Nonsymmetric Macdonald Superpolynomials

Nonsymmetric Macdonald Superpolynomials

Charles F. DUNKL

C.F. Dunkl

Department of Mathematics, University of Virginia,
PO Box 400137, Charlottesville VA 22904-4137, USA<EMAIL_ADDRESS>https://uva.theopenscholar.com/charles-dunkl

Received November 16, 2020, in final form May 13, 2021; Published online May
23, 2021

There are representations of the type-A Hecke algebra on spaces of polynomials
in anti-commuting variables. Luque and the author [Sém. Lothar. Combin. 66
(2012), Art. B66b, 68 pages, arXiv:1106.0875] constructed nonsymmetric
Macdonald polynomials taking values in arbitrary modules of the Hecke algebra.
In this paper the two ideas are combined to define and study nonsymmetric
Macdonald polynomials taking values in the aforementioned anti-commuting
polynomials, in other words, superpolynomials. The modules, their orthogonal
bases and their properties are first derived. In terms of the standard Young
tableau approach to representations these modules correspond to hook tableaux.
The details of the Dunkl–Luque theory and the particular application are
presented. There is an inner product on the polynomials for which the
Macdonald polynomials are mutually orthogonal. The squared norms for this
product are determined. By using techniques of Baker and Forrester [Ann. Comb.
3 (1999), 159–170, arXiv:q-alg/9707001] symmetric Macdonald polynomials are
built up from the nonsymmetric theory. Here “symmetric” means in the Hecke
algebra sense, not in the classical group sense. There is a concise formula
for the squared norm of the minimal symmetric polynomial, and some formulas
for anti-symmetric polynomials. For both symmetric and anti-symmetric
polynomials there is a factorization when the polynomials are evaluated at
special points.

superpolynomials; Hecke algebra; symmetrization; norms

33D56; 20C08; 05E05

##  1 Introduction

Nonsymmetric Macdonald [13] polynomials are simultaneous eigenfunctions of a
set of mutually commuting operators derived from an action of the type-$A$
Hecke algebra on the space of polynomials in $N$ variables. They are
significantly different from the symmetric Macdonald polynomials in the
technique of their respective definitions and yet Baker and Forrester [1]
established a strong relation between them. In the analogous theory of
nonsymmetric Jack polynomials Griffeth [11] constructed such polynomials which
take values in modules of the underlying groups, specifically the complex
reflection groups in the infinite family $G(\ell,p,N)$. These polynomials
constitute a standard module of the rational Cherednik algebra. Luque and the
author [9] extended the theory of nonsymmetric Macdonald polynomials in the
direction suggested by Griffeth’s work by studying polynomials taking values
in modules of the Hecke algebra. The development relies on exploiting standard
Young tableaux and the Yang–Baxter graph technique of Lascoux [12].

The superpolynomials considered here are generated by $N$ anti-commuting and
$N$ commuting variables. By defining representations of the Hecke algebra on
anti-commuting variables the theory of vector-valued nonsymmetric Macdonald
polynomials is applied to define and analyze superpolynomials. There is a
theory of symmetric Macdonald superpolynomials initiated by Blondeau-Fournier,
Desrosiers, Lapointe, and Mathieu [3] with further developments on norm and
special point values by González and Lapointe [10]. Their approach and
definitions are based on differential operators and linear combinations of the
classical nonsymmetric Macdonald polynomials, whose coefficients involve anti-
commuting variables. The theory developed in the present paper is different
due to the method of using anti-commuting variables to form Hecke algebra
modules.

Nonsymmetric Macdonald polynomials associated with general root systems were
intensively studied by Cherednik [5]. By specializing to root systems of type
$A$ it becomes possible to develop more detailed relations, formulas and
structure. In particular, the papers of Noumi and Mimachi [14], Baker and
Forrester [1] provide important background for the present paper. Note that
some authors use different axioms for the quadratic relations of the Hecke
algebra, such as $\big{(}T-t^{1/2}\big{)}\big{(}T+t^{-1/2}\big{)}=0$, rather
than $(T-t)(T+1)=0$.

The theory of Hecke algebras of type $A$ and their representations is briefly
described in Section 2 and then applied to modules of polynomials in anti-
commuting variables. In general the irreducible representations are
constructed as spans of standard Young tableaux whose shape corresponds to a
fixed partition of $N$. In the present situation it is the hook tableaux which
arise. The basis vectors are constructed and the important transformation
formulas are stated. There is an inner product in which the generators of the
Hecke algebra are self-adjoint which leads to evaluation of the squared norms
of the basis elements.

In Section 3 the theory of vector-valued nonsymmetric Macdonald polynomials
developed in [9] is applied to produce superpolynomials, considered as
polynomials taking values in modules of anti-commuting variables. The main
results are stated without proofs but some important details are carefully
worked out. In [8] the author constructed an inner product in which the
nonsymmetric Macdonald polynomials are mutually orthogonal, in the general
vector-valued situation. This structure is worked out for the superpolynomials
in Section 3.3 and the squared norms are computed. In Section 4 the techniques
of Baker and Forrester [1] are used to produce supersymmetric Macdonald
polynomials, and the squared norms. From results of [9] the labels of these
polynomials correspond to the superpartitions of Desrosiers, Lapointe, and
Mathieu [6]. It has to be emphasized that in this paper the meaning of
symmetric is with respect to the Hecke algebra, not the symmetric group. Also
the squared norm of the lowest degree supersymmetric polynomial is determined
– the formula is more elegant than the general formula; its calculation is
able to use telescoping arguments for simplifications. There is a derivation
of formulas for antisymmetric Macdonald polynomials in Section 4.5. In the
conclusion some further topics of investigation, such as evaluation at special
points, are discussed.

##  2 The Hecke algebra of type $\boldsymbol{A}$

###  2.1 Definitions and Jucys–Murphy elements

The Hecke algebra $\mathcal{H}_{N}(t)$ of type $A_{N-1}$ with parameter $t$ is
the associative algebra over an extension field of $\mathbb{Q}$, generated by
$\big{\\{}T_{1},\dots,T_{N-1}\big{\\}}$ subject to the braid relations

$\displaystyle T_{i}T_{i+1}T_{i}=T_{i+1}T_{i}T_{i+1},\qquad 1\leq i<N-1,$
(2.1a) $\displaystyle T_{i}T_{j}=T_{j}T_{i},\qquad|i-j|\geq 2,$ (2.1b) and the
quadratic relations  $\displaystyle(T_{i}-t)(T_{i}+1)=0,\qquad 1\leq i<N,$
(2.2)

where $t$ is a generic parameter (this means $t^{n}\neq 1$ for $2\leq n\leq
N$). The quadratic relation implies $T_{i}^{-1}=\frac{1}{t}(T_{i}+1-t)$. There
is a commutative set in $\mathcal{H}_{N}(t)$ of Jucys–Murphy elements defined
by $\omega_{N}=1$, $\omega_{i}=t^{-1}T_{i}\omega_{i+1}T_{i}$ for $1\leq i<N$,
that is,

$\displaystyle\omega_{i}=t^{i-N}T_{i}T_{i+1}\cdots T_{N-1}T_{N-1}T_{N-2}\cdots
T_{i}.$

Simultaneous eigenvectors of $\\{\omega_{i}\\}$ form bases of irreducible
representations of the algebra. The symmetric group $\mathcal{S}_{N}$ is the
group of permutations of $\\{1,2,\dots,N\\}$ and is generated by the simple
reflections (adjacent transpositions) $\\{s_{i}\colon 1\leq i<N\\}$, where
$s_{i}$ interchanges $i$, $i+1$ and fixes the other points (the $s_{i}$
satisfy the braid relations and $s_{i}^{2}=1$). There is a linear isomorphism
$\mathbb{Z}\mathcal{S}_{N}\rightarrow\mathcal{H}_{N}(t)$ given by
$\sum_{u\in\mathcal{S}_{N}}a_{u}u\rightarrow\sum_{u\in\mathcal{S}_{N}}a_{u}T(u)$,
where $T(u)=T_{i_{1}}\cdots T_{i_{\ell}}$ with $u=s_{i_{1}}\cdots
s_{i_{\ell}}$ being a shortest expression for $u$ (in fact
$\ell=\\#\\{(i,j)\colon i<j,\,u(i)>u(j)\\}$); $T(u)$ is well-defined because
of the braid relations (see [7]).

###  2.2 Modules of anti-commuting variables

Consider polynomials in $N$ anti-commuting (fermionic) variables
$\theta_{1},\theta_{2},\dots,\theta_{N}$. They satisfy $\theta_{i}^{2}=0$ and
$\theta_{i}\theta_{j}+\theta_{j}\theta_{i}=0$ for $i\neq j$. The basis for
these polynomials consists of monomials labeled by subsets of
$\\{1,2,\dots,N\\}$:

$\displaystyle\phi_{E}:=\theta_{i_{1}}\cdots\theta_{i_{m}},\qquad
E=\\{i_{1},i_{2},\dots,i_{m}\\},\qquad 1\leq i_{1}<i_{2}<\cdots<i_{m}\leq N.$

The polynomials have coefficients in an extension field of $\mathbb{Q}(q,t)$
with transcendental $q$, $t$, or generic $q$, $t$ satisfying $q,t\neq 0$,
$q^{a}\neq 1$, $q^{a}t^{n}\neq 1$ for $a\in\mathbb{Z}$ and $n\neq
2,3,\dots,N$.

######  Definition 2.1.

$\mathcal{P}:=\operatorname{span}\big{\\{}\phi_{E}\colon
E\subset\\{1,\dots,N\\}\big{\\}}$ and
$\mathcal{P}_{m}:=\operatorname{span}\big{\\{}\phi_{E}\colon\\#E=m\big{\\}}$
for $0\leq m$ $\leq N$. The fermionic degree of $\phi_{E}$ is $\\#E$.

Some utility formulas are used for working with $\\{\phi_{E}\\}$.

######  Definition 2.2.

For a subset $E\subset\\{1,2,\dots,N\\}$ and $1\leq i<N$ let

$\displaystyle E^{C}:=,2,\dots,N\\}\backslash E,$
$\displaystyle\operatorname{inv}(E):=\\#\big{\\{}(i,j)\in E\times E^{C}\colon
i<j\big{\\}},$ $\displaystyle
s_{i}\phi_{E}=\phi_{s_{i}E}=\phi_{(E\backslash\\{i\\})\cup\\{i+1\\}},\qquad(i,i+1)\in
E\times E^{C},$ $\displaystyle
s_{i}\phi_{E}=\phi_{s_{i}E}=\phi_{(E\backslash\\{i+1\\})\cup\\{i\\}},\qquad(i,i+1)\in
E^{C}\times E.$

(When $\\{i,i+1\\}\subset E$ or $\subset E^{C}$ then $s_{i}E=E$ and
$s_{i}\phi_{E}=\phi_{E}$.) Introduce a representation of $\mathcal{H}_{N}(t)$
on $\mathcal{P}$.

######  Definition 2.3.

For $1\leq i<N$

$\displaystyle T_{i}\phi_{E}=\begin{cases}-\phi_{E},&\\{i,i+1\\}\subset E,\\\
t\phi_{E},&\\{i,i+1\\}\subset E^{C},\\\ s_{i}\phi_{E},&(i,i+1)\in E\times
E^{C},\\\ (t-1)\phi_{E}+ts_{i}\phi_{E},&(i,i+1)\in E^{C}\times E.\end{cases}$

######  Proposition 2.4.

The operators $\\{T_{i}\\}$ satisfy the braid and quadratic relations (2.1a)
and (2.2).

###### Proof.

It suffices to verify that $T_{1}T_{2}T_{1}=T_{2}T_{1}T_{2}$ and
$(T_{1}-t)(T_{1}+1)=0$ on the spaces
$\operatorname{span}\\{\theta_{1},\theta_{2},\theta_{3}\\}$ and
$\operatorname{span}\\{\theta_{2}\theta_{3},\theta_{1}\theta_{3},\theta_{1}\theta_{2}\\}$.
The relations are trivially satisfied on $\operatorname{span}\\{1\\}$ and
$\operatorname{span}\\{\theta_{1}\theta_{2}\theta_{3}\\}$. ∎

######  Remark 2.5.

For symbolic computation and to verify the previous proposition use

$\displaystyle
T_{i}f(\theta_{1},\dots,\theta_{N})=tf+(t\theta_{i}-\theta_{i+1})\bigg{(}\frac{\partial}{\partial\theta_{i+1}}-\frac{\partial}{\partial\theta_{i}}\bigg{)}f-\big{(}t\theta_{i}^{2}+\theta_{i+1}^{2}\big{)}\frac{\partial^{2}}{\partial\theta_{i}\partial\theta_{i+1}}f,$
$\displaystyle
T_{i}^{-1}f(\theta_{1},\dots,\theta_{N})=\frac{1}{t}f+\bigg{(}\theta_{i}-\frac{1}{t}\theta_{i+1}\bigg{)}\bigg{(}\frac{\partial}{\partial\theta_{i+1}}-\frac{\partial}{\partial\theta_{i}}\bigg{)}f-\bigg{(}\theta_{i}^{2}+\frac{1}{t}\theta_{i+1}^{2}\bigg{)}\frac{\partial^{2}}{\partial\theta_{i}\partial\theta_{i+1}}f,$

with the partial derivatives being formal (the order of variables is ignored).

There is a symmetric bilinear form on $\mathcal{P}$ which is positive-definite
for $t>0$ and in which $T_{i}$ is self-adjoint for $1\leq i<N$. The purpose of
the form is to make the simultaneous eigenvectors of $\\{\omega_{i}\\}$
mutually perpendicular.

######  Definition 2.6.

For $E,F\subset\\{1,2,\dots,N\\}$ define
$\left\langle\phi_{E},\phi_{F}\right\rangle=\delta_{E,F}t^{-\operatorname{inv}(E)}$
and extend the form to $\mathcal{P}$ by linearity.

######  Proposition 2.7.

Suppose $f,g\in\mathcal{P}$ and $1\leq i<N$ then $\left\langle
T_{i}f,g\right\rangle=\left\langle f,T_{i}g\right\rangle$.

###### Proof.

It suffices to consider $T_{1}$. Let $F\subset\left\\{3,4,\dots,N\right\\}$.
Then $T_{1}\phi_{F}=t\phi_{F}$ and $T_{1}\phi_{\left\\{1,2\right\\}\cup
F}=-\phi_{\left\\{1,2\right\\}\cup F}$. Let $F_{i}=F\cup\\{i\\}$ for $i=1,2$
and $T_{1}\phi_{F_{1}}=\phi_{F_{2}}$ and
$T_{1}\phi_{F_{2}}=(t-1)\phi_{F_{2}}+t\phi_{F_{1}}$ so that

$\displaystyle\left\langle
T_{1}\phi_{F_{1}},\phi_{F_{2}}\right\rangle=\left\langle\phi_{F_{2}},\phi_{F_{2}}\right\rangle=t^{-\operatorname{inv}\left(F_{2}\right)},$
$\displaystyle\left\langle\phi_{F_{1}},T_{1}\phi_{F_{2}}\right\rangle=\left\langle\phi_{F_{1}},(t-1)\phi_{F_{2}}+t\phi_{F_{1}}\right\rangle=t\left\langle\phi_{F_{1}},\phi_{F_{1}}\right\rangle=t^{1-\operatorname{inv}\left(F_{1}\right)},$

and $\operatorname{inv}(F_{1})=\operatorname{inv}(F_{2})+1$ by counting the
pair $(1,2)\in F_{1}\times F_{1}^{C}$. ∎

######  Corollary 2.8.

If $f,g\in\mathcal{P}$ and $1\leq i\leq N$ then
$\left\langle\omega_{i}f,g\right\rangle=\left\langle
f,\omega_{i}g\right\rangle$.

###### Proof.

This follows from $\omega_{N}=1$ and $\omega_{i}=t^{-1}T_{i}\omega_{i+1}T_{i}$
for $i<N$. ∎

There are two degree-changing linear maps which commute with the Hecke algebra
action.

######  Definition 2.9.

For $n\in\mathbb{Z}$ set $\sigma(n):=(-1)^{n}$ and for
$E\subset\\{1,2,\dots,N\\}$, $1\leq i\leq N$ set $s(i,E):=\\#\\{j\in E\colon
j<i\\}$. Define the operators $\partial_{i}$ and $\widehat{\theta}_{i}$ by
$\partial_{i}\theta_{i}\phi_{E}=\phi_{E}$, $\partial_{i}\phi_{E}=0$ and
$\widehat{\theta}_{i}\phi_{E}=\theta_{i}\phi_{E}=\sigma(s(i,E))\phi_{E\cup\\{i\\}}$
for $i\notin E$, while $\widehat{\theta}_{i}\phi_{E}=0$ for $i\in E$
$\big{(}$also $i\in E$ implies
$\phi_{E}=\sigma(s(i,E))\theta_{i}\phi_{E\backslash\\{i\\}}$ and
$\partial_{i}\phi_{E}=\sigma(s(i,E))\phi_{E\backslash\\{i\\}}\big{)}$. Define
$M:=\sum_{i=1}^{N}\widehat{\theta}_{i}$ and
$D:=\sum_{i=1}^{N}t^{i-1}\partial_{i}$.

By direct computation one can show that
$\widehat{\theta}_{i}\partial_{j}=-\partial_{j}\widehat{\theta}_{i}$ for
$i\neq j$.

######  Proposition 2.10.

$M$ and $D$ commute with $T_{i}$ for $1\leq i<N.$

###### Proof.

It follows from the definitions that $\partial_{i}$ and $\widehat{\theta}_{j}$
commute with $T_{i}$ when $j<i$ or $j>i+1$. It suffices to show
$\partial_{1}+t\partial_{2}$ and $\widehat{\theta}_{1}+\widehat{\theta}_{2}$
commute with $T_{1}$ applied to $p_{1}:=\phi_{F}$,
$p_{2}:=(\theta_{1}+\theta_{2})\phi_{F}$,
$~{}p_{3}:=(t\theta_{1}-\theta_{2})\phi_{F}$,
$p_{4}:=\theta_{1}\theta_{2}\phi_{F}$ with $1,2\notin F$. Then
$T_{1}p_{i}=tp_{i}$ for $i=1,2$, $T_{1}p_{i}=-p_{i}$ for $i=3,4$ and

$\displaystyle(\partial_{1}+t\partial_{2})[p_{1},p_{2},p_{3},p_{4}]=[0,(t+1)p_{1},0,-p_{3}],$
$\displaystyle(\widehat{\theta}_{1}+\widehat{\theta}_{2})[p_{1},p_{2},p_{3},p_{4}]=[p_{2},0,-(t+1)p_{4},0].$

This concludes the proof. ∎

It is clear that $D^{2}=0=M^{2}$. For $n=0,1,2,\dots$ let
$[n]_{t}:=\frac{1-t^{n}}{1-t}$ and $[n]_{t}!:=[1]_{t}[2]_{t}\cdots[n]_{t}$.

######  Proposition 2.11.

$MD+DM=[N]_{t}$.

###### Proof.

Fix $\phi_{E}$, $\\#E=m$;
$\partial_{j}\phi_{E}=\sigma(s(j,E))\phi_{E\backslash\\{j\\}}$, then
$\widehat{\theta}_{j}\partial_{j}\phi_{E}=\sigma(s(j,E))\theta_{j}\phi_{E\backslash\\{j\\}}=\phi_{E}$
thus the coefficient of $\phi_{E}$ in $MD$ is $\sum_{j\in E}t^{j-1}$. Also
$M\phi_{E}=\sum_{i\notin E}\theta_{i}\phi_{E}$ and the coefficient of
$\phi_{E}$ in $DM\phi_{E}$ is $\sum_{i\notin E}t^{i-1}$ so that the
coefficient of $\phi_{E}$ in $MD\\!+\\!DM$ is
$\sum_{j=1}^{N}\\!t^{j-1}=[N]_{t}$. Suppose $i\in E$, $j\notin E$ then
$\widehat{\theta}_{j}\phi_{E\backslash\\{i\\}}$ appears in $MD$ with
coefficient $t^{i-1}\sigma(s(i,E))$ while
$t^{i-1}\partial_{i}\theta_{j}\phi_{E}=\sigma(s(i,E))t^{i-1}\partial_{i}\theta_{j}\theta_{i}\phi_{E\backslash\\{i\\}}=-\sigma(s(i,E))t^{i-1}\widehat{\theta}_{j}\phi_{E\backslash\\{i\\}}$,
and this term is canceled out in $MD+DM$. ∎

###  2.3 Representations of $\boldsymbol{\mathcal{H}_{N}(t)}$

These representations correspond to partitions of $N$, namely
$\lambda=\left(\lambda_{1},\dots,\lambda_{N}\right)\in\mathbb{N}_{0}^{N}$ with
$\lambda_{1}\geq\lambda_{2}\geq\cdots\geq\lambda_{N}$ and
$\sum_{i=1}^{N}\lambda_{i}=N$. The length of $\lambda$ is
$\ell(\lambda)=\max\\{i\colon\lambda_{i}\geq 1\\}$. There is a graphical
device to picture $\lambda$, called the Ferrers diagram, which has boxes at
$\\{(i,j)\colon 1\leq i\leq\ell(\lambda)\,,1\leq j\leq\lambda_{i}\\}$ (integer
points). A reverse standard tableau (RSYT) is a filling of the Ferrers diagram
with the numbers $\\{1,2,\dots,N\\}$ such that the entries decrease in each
row and in each column. The relevant representation of $\mathcal{H}_{N}(t)$ is
defined on the span of the RSYT’s of shape $\lambda$ in such a way that
$\omega_{i}Y=t^{c(i,Y)}Y$ for $1\leq i\leq N$, where $Y[a,b]=i$, $c(i,Y)=b-a$
($b-a$ is called the content of $[a,b]$), and $Y$ is a RSYT of shape
$\lambda$. In the present work only hook tableaux will occur, namely
partitions of the form $\lambda=(N-n,1^{n})$ (the part $1$ is repeated $n$
times), so that $\ell(\lambda)=n+1$.

We will show that $\mathcal{P}_{m}$ is a direct sum of the
$\mathcal{H}_{N}(t)$-modules corresponding to $(N-m,1^{m})$ and
$(N+1-m,1^{m-1})$. Here is a structure for labeling the $\phi_{E}$ of
interest.

######  Definition 2.12.

Let $\mathcal{Y}_{0}:=\\{E\colon\\#E=m+1,\,N\in E\\}$ and
$\mathcal{Y}_{1}:=\\{E\colon\\#E=m-1,\,N\notin E\\}$.

These sets are associated to RSYT’s of shape $\big{(}N-m,1^{m}\big{)}$ and
$\big{(}N-m+1,1^{m-1}\big{)}$ respectively, and this correspondence will be
used to define content vectors for $E$.

