# Bispectrality of $AG_{2}$ Calogero–Moser–Sutherland system

Misha Feigin  Martin Vrabec

###### Abstract

We consider the generalised Calogero–Moser–Sutherland quantum integrable
system associated to the configuration of vectors $AG_{2}$, which is a union
of the root systems $A_{2}$ and $G_{2}$. We establish the existence of and
construct a suitably defined Baker–Akhiezer function for the system, and we
show that it satisfies bispectrality. We also find two corresponding dual
difference operators of rational Macdonald–Ruijsenaars type in an explicit
form.

##  1 Introduction

The Calogero–Moser–Sutherland (CMS) systems form an important and rich class
of integrable systems with deep relations to geometry, algebra, combinatorics,
and other areas. The original works of Calogero, Sutherland and Moser dealt
with Hamiltonians describing a pairwise interaction of systems of particles on
a circle or a line [2, 20, 28]. This case was related with the root system
$A_{n}$ by Olshanetsky and Perelomov, who extended CMS Hamiltonians to the
case of any root system [21]. In the quantum case, these Hamiltonians are
closely related to the radial parts of Laplace–Beltrami operators on symmetric
spaces [1, 22].

Ruijsenaars and Schneider introduced a relativistic version of the CMS system
in the classical case in [23]. The quantum Hamiltonian and its integrals,
which are difference operators, were studied by Ruijsenaars in [24]. For an
arbitrary (reduced) root system, the corresponding commuting operators were
introduced by Macdonald [19].

Ruijsenaars established a duality relation between the classical
(trigonometric) CMS system and its (rational) relativistic version which
essentially swaps action and angle variables of the two systems [25]. He also
conjectured a duality relation in the quantum case [26]. In this case, there
exists a function $\psi$ of two sets of variables $x$ and $z$ such that

$\displaystyle L\psi$ $\displaystyle=\lambda\psi,$ (1.1) $\displaystyle M\psi$
$\displaystyle=\mu\psi,$

where $L=L(x,\partial_{x})$ is the CMS Hamiltonian or its quantum integral,
$M$ is Ruijsenaars’ relativistic CMS operator acting in the variables $z$, and
$\lambda=\lambda(z)$, $\mu=\mu(x)$ are the corresponding eigenvalues. The
relations (1.1) may be viewed as a higher-dimensional differential-difference
version of the one-dimensional differential-differential bispectrality
relation studied by Duistermaat and Grünbaum [11].

The bispectrality relation (1.1) for the CMS Hamiltonian $L$ related to any
root system and for the corresponding Macdonald–Ruijsenaars operators $M$ was
established by Chalykh in [8] in the case of integer coupling parameters. In
this case, the function $\psi$ can be taken to be a multi-dimensional
Baker–Akhiezer (BA) function [3, 4, 7]. It has the form

$\psi(z,x)=Pe^{\langle x,z\rangle},$ (1.2)

where $P$ is a polynomial in $z$ which also depends on $x$, and $\psi$ can be
characterised by its properties as a function of the variables $z$. The
relation (1.1) for the root system $A_{n}$, another function $\psi$, and an
arbitrary coupling parameter was established recently by Kharchev and
Khoroshkin in [17]. We refer to [16] and references therein for aspects of
this and various other dualities in the broad realm of integrable systems.

In general, the eigenfunctions of quantum integrable systems may be
complicated. However, the multi-dimensional BA functions turn out to be
manageable algebraic functions. An axiomatic definition of BA functions was
formulated by Chalykh, Veselov and Styrkas for an arbitrary finite collection
of non-collinear vectors with integer multiplicities in [3, 4] (see also [13]
for a weaker version of the function; we also refer to [18] for one-
dimensional BA functions). The key properties are the quasi-invariant
conditions of the form

$\psi(z+s\alpha,x)=\psi(z-s\alpha,x)$

which are satisfied at $\langle\alpha,z\rangle=0$ for any vector $\alpha$ from
the configuration, and $s$ takes special values.

It was shown that the BA function exists for very special configurations only,
which includes positive subsystems of reduced root systems [4]. With a slight
modification, one can also formulate an axiomatic definition of the BA
function in the case of the (only) non-reduced root systems $BC_{n}$ [8].

The BA function is known to exist for certain generalisations of CMS systems
related to special configurations of vectors which are not root systems. The
known examples are deformations $A_{n,1}(m)$, $C_{n}(m,l)$ of the roots
systems $A_{n}$ and $C_{n}$, respectively, found by Chalykh, Veselov and one
of the authors in [6, 7]; and another deformation $A_{n,2}(m)$ of the root
system $A_{n+1}$ [9], which satisfies a weaker axiomatics [13]. The BA
function for the configuration $A_{n,1}(m)$ was constructed in [8], and for
the cases $C_{n}(m,l)$, $A_{n,2}(m)$ it was constructed in [13]. It was also
shown in the papers [8, 13] that the bispectrality relation (1.1) holds for
the corresponding generalisations of Macdonald–Ruijsenaars operators. An
important property of these operators is the preservation of a space of quasi-
invariant analytic functions.

In the process of classification of algebraically integrable two-dimensional
generalised CMS operators, the configuration of vectors $AG_{2}$ emerged in
the work of Fairley and one of the authors [12]. We studied the corresponding
generalised CMS operator in [14]. There we found its eigenfunction $\psi$ of
the form (1.2) as $\psi=\mathcal{D}\psi_{0}$ where $\psi_{0}$ is the BA
function for the root system $G_{2}$ and $\mathcal{D}$ is an explicit
differential operator of order three.

In this paper, we study the function $\psi$ further. We describe it as a BA
function by finding the conditions in the variables $z$ which this function
satisfies. We find a pair of explicit difference operators which have $\psi$
as their common eigenfunction. Thus we establish bispectral duality for the
generalised CMS system of type $AG_{2}$. We also obtain an expression for the
function $\psi$ by iterations of the action of the difference operators, in
analogy with formulas from the works [8, 13] for other configurations.

Let us now introduce more precisely the generalised CMS operators and the
configuration of vectors $AG_{2}$. Let $\langle\cdot,\cdot\rangle$ denote the
standard Euclidean inner product on $\mathbb{R}^{n}$. We consider the
complexification of the real Euclidean space $\mathbb{R}^{n}$ and extend
$\langle\cdot,\cdot\rangle$ bilinearly. Let
$x=(x_{1},\dots,x_{n})\in\mathbb{C}^{n}$.

Let $A\subset\mathbb{C}^{n}$ be a finite collection of non-isotropic vectors.
A multiplicity map is a function $m\colon A\to\mathbb{C}$, $\alpha\mapsto
m_{\alpha}=m(\alpha)$. One calls $m_{\alpha}$ the multiplicity of $\alpha$.
One denotes such collections of vectors with multiplicities by
$\mathcal{A}=(A,m)$, or just $A$ if the intended multiplicity map is clear
from the context. The corresponding generalised CMS operator has the form

$L=-\Delta+\sum_{\alpha\in A}g_{\alpha}\sinh^{-2}\langle\alpha,x\rangle,$
(1.3)

where $\Delta=\sum_{i=1}^{n}\partial_{x_{i}}^{2}$ is the Laplacian on
$\mathbb{C}^{n}$ and we have the convention of writing the constants
$g_{\alpha}$ as

$g_{\alpha}=m_{\alpha}(m_{\alpha}+2m_{2\alpha}+1)\langle\alpha,\alpha\rangle$
(1.4)

with $m_{2\alpha}=0$ if $2\alpha\notin A$. The convention (1.4) is used in
symmetric spaces theory (see e.g. [27]).

The Olshanetsky–Perelomov operators correspond to the case when $A$ is a
positive half of a root system and the multiplicity function $m$ is Weyl-
invariant. In that case, if $m_{\alpha}\in\mathbb{N}$ for all $\alpha$ then
the operator (1.3) admits additional integrals and is algebraically integrable
(see [3, 4]).

The configuration of vectors $AG_{2}$ is a non-reduced configuration in
$\mathbb{C}^{2}$ obtained as a union of the root systems $G_{2}$ and $A_{2}$.
A positive half $AG_{2,+}$ is shown in Figure 1, where
${G_{2,+}}=\\{\alpha_{i},\beta_{i}\colon i=1,2,3\\}$, and
${A_{2,+}}=\\{2\beta_{i}\colon i=1,2,3\\}$. There are in total $18$ vectors in
the configuration $AG_{2}$. The subscripts on the $\alpha_{i}$’s are assigned
in such a way that $\langle\alpha_{i},\beta_{i}\rangle=0$ for all $i=1,2,3$.
The ratio of the squared lengths of the long roots $\alpha_{i}$ relative to
the short roots $\beta_{i}$ from the root system $G_{2}$ is
$\alpha_{i}^{2}/\beta_{i}^{2}=3$ for all $i=1,2,3$. The vectors $\alpha_{i}$,
$\beta_{i}$ and $2\beta_{i}$ are assigned respectively the multiplicities $m$,
$3m$ and $1$, where $m\in\mathbb{C}$ is a parameter.

We adopt a coordinate system where the vectors take the form

$\alpha_{1}=\omega\big{(}0,\sqrt{3}\big{)},\quad\alpha_{2}=\omega\left({-}\tfrac{3}{2},\tfrac{\sqrt{3}}{2}\right),\quad\alpha_{3}=\omega\left(\tfrac{3}{2},\tfrac{\sqrt{3}}{2}\right)$

and

$\beta_{1}=\omega\big{(}1,0\big{)},\quad\beta_{2}=\omega\left(\tfrac{{1}}{2},\tfrac{\sqrt{3}}{2}\right),\quad\beta_{3}=\omega\left({-}\tfrac{{1}}{2},\tfrac{\sqrt{3}}{2}\right)$

for some scaling $\omega\in\mathbb{C}^{\times}$.

Figure 1: A positive half of the configuration $AG_{2}$ [14].

We have $\beta_{1}+\beta_{3}=\beta_{2}$, $\alpha_{2}+\alpha_{3}=\alpha_{1}$,
and also

$\displaystyle\beta_{1}=2\beta_{2}-\alpha_{1}=\alpha_{1}-2\beta_{3}=\alpha_{3}-\beta_{2}=\beta_{3}-\alpha_{2},$
$\displaystyle\alpha_{1}=\tfrac{3}{2}\beta_{2}+\tfrac{1}{2}\alpha_{2}=\tfrac{3}{2}\beta_{3}+\tfrac{1}{2}\alpha_{3},\text{
and
}\beta_{1}=\tfrac{1}{2}\beta_{2}-\tfrac{1}{2}\alpha_{2}=-\tfrac{1}{2}\beta_{3}+\tfrac{1}{2}\alpha_{3}.$

The configuration $AG_{2}$ is contained in the lattice
$\mathbb{Z}\beta_{1}\oplus\mathbb{Z}\alpha_{2}\subset\mathbb{C}^{2}$. It is
invariant under the Weyl group of type $G_{2}$, but it is not a
crystallographic root system because, for instance, the vectors $\beta_{1}$
and $2\beta_{2}$ have $2\langle\beta_{1},2\beta_{2}\rangle/\langle
2\beta_{2},2\beta_{2}\rangle=\frac{1}{2}\notin\mathbb{Z}$.

The corresponding generalised CMS quantum Hamiltonian (1.3) with $A=AG_{2,+}$
has recently been shown to be quantum integrable for any value of the
parameter $m\in\mathbb{C}$ [14]. It is moreover algebraically integrable for
$m\in\mathbb{N}$ by virtue of $AG_{2}$ being a locus configuration [12], as a
consequence of the general results presented in [10] (see also [14]). In this
paper, we continue investigating the generalised CMS operator for $AG_{2}$ in
the special case when $m\in\mathbb{N}$.

The structure of the paper is as follows. In Section 2, we recall the notion
of a multi-dimensional BA function associated with a configuration
$\mathcal{A}$ following [3, 4], and we generalise it to a case when $A$ has
proportional vectors. We show that if such a function exists then it is an
eigenfunction for the generalised CMS operator (1.3). In Section 3, we give a
general ansatz for the dual generalised Macdonald–Ruijsenaars difference
operators with rational coefficients, and we find sufficient conditions for
these operators to preserve a space of quasi-invariant analytic functions. In
Section 4, we find a difference operator ${\mathcal{D}}_{1}$ related to the
configuration $AG_{2}$ which satisfies the conditions from Section 3. In
Section 5, we use this operator to show that the BA function for the
configuration $AG_{2}$ exists, and we express this function by iterated action
of the operator. We also show that the BA function is an eigenfunction of the
operator ${\mathcal{D}}_{1}$, thus establishing bispectral duality. In Section
6, we present another dual Macdonald–Ruijsenaars operator ${\mathcal{D}}_{2}$
for the configuration $AG_{2}$, and we establish the corresponding statements
for this operator analogous to the ones from Sections 4 and 5.

In Section 7, we consider the operator ${\mathcal{D}}_{1}$ from Section 4 at
$m=0$, which gives a Macdonald–Ruijsenaars operator for the root system
$A_{2}$ with multiplicity $1$. We also consider a version of this operator for
the root system $A_{1}$ and decompose it into a sum of two non-symmetric
commuting difference operators. We relate these operators with the standard
Macdonald–Ruijsenaars operator for the minuscule weight of the root system
$A_{1}$.

##  2 Baker–Akhiezer functions

We propose a modified definition of multi-dimensional BA functions so as to
extend the definitions from papers [3, 4, 8] to include the configuration
$AG_{2}$. We formulate the definition in such a way that it naturally
generalises the case of reduced root systems as well as the case of the root
system $BC_{n}$.

Let $R\subset\mathbb{C}^{n}$ be a (possibly non-reduced) finite collection of
non-isotropic vectors. We assume there is a subset $R_{+}\subset R$ such that
any collinear vectors in $R_{+}$ are of the form $\alpha$, $2\alpha$ and
$R=R_{+}\sqcup(-R_{+})$. Let $R^{r}=\\{\alpha\in
R\colon\frac{1}{2}\alpha\notin R\\}$ and $R^{r}_{+}=R^{r}\cap R_{+}$. Let
$m\colon R\to\mathbb{Z}_{\geq 0}$ be a multiplicity map with
$m(R^{r})\subset\mathbb{N}$. We extend it to a map $m\colon R\cup
2R^{r}\to\mathbb{Z}_{\geq 0}$ by putting $m_{2\alpha}=0$ if $2\alpha\notin R$,
$\alpha\in R^{r}$.

######  Definition 2.1.

A function $\psi(z,x)$ $(z,x\in\mathbb{C}^{n})$ is a BA function for $(R,m)$
if

  1. 1.

$\psi(z,x)=P(z,x)\exp\langle z,x\rangle$ for some polynomial $P$ in $z$ with
highest order term $\prod_{\gamma\in
R_{+}}\langle\gamma,z\rangle^{m_{\gamma}}$;

  2. 2.

$\psi(z+s\alpha,x)=\psi(z-s\alpha,x)$ at $\langle z,\alpha\rangle=0$ for
$s=1,2,3,\dots,m_{\alpha},m_{\alpha}+2,\dots,m_{\alpha}+2m_{2\alpha}$ for all
$\alpha\in R^{r}_{+}$.

The multiplicities of the vectors in $AG_{2}$ are $m_{\beta_{i}}=3m$,
$m_{2\beta_{i}}=1$, $m_{\alpha_{i}}=m$, and we put $m_{2\alpha_{i}}=0$ for all
$i=1,2,3$, where $m\in\mathbb{N}$. When we apply Definition 2.1 to this
configuration $R=AG_{2}$, we get the notion of the BA function for $AG_{2}$.

For $\gamma\in\mathbb{C}^{n}$, we denote by $\delta_{\gamma}$ the operator
that acts on functions $f(z,x)$ by

$\delta_{\gamma}f(z,x)=f(z+\gamma,x)-f(z-\gamma,x).$

The condition 2 in Definition 2.1 admits the following equivalent
characterisation.

######  Lemma 2.2.

Let $\alpha\in\mathbb{C}^{n}$ be non-isotropic, $m_{\alpha}\in\mathbb{N}$, and
$m_{2\alpha}\in\mathbb{Z}_{\geq 0}$. A function $\psi(z,x)$
$(z,x\in\mathbb{C}^{n})$ analytic in $z$ satisfies
$\psi(z+s\alpha,x)=\psi(z-s\alpha,x)\text{ at }\langle
z,\alpha\rangle=0\textnormal{ for
}s=1,2,3,\dots,m_{\alpha},m_{\alpha}+2,\dots,m_{\alpha}+2m_{2\alpha}$ if and
only if

$\bigg{(}\delta_{\alpha}\circ\frac{1}{\langle
z,\alpha\rangle}\bigg{)}^{s-1}\delta_{\alpha}\psi(z,x)=0\text{ at }\langle
z,\alpha\rangle=0,s=1,\dots,m_{\alpha}$ (2.1)

and

$\bigg{(}\delta_{2\alpha}\circ\frac{1}{\langle
z,\alpha\rangle}\bigg{)}^{t}\circ\bigg{(}\delta_{\alpha}\circ\frac{1}{\langle
z,\alpha\rangle}\bigg{)}^{m_{\alpha}-1}\delta_{\alpha}\psi(z,x)=0\text{ at
}\langle z,\alpha\rangle=0,t=1,\dots,m_{2\alpha}.$ (2.2)

The proof will follow from the one-dimensional considerations in Lemma 2.4
below (see also [4] where the corresponding statements in the case
$m_{2\alpha}=0$ were stated). Let $\delta_{r}$ ($r\in\mathbb{N}$) be the
difference operator that acts on functions $F(k)$ ($k\in\mathbb{C}$) by
$\delta_{r}F(k)=F(k+r)-F(k-r)$. Write $\delta=\delta_{1}$ for short. The next
Lemma will be used in the proof of Lemma 2.4.

