# Fusion for the Yang-Baxter equation and the braid group

L. Poulain d’Andecy111Laboratoire de Mathématiques de Reims UMR 9008,
Université de Reims Champagne-Ardenne, Moulin de la Housse BP 1039, 51100
Reims, France<EMAIL_ADDRESS>
###### Abstract

These are the extended notes of a mini-course given at the school
_WinterBraids X_. We discuss algebras simultaneously related to: the braid
group, the Yang–Baxter equation and the representation theory of quantum
groups. The main goal is to explain the idea of the fusion procedure for the
Yang–Baxter equation and to show how it leads to new examples of such
algebras: the fused Hecke algebras.

$1$$2$$3$$u_{12}$$u_{13}$$u_{23}$$=$$1$$2$$3$$u_{23}$$u_{13}$$u_{12}$

$=$

_The Yang–Baxter equation and the braid relation._

###### Contents

  1. Introduction
  2. 1 The Braid Group and the Yang–Baxter Equation
    1. 1.1 The Braid group
    2. 1.2 The Yang–Baxter equation
  3. 2 The Guiding Examples: Hecke and Birman–Murakami–Wenzl
    1. 2.1 The Hecke algebra
    2. 2.2 The Birman–Murakami–Wenzl algebra
    3. 2.3 Baxterization
  4. 3 Quantum Groups
    1. 3.1 Some properties of quantum groups
    2. 3.2 Local representations of the braid group
    3. 3.3 Solutions of the YB equation
    4. 3.4 Example of $U_{q}(sl_{2})$
  5. 4 Centralisers of Tensor Representations of Quantum Groups
    1. 4.1 Properties of centralisers
    2. 4.2 Schur–Weyl duality
  6. 5 Fusion Procedure for the Yang–Baxter Equation
    1. 5.1 Fusion procedure for matrix solutions
    2. 5.2 Fusion procedure, the braided version
    3. 5.3 Fusion procedure in the Hecke algebra
    4. 5.4 An example
  7. 6 Fused Braids and the Fused Hecke Algebra
    1. 6.1 Objects and multiplication
    2. 6.2 Fused permutations and standard basis
    3. 6.3 Conclusion

## Introduction

The main purpose of these lectures is to discuss algebras with the following
features:

  * $\bullet$

they are quotients of the braid group algebra;

  * $\bullet$

they contain abstract solutions of the Yang–Baxter equation;

  * $\bullet$

they admit representations on vector spaces of the form $V^{\otimes n}$.

This “wish list” of properties is fulfilled in three famous situations, the
Temperley–Lieb algebra, the Hecke algebra and the Birman–Murakami–Wenzl
algebra. These algebras are well-known quotients of the algebra of the braid
group; in fact they are the algebras behind the following invariants of links:
the Jones polynomial, the Homfly-pt polynomial and the Kauffman polynomial.

It is also well-know that they contain solutions of the Yang–Baxter equation
(so-called Baxterization formulas). We call them abstract solutions, since
they are solutions in some algebras, as opposed to solutions acting on vector
spaces (matrix solutions) as genuine solutions of the Yang–Baxter equation
should do. To obtain genuine matrix solutions, we need to look for
representations of some special form, and these three algebras indeed have
such representations on spaces $V^{\otimes n}$.

The third item on our wish list, the existence of representations on tensor
spaces $V^{\otimes n}$, is the key to unifying these examples. In fact, the
Temperley–Lieb algebra, the Hecke algebra and the Birman–Murakami–Wenzl
algebra (through their actions on tensor spaces) can be seen as centralisers
of representations of some interesting algebras: quantum groups. The first
goal of these notes is to discuss briefly how the quantum groups (through
their centralisers) are nicely designed to produce algebras fulfilling our
wish list above.

Once we know that our sought-after algebras can be found among the
centralisers of representations of quantum groups, we would like to be able to
describe and study these centralisers. The second goal of these notes is to
explain how to fit the fusion procedure in this picture. The fusion procedure
was designed at the matrix level to produce new solutions of the Yang–Baxter
equation. Here we would like to use it in order to understand better the
centralisers. This leads us to introduce topological objects that we call
fused braids, and to define from them new algebras, called fused Hecke
algebras.

We proceed as follows. The first two sections are introductory to the braid
group and the Yang–Baxter equation, and describe in details our favourite
examples: the Hecke algebra and the Birman–Murakami–Wenzl algebra. Sections 3
and 4 discuss quantum groups and their centralisers. Without full details,
they are intended to give an idea of how these objects (quantum groups) are
constructed, how they behave and how they are relevant to our subject.

In Section 5, we present the fusion procedure and we revert to a more rigorous
presentation for this part. Finally, the last section, presents the fused
Hecke algebras. After discussing the braid-like description of these algebras,
we go over our wish list, and we check that we indeed produced algebras
fulfilling every wishes on the list, and without too much surprise, that these
algebras are related to the centralisers of some quantum groups
representations.

The lectures were intended to PhD students and young researchers studying
braids and their different aspects and applications. In these extended notes,
more details are given than during the lectures, but many are still omitted or
hidden, and full mathematical precision is not always the priority. Some
background in basic algebra and representation theory is assumed.

##### Aknowledgements.

It is a pleasure to thank the organisers Paolo Bellingeri, Vincent Florens, JB
Meilhan, Emmanuel Wagner of the Winterbraids series of schools and the
organiser Filippo Callegaro of the session in Pisa, along with all
participants of the tenth edition. The work of the author is supported by
Agence Nationale de la Recherche Projet AHA ANR-18-CE40-0001.

##  1 The Braid Group and the Yang–Baxter Equation

###  1.1 The Braid group

For precise definitions and more details on the braid group, we refer to [46].
Here is an example of a braid (with $5$ strands):

$\scriptstyle{\sigma_{1}}$$\scriptstyle{\sigma_{1}}$$\scriptstyle{\sigma_{2}}$$\scriptstyle{\sigma_{3}}$$\scriptstyle{\sigma_{4}^{-1}}$$\scriptstyle{\sigma_{2}}$$\scriptstyle{\sigma_{2}^{-1}}$$\scriptstyle{\sigma_{4}^{-1}}$$\scriptstyle{\sigma_{4}}$$\scriptstyle{\sigma_{3}}$

We fix two horizontal parallel lines each having $5$ fixed points, and we
connect bijectively each point on the top line to a point on the bottom line
by a “strand”.

This is an object in a three-dimensional space, so the strands can pass “over”
or “under” other strands. For now, we may ignore the labels $\sigma_{i}^{\pm
1}$ next to the crossings.

Braids are considered up to isotopy, meaning that we can move continuously the
strands while leaving their end points fixed, and this is still the same
braid.

As shown by the example, a braid with $n$ strands is drawn in a rectangular
strip with a top line of $n$ fixed dots and a bottom line of $n$ fixed dots.
We connect bijectively each top dot to a bottom dot by a strand inside the
strip. A strand is a continuous line going from a top point to a bottom point.
By convention, we assume that the vertical coordinate is always decreasing
along the line (no strand is allowed to go back towards the top).

At each point of the strip at most two strands are crossing each other, and at
each crossing, we indicate which strand passes over the other one. We call a
crossing positive (resp. negative) when the strand coming from the left passes
over (resp. under) the strand coming from the right. Such diagram is called a
braid with $n$ strands and braids are considered up to isotopy (continuous
moves of the strands with fixed end points).

The set of all braids with $n$ strands forms a group, denoted $B_{n}$, the
multiplication being simply the vertical concatenation of diagrams. If
$\alpha,\beta\in B_{n}$, to perform the product $\alpha\beta$, we place the
diagram of $\alpha$ on top of the diagram of $\beta$, we identify the bottom
dots of $\alpha$ with the top dots of $\beta$, thereby connecting the strands,
and we delete these middle dots.

The identity element $1$ of $B_{n}$ is the braid where the $n$ strands are
vertical and parallel. It might not be completely obvious at first sight that
the braids form a group. In fact, to find the inverse of a given braid,
reflect it through the bottom horizontal line and then invert all the
crossings. Up to isotopy, the concatenation with the given braid reduces to
the identity element.

The following elements are called elementary braidings:

$\sigma_{i}=$$1$$\dots$$i-1$$i$$i+1$$i+2$$\dots$$n$$i\in\\{1,\dots,n-1\\}$;

We can see that sometimes we number the dots $1,...,n$ from left to right to
be able to speak of the “$i$-th dot” and to help visualize where a crossing
is. The previous braid provides an example of positive crossing. As an example
of the multiplication rule and of the use of isotopy, one can check that the
inverse $\sigma_{i}^{-1}$ of $\sigma_{i}$ is:

$\sigma_{i}^{-1}=$$1$$\dots$$i-1$$i$$i+1$$i+2$$\dots$$n$$i\in\\{1,\dots,n-1\\}$;

It is rather easy to accept that the braid group $B_{n}$ is generated by the
elements $\sigma_{1},\dots,\sigma_{n-1}$, meaning that any braid can be
written as a product of elementary braidings $\sigma_{i}$ and their inverses.
In the braid pictured above, the elementary braiding are indicated next to
each crossing, and we see that this braid is:

$\sigma_{1}\sigma_{3}\sigma_{4}^{-1}\sigma_{1}\sigma_{2}\sigma_{4}^{-1}\sigma_{2}\sigma_{3}\sigma_{4}\sigma_{2}^{-1}\
.$

As we can see the decomposition of a braid as a product of generators is not
unique (for example, we could have ended the word by
$\sigma_{2}^{-1}\sigma_{4}$). This comes from the fact that some relations are
satisfied by the generators, reflecting the invariance under isotopy. Namely,
in addition to the trivial relations
$\sigma_{i}\sigma_{i}^{-1}=\sigma_{i}^{-1}\sigma_{i}=1$, it is easy to see
that the following relations are satisfied:

$\sigma_{i}\sigma_{j}=\sigma_{j}\sigma_{i}\ \ \ \ \ \text{if $|i-j|>1$,}$

since if $|i-j|>1$, then the pairs of strands $i,i+1$ and $j,j+1$ can be
manipulated independently. It is also immediate that the following relations
are satisfied: $i$$i\\!+\\!1$$i\\!+\\!2$$=$$i$$i\\!+\\!1$$i\\!+\\!2$ Indeed,
one simply has to move the middle strand. In algebraic terms, we have:
$\sigma_{i}\sigma_{i+1}\sigma_{i}=\sigma_{i+1}\sigma_{i}\sigma_{i+1}\,,\ \ \ \
\text{for $i=1,\dots,n-2$.}$ It is remarkable that only the two sorts of
relations above are enough to completely characterise the braid group. This is
the well known following theorem.

######  Theorem 1.1 (Artin [1, 2, 8]).

The braid group $B_{n}$ is generated by $\sigma_{1},\dots,\sigma_{n-1}$ with
defining relations:

$\begin{array}[]{ll}\sigma_{i}\sigma_{i+1}\sigma_{i}=\sigma_{i+1}\sigma_{i}\sigma_{i+1}\,,&\text{for
$i\in\\{1,\dots,n-2\\}$}\,,\\\\[1.99997pt]
\sigma_{i}\sigma_{j}=\sigma_{j}\sigma_{i}\,,&\text{for
$i,j\in\\{1,\dots,n-1\\}$ such that $|i-j|>1$}\,.\end{array}$ (1)

The phraseology “defining relations” means that any other relation involving
braids in $B_{n}$ is implied by these ones (and the trivial ones
$\sigma_{i}\sigma_{i}^{-1}=\sigma_{i}^{-1}\sigma_{i}=1$). Equivalently, it
means that any other group with generators satisfying the relations (1) must
be a quotient of the braid group $B_{n}$.

##### Local representations of $B_{n}$.

Let $V$ be a vector space and form the tensor product $V^{\otimes n}$ of $n$
copies of $V$. We take an invertible element $R\in\text{End}(V\otimes V)$ and
we define elements $R_{1},\dots,R_{n-1}$ like this:

$\text{Operators $R_{1},\dots,R_{n-1}$ on $V^{\otimes n}$:}\qquad\ \ \ \hbox
to0.0pt{$\overbrace{\phantom{V\otimes
V}}^{R_{1}}$\hss}V\otimes\overbrace{V\otimes
V}^{R_{2}}\otimes\dots\overbrace{\ldots\otimes V}^{R_{n-1}}$

More precisely, for $i=1,\dots,n-1$, the operator $R_{i}$ acts on $V^{\otimes
n}$ non-trivially only in the copies $i,i+1$ of $V$, where it acts by $R$.
That is, it is defined by $R_{i}=\text{Id}_{V^{\otimes i-1}}\otimes
R\otimes\text{Id}_{V^{\otimes n-i-1}}$.

Now we say that $R$ defines a local representation of the braid group $B_{n}$,
or that the following map

$\begin{array}[]{lcrcl}\rho&:&\ B_{n}&\to&\text{End}(V^{\otimes n})\\\\[5.0pt]
&&\sigma_{i}&\mapsto&R_{i}\end{array}$

is a local representation of $B_{n}$ on the tensor space $V^{\otimes n}$, if
the braid relation is satisfied

$R_{1}R_{2}R_{1}=R_{2}R_{1}R_{2}\ \quad\ \ \ \ \text{on $V\otimes V\otimes
V$.}$ (2)

This immediately implies all the other defining relations of $B_{n}$, thereby
making the map $\rho$ to be a representation of the group $B_{n}$. Note that
the other relation $R_{i}R_{j}=R_{j}R_{i}$ if $|i-j|>1$ follows from the
locality of the operators $R_{i},R_{j}$: they do not act on the same copies of
$V$ if $|i-j|>1$.

A local representation is thus a representation of $B_{n}$ on a tensor space
$V^{\otimes n}$ with the special (local) form explained above for the action
of the generators $\sigma_{i}$. It depends only on the element
$R\in\text{End}(V\otimes V)$ satisfying (2) (and invertible in order to have a
representation of a group).

######  Remark 1.2.

A direct, or numerical, approach to find local representation of the braid
group $B_{n}$ is quite difficult. Indeed if $\dim(V)=n$ then one has to solve
cubic equations in $n^{4}$ variables. ∎

######  Example 1.3.

Let $P\in\text{End}(V\otimes V)$ be the permutation operator sending $x\otimes
y$ to $y\otimes x$. Then it provides a local representation of the braid
group. In fact, it factors through the natural permutation representation of
the symmetric group on $V^{\otimes n}$. All this amounts to the simple facts
that $P^{2}=\text{Id}$ and that the following equality is true when one
composes transpositions: $(1,2)(2,3)(1,2)=(2,3)(1,2)(2,3)$.

