} -1} t^{N -m +c \left (m ,S\right )} \eta  \left (\beta  ,S\right )\text{,} \\
\eta  \left (\lambda  ,S\right ) &  =\eta  \left (\alpha  ,S\right ) =q^{\lambda _{m} -1} t^{N -m +c \left (m ,S\right )} \eta  \left (\gamma  ,S\right )\text{.} \nonumber \end{align}Compute
\begin{align}\mathcal{E} \left (\alpha  ,S\right ) &  =\prod \limits _{j =m +1}^{N}u \left (q^{\lambda _{m}} t^{c \left (m ,S\right ) -c \left (j ,S\right )}\right )\text{,} \\
\mathcal{E} \left (\beta  ,S\right ) &  =\prod \limits _{i =1}^{m -1}u \left (q^{\lambda _{i} -\lambda _{m} +1} t^{c \left (i ,S\right ) -c \left (m ,S\right )}\right )\text{.} \nonumber \end{align}

The argument also shows that $\eta  \left (\lambda  ,S\right ) =q^{\Sigma _{1} \left (\lambda \right )} t^{\Sigma _{2} \left (\lambda  ,S\right )}$ where\break $\Sigma _{1} \left (\lambda \right ):=\frac{1}{2} \sum _{i =1}^{N}\lambda _{i} \left (\lambda _{i} -1\right )$ and $\Sigma _{2} \left (\lambda  ,S\right ) =\sum _{i =1}^{N}\lambda _{i} \left (N -i +c \left (i ,S\right )\right )$. Recall $k \left (\lambda \right ) =\sum _{i =1}^{N}\left (N -2 i +1\right ) \lambda _{i}$ for $\lambda  \in \mathbb{N}_{0}^{N , +}$. 


\begin{theorem}
Suppose {\em(BF2)} holds, $\lambda  \in \mathbb{N}_{0}^{N , +}$ and $S \in \mathcal{Y} \left (\tau \right )$. Then
\begin{multline}\left \Vert M_{\lambda  ,S}\right \Vert ^{2}   =t^{k \left (\lambda \right )} \left \Vert S\right \Vert _{0}^{2}\; \left (1 -q\right )^{ -\left \vert \lambda \right \vert } \prod \limits _{i =1}^{N}\left (q t^{c \left (i ,S\right )} ;q\right )_{\lambda _{i}} \label{normdef2} \\
   \times \prod _{1 \leq i <j \leq N}\frac{\left (q t^{c \left (i ,S\right ) -c \left (j ,S\right ) -1} ;q\right )_{\lambda _{i} -\lambda _{j}} \left (q t^{c \left (i ,S\right ) -c \left (j ,S\right ) +1} ;q\right )_{\lambda _{i} -\lambda _{j}}}{\left (q t^{c \left (i ,S\right ) -c \left (j ,S\right )} ;q\right )_{\lambda _{i} -\lambda _{j}}^{2}} . \nonumber \end{multline}
\end{theorem}


\begin{proof}
Denote the $\left (i ,j\right )$-product by $\Pi _{\lambda }$. Suppose $\lambda _{m} >0 =\lambda _{m +1}$ and $\gamma  =\left (\lambda _{1} ,\ldots  ,\lambda _{m -1} ,\lambda _{m} -1 ,0 ,\ldots \right )$.with $\alpha  ,\beta $ as in \eqref{abgamma}. Then
\begin{align}\frac{\Pi _{\lambda }}{\Pi _{\gamma }} &  =\prod \limits _{i =1}^{m -1}\frac{\left (1 -q^{\lambda _{i} -\lambda _{m} +1} t^{c \left (i ,S\right ) -c \left (m ,S\right )}\right )^{2}}{\left (1 -q^{\lambda _{i} -\lambda _{m} +1} t^{c \left (i ,S\right ) -c \left (m ,S\right ) -1}\right ) \left (1 -q^{\lambda _{i} -\lambda _{m} +1} t^{c \left (i ,S\right ) -c \left (m ,S\right ) +1}\right )} \\
 &\kern1cm  \times \prod \limits _{j =m +1}^{N}\frac{\left (1 -q^{\lambda _{m}} t^{c \left (m ,S\right ) -c \left (j ,S\right ) -1}\right ) \left (1 -q^{\lambda _{m}} t^{c \left (m ,S\right ) -c \left (j ,S\right ) +1}\right )}{\left (1 -q^{\lambda _{m}} t^{c \left (m ,S\right ) -c \left (j ,S\right )}\right )^{2}} \nonumber  \\
 &  =t^{2 m -1 -N} \prod \limits _{i =1}^{m -1}u \left (q^{\lambda _{i} -\lambda _{m} +1} t^{c \left (i ,S\right ) -c \left (m ,S\right )}\right )^{ -1} \prod \limits _{j =m +1}^{N}u \left (q^{\lambda _{m}} t^{c \left (m ,S\right ) -c \left (j ,S\right )}\right ) \nonumber  \\
 &  =t^{2 m -1 -N} \mathcal{E} \left (\alpha  ,S\right )/\mathcal{E} \left (\beta  ,S\right )\text{.} \nonumber \end{align}Also $\prod \limits _{i =1}^{N}\left (q t^{c \left (i ,S\right )} ;q\right )_{\lambda _{i}}/\prod \limits _{i =1}^{N} \left (q t^{c \left (i ,S\right )} ;q\right )_{\gamma _{i}} =1 -q^{\lambda _{m}} t^{c \left (m ,S\right )}$. The formula satisfies the relation $\left \Vert M_{\lambda  ,S}\right \Vert ^{2} =\frac{1 -q^{\lambda _{m}} t^{c \left (m ,S\right )}}{1 -q} \frac{\mathcal{E} \left (\alpha  ,S\right )}{\mathcal{E} \left (\beta  ,S\right )} \left \Vert M_{\gamma  ,S}\right \Vert ^{2}$ and is valid at $\lambda  =\mathbf{0}$ since $M_{\mathbf{0} ,S} =1 \otimes S$ and $\left \Vert 1 \otimes S\right \Vert ^{2} =\left \Vert S\right \Vert _{0}^{2}$. 
\end{proof}