######  Lemma 2.3.

Suppose $\delta_{r}F=k\widehat{F}$ for analytic functions $F(k)$ and
$\widehat{F}(k)$. Suppose also that $F(s-r)=F(r-s)$ for some $s$. Then
$\widehat{F}(s)=\widehat{F}(-s)$ if and only if $F(s+r)=F(-s-r)$.

###### Proof.

The statement follows by taking $k=s$ and $k=-s$ in the equality
$\delta_{r}F=k\widehat{F}$. ∎

######  Lemma 2.4.

The following two properties are equivalent for any analytic function $F(k)$
$(k\in\mathbb{C})$ and any $n\in\mathbb{N}$, $m\in\mathbb{Z}_{\geq 0}$.

  1. 1.

For all $s=1,\dots,n,$

$\bigg{(}\delta\circ\frac{1}{k}\bigg{)}^{s-1}\delta
F(k)\bigg{\rvert}_{k=0}=0,$ (2.3)

and for all $t=1,\dots,m,$

$\bigg{(}\delta_{2}\circ\frac{1}{k}\bigg{)}^{t}\circ\bigg{(}\delta\circ\frac{1}{k}\bigg{)}^{n-1}\delta
F(k)\bigg{\rvert}_{k=0}=0.$ (2.4)

  2. 2.

$F(s)=F(-s)$ for all $s=1,2,3,\dots,n,n+2,\dots,n+2m$.

###### Proof.

Let us firstly prove by induction on $n$ that conditions (2.3) are equivalent
to the existence of analytic functions $G_{1},\ldots,G_{n}$ such that $\delta
G_{q}(k)=kG_{q+1}(k)$ for $0\leq q\leq n-1$, $G_{0}=F$, with the conditions
$G_{q}(s)=G_{q}(-s)$ for $1\leq s\leq n-q$.

For the base case $n=1$, the condition (2.3) is equivalent to $F(1)=F(-1)$,
which is equivalent to the existence of an analytic function $G_{1}$ such that
$\delta F=kG_{1}$. Thus the base of induction holds.

Let us now assume that the induction hypothesis is satisfied for some
$n\in\mathbb{N}$. The relation (2.3) for $s=n+1$ is equivalent to $\delta
G_{n}(k)=0$ at $k=0$, that is $G_{n}(1)=G_{n}(-1)$, which is also equivalent
to the existence of an analytic function $G_{n+1}$ such that $\delta
G_{n}=kG_{n+1}$.

We also have $kG_{n}=\delta G_{n-1}$ by the induction hypothesis. By applying
Lemma 2.3 with $s=1$ we get $G_{n-1}(2)=G_{n-1}(-2)$. By considering
$kG_{n-1}=\delta G_{n-2}$ and using that by induction hypothesis
$G_{n-2}(1)=G_{n-2}(-1)$, we obtain by Lemma 2.3 with $s=2$ that
$G_{n-2}(3)=G_{n-2}(-3)$. Similarly we obtain $G_{q}(n-q+1)=G_{q}(-n+q-1)$
successively for $q=n-3,\ldots,0$, which completes the proof of the claim by
induction.

For any fixed $n$ let us now prove by induction on $m$ that relations (2.3),
(2.4) are equivalent to the existence of functions $G_{1},\ldots,G_{n}$ as in
the first paragraph of the proof with the additional conditions
$G_{q}(s)=G_{q}(-s)$ for $s=n-q+2t$ with $1\leq t\leq m$, $0\leq q\leq n-1$,
together with the existence of analytic functions $H_{1},\ldots,H_{m}$
satisfying $\delta_{2}H_{q}=kH_{q+1}$ for $0\leq q\leq m-1$, $H_{0}=G_{n}$,
and $H_{q}(2s)=H_{q}(-2s)$ for $1\leq s\leq m-q$.

The base case $m=0$ is already done. Let us now assume that the induction
hypothesis is satisfied for some $m\in\mathbb{Z}_{\geq 0}$. The relation (2.4)
for $t=m+1$ is equivalent to $\delta_{2}H_{m}(k)=0$ at $k=0$, that is
$H_{m}(2)=H_{m}(-2)$, which is also equivalent to the existence of an analytic
function $H_{m+1}$ such that $\delta_{2}H_{m}=kH_{m+1}$.

We also have $kH_{m}=\delta_{2}H_{m-1}$ by the induction hypothesis. By
applying Lemma 2.3, we obtain $H_{q}(2(m-q+1))=H_{q}(-2(m-q+1))$ successively
for $q=m-1,\ldots,0$.

By considering $kH_{0}=kG_{n}=\delta G_{n-1}$ and using that by induction
hypothesis $G_{n-1}(2m+1)=G_{n-1}(-2m-1)$, we obtain by Lemma 2.3 for
$s=2(m+1)$ that $G_{n-1}(2m+3)=G_{n-1}(-2m-3)$. Similarly we obtain
$G_{q}(n-q+2m+2)=G_{q}(-n+q-2m-2)$ successively for $q=n-2,\ldots,0$, which
completes the proof by induction on $m$.

It follows that condition $1$ in Lemma 2.4 implies condition $2$.

Let us now show that condition $2$ implies condition $1$. Firstly, let us
prove it in the case $m=0$ by induction on $n$. The base case $n=1$ is clear.
Let us assume it for some $n\in\mathbb{N}$. Suppose now that $F(s)=F(-s)$ for
$s=1,\dots,n,n+1$. By the induction hypothesis and the above analysis in the
first part of the proof, we know that there exist functions
$G_{1},\dots,G_{n}$ with properties as stated in the first paragraph of the
proof. By using $F(n\pm 1)=F(-n\mp 1)$ and $\delta F=kG_{1}$ and applying
Lemma 2.3 with $s=n$, we get $G_{1}(n)=G_{1}(-n)$. Similarly we obtain
$G_{q}(n+1-q)=G_{q}(-n-1+q)$ successively for $q=2,\dots,n$. In particular,
$G_{n}(1)=G_{n}(-1)$, which implies that the relation (2.3) holds for $s=n+1$,
completing the induction on $n$.

Suppose that condition $2$ implies condition $1$ for some
$m\in\mathbb{Z}_{\geq 0}$. Now suppose $F$ satisfies condition $2$ and
$F(n+2m+2)=F(-n-2m-2)$. By the induction hypothesis and the above analysis in
the first part of the proof, we have existence of functions
$G_{1},\dots,G_{n}$ and $H_{1},\dots,H_{m}$ with properties as stated in the
fifth paragraph of the proof. By using the assumptions on $F$, the fact that
$\delta F=kG_{1}$ and applying Lemma 2.3 for $s=n+2m+1$, we get that
$G_{1}(n+2m+1)=G_{1}(-n-2m-1)$. Similarly we obtain
$G_{q}(n+2m+2-q)=G_{q}(-n-2m-2+q)$ successively for $q=2,\dots,n$, and
$H_{q}(2m+2-2q)=H_{q}(-2m-2+2q)$ successively for $q=0,\dots,m$. In
particular, $H_{m}(2)=H_{m}(-2)$, which implies that the relation (2.4) holds
for $t=m+1$, which completes the proof. ∎

The following Lemma is a generalisation to the present context of [13, Lemma
1] (see also [4, Proposition 1]). In its proof we use that for a polynomial
$p(z)$ we have $p(z+\gamma)=p(z)+\text{l.o.t.}$ by the binomial theorem, where
“l.o.t.” denotes terms of lower order in $z$ than $\deg p(z)$. This Lemma will
be used below to prove uniqueness of the BA function whenever such a function
exists.

######  Lemma 2.5.

Let $\psi(z,x)=P(z,x)\exp\langle z,x\rangle$ $(z,x\in\mathbb{C}^{n})$, where
$P(z,x)$ is a polynomial in $z$. Suppose that $\psi$ satisfies conditions
(2.1) and (2.2) for some non-zero $\alpha\in\mathbb{C}^{n}$,
$m_{\alpha}\in\mathbb{N}$, $m_{2\alpha}\in\mathbb{Z}_{\geq 0}$. Then
$\langle\alpha,z\rangle^{m_{\alpha}+m_{2\alpha}}$ divides the highest order
term $P_{0}(z,x)$ of $P(z,x)$.

###### Proof.

Firstly, the condition (2.1) with $s=1$ gives

$\displaystyle 0=\delta_{\alpha}\psi(z,x)$
$\displaystyle=\big{(}P_{0}(z,x)(\exp{\langle\alpha,x\rangle}-\exp{\langle-\alpha,x\rangle})+\text{l.o.t.}\big{)}\exp{\langle
z,x\rangle}$ $\displaystyle=\big{(}2\sinh\langle\alpha,x\rangle
P_{0}(z,x)+\text{l.o.t.}\big{)}\exp{\langle z,x\rangle}$

at $\langle z,\alpha\rangle=0$ for all $x\in\mathbb{C}^{n}$. This implies that
$P_{0}(z,x)$ and the l.o.t. must be divisible by $\langle z,\alpha\rangle$.
Let $P_{0}(z,x)=\langle z,\alpha\rangle P_{0}^{(1)}(z,x)$, and let

$\psi^{(1)}(z,x)=\frac{\delta_{\alpha}\psi(z,x)}{\langle
z,\alpha\rangle}=\big{(}2\sinh\langle\alpha,x\rangle
P_{0}^{(1)}(z,x)+\text{l.o.t.}\big{)}\exp{\langle z,x\rangle}.$

The condition (2.1) with $s=2$ gives

$0=\delta_{\alpha}\psi^{(1)}(z,x)=\big{(}4\sinh^{2}\langle\alpha,x\rangle
P_{0}^{(1)}(z,x)+\text{l.o.t.}\big{)}\exp{\langle z,x\rangle}$

at $\langle z,\alpha\rangle=0$ for all $x\in\mathbb{C}^{n}$. This implies that
$\langle z,\alpha\rangle$ divides $P_{0}^{(1)}(z,x)$. In particular,
$P_{0}(z,x)$ is divisible by $\langle z,\alpha\rangle^{2}$: $P_{0}(z,x)\langle
z,\alpha\rangle^{-2}=P_{0}^{(2)}(z,x)$ for some polynomial $P_{0}^{(2)}$. We
let $\psi^{(2)}(z,x)={\langle
z,\alpha\rangle}^{-1}\delta_{\alpha}\psi^{(1)}(z,x).$ By iterating for
$s=3,\dots,m_{\alpha}$, we get that $P_{0}(z,x)$ is divisible by $\langle
z,\alpha\rangle^{m_{\alpha}}$, and we recursively obtain polynomials
$P_{0}^{(s)}(z,x)=P_{0}^{(s-1)}(z,x)\langle
z,\alpha\rangle^{-1}=P_{0}(z,x)\langle z,\alpha\rangle^{-s}$ and functions

$\psi^{(s)}(z,x)=\frac{\delta_{\alpha}\psi^{(s-1)}(z,x)}{\langle
z,\alpha\rangle}=\big{(}2^{s}\sinh^{s}\langle\alpha,x\rangle
P_{0}^{(s)}(z,x)+\text{l.o.t.}\big{)}\exp{\langle z,x\rangle}.$

The condition (2.2) with $t=1$ then gives

$0=\delta_{2\alpha}\psi^{m_{\alpha}}(z,x)=\big{(}2^{m_{\alpha}+1}\sinh^{m_{\alpha}}\langle\alpha,x\rangle\sinh\langle
2\alpha,x\rangle P_{0}^{(m_{\alpha})}(z,x)+\text{l.o.t.}\big{)}\exp{\langle
z,x\rangle}$

at $\langle z,\alpha\rangle=0$. This implies that $\langle z,\alpha\rangle$
divides $P_{0}^{(m_{\alpha})}(z,x)$ and the l.o.t., in particular,
$P_{0}(z,x)$ is divisible by $\langle z,\alpha\rangle^{m_{\alpha}+1}$.
Continuing this iteratively for $t=2,\dots,m_{2\alpha}$ completes the proof. ∎

Lemma 2.5 has the following consequence.

######  Lemma 2.6.

Let $\psi(z,x)=P(z,x)\exp\langle z,x\rangle$ $(z,x\in\mathbb{C}^{n})$ satisfy
condition 2 in Definition 2.1, where $P(z,x)$ is a polynomial in $z$ with
highest order term $P_{0}(z,x)$. Then $\prod_{\gamma\in
R_{+}}\langle\gamma,z\rangle^{m_{\gamma}}$ divides $P_{0}(z,x)$.

###### Proof.

Lemma 2.5 gives that $P_{0}(z,x)$ is divisible by $\langle
z,\alpha\rangle^{m_{\alpha}+m_{2\alpha}}$ for all $\alpha\in R^{r}_{+}$. This
is a constant multiple of $\langle z,\alpha\rangle^{m_{\alpha}}\langle
z,2\alpha\rangle^{m_{2\alpha}}$. The statement follows since we are assuming
that collinear vectors in $R_{+}$ are only of the form $\alpha$, $2\alpha$. ∎

Lemma 2.6 leads to the following uniqueness statement analogous to [4,
Proposition 1] (cf. also [13, Proposition 1]) with a completely analogous
proof.

######  Proposition 2.7.

If a BA function satisfying Definition 2.1 exists, then it is unique. In
particular, if the BA function for $AG_{2}$ exists then it is unique.

The next Theorem generalises Theorem [4, Theorem 1] to the present context,
and it is proved analogously to how that Theorem is proved. It states that if
the BA function satisfying Definition 2.1 exists, then it is a joint
eigenfunction of a commutative ring of differential operators in the variables
$x$. Let us first define an isomorphic ring of polynomials.

Let $\mathcal{R}$ be the ring of polynomials
$p(z)\in\mathbb{C}[z_{1},\ldots,z_{n}]$ satisfying

$\displaystyle p(z+s\alpha)=p(z-s\alpha)\text{ at }\langle
z,\alpha\rangle=0,s=1,2,3,\dots,m_{\alpha},m_{\alpha}+2,\dots,m_{\alpha}+2m_{2\alpha}$

for all $\alpha\in R^{r}_{+}$ (we remark that this is similar to condition 2
in Definition 2.1). We have $z^{2}\in\mathcal{R}$. Indeed, for any
$\gamma\in\mathbb{C}^{n}$, $s\in\mathbb{N}$, we have $(z\pm
s\gamma)^{2}=z^{2}\pm 2s\langle
z,\gamma\rangle+s^{2}\gamma^{2}=z^{2}+s^{2}\gamma^{2}$ at $\langle
z,\gamma\rangle=0$.

For a polynomial $p(z)=p(z_{1},\dots,z_{n})$, by $p(\partial_{x})$ we will
mean $p(\partial_{x_{1}},\dots,\partial_{x_{n}})$. For example, if
$p(z)=z^{2}=z_{1}^{2}+\dots+z_{n}^{2}$, then $p(\partial_{x})=\Delta_{x}$ is
the Laplace operator in $n$ dimensions acting in the variables $x$.

The following statement takes places.

######  Theorem 2.8.

If the BA function $\psi(z,x)$ satisfying Definition 2.1 exists, then for any
$p(z)\in\mathcal{R}$ there is a differential operator $L_{p}(x,\partial_{x})$
with highest order term $p(\partial_{x})$ such that

$L_{p}(x,\partial_{x})\psi(z,x)=p(z)\psi(z,x).$

For any $p,q\in\mathcal{R}$, the operators $L_{p}$ and $L_{q}$ commute.

The next Lemma will be used below to prove that the differential operator
$L_{z^{2}}$ from Theorem 2.8 corresponding to the element $z^{2}$ of
$\mathcal{R}$ coincides with the generalised CMS Hamiltonian (1.3) associated
to the configuration $A=R_{+}$. The next Lemma is a generalisation of [13,
Lemma 2] (see also [4]).

######  Lemma 2.9.

Suppose $\psi(z,x)=P(z,x)\exp\langle z,x\rangle$ satisfies Definition 2.1. Let
$N=\sum_{\gamma\in R_{+}}m_{\gamma}$. Write $P(z,x)=\sum_{i=0}^{N}P_{i}(z,x)$
where $P_{0}(z,x)=\prod_{\gamma\in R_{+}}\langle\gamma,z\rangle^{m_{\gamma}}$
and $P_{i}$ are polynomials homogeneous in $z$ with $\deg P_{i}=N-i$. Then

$\frac{P_{1}(z,x)}{P_{0}(z,x)}=-\sum_{\gamma\in
R_{+}}\frac{m_{\gamma}(m_{\gamma}+2m_{2\gamma}+1)\gamma^{2}}{2\langle\gamma,z\rangle}\coth\langle\gamma,x\rangle.$
(2.5)

###### Proof.

We introduce the following notations

$A(z)=P_{0}(z,x),\quad A_{\alpha}(z)=\frac{A(z)}{\langle\alpha,z\rangle},\quad
A_{\alpha\beta}(z)=\frac{A(z)}{\langle\alpha,z\rangle\langle\beta,z\rangle}=A_{\beta\alpha}(z),\quad\textnormal{etc.}$

As a shorthand, we will write $A_{\alpha^{2}}$ for $A_{\alpha\alpha}$, etc. In
this notation, the equality (2.5) can be written as $P_{1}=-\sum_{\gamma\in
R_{+}}\frac{1}{2}m_{\gamma}(m_{\gamma}+2m_{2\gamma}+1)\gamma^{2}A_{\gamma}(z)\coth\langle\gamma,x\rangle$.