###  1.2 The Yang–Baxter equation

Let $V$ be a finite-dimensional vector space. We introduce a bit more
notations for operators on $V^{\otimes n}$. If $S\in\text{End}(V\otimes V)$,
we use the notation $S_{ij}$ for the operator on $V^{\otimes n}$ acting as $S$
on copies $i$ and $j$ and trivially otherwise. More formally, write
$S=\sum_{a}s_{a}\otimes t_{a}$ , where $s_{a},t_{a}\in\text{End}(V)$. Then by
definition, we have:

$S_{ij}=\sum_{a}\text{Id}_{V}\otimes\dots\otimes\text{Id}_{V}\otimes
s_{a}\otimes\text{Id}_{V}\otimes\dots\otimes\text{Id}_{V}\otimes
t_{a}\otimes\text{Id}_{V}\otimes\dots\otimes\text{Id}_{V}\ ,$

where $s_{a}$ is in position $i$ and $t_{a}$ is in position $j$. We use this
notation in particular for the permutation operator $P$ of $V\otimes V$.
Explicitly, we have:

$P_{ij}(v_{1}\otimes\dots\otimes v_{i}\otimes\dots\otimes
v_{j}\otimes\dots\otimes v_{n})=v_{1}\otimes\dots\otimes
v_{j}\otimes\dots\otimes v_{i}\otimes\dots\otimes v_{n}\ .$

The basic property about all these notations that we use repeatedly is:

$S_{ab}P_{ij}=P_{ij}S_{\pi_{i,j}(a)\pi_{i,j}(b)}\,,\ \ \ \ \ \ \ \text{where
$\pi_{i,j}$ is the transposition $(i,j)$.}$

For example, $S_{13}P_{12}=P_{12}S_{23}$ or $S_{13}=P_{23}S_{12}P_{23}$, etc.

Now we consider a function

$\begin{array}[]{lcrcl}R&:&\ \mathbb{C}^{2}&\to&\text{End}(V\otimes
V)\\\\[5.0pt] &&(u,v)&\mapsto&R(u,v)\end{array}\ .$

The Yang–Baxter equation (YB equation for short) is a functional equation for
such a function $R$:

$R_{12}(u_{1},u_{2})R_{13}(u_{1},u_{3})R_{23}(u_{2},u_{3})=R_{23}(u_{2},u_{3})R_{13}(u_{1},u_{3})R_{12}(u_{1},u_{2})\
\quad\ \ \ \ \text{on $V\otimes V\otimes V$.}$ (3)

In this context, the variables $u_{1},u_{2},u_{3}$ are often called _spectral
parameters_ , or _spectral variables_.

In our perspective, the so-called braided version of this equation is more
relevant. From $R$ define another function:

$\begin{array}[]{lcrcl}\check{R}&:&\ \mathbb{C}^{2}&\to&\text{End}(V\otimes
V)\\\\[5.0pt] &&(u,v)&\mapsto&PR(u,v)\end{array}\ ,$

where $P$ is the permutation operator sending $x\otimes y$ to $y\otimes x$. To
see better the connections with the braid relation, set
$\check{R}_{1}(u,v):=\check{R}_{12}(u,v)$ and
$\check{R}_{2}(u,v):=\check{R}_{23}(u,v)$. Then an easy manipulation shows
that the YB equation is equivalent to the following equation only involving
the function $\check{R}$:

$\check{R}_{1}(u_{1},u_{2})\check{R}_{2}(u_{1},u_{3})\check{R}_{1}(u_{2},u_{3})=\check{R}_{2}(u_{2},u_{3})\check{R}_{1}(u_{1},u_{3})\check{R}_{2}(u_{1},u_{2})\
\quad\ \ \ \ \text{on $V\otimes V\otimes V$.}$ (4)

When precision is needed, we will refer to this version as the braided YB
equation.

It happens in many situations that the function $\check{R}$ (or equivalently
$R$) depends only on the ratio $u/v$. In this particular case (set
$u=u_{1}/u_{2}$ and $v=u_{2}/u_{3}$), the braided YB equation becomes:

$\check{R}_{1}(u)\check{R}_{2}(uv)\check{R}_{1}(v)=\check{R}_{2}(v)\check{R}_{1}(uv)\check{R}_{2}(u)\
\quad\ \ \ \ \text{on $V\otimes V\otimes V$.}$ (5)

Instead of a multiplicative version, we sometimes also have the additive one,
where the function $\check{R}$ depends only on the difference $u-v$:

$\check{R}_{1}(u)\check{R}_{2}(u+v)\check{R}_{1}(v)=\check{R}_{2}(v)\check{R}_{1}(u+v)\check{R}_{2}(u)\
\quad\ \ \ \ \text{on $V\otimes V\otimes V$.}$ (6)

######  Example 1.4 (constant solution, Yang solution).

$\bullet$ For a first encounter with the YB equation, one may ask for constant
solutions. The equation (4) for a constant function is simply the braid
relation. Thus, a constant solution of the braided YB equation is equivalent
to a local representation of the braid group, as defined previously.

$\bullet$ One can check by hand that the following function satisfies (3):

$R(u,v)=\text{Id}_{V\otimes V}+\frac{P}{u-v}\,,$

or equaivalently that the braided YB equation (4) is satisfied by:

$\check{R}(u,v)=PR(u,v)=P+\frac{\text{Id}_{V\otimes V}}{u-v}\,.$

Note that in this verification, one uses that $P$ satisfies the braid relation
(this is an example of _Baxterization_ , as we will see later). This solution
is an example depending on the spectral parameters only through their
difference. Namely we have that $\check{R}(u)=P+\frac{\text{Id}_{V\otimes
V}}{u}$ satisfies (6).

####  1.2.1 The Yang–Baxter equation in Physics

The Yang–Baxter equation is certainly one of the fundamental equations in
theoretical and mathematical physics. Its history is complex and naturally
most of it requires some background in physics that we are not ready to
discuss here. So let us only give some hints and vague indications which might
be enough to get a feeling of its importance and to feel a little motivated
for its study. Some references among the huge literature on this subject are
[3, 6, 23, 24, 29, 38, 39, 40, 49]. Though the YB equation belongs to the
theoretical and mathematical side of Physics, it is interesting to note that
connections with more experimental physics can be investigated today, see for
example [4, 5, 32].

##### Factorization of interaction processes.

Probably the most mathematicians-friendly interpretation of the YB equation in
physics is as a factorisation property. Let us depict the interaction of two
particles called $1$ and $2$ like this:

$\leavevmode\hbox to27.24pt{\vbox
to28.85pt{\pgfpicture\makeatletter\hbox{\hskip 5.08301pt\lower-17.47182pt\hbox
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}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{\scriptsize{$1$}}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
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}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{\scriptsize{$2$}}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
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}{}\pgfsys@moveto{17.07182pt}{0.0pt}\pgfsys@lineto{0.0pt}{-17.07182pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}{}\pgfsys@moveto{5.1216pt}{-5.1216pt}\pgfsys@lineto{11.95021pt}{-5.1216pt}\pgfsys@stroke\pgfsys@invoke{
} {{}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{}{{ {}{}}}{ {}{}}
{{}{{}}}{{}{}}{}{{}{}} { }{{{{}}\pgfsys@beginscope\pgfsys@invoke{
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}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
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}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
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\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}\ \ \ \ :\ \ \ \
\text{space-time trajectories for the interactions of two particles,}$ (7)

where we think of the vertical direction as time and of the horizontal one as
space. The parameter $u$ is the difference of rapidity of the two
particles222In relativity, the rapidity corresponding to the speed $v$ is
$\text{tanh}^{-1}(\frac{v}{c})$ and is additive for one-dimensional motion
($c$ is the speed of light).. If we think of the particles as being described
by a vector space $V$ (their internal state space), the interaction is then
controlled by an operator, say $S(u)$, in $\text{End}(V\otimes V)$.

Now consider $n$ particles on the line (numbered from $1$ to $n$) and they all
interact with each other. We imagine our $n$ particles coming closer and
closer to each other, interacting when they cross each other (controlled by
the scattering operator $S(u)$), and after having all interacted once with
each other, their order is completely reversed. As an example, if we consider
three particles, then we find two different possibilities for ordering these
interactions. The space-time trajectories are shown below:

$1$$2$$3$$u_{12}$$u_{13}$$u_{23}$$=$$1$$2$$3$$u_{23}$$u_{13}$$u_{12}$

where $u_{ij}=u_{i}-u_{j}$ are the differences of the rapidities. The
Yang–Baxter equation is simply the hypothesis that these two interacting
processes lead to the same result. Namely, in operator notation, we find:

$S_{12}(u_{12})S_{13}(u_{13})S_{23}(u_{23})=S_{23}(u_{23})S_{13}(u_{13})S_{12}(u_{12})\
,$

which is the Yang–Baxter equation for the function $S\ :\ (u,v)\mapsto
S(u-v)$.

The fundamental fact about this is the following. Once we assume the YB
equation for the interaction of three particles, then it follows that the
interaction of $n$ particles, decomposed as a sequence of $2$-particles
interactions, is independent of the chosen sequence. In other words, the $n$
particles interaction is unambiguously given by the two particles interaction.
The YB equation is a compatibility condition for the factorization of the
$n$-body interaction in terms of the two-body interaction. This originates in
the papers [55, 68, 69, 70] and accounts for the first half of the name of the
equation.

######  Remark 1.5.

The above interpretation of the YB equation is very similar to the following
classical fact about the braid group and the symmetric group. The longest
element of the symmetric group (reversing the order of $1,\dots,n$) can be
factorised in different ways as a minimal-length product of simple
transpositions, and all these different factorisations are seen to be equal
just by assuming the braid relation. The YB equation above plays the same role
as the braid equation here. ∎

##### 2-dimensional statistical physics.

The YB equation has a long history in 2-dimensional statistical models, and we
shall not attempt to describe it. Suffices it to say that it seems to have
made its first apparition (somewhat hidden) in the solution of the Ising model
by Onsager in the 40’s [61] and its importance was gradually recognized,
culminating in the work of Baxter in the 70’s [6]. It is instructing to search
for explicit apparitions of the YB equation in the classical book by Baxter
[6]. The first two are located near the end of the discussions on the Ising
model (§7.13) and on the ice model (§9.7). Remarkably, the next one (§10.4)
appears at the beginning of the discussion of the 8-vertex model. The meaning
is clear: the decisive step of promoting the YB equation as the key to
solvability of these models was done. At that time, the YB equation was called
the star-triangle relation, or the parametrized star-triangle relation to
emphasize the presence of the spectral parameter.

In these statistical models, there is a partition function which is the thing
one would like to calculate. The partition function can be expressed as the
trace of some powers of an object called the transfer matrix $t(u)$ depending
on a parameter. So of course, one natural approach is to try to diagonalize
this matrix. Then the key property that one wishes for this transfer matrix is
its commutation property:

$[t(u),t(v)]=0\ \ \ \ \ \forall u,v\,,$

since it provides an infinite number of commuting operators. The YB equation
(for which operator? we shall see below) was identified as a condition
ensuring the commutativity of the transfer matrices.

After this very brief summary, let us at least explain how an operator in
$\text{End}(V\otimes V)$ arises in this setting. A vertex model in
2-dimensional statistical physics starts with a lattice in the plane, say a
square lattice for definiteness. Each edge can be in a certain state $s$ taken
in a given finite set $S$. For example $S$ can be of cardinal $2$,
corresponding to the two orientations of a spin, or two states “empty” and
“occupied”. The interactions happen at the vertices (hence the name “vertex
models”) and at each vertex, there is a Boltzmann weight333In statistical
physics, the Boltzmann weight of a state is equal to $e^{-E/k_{B}T}$ where $E$
is the energy, $T$ is the temperature and $k_{B}$ is the Boltzmann constant.
The Boltzmann weight expresses the probability for a system to be in a state
with energy $E$. depending on the states of the edges connected to this
vertex. To define the partition function, take a configuration of the lattice
(fix a state for each edge) and take the product of the Boltzmann weights of
all vertices. Then sum over all the possible configurations of the lattice.

Thus to define the model, it suffices to give the Boltzmann weight of a vertex
for each configuration of the edges. If we organise these weights in an array
as follows:

$\scriptstyle{i}$$\scriptstyle{j}$$\scriptstyle{l}$$\scriptstyle{k}$$\leadsto$$\bigl{(}R_{ik}^{jl}\bigr{)}_{i,j,k,l\in
S}$

we realise that it is an array of numbers with $4$ indices in $S$. Thus this
can be seen as an element of $\text{End}(V\otimes V)$, where $V$ is a vector
space with basis indexed by $S$. The appearance of the spectral parameter $u$
in this context is not easily explained, it can be seen as a clever way of
parametrizing the Boltzmann weights (one can start from the occurrences of the
YB equation in [6] previously indicated, and read backwards to see where the
spectral parameter came from). Some symmetries can be required, restricting
the values of the various weights. For example if $\dim(V)=2$, for the
6-vertex model (respectively, 8-vertex model), only 6 entries (respectively, 8
entries) of the $R$-matrix are non-zero.

If it happens that this operator compiling the various Boltzmann weights
satisfies the YB equation, then the commuting property of the transfer matrix
is ensured.

##### Quantum spin chains.

A quantum spin chain is a quantum-mechanical model of interacting particles on
a line. The Hilbert space of states is the tensor product $V^{\otimes n}$ ($n$
particles, each having $V$ as its Hilbert space of states), and there is a
Hamiltonian describing the interactions. For example, $V$ can be
$\mathbb{C}^{2}$ and the Hamiltonian can be:

$\sum_{i=1}^{n}(J_{x}\sigma_{i}^{x}\sigma_{i+1}^{x}+J_{y}\sigma_{i}^{y}\sigma_{i+1}^{y}+J_{z}\sigma_{i}^{z}\sigma_{i+1}^{z})\,,$

where $\sigma^{x},\sigma^{y},\sigma^{z}$ are the famous Pauli matrices and the
indices $i,i+1$ indicate the positions in the tensor product $V^{\otimes n}$
($n+1$ is understood as $1$). This is called the XYZ model. It reduces to the
XXZ model if we take $J_{x}=J_{y}$ and further to the $XXX$ model (or
Heisenberg spin chain) if $J_{x}=J_{y}=J_{z}$.