######  Definition 2.13.

Suppose $E\in\mathcal{Y}_{0}$ and $E=\\{i_{1},\dots,i_{m},i_{m+1}\\}$,
$E^{C}=\\{j_{1},\dots,j_{N-m-1}\\}$ with $i_{1}<i_{2}<\cdots<i_{m+1}=N$ and
$j_{1}<j_{2}<\cdots$ then $Y_{E}$ is the RSYT of shape $(N-m,1^{m})$ given by
$Y_{E}[k,1]=i_{m+2-k}$ for $1\leq k\leq m+1$, and $Y_{E}[1,k]=j_{N-m+1-k}$ for
$2\leq k\leq N-m$. Suppose $E\in\mathcal{Y}_{1}$ and
$E=\\{i_{1},\dots,i_{m-1}\\}$, $E^{C}=\\{j_{1},\dots,j_{N-m+1}\\}$ with
$i_{1}<i_{2}<\cdots$ and $j_{1}<j_{2}<\cdots<j_{N-m+1}=N$ then $Y_{E}$ is the
RSYT of shape $(N-m+1,1^{m-1})$ given by $Y_{E}[k,1]=i_{m+1-k}$ for $2\leq
k\leq m$, $Y_{E}[1,k]=j_{N-m+2-k}$ for $1\leq k\leq N-m+1$. In both cases
define the content vector $c(i,E)=c\big{(}i,Y_{E}\big{)}$ for $1\leq i\leq N$.

For space-saving convenience the RSYT’s are displayed in two rows, with the
second row consisting of the entries $Y_{E}[2,1],\,Y_{E}[3,1],\dots$ . Recall
the content of cell $[i,j]$ is $j-i$.

As example let $N=8$, $m=3$, $E=\\{2,5,7,8\\}$ then

$\displaystyle Y_{E}=\begin{bmatrix}8&6&4&3&1\\\ \cdot&7&5&2&\end{bmatrix}$

and $[c(i,E)]_{i=1}^{8}=[4,-3,3,2,-2,1,-1,0]$.

We will construct for each $E\in\mathcal{Y}_{0}\cup\mathcal{Y}_{1}$ a
polynomial $\tau_{E}\in\mathcal{P}_{m}$ such that
$\omega_{i}\tau_{E}=t^{c(i,E)}\tau_{E}$ for $1\leq i\leq N$. To start let
$E_{0}:=\\{N-m,N-m+1,\dots,N\\}\in\mathcal{Y}_{0}$. Then

$\displaystyle Y_{E_{0}}=\begin{bmatrix}N&N-m-1&N-m-2&\cdots&\cdots&1\\\
\cdot&N-1&N-2&\cdots&N-m&\end{bmatrix}\\!,$
$\displaystyle\left[c(i,E_{0})\right]_{i=1}^{N}=[N-m-1,N-m-2,\dots,1,-m,1-m,\dots,-1,0].$

######  Theorem 2.14.

Let $\psi_{0}=D\phi_{E_{0}}\in\ker D\cap\mathcal{P}_{m}$, then
$\omega_{i}\psi_{0}=t^{c(i,E_{0})}\psi_{0}$ for $1\leq i\leq N$.

###### Proof.

If $N-m\leq i<N$ then $T_{i}\phi_{E_{0}}=-\phi_{E_{0}}$ and so
$T_{i}\psi_{0}=-\psi_{0}$, because $T_{i}D=DT_{i}$ and
$\psi_{0}=D\phi_{E_{0}}$, thus $\omega_{i}\psi_{0}=t^{i-N}\psi_{0}$. It is
clear that $T_{i}\psi_{0}=t\psi_{0}$ for $1\leq i<N-m-1$ and so it remains to
prove $\omega_{N-m-1}\psi_{0}=t\psi_{0}$. (The remaining part of the argument
is straightforward, and is at the end of this proof; for example
$\omega_{N-m-2}\psi_{0}=t^{-1}T_{N-m-2}\omega_{N-m-1}T_{N-m-2}\psi_{0}=T_{N-m-2}\omega_{N-m-1}\psi_{0}=t^{2}\psi_{0}$.)
Let $F:=\\{N-m-1,N-m,\dots,N\\}$ and $F_{j}:=F\backslash\\{j\\}$,
$p_{j}=\phi_{F_{j}}$, (so that $p_{N-m-1}=\phi_{E_{0}}$) then
$T_{i}p_{j}=-p_{j}$ if $i>j$ or $N-m-1\leq i<j-1$,
$T_{j}p_{j}=(1-t)p_{j}+tp_{j+1}$ and $T_{j}p_{j+1}=p_{j}$. To set up an
induction argument let $U_{N-m-1}=T_{N-m-1}$ and $U_{i+1}=T_{i+1}U_{i}$ for
$i<N-1$. We claim

$U_{i}\phi_{E_{0}}=t^{i-N+m+2}p_{i+1}+(t-1)\sum_{j=N-m-1}^{i}(-1)^{i-j}t^{j-N+m+1}p_{j}.$

At the start of the induction $T_{N-m-1}p_{N-m-1}=tp_{N-m}+(t-1)p_{N-m-1}$.
Suppose the formula holds for $i$ then

$\displaystyle
U_{i+1}\phi_{E_{0}}=t^{i-N+m+2}T_{i+1}p_{i+1}+(t-1)\sum_{j=N-m-1}^{i}(-1)^{i-j}t^{j-N+m+1}T_{i+1}p_{j}$
$\displaystyle\hphantom{U_{i+1}\phi_{E_{0}}}{}=t^{i-N+m+2}(tp_{i+2}+(t-1)p_{i+1})-(t-1)\sum_{j=N-m-1}^{i}(-1)^{i-j}t^{j-N+m+1}p_{j}$
$\displaystyle\hphantom{U_{i+1}\phi_{E_{0}}}{}=t^{i-N+m+3}p_{i+2}+(t-1)\sum_{j=N-m}^{i+1}(-1)^{j-i+1}t^{j-N+m+1}p_{j},$

as is to be shown. We also need

$D\phi_{F}=\sum_{j=N-m-1}^{N-1}(-1)^{j-N+m+1}t^{j-1}p_{j}+(-1)^{m+1}t^{N-1}p_{N}.$

Thus

$\displaystyle
U_{N-1}\phi_{E_{0}}=t^{m+1}p_{N}+(t-1)\sum_{j=N-m-1}^{N-1}(-1)^{N-1-j}t^{j-N+m+1}p_{j}$
$\displaystyle\hphantom{U_{N-1}\phi_{E_{0}}}{}=t^{m+1}p_{N}+(t-1)(-1)^{m}t^{-N+m+2}\big{\\{}D\phi_{F}-(-1)^{m+1}t^{N-1}p_{N}\big{\\}}$
$\displaystyle\hphantom{U_{N-1}\phi_{E_{0}}}{}=t^{m+2}p_{N}+(t-1)(-1)^{m}t^{-N+m+2}D\phi_{F}.$

Then $U_{N-1}\psi_{0}=DU_{N-1}\phi_{E_{0}}=t^{m+2}Dp_{N}$ and
$T_{N-m-1}T_{N-m}\cdots T_{N-1}U_{N-1}\psi_{0}=t^{m+2}Dp_{N-m-1}$
$=t^{m+2}\psi_{0}$ since $T_{j}p_{j+1}=p_{j}$ for $N-m-1\leq j<N$. Hence
$\omega_{N-m-1}\psi_{0}=t^{N-m-1-N}t^{m+2}\psi_{0}$ $=t\psi_{0}$. It follows
that

$\displaystyle\omega_{i}\psi_{0}$ $\displaystyle=t^{i-N+m+1}T_{i}\cdots
T_{N-m-2}\omega_{N-m-1}T_{N-m-2}\cdots T_{i}\psi_{0}$
$\displaystyle=t^{i-N+m+1}t^{1+2(N-m-1-i)}\psi_{0}=t^{N-m-i}\psi_{0},$

for $1\leq i\leq N-m-1$. ∎

Turning to the isotype $\left(N-m+1,1^{m-1}\right)$, let
$E_{1}:=\\{1,2,\dots,m-1\\}\in\mathcal{Y}_{1}$ so that

$\displaystyle Y_{E_{1}}=\begin{bmatrix}N&N-1&N-2&\cdots&\cdots&m\\\
\cdot&m-1&m-2&\cdots&1&\end{bmatrix}\\!,$
$\displaystyle[c(i,E_{1})]_{i=1}^{N}=[1-m,2-m,\dots,-1,N-m,N-m-1,\dots,1,0].$

######  Theorem 2.15.

Let $\eta_{E_{1}}=M\phi_{E_{1}}\in\ker M\cap\mathcal{P}_{m}$. Then
$\omega_{i}\eta_{E_{1}}=t^{c(i,E_{1})}\eta_{E_{1}}$ for $1\leq i\leq N$.

###### Proof.

Since $T_{i}\phi_{E_{1}}=t\phi_{E_{1}}$ for $m\leq i<N$ it follows that
$\omega_{i}M\phi_{E_{1}}=t^{N-i}M\phi_{E_{1}}$. Also
$T_{i}\phi_{E_{1}}=-\phi_{E_{1}}$ and $T_{i}M\phi_{E_{1}}=-M\phi_{E_{1}}$ for
$1\leq i<m-1$ and so it remains to show
$\omega_{m-1}\eta_{E_{1}}=t^{-1}\eta_{E_{1}}$. Let $F=\\{1,2,\dots,m-2\\}$ and
for $m-1\leq j\leq N$ let $F_{j}=F\cup\\{j\\}$ and $p_{j}=\phi_{F_{j}}$
($\phi_{E_{1}}=p_{m-1}$). Then $T_{j}p_{j}=p_{j+1}$ so that
$T_{N-1}T_{N-2}\cdots T_{m-1}p_{m-1}=p_{N}$. Also
$T_{j}p_{j+1}=tp_{j}+(t-1)p_{j+1}$ and $T_{j}p_{i}=tp_{i}$ for $i>j+1$. By
induction we prove that

$T_{i}T_{i+1}\cdots
T_{N-1}p_{N}=t^{N-i}p_{i}+(t-1)t^{N-i-1}\sum_{j=i+1}^{N}p_{j}.$

The formula is valid for $i=N-1$ and assuming it is true for $i$ apply
$T_{i-1}$ to both sides, then the first term becomes
$t^{N-i}\left(p_{i-1}+(t-1)p_{i}\right)$ and the second term is multiplied by
$t$. Substitute $M\phi_{F}=\sum_{j=m-1}^{N}p_{j}$ in the formula with $i=m-1$
to obtain

$\displaystyle T_{m-1}\cdots
T_{N-1}p_{N}=t^{N-m+1}p_{m-1}+(t-1)t^{N-m}\big{\\{}M\phi_{F}-p_{m-1}\big{\\}}$
$\displaystyle\hphantom{T_{m-1}\cdots
T_{N-1}p_{N}}{}=t^{N-m}p_{m-1}+(t-1)t^{N-m}M\phi_{F}.$

Thus $\omega_{m-1}\eta_{E_{1}}=t^{m-1-N}t^{N-m}Mp_{m-1}=t^{-1}\eta_{E_{1}}$
(since $M^{2}=0$). From $T_{i}\eta_{E_{1}}=-\eta_{E_{1}}$ for $1\leq i<m-1$ it
follows that $\omega_{i}\eta_{E_{1}}=t^{i-m}\eta_{E_{1}}$. Thus
$\omega_{i}\eta_{E_{1}}=t^{c(i,E_{1})}\eta_{E_{1}}$ for $1\leq i\leq N$. ∎

###  2.4 Steps

Having found two polynomials which are $\\{\omega_{i}\\}$ simultaneous
eigenfunctions we describe the method for constructing for each
$E\in\mathcal{Y}_{0}\cup\mathcal{Y}_{1}$ a polynomial
$\tau_{E}\in\mathcal{P}_{m}$ such that $\omega_{i}\tau_{E}=t^{c(i,E)}\tau_{E}$
for $1\leq i\leq N$. Recall the standard properties
$T_{i}\omega_{j}=\omega_{j}T_{i}$ for $i<j-1$ (obvious) and for $i>j$ (suppose
$i=j+1$) then

$\displaystyle T_{j+1}\omega_{j}$
$\displaystyle=t^{-2}T_{j+1}T_{j}T_{j+1}\omega_{j+2}T_{j+1}T_{j}=t^{-2}T_{j}T_{j+1}T_{j}\omega_{j+2}T_{j+1}T_{j}=t^{-2}T_{j}T_{j+1}\omega_{j+2}T_{j}T_{j+1}T_{j}$
$\displaystyle=t^{-2}T_{j}T_{j+1}\omega_{j+2}T_{j=1}T_{j}T_{j+1}=\omega_{j}T_{j+1},$

by the braid relations; and

$\displaystyle T_{j}\omega_{j}$
$\displaystyle=t^{-1}T_{j}^{2}\omega_{j+1}T_{j}=t^{-1}\\{(t-1)T_{j}+t\\}\omega_{j+1}T_{j}=(t-1)\omega_{j}+\omega_{j+1}T_{j},$
$\displaystyle\omega_{j}T_{j}$
$\displaystyle=T_{j}\omega_{j+1}+(t-1)\omega_{j}.$ (2.3)

######  Proposition 2.16.

Suppose $\omega_{j}f=\lambda_{j}f$ for $1\leq j\leq N$ $(f\neq 0)$,
$\lambda_{i}\neq\lambda_{i+1}$ and

$g:=T_{i}f+\dfrac{(t-1)\lambda_{i}}{\lambda_{i+1}-\lambda_{i}}f$

then $\omega_{j}g=\lambda_{j}g$ for all $j\neq i,i+1$ and
$\omega_{i}g=\lambda_{i+1}g$, $\omega_{i+1}g=\lambda_{i}g$. If
$\lambda_{i+1}\neq t^{\pm 1}\lambda_{i}$ then $g\neq 0$.

###### Proof.

If $j>i+1$ or $j<i$ then $\omega_{j}T_{i}=T_{i}\omega_{j}$ and thus
$\omega_{j}g=\lambda_{j}g$. By (2.3)

$\displaystyle\omega_{i}g$
$\displaystyle=\omega_{i}T_{i}f+\lambda_{i}\dfrac{(t-1)\lambda_{i}}{\lambda_{i+1}-\lambda_{i}}f=T_{i}\omega_{i+1}f+\bigg{\\{}t-1+\dfrac{(t-1)\lambda_{i}}{\lambda_{i+1}-\lambda_{i}}\bigg{\\}}\lambda_{i}f$
$\displaystyle=\lambda_{i+1}T_{i}f+\frac{(t-1)\lambda_{i+1}\lambda_{i}}{\lambda_{i+1}-\lambda_{i}}f=\lambda_{i+1}g.$

A similar calculation using
$\omega_{i+1}T_{i}=T_{i}\omega_{i}-(t-1)\omega_{i}$ shows that
$\omega_{i+1}g=\lambda_{i}g.$ Since $T_{i}^{2}=(t-1)T_{i}+t$

$\bigg{(}T_{i}+\dfrac{(t-1)\lambda_{i+1}}{\lambda_{i}-\lambda_{i+1}}\bigg{)}\bigg{(}T_{i}+\dfrac{(t-1)\lambda_{i}}{\lambda_{i+1}-\lambda_{i}}\bigg{)}=\frac{(\lambda_{i}t-\lambda_{i+1})(\lambda_{i}-t\lambda_{i+1})}{(\lambda_{i}-\lambda_{i+1})^{2}},$

thus $\lambda_{i+1}\neq t^{\pm 1}\lambda_{i}$ implies $g\neq 0$. ∎

Given the hypotheses of the proposition and the self-adjointness of
$\omega_{i}$ (Corollary 2.8) it follows that $\langle f,g\rangle=0$
($\lambda_{i}\langle f,g\rangle=\langle\omega_{i}f,g\rangle=\langle
f,\omega_{i}g\rangle=\lambda_{i+1}\langle f,g\rangle$).

######  Lemma 2.17.

Suppose $g=(T_{i}+b)f$ and $\langle f,g\rangle=0$ then
$\|g\|^{2}=(1-b)(t+b)\|f\|^{2}$.

###### Proof.

It follows from $T_{i}$ being self-adjoint that $\langle
T_{i}f,T_{i}f\rangle=\left\langle T_{i}^{2}f,f\right\rangle=(t-1)\langle
T_{i}f,f\rangle+t\|f\|^{2}$ and $\langle f,g\rangle=0$ implies $\langle
T_{i}f,f\rangle+b\|f\|^{2}=0$. Thus

$\displaystyle\|g\|^{2}=\|T_{i}f\|^{2}+2b\langle
T_{i}f,f\rangle+b^{2}\|f\|^{2}=(t-1+2b)\langle
T_{i}f,f\rangle+(t+b^{2})\|f\|^{2}$
$\displaystyle\hphantom{\|g\|^{2}}{}=\big{\\{}{-}b(t-1+2b)+t+b^{2}\big{\\}}\|f\|^{2}=\big{(}{-}b^{2}-b(t-1)+t\big{)}\|f\|^{2}=(1-b)(t+b)\|f\|^{2}.\\!\\!\\!\\!\\!\\!$
∎

######  Corollary 2.18.

Suppose $g=(T_{i}+b)f$, $\langle g,f\rangle=0$ and $b=1$ or $b=-t$ then $g=0$.

####  2.4.1 Isotype $\boldsymbol{(N-m,1^{m})}$

This concerns the polynomials in $\mathcal{P}_{m,0}=\ker
D\cap\mathcal{P}_{m}=D\mathcal{P}_{m+1}.$ Recall
$\mathcal{Y}_{0}=\\{E\colon\\#E=m+1$, $N\in E\\}$.

######  Definition 2.19.

For $\\#E=m+1$ define $\psi_{E}=D\phi_{E}$.

The set $\big{\\{}\psi_{E}\colon E\in\mathcal{Y}_{0}\big{\\}}$ spans
$\mathcal{P}_{m,0}$; for suppose $N\notin E$ then $\theta_{N}D\phi_{E}$ is a
linear combination of $\phi_{F}$ with $F\in\mathcal{Y}_{0}$ and
$t^{1-N}D\theta_{N}D\phi_{E}=D\phi_{E}=\psi_{E}$. The map
$p\big{(}\theta_{1},\dots,\theta_{N}\big{)}\rightarrow
p\big{(}\theta_{1},\dots,\theta_{N-1},0\big{)}$ takes $\psi_{E}$ to
$t^{N-1}\phi_{E\backslash\\{N\\}}$; thus
$\dim\mathcal{P}_{m,0}=\binom{N-1}{m}$. The function $\operatorname{inv}(E)$
provides a partial order on $\mathcal{Y}_{0}$.

######  Definition 2.20.

For $0\leq n\leq m(N-1-m)$ let

$\mathcal{P}_{m,0}^{(n)}:=\operatorname{span}\big{\\{}\psi_{E}\colon
E\in\mathcal{Y}_{0},\,\operatorname{inv}(E)\leq n\big{\\}}.$

The extreme cases are $\operatorname{inv}(\\{N-m,\dots,N\\})=0$ and
$\operatorname{inv}(\\{1,2,\dots,m,N\\})=m(N-1-m)$.

######  Lemma 2.21.

Suppose $E\in\mathcal{Y}_{0}$ and $\operatorname{inv}(E)=n$. If
$\\{i,i+1\\}\subset E$ then $T_{i}\psi_{E}=-\psi_{E}$, or if $\\{i,i+1\\}\cap
E=\varnothing$ then $T_{i}\psi_{E}=t\psi_{E}$. If $(i,i+1)\in E\times E^{C}$
then $\operatorname{inv}(s_{i}E)=n-1$ and $T_{i}\psi_{E}=\psi_{s_{i}E}$. If
$i(i,i+1)\in E^{C}\times E$ then $\operatorname{inv}(s_{i}E)=n+1$ and
$T_{i}\psi_{E}=(t-1)\psi_{E}+t\psi_{s_{i}E}\in\mathcal{P}_{m,0}^{(n+1)}$. That
is, $T_{i}\mathcal{P}_{m,0}^{(n)}\subset\mathcal{P}_{m,0}^{(n+1)}$.

###### Proof.

The transformation rules follow from Definition 2.3 and $DT_{i}=T_{i}D$ (see
Proposition 2.10). ∎

######  Theorem 2.22.

Suppose for some $n$ and for each $E\in\mathcal{Y}_{0}$ with
$\operatorname{inv}(E)=n$ there is a polynomial $\tau_{E}=t^{n}\psi_{E}+p_{E}$
with $p_{E}\in\mathcal{P}_{m,0}^{(n-1)}$ such that
$\omega_{i}\tau_{E}=t^{c(i,E)}\tau_{E}$ for all $i$ then this property holds
for $n+1$.

###### Proof.

Suppose $E\in\mathcal{Y}_{0}$ with $\operatorname{inv}(E)=n+1$ then for some
$i$ it holds that $(i,i+1)\in E\times E^{C}$ (otherwise
$E=E_{0}=\\{N-m,N-m+1,\dots,N\\}$ and $\operatorname{inv}(E_{0})=0$), then let
$F=s_{i}E$ so that $\operatorname{inv}(F)=n$. Then $c(j,E)=c(j,F)$ for $j\neq
i,i+1$ and $c(i,E)=c(i+1,F)\leq-1$, $c(i+1,E)=c(i,F)\geq 1$. By Proposition
2.16 let

$\tau_{E}=\bigg{(}T_{i}+\dfrac{(t-1)t^{c(i,F)}}{t^{c(i+1,F)}-t^{c(i,F)}}\bigg{)}\tau_{F}.$

Then $\omega_{j}\tau_{E}=t^{c(j,E)}\tau_{E}$ for all $j$; and
$\tau_{E}=(T_{i}+b)\big{(}t^{n}\psi_{F}+p_{F}\big{)}=t^{n+1}\psi_{E}+t^{n}(t-1+b)\psi_{F}+(T_{i}+b)p_{E}$
(for a constant $b$). By the lemma
$t^{n}(t-1+b)\psi_{F}+(T_{i}+b)p_{E}\in\mathcal{P}_{m,0}^{(n)}$. ∎

######  Corollary 2.23.

For each $E\in\mathcal{Y}_{0}$ and $\operatorname{inv}(E)=n$ there is a unique
$\tau_{E}\in\mathcal{P}_{m,0}$ with $\tau_{E}=t^{n}\psi_{E}+p_{E}$ and
$p_{E}\in\mathcal{P}_{m,0}^{(n-1)}$ such that
$\omega_{i}\tau_{E}=t^{c(i,E)}\tau_{E}$ for all $i$.