The strategy is as follows. Applying condition 2 in Definition 2.1 and
equating the homogeneous terms of each degree on the right- and left-hand
sides, one gets equations for $P_{1},\dots,P_{N}$. We are assuming that $\psi$
satisfies the definition, hence a solution exists. We are only interested in
$P_{1}$ here.

Fix any $\alpha\in R^{r}_{+}$. We will omit arguments of functions whenever
convenient, and write $e^{\alpha}$ for $\exp\langle\alpha,x\rangle$. By the
binomial theorem $A(z\pm\alpha)$ has the following homogeneous components

$A(z\pm\alpha)=A(z)\pm\sum_{\beta\in R_{+}}m_{\beta}\langle\alpha,\beta\rangle
A_{\beta}(z)+\text{l.o.t.}$

and similarly $P_{1}(z\pm\alpha,x)=P_{1}(z,x)+\text{l.o.t.}$

We will use the fact that for this fixed $\alpha$ condition 2 in Definition
2.1 admits the equivalent characterisation given in Lemma 2.2. We will apply
condition (2.1) for $s=1,\dots,m_{\alpha}$, and then if $m_{2\alpha}>0$ also
condition (2.2) for $t=1,\dots,m_{2\alpha}$, in order to see what conditions
this places on $P_{1}$. Applying condition (2.1) with $s=1$ gives
$0=\delta_{\alpha}\psi(z,x)=(P(z+\alpha,x)e^{\alpha}-P(z-\alpha,x)e^{-\alpha})\exp\langle
z,x\rangle$ at $\langle\alpha,z\rangle=0$. This equation can be rearranged
(after dividing through by $\exp\langle z,x\rangle$) as

$\displaystyle 0=A(z)(e^{\alpha}-e^{-\alpha})+\sum_{\beta\in
R_{+}}m_{\beta}\langle\alpha,\beta\rangle
A_{\beta}(z)(e^{\alpha}+e^{-\alpha})+P_{1}(z,x)(e^{\alpha}-e^{-\alpha})+\text{l.o.t.}$
(2.6)

Note that for $\beta\neq\alpha$ and $\beta\neq 2\alpha$ we have
$A_{\beta}(z)=0$ at $\langle\alpha,z\rangle=0$. By identifying the homogeneous
terms of degree $N-1$ in (2.6), we get at $\langle\alpha,z\rangle=0$ that

$\displaystyle 0$
$\displaystyle=\left(m_{\alpha}\alpha^{2}A_{\alpha}+2m_{2\alpha}\alpha^{2}A_{2\alpha}\right)(e^{\alpha}+e^{-\alpha})+P_{1}(e^{\alpha}-e^{-\alpha})$
$\displaystyle=(m_{\alpha}+m_{2\alpha})\alpha^{2}A_{\alpha}(e^{\alpha}+e^{-\alpha})+P_{1}(e^{\alpha}-e^{-\alpha}).$
(2.7)

Notice that this can be rearranged as
$P_{1}=-(m_{\alpha}+m_{2\alpha})\alpha^{2}A_{\alpha}\coth\langle\alpha,x\rangle$
at $\langle\alpha,z\rangle=0$.

Assume that $m_{\alpha}>1$. Then $A_{\alpha}(z)=0$ at
$\langle\alpha,z\rangle=0$, which forces $P_{1}=0$ at
$\langle\alpha,z\rangle=0$. This is equivalent to divisibility of $P_{1}$ by
$\langle\alpha,z\rangle$. We are next going to consider the condition (2.1)
from Lemma 2.2 with $s=2$, and we are now interested in the degree $N-2$ terms
in the polynomial part of
$(\delta_{\alpha}\circ\langle\alpha,z\rangle^{-1})\delta_{\alpha}\psi(z,x)$.
These terms arise in two ways: firstly they come from the degree $N-2$ terms
in the polynomial part of $\langle\alpha,z\rangle^{-1}\delta_{\alpha}\psi$ and
further action of $\delta_{\alpha}$ on $\exp\langle\alpha,z\rangle$; secondly
they come from application of $\delta_{\alpha}$ onto the terms in
$\langle\alpha,z\rangle^{-1}\delta_{\alpha}\psi$ whose polynomial part has
degree $N-1$. Let us examine these two possibilities individually. From (2.6)
we see that the terms of degree $N-2$ in
$\exp\langle-z,x\rangle\langle\alpha,z\rangle^{-1}\delta_{\alpha}\psi$ are

$\displaystyle\sum_{\beta\in R_{+}}m_{\beta}\langle\alpha,\beta\rangle
A_{\alpha\beta}(z)(e^{\alpha}+e^{-\alpha})+\frac{P_{1}(z,x)}{\langle\alpha,z\rangle}(e^{\alpha}-e^{-\alpha})$
(2.8)
$\displaystyle=(m_{\alpha}+m_{2\alpha})\alpha^{2}A_{\alpha^{2}}(z)(e^{\alpha}+e^{-\alpha})+\sum_{\beta\in
R_{+}\setminus\\{\alpha,2\alpha\\}}m_{\beta}\langle\alpha,\beta\rangle
A_{\alpha\beta}(z)(e^{\alpha}+e^{-\alpha})+\frac{P_{1}(z,x)}{\langle\alpha,z\rangle}(e^{\alpha}-e^{-\alpha}).$

And from (2.6) we see that the term of degree $N-1$ in
$\exp\langle-z,x\rangle\langle\alpha,z\rangle^{-1}\delta_{\alpha}\psi$ is
$A_{\alpha}(z)(e^{\alpha}-e^{-\alpha})$. Since we need to know how
$\delta_{\alpha}$ acts on it, we note here that

$\displaystyle
A_{\alpha}(z\pm\alpha)=A_{\alpha}(z)\pm(m_{\alpha}+m_{2\alpha}-1)\alpha^{2}A_{\alpha^{2}}(z)\pm\sum_{\beta\in
R_{+}\setminus\\{\alpha,2\alpha\\}}m_{\beta}\langle\alpha,\beta\rangle
A_{\alpha\beta}(z)+\text{l.o.t.}$ (2.9)

Let us now consider condition (2.1) from Lemma 2.2 with $s=2$, namely
$0=(\delta_{\alpha}\circ\langle\alpha,z\rangle^{-1})\delta_{\alpha}\psi(z,x)$
at $\langle\alpha,z\rangle=0$. By making use of formulas (2.8) and (2.9) this
condition can be rearranged as

$\displaystyle 0=A_{\alpha}(z)(e^{\alpha}-e^{-\alpha})^{2}$
$\displaystyle+(m_{\alpha}+m_{2\alpha}-1)\alpha^{2}A_{\alpha^{2}}(z)(e^{\alpha}-e^{-\alpha})(e^{\alpha}+e^{-\alpha})$
(2.10)
$\displaystyle+(m_{\alpha}+m_{2\alpha})\alpha^{2}A_{\alpha^{2}}(z)(e^{\alpha}+e^{-\alpha})(e^{\alpha}-e^{-\alpha})$
$\displaystyle+\sum_{\beta\in
R_{+}\setminus\\{\alpha,2\alpha\\}}2m_{\beta}\langle\alpha,\beta\rangle
A_{\alpha\beta}(z)(e^{\alpha}+e^{-\alpha})(e^{\alpha}-e^{-\alpha})$
$\displaystyle+\frac{P_{1}(z,x)}{\langle\alpha,z\rangle}(e^{\alpha}-e^{-\alpha})^{2}+\text{l.o.t.}$

Note that for $\beta\neq\alpha$ and $\beta\neq 2\alpha$ we have
$A_{\alpha\beta}(z)=0$ at $\langle\alpha,z\rangle=0$. By identifying the
homogeneous terms of degree $N-2$ in (2.10) we get at
$\langle\alpha,z\rangle=0$ that

$\displaystyle
0=\left((m_{\alpha}+m_{2\alpha})+(m_{\alpha}+m_{2\alpha}-1)\right)\alpha^{2}A_{\alpha^{2}}(z)(e^{\alpha}-e^{-\alpha})(e^{\alpha}+e^{-\alpha})+\frac{P_{1}}{\langle\alpha,z\rangle}(e^{\alpha}-e^{-\alpha})^{2}.$
(2.11)

If $m_{\alpha}>2$, then $A_{\alpha^{2}}(z)=0$ at $\langle\alpha,z\rangle=0$,
so (2.11) gives divisibility of $\langle\alpha,z\rangle^{-1}P_{1}$ by
$\langle\alpha,z\rangle$. Continuing iteratively for $s=3,\dots,m_{\alpha}$
the process above, after the $s=m_{\alpha}$ step we get that at
$\langle\alpha,z\rangle=0$

$\displaystyle 0=$
$\displaystyle\left((m_{\alpha}+m_{2\alpha})+(m_{\alpha}+m_{2\alpha}-1)+\dots+(1+m_{2\alpha})\right)\alpha^{2}A_{\alpha^{m_{\alpha}}}(z)(e^{\alpha}-e^{-\alpha})^{m_{\alpha}-1}(e^{\alpha}+e^{-\alpha})$
(2.12)
$\displaystyle+\frac{P_{1}}{\langle\alpha,z\rangle^{m_{\alpha}-1}}(e^{\alpha}-e^{-\alpha})^{m_{\alpha}}.$

We have

$\displaystyle(m_{\alpha}+m_{2\alpha})+(m_{\alpha}+m_{2\alpha}-1)+\dots+(1+m_{2\alpha})$
$\displaystyle=m_{\alpha}m_{2\alpha}+(1+\dots+m_{\alpha})=\frac{m_{\alpha}(m_{\alpha}+2m_{2\alpha}+1)}{2}.$

Therefore (2.12) can be rearranged as

$\frac{P_{1}}{\langle\alpha,z\rangle^{m_{\alpha}-1}}=-\frac{m_{\alpha}(m_{\alpha}+2m_{2\alpha}+1)}{2}\alpha^{2}A_{\alpha^{m_{\alpha}}}(z)\coth\langle\alpha,x\rangle\textnormal{
at }\langle\alpha,z\rangle=0.$ (2.13)

If $m_{2\alpha}=0$ for all $\alpha\in R_{+}$ for which $\frac{\alpha}{2}\notin
R_{+}$, then for each such $\alpha$ there are no more conditions in Definition
2.1 to be satisfied, and $P_{1}$ given by (2.5) satisfies (2.13). This
completes the proof in this case, by uniqueness. Otherwise we continue and
consider condition (2.2). Note that the highest order term that we are
carrying forward is $A_{\alpha^{m_{\alpha}-1}}(z)$.

Assume $m_{2\alpha}>0$. Then $A_{\alpha^{m_{\alpha}}}(z)=0$ at
$\langle\alpha,z\rangle=0$, so (2.13) implies divisibility of
$\langle\alpha,z\rangle^{1-m_{\alpha}}P_{1}$ by $\langle\alpha,z\rangle$.
Consider the condition (2.2) from Lemma 2.2 with $t=1$. Note that

$A_{\alpha^{m_{\alpha}}}(z\pm 2\alpha)=A_{\alpha^{m_{\alpha}}}(z)\pm
m_{2\alpha}(2\alpha)^{2}A_{(2\alpha)\alpha^{m_{\alpha}}}+f(z)$

where the term $f(z)$ vanishes at $\langle\alpha,z\rangle=0$. Thus at this
step we obtain

$\displaystyle 0=$ $\displaystyle
m_{2\alpha}(2\alpha)^{2}A_{(2\alpha)\alpha^{m_{\alpha}}}(z)(e^{2\alpha}+e^{-2\alpha})(e^{\alpha}-e^{-\alpha})^{m_{\alpha}}$
(2.14)
$\displaystyle+\frac{m_{\alpha}(m_{\alpha}+2m_{2\alpha}+1)\alpha^{2}}{2}A_{\alpha^{m_{\alpha}+1}}(z)(e^{2\alpha}-e^{-2\alpha})(e^{\alpha}-e^{-\alpha})^{m_{\alpha}-1}(e^{\alpha}+e^{-\alpha})$
$\displaystyle+\frac{P_{1}}{\langle\alpha,z\rangle^{m_{\alpha}}}(e^{2\alpha}-e^{-2\alpha})(e^{\alpha}-e^{-\alpha})^{m_{\alpha}}$

at $\langle\alpha,z\rangle=0$. This can be rearranged as

$\frac{P_{1}}{\langle\alpha,z\rangle^{m_{\alpha}}}=-m_{2\alpha}(2\alpha)^{2}A_{(2\alpha)\alpha^{m_{\alpha}}}(z)\coth\langle
2\alpha,x\rangle-\frac{m_{\alpha}(m_{\alpha}+2m_{2\alpha}+1)\alpha^{2}}{2}A_{\alpha^{m_{\alpha}+1}}(z)\coth\langle\alpha,x\rangle.$
(2.15)

Assume $m_{2\alpha}=1$, then one can check that $P_{1}$ given by (2.5)
satisfies (2.15). Otherwise $m_{2\alpha}>1$ and we continue recursively. For
$t=2$, for example, to treat the highest order term we need to consider

$A_{\alpha^{m_{\alpha}+1}}(z\pm
2\alpha)=A_{\alpha^{m_{\alpha}+1}}(z)\pm(m_{2\alpha}-1)(2\alpha)^{2}A_{(2\alpha)\alpha^{m_{\alpha}+1}}+g(z)$

where the term $g(z)$ vanishes at $\langle\alpha,z\rangle=0$.

We thus iterate for $t=2,\dots,m_{2\alpha}$. One has to use that
$1+\dots+m_{2\alpha}=\frac{1}{2}m_{2\alpha}(m_{2\alpha}+1)$, and at the end we
get

$\displaystyle\frac{P_{1}}{\langle\alpha,z\rangle^{m_{\alpha}+m_{2\alpha}-1}}=$
$\displaystyle-\frac{m_{2\alpha}(m_{2\alpha}+1)(2\alpha)^{2}}{2}A_{(2\alpha)\alpha^{m_{\alpha}+m_{2\alpha}-1}}\coth\langle
2\alpha,x\rangle$ (2.16)
$\displaystyle-\frac{m_{\alpha}(m_{\alpha}+2m_{2\alpha}+1)\alpha^{2}}{2}A_{\alpha^{m_{\alpha}+m_{2\alpha}}}\coth\langle\alpha,x\rangle$

at $\langle\alpha,z\rangle=0$. It is straightforward to check that $P_{1}$
given by (2.5) satisfies (2.16) for all $\alpha\in R_{+}$ such that
$\frac{\alpha}{2}\notin R_{+}$.

If there existed some other $\widetilde{P}_{1}$ of degree $N-1$ satisfying
(2.16), then (2.16) would imply that $P_{1}-\widetilde{P}_{1}$ is divisible by
$\langle\alpha,z\rangle^{m_{\alpha}+m_{2\alpha}}$ for all $\alpha\in R_{+}$
with $\frac{\alpha}{2}\notin R_{+}$. But unless $P_{1}-\widetilde{P}_{1}=0$
this would mean that $P_{1}-\widetilde{P}_{1}$ has degree at least $N$ which
is not possible. This completes the proof. ∎

The next Propostion has a completely analogous proof to [13, Proposition 2],
it just uses Lemma 2.9 in place of [13, Lemma 2] (see also [4]).

######  Proposition 2.10.

With notations and assumptions as in Theorem 2.8, the element $p(z)=z^{2}$ of
$\mathcal{R}$ corresponds to the differential operator

$L_{z^{2}}=\Delta-\sum_{\gamma\in
R_{+}}\frac{m_{\gamma}(m_{\gamma}+2m_{2\gamma}+1)\gamma^{2}}{\sinh^{2}\langle\gamma,x\rangle},$

which coincides (up to sign) with the generalised CMS operator (1.3) for
$\mathcal{A}=(R_{+},m)$.

This then implies integrability of the Hamiltonian and provides an integral of
motion $L_{p}$ for each $p(z)\in\mathcal{R}$. The following statement is one
of the main results of this paper. The proof will be presented below.

######  Theorem 2.11.

There exists a BA function for the configuration $AG_{2}$.

##  3 Generalised Macdonald–Ruijsenaars operators

In Section 4, we will utilise a method for explicit construction of BA
functions which was proposed by Chalykh [8] (see also [13] for further
examples where this method is applied, and [5, 7] for the differential case).
The construction uses (generalised) Macdonald–Ruijsenaars difference
operators. The key element of the method is the preservation of a space of
quasi-invariant analytic functions under the action of the
Macdonald–Ruijsenaars operators. In this Section, we find sufficient
conditions for a sufficiently general difference operator to preserve such a
ring of quasi-invariants.

Let $W=\langle s_{\alpha}\colon\alpha\in R\rangle$, where $s_{\alpha}$ is the
orthogonal reflection about the hyperplane $\langle\alpha,x\rangle=0$. We
assume now that the collection $R$ is $W$-invariant, that is $w(R)=R$ for all
$w\in W$, and that the multiplicity map $m$ is $W$-invariant, too. Let
$u^{\vee}=2u/\langle u,u\rangle$ for any $u\in\mathbb{C}^{n}$ such that
$\langle u,u\rangle\neq 0$.