In quantum mechanics, the goal is clear: we want the eigenvalues and
eigenvectors of the Hamiltonian. The connections with the seemingly unrelated
statistical models of the previous paragraph is as follows. The transfer
matrix of a vertex model is also an operator on some tensor product
$V^{\otimes n}$. And it was found that for the ice model, the eigenvectors of
the transfer matrix are the same as for the $XXZ$ Hamiltonian. In fact, it
turned out that the Hamiltonian commutes with the transfer matrix $t(u)$, thus
explaining this coincidence. Even better, it was finally realized that the
Hamiltonian is included in a sense in the transfer matrix (the Hamiltonian can
be recovered from $t(u)$), and this already for the $XYZ$ model (the
corresponding vertex model is the 8-vertex model). So in fact quantum spin
chains also fall in the realm of models governed by solutions of the YB
equation.

##### Algebraic Bethe Ansatz.

In the end of the 70s, a general method to construct and study integrable
systems was developed under the name of Algebraic Bethe Ansatz (see [23, 64]
and references therein). An essential ingredient taken as a starting point is
an $R$-matrix, namely a solution of the YB equation. Long story short, an $L$
matrix, a monodromy matrix and a transfer matrix are successively constructed
from it, and ultimately, the commutation relation for the transfer matrix
follows from the YB equation.

This approach put forward the algebraic relations between the various
operators involved (including of course the YB equation), and it naturally led
to wonder about abstract and general algebraic structures behind all this.
This was the birth of quantum groups, and the moment in time where the YB
equation started to diffuse into mathematics. Quantum groups are one of our
next subjects of discussions. It is remarkable that these structures (quantum
groups) originating from considerations in mathematical physics turn out to be
very important in modern representation theory, for example for the symmetric
group in positive characteristic. We will not talk about that.

##  2 The Guiding Examples: Hecke and Birman–Murakami–Wenzl

###  2.1 The Hecke algebra

There are several possible equivalent definitions for the Hecke algebra, in
quite different contexts. The most natural here is as a quotient of the
algebra of the braid group. The idea is as follows. The braid group is a
fairly complicated group algebraically since we can make an arbitrary number
of crossing between strands. There is no way to reduce this number of
crossings (except in the trivial situation when a positive crossing meets a
negative crossing on the same two strands, that is, when we can use the
relation $\sigma_{i}\sigma_{i}^{-1}=1$). So the naive idea is to add an
algebraic relation allowing to deal with all these crossings. One of the
simplest way to do this results in the Hecke algebra.

More precisely, we add to the definition of the braid group the following
relation:

$=$$-\,(q-q^{-1})$ (8)

where $q$ is a parameter, which can be seen as an indeterminate or as a non-
zero complex number. This relation is a local relation, meaning that for any
braid and any of its crossing, the braid is equal to a sum of two terms: the
braid obtained by replacing the crossing by its opposite and $\pm(q-q^{-1})$
(depending on the sign of the original crossing) times the braid obtained by
replacing the crossing by two pieces of vertical strands. In particular this
allows one to transform all the negative crossings into positive ones.

Algebraically, the new relation is equivalent to imposing
$\sigma_{i}^{-1}=\sigma_{i}-(q-q^{-1})$ for all generators $\sigma_{i}$. Of
course, if $q^{2}\neq 1$ we leave the realm of groups with such a relation,
and we end up with an algebra. So the algebraic definition of the Hecke
algebra $H_{n}(q)$ goes by defining it as the algebra generated by elements
$\sigma_{1},\dots,\sigma_{m-1}$ with defining relations:

$\begin{array}[]{ll}\sigma_{i}\sigma_{i+1}\sigma_{i}=\sigma_{i+1}\sigma_{i}\sigma_{i+1}\,,&\text{for
$i\in\\{1,\dots,n-2\\}$}\,,\\\\[1.99997pt]
\sigma_{i}\sigma_{j}=\sigma_{j}\sigma_{i}\,,&\text{for
$i,j\in\\{1,\dots,n-1\\}$ such that $|i-j|>1$}\,,\\\\[1.99997pt]
\sigma_{i}^{2}=1+(q-q^{-1})\sigma_{i}\,,&\text{for
$i\in\\{1,\dots,n-1\\}$}\,.\end{array}$ (9)

It is an algebra over $\mathbb{C}$ if $q$ is a complex number, and can be
defined also as an algebra over $\mathbb{C}[q,q^{-1}]$ if $q$ is an
indeterminate. Comparing with the algebraic presentation of the braid group
$B_{n}$ in Theorem 1.1, we see that the Hecke algebra $H_{n}(q)$ is a quotient
of the group algebra of the braid group $B_{n}$ (the algebra $\mathbb{C}B_{n}$
consisting of linear combinations of elements of $B_{n}$). The characteristic
equation of degree 2 for the generators is:

$(\sigma_{i}-q)(\sigma_{i}+q^{-1})=0\ .$

If $q^{2}=1$ the relations (9) are defining relations for the symmetric group
$S_{n}$, and the Hecke algebra $H_{n}(1)$ is thus the group algebra
$\mathbb{C}S_{n}$. In this case, the generators $\sigma_{i}$ correspond to the
transpositions $(i,i+1)$. It should feel natural since if $q^{2}=1$, the local
Hecke relation says that we can forget about the signs of the crossings, and
so all topological information is lost, and what remains of a braid is simply
the permutation of the $n$ points induced by the strands.

Thus the Hecke algebra is a deformation of the group algebra of the symmetric
group $S_{n}$. The symmetric group can be seen as a “simplification” of the
braid group, but it is in some sense “too simple”. The Hecke algebra retains
more information (in particular, topological) because of the additional
freedom given by the deformation parameter $q$.

######  Remark 2.1 (Jones polynomial).

A breakthrough in the theory of invariants of links was the discovery of the
Jones polynomial (and its generalisation, the so-called HOMFLY-PT polynomial)
[26, 41, 63]. In algebraic terms, this invariant is obtained from a Markov
trace on the chain of Hecke algebras (see for example [28, §4.5]). ∎

######  Remark 2.2.

There are several related algebraic structures called Hecke algebras:
$\bullet$ One purely algebraic definition is as deformations of Coxeter
groups. Our example corresponds to a finite Coxeter group of type $A$,
_a.k.a._ the symmetric group.
$\bullet$ There is also a very general definition of Hecke algebras
(explaining the name “Hecke”) as endomorphism algebras of induced
representations. We know of several coincidences of the two definitions. Let
us only indicate that if we take $G$ the group of invertible matrices over a
finite field with $q$ elements, and induce to $G$ the trivial representation
of the subgroup consisting of upper-diagonal matrices, then the endomorphism
algebra turns out to be the Hecke algebra $H_{n}(q)$.
$\bullet$ In our case of $H_{n}(q)$, there is another possible definition as
the centraliser of the action of a quantum group in a tensor product of
representations. We will come back to this later. ∎

##### The YB equation.

For any $i\in\\{1,\dots,n-1\\}$, define the following function taking values
in the algebra $H_{n}(q)$:

$\sigma_{i}(u)=\sigma_{i}+(q-q^{-1})\frac{1}{u-1}\ .$ (10)

Then a straightforward calculation using the relations in $H_{n}(q)$ shows
that:

$\sigma_{i}(u)\sigma_{i+1}(uv)\sigma_{i}(v)=\sigma_{i+1}(v)\sigma_{i}(uv)\sigma_{i+1}(u)\
,$

namely, the braided YB equation is satisfied in $H_{n}(q)$ by these functions.

So we have a solution of the braided YB equation inside the Hecke algebra
$H_{n}(q)$, given by a rather simple formula. However this is not a _genuine_
solution of the YB equation since it is not yet an operator on a tensor
product of vector spaces. The ingredient one has to add is a little bit of
representation theory, namely a representation of $H_{n}(q)$ on a tensor space
$V^{\otimes n}$. More precisely, we want a local representation of $H_{n}(q)$,
that is, a representation of the form:

$\sigma_{i}\mapsto\check{R}_{i}=\text{Id}_{V^{\otimes
i-1}}\otimes\check{R}\otimes\text{Id}_{V^{\otimes n-i-1}}\,,$

for some operator $\check{R}\in\text{End}(V\otimes V)$. Indeed, assume that we
have such a representation and set:

$\check{R}(u)=\check{R}+(q-q^{-1})\frac{\text{Id}_{V\otimes V}}{u-1}\ .$

Then it follows from (10) that $\check{R}(u)$ satisfies the braided YB
equation.

This is all very nice, only if we can find local representations of the Hecke
algebra $H_{n}(q)$. It turns out that there are some. In fact, take any vector
space $V$, then we can construct a local representation of $H_{n}(q)$ on
$V^{\otimes n}$. The rough idea is that for the symmetric group $S_{n}$, there
is one local representation which is simply by permuting the $n$ components in
the tensor product. As the Hecke algebra $H_{n}(q)$ is a deformation of
$S_{n}$, one may try to deform the permutation representation. It works.

More precisely, fix a basis $(e_{1},\dots,e_{N})$ of $V$ and define a linear
operator on $V\otimes V$ by:

$\check{R}(e_{a}\otimes e_{b}):=\left\\{\begin{array}[]{ll}q\,e_{a}\otimes
e_{b}&\text{if $a=b$,}\\\\[8.00003pt] e_{b}\otimes
e_{a}+(q-q^{-1})\,e_{a}\otimes e_{b}&\text{if $a<b$,}\\\\[3.99994pt]
e_{b}\otimes e_{a}&\text{if $a>b$.}\end{array}\right.\ \ \ \text{where
$a,b=1,\dots,N$.}$ (11)

One can check by hand that this provides a local representation of the braid
group:
$\check{R}_{i}\check{R}_{i+1}\check{R}_{i}=\check{R}_{i+1}\check{R}_{i}\check{R}_{i+1}$;
and moreover, that it factors through the Hecke algebra since the Hecke
relation $\check{R}^{2}=\text{Id}_{V\otimes V}+(q-q^{-1})\check{R}$ is
satisfied. So finally we get a solution of the braided YB equation on any
vector space $V$ by the above procedure.

######  Example 2.3.

For example, if $\dim(V)=2$, in the lexicographic ordering of the basis of
$V\otimes V$, we find:

$\check{R}=\left(\begin{array}[]{cccc}q&\cdot&\cdot&\cdot\\\
\cdot&q-q^{-1}&1&\cdot\\\ \cdot&1&0&\cdot\\\
\cdot&\cdot&\cdot&q\end{array}\right)\ ,\ \ \ \ \ \
\check{R}(u)=\left(\begin{array}[]{cccc}\displaystyle\frac{qu-q^{-1}}{u-1}&\cdot&\cdot&\cdot\\\
\cdot&\displaystyle\frac{(q-q^{-1})u}{u-1}&1&\cdot\\\
\cdot&1&\displaystyle\frac{q-q^{-1}}{u-1}&\cdot\\\
\cdot&\cdot&\cdot&\displaystyle\frac{qu-q^{-1}}{u-1}\end{array}\right)\ .$

The solution $\check{R}(u)$ coming from the Hecke algebra is a deformation of
the Yang solution $\check{R}(\alpha)=P+\frac{\text{Id}}{\alpha}$ from Example
1.4 in the following sense: set $u=q^{2\alpha}$ and take the limit $q\to 1$.

###  2.2 The Birman–Murakami–Wenzl algebra

The Hecke algebra is more or less the most general quotient of the algebra of
the braid group by a quadratic characteristic equation for the generators. So
following the idea of simplifying the braid group, it is natural to turn our
attention to quotients by a cubic characteristic equation.

##### On quotients of the braid group.

A naive idea for defining quotients of the braid group algebra
$\mathbb{C}B_{n}$ is to add the following relations:

$\sigma_{i}^{m}=a_{0}+a_{1}\sigma_{i}+\dots+a_{m-1}\sigma_{i}^{m-1}\ \ \
\text{for $i=1,\dots,n-1$.}$

In this context, we would like the generators $\sigma_{i}$ to stay invertible
so that, up to a renormalisation of the generators $\sigma_{i}$ which does not
change the braid relations, we can assume that $a_{0}=1$. Note that all
generators $\sigma_{i}$ are conjugated in the braid group so they must satisfy
the same characteristic equation (hence the same relation for all
$i=1,\dots,n-1$). If $m=2$, up to some conventions, the resulting algebra is
the Hecke algebra $H_{n}(q)$.

A first natural step is to look at the group we obtain if we consider the
relation $\sigma_{i}^{m}=1$. Then we could try to see the algebra resulting
from the relations above as deformations of these groups. This would certainly
be nice if the groups (respectively, the algebras) were finite (respectively,
finite-dimensional). This question has been settled by Coxeter [16] for the
groups resulting in the following remarkable result: if we make the quotient
of the braid group $B_{n}$ on $n$ strands by the relation $\sigma_{i}^{m}=1$
then the resulting group is finite if and only if
$\frac{1}{n}+\frac{1}{m}>\frac{1}{2}$. This gives the following possibilities:

$(n,m)\in\bigl{\\{}\ (2,m)\,,\ \ (n,2)\,,\ \ (3,3)\,,\ \ (3,4)\,,\ \ (3,5)\,\
\ (4,3)\,,\ \ (5,3)\ \bigr{\\}}\ .$

The first two families correspond to cyclic groups and symmetric groups. Aside
from these groups, there are 5 other situations giving a finite group.

The second step is to try to deform the above finite groups by considering an
arbitrary characteristic equation as above. It turns out that there is such a
deformation theory for these groups, since they all belong to the family of
finite complex reflection groups. The resulting algebra are called cyclotomic
Hecke algebras, and they are flat deformations in the sense that their
dimensions remains equal to the order of the corresponding group. We refer to
[53, 13].

##### An example of a cubic quotient: the BMW algebra.

As we have discussed just above, a generic cubic quotient of the braid group
algebra does not have to be finite-dimensional (as soon as $n>5$). So an idea
is to again add relations to make it finite-dimensional. Complete study of
such quotients turns out to be quite involved, and it seems to be still an
open question what would be the most generic finite-dimensional quotient of
the braid group algebra including a cubic characteristic equation for the
generators, see for example [54]. The Birman–Murakami–Wenzl (BMW) algebra [7,
58] is the most well-known particular case of such a quotient.

The topological definition of the BMW algebra uses the notion of tangles,
generalising braids. We keep the rectangular strip with the two lines of $n$
dots, and we still connect each dot to another dot by a strand inside the
strip. However now we are allowed to connect a top dot to another top dot (and
similarly for bottom dots). An example is:

We can multiply such objects by vertical concatenation as before. The BMW
algebra is generated by such objects, with an adequate notion of isotopy and
adding some local relations. Note that closed links (possibly intertwined with
the strands) can live inside the rectangular strip, due to the desired
stability by concatenation. One important local relation is the Kauffman skein
relation:

$-$$=-(q-q^{-1})$$\Bigl{(}$$-$$\Bigr{)}$ (12)

allowing in some sense to “resolve” the crossings, in a way similar to the
Hecke algebra situation, though more involved. We will not give the precise
topological definition of the BMW algebra (see for example [57]), and we shall
be happy with an algebraic description by generators and relations.