###### Proof.

The existence follows from induction starting with $\tau_{E_{0}}=\psi_{E_{0}}$
and Theorem 2.14. Uniqueness follows from the leading term. The
$\\{\omega_{i}\\}$-eigenvalues of $\tau_{E}$ determine $E$ uniquely. ∎

######  Corollary 2.24.

Suppose $E\in\mathcal{Y}_{0}$, if $\\{i,i+1\\}\subset E$ then
$T_{i}\tau_{E}=-\tau_{E}$ and if $\\{i,i+1\\}\cap E=\varnothing$ or
$i=N-1\notin E$ then $T_{i}\tau_{E}=t\tau_{E}$.

###### Proof.

Let
$b=\frac{(t-1)t^{c(i,E)}}{t^{c(i+1,E)}-t^{c(i,E)}}=\frac{t-1}{t^{c(i+1,E)-c(i,E)}-1}$.
If $\\{i,i+1\\}\subset E$ then $c(i+1,E)=1+c(i,E)$ and $b=1$; thus
$\left\langle(T_{i}+1)\tau_{E},\tau_{E}\right\rangle=0$ (by the comment after
Proposition 2.16) and $\|(T_{i}+1)\tau_{E}\|^{2}$ $=0$ by Corollary 2.18. If
$\\{i,i+1\\}\cap E=\varnothing$ (or $i=N-1\notin E$) then $c(i+1,E)=c(i,E)-1$
and $b=\frac{t-1}{t^{-1}-1}=-t$; thus
$\left\|(T_{i}-t)\tau_{E}\right\|^{2}=0$. ∎

######  Definition 2.25.

Let $u(z):=\frac{(t-z)(1-tz)}{(1-z)^{2}}$. Suppose $v\in\mathbb{Z}^{N}$ and
$v_{j}\neq 0$ for $j<N$, $v_{N}=0$, then

$\displaystyle\mathcal{C}(v):=\prod\limits_{1\leq
i<j<N}\big{\\{}u\big{(}t^{v_{i}-v_{j}}\big{)}\colon v_{i}<0<v_{j}\big{\\}}.$
(2.4)

######  Proposition 2.26.

Suppose $E\in\mathcal{Y}_{0}$ then

$\|\tau_{E}\|^{2}=t^{2(N-m-1)}[m+1]_{t}\mathcal{C}\big{(}[c(i,E)]_{i=1}^{N}\big{)}.$

###### Proof.

By definition
$\tau_{E_{0}}=\sum_{j=N-m}^{N}t^{j-1}(-1)^{N-m-j}\phi_{E_{0}\backslash\\{j\\}}$
and $\left\|\tau_{E_{0}}\right\|^{2}=\sum_{j=N-m}^{N}t^{2j-2}t^{-i_{j}}$,
where $i_{j}=\operatorname{inv}\big{(}E_{0}\backslash\\{j\\}\big{)}=j-N+m$;
thus $\|\tau_{E_{0}}\|^{2}=t^{2(N-m-1)}\sum_{j=0}^{m}t^{j}$. Suppose the
formula is valid for all $E$ with $\operatorname{inv}(E)\leq n$ and
$\operatorname{inv}(E)=n+1$. Then $E=s_{i}F$ for some $i\in E$ with $i+1\notin
E$ and $\operatorname{inv}(F)=n$ (so $i+1\in F$, $i\notin F$). By Lemma 2.17
$\|\tau_{E}\|^{2}=(1-b)(t+b)\|\tau_{F}\|^{2}$, where
$b=\frac{t-1}{t^{c(i+1,F)-c(i,F)}-1}$. Write
$z=t^{c(i+1,F)-c(i,F)}=t^{c(i,E)-c(i+1,E)}$ then

$\|\tau_{E}\|^{2}=\frac{(t-z)(1-tz)}{(1-z)^{2}}\|\tau_{F}\|^{2}=u(z)\|\tau_{F}\|^{2}.$

In the product for $\|\tau_{F}\|^{2}$ the factors for pairs $(i)$ $(k,\ell)$
with $\\{k,\ell\\}\cap\\{i,i+1\\}=\varnothing$, $(ii)$ $(k,i)$, $k\in F$,
$k<i$, $(iii)$ $(i+1,\ell)$, $\ell\notin F,\ell>i+1$ are the same in the
product for $\|\tau_{E}\|^{2}$ for the pairs $(i)$ $(k,\ell)$, $(ii)$
$(k,i+1)$, $(iii)$ $(i,\ell)$, respectively. The extra factor in the product
for $\|\tau_{E}\|^{2}$ has the desired value. ∎

####  2.4.2 Isotype $\boldsymbol{(N-m+1,1^{m-1})}$

This concerns the polynomials in $\mathcal{P}_{m,1}=\ker
M\cap\mathcal{P}_{m}=M\mathcal{P}_{m-1}$.

######  Definition 2.27.

For $\\#E=m-1$ define $\eta_{E}:=M\phi_{E}$. The set
$\mathcal{Y}_{1}=\big{\\{}E\colon\\#E=m-1,N\notin E\big{\\}}$.

The set $\big{\\{}\eta_{E}\colon E\in\mathcal{Y}_{1}\big{\\}}$ spans
$\mathcal{P}_{m,1}$; for suppose $N\in E$ then $\partial_{N}M\phi_{E}$ is a
linear combination of $\phi_{F}$ with $F\in\mathcal{Y}_{1}$ and
$M\big{(}\partial_{N}M\phi_{E}\big{)}=M\phi_{E}=\eta_{E}$. Furthermore
$\partial_{N}\eta_{E}=\phi_{E}$ and thus
$\dim\mathcal{P}_{m,1}=\binom{N-1}{m-1}$. The function $\operatorname{inv}(E)$
provides a partial order on $\mathcal{Y}_{1}$. The extreme cases are
$\operatorname{inv}(\\{N-m+1,\dots,N-1\\})=m-1$ and
$\operatorname{inv}(\\{1,2,\dots,m-1\\})=(m-1)(N-m+1)$.

######  Definition 2.28.

For $0\leq n\leq(m-1)(N-m+1)$ let

$\mathcal{P}_{m,1}^{(n)}:=\operatorname{span}\big{\\{}\eta_{E}\colon
E\in\mathcal{Y}_{1},\,\operatorname{inv}(E)\geq n\big{\\}}.$

######  Lemma 2.29.

Suppose $E\in\mathcal{Y}_{1}$ and $\operatorname{inv}(E)=n$. If
$\\{i,i+1\\}\subset E$ then $T_{i}\eta_{E}=-\eta_{E}$, or if $\\{i,i+1\\}\cap
E=\varnothing$ then $T_{i}\eta_{E}=t\eta_{E}$. If $i\notin E$, $i+1\in E$ then
$\operatorname{inv}(s_{i}E)=n+1$ and
$T_{i}\eta_{E}=(t-1)\eta_{E}+t\eta_{s_{i}E}\in\mathcal{P}_{m,1}^{(n)}$. If
$i\in E$, $i+1\notin E$ then $\operatorname{inv}(s_{i}E)=n-1$ and
$T_{i}\eta_{E}=\eta_{s_{i}E}\in\mathcal{P}_{m,1}^{(n-1)}$.

###### Proof.

The transformation rules follow from Definition 2.3 and $MT_{i}=T_{i}M$. ∎

######  Theorem 2.30.

Suppose for some $n$ and for each $E\in\mathcal{Y}_{1}$ with
$\operatorname{inv}(E)=n$ there is a polynomial $\tau_{E}=\eta_{E}+p_{E}$ with
$p_{E}\in\mathcal{P}_{m,1}^{(n+1)}$ such that
$\omega_{i}\tau_{E}=t^{c(i,E)}\tau_{E}$ for all $i$ then this property holds
for $n-1$.

###### Proof.

Suppose $E\in\mathcal{Y}_{1}$ with $\operatorname{inv}(E)=n-1$ then for some
$i$ it holds that $i\notin E,i+1\in E$ (otherwise $E=E_{1}=\\{1,\dots,m-1\\}$
and $\operatorname{inv}(E_{1})=(m-1)(N-m+1)$), then let $F=s_{i}E$ so that
$\operatorname{inv}(F)=n$. Then $c(j,E)=c(j,F)$ for $j\neq i,i+1$ and
$c(i,E)=c(i+1,F)\geq 1$, $c(i+1,E)=c(i,F)\leq-1$. By Proposition 2.16 let

$\tau_{E}=\bigg{(}T_{i}+\dfrac{(t-1)t^{c(i,F)}}{t^{c(i+1,F)}-t^{c(i,F)}}\bigg{)}\tau_{F}.$

Then $\omega_{j}\tau_{E}=t^{c(j,E)}\tau_{E}$ for all $j$; and
$\tau_{E}=(T_{i}+b)\big{(}\eta_{F}+p_{F}\big{)}=\eta_{E}+b\eta_{F}+(T_{i}+b)p_{F}$
(for a constant $b$). By the previous lemma
$b\eta_{F}+(T_{i}+b)p_{E}\in\mathcal{P}_{m,1}^{(n)}$. ∎

######  Corollary 2.31.

For each $E\in\mathcal{Y}_{1}$ and $\operatorname{inv}(E)=n\leq(m-1)(N-m+1)$
there is a unique $\tau_{E}\in\mathcal{P}_{m,1}$ with
$\tau_{E}=\eta_{E}+p_{E}$ and $p_{E}\in\mathcal{P}_{m,1}^{(n+1)}$ such that
$\omega_{i}\tau_{E}=t^{c(i,E)}\tau_{E}$ for all $i$.

###### Proof.

The existence follows from induction starting with
$\tau_{E_{1}}=M\phi_{E_{1}}=\eta_{E_{1}}$ and Theorem 2.15. Uniqueness follows
from the leading term. The eigenvalues of $\tau_{E}$ determine $E$ uniquely. ∎

######  Corollary 2.32.

Suppose $E\in\mathcal{Y}_{1}$, if $\\{i,i+1\\}\in E$ then
$T_{i}\tau_{E}=-\tau_{E}$ and if $\\{i,i+1\\}\cap E=\varnothing$ then
$T_{i}\tau_{E}=t\tau_{E}$.

###### Proof.

This has the same proof as Corollary 2.24. ∎

######  Proposition 2.33.

Suppose $E\in\mathcal{Y}_{1}$ then

$\|\tau_{E}\|^{2}=t^{-m(N-m)}[N-m+1]_{t}~{}\mathcal{C}\big{(}[-c(i,E)]_{i=1}^{N}\big{)}.$

###### Proof.

By definition $\tau_{E_{1}}=(-1)^{m-1}\sum_{j=m}^{N}\phi_{E_{1}\cup\\{j\\}}$
and $\|\tau_{E_{1}}\|^{2}=\sum_{j=m}^{N}t^{-i_{j}}$, where
$i_{j}=\operatorname{inv}\big{(}E_{1}\cup\\{j\\}\big{)}=m(N+1-m)-j$; thus
$\|\tau_{E_{1}}\|^{2}=t^{-m(N-m)}\sum_{j=0}^{N-m}t^{j}$. Suppose the formula
is valid for all $E$ with $\operatorname{inv}(E)\geq n$ and
$\operatorname{inv}(E)=n-1$. Then $E=s_{i}F$ for some $i\notin E$ with $i+1\in
E$ and $\operatorname{inv}(F)=n$ (so $(i,i+1)\in F\times F^{C}$). By Lemma
2.17 $\|\tau_{E}\|^{2}=(1-b)(t+b)\|\tau_{F}\|^{2}$, where
$b=\frac{t-1}{t^{c(i+1,F)-c(i,F)}-1}$. Write
$z=t^{c(i+1,F)-c(i,F)}=t^{c(i,E)-c(i+1,E)}$ then

$\|\tau_{E}\|^{2}=\frac{(t-z)(1-tz)}{(1-z)^{2}}\|\tau_{F}\|^{2}=u(z)\|\tau_{F}\|^{2}.$

In the product for $|\tau_{F}|^{2}$ the factors for pairs $(i)$ $(k,\ell)$
with $\\{k,\ell\\}\cap\\{i,i+1\\}=\varnothing$, $(ii)$ $(k,i)$, $k\notin F$,
$k<i$, $(iii)$ $(i+1,\ell)$, $\ell\in F$, $\ell>i+1$ are the same in the
product for $|\tau_{E}|^{2}$ for the pairs $(i)$ $(k,\ell)$, $(ii)$ $(k,i+1)$,
$(iii)$ $(i,\ell)$, respectively. The extra factor in the product for
$|\tau_{E}|^{2}$ has the desired value. ∎

######  Proposition 2.34.

Let $F_{0}=\\{1,\dots,m,N\\}$, $F_{1}=\\{1,\dots,m\\}$ then
$\tau_{F_{0}}\in\mathcal{P}_{m,0}$ and $\tau_{F_{1}}\in\mathcal{P}_{m+1,1}$
have the same $\\{\omega_{i}\\}$-eigenvalues and

$\displaystyle\|\tau_{F_{0}}\|^{2}=t^{(m+2)(N-m-1)}\frac{[N]_{t}}{[N-m]_{t}},\qquad\|\tau_{F_{1}}\|^{2}=t^{-(m+1)(N-m-1)}[N-m]_{t},$
$\displaystyle\|\tau_{F_{0}}\|^{2}=t^{(2m+3)(N-m-1)}[N]_{t}[N-m]_{t}^{-2}\|\tau_{F_{1}}\|^{2},$
$\displaystyle
D\tau_{F_{1}}=t^{-(m+1)(N-m-1)}[N-m]_{t}\tau_{F_{0}},\qquad\|D\tau_{F_{1}}\|^{2}=t^{N-m-1}[N]_{t}\|\tau_{F_{1}}\|^{2}.$

###### Proof.

The content vector for $F_{0}$ is $[-m,1-m,\dots,-1,N-m-1,\dots,1,0]$ (the
same as $F_{1}$) so that

$\displaystyle\mathcal{C}\big{(}[c(i,F_{0})]_{i=1}^{N}\big{)}=\prod\limits_{i=1}^{m}\prod\limits_{j=1}^{N-m-1}\frac{(t-t^{-i-j})(1-t^{1-i-j})}{(1-t^{-i-j})^{2}}$
$\displaystyle\phantom{\mathcal{C}\big{(}[c(i,F_{0})]_{i=1}^{N}\big{)}}{}=t^{m(N-m-1)}\prod\limits_{i=1}^{m}\prod\limits_{j=1}^{N-m-1}\bigg{(}\frac{t^{i+j+1}-1}{t^{i+j}-1}\bigg{)}\bigg{(}\frac{t^{i+j-1}-1}{t^{i+j}-1}\bigg{)}$
$\displaystyle\phantom{\mathcal{C}\big{(}[c(i,F_{0})]_{i=1}^{N}\big{)}}{}=t^{m(N-m-1)}\prod\limits_{j=1}^{N-m-1}\bigg{(}\frac{t^{m+1+j}-1}{t^{1+j}-1}\bigg{)}\bigg{(}\frac{t^{j}-1}{t^{m+j}-1}\bigg{)}$
$\displaystyle\phantom{\mathcal{C}\big{(}[c(i,F_{0})]_{i=1}^{N}\big{)}}{}=t^{m(N-m-1)}\bigg{(}\frac{t-1}{t^{N-m}-1}\bigg{)}\bigg{(}\frac{t^{N}-1}{t^{m+1}-1}\bigg{)}=\frac{t^{m(N-m-1)}[N]_{t}}{[N-m]_{t}[m+1]_{t}}$

by use of telescoping arguments. Thus
$\|\tau_{F_{0}}\|^{2}=t^{(m+2)(N-m-1)}\frac{[N]_{t}}{[N-m]_{t}}$ by
Proposition 2.26. The value of $\|\tau_{F_{1}}\|^{2}$ is from Proposition
2.33. By definition
$\tau_{F_{1}}=M\phi_{F_{1}}=(\theta_{m+1}+\cdots+\theta_{N})\times\theta_{1}\theta_{2}\cdots\theta_{m}$
and the coefficient of $\phi_{F_{1}}$ in $D\tau_{F_{1}}$ is
$\sum_{i=m+1}^{N}t^{i-1}=t^{m}[N-m]_{t}$. The coefficient of $\phi_{F_{1}}$ in
$\tau_{F_{0}}=D\phi_{F_{1}}$ is $(-1)^{m}t^{m(N-1-m)+N-1}$(see Theorem 2.22).
From $D\tau_{F_{1}}$ and $\tau_{F_{0}}$ having the same
$\\{\omega_{i}\\}$-eigenvalues it follows that $D\tau_{F_{1}}=a\tau_{F_{0}}$
for some constant. ∎

###  2.5 Isomorphisms

This section concerns the action of the maps $M$, $D$ on the irreducible
$\mathcal{H}_{N}(t)$-modules. The following is a version of Schur’s lemma for
irreducible representations.

######  Lemma 2.35.

Suppose $\mu$ is a linear isomorphism $V_{1}\rightarrow V_{2}$ of irreducible
$\mathcal{H}_{N}(t)$-modules such that $T_{i}\mu=\mu T_{i}$ for $1\leq i<N$
and $V_{1}$, $V_{2}$ are equipped with inner products in which each $T_{i}$ is
self-adjoint, then $\|\mu f\|^{2}/\|f\|^{2}$ is constant for $f\in V_{1}$.

###### Proof.

The argument is based on orthogonal bases defined in the previous sections. By
hypothesis $V_{1}$ has an orthogonal basis consisting of
$\\{\omega_{i}\\}$-eigenfunctions. The image of this basis under $\mu$ has the
same property. For a typical basis element $f\in V_{1}$ suppose
$\omega_{j}f=\lambda_{j}f$ for all $j$ and $\lambda_{i+1}\neq t^{\pm
1}\lambda_{i}$ then $g=(T_{i}+b)f$ satisfies
$\omega_{i}g=\lambda_{i+1}g,\omega_{i+1}g=\lambda_{i}g$ for
$b=\frac{(t-1)\lambda_{i}}{\lambda_{i+1}-\lambda_{i}}$ and
$\|g\|^{2}=(1-b)(t+b)\|f\|^{2}$ (this equation follows from $T_{i}$ being
self-adjoint and $\langle f,g\rangle=0$). By hypothesis $\omega_{j}\mu
f=\lambda_{j}\mu f$ for all $j$ and $\mu g=(T_{i}+b)\mu f$ satisfies
$\omega_{i}\mu g=\lambda_{i+1}\mu g$, $\omega_{i+1}\mu g=\lambda_{i}\mu g$. By
Lemma 2.17 $\|\mu g\|^{2}=(1-b)(t+b)\|\mu f\|^{2}$ and so $\gamma:=\left\|\mu
g\right\|^{2}/\|g\|^{2}=\|\mu f\|^{2}/\|f\|^{2}$. By the step constructions
$\|\mu f\|^{2}/\|f\|^{2}=\gamma$ holds for every basis vector of $V_{1}$. ∎

The relation $MD+DM=[N]_{t}$ (Proposition 2.11) implies that $\mathcal{P}_{m}$
is a direct sum of $\mathcal{P}_{m,0}=\mathcal{P}_{m}\cap\ker D$ and
$\mathcal{P}_{m,1}=\mathcal{P}_{m}\cap\ker M$.

######  Theorem 2.36.

The maps $M$, $D$ are linear isomorphisms
$\mathcal{P}_{m,0}\rightarrow\mathcal{P}_{m+1,1}$,
$\mathcal{P}_{m+1,1}\rightarrow\mathcal{P}_{m,0}$ respectively, of
$\mathcal{H}_{N}(t)$-modules, $\|Mf\|^{2}=t^{m+1-N}[N]_{t}\|f\|^{2}$ for
$f\in\mathcal{P}_{m,0}$ and $\|Dg\|^{2}=t^{N-m-1}\times[N]_{t}\|g\|^{2}$ for
$g\in\mathcal{P}_{m+1,1}$.

###### Proof.

$M$ and $D$ commute with each $T_{i}$ and hence with each $\omega_{i}$.
Furthermore if $f\in\mathcal{P}_{m,0}$ then $(MD+DM)f=DMf=[N]_{t}f$ (by
Proposition 2.11) and thus $M$ is one-to-one on $\mathcal{P}_{m,0}$. Similarly
if $g\in\mathcal{P}_{m+1,1}$ then $[N]_{t}g=(MD+DM)g=MDg$ and $D$ is one-to-
one. By the lemma there are constants $\gamma_{1},\gamma_{2}$ such that
$\|Mf\|^{2}=\gamma_{1}\|f\|^{2}$ and $\|Dg\|^{2}=\gamma_{2}\|g\|^{2}$. From
$(MD+DM)f=DMf$ it follows that
$\|DMf\|^{2}=[N]_{t}^{2}\|f\|^{2}=\gamma_{2}\|Mf\|^{2}$ and
$\gamma_{1}=[N]_{t}^{2}/\gamma_{2}$. By Proposition 2.34
$\gamma_{2}=t^{N-m-1}[N]_{t}$ and thus $\gamma_{1}=t^{m+1-N}[N]_{t}$. ∎

##  3 Nonsymmetric Macdonald polynomials

###  3.1 Operators on polynomials

The following presents the key concepts for our constructions: the definition
of the action of $\mathcal{H}_{N}(t)$ on superpolynomials and the ingredients
necessary to define the Cherednik operators whose simultaneous eigenvectors
are the nonsymmetric Macdonald superpolynomials. Here we extend the
polynomials in $\\{\theta_{i}\\}$ by adjoining $N$ commuting variables
$x_{1},\dots,x_{N}$ (that is $[x_{i},x_{j}]=0$, $[x_{i},\theta_{j}]=0$,
$\theta_{i}\theta_{j}=-\theta_{j}\theta_{i}$ for all $i$, $j$). Each
polynomial is a sum of monomials $x^{\alpha}\phi_{E}$, where
$E\subset\\{1,2,\dots,N\\}$ and $\alpha\in\mathbb{N}_{0}^{N}$,
$x^{\alpha}:=\prod_{i=1}^{N}x_{i}^{\alpha_{i}}$. The partitions in
$\mathbb{N}_{0}^{N}$ are denoted by $\mathbb{N}_{0}^{N,+}$
($\lambda\in\mathbb{N}_{0}^{N,+}$ if and only if
$\lambda_{1}\geq\lambda_{2}\geq\cdots\geq\lambda_{N}$). The fermionic degree
of this monomial is $\\#E$ and the bosonic degree is
$|\alpha|:=\sum_{i=1}^{N}\alpha_{i}$. Let
$s\mathcal{P}_{m}:=\operatorname{span}\big{\\{}x^{\alpha}\phi_{E}\colon\alpha\in\mathbb{N}_{0}^{N},\,\\#E=m\big{\\}}$.
Then using the decomposition
$\mathcal{P}_{m}=\mathcal{P}_{m,0}\oplus\mathcal{P}_{m,1}$ let

$\displaystyle
s\mathcal{P}_{m,0}=\operatorname{span}\big{\\{}x^{\alpha}\psi_{E}\colon\alpha\in\mathbb{N}_{0}^{N},\,E\in\mathcal{Y}_{0}\big{\\}},$
$\displaystyle
s\mathcal{P}_{m,1}=\operatorname{span}\big{\\{}x^{\alpha}\eta_{E}\colon\alpha\in\mathbb{N}_{0}^{N},\,E\in\mathcal{Y}_{1}\big{\\}}.$

The Hecke algebra $\mathcal{H}_{N}(t)$ is represented on $s\mathcal{P}_{m}$.
This allows us to apply the theory of nonsymmetric Macdonald polynomials
taking values in $\mathcal{H}_{N}(t)$-modules (see [8, 9]).