\begin{definition}
The symmetric bilinear form is given by \eqref{normdef2} for $\lambda  \in \mathbb{N}_{0}^{N , +} ,S \in \mathcal{Y} \left (\tau \right )$, by $\left \Vert M_{\alpha  ,S}\right \Vert ^{2} =\mathcal{E} \left (\alpha  ,S\right )^{ -1} \left \Vert M_{\alpha ^{ +} ,S}\right \Vert ^{2}$ for $\alpha  \in \mathbb{N}_{0}^{N}$ and by $ \langle M_{\alpha  ,S} ,M_{\beta  ,S^{ \prime }} \rangle  =0$ for $\left (\alpha  ,S\right ) \neq \left (\beta  ,S^{ \prime }\right )\text{.}$ 
\end{definition}

Next we show that the definition satisfies the hypotheses (BF2). 

The
step $s_{i}$ with $\alpha _{i} <\alpha _{i +1}$ satisfies \eqref{BF2.2} because of the value $\frac{\mathcal{E} \left (\alpha  .s_{i} ,S\right )}{\mathcal{E} \left (\alpha  ,S\right )}$. It remains to check the step with $\alpha _{i} =\alpha _{i +1}$ and the affine step. The $\left (i ,j\right )$-product in \eqref{normdef2}
can be written as (note $t^{ -1} u \left (z\right ) =\frac{\left (1 -z/t\right ) \left (1 -t z\right )}{\left (1 -z\right )^{2}}$)\begin{equation*}\prod _{1 \leq i <j \leq N}t^{\lambda _{j} -\lambda _{i}} \prod \limits _{l =1}^{\lambda _{i} -\lambda _{j}}u \left (q^{l} t^{c \left (i ,S\right ) -c \left (j ,S\right )}\right )\text{.}
\end{equation*} Suppose $\alpha  \in \mathbb{N}_{0}^{N}$ and $\lambda :=\alpha ^{ +}$; in the formula for $\mathcal{E} \left (\alpha  ,S\right )$ the condition $\left (i <j\right ) \&\left (\alpha _{i} <\alpha _{j}\right )$ is equivalent to $\left (i <j\right ) \&\left (r_{\alpha } \left (i\right ) >r_{\alpha } \left (j\right )\right )$. Let $v_{\alpha } =r_{\alpha }^{ -1}$ so that $\lambda _{i} =\alpha _{v_{\alpha } \left (i\right )}$. Then the product can be indexed by $\left (v_{\alpha } \left (i^{ \prime }\right ) <v_{\alpha } \left (j^{ \prime }\right )\right ) \&\left (i^{ \prime } >j^{ \prime }\right )$ (where $i^{ \prime } =r_{\alpha } \left (i\right ) $, $j^{ \prime } =r_{\alpha } \left (j\right )$). Thus \begin{equation*}\mathcal{E} \left (\alpha  ,S\right ) =\prod _{1 \leq j^{ \prime } <i^{ \prime } \leq N ,v_{\alpha } \left (i^{ \prime }\right ) <v_{\alpha } \left (j^{ \prime }\right )}u \left (q^{\lambda _{i^{ \prime }} -\lambda _{j^{ \prime }}} t^{c \left (i^{ \prime } ,S\right ) -c \left (j^{ \prime } ,S\right )}\right )\text{.}
\end{equation*} 


\begin{proposition}
Suppose $\alpha _{i} =\alpha _{i +1} $, $j =r_{\alpha } \left (i\right )$ and $m =c \left (j ,S\right ) -c \left (j +1 ,S\right ) \geq 2$. Then $\left \Vert M_{\alpha  ,S^{\left (j\right )}}\right \Vert ^{2} =\frac{\left (1 -t^{1 -m}\right ) \left (t -t^{ -m}\right )}{\left (1 -t^{ -m}\right )^{2}} \left \Vert M_{\alpha  ,S}\right \Vert ^{2}$. 
\end{proposition}