The BMW algebra $BMW_{n}(a,q)$ is the quotient of the braid group algebra
$\mathbb{C}B_{n}$ by the relations:

$\begin{array}[]{ll}e_{i}\sigma_{i}=ae_{i}&\text{for
$i=1,\dots,n-1$,}\\\\[3.99994pt] e_{i}\sigma_{i+1}^{\pm 1}e_{i}=a^{\mp
1}e_{i}&\text{for $i=1,\dots,n-2$,}\end{array}$

where we have set $e_{i}=1-\frac{\sigma_{i}-\sigma_{i}^{-1}}{q-q^{-1}}$. Here
$q$ and $a$ are two non-zero complex numbers and $q^{2}\neq 1$. In the
realisation as an algebra of tangles, the element $e_{i}$ is

$e_{i}=$$1$$\dots$$i-1$$i$$i+1$$i+2$$\dots$$n$$i\in\\{1,\dots,n-1\\}$,

as can be seen from the local Kauffman relation. The first relation, when
written only in terms of $\sigma_{i}$, is:

$(\sigma_{i}-q)(\sigma_{i}+q^{-1})(\sigma_{i}-a)=0\ ,$

so that we have a cubic characteristic equation for the generators. The other
relation involves three different strands. It is clear from the algebraic
presentation that if we set $e_{i}=0$ we recover the Hecke algebra $H_{n}(q)$.

######  Remark 2.4 (Kauffman polynomial).

The chain of BMW algebras supports a Markov trace, different from the one on
the chain of Hecke algebras, resulting in an invariant of links called the
Kauffman polynomial [47]. ∎

##### The YB equation.

For any $i\in\\{1,\dots,n-1\\}$, define the following function taking values
in the algebra $BMW_{n}(q,a)$:

$\sigma_{i}(u)=\sigma_{i}+(q-q^{-1})\frac{1}{u-1}+(q-q^{-1})\frac{1}{a^{-1}qu+1}e_{i}\
.$ (13)

It turns out that the braided YB equation is satisfied in the BMW algebra by
these functions [42]:

$\sigma_{i}(u)\sigma_{i+1}(uv)\sigma_{i}(v)=\sigma_{i+1}(v)\sigma_{i}(uv)\sigma_{i+1}(u)\
.$

Following the same steps as for the Hecke algebra, it remains to discuss
whether we can find some local representation of the BMW algebra, in order to
obtain genuine matrix solutions of the YB equation. As for the Hecke algebra,
it turns out to be possible. We do not give the details and refer to [59, 66]
for explicit formulas. In both cases (Hecke and BMW), the existence of a local
representation has a far-reaching significance in the context of Schur–Weyl
dualities. We will discuss this later in Section 4.

###  2.3 Baxterization

The two preceding subsections provide examples of Baxterization formulas
(inside an algebra). We will quickly review this notion, very natural in our
discussion. The terminology is due to V. Jones [43]. Some references for
Baxterization are [9, 14, 17, 19, 35, 50, 71].

Start with a local representation of the braid group given by an operator
$\check{R}\in\text{End}(V\otimes V)$. Baxterizing this solution is, roughly
speaking, to find a way to add the spectral parameters. More precisely, one
looks for an expression $\check{R}(u)$ depending on a parameter $u$, and built
out of the matrix $\check{R}$, such that the braided YB equation is satisfied:

$\check{R}_{1}(u)\check{R}_{2}(uv)\check{R}_{1}(u)=\check{R}_{2}(v)\check{R}_{1}(uv)\check{R}_{2}(u)\
\ \ \ \ \text{on $V\otimes V\otimes V$}.$

The expression for $\check{R}(u)$ is seen as a function of $u$ taking values
in $\text{End}(V\otimes V)$. Typically it takes values in the subalgebra of
$\text{End}(V\otimes V)$ generated by $\check{R}$. For example, if $\check{R}$
is diagonalisable, one can look at an expression of the form:

$\check{R}(u)=\sum_{i}f_{i}(u)P_{i}\ ,$

where $P_{i}$ are the projectors on eigenspaces of $\check{R}$, and one tries
to find expressions for the functions $f_{i}(u)$ such that the YB equation is
satisfied. usually, there is a way ro recover the operator $\check{R}$ by
taking some limiting values of the parameters $u$.

At the level of an algebra $A$, a Baxterization formula often refers to the
following construction. Assume for simplicity that $A$ is generated by
elements $\sigma_{1},\dots,\sigma_{n-1}$ satisfying the braid relations. Then
a Baxterization formula in the algebra $A$ is an explicit formula for a
function $\sigma_{i}(u)$ with values in the algebra $A$ (or more precisely, in
the subalgebra generated by $\sigma_{i}$), such that the braided YB equation
is satisfied inside $A$:

$\sigma_{i}(u)\sigma_{i+1}(uv)\sigma_{i}(v)=\sigma_{i+1}(v)\sigma_{i}(uv)\sigma_{i+1}(u)\
.$

The expression of $\sigma_{i}(u)$ in terms of $\sigma_{i}$ should be the same
for every $i$.

If we have a Baxterization for an algebra $A$, then we should look for local
representations of $A$, that is, representations of $A$ in
$\text{End}(V^{\otimes n})$ for some vector space $V$ given in the form:

$\sigma_{i}\mapsto\text{Id}^{\otimes
i-1}\otimes\check{R}\otimes\text{Id}^{\otimes n-i-1}\ ,$

where $\check{R}\in\text{End}(V\otimes V)$. Applying the representation, the
function $\sigma_{i}(u)$ is sent to a function with values in
$\text{End}(V^{\otimes n})$ satisfying the YB equation for matrices.

We have already seen three Baxterizations formulas, one for the Hecke algebra
(10), one for the BMW algebra (13) and one for the symmetric group in Example
1.4. The Baxterization formula for the symmetric group (more accurately, its
group algebra) is:

$\check{R}(u)=\sigma_{i}+\frac{1}{u}\,,\ \ \ \ \ \text{(where
$\sigma_{i}=(i,i+1)$)}.$

##  3 Quantum Groups

We seek for a common ground to interpret both the Hecke algebra and the BMW
algebra, and if possible explaining that both of them admit a Baxterization
formula. To do so, we make a little detour through quantum groups.

Quantum groups were introduced in the mid 80s by Drinfeld and Jimbo [20, 36].
Some constructions in special cases were already found in [51, 65]. Some
general references on quantum groups are [11, 21, 44, 45, 48].

###  3.1 Some properties of quantum groups

Let $\mathfrak{g}$ be a complex simple Lie algebra (for example,
$\mathfrak{g}=sl_{N}$). To keep these notes reasonable in size, we will not
give the definitions of quasi-triangular Hopf algebras of which the quantum
groups are famous examples (we refer the interested reader to [11, §4], [44,
§3&8], [45, §2] and [48, §1]). We will only pick up from this whole theory the
properties especially relevant for us and briefly discuss them. This section
and the following are only meant to roughly indicate, not going into details,
that there exists interesting algebraic objects (quantum groups) which provide
us with representations of the braid group and even more, with solutions of
the Yang–Baxter equation.

To get a first feeling of these objects, let us mention that the quantum group
that we denote $U_{q}(\mathfrak{g})$ is an associative algebra444Regarding the
terminology “quantum group”, it may be satisfying enough to note that the
quantum groups $U_{q}(\mathfrak{g})$ are deformations of the classical objects
$U(\mathfrak{g})$ associated to Lie groups and moreover, during the
deformation, something which was commutative (the coproduct) becomes non-
commutative. More convincing explanations can be found in [21]., which can be
defined explicitly by generators and relations. As a vector space, it looks
like the universal enveloping algebra $U(\mathfrak{g})$ of the Lie algebra
$\mathfrak{g}$. However, the way of multiplying elements is different. It is a
deformation of the multiplication in $U(\mathfrak{g})$, in the sense that
there is a certain way to send the parameter $q$ to $1$ which recovers the
algebra $U(\mathfrak{g})$. The example of $\mathfrak{g}=sl_{2}$ will be
treated after the general discussion.

To summarize, the main properties that we are going to discuss here are555The
existence of a trivial representation and of contragredient (or dual)
representations are important properties of quantum groups that we omit in our
discussion.:

  * •

the fact that we can make tensor products of representations of
$U_{q}(\mathfrak{g})$;

  * •

the fact that $V\otimes W$ and $W\otimes V$ are isomorphic as representations
of $U_{q}(\mathfrak{g})$, and that the isomorphisms provide local
representations of the braid group;

  * •

and finally, the fact that we can upgrade this picture, using affine quantum
groups, to get solutions of the Yang–Baxter equation.

######  Remark 3.1.

From general deformation theory, it is known that there is no non-trivial
deformation of the algebra $U(\mathfrak{g})$ for a simple Lie algebra
$\mathfrak{g}$ (see [44]). It means that in a sense $U_{q}(\mathfrak{g})$ and
$U(\mathfrak{g})$ are isomorphic as algebras (even though the explicit
isomorphism is not obvious at all). At first, this remark can be somewhat
disturbing for we realise that the new multiplication of $U_{q}(\mathfrak{g})$
is more or less equivalent to the usual one of $U(\mathfrak{g})$. However, one
has to keep in mind that $U_{q}(\mathfrak{g})$ and $U(\mathfrak{g})$ are not
isomorphic as Hopf algebras, meaning that the coproduct (see below) of
$U_{q}(\mathfrak{g})$ is really different from the usual one of
$U(\mathfrak{g})$. Roughly speaking, we could say that going from
$U(\mathfrak{g})$ to $U_{q}(\mathfrak{g})$ does not change much the algebra
structure, and in particular the representation theory, but it changes non-
trivially the way to perform tensor products of representations. ∎

Let $V_{1}$ and $V_{2}$ be two representations of $U_{q}(\mathfrak{g})$. The
vector space $V_{1}\otimes V_{2}$ can be made a representation of
$U_{q}(\mathfrak{g})$. This is made possible by the existence of a so-called
coproduct $\Delta$, which is a morphism of algebras:

$\Delta\ :\ \ U_{q}(\mathfrak{g})\ \longrightarrow\ U_{q}(\mathfrak{g})\otimes
U_{q}(\mathfrak{g})\ .$

Note that the vector space $V_{1}\otimes V_{2}$ naturally carries a
representation of $U_{q}(\mathfrak{g})\otimes U_{q}(\mathfrak{g})$. Then the
way to construct a representation of $U_{q}(\mathfrak{g})$ on $V_{1}\otimes
V_{2}$ is simply by precomposing with the coproduct.

More explicitly, denote $\rho_{1}$ the morphism
$U_{q}(\mathfrak{g})\to\text{End}(V_{1})$ and similarly for $\rho_{2}$. Then,
explicitly, the representation of $U_{q}(\mathfrak{g})$ on $V_{1}\otimes
V_{2}$ is defined as:

$\begin{array}[]{ccccc}U_{q}(\mathfrak{g})&\stackrel{{\scriptstyle\Delta}}{{\longrightarrow}}&U_{q}(\mathfrak{g})\otimes
U_{q}(\mathfrak{g})&\stackrel{{\scriptstyle\rho_{1}\otimes\rho_{2}}}{{\longrightarrow}}&\text{End}(V_{1}\otimes
V_{2})\\\\[5.0pt] a&\mapsto&\Delta(a)=\sum a^{\prime}\otimes
a^{\prime\prime}&\mapsto&\sum\rho_{1}(a^{\prime})\otimes\rho_{2}(a^{\prime\prime})\end{array}$

Here and in some places below, we use the useful Sweedler notation
$\Delta(a)=\sum a^{\prime}\otimes a^{\prime\prime}$, where the summation index
is omitted. In rigorous mathematical notations, it would have been
$\Delta(a)=\sum_{i=1}^{N}a^{\prime}_{i}\otimes a^{\prime\prime}_{i}$ for some
$N>0$ and some $a^{\prime}_{i},a^{\prime\prime}_{i}\in U_{q}(\mathfrak{g})$.

######  Example 3.2.

For a group algebra $\mathbb{C}G$, the map $\Delta$ from $\mathbb{C}G$ to
$\mathbb{C}G\otimes\mathbb{C}G$ defined by $\Delta(g)=g\otimes g$ for any
$g\in G$ extends to a morphism of algebras and leads to the standard way of
performing tensor products of representations of a group.

For a Lie algebra $\mathfrak{g}$, the map $\Delta$ from $U(\mathfrak{g})$ to
$U(\mathfrak{g})\otimes U(\mathfrak{g})$ defined by $\Delta(x)=x\otimes
1+1\otimes x$ for any $x\in\mathfrak{g}$ extends to a morphism of algebras and
leads to the standard way of performing tensor products of representations of
a Lie algebra.

Not any morphism of algebras $U_{q}(\mathfrak{g})\to
U_{q}(\mathfrak{g})\otimes U_{q}(\mathfrak{g})$ will work as a nice coproduct.
One property which is usually required is the coassociativity property. In
formulas, it reads

$(\Delta\otimes\text{Id})\circ\Delta=(\text{Id}\otimes\Delta)\circ\Delta\ .$

This is an equality for applications from $U_{q}(\mathfrak{g})$ to
$U_{q}(\mathfrak{g})^{\otimes 3}$. Its meaning is the following. At this
point, there are two ways of performing the tensor product of three
representations: $(V_{1}\otimes V_{2})\otimes V_{3}$ and
$V_{1}\otimes(V_{2}\otimes V_{3})$ and the resulting representations on
$V_{1}\otimes V_{2}\otimes V_{3}$ do not have to be isomorphic. The
coassociativity property ensures that the two representations in fact simply
coincide. Thanks to the coassociativity, we do not need to worry about putting
parentheses in a tensor product, since every possible ways of performing it
lead to the same result.

It might be worth emphasizing that, in general, for an arbitrary associative
algebra, there is no natural coproduct and no natural way of performing tensor
product of representations (for example, for the Hecke algebra $H_{n}(q)$). So
the existence of a coproduct is a first remarkable property of quantum groups.

