Higher Rank Relations for the Askey–Wilson and $q$-Bannai–Ito Algebra

Higher Rank Relations for the Askey–Wilson
and $\boldsymbol{q}$-Bannai–Ito Algebra

Hadewijch DE CLERCQ

H. De Clercq

Department of Electronics and Information Systems, Faculty of Engineering and
Architecture,
Ghent University, Belgium<EMAIL_ADDRESS>
Received September 03, 2019, in final form December 13, 2019; Published online
December 19, 2019

The higher rank Askey–Wilson algebra was recently constructed in the $n$-fold
tensor product of $U_{q}(\mathfrak{sl}_{2})$. In this paper we prove a class
of identities inside this algebra, which generalize the defining relations of
the rank one Askey–Wilson algebra. We extend the known construction algorithm
by several equivalent methods, using a novel coaction. These allow to simplify
calculations significantly. At the same time, this provides a proof of the
corresponding relations for the higher rank $q$-Bannai–Ito algebra.

Askey–Wilson algebra; Bannai–Ito algebra

16T05; 16T15; 17B37; 81R50

##  1 Introduction

The Askey–Wilson algebra was introduced in [33] as an algebraic foundation for
the bispectral problem of the Askey–Wilson orthogonal polynomials [22]. More
precisely, Zhedanov defined this algebra by generators and relations, which
turned out to be satisfied when realizing the generators as the Askey–Wilson
$q$-difference operator on the one hand, and the multiplication with the
variable on the other. A central extension, which allows a
$\mathbb{Z}_{3}$-symmetric presentation, was defined and studied by
Terwilliger [29]. He calls this central extension the universal Askey–Wilson
algebra. We will denote it by ${\rm AW}(3)$.

The irreducible finite-dimensional representations of the Askey–Wilson algebra
have been classified by Leonard pairs [26, 31] and a similar classification
for the universal Askey–Wilson algebra ${\rm AW}(3)$ has appeared in [18].
Further applications arise in the theory of special functions [4, 24],
tridiagonal and Leonard pairs [25, 28, 32], superintegrable quantum systems
[3] and the reflection equation [2]. Its use to quantum mechanics is further
emphasized by the identification of the Askey–Wilson algebra as a quotient of
the $q$-Onsager algebra [27], which originates from statistical mechanics.
This connection was later extended to the universal Askey–Wilson algebra [30].
Furthermore, an explicit homomorphism from the original Askey–Wilson algebra
to the double affine Hecke algebra of type $\big{(}C_{1}^{\vee},C_{1}\big{)}$
has been constructed in [21, 23]. Recently, connections with $q$-Higgs
algebras [13] and Howe dual pairs [14] have been obtained.

The original Askey–Wilson algebra [33] was realized inside the quantum group
$U_{q}(\mathfrak{sl}_{2})$ in [17]. A different realization in the three-fold
tensor product of $U_{q}(\mathfrak{sl}_{2})$ was given in [16], and was later
extended to the universal Askey–Wilson algebra ${\rm AW}(3)$ of [29] in [19].
In the latter reference, ${\rm AW}(3)$ is embedded in the threefold tensor
product of $U_{q}(\mathfrak{sl}_{2})$: if one denotes by $\Lambda$ the
quadratic Casimir element of $U_{q}(\mathfrak{sl}_{2})$ and by $\Delta$ its
coproduct, and defines

$\displaystyle\Lambda_{\\{1\\}}=\Lambda\otimes 1\otimes
1,\qquad\Lambda_{\\{2\\}}=1\otimes\Lambda\otimes
1,\qquad\Lambda_{\\{3\\}}=1\otimes 1\otimes\Lambda,$
$\displaystyle\Lambda_{\\{1,2\\}}=\Delta(\Lambda)\otimes
1,\qquad\Lambda_{\\{2,3\\}}=1\otimes\Delta(\Lambda),\qquad\Lambda_{\\{1,2,3\\}}=(1\otimes\Delta)\Delta(\Lambda),$
(1.1)

then these elements generate the universal Askey–Wilson algebra. Indeed, they
satisfy the $q$-commutation relations

$\displaystyle[\Lambda_{\\{1,2\\}},\Lambda_{\\{2,3\\}}]_{q}=\big{(}q^{-2}-q^{2}\big{)}\Lambda_{\\{1,3\\}}+\big{(}q-q^{-1}\big{)}\big{(}\Lambda_{\\{2\\}}\Lambda_{\\{1,2,3\\}}+\Lambda_{\\{1\\}}\Lambda_{\\{3\\}}\big{)},$
(1.2)
$\displaystyle[\Lambda_{\\{2,3\\}},\Lambda_{\\{1,3\\}}]_{q}=\big{(}q^{-2}-q^{2}\big{)}\Lambda_{\\{1,2\\}}+\big{(}q-q^{-1}\big{)}\big{(}\Lambda_{\\{3\\}}\Lambda_{\\{1,2,3\\}}+\Lambda_{\\{1\\}}\Lambda_{\\{2\\}}\big{)},$
(1.3)
$\displaystyle[\Lambda_{\\{1,3\\}},\Lambda_{\\{1,2\\}}]_{q}=\big{(}q^{-2}-q^{2}\big{)}\Lambda_{\\{2,3\\}}+\big{(}q-q^{-1}\big{)}\big{(}\Lambda_{\\{1\\}}\Lambda_{\\{1,2,3\\}}+\Lambda_{\\{2\\}}\Lambda_{\\{3\\}}\big{)},$
(1.4)

where $\Lambda_{\\{1,3\\}}$ is defined through relation (1.2), and
$\Lambda_{\\{1,2,3\\}}$ and all $\Lambda_{\\{i\\}}$ are central. This
coincides with the presentation for ${\rm AW}(3)$ given in [29]. This
illustrates that there exists an algebra homomorphism from ${\rm AW}(3)$ to
$U_{q}(\mathfrak{sl}_{2})^{\otimes 3}$ and this map turns out to be injective,
as shown in [19, Theorem 4.8].

In [8] this approach was generalized to $n$-fold tensor products for arbitrary
$n$, which leads to an extension of the universal Askey–Wilson algebra to
higher rank, denoted ${\rm AW}(n)$. The same algorithm allows to construct a
higher rank extension of the $q$-Bannai–Ito algebra, which is isomorphic to
${\rm AW}(3)$ under a transformation $q\to-q^{2}$ and allows a similar
embedding in $\mathfrak{osp}_{q}(1|2)^{\otimes 3}$ [15]. Also the limiting
cases $q=1$ provide interesting algebras, as summarized graphically below.

Askey–Wilson algebra Racah algebra $q$-Bannai–Ito algebra Bannai–Ito algebra
$q\to 1$ $q\to-q^{2}$ $q\to 1$

Such higher rank algebras are motivated by their role as symmetry algebras for
superintegrable quantum systems of higher dimension. This has been confirmed
in the limiting case $q=1$ [9, 10, 11], and later also for general $q$ [8]. In
both cases, the Hamiltonians under consideration are built from Dunkl
operators with $\mathbb{Z}_{2}^{n}$ symmetry [12], possibly $q$-deformed [5].
Moreover, these higher rank algebras allow to extend known connections with
orthogonal polynomials to multiple variables. This was achieved in [7] for the
$q$-Bannai–Ito algebra. To be concrete, an action of the higher rank
$q$-Bannai–Ito algebra on an abstract vector space was considered, leading to
various orthonormal bases for this space. The overlap coefficients between
such bases turned out to be multivariable $(-q)$-Racah polynomials, the
truncated analogs of Askey–Wilson polynomials. This has allowed to construct a
realization of the higher rank $q$-Bannai–Ito algebra with Iliev’s
$q$-difference operators [20], which have thereby obtained an algebraic
interpretation.

The construction of ${\rm AW}(n)$ is rather intricate: in [8] we have outlined
an algorithm which repeatedly applies the $U_{q}(\mathfrak{sl}_{2})$-coproduct
$\Delta$ and a coaction $\tau$ to the Casimir element $\Lambda$, in a specific
order. This way we construct, as an extension of (1.2)–(1.4), an element
$\Lambda_{A}\in U_{q}(\mathfrak{sl}_{2})^{\otimes n}$ for each
$A\subseteq\\{1,\dots,n\\}$. In this paper we rephrase this extension
algorithm in more accessible notation and provide alternative construction
methods for the elements $\Lambda_{A}$ which use a novel coaction. This new
approach is of major use to derive algebraic identities in ${\rm AW}(n)$, as
we showcase in Theorems 3.1 and 3.2 by significantly generalizing the
algebraic relations given in [8]. The main achievement of this paper is hence
a general criterion for two generators $\Lambda_{A}$ and $\Lambda_{B}$ of
${\rm AW}(n)$ to commute or to satisfy a relation of the form

$\displaystyle[\Lambda_{A},\Lambda_{B}]_{q}=\big{(}q^{-2}-q^{2}\big{)}\Lambda_{(A\cup
B)\setminus(A\cap B)}+\big{(}q-q^{-1}\big{)}\big{(}\Lambda_{A\cap
B}\Lambda_{A\cup B}+\Lambda_{A\setminus(A\cap B)}\Lambda_{B\setminus(A\cap
B)}\big{)}.\\!\\!\\!$ (1.5)

More precisely, we show in Theorem 3.1 that

$[\Lambda_{A},\Lambda_{B}]=0\qquad\mathrm{if}\quad B\subseteq A,$

and in Theorem 3.2 we prove that the relation (1.5) is satisfied for

$\displaystyle A=A_{1}\cup A_{2}\cup A_{4},\qquad$ $\displaystyle B=A_{2}\cup
A_{3},$ $\displaystyle A=A_{2}\cup A_{3},\qquad$ $\displaystyle B=A_{1}\cup
A_{3}\cup A_{4},$ $\displaystyle A=A_{1}\cup A_{3}\cup A_{4},\qquad$
$\displaystyle B=A_{1}\cup A_{2}\cup A_{4},$ (1.6)

where $A_{1},A_{2},A_{3},A_{4}\subseteq\\{1,\dots,n\\}$ are such that for each
$i\in\\{1,2,3\\}$ one has either $\max(A_{i})<\min(A_{i+1})$ or
$A_{i}=\varnothing$ or $A_{i+1}=\varnothing$.

It is not clear at this point whether these relations define the algebra ${\rm
AW}(n)$ abstractly, or, in case the answer turns to be negative, which
supplementary relations should be added in order to attain this purpose.
However, calculations with computer algebra packages suggest that the
condition (1.6) describes the most general situation for the relations (1.5)
to be satisfied.

Our methods are elementary and intrinsic: they are independent of the
expressions for the coactions and the $U_{q}(\mathfrak{sl}_{2})$-Casimir
$\Lambda$, and only recur to natural algebraic properties like coassociativity
and the cotensor product property. As a consequence, the results of this paper
are equally applicable to the higher rank $q$-Bannai–Ito algebra of [8],
without any modification.

The paper is organized as follows. In Section 2 we construct the higher rank
Askey–Wilson algebra ${\rm AW}(n)$ as a subalgebra of
$U_{q}(\mathfrak{sl}_{2})^{\otimes n}$ through different extension processes,
which we prove to be equivalent. Section 3 lists the main results of this
paper and explains the general strategy of proof. Consequently, in Sections 4
and 5 we prove some intermediate results which will be relied on in Sections 6
and 7, where we prove Theorems 3.1 and 3.2. Finally in Section 8 we introduce
similar extension processes to construct a higher rank $q$-Bannai–Ito algebra
as a subalgebra of $\mathfrak{osp}_{q}(1|2)^{\otimes n}$. We state two
explicit theorems describing their algebraic relations.

##  2 Defining the higher rank generators

Throughout this paper, we will work with the quantum group
$U_{q}(\mathfrak{sl}_{2})$, which can be presented as the associative algebra
over a field $\mathbb{K}$ with generators $E$, $F$, $K$ and $K^{-1}$, and
relations

$KK^{-1}=K^{-1}K=1,\qquad KE=q^{2}EK,\qquad
KF=q^{-2}FK,\qquad[E,F]=\frac{K-K^{-1}}{q-q^{-1}},$

where $q$ is a fixed parameter in $\mathbb{K}$, assumed not to be a root of
unity. A Casimir element, which commutes with all elements of
$U_{q}(\mathfrak{sl}_{2})$, is given by

$\displaystyle\Lambda=\big{(}q-q^{-1}\big{)}^{2}EF+q^{-1}K+qK^{-1}.$ (2.1)

The quantum group $U_{q}(\mathfrak{sl}_{2})$ has the structure of a bialgebra:
it is equipped with a coproduct $\Delta\colon U_{q}(\mathfrak{sl}_{2})\to
U_{q}(\mathfrak{sl}_{2})^{\otimes 2}$, which satisfies the coassociativity
property $(1\otimes\Delta)\Delta=(\Delta\otimes 1)\Delta$, and a counit
$\epsilon\colon U_{q}(\mathfrak{sl}_{2})\to\mathbb{K}$ satisfying
$(1\otimes\epsilon)\Delta=(\epsilon\otimes 1)\Delta=1$, where $1$ denotes the
identity mapping on $U_{q}(\mathfrak{sl}_{2})$. Explicitly, they are given by

$\displaystyle\Delta(E)=E\otimes 1+K\otimes E,\qquad\\!\Delta(F)=F\otimes
K^{-1}+1\otimes F,\qquad\\!\Delta\big{(}K^{\pm 1}\big{)}=K^{\pm 1}\otimes
K^{\pm 1},\\!\\!$ (2.2)
$\displaystyle\epsilon(E)=\epsilon(F)=0,\qquad\epsilon(K)=\epsilon\big{(}K^{-1}\big{)}=1.$

A binary operation we will often use is the so-called $q$-commutator. For
$X,Y\in U_{q}(\mathfrak{sl}_{2})$ we write

$[X,Y]_{q}=qXY-q^{-1}YX.$

For $i$ and $j$ natural numbers with $i\leq j$, we will write $[i;j]$ to
denote the discrete interval $\\{i,i+1,\dots,j-1,j\\}$. If we consider
disjoint unions of discrete intervals, often denoted by
$[i_{1};j_{1}]\cup[i_{2};j_{2}]\cup\dots\cup[i_{k};j_{k}]$, it is always
understood that $i_{\ell}\leq j_{\ell}<i_{\ell+1}-1$ for all $\ell$. Note that
this implies that $i_{k}\geq i_{1}+2k-2$. Moreover, if $B$ is any set of
natural numbers and $a\in\mathbb{N}$, we will write $B-a$ for the set
$\\{b-a\colon b\in B\\}$.

###  2.1 Coideals and comodules

In [8] we have introduced the following $U_{q}(\mathfrak{sl}_{2})$-subalgebra.

######  Definition 2.1.

We denote by $\mathcal{I}_{R}$ the subalgebra of $U_{q}(\mathfrak{sl}_{2})$
generated by $EK^{-1}$, $F$, $K^{-1}$ and $\Lambda$, and we define the algebra
morphism $\tau_{R}\colon\mathcal{I}_{R}\to
U_{q}(\mathfrak{sl}_{2})\otimes\mathcal{I}_{R}$ through its action on the
generators:

$\displaystyle\tau_{R}\big{(}EK^{-1}\big{)}=K^{-1}\otimes EK^{-1},$
$\displaystyle\tau_{R}(F)=K\otimes
F-q^{-3}\big{(}q-q^{-1}\big{)}^{2}F^{2}K\otimes
EK^{-1}+q^{-1}\big{(}q+q^{-1}\big{)}FK\otimes K^{-1}-q^{-1}FK\otimes\Lambda,$
$\displaystyle\tau_{R}\big{(}K^{-1}\big{)}=1\otimes
K^{-1}-q^{-1}\big{(}q-q^{-1}\big{)}^{2}F\otimes EK^{-1},$
$\displaystyle\tau_{R}(\Lambda)=1\otimes\Lambda.$

It is readily checked that these definitions comply with the algebra relations
in $\mathcal{I}_{R}$. The following important observation was made in [8,
Proposition 3].

######  Proposition 2.1.

The algebra $\mathcal{I}_{R}$ is a left coideal subalgebra of
$U_{q}(\mathfrak{sl}_{2})$ and a left $U_{q}(\mathfrak{sl}_{2})$-comodule with
coaction $\tau_{R}$. This means that $\Delta(\mathcal{I}_{R})\subset
U_{q}(\mathfrak{sl}_{2})\otimes\mathcal{I}_{R}$ and that one has

$\displaystyle(1\otimes\tau_{R})\tau_{R}=(\Delta\otimes 1)\tau_{R},$ (2.3)
$\displaystyle(\epsilon\otimes 1)\tau_{R}=1.$

An interpretation of the coaction $\tau_{R}$ in terms of the universal
$R$-matrix for $U_{q}(\mathfrak{sl}_{2})$ was recently given in [6]. This
coaction was constructed so as to satisfy the identity

$\displaystyle\Lambda_{\\{1,3\\}}=(1\otimes\tau_{R})\Delta(\Lambda),$ (2.4)

with $\Lambda_{\\{1,3\\}}$ defined through (1.1) and (1.2). A similar mapping
$\tau_{L}$ can be constructed by demanding that

$\displaystyle\Lambda_{\\{1,3\\}}=(\tau_{L}\otimes 1)\Delta(\Lambda).$ (2.5)

This suggests the following definition.

######  Definition 2.2.

We denote by $\mathcal{I}_{L}$ the subalgebra of $U_{q}(\mathfrak{sl}_{2})$
generated by $E$, $FK$, $K$ and $\Lambda$, and we define the algebra morphism
$\tau_{L}\colon\mathcal{I}_{L}\to\mathcal{I}_{L}\otimes
U_{q}(\mathfrak{sl}_{2})$ through its action on the generators:

$\displaystyle\tau_{L}(E)=E\otimes K,$ $\displaystyle\tau_{L}(FK)=FK\otimes
K^{-1}-q^{-1}\big{(}q-q^{-1}\big{)}^{2}E\otimes
F^{2}K+q\big{(}q+q^{-1}\big{)}K\otimes F-q\Lambda\otimes F,$
$\displaystyle\tau_{L}(K)=K\otimes 1-q^{-1}\big{(}q-q^{-1}\big{)}^{2}E\otimes
FK,$ $\displaystyle\tau_{L}(\Lambda)=\Lambda\otimes 1.$

This subalgebra behaves in a similar fashion.

######  Proposition 2.2.

The algebra $\mathcal{I}_{L}$ is a right coideal subalgebra of
$U_{q}(\mathfrak{sl}_{2})$ and a right $U_{q}(\mathfrak{sl}_{2})$-comodule
with coaction $\tau_{L}$.

###### Proof.

It suffices to check explicitly on each of the generators that

  1. 1)

$\mathcal{I}_{L}$ is a right coideal of $U_{q}(\mathfrak{sl}_{2})$:
$\Delta(\mathcal{I}_{L})\subset\mathcal{I}_{L}\otimes
U_{q}(\mathfrak{sl}_{2})$,

  2. 2)

$\tau_{L}$ is a right coaction: it preserves the algebra relations in
$\mathcal{I}_{L}$ and satisfies

$\displaystyle(\tau_{L}\otimes 1)\tau_{L}=(1\otimes\Delta)\tau_{L},$ (2.6)
$\displaystyle(1\otimes\epsilon)\tau_{L}=1.$ ∎

It is readily checked that the element $\Delta(\Lambda)$ lies in
$\mathcal{I}_{L}\otimes\mathcal{I}_{R}$. Indeed, it follows immediately from
(2.1) and (2.2) that one has

$\Delta(\Lambda)=\Lambda\otimes
K^{-1}+K\otimes\Lambda-\big{(}q+q^{-1}\big{)}K\otimes
K^{-1}+\big{(}q-q^{-1}\big{)}^{2}\big{(}E\otimes F+q^{-2}FK\otimes
EK^{-1}\big{)}.$

In the language of category theory [1], the equality of (2.4) and (2.5) can be
phrased as follows.