Let $\mathcal{R}^{a}$ be the ring of analytic functions $p(z)$ such that

$p(z+t\alpha)=p(z-t\alpha)\textnormal{ at
}\langle\alpha,z\rangle=0\textnormal{ for }t\in A_{\alpha}$

for all $\alpha\in R^{r}_{+}$, where $A_{\alpha}\subset\mathbb{N}$ specifies
the axiomatics that one wants to consider. For instance, it can be
$A_{\alpha}=\\{1,2,3,\dots,m_{\alpha},m_{\alpha}+2,m_{\alpha}+4,\dots,m_{\alpha}+2m_{2\alpha}\\}$.
We assume that $A_{|w\alpha|}=A_{\alpha}$ for all $w\in W$, $\alpha\in
R^{r}_{+}$, where $|w\alpha|=w\alpha$ if $w\alpha\in R^{r}_{+}$ and
$|w\alpha|=-w\alpha$ if $w\alpha\in-R^{r}_{+}$. For $\alpha\in R$, we let
$\operatorname{sgn}\alpha=1$ if $\alpha\in R_{+}$ and
$\operatorname{sgn}\alpha=-1$ if $\alpha\in-R_{+}$.

Let $S\subset\mathbb{C}^{n}\setminus\\{0\\}$ be a $W$-invariant finite
collection of vectors. Let $z\in\mathbb{C}^{n}$ and for any
$\gamma\in\mathbb{C}^{n}$ let $T_{\gamma}$ be the shift operator that acts on
functions $f(z)$ by $T_{\gamma}f(z)=f(z+\gamma)$. We are interested in
difference operators $D$ of the form

$D=\sum_{\tau\in S}a_{\tau}(z)(T_{\tau}-1),$ (3.1)

where $a_{\tau}$ are rational functions with the following three properties:

  1. ($D_{1}$)

$\deg a_{\tau}=0$.

  2. ($D_{2}$)

$a_{\tau}(z)$ has a simple pole at $\langle\alpha,z\rangle=c\alpha^{2}$ for
$\alpha\in R^{r}_{+}$, $c\in\mathbb{C}$ if and only if
$\lambda=s_{\alpha}(\tau)-2c\alpha\in S\cup\\{0\\}$ and

$|c+\langle\tau,(2\alpha)^{\vee}\rangle|=|\langle\tau+c\alpha,\alpha\rangle/\alpha^{2}|\in
A_{\alpha}.$

There are no other singularities in $a_{\tau}$. Denote the set of all such
$(\alpha,c)$ for this $\tau$ by $S_{\tau}$.

  3. ($D_{3}$)

$wa_{\tau}=a_{w\tau}$ for all $w\in W$.

The condition ($D_{2}$) implies that if $a_{\tau}$ has a singularity
$\langle\alpha,z\rangle=c\alpha^{2}$, then for any such $z$ the vectors
$z+\tau$ and $z+\lambda$ are of the form $z+\tau=\widetilde{z}+t\alpha$ and
$z+\lambda=\widetilde{z}-t\alpha$ for some $\widetilde{z}$ with
$\langle\alpha,\widetilde{z}\rangle=0$ and
$|t|=|c+\langle\tau,(2\alpha^{\vee})\rangle|\in A_{\alpha}$. We note also that
$\lambda\neq\tau$ since $0\notin A_{\alpha}$.

Note that if $a_{\tau}$ has a singularity $\langle\alpha,z\rangle=c\alpha^{2}$
and the corresponding $\lambda\neq 0$, then by condition $(D_{2})$
$a_{\lambda}(z)$ necessarily also has a singularity at
$\langle\alpha,z\rangle=c\alpha^{2}$, since
$s_{\alpha}(\lambda)-2c\alpha=\tau\in S$ and
$\langle\lambda+c\alpha,\alpha\rangle=\langle
s_{\alpha}(\tau+c\alpha),\alpha\rangle=-\langle\tau+c\alpha,\alpha\rangle$. In
other words, by condition $(D_{2})$ we have $(\alpha,c)\in S_{\tau}$ if and
only if $(\alpha,c)\in S_{\lambda}$ provided that both $\tau,\lambda\neq 0$.
We additionally observe the following.

######  Lemma 3.1.

For any $w\in W$, $(\alpha,c)\in S_{\tau}$ if and only if
$(|w\alpha|,\operatorname{sgn}(w\alpha)c)\in S_{w\tau}$.

###### Proof.

Let $\varepsilon=\operatorname{sgn}(w\alpha)$. Since
$s_{|w\alpha|}=ws_{\alpha}w^{-1}$ and $\varepsilon|w\alpha|=w\alpha$, we get
that $s_{|w\alpha|}(w\tau)-2\varepsilon
c|w\alpha|=w(s_{\alpha}(\tau)-2c\alpha)$ belongs to $S\cup\\{0\\}$ if and only
if $s_{\alpha}(\tau)-2c\alpha\in S\cup\\{0\\}$, by $W$-invariance of $S$.
Furthermore, $|\langle w\tau+\varepsilon
c|w\alpha|,|w\alpha|\rangle/|w\alpha|^{2}|=|\langle\tau+c\alpha,\alpha\rangle/\alpha^{2}|$,
and $A_{\alpha}=A_{|w\alpha|}$, by assumption. ∎

More explicitly, we are looking at operators of the form

$D=\sum_{\tau\in S}P_{\tau}(z)\left(\prod_{(\alpha,c)\in
S_{\tau}}\left(\langle\alpha,z\rangle-c\alpha^{2}\right)^{-1}\right)(T_{\tau}-1)$

for some polynomials $P_{\tau}(z)$ of degree $|S_{\tau}|$ so that $\deg
a_{\tau}=0$ and such that the condition ($D_{3}$) holds. We want to find some
sufficient conditions that would ensure that $D$ preserves the ring
$\mathcal{R}^{a}$.

######  Theorem 3.2.

Suppose the operator (3.1) satisfies conditions ($D_{2}$) and ($D_{3}$). Then
for any $\alpha\in R^{r}_{+}$ and arbitrary $p(z)\in\mathcal{R}^{a}$, we have
the following two properties.

  1. 1.

$Dp(z)$ is non-singular at $\langle\alpha,z\rangle=0$. Moreover, for any
$c\neq 0$, provided that for all $\tau\in S$ such that $(\alpha,c)\in
S_{\tau}$ and such that $\lambda=s_{\alpha}(\tau)-2c\alpha\neq 0$ we have

$\operatorname{res}_{\langle\alpha,z\rangle=c\alpha^{2}}(a_{\tau}+a_{\lambda})=0,$

then $Dp(z)$ is non-singular at $\langle\alpha,z\rangle=c\alpha^{2}$.

  2. 2.

Suppose, in addition to assumptions of part 1, that for all $\tau\in S$ and
any $t\in A_{\alpha}$, the following is satisfied whenever
$|t+\langle\tau,(2\alpha)^{\vee}\rangle|\notin A_{\alpha}\cup\\{0\\}$:

    1. (a)

$a_{\tau}(z+t\alpha)=0\textnormal{ at }\langle\alpha,z\rangle=0$
(equivalently, $P_{\tau}(z)$ has a factor of
$\langle\alpha,z\rangle-t\alpha^{2}$),

or

    1. (b)

$\lambda=s_{\alpha}(\tau)-2t\alpha\in S$ and
$a_{\lambda}(z+t\alpha)=a_{\tau}(z+t\alpha)$ at $\langle\alpha,z\rangle=0$.

Then $Dp(z+t\alpha)=Dp(z-t\alpha)\textnormal{ at
}\langle\alpha,z\rangle=0\textnormal{ for all }t\in A_{\alpha}.$

###### Proof.

1\. Let $c\in\mathbb{C}$. We want to show that the residue at
$\langle\alpha,z\rangle=c\alpha^{2}$ of $Dp(z)$ is zero. Take any $\tau\in S$
such that $(\alpha,c)\in S_{\tau}$. Write $\tau+c\alpha=a\alpha+\gamma$ where
$\langle\gamma,\alpha\rangle=0$ and
$a=\langle\tau+c\alpha,\alpha\rangle/\alpha^{2}$. Let
$\lambda=s_{\alpha}(\tau)-2c\alpha$. Then
$\lambda+c\alpha=s_{\alpha}(\tau+c\alpha)=-a\alpha+\gamma$. At
$\langle\alpha,z\rangle=c\alpha^{2}$ we thus have

$\displaystyle p(z+\tau)$
$\displaystyle=p(z-c\alpha+c\alpha+\tau)=p((z-c\alpha+\gamma)+a\alpha)=p((z-c\alpha+\gamma)-a\alpha)=p(z+\lambda)$

as $\langle z-c\alpha+\gamma,\alpha\rangle=0$ and $|a|\in A_{\alpha}$ by
assumption ($D_{2}$). So, if $\lambda=0$ then the simple pole at
$\langle\alpha,z\rangle=c\alpha^{2}$ present in $a_{\tau}(z)$ is cancelled by
$(T_{\tau}-1)[p(z)]=p(z+\tau)-p(z)$. And if $\lambda\neq 0$, then the sum

$a_{\tau}(z)(p(z+\tau)-p(z))+a_{\lambda}(z)(p(z+\lambda)-p(z))$

contributes zero to the residue provided that the residue of
$a_{\tau}+a_{\lambda}$ is zero. For $c\neq 0$ the latter is satisfied by
assumption. In the case of $c=0$, we have $\lambda=s_{\alpha}(\tau)$, hence
$a_{\tau}(s_{\alpha}(z))=a_{\lambda}(z)$ by the symmetry ($D_{3}$) of the
operator, so we get

$\lim_{\langle\alpha,z\rangle\to 0}\langle\alpha,z\rangle
a_{\tau}(z)=\lim_{\langle\alpha,z\rangle\to
0}\langle\alpha,s_{\alpha}(z)\rangle
a_{\tau}(s_{\alpha}(z))=-\lim_{\langle\alpha,z\rangle\to
0}\langle\alpha,z\rangle a_{\lambda}(z),$

that is, the residue of $a_{\tau}(z)$ at $\langle\alpha,z\rangle=0$ is minus
that of $a_{\lambda}(z)$, as needed.

2\. Fix $t\in A_{\alpha}$. By symmetry ($D_{3}$) of the operator, at
$\langle\alpha,z\rangle=0$ we have
$a_{\mu}(z+s\alpha)=a_{s_{\alpha}(\mu)}(z-s\alpha)$ for all generic
$s\in\mathbb{C}$ and generic $z\in\mathbb{C}^{2}$ with
$\langle\alpha,z\rangle=0$, $\mu\in S$. Using that $s_{\alpha}(S)=S$ we can
thus write $Dp(z+t\alpha)-Dp(z-t\alpha)$ as

$\lim_{s\to t}\sum_{\mu\in
S}a_{\mu}(z+s\alpha)\bigg{(}p(z+s\alpha+\mu)-p(z-s\alpha+s_{\alpha}(\mu))-p(z+s\alpha)+p(z-s\alpha)\bigg{)}.$
(3.2)

Firstly, let us consider any $\tau\in S$ for which $a_{\tau}(z+t\alpha)$ is
non-singular at $\langle\alpha,z\rangle=0$ (for generic $z$). Then the
corresponding $\mu=\tau$ term in the sum (3.2) can be simplified to

$a_{\tau}(z+t\alpha)\left(p(z+t\alpha+\tau)-p(z-t\alpha+s_{\alpha}(\tau)\right)$
(3.3)

as $p(z)\in\mathcal{R}^{a}$. Let $\tau=b\alpha+\delta$, where
$\langle\delta,\alpha\rangle=0$ and $b=\langle\tau,(2\alpha)^{\vee}\rangle$.
Then $s_{\alpha}(\tau)=-b\alpha+\delta$. Thus

$p(z+t\alpha+\tau)-p(z-t\alpha+s_{\alpha}(\tau))=p(z+\delta+(t+b)\alpha)-p(z+\delta-(t+b)\alpha),$
(3.4)

where $\langle\alpha,z+\delta\rangle=0$. Hence if $|t+b|\in
A_{\alpha}\cup\\{0\\}$, then (3.4) equals zero, and the whole term (3.3)
vanishes. Else, we have by assumption two possibilities. If
$a_{\tau}(z+t\alpha)=0$ at $\langle\alpha,z\rangle=0$ then (3.3) vanishes, and
if $a_{\tau}(z+t\alpha)\neq 0$ then $\lambda=s_{\alpha}(\tau)-2t\alpha\in
S\setminus\\{\tau\\}$ and $a_{\lambda}(z+t\alpha)=a_{\tau}(z+t\alpha)$. (Note
that $|t+\langle\tau,(2\alpha)^{\vee}\rangle|\notin A_{\alpha}\cup\\{0\\}$
implies that $\lambda\neq 0,\tau$, and due to ($D_{2}$) also that
$a_{\lambda}(z+t\alpha)$ and $a_{\tau}(z+t\alpha)$ are well-defined at
$\langle\alpha,z\rangle=0$ for generic $z$). In the latter case, the term
corresponding to $\mu=\lambda$ in the sum (3.2) can be simplified to

$a_{\lambda}(z+t\alpha)\left(p(z+t\alpha+\lambda)-p(z-t\alpha+s_{\alpha}(\lambda)\right)=a_{\tau}(z+t\alpha)\left(p(z-t\alpha+s_{\alpha}(\tau))-p(z+t\alpha+\tau)\right),$

which is the negative of (3.3), hence the terms corresponding to $\mu=\tau$
and $\mu=\lambda$ in (3.2) cancel out.

Secondly, let us consider any $\tau\in S$ for which $a_{\tau}(z+t\alpha)$ is
singular at $\langle\alpha,z\rangle=0$. Equivalently,
$a_{\tau}(\widetilde{z})$ is singular at
$\langle\alpha,\widetilde{z}\rangle=t\alpha^{2}$. Hence $(\alpha,t)\in
S_{\tau}$, and $|t+b|\in A_{\alpha}$ by assumption ($D_{2}$). It follows that
the expression (3.4) vanishes. We can restate this as

$p(z+s\alpha+\tau)-p(z-s\alpha+s_{\alpha}(\tau))=(s-t)q(s)$

for some analytic function $q(s)$ ($s\in\mathbb{C}$). Similarly,
$p(z+t\alpha)=p(z-t\alpha)$ can be restated as

$p(z+s\alpha)-p(z-s\alpha)=(s-t)r(s)$ (3.5)

for some analytic function $r(s)$. Moreover, we also have

$p(z+s\alpha+\lambda)-p(z-s\alpha+s_{\alpha}(\lambda))=p(z-(2t-s)\alpha+s_{\alpha}(\tau))-p(z+(2t-s)\alpha+\tau)=(s-t)q(2t-s).$
(3.6)

Suppose firstly that $\lambda\neq 0$. Then in the sum (3.2) the two terms
corresponding to $\mu=\tau$ and $\mu=\lambda$ cancel out. Indeed, they equal

$\displaystyle\lim_{s\to
t}a_{\tau}(z+s\alpha)(s-t)(q(s)-r(s))+a_{\lambda}(z+s\alpha)(s-t)(q(2t-s)-r(s))$
$\displaystyle\quad=\operatorname{res}_{\langle
z,\alpha\rangle=t\alpha^{2}}(a_{\tau}+a_{\lambda})(q(t)-r(t))=0,$

because $\operatorname{res}_{\langle
z,\alpha\rangle=t\alpha^{2}}(a_{\tau}+a_{\lambda})=0$ by assumptions of part 1
with $c=t$. Suppose now that $\lambda=0$, then $r(s)=q(2t-s)$ by equalities
(3.5) and (3.6). Therefore the term corresponding to $\mu=\tau$ in the sum
(3.2) equals $\lim_{s\to
t}a_{\tau}(z+s\alpha)(s-t)\left(q(s)-q(2t-s)\right)=0.$

It follows that the sum (3.2) vanishes, as required. ∎

It follows that if the assumptions of both parts 1 and 2 of Theorem 3.2 are
satisfied for all $\alpha\in R^{r}_{+}$, then $D$ preserves the ring
$\mathcal{R}^{a}$, that is $Dp(z)\in\mathcal{R}^{a}$ if
$p(z)\in\mathcal{R}^{a}$.

Additionally, we can use the symmetry assumption ($D_{3}$) to reduce the
number of conditions that we have to consider in Theorem 3.2. The following
statements take place.

######  Lemma 3.3.

Suppose that condition ($D_{3}$) holds. If $a_{\tau}+a_{\lambda}$ has zero
residue at $\langle\alpha,z\rangle=c\alpha^{2}$, then $a_{w\tau}+a_{w\lambda}$
has zero residue at
$\langle|w\alpha|,z\rangle=\operatorname{sgn}(w\alpha)c|w\alpha|^{2}$ (or,
equivalently, at $\langle\alpha,w^{-1}z\rangle=c\alpha^{2}$) for any $w\in W$.

###### Proof.

By the property ($D_{3}$) we have
$a_{w\tau}(z)+a_{w\lambda}(z)=a_{\tau}(w^{-1}z)+a_{\lambda}(w^{-1}z)$,
therefore

$\displaystyle\operatorname{res}_{\langle\alpha,w^{-1}z\rangle=c\alpha^{2}}(a_{w\tau}(z)+a_{w\lambda}(z))=\lim_{\langle\alpha,w^{-1}z\rangle\to
c\alpha^{2}}(\langle\alpha,w^{-1}z\rangle-c\alpha^{2})(a_{w\tau}(z)+a_{w\lambda}(z))$
$\displaystyle=\lim_{\langle\alpha,w^{-1}z\rangle\to
c\alpha^{2}}(\langle\alpha,w^{-1}z\rangle-c\alpha^{2})(a_{\tau}(w^{-1}z)+a_{\lambda}(w^{-1}z))=\operatorname{res}_{\langle\alpha,\widetilde{z}\rangle=c\alpha^{2}}(a_{\tau}(\widetilde{z})+a_{\lambda}(\widetilde{z}))=0.\qed$

By combining Lemmas 3.1, 3.3 and Theorem 3.2, we obtain the following.