Now, for our discussion, one crucial property of tensor products of
representations of $U_{q}(\mathfrak{g})$ is the following: for any two
representations $V,W$ of $U_{q}(\mathfrak{g})$, the representations $V\otimes
W$ and $W\otimes V$ are isomorphic, that is, we have an invertible linear
operator $\check{R}_{V,W}$ between $V\otimes W$ and $W\otimes V$ which
commutes with the action of $U_{q}(\mathfrak{g})$:

$\check{R}_{V,W}\ :\ \ V\otimes W\ \stackrel{{\scriptstyle\sim}}{{\to}}\
W\otimes V\ ,\ \ \ \ \ \ \ \text{isomorphism of
$U_{q}(\mathfrak{g})$-representations.}$

######  Remark 3.3.

The permutation operator $\text{P}_{V,W}$ sending $v\otimes w$ to $w\otimes
v$, is the natural isomorphism of vector spaces between $V\otimes W$ and
$W\otimes V$. For $U(\mathfrak{g})$, the permutation provides the isomorphism
of representations. This is not so for $U_{q}(\mathfrak{g})$ (that is,
$\check{R}_{V,W}\neq\text{P}_{V,W}$) and this is one of the important points
about quantum groups. ∎

The existence of the isomorphisms $\check{R}_{V,W}$ is not enough for our
purpose, we need moreover the fact that these isomorphisms satisfy a
compatibility condition when considering a tensor product of three
representations:

$V_{1}\otimes V_{2}\otimes V_{3}$$V_{2}\otimes V_{1}\otimes
V_{3}$$V_{1}\otimes V_{3}\otimes V_{2}$$V_{2}\otimes V_{3}\otimes
V_{1}$$V_{3}\otimes V_{1}\otimes V_{2}$$V_{3}\otimes V_{2}\otimes V_{1}$

This diagram shows the two possible paths from $V_{1}\otimes V_{2}\otimes
V_{3}$ to $V_{3}\otimes V_{2}\otimes V_{1}$ by applying the isomorphisms
$\check{R}_{V_{i},V_{j}}$. The compatibility condition is that these two paths
coincide. This condition writes as an equality of linear operator from
$V_{1}\otimes V_{2}\otimes V_{3}$ to $V_{3}\otimes V_{2}\otimes V_{1}$:

$\bigl{(}\check{R}_{V_{1},V_{2}}\otimes\text{Id}_{V_{3}}\bigr{)}\circ\bigl{(}\text{Id}_{V_{2}}\otimes\check{R}_{V_{1},V_{3}}\bigr{)}\circ\bigl{(}\check{R}_{V_{2},V_{3}}\otimes\text{Id}_{V_{1}}\bigr{)}=\bigl{(}\text{Id}_{V_{1}}\otimes\check{R}_{V_{2},V_{3}}\bigr{)}\circ\bigl{(}\check{R}_{V_{1},V_{3}}\otimes\text{Id}_{V_{2}}\bigr{)}\circ\bigl{(}\text{Id}_{V_{3}}\otimes\check{R}_{V_{1},V_{2}}\bigr{)}\
.$ (14)

As we will see just below, this is at this step of the story that the braid
group makes its appearance.

###  3.2 Local representations of the braid group

If we consider the situation where $V_{1}=V_{2}=V_{3}=V$ and we denote simply
by $\check{R}$ the isomorphism $\check{R}_{V,V}$ which is an invertible
element in $\text{End}(V\otimes V)$, then the compatibility condition above
becomes simply the braid relation:

$\check{R}_{1}\check{R}_{2}\check{R}_{1}=\check{R}_{2}\check{R}_{1}\check{R}_{2}\
\quad\ \ \ \ \text{on $V\otimes V\otimes V$.}$

The operator $\check{R}$ on $V\otimes V$ is called the $R$-matrix associated
to the representation $V$ of the quantum group $U_{q}(\mathfrak{g})$.

######  Conclusion 1.

For any quantum group $U_{q}(\mathfrak{g})$ and any representation $V$ of
$U_{q}(\mathfrak{g})$, we have a local representation of the braid group on
$V^{\otimes n}$.

Explicitly, that is to say that we have elements
$\check{R}_{1},\dots,\check{R}_{n-1}$ in $\text{End}(V^{\otimes n})$, which
are constructed from the element $\check{R}\in\text{End}(V\otimes V)$ like
this:

$\hbox to0.0pt{$\overbrace{\phantom{V\otimes
V}}^{\check{R}_{1}}$\hss}V\otimes\overbrace{V\otimes
V}^{\check{R}_{2}}\otimes\dots\overbrace{\ldots\otimes V}^{\check{R}_{n-1}}$

and which satisfy the braid relations.

##### Universal $R$-matrix.

For an algebra $A$ admitting a coproduct $\Delta$, we denote $\Delta^{op}$ the
composition of $\Delta$ with the permutation of $A\otimes A$ ($a\otimes
b\mapsto b\otimes a$). We say that the coproduct is cocommutative when the
images $\Delta(a)$ are invariant under the transposition of the two
components, that is, when $\Delta=\Delta^{op}$. Note that the two classical
examples of coproducts for groups and for Lie algebras (Example 3.2) are
cocommutative. For an algebra with a cocommutative coproduct, it is easy to
see that the permutation operator from $V\otimes W$ to $W\otimes V$ provides
an isomorphism of representations. So in this case, the obtained local
representation of the braid group is always the trivial one given by the
permutation operator.

One of the main interest of quantum groups is that their coproduct is not
cocommutative, and thus the obtained local representations of the braid group
are not trivial. So regarding the construction of local representations of the
braid group, the quantum groups $U_{q}(\mathfrak{g})$ are a considerable
improvement compared to their classical analogue $U(\mathfrak{g})$.

For a non-cocommutative coproduct on an algebra $A$, there is a natural
condition weakening the cocommutativity that ensures the isomorphisms of
representations $V\otimes W$ and $W\otimes V$. This is the existence of an
invertible element $\mathcal{R}$ in $A\otimes A$ such that:

$\mathcal{R}\cdot\Delta(a)=\Delta^{op}(a)\cdot\mathcal{R}\ \ \ \ \forall a\in
A\ .$ (15)

It follows immediately from this that the following element provides an
isomorphism $\check{R}_{V,W}$ of representation between $V\otimes W$ and
$W\otimes V$:

$\check{R}_{V,W}=P_{V,W}\cdot\rho_{V\otimes W}(\mathcal{R})\,,$

where $\rho_{V\otimes W}(\mathcal{R})$ is the image of the element
$\mathcal{R}$ of $A\otimes A$ in the representation on $V\otimes W$; it
belongs to $\text{End}(V\otimes W)$. And $P_{V,W}$ is the permutation operator
from $V\otimes W$ to $W\otimes V$.

At this point, there is no reason for the compatibility condition (14) to be
satisfied. A way to get it is obtained through the so-called
quasitriangularity conditions, which read as:

$(\Delta\otimes\text{Id}_{A})(\mathcal{R})=\mathcal{R}_{13}\mathcal{R}_{23}\ \
\ \text{and}\ \ \
(\text{Id}_{A}\otimes\Delta)(\mathcal{R})=\mathcal{R}_{13}\mathcal{R}_{12}\ .$
(16)

These two conditions, which are equations in $A\otimes A\otimes A$, imply in
particular the constant YB equation
$\mathcal{R}_{12}\mathcal{R}_{13}\mathcal{R}_{23}=\mathcal{R}_{23}\mathcal{R}_{13}\mathcal{R}_{12}$
directly in $A\otimes A\otimes A$. From this, the compatibility condition for
all representations as in (14) is ensured, and in particular the local
representations of the braid group as in Conclusion 1 are obtained.

Such an element $\mathcal{R}$ satisfying (15) and (16) is called a universal
$R$-matrix. It turns out that the quantum group $U_{q}(\mathfrak{g})$ admits a
universal $R$-matrix satisfying all the required properties. We note that an
appropriate definition of $U_{q}(\mathfrak{g})$ and
$U_{q}(\mathfrak{g})\otimes U_{q}(\mathfrak{g})$ is needed, involving formal
power series and suitable completions (see the example of $sl_{2}$ below).

###  3.3 Solutions of the YB equation

The quantum groups were originally designed for being useful in the study of
the YB equation, but so far we have only discussed how they are related to
local representations of the braid group. In short we must explain how to add
the spectral parameters. This requires increasing by one (rather big) step the
technical difficulties by going to the so-called “affine” Lie algebras. We are
thus going to be even more sketchy than before. We note that the use of affine
quantum groups for the YB equation was present since the origin of the theory
of quantum groups [37]. References for this section are [10, 11, 22, 25, 29].
The YB equation is now connected with many more mathematical structures than
only quantum groups, see for example [33] and references therein.

The property we would like to discuss is the existence of solutions of the YB
equation on representations of $U_{q}(\mathfrak{g})$. We will only try to
convey the idea that this remarkable property comes from the existence of
affine quantum groups with good properties. We organise the discussion in a
series of steps towards the main conclusion.

$\bullet$ The affine Lie algebra $\hat{\mathfrak{g}}$ is a central extension
of the Lie algebra $\mathfrak{g}\otimes\mathbb{C}[t,t^{-1}]$ of Laurent
polynomials with coefficients in $\mathfrak{g}$ (we will forget the central
extension in our discussion). If $V$ is a representation of $\mathfrak{g}$
then for any non-zero $a\in\mathbb{C}$, there is a representation $V(a)$ of
$\hat{\mathfrak{g}}$ called evaluation representation. As a representation of
$\mathfrak{g}$, $V(a)$ is simply $V$, while the generator $t$ is evaluated to
the number $a$. The new parameter $a$ in the fixed representation $V$ of
$\mathfrak{g}$ will play the role of the spectral parameter in the YB
equation. Only problem is we need to do the same thing for representations of
$U_{q}(\mathfrak{g})$. This is where a quantum version of $\hat{\mathfrak{g}}$
is required.

$\bullet$ The affine Lie algebra $\hat{\mathfrak{g}}$ is not a simple Lie
algebra. However, it belongs to the family of Kac–Moody Lie algebra. In a few
words, if we cast the algebraic definition of $\mathfrak{g}$ such that it
depends only on its Cartan matrix, then we realise that we can use a similar
definition for generalised Cartan matrices. This results in the Kac–Moody Lie
algebras, including the affine Lie algebras. It turns out that the definition
of the quantum group $U_{q}(\mathfrak{g})$ that we have avoided to give also
only depends on the Cartan matrix. And thus it is perfectly generalisable to
the generalised Cartan matrices leading to quantum groups associated to
Kac–Moody Lie algebras. The point is that there is a definition of affine
quantum groups $U_{q}(\hat{\mathfrak{g}})$ similar to the definition of
quantum groups $U_{q}(\mathfrak{g})$ (the resulting algebras are quite more
complicated though).

As in the preceding discussion for $U_{q}(\mathfrak{g})$, there is an
analogous story for tensor products of representations of
$U_{q}(\hat{\mathfrak{g}})$. The algebra $U_{q}(\hat{\mathfrak{g}})$ still
admits a coproduct, so we can make tensor products of representations, and
there is in some completion a universal $R$-matrix allowing, roughly speaking,
to reproduce the discussion before Formula (14).

$\bullet$ Generalising the natural embedding of $\mathfrak{g}$ in
$\hat{\mathfrak{g}}$, one may regard $U_{q}(\mathfrak{g})$ as a subalgebra of
$U_{q}(\hat{\mathfrak{g}})$. This allows to restrict representations of
$U_{q}(\hat{\mathfrak{g}})$ to representations of $U_{q}(\mathfrak{g})$. Let
us assume for now that for a representation $V$ of $U_{q}(\mathfrak{g})$,
there exists an analogue of evaluation representations. Namely that we have
representations $V(a)$ of $U_{q}(\hat{\mathfrak{g}})$, depending on a
parameter $a$, such that the restriction to $U_{q}(\mathfrak{g})$ is the
representation $V$ we started with.

For two “evaluation” representations $V(a)$ and $V(b)$ of
$U_{q}(\hat{\mathfrak{g}})$, we can find for generic values of $a$ and $b$ an
operator intertwining the representations $V(a)\otimes V(b)$ and $V(b)\otimes
V(a)$. This operator, denoted $\check{R}(a,b)$, thus lives in
$\text{End}(V\otimes V)$ since the vector space underlying both
representations $V(a)$ and $V(b)$ is $V$. In fact, it turns out that it can be
seen as a rational function in the parameters $a,b$:

$\begin{array}[]{lcrcl}\check{R}&:&\ \mathbb{C}^{2}&\to&\text{End}(V\otimes
V)\\\\[5.0pt] &&(a,b)&\mapsto&\check{R}(a,b)\end{array}\ ,$

The compatibility condition applied to $V(a)\otimes V(b)\otimes V(c)$ looks
like this:

$V(a)\otimes V(b)\otimes V(c)$$V(b)\otimes V(a)\otimes V(c)$$V(a)\otimes
V(c)\otimes V(b)$$V(b)\otimes V(c)\otimes V(a)$$V(c)\otimes V(a)\otimes
V(b)$$V(c)\otimes V(b)\otimes V(a)$

For the function $\check{R}(a,b)$, this reads:

$\check{R}_{12}(b,c)\check{R}_{23}(a,c)\check{R}_{12}(a,b)=\check{R}_{23}(a,b)\check{R}_{12}(a,c)\check{R}_{23}(b,c)\
\quad\ \ \ \ \text{on $V\otimes V\otimes V$.}$

This is the braided YB equation. Note that in the end the spectral parameters
came from a family of representations of $U_{q}(\hat{\mathfrak{g}})$ extending
the given representation of $U_{q}(\mathfrak{g})$. So it would not be too far
from truth to say that the YB equation is the braid relation applied to some
representations of affine quantum groups.

At last we can draw our second conclusion regarding the uses of quantum groups
in our context.

######  Conclusion 2.

For a quantum group $U_{q}(\mathfrak{g})$ and some of its representations $V$,
we have a solution of the YB equation on $V\otimes V$.

Let us admit for the sake of simplicity that the function $\check{R}(a,b)$
depends only on the ratio of the two parameters, and therefore is equivalent
to a function of one parameter $\check{R}(u)$. The conclusion says that we
have functions $\check{R}_{1}(u),\dots,\check{R}_{n-1}(u)$ taking values in
$\text{End}(V^{\otimes n})$, which are constructed from the function
$\check{R}(u)$ like this:

$\hbox to0.0pt{$\overbrace{\phantom{V\otimes
V}}^{\check{R}_{1}(u)}$\hss}V\otimes\overbrace{V\otimes
V}^{...}\otimes\dots\overbrace{\ldots\otimes V}^{\check{R}_{n-1}(u)}$

and which satisfy the braided YB equation:

$\check{R}_{i}(u)\check{R}_{i+1}(uv)\check{R}_{i}(v)=\check{R}_{i+1}(v)\check{R}_{i}(uv)\check{R}_{i+1}(u)\
.$

##### On the validity of Conclusion 2.

In the above statement we had to reduce the generality by writing “some of its
representations”. This is because we have assumed that for a representation
$V$ of $U_{q}(\mathfrak{g})$, we had a family of “evaluation” representations
$V(a)$ of $U_{q}(\hat{\mathfrak{g}})$. This assumption is not always satisfied
and we will discuss briefly its validity.