######  Corollary 2.1.

The element $\Delta(\Lambda)\in\mathcal{I}_{L}\otimes\mathcal{I}_{R}$, with
$\Lambda$ defined in (2.1), belongs to the cotensor product of the coideal
comodule subalgebras $\mathcal{I}_{L}$ and $\mathcal{I}_{R}$ of
$U_{q}(\mathfrak{sl}_{2})$:

$\displaystyle(1\otimes\tau_{R})\Delta(\Lambda)=(\tau_{L}\otimes
1)\Delta(\Lambda).$ (2.7)

###  2.2 The right extension process

Our goal in this section will be to associate to each set $A\subseteq[1;n]$ an
element $\Lambda_{A}\in U_{q}(\mathfrak{sl}_{2})^{\otimes n}$, which will
serve as a generator for the higher rank Askey–Wilson algebra ${\rm AW}(n)$.
For the empty set, this will simply be the scalar

$\displaystyle\Lambda_{\varnothing}=q+q^{-1}.$ (2.8)

For general $A$, a construction algorithm was given in [8]. We will repeat it
here in a more accessible notation.

######  Definition 2.3.

For any set $A=\\{a_{1},\dots,a_{m}\\}\subseteq[1;n]$, ordered such that
$a_{i}<a_{i+1}$ for all $i$, we define $\Lambda_{A}\in
U_{q}(\mathfrak{sl}_{2})^{\otimes n}$ by

$\Lambda_{A}=1^{\otimes(a_{1}-1)}\otimes\left(\overrightarrow{\prod_{i=2}^{m}}\,\mu^{A}_{i}\right)(\Lambda)\otimes
1^{\otimes(n-a_{m})},$

with

$\displaystyle\mu_{i}^{A}=\left(\overrightarrow{\prod_{\ell=a_{i-1}-a_{1}+1}^{a_{i}-a_{1}-1}}\big{(}1^{\otimes\ell}\otimes\tau_{R}\big{)}\right)\big{(}1^{\otimes(a_{i-1}-a_{1})}\otimes\Delta\big{)},$
(2.9)

where it is understood that the term between brackets in (2.9) is absent if
$a_{i}=a_{i-1}+1$.

######  Example 2.1.

If $n=9$ and $A=\\{2,4,5,8\\}$, then we have

$\Lambda_{\\{2,4,5,8\\}}=1\otimes\big{(}\mu_{4}^{\\{2,4,5,8\\}}\mu_{3}^{\\{2,4,5,8\\}}\mu_{2}^{\\{2,4,5,8\\}}\big{)}(\Lambda)\otimes
1,$

with $\mu_{2}^{\\{2,4,5,8\\}}=(1\otimes\tau_{R})\Delta$,
$\mu_{3}^{\\{2,4,5,8\\}}=1^{\otimes 2}\otimes\Delta$ and
$\mu_{4}^{\\{2,4,5,8\\}}=\big{(}1^{\otimes
5}\otimes\tau_{R}\big{)}\big{(}1^{\otimes
4}\otimes\tau_{R}\big{)}\big{(}1^{\otimes 3}\otimes\Delta\big{)}$.

The rationale behind this construction is that each element of $A$, except for
its minimum $a_{1}$, corresponds to an application of $\Delta$, whereas the
coaction $\tau_{R}$ is used to _create the gaps_ between the elements of $A$.
In the example above, $\tau_{R}$ is applied first once and then twice,
corresponding to a _hole_ of size 1 between 2 and 4 and one of size 2 between
5 and 8. The improvement with respect to the notation of [8] lies in the fact
that here we iterate over the elements of the set $A$ rather than over all
elements of $[1;n]$, such that we can avoid distinguishing several cases as in
[8].

######  Definition 2.4.

The Askey–Wilson algebra of rank $n-2$, denoted ${\rm AW}(n)$, is the
subalgebra of $U_{q}(\mathfrak{sl}_{2})^{\otimes n}$ generated by all
$\Lambda_{A}$ with $A\subseteq[1;n]$.

We will refer to the algorithm described in Definition 2.3 as the right
extension process, as to make the distinction with the following, alternative
construction method.

###  2.3 The left and mixed extension processes

An alternative method to associate to each $A\subseteq[1;n]$ an element of
$U_{q}(\mathfrak{sl}_{2})^{\otimes n}$ uses the left coideal comodule
subalgebra $\mathcal{I}_{L}$ and its coaction $\tau_{L}$, as introduced in
Definition 2.2.

######  Definition 2.5.

For any set $A=\\{a_{1},\dots,a_{m}\\}\subseteq[1;n]$, ordered such that
$a_{i}<a_{i+1}$ for all $i$, we define $\widehat{\Lambda}_{A}\in
U_{q}(\mathfrak{sl}_{2})^{\otimes n}$ by

$\widehat{\Lambda}_{A}=1^{\otimes(a_{1}-1)}\otimes\left(\overleftarrow{\prod_{i=1}^{m-1}}\widehat{\mu}_{i}^{A}\right)(\Lambda)\otimes
1^{\otimes(n-a_{m})},$

with

$\displaystyle\widehat{\mu}_{i}^{A}=\left(\overrightarrow{\prod_{\ell=a_{m}-a_{i+1}+1}^{a_{m}-a_{i}-1}}\big{(}\tau_{L}\otimes
1^{\otimes\ell}\big{)}\right)\big{(}\Delta\otimes
1^{\otimes(a_{m}-a_{i+1})}\big{)},$ (2.10)

where of course the term between brackets is absent if $a_{i+1}=a_{i}+1$.

######  Example 2.2.

As before we take $n=9$ and $A=\\{2,4,5,8\\}$, and find

$\widehat{\Lambda}_{\\{2,4,5,8\\}}=1\otimes\big{(}\widehat{\mu}_{1}^{\\{2,4,5,8\\}}\widehat{\mu}_{2}^{\\{2,4,5,8\\}}\widehat{\mu}_{3}^{\\{2,4,5,8\\}}\big{)}(\Lambda)\otimes
1,$

with $\widehat{\mu}_{3}^{\\{2,4,5,8\\}}=\big{(}\tau_{L}\otimes 1^{\otimes
2}\big{)}(\tau_{L}\otimes 1)\Delta$,
$\widehat{\mu}_{2}^{\\{2,4,5,8\\}}=\Delta\otimes 1^{\otimes 3}$ and
$\widehat{\mu}_{1}^{\\{2,4,5,8\\}}=\big{(}\tau_{L}\otimes 1^{\otimes
5}\big{)}\big{(}\Delta\otimes 1^{\otimes 4}\big{)}$.

Again, the idea is that each element of $A$ but the maximum $a_{m}$
corresponds to an application of $\Delta$, whereas $\tau_{L}$ _creates the
holes_. However, as opposed to Definition 2.3, we now run through the elements
of $A$ in decreasing order, from right to left. This is why we refer to the
algorithm of Definition 2.5 as the left extension process.

Our first major task will be to prove the equivalence of the right and left
extension processes, i.e., to show that they produce the same elements, for
each set $A$. To do so, it will often be needed to switch the order of certain
algebra morphisms which act on mutually disjoint tensor product positions.
More precisely, if $X\in U_{q}(\mathfrak{sl}_{2})^{\otimes 2}$ and
$\varphi,\psi\colon U_{q}(\mathfrak{sl}_{2})\to
U_{q}(\mathfrak{sl}_{2})^{\otimes 2}$, then we have the following basic
property:

$\displaystyle(1\otimes 1\otimes\psi)(\varphi\otimes 1)X=(\varphi\otimes
1\otimes 1)(1\otimes\psi)X.$ (2.11)

######  Remark 2.1.

This property also allows to replace the definitions (2.9) and (2.10) of
$\mu_{i}^{A}$ and $\widehat{\mu}_{i}^{A}$ by certain equivalent expressions.
For example, $\mu_{2}^{\\{1,5\\}}=\big{(}1^{\otimes
3}\otimes\tau_{R}\big{)}\big{(}1^{\otimes
2}\otimes\tau_{R}\big{)}(1\otimes\tau_{R})\Delta$. Invoking (2.3) and (2.11),
we have

$\displaystyle\big{(}1^{\otimes
2}\otimes\tau_{R}\big{)}(1\otimes\tau_{R})\tau_{R}=\big{(}1^{\otimes
2}\otimes\tau_{R}\big{)}(\Delta\otimes 1)\tau_{R}$
$\displaystyle\hphantom{\big{(}1^{\otimes
2}\otimes\tau_{R}\big{)}(1\otimes\tau_{R})\tau_{R}}{}=\big{(}\Delta\otimes
1^{\otimes 2}\big{)}(1\otimes\tau_{R})\tau_{R}=\big{(}\Delta\otimes 1^{\otimes
2}\big{)}(\Delta\otimes 1)\tau_{R},$

which allows to rewrite $\mu_{2}^{\\{1,5\\}}$, and of course this applies to
any morphism of the form $\mu_{i}^{A}$ with $a_{i}-a_{i-1}>2$. Similarly, one
finds from (2.6) and (2.11) that

$\big{(}\tau_{L}\otimes 1^{\otimes 2}\big{)}(\tau_{L}\otimes
1)\tau_{L}=\big{(}1^{\otimes 2}\otimes\Delta\big{)}(1\otimes\Delta)\tau_{L}$

and its generalizations for any $\widehat{\mu}_{i}^{A}$ with
$a_{i+1}-a_{i}>2$.

This observation will now help us show the equivalence of both extension
processes.

######  Proposition 2.3.

The right and left extension processes produce exactly the same generators:
for each $A\subseteq[1;n]$ one has $\Lambda_{A}=\widehat{\Lambda}_{A}$.

###### Proof.

Without loss of generality, we can assume that $\min(A)=1$ and $\max(A)=n$,
since otherwise it suffices to add $1$ in the remaining positions. We write
$A=[1;j_{1}]\cup[i_{2};j_{2}]\cup\dots\cup[i_{m};n]$ and proceed by induction
on $m$. The case $m=1$ follows from coassociativity:

$\Lambda_{[1;n]}=\big{(}1^{\otimes(n-2)}\otimes\Delta\big{)}\cdots(1\otimes\Delta)\Delta(\Lambda)=\big{(}\Delta\otimes
1^{\otimes(n-2)}\big{)}\cdots(\Delta\otimes
1)\Delta(\Lambda)=\widehat{\Lambda}_{[1;n]}.$

Suppose now the claim is true for all sets consisting of $m-1$ discrete
intervals, including
$\widehat{A}=[1;j_{1}]\cup[i_{2};j_{2}]\cup\dots\cup[i_{m-1};j_{m-1}+1]$:
$\Lambda_{\widehat{A}}=\widehat{\Lambda}_{\widehat{A}}$. The right extension
process asserts

$\displaystyle\Lambda_{A}=\big{(}1^{\otimes(n-2)}\otimes\Delta\big{)}\cdots\big{(}1^{\otimes(i_{m}-1)}\otimes\Delta\big{)}\big{(}1^{\otimes(i_{m}-2)}\otimes\tau_{R}\big{)}\cdots\big{(}1^{\otimes
j_{m-1}}\otimes\tau_{R}\big{)}\Lambda_{\widehat{A}}.$ (2.12)

On the other hand, by the left extension process we have

$\displaystyle\Lambda_{\widehat{A}}=\widehat{\Lambda}_{\widehat{A}}=\big{(}\alpha_{j_{m-1}-1}\otimes
1^{\otimes(j_{m-1}-1)}\big{)}\cdots(\alpha_{1}\otimes 1)\Delta(\Lambda),$
(2.13)

with each $\alpha_{i}\in\\{\Delta,\tau_{L}\\}$. When combining (2.12) and
(2.13), it is clear that the second tensor product position in
$\Delta(\Lambda)$ is left invariant by all the $\alpha_{i}$. Hence by (2.11)
we may shift the morphisms in (2.12) through those in (2.13), such that

$\displaystyle\Lambda_{A}=\big{(}\alpha_{j_{m-1}-1}\otimes
1^{\otimes(n-2)}\big{)}\cdots\big{(}\alpha_{1}\otimes
1^{\otimes(n-j_{m-1})}\big{)}\Lambda_{B},$ (2.14)

with

$\Lambda_{B}=\big{(}1^{\otimes(n-j_{m-1}-1)}\otimes\Delta\big{)}\cdots\big{(}1^{\otimes(i_{m}-j_{m-1})}\otimes\Delta\big{)}\big{(}1^{\otimes(i_{m}-j_{m-1}-1)}\otimes\tau_{R}\big{)}\cdots(1\otimes\tau_{R})\Delta(\Lambda).$

By (2.7) and (2.11) we have

$(1\otimes 1\otimes\tau_{R})(1\otimes\tau_{R})\Delta(\Lambda)=(\tau_{L}\otimes
1\otimes 1)(1\otimes\tau_{R})\Delta(\Lambda)=(\tau_{L}\otimes 1\otimes
1)(\tau_{L}\otimes 1)\Delta(\Lambda).$

Repeating this, we find

$\Lambda_{B}=\big{(}1^{\otimes(n-j_{m-1}-1)}\otimes\Delta\big{)}\cdots\big{(}1^{\otimes(i_{m}-j_{m-1})}\otimes\Delta\big{)}\big{(}\tau_{L}\otimes
1^{\otimes(i_{m}-j_{m-1}-1)}\big{)}\cdots(\tau_{L}\otimes 1)\Delta(\Lambda).$

Invoking (2.11) again, we may shift all the $1^{\otimes\ell}\otimes\Delta$
through all the $\tau_{L}\otimes 1^{\otimes m}$, which, after using
coassociativity, gives

$\displaystyle\Lambda_{B}=\big{(}\tau_{L}\otimes
1^{\otimes(n-j_{m-1}-1)}\big{)}\cdots\big{(}\tau_{L}\otimes
1^{\otimes(n-i_{m}+1)}\big{)}\big{(}\Delta\otimes
1^{\otimes(n-i_{m})}\big{)}\cdots(\Delta\otimes 1)\Delta(\Lambda)$
$\displaystyle\hphantom{\Lambda_{B}}{}=\widehat{\Lambda}_{\\{1\\}\cup[i_{m}-j_{m-1}+1;n-j_{m-1}+1]}.$

Moreover, the left extension process and (2.13) imply that

$\widehat{\Lambda}_{A}=\big{(}\alpha_{j_{m-1}-1}\otimes
1^{\otimes(n-2)}\big{)}\cdots\big{(}\alpha_{1}\otimes
1^{\otimes(n-j_{m-1})}\big{)}\widehat{\Lambda}_{\\{1\\}\cup[i_{m}-j_{m-1}+1;n-j_{m-1}+1]}.$

The statement now follows from (2.14). ∎

The reasoning established in the proof of Proposition 2.3 suggests the
existence of several other, so-called mixed extension processes, which produce
the same elements $\Lambda_{A}$. To be precise, one can split the set $A$ at
any $a_{j}$ and perform the right extension process for the subset
$\\{a_{j+1},\dots,a_{m}\\}$ and consequently the left extension process for
$\\{a_{1},\dots,a_{j-1}\\}$. This is described in the following definition.

######  Definition 2.6.

For any set $A=\\{a_{1},\dots,a_{m}\\}\subseteq[1;n]$, ordered such that
$a_{i}<a_{i+1}$ for all $i$, we define $\Lambda_{A}^{(j)}\in
U_{q}(\mathfrak{sl}_{2})^{\otimes n}$ by

$\Lambda_{A}^{(j)}=1^{\otimes(a_{1}-1)}\otimes\left(\overleftarrow{\prod_{i=1}^{j-1}}\widehat{\mu}_{i}^{A}\right)\left(\overrightarrow{\prod_{i=j+1}^{m}\mu_{i,j}^{A}}\right)(\Lambda)\otimes
1^{\otimes(n-a_{m})},$

with $\widehat{\mu}_{i}^{A}$ as in (2.10) and

$\mu_{i,j}^{A}=\left(\overrightarrow{\prod_{\ell=a_{i-1}-a_{j}+1}^{a_{i}-a_{j}-1}}\big{(}1^{\otimes\ell}\otimes\tau_{R}\big{)}\right)\big{(}1^{\otimes(a_{i-1}-a_{j})}\otimes\Delta\big{)}.$

By definition, we have $\Lambda_{A}^{(1)}=\Lambda_{A}$ and
$\Lambda_{A}^{(m)}=\widehat{\Lambda}_{A}$ and following the proof of
Proposition 2.3, we even have $\Lambda_{A}^{(j)}=\Lambda_{A}$ for all
$j\in\\{1,\dots,m\\}$.

######  Example 2.3.

For $n=9$, $A=\\{2,4,5,8\\}$ and $j=2$ we have

$\Lambda_{\\{2,4,5,8\\}}^{(2)}=1\otimes\big{(}\widehat{\mu}_{1}^{\\{2,4,5,8\\}}\mu_{4,2}^{\\{2,4,5,8\\}}\mu_{3,2}^{\\{2,4,5,8\\}}\big{)}(\Lambda)\otimes
1,$

with $\mu_{3,2}^{\\{2,4,5,8\\}}=\Delta$,
$\mu_{4,2}^{\\{2,4,5,8\\}}=\big{(}1^{\otimes
3}\otimes\tau_{R}\big{)}\big{(}1^{\otimes
2}\otimes\tau_{R}\big{)}(1\otimes\Delta)$ and
$\widehat{\mu}_{1}^{\\{2,4,5,8\\}}=\big{(}\tau_{L}\otimes 1^{\otimes
5}\big{)}\big{(}\Delta\otimes 1^{\otimes 4}\big{)}$.

###  2.4 Another layer of freedom

In the extension processes described above, the order in which we apply the
different morphisms is of high importance. Nevertheless, there is some
additional freedom in this order, which will come in handy in many of the
following proofs. More precisely, the upcoming Proposition 2.4 asserts that
when constructing $\Lambda_{A}$, it suffices to first create all the holes
between elements of $A$ in order of appearance and then enlarge all holes and
all intervals by applying repeatedly the coproduct $\Delta$.

######  Proposition 2.4.

Let $A=[1;j_{1}]\cup[i_{2};j_{2}]\cup\dots\cup[i_{k};j_{k}]$, then one has

$\displaystyle\Lambda_{A}=\left(\overleftarrow{\prod_{n=0}^{2k-2}}\overrightarrow{\prod_{\ell=\alpha_{n,\mathbf{i},\mathbf{j}}}^{\beta_{n,\mathbf{i},\mathbf{j}}}}(1^{\otimes
n}\otimes\Delta\otimes
1^{\otimes\ell})\right)\Lambda_{\\{1,3,5,\dots,2k-1\\}},$ (2.15)

where

$\alpha_{n,\mathbf{i},\mathbf{j}}=\begin{cases}j_{k}-j_{m}&\text{if}\ n=2m-2\
\text{is even},\\\ j_{k}-i_{m}+1&\text{if}\ n=2m-3\ \text{is odd}\end{cases}$

and

$\beta_{n,\mathbf{i},\mathbf{j}}=\begin{cases}j_{k}-i_{m}-1&\text{if}\ n=2m-2\
\text{is even},\\\ j_{k}-j_{m-1}-2&\text{if}\ n=2m-3\ \text{is
odd}\end{cases}$

and where we set $\mathbf{i}=(1,i_{2},\dots,i_{k})$ and
$\mathbf{j}=(j_{1},j_{2},\dots,j_{k})$.