######  Corollary 3.4.

Suppose the operator (3.1) satisfies conditions ($D_{2}$) and ($D_{3}$). If
the assumptions of part 1 of Theorem 3.2 are satisfied for some $\alpha\in
R^{r}_{+}$ then $Dp(z)$ is non-singular at
$\langle|w\alpha|,z\rangle=\operatorname{sgn}(w\alpha)c|w\alpha|^{2}$ for all
$w\in W$.

###### Proof.

By Theorem 3.2 part 1, it suffices to check that for any $\widetilde{\tau}\in
S$ and $c\neq 0$ such that $(|w\alpha|,\operatorname{sgn}(w\alpha)c)\in
S_{\widetilde{\tau}}$ and such that
$\widetilde{\lambda}=s_{|w\alpha|}(\widetilde{\tau})-2cw\alpha\neq 0$ we have
that the residue of $a_{\widetilde{\tau}}+a_{\widetilde{\lambda}}$ at
$\langle|w\alpha|,z\rangle=\operatorname{sgn}(w\alpha)c|w\alpha|^{2}$ is zero.
Since $S$ is invariant, we can write $\widetilde{\tau}=w\tau$ for some
$\tau\in S$. Lemma 3.1 then gives $(\alpha,c)\in S_{\tau}$. Note that
$\widetilde{\lambda}=w\lambda$ for $\lambda=s_{\alpha}(\tau)-2c\alpha$ (in
particular, $\lambda\neq 0$ as $\widetilde{\lambda}\neq 0$). By assumption,
part 1 of Theorem 3.2 holds for this $\alpha$ and this $c$, that is
$\operatorname{res}_{\langle\alpha,z\rangle=c\alpha^{2}}(a_{\tau}+a_{\lambda})=0$.
Lemma 3.3 now gives what we need. ∎

######  Lemma 3.5.

Suppose the operator (3.1) satisfies conditions ($D_{2}$) and ($D_{3}$). If
the assumptions of parts 1 and 2 of Theorem 3.2 are satisfied for some
$\alpha\in R^{r}_{+}$, then the assumptions of part 2 are also satisfied for
$w\alpha$ for all $w\in W$ such that $w\alpha\in R^{r}_{+}$.

###### Proof.

Note that $A_{w\alpha}=A_{\alpha}$. Thus we need to prove that whenever for
some $t\in A_{\alpha}$ and $\widetilde{\tau}\in S$ we have
$|t+\langle\widetilde{\tau},(2w\alpha)^{\vee}\rangle|\notin
A_{\alpha}\cup\\{0\\}$, then either $a_{\widetilde{\tau}}(z+tw\alpha)=0$ at
$\langle w\alpha,z\rangle=0$, or else
$\widetilde{\lambda}=s_{\alpha}(\widetilde{\tau})-2tw\alpha\in S$ and
$a_{\widetilde{\lambda}}(z+tw\alpha)=a_{\widetilde{\tau}}(z+tw\alpha)$ at
$\langle w\alpha,z\rangle=0$.

So suppose that $|t+\langle\widetilde{\tau},(2w\alpha)^{\vee}\rangle|\notin
A_{\alpha}\cup\\{0\\}$. Since $S$ is invariant, we can write
$\widetilde{\tau}=w\tau$ for some $\tau\in S$. Note that then
$\widetilde{\lambda}=w(s_{\alpha}(\tau)-2t\alpha)=w\lambda$. Note also that
$(2w\alpha)^{\vee}=w(2\alpha)^{\vee}$. Therefore
$|t+\langle\tau,(2\alpha)^{\vee}\rangle|=|t+\langle\widetilde{\tau},(2w\alpha)^{\vee}\rangle|\notin
A_{\alpha}\cup\\{0\\}$. By assumption, part 2 of Theorem 3.2 holds for this
$\alpha$. Suppose first that $a_{\tau}(\widetilde{z}+t\alpha)=0$ at
$\langle\alpha,\widetilde{z}\rangle=0$. By symmetry ($D_{3}$) of the operator,
at $\langle w\alpha,z\rangle=0$ (or, equivalently at
$\langle\alpha,w^{-1}z\rangle=0$) we thus get
$a_{\widetilde{\tau}}(z+tw\alpha)=a_{\tau}(w^{-1}z+t\alpha)=0$, as required.
Otherwise, $\lambda\in S$ so $\widetilde{\lambda}=w\lambda\in S$ by
invariance, and at $\langle w\alpha,z\rangle=0$ property ($D_{3}$) gives
$a_{\widetilde{\lambda}}(z+tw\alpha)-a_{\widetilde{\tau}}(z+tw\alpha)=a_{\lambda}(w^{-1}z+t\alpha)-a_{\tau}(w^{-1}z+t\alpha)=0$,
as needed. ∎

######  Remark 3.6.

Let $\alpha\in R^{r}_{+}$. Suppose $w\in W$ satisfies $w\alpha=\alpha$. Then,
for any $\tau\in S$, in part 2 of Theorem 3.2 it suffices to check the given
conditions for either $\tau$ or $w\tau$, as the other one then follows.
Indeed, for any $t\in A_{\alpha}$ we have $|t+\langle
w\tau,(2\alpha)^{\vee}\rangle|=|t+\langle\tau,(2\alpha)^{\vee}\rangle|$. Also
$s_{\alpha}(w\tau)-2t\alpha=w\lambda$, and at $\langle\alpha,z\rangle=0$
(equivalently, at $\langle\alpha,w^{-1}z\rangle=0$) by the symmetry ($D_{3}$)
we have $a_{w\tau}(z+t\alpha)=a_{\tau}(w^{-1}z+t\alpha)$ and in case (b) also
$a_{w\lambda}(z+t\alpha)=a_{\lambda}(w^{-1}z+t\alpha)$.

##  4 Bispectral dual difference operator for $AG_{2}$

In this Section, we find a difference operator ${\mathcal{D}}_{1}$ satisfying
the conditions of Theorem 3.2 for the configuration $AG_{2}$. We define a
difference operator acting in the variable $z\in\mathbb{C}^{2}$ of the form

${\mathcal{D}}_{1}=\sum_{\tau\colon\frac{1}{2}\tau\in
G_{2}}a_{\tau}(z)(T_{\tau}-1).$ (4.1)

Let $W$ be the Weyl group of the root system $G_{2}$. For
$\tau=2\varepsilon\beta_{j},\varepsilon\in\\{\pm 1\\}$, we define

$\displaystyle a_{2\varepsilon\beta_{j}}(z)$
$\displaystyle=3\prod_{\begin{subarray}{c}\gamma\in W\beta_{1}\\\ \langle
2\varepsilon\beta_{j},(2\gamma)^{\vee}\rangle=1\end{subarray}}\bigg{(}1-\frac{(3m+2)\gamma^{2}}{\langle\gamma,z\rangle}\bigg{)}\bigg{(}1+\frac{3m\gamma^{2}}{\langle\gamma,z\rangle+2\gamma^{2}}\bigg{)}\bigg{(}1-\frac{(3m-1)\gamma^{2}}{\langle\gamma,z\rangle-\gamma^{2}}\bigg{)}$
(4.2) $\displaystyle\times\prod_{\begin{subarray}{c}\gamma\in W\alpha_{1}\\\
\langle
2\varepsilon\beta_{j},(2\gamma)^{\vee}\rangle=1\end{subarray}}\bigg{(}1-\frac{m\gamma^{2}}{\langle\gamma,z\rangle}\bigg{)}\times\bigg{(}1-\frac{(3m+2)\beta_{j}^{2}}{\langle\varepsilon\beta_{j},z\rangle}\bigg{)}\bigg{(}1-\frac{3m\beta_{j}^{2}}{\langle\varepsilon\beta_{j},z\rangle+\beta_{j}^{2}}\bigg{)}.$

For $\tau=2\varepsilon\alpha_{j}$, we define

$\displaystyle a_{2\varepsilon\alpha_{j}}(z)$
$\displaystyle=\prod_{\begin{subarray}{c}\gamma\in W\beta_{1}\\\ \langle
2\varepsilon\alpha_{j},(2\gamma)^{\vee}\rangle=3\end{subarray}}\bigg{(}1-\frac{(3m+2)\gamma^{2}}{\langle\gamma,z\rangle}\bigg{)}\bigg{(}1-\frac{(3m+1)\gamma^{2}}{\langle\gamma,z\rangle+\gamma^{2}}\bigg{)}\bigg{(}1-\frac{3m\gamma^{2}}{\langle\gamma,z\rangle+2\gamma^{2}}\bigg{)}$
(4.3) $\displaystyle\times\prod_{\begin{subarray}{c}\gamma\in W\alpha_{1}\\\
\langle
2\varepsilon\alpha_{j},(2\gamma)^{\vee}\rangle=1\end{subarray}}\bigg{(}1-\frac{m\gamma^{2}}{\langle\gamma,z\rangle}\bigg{)}\times\bigg{(}1-\frac{m\alpha_{j}^{2}}{\langle\varepsilon\alpha_{j},z\rangle}\bigg{)}\bigg{(}1-\frac{m\alpha_{j}^{2}}{\langle\varepsilon\alpha_{j},z\rangle+\alpha_{j}^{2}}\bigg{)}.$

The following Lemma shows that these functions $a_{\tau}(z)$ have $G_{2}$
symmetry.

######  Lemma 4.1.

Let $a_{\tau}(z)$ be defined as in (4.2) and (4.3). Then for all $w\in W$ we
have $wa_{\tau}=a_{w\tau}$.

###### Proof.

For any $w\in W$ we have $w(W\alpha_{1})=W\alpha_{1}$,
$w(W\beta_{1})=W\beta_{1}$, and $\langle
w\tau,w\gamma\rangle=\langle\tau,\gamma\rangle$ for all
$\gamma,\tau\in\mathbb{C}^{2}$. The multiplicities are invariant, too. The
statement follows. ∎

Define the ring $\mathcal{R}^{a}_{AG_{2}}$ of analytic functions $p(z)$
satisfying conditions

$\displaystyle p(z+s\alpha_{j})=p(z-s\alpha_{j})\textnormal{ at
}\langle\alpha_{j},z\rangle=0,s=1,2,\dots,m,$ (4.4) $\displaystyle
p(z+s\beta_{j})=p(z-s\beta_{j})\textnormal{ at
}\langle\beta_{j},z\rangle=0,s=1,2,\dots,3m,3m+2$

for all $j=1,2,3$.

######  Theorem 4.2.

The operator (4.1) preserves the ring $\mathcal{R}^{a}_{AG_{2}}$.

###### Proof.

One can check that this operator has the property ($D_{2}$) where $S=2G_{2}$.
Let $p(z)\in\mathcal{R}^{a}_{AG_{2}}$ be arbitrary. Without loss of
generality, we put $\omega=\sqrt{2}$. We introduce new coordinates $(A,B)$ on
$\mathbb{C}^{2}$ given by $A=\langle\alpha_{1},z\rangle$ and
$B=\langle\beta_{1},z\rangle$.

If $B=4$ (equivalently, $\langle\beta_{1},z\rangle=2\beta_{1}^{2}$), then
$\langle\beta_{2},z\rangle=2+\frac{1}{2}A$,
$\langle\beta_{3},z\rangle=-2+\frac{1}{2}A$,
$\langle\alpha_{2},z\rangle=-6+\frac{1}{2}A$ and
$\langle\alpha_{3},z\rangle=6+\frac{1}{2}A$. The only terms singular at $B=4$
are $a_{-2\beta_{2}}$, $a_{-2\alpha_{3}}$, $a_{2\beta_{3}}$ and
$a_{2\alpha_{2}}$. Note that
$s_{\beta_{1}}(-2\beta_{2})-4\beta_{1}=-2\alpha_{3}$, and we compute that
$\operatorname{res}_{B=4}(a_{-2\beta_{2}})=-\operatorname{res}_{B=4}(a_{-2\alpha_{3}})$
equals

$\displaystyle-\tfrac{3m(3m+2)(3m+4)(A-12-12m)(A+6m)(A-4+12m)(A+12m)(A+4+12m)(A+12+12m)^{2}}{(A-12)(A-4)A^{3}(A+4)(A+12)}.$

As $s_{\alpha_{1}}(-2\beta_{2})=2\beta_{3}$ and
$s_{\alpha_{1}}(-2\alpha_{3})=2\alpha_{2}$, by Lemma 3.3 with
$w=s_{\alpha_{1}}$ we get that $a_{2\beta_{3}}+a_{2\alpha_{2}}$ has zero
residue at $B=4$, too. By Theorem 3.2 part 1, there is thus no singularity at
$B=4$ in ${\mathcal{D}}_{1}p(z)$.

If $B=2$ (equivalently, $\langle\beta_{1},z\rangle=\beta_{1}^{2}$), then
$\langle\beta_{2},z\rangle=1+\frac{1}{2}A$,
$\langle\beta_{3},z\rangle=-1+\frac{1}{2}A$,
$\langle\alpha_{2},z\rangle=-3+\frac{1}{2}A$ and
$\langle\alpha_{3},z\rangle=3+\frac{1}{2}A$. The only $\tau\in 2G_{2}$ for
which $a_{\tau}$ is singular at $B=2$ and for which the corresponding
$\lambda=s_{\beta_{1}}(\tau)-2\beta_{1}\neq 0$ are
$\tau=2\beta_{2},2\alpha_{2},-2\beta_{3},-2\alpha_{3}$. Note that
$s_{\beta_{1}}(2\beta_{2})-2\beta_{1}=2\alpha_{2}$, and we compute that
$\operatorname{res}_{B=2}(a_{2\beta_{2}})=-\operatorname{res}_{B=2}(a_{2\alpha_{2}})$
equals

$\displaystyle\tfrac{6(m+1)(3m-1)(3m+1)(A-10-12m)(A-6-12m)(A-2-12m)(A+6-12m)^{2}(A-6m)(A+6+12m)}{(A-6)(A-2)A(A+2)(A+6)^{3}}.$

As $s_{\alpha_{1}}(2\beta_{2})=-2\beta_{3}$ and
$s_{\alpha_{1}}(2\alpha_{2})=-2\alpha_{3}$, by Lemma 3.3 we get that
$a_{-2\beta_{3}}+a_{-2\alpha_{3}}$ has zero residue at $B=2$, too. By Theorem
3.2 part 1, there is thus no singularity at $B=2$ in ${\mathcal{D}}_{1}p(z)$,
nor at $B=0$.

It follows from the above analysis and from the form of the coefficient
functions (4.2) and (4.3) that there are no singularities in
${\mathcal{D}}_{1}p(z)$ at $B=c$ for all $c\geq 0$. By Corollary 3.4, there is
also no singularity in ${\mathcal{D}}_{1}p(z)$ at
$\langle\beta_{i},z\rangle=c$ for all $i=1,2,3$ and all $c\in\mathbb{C}$.

The only singularity at $A=\mathrm{const}>0$ present in the coefficients
$a_{\tau}$ for some $\tau$ is at $A=6$ (equivalently,
$\langle\alpha_{1},z\rangle=\alpha_{1}^{2}$) when $\tau=-2\alpha_{1}$. This
singularity cancels in ${\mathcal{D}}_{1}p(z)$ by Theorem 3.2 part 1, since
the corresponding $\lambda=s_{\alpha_{1}}(-2\alpha_{1})-2\alpha_{1}=0$. By
Corollary 3.4, there is also no singularity in ${\mathcal{D}}_{1}p(z)$ at
$\langle\alpha_{i},z\rangle=c$ for all $i=1,2,3$ and for all $c\in\mathbb{R}$.
This completes the proof that ${\mathcal{D}}_{1}p(z)$ is analytic.

Let us now show ${\mathcal{D}}_{1}p(z)$ satisfies the axiomatics of
$\mathcal{R}^{a}_{AG_{2}}$. We have $A_{\beta_{i}}=\\{1,2,3,\dots,3m,3m+2\\}$
and $A_{\alpha_{i}}=\\{1,2,\dots,m\\}$ ($i=1,2,3$). Let us show firstly that
${\mathcal{D}}_{1}p(z+t\beta_{1})={\mathcal{D}}_{1}p(z-t\beta_{1})$ at
$\langle\beta_{1},z\rangle=0$ for all $t\in A_{\beta_{1}}$. To do so we will
check condition 2 in Theorem 3.2 for all $\tau\in 2G_{2}$.

Note that $(2\beta_{1})^{\vee}=\frac{1}{2}\beta_{1}$. Let $\tau=2\beta_{1}$.
Then $|t+\langle\tau,(2\beta_{1})^{\vee}\rangle|=t+2$ which does not belong to
$A_{\beta_{1}}\cup\\{0\\}$ if and only if $t=3m-1$ or $t=3m+2$. But

$a_{2\beta_{1}}(z+(3m+2)\beta_{1})=a_{2\beta_{1}}(z+(3m-1)\beta_{1})=0\textnormal{
at }\langle\beta_{1},z\rangle=0$

because $a_{2\beta_{1}}(z)$ contains the factors
$(1-\frac{(3m+2)\beta_{1}^{2}}{\langle\beta_{1},z\rangle})(1-\frac{3m\beta_{1}^{2}}{\langle\beta_{1},z\rangle+\beta_{1}^{2}})$.

Let now $\tau=-2\beta_{1}$. Then
$|t+\langle\tau,(2\beta_{1})^{\vee}\rangle|=|t-2|\in A_{\beta_{1}}\cup\\{0\\}$
for all $t\in A_{\beta_{1}}$, as needed.