######  Definition 3.1.

Suppose $p\in s\mathcal{P}_{m}$ and $1\leq i<N$ then set

$\boldsymbol{T}_{i}p(x;\theta):=(1-t)x_{i+1}\frac{p(x;\theta)-p(xs_{i};\theta)}{x_{i}-x_{i+1}}+T_{i}p(xs_{i};\theta).$

Note that $T_{i}$ acts on the $\theta$ variables according to Definition 2.3.

######  Theorem 3.2 ([9, Proposition 3.5]).

Suppose $1\leq i<N-1$ then
$\boldsymbol{T}_{i}\boldsymbol{T}_{i+1}\boldsymbol{T}_{i}=\boldsymbol{T}_{i+1}\boldsymbol{T}_{i}\boldsymbol{T}_{i+1}$,
if $1\leq i<N$ then $(\boldsymbol{T}_{i}+1)(\boldsymbol{T}_{i}-t)=0$ and if
$1\leq i<j-1\leq N-2$ then
$\boldsymbol{T}_{i}\boldsymbol{T}_{j}=\boldsymbol{T}_{j}\boldsymbol{T}_{i}$.

We also use

$\boldsymbol{T}_{i}^{-1}p(x;\theta)=\bigg{(}\frac{1-t}{t}\bigg{)}x_{i}\frac{p(x;\theta)-p(xs_{i};\theta)}{x_{i}-x_{i+1}}+T_{i}^{-1}p(xs_{i};\theta).$

######  Definition 3.3.

Let $T^{(N)}=T_{N-1}T_{N-2}\cdots T_{1}$ and for $p\in s\mathcal{P}_{m}$ and
$1\leq i\leq N$

$\displaystyle\boldsymbol{w}p(x;\theta):=T^{(N)}p(qx_{N},x_{1},x_{2},\dots,x_{N-1};\theta),$
$\displaystyle\xi_{i}p(x;\theta):=t^{i-N}\boldsymbol{T}_{i}\boldsymbol{T}_{i+1}\cdots\boldsymbol{T}_{N-1}\boldsymbol{wT}_{1}^{-1}\boldsymbol{T}_{2}^{-1}\cdots\boldsymbol{T}_{i-1}^{-1}p(x;\theta).$

The operators $\xi_{i}$ are Cherednik operators, defined by Baker and
Forrester [1] (see Braverman et al. [4] for the significance of these
operators in double affine Hecke algebras). They mutually commute (the proof
in the vector-valued situation is in [9, Theorem 3.8]). Observe
$\xi_{i-1}=t^{-1}\boldsymbol{T}_{i-1}\xi_{i}\boldsymbol{T}_{i-1}$. Their key
properties are

$\displaystyle T^{(N)}T_{i+1}=T_{N-1}\cdots T_{i+1}T_{i}T_{i+1}T_{i-1}\cdots
T_{1}=T_{N-1}\cdots T_{i+2}T_{i}T_{i+1}T_{i}\cdots T_{1}=T_{i}T^{(N)},$
$\displaystyle\big{(}T^{(N)}\big{)}^{2}=T^{(N)}\big{(}T_{N-1}\cdots
T_{2}\big{)}T_{1}=\big{(}T_{N-2}\cdots
T_{1}\big{)}T^{(N)}T_{1}=T_{N-1}^{-1}\big{(}T^{(N)}\big{)}^{2}T_{1},$
$\displaystyle\big{(}T^{(N)}\big{)}^{2}T_{1}=T_{N-1}\big{(}T^{(N)}\big{)}^{2},$
$\displaystyle\boldsymbol{wT}_{i+1}=\boldsymbol{T}_{i}\boldsymbol{w},\qquad\boldsymbol{w}^{2}\boldsymbol{T}_{1}=\boldsymbol{T}_{N-1}\boldsymbol{w}^{2},$
$\displaystyle\boldsymbol{w}^{-1}p(x;\theta)=T_{1}^{-1}T_{2}^{-1}\cdots
T_{N-1}^{-1}p(x_{2},x_{3},\dots,x_{N},q^{-1}x_{1};\theta).$ (3.1)

There is a basis of $s\mathcal{P}_{m}$ consisting of simultaneous eigenvectors
of $\\{\xi_{i}\\}$ and these are the nonsymmetric Macdonald superpolynomials
(henceforth abbreviated to “NSMP”).

Suppose $p(\theta)$ is independent of $x$ then $\boldsymbol{T}_{i}p=T_{i}p$
and

$\displaystyle\xi_{i}p(\theta)$ $\displaystyle=t^{i-N}T_{i}T_{i+1}\cdots
T_{N-1}\big{(}T_{N-1}\cdots T_{2}T_{1}\big{)}T_{1}^{-1}T_{2}^{-1}\cdots
T_{i-1}^{-1}p(\theta)$ $\displaystyle=t^{i-N}T_{i}T_{i+1}\cdots
T_{N-1}T_{N-1}\cdots T_{i}p(\theta)=\omega_{i}p(\theta),$

that is $\xi_{i}$ agrees with $\omega_{i}$ on polynomials of bosonic degree
0\. Also $\boldsymbol{wT}_{i+1}=\boldsymbol{T}_{i}\boldsymbol{w}$. Suppose
$j>i+1$ then

$\displaystyle\xi_{i}\boldsymbol{T}_{j}$
$\displaystyle=t^{i-N}\boldsymbol{T}_{i}\boldsymbol{T}_{i+1}\cdots\boldsymbol{T}_{N-1}w\boldsymbol{T}_{j}\boldsymbol{T}_{1}^{-1}\boldsymbol{T}_{2}^{-1}\cdots\boldsymbol{T}_{i-1}^{-1}$
$\displaystyle=t^{i-N}\boldsymbol{T}_{i}\boldsymbol{T}_{i+1}\cdots\boldsymbol{T}_{N-1}\boldsymbol{T}_{j-1}w\boldsymbol{T}_{1}^{-1}\boldsymbol{T}_{2}^{-1}\cdots\boldsymbol{T}_{i-1}^{-1}$
$\displaystyle=t^{i-N}\boldsymbol{T}_{i}\cdots(\boldsymbol{T}_{j-1}\boldsymbol{T}_{j}\boldsymbol{T}_{j-1})\cdots\boldsymbol{T}_{N-1}w\boldsymbol{T}_{1}^{-1}\boldsymbol{T}_{2}^{-1}\cdots\boldsymbol{T}_{i-1}^{-1}$
$\displaystyle=t^{i-N}\boldsymbol{T}_{i}\cdots(\boldsymbol{T}_{j}\boldsymbol{T}_{j-1}\boldsymbol{T}_{j})\cdots\boldsymbol{T}_{N-1}w\boldsymbol{T}_{1}^{-1}\boldsymbol{T}_{2}^{-1}\cdots\boldsymbol{T}_{i-1}^{-1}=\boldsymbol{T}_{j}\xi_{i}.$

A similar argument shows $\xi_{i}\boldsymbol{T}_{j}=\boldsymbol{T}_{j}\xi_{i}$
when $j<i-1$, by using
$\boldsymbol{T}_{j}^{-1}\boldsymbol{T}_{j+1}^{-1}\boldsymbol{T}_{j}=\boldsymbol{T}_{j+1}\boldsymbol{T}_{j}^{-1}\boldsymbol{T}_{j+1}^{-1}$.

###  3.2 Properties of nonsymmetric Macdonald polynomials

They have a triangularity property with respect to the partial order $\rhd$ on
the compositions $\mathbb{N}_{0}^{N}$, which is derived from the dominance
order:

$\displaystyle\alpha\prec\beta\Longleftrightarrow\sum_{j=1}^{i}\alpha_{j}\leq\sum_{j=1}^{i}\beta_{j},\qquad
1\leq i\leq N,\qquad\alpha\neq\beta,$
$\displaystyle\alpha\lhd\beta\Longleftrightarrow(|\alpha|=|\beta|)\wedge\big{[}\big{(}\alpha^{+}\prec\beta^{+}\big{)}\vee\big{(}\alpha^{+}=\beta^{+}\wedge\alpha\prec\beta\big{)}\big{]}.$

The rank function on compositions is involved in the formula for an NSMP.

######  Definition 3.4.

For $\alpha\in\mathbb{N}_{0}^{N}$, $1\leq i\leq N$

$r_{\alpha}(i):=\\#\\{j\colon\alpha_{j}>\alpha_{i}\\}+\\#\\{j\colon 1\leq
j\leq i,\,\alpha_{j}=\alpha_{i}\\},$

then $r_{\alpha}\in\mathcal{S}_{N}$. There is a shortest expression
$r_{\alpha}=s_{i_{1}}s_{i_{2}}\cdots s_{i_{k}}$ and
$R_{\alpha}:=(T_{i_{1}}T_{i_{2}}\cdots T_{i_{k}})^{-1}\in\mathcal{H}_{N}(t)$
(that is, $R_{\alpha}=T(r_{\alpha})^{-1}$).

A consequence is that $r_{\alpha}\alpha=\alpha^{+}$, the nonincreasing
rearrangement of $\alpha$, for any $\alpha\in\mathbb{N}_{0}^{N}$. For example
if $\alpha=(1,2,0,5,4,5)$ then $r_{\alpha}=[5,4,6,1,3,2]$ and
$r_{\alpha}\alpha=\alpha^{+}=(5,5,4,2,1,0)$ (recall
$(u\alpha)_{i}=\alpha_{u^{-1}(i)}$). Also $r_{\alpha}=I$ if and only if
$\alpha\in\mathbb{N}_{0}^{N,+}$.

######  Theorem 3.5 ([9, Theorem 4.12]).

Suppose $\alpha\in\mathbb{N}_{0}^{N}$ and $E\in\mathcal{Y}_{k}$, $k=0,1$ then
there exists a $(\xi_{i})$-simultaneous eigenfunction

$\displaystyle
M_{\alpha,E}(x;\theta)=t^{e(\alpha^{+})}q^{b(\alpha)}x^{\alpha}R_{\alpha}(\tau_{E}(\theta))+\sum_{\beta\vartriangleleft\alpha}x^{\beta}v_{\alpha,\beta,E}(\theta;q,t),$
(3.2)

where $v_{\alpha,\beta,E}(\theta;q,t)\in\mathcal{P}_{m,k}$ and its
coefficients are rational functions of $q$, $t$. Also
$\xi_{i}M_{\alpha,E}(x;\theta)=\zeta_{\alpha,E}(i)M_{\alpha,E}(x;\theta)$,
where $\zeta_{\alpha,E}(i)=q^{\alpha_{i}}t^{c(r_{\alpha}(i),E)}$ for $1\leq
i\leq N$. The exponent $b(\alpha):=\sum_{i=1}^{N}\binom{\alpha_{i}}{2}$ and
$e(\alpha^{+}):=\sum_{i=1}^{N}\alpha_{i}^{+}(N-i+c(i,E))$.

The spectral vector is $[\zeta_{\alpha,E}(i)]_{i=1}^{N}$. Note that the
leading term involves the element $R_{\alpha}(\tau_{E}(\theta))$ of
$\mathcal{H}_{N}(t)$ acting on fermionic variables. The explanation for the
exponents $e(\alpha^{+})$ and $b(\alpha)$ is in Proposition 3.10 below. The
relations (2.3) hold when $\omega_{i}$, $T_{i}$ is replaced by $\xi_{i}$,
$\boldsymbol{T}_{i}$ respectively

$\displaystyle\boldsymbol{T}_{j}\xi_{j}=(t-1)\xi_{j}+\xi_{j+1}\boldsymbol{T}_{j},$
$\displaystyle\xi_{j}\boldsymbol{T}_{j}=\boldsymbol{T}_{j}\xi_{j+1}+(t-1)\xi_{j},$

and this leads to the following, which has the same proof as Proposition 2.16:

######  Proposition 3.6.

Suppose $\xi_{j}f=\lambda_{j}f$ for $1\leq j\leq N$ $(f\neq 0$ and $f\in
s\mathcal{P}_{m})$ and
$g:=\boldsymbol{T}_{i}f+\frac{t-1}{\lambda_{i+1}/\lambda_{i}-1}f$ then
$\xi_{j}g=\lambda_{j}g$ for all $j\neq i,i+1$ and $\xi_{i}g=\lambda_{i+1}g$,
$\xi_{i+1}g=\lambda_{i}g$. If $\lambda_{i+1}\neq t^{\pm 1}\lambda_{i}$ then
$g\neq 0$.

This together with a degree-raising operation provides the method for
constructing the Macdonald polynomials.

Suppose $\alpha\in\mathbb{N}_{0}^{N}$,
$E\in\mathcal{Y}_{0}\cup\mathcal{Y}_{1}$ and $\alpha_{i}\neq\alpha_{i+1}$ then
let $z=\zeta_{\alpha,E}(i+1)/\zeta_{\alpha,E}(i)$ and

$\displaystyle
M_{s_{i}\alpha,E}=c_{\alpha,E}\bigg{(}\boldsymbol{T}_{i}+\frac{t-1}{z-1}\bigg{)}M_{\alpha,E},$
(3.3)

where $c_{\alpha,E}=1$ if $\alpha_{i}<\alpha_{i+1}$ and $c_{\alpha,E}=u(z)$ if
$\alpha_{i}>\alpha_{i+1}$. The spectral vector
$\zeta_{s_{i}\alpha,E}=s_{i}\zeta_{\alpha,E}$.

Suppose $\alpha\in\mathbb{N}_{0}^{N}$ and
$E\in\mathcal{Y}_{0}\cup\mathcal{Y}_{1}$ and $\alpha_{i}=\alpha_{i+1}$, then
let $j=r_{\alpha}(i)$. If $\\{j,j+1\\}\in E$ or $j=N-1\in E$ then
$\boldsymbol{T}_{i}M_{\alpha,E}=-M_{\alpha,E}$. If $\\{j,j+1\\}\cap
E=\varnothing$ or $j=N-1\notin E$ then
$\boldsymbol{T}_{i}M_{\alpha,E}=tM_{\alpha,E}$. If $j<N-2$ and $(j,j+1)\in
E\times E^{C}$ or $(j,j+1)\in E^{C}\times E$ then
$z=\zeta_{\alpha,E}(i+1)/\zeta_{\alpha,E}(i)=t^{c\left(j+1,E\right)-c(j,E)}$
and

$\displaystyle
M_{\alpha,s_{j}E}=c_{\alpha,E}\bigg{(}\boldsymbol{T}_{i}+\frac{t-1}{z-1}\bigg{)}M_{\alpha,E},$

where $c_{\alpha,E}=1$ if (1) $E\in\mathcal{Y}_{0}$ and $(j,j+1)\in
E^{C}\times E$ or if (2) $E\in\mathcal{Y}_{1}$ and $(j,j+1)\in E\times E^{C}$,
or $c_{\alpha,E}=u(z)^{-1}$ if (1) $E\in\mathcal{Y}_{0}$ and $(j,j+1)\in
E\times E^{C}$ or if (2) $E\in\mathcal{Y}_{1}$ and $(j,j+1)\in E^{C}\times E$.
In all cases $\zeta_{\alpha,s_{j}E}=s_{i}\zeta_{\alpha,E}.$

The above equations are implicit formulas for
$\boldsymbol{T}_{i}M_{\alpha,E}$. Formula (3.3) is the same as that for the
scalar case, as in [1, 14].

The affine step is defined as follows: for $\alpha\in\mathbb{N}_{0}^{N}$,
$E\in\mathcal{Y}_{0}\cup\mathcal{Y}_{1}$

$\displaystyle\Phi\alpha=(\alpha_{2},\alpha_{3},\dots,\alpha_{N},\alpha_{1}+1),$
$\displaystyle\zeta_{\Phi\alpha,E}=\big{[}\zeta_{\alpha,E}(2),\zeta_{\alpha,E}(3),\dots,\zeta_{\alpha,E}(N),q\zeta_{\alpha,E}(1)\big{]},$
$\displaystyle M_{\Phi\alpha,E}=x_{N}\boldsymbol{w}M_{\alpha,E}.$

This is based on the relations
$\xi_{N}x_{N}\boldsymbol{w}=qx_{N}\boldsymbol{w}\xi_{1}$ and
$\xi_{i}x_{N}\boldsymbol{w}=x_{N}\boldsymbol{w}\xi_{i+1}$ for $1\leq i<N$.

Denote $\boldsymbol{0}=(0,\dots,0)\in\mathbb{N}_{0}^{N}$ and recall
$E_{0}=\\{N-m,\dots,N\\}$, $E_{1}=\\{1,2,\dots,m-1\\}$. In [9, Section 4.1] a
Yang–Baxter directed graph method is used to inductively construct the
$M_{\alpha,E}$ (this technique is due to Lascoux [12]). Label the nodes
$(\alpha,E)$; for $\mathcal{Y}_{0}$ the root is
$\left(\boldsymbol{0,}E_{0}\right)$ with
$M_{\boldsymbol{0,}E_{0}}=\tau_{E_{0}}=D\phi_{E_{0}}$; and for
$\mathcal{Y}_{1}$ the root is $\left(\boldsymbol{0},E_{1}\right)$ with
$M_{\boldsymbol{0,}E_{1}}=\tau_{E_{1}}=M\phi_{E_{1}}$. The equations have been
set up so that $M_{\boldsymbol{0},E}=\tau_{E}$. The arrows in the graph point
from $(\alpha,E)$ to $(\Phi\alpha,E)$, and from $(\alpha,E)$ to
$(s_{i}\alpha,E)$ ($\alpha_{i}<\alpha_{i+1}$) or to $(\alpha,s_{j}E)$ when
$j=r_{\alpha}(E)$ and $c_{\alpha,E}=1$ in the cases described above.

Here is a brief discussion of the effect of $\boldsymbol{T}_{i}$ on
$x^{\alpha}R_{\alpha}\tau_{E}$ for the $c_{\alpha,E}=1$ cases. For
$\alpha\in\mathbb{N}_{0}^{N}$ let

$\operatorname{inv}(\alpha):=\\#\big{\\{}(i,j)\colon
i<j,\,\alpha_{i}<\alpha_{j}\big{\\}}$

then $r_{\alpha}\alpha=\alpha^{+}$ and $r_{\alpha}=s_{i_{1}}\cdots
s_{i_{\ell}}$, where $\ell=\operatorname{inv}(\alpha)$. Recall
$R_{\alpha}=T_{i_{\ell}}^{-1}\cdots T_{i_{1}}^{-1}$ and the value of
$R_{\alpha}$ is independent of the expressions for $r_{\alpha}$ of length
$\ell$. Suppose $\alpha_{i}<\alpha_{i+1}$ then
$\operatorname{inv}(\alpha)=\operatorname{inv}(s_{i}\alpha)+1$; write
$s_{i}\alpha=r_{s_{i}\alpha}^{-1}\alpha_{+}$ and
$r_{s_{i}\alpha}=s_{i_{1}}\cdots s_{i_{\ell}}$ with
$\ell=\operatorname{inv}(s_{i}\alpha)$. Thus
$r_{\alpha}^{-1}=s_{i}s_{i_{1}}\cdots s_{i_{\ell}}$ and
$R_{\alpha}=T_{i}^{-1}R_{s_{i}\alpha}$ and so
$\boldsymbol{T}_{i}x^{\alpha}R_{\alpha}\tau_{E}=x^{s_{i}\alpha}R_{s_{i}\alpha}\tau_{E}+p(x;\theta)$,
where $p$ is a sum of $x^{\beta}p^{\prime}(\theta)$ with
$s_{i}\alpha\vartriangleright\beta$.

######  Proposition 3.7.

Suppose $\alpha_{i}=\alpha_{i+1}$ with $j=r_{\alpha}(i)$ then
$\boldsymbol{T}_{i}x^{\alpha}R_{\alpha}\tau_{E}=x^{\alpha}T_{i}R_{\alpha}\tau_{E}$
and $T_{i}R_{\alpha}=R_{\alpha}T_{j}$.

###### Proof.

Let $r_{\alpha}=s_{i_{1}}\cdots s_{i_{\ell}}$ with
$\ell=\operatorname{inv}(\alpha)$. Let $\beta=\alpha$ except
$\beta_{i+1}=\alpha_{i}+\frac{1}{2}$ then $\operatorname{inv}(\beta)=\ell+1.$
From $r_{\alpha}\alpha=\alpha^{+}$ it follows that
$r_{\alpha}s_{i}\beta=\beta^{+}$. By definition of $j$ we obtain
$\beta_{j}^{+}=\alpha_{i}+\frac{1}{2},\beta_{j+1}^{+}=\alpha_{i}$ and
$r_{\alpha}^{-1}s_{j}\beta^{+}=\beta$. Now $r_{\alpha}s_{i}\beta=\beta^{+}$
and $s_{j}r_{\alpha}\beta=\beta^{+}$ and $r_{\alpha}s_{i},s_{j}r_{\alpha}$
have (at least) $\ell+1$ factors $s_{i_{n}}$ and
$\operatorname{inv}(\beta)=\ell+1$. This implies
$r_{\alpha}s_{i}=s_{j}r_{\alpha}$ and $T_{i_{1}}T_{i_{2}}\cdots
T_{i_{\ell}}T_{i}=T_{j}T_{i_{1}}T_{i_{2}}\cdots T_{i_{\ell}}$, that is,
$R_{\alpha}^{-1}T_{i}=T_{j}R_{\alpha}^{-1}$. ∎

Thus in the case $\alpha_{i}=\alpha_{i+1}$ the transformation laws (Theorem
2.22 and Corollary 2.24) apply to
$\big{(}T_{j}+\frac{t-1}{z-1}\big{)}\tau_{E}$ with $z=t^{c(j+1,E)-c(j,E)}$
(with the possible results $-\tau_{E},t\tau_{E},\tau_{s_{j}E},\dots$).

Hence only the affine step changes the power of $t$ in the coefficient of
$x^{\alpha}$. It remains to consider
$x^{N}\boldsymbol{w}x^{\alpha}R_{\alpha}\tau_{E}$.