###### Proof.

By induction on $k$, the case $k=1$ being trivial by coassociativity. Suppose
hence the claim holds for all sets consisting of $k-1$ discrete intervals,
including
$\widehat{A}=[1;j_{1}]\cup[i_{2};j_{2}]\cup\dots\cup[i_{k-1};j_{k-1}]$:

$\displaystyle\Lambda_{\widehat{A}}=\left(\overleftarrow{\prod_{n=0}^{2k-4}}\overrightarrow{\prod_{\ell=\alpha_{n,\widetilde{\mathbf{i}},\widetilde{\mathbf{j}}}}^{\beta_{n,\widetilde{\mathbf{i}},\widetilde{\mathbf{j}}}}}\big{(}1^{\otimes
n}\otimes\Delta\otimes
1^{\otimes\ell}\big{)}\right)\Lambda_{\\{1,3,5,\dots,2k-3\\}},$ (2.16)

where $\widetilde{\mathbf{i}}=(1,i_{2},\dots,i_{k-1})$ and
$\widetilde{\mathbf{j}}=(j_{1},j_{2},\dots,j_{k-1})$. From the right extension
process, using coassociativity and Remark 2.1, it is clear that

$\displaystyle\Lambda_{A}=\big{(}1^{\otimes(i_{k}-1)}\otimes\Delta\otimes
1^{\otimes(j_{k}-i_{k}-1)}\big{)}\cdots\big{(}1^{\otimes(i_{k}-1)}\otimes\Delta\big{)}\big{(}1^{\otimes
j_{k-1}}\otimes\Delta\otimes 1^{\otimes(i_{k}-j_{k-1}-2)}\big{)}\cdots$
$\displaystyle\hphantom{\Lambda_{A}=}{}\times\big{(}1^{\otimes
j_{k-1}}\otimes\Delta\otimes 1\big{)}\big{(}1^{\otimes
j_{k-1}}\otimes\tau_{R}\big{)}\big{(}1^{\otimes(j_{k-1}-1)}\otimes\Delta\big{)}\Lambda_{\widehat{A}}.$
(2.17)

Observe that when combining (2.16) and (2.17), the morphisms in (2.16) with
$n\leq 2k-5$ will act on tensor product positions in
$\Lambda_{\\{1,3,5,\dots,2k-3\\}}$ of lower index than the morphisms in the
second line in (2.17). We may hence switch their order as in (2.11):

$\displaystyle\big{(}1^{\otimes
j_{k-1}}\otimes\tau_{R}\big{)}\big{(}1^{\otimes(j_{k-1}-1)}\otimes\Delta\big{)}\Lambda_{\widehat{A}}=\left(\overleftarrow{\prod_{n=0}^{2k-5}}\overrightarrow{\prod_{\ell=\alpha_{n,\widetilde{\mathbf{i}},\widetilde{\mathbf{j}}}+2}^{\beta_{n,\widetilde{\mathbf{i}},\widetilde{\mathbf{j}}}+2}}\big{(}1^{\otimes
n}\otimes\Delta\otimes 1^{\otimes\ell}\big{)}\right)\Lambda_{B},$ (2.18)

where $\Lambda_{B}$ is defined as $\chi(\Lambda_{\\{1,3,5,\dots,2k-3\\}})$,
where $\chi$ is the morphism

$\displaystyle\big{(}1^{\otimes(j_{k-1}-i_{k-1}+2k-3)}\otimes\tau_{R}\big{)}\big{(}1^{\otimes(j_{k-1}-i_{k-1}+2k-4)}\otimes\Delta\big{)}\left(\overrightarrow{\prod_{\ell=0}^{j_{k-1}-i_{k-1}-1}}\big{(}1^{\otimes(2k-4)}\otimes\Delta\otimes
1^{\otimes\ell}\big{)}\right).$

Here, the term between brackets corresponds to $n=2k-4$ in (2.16), since

$\displaystyle\alpha_{2k-4,\widetilde{\mathbf{i}},\widetilde{\mathbf{j}}}=0\qquad\mathrm{and}\qquad\beta_{2k-4,\widetilde{\mathbf{i}},\widetilde{\mathbf{j}}}=j_{k-1}-i_{k-1}-1.$
(2.19)

Using coassociativity, separating the term for $\ell=0$ and relying on the
right extension process, we have

$\displaystyle\Lambda_{B}=\big{(}1^{\otimes(j_{k-1}-i_{k-1}+2k-3)}\otimes\tau_{R}\big{)}\left(\overrightarrow{\prod_{\ell=1}^{j_{k-1}-i_{k-1}}}\big{(}1^{\otimes(2k-4)}\otimes\Delta\otimes
1^{\otimes\ell}\big{)}\right)\Lambda_{\\{1,3,5,\dots,2k-3,2k-2\\}}.$

Now it is manifest that all $\Delta$ in the product between brackets act on
the tensor product position $2k-3$ in $\Lambda_{\\{1,3,5,\dots,2k-3,2k-2\\}}$,
whereas $\tau_{R}$ in fact acts on the last position $2k-2$. Hence we may
again apply (2.11) to switch the order:

$\Lambda_{B}=\left(\overrightarrow{\prod_{\ell=2}^{j_{k-1}-i_{k-1}+1}}\big{(}1^{\otimes(2k-4)}\otimes\Delta\otimes
1^{\otimes\ell}\big{)}\right)\Lambda_{\\{1,3,5,\dots,2k-3,2k-1\\}}.$

Note that the lower and upper bounds in the product between brackets equal
$\alpha_{2k-4,\widetilde{\mathbf{i}},\widetilde{\mathbf{j}}}+2$ and
$\beta_{2k-4,\widetilde{\mathbf{i}},\widetilde{\mathbf{j}}}+2$ respectively,
by (2.19).

Combined with (2.17) and (2.18), this yields

$\displaystyle\Lambda_{A}=\big{(}1^{\otimes(i_{k}-1)}\otimes\Delta\otimes
1^{\otimes(j_{k}-i_{k}-1)}\big{)}\cdots\big{(}1^{\otimes(i_{k}-1)}\otimes\Delta\big{)}\big{(}1^{\otimes
j_{k-1}}\otimes\Delta\otimes 1^{\otimes(i_{k}-j_{k-1}-2)}\big{)}\cdots$
$\displaystyle\hphantom{\Lambda_{A}=}{}\times\big{(}1^{\otimes
j_{k-1}}\otimes\Delta\otimes
1\big{)}\left(\overleftarrow{\prod_{n=0}^{2k-4}}\overrightarrow{\prod_{\ell=\alpha_{n,\widetilde{\mathbf{i}},\widetilde{\mathbf{j}}}+2}^{\beta_{n,\widetilde{\mathbf{i}},\widetilde{\mathbf{j}}}+2}}\big{(}1^{\otimes
n}\otimes\Delta\otimes
1^{\otimes\ell}\big{)}\right)\Lambda_{\\{1,3,5,\dots,2k-3,2k-1\\}}.$

Moreover, all morphisms in the first line act on the tensor product positions
$2k-2$ or $2k-1$ of $\Lambda_{\\{1,3,5,\dots,2k-3,2k-1\\}}$, whereas those on
the second line act on positions 1 to $2k-3$. The order can hence be switched
again by (2.11), leading first to

$\displaystyle\Lambda_{A}=\left(\overleftarrow{\prod_{n=0}^{2k-4}}\overrightarrow{\prod_{\ell=\alpha_{n,\widetilde{\mathbf{i}},\widetilde{\mathbf{j}}}+j_{k}-j_{k-1}}^{\beta_{n,\widetilde{\mathbf{i}},\widetilde{\mathbf{j}}}+j_{k}-j_{k-1}}}\big{(}1^{\otimes
n}\otimes\Delta\otimes 1^{\otimes\ell}\big{)}\right)$
$\displaystyle\hphantom{\Lambda_{A}=}{}\times\big{(}1^{\otimes(i_{k}-j_{k-1}+2k-4)}\otimes\Delta\otimes
1^{\otimes(j_{k}-i_{k}-1)}\big{)}\cdots\big{(}1^{\otimes(i_{k}-j_{k-1}+2k-4)}\otimes\Delta\big{)}$
$\displaystyle\hphantom{\Lambda_{A}=}{}\times\big{(}1^{\otimes(2k-3)}\otimes\Delta\otimes
1^{\otimes(i_{k}-j_{k-1}-2)}\big{)}\cdots\big{(}1^{\otimes(2k-3)}\otimes\Delta\otimes
1\big{)}\Lambda_{\\{1,3,5,\dots,2k-3,2k-1\\}}$

and then, after switching the morphisms on the second with those on the third
line, to

$\displaystyle\Lambda_{A}=\left(\overleftarrow{\prod_{n=0}^{2k-4}}\overrightarrow{\prod_{\ell=\alpha_{n,\widetilde{\mathbf{i}},\widetilde{\mathbf{j}}}+j_{k}-j_{k-1}}^{\beta_{n,\widetilde{\mathbf{i}},\widetilde{\mathbf{j}}}+j_{k}-j_{k-1}}}\big{(}1^{\otimes
n}\otimes\Delta\otimes 1^{\otimes\ell}\big{)}\right)$
$\displaystyle\hphantom{\Lambda_{A}=}{}\times\big{(}1^{\otimes(2k-3)}\otimes\Delta\otimes
1^{\otimes(j_{k}-j_{k-1}-2)}\big{)}\cdots\big{(}1^{\otimes(2k-3)}\otimes\Delta\otimes
1^{\otimes(j_{k}-i_{k}+1)}\big{)}$
$\displaystyle\hphantom{\Lambda_{A}=}{}\times\big{(}1^{\otimes(2k-2)}\otimes\Delta\otimes
1^{\otimes(j_{k}-i_{k}-1)}\big{)}\cdots\big{(}1^{\otimes(2k-2)}\otimes\Delta\big{)}\Lambda_{\\{1,3,5,\dots,2k-3,2k-1\\}}.$

Now observe that for every $n\in\\{0,1,\dots,2k-4\\}$ one has
$\alpha_{n,\mathbf{i},\mathbf{j}}=\alpha_{n,\widetilde{\mathbf{i}},\widetilde{\mathbf{j}}}+j_{k}-j_{k-1}$
and
$\beta_{n,\mathbf{i},\mathbf{j}}=\beta_{n,\widetilde{\mathbf{i}},\widetilde{\mathbf{j}}}+j_{k}-j_{k-1}$,
and moreover we have $\alpha_{2k-3,\mathbf{i},\mathbf{j}}=j_{k}-i_{k}+1$,
$\beta_{2k-3,\mathbf{i},\mathbf{j}}=j_{k}-j_{k-1}-2$,
$\alpha_{2k-2,\mathbf{i},\mathbf{j}}=0$ and
$\beta_{2k-2,\mathbf{i},\mathbf{j}}=j_{k}-i_{k}-1$. Hence the expression above
coincides with (2.15). ∎

##  3 Main results and strategy of proof

In this section, we formulate the main results of this paper: the algebraic
relations satisfied in the higher rank Askey–Wilson algebra ${\rm AW}(n)$. As
in the rank one case, these will be of the form

$\displaystyle[\Lambda_{A},\Lambda_{B}]_{q}=\big{(}q^{-2}-q^{2}\big{)}\Lambda_{(A\cup
B)\setminus(A\cap B)}+\big{(}q-q^{-1}\big{)}\big{(}\Lambda_{A\cap
B}\Lambda_{A\cup B}+\Lambda_{A\setminus(A\cap B)}\Lambda_{B\setminus(A\cap
B)}\big{)},$ ($\ast$)

or

$\displaystyle[\Lambda_{A},\Lambda_{B}]=0,$ ($\Delta$)

under suitable conditions on the sets $A$ and $B$. In this section, we present
these conditions, which, based on extensive computer calculations, we believe
to be minimal. They can be stated as follows.

######  Theorem 3.1.

Let $A,B\subseteq[1;n]$ be such that $B\subseteq A$, then $\Lambda_{A}$ and
$\Lambda_{B}$ commute.

Section 6 will be devoted to the proof of this theorem.

######  Definition 3.1.

For $A,B\subseteq[1;n]$ we write $A\prec B$ if $\max(A)<\min(B)$ or if either
$A$ or $B$ is empty.

######  Theorem 3.2.

Let $A_{1}$, $A_{2}$, $A_{3}$ and $A_{4}$ be $($potentially empty$)$ subsets
of $[1;n]$ satisfying

$A_{1}\prec A_{2}\prec A_{3}\prec A_{4}.$

The standard relation

$[\Lambda_{A},\Lambda_{B}]_{q}=\big{(}q^{-2}-q^{2}\big{)}\Lambda_{(A\cup
B)\setminus(A\cap B)}+\big{(}q-q^{-1}\big{)}\big{(}\Lambda_{A\cap
B}\Lambda_{A\cup B}+\Lambda_{A\setminus(A\cap B)}\Lambda_{B\setminus(A\cap
B)}\big{)}$

is satisfied for $A$ and $B$ defined by one of the following relations:

$\displaystyle A=A_{1}\cup A_{2}\cup A_{4},\qquad$ $\displaystyle B=A_{2}\cup
A_{3},$ (3.1) $\displaystyle A=A_{2}\cup A_{3},\qquad$ $\displaystyle
B=A_{1}\cup A_{3}\cup A_{4},$ (3.2) $\displaystyle A=A_{1}\cup A_{3}\cup
A_{4},\qquad$ $\displaystyle B=A_{1}\cup A_{2}\cup A_{4}.$ (3.3)

This will be shown in Section 7.

Our general strategy to prove a relation of the form ($\ast$ ‣ 3) will be as
follows. First we will construct an operator $\chi$, by combining morphisms of
the form $1^{\otimes n}\otimes\alpha\otimes 1^{\otimes m}$,
$\alpha\in\\{\Delta,\tau_{R},\tau_{L}\\}$ and $n,m\in\mathbb{N}$, such that

$\chi(\Lambda_{A^{\prime}})=\Lambda_{A},\qquad\chi(\Lambda_{B^{\prime}})=\Lambda_{B},\qquad\chi(\Lambda_{(A^{\prime}\cup
B^{\prime})\setminus(A^{\prime}\cap B^{\prime})})=\Lambda_{(A\cup
B)\setminus(A\cap B)},\qquad\dots$

for certain (less complicated) sets $A^{\prime}$, $B^{\prime}$. To prove
($\ast$ ‣ 3) it now suffices to show

$\displaystyle[\Lambda_{A^{\prime}},\Lambda_{B^{\prime}}]_{q}=\big{(}q^{-2}-q^{2}\big{)}\Lambda_{(A^{\prime}\cup
B^{\prime})\setminus(A^{\prime}\cap B^{\prime})}$
$\displaystyle\hphantom{[\Lambda_{A^{\prime}},\Lambda_{B^{\prime}}]_{q}=}{}+\big{(}q-q^{-1}\big{)}\big{(}\Lambda_{A^{\prime}\cap
B^{\prime}}\Lambda_{A^{\prime}\cup
B^{\prime}}+\Lambda_{A^{\prime}\setminus(A^{\prime}\cap
B^{\prime})}\Lambda_{B^{\prime}\setminus(A^{\prime}\cap B^{\prime})}\big{)},$
($\ast\ast$)

apply the operator $\chi$ to both sides of the equation and use its linearity
and multiplicativity. In this case we will write “($\ast$ ‣ 3) follows from
($\ast\ast$ ‣ 3) by $\chi$”. The same strategy applies to relations of the
form ($\Delta$). In this respect, the relations of Theorem 3.2 can in fact be
derived from the following 9 fundamental cases.

######  Proposition 3.1.

The standard relation ($\ast$ ‣ 3) holds for the following combinations of
sets $A$ and $B$:

$\displaystyle\begin{split}&A=\\{1,2,4,6,\dots,2k\\},\\\
&B=\\{2,4,6,\dots,2k,2k+1\\};\end{split}\vspace{2mm}$ (C1)
$\displaystyle\begin{split}&A=\\{2,4,6,\dots,2k,2k+1\\},\\\
&B=\\{1,2k+1\\};\end{split}\vspace{2mm}$ (C2)
$\displaystyle\begin{split}&A=\\{1,2k+1\\},\\\
&B=\\{1,2,4,6,\dots,2k\\};\end{split}\vspace{2mm}$ (C3)
$\displaystyle\begin{split}&A=\\{1,2,4,6,\dots,2k,2k+2\ell+2\\},\\\
&B=\\{2,4,6,\dots,2k,2k+1,2k+3,\dots,2k+2\ell+1\\};\end{split}\vspace{2mm}$
(C4) $\displaystyle\begin{split}&A=\\{1,2,4,6,\dots,2k,2k+2\ell+3\\},\\\
&B=\\{2,4,6,\dots,2k,2k+2,2k+4,\dots,2k+2\ell+2\\};\end{split}\vspace{2mm}$
(C4′)
$\displaystyle\begin{split}&A=\\{2,4,6,\dots,2k,2k+1,2k+3,\dots,2k+2\ell+1\\},\\\
&B=\\{1,2k+1,2k+3,\dots,2k+2\ell+1,2k+2\ell+2\\};\end{split}\vspace{2mm}$ (C5)
$\displaystyle\begin{split}&A=\\{2,4,6,\dots,2k,2k+2,2k+4,\dots,2k+2\ell+2\\},\\\
&B=\\{1,2k+2,2k+4,\dots,2k+2\ell+2,2k+2\ell+3\\};\end{split}\vspace{2mm}$
(C5′)
$\displaystyle\begin{split}&A=\\{1,2k+1,2k+3,\dots,2k+2\ell+1,2k+2\ell+2\\},\\\
&B=\\{1,2,4,6,\dots,2k,2k+2\ell+2\\};\end{split}\vspace{2mm}$ (C6)
$\displaystyle\begin{split}&A=\\{1,2k+2,2k+4,\dots,2k+2\ell+2,2k+2\ell+3\\},\\\
&B=\\{1,2,4,6,\dots,2k,2k+2\ell+3\\},\end{split}$ (C6′)

with $k,\ell\in\mathbb{N}$.

In the next section, we will provide an elaborate proof for each of these
fundamental relations. They will serve as the building blocks to prove similar
fundamental commutation relations in Section 5 and consequently to prove
Theorems 3.1 and 3.2 in Sections 6 and 7 respectively.

##  4 Proof of Proposition 3.1

###  4.1 Some basic commutation relations

Throughout the whole paper, it will turn out useful to switch orders in nested
$q$-commutators. A straightforward calculation gives the following:

######  Lemma 4.1.

Let $\mathcal{A}$ be any algebra and $\alpha$, $\beta$, $\gamma$, $\delta$
elements of $\mathcal{A}$, then one has

$\displaystyle[\alpha,[\gamma,\delta]_{q}]_{q}=[[\alpha,\gamma]_{q},\delta]_{q}\qquad\mathrm{if}\quad[\alpha,\delta]=0,$
(4.1)
$\displaystyle[\alpha,[\gamma,\delta]_{q}]_{q}=[\gamma,[\alpha,\delta]_{q}]_{q}\qquad\mathrm{if}\quad[\alpha,\gamma]=0,$
(4.2)
$\displaystyle[[\gamma,\delta]_{q},\beta]_{q}=[[\gamma,\beta]_{q},\delta]_{q}\qquad\mathrm{if}\quad[\beta,\delta]=0,$
(4.3)
$\displaystyle[[\gamma,\delta]_{q},\beta]_{q}=[\gamma,[\delta,\beta]_{q}]_{q}\qquad\mathrm{if}\quad[\beta,\gamma]=0.$
(4.4)

The following commutation relations are so natural that they will often be
relied on in proofs in Subsection 4.2 without explicit reference.

######  Lemma 4.2.

For any $i\in[1;n]$ and any subset $A\subseteq[1;n]$ one has

$[\Lambda_{A},\Lambda_{\\{i\\}}]=0.$

###### Proof.

This is immediate by the fact that
$\Lambda_{\\{i\\}}=1^{\otimes(i-1)}\otimes\Lambda\otimes 1^{\otimes(n-i)}$ and
that $\Lambda$ is the Casimir operator of $U_{q}(\mathfrak{sl}_{2})$. ∎

######  Lemma 4.3.

Let $A,A^{\prime}\subseteq[1;n]$ be such that $A\prec A^{\prime}$, where we
use Definition 3.1, then one has

$[\Lambda_{A},\Lambda_{A^{\prime}}]=0,\qquad[\Lambda_{A},\Lambda_{A\cup
A^{\prime}}]=0,\qquad[\Lambda_{A^{\prime}},\Lambda_{A\cup A^{\prime}}]=0.$

###### Proof.

The first statement is trivial, since $\Lambda_{A}$ and $\Lambda_{A^{\prime}}$
live in disjoint tensor product positions. For the second claim, writing

$A=[1;j_{1}]\cup[i_{2};j_{2}]\cup\cdots\cup[i_{k};j_{k}],\qquad
A^{\prime}=[i_{k+1};j_{k+1}]\cup\cdots\cup[i_{k+\ell};j_{k+\ell}],$

the statement follows from $[\Lambda_{\\{1\\}},\Lambda_{\\{1,2\\}}]=0$ by a
suitable morphism of the form

$\displaystyle\chi=\left(\overrightarrow{\prod_{m=i_{k+1}-1}^{j_{k+\ell}-2}}\big{(}1^{\otimes
m}\otimes\beta_{m}\big{)}\right)\left(\overrightarrow{\prod_{m=1}^{i_{k+1}-2}}\big{(}\alpha_{m}\otimes
1^{\otimes m}\big{)}\right),$

where each $\alpha_{m}\in\\{\Delta,\tau_{L}\\}$ and each
$\beta_{m}\in\\{\Delta,\tau_{R}\\}$. The third statement follows analogously.
∎

The next lemma provides a first generalization of the relations (1.2)–(1.4).