Let now $\tau=2\beta_{2}$. Then
$|t+\langle\tau,(2\beta_{1})^{\vee}\rangle|=t+1$ which does not belong to
$A_{\beta_{1}}\cup\\{0\\}$ if and only if $t=3m$ or $t=3m+2$. But

$a_{2\beta_{2}}(z+(3m+2)\beta_{1})=a_{2\beta_{2}}(z+3m\beta_{1})=0\textnormal{
at }\langle\beta_{1},z\rangle=0$

because $a_{2\beta_{2}}(z)$ contains the factors
$(1-\frac{(3m+2)\beta_{1}^{2}}{\langle\beta_{1},z\rangle})(1-\frac{(3m-1)\beta_{1}^{2}}{\langle\beta_{1},z\rangle-\beta_{1}^{2}})$.

Let now $\tau=-2\beta_{2}$. Then
$|t+\langle\tau,(2\beta_{1})^{\vee}\rangle|=t-1$ which does not belong to
$A_{\beta_{1}}\cup\\{0\\}$ if and only if $t=3m+2$. But
$a_{-2\beta_{2}}(z+(3m+2)\beta_{1})=0$ at $\langle\beta_{1},z\rangle=0$
because $a_{-2\beta_{2}}$ contains the factor
$(1+\frac{3m\beta_{1}^{2}}{-\langle\beta_{1},z\rangle+2\beta_{1}^{2}})$. Since
$s_{\alpha_{1}}(\beta_{1})=\beta_{1}$, by Remark 3.6 there is nothing to check
for $\tau=\pm 2\beta_{3}=s_{\alpha_{1}}(\mp 2\beta_{2})$.

For $\tau=\pm 2\alpha_{1}$, we get
$|t+\langle\tau,(2\beta_{1})^{\vee}\rangle|=t\in A_{\beta_{1}}$, as needed.
Similarly for $\tau=2\alpha_{2}$, $|t+\langle\tau,(2\beta_{1})^{\vee}\rangle|$
$=|t-3|\in A_{\beta_{1}}\cup\\{0\\}$ for all $t\in A_{\beta_{1}}$, as needed.

Finally, let $\tau=-2\alpha_{2}$. Then
$|t+\langle\tau,(2\beta_{1})^{\vee}\rangle|=t+3\notin
A_{\beta_{1}}\cup\\{0\\}$ if and only if $t=3m+2$, $t=3m$ or $t=3m-2$, but
$a_{-2\alpha_{2}}(z+t\beta_{1})=0$ at $\langle\beta_{1},z\rangle=0$ for those
$t$ because $a_{-2\alpha_{2}}$ contains the factors
$(1-\frac{(3m+2)\beta_{1}^{2}}{\langle\beta_{1},z\rangle})(1-\frac{(3m+1)\beta_{1}^{2}}{\langle\beta_{1},z\rangle+\beta_{1}^{2}})(1-\frac{3m\beta_{1}^{2}}{\langle\beta_{1},z\rangle+2\beta_{1}^{2}})$.
By Remark 3.6, there is nothing to check for $\tau=\pm
2\alpha_{3}=s_{\alpha_{1}}(\mp 2\alpha_{2})$.

Let us show next that
${\mathcal{D}}_{1}p(z+t\alpha_{1})={\mathcal{D}}_{1}p(z-t\alpha_{1})$ at
$\langle\alpha_{1},z\rangle=0$ for all $t\in A_{\alpha_{1}}$. By Theorem 3.2
it is sufficient to check its condition 2 for all $\tau\in 2G_{2}$.

Let $\tau=\pm 2\beta_{1}$. Then
$|t+\langle\tau,(2\alpha_{1})^{\vee}\rangle|=t\in A_{\alpha_{1}}$, as needed.

Let now $\tau=2\beta_{2}$. Note that
$(2\alpha_{1})^{\vee}=\frac{1}{6}\alpha_{1}$. Then
$|t+\langle\tau,(2\alpha_{1})^{\vee}\rangle|=t+1\notin
A_{\alpha_{1}}\cup\\{0\\}$ if and only if $t=m$. But
$a_{2\beta_{2}}(z+m\alpha_{1})=0$ at $\langle\alpha_{1},z\rangle=0$ because
$a_{2\beta_{2}}$ contains the factor
$(1-\frac{m\alpha_{1}^{2}}{\langle\alpha_{1},z\rangle})$.

Let now $\tau=-2\beta_{2}$. Then
$|t+\langle\tau,(2\alpha_{1})^{\vee}\rangle|=t-1\in A_{\alpha_{1}}\cup\\{0\\}$
for all $t\in A_{\alpha_{1}}$, as needed. Since
$s_{\beta_{1}}(\alpha_{1})=\alpha_{1}$, by Remark 3.6 there is nothing to
check for $\tau=\pm 2\beta_{3}=s_{\beta_{1}}(\pm 2\beta_{2})$.

Let now $\tau=2\alpha_{1}$. Then
$|t+\langle\tau,(2\alpha_{1})^{\vee}\rangle|=t+2\notin
A_{\alpha_{1}}\cup\\{0\\}$ if and only if $t=m$ or $t=m-1$. But
$a_{2\alpha_{1}}(z+t\alpha_{1})=0$ at $\langle\alpha_{1},z\rangle=0$ for those
$t$ because $a_{2\alpha_{1}}$ contains the factors
$(1-\frac{m\alpha_{1}^{2}}{\langle\alpha_{1},z\rangle})(1-\frac{m\alpha_{1}^{2}}{\langle\alpha_{1},z\rangle+\alpha_{1}^{2}})$.

Let now $\tau=-2\alpha_{1}$. Then
$|t+\langle\tau,(2\alpha_{1})^{\vee}\rangle|=|t-2|\in
A_{\alpha_{1}}\cup\\{0\\}$ for all $t\in A_{\alpha_{1}}$, as needed.

Let now $\tau=2\alpha_{2}$. Then
$|t+\langle\tau,(2\alpha_{1})^{\vee}\rangle|=t+1\notin
A_{\alpha_{1}}\cup\\{0\\}$ if and only if $t=m$. But
$a_{2\alpha_{2}}(z+m\alpha_{1})=0$ at $\langle\alpha_{1},z\rangle=0$ because
$a_{2\alpha_{2}}$ contains the factor
$(1-\frac{m\alpha_{1}^{2}}{\langle\alpha_{1},z\rangle})$.

Finally, for $\tau=-2\alpha_{2}$ we get
$|t+\langle\tau,(2\alpha_{1})^{\vee}\rangle|=t-1\in A_{\alpha_{1}}\cup\\{0\\}$
for all $t\in A_{\alpha_{1}}$. And by Remark 3.6 there is nothing to check for
$\tau=\pm 2\alpha_{3}=s_{\beta_{1}}(\pm 2\alpha_{2})$.

Since all the vectors $\alpha_{i},\beta_{i}$ are in the $W$-orbit of
$\alpha_{1}\cup\beta_{1}$ the statement follows by Lemma 3.5. ∎

Let us now look at the expansion of the operator (4.1) as $\omega\to 0$. It
produces the rational CMS operator for the root system of type $G_{2}$ with
multiplicity $m$ for the long roots and multiplicity $3m+1$ for the short
roots, as the next Proposition shows. Let
$\widetilde{\beta}_{j}=\omega^{-1}\beta_{j}$ and
$\widetilde{\alpha}_{j}=\omega^{-1}\alpha_{j}$ $(j=1,2,3)$ with the same
multiplicities as $\beta_{j}$ and $\alpha_{j}$, respectively.

######  Proposition 4.3.

We have

$\lim_{\omega\to
0}\frac{{\mathcal{D}}_{1}}{72\omega^{2}}=\Delta-\sum_{\gamma\in\\{\widetilde{\beta_{i}},\widetilde{\alpha}_{i}\colon
i=1,2,3\\}}\frac{2(m_{\gamma}+m_{2\gamma})}{\langle\gamma,z\rangle}\partial_{\gamma},$

where $\Delta=\partial^{2}_{z_{1}}+\partial^{2}_{z_{2}}$.

###### Proof.

We have that $T_{\pm
2\beta_{j}}-1=\pm\omega\partial_{2\widetilde{\beta}_{j}}+\frac{1}{2}\omega^{2}\partial_{2\widetilde{\beta}_{j}}^{2}+\dots$,
and similarly for the other shifts. The terms at $\omega$ in the expansion
$\omega\to 0$ of the operator ${\mathcal{D}}_{1}$ vanish. The terms that are
second order in derivatives in the coefficient at $\omega^{2}$ in the
expansion $\omega\to 0$ of the operator ${\mathcal{D}}_{1}$ are

$\displaystyle
3\sum_{j=1}^{3}\partial_{2\widetilde{\beta}_{j}}^{2}+\sum_{j=1}^{3}\partial_{2\widetilde{\alpha}_{j}}^{2}=12\sum_{j=1}^{3}\partial_{\widetilde{\beta}_{j}}^{2}+4\sum_{j=1}^{3}\partial_{\widetilde{\alpha}_{j}}^{2}=72\Delta.$

Let us now consider the terms that are first order in derivatives in the
coefficient at $\omega^{2}$. It is easy to see that such terms containing
$\langle\widetilde{\beta}_{1},z\rangle^{-1}$ are

$\displaystyle-12(3m+1)\left(2\partial_{2\widetilde{\beta}_{1}}+\partial_{2\widetilde{\beta}_{2}}-\partial_{2\widetilde{\beta}_{3}}+\partial_{2\widetilde{\alpha}_{3}}-\partial_{2\widetilde{\alpha}_{2}}\right)=-144(m_{\beta_{1}}+m_{2\beta_{1}})\partial_{\widetilde{\beta}_{1}}.$

Altogether, the term at $\omega^{2}$ in the expansion of the operator
${\mathcal{D}}_{1}$ is as required. ∎

##  5 Construction of the Baker–Akhiezer function for $AG_{2}$

In this Section, we employ the method from [8] to give a construction of the
BA function for the configuration $AG_{2}$. The BA function will be an
eigenfunction for the difference operator from Section 4, which establishes
bispectrality of the CMS $AG_{2}$ Hamiltonian.

Consider the operator ${\mathcal{D}}_{1}$ given by (4.1) with functions
$a_{\tau}$ as given in (4.2) and (4.3). The following Lemma gives a useful way
of expanding the functions $a_{\tau}$.

######  Lemma 5.1.

Let $a_{\tau}(z)$ be defined as in (4.2) and (4.3). Then

$a_{\tau}(z)=\kappa_{\tau}-\kappa_{\tau}\sum_{\gamma\in
G_{2,+}}\frac{\langle\tau,\gamma\rangle(m_{\gamma}+m_{2\gamma})}{\langle\gamma,z\rangle}+R_{\tau}(z)$
(5.1)

where $\kappa_{2\varepsilon\beta_{j}}=3$ and
$\kappa_{2\varepsilon\alpha_{j}}=1$, and $R_{\tau}(z)$ is a rational function
with $\deg R_{\tau}\leq-2$.

###### Proof.

For the factors in $a_{\tau}$ with shifted singularities at
$\langle\gamma,z\rangle+c=0$ for $c\neq 0$ we can use that

$\frac{1}{\langle\gamma,z\rangle+c}=\frac{1}{\langle\gamma,z\rangle}-\frac{c}{(\langle\gamma,z\rangle+c)\langle\gamma,z\rangle},$

which differs from $\langle\gamma,z\rangle^{-1}$ only by a rational function
of degree $-2$ which cannot affect the coefficient at
$\langle\gamma,z\rangle^{-1}$. The relation (5.1) is then obtained by
multiplying out the factors in each of the $a_{\tau}$. ∎

The next Lemma is proved by a direct computation that uses Lemma 5.1. We will
apply it in the proof of Theorem 5.3 below.

######  Lemma 5.2.

For $\gamma\in G_{2,+}$ let $n_{\gamma}\in\mathbb{N}$ be arbitrary. Let
$N=\sum_{\gamma\in G_{2,+}}n_{\gamma}$. Let

$\mu(x)=\sum_{\tau\colon\frac{1}{2}\tau\in G_{2}}\kappa_{\tau}(\exp\langle
x,\tau\rangle-1),$

where $\kappa_{\tau}$ are as in Lemma 5.1. Let $A(z)=\prod_{\gamma\in
G_{2,+}}\langle\gamma,z\rangle^{n_{\gamma}}$. Write
$({\mathcal{D}}_{1}-\mu(x))[A(z)\exp{\langle x,z\rangle}]=R(x,z)\exp\langle
x,z\rangle$ for some rational function $R(x,z)$ in $z$, which has degree less
than or equal to $N$. Then

$\displaystyle R(x,z)=\sum_{\gamma\in
G_{2,+}}\big{(}n_{\gamma}-(m_{\gamma}+m_{2\gamma})\big{)}\left(\sum_{\tau\colon\frac{1}{2}\tau\in
G_{2}}\kappa_{\tau}\langle\tau,\gamma\rangle\exp{\langle
x,\tau\rangle}\right)A(z)\langle\gamma,z\rangle^{-1}+S(x,z)$

for some rational function $S(x,z)$ in $z$ with degree less than or equal to
$N-2$.

###### Proof.

By making use of the expression for $a_{\tau}(z)$ given in Lemma 5.1 we get

$\displaystyle{\mathcal{D}}_{1}[A(z)\exp{\langle
x,z\rangle}]=\sum_{\tau\colon\frac{1}{2}\tau\in
G_{2}}a_{\tau}(z)(T_{\tau}-1)[A(z)\exp{\langle x,z\rangle}]$
$\displaystyle=\exp{\langle x,z\rangle}\sum_{\tau\colon\frac{1}{2}\tau\in
G_{2}}a_{\tau}(z)\bigg{(}\exp{\langle x,\tau\rangle}\prod_{\gamma\in
G_{2,+}}(\langle\gamma,z\rangle+\langle\tau,\gamma\rangle)^{n_{\gamma}}-A(z)\bigg{)}$
$\displaystyle=A(z)\exp{\langle x,z\rangle}\sum_{\tau\colon\frac{1}{2}\tau\in
G_{2}}\kappa_{\tau}\left(1-\sum_{\gamma\in
G_{2,+}}\langle\tau,\gamma\rangle(m_{\gamma}+m_{2\gamma})\langle\gamma,z\rangle^{-1}+\text{l.o.t.}\right)$
$\displaystyle\hskip 140.00021pt\times\left(\exp{\langle
x,\tau\rangle}\bigg{(}1+\sum_{\gamma\in
G_{2,+}}n_{\gamma}\langle\tau,\gamma\rangle\langle\gamma,z\rangle^{-1}+\text{l.o.t.}\bigg{)}-1\right)$
$\displaystyle=A(z)\exp\langle x,z\rangle\left(\mu(x)+\sum_{\gamma\in
G_{2,+}}\big{(}n_{\gamma}-(m_{\gamma}+m_{2\gamma})\big{)}\left(\sum_{\tau\colon\frac{1}{2}\tau\in
G_{2}}\kappa_{\tau}\langle\tau,\gamma\rangle\exp{\langle
x,\tau\rangle}\right)\langle\gamma,z\rangle^{-1}+\text{l.o.t.}\right),$

where l.o.t. denotes lower degree terms in $z$, and where we used that
$\sum_{\tau\colon\frac{1}{2}\tau\in
G_{2}}\kappa_{\tau}\langle\tau,\gamma\rangle=0$ for all $\gamma\in G_{2,+}$
since if $\frac{1}{2}\tau\in G_{2}$ then also $-\frac{1}{2}\tau\in G_{2}$ and
$\kappa_{\tau}=\kappa_{-\tau}$. ∎

We are ready to give the main result of this Section.

######  Theorem 5.3.

Let $M=\sum_{\gamma\in AG_{2,+}}m_{\gamma}=12m+3.$ For $x\in\mathbb{C}^{2}$,
let

$\mu(x)=\sum_{\tau\colon\frac{1}{2}\tau\in G_{2}}\kappa_{\tau}(\exp\langle
x,\tau\rangle-1),$

and

$c(x)=\frac{M!}{8}\prod_{\gamma\in
G_{2,+}}\left(\sum_{\tau\colon\frac{1}{2}\tau\in
G_{2}}\kappa_{\tau}\langle\tau,\gamma\rangle\exp{\langle
x,\tau\rangle}\right)^{m_{\gamma}+m_{2\gamma}}.$ (5.2)

We also define for $z\in\mathbb{C}^{2}$ the polynomial belonging to
$\mathcal{R}^{a}_{AG_{2}}$ given by

$Q(z)=\prod_{\begin{subarray}{c}\gamma\in G_{2,+}\\\ s\in
A_{\gamma}\end{subarray}}\left(\langle\gamma,z\rangle^{2}-s^{2}\gamma^{2}\right),$

where we recall
$A_{\gamma}=\\{1,2,3,\dots,m_{\gamma},m_{\gamma}+2m_{2\gamma}\\}$ for all
$\gamma\in G_{2,+}$. Then the function

$\psi(z,x)=c^{-1}(x)({\mathcal{D}}_{1}-\mu(x))^{M}[Q(z)\exp\langle
z,x\rangle]$ (5.3)

is the BA function for $AG_{2}$. Moreover, $\psi$ is also an eigenfunction of
the operator ${\mathcal{D}}_{1}$ with ${\mathcal{D}}_{1}\psi=\mu(x)\psi$, thus
bispectrality holds.