######  Proposition 3.8.

Suppose $\alpha\in\mathbb{N}_{0}^{N}$ and $r_{\alpha}(1)=j$ then
$T^{(N)}R_{\alpha}=t^{N-j}R_{\Phi\alpha}\omega_{j}$.

###### Proof.

Let $\ell=\\#\\{(i,j)\colon 1<i<j\leq N,\,\alpha_{i}<\alpha_{j}\\}$. Then
there is a product $u=s_{i_{1}}s_{i_{2}}\cdots s_{i_{\ell}}$ such that
$u\alpha=\big{(}\alpha_{1},\alpha_{1}^{+},\dots,\alpha_{j-1}^{+},\alpha_{j+1}^{+},\dots,\alpha_{N}^{+}\big{)}$,
and each $i_{k}>1$. By definition $i>j$ implies
$\alpha_{1}^{+}\leq\alpha_{i}<\alpha_{1}+1$, and $i<j$ implies
$\alpha_{i}^{+}\geq\alpha_{1}+1$. Then $s_{j-1}s_{j-2}\cdots
s_{1}u\alpha=\alpha^{+}$ and $R_{\alpha}^{-1}=\big{(}T_{j-1}\cdots
T_{1}\big{)}U$, where $U=T_{i_{1}}\cdots T_{i_{\ell}}$. Let
$u^{\prime}=s_{i_{1}-1}\cdots s_{i_{\ell}-1}$ then

$u^{\prime}(\Phi\alpha)=\big{(}\alpha_{1}^{+},\dots,\alpha_{j-1}^{+},\alpha_{j+1}^{+},\dots,\alpha_{N}^{+},\alpha_{1}+1\big{)}$

and $s_{j}s_{j+1}\cdots s_{N-1}u^{\prime}(\Phi\alpha)=(\Phi\alpha)^{+}$. Thus
$R_{\Phi\alpha}^{-1}=\big{(}T_{j}\cdots T_{N-1}\big{)}U^{\prime}$ and
$U^{\prime}=T_{i_{1}-1}\cdots T_{i_{\ell}-1}$. By (3.1)
$T^{(N)}U=U^{\prime}T^{(N)}$ and

$\displaystyle R_{\Phi\alpha}^{-1}T^{(N)}$ $\displaystyle=\big{(}T_{j}\cdots
T_{N-1}\big{)}U^{\prime}T^{(N)}=\big{(}T_{j}\cdots T_{N-1}\big{)}T^{(N)}U$
$\displaystyle=\big{(}T_{j}\cdots T_{N-1}T_{N-1}\cdots
T_{j}\big{)}T_{j-1}\cdots T_{1}U=t^{N-j}\omega_{j}R_{\alpha}^{-1}.$ ∎

As a consequence
$x_{N}\boldsymbol{w}x^{\alpha}R_{\alpha}\tau_{E}=q^{\alpha_{1}}t^{N-j+c(j,E)}x^{\Phi\alpha}R_{\Phi\alpha}\tau_{E}$
with $j=r_{\alpha}(1)$. Denote $(\alpha_{1}+1$,
$\alpha_{2}+1,\cdots,\alpha_{N}+1)$ by $\alpha+\boldsymbol{1}$.

######  Corollary 3.9.

$(x_{N}\boldsymbol{w})^{N}M_{\alpha,E}=M_{\alpha+\boldsymbol{1},E}=q^{|\alpha|}t^{v}(x_{1}x_{2}\cdots
x_{N})M_{\alpha,E}$, where $v=\frac{N(N-1)}{2}+\sum_{i=1}^{N}c(i,E)$.

###### Proof.

Suppose the coefficient of $x^{\alpha}R_{\alpha}\left(\tau_{E}(\theta)\right)$
in $M_{\alpha,E}$ is $q^{a}t^{b}$ then

$x_{N}\boldsymbol{w}M_{\alpha,E}=M_{\Phi\alpha,E}=q^{a+\alpha_{1}}t^{N-r_{\alpha}(1)+c(r_{\alpha}(1),E)}x^{\Phi\alpha}R_{\Phi\alpha}(\tau_{E})+\cdots.$

Then $(x_{N}\boldsymbol{w})^{2}M_{\alpha,E}$ involves
$r_{\Phi\alpha}(1)=r_{\alpha}(2)$. Repeating this process yields a factor of
$q^{|\alpha|}$ and the $t$-exponent
$\sum_{i=1}^{N}(N-r_{\alpha}(i)+c(r_{\alpha}(i),E))=\frac{N(N-1)}{2}+\sum_{i=1}^{N}c(i,E).$
Furthermore $\Phi^{N}\alpha=\alpha+\boldsymbol{1}$ and
$R_{\alpha+\boldsymbol{1}}=R_{\alpha}$. ∎

If $E\in\mathcal{Y}_{0},\mathcal{Y}_{1}$ then
$\sum_{i=1}^{N}c(i,E)=\frac{N(N-1)}{2}-Nm$, $\frac{N(N-1)}{2}-N(m-1)$
respectively (so $v=N(N-m-1)$ or $N(N-m)$).

######  Proposition 3.10.

The exponent on $q$ in (3.2) is
$b(\alpha)=\sum_{i=1}^{N}\binom{\alpha_{i}}{2}$. The exponent on $t$ in (3.2)
is $\sum_{i=1}^{N}\lambda_{i}(N-i+c(i,E))$.

###### Proof.

For the value of $b(\alpha)$ we use induction on $|\alpha|$. The statement
$b(\boldsymbol{0})=0$ is true since $M_{\boldsymbol{0},E}=\tau_{E}$. The steps
at $\alpha_{i}<\alpha_{i+1}$ taking $x^{\alpha}$ to $x^{s_{i}\alpha}$ do not
involve $q$ (which does affect the other terms of the polynomial), and indeed
$b(\alpha)$ is invariant under $s_{i}$. The affine step takes $x^{\alpha}$ to
$x^{\Phi\alpha}$ and multiplies by $q^{\alpha_{1}}$, that is,
$b(\Phi\alpha)=b(\alpha)+\alpha_{1}$, and
$\binom{\alpha_{1}}{2}+\alpha_{1}=\binom{\alpha_{1}+1}{2}$. Induct on
$\ell(\lambda)$ for the $t$-exponent. Note that only the affine step affects
the exponent so it depends only on $\lambda=\alpha^{+}$. Suppose
$\lambda_{k}\geq 1$ and $\lambda_{i}=0$ for $i>k$. The affine step from
$\lambda^{(k)}$ with $\lambda_{i}^{(k)}=\lambda_{i}$ except
$\lambda_{k}^{(k)}=\lambda_{k}-1$ proceeds by
$\lambda^{(k)}\rightarrow(\lambda_{k}-1,\lambda_{1},\dots,\lambda_{k-1},0,\dots,0)\rightarrow(\lambda_{1},\dots,\lambda_{k=1},0,\dots,0,\lambda_{k})\rightarrow\lambda$
and multiplies by $t^{N-k+c(k,E)}$. Thus passing from
$(\lambda_{1},\dots,\lambda_{k-1},0,\dots,0)$ to $\lambda$ contributes a
factor of $t^{\lambda_{k}(N-k+c(k,E))}$. ∎

######  Example 3.11.

Explicit formulas for $M_{\alpha,E}$ tend to be complicated; here is a fairly
simple one. Let $N=5$, $m=2$, $E=\\{3,4,5\\}$ and $\alpha=(0,0,1,0,0)$:

$\displaystyle
M_{\alpha,E}=t^{6}x_{3}(t^{3}\theta_{2}\theta_{4}-t^{2}\theta_{2}\theta_{4}+\theta_{3}\theta_{4})$
$\displaystyle\hphantom{M_{\alpha,E}=}{}+\frac{(t-1)t^{9}q}{qt^{3}-1}\big{\\{}x_{4}\big{(}t^{3}\theta_{2}\theta_{3}-t\theta_{2}\theta_{5}+\theta_{3}\theta_{5}\big{)}-x_{5}\big{(}t^{2}\theta_{2}\theta_{3}-t\theta_{2}\theta_{4}+\theta_{3}\theta_{4}\big{)}\big{\\}}.$

The spectral vector is $\big{[}t,t^{-2},qt^{2},t^{-1},1\big{]}$ and
$\boldsymbol{T}_{4}M_{\alpha,E}=-M_{\alpha,E}$.

There is a pairwise commuting set of (bosonic) degree lowering operations,
namely the Dunkl operators defined by Baker and Forrester [1].

######  Definition 3.12.

Suppose $f\in s\mathcal{P}_{m}$ then
$\boldsymbol{D}_{N}f:=\frac{1}{x_{N}}(f-\xi_{N}f)$ and if $i<N$ then
$\boldsymbol{D}_{i}f=\frac{1}{t}\boldsymbol{T}_{i}\boldsymbol{D}_{i+1}\boldsymbol{T}_{i}f$.

Assuming the existence of the nonsymmetric Macdonald polynomials
$M_{\alpha,E}$ the argument for showing that $\boldsymbol{D}_{i}f$ is a
polynomial is the following:

######  Proposition 3.13.

Suppose $\alpha\in\mathbb{N}_{0}^{N}$ and
$E\in\mathcal{Y}_{0}\cup\mathcal{Y}_{1}$; if $\alpha_{N}=0$ then
$\boldsymbol{D}_{N}M_{\alpha,E}=0$ or if $\alpha_{N}\geq 1$ then
$\boldsymbol{D}_{N}M_{\alpha,E}=(1-\zeta_{\alpha,E}(N))\boldsymbol{w}M_{\beta,E}$,
where $\alpha=\Phi\beta$. Also
$\boldsymbol{D}_{N}M_{\Phi\alpha,E}=(1-q\zeta_{\alpha}(1))\boldsymbol{w}M_{\alpha,E}$.

###### Proof.

If $\alpha_{N}=0$ then $r_{\alpha}(N)=N,c(N,E)=0$ and
$\xi_{N}M_{\alpha,E}=M_{\alpha,E}$ and $(1-\xi_{N})M_{\alpha,E}=0$. If
$\alpha_{N}\geq 1$ then $\alpha=\Phi\beta$ with $|\beta|=|\alpha|-1$ and
$(1-\xi_{N})M_{\alpha,E}=(1-\zeta_{\alpha,E}(N))M_{\alpha,E}=(1-\zeta_{\alpha,E}(N))x_{N}\boldsymbol{w}M_{\beta,E}$,
and thus
$\boldsymbol{D}_{N}M_{\alpha,E}=(1-\zeta_{\alpha,E}(N))\boldsymbol{w}M_{\beta,E}$.
For the other statement note that
$\zeta_{\Phi\alpha,E}(N)=q\zeta_{\alpha,E}(1)$. ∎

######  Remark 3.14.

We conjecture that there are evaluation formulas for the special case
$E=\\{1,2,\dots,m,N\\}$, $\alpha_{i}=0$ for $i>m$ and
$x_{0}=\big{(}t^{N-1},t^{N-2},\dots,t,1\big{)}$. Here are some small examples
in isotype $(2,1,1)$

$\displaystyle
J_{(1,1,0,0),E}(x_{0})=\frac{t^{5}\big{(}q-t^{4}\big{)}}{q-t^{2}}g_{12}(\theta),$
$\displaystyle
J_{(1,2,0,0),E}(x_{0})=\frac{t^{7}\big{(}q-t^{4}\big{)}\big{(}q^{2}-t^{4}\big{)}}{\big{(}q-t^{2}\big{)}\big{(}q^{2}-t^{3}\big{)}}g_{12}(\theta),$
$\displaystyle
J_{(2,1,0,0),E}(x_{0})=\frac{t^{7}\big{(}q-t^{2}\big{)}\big{(}q-t^{4}\big{)}\big{(}q^{2}-t^{4}\big{)}}{(q-t)\big{(}q-t^{2}\big{)}\big{(}q^{2}-t^{3}\big{)}}g_{12}(\theta),$
$\displaystyle
g_{12}(\theta)=\theta_{1}\theta_{2}-\dfrac{1}{t^{2}(1+t)}(t\theta_{1}-\theta_{2})(\theta_{3}+\theta_{4}).$

Replace $g_{12}(\theta)$ by
$(\theta_{1}\theta_{2}\theta_{3}+\theta_{1}\theta_{2}\theta_{4})$ for the
$\mathcal{P}_{3,1}$ version (by applying $M$).

###  3.3 Symmetric bilinear form

In this section we define an inner product (symmetric bilinear form) on
$s\mathcal{P}_{m}$ in which $\boldsymbol{T}_{i}$, $\xi_{i}$ are self-adjoint,
the Macdonald polynomials are pairwise orthogonal and it is positive-definite
for $t,q>0$, $q\neq 1$ and
$\min\big{(}q^{1/N},q^{-1/N}\big{)}<t<\max\big{(}q^{1/N},q^{-1/N}\big{)}$. The
background and proofs for this section are in [8]. The hypotheses
$\langle\boldsymbol{T}_{i}f,g\rangle=\langle f,\boldsymbol{T}_{i}g\rangle$ for
$1\leq i<N$ and $\langle\xi_{N}f,g\rangle=\langle f,\xi_{N}g\rangle$ already
imply that $\langle\xi_{i}f,g\rangle=\langle f,\xi_{i}g\rangle$ for all $i$
since $\xi_{i}=t^{-1}\boldsymbol{T}_{i}\xi_{i+1}\boldsymbol{T}_{i}$ and thus
$\langle M_{\alpha,E},M_{\beta,F}\rangle=0$ if $(\alpha,E)\neq(\beta,F)$ (at
least one different $\\{\xi_{i}\\}$-eigenvalue). Denote $\langle
f,f\rangle=\|f\|^{2}$, even if possibly nonpositive. The aim is to determine a
formula for $\|M_{\alpha,E}\|^{2}$ which, other than leading coefficients
$q^{\ast}t^{\ast}$, involves only linear factors of the form $(1-q^{q}t^{b})$
with $a\in\mathbb{N}_{0}$, $b\in\mathbb{Z}$, $|b|\leq N$. Recall
$u(z):=\frac{(t-z)(1-tz)}{(1-z)^{2}}$ from Definition 2.25.

######  Proposition 3.15.

Suppose there is a symmetric bilinear form on $s\mathcal{P}$ in which each
$\boldsymbol{T}_{i}$ and $\xi_{i}$ is self-adjoint and suppose
$E\in\mathcal{Y}_{0}\cup\mathcal{Y}_{1}$, $\alpha\in\mathbb{N}_{0}^{N}$ and
$\alpha_{i}<\alpha_{i+1}$ for some $i$ then

$\|M_{s_{i}\alpha,E}\|^{2}=u\big{(}q^{\alpha_{i+1}-\alpha_{i}}t^{c(r_{\alpha}(i+1),E)-c(r_{\alpha}(i),E)}\big{)}\|M_{\alpha,E}\|^{2}.$

###### Proof.

This is the same argument used in Lemma 2.17. ∎

We introduce a product for expressing $\|M_{\alpha,E}\|^{2}$ in terms of
$\|M_{\alpha^{+},E}\|^{2}$.

######  Definition 3.16.

For $E\in\mathcal{Y}_{0}\cup\mathcal{Y}_{1}$, $\alpha\in\mathbb{N}_{0}^{N}$
let

$\mathcal{R}(\alpha,E):=\prod\limits_{1\leq i<j\leq
N,\,\alpha_{i}<\alpha_{j}}u\big{(}q^{\alpha_{j}-\alpha_{i}}t^{c(r_{\alpha}(j),E)-c(r_{\alpha}(i),E)}\big{)}.$

There are $\operatorname{inv}(\alpha)$ terms in the product. The next
proposition assumes the same hypotheses on the bilinear form.

######  Proposition 3.17.

Suppose $E\in\mathcal{Y}_{0}\cup\mathcal{Y}_{1}$,
$\alpha\in\mathbb{N}_{0}^{N}$ then

$\|M_{\alpha^{+},E}\|^{2}=\mathcal{R}(\alpha,E)\|M_{\alpha,E}\|^{2}.$

###### Proof.

With the same argument as in Proposition 2.26 one shows
$\alpha_{i}<\alpha_{i+1}$ implies

$\displaystyle\frac{\mathcal{R}(\alpha,E)}{\mathcal{R}(s_{i}\alpha,E)}=u\big{(}q^{\alpha_{i+1}-\alpha_{i}}t^{c(r_{\alpha}(i+1),E)-c(r_{\alpha}(i),E)}\big{)}.$
∎

Another hypothesis is required to define the inner product for all polynomials
starting with bosonic degree $0$ ($M_{\boldsymbol{0},E}=\tau_{E}$). The
approach of making $\boldsymbol{D}_{i}$ the adjoint of multiplication by
$x_{i}$, or making an isometry out of the latter (torus norm) as is done in
the Jack polynomial situation, does not work here without a modification.

######  Theorem 3.18 ([8, Section 3.3]).

There is a unique symmetric bilinear form on $s\mathcal{P}_{m}$ which extends
the form in Definition 2.6 and satisfies $($for $f,g\in s\mathcal{P}_{m}$ and
$1\leq i<N)$

$\displaystyle\langle\boldsymbol{T}_{i}f,g\rangle=\langle
f,\boldsymbol{T}_{i}g\rangle,$ (3.4)
$\displaystyle\langle\xi_{N}f,g\rangle=\langle f,\xi_{N}g\rangle,$ (3.5)
$\displaystyle\langle\boldsymbol{w}^{-1}\boldsymbol{D}_{N}f,g\rangle=(1-q)\langle
f,x_{N}\boldsymbol{w}g\rangle.$ (3.6)

It follows from $\xi_{i}=t^{-1}\boldsymbol{T}_{i}\xi_{i+1}\boldsymbol{T}_{i}$
that $\langle\xi_{i}f,g\rangle=\langle f,\xi_{i}g\rangle$ for all $i$. The
reason for the factor $(1-q)$ is to allow the limit as $t\rightarrow 1$ when
$q=t^{1/\kappa}$ to obtain nonsymmetric Jack polynomials.

In [8] hypothesis (3.6) is stated in the equivalent form

$\langle\boldsymbol{D}_{N}f,g\rangle=(1-q)\langle
f,x_{N}\boldsymbol{ww}^{\ast}g\rangle,$

where

$\displaystyle\boldsymbol{w}^{\ast}=\boldsymbol{T}_{1}\boldsymbol{T}_{2}\cdots\boldsymbol{T}_{N-1}\boldsymbol{wT}_{1}^{-1}\boldsymbol{T}_{2}^{-1}\cdots\boldsymbol{T}_{N-1}^{-1},$
$\displaystyle\langle\boldsymbol{w}f,g\rangle=\langle
f,\boldsymbol{w}^{\ast}g\rangle,$

and this expression follows from
$\boldsymbol{w}=t^{N-1}\boldsymbol{T}_{N-1}^{-1}\cdots\boldsymbol{T}_{2}^{-1}\boldsymbol{T}_{1}^{-1}\xi_{1}$.
Next we use hypothesis (3.6) to relate norms for polynomials of different
bosonic degrees.

######  Proposition 3.19.

Suppose $E\in\mathcal{Y}_{0}\cup\mathcal{Y}_{1}$,
$\alpha\in\mathbb{N}_{0}^{N}$ then

$\|M_{\Phi\alpha,E}\|^{2}=\frac{1-q^{\alpha_{1}+1}t^{c(r_{\alpha}(1),E)}}{1-q}\|M_{\alpha,E}\|^{2}.$

###### Proof.

In (3.6) set $g=M_{\alpha,E}$ and $f=M_{\Phi\alpha,E}$ then $(1-q)\langle
f,x_{N}\boldsymbol{w}g\rangle=(1-q)\|M_{\Phi\alpha,E}\|^{2}$. On the other
hand

$\displaystyle\boldsymbol{D}_{N}f=\frac{1}{x_{N}}(1-\xi_{N})f=\frac{1}{x_{N}}(1-\zeta_{\Phi\alpha,E}(N))M_{\Phi\alpha,E}=(1-\zeta_{\Phi\alpha,E}(N))\boldsymbol{w}M_{\alpha,E},$
$\displaystyle\langle\boldsymbol{w}^{-1}\boldsymbol{D}_{N}f,g\rangle=(1-\zeta_{\Phi\alpha,E}(N))\langle
M_{\alpha,E},M_{\alpha,E}\rangle,$

thus
$\|M_{\Phi\alpha,E}\|^{2}=\frac{1-\zeta_{\Phi\alpha,E}(N)}{1-q}\|M_{\alpha,E}\|^{2}$
and
$\zeta_{\Phi\alpha,E}(N)=q\zeta_{\alpha,E}(1)=q^{\alpha_{1}+1}t^{c(r_{\alpha}(1),E)}$.
∎

With this formula and Proposition 3.17 we can use induction to find
$\|M_{\alpha,E}\|^{2}$ for any $\alpha$. The first step uses
$\alpha=\boldsymbol{0}$ and any $E\in\mathcal{Y}_{0}\cup\mathcal{Y}_{1}$,
where $M_{\boldsymbol{0},E}=\tau_{E}$ and the spectral vector
$[\zeta_{\boldsymbol{0},E}(i)]_{i=1}^{N}=\big{[}t^{c(i,E)}\big{]}_{i=1}^{N}$.
Then $\Phi\alpha=(0,\dots,0,1)$ and $M_{\Phi\alpha,E}=x_{N}T^{(N)}\tau_{E}$,
$\|M_{\Phi\alpha,E}\|^{2}=\frac{1-qt^{c(1,E)}}{1-q}\|\tau_{E}\|^{2}$ (see
Propositions 2.26 and 2.33 for this value).

The argument for establishing the formula for $\|M_{\lambda,E}\|^{2}$, where
$\lambda\in\mathbb{N}_{0}^{N,+}$ uses the following steps, starting with the
assumption that $\lambda_{k}\geq 1$ and $\lambda_{j}=0$ for $k<j\leq N$.
Throughout $E$ is fixed. Let

$\displaystyle\mu=(\lambda_{1},\dots,\lambda_{k-1},\lambda_{k}-1,0..,0)=\alpha^{+},$
$\displaystyle\alpha=(\lambda_{k}-1,\lambda_{1},\dots,\lambda_{k-1},0,\dots,0),$
$\displaystyle\beta=(\lambda_{1},\dots,\lambda_{k-1},0,\dots,0,\lambda_{k})=\Phi\alpha$

so that $\|M_{\alpha,E}\|^{2}=\mathcal{R}(\alpha,E)^{-1}\|M_{\mu,E}\|^{2}$,
$\|M_{\beta,E}\|^{2}=\frac{1-q^{\lambda_{k}}t^{c(k,E)}}{1-q}\|M_{\alpha,E}\|^{2}$
(since $r_{\alpha}(1)=k$) and
$\|M_{\lambda,E}\|^{2}=\mathcal{R}(\beta,E)\|M_{\beta,E}\|^{2}$. In the
resulting formula we use a slightly different expression for
$u(z)=\frac{(t-z)(1-tz)}{(z-1)^{2}}=t\frac{(1-z/t)(1-tz)}{(z-1)^{2}}$ because
$u(z)/t$ is invariant under $t\rightarrow\frac{1}{t}$,
$z\rightarrow\frac{1}{z}$, but this causes a power of $t$ to appear in the
form $k(\lambda):=\sum_{i=1}^{N}(N-2i+1)\lambda_{i}$ for
$\lambda\in\mathbb{N}_{0}^{N,+}$. The shifted $q$-factorial
$(z;q)_{0}=1,(z;q)_{n+1}=(z;q)_{n}(1-zq^{n})$, $n=0,1,2,\dots$ is used.