######  Lemma 4.4.

Let $i\in[1;n]$ and $A_{1},A_{2}\subseteq[1;n]$ be such that
$A_{1}\prec\\{i\\}\prec A_{2}$, then the relation ($\ast$ ‣ 3) holds for
$(A,B)$ one of the couples

$(A_{1}\cup\\{i\\},\\{i\\}\cup A_{2}),\qquad(\\{i\\}\cup A_{2},A_{1}\cup
A_{2}),\qquad(A_{1}\cup A_{2},A_{1}\cup\\{i\\}).$

###### Proof.

Let $A_{1}\cup\\{i\\}\cup A_{2}=\\{a_{1},\dots,a_{m}\\}$, ordered such that
$a_{\ell}<a_{\ell+1}$ for all $\ell$ and let $j$ be such that $a_{j}=i$. The
mixed extension process with parameter $j$ asserts the existence of a morphism
$\chi$ which sends

$\displaystyle\\{1\\}\mapsto
A_{1},\qquad\\{2\\}\mapsto\\{i\\},\qquad\\{3\\}\mapsto A_{2}.$

Hence the statements follow from (1.2)–(1.4) by $\chi$. ∎

A final immediate commutation relation is the following.

######  Lemma 4.5.

For any $k\in\mathbb{N}$ one has

$[\Lambda_{\\{2,4,\dots,2k\\}},\Lambda_{\\{1,2k+1\\}}]=0.$

###### Proof.

By induction on $k$. The case $k=1$ is trivial by Lemma 4.2. Suppose hence the
claim has been proven for $k-1$. Lemma 4.4 asserts

$\Lambda_{\\{2,4,6,\dots,2k\\}}=\frac{[\Lambda_{\\{2,3\\}},\Lambda_{\\{3,4,6,\dots,2k\\}}]_{q}}{q^{-2}-q^{2}}+\frac{\Lambda_{\\{3\\}}\Lambda_{\\{2,3,4,6,\dots,2k\\}}+\Lambda_{\\{2\\}}\Lambda_{\\{4,6,\dots,2k\\}}}{q+q^{-1}},$

hence it suffices to show that $\Lambda_{\\{1,2k+1\\}}$ commutes with each
term in the right-hand side. For $\Lambda_{\\{2,3\\}}$ this follows from
$[\Lambda_{\\{1,3\\}},\Lambda_{\\{2\\}}]=0$ by
$\chi=\big{(}1^{\otimes(2k-1)}\otimes\tau_{R})\cdots\big{(}1^{\otimes
3}\otimes\tau_{R}\big{)}(1\otimes\Delta\otimes 1)$. The other nontrivial
commutation relations follow from the induction hypothesis, by
$\chi=\big{(}\tau_{L}\otimes
1^{\otimes(2k-1)}\big{)}\big{(}1\otimes\Delta\otimes
1^{\otimes(2k-3)}\big{)}$, $\chi=\big{(}1\otimes\Delta\otimes
1^{\otimes(2k-2)}\big{)}\big{(}1\otimes\Delta\otimes 1^{\otimes(2k-3)}\big{)}$
and $\chi=\big{(}\tau_{L}\otimes
1^{\otimes(2k-3)}\big{)}\big{(}\tau_{L}\otimes 1^{\otimes(2k-2)}\big{)}$
respectively. ∎

###  4.2 The fundamental cases (C1)–(C6′)

In this section we will prove that the relation ($\ast$ ‣ 3) is satisfied for
the combinations of sets (C1)–(C6′). We will work out the proof for three of
these cases in detail, and describe concisely how one can show the remaining
cases.

######  Lemma 4.6.

The relation ($\ast$ ‣ 3) holds for the sets (C2) and (C5) with $\ell=0$.

###### Proof.

We will prove both claims together in one single induction on $k$. For $k=1$
the first claim coincides with (1.3) and the second follows by direct
calculation.

Our induction hypothesis states that ($\ast$ ‣ 3) holds for

$\displaystyle A=\\{2,4,\dots,2k-2,2k-1\\},\qquad B=\\{1,2k-1\\}$ (4.5)

and

$\displaystyle A=\\{2,4,\dots,2k-2,2k-1\\},\qquad B=\\{1,2k-1,2k\\}.$ (4.6)

By $\chi=\big{(}1^{\otimes(2k-3)}\otimes\Delta\otimes 1\otimes
1\big{)}\big{(}1^{\otimes(2k-3)}\otimes\Delta\otimes 1\big{)}$, (4.5) implies
($\ast$ ‣ 3) for

$\displaystyle A=\\{2,4,\dots,2k-2,2k-1,2k,2k+1\\},\qquad B=\\{1,2k+1\\}$
(4.7)

and by $\chi=1^{\otimes(2k-1)}\otimes\Delta$, (4.6) gives rise to ($\ast$ ‣ 3)
for

$\displaystyle A=\\{2,4,\dots,2k-2,2k-1\\},\qquad B=\\{1,2k-1,2k,2k+1\\}.$
(4.8)

We will first compute

$\displaystyle[\Lambda_{\\{2,4,6,\dots,2k,2k+1\\}},\Lambda_{\\{1,2k+1\\}}]_{q}.$
(4.9)

By Lemma 4.5 we have ($\Delta$) for $A=\\{1,2k-1\\}$,
$B=\\{2,4,\dots,2k-2\\}$, which, upon applying
$\chi=\big{(}1^{\otimes(2k-1)}\otimes\tau_{R}\big{)}\big{(}1^{\otimes(2k-3)}\otimes\Delta\otimes
1\big{)}$, yields ($\Delta$) for

$\displaystyle A=\\{1,2k+1\\},\qquad B=\\{2,4,\dots,2k-2,2k-1\\},$ (4.10)

whereas by
$\chi=\big{(}1^{\otimes(2k-1)}\otimes\tau_{R}\big{)}\big{(}1^{\otimes(2k-2)}\otimes\tau_{R}\big{)}$
we have ($\Delta$) for

$\displaystyle A=\\{1,2k+1\\},\qquad B=\\{2,4,\dots,2k-2\\}.$ (4.11)

By Lemma 4.4 we may write

$\displaystyle\Lambda_{\\{2,4,\dots,2k,2k+1\\}}=\frac{[\Lambda_{\\{2,4,\dots,2k-2,2k-1\\}},\Lambda_{\\{2k-1,2k,2k+1\\}}]_{q}}{q^{-2}-q^{2}}$
$\displaystyle\hphantom{\Lambda_{\\{2,4,\dots,2k,2k+1\\}}=}{}+\frac{\Lambda_{\\{2k-1\\}}\Lambda_{\\{2,4,\dots,2k-2,2k-1,2k,2k+1\\}}+\Lambda_{\\{2,4,\dots,2k-2\\}}\Lambda_{\\{2k,2k+1\\}}}{q+q^{-1}}.$

Substituting this in (4.9) and using (4.4) by (4.10), (4.9) becomes

$\displaystyle\frac{[\Lambda_{\\{2,4,\dots,2k-2,2k-1\\}},[\Lambda_{\\{2k-1,2k,2k+1\\}},\Lambda_{\\{1,2k+1\\}}]_{q}]_{q}}{q^{-2}-q^{2}}$
$\displaystyle\quad{}+\frac{\Lambda_{\\{2k-1\\}}[\Lambda_{\\{2,4,\dots,2k-2,2k-1,2k,2k+1\\}},\Lambda_{\\{1,2k+1\\}}]_{q}+\Lambda_{\\{2,4,\dots,2k-2\\}}[\Lambda_{\\{2k,2k+1\\}},\Lambda_{\\{1,2k+1\\}}]_{q}}{q+q^{-1}},$

where we have used (4.11) and Lemma 4.3.

The relation ($\ast$ ‣ 3) holds for $A=\\{2k-1,2k,2k+1\\}$, $B=\\{1,2k+1\\}$
and for $A=\\{2k,2k+1\\}$, $B=\\{1,2k+1\\}$, as one sees by applying
$\chi=\big{(}\tau_{L}\otimes
1^{\otimes(2k-1)}\big{)}\cdots\big{(}\tau_{L}\otimes 1^{\otimes
3}\big{)}(1\otimes\Delta\otimes 1)$ resp. $\chi=\big{(}\tau_{L}\otimes
1^{\otimes(2k-1)}\big{)}\cdots\big{(}\tau_{L}\otimes 1^{\otimes 2}\big{)}$ to
(1.3). With this and (4.7), (4.9) becomes

$\displaystyle[\Lambda_{\\{2,4,\dots,2k-2,2k-1\\}},\Lambda_{\\{1,2k-1,2k\\}}]_{q}$
$\displaystyle{}-\frac{\Lambda_{\\{2k+1\\}}[\Lambda_{\\{2,4,\dots,2k-2,2k-1\\}},\Lambda_{\\{1,2k-1,2k,2k+1\\}}]_{q}+\Lambda_{\\{1\\}}[\Lambda_{\\{2,4,\dots,2k-2,2k-1\\}},\Lambda_{\\{2k-1,2k\\}}]_{q}}{q+q^{-1}}$
$\displaystyle{}-\big{(}q-q^{-1}\big{)}\Lambda_{\\{2k-1\\}}\Lambda_{\\{1,2,4,\dots,2k-2,2k-1,2k\\}}$
$\displaystyle{}+\frac{q-q^{-1}}{q+q^{-1}}\Lambda_{\\{2k-1\\}}\big{(}\Lambda_{\\{2k+1\\}}\Lambda_{\\{1,2,4,\dots,2k-2,2k-1,2k,2k+1\\}}+\Lambda_{\\{1\\}}\Lambda_{\\{2,4,\dots,2k-2,2k-1,2k\\}}\big{)}$
$\displaystyle{}-\big{(}q-q^{-1}\big{)}\Lambda_{\\{2,4,\dots,2k-2\\}}\Lambda_{\\{1,2k\\}}+\frac{q-q^{-1}}{q+q^{-1}}\Lambda_{\\{2,4,\dots,2k-2\\}}\big{(}\Lambda_{\\{2k+1\\}}\Lambda_{\\{1,2k,2k+1\\}}+\Lambda_{\\{1\\}}\Lambda_{\\{2k\\}}\big{)}.$

The first $q$-commutator can be expanded by (4.6), the second by (4.8), the
third by Lemma 4.4. Writing everything down, a lot of common terms will
cancel, eventually leading to

$\displaystyle[\Lambda_{\\{2,4,6,\dots,2k,2k+1\\}},\Lambda_{\\{1,2k+1\\}}]_{q}$
$\displaystyle\quad=\big{(}q^{-2}-q^{2}\big{)}\Lambda_{\\{1,2,4,\dots,2k\\}}+\big{(}q-q^{-1}\big{)}\big{(}\Lambda_{\\{2k+1\\}}\Lambda_{\\{1,2,4,\dots,2k,2k+1\\}}+\Lambda_{\\{1\\}}\Lambda_{\\{2,4,\dots,2k\\}}\big{)}.\\!\\!\\!$
(4.12)

This proves the first part of the claim.

By $\chi=1^{\otimes(2k-1)}\otimes\Delta\otimes 1$ respectively
$\chi=1^{\otimes 2k}\otimes\tau_{R}$, (4.12) implies that ($\ast$ ‣ 3) holds
for

$\displaystyle A=\\{2,4,\dots,2k,2k+1,2k+2\\},\qquad B=\\{1,2k+2\\}$ (4.13)

and

$\displaystyle A=\\{2,4,\dots,2k,2k+2\\},\qquad B=\\{1,2k+2\\}.$ (4.14)

Applying $\chi=\big{(}\tau_{L}\otimes 1^{\otimes
2k}\big{)}\cdots\big{(}\tau_{L}\otimes 1^{\otimes 2}\big{)}$ to (1.3), we may
write

$\displaystyle\Lambda_{\\{1,2k+1\\}}=\frac{[\Lambda_{\\{2k+1,2k+2\\}},\Lambda_{\\{1,2k+2\\}}]_{q}}{q^{-2}-q^{2}}+\frac{\Lambda_{\\{2k+2\\}}\Lambda_{\\{1,2k+1,2k+2\\}}+\Lambda_{\\{2k+1\\}}\Lambda_{\\{1\\}}}{q+q^{-1}}.$
(4.15)

We have ($\Delta$) for $A=\\{1,2k+2\\}$ and $B=\\{2,4,\dots,2k,2k+1\\}$, as
follows from Lemma 4.5 by $\chi=1^{\otimes(2k-1)}\otimes\Delta\otimes 1$.
Substituting (4.15) in (4.9) and using (4.1), (4.9) becomes

$\displaystyle\frac{[[\Lambda_{\\{2,4,\dots,2k,2k+1\\}},\Lambda_{\\{2k+1,2k+2\\}}]_{q},\Lambda_{\\{1,2k+2\\}}]_{q}}{q^{-2}-q^{2}}$
$\displaystyle\quad{}+\frac{\Lambda_{\\{2k+2\\}}[\Lambda_{\\{2,4,\dots,2k,2k+1\\}},\Lambda_{\\{1,2k+1,2k+2\\}}]_{q}+\big{(}q-q^{-1}\big{)}\Lambda_{\\{1\\}}\Lambda_{\\{2k+1\\}}\Lambda_{\\{2,4,\dots,2k,2k+1\\}}}{q+q^{-1}}.$

The relation ($\ast$ ‣ 3) holds for $A=\\{2,4,\dots,2k,2k+1\\}$ and
$B=\\{2k+1,2k+2\\}$ by Lemma 4.4. Hence (4.9) becomes

$\displaystyle[\Lambda_{\\{2,4,\dots,2k,2k+2\\}},\Lambda_{\\{1,2k+2\\}}]_{q}$
$\displaystyle{}-\frac{\Lambda_{\\{2k+1\\}}[\Lambda_{\\{2,4,\dots,2k,2k+1,2k+2\\}},\Lambda_{\\{1,2k+2\\}}]_{q}+\big{(}q-q^{-1}\big{)}\Lambda_{\\{2k+2\\}}\Lambda_{\\{2,4,\dots,2k\\}}\Lambda_{\\{1,2k+2\\}}}{q+q^{-1}}$
$\displaystyle{}+\frac{\Lambda_{\\{2k+2\\}}[\Lambda_{\\{2,4,\dots,2k,2k+1\\}},\Lambda_{\\{1,2k+1,2k+2\\}}]_{q}+\big{(}q-q^{-1}\big{)}\Lambda_{\\{1\\}}\Lambda_{\\{2k+1\\}}\Lambda_{\\{2,4,\dots,2k,2k+1\\}}}{q+q^{-1}},$
(4.16)

where we have used ($\Delta$) for $A=\\{2,4,\dots,2k\\}$, $B=\\{1,2k+2\\}$,
which follows from Lemma 4.5 by $\chi=1^{\otimes 2k}\otimes\tau_{R}$. The
first $q$-commutator can be expanded by (4.14), the second by (4.13). On the
other hand, we already know an expression for (4.9), namely (4.12). Comparing
these, the only remaining $q$-commutator in (4.16) can be expanded as

$\displaystyle[\Lambda_{\\{2,4,\dots,2k,2k+1\\}},\Lambda_{\\{1,2k+1,2k+2\\}}]_{q}=\big{(}q^{-2}-q^{2}\big{)}\Lambda_{\\{1,2,4,\dots,2k,2k+2\\}}$
$\displaystyle\quad{}+\big{(}q-q^{-1}\big{)}\big{(}\Lambda_{\\{2k+1\\}}\Lambda_{\\{1,2,4,\dots,2k,2k+1,2k+2\\}}+\Lambda_{\\{1,2k+2\\}}\Lambda_{\\{2,4,\dots,2k\\}}\big{)}.$

This concludes the induction. ∎

By a completely analogous inductive proof, one can show the following.

######  Lemma 4.7.

The relation ($\ast$ ‣ 3) holds for the sets (C3) and (C4) with $k=1$.

Somewhat different is our strategy to prove the relation ($\ast$ ‣ 3) for the
sets (C4) and (C4′).

######  Lemma 4.8.

The standard relation ($\ast$ ‣ 3) holds for the sets (C4) and (C4′), i.e.,
for

$\displaystyle A=\\{1,2,4,6,\dots,2k,2k+2\ell+2+\delta\\},$ $\displaystyle
B=\\{2,4,6,\dots,2k,2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta\\},$

with $k,\ell\in\mathbb{N}$ and $\delta\in\\{0,1\\}$.

###### Proof.

We need to rewrite

$\displaystyle[\Lambda_{\\{1,2,4,6,\dots,2k,2k+2\ell+2+\delta\\}},\Lambda_{\\{2,4,6,\dots,2k,2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta\\}}]_{q}.$
(4.17)

First observe that

$\displaystyle[\Lambda_{\\{1,2k+2\ell+3+\delta\\}},\Lambda_{\\{2,4,6,\dots,2k,2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta\\}}]=0.$
(4.18)

This follows from Lemma 4.5 with $k+\ell+\delta$ instead of $k$, by

$\chi=\big{(}1^{\otimes(2k-1)}\otimes\Delta\otimes
1^{\otimes(2\ell+2)}\big{)}^{1-\delta}\big{(}1^{\otimes(2k+2\ell+2\delta)}\otimes\tau_{R}\big{)}.$

The relation ($\ast$ ‣ 3) for (C2) with $k+1$ instead of $k$, acted upon with

$\chi=\big{(}1^{\otimes 2k}\otimes\Delta\otimes
1^{\otimes(2\ell+1+\delta)}\big{)}\cdots\big{(}1^{\otimes
2k}\otimes\Delta\otimes 1^{\otimes 2}\big{)},$

gives rise to the identity

$\displaystyle\Lambda_{\\{1,2,4,6,\dots,2k,2k+2\ell+2+\delta\\}}=\frac{[\Lambda_{\\{2,4,6,\dots,2k,2k+2\ell+2+\delta,2k+2\ell+3+\delta\\}},\Lambda_{\\{1,2k+2\ell+3+\delta\\}}]_{q}}{q^{-2}-q^{2}}$
$\displaystyle\qquad{}+\frac{\Lambda_{\\{2k+2\ell+3+\delta\\}}\Lambda_{\\{1,2,4,\dots,2k,2k+2\ell+2+\delta,2k+2\ell+3+\delta\\}}+\Lambda_{\\{1\\}}\Lambda_{\\{2,4,\dots,2k,2k+2\ell+2+\delta\\}}}{q+q^{-1}}.$

Using (4.3) by (4.18), we may write (4.17) as

$\displaystyle\frac{[[\Lambda_{\\{2,4,\dots,2k,2k+2\ell+2+\delta,2k+2\ell+3+\delta\\}},\Lambda_{\\{2,4,\dots,2k,2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta\\}}]_{q},\Lambda_{\\{1,2k+2\ell+3+\delta\\}}]_{q}}{q^{-2}-q^{2}}$
$\displaystyle{}+\frac{\Lambda_{\\{2k+2\ell+3+\delta\\}}[\Lambda_{\\{1,2,4,\dots,2k,2k+2\ell+2+\delta,2k+2\ell+3+\delta\\}},\Lambda_{\\{2,4,\dots,2k,2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta\\}}]_{q}}{q+q^{-1}}$
$\displaystyle{}+\frac{\Lambda_{\\{1\\}}[\Lambda_{\\{2,4,\dots,2k,2k+2\ell+2+\delta\\}},\Lambda_{\\{2,4,\dots,2k,2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta\\}}]_{q}}{q+q^{-1}}.$
(4.19)

The relation ($\ast$ ‣ 3) is satisfied for

$\displaystyle A=\\{2,4,6,\dots,2k,2k+2\ell+2+\delta\\},$ $\displaystyle
B=\\{2,4,6,\dots,2k,2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta\\},$

as follows from (C3) with $\ell+1$ instead of $k$, by

$\chi=\left(\overrightarrow{\prod_{n=2\ell+2+\delta}^{2k+2\ell-2+\delta}}\big{(}\tau_{L}\otimes
1^{\otimes(n+1)}\big{)}\big{(}\Delta\otimes 1^{\otimes
n}\big{)}\right)\big{(}\tau_{L}\otimes 1^{\otimes(2\ell+2)}\big{)}^{\delta}$

and putting a factor $1\otimes$ in front. By
$\chi=1^{\otimes(2k+2\ell+1+\delta)}\otimes\Delta$, we also have ($\ast$ ‣ 3)
for