###### Proof.

The operator ${\mathcal{D}}_{1}$ preserves the ring $\mathcal{R}^{a}_{AG_{2}}$
by Theorem 4.2. The function $Q(z)\exp\langle z,x\rangle$ belongs to
$\mathcal{R}^{a}_{AG_{2}}$ since it is analytic and satisfies conditions (4.4)
given that $Q(z+s\gamma)=Q(z-s\gamma)=0$ at $\langle\gamma,z\rangle=0$, $s\in
A_{\gamma}$, for all $\gamma\in G_{2,+}$. Since ${\mathcal{D}}_{1}$ preserves
$\mathcal{R}^{a}_{AG_{2}}$, so does ${\mathcal{D}}_{1}-\mu(x)$, hence
$\psi(z,x)$ given by (5.3) belongs to $\mathcal{R}^{a}_{AG_{2}}$. Its
analyticity and the form of the functions $a_{\tau}$ imply that it equals
$c^{-1}(x)P(z,x)\exp\langle z,x\rangle$ for some polynomial $P(z,x)$ in $z$.
To prove that $\psi(z,x)$ satisfies the definition of the BA function, it thus
suffices to calculate the highest degree term in $P(z,x)$.

The highest degree term in $Q(z)$ is $Q_{0}(z)=\prod_{\gamma\in
G_{2,+}}\langle\gamma,z\rangle^{2(m_{\gamma}+m_{2\gamma})}$ and $\deg
Q_{0}=2M$. For all $k\in\mathbb{N}$ with $k\leq M$, an analogous argument as
above gives that $({\mathcal{D}}_{1}-\mu(x))^{k}[Q(z)\exp\langle z,x\rangle]$
belongs to $\mathcal{R}^{a}_{AG_{2}}$ and is of the form
$Q^{(k)}(z,x)\exp\langle z,x\rangle$ for some polynomial $Q^{(k)}(z,x)$ in
$z$. Let its highest-degree homogeneous component be $Q_{0}^{(k)}(z,x)$. Lemma
5.2 allows to compute $Q_{0}^{(k)}(z,x)$.

Lemma 5.2 gives that after the first application of ${\mathcal{D}}_{1}-\mu(x)$
onto $Q(z)\exp\langle z,x\rangle$ we get

$Q_{0}^{(1)}=\sum_{\gamma\in
G_{2,+}}(m_{\gamma}+m_{2\gamma})\left(\sum_{\tau\colon\frac{1}{2}\tau\in
G_{2}}\kappa_{\tau}\langle\tau,\gamma\rangle\exp{\langle
x,\tau\rangle}\right)\langle\gamma,z\rangle^{-1}Q_{0}(z).$

The second application gives

$\displaystyle Q_{0}^{(2)}=\sum_{\gamma\in
G_{2,+}}(m_{\gamma}+m_{2\gamma})(m_{\gamma}+m_{2\gamma}-1)\left(\sum_{\tau\colon\frac{1}{2}\tau\in
G_{2}}\kappa_{\tau}\langle\tau,\gamma\rangle\exp{\langle
x,\tau\rangle}\right)^{2}\langle\gamma,z\rangle^{-2}Q_{0}(z)$
$\displaystyle+\sum_{\begin{subarray}{c}\gamma\neq\delta\\\ \gamma,\delta\in
G_{2,+}\end{subarray}}(m_{\gamma}+m_{2\gamma})(m_{\delta}+m_{2\delta})\left(\sum_{\tau\colon\frac{1}{2}\tau\in
G_{2}}\kappa_{\tau}\langle\tau,\gamma\rangle\exp{\langle
x,\tau\rangle}\right)$ $\displaystyle\hskip
50.00008pt\times\left(\sum_{\tau\colon\frac{1}{2}\tau\in
G_{2}}\kappa_{\tau}\langle\tau,\delta\rangle\exp{\langle
x,\delta\rangle}\right)\langle\gamma,z\rangle^{-1}\langle\delta,z\rangle^{-1}Q_{0}(z).$

By repeatedly applying Lemma 5.2 we get

$Q_{0}^{(k)}=\sum_{\mathbf{n}}f_{\mathbf{n}}(x)Q_{0}(z)\prod_{\gamma\in
G_{2,+}}\langle\gamma,z\rangle^{-n_{\gamma}}$

where $\mathbf{n}=(n_{\gamma})_{\gamma\in G_{2,+}}$ for
$n_{\gamma}\in\mathbb{Z}_{\geq 0}$ such that $n_{\gamma}$ add up to $k$ and
where $f_{\bf{n}}(x)$ is non-zero only if $n_{\gamma}\leq
m_{\gamma}+m_{2\gamma}$ for all $\gamma$. It follows that $\deg P\leq M$ and
that the highest degree term of $P(z,x)$ is

$d(x)\prod_{\gamma\in
G_{2,+}}\langle\gamma,z\rangle^{m_{\gamma}+m_{2\gamma}}=\frac{1}{8}d(x)\prod_{\gamma\in
AG_{2,+}}\langle\gamma,z\rangle^{m_{\gamma}}$

for some function $d(x)$. It also implies that the polynomial part of
$({\mathcal{D}}_{1}-\mu(x))^{M+1}[Q(z)\exp\langle z,x\rangle]$ has degree less
than $M$ hence vanishes as a consequence of Lemma 2.6, giving
${\mathcal{D}}_{1}\psi=\mu(x)\psi$. So to complete the proof we just need to
verify that $c(x)$ given by (5.2) equals $\frac{1}{8}d(x)$.

To arrive at $\prod_{\gamma\in
G_{2,+}}\langle\gamma,z\rangle^{m_{\gamma}+m_{2\gamma}}$ starting from
$Q_{0}(z)$ we overall need to reduce the power on each of the factors
$\langle\gamma,z\rangle$ by $m_{\gamma}+m_{2\gamma}$ and we do this by
reducing the power of one of them by one at each step. The total number of
possible orderings of doing that corresponds to the number of words of length
$M$ in the alphabet $G_{2,+}$ such that $\gamma$ appears in the word
$m_{\gamma}+m_{2\gamma}$ times for each $\gamma\in G_{2,+}$. This gives

$\frac{M!}{\prod_{\gamma\in G_{2,+}}(m_{\gamma}+m_{2\gamma})!}$

possibilities, and for each of them the total proportionality factor picked up
equals by Lemma 5.2

$\prod_{\gamma\in
G_{2,+}}(m_{\gamma}+m_{2\gamma})!\left(\sum_{\tau\colon\frac{1}{2}\tau\in
G_{2}}\kappa_{\tau}\langle\tau,\gamma\rangle\exp{\langle
x,\tau\rangle}\right)^{m_{\gamma}+m_{2\gamma}}.$

It follows that $c(x)$ has the required form. ∎

##  6 Another dual operator

In this Section, we present another difference operator for the configuration
$AG_{2}$ which preserves the quasi-invariants. We also give the corresponding
second construction of the BA function.

We define a difference operator acting in the variable $z\in\mathbb{C}^{2}$ of
the form

${\mathcal{D}}_{2}=\sum_{\tau\colon\frac{1}{2}\tau\in
AG_{2}}a_{\tau}(z)(T_{\tau}-1).$ (6.1)

We now proceed to specify the functions $a_{\tau}(z)$. Let
$\lambda_{\tau}=g_{\tau/2}$ where $g$ is defined in terms of the multiplicity
map of $AG_{2}$ in accordance with the convention (1.4) for couplings. That
is, we set

$\lambda_{\tau}=\frac{1}{4}m_{\frac{1}{2}\tau}(m_{\frac{1}{2}\tau}+2m_{\tau}+1)\tau^{2}.$

Recall that the multiplicities are $m_{\varepsilon\beta_{j}}=3m$,
$m_{2\varepsilon\beta_{j}}=1$ and $m_{\varepsilon\alpha_{j}}=m$ ($j=1,2,3$,
$\varepsilon\in\\{\pm 1\\}$, $m\in\mathbb{N}$), and we put
$m_{2\varepsilon\alpha_{j}}=0$. That means
$\lambda_{4\varepsilon\beta_{j}}=8\beta_{j}^{2}$,
$\lambda_{2\varepsilon\beta_{j}}=9m(m+1)\beta_{j}^{2}$ and
$\lambda_{2\varepsilon\alpha_{j}}=m(m+1)\alpha_{j}^{2}$. For
$\tau=4\varepsilon\beta_{j}$ we define

$\displaystyle
a_{4\varepsilon\beta_{j}}(z)=\lambda_{4\varepsilon\beta_{j}}\prod_{\begin{subarray}{c}\gamma\in
W\alpha_{1}\\\ \langle
4\varepsilon\beta_{j},(2\gamma)^{\vee}\rangle=2\end{subarray}}\bigg{(}1-\frac{m\gamma^{2}}{\langle\gamma,z\rangle}\bigg{)}\bigg{(}1-\frac{m\gamma^{2}}{\langle\gamma,z\rangle+\gamma^{2}}\bigg{)}$
(6.2) $\displaystyle\times\prod_{\begin{subarray}{c}\gamma\in W\beta_{1}\\\
\langle
4\varepsilon\beta_{j},(2\gamma)^{\vee}\rangle=2\end{subarray}}\bigg{(}1-\frac{(3m+2)\gamma^{2}}{\langle\gamma,z\rangle}\bigg{)}\bigg{(}1-\frac{3m\gamma^{2}}{\langle\gamma,z\rangle+\gamma^{2}}\bigg{)}$
$\displaystyle\times\bigg{(}1-\frac{(3m+2)\beta_{j}^{2}}{\langle\varepsilon\beta_{j},z\rangle}\bigg{)}\bigg{(}1-\frac{3m\beta_{j}^{2}}{\langle\varepsilon\beta_{j},z\rangle+\beta_{j}^{2}}\bigg{)}\bigg{(}1-\frac{(3m+2)\beta_{j}^{2}}{\langle\varepsilon\beta_{j},z\rangle+2\beta_{j}^{2}}\bigg{)}\bigg{(}1-\frac{3m\beta_{j}^{2}}{\langle\varepsilon\beta_{j},z\rangle+3\beta_{j}^{2}}\bigg{)}.$

For $\tau=2\varepsilon\beta_{j}$ we define

$\displaystyle
a_{2\varepsilon\beta_{j}}(z)=\lambda_{2\varepsilon\beta_{j}}\prod_{\begin{subarray}{c}\gamma\in
W\alpha_{1}\\\ \langle
2\varepsilon\beta_{j},(2\gamma)^{\vee}\rangle=0\end{subarray}}\bigg{(}1-\frac{\frac{2}{3}\gamma^{2}}{\langle\gamma,z\rangle-\gamma^{2}}\bigg{)}\prod_{\begin{subarray}{c}\gamma\in
W\alpha_{1}\\\ \langle
2\varepsilon\beta_{j},(2\gamma)^{\vee}\rangle=1\end{subarray}}\bigg{(}1-\frac{m\gamma^{2}}{\langle\gamma,z\rangle}\bigg{)}$
(6.3) $\displaystyle\times\prod_{\begin{subarray}{c}\gamma\in W\beta_{1}\\\
\langle
2\varepsilon\beta_{j},(2\gamma)^{\vee}\rangle=1\end{subarray}}\bigg{(}1-\frac{(3m+2)\gamma^{2}}{\langle\gamma,z\rangle}\bigg{)}\bigg{(}1+\frac{3m\gamma^{2}}{\langle\gamma,z\rangle+2\gamma^{2}}\bigg{)}\bigg{(}1-\frac{(3m-1)\gamma^{2}}{\langle\gamma,z\rangle-\gamma^{2}}\bigg{)}$
$\displaystyle\times\bigg{(}1-\frac{(3m+2)\beta_{j}^{2}}{\langle\varepsilon\beta_{j},z\rangle}\bigg{)}\bigg{(}1-\frac{3m\beta_{j}^{2}}{\langle\varepsilon\beta_{j},z\rangle+\beta_{j}^{2}}\bigg{)}\bigg{(}1+\frac{4\beta_{j}^{2}}{\langle\varepsilon\beta_{j},z\rangle+3\beta_{j}^{2}}\bigg{)}\bigg{(}1-\frac{4\beta_{j}^{2}}{\langle\varepsilon\beta_{j},z\rangle-\beta_{j}^{2}}\bigg{)}.$

For $\tau=2\varepsilon\alpha_{j}$ we define

$\displaystyle a_{2\varepsilon\alpha_{j}}(z)$
$\displaystyle=\lambda_{2\varepsilon\alpha_{j}}\prod_{\begin{subarray}{c}\gamma\in
W\beta_{1}\\\ \langle
2\varepsilon\alpha_{j},(2\gamma)^{\vee}\rangle=3\end{subarray}}\bigg{(}1-\frac{(3m+2)\gamma^{2}}{\langle\gamma,z\rangle}\bigg{)}\bigg{(}1-\frac{(3m+1)\gamma^{2}}{\langle\gamma,z\rangle+\gamma^{2}}\bigg{)}\bigg{(}1-\frac{3m\gamma^{2}}{\langle\gamma,z\rangle+2\gamma^{2}}\bigg{)}$
(6.4) $\displaystyle\times\prod_{\begin{subarray}{c}\gamma\in W\beta_{1}\\\
\langle
2\varepsilon\alpha_{j},(2\gamma)^{\vee}\rangle=0\end{subarray}}\bigg{(}1-\frac{6\gamma^{2}}{\langle\gamma,z\rangle-\gamma^{2}}\bigg{)}\prod_{\begin{subarray}{c}\gamma\in
W\alpha_{1}\\\ \langle
2\varepsilon\alpha_{j},(2\gamma)^{\vee}\rangle=1\end{subarray}}\bigg{(}1-\frac{m\gamma^{2}}{\langle\gamma,z\rangle}\bigg{)}$
$\displaystyle\times\bigg{(}1-\frac{m\alpha_{j}^{2}}{\langle\varepsilon\alpha_{j},z\rangle}\bigg{)}\bigg{(}1-\frac{m\alpha_{j}^{2}}{\langle\varepsilon\alpha_{j},z\rangle+\alpha_{j}^{2}}\bigg{)}.$

Lemma 6.1 below shows that the functions $a_{\tau}(z)$ have $G_{2}$ symmetry.

######  Lemma 6.1.

Let $a_{\tau}(z)$ be defined as above. Then for all $w\in W$ we have
$wa_{\tau}=a_{w\tau}$.

###### Proof.

For any $w\in W$ we have $w(W\alpha_{1})=W\alpha_{1}$,
$w(W\beta_{1})=W\beta_{1}$, and $\langle
w\tau,w\gamma\rangle=\langle\tau,\gamma\rangle$ for all
$\gamma,\tau\in\mathbb{C}^{2}$. The multiplicities are invariant, too, and
$\lambda_{w\tau}=\lambda_{\tau}$ for any $\tau$ such that $\frac{1}{2}\tau\in
AG_{2}$. The statement follows. ∎

######  Theorem 6.2.

The operator (6.1) preserves the ring $\mathcal{R}^{a}_{AG_{2}}$.

###### Proof.

One can check that the operator satisfies condition ($D_{2}$). Let
$p(z)\in\mathcal{R}^{a}_{AG_{2}}$ be arbitrary. Without loss of generality, we
put $\omega=\sqrt{2}$. We introduce new coordinates $(A,B)$ on
$\mathbb{C}^{2}$ given by $A=\langle\alpha_{1},z\rangle$ and
$B=\langle\beta_{1},z\rangle$.

It follows from the form of the coefficient functions (6.2), (6.3) and (6.4)
and Theorem 3.2 that there are no singularities in ${\mathcal{D}}_{2}p(z)$ at
$B=c$ for $c\geq 0$ unless $B=2,4,6$. Let us consider these cases.

If $B=6$ (equivalently, $\langle\beta_{1},z\rangle=3\beta_{1}^{2}$), then
$\langle\beta_{2},z\rangle=3+\frac{1}{2}A$,
$\langle\beta_{3},z\rangle=-3+\frac{1}{2}A$,
$\langle\alpha_{2},z\rangle=-9+\frac{1}{2}A$ and
$\langle\alpha_{3},z\rangle=9+\frac{1}{2}A$. The only terms singular at $B=6$
are $a_{-4\beta_{1}}$ and $a_{-2\beta_{1}}$. We note that
$s_{\beta_{1}}(-4\beta_{1})-6\beta_{1}=-2\beta_{1}$, and we compute that
$\operatorname{res}_{B=6}(a_{-4\beta_{1}})=-\operatorname{res}_{B=6}(a_{-2\beta_{1}})$
equals

$\displaystyle\tfrac{48m(m+1)(3m+2)(3m+5)(A-2-12m)(A-6-12m)(A-14-12m)(A-18-12m)(A+2+12m)(A+6+12m)(A+14+12m)(A+18+12m)}{(A-18)(A-6)^{2}(A-2)(A+2)(A+6)^{2}(A+18)}.$

Therefore, by Theorem 3.2 part 1, there is no singularity at $B=6$ in
${\mathcal{D}}_{2}p(z)$.