First, for the simplest situation $\mathfrak{g}=sl_{N}$, this assumption is
valid for any representation $V$ of $U_{q}(\mathfrak{g})$. Indeed, for
$\mathfrak{g}=sl_{N}$, there is a morphism from $U_{q}(\hat{\mathfrak{g}})$ to
$U_{q}(\mathfrak{g})$ depending on a parameter $a$, which is the analogue of
the evaluation morphism from $\hat{\mathfrak{g}}$ to $\mathfrak{g}$. This
allows immediately to construct the “evaluation” representations $V(a)$ of
$U_{q}(\hat{\mathfrak{g}})$ from a representation $V$ of
$U_{q}(\mathfrak{g})$.

For other simple Lie algebras $\mathfrak{g}$, such an analogue of evaluation
morphism does not exist. However, what is true in general is that there is a
family of automorphisms of $U_{q}(\hat{\mathfrak{g}})$ depending on a
parameter $a$. So assume that a representation $V$ of $U_{q}(\mathfrak{g})$
can be extended to a representation of $U_{q}(\hat{\mathfrak{g}})$. Then by
twisting this representation by the automorphisms, one get a family of
representations $V(a)$ of $U_{q}(\hat{\mathfrak{g}})$ extending $V$ (they do
not come anymore from a morphism $U_{q}(\hat{\mathfrak{g}})\to
U_{q}(\mathfrak{g})$ but they still serve for finding solutions of YB equation
on $V$).

This leaves us with the following conclusion: we have a solution of the YB
equation on representations $V$ of $U_{q}(\mathfrak{g})$ which can be extended
to $U_{q}(\hat{\mathfrak{g}})$. To insist once more, for
$\mathfrak{g}=sl_{N}$, all representations can be extended but for arbitrary
$\mathfrak{g}$ this is not true. This leads to the natural concept of “minimal
affinization” of a representation [12]. Note that reversing the point of view,
one could also say: we have a solution of the YB equation on representations
$V$ of $U_{q}(\mathfrak{g})$ which can be obtained from restrictions of
representations of $U_{q}(\hat{\mathfrak{g}})$.

######  Remark 3.4.

Example 2.3 is associated to $U_{q}(sl_{2})$ and the fundamental
representation. However, before that, we had the example 1.4, the Yang
solution, which is a certain limit of the one from $U_{q}(sl_{2})$. In fact,
this example also comes from an algebra, sharing many properties with
$U_{q}(\hat{\mathfrak{g}})$, which is called the Yangian $Y(\mathfrak{g})$
(see [11]). This algebra, with no parameter $q$, can be seen a certain limit
of $U_{q}(\hat{\mathfrak{g}})$ and also provides solutions of the YB equation
along the same lines as outlined above. ∎

###  3.4 Example of $U_{q}(sl_{2})$

This is about time for an explicit example, so we will take the simplest
situation $\mathfrak{g}=sl_{2}$, see [44]. As a vector space, $sl_{2}$ is
generated by:

$h=\left(\begin{array}[]{cc}1&0\\\ 0&-1\end{array}\right)\,,\ \ \
x=\left(\begin{array}[]{cc}0&1\\\ 0&0\end{array}\right)\,,\ \ \
y=\left(\begin{array}[]{cc}0&0\\\ 1&0\end{array}\right)\,.$

Its Lie algebra structure is given by:

$[h,x]=2x\,,\ \ [h,y]=-2y\,,\ \ [x,y]=h\ .$ (17)

Thus, as an algebra, $U(sl_{2})$ is generated by elements $h,x,y$ with
relations (17) as defining relations. Note that we abuse notations by keeping
the same names $x,y,h$ for the generators of $U(sl_{2})$. In $U(sl_{2})$, the
bracket $[a,b]$ simply means $ab-ba$.

##### The algebra $U_{q}(sl_{2})$.

The quantum group $U_{q}(sl_{2})$ is often defined as the algebra generated by
$X,Y,K$ with $K$ required to be invertible, and with defining relations:

$KXK^{-1}=q^{2}X\,,\ \ \ KYK^{-1}=q^{-2}Y\,,\ \ \ XY-
YX=\frac{K-K^{-1}}{q-q^{-1}}\ .$ (18)

In this version, it is not so clear how to relate it to $U(sl_{2})$ (at this
point, the limit $q\to 1$ does not really make sense) and it would also
require some completions to be able to find a universal $R$-matrix.

The alternative standard definition of $U_{q}(sl_{2})$, the one which we are
going to consider from now is the following one. As a vector space, we define
$U_{q}(sl_{2})$ to be

$U(sl_{2})[[\alpha]]=\\{c_{0}+c_{1}\alpha+\dots\,,\ c_{i}\in U(sl_{2})\\}\,,$

the vector space of formal power series in $\alpha$ with coefficients in
$U(sl_{2})$. We abuse again notation and keep the names $x,y,h$ for the
generators of $U(sl_{2})$ but we insist that we mean here only the vector
space $U(sl_{2})$; the multiplication will be different. The multiplication of
elements of $U_{q}(sl_{2})$ is the usual multiplication of formal series,
together with the following (new) defining relations between the elements
$x,y,h$:

$hx-xh=2x\,,\ \ hy-yh=-2y\,,\ \ \ xy-yx=\frac{e^{\alpha h}-e^{-\alpha
h}}{e^{\alpha}-e^{-\alpha}}\ .$ (19)

One has to notice that the right hand side of the last relation is indeed a
power series in $\alpha$. We should insist that as a vector space,
$U(sl_{2})[[\alpha]]$ is not the same as $U(sl_{2})$ with coefficients in
$\mathbb{C}[[\alpha]]$. This is because $U(sl_{2})$ is not finite dimensional.
For example, $e^{\alpha h}$ is in $U(sl_{2})[[\alpha]]$.

One can see the connections with the previous definition of $U_{q}(sl_{2})$ as
follows: if one sets $q=e^{\alpha}$ and $K=e^{\alpha h}$ then Relations (18)
are satisfied. One advantage of the definition of $U_{q}(sl_{2})$ involving
power series in $\alpha$ is that the connection with $U(sl_{2})$ is quite
transparent. The limit $\alpha=0$ is well-defined and one recovers the algebra
$U(sl_{2})$ as can be seen immediately in the defining relations.

##### Coproduct and $R$-matrix for $U_{q}(sl_{2})$.

We start by defining the tensor product
$U_{q}(sl_{2})\hat{\otimes}U_{q}(sl_{2})$ as follows (and similarly for
$U_{\alpha}(sl_{2})^{\hat{\otimes}L}$ for any $L$). As a vector space, this
is:

$\bigl{(}U(sl_{2})\otimes U(sl_{2})\bigr{)}[[\alpha]]\ .$

The power series are multiplied as usual, and the elements $x,y,h$ still
satisfy the relations (19) in each factor of the tensor product; moreover
elements in different factors commute.

Then, the coproduct is defined on the generators by:

$\Delta(x)=x\otimes e^{-\alpha h/2}+e^{\alpha h/2}\otimes x\,,\ \ \
\Delta(y)=y\otimes e^{-\alpha h/2}+e^{\alpha h/2}\otimes y\,,\ \ \
\Delta(h)=h\otimes 1+1\otimes h\,,$ (20)

and extended to $U_{q}(sl_{2})$ by $\Delta(\sum
c_{i}\alpha^{i})=\sum\Delta(c_{i})\alpha^{i}$. One can check that $\Delta$
extends to an algebra homomorphism from $U_{q}(sl_{2})$ to
$U_{q}(sl_{2})\hat{\otimes}U_{q}(sl_{2})$. It is immediate to see that the
limit $\alpha=0$ gives back the usual coproduct on $U(sl_{2})$.

Note that $U_{q}(sl_{2})\hat{\otimes}U_{q}(sl_{2})$ is not the same vector
space as $U_{q}(sl_{2})\otimes U_{q}(sl_{2})$. For example $e^{\alpha(h\otimes
h)}$ is in the former but not in the latter. At this point, it may seem
superfluous to consider the completed tensor product $\hat{\otimes}$, since
the coproduct takes values in the usual tensor product. However, the completed
tensor product is relevant because it turns out that there is an universal
$R$-matrix $\mathcal{R}$ with the required properties in
$U_{q}(sl_{2})\hat{\otimes}U_{q}(sl_{2})$. An explicit formula is:

$\mathcal{R}=e^{\alpha(h\otimes h)/2}\sum_{n\geq
0}\frac{(q-q^{-1})^{n}}{[n]_{q}!}q^{n(n-1)/2}\Bigl{(}e^{-\alpha h/2}y\otimes
xe^{\alpha h/2}\Bigr{)}^{n}\,,$ (21)

where we have set $q=e^{\alpha h}$, $[n]_{q}!=[2]_{q}\cdot[3]_{q}\dots[n]_{q}$
where $[k]_{q}=\frac{q^{k}-q^{-k}}{q-q^{-1}}$. The term in the sum being a
multiple of $\alpha^{n}$, it follows that the element $\mathcal{R}$ is a well-
defined element of $U_{q}(sl_{2})\hat{\otimes}U_{q}(sl_{2})$.

##### Example of representations of $U_{q}(sl_{2})$.

By checking directly the defining relations, we find that:

$h\mapsto\left(\begin{array}[]{cc}1&0\\\ 0&-1\end{array}\right)\,,\ \ \
x\mapsto\left(\begin{array}[]{cc}0&1\\\ 0&0\end{array}\right)\,,\ \ \
y\mapsto\left(\begin{array}[]{cc}0&0\\\ 1&0\end{array}\right)\,,$

defines a representation of $U_{q}(sl_{2})$ on a vector space $V$ of dimension
2. It is called the vector representation (it looks exactly the same as the
natural representation of $sl_{2}$, but look at the next example). To
calculate the $R$-matrix for this representation, one notes that the images of
$x$ and $y$ are nilpotent of index 2, and thus it is enough to take only
$n=0,1$ in the sum in (21). Say $(v_{1},v_{2})$ was the basis of $V$ used to
give the matrices above, then in the basis $(v_{i}\otimes v_{j})_{i,j=1,2}$
ordered lexicographically, we easily find the image of $\mathcal{R}$ and in
turn the $R$-matrix:

$\rho_{V\otimes
V}(\mathcal{R})=\left(\begin{array}[]{cccc}e^{\alpha/2}&\cdot&\cdot&\cdot\\\
\cdot&e^{-\alpha/2}&\cdot&\cdot\\\
\cdot&e^{-\alpha/2}(q-q^{-1})&e^{-\alpha/2}&\cdot\\\
\cdot&\cdot&\cdot&e^{\alpha/2}\end{array}\right)\ \ \ \ \Rightarrow\ \ \
\check{R}_{V,V}=e^{-\alpha/2}\left(\begin{array}[]{cccc}q&\cdot&\cdot&\cdot\\\
\cdot&q-q^{-1}&1&\cdot\\\ \cdot&1&0&\cdot\\\
\cdot&\cdot&\cdot&q\end{array}\right)\,,$

where we recall that the matrix $\check{R}$ is obtained by multiplying by the
permutation operator. Up to a global factor, we recover the solution in
Example 2.3 coming from the Hecke algebra for a vector space of dimension 2.

As a first instance of a procedure whose generalisation we are going to
discuss later, we note that the $R$-matrix $\check{R}_{V,V}$ commutes with the
action of $U_{q}(sl_{2})$ on $V\otimes V$. Namely, it commutes with the image
of $\Delta(h),\Delta(x),\Delta(y)$ in $\text{End}(V\otimes V)$. This can be
easily checked here with a direct calculation, but as we will see, it follows
by construction of $R$-matrices.

Therefore, the action of $U_{q}(sl_{2})$ leaves invariant the eigenspaces of
$\check{R}_{V,V}$. A basis of the eigenspace for the eigenvalue $q$ is
$(v_{1}\otimes v_{1},\,e^{-\alpha/2}v_{1}\otimes
v_{2}+e^{\alpha/2}v_{2}\otimes v_{1},\,v_{2}\otimes v_{2})$. This is the
deformation of the symmetric square of $V$. On this subspace, we find the
following representation of $U_{q}(sl_{2})$:

$h\mapsto\left(\begin{array}[]{ccc}2&0&0\\\ 0&0&0\\\
0&0&-2\end{array}\right)\,,\ \ \
x\mapsto\left(\begin{array}[]{ccc}0&q+q^{-1}&0\\\ 0&0&1\\\
0&0&0\end{array}\right)\,,\ \ \ y\mapsto\left(\begin{array}[]{ccc}0&0&0\\\
1&0&0\\\ 0&q+q^{-1}&0\end{array}\right)\,.$

This is the deformation for $U_{q}(sl_{2})$ of the three-dimensional
representation of $sl_{2}$. The $R$-matrix associated to this representation
can also be calculated directly from the formula (21), we leave the details
for the reader.

We note that the $R$-matrix for this representation does not come anymore from
the Hecke algebra, and one goal of these notes is to discuss the algebras
controlling this solution and the analogues for higher symmetric powers.