$\displaystyle A=\\{2,4,6,\dots,2k,2k+2\ell+2+\delta,2k+2\ell+3+\delta\\},$
$\displaystyle
B=\\{2,4,6,\dots,2k,2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta\\}.$

This helps us expand the first and third line of (4.19), such that (4.17)
becomes

$\displaystyle[\Lambda_{\\{2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta,2k+2\ell+2+\delta,2k+2\ell+3+\delta\\}},\Lambda_{\\{1,2k+2\ell+3+\delta\\}}]_{q}$
$\displaystyle{}-\frac{\Lambda_{\\{2,4\dots,2k\\}}}{q+q^{-1}}[\Lambda_{\\{2,4,\dots,2k,2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta,2k+2\ell+2+\delta,2k+2\ell+3+\delta\\}},\Lambda_{\\{1,2k+2\ell+3+\delta\\}}]_{q}$
$\displaystyle{}-\frac{\Lambda_{\\{2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta\\}}}{q+q^{-1}}[\Lambda_{\\{2k+2\ell+2+\delta,2k+2\ell+3+\delta\\}},\Lambda_{\\{1,2k+2\ell+3+\delta\\}}]_{q}$
$\displaystyle{}+\frac{\Lambda_{\\{2k+2\ell+3+\delta\\}}}{q+q^{-1}}[\Lambda_{\\{1,2,4,\dots,2k,2k+2\ell+2+\delta,2k+2\ell+3+\delta\\}},\Lambda_{\\{2,4,\dots,2k,2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta\\}}]_{q}$
$\displaystyle{}-\big{(}q-q^{-1}\big{)}\Lambda_{\\{1\\}}\Lambda_{\\{2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta,2k+2\ell+2+\delta\\}}$
$\displaystyle{}+\frac{q-q^{-1}}{q+q^{-1}}\Lambda_{\\{1\\}}\Lambda_{\\{2,4,\dots,2k\\}}\Lambda_{\\{2,4,\dots,2k,2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta,2k+2\ell+2+\delta\\}}$
$\displaystyle{}+\frac{q-q^{-1}}{q+q^{-1}}\Lambda_{\\{1\\}}\Lambda_{\\{2k+2\ell+2+\delta\\}}\Lambda_{\\{2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta\\}}.$
(4.20)

Here we have used the commutation relations

$\displaystyle[\Lambda_{\\{2,4,\dots,2k\\}},\Lambda_{\\{1,2k+2\ell+3+\delta\\}}]=0,$
$\displaystyle[\Lambda_{\\{2,4,\dots,2k\\}},\Lambda_{\\{2,4,\dots,2k,2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta,2k+2\ell+2+\delta,2k+2\ell+3+\delta\\}}]=0,$
$\displaystyle[\Lambda_{\\{2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta\\}},\Lambda_{\\{2k+2\ell+2+\delta,2k+2\ell+3+\delta\\}}]=0,$
$\displaystyle[\Lambda_{\\{2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta\\}},\Lambda_{\\{1,2k+2\ell+3+\delta\\}}]=0,$

which follow from Lemmas 4.3 and 4.5.

The relation ($\ast$ ‣ 3) holds for the sets

$\displaystyle\begin{split}&A=\\{2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta,2k+2\ell+2+\delta,2k+2\ell+3+\delta\\},\\\
&B=\\{1,2k+2\ell+3+\delta\\};\end{split}$ (4.21)
$\displaystyle\begin{split}&A=\\{2k+2\ell+2+\delta,2k+2\ell+3+\delta\\},\\\
&B=\\{1,2k+2\ell+3+\delta\\};\end{split}\vspace{2mm}$ (4.22)
$\displaystyle\begin{split}&A=\\{2,4,\dots,2k,2k+1+\delta,2k+3+\delta,\dots,\\\
&\hphantom{A=\\{}{}2k+2\ell+1+\delta,2k+2\ell+2+\delta,2k+2\ell+3+\delta\\},\\\
&B=\\{1,2k+2\ell+3+\delta\\}.\end{split}$ (4.23)

For (4.21), this follows from (C2) with $\ell+1$ instead of $k$, upon applying

$\chi=\left(\overrightarrow{\prod_{n=2\ell+3}^{2k+2\ell+1+\delta}}\big{(}\tau_{L}\otimes
1^{\otimes n}\big{)}\right)\big{(}1^{\otimes(2\ell+1)}\otimes\Delta\otimes
1\big{)}.$

The claim for (4.22) follows from (1.3), by
$\chi=\overrightarrow{\prod\limits_{n=2}^{2k+2\ell+1+\delta}}\big{(}\tau_{L}\otimes
1^{\otimes n}\big{)}$, and finally (4.23) follows from (C2) with
$k+\ell+\delta$ instead of $k$, by
$\chi=\big{(}1^{\otimes(2k-1)}\otimes\Delta\otimes
1^{\otimes(2\ell+2)}\big{)}^{1-\delta}\big{(}1^{\otimes(2k+2\ell-1+2\delta)}\otimes\Delta\otimes
1\big{)}$. Bringing the fourth line of (4.20) to the left-hand side and
expanding everything as explained, we find

$\displaystyle[\Lambda_{\\{1,2,4,6,\dots,2k,2k+2\ell+2+\delta\\}},\Lambda_{\\{2,4,6,\dots,2k,2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta\\}}]_{q}$
$\displaystyle-\frac{\Lambda_{\\{2k+2\ell+3+\delta\\}}}{q+q^{-1}}[\Lambda_{\\{1,2,4,\dots,2k,2k+2\ell+2+\delta,2k+2\ell+3+\delta\\}},\Lambda_{\\{2,4,\dots,2k,2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta\\}}]_{q}$
$\displaystyle=\big{(}q^{-2}-q^{2}\big{)}\Lambda_{\\{1,2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta,2k+2\ell+2+\delta\\}}$
$\displaystyle+\big{(}q-q^{-1}\big{)}\Lambda_{\\{2,4,\dots,2k\\}}\Lambda_{\\{1,2,4,\dots,2k,2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta,2k+2\ell+2+\delta\\}}$
$\displaystyle+\big{(}q-q^{-1}\big{)}\Lambda_{\\{1,2k+2\ell+2+\delta\\}}\Lambda_{\\{2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta\\}}$
$\displaystyle+\big{(}q-q^{-1}\big{)}\Lambda_{\\{2k+2\ell+3+\delta\\}}\Lambda_{\\{1,2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta,2k+2\ell+2+\delta,2k+2\ell+3+\delta\\}}$
$\displaystyle-\frac{q-q^{-1}}{q+q^{-1}}\Lambda_{\\{2k+2\ell+3+\delta\\}}\Lambda_{\\{2,4,\dots,2k\\}}\Lambda_{\\{1,2,4,\dots,2k,2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta,2k+2\ell+2+\delta,2k+2\ell+3+\delta\\}}$
$\displaystyle-\frac{q-q^{-1}}{q+q^{-1}}\Lambda_{\\{2k+2\ell+3+\delta\\}}\Lambda_{\\{1,2k+2\ell+2+\delta,2k+2\ell+3+\delta\\}}\Lambda_{\\{2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta\\}}.$

Now let us denote by $\Theta$ and $\Xi$ respectively the expressions

$\displaystyle\Theta=[\Lambda_{\\{1,2,4,6,\dots,2k,2k+2\ell+2+\delta\\}},\Lambda_{\\{2,4,6,\dots,2k,2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta\\}}]_{q},$
$\displaystyle\Xi=\big{(}q^{-2}-q^{2}\big{)}\Lambda_{\\{1,2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta,2k+2\ell+2+\delta\\}}$
$\displaystyle\hphantom{\Xi=}{}+\big{(}q-q^{-1}\big{)}\Lambda_{\\{2,4,\dots,2k\\}}\Lambda_{\\{1,2,4,\dots,2k,2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta,2k+2\ell+2+\delta\\}}$
$\displaystyle\hphantom{\Xi=}{}+\big{(}q-q^{-1}\big{)}\Lambda_{\\{1,2k+2\ell+2+\delta\\}}\Lambda_{\\{2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta\\}},$

both in the $(2k+2\ell+2+\delta)$-fold tensor product. Then the relation above
becomes

$\displaystyle\Theta\otimes
1-\frac{1}{q+q^{-1}}\Lambda_{\\{2k+2\ell+3+\delta\\}}\big{(}1^{\otimes(2k+2\ell+1+\delta)}\otimes\Delta\big{)}\Theta$
$\displaystyle\qquad{}=\Xi\otimes
1-\frac{1}{q+q^{-1}}\Lambda_{\\{2k+2\ell+3+\delta\\}}\big{(}1^{\otimes(2k+2\ell+1+\delta)}\otimes\Delta\big{)}\Xi,$

or equivalently

$\displaystyle(\Theta-\Xi)\otimes
1=\frac{1}{q+q^{-1}}\Lambda_{\\{2k+2\ell+3+\delta\\}}\big{(}1^{\otimes(2k+2\ell+1+\delta)}\otimes\Delta\big{)}(\Theta-\Xi).$
(4.24)

We want to show that $\Theta=\Xi$.

Let us start by writing $\Theta-\Xi$ explicitly as a tensor product, i.e.,

$\Theta-\Xi=\sum_{i\in I}a_{1}^{(i)}\otimes\cdots\otimes
a_{2k+2\ell+2+\delta}^{(i)},$

for certain $a_{j}^{(i)}\in U_{q}(\mathfrak{sl}_{2})$ and some finite index
set $I$, and where we have grouped the elements in such a way that the set

$\displaystyle\big{\\{}a_{1}^{(i)}\otimes\dots\otimes
a_{2k+2\ell+1+\delta}^{(i)}\colon i\in I\big{\\}}$ (4.25)

is linearly independent. The equality (4.24) then asserts

$\sum_{i\in I}a_{1}^{(i)}\otimes\dots\otimes
a_{2k+2\ell+1+\delta}^{(i)}\otimes\left(a_{2k+2\ell+2+\delta}^{(i)}\otimes
1-\frac{1}{q+q^{-1}}(1\otimes\Lambda)\Delta\big{(}a_{2k+2\ell+2+\delta}^{(i)}\big{)}\right)=0,$

which by our convention (4.25) implies

$a_{2k+2\ell+2+\delta}^{(i)}\otimes
1=\frac{1}{q+q^{-1}}(1\otimes\Lambda)\Delta\big{(}a_{2k+2\ell+2+\delta}^{(i)}\big{)}$

for every $i\in I$. But this now implies that every
$a_{2k+2\ell+2+\delta}^{(i)}$ vanishes, since if not, then
$\Delta\big{(}a_{2k+2\ell+2+\delta}^{(i)}\big{)}=\sum\limits_{j\in
J}b_{1}^{(j)}\otimes b_{2}^{(j)}$ is nontrivial and must be such that

$a_{2k+2\ell+2+\delta}^{(i)}\otimes 1=\frac{1}{q+q^{-1}}\sum_{j\in
J}b_{1}^{(j)}\otimes\Lambda b_{2}^{(j)}.$

But by comparing degrees in the generators $E$ and $F$, it is clear that
$\Lambda b_{2}^{(j)}\neq 1$ for every possible $b_{2}^{(j)}\in
U_{q}(\mathfrak{sl}_{2})$, i.e., $\Lambda$ has no inverse. So we have
$a_{2k+2\ell+2+\delta}^{(i)}=0$ for every $i$. Hence we have shown that indeed
$\Theta=\Xi$. This concludes the proof. ∎

The two preceding proofs manifest the general strategy that can be used to
show Proposition 3.1. Therefore, and for the sake of brevity, we will only
concisely mention how one can show the remaining relations.

######  Lemma 4.9.

The standard relation ($\ast$ ‣ 3) holds for the sets (C1).

###### Sketch of the proof.

Use ($\ast$ ‣ 3) for $A=\\{2,4,\dots,2k,2k+2\\}$, $B=\\{1,2k+2\\}$, as follows
from (C2), to rewrite $\Lambda_{\\{1,2,4,\dots,2k\\}}$ in

$[\Lambda_{\\{1,2,4,\dots,2k\\}},\Lambda_{\\{2,4,\dots,2k,2k+1\\}}]_{q}.$

Use (4.3) to switch the order of the nested $q$-commutators. Use (C4) with
$\ell=0$ and (1.4) acted upon with
$\chi=\overrightarrow{\prod\limits_{\ell=1}^{k-1}}\big{(}\tau_{L}\otimes
1^{\otimes(2\ell+1)}\big{)}\big{(}\Delta\otimes 1^{\otimes 2\ell}\big{)}$ to
expand further. The remaining $q$-commutators can be worked out from (1.3) and
(C2). ∎

######  Lemma 4.10.

The standard relation ($\ast$ ‣ 3) holds for the sets (C5) and (C5′).

###### Sketch of the proof.

Use ($\ast$ ‣ 3) for (C3) with $\ell+1$ instead of $k$, acted upon with

$\chi=\big{(}\tau_{L}\otimes
1^{\otimes(2k+2\ell+\delta)}\big{)}\cdots\big{(}\tau_{L}\otimes
1^{\otimes(2\ell+2)}\big{)},$

to rewrite $\Lambda_{B^{\prime}}$ in
$[\Lambda_{A^{\prime}},\Lambda_{B^{\prime}}]_{q}$ with

$\displaystyle
A^{\prime}=\\{2,4,\dots,2k,2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta\\},$
$\displaystyle
B^{\prime}=\\{2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta,2k+2\ell+2+\delta\\},$

where $\delta\in\\{0,1\\}$. Use (4.2) and expand the inner $q$-commutator by
(C2) acted upon with

$\displaystyle\chi=\big{(}1^{\otimes(2k+2\ell-1+\delta)}\otimes\tau_{R}\big{)}\big{(}1^{\otimes(2k+2\ell-2+\delta)}\otimes\Delta\big{)}\cdots$
$\displaystyle\hphantom{\chi=}{}\times\big{(}1^{\otimes(2k+1+\delta)}\otimes\tau_{R}\big{)}\big{(}1^{\otimes(2k+\delta)}\otimes\Delta\big{)}\big{(}1^{\otimes
2k}\otimes\tau_{R}\big{)}^{\delta}.$

Apply the relation ($\ast$ ‣ 3) for the sets

$\displaystyle\begin{split}&A=\\{1,2k+2\ell+2+\delta\\},\\\
&B=\\{1,2,4,\dots,2k\\};\end{split}\vspace{2mm}$
$\displaystyle\begin{split}&A=\\{1,2k+2\ell+2+\delta\\},\\\
&B=\\{1,2,4,\dots,2k,2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta\\},\end{split}$

as follows from (C3) by
$\chi=\big{(}1^{\otimes(2k+2\ell+\delta)}\otimes\tau_{R}\big{)}\cdots\big{(}1^{\otimes
2k}\otimes\tau_{R}\big{)}$ respectively from (C3) with $k+\ell+\delta$ instead
of $k$, acted upon with $\chi=\big{(}1^{\otimes(2k-1)}\otimes\Delta\otimes
1^{\otimes(2\ell+1)}\big{)}^{1-\delta}$. In our expression for
$[\Lambda_{A^{\prime}},\Lambda_{B^{\prime}}]_{q}$, there is now one term left
containing a $q$-commutator, namely

$\displaystyle\frac{1}{q+q^{-1}}\Lambda_{\\{1\\}}[\Lambda_{\\{2,4,\dots,2k,2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta\\}},$
$\displaystyle\hphantom{\frac{1}{q+q^{-1}}\Lambda_{\\{1\\}}[}{}\Lambda_{\\{1,2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta,2k+2\ell+2+\delta\\}}]_{q}.$
(4.26)

On the other hand, we already know how to expand
$[\Lambda_{A^{\prime}},\Lambda_{B^{\prime}}]_{q}$ from (C1) with $\ell+1$
instead of $k$, acted upon with

$\displaystyle\chi=\big{(}\tau_{L}\otimes
1^{\otimes(2k+2\ell-1+\delta)}\big{)}\big{(}\Delta\otimes
1^{\otimes(2k+2\ell-2+\delta)}\big{)}\cdots$
$\displaystyle\hphantom{\chi=}{}\times\big{(}\tau_{L}\otimes
1^{\otimes(2\ell+3+\delta)}\big{)}\big{(}\Delta\otimes
1^{\otimes(2\ell+2+\delta)}\big{)}\big{(}\tau_{L}\otimes
1^{\otimes(2\ell+2)}\big{)}^{\delta},$

and adding a factor $1\otimes$ in front. This leads to the sought expression
for the remaining $q$-commutator in (4.26). ∎

To show the final cases (C6) and (C6′), we will need another commutation
relation.

######  Lemma 4.11.

For any $k\in\mathbb{N}$ one has

$[\Lambda_{\\{1,2k+1\\}},\Lambda_{\\{1,2,4,\dots,2k,2k+1\\}}]=0.$

###### Proof.

By induction on $k$, the case $k=1$ being obvious from a direct calculation.
Suppose the claim has been proven for $k-1$. Since ($\ast$ ‣ 3) holds for
(C5), with $k-1$ instead of $k$ and with $\ell=0$, we have by
$\chi=1^{\otimes(2k-1)}\otimes\Delta$

$\displaystyle\Lambda_{\\{1,2,4,\dots,2k,2k+1\\}}=\frac{[\Lambda_{\\{2,4,\dots,2k-2,2k-1\\}},\Lambda_{\\{1,2k-1,2k,2k+1\\}}]_{q}}{q^{-2}-q^{2}}$
$\displaystyle\hphantom{\Lambda_{\\{1,2,4,\dots,2k,2k+1\\}}=}{}+\frac{\Lambda_{\\{2k-1\\}}\Lambda_{\\{1,2,4,\dots,2k-2,2k-1,2k,2k+1\\}}+\Lambda_{\\{2,4,\dots,2k-2\\}}\Lambda_{\\{1,2k,2k+1\\}}}{q+q^{-1}}.$

It suffices now to show that each of the generators in the right-hand side
commutes with $\Lambda_{\\{1,2k+1\\}}$. For
$\Lambda_{\\{2,4,\dots,2k-2,2k-1\\}}$ and $\Lambda_{\\{2,4,\dots,2k-2\\}}$
this follows from Lemma 4.5 with $k-1$ instead of $k$, upon applying
$\chi=\big{(}1^{\otimes(2k-1)}\otimes\tau_{R}\big{)}\big{(}1^{\otimes(2k-3)}\otimes\Delta\otimes
1\big{)}$ and
$\chi=\big{(}1^{\otimes(2k-1)}\otimes\tau_{R}\big{)}\big{(}1^{\otimes(2k-2)}\otimes\tau_{R}\big{)}$
respectively. For $\Lambda_{\\{1,2,4,\dots,2k-2,2k-1,2k,2k+1\\}}$ we apply
$\chi=\big{(}1^{\otimes(2k-3)}\otimes\Delta\otimes 1\otimes
1\big{)}\big{(}1^{\otimes(2k-3)}\otimes\Delta\otimes 1\big{)}$ to the
induction hypothesis. The remaining nontrivial commutation relations follow
from $[\Lambda_{\\{1,3\\}},\Lambda_{\\{1,2,3\\}}]=0$ upon applying
respectively

$\chi=\big{(}\tau_{L}\otimes
1^{\otimes(2k-1)}\big{)}\cdots\big{(}\tau_{L}\otimes 1^{\otimes
4}\big{)}\big{(}\tau_{L}\otimes 1^{\otimes 3}\big{)}(1\otimes\Delta\otimes 1)$

and

$\displaystyle\chi=\big{(}\tau_{L}\otimes
1^{\otimes(2k-1)}\big{)}\cdots\big{(}\tau_{L}\otimes 1^{\otimes
3}\big{)}\big{(}\tau_{L}\otimes 1^{\otimes 2}\big{)}.$ ∎

######  Corollary 4.1.

$\Lambda_{\\{1,2k\\}}$ commutes with $\Lambda_{\\{1,2,4,\dots,2k\\}}$ and
$\Lambda_{\\{1,3,5,\dots,2k-1,2k\\}}$.