If $B=4$ (equivalently, $\langle\beta_{1},z\rangle=2\beta_{1}^{2}$), then
$\langle\beta_{2},z\rangle=2+\frac{1}{2}A$,
$\langle\beta_{3},z\rangle=-2+\frac{1}{2}A$,
$\langle\alpha_{2},z\rangle=-6+\frac{1}{2}A$ and
$\langle\alpha_{3},z\rangle=6+\frac{1}{2}A$. The only $\tau\in AG_{2}$ for
which $a_{\tau}$ is singular at $B=4$ and for which the corresponding
$\lambda=s_{\beta_{1}}(\tau)-4\beta_{1}\neq 0$ are
$\tau=-2\beta_{2},-2\alpha_{3},2\beta_{3}$ and $2\alpha_{2}$. We note that
$s_{\beta_{1}}(-2\beta_{2})-4\beta_{1}=-2\alpha_{3}$, and we compute that
$\operatorname{res}_{B=4}(a_{-2\beta_{2}})=-\operatorname{res}_{B=4}(a_{-2\alpha_{3}})$
equals

$\displaystyle-\tfrac{18m^{2}(m+1)(3m+2)(3m+4)(A-32)(A+24)(A-12-12m)(A+6m)(A-4+12m)(A+12m)(A+4+12m)(A+12+12m)^{2}}{(A-12)(A-8)(A-4)A^{4}(A+4)(A+12)}.$

Since $s_{\alpha_{1}}(-2\beta_{2})=2\beta_{3}$ and
$s_{\alpha_{1}}(-2\alpha_{3})=2\alpha_{2}$, by Lemma 3.3 the residue of
$a_{2\beta_{3}}+a_{2\alpha_{2}}$ at $B=4$ is also zero. Thus, by Theorem 3.2
part 1, there is no singularity at $B=4$ in ${\mathcal{D}}_{2}p(z)$.

If $B=2$ (equivalently, $\langle\beta_{1},z\rangle=\beta_{1}^{2}$), then
$\langle\beta_{2},z\rangle=1+\frac{1}{2}A$,
$\langle\beta_{3},z\rangle=-1+\frac{1}{2}A$,
$\langle\alpha_{2},z\rangle=-3+\frac{1}{2}A$ and
$\langle\alpha_{3},z\rangle=3+\frac{1}{2}A$. The only $\tau\in AG_{2}$ for
which $a_{\tau}$ is singular at $B=2$ and for which the corresponding
$\lambda=s_{\beta_{1}}(\tau)-2\beta_{1}\neq 0$ are
$\tau=-4\beta_{1},2\beta_{1},-4\beta_{2},4\beta_{3},\pm
2\alpha_{1},2\beta_{2},2\alpha_{2},-2\beta_{3}$ and $-2\alpha_{3}$. We note
that $s_{\beta_{1}}(-4\beta_{1})-2\beta_{1}=2\beta_{1}$, and we compute that
$\operatorname{res}_{B=2}(a_{-4\beta_{1}})=-\operatorname{res}_{B=2}(a_{2\beta_{1}})$
equals

$\displaystyle\tfrac{144m(m+1)(3m-2)(3m+1)(A+6-12m)(A+2-12m)(A-6-12m)(A-10-12m)(A-6+12m)(A-2+12m)(A+6+12m)(A+10+12m)}{(A-6)^{2}(A-2)^{2}(A+2)^{2}(A+6)^{2}}.$

Similarly, we note that $s_{\beta_{1}}(-4\beta_{2})-2\beta_{1}=-2\alpha_{1}$,
and we compute that
$\operatorname{res}_{B=2}(a_{-4\beta_{2}})=-\operatorname{res}_{B=2}(a_{-2\alpha_{1}})$
equals

$\displaystyle\tfrac{288m(m+1)(A-6+6m)(A+6m)(A-10+12m)(A-6+12m)^{2}(A-2+12m)(A+2+12m)(A+6+12m)^{2}(A+10+12m)}{(A-10)(A-6)^{4}(A-2)^{2}A(A+2)(A+6)}.$

Since $s_{\alpha_{1}}(-4\beta_{2})=4\beta_{3}$ and
$s_{\alpha_{1}}(-2\alpha_{1})=2\alpha_{1}$, it follows by Lemma 3.3 that the
residue of $a_{4\beta_{3}}+a_{2\alpha_{1}}$ at $B=2$ is also zero. Next we
note that $s_{\beta_{1}}(2\beta_{2})-2\beta_{1}=2\alpha_{2}$, and we compute
that
$\operatorname{res}_{B=2}(a_{2\beta_{2}})=-\operatorname{res}_{B=2}(a_{2\alpha_{2}})$
equals

$\displaystyle\tfrac{36m(m+1)^{2}(3m-1)(3m+1)(A-26)(A+30)(A-10-12m)(A-6-12m)(A-2-12m)(A+6-12m)^{2}(A-6m)(A+6+12m)}{(A-6)(A-2)^{2}A(A+2)(A+6)^{4}}.$

Since $s_{\alpha_{1}}(2\beta_{2})=-2\beta_{3}$ and
$s_{\alpha_{1}}(2\alpha_{2})=-2\alpha_{3}$, it follows by Lemma 3.3 that the
residue of $a_{-2\beta_{3}}+a_{-2\alpha_{3}}$ at $B=2$ is also zero. Thus, by
Theorem 3.2 part 1 there is no singularity at $B=2$ in
${\mathcal{D}}_{2}p(z)$.

Let us now consider possible singularities in ${\mathcal{D}}_{2}p(z)$ at
$A=c\geq 0$. By Theorem 3.2 part 1 and the form of the coefficients (6.2) –
(6.4) it is sufficient to consider the case $A=6$ (equivalently,
$\langle\alpha_{1},z\rangle=\alpha_{1}^{2}$). In this case
$\langle\beta_{2},z\rangle=\frac{1}{2}B+3$,
$\langle\beta_{3},z\rangle=-\frac{1}{2}B+3$,
$\langle\alpha_{2},z\rangle=-\frac{3}{2}B+3$ and
$\langle\alpha_{3},z\rangle=\frac{3}{2}B+3$. The only $\tau\in AG_{2}$ for
which $a_{\tau}$ is singular at $A=6$ and for which the corresponding
$\lambda=s_{\alpha_{1}}(\tau)-2\alpha_{1}\neq 0$ are
$\tau=-4\beta_{2},-4\beta_{3}$ and $\pm 2\beta_{1}$. We note that
$s_{\alpha_{1}}(-4\beta_{2})-2\alpha_{1}=-2\beta_{1}$, and we compute that
$\operatorname{res}_{A=6}(a_{-4\beta_{2}})=-\operatorname{res}_{A=6}(a_{-2\beta_{1}})$
equals

$\displaystyle\tfrac{96m(m+1)(B-14-12m)(B-2-12m)(B-2+4m)(B+2+4m)(B-2+6m)(B+4+6m)(B-6+12m)(B+2+12m)(B+6+12m)(B+14+12m)}{(B-6)^{2}(B-2)^{4}B(B+2)^{2}(B+6)}.$

Since $s_{\beta_{1}}(-4\beta_{2})=-4\beta_{3}$ and
$s_{\beta_{1}}(-2\beta_{1})=2\beta_{1}$, it follows by Lemma 3.3 that the
residue of $a_{-4\beta_{3}}+a_{2\beta_{1}}$ at $A=6$ is also zero. By Theorem
3.2 part 1 there is thus no singularity at $A=6$ in ${\mathcal{D}}_{2}p(z)$.

By Corollary 3.4 it follows that ${\mathcal{D}}_{2}p(z)$ has no singularities.
The proof that ${\mathcal{D}}_{2}p(z)$ belongs to $\mathcal{R}^{a}_{AG_{2}}$
can be completed in an analogous way to how it was done for the operator (4.1)
in the proof of Theorem 4.2. ∎

We now give a second construction of the Baker–Akhiezer function for $AG_{2}$.

######  Theorem 6.3.

Let $M=\sum_{\gamma\in AG_{2,+}}m_{\gamma}=12m+3.$ For $x\in\mathbb{C}^{2}$,
let

$\mu(x)=\sum_{\tau\colon\frac{1}{2}\tau\in AG_{2}}\lambda_{\tau}(\exp\langle
x,\tau\rangle-1),$

and

$c(x)=\frac{M!}{8}\prod_{\gamma\in
G_{2,+}}\left(\sum_{\tau\colon\frac{1}{2}\tau\in
AG_{2}}\lambda_{\tau}\langle\tau,\gamma\rangle\exp{\langle
x,\tau\rangle}\right)^{m_{\gamma}+m_{2\gamma}}.$ (6.5)

We also define for $z\in\mathbb{C}^{2}$ the polynomial belonging to
$\mathcal{R}^{a}_{AG_{2}}$ given by

$Q(z)=\prod_{\begin{subarray}{c}\gamma\in G_{2,+}\\\ s\in
A_{\gamma}\end{subarray}}\left(\langle\gamma,z\rangle^{2}-s^{2}\gamma^{2}\right),$

where we recall
$A_{\gamma}=\\{1,2,3,\dots,m_{\gamma},m_{\gamma}+2m_{2\gamma}\\}$ for all
$\gamma\in G_{2,+}$. Then the function

$\psi(z,x)=c^{-1}(x)({\mathcal{D}}_{2}-\mu(x))^{M}[Q(z)\exp\langle
z,x\rangle]$ (6.6)

is the BA function for $AG_{2}$. Moreover, $\psi$ is also an eigenfunction of
the operator ${\mathcal{D}}_{2}$ with ${\mathcal{D}}_{2}\psi=\mu(x)\psi$, thus
bispectrality holds.

The proof is similar to the proof of Theorem 5.3 and it can be found in [30].

######  Remark 6.4.

One can show that operators (4.1) – (4.3), (6.1) – (6.4) commute:
$[{\mathcal{D}}_{1},{\mathcal{D}}_{2}]=0$. This can be proven by taking the
rational limit of the more general trigonometric versions of these operators,
which also commute [15].

##  7 Relation with $A_{2}$ and $A_{1}$ Macdonald–Ruijsenaars systems

In the case when $m=0$, the configuration $AG_{2}$ reduces to the root system
$\\{\pm 2\beta_{i}\colon i=1,2,3\\}$ of type $A_{2}$ with multiplicity $1$ for
all vectors. In this limit, the operator (6.1) reduces to the quasiminuscule
operator for (twice) this root system. Let us now consider the $m=0$ limit of
the operator (4.1). After a rescaling, this gives an operator of the form

$D_{0}=\sum_{\tau\in G_{2}}a_{\tau,0}(z)(T_{\tau}-1)=-24+\sum_{\tau\in
G_{2}}a_{\tau,0}(z)T_{\tau},$ (7.1)

where for $\tau=\varepsilon\beta_{j}$, $\varepsilon\in\\{\pm 1\\}$, $j=1,2,3,$
we have

$a_{\tau,0}(z)=3\prod_{\begin{subarray}{c}\gamma\in W\beta_{1}\\\
\langle\tau,(2\gamma)^{\vee}\rangle=\frac{1}{2}\end{subarray}}\left(1-\frac{\frac{1}{2}\gamma^{2}}{\langle\gamma,z\rangle-\frac{1}{2}\gamma^{2}}\right)\prod_{\begin{subarray}{c}\gamma\in
W\beta_{1}\\\
\langle\tau,(2\gamma)^{\vee}\rangle=1\end{subarray}}\left(1-\frac{\gamma^{2}}{\langle\gamma,z\rangle}\right),$

and for $\tau=\varepsilon\alpha_{j}$ we have

$a_{\tau,0}(z)=\prod_{\begin{subarray}{c}\gamma\in W\beta_{1}\\\
\langle\tau,(2\gamma)^{\vee}\rangle=\frac{3}{2}\end{subarray}}\left(1-\frac{\frac{3}{2}\gamma^{2}}{\langle\gamma,z\rangle+\frac{1}{2}\gamma^{2}}\right).$

######  Proposition 7.1.

The operator (7.1) preserves the ring of analytic functions $p(z)$ satisfying
$p(z+\beta_{i})$ $=p(z-\beta_{i})$ at $\langle\beta_{i},z\rangle=0$ for all
$i=1,2,3$.

The proof is parallel to the proof of Theorem 4.2. In this case, though, the
condition 2(b) of Theorem 3.2 is needed while it does not play a role in the
proofs of Theorems 4.2 and 6.2.

Let us rewrite the operator $D_{0}$ for the more standard realisation of the
root system $A_{2}$ given by $A_{2}=\\{e_{i}-e_{j}\colon 1\leq i\neq j\leq
3\\}\subset\mathbb{R}^{3}$, where $e_{i}$ are the standard basis vectors.

######  Proposition 7.2.

Define the set $S=S_{1}\cup S_{2}$, where

$S_{1}=\\{3e_{i}\colon i=1,2,3\\}\cup\\{2e_{i}+2e_{j}-e_{k}:1\leq i<j\neq
k\leq 3,i\neq k\\}\textnormal{ and }S_{2}=\\{2e_{i}+e_{j}\colon 1\leq i\neq
j\leq 3\\}.$

Then the operator acting in the variable
$z=(z_{1},z_{2},z_{3})\in\mathbb{C}^{3}$ given by

$\displaystyle\widetilde{D}_{0}$ $\displaystyle=3\sum_{\tau\in
S_{2}}\left(\prod_{\begin{subarray}{c}i\neq j\\\
\langle\tau,e_{i}-e_{j}\rangle=1\end{subarray}}\left(1-\frac{1}{z_{i}-z_{j}-1}\right)\prod_{\begin{subarray}{c}i\neq
j\\\
\langle\tau,e_{i}-e_{j}\rangle=2\end{subarray}}\left(1-\frac{2}{z_{i}-z_{j}}\right)\right)T_{\tau}$
(7.2) $\displaystyle+\sum_{\tau\in S_{1}}\left(\prod_{\begin{subarray}{c}i\neq
j\\\
\langle\tau,e_{i}-e_{j}\rangle=3\end{subarray}}\left(1-\frac{3}{z_{i}-z_{j}+1}\right)\right)T_{\tau}.$

preserves the ring of analytic functions $p(z)$ satisfying
$p(z+e_{i}-e_{j})=p(z-e_{i}+e_{j})$ at $z_{i}=z_{j}$ for all $i,j=1,2,3$.

Note that $S_{1}=\\{\tau\in
S\colon|\langle\tau,e_{i}-e_{j}\rangle|\in\\{0,3\\}\textnormal{ for all
}i,j=1,2,3\\}$ and $S_{2}=\\{\tau\in
S\colon|\langle\tau,e_{i}-e_{j}\rangle|\in\\{0,1,2\\}\textnormal{ for all
}i,j=1,2,3\\}$.

Let us now consider a version of the operator (7.2) for the root system
$A_{1}$. Let $\sim$ denote equality of operators when acting on functions
constant along the direction normal to the hyperplane $z_{1}+z_{2}=0$.

######  Proposition 7.3.

Let $S_{1}^{\prime}=\\{3e_{1},3e_{2}\\}$ and
$S_{2}^{\prime}=\\{2e_{1}+e_{2},e_{1}+2e_{2}\\}.$ Then formula (7.2) after
replacement of $S_{i}$ with $S_{i}^{\prime}$, $i=1,2$, gives an operator
$\widehat{D}_{0}$ acting in the variable $z=(z_{1},z_{2})\in\mathbb{C}^{2}$
that preserves the ring $\mathcal{R}^{a}_{A_{1}}$ of analytic functions $p(z)$
satisfying $p(z+e_{1}-e_{2})=p(z-e_{1}+e_{2})$ at $z_{1}=z_{2}$. Moreover, if
we split the operator $\widehat{D}_{0}=D_{1}+D_{2}$, where

$D_{1}=3\left(1-\frac{1}{z_{2}-z_{1}-1}\right)T_{e_{1}+2e_{2}}+\left(1-\frac{3}{z_{1}-z_{2}+1}\right)T_{3e_{1}}$

and

$D_{2}=3\left(1-\frac{1}{z_{1}-z_{2}-1}\right)T_{2e_{1}+e_{2}}+\left(1-\frac{3}{z_{2}-z_{1}+1}\right)T_{3e_{2}},$

then $D_{i}(\mathcal{R}^{a}_{A_{1}})\subseteq\mathcal{R}^{a}_{A_{1}}$ for
$i=1,2$. The operators $D_{i}$ satisfy commutativity relations

$[D_{1},D_{2}]=[D_{1},D^{msl}]=[D_{2},D^{msl}]=0,$

where $D^{msl}$ is the operator for the minuscule weight $2e_{1}$ of the root
system $2A_{1}$ with multiplicity 1 given by

$D^{msl}=\left(1-\frac{2}{z_{1}-z_{2}}\right)T_{2e_{1}}+\left(1-\frac{2}{z_{2}-z_{1}}\right)T_{2e_{2}}.$

We also have ${\widehat{D}_{0}}^{2}\sim(D^{msl}+2)^{3}$, and $D_{1}D_{2}\sim
3D^{qm}+16\sim 3(D^{msl})^{2}+4$, where

$\displaystyle D^{qm}$
$\displaystyle=\left(1-\frac{2}{z_{1}-z_{2}}\right)\left(1-\frac{2}{z_{1}-z_{2}+2}\right)(T_{4e_{1}}-1)+\left(1-\frac{2}{z_{2}-z_{1}}\right)\left(1-\frac{2}{z_{2}-z_{1}+2}\right)(T_{4e_{2}}-1)$

is the operator for the quasiminuscule weight $4e_{1}$.

We note that the operators $D_{1}$ and $D_{2}$ are not symmetric under the
swap of the variables $z_{1},z_{2}$. The operator $D^{msl}$ is symmetric and
all three operators commute.

Acknowledgments. M.V. acknowledges support received from the Institute of
Mathematics and its Applications through their Small Grant Scheme, as well as
support from the School of Mathematics and Statistics of the University of
Glasgow for which we would like to thank Ian Strachan. The work of M.V. was
also partially supported by a Carnegie PhD Scholarship from the Carnegie Trust
for the Universities of Scotland.

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School of Mathematics and Statistics, University of Glasgow, University Place,
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