##  4 Centralisers of Tensor Representations of Quantum Groups

###  4.1 Properties of centralisers

##### Generalities.

Let $\rho\ :\ A\to\text{End}(E)$ be a representation of an algebra $A$ on a
vector space $E$. The centraliser of the representation is a classical object
in linear algebra. In words, it is the commutant in $\text{End}(E)$ of the
family of endomorphisms $\\{\rho(a)\\}_{a\in A}$ . More formally, we define:

$\text{End}_{A}(E):=\\{x\in\text{End}(E)\ |\ \rho(a)x=x\rho(a)\,,\ \forall
a\in A\\}\ .$

The meaning of the centraliser in representation theory is quickly understood
when one considers a semisimple representation $E$, that is a representation
$E$ which decomposes as a direct sum of irreducible representations:

$E=\bigoplus_{i}E_{i}^{\oplus m_{i}}\,,$

where $m_{i}\in\mathbb{Z}_{>0}$ are the multiplicities of the irreducible
representations $E_{i}$ in $E$. Using Schur Lemma, the centraliser
$\text{End}_{A}(E)$ is described as follows. First it leaves invariant the
subspaces $E_{i}^{\oplus m_{i}}$. Besides, for a given $i$, choose a basis for
the representation $E_{i}$ and denote by $\rho_{i}(a)$ the matrices of
elements of $A$ acting in this basis of $E_{i}$. Form a basis of
$E_{i}^{\oplus m_{i}}$ by concatenating $m_{i}$ times this same basis. Then on
$E_{i}^{\oplus m_{i}}$, the operators of $A$ and of the centralisers look as
follows:

$A\ni a\mapsto\left(\begin{array}[]{cccc}\rho_{i}(a)&0&\dots&0\\\
0&\rho_{i}(a)&\ddots&\vdots\\\ \vdots&\ddots&\ddots&0\\\
0&\ldots&0&\rho_{i}(a)\\\ \end{array}\right)\,,\ \ \ \ \ \
\text{End}_{A}(E)\ni
x\mapsto\left(\begin{array}[]{cccc}x_{11}\text{Id}&x_{12}\text{Id}&\dots&x_{1m_{i}}\text{Id}\\\
x_{21}\text{Id}&x_{22}\text{Id}&\ddots&\vdots\\\
\vdots&\ddots&\ddots&\vdots\\\
x_{m_{i}1}\text{Id}&\ldots&\ldots&x_{m_{i}m_{i}}\text{Id}\\\
\end{array}\right)\,,$

as block-matrices with $m_{i}\times m_{i}$ blocks. Thus we see that the
centraliser $\text{End}_{A}(E)$ is isomorphic to a direct sum of matrix
algebras:

$\text{End}_{A}(E)\cong\bigoplus_{i}\text{Mat}_{m_{i}}(\mathbb{C})\ ,$

the sizes of which correspond to the multiplicities.

One can go further and note that $E$ can also be seen as a representation of
the centraliser $\text{End}_{A}(E)$, and that for this representation, the
centraliser is the image of $A$ in $\text{End}(E)$. This is an instance of a
double centralising theorem (note that the assumption that the representation
of $A$ we started with is semisimple is important). Finally, as the images
commute, $E$ can also be seen as a representation of the algebra
$A\otimes\text{End}_{A}(E)$ (the multiplication is performed independently in
each factor).

To summarise, we have the following decompositions of $E$ as a representation,
respectively, of $A$, of $\text{End}_{A}(E)$ and finally of
$A\otimes\text{End}_{A}(E)$:

$E=\bigoplus_{i}E_{i}^{\oplus m_{i}}\,,\quad\ \ \
E=\bigoplus_{i}\bigl{(}\mathbb{C}^{m_{i}}\bigr{)}^{\oplus\dim(E_{i})}\,,\quad\
\ \ E=\bigoplus_{i}E_{i}\otimes\mathbb{C}^{m_{i}}\,.$ (22)

##### Centralisers of $U_{q}(\mathfrak{g})$.

Now, we are mainly interested here in the case of $A=U_{q}(\mathfrak{g})$ a
quantum group as in Section 3 and $E$ is a tensor product of representations.
More precisely, we take a representation $V$ of $U_{q}(\mathfrak{g})$ and we
form the tensor product $V^{\otimes n}$. We have explained the fundamental
property of quantum groups that this tensor product $V^{\otimes n}$ is also a
representation of $U_{q}(\mathfrak{g})$. The objects of main interest to us
are the centralisers of such representations:

$\text{End}_{U_{q}(\mathfrak{g})}(V^{\otimes n})\ .$

Now it is time to draw on our previous discussion on quantum groups and
fulfill the objective set up in the introduction. Recall from Section 3.2
(Conclusion 1) that we have elements $\check{R}_{1},\dots,\check{R}_{n-1}$ of
$\text{End}(V^{\otimes n})$, satisfying the braid relations.

Moreover recall from Section 3.3 that, assuming the representation $V$ is such
that Conclusion 2 is valid, we have functions
$\check{R}_{1}(u),\dots,\check{R}_{n-1}(u)$ taking values
$\text{End}(V^{\otimes n})$ satisfying the braided YB equation. The quantum
groups are so cleverly designed that the following holds.

######  Conclusion 3.

  1. 1.

The elements $\check{R}_{1},\dots,\check{R}_{n-1}$ belong to the centraliser
$\text{End}_{U_{q}(\mathfrak{g})}(V^{\otimes n})$.

  2. 2.

The functions $\check{R}_{1}(u),\dots,\check{R}_{n-1}(u)$ take values in the
centraliser $\text{End}_{U_{q}(\mathfrak{g})}(V^{\otimes n})$.

###### Proof of Conclusion 3.

The facts stated above being clearly fundamental in our discussion, this is a
good point to give one proof in these notes. It may help to see a little bit
better how the subtle machinery of quantum groups works. So let us prove item
1.

Denote $\rho\ :\ U_{q}(\mathfrak{g})\to\text{End}(V)$ the representation. Let
$a$ be an arbitrary element of $U_{q}(\mathfrak{g})$ and denote
$\Delta(a)=\sum a^{\prime}\otimes a^{\prime\prime}$ its coproduct.

We discussed in Section 3 the coassociativity property of the coproduct and
how it allows to make tensor products without worrying about parentheses. Let
us see more explicitly how this happens. Recall that the representation
$\rho^{(2)}\ :\ U_{q}(\mathfrak{g})\to\text{End}(V\otimes V)$ is constructed
using the coproduct and is given by

$\rho^{(2)}(a)=(\rho\otimes\rho)\bigl{(}\Delta(a)\bigr{)}=\sum\rho(a^{\prime})\otimes\rho(a^{\prime\prime})\
.$

To construct a representation on $V^{\otimes n}$, one should first put
parentheses in $V^{\otimes n}$ and then perform a sequence of tensor products
of two spaces. For $n=3$, we have two possibilities:

$V\otimes(V\otimes V)\ \ \ \ \text{or}\ \ \ \ (V\otimes V)\otimes V\ .$

These two possibilities correspond to representations on $V^{\otimes 3}$ given
by:

$\text{applying $\rho\otimes\rho\otimes\rho$ on:}\quad\ \
\bigl{(}(\text{Id}\otimes\Delta)\circ\Delta\bigr{)}(a)\quad\ \ \
\text{or}\quad\ \ \ \bigl{(}(\Delta\otimes\text{Id})\circ\Delta\bigr{)}(a)\ .$

The coassociativity condition
$(\text{Id}\otimes\Delta)\circ\Delta=(\Delta\otimes\text{Id})\circ\Delta$
ensures that these two possibilities give the same representation. Now,
similarly, any way of putting parentheses on $V^{\otimes n}$ also corresponds
to a certain composition of coproducts. For example, if $n=5$, here is one
possibility:

$V\ \ \leadsto\ \ V\otimes V\ \ \leadsto\ \ (V\otimes V)\otimes V\ \ \leadsto\
\ (V\otimes(V\otimes V))\otimes V\ \ \leadsto\ \ (V\otimes(V\otimes
V))\otimes(V\otimes V)\,,$

corresponding to the following composition:
$(\text{Id}\otimes\text{Id}\otimes\text{Id}\otimes\Delta)\circ(\text{Id}\otimes\Delta\otimes\text{Id})\circ(\Delta\otimes\text{Id})\circ\Delta$.
So to check that we really obtain always the same representation on
$V^{\otimes n}$, we first consider one way of doing it, for example by
defining $\Delta^{(2)}=\Delta$ and for $n>2$:

$\Delta^{(n)}:=(\Delta\otimes\text{Id}\otimes\dots\otimes\text{Id})\circ\Delta^{(n-1)}\,,$

which corresponds to the parenthesized product
$\Bigl{(}\\!\\!...\\!\bigl{(}(V\otimes V)\otimes V\bigr{)}\otimes\dots\otimes
V\Bigr{)}$. And then it is easy to check by induction on $n$ that it coincides
with any other possibilities:

$\Delta^{(n)}=(\text{Id}\otimes\dots\otimes\text{Id}\otimes\Delta\otimes\text{Id}\otimes\dots\otimes\text{Id})\circ\Delta^{(n-1)}\,,\
\ \ \ \ \text{for any position of $\Delta$ among the Id's}\ .$

So finally there is only one representation on
$\rho^{(n)}:U_{q}(\mathfrak{g})\to\text{End}(V^{\otimes n})$, and it is given
by the following recursive formula for any choice of $i$:

$\rho^{(n)}(a)=\sum\rho^{(n-i)}(a^{\prime})\otimes\rho^{(i)}(a^{\prime\prime})\
\ \ \ \ \text{for $i\in\\{1,\dots,n-1\\}$.}$

$\bullet$ $n=2$: After all this preparation, we are ready for the proof at
last. Consider the element $\mathcal{R}$ in $U_{q}(\mathfrak{g})\otimes
U_{q}(\mathfrak{g})$ satisfying:

$\mathcal{R}\Delta(a)=\Delta^{op}(a)\mathcal{R}\,,$

and recall that the element $\check{R}\in\text{End}(V\otimes V)$ comes from
$\mathcal{R}$ in the sense: $\check{R}=P\rho\otimes\rho(\mathcal{R})$ (where
$P$ is the permutation operator of $V\otimes V$). So applying the
representation $\rho\otimes\rho$ to the equality above, we find:

$P\check{R}\,\Bigl{(}\sum\rho(a^{\prime})\otimes\rho(a^{\prime\prime})\Bigr{)}=\Bigl{(}\sum\rho(a^{\prime\prime})\otimes\rho(a^{\prime})\Bigr{)}\,P\check{R}\
.$

Now it is clear that, moving the permutation operator to the left and removing
it, we have:

$\check{R}\rho^{(2)}(a)=\rho^{(2)}(a)\check{R}\ ,$

which is the statement that $\check{R}$ belongs to
$\text{End}_{U_{q}(\mathfrak{g})}(V\otimes V)$.

$\bullet$ $n>2$: By induction on $n$, using respectively the formulas:

$\rho^{(n)}(a)=\sum\rho^{(n-1)}(a^{\prime})\otimes\rho(a^{\prime\prime})\ \ \
\ \text{and}\ \ \ \ \
\rho^{(n)}(a)=\sum\rho(a^{\prime})\otimes\rho^{(n-1)}(a^{\prime\prime})\,,$

we obtain first that $\check{R}_{1},\dots,\check{R}_{n-2}$ commutes with
$\rho^{(n)}(a)$ and second that $\check{R}_{n-1}$ commutes with
$\rho^{(n)}(a)$ as well.

For item 2, we can follow the same reasoning for $\check{R}(u)$ in the
suitable tensor product of “evaluation” representations $V(a)$ of the affine
quantum group $U_{q}(\hat{\mathfrak{g}})$. Restricted to
$U_{q}(\mathfrak{g})$, the dependance on the “evaluation” parameter disappears
and the tensor product is simply $V^{\otimes n}$. We skip the details here. ∎

######  Example 4.1.

Let us discuss a classical example of centralisers. Consider the universal
enveloping algebra $U(sl_{N})$ (so this is the classical limit of the
preceding situation). The Lie algebra $sl_{N}$ is the Lie algebra of $N\times
N$ matrices with zero trace. As such there is a natural representation of
dimension $N$ (send a matrix to itself). Call it $V$ and construct the
representation $V^{\otimes n}$. Recall that for a Lie algebra, the action of
an element $g$ on $V^{\otimes n}$ is the sum of the action on each factor:

$g\otimes\text{Id}\otimes\dots\otimes\text{Id}+\text{Id}\otimes
g\otimes\text{Id}\otimes\dots\otimes\text{Id}+\dots+\text{Id}\otimes\dots\otimes\text{Id}\otimes
g\ .$

Now it is clear that any permutation of the factors in $V^{\otimes n}$
commutes with the action of $U(sl_{N})$. In particular, take $P_{i}$ to be the
transposition of positions $i$ and $i+1$, then we have that
$P_{1},\dots,P_{n-1}$ belong to $\text{End}_{U(sl_{N})}(V^{\otimes n})$. These
operators satisfy the braid relations, so we just described the analogue of
item 1 of Conclusion 3 in this case. Concerning item 2, the solutions of the
braided YB equation (with additive spectral parameters) are given by the Yang
formula, and they obviously take values in $\text{End}_{U(sl_{N})}(V^{\otimes
n})$:

$\check{R}_{i}=P_{i}+\frac{Id_{V^{\otimes n}}}{u}\ .$

Moreover, in this case, we can obtain a complete description of the
centraliser $\text{End}_{U(sl_{N})}(V^{\otimes n})$ since we can show that it
is actually generated by the permutations (so it is a quotient of the algebra
of the symmetric group). This is the classical Schur–Weyl duality, which we
will describe for $U_{q}(sl_{N})$ in the next subsection.

###  4.2 Schur–Weyl duality

At this point, we should be convinced that the centralisers
$\text{End}_{U_{q}(\mathfrak{g})}(V^{\otimes n})$ are the algebras we were
looking for, in the sense that they satisfy the properties discussed in the
introduction: they contain a quotient of the braid group algebra and they
contain solutions of the YB equation. About a representation on a tensor space
$V^{\otimes n}$, well, they are defined as subalgebras of
$\text{End}(V^{\otimes n})$, so they admit one by definition.

However, we are far from an explicit algebraic (or any other) description of
$\text{End}_{U_{q}(\mathfrak{g})}(V^{\otimes n})$ for any $\mathfrak{g}$ and
$V$. So the conclusion above is more a general statement indicating that this
should be very interesting for us to study these algebras. Moreover, the
claimed objective that we were going to “explain” why the Hecke (and BMW)
algebras fit so nicely in our picture is still not achieved. This is what we
are going to do now.

##### Quantum group $U_{q}(sl_{N})$.

So we take $\mathfrak{g}=sl_{N}$ for some $N>0$ and, for the representation
$V$, we take the analogue for $U_{q}(sl_{N})$ of the natural representation of
$sl_{N}$. From the general representation theory of quantum groups, we know
that any (finite-dimensional irreducible) representation of $\mathfrak{g}$ can
be deformed to a representation of $U_{q}(\mathfrak{g})$. However, let us be
more explicit than that.

First, the quantum group $U_{q}(sl_{N})$ should be described a bit more. We
build on our previous description of $U_{q}(sl_{2})$ in Section 3.4 involving
power series in $\alpha$ and so on, so we do not repeat our discussion ($q$ is
still $e^{\alpha}$). Natural generators of $sl_{N}$ are the elements
$h_{i},x_{i},y_{i}$, with $i=1,\dots,N-1$. The triplet of elements
$h_{i},x_{i},y_{i}$ corresponds to the following matrices of $sl_{N}$: take
the matrices $\left(\begin{array}[]{cc}1&0\\\ 0&-1\end{array}\right)$,
$\left(\begin{array}[]{cc}0&1\\\ 0&0\end{array}\right)$,
$\left(\begin{array}[]{cc}0&0\\\ 1&0\end{array}\right)$ as for $sl_{2}$ and
plug them in lines and columns $i$ and $i+1$.