###### Proof.

Act on the relation of Lemma 4.11, with $k-1$ instead of $k$, with
$\chi=1^{\otimes(2k-2)}\otimes\tau_{R}$ and $\chi=\tau_{L}\otimes
1^{\otimes(2k-2)}$ respectively. ∎

######  Lemma 4.12.

The standard relation ($\ast$ ‣ 3) holds for the sets (C6) and (C6′), i.e.,
for

$\displaystyle
A=\\{1,2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta,2k+2\ell+2+\delta\\},$
$\displaystyle B=\\{1,2,4,\dots,2k,2k+2\ell+2+\delta\\}$ (4.27)

with $k,\ell\in\mathbb{N}$ and $\delta\in\\{0,1\\}$.

###### Sketch of the proof.

By induction on $k$. Already the induction basis $k=1$, i.e., to prove ($\ast$
‣ 3) for

$\displaystyle
A=\\{1,3+\delta,5+\delta,\dots,2\ell+3+\delta,2\ell+4+\delta\\},\qquad
B=\\{1,2,2\ell+4+\delta\\},$ (4.28)

is difficult and needs to be shown by induction on $\ell$, the case $\ell=0$
being trivial. Suppose thus that the claim holds for $\ell-1$, i.e., the
relation ($\ast$ ‣ 3) holds for

$\displaystyle
A=\\{1,3+\delta,5+\delta,\dots,2\ell+1+\delta,2\ell+2+\delta\\},\qquad
B=\\{1,2,2\ell+2+\delta\\}.$ (4.29)

We will find an expression for

$\displaystyle[\Lambda_{\\{1,3+\delta,5+\delta,\dots,2\ell+3+\delta,2\ell+4+\delta\\}},\Lambda_{\\{1,2,2\ell+4+\delta\\}}]_{q}.$
(4.30)

Rewrite
$\Lambda_{\\{1,3+\delta,5+\delta,\dots,2\ell+3+\delta,2\ell+4+\delta\\}}$ in
(4.30) by (C4) with $k=1$ and $\ell=0$, acted upon by

$\displaystyle\begin{split}&\chi=\big{(}\tau_{L}\otimes
1^{\otimes(2\ell+3)}\big{)}^{\delta}\big{(}\tau_{L}\otimes
1^{\otimes(2\ell+2)}\big{)}\big{(}\Delta\otimes
1^{\otimes(2\ell+1)}\big{)}\cdots\\\
&\hphantom{\chi=}{}\times\big{(}\tau_{L}\otimes 1^{\otimes
6}\big{)}\big{(}\Delta\otimes 1^{\otimes 5}\big{)}\big{(}\tau_{L}\otimes
1^{\otimes 4}\big{)}\big{(}\Delta\otimes 1^{\otimes 3}\big{)}.\end{split}$

Use (4.3) and expand further by ($\ast$ ‣ 3) for

$\displaystyle\begin{split}&A=\\{1,3+\delta,5+\delta,\dots,2\ell+1+\delta,2\ell+2+\delta,2\ell+4+\delta\\},\\\
&B=\\{1,2,2\ell+4+\delta\\};\end{split}\vspace{2mm}$
$\displaystyle\begin{split}&A=\\{1,3+\delta,5+\delta,\dots,2\ell+1+\delta,2\ell+2+\delta,2\ell+3+\delta,2\ell+4+\delta\\},\\\
&B=\\{1,2,2\ell+4+\delta\\};\end{split}\vspace{2mm}$
$\displaystyle\begin{split}&A=\\{1,3+\delta,5+\delta,\dots,2\ell+1+\delta,2\ell+4+\delta\\},\\\
&B=\\{1,2,2\ell+4+\delta\\},\end{split}$

as follows from the induction hypothesis (4.29) by respectively

$\displaystyle\chi=\big{(}1^{\otimes(2\ell+\delta)}\otimes\Delta\otimes
1^{\otimes 2}\big{)}\big{(}1^{\otimes(2\ell+1+\delta)}\otimes\tau_{R}\big{)},$
$\displaystyle\chi=\big{(}1^{\otimes(2\ell+\delta)}\otimes\Delta\otimes
1^{\otimes 2}\big{)}\big{(}1^{\otimes(2\ell+\delta)}\otimes\Delta\otimes
1\big{)},$
$\displaystyle\chi=\big{(}1^{\otimes(2\ell+2+\delta)}\otimes\tau_{R}\big{)}\big{(}1^{\otimes(2\ell+1+\delta)}\otimes\tau_{R}\big{)}.$

Use the relation
$[\Lambda_{\\{1,2\ell+4+\delta\\}},\Lambda_{\\{1,2,3+\delta,5+\delta,\dots,2\ell+1+\delta,2\ell+2+\delta,2\ell+4+\delta\\}}]=0$,
as follows from Lemma 4.11 with $\ell+\delta$ instead of $k$, after applying

$\chi=\big{(}1\otimes\Delta\otimes
1^{\otimes(2\ell+1)}\big{)}^{1-\delta}\big{(}1^{\otimes(2\ell-1+2\delta)}\otimes\Delta\otimes
1^{\otimes 2}\big{)}\big{(}1^{\otimes(2\ell+2\delta)}\otimes\tau_{R}\big{)}.$

The remaining $q$-commutators can now be expanded using Lemma 4.4 and (C4)
with $k=1$ and $\ell=0$, acted upon by

$\displaystyle\chi=\big{(}\Delta\otimes
1^{\otimes(2\ell+2+\delta)}\big{)}\big{(}\tau_{L}\otimes
1^{\otimes(2\ell+2)}\big{)}^{\delta}\big{(}\Delta\otimes
1^{\otimes(2\ell+1)}\big{)}\big{(}\tau_{L}\otimes 1^{\otimes
2\ell}\big{)}\big{(}\Delta\otimes 1^{\otimes(2\ell-1)}\big{)}\cdots$
$\displaystyle\hphantom{\chi=}{}\times\big{(}\tau_{L}\otimes 1^{\otimes
4}\big{)}\big{(}\Delta\otimes 1^{\otimes 3}\big{)}.$

This leads to ($\ast$ ‣ 3) for (4.28), which will serve as the basis for the
induction on $k$ we perform in our global proof, i.e., to show ($\ast$ ‣ 3)
for (4.27).

Suppose now the claim holds for $k-1$, i.e., the relation ($\ast$ ‣ 3) holds
for

$\displaystyle
A=\\{1,2k-1+\delta,2k+1+\delta,\dots,2k+2\ell-1+\delta,2k+2\ell+\delta\\},$
$\displaystyle B=\\{1,2,4,\dots,2k-2,2k+2\ell+\delta\\},$ (4.31)

for arbitrary $\ell\in\mathbb{N}$. We will find an expression for

$\displaystyle[\Lambda_{\\{1,2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta,2k+2\ell+2+\delta\\}},\Lambda_{\\{1,2,4,\dots,2k,2k+2\ell+2+\delta\\}}]_{q}.$
(4.32)

Use (C5) with $k=1$ and $\ell=0$, acted upon with

$\chi=\left(\overrightarrow{\prod_{m=2k+1}^{2k+2\ell+\delta}}\big{(}1^{\otimes
m}\otimes\tau_{R}\big{)}\right)\left(\overrightarrow{\prod_{m=2}^{k}}\big{(}1^{\otimes
2m}\otimes\tau_{R}\big{)}\big{(}1^{\otimes(2m-1)}\otimes\Delta\big{)}\right),$

to rewrite $\Lambda_{\\{1,2,4,\dots,2k,2k+2\ell+2+\delta\\}}$ in (4.32) and
apply (4.2). Expand further by ($\ast$ ‣ 3) for
$A=\\{1,2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta,2k+2\ell+2+\delta\\}$
and

$\displaystyle B=\\{1,3,4,6,\dots,2k-2,2k,2k+2\ell+2+\delta\\},$
$\displaystyle B=\\{1,2,3,4,6,\dots,2k-2,2k,2k+2\ell+2+\delta\\},$
$\displaystyle B=\\{1,4,6,\dots,2k-2,2k,2k+2\ell+2+\delta\\},$

as follows from the induction hypothesis (4.31) by applying respectively

$\displaystyle\chi=\big{(}\tau_{L}\otimes
1^{\otimes(2k+2\ell+\delta)}\big{)}\big{(}1\otimes\Delta\otimes
1^{\otimes(2k+2\ell-2+\delta)}\big{)},$
$\displaystyle\chi=\big{(}1\otimes\Delta\otimes
1^{\otimes(2k+2\ell-1+\delta)}\big{)}\big{(}1\otimes\Delta\otimes
1^{\otimes(2k+2\ell-2+\delta)}\big{)},$
$\displaystyle\chi=\big{(}\tau_{L}\otimes
1^{\otimes(2k+2\ell+\delta)}\big{)}\big{(}\tau_{L}\otimes
1^{\otimes(2k+2\ell-1+\delta)}\big{)}.$

Apply the commutation relation

$[\Lambda_{\\{1,2k+2\ell+2+\delta\\}},\Lambda_{\\{1,3,4,6,\dots,2k,2k+1+\delta,2k+3+\delta,\dots,2k+2\ell+1+\delta,2k+2\ell+2+\delta\\}}]=0,$

which follows from Lemma 4.11. The remaining $q$-commutators can now be
expanded using Lemma 4.4 and (C5) with $k=1$ and $\ell=0$. This leads to the
anticipated expression for (4.32). ∎

##  5 More commutation relations

In this section we derive several criteria on the sets $A$ and $B$ for the
elements $\Lambda_{A}$ and $\Lambda_{B}$ to commute. Eventually, it will be
our aim to show that it suffices that $B\subseteq A$ in order to have
$[\Lambda_{A},\Lambda_{B}]=0$. This result, stated in Theorem 3.1, will be
proven in Section 6 and its proof will rely on the results of the present
section.

######  Lemma 5.1.

The element $\Lambda_{\\{1,3,5,\dots,2k-1\\}}$ commutes with
$\Lambda_{[1;2k]}$ and $\Lambda_{[1;2k-1]}$, and
$\Lambda_{\\{2,4,6,\dots,2k\\}}$ commutes with $\Lambda_{[1;2k]}$ and
$\Lambda_{[1;2k+1]}$.

###### Proof.

We will only show the first relation, the others follow in complete analogy.
We proceed by induction on $k$, the case $k=1$ being trivial. Suppose hence
that the claim holds for $k-1$, i.e., $\Lambda_{A}$ commutes with
$\Lambda_{B}$ for

$\displaystyle A=\\{1,3,5,\dots,2k-3\\},\qquad B=[1;2k-2].$ (5.1)

By Lemma 4.4 we may write

$\displaystyle\Lambda_{\\{1,3,5,\dots,2k-1\\}}=\frac{[\Lambda_{\\{1,3,5,\dots,2k-3,2k-2\\}},\Lambda_{\\{2k-2,2k-1\\}}]_{q}}{q^{-2}-q^{2}}$
$\displaystyle\hphantom{\Lambda_{\\{1,3,5,\dots,2k-1\\}}=}{}+\frac{\Lambda_{\\{2k-2\\}}\Lambda_{\\{1,3,5,\dots,2k-3,2k-2,2k-1\\}}+\Lambda_{\\{2k-1\\}}\Lambda_{\\{1,3,5,\dots,2k-3\\}}}{q+q^{-1}}.$

Hence it suffices to show that $\Lambda_{[1;2k]}$ commutes with each of the
terms in the right-hand side. For $\Lambda_{\\{1,3,5,\dots,2k-3,2k-2\\}}$,
$\Lambda_{\\{1,3,5,\dots,2k-3,2k-2,2k-1\\}}$ and
$\Lambda_{\\{1,3,5,\dots,2k-3\\}}$ this follows from the induction hypothesis
(5.1) by
$\chi=\big{(}1^{\otimes(2k-2)}\otimes\Delta\big{)}\big{(}1^{\otimes(2k-4)}\otimes\Delta\otimes
1\big{)}$, $\chi=\big{(}1^{\otimes(2k-3)}\otimes\Delta\otimes
1\big{)}\big{(}1^{\otimes(2k-4)}\otimes\Delta\otimes 1)$ and
$\chi=\big{(}1^{\otimes(2k-2)}\otimes\Delta\big{)}\big{(}1^{\otimes(2k-3)}\otimes\Delta\big{)}$
respectively. For $\Lambda_{\\{2k-2,2k-1\\}}$ this follows from
$[\Lambda_{\\{2\\}},\Lambda_{\\{1,2,3\\}}]=0$ acted upon with
$1\otimes\Delta\otimes 1$ and iterations of $\Delta\otimes 1^{\otimes n}$. ∎

Another useful commutation relation relies on Lemma 4.12. The proof is similar
to the one above, we will hence just sketch it.

######  Lemma 5.2.

The element $\Lambda_{\\{2,4,\dots,2k\\}}$ commutes with
$\Lambda_{\\{1,2,4,\dots,2k,2k+1\\}}$.

###### Sketch of the proof.

By induction on $k$. Rewrite $\Lambda_{\\{2,4,\dots,2k\\}}$ by ($\ast$ ‣ 3)
for (C6′), with $k-1$ instead of $k$ and $\ell=0$. Check that
$\Lambda_{\\{1,2,4,\dots,2k,2k+1\\}}$ commutes with all generators in the new
expression. This requires us to use Lemma 4.11, the first statement of
Corollary 4.1 acted upon with $1^{\otimes(2k-1)}\otimes\Delta$, the induction
hypothesis acted upon with
$\big{(}1^{\otimes(2k-1)}\otimes\Delta\big{)}\big{(}1^{\otimes(2k-2)}\otimes\tau_{R}\big{)}$
and the relation $[\Lambda_{\\{1,3\\}},\Lambda_{\\{1,2,3\\}}]=0$ acted upon
with iterations of $\big{(}\Delta\otimes
1^{\otimes(n+1)}\big{)}\big{(}\tau_{L}\otimes 1^{\otimes n}\big{)}$. ∎

In the following commutation relations, the indexing subsets will no longer be
defined by integers $k$ and $\ell$, but rather by more general conditions.

######  Lemma 5.3.

Let $A$ be a set of consecutive integers and $B\subseteq A$. Then
$\Lambda_{A}$ and $\Lambda_{B}$ commute.

###### Proof.

Without loss of generality, we may assume $A=[1;n]$ and write $B$ as

$B=[i_{1};j_{1}]\cup\cdots\cup[i_{k};j_{k}].$

The claim follows upon applying

$\left(\overrightarrow{\prod_{\ell=n-i_{1}+1}^{n-2}}\big{(}\Delta\otimes
1^{\otimes\ell}\big{)}\right)^{1-\delta_{i_{1},1}}\left(\overrightarrow{\prod_{\ell=j_{k}-i_{1}+2-\delta_{i_{1},1}}^{n-i_{1}-\delta_{i_{1},1}}}\big{(}1^{\otimes\ell}\otimes\Delta\big{)}\right)^{1-\delta_{j_{k},n}}\chi^{\prime},$

with

$\chi^{\prime}=\overleftarrow{\prod_{m=1-\delta_{i_{1},1}}^{2k-1-\delta_{i_{1},1}}}\overrightarrow{\prod_{\ell=\alpha_{n,\mathbf{i},\mathbf{j}}+1-\delta_{j_{k},n}}^{\beta_{n,\mathbf{i},\mathbf{j}}+1-\delta_{j_{k},n}}}\big{(}1^{\otimes
m}\otimes\Delta\otimes 1^{\otimes\ell}\big{)}$

to the commutation relation ($\Delta$) for

$A=\\{2-\delta_{i_{1},1},4-\delta_{i_{1},1},\dots,2k-\delta_{i_{1},1}\\},\qquad
B=[1;2k+1-\delta_{i_{1},1}-\delta_{j_{k},n}],$

which follows from Lemma 5.1. ∎

It can easily be checked explicitly that one has

$\displaystyle[\Lambda_{\\{1,3,4\\}},\Lambda_{\\{1,3\\}}]=0,\qquad$
$\displaystyle[\Lambda_{\\{1,3,4\\}},\Lambda_{\\{1,4\\}}]=0,\qquad$
$\displaystyle[\Lambda_{\\{1,3,4,5\\}},\Lambda_{\\{1,4\\}}]=0,$
$\displaystyle[\Lambda_{\\{1,2,4\\}},\Lambda_{\\{1,4\\}}]=0,\qquad$
$\displaystyle[\Lambda_{\\{1,2,4,5\\}},\Lambda_{\\{1,4\\}}]=0,\qquad$
$\displaystyle[\Lambda_{\\{1,2,4,5\\}},\Lambda_{\\{1,5\\}}]=0,$
$\displaystyle[\Lambda_{\\{1,2,4,5,6\\}},\Lambda_{\\{1,5\\}}]=0,\qquad$
$\displaystyle[\Lambda_{\\{1,2,4\\}},\Lambda_{\\{2,4\\}}]=0,\qquad$
$\displaystyle[\Lambda_{\\{1,2,4,5\\}},\Lambda_{\\{2,4\\}}]=0,$
$\displaystyle[\Lambda_{\\{1,2,4,5\\}},\Lambda_{\\{2,5\\}}]=0,\qquad$
$\displaystyle[\Lambda_{\\{1,2,4,5,6\\}},\Lambda_{\\{2,5\\}}]=0,\qquad$
$\displaystyle[\Lambda_{\\{1,2,3,5\\}},\Lambda_{\\{2,5\\}}]=0,$
$\displaystyle[\Lambda_{\\{1,2,3,5,6\\}},\Lambda_{\\{2,5\\}}]=0,\qquad$
$\displaystyle[\Lambda_{\\{1,2,3,5,6\\}},\Lambda_{\\{2,6\\}}]=0,\qquad$
$\displaystyle[\Lambda_{\\{1,2,3,5,6,7\\}},\Lambda_{\\{2,6\\}}]=0.$ (5.2)

These relations are needed in order to show the following result.

######  Lemma 5.4.

Let $A=[1;j_{1}]\cup[i_{2};j_{2}]$ and $B$ be such that $B\subseteq A$ and
$|B|=2$, then $\Lambda_{A}$ and $\Lambda_{B}$ commute.

###### Sketch of the proof.

If $B\subseteq[1;j_{1}]$, then by acting with morphisms
$1^{\otimes\ell}\otimes\tau_{R}$ and $1^{\otimes m}\otimes\Delta$, the claim
follows from $[\Lambda_{[1;j_{1}+1]},\Lambda_{B}]=0$, as asserted by Lemma
5.3. Similarly if $B\subseteq[i_{2};j_{2}]$.

Now suppose $B\nsubseteq[1;j_{1}]$ and $B\nsubseteq[i_{2};j_{2}]$, then
$B\cap[1;j_{1}]$ consists of a single element $n_{1}$ and $B\cap[i_{2};j_{2}]$
of a single element $n_{2}$. One can distinguish 16 cases: each of the 4 cases
$n_{1}=1=j_{1}$, $n_{1}=1\neq j_{1}$, $n_{1}=j_{1}\neq 1$, $j_{1}\neq
n_{1}\neq 1$ can be combined with each of the 4 cases determined similarly by
the mutual equality of $n_{2}$, $j_{2}$ and $i_{2}$. Each of those cases
follows from one of the relations in (5.2) upon applying $\Delta$ on suitable
tensor product positions. ∎

The following lemma aims to remove part of the restrictions on the set $B$.

######  Lemma 5.5.

Let $A=[1;j_{1}]\cup[i_{2};j_{2}]$, and $B$ be such that $B\subseteq A$ and
$|B\cap[i_{2};j_{2}]|=1$, then $\Lambda_{A}$ and $\Lambda_{B}$ commute.

###### Proof.

By induction on $k=|B\cap[1;j_{1}]|$, the case $k=0$ being trivial by Lemma
4.2 and $k=1$ following from Lemma 5.4. So let $k\geq 2$ and suppose the claim
holds for all sets $B^{\prime}\subset A$ with $|B^{\prime}\cap[1;j_{1}]|$
strictly less than $k$.