\begin{proof}
By hypothesis $\zeta _{\alpha  ,S} \left (i\right ) =q^{\alpha _{i}} t^{c \left (j ,S\right )}$ and $\zeta _{\alpha  ,S} \left (i +1\right ) =q^{\alpha _{i}} t^{c \left (j +1 ,S\right )}$ so that $\zeta _{\alpha  ,S} \left (i +1\right )/\zeta _{\alpha  ,S} \left (i\right ) =t^{ -m}$. Also by Proposition~\ref{Sform} we have $\left \Vert S^{\left (j\right )}\right \Vert _{0}^{2} =u \left (t^{ -m}\right ) \left \Vert S\right \Vert _{0}^{2}$. Suppose first that $\alpha  \in \mathbb{N}_{0}^{N , +}$. Then $j =i$. In the formula for $\left \Vert M_{\alpha  ,S}\right \Vert ^{2}$ the first product does not change when $S$ is replaced by $S^{\left (j\right )}$; the factors $\left (q t^{c \left (i ,S\right )} ;q\right )_{\lambda _{i}}$,$\left (q t^{c \left (i +1 ,S\right )} ;q\right )_{\lambda _{i}}$ trade places. By a similar argument the $\left (i ,j\right )$-product also does not change, and $\left \Vert M_{\alpha  ,S^{\left (j\right )}}\right \Vert ^{2}/\left \Vert S^{\left (j\right )}\right \Vert _{0}^{2} =\left \Vert M_{\alpha  ,S}\right \Vert ^{2}/\left \Vert S\right \Vert _{0}^{2}$. Otherwise $\alpha  \neq \alpha ^{ +}$ and
\begin{align}\frac{\left \Vert M_{\alpha  ,S^{\left (j\right )}}\right \Vert ^{2}}{\left \Vert S^{\left (j\right )}\right \Vert _{0}^{2}} &=\frac{\left \Vert M_{\alpha ^{ +} ,S^{\left (j\right )}}\right \Vert ^{2}}{\mathcal{E} \left (\alpha  ,S^{\left (j\right )}\right ) \left \Vert S^{\left (j\right )}\right \Vert _{0}^{2}} =\frac{\left \Vert M_{\alpha ^{ +} ,S}\right \Vert ^{2}}{\mathcal{E} \left (\alpha  ,S^{\left (j\right )}\right ) \left \Vert S\right \Vert _{0}^{2}} \\
& =\frac{\mathcal{E} \left (\alpha  ,S\right )}{\mathcal{E} \left (\alpha  ,S^{\left (j\right )}\right )} \frac{\left \Vert M_{\alpha  ,S}\right \Vert ^{2}}{\left \Vert S\right \Vert _{0}^{2}}\text{.} \nonumber \end{align} Recall $\mathcal{E} \left (\alpha  ,S\right ) =\prod _{1 \leq l <n \leq N ,\,\alpha _{l} <\alpha _{n}}u \left (q^{\alpha _{n} -\alpha _{l}} t^{c \left (r_{\alpha } \left (n\right ) ,S\right ) -c \left (r_{\alpha } \left (l\right ) ,S\right )}\right )$ and the product does not change when $S$ is replaced by $S^{\left (j\right )}$ (the factors involving $l =i$ or $n =i$ are interchanged with those involving $l =i +1$ or $n =i +1$). Thus $\mathcal{E} \left (\alpha  ,S^{\left (j\right )}\right ) =\mathcal{E} \left (\alpha  ,S\right )$. 
\end{proof}


\begin{proposition}
Suppose $\alpha  \in \mathbb{N}_{0}^{N} ,S \in \mathcal{Y} \left (\tau \right )$. Then\begin{equation*}\left \Vert M_{\alpha  \Phi  ,S}\right \Vert ^{2} =\dfrac{1 -q^{\alpha _{1} +1} t^{c \left (r_{\alpha } \left (1\right ) ,S\right )}}{1 -q} \left \Vert M_{\alpha  ,S}\right \Vert ^{2}\text{.}
\end{equation*} 
\end{proposition}