The multiplication is deformed as follows. First each triplet
$h_{i},x_{i},y_{i}$ satisfies the relations of $U_{q}(sl_{2})$ in (19),
replacing $x$ to $x_{i}$, $y$ to $y_{i}$ and $h$ to $h_{i}$. Then we keep the
following usual relations of matrices:

$[h_{i},x_{i\pm 1}]=-x_{i\pm 1}\,,\ \ \ \ \ [h_{i},y_{i\pm 1}]=y_{i\pm 1}\,,\
\ \ \ \ [h_{i},x_{j}]=[h_{i},y_{j}]=0\ \ (|i-j|>1)\,,$ $[h_{a},h_{b}]=0\ \
(\forall a,b)\,,\ \ \ \ \ \ [x_{i},y_{j}]=0\ \ (i\neq j)\ .$

And finally we add the deformation of the so-called Serre relations (which is
really the new ingredient compared to $sl_{2}$):

$x_{j}x_{i}^{2}-(q+q^{-1})x_{i}x_{j}x_{i}+x_{i}^{2}x_{j}=y_{j}y_{i}^{2}-(q+q^{-1})y_{i}y_{j}y_{i}+y_{i}^{2}y_{j}=0\
\ \ \ \text{ if $|i-j|=1$\,.}$

For each triplet $h_{i},x_{i},y_{i}$, the coproduct is given by the same
formulas as for $U_{q}(sl_{2})$ in (20). That is it, we have defined
$U_{q}(sl_{N})$ and how to make tensor products of its representations.

As for $sl_{2}$, the standard representation $V$ of $U_{q}(sl_{N})$ is simply
given by assigning to each $h_{i},x_{i},y_{i}$ its natural $N\times N$ matrix,
as if we were dealing with $sl_{N}$. One can be surprised maybe, but all
defining relations of $U_{q}(sl_{N})$ are indeed satisfied. So again, and
especially in the standard representation, we are not changing much the
algebra structure, but the “quantum” novelty of $U_{q}(sl_{N})$ compared to
classical $U(sl_{N})$ is that we have a different way to construct a
representation on $V^{\otimes n}$. Comparing to Example 4.1, now the
permutations do not belong to the centraliser
$\text{End}_{U_{q}(sl_{N})}(V^{\otimes n})$. This is where we meet again, at
last, with the Hecke algebra.

##### Hecke algebra $H_{n}(q)$ and Jimbo–Schur–Weyl duality.

Informations on the classical Schur–Weyl duality can be found in [27, 30, 67],
and on the quantum version in [11, 37, 48]. Recall that we have a
representation of the Hecke algebra $H_{n}(q)$ on the tensor product
$V^{\otimes n}$. This was given explicitly in (11). So our situation is that
we have two algebras represented on the same vector space, and we can picture
it like this:

$U_{q}(sl_{N})\ \ \stackrel{{\scriptstyle\rho^{(n)}}}{{\longrightarrow}}\ \ \
\ \text{End}(V^{\otimes n})\ \ \ \
\stackrel{{\scriptstyle\pi}}{{\longleftarrow}}\ \ H_{n}(q)\ $

Now we can state the (first part of) the Schur–Weyl duality.

######  Theorem 4.2 (Schur–Weyl I).

The centraliser $\text{End}_{U_{q}(sl_{N})}(V^{\otimes n})$ is the image of
the Hecke algebra $H_{n}(q)$:

$\text{End}_{U_{q}(sl_{N})}(V^{\otimes n})=\pi\bigl{(}H_{n}(q)\bigr{)}\ .$

Since we are in the semisimple situation, from the general consideration on
centralisers sketched above, we have actually that
$\pi\bigl{(}H_{n}(q)\bigr{)}$ and $\rho^{(n)}\bigl{(}U_{q}(sl_{N})\bigr{)}$
are the mutual centralisers of each other. This is why we call it a duality.

From the generalities discussed above, see around Formulas (22), we know that
the centraliser is related to various decompositions of the representation
$V^{\otimes n}$ into irreducible summands. To describe this in our particular
situation, we need some notations about irreducible representations of
$U_{q}(sl_{N})$ and of $H_{n}(q)$.

A partition $\lambda$ of $n$ is a family of integers
$\lambda=(\lambda_{1},\dots,\lambda_{l})$ such that
$\lambda_{1}\geq\lambda_{2}\geq\dots\geq\lambda_{l}\geq 0$ and
$\lambda_{1}+\dots+\lambda_{l}=n$. We note $\lambda\vdash n$ and we say that
$\lambda$ is a partition of size $n$. The number $\ell(\lambda)$ of non-zero
parts is called the length of $\lambda$. Then we have:

  * •

The quantum group $U_{q}(sl_{N})$ has irreducible representations indexed by
highest weights, which are here identified with partitions $\lambda$ such that
$\ell(\lambda)\leq N$. Let us denote them by $L_{\lambda}^{N}$. Note that here
the size of $\lambda$ can be arbitrary. We put a $N$ in the notation because
the same partition can index a representation of $U_{q}(sl_{N})$ for various
$N$. The standard representation $V$ corresponds to $\lambda=(1)$.

  * •

The Hecke algebra $H_{n}(q)$ has irreducible representations indexed by
partitions $\lambda$ of size $n$. Let us denote them by $S_{\lambda}$ (our
convention is such that $S_{(n)}$ is the one-dimensional representation
$\sigma_{i}\mapsto q$).

Now the decomposition of $V^{\otimes n}$, respectively, as a representation of
$U_{q}(sl_{N})$, as a representation of $H_{n}(q)$ and finally as a
representation of $U_{q}(sl_{N})\otimes H_{n}(q)$ is:

$V^{\otimes n}=\bigoplus_{\begin{subarray}{c}\lambda\vdash n\\\\[1.39998pt]
\ell(\lambda)\leq
N\end{subarray}}(L_{\lambda}^{N})^{\oplus\dim(S_{\lambda})}\,,\quad\ \ \
V^{\otimes n}=\bigoplus_{\begin{subarray}{c}\lambda\vdash n\\\\[1.39998pt]
\ell(\lambda)\leq
N\end{subarray}}S_{\lambda}^{\oplus\dim(L_{\lambda}^{N})}\,,\quad\ \ \
V^{\otimes n}=\bigoplus_{\begin{subarray}{c}\lambda\vdash n\\\\[1.39998pt]
\ell(\lambda)\leq N\end{subarray}}L_{\lambda}^{N}\otimes S_{\lambda}\,.$ (23)

The duality here is apparent. Note that the knowledge of the representations
$S_{\lambda}$ of $H_{n}(q)$ can be used to construct the representations
$L_{\lambda}^{N}$ of $U_{q}(sl_{N})$ (this is the natural point of view in
these notes), or the other way around.

##### Kernel in the Schur–Weyl duality.

The first step of the Schur–Weyl duality is a great step in the understanding
of $\text{End}_{U_{q}(sl_{N})}(V^{\otimes n})$. Indeed Theorem 4.2 asserts
that it is generated by the elements $\check{R}_{1},\dots,\check{R}_{n-1}$,
and that these elements, in addition to satisfying the braid relations, also
satisfy the Hecke relations. However, we can not say that the centraliser _is_
the Hecke algebra $H_{n}(q)$. It only says that the centraliser is the image
of the map $\pi$, and this map may have a kernel. So the second part of the
Schur–Weyl duality shall be the description of this kernel. It turns out that
there is a quite simple description of the kernel. The kernel was actually
implicitly described in (23) but let us not worry about that and just give
directly its algebraic description.

We need to go a little bit into the algebraic structure of the Hecke algebra
$H_{n}(q)$. For any element $w$ of the symmetric group $\mathfrak{S}_{n}$, let
$w=s_{a_{1}}\dots s_{a_{k}}$ be a reduced (_i.e._ minimal length) expression
for $w$ in terms of the generators $s_{i}=(i,i+1)$, and denote $\ell(w)=k$.
Then define $\sigma_{w}:=\sigma_{a_{1}}\dots\sigma_{a_{k}}\in H_{n}(q)$. This
definition does not depend on the reduced expression for $w$ and the set
$\\{\sigma_{w}\\}_{w\in\mathfrak{S}_{n}}$ forms a basis of $H_{n}(q)$.

For $L\in\mathbb{Z}_{\geq 0}$, we define the $q$-numbers as follows:

$[L]_{q}:=\frac{q^{L}-q^{-L}}{q-q^{-1}}=q^{L-1}+q^{L-3}+\dots+q^{-(L-1)}\ \ \
\ \ \text{and}\ \ \ \ \ \ \ [L]_{q}!:=[1]_{q}[2]_{q}\dots[L]_{q}\ .$ (24)

Then the $q$-symmetriser in $H_{n}(q)$ is the following element:

$P_{n}=\frac{\sum_{w\in\mathfrak{S}_{n}}q^{\ell(w)}\sigma_{w}}{\sum_{w\in\mathfrak{S}_{n}}q^{2\ell(w)}}=\frac{q^{-n(n-1)/2}}{[n]_{q}!}\sum_{w\in\mathfrak{S}_{n}}q^{\ell(w)}\sigma_{w}\
.$ (25)

and the $q$-antisymmetriser in $H_{n}(q)$ is the following element:

$P^{\prime}_{n}=\frac{\sum_{w\in\mathfrak{S}_{n}}(-q^{-1})^{\ell(w)}\sigma_{w}}{\sum_{w\in\mathfrak{S}_{n}}q^{-2\ell(w)}}=\frac{q^{n(n-1)/2}}{[n]_{q}!}\sum_{w\in\mathfrak{S}_{n}}(-q^{-1})^{\ell(w)}\sigma_{w}\
.$ (26)

We take this opportunity to emphasize the nice formula
$\sum_{w\in\mathfrak{S}_{n}}q^{2\ell(w)}=q^{n(n-1)/2}[n]_{q}!$.

######  Remark 4.3.

For $q=1$, the $q$-symmetriser becomes the usual symmetriser in the group
algebra $\mathbb{C}\mathfrak{S}_{n}$: the sum of all elements of
$\mathfrak{S}_{n}$ divided by $n!$. Similarly, the $q$-antisymmetriser becomes
the usual antisymmetriser: the sum of all elements of $\mathfrak{S}_{n}$
multiplied by their signature and divided by $n!$. ∎

We remark that up to now the description of the centraliser does not
explicitly depend on the dimension $N$ of $V$, since the Hecke algebra does
not depend on $N$. Of course, this dependence is hidden in the map $\pi$ and
in its kernel. So it is only natural that it appears explicitly now.

######  Theorem 4.4 (Schur–Weyl II).

  * •

If $n\leq N$ then the kernel of the map $\pi$ is $\\{(0)\\}$

  * •

If $n>N$ then the kernel of the map $\pi$ is generated in $H_{n}(q)$ by the
element $P^{\prime}_{N+1}$.

Note that we see the element $P^{\prime}_{N+1}$ as an element of $H_{n}(q)$ by
the natural inclusion of $H_{N+1}(q)$ in $H_{n}(q)$ (if $n>N$). In words, to
obtain the complete description of the centraliser, we start from the Hecke
algebra $H_{n}(q)$ and, if $n>N$, we cancel the $q$-antisymmetriser on $N+1$
letters.

Explicitly, the centraliser $\text{End}_{U_{q}(sl_{N})}(V^{\otimes n})$ is
isomorphic to the algebra generated by $\sigma_{1},\dots,\sigma_{n-1}$ with
defining relations:

$\begin{array}[]{ll}\sigma_{i}\sigma_{i+1}\sigma_{i}=\sigma_{i+1}\sigma_{i}\sigma_{i+1}\,,&\text{for
$i\in\\{1,\dots,n-2\\}$}\,,\\\\[1.99997pt]
\sigma_{i}\sigma_{j}=\sigma_{j}\sigma_{i}\,,&\text{for
$i,j\in\\{1,\dots,n-1\\}$ such that $|i-j|>1$}\,,\\\\[1.99997pt]
\sigma_{i}^{2}=1+(q-q^{-1})\sigma_{i}\,,&\text{for
$i\in\\{1,\dots,n-1\\}$}\,,\\\\[1.99997pt] P^{\prime}_{N+1}=0&\text{if
$n>N$.}\end{array}$

######  Example 4.5 (Temperley–Lieb algebra).

Let $N=2$. In this case, the centraliser
$\text{End}_{U_{q}(sl_{2})}(V^{\otimes n})$ is called the Temperley–Lieb
algebra. The additional relation $P^{\prime}_{3}=0$ reads:

$1-q^{-1}(\sigma_{1}+\sigma_{2})+q^{-2}(\sigma_{1}\sigma_{2}+\sigma_{2}\sigma_{1})-q^{-3}\sigma_{1}\sigma_{2}\sigma_{1}=0\
.$

One can show that it implies the same relation with indices $i,i+1$ for all
$i=1,\dots,n-2$. Then setting $\tau_{i}:=\sigma_{i}-q$, one recovers the other
standard presentation of the Temperley–Lieb algebra:

$\tau_{i}^{2}=-(q+q^{-1})\tau_{i}\,,\ \ \ \
\tau_{i}\tau_{i+1}\tau_{i}=\tau_{i}\,,\ \ \ \
\tau_{i+1}\tau_{i}\tau_{i+1}=\tau_{i+1}\ \ \ \text{and}\ \ \
\tau_{i}\tau_{j}=\tau_{j}\tau_{i}\ \text{if $|i-j|>1$.}$

##### And the BMW algebra?

Now we explained, at length, how the Hecke algebra fits in the story of
centralisers of quantum groups representations. This beautiful story has its
counterpart for the BMW algebra. Indead instead of $sl_{N}$ we can consider
the other classical Lie algebras $so_{N}$ or $sp_{N}$. We can form the quantum
groups $U_{q}(so_{N})$ or $U_{q}(sp_{N})$, and we can consider the tensor
product $V^{\otimes n}$, where $V$ is the (analogue of the) vector
representation for $U_{q}(so_{N})$ or $U_{q}(sp_{N})$. Then the centralisers
of $V^{\otimes n}$ are described in a way similar to above, where the Hecke
algebra is replaced by another algebra. This other algebra is the BMW algebra
(with the parameter $a$ specialised to a