If $B\cap[1;j_{1}]$ is a set of consecutive integers, the statement follows
from Lemma 5.4 upon applying $\Delta$ on suitable positions as in Proposition
2.4. If not, then $B\cap[1;j_{1}]$ contains at least one hole, say between the
elements $x_{1}$ and $x_{2}$. Let us write $B_{1}=\\{b\in B\colon b<x_{1}\\}$
and $B_{2}=\\{b\in B\colon b>x_{2}\\}$. By Lemma 4.4 we may write

$\displaystyle\Lambda_{B}=\frac{[\Lambda_{B_{1}\cup[x_{1};x_{2}-1]},\Lambda_{[x_{1}+1;x_{2}]\cup
B_{2}}]_{q}}{q^{-2}-q^{2}}+\frac{\Lambda_{[x_{1}+1;x_{2}-1]}\Lambda_{B_{1}\cup[x_{1};x_{2}]\cup
B_{2}}+\Lambda_{B_{1}\cup\\{x_{1}\\}}\Lambda_{\\{x_{2}\\}\cup
B_{2}}}{q+q^{-1}}\\!\\!\\!$ (5.3)

and hence it suffices to show that $\Lambda_{A}$ commutes with each of the
terms in the right-hand side. For $\Lambda_{B_{1}\cup[x_{1};x_{2}-1]}$,
$\Lambda_{[x_{1}+1;x_{2}-1]}$ and $\Lambda_{B_{1}\cup\\{x_{1}\\}}$ this
follows from Lemma 5.3 with $A=[1;j_{1}+1]$, upon applying repeatedly
$1^{\otimes m}\otimes\tau_{R}$ and $1^{\otimes\ell}\otimes\Delta$. For
$\Lambda_{[x_{1}+1;x_{2}]\cup B_{2}}$, $\Lambda_{B_{1}\cup[x_{1};x_{2}]\cup
B_{2}}$ and $\Lambda_{\\{x_{2}\\}\cup B_{2}}$ this follows from the induction
hypothesis. This concludes the proof. ∎

We can now also remove the constraint on $|B\cap[i_{2};j_{2}]|$. The proof is
completely similar to the previous one, we will hence only sketch it.

######  Proposition 5.1.

Let $A=[1;j_{1}]\cup[i_{2};j_{2}]$ and $B$ be such that $B\subseteq A$, then
$\Lambda_{A}$ commutes with $\Lambda_{B}$.

###### Sketch of the proof.

If $B\cap[1;j_{1}]$ is empty, the result follows from Lemma 5.3. If it is not,
then we will proceed by induction on $k=|B\cap[i_{2};j_{2}]|$. The case $k=0$
follows from Lemma 5.3 and $k=1$ from Lemma 5.5. Let hence $k\geq 2$ and
suppose the claim holds for all values of $|B\cap[i_{2};j_{2}]|$ strictly less
than $k$. We again distinguish between $B\cap[i_{2};j_{2}]$ without holes, in
which case Lemma 5.5 suffices, and $B\cap[i_{2};j_{2}]$ containing holes, in
which case we can rewrite $\Lambda_{B}$ using (5.3) and apply Lemma 5.3 and
the induction hypothesis. ∎

Let us now take a look at the case where the set $A$ consists of more than
just 2 discrete intervals.

######  Lemma 5.6.

Let
$A=[1;j_{1}]\cup[i_{2};j_{2}]\cup\dots\cup[i_{k-1};j_{k-1}]\cup[i_{k};j_{k}]$,
with $k\geq 3$. Then $\Lambda_{A}$ commutes with $\Lambda_{B}$, where $B$ is
any of the sets

$\displaystyle
B=[i_{1}^{\prime};j_{1}]\cup[i_{2};j_{2}]\cup\dots\cup[i_{k-1};j_{k-1}]\cup[i_{k};j_{k}^{\prime}],\quad\
$ $\displaystyle
B=[i_{1}^{\prime};j_{1}]\cup[i_{2};j_{2}]\cup\dots\cup[i_{k-1};j_{k-1}],$
$\displaystyle
B=[i_{2};j_{2}]\cup\dots\cup[i_{k-1};j_{k-1}]\cup[i_{k};j_{k}^{\prime}],\quad$
$\displaystyle B=[i_{2};j_{2}]\cup\dots\cup[i_{k-1};j_{k-1}],$

where $1\leq i_{1}^{\prime}\leq j_{1}$ and $i_{k}\leq j_{k}^{\prime}\leq
j_{k}$.

###### Proof.

Let us start with the first claim. If $i_{1}^{\prime}=1$ or
$j_{k}^{\prime}=j_{k}$, this follows from Lemma 4.3. So suppose
$1<i_{1}^{\prime},j_{k}^{\prime}<j_{k}$. Then the statement follows upon
applying

$\displaystyle\chi=\left(\overrightarrow{\prod_{\ell=j_{k}-i_{1}^{\prime}+1}^{j_{k}-2}}\big{(}\Delta\otimes
1^{\otimes\ell}\big{)}\right)\left(\overrightarrow{\prod_{\ell=j_{k}-j_{1}}^{j_{k}-i_{1}^{\prime}-1}}\big{(}1\otimes\Delta\otimes
1^{\otimes\ell}\big{)}\right)\left(\overleftarrow{\prod_{n=1}^{2k-3}}\overrightarrow{\prod_{\ell=\alpha_{n,\mathbf{i},\mathbf{j}}}^{\beta_{n,\mathbf{i},\mathbf{j}}}}\big{(}1^{\otimes(n+1)}\otimes\Delta\otimes
1^{\otimes\ell}\big{)}\right)$
$\displaystyle\hphantom{\chi=}{}\times\left(\overrightarrow{\prod_{\ell=j_{k}-j_{k}^{\prime}}^{j_{k}-i_{k}-1}}\big{(}1^{\otimes(2k-1)}\otimes\Delta\otimes
1^{\otimes\ell}\big{)}\right)\left(\overrightarrow{\prod_{\ell=0}^{j_{k}-j_{k}^{\prime}-2}}\big{(}1^{\otimes
2k}\otimes\Delta\otimes 1^{\otimes\ell}\big{)}\right)$

to the statement of Lemma 5.2, as asserted by Proposition 2.4. The second
claim follows from the first with $A$ replaced by
$[1;j_{1}]\cup[i_{2};j_{2}]\cup\dots\cup[i_{k-1};j_{k-1}+1]$, acted upon by
iterations of $1^{\otimes\ell}\otimes\tau_{R}$ and $1^{\otimes
m}\otimes\Delta$. Similarly for the other statements. ∎

For any $m\in\mathbb{N}$ one can define the following operators:

$\displaystyle\chi_{m}=\overleftarrow{\prod_{n=1}^{m}}\left(\left(\overrightarrow{\prod_{\ell=j_{k}-j_{n}}^{j_{k}-i_{n}-\delta_{n,1}}}\big{(}\Delta\otimes
1^{\otimes\ell}\big{)}\right)\left(\overrightarrow{\prod_{\ell=j_{k}-i_{n+1}+1}^{j_{k}-j_{n}-1}}\big{(}\tau_{L}\otimes
1^{\otimes\ell}\big{)}\right)\right),$ (5.4)
$\displaystyle\widetilde{\chi}_{m}=\overrightarrow{\prod_{n=m}^{k}}\left(\left(\overrightarrow{\prod_{\ell=i_{n}-1}^{j_{n}-1-\delta_{n,k}}}\big{(}1^{\otimes\ell}\otimes\Delta\big{)}\right)\left(\overrightarrow{\prod_{\ell=j_{n-1}}^{i_{n}-2}}\big{(}1^{\otimes\ell}\otimes\tau_{R}\big{)}\right)\right).$
(5.5)

These will be of use to show the following commutation relation.

######  Lemma 5.7.

Let
$A=[1;j_{1}]\cup[i_{2};j_{2}]\cup\dots\cup[i_{k-1};j_{k-1}]\cup[i_{k};j_{k}]$
and let $B$ be any of the sets

$B=[i_{1}^{\prime};j_{1}]\cup[i_{2};j_{2}]\cup\dots\cup[i_{k-1};j_{k-1}]\cup
B_{k},\qquad B=[i_{2};j_{2}]\cup\dots\cup[i_{k-1};j_{k-1}]\cup B_{k},$

where $k\geq 3$, $1\leq i_{1}^{\prime}\leq j_{1}$ and where
$B_{k}\subseteq[i_{k};j_{k}]$. Then $\Lambda_{A}$ and $\Lambda_{B}$ commute.

###### Proof.

We will start with the first claim. If $B_{k}$ is empty or of the form
$[i_{k};j_{k}^{\prime}]$ with $i_{k}\leq j_{k}^{\prime}\leq j_{k}$, the
statement follows from Lemma 5.6. So suppose $B_{k}$ is nonempty and not of
that particular form. We will prove the special case $i_{1}^{\prime}=j_{1}=2$,
the general case then follows upon applying $\Delta$ on tensor product
positions 1 and 2. Let $x_{1}=\min([i_{k};j_{k}]\setminus B_{k})$. Define the
sets

$\displaystyle B_{1}=\\{b\in B\colon
b<x_{1}\\}=\\{2\\}\cup[i_{2};j_{2}]\cup\dots\cup[i_{k-1};j_{k-1}]\cup[i_{k};x_{1}-1],$
$\displaystyle B_{2}=\\{b\in B_{k}\colon b>x_{1}\\},$

where we interpret $[i_{k};x_{1}-1]$ as the empty set in case $x_{1}=i_{k}$.
Note that both sets are nonempty, by the assumptions on $B_{k}$. By Lemma 4.4
we have

$\displaystyle\Lambda_{B}=\frac{[\Lambda_{B_{1}\cup\\{x_{1}\\}},\Lambda_{\\{x_{1}\\}\cup
B_{2}}]_{q}}{q^{-2}-q^{2}}+\frac{\Lambda_{\\{x_{1}\\}}\Lambda_{B\cup\\{x_{1}\\}}+\Lambda_{B_{1}}\Lambda_{B_{2}}}{q+q^{-1}}.$
(5.6)

Both $\Lambda_{B_{1}}$ and $\Lambda_{B_{1}\cup\\{x_{1}\\}}$ commute with
$\Lambda_{A}$, as follows from Lemma 5.6 and the form of $B_{1}$. Moreover,
Lemma 5.3 asserts

$[\Lambda_{[1;j_{k}-i_{k}+2]},\Lambda_{B_{2}-(i_{k}-2)}]=0,\qquad[\Lambda_{[1;j_{k}-i_{k}+2]},\Lambda_{(\\{x_{1}\\}\cup
B_{2})-(i_{k}-2)}]=0.$

Acted upon with $\chi_{k-1}$, defined in (5.4), this implies $\Lambda_{A}$
commutes with $\Lambda_{B_{2}}$ and $\Lambda_{\\{x_{1}\\}\cup B_{2}}$. Hence
it follows from (5.6) that

$[\Lambda_{A},\Lambda_{B}]=\frac{1}{q+q^{-1}}\Lambda_{\\{x_{1}\\}}[\Lambda_{A},\Lambda_{B\cup\\{x_{1}\\}}].$

We now repeat our reasoning. Either $B_{k}\cup\\{x_{1}\\}$ is of the form
$[i_{k};j_{k}^{\prime}]$, in which case the statement follows from Lemma 5.6.
If not, then upon defining
$x_{2}=\min([i_{k};j_{k}]\setminus(B_{k}\cup\\{x_{1}\\}))$, we find by the
same arguments

$[\Lambda_{A},\Lambda_{B\cup\\{x_{1}\\}}]=\frac{1}{q+q^{-1}}\Lambda_{\\{x_{2}\\}}[\Lambda_{A},\Lambda_{B\cup\\{x_{1},x_{2}\\}}].$

If we continue this process, then at some point the set
$B_{k}\cup\\{x_{1},x_{2},\dots,x_{m}\\}$ will inevitably be of the form
$[i_{k};j_{k}^{\prime}]$, with $j_{k}^{\prime}=\max(B)$, namely when we have
_filled up all the holes_ in the set $B_{k}$. At this point we have

$[\Lambda_{A},\Lambda_{B}]=\frac{1}{\big{(}q+q^{-1}\big{)}^{m}}\left(\prod_{\ell=1}^{m}\Lambda_{\\{x_{\ell}\\}}\right)[\Lambda_{A},\Lambda_{B\cup\\{x_{1},\dots,x_{m}\\}}]=0,$

where the last step uses Lemma 5.6. This shows the claim for the first given
form of $B$.

For the second form, our first claim asserts

$[\Lambda_{([i_{2}-1;j_{2}]\cup\dots\cup[i_{k-1};j_{k-1}]\cup[i_{k};j_{k}])-(i_{2}-2)},\Lambda_{([i_{2};j_{2}]\cup\dots\cup[i_{k-1};j_{k-1}]\cup
B_{k})-(i_{2}-2)}]=0,$

which yields the anticipated result by the left extension process. ∎

We can now also replace $[i_{1}^{\prime};j_{1}]$ in the previous lemma by an
arbitrary set. The proof will be completely parallel to the previous one.

######  Proposition 5.2.

Let
$A=[1;j_{1}]\cup[i_{2};j_{2}]\cup\dots\cup[i_{k-1};j_{k-1}]\cup[i_{k};j_{k}]$,
with $k\geq 3$, and let $B$ be such that $B\subseteq A$ and
$[i_{2};j_{2}]\cup\dots\cup[i_{k-1};j_{k-1}]\subseteq B$, then $\Lambda_{A}$
commutes with $\Lambda_{B}$.

###### Sketch of the proof.

If $B\cap[1;j_{1}]$ is empty or of the form $[i_{1}^{\prime};j_{1}]$, the
claim follows from Lemma 5.7. If not, then with $x_{1}=\max([1;j_{1}]\setminus
B)$, $B_{1}=\\{b\in B\colon b<x_{1}\\}$ and $B_{2}=\\{b\in B\colon
b>x_{1}\\}=[x_{1}+1;j_{1}]\cup[i_{2};j_{2}]\cup\dots\cup[i_{k-1};j_{k-1}]\cup(B\cap[i_{k};j_{k}])$,
we have again the relation (5.6). Each of the terms in the right-hand side of
(5.6) commutes with $\Lambda_{A}$, by Lemmas 5.3 and 5.7, except for
$\Lambda_{B\cup\\{x_{1}\\}}$. Defining recursively
$x_{i}=\max([1;j_{1}]\setminus(B\cup\\{x_{1},\dots,x_{i-1}\\}))$, we find

$[\Lambda_{A},\Lambda_{B}]=\frac{1}{q+q^{-1}}\Lambda_{\\{x_{1}\\}}[\Lambda_{A},\Lambda_{B\cup\\{x_{1}\\}}]=\frac{1}{\big{(}q+q^{-1}\big{)}^{2}}\Lambda_{\\{x_{1}\\}}\Lambda_{\\{x_{2}\\}}[\Lambda_{A},\Lambda_{B\cup\\{x_{1},x_{2}\\}}]=\cdots,$

which eventually becomes zero, since at some point the set
$(B\cup\\{x_{1},\dots,x_{m}\\})\cap[1;j_{1}]$ will inevitably be of the form
$[i_{1}^{\prime};j_{1}]$, with $i_{1}^{\prime}=\min(B)$, such that Lemma 5.7
will be applicable. ∎

##  6 Proof of Theorem 3.1

###### Proof.

Without loss of generality, we may assume $\min(A)=1$ and thus we can write
$A=[1;j_{1}]\cup[i_{2};j_{2}]\cup\dots\cup[i_{k};j_{k}]$. For ease of
notation, we will write $A_{\ell}$ for the discrete interval
$[i_{\ell};j_{\ell}]$. We will proceed by induction on $k$.

For $k=1$ the statement follows from Lemma 5.3. For $k=2$ we may invoke
Proposition 5.1. Suppose hence that $k\geq 3$ and that the statement holds for
any set $A^{\prime}$ consisting of strictly less than $k$ discrete intervals,
and for any set $B^{\prime}$ contained in $A^{\prime}$. We distinguish four
cases.

Case 1: $A_{2}\cup\dots\cup A_{k-1}\subseteq B$.

This case follows immediately from Proposition 5.2.

Case 2: $B\cap A_{k}=\varnothing$.

In this case the induction hypothesis asserts $[\Lambda_{A_{1}\cup\dots\cup
A_{k-1}\cup\\{j_{k-1}+1\\}},\Lambda_{B}]\allowbreak=0$, which implies our
claim upon acting repeatedly with $1^{\otimes\ell}\otimes\tau_{R}$ and
$1^{\otimes m}\otimes\Delta$.

Case 3: $B\cap A_{1}=\varnothing$.

Similarly, using the induction hypothesis and the left extension process.

Case 4: $A_{2}\cup\dots\cup A_{k-1}\nsubseteq B$, $B\cap A_{1}\neq\varnothing$
and $B\cap A_{k}\neq\varnothing$.

Let $x_{1}\in A\setminus B$ be such that $x_{1}\in A_{i}$ with
$i\in\\{2,\dots,k-1\\}$. Let us define the sets $B_{1}=\\{b\in B\colon
b<x_{1}\\}$ and $B_{2}=\\{b\in B\colon b>x_{1}\\}$. Then it follows from Lemma
4.4 that

$\displaystyle\Lambda_{B}=\frac{[\Lambda_{B_{1}\cup\\{x_{1}\\}},\Lambda_{\\{x_{1}\\}\cup
B_{2}}]_{q}}{q^{-2}-q^{2}}+\frac{\Lambda_{\\{x_{1}\\}}\Lambda_{B\cup\\{x_{1}\\}}+\Lambda_{B_{1}}\Lambda_{B_{2}}}{q+q^{-1}}.$
(6.1)

Note that none of the indexing sets above is empty, by our assumptions on $B$.
Observe that $B_{1}$ and $B_{1}\cup\\{x_{1}\\}$ are contained in
$A_{1}\cup\dots\cup A_{i}\cup\\{j_{i}+1\\}$, which contains strictly fewer
discrete intervals than $k$. Hence the induction hypothesis asserts

$\displaystyle[\Lambda_{B_{1}},\Lambda_{A_{1}\cup\dots\cup
A_{i}\cup\\{j_{i}+1\\}}]=0,\qquad[\Lambda_{B_{1}\cup\\{x_{1}\\}},\Lambda_{A_{1}\cup\dots\cup
A_{i}\cup\\{j_{i}+1\\}}]=0,$

which by $\widetilde{\chi}_{i+1}$, as defined in (5.5), asserts that
$\Lambda_{B_{1}}$ and $\Lambda_{B_{1}\cup\\{x_{1}\\}}$ commute with
$\Lambda_{A}$. A similar reasoning using the induction hypothesis and the
morphism $\chi_{i-1}$, defined in (5.4), shows that $\Lambda_{A}$ commutes
with $\Lambda_{B_{2}}$ and $\Lambda_{\\{x_{1}\\}\cup B_{2}}$. The expression
(6.1) now implies

$[\Lambda_{A},\Lambda_{B}]=\frac{1}{q+q^{-1}}\Lambda_{\\{x_{1}\\}}[\Lambda_{A},\Lambda_{B\cup\\{x_{1}\\}}].$

We may now repeat this reasoning. Either $B\cup\\{x_{1}\\}$ contains
$A_{2}\cup\dots\cup A_{k-1}$, in which case
$[\Lambda_{A},\Lambda_{B\cup\\{x_{1}\\}}]=0$ by Proposition 5.2. If not, then
as established before $A\setminus B$ contains an element $x_{2}$ contained in
$A_{j}$ for a certain $j\in\\{2,\dots,k-1\\}$, and as before this implies that

$[\Lambda_{A},\Lambda_{B\cup\\{x_{1}\\}}]=\frac{1}{q+q^{-1}}\Lambda_{\\{x_{2}\\}}[\Lambda_{A},\Lambda_{B\cup\\{x_{1},x_{2}\\}}].$

If we continue this process, then at some point the set
$B\cup\\{x_{1},\dots,x_{m}\\}$ must inevitably contain $A_{2}\cup\dots\cup
A_{k-1}$. At this point we have

$[\Lambda_{A},\Lambda_{B}]=\frac{1}{\big{(}q+q^{-1}\big{)}^{m}}\left(\prod_{\ell=1}^{m}\Lambda_{\\{x_{\ell}\\}}\right)[\Lambda_{A},\Lambda_{B\cup\\{x_{1},\dots,x_{m}\\}}]=0,$

where in the last step we have used Proposition 5.2. This concludes the proof.
∎

##  7 Proof of Theorem 3.2

###### Proof.

We will start by proving the case $A_{1},A_{2},A_{3}\neq\varnothing$ and
$A_{4}=\varnothing$. It suffices to do this in the situation where
$A_{1}=\\{1\\}$, $\min(A_{2})=2$ and $A_{3}=\\{\max(A_{2})+1\\}$, the general
case then follows upon applying suitable morphisms of the form $\Delta\otimes
1^{\otimes n}$, $\tau_{L}\otimes 1^{\otimes n}$, $1^{\otimes n}\otimes\Delta$
and $1^{\otimes n}\otimes\tau_{R}$. Let us write $A_{2}$ in the form