######  Theorem 3.20 ([8]).

Suppose $\lambda\in\mathbb{N}_{0}^{N,+}$, $\alpha,\beta\in\mathbb{N}_{0}^{N}$
and $E,F\in\mathcal{Y}_{0}\cup\mathcal{Y}_{1}$ then the following satisfy the
hypotheses (3.4), (3.5), and (3.6)$:$

$\displaystyle\langle
M_{\alpha,E},M_{\beta,F}\rangle=0,\qquad(\alpha,E)\neq(\beta,F),$
$\displaystyle\|M_{\alpha,E}\|^{2}=\mathcal{R}(\alpha,E)^{-1}\|M_{\alpha^{+},E}\|^{2},$
$\displaystyle\|M_{\lambda,E}\|^{2}=t^{k(\lambda)}\|\tau_{E}\|^{2}(1-q)^{-|\lambda|}\prod\limits_{i=1}^{N}\big{(}qt^{c(i,E)};q\big{)}_{\lambda_{i}}$
$\displaystyle\hphantom{\|M_{\lambda,E}\|^{2}=}{}\times\prod\limits_{1\leq
i<j\leq
N}\frac{\big{(}qt^{c(i,E)-c(j,E)-1};q\big{)}_{\lambda_{i}-\lambda_{j}}\big{(}qt^{c(i,E)-c(j,E)+1};q\big{)}_{\lambda_{i}-\lambda_{j}}}{\big{(}qt^{c(i,E)-c(j,E)};q\big{)}_{\lambda_{i}-\lambda_{j}}^{2}}.$

As Griffeth [11] pointed out there is not much cancellation between successive
terms in general; there is a certain amount for the extreme cases
$E_{0}=\\{N-m,N-m+1,\dots,N\\}$ and $E_{1}=\\{1,2,\dots,m-1\\}$. By [8,
Proposition 11] $\|M_{\alpha,E}\|^{2}>0$ if $q>0$ and
$\min\big{(}q^{-1/N},q^{1/N}\big{)}<t<\max\big{(}q^{-1/N},q^{1/N}\big{)}$.

##  4 Symmetric Macdonald superpolynomials

###  4.1 From nonsymmetric to symmetric

This concerns polynomials $p$ in $s\mathcal{P}_{m}$ which satisfy
$\boldsymbol{T}_{i}p=tp$ for $1\leq i<N$ and which are eigenfunctions of
$\sum_{i=1}^{N}\xi_{i}$. We call polynomials satisfying
$\boldsymbol{T}_{i}p=tp$ for all $i$, or $\boldsymbol{T}_{i}p=-p$ for all $i$,
symmetric or antisymmetric, respectively. (The meaning of symmetric here is
not the same as for the symmetric group situation, as will be shown by
example.) There are two approaches to producing symmetric polynomials. One way
is to identify a set of $M_{\alpha,E}$ which is closed under the steps
$f\rightarrow(\boldsymbol{T}_{i}+b)f$ of the type described in Proposition 3.6
and then to apply the symmetry conditions to a linear combination of these
polynomials with undetermined coefficients. The other way is to apply a
symmetrization operator to one polynomial. The original idea for these
approaches comes from Baker and Forrester [2].

######  Definition 4.1.

For $\alpha\in\mathbb{N}_{0}^{N}$ and $E\in\mathcal{Y}_{0}\cup\mathcal{Y}_{1}$
let $\lfloor\alpha,E\rfloor$ denote the tableau obtained from $Y_{E}$ by
replacing $i$ by $\alpha_{i}^{+}$ for $1\leq i\leq N$. Let
$\mathcal{M}(\alpha,E)=\operatorname{span}\\{M_{\beta,F}\colon\lfloor\beta,F\rfloor=\lfloor\alpha,E\rfloor\\}$.

Example: let $N=9$, $m=4$, $E=\\{2,3,6,8,9\\}$, $\alpha=(3,5,6,2,2,1,4,4,6)$,
$\alpha^{+}=(6,6,5,4,4,$ $3,2,2,1)$ and

$Y_{E}=\begin{bmatrix}9&7&5&4&1\\\
\cdot&8&6&3&2\end{bmatrix}\\!,\qquad\lfloor\alpha,E\rfloor=\begin{bmatrix}1&2&4&4&6\\\
\cdot&2&3&5&6\end{bmatrix}\\!.$

######  Theorem 4.2 ([9, Proposition 5.2]).

Suppose $\alpha\in\mathbb{N}_{0}^{N}$ and
$E\in\mathcal{Y}_{0}\cup\mathcal{Y}_{1}$, then there is a series of
transformations of the form $a(\boldsymbol{T}_{i}+b)$ mapping $M_{\alpha,E}$
to $M_{\beta,F}$ if and only if
$\lfloor\beta,F\rfloor=\lfloor\alpha,E\rfloor$.

It is a consequence of the transformation rules that if
$\lfloor\beta,F\rfloor=\lfloor\alpha,E\rfloor$ then the spectral vector
$\zeta_{\beta,F}$ is a permutation of $\zeta_{\alpha,E}$. Furthermore
$\mathcal{M}(\alpha,E)$ is an $\mathcal{H}_{N}(t)$-module.

######  Theorem 4.3 ([9, Theorem 5.27]).

Suppose $\alpha\in\mathbb{N}_{0}^{N}$ and
$E\in\mathcal{Y}_{0}\cup\mathcal{Y}_{1}$ and $\lfloor\alpha,E\rfloor$ is
column-strict $($the entries in column 1 are strictly decreasing$)$ then there
is a unique symmetric polynomial $($up to multiplication by a constant$)$ in
$\mathcal{M}(\alpha,E)$ otherwise there is no nonzero symmetric polynomial.

In [6] the authors defined a superpartition with $N$ parts and fermionic
degree $m$ as an $N$-tuple
$\big{(}\Lambda_{1},\dots,\Lambda_{m};\Lambda_{m+1},\dots,\Lambda_{N}\big{)}$
which satisfies $\Lambda_{1}>\Lambda_{2}>\cdots>\Lambda_{m}$ and
$\Lambda_{m+1}\geq\Lambda_{m+2}\geq\cdots\geq\Lambda_{N}$. Suppose
$\lambda\in\mathbb{N}_{0}^{N,+}$, $E\in\mathcal{Y}_{0}$ and
$\lfloor\lambda,E\rfloor$ is column strict, then
$\Lambda_{i}=\lfloor\lambda,E\rfloor[m+2-i,1]$ for $1\leq i$ $\leq m$ and
$\Lambda_{i}=\lfloor\lambda,E\rfloor[1,N+1-i]$ for $m+1\leq i\leq N$, and also
$\Lambda_{m}>\Lambda_{N}$. Alternatively suppose
$\lambda\in\mathbb{N}_{0}^{N,+}$, $E\in\mathcal{Y}_{1}$ and
$\lfloor\lambda,E\rfloor$ is column strict, then
$\Lambda_{i}=\lfloor\lambda,E\rfloor[m+1-i,1]$ for $1\leq i\leq m$ and
$\Lambda_{i}=\lfloor\lambda,E\rfloor[1,N+2-i]$ for $m+1\leq i\leq N$, and also
$\Lambda_{m}\leq\Lambda_{N}$ (because
$\Lambda_{m}=\lfloor\lambda,E\rfloor[1,1]$ and
$\Lambda_{N}=\lfloor\lambda,E\rfloor[1,2]$). Thus the inequalities
$\Lambda_{m}>\Lambda_{N}$ and $\Lambda_{m}\leq\Lambda_{N}$ distinguish
$\mathcal{Y}_{0}$ from $\mathcal{Y}_{1}$.

As a standardization for the labels use $\lambda=\alpha^{+}$ and for $E$ use
the root $E_{R}$ or the sink $E_{S}$

######  Definition 4.4.

Suppose $E\in\mathcal{Y}_{0},\lambda\in\mathbb{N}_{0}^{N,+}$ then the root
$E_{R}$ and the sink $E_{S}$ (which implicitly depend on $\lambda$) satisfy

$\displaystyle\operatorname{inv}(E_{R})=\min\big{\\{}\operatorname{inv}(F)\colon\lfloor\alpha,F\rfloor=\lfloor\lambda,E\rfloor\big{\\}},$
$\displaystyle\operatorname{inv}(E_{S})=\max\big{\\{}\operatorname{inv}(F)\colon\lfloor\alpha,F\rfloor=\lfloor\lambda,E\rfloor\big{\\}}.$

The root and the sink are produced by minimizing the entries of $F$ in row 1,
respectively minimizing the entries of $F$ in column 1. For
$E\in\mathcal{Y}_{1}$ the definitions of $E_{R}$ and $E_{S}$ are reversed.

So in the above example $E_{R}=E$ and $E_{S}=\\{1,3,6,7,9\\}$ and there are
four sets $F$ such that $\lfloor\lambda,F\rfloor=\lfloor\lambda,E_{S}\rfloor$.

Consider
$p=\sum_{\lfloor\beta,F\rfloor=\lfloor\lambda,E_{R}\rfloor}A(\beta,F)M_{\beta,F}$
then the action of $\boldsymbol{T}_{i}$ decomposes the sum into pairs and
singletons. Suppose
$\lfloor\beta,F\rfloor=\big{\lfloor}\lambda,E_{R}\big{\rfloor}$ for some
$\beta$ with $\beta_{i}<\beta_{i+1}$, some $i$. Let
$z=\zeta_{\beta,F}(i+1)/\zeta_{\beta,F}(i)$ then

$\displaystyle\bigg{(}\boldsymbol{T}_{i}+\frac{t-1}{z-1}\bigg{)}M_{\beta,F}=M_{s_{i}\beta,F},\qquad\boldsymbol{T}_{i}M_{\beta,F}=-\frac{t-1}{z-1}M_{\beta,F}+M_{s_{i}\beta,F},$
$\displaystyle\bigg{(}\boldsymbol{T}_{i}+\frac{t-1}{z^{-1}-1}\bigg{)}M_{s_{i}\beta,F}=u(z)M_{\beta,F},\qquad\boldsymbol{T}_{i}M_{s_{i}\beta,F}=u(z)M_{\beta,F}-\frac{t-1}{z^{-1}-1}M_{s_{i}\beta,F},$

and

$(\boldsymbol{T}_{i}-t)\big{(}A(\beta,F)M_{\beta,F}+A(s_{i}\beta,F)M_{s_{i}\beta,F}\big{)}=0$

implies $A(\beta,F)=\frac{t-z}{1-z}A(s_{i}\beta,F)$.

######  Definition 4.5.

Let $u_{0}(z):=\frac{t-z}{1-z}$, $u_{1}(z):=\frac{1-tz}{1-z}$ and for
$\beta\in\mathbb{N}_{0}^{N}$, $F\in\mathcal{Y}_{0}\cup\mathcal{Y}_{1}$,
$k=1,2$ let

$\mathcal{R}_{k}(\beta,F)=\prod\limits_{1\leq i<j\leq
N,\,\beta_{i}<\beta_{j}}u_{k}\big{(}q^{\beta_{j}-\beta_{i}}t^{c(r_{\beta}(j),F)-c(r_{\beta}(i),F)}\big{)}.$

Thus $\mathcal{R}(\beta,F)=\mathcal{R}_{0}(\beta,F)\mathcal{R}_{1}(\beta,F)$
(see Definition 3.16).

######  Lemma 4.6.

If $\boldsymbol{T}_{i}p=tp$ for $1\leq i<N$ then

$A(\beta,F)=\mathcal{R}_{0}(\beta,F)A\big{(}\beta^{+},F\big{)}.$

###### Proof.

Suppose $\beta_{i}>\beta_{i+1}$ then

$\frac{\mathcal{R}_{0}(s_{i}\beta,F)}{\mathcal{R}_{0}(\beta,F)}=u_{0}\bigg{(}\frac{\zeta_{\beta,F}(i)}{\zeta_{\beta,F}(i+1)}\bigg{)}.$

The same argument as in Proposition 2.26 applies. ∎

######  Definition 4.7.

For $k=0,1$ let $\mathcal{C}_{k}(E):=\prod\limits_{\begin{subarray}{c}1\leq
i<j<N,\\\
c(i,E)<0<c(j,E)\end{subarray}}u_{k}\big{(}t^{v(i,E)-c(ij,E)}\big{)}$.

Thus
$\mathcal{C}\big{[}c(i,E)]_{i=1}^{N}\big{)}=\mathcal{C}_{0}(E)\mathcal{C}_{1}(E)$
(see (2.4)).

######  Lemma 4.8.

$\dfrac{A(\beta,F)}{\mathcal{C}_{0}(F)}=\dfrac{A\left(\beta,E_{S}\right)}{\mathcal{C}_{0}\left(E_{S}\right)}$.

###### Proof.

Consider the possibilities when $\beta_{i}=\beta_{i+1}$ and $j=r_{\beta}(i)$:
if $c(j,F)=c(j+1,F)+1$, that is, $j$ and $j+1$ are in adjacent cells of row 1
of $Y_{F}$ then $\boldsymbol{T}_{i}M_{\beta,F}=tM_{\beta,F}$, imposing no
conditions on $A(\beta,F)$; if $c(j,F)=c(j+1,F)-1$ then
$\boldsymbol{T}_{i}M_{\beta,F}=-M_{\beta,F}$ but this occurs only if there are
adjacent equal values ($\beta_{i}$) in column 1 of
$\lfloor\lambda,E_{R}\rfloor$, ruled out by hypothesis; $c(j,F)<0<c(j+1,F)$.
In this case we relate $M_{\beta,F}$ to $M_{\beta,s_{j}F}$, where
$\operatorname{inv}(s_{j}F)=\operatorname{inv}(F)-1$: the formulas similar to
(3.2) with
$z=\frac{\zeta_{\beta,F}(i)}{\zeta_{\beta,F}(i+1)}=t^{c(j,F)-c(j+1,F)}$ appear
here:

$\displaystyle\boldsymbol{T}_{i}M_{\beta,s_{j}F}=-\frac{t-1}{z-1}M_{\beta,s_{j}F}+M_{\beta,F},$
$\displaystyle\boldsymbol{T}_{i}M_{\beta,F}=u(z)M_{\beta,s_{j}F}-\frac{t-1}{z^{-1}-1}M_{\beta,F},$

then
$(\boldsymbol{T}_{i}-t)\big{(}A(\beta,F)M_{\beta,F}+A(\beta,s_{j}F)M_{\beta,s_{j}F}\big{)}=0$
implies $A\big{(}\beta,s_{j}F\big{)}\allowbreak=\frac{t-z}{1-z}A(\beta,F)$. ∎

So
$A(\beta,F)=\mathcal{R}_{0}(\beta,F)A(\beta^{+},F)=A(\lambda,E_{S})\dfrac{\mathcal{R}_{0}(\beta,F)\mathcal{C}_{0}(E_{S})}{\mathcal{C}_{0}(F)}$.

######  Theorem 4.9.

Suppose $\lambda\in\mathbb{N}_{0}^{N,+}$, $E\in\mathcal{Y}_{0}$, and
$\lfloor\lambda,E\rfloor$ is column-strict then

$p_{\lambda,E}=\sum_{\left\lfloor\alpha,F\right\rfloor=\lfloor\lambda,E\rfloor}\frac{\mathcal{C}_{0}\left(E_{S}\right)\mathcal{R}_{0}\left(\alpha,F\right)}{\mathcal{C}_{0}(F)}M_{\alpha,F}$

is the supersymmetric polynomial in $\mathcal{M}(\lambda,E)$, unique when the
coefficient of $M_{\lambda,E_{S}}$ is $1$.

To show that this meaning of symmetric is different from the group case
consider $N=4$, $E=\\{1,2\\}$, $m=3$, $\lambda=(2,1,0,0)$ and the
corresponding symmetric polynomial (too large to display here) begins:

$p=x_{1}^{2}x_{2}\theta_{1}\theta_{2}(\theta_{3}+\theta_{4})-x_{1}^{2}x_{3}\theta_{1}(t\theta_{2}\theta_{3}+(t-1)\theta_{2}\theta_{4}-\theta_{3}\theta_{4})-tx_{1}^{2}x_{4}\theta_{1}(\theta_{2}+\theta_{3})\theta_{4}+\cdots\,.$

###  4.2 Symmetrization operator and norms

The symmetrization operator is defined analogously to the group case.

######  Definition 4.10.

For $n\geq 1$ let $X_{0}=1$ and $X_{n}=1+\boldsymbol{T}_{n}X_{n-1}$, and
$S^{(n)}=X_{1}X_{2}\cdots X_{n}$.

Equivalently
$X_{n}=1+\boldsymbol{T}_{n}+\boldsymbol{T}_{n}\boldsymbol{T}_{n-1}+\dots+\boldsymbol{T}_{n}\cdots\boldsymbol{T}_{2}\boldsymbol{T}_{1}$.

######  Theorem 4.11.

If $1\leq j\leq n$ then $\big{(}\boldsymbol{T}_{j}-t\big{)}S^{(n)}=0$.

###### Proof.

Consider the same formulas with $\boldsymbol{T}_{i}$ replaced by $s_{i}$ and
denote $\widetilde{X}_{n}=1+s_{n}\widetilde{X}_{n-1}$. In the full expansion
there are $(n+1)!$ terms and the coefficient of $t^{k}$ in $[n+1]_{t}!$ is the
number of terms with $k$ factors. Claim that
$\widetilde{S}^{(n)}=\widetilde{X}_{1}\widetilde{X}_{2}\cdots\widetilde{X}_{n}=\sum_{u\in\mathcal{S}_{n+1}}u\in\mathbb{Z}\mathcal{S}_{n+1}$;
proceeding by induction the statement is true for $n=1$, where
$\widetilde{X}_{1}=1+s_{1}$ and now suppose it is true for $n$ and consider
$\sum_{u\in\mathcal{S}_{n+1}}u(1+s_{n+1}+s_{n+1}s_{n}+\dots+s_{N+1}\cdots
s_{1})$ acting on $\gamma=(\gamma_{1},\dots,\gamma_{n+2})$; then
$s_{n+1}\cdots
s_{j}\gamma=(\gamma_{1},\dots,\gamma_{i-1},\gamma_{i+1},\dots,\gamma_{n+2},\gamma_{i})$.
Thus $\sum_{u\in\mathcal{S}_{n+1}}us_{n+1}\cdots s_{j}$ is the sum of all
$u^{(i)}$ such that $\big{(}u^{(i)}\gamma\big{)}_{n+2}=\gamma_{i}$. This shows
$\widetilde{S}^{(n+1)}=\sum_{u\in\mathcal{S}_{n+2}}u$. Since the number of
terms with $k$ factors in $\widetilde{S}^{(n)}$ is the same as the number of
$u$ of length $k$ each term is of minimum length (the shortest expression of
$u$ as a product of $\\{s_{i}\\}$). Thus replacing each $s_{i}$ by
$\boldsymbol{T}_{i}$ shows that
$S^{(n)}=\sum_{u\in\mathcal{S}_{n+1}}\boldsymbol{T}(u)$.

Replacing $\boldsymbol{T}_{i}$ by $\boldsymbol{T}_{n+1-i}$ for $1\leq i\leq n$
in $S^{(n)}$ does not affect the sum (implicitly the braid relations are
used). Given $j\leq n$ apply the map
$\boldsymbol{T}_{i}\rightarrow\boldsymbol{T}_{j+1-i}$ in $X_{1}X_{2}\cdots
X_{j}$ to obtain

$S^{(n)}=\big{(}1+\boldsymbol{T}_{j}\big{)}\big{(}1+\boldsymbol{T}_{j-1}+\boldsymbol{T}_{j-1}\boldsymbol{T}_{j}\big{)}\cdots\big{(}1+\boldsymbol{T}_{1}+\dots+\boldsymbol{T}_{1}\cdots\boldsymbol{T}_{j}\big{)}X_{j+1}\cdots
X_{n}$

and it is now obvious that $\big{(}\boldsymbol{T}_{j}-t\big{)}S^{(n)}=0$. ∎

######  Corollary 4.12.

Suppose $f\in s\mathcal{P}_{m}$ then
$\boldsymbol{T}_{j}\big{(}S^{(N-1)}f\big{)}=tS^{(N-1)}f$ for $1\leq j<N$.

######  Corollary 4.13.

$S^{(n)}S^{(n)}=[n+1]_{t}!S^{(n)}$.

###### Proof.

The effect of $X_{j}$ on an invariant polynomial is to multiply by
$1+t+t^{2}+\dots+t^{j}$. ∎

######  Corollary 4.14.

Suppose $f,g\in s\mathcal{P}_{m}$ then
$\big{\langle}S^{(N-1)}f,g\big{\rangle}=\big{\langle}f,S^{(N-1)}g\big{\rangle}$.