\begin{proof}
We need to compute various ratios of $\mathcal{E} \left (\alpha  ,S\right ) $, $\left \Vert \text{}M_{\alpha ^{ +} ,S}\right \Vert ^{2} $,\break 
$\mathcal{E} \left (\alpha  \Phi  ,S\right )$, $\left \Vert M_{\left (\alpha  \Phi \right )^{ +} ,S}\right \Vert ^{2}$. Also $r_{\alpha } \left (i +1\right ) =r_{\alpha  \Phi } \left (i\right )$ for $1 \leq i <N $, $r_{\alpha } \left (1\right ) =r_{\alpha  \Phi } \left (N\right )$. Let $\lambda :=\alpha ^{ +}$. Then $\lambda _{r_{\alpha } \left (i\right )} =\alpha _{i}$ for all $i$. Let $m:=r_{\alpha } \left (1\right )$. This implies $\# \left \{i :\alpha _{i} >\alpha _{1}\right \} =m -1$, thus $\lambda _{m -1} >\lambda _{m}$ and $\left (\alpha  \Phi \right )_{m}^{ +} =\lambda _{m} +1$. Also $k\left (\left (\alpha \Phi \right )^{ +}\right ) -k\left (\lambda \right ) =N -2m +1 $ . This implies
\begin{align*}\frac{\left \Vert M_{\left (\alpha  \Phi \right )^{ +} ,S}\right \Vert ^{2}}{\left \Vert M_{\lambda  ,S}\right \Vert ^{2}} &=t^{N -2 m +1} \frac{1 -q^{\lambda _{m} +1} t^{c \left (m ,S\right )}}{1 -q}  \\
&\kern1cm \times t^{m -1} \prod \limits _{i =1}^{m -1}u \left (q^{\lambda _{i} -\lambda _{m}} t^{c \left (i ,S\right ) -c \left (m ,S\right )}\right )^{ -1} \\
&\kern1cm \times t^{m -N} \prod \limits _{j =m +1}^{N}u \left (q^{\lambda _{m} +1 -\lambda _{i}} t^{c \left (m ,S\right ) -c \left (j ,S\right )}\right )\text{.}\end{align*}Let $\mu :=\left (\alpha  \Phi \right )^{ +}$. Then 
$$\mathcal{E} \left (\alpha  \Phi  ,S\right ) =\underset{v_{\Phi } \left (i\right ) >v_{\alpha  \Phi } \left (j\right )}{\prod _{i <j}}u \left (q^{\mu _{i} -\mu _{j}} t^{c \left (i ,S\right ) -c \left (j ,S\right )}\right )$$ 
and 
$$\mathcal{E} \left (\alpha  ,S\right ) =\underset{v_{\alpha } \left (i\right ) >v_{\alpha } \left (j\right )}{\prod _{i <j}}u \left (q^{\lambda _{i} -\lambda _{j}} t^{c \left (i ,S\right ) -c \left (j ,S\right )}\right )\text{.}$$ 
From $v_{\alpha } \left (i\right ) = 
v_{\alpha  \Phi } \left (i\right ) +1$ except $v_{\alpha } \left (m\right ) =1 $, $v_{\alpha  \Phi } \left (m\right ) =N$ the inversions  
$\left \{\left (i ,j\right ) :i <j ,v_{\alpha }\left (i\right ) >v_{\alpha \left .\right .}\left (j\right )\right \}$ occur in both the products provided that $j \neq m$ in which case the pairs $\left \{\left (i ,m\right ) :1 \leq i \leq m -1\right \}$ do not occur in $\mathcal{E} \left (\alpha  \Phi  ,S\right )$, or if $i =m$ and the pairs $\left \{\left (m ,j\right ) :m <j \leq N\right \}$ do not occur in $\mathcal{E} \left (\alpha  ,S\right )$. Also $\mu _{i} =\lambda _{i}$ for all $i$ except $\mu _{m} =\lambda _{m} +1$. Thus
\begin{multline*}\frac{\mathcal{E} \left (\alpha  ,S\right )}{\mathcal{E} \left (\alpha  \Phi  ,S\right )} =\prod \limits _{i =1}^{m -1}u \left (q^{\lambda _{i} -\lambda _{m}} t^{c \left (i ,S\right ) -c \left (m ,S\right )}\right ) \\
\times\prod \limits _{j =m +1}^{N}u \left (q^{\lambda _{m} +1 -\lambda _{j}} t^{c \left (m ,S\right ) -c \left (j ,S\right )}\right )^{ -1}\text{,}
\end{multline*} and\begin{equation*}\frac{\left \Vert M_{\alpha  \Phi  ,S}\right \Vert ^{2}}{\left \Vert M_{\alpha  ,S}\right \Vert ^{2}} =\frac{\left \Vert M_{\left (\alpha  \Phi \right )^{ +} ,S}\right \Vert ^{2}}{\left \Vert M_{\lambda  ,S}\right \Vert ^{2}} \frac{\mathcal{E} \left (\alpha  ,S\right )}{\mathcal{E} \left (\alpha  \Phi  ,S\right )} =\frac{1 -q^{\lambda _{m} +1} t^{c \left (m ,S\right )}}{1 -q}\text{.}
\end{equation*} Finally $\zeta _{\alpha  ,S} \left (1\right ) =q^{\alpha _{1}} t^{c \left (r_{\alpha } \left (1\right ) ,S\right )} =q^{\lambda _{m}} t^{c \left (m ,S\right )}$. 
\end{proof}