$A_{2}=[2;j_{1}]\cup[i_{2};j_{2}]\cup\dots\cup[i_{k};j_{k}].$

Cases (3.1), (3.2) and (3.3) then follow from (C1), (C2) and (C3)
respectively, by

$\displaystyle\chi=\left(\overrightarrow{\prod_{\ell=j_{k}-j_{1}+1}^{j_{k}-2}}\big{(}1\otimes\Delta\otimes
1^{\otimes\ell}\big{)}\right)\left(\overleftarrow{\prod_{n=1}^{2k-2}}\overrightarrow{\prod_{\ell=\alpha_{n,\mathbf{i},\mathbf{j}}+1}^{\beta_{n,\mathbf{i},\mathbf{j}}+1}}\big{(}1^{\otimes(n+1)}\otimes\Delta\otimes
1^{\otimes\ell}\big{)}\right).$

Indeed, it follows from Proposition 2.4, after separating the terms
corresponding to $n=0$ and applying coassociativity, that

$\displaystyle\Lambda_{A}=\left(\overrightarrow{\prod_{\ell=j_{k}-j_{1}+1}^{j_{k}-2}}\big{(}1\otimes\Delta\otimes
1^{\otimes\ell}\big{)}\right)\big{(}\Delta\otimes
1^{\otimes(j_{k}-j_{1}+1)}\big{)}$
$\displaystyle\hphantom{\Lambda_{A}=}{}\times\left(\overleftarrow{\prod_{n=1}^{2k-2}}\overrightarrow{\prod_{\ell=\alpha_{n,\mathbf{i},\mathbf{j}}+1}^{\beta_{n,\mathbf{i},\mathbf{j}}+1}}\big{(}1^{\otimes
n}\otimes\Delta\otimes
1^{\otimes\ell}\big{)}\right)\Lambda_{\\{1,3,5,\dots,2k-1\\}},$

where $\Lambda_{\\{1,3,5,{\dots},2k{-}1\\}}\\!$ is considered inside a
$(2k)$-fold tensor product. The morphism
$\Delta{\otimes}1^{\\!{\otimes}(j_{k}{-}j_{1}{+}1)\\!}\\!$ can be shifted
through the subsequent ones by (2.11), thus acting directly on
$\Lambda_{\\{1,3,5,\dots,2k-1\\}}$ by sending it to
$\Lambda_{\\{1,2,4,6,\dots,2k\\}}$. Hence we have
$\Lambda_{A}=\chi(\Lambda_{\\{1,2,4,6,\dots,2k\\}})$. Similarly, $\chi$ sends
$\Lambda_{\\{2,4,6,\dots,2k,2k+1\\}}$ to $\Lambda_{B}$ and so on.

The case $A_{1}=\varnothing$ and $A_{2},A_{3},A_{4}\neq\varnothing$ is
identical.

Now consider the case $A_{2}=\varnothing$ and
$A_{1},A_{3},A_{4}\neq\varnothing$. Then (3.2) and (3.3) follow from Theorem
3.1 and (2.8). To show (3.1), it again suffices to prove
$[\Lambda_{A},\Lambda_{B}]=0$, by (2.8), and to take $A_{1}=\\{1\\}$,
$\min(A_{3})=2$ and $A_{4}=\\{\max(A_{3})+1\\}$. Let us write $A_{3}$ in the
form

$A_{3}=[2;j_{1}]\cup[i_{2};j_{2}]\cup\dots\cup[i_{k};j_{k}].$

By

$\chi=\overleftarrow{\prod_{n=0}^{2k-2}}\overrightarrow{\prod_{\ell=\alpha_{n,\mathbf{i},\mathbf{j}}}^{\beta_{n,\mathbf{i},\mathbf{j}}}}\big{(}1^{\otimes(n+1)}\otimes\Delta\otimes
1^{\otimes(\ell+1)}\big{)}$

with $i_{1}=2$, this follows from Lemma 4.5.

The case $A_{3}=\varnothing$ and $A_{1},A_{2},A_{4}\neq\varnothing$ is
identical.

If two or more of the $A_{i}$ are empty, then again the statements follow from
Theorem 3.1 and (2.8).

Finally, suppose none of the $A_{i}$ is empty. Again, it suffices to consider
the special case where $A_{1}=\\{1\\}$, $\min(A_{2})=2$ and
$A_{4}=\\{\max(A_{3})+1\\}$. We will write $A_{2}$ and $A_{3}$ as disjoint
unions of discrete intervals. If $\min(A_{3})=\max(A_{2})+1$, then

$\displaystyle A_{2}=[2;j_{1}]\cup[i_{2};j_{2}]\cup\dots\cup[i_{k};j_{k}],$
$\displaystyle
A_{3}=[j_{k}+1;j_{k+1}]\cup[i_{k+2};j_{k+2}]\dots\cup[i_{k+\ell+1};j_{k+\ell+1}].$

Cases (3.1), (3.2) and (3.3) then follow from (C4), (C5) and (C6)
respectively, by

$\displaystyle\chi=\left(\overleftarrow{\prod_{n=0}^{2k-2}}\overrightarrow{\prod_{\ell=\alpha_{n,\mathbf{i},\mathbf{j}}+1}^{\beta_{n,\mathbf{i},\mathbf{j}}+1}}\big{(}1^{\otimes(n+1)}\otimes\Delta\otimes
1^{\otimes\ell}\big{)}\right)\left(\overrightarrow{\prod_{\ell=j_{k+\ell+1}-j_{k+1}+1}^{j_{k+\ell+1}-j_{k}-1}}\big{(}1^{\otimes
2k}\otimes\Delta\otimes 1^{\otimes\ell}\big{)}\right)$
$\displaystyle\hphantom{\chi=}{}\times\left(\overleftarrow{\prod_{n=2k+1}^{2k+2\ell}}\overrightarrow{\prod_{\ell=\alpha_{n,\mathbf{i},\mathbf{j}}+1}^{\beta_{n,\mathbf{i},\mathbf{j}}+1}}\big{(}1^{\otimes
n}\otimes\Delta\otimes 1^{\otimes\ell}\big{)}\right),$

where $\mathbf{i}=(2,i_{2},\dots,i_{k},j_{k}+1,i_{k+2},\dots,i_{k+\ell+1})$
and $\mathbf{j}=(j_{1},\dots,j_{k+\ell+1})$.

If $\min(A_{3})>\max(A_{2})+1$, then we have

$\displaystyle
A_{2}=[2;j_{1}]\cup[i_{2};j_{2}]\cup\dots\cup[i_{k};j_{k}],\qquad
A_{3}=[i_{k+1};j_{k+1}]\cup[i_{k+2};j_{k+2}]\dots\cup[i_{k+\ell+1};j_{k+\ell+1}]$

and by

$\displaystyle\chi=\overleftarrow{\prod_{n=0}^{2k+2\ell}}\overrightarrow{\prod_{\ell=\alpha_{n,\mathbf{i},\mathbf{j}}+1}^{\beta_{n,\mathbf{i},\mathbf{j}}+1}}\big{(}1^{\otimes(n+1)}\otimes\Delta\otimes
1^{\otimes\ell}\big{)},$

with $\mathbf{i}=(2,i_{2},\dots,i_{k+\ell+1})$ and
$\mathbf{j}=(j_{1},\dots,j_{k+\ell+1})$, the cases (3.1), (3.2) and (3.3)
follow from (C4′), (C5′) and (C6′) respectively. ∎

##  8 Higher rank $\boldsymbol{q}$-Bannai–Ito relations

Each of the results and proofs outlined in this paper carries over to the
higher rank $q$-Bannai–Ito algebra, which is canonically isomorphic to the
higher rank Askey–Wilson algebra. In this final section we will apply the
methods of Section 2 to construct this higher rank extension of the
$q$-Bannai–Ito algebra and we will argue why the Theorems 3.1 and 3.2 have
natural $q$-Bannai–Ito analogs.

Let us start by the defining the quantum superalgebra
$\mathfrak{osp}_{q}(1|2)$ as the $\mathbb{Z}_{2}$-graded associative algebra
over a field $\mathbb{K}$ generated by elements $A_{+}$, $A_{-}$, $K$,
$K^{-1}$ and $P$ subject to the relations

$\displaystyle KA_{+}K^{-1}=q^{1/2}A_{+},\qquad
KA_{-}K^{-1}=q^{-1/2}A_{-},\qquad\\{A_{+},A_{-}\\}=\frac{K^{2}-K^{-2}}{q^{1/2}-q^{-1/2}},$
$\displaystyle\\{P,A_{\pm}\\}=0,\qquad[P,K]=0,\qquad[P,K^{-1}]=0,\qquad
KK^{-1}=K^{-1}K=1,\qquad P^{2}=1.$

A Casimir element is given by

$\displaystyle\Gamma^{q}=\left(-A_{+}A_{-}+\frac{q^{-1/2}K^{2}-q^{1/2}K^{-2}}{q-q^{-1}}\right)P.$

Its coproduct
$\Delta\colon\mathfrak{osp}_{q}(1|2)\to\mathfrak{osp}_{q}(1|2)^{\otimes 2}$
and counit $\epsilon\colon\mathfrak{osp}_{q}(1|2)\to\mathbb{K}$ are defined by

$\displaystyle\Delta(A_{\pm})=A_{\pm}\otimes KP+K^{-1}\otimes A_{\pm},\quad\ \
\\!$ $\displaystyle\Delta\big{(}K^{\pm 1}\big{)}=K^{\pm 1}\otimes K^{\pm
1},\quad\ \ \\!$ $\displaystyle\Delta(P)=P\otimes P,$ (8.1)
$\displaystyle\epsilon(A_{\pm})=0,\qquad$ $\displaystyle\epsilon\big{(}K^{\pm
1}\big{)}=1,\qquad$ $\displaystyle\epsilon(P)=1.$

We will also need the definition of the $q$-anticommutator, given by

$\\{X,Y\\}_{q}=q^{1/2}XY+q^{-1/2}YX.$

Now let $\mathcal{I}_{R}$ be the subalgebra of $\mathfrak{osp}_{q}(1|2)$
generated by $A_{+}K$, $A_{-}K$, $K^{2}P$ and $\Gamma^{q}$ and similarly write
$\mathcal{I}_{L}$ for the $\mathfrak{osp}_{q}(1|2)$-subalgebra generated by
the elements $A_{+}K^{-1}P$, $A_{-}K^{-1}P$, $K^{-2}P$ and $\Gamma^{q}$. If
one defines the algebra morphism
$\tau_{R}\colon\mathcal{I}_{R}\to\mathfrak{osp}_{q}(1|2)\otimes\mathcal{I}_{R}$
by

$\displaystyle\tau_{R}(A_{-}K)=K^{2}P\otimes A_{-}K,$
$\displaystyle\tau_{R}(A_{+}K)=\big{(}K^{-2}P\otimes
A_{+}K\big{)}+q^{-1/2}\big{(}q-q^{-1}\big{)}\big{(}A_{+}^{2}P\otimes
A_{-}K\big{)}$
$\displaystyle\hphantom{\tau_{R}(A_{+}K)=}{}+q^{-1/2}\big{(}q^{1/2}-q^{-1/2}\big{)}\big{(}A_{+}K^{-1}P\otimes
K^{2}P\big{)}+q^{-1/2}\big{(}q-q^{-1}\big{)}\big{(}A_{+}K^{-1}P\otimes\Gamma^{q}\big{)},$
$\displaystyle\tau_{R}\big{(}K^{2}P\big{)}=1\otimes
K^{2}P-\big{(}q-q^{-1}\big{)}(A_{+}K\otimes A_{-}K),$
$\displaystyle\tau_{R}\big{(}\Gamma^{q}\big{)}=1\otimes\Gamma^{q},$ (8.2)

then $\mathcal{I}_{R}$ is readily checked to be a left coideal subalgebra of
$\mathfrak{osp}_{q}(1|2)$ and a left $\mathfrak{osp}_{q}(1|2)$-comodule with
coaction $\tau_{R}$. Similarly, the subalgebra $\mathcal{I}_{L}$ is a right
coideal subalgebra of $\mathfrak{osp}_{q}(1|2)$ and a right
$\mathfrak{osp}_{q}(1|2)$-comodule with coaction
$\tau_{L}\colon\mathcal{I}_{L}\to\mathcal{I}_{L}\otimes\mathfrak{osp}_{q}(1|2)$
defined by

$\displaystyle\tau_{L}\big{(}A_{-}K^{-1}P\big{)}=A_{-}K^{-1}P\otimes K^{-2}P,$
$\displaystyle\tau_{L}\big{(}A_{+}K^{-1}P\big{)}=A_{+}K^{-1}P\otimes
K^{2}P-q^{1/2}\big{(}q-q^{-1}\big{)}A_{-}K^{-1}P\otimes A_{+}^{2}P$
$\displaystyle\hphantom{\tau_{L}\big{(}A_{+}K^{-1}P\big{)}=}{}-q^{1/2}\big{(}q^{1/2}-q^{-1/2}\big{)}K^{-2}P\otimes
A_{+}K-q^{1/2}\big{(}q-q^{-1}\big{)}\Gamma^{q}\otimes A_{+}K,$
$\displaystyle\tau_{L}\big{(}K^{-2}P\big{)}=K^{-2}P\otimes
1-\big{(}q-q^{-1}\big{)}A_{-}K^{-1}P\otimes A_{+}K^{-1}P,$
$\displaystyle\tau_{L}\big{(}\Gamma^{q}\big{)}=\Gamma^{q}\otimes 1.$ (8.3)

In analogy to Corollary 2.1, one can easily verify that the element
$\Delta(\Gamma^{q})\in\mathcal{I}_{L}\otimes\mathcal{I}_{R}$ lies in the
cotensor product of $\mathcal{I}_{L}$ and $\mathcal{I}_{R}$, i.e.,

$(1\otimes\tau_{R})\Delta\big{(}\Gamma^{q}\big{)}=(\tau_{L}\otimes
1)\Delta\big{(}\Gamma^{q}\big{)}.$

This observation makes it possible to repeat the extension processes outlined
in Section 2 for the $q$-Bannai–Ito case, and in particular to state the
following definition.

######  Definition 8.1.

The $q$-Bannai–Ito algebra of rank $n-2$ is the subalgebra of
$\mathfrak{osp}_{q}(1|2)^{\otimes n}$ generated by the elements
$\Gamma_{A}^{q}$ with $A\subseteq[1;n]$, which are constructed by the left,
right or mixed extension processes of Definitions 2.3, 2.5 and 2.6, upon
replacing $\Lambda\in U_{q}(\mathfrak{sl}_{2})$ by
$\Gamma^{q}\in\mathfrak{osp}_{q}(1|2)$ and $\Delta$, $\tau_{R}$ and $\tau_{L}$
by the morphisms (8.1), (8.2) and (8.3).

It was explained in [8, Section 2.3] that the rank $n-2$ $q$-Bannai–Ito
algebra is isomorphic to the Askey–Wilson algebra of the same rank. The
isomorphism is explicitly given by

$\Lambda_{A}\mapsto-i\big{(}q-q^{-1}\big{)}\Gamma_{A}^{q},\qquad q\mapsto
iq^{1/2},$

where $i\in\mathbb{K}$ is a square root of $-1$.

Looking back at the Sections 3 to 7, it is clear that the Theorems 3.1 and 3.2
were proven based solely on the coassociativity of the coproduct, the cotensor
product property and the rank one relations (1.2)–(1.4). The whole strategy of
proof can thus be repeated to extend the rank one $q$-Bannai–Ito relations

$\displaystyle\big{\\{}\Gamma_{\\{1,2\\}}^{q},\Gamma_{\\{2,3\\}}^{q}\big{\\}}_{q}=\Gamma_{\\{1,3\\}}^{q}+\big{(}q^{1/2}+q^{-1/2}\big{)}\big{(}\Gamma_{\\{2\\}}^{q}\Gamma_{\\{1,2,3\\}}^{q}+\Gamma_{\\{1\\}}^{q}\Gamma_{\\{3\\}}^{q}\big{)},$
$\displaystyle\big{\\{}\Gamma_{\\{2,3\\}}^{q},\Gamma_{\\{1,3\\}}^{q}\big{\\}}_{q}=\Gamma_{\\{1,2\\}}^{q}+\big{(}q^{1/2}+q^{-1/2}\big{)}\big{(}\Gamma_{\\{3\\}}^{q}\Gamma_{\\{1,2,3\\}}^{q}+\Gamma_{\\{1\\}}^{q}\Gamma_{\\{2\\}}^{q}\big{)},$
$\displaystyle\big{\\{}\Gamma_{\\{1,3\\}}^{q},\Gamma_{\\{1,2\\}}^{q}\big{\\}}_{q}=\Gamma_{\\{2,3\\}}^{q}+\big{(}q^{1/2}+q^{-1/2}\big{)}\big{(}\Gamma_{\\{1\\}}^{q}\Gamma_{\\{1,2,3\\}}^{q}+\Gamma_{\\{2\\}}^{q}\Gamma_{\\{3\\}}^{q}\big{)},$

as shown in [15], to higher rank. This leads to the following analog of
Theorems 3.1 and 3.2.

######  Theorem 8.1.

Let $A,B\subseteq[1;n]$ be such that $B\subseteq A$, then $\Gamma_{A}^{q}$ and
$\Gamma_{B}^{q}$ commute.

######  Theorem 8.2.

Let $A_{1}$, $A_{2}$, $A_{3}$ and $A_{4}$ be $($potentially empty$)$ subsets
of $[1;n]$ satisfying

$A_{1}\prec A_{2}\prec A_{3}\prec A_{4},$

where we have used Definition 3.1. The relation

$\big{\\{}\Gamma_{A}^{q},\Gamma_{B}^{q}\big{\\}}_{q}=\Gamma_{(A\cup
B)\setminus(A\cap B)}^{q}+\big{(}q^{1/2}+q^{-1/2}\big{)}\big{(}\Gamma_{A\cap
B}^{q}\Gamma_{A\cup B}^{q}+\Gamma_{A\setminus(A\cap
B)}^{q}\Gamma_{B\setminus(A\cap B)}^{q}\big{)}$

is satisfied for $A$ and $B$ defined by one of the relations (3.1)–(3.3).

##  9 Conclusions and outlook

In this paper, we have presented several construction techniques for the
higher rank Askey–Wilson algebra ${\rm AW}(n)$, equivalent with the algorithm
given in [8]. The key observation was the existence of a novel, right coideal
comodule subalgebra of $U_{q}(\mathfrak{sl}_{2})$. We have proven a large
class of algebraic identities inside ${\rm AW}(n)$, culminating in Theorems
3.1 and 3.2, by elementary and intrinsic methods. Each of the proofs also
applies, mutatis mutandis, to the higher rank $q$-Bannai–Ito algebra, which is
isomorphic to ${\rm AW}(n)$.

At present, it is still an open question whether the relations obtained in
Theorems 3.1 and 3.2 define the algebras ${\rm AW}(n)$ abstractly. Using
computer algebra packages, we have obtained all subsets
$A\subseteq\\{1,\dots,n\\}$ for which the relation ($\ast$ ‣ 3) is satisfied,
for several values of $n$, and each of them turned out to be of the form
(3.1), (3.2) or (3.3). Hence if supplementary relations are needed to obtain a
complete set of defining relations for the algebras ${\rm AW}(n)$, they will
most likely not be of the form ($\ast$ ‣ 3). Other related problems are the
construction of similar algebras for more general quantum groups
$U_{q}(\mathfrak{g})$ and the behavior of the obtained algebras at $q$ a root
of unity different from 1. We plan to look into these questions in further
research.

### Acknowledgements

HDC is a PhD Fellow of the Research Foundation Flanders (FWO). This work was
also supported by FWO Grant EOS 30889451. The author wishes to thank the
anonymous referees for their valuable suggestions and comments.

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