###### Proof.

Suppose $u\in\mathcal{S}_{N}$ and $u=s_{i_{1}}\cdots s_{i_{\ell}}$ is a
shortest expression for $u$ so that
$\boldsymbol{T}(u)=\boldsymbol{T}_{i_{1}}\cdots$
${}\times\boldsymbol{T}_{i_{\ell}}$ then
$\langle\boldsymbol{T}(u)f,g\rangle=\langle
f,\boldsymbol{T}_{i_{\ell}}\cdots\boldsymbol{T}_{i_{1}}g\rangle=\big{\langle}f,\boldsymbol{T}(u^{-1})g\big{\rangle}$.
Since
$\sum_{u\in\mathcal{S}_{N}}\\!\boldsymbol{T}(u)=\sum_{u\in\mathcal{S}_{N}}\\!\boldsymbol{T}(u^{-1})$
this completes the proof. ∎

There is a summation-free formula for $\|p_{\lambda,E}\|^{2}$, derived as
follows:

Suppose $\lfloor\alpha,F\rfloor=\lfloor\lambda,E\rfloor$ then
$S^{(N-1)}M_{\alpha,F}=cp_{\lambda,E}$ for some constant $c$, because of the
uniqueness of $p_{\lambda,E}$ in $\mathcal{M}(\lambda,E)$. Then

$\displaystyle\big{\langle}p_{\lambda,E},S^{(N-1)}M_{\alpha,F}\big{\rangle}=c\langle
p_{\lambda,E},p_{\lambda,E}\rangle=\big{\langle}S^{(N-1)}p_{\lambda,E},M_{\alpha,F}\big{\rangle}$
$\displaystyle\hphantom{\big{\langle}p_{\lambda,E},S^{(N-1)}M_{\alpha,F}\big{\rangle}}{}=[N]_{t}!\langle
p_{\lambda,E},M_{\alpha,F}\rangle=[N]_{t}!\frac{\mathcal{C}_{0}\big{(}E_{S}\big{)}\mathcal{R}_{0}(\alpha,F)}{\mathcal{C}_{0}(F)}\|M_{\alpha,F}\|^{2}.$
(4.1)

The evaluation depends on determining $c$, which can be done by using
$M_{\lambda^{-},E_{R}}$, where $\lambda^{-}$ is the nondecreasing
rearrangement of $\lambda$. For each $i\leq\lambda_{1}$ let
$m_{i}=\\#\big{\\{}j\colon\big{\lfloor}\lambda,E_{S}\big{\rfloor}[1,j]=i\big{\\}}$
(the multiplicity of $i$ in row 1 of
$\big{\lfloor}\lambda,E_{S}\big{\rfloor}$). We will show that the coefficient
of $M_{\lambda,E_{S}}$ in $S^{(N-1)}M_{\lambda_{-},E_{R}}$ is $[m_{i}]_{t}!$.
(This was shown in [9, Theorem 5.39]; we are outlining a proof here with
simplifications due to the simple hook shape $(N-m,1^{m})$, also to
accommodate the different notation.) Here is an illustration of the following
theorem and the method of proof. Suppose $\lambda=(3,2,2,2,1,0)$ and

$Y_{E_{R}}=\begin{bmatrix}6&5&3&2\\\
\cdot&4&1&\end{bmatrix}\\!,\qquad\big{\lfloor}\lambda,E_{S}\big{\rfloor}=\begin{bmatrix}0&1&2&2\\\
\cdot&2&3&\end{bmatrix}\\!,\qquad Y_{E_{S}}=\begin{bmatrix}6&5&4&3\\\
\cdot&2&1&\end{bmatrix}\\!.$

We demonstrate the effect on the significant terms by using the spectral
vectors:

$\displaystyle\big{(}\lambda^{-},E_{R}\big{)}\simeq\big{(}1,qt,q^{2}t^{3},q^{2}t^{2},q^{2}t^{-1},q^{3}t^{-2}\big{)},$
$\displaystyle X_{5}\colon\
\big{(}qt,q^{2}t^{3},q^{2}t^{2},q^{2}t^{-1},q^{3}t^{-2},1\big{)},$
$\displaystyle X_{4}\colon\
\big{(}q^{2}t^{3},q^{2}t^{2},q^{2}t^{-1},q^{3}t^{-2},qt,1\big{)},$
$\displaystyle X_{3}\colon\
(1+t)\times\big{(}q^{2}t^{3},q^{2}t^{-1},q^{3}t^{-2},q^{2}t^{2},qt,1\big{)},$
$\displaystyle X_{2}\colon\
(1+t)\times\big{(}q^{2}t^{-1},q^{3}t^{-2},q^{2}t^{3},q^{2}t^{2},qt,1\big{)},$
$\displaystyle X_{1}\colon\
(1+t)\times\big{(}q^{3}t^{-2},q^{2}t^{-1},q^{2}t^{3},q^{2}t^{2},qt,1\big{)},$

and this is the spectral vector of $\big{(}\lambda,E_{S}\big{)}$.

######  Theorem 4.15.

For each $M_{\beta,F}\in\mathcal{M}\big{(}\lambda,E_{S}\big{)}$ with
$(\beta,F)\neq\big{(}\lambda,E_{S}\big{)}$ there is a constant $c_{\beta,F}$
such that

$S^{(N-1)}M_{\lambda^{-},E_{R}}=\prod\limits_{i=1}^{\lambda_{1}}[m_{i}]_{t}!M_{\lambda,E_{S}}+\sum_{\lfloor\beta,F\rfloor=\left\lfloor\lambda,E_{S}\right\rfloor}\big{\\{}c_{\beta,F}M_{\beta,F}\colon(\beta,F)\neq\big{(}\lambda,E_{S}\big{)}\big{\\}}.$

###### Proof.

The proof relies on identifying the intermediate steps in transforming
$x^{\lambda^{-}}R_{\lambda_{-}}(\tau_{E_{R}})$ to $x^{\lambda}\tau_{E_{S}}$.
Roughly the action of $X_{N-1-i}$ transforms $M_{\alpha(i),E}$ to
$M_{\alpha(i+1),E}$ by means of $\boldsymbol{T}_{N-1-i}$
${}\times\cdots\boldsymbol{T}_{1}$, where $\alpha(0)=\lambda^{-}$ and for
$i\geq 1$

$\alpha(i)=\big{(}\lambda_{N-i},\lambda_{N-2},\dots,\lambda_{1},\lambda_{N-i+1},\dots,\lambda_{N}\big{)}$

however the situation is not this simple because repeated values of
$\lambda_{j}$ have to taken into account. Note that $X_{j}$ does not affect
the variables $x_{k}$ for $k>j+1$. It (almost) suffices to consider the
coefficient of $x^{\alpha(1)}$ in $X_{N-1}M_{\lambda_{-},E_{R}}$. (Throughout
we use $\Sigma$ to denote a linear combination of terms $M_{\beta,F}$ which
can not be transformed into $M_{\lambda,E_{S}}$ by the operators $X_{1}\cdots
X_{j}$.) Suppose that $\lambda_{N}<\lambda_{N-1}$ (that is,
$\lambda_{1}^{-}<\lambda_{2}^{-}$) then
$\boldsymbol{T}_{N-1-i}\cdots\boldsymbol{T}_{1}M_{\lambda^{-},E_{R}}=M_{\alpha(1),E_{R}}+\Sigma$,
and the process is repeated with $M_{\alpha(1),E_{R}}.$ The other possibility
is that $\lambda_{N-k}>\lambda_{N-k+1}=\cdots=\lambda_{N}$ for some $k\geq 2$.
This implies $r_{\lambda^{-}}(i)=N-k+i$ for $1\leq i\leq k$ and
$c\left(r_{\lambda^{-}}(i),E_{R}\right)=k-i$, because the entries
$N-k+1,\dots,N$ are adjacent in row 1 of $Y_{E_{R}}$ (by hypothesis
$\lfloor\lambda,E\rfloor[2,1]>\lambda_{N}$ and so $\lambda_{N}=\lambda_{N-1}$
implies $\lfloor\lambda,E\rfloor[1,2]=N-1$). Thus
$\boldsymbol{T}_{i}M_{\lambda^{-},E_{R}}=tM_{\lambda^{-},E_{R}}$ for $1\leq
i\leq k-1$ and

$\displaystyle\boldsymbol{T}_{N-1}\cdots\boldsymbol{T}_{k}(1+\boldsymbol{T}_{k-1}+\boldsymbol{T}_{k-1}\boldsymbol{T}_{k-2}+\dots+\boldsymbol{T}_{k-1}\cdots\boldsymbol{T}_{1})M_{\lambda^{-},E_{R}}$
$\displaystyle\qquad{}=\big{(}1+t+t^{2}+\dots+t^{k-1}\big{)}\boldsymbol{T}_{N-1}\cdots\boldsymbol{T}_{k}M_{\lambda^{-},E_{R}}=[k]_{t}\big{\\{}M_{\alpha(1),E_{R}}+\Sigma\big{\\}}.$

Then
$\alpha(1)=\big{(}\lambda_{N-1},\dots,\lambda_{N-k+1},\lambda_{N-k},\dots,\lambda_{1},\lambda_{N}\big{)}$
and the previous argument applies with $k-1$ replacing $k$. The result of
applying $X_{N-k}\cdots X_{N-1}$ is $[k]_{t}M_{\alpha(k),E^{r}}+\Sigma$. Now
$\alpha(k)=\big{(}\lambda_{N-k},\lambda_{N-k+1},\cdots\big{)}$ and the
$\lambda_{N}<\lambda_{N-1}$ type process applies.

The last case to consider is
$\lambda_{N-i-1-k}>\lambda_{N-i-k}=\dots=\lambda_{N-i}>\lambda_{N-i+1}$, where
$Y_{E_{R}}[\ell,1]=N-i$ and the entries $N-i-1,N-i-2,\dots,N-i-k$ are adjacent
in row 1 of $Y_{E_{R}}$. Then

$\displaystyle\alpha(i)=\big{(}\lambda_{N-i},\dots,\lambda_{N-i-k},\lambda_{N-i-k-1}\big{)},$

and similarly to the previous case
$\boldsymbol{T}_{i}M_{\alpha(i),E^{\prime}}=tM_{\alpha(i),E^{\prime}}$ for
$1\leq i\leq k-1$. The set $E^{\prime}$ is an intermediate step in a series of
transpositions transforming $E_{R}$ to $E_{S}$, at this stage using only
$s_{j}$ with $i>N-i$. Similarly

$\displaystyle\boldsymbol{T}_{N-i-1}\cdots\boldsymbol{T}_{k}(1+\boldsymbol{T}_{k-1}+\boldsymbol{T}_{k-1}\boldsymbol{T}_{k-2}+\dots+\boldsymbol{T}_{k-1}\cdots\boldsymbol{T}_{1})M_{\alpha(i),E^{\prime}}$
$\displaystyle\qquad=\big{(}1+t+t^{2}+\dots+t^{k-1}\big{)}\boldsymbol{T}_{N-i-1}\cdots\boldsymbol{T}_{k}M_{\alpha(i),E^{\prime}}=[k]_{t}\big{\\{}M_{\alpha(i+1),E"}+\Sigma\big{\\}}.$

Here $\boldsymbol{T}_{k}$ transforms $M_{\alpha(i),E^{\prime}}$ to
$M_{\alpha(i),E^{\prime\prime}}$, where $E^{\prime\prime}=s_{N-i-1}E^{\prime}$
(since $c(N-i,E^{\prime})<0<c(N-i-1,E^{\prime})$ and
$\operatorname{inv}(E^{\prime\prime})=\operatorname{inv}(E^{\prime})+1$.
Eventually these steps transform $E_{R}$ to $E_{S}$ and $\lambda^{-}$ to
$\lambda$. Each set of $m_{i}$ (contiguous) $\lambda_{i}$ values
$\lambda_{j}=i$ in row 1 of $\big{\lfloor}\lambda,E_{S}\big{\rfloor}$
contributes a factor of $[m_{i}]_{t}!$. By beginning with $E_{R}$ the factors
appearing in $(\boldsymbol{T}_{i}+b)M_{\beta,F}$ are always $1$ (see (3.3) and
(3.2)). ∎

######  Lemma 4.16.

Suppose

$F(\alpha,E):=\prod\limits_{\begin{subarray}{c}1\leq i<j\leq N\\\
\alpha_{i}<\alpha_{j},\end{subarray}}g\big{(}\alpha_{j}-\alpha_{i},c(r_{\alpha}(j),E)-c(r_{\alpha}(i,E))\big{)}$

for some function $g$ then

$F(\lambda^{-},E)=\prod\limits_{\lambda_{i}>\lambda_{j}}g\big{(}\lambda_{i}-\lambda_{j},c(i,E)-c(j,E)\big{)}.$

###### Proof.

It is clear that $\lambda_{i}>\lambda_{j}$ if and only if
$\lambda_{N+1-i}^{-}>\lambda_{N+1-j}^{-}$. If
$\lambda_{a-1}>\lambda_{a}=\cdots=\lambda_{a+k-1}>\lambda_{a+k}$ then
$[r_{\lambda^{-}}(b+i)]_{i=1}^{k}=[a,a+1,\dots,a+k-1]$ and
$\lambda_{b+i}^{-}=\lambda_{a}$ for $b=N+1-a-k,1\leq i\leq k$. The
corresponding contents $[c(r_{\lambda^{-}}(b+i)),E]_{i=1}^{k}$ are the same as
$[c(a+i-1),E]_{i=1}^{k}$. So each term in $F_{\lambda^{-},E}$ matches one in
the stated $\lambda$-product. ∎

######  Theorem 4.17.

Suppose $\lfloor\lambda,E_{S}\rfloor$ is column-strict and
$m_{i}=\\#\big{\\{}j\colon\lfloor\lambda,E_{S}\rfloor[1,j]=i\big{\\}}$ for
$0\leq i$ $\leq\lambda_{1}$ then

$\displaystyle\|p_{\lambda,E_{S}}\|^{2}=t^{2(N-m-1)+k(\lambda)}[m+1]_{t}(1-q)^{-|\lambda|}\prod\limits_{i=1}^{N}\big{(}qt^{c\left(i,E_{S}\right)};q\big{)}_{\lambda_{i}}$
$\displaystyle\hphantom{\|p_{\lambda,E_{S}}\|^{2}=}{}\times\prod\limits_{1\leq
i<j\leq
N}\frac{\big{(}qt^{c\left(i,E_{S}\right)-c\left(j,E_{S}\right)-1};q\big{)}_{\lambda_{i}-\lambda_{j}}\big{(}qt^{c\left(i,E_{S}\right)-c\left(j,E_{S}\right)+1};q\big{)}_{\lambda_{i}-\lambda_{j}-1}}{\big{(}1-q^{\lambda_{i}-\lambda_{j}}t^{c\left(i,E_{S}\right)-c\left(j,E_{S}\right)}\big{)}\big{(}qt^{c\left(i,E_{S}\right)-c\left(j,E_{S}\right)};q\big{)}_{\lambda_{i}-\lambda_{j}-1}^{2}}$
$\displaystyle\hphantom{\|p_{\lambda,E_{S}}\|^{2}=}{}\times\frac{[N]_{t}!}{\prod\limits_{i\geq
0}[m_{i}]_{t}!}\mathcal{C}_{0}\big{(}E_{S}\big{)}\mathcal{C}_{1}\big{(}E_{R}\big{)}.$

###### Proof.

By (4.1) and Theorem 4.15 $c=\prod_{i\geq 0}[m_{i}]_{t}!$,

$\displaystyle\|p_{\lambda,E_{S}}\|^{2}$
$\displaystyle=\frac{[N]_{t}!}{\prod\limits_{i\geq
0}[m_{i}]_{t}!}\frac{\mathcal{C}_{0}\big{(}E_{S}\big{)}\mathcal{R}_{0}\big{(}\lambda^{-},E_{R}\big{)}}{\mathcal{C}_{0}\big{(}E_{R}\big{)}}\big{\|}M_{\lambda^{-},E_{R}}\big{\|}^{2}$
$\displaystyle=\frac{[N]_{t}!}{\prod\limits_{i\geq
0}[m_{i}]_{t}!}\frac{\mathcal{C}_{0}\big{(}E_{S}\big{)}\mathcal{R}_{0}\big{(}\lambda^{-},E_{R}\big{)}}{\mathcal{C}_{0}\big{(}E_{R}\big{)}\mathcal{E}\big{(}\lambda^{-},E_{R}\big{)}}\big{\|}M_{\lambda,E_{R}}\big{\|}^{2}$

and

$\displaystyle\big{\|}M_{\lambda,E_{R}}\big{\|}^{2}=t^{\ell(\lambda)}\|\tau_{E_{R}}\|^{2}(1-q)^{-|\lambda|}\prod\limits_{i=1}^{N}\big{(}qt^{c\left(i,E_{R}\right)};q\big{)}_{\lambda_{i}}$
$\displaystyle\hphantom{\big{\|}M_{\lambda,E_{R}}\big{\|}^{2}=}{}\times\prod\limits_{1\leq
i<j\leq
N}\frac{\big{(}qt^{c\left(i,E_{R}\right)-c\left(j,E_{R}\right)-1};q\big{)}_{\lambda_{i}-\lambda_{j}}\big{(}qt^{c\left(i,E_{R}\right)-c\left(j,E_{R}\right)+1};q\big{)}_{\lambda_{i}-\lambda_{j}}}{\big{(}qt^{c\left(i,E_{R}\right)-c\left(j,E_{R}\right)};q\big{)}_{\lambda_{i}-\lambda_{j}}^{2}}.$

Also
$\|\tau_{E_{R}}\|^{2}=t^{2(N-m-1)}[m+1]_{t}\mathcal{C}\big{(}[c(i,E_{R})]_{i=1}^{N}\big{)}=t^{2(N-m-1)}[m+1]_{t}\mathcal{C}_{0}\big{(}E_{R}\big{)}\mathcal{C}_{1}\big{(}E_{R}\big{)}$.
Since
$\mathcal{E}\big{(}\lambda^{-},E_{R}\big{)}=\mathcal{R}_{0}\big{(}\lambda^{-},E_{R}\big{)}\mathcal{R}_{1}\big{(}\lambda^{-},E_{R}\big{)}$
the terms $\mathcal{R}_{0}\big{(}\lambda^{-},E_{R}\big{)}$ and
$\mathcal{C}_{0}\big{(}E_{R}\big{)}$ cancel out. By the lemma

$\displaystyle\mathcal{R}_{1}\big{(}\lambda^{-},E_{S}\big{)}=\prod\limits_{1\leq
i<j\leq
N}u_{1}\big{(}q^{\lambda_{i}-\lambda_{j}}t^{c\left(i,E_{R}\right)-c\left(j,E_{R}\right)}\big{)}=\prod\limits_{1\leq
i<j\leq
N}\frac{1-q^{\lambda_{i}-\lambda_{j}}t^{c\left(i,E_{R}\right)-c\left(j,E_{R}\right)+1}}{1-q^{\lambda_{i}-\lambda_{j}}t^{c\left(i,E_{R}\right)-c\left(j,E_{R}\right)}}$

and dividing by this product changes two of the
$\left(\ast;q\right)_{\lambda_{i}-\lambda j}$ terms to
$\left(\ast;q\right)_{\lambda_{i}-\lambda j-1}$. ∎

Note that $E_{R}$ can be replaced by $E_{S}$ in the first two lines of the
formula for $\|p_{\lambda,E_{S}}\|^{2}.$ By using the $M$ map the formulas
produce supersymmetric polynomials in $\mathcal{P}_{m,1}$: consider the
polynomials $M(p_{\lambda,E_{S}})$, where
$p_{\lambda,E_{S}}\in\mathcal{P}_{m-1,0}$. This is why we do not go into
detail about the $E\in\mathcal{E}_{1}$ case. The norm formula implies the
identity

$\sum_{\lfloor\alpha,F\rfloor=\lfloor\lambda,E\rfloor}\frac{\mathcal{C}_{1}(F)\mathcal{R}_{0}(\alpha,F)}{\mathcal{C}_{0}(F)\mathcal{R}_{1}(\alpha,F)}=\frac{[N]_{t}!}{\prod\limits_{i\geq
0}[m_{i}]_{t}!}\frac{\mathcal{C}_{1}\big{(}E_{R}\big{)}}{\mathcal{C}_{0}\big{(}E_{S}\big{)}\mathcal{R}_{1}\big{(}\lambda^{-},E_{R}\big{)}}.$

This formula was checked by computer algebra for a “small” example, $N=5$,
$m=2$, $\lambda=(2,2,1,1,0)$ with

$\big{\lfloor}\lambda,E_{S}\big{\rfloor}=\begin{bmatrix}0&1&2\\\
\cdot&1&2\end{bmatrix}\\!,\qquad E_{R}=\\{2,4,5\\},\qquad E_{S}=\\{1,3,5\\};$

there are $120$ labels $(\beta,F)$ with
$\lfloor\beta,F\rfloor=\big{\lfloor}\lambda,E_{S}\big{\rfloor}$, that is
$\dim\mathcal{M}\big{(}\lambda,E_{S}\big{)}=120$.

###  4.3 Special values

In the scalar Macdonald polynomial situation there are formulas for special
values. There is one such fairly simple formula here. Let
$F=\\{1,2,\dots,m,N\\}$. (In the $\mathcal{Y}_{1}$-case use $E_{1}$.) Refer to
Proposition 2.34 for useful facts about $\tau_{F}$ and $\|\tau_{F}\|^{2}$.

######  Proposition 4.18.

Suppose $p\in s\mathcal{P}_{m}$ is symmetric and

$z:=\big{(}z_{1},z_{2},\dots,z_{m},t^{N-m-1},\dots,t^{2},t,1\big{)}$

then

$p(z;\theta)=\prod\limits_{1\leq i<j\leq
m}(z_{i}-tz_{j})\prod\limits_{k=1}^{m}\big{(}z_{k}-t^{N-m}\big{)}p_{0}(z_{1},\dots,z_{m})\tau_{F},$

where $p_{0}$ is $\mathcal{S}_{m}$-symmetric.

###### Proof.

First we show that if $\boldsymbol{T}_{i}p(x;\theta)=tp(x;\theta)$ then
$T_{i}p\big{(}x^{(i)};\theta\big{)}=tp\big{(}x^{(i)};\theta\big{)}$, where
$x_{i}^{(i)}=tx_{i+1}^{(i)}$. By hypothesis

$\displaystyle(1+\boldsymbol{T}_{i})p(x;\theta)=(1+t)p(x;\theta)=p(x;\theta)+(1-t)x_{i+1}\frac{p(x;\theta)-p(xs_{i};\theta)}{x_{i}-x_{i+1}}+T_{i}p(xs_{i};\theta).$

Substitute $x_{i}=tx_{i+1}$ in the equations:

$\displaystyle(1+t)p(x;\theta)=p(x;\theta)-(p(x;\theta)-p(xs_{i};\theta))+T_{i}p(xs_{i};\theta)=(1+T_{i})p(xs_{i};\theta)$

and this shows $(t-T_{i})p(x;\theta)=0$ at $x=x^{(i)}$. By hypothesis on $z$
this shows $T_{i}p(z;\theta)=tp(z;\theta)$ for $m+1\leq i<N$ and this implies
$\omega_{i}p(z;\theta)=t^{N-i}p(z;\theta)$ for $m+1\leq i\leq N$. There is
only one $\tau_{E}$ which has these $\\{\omega_{i}\\}$-eigenvalues, namely
$\tau_{F}$. Thus $p(z;\theta)=\widetilde{p}(z_{1},\dots,z_{m})\tau_{F}$ for
some polynomial $\widetilde{p}$ and
$\boldsymbol{T}_{i}\widetilde{p}(z)\tau_{F}=t\widetilde{p}(z)\tau_{F}$ for
$1\leq i<m$. In this range $T_{i}\tau_{F}=-\tau_{F}$ thus $\widetilde{p}$
satisfies the equation

$\displaystyle
t\widetilde{p}(z)\tau_{E}=(1-t)z_{i+1}\frac{\widetilde{p}(z)-\widetilde{p}(zs_{i})}{z_{i}-z_{i+1}}\tau_{E}-\widetilde{p}(zs_{i})\tau_{E},$
$\displaystyle\widetilde{p}(zs_{i})=\frac{z_{i+1}-tz_{i}}{z_{i}-tz_{i+1}}\widetilde{p}(z).$

Thus $z_{i}-tz_{i+1}$ is a factor of $\widetilde{p}(z)$ because
$\widetilde{p}(zs_{i})$ is polynomial. Furthermore
$\widetilde{p}(z)/(z_{i}-tz_{i+1})$ is $s_{i}$-invariant. We claim by
induction that $(z_{i}-tz_{i+k})$ is a factor of $\widetilde{p}(z)$ for $1\leq
i<i+k\leq m$: this is valid for $k=1$ so consider that $(z_{i}-tz_{i+k})$ is a
factor of $\widetilde{p}(z)$ and $\widetilde{p}(z)/(z_{i+k}-tz_{i+k+1})$ is
$s_{i+k}$-invariant thus $(z_{i}-tz_{i+k+1})$ is a factor (where $i+k+1\leq
m$).