\begin{corollary}
The bilinear form satisfies \eqref{BF2.4}. 
\end{corollary}


\begin{proof}
By Lemma \ref{Mdiff2} we have
\begin{align*} \langle M_{\alpha  \Phi  ,S} \mathcal{D}_{N} ,M_{\alpha  ,S} \left (\boldsymbol{w}^{ \ast }\right )^{ -1} \rangle  &  =\left (1 -q \zeta _{\alpha  ,S} \left (1\right )\right )  \langle M_{\alpha  ,S} \boldsymbol{w}\text{,}M_{\alpha  ,S} \left (\boldsymbol{w}^{ \ast }\right )^{ -1} \rangle  \\
 &  =\left (1 -q \zeta _{\alpha  ,S} \left (1\right )\right ) \left \Vert M_{\alpha  ,S}\right \Vert ^{2}
\end{align*}
and
\begin{align*}
 \langle M_{\alpha  \Phi  ,S} ,x_{N} \left (M_{\alpha  ,S} \left (\boldsymbol{w}^{ \ast }\right )^{ -1}\right ) \boldsymbol{w}^{ \ast } \boldsymbol{w} \rangle  &  = \langle M_{\alpha  \Phi  ,S} ,M_{\alpha  \Phi  ,S} \rangle  \\
 &  =\dfrac{1 -q \zeta _{\alpha  ,S} \left (1\right )}{1 -q} \left \Vert M_{\alpha  ,S}\right \Vert ^{2}\end{align*}by the proposition, thus $\left (1 -q\right ) \langle M_{\alpha  \Phi  ,S} ,x_{N} g \boldsymbol{w}^{ \ast } \boldsymbol{w} \rangle  = \langle M_{\alpha  \Phi  ,S} \mathcal{D}_{N} ,g \rangle $ when $g =M_{\alpha  ,S} \left (\boldsymbol{w}^{ \ast }\right )^{ -1}$. It suffices to prove 
$$ \langle f \mathcal{D}_{N} ,g \rangle  = 
\left (1 -q\right ) \langle f ,x_{N} \left (g \boldsymbol{w}^{ \ast } \boldsymbol{w}\right ) \rangle $$ 
for $f =M_{\gamma  ,S}$ and $g \boldsymbol{w}^{ \ast } =M_{\beta  ,S^{ \prime }}$ with $\left \vert \gamma \right \vert  =\left \vert \beta \right \vert  +1$. If $\gamma _{N} =0$ then $M_{\gamma  ,S} \mathcal{D}_{N} =0$ and $ \langle M_{\gamma  ,S} \mathcal{D}_{N} ,M_{\beta  .S^{ \prime }} \left (\boldsymbol{w}^{ \ast }\right )^{ -1} \rangle  =0$ while 
$$ \langle M_{\gamma  ,S} ,x_{N} \left (M_{\beta  ,S^{ \prime }} \boldsymbol{w}\right ) \rangle  = \langle M_{\gamma  ,S} ,M_{\beta  \Phi  ,S^{ \prime }} \rangle  =0$$ 
because $\gamma  \neq \beta  \Phi $. If $\gamma  =\alpha  \Phi $ for some $\alpha $ with $\left (\alpha  ,S\right ) \neq \left (\beta  ,S^{ \prime }\right )$ then 
$$ \langle M_{\alpha  \Phi  ,S} ,x_{N} \left (M_{\beta  ,S^{ \prime }} \boldsymbol{w}\right ) \rangle  = \langle M_{\alpha  \Phi  ,S} ,M_{\beta  \Phi  ,S^{ \prime }} \rangle  =0$$ and 
\begin{align*} 
\langle M_{\alpha  \Phi  ,S} \mathcal{D}_{N} ,M_{\beta  .S^{ \prime }} \left (\boldsymbol{w}^{ \ast }\right )^{ -1} \rangle  &=\left (1 -q \zeta _{\alpha } \left (1\right )\right )  
 \langle M_{\alpha  ,S} \boldsymbol{w}\text{,}M_{\beta  .S^{ \prime }} \left (\boldsymbol{w}^{ \ast }\right )^{ -1} \rangle  \\
&=\left (1 -q \zeta _{\alpha } \left (1\right )\right )  \langle M_{\alpha  ,S}\text{,}M_{\beta  .S^{ \prime }} \rangle  =0\text{.}
\end{align*}
The case $\left (\alpha  ,S\right ) =\left (\beta  ,S^{ \prime }\right )$ is already done. 
\end{proof}


\begin{proposition}
\label{regpos2}Suppose $\dim  V_{\tau } \geq 2$, $q ,t >0$ and $q \neq 1$. Then the form $ \langle  \cdot  , \cdot  \rangle $ is positive-definite provided $0 <q <\min \left (t^{ -h_{\tau }} ,t^{h_{\tau }}\right )$ or $q >\max \left (t^{ -h_{\tau }} ,t^{h_{\tau }}\right )$, that is, $\min \left (q^{ -1/h_{\tau }} ,q^{1/h_{\tau }}\right ) <t <\max \left (q^{ -1/h_{\tau }} ,q^{1/h_{\tau }}\right )$. 
\end{proposition}


\begin{proof}
In the definition of $ \langle M_{\alpha  ,S} ,M_{\alpha  ,S} \rangle $ there is an even number of factors of the form
$1 -q^{a} t^{b}$ where $a =1 ,2 ,3 ,\ldots $ and $b$ is one of $c \left (i ,S\right )$, $c \left (i ,S\right ) -c \left (j ,S\right )$, or $c \left (i ,S\right ) -c \left (j ,S\right ) \pm 1$. The $c \left (i ,S\right )$ values lie in $\left [1 -\ell  \left (\tau \right ) ,\tau _{1} -1\right ]$; thus $ -h_{\tau } \leq b \leq h_{\tau }$ where $h_{\tau } =\tau _{1} +\ell  \left (\tau \right ) -1$, the maximum hook length in the Ferrers diagram $\lambda $. Consider the four cases 
\begin{enumerate}
\item $0 <q <1 $, $0 <t <1$. Then $q^{a} t^{b} \leq q t^{ -h_{\tau }} <1$ provided $q <t^{h_{\tau }}$.
\item $0 <q <1 $, $t \geq 1$. Then $q^{a} t^{b} \leq q t^{h_{\tau }} <1$ provided $q <t^{ -h_{\tau }}$.
\item $q >1 $, $0 <t <1$. Then $q^{a} t^{b} \geq q t^{h_{\tau }} >1$ provided $q >t^{ -h_{\tau }}$.
\item $q >1 $, $t \geq 1$. Then $q^{q} t^{b} \geq q t^{ -h_{\gamma }} >1$ provided $q >t^{h_{\tau }}$.
\end{enumerate} 
Thus $\left \Vert M_{\alpha  ,S}\right \Vert ^{2} >0$ if $\min \left (q^{ -1/h_{\tau }} ,q^{1/h_{\tau }}\right ) <t <\max \left (q^{ -1/h_{\tau }} ,q^{1/h_{\tau }}\right )$. 
\end{proof}

There is an illustration in Figure \ref{t20} with $h_{\tau } =3$ (for $\tau  =\left (2 ,1\right )$ or $\tau  =\left (2 ,2\right )$). 

From a similar argument it follows that the transformation formulas for Macdonald polynomials have no poles
when $\min \left (q^{ -1/k} ,q^{1/k}\right ) <t <\max \left (q^{ -1/k} ,q^{1/k}\right )$ with $k =h_{\tau } -1$.

      
\begin{figure}\centering 
\setlength\fboxrule{0in}\setlength\fboxsep{0.2in}\fcolorbox[HTML]{000000}{FFFFFF}{\includegraphics[ width=3.1514in, height=3.1514in,]{posreg2}
}
\caption{\label{t20}Logarithmic coordinates, $h = 3$}\end{figure}


\subsection{Singular polynomials}
A singular polynomial $f \in \mathcal{P}_{\tau }$ is one which satisfies $f \mathcal{D}_{i} =0$ for all $i$ when $\left (q ,t\right )$ are specialized to some specific relation of the form $q^{a} t^{b} =1$. By Proposition~\ref{fDxg} the polynomial $f$ satisfies $ \langle f ,g \rangle  =0$ for all $g \in \mathcal{P}_{\tau }$, and in particular $ \langle f ,f \rangle  =0$. Thus the singular polynomial phenomenon can not occur in the $\left (q ,t\right )$-region of positivity. The boundary of the region does allow singular
polynomials. There are Macdonald polynomials which are singular when specialized to $q =t^{h_{\tau }}$ or $q =t^{ -h_{\tau }}$. These values do not produce poles in the polynomial coefficients as remarked above, since $\frac{1}{h_{\tau }} <\frac{1}{h_{\tau } -1}$. 


\begin{proposition}
\label{zerodiff}Suppose $\alpha  \in \mathbb{N}_{0}^{N} $, $S \in \mathcal{Y} \left (\tau \right )$ and $\alpha _{i} =0$ for $m <i \leq N$. Then $M_{\alpha  ,S} \mathcal{D}_{j} =0$ for $m <j \leq N$. 
\end{proposition}


\begin{proof}
Arguing by induction the start is 
$$M_{\alpha  ,S} \mathcal{D}_{N} =\frac{1}{x_{N}}M_{\alpha  ,S}\left (1 -\xi _{N}\right ) =\frac{1}{x_{N}}\left (1 -\zeta _{\alpha  ,S}\left (N\right )\right )M_{\alpha  ,S} =0\text{,}$$ 
since $\zeta _{\alpha  ,S} \left (N\right ) =1$. Suppose now that $\beta _{i} =0$ for $i \geq k +1$ implies $M_{\beta  ,S^{ \prime }} \mathcal{D}_{j} =0$ for $j \geq k +1$ and any $\left (\beta  ,S^{ \prime }\right )$. Suppose $\alpha _{i} =0$ for $i \geq k$. Then $r_{\alpha } \left (i\right ) =i$ and $\zeta _{\alpha  ,S} \left (i\right ) =t^{r \left (i ,S\right )}$ for $i \geq k$ and $M_{\alpha  ,S} \boldsymbol{T}_{k}$ is one of $t M_{\alpha  ,S}$, $ -M_{\alpha  ,S}$, $M_{\alpha  ,S^{\left (k\right )}} -\frac{t -1}{\rho  -1} M_{\alpha  ,S}$, $\frac{\left (1 -t \rho \right ) \left (t -\rho \right )}{\left (1 -\rho \right )^{2}} M_{\alpha  ,S^{\left (k\right )}} -\frac{t -1}{\left (1 -\rho \right )} M_{\alpha  ,S}$ depending on $c \left (k +1 ,S\right ) -c \left (k ,S\right ) =1$, $ = -1$, $ \geq 2$, $ \leq  -2$ respectively and $\rho  =t^{c \left (k +1 ,S\right ) -c \left (k ,S\right )}$. Then $M_{\alpha  ,S} \mathcal{D}_{k} =\frac{1}{t} \left (M_{\alpha  ,S} \boldsymbol{T}_{k}\right ) \mathcal{D}_{k +1} \boldsymbol{T}_{k}$ and $\left (M_{\alpha  ,S} \boldsymbol{T}_{k}\right ) \mathcal{D}_{k +1} =0$ by the inductive hypothesis. 
\end{proof}