Suppose $z_{m}^{\prime}=t^{N-m}=tz_{N-m+1}^{\prime}$ then
$T_{m}\widetilde{p}(z^{\prime})\tau_{F}=t\widetilde{p}(z^{\prime})\tau_{F}$
but this implies $\widetilde{p}(z^{\prime})=0$ or
$\omega_{m}\tau_{F}=t^{N-m}\tau_{F}$ which is impossible. Thus
$\big{(}z_{m}-t^{N-m}\big{)}$ is a factor of $\widetilde{p}(z)$. The symmetry
properties imply $\big{(}z^{i}-t^{N-m}\big{)}$ is a factor of
$\widetilde{p}(z)$ for $1\leq i\leq m$. ∎

González and Lapointe [10] proved an evaluation formula for the version of
supersymmetric Macdonald polynomials constructed in [3], with

$z=\big{(}t^{N-1}q^{-m},t^{N-2}q^{1-m},\dots,t^{N-m}q^{-1},t^{N-m-1},\dots,t,1\big{)}.$

An example appears to show there is no such general result in our version.
However there may be one for the special case where
$\lfloor\lambda,E_{S}\rfloor[1,j]=0$ for $1\leq j\leq N-m$. At this point we
offer no conjecture, but some very small examples with $N=3$, $4$ and
$|\lambda|\leq 4$ suggest there is something to be found.

###  4.4 Minimal symmetric polynomial

For given $N,m$, isotype $(N-m,1^{m})$ there is a unique column-strict tableau
with minimum sum of entries, namely

$\lfloor\lambda,E\rfloor=\begin{bmatrix}0&0&0&\cdots&0\\\
\cdot&1&2&\cdots&m\end{bmatrix}\\!,$

thus $\lambda=(m,m-1,\dots,2,1,0,\dots,0)$ and
$E_{R}=E_{S}=\\{1,2,\dots,m,N\\}$. There is non-trivial multiplicity
$m_{0}=N-m$. Thus $p_{\lambda,E}$ is the symmetric polynomial in
$\mathcal{P}_{m,0}$ of minimum bosonic degree $\frac{m(m+1)}{2}$ There is a
concise formula for $\|p_{\lambda,E}\|^{2}$.

######  Theorem 4.19.

Suppose $\lambda=(m,m-1,\dots,1,0,\dots,0)\in\mathbb{N}_{0}^{N,+}$ and
$E=\\{1,2,\dots,m,N\\}$ then $\lfloor\lambda,E\rfloor$ is column-strict and

$\|p_{\lambda,E}\|^{2}=\frac{[N]_{t}![N]_{t}t^{\gamma}}{[N-m]_{t}![N-m]_{t}}\big{(}qt^{-N};q\big{)}_{m}\prod\limits_{j=2}^{m}\big{(}qt^{-j};q\big{)}_{j-1},$

with
$\gamma=(N-m-1)\big{(}1+\frac{(m+1)(m+2)}{2}\big{)}+\frac{m(m+1)(m+2)}{6}$.

###### Proof.

The exponent on $t$ is

$\displaystyle\gamma$ $\displaystyle=2(N-m-1)+k(\lambda)+m(N-m-1)$
$\displaystyle=(m+2)(N-m-1)+\sum_{i=1}^{m}(N-2i+1)(m+1-i).$

The spectral vector is
$\zeta_{\lambda,E}=\big{[}q^{m}t^{-m},q^{m-1}t^{1-m},\dots,qt^{-1},t^{N-m-1},\dots,t,1\big{]}$.
Consider the content product (part of $\mathcal{C}$) $\Pi_{k}$ for the pairs
$(m+1-k,m+j)$ with $1\leq k\leq m$ and $j=1,\dots,N-m-1$:

$\displaystyle\Pi_{k}=\\!\\!\\!\prod\limits_{j=1}^{N-m-1}\\!\\!\\!u\big{(}t^{-k-j}\big{)}=\\!\\!\\!\prod\limits_{j=1}^{N-m-1}\\!\frac{1\\!-\\!t^{-k-j+1}}{1\\!-t^{-k-j}}\frac{t\\!-t^{-k-j}}{1\\!-t^{-k-j}}=t^{N-m-1}\frac{1\\!-\\!t^{-k}}{1\\!-t^{-k-N+m+1}}\frac{1\\!-\\!t^{-k-N+m}}{1\\!-t^{-k-1}},$

by telescoping the products (and $t-t^{-k-j}=t(1-t^{-k-j-1})$) then

$\prod\limits_{k=1}^{m}\Pi_{k}=t^{m(N-m-1)}\frac{1-t^{-1}}{1-t^{-m-1}}\frac{1-t^{-N}}{1-t^{-N+m}}=t^{m(N-m-1)}\frac{[N]_{t}}{[m+1]_{t}[N-m]_{t}}.$

Consider the $q$-factors for the pairs $(m+1-k,m+j)$ with $1\leq k\leq m$ and
$j=1,\dots,N-m-1$ (use telescoping)

$\displaystyle
P_{1,k}=\prod\limits_{j=1}^{N-m-1}\frac{(qt^{-k-j-1};q)_{k}\big{(}qt^{-k-j+1};q\big{)}_{k-1}}{\big{(}qt^{-k-j};q\big{)}_{k}\big{(}qt^{-k-j};q\big{)}_{k-1}}=\frac{\big{(}qt^{-k-N+m};q\big{)}_{k}\big{(}qt^{-k};q\big{)}_{k-1}}{(qt^{-m-1};q)_{m}\big{(}qt^{-k-N+m+1};q\big{)}_{k-1}},$

then

$P_{1}:=\prod\limits_{k=1}^{m}P_{1,k}=\frac{\big{(}qt^{-N};q\big{)}_{m}}{\big{(}qt^{-m-1};q\big{)}_{m}}.$

Next consider $(m+1-k,m+1-i)$ with $k>i\geq 1$ together with $(m+1-k,N)$:

$P_{2,k}=\prod\limits_{i=0}^{k-1}\frac{\big{(}qt^{-k+i-1};q\big{)}_{k-i}\big{(}qt^{-k+i+1};q\big{)}_{k-i-1}}{\big{(}qt^{-k+i};q\big{)}_{k-i}\big{(}qt^{-k+i};q\big{)}_{k-i-1}}=\frac{\big{(}qt^{-k-1};q\big{)}_{k}}{\big{(}qt^{-k};q\big{)}_{k}}.$

Combine

$\displaystyle\prod\limits_{i=1}^{N}\big{(}qt^{c(i,E)};q\big{)}_{\lambda_{i}}P_{1}\prod\limits_{k=1}^{m}P_{2,k}$
$\displaystyle=\frac{\big{(}qt^{-N};q\big{)}_{m}}{\big{(}qt^{-m-1};q\big{)}_{m}}\prod\limits_{k=1}^{m}\frac{\big{(}qt^{-k};q\big{)}_{k}\big{(}qt^{-k-1};q\big{)}_{k}}{\big{(}qt^{-k};q\big{)}_{k}}$
$\displaystyle=\big{(}qt^{-N};q\big{)}_{m}\prod\limits_{k=1}^{m-1}\big{(}qt^{-k-1};q\big{)}_{k}.$

This concludes the proof. ∎

The formula has a hook length interpretation: $\prod_{(i,j)\in
Y_{E}}\\!\big{(}qt^{-\mathrm{hook}(i,j)};q\big{)}_{\mathrm{leg}(i,j)}$; here
$\mathrm{leg}(1,j)\\!=\\!0$ for $2\leq j\leq N-m$; $\mathrm{hook}(i,1)=m+2-i$,
$\mathrm{leg}(i,1)=m+1-i$, $2\leq i\leq m+1$ and $\mathrm{hook}(1,1)=N$,
$\mathrm{leg}(1,1)=m$ (see [9, Theorem 6.22]).

###  4.5 Antisymmetric superpolynomials

Consider
$p_{a}=\sum_{\lfloor\beta,F\rfloor=\lfloor\lambda,E_{R}\rfloor}A(\beta,F)M_{\beta,F}$,
where $\lfloor\lambda,E\rfloor$ is row-strict and
$\boldsymbol{T}_{i}p_{a}=-p_{a}$ for $1\leq i<N$. Recall the action of
$\boldsymbol{T}_{i}$ on the sum, which decomposes into pairs and singletons.
Suppose $\lfloor\beta,F\rfloor=\lfloor\lambda,E_{R}\rfloor$ for some $\beta$
with $\beta_{i}<\beta_{i+1}$, some $i$. Let
$z=\zeta_{\beta,F}(i+1)/\zeta_{\beta,F}(i)$ then by (3.3)

$(\boldsymbol{T}_{i}+1)\big{(}A(\beta,F)M_{\beta,F}+A(s_{i}\beta,F)M_{s_{i}\beta,F}\big{)}=0$

implies $A(\beta,F)=-\frac{1-zt}{1-z}A(s_{i}\beta,F)$. Recall
$\sigma(n)=(-1)^{n}$ (Definition 2.9).

######  Lemma 4.20.

If $\boldsymbol{T}_{i}p_{a}=-p_{a}$ for $1\leq i<N$ then

$A(\beta,F)=\sigma(\operatorname{inv}(\beta))\mathcal{R}_{1}(\beta,F)A(\beta^{+},F).$

###### Proof.

Suppose $\beta_{i}>\beta_{i+1}$ then

$\displaystyle\frac{\mathcal{R}_{1}(s_{i}\beta,F)}{\mathcal{R}_{1}(\beta,F)}=u_{1}\bigg{(}\frac{\zeta_{\beta,F}(i)}{\zeta_{\beta,F}(i+1)}\bigg{)}.$
∎

Consider the possibilities when $\beta_{i}=\beta_{i+1}$ and $j=r_{\beta}(i)$:
$(i)$ if $c(j,F)=c(j+1,F)-1$, that is, $j$ and $j+1$ are in adjacent cells of
column 1 of $Y_{F}$ then $\boldsymbol{T}_{i}M_{\beta,F}=-M_{\beta,F}$,
imposing no conditions on $A(\beta,F)$; $(ii)$ if $c(j,F)=c(j+1,F)+1$ then
$\boldsymbol{T}_{i}M_{\beta,F}=tM_{\beta,F}$ but this occurs only if there are
adjacent equal values ($\beta_{i}$) in row 1 of $\lfloor\lambda,E_{R}\rfloor$,
ruled out by hypothesis; $(iii)$ $c(j,F)<0<c(j+1,F)$. In this case we relate
$M_{\beta,F}$ to $M_{\beta,s_{j}F}$, where
$\operatorname{inv}(s_{j}F)=\operatorname{inv}(F)-1$: the formulas similar to
(3.2) with $z=\zeta_{\beta,F}(i)/\zeta_{\beta,F}(i+1)=t^{c(j,F)-c(j+1,F)}$
appear here: then

$(\boldsymbol{T}_{i}+1)\big{(}A(\beta,F)M_{\beta,F}+A(\beta,s_{j}F)M_{\beta,s_{j}F}\big{)}=0$

implies $A\big{(}\beta,s_{j}F\big{)}=-\frac{1-tz}{1-z}A(\beta,F)$.

######  Lemma 4.21.

$\sigma(\operatorname{inv}(F))\dfrac{A(\beta,F)}{\mathcal{C}_{1}(F)}=\sigma\big{(}\operatorname{inv}(E_{S})\big{)}\dfrac{A\big{(}\beta,E_{S}\big{)}}{\mathcal{C}_{1}\big{(}E_{S}\big{)}}$.

Thus

$\displaystyle A(\beta,F)$
$\displaystyle=\sigma(\operatorname{inv}(\beta))\mathcal{R}_{1}(\beta,F)A(\beta^{+},F)$
$\displaystyle=\sigma\big{(}\operatorname{inv}(\beta)+\operatorname{inv}(F)+\operatorname{inv}\big{(}E_{S}\big{)}\big{)}A\big{(}\lambda,E_{S}\big{)}\dfrac{\mathcal{R}_{1}(\beta,F)\mathcal{C}_{1}\big{(}E_{S}\big{)}}{\mathcal{C}_{1}(F)}.$

######  Theorem 4.22.

Suppose $\lambda\in\mathbb{N}_{0}^{N,+}$, $E\in\mathcal{Y}_{0}$, and
$\lfloor\lambda,E\rfloor$ is row-strict then

$p_{\lambda,E}^{a}=\sum_{\lfloor\alpha,F\rfloor\in\mathcal{T}(\lambda,E)}\sigma\big{(}\operatorname{inv}(\beta)+\operatorname{inv}(F)+\operatorname{inv}\big{(}E_{S}\big{)}\big{)}\frac{\mathcal{C}_{1}\big{(}E_{S}\big{)}\mathcal{R}_{1}(\alpha,F)}{\mathcal{C}_{1}(F)}M_{\alpha,F}$

is the antisymmetric polynomial in $\mathcal{M}(\lambda,E)$, unique when the
coefficient of $M_{\lambda,E_{S}}$ is $1$.

The antisymmetrizing operator is defined analogously to $S$.

######  Definition 4.23.

For $n\geq 1$ let $X_{0}^{a}=1$ and
$X_{n}^{a}=1-\frac{1}{t}\boldsymbol{T}_{n}X_{n-1}$, and
$A^{(n)}=X_{1}^{a}X_{2}^{a}\cdots X_{n}^{a}$.

Equivalently
$X_{n}^{a}=1-\frac{1}{t}\boldsymbol{T}_{n}+\frac{1}{t^{2}}\boldsymbol{T}_{n}\boldsymbol{T}_{n-1}+\dots+\frac{(-1)^{n}}{t^{n}}\boldsymbol{T}_{n}\cdots\boldsymbol{T}_{2}\boldsymbol{T}_{1}$.

######  Theorem 4.24.

If $1\leq j\leq n$ then $(\boldsymbol{T}_{j}+1)A^{(n)}=0$.

###### Proof.

The operators $\big{\\{}{-}\frac{1}{t}\boldsymbol{T}_{i}\big{\\}}$ satisfy the
braid relations so the same approach as in Theorem 4.11 works here, and the
proof then follows from
$(\boldsymbol{T}_{i}+1)\big{(}1-\frac{1}{t}\boldsymbol{T}_{i}\big{)}=0$. ∎

Similarly to Corollary 4.13 one can show that

$A^{(N-1)}A^{(N-1)}=t^{-N(N-1)/2}[N]_{t}!A^{(N-1)}.$

There is a result analogous to Proposition 4.18.

######  Lemma 4.25.

Suppose for some $i$ that $\boldsymbol{T}_{i}p(x;\theta)=-p(x;\theta)$ then
$T_{i}p\left(x^{(i)};\theta\right)=-p\left(x^{(i)};\theta\right)$, where
$x_{i+1}^{(i)}=tx_{i}^{(i)}$.

###### Proof.

By hypothesis

$\displaystyle(t-\boldsymbol{T}_{i})p(x;\theta)=(t+1)p(x;\theta)=tp(x;\theta)-(1-t)x_{i+1}\frac{p(x;\theta)-p(xs_{i};\theta)}{x_{i}-x_{i+1}}T_{i}p(xs_{i};\theta).$

Substitute $x_{i+1}=tx_{i}$ in the equations:

$\displaystyle(t+1)p(x;\theta)=tp(x;\theta)-t(p(x;\theta)-p(xs_{i};\theta))-T_{i}p(xs_{i};\theta)=(t-T_{i})p(xs_{i};\theta)$

and this shows $(1+T_{i})p(x;\theta)=0$ at $x=x^{(i)}$. ∎

Suppose $p_{\lambda,E}^{a}$ is antisymmetric and
$E_{0}=\\{N-m,N-m+1,\dots,N\\}$ and consider $p_{\lambda,E}^{a}(z)$, where
$z=\big{(}z_{1},z_{2},\dots,z_{N-m-1},t^{-m},\dots,t^{-2},t^{-1},1\big{)}$,
then by the lemma $T_{i}p(z;\theta)=-p(z;\theta)$ for $N-m\leq i<N$ which
$\omega_{i}p(z;\theta)=t^{i-N}p(z;\theta)$ for $N-m\leq i\leq N$. The
eigenvalues determine $\tau_{E_{0}}$ and thus
$p(z;\theta)=\widetilde{p}(z)\tau_{E_{0}}(\theta)$. If range $1\leq i\leq
N-m-2$ then $T_{i}\tau_{F}=t\tau_{F}$ thus $\widetilde{p}$ satisfies the
equation

$\displaystyle-\widetilde{p}(z)\tau_{E_{0}}=(1-t)z_{i+1}\frac{\widetilde{p}(z)-\widetilde{p}(zs_{i})}{z_{i}-z_{i+1}}\tau_{E_{0}}+t\widetilde{p}(zs_{i})\tau_{E_{0}},$
$\displaystyle\widetilde{p}(zs_{i})=\frac{z_{i}-tz_{i+1}}{z_{i+1}-tz_{i}}\widetilde{p}(z).$

This implies $(z_{i+1}-tz_{i})$ is a factor of $\widetilde{p}(z)$ and
$\frac{\widetilde{p}(z)}{z_{i+1}-tz_{i}}$ is $s_{i}$-invariant. Furthermore
$(tz_{i}-z_{j})$ is a factor of $\widetilde{p}(z)$ for $1\leq i<j\leq N-m-1$.
Also $z_{N-m-1}=t^{-m-1}$ implies $\widetilde{p}(z)=0$ (or else
$\omega_{N-m-1}\tau_{E_{0}}=t^{-m-1}\tau_{E_{0}}$, contra) and so
$\big{(}t^{m+1}z_{N-m-1}-1\big{)}$ is a factor of $\widetilde{p}(z)$. Thus

$p^{a}(z;\theta)=\prod\limits_{1\leq i<j\leq
N-m-1}(tz_{i}-z_{j})\prod\limits_{k=1}^{N-m-1}\big{(}t^{m+1}z_{k}-1\big{)}p_{0}(z_{1},\dots,z_{N-m-1})\tau_{E_{0}}$

and $p_{0}$ is $\mathcal{S}_{N-m-1}$-symmetric. With methods similar to those
of Theorem 4.19 and by use of the antisymmetrizing operator $A^{(N-1)}$ one
can derive a formula for $\|p_{\lambda,E}^{a}\|^{2}$.

##  5 Conclusion

We constructed a representation of the Hecke algebra $\mathcal{H}_{N}(t)$ on
superpolynomials and applied the theory of vector-valued nonsymmetric
Macdonald polynomials to this situation. The basic facts such as orthogonal
bases for irreducible representations on fermionic variables, the partial
order on compositions used in expressions for the Macdonald polynomials, and a
sketch of the Yang–Baxter graph technique for constructing the polynomials
starting from degree zero were presented. The polynomials are mutually
orthogonal with respect to a bilinear form in which the generators of the
Hecke algebra are self-adjoint. The ideas of Baker and Forrester were used to
construct symmetric polynomials and to determine their squared norms.

There are some topics which deserve further investigation. What can be proven
about values of the nonsymmetric Macdonald polynomials at special points such
as $\big{(}t^{N-1},t^{N-2},\dots,t,1\big{)}$ (see Remark 3.14)? Are there
special values of the symmetric and anti-symmetric polynomials? A minimal
factorization was proven in Proposition 4.18.

Characterizing singular values of the parameters $q$, $t$ and the
corresponding polynomials is another important problem: this means that for a
specific value of $\left(q,t\right)$ there is a polynomial annihilated by
$\boldsymbol{D}_{i}$ for $1\leq i\leq N$ (see Definition 3.12). This problem
is connected with the existence of maps between different modules. Also there
should be interesting factorizations. Here are two examples with $N=5$.

Let $\alpha=(2,0,0,0,0)$, $m=2$, $E=\\{3,4,5\\}\in\mathcal{Y}_{0}$ then
$\boldsymbol{D}_{i}M_{\alpha,E}=0$ for $1\leq i\leq 5$ when $q^{2}t^{5}=1$ or
$qt=-1$ (that is $q^{2}t^{2}=1$, $qt\neq 1$), and

$\displaystyle
M_{\alpha,E}\big{(}x_{1},x_{2},tx_{2},t^{2}x_{2},t^{3}x_{2}\big{)}=t^{10}(tx_{1}-x_{2})(qtx_{1}-x_{2})\tau_{E},$
$\displaystyle\tau_{E}=t^{4}\theta_{3}\theta_{4}-t^{3}\theta_{3}\theta_{5}+t^{2}\theta_{4}\theta_{5},$

when $(q,t)$ takes on a singular value (note if $q=-1/t$ then
$qtx_{1}-x_{2}=-(x_{1}+x_{2})$).

Let $\alpha=(2,0,0,0,0)$, $m=3$, $E=\\{1,2\\}\in\mathcal{Y}_{1}$ then
$\boldsymbol{D}_{i}M_{\alpha,E}=0$ for $1\leq i\leq 5$ when $q^{2}t^{-5}=1$ or
$q=-t$ (that is $q^{2}t^{-2}=1$ and $qt^{-1}\neq 1$), and

$\displaystyle
M_{\alpha,E}\big{(}x_{1},x_{2},t^{-1}x_{2},t^{-2}x_{2},t^{-3}x_{2}\big{)}=t^{6}\big{(}t^{-1}x_{1}-x_{2}\big{)}\big{(}qt^{-1}x_{1}-x_{2}\big{)}\tau_{E},$
$\displaystyle\tau_{E}=\theta_{1}\theta_{2}(\theta_{3}+\theta_{4}+\theta_{5}),$

when $q^{2}=t^{5}$ (set $q=u^{5}$, $t=u^{2}$) or $q=-t$.

Obviously there are delicate interactions among $\alpha$, $E$, $q$, $t$, $x$
for such factorizations to hold.

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