\begin{lemma} \label{lemma}
Suppose $\alpha  =\left (\alpha _{1} ,\ldots  ,\alpha _{m -1} ,1 ,0 ,\ldots \right )$ with $\alpha _{i} \geq 1$ for $i \leq m$ and $S \in \mathcal{Y} \left (\tau \right )$. Then \begin{equation*}M_{\alpha  ,S} \mathcal{D}_{m} =t^{m -N} \prod \limits _{j =m}^{N -1}u \left (q t^{c \left (m ,S\right ) -c \left (j +1 ,S\right )}\right ) M_{\alpha ^{\left (N\right )} ,S} \mathcal{D}_{N} \boldsymbol{T}_{N -1} \cdots  \boldsymbol{T}_{m}\text{,}
\end{equation*} where $\alpha ^{\left (N\right )} =\left (\alpha _{1} ,\ldots  ,\alpha _{m -1} ,0 ,0 ,\ldots  ,1\right )$. 
\end{lemma}


\begin{proof}
For $m \leq j \leq N$ let $\alpha ^{\left (j\right )} =\left (\alpha _{1} ,\ldots  ,\alpha _{m -1} ,0 ,\ldots  ,\overset{j}{1} ,0 \ldots \right )$ so that $\alpha _{i}^{\left (j\right )} =\alpha _{i}$ except $\alpha _{j}^{\left (j\right )} =1$ and $\alpha _{m}^{\left (j\right )} =0$ (when $j \neq m$). Then $\zeta _{\alpha ^{\left (j\right )} ,S} \left (j\right ) =q t^{c \left (m ,S\right )}$ and $\zeta _{\alpha ^{\left (j\right )} ,S} \left (j +1\right ) =t^{c \left (j +1 ,S\right )}$ (since $r_{\alpha ^{\left (m\right )}} \left (j\right ) =m$) and \begin{equation*}M_{\alpha ^{\left (j\right )} ,S} \boldsymbol{T}_{j} =\frac{\left (1 -t \rho \right ) \left (t -\rho \right )}{\left (1 -\rho \right )^{2}} M_{\alpha ^{\left (j +1\right )} ,S} +\frac{\rho  \left (1 -t\right )}{\left (1 -\rho \right )} M_{\alpha ^{\left (j\right )} ,S}
\end{equation*} from \eqref{MT<i} with $\rho  =q t^{c \left (m ,S\right ) -c \left (j +1 ,S\right )}$. Thus 
\begin{align*}M_{\alpha ^{\left (j\right )} ,S} \mathcal{D}_{j} &=\frac{1}{t} M_{\alpha ^{\left (j\right )} ,S} \boldsymbol{T}_{j} \mathcal{D}_{j +1} \boldsymbol{T}_{j} \\
&=\frac{1}{t} u \left (q t^{c \left (m ,S\right ) -c \left (j +1 ,S\right )}\right ) M_{\alpha ^{\left (j +1\right )} ,S} \mathcal{D}_{j +1} \boldsymbol{T}_{j}\text{,}
\end{align*} 
because $M_{\alpha ^{\left (j\right )} ,S} \mathcal{D}_{j +1} =0$. Iterate this formula starting with $j =m$ and $\alpha ^{\left (m\right )} =\alpha $, ending with $j =N -1$ to obtain the stated formula. 
\end{proof}

Recall that $S_{1}$ is the {\it inv}-minimal RSYT with the numbers $N ,N -1 ,N -2 ,\ldots  ,1$ entered row-by-row and let $l =\ell  \left (\tau \right )$, $\alpha  =\left (1^{\tau _{l}} ,0^{N -\tau _{l}}\right )$. Thus the entry at $\left (l ,1\right )$ is $\tau _{l}$ and $c \left (\tau _{l} ,S_{1}\right ) =1 -l$. The entry at $\left (1 ,\tau _{1}\right )$ is $N -\tau _{1} +1$ and $c \left (N -\tau _{1} +1 ,S_{1}\right ) =\tau _{1} -1$. 


\begin{proposition}
$M_{\alpha  ,S_{1}}$ is singular for $q =t^{h_{\tau }}$. 
\end{proposition}


\begin{proof}
By the lemma with $m =\tau _{l}$, we have \begin{equation*}M_{\alpha  ,S_{1}} \mathcal{D}_{\tau _{l}} =t^{\tau _{l} -N} \prod \limits _{j =\tau _{l}}^{N -1}u \left (q t^{1 -l -c \left (j +1 ,S_{1}\right )}\right ) M_{\alpha ^{\left (N\right )} ,S_{1}} \mathcal{D}_{N} \boldsymbol{T}_{N -1} \cdots  \boldsymbol{T}_{\tau _{l}}\text{.}
\end{equation*} The factors in the denominator of the product are of the form $1 -q t^{1 -l -c \left (j +1 ,S_{1}\right )}$ with $c \left (j +1 ,S_{1}\right ) \leq \tau _{1} -1$ so that $1 -l -c \left (j +1 ,S_{1}\right ) \geq 2 -l -\tau _{1} =1 -h_{\tau } >h_{\tau }$. Furthermore the numerator factor at $j =N -\tau _{1}$ is $\left (t -q t^{2 -l -\tau _{1}}\right ) \left (1 -q t^{3 -l -\tau _{1}}\right )$ which vanishes at $q t^{ -h_{\tau }} =1$. By Proposition~\ref{zerodiff} we have $M_{\alpha  ,S_{1}} \mathcal{D}_{i} =0$ for $i >\tau _{l}$. If $1 \leq i <\tau _{l}$ then $M_{\alpha  ,S_{1}} \boldsymbol{T}_{i} =t M_{\alpha  ,S_{1}}$ (because $i ,i +1$ are in the same row of $S_{1})$, thus
\begin{align*}M_{\alpha  ,S_{1}} \mathcal{D}_{i} &  =t^{i -\tau _{l}} M_{\alpha  ,S_{i}} \boldsymbol{T}_{i} \boldsymbol{T}_{i +1} \cdots  \boldsymbol{T}_{\tau _{l} -1} \mathcal{D}_{\tau _{l}} \boldsymbol{T}_{\tau _{l} -1} \cdots  \boldsymbol{T}_{i} \\
 &  =M_{\alpha  ,S_{i}} \mathcal{D}_{\tau _{l}} \boldsymbol{T}_{\tau _{l} -1} \cdots  \boldsymbol{T}_{i} =0\end{align*}when $q =t^{h_{\tau }}$. 
\end{proof}

We apply the same argument to $S_{0}$ where the numbers $N ,N -1 ,\ldots  ,1$ are entered column-by-column. Let $m =\tau _{\tau _{1}}^{ \prime }$, that is, the length of the last column of $\tau $. Then the entry at $\left (\tau _{1} ,1\right )$ is $m$ and $c \left (m ,S_{0}\right ) =\tau _{1} -1$. Also the entry at $\left (l ,1\right )$ is $N -l +1$ and $c \left (N -l +1\right ) =1 -l$. 


\begin{proposition}
Set $\alpha  =\left (1^{m} ,0^{N -m}\right )$. Then $M_{\alpha  ,S_{0}}$ is singular for $q =t^{ -h_{\tau }}$. 
\end{proposition}


\begin{proof}
By Lemma~\ref{lemma} we have\begin{equation*}M_{\alpha  ,S_{0}} \mathcal{D}_{m} =t^{m -N} \prod \limits _{j =m}^{N -1}u \left (q t^{\tau _{1} -1 -c \left (j +1 ,S_{1}\right )}\right ) M_{\alpha ^{\left (N\right )} ,S_{0}} \mathcal{D}_{N} \boldsymbol{T}_{N -1} \cdots  \boldsymbol{T}_{m}\text{.}
\end{equation*} The factors in the denominator of the product are of the form $1 -q t^{\tau _{1} -1 -c \left (j +1 ,S_{1}\right )}$ with $c \left (j +1 ,S_{1}\right ) \geq 1 -l$ so that $\tau _{1} -1 -c \left (j +1 ,S_{1}\right ) \leq \tau _{1} +l -2 <h_{\tau }$. Furthermore the numerator factor at $j =N -l$ is $\left (t -q t^{\tau _{1} +l -2}\right ) \left (1 -q t^{\tau _{1} +l -1}\right )$ which vanishes at $q t^{h_{\tau }} =1$. The rest of the argument is as in the previous proposition with the difference that $M_{\alpha  ,S_{0}} \boldsymbol{T}_{i} = -M_{\alpha  ,S_{0}}$ for $1 \leq i <m$. 
\end{proof}

In conclusion we have constructed a symmetric bilinear form on $\mathcal{P}_{\tau }$ for which the operators $\boldsymbol{T}_{i}$ and $\mathbf{\xi }_{i}$ are self-adjoint, the Macdonald polynomials $M_{\alpha  ,S}$ are mutually orthogonal, and the form is positive-definite for $q >0 ,q \neq 1$ and $\min \left (q^{ -1/h_{\tau }} ,q^{1/h_{\tau }}\right ) <t <\max \left (q^{ -1/h_{\tau }} ,q^{1/h_{\tau }}\right )$) where $h_{\tau } =\tau _{1} +\ell  \left (\tau \right ) -1$. The bound is sharp, as demonstrated by the existence of singular polynomials for $q =t^{ \pm h_{\tau }}$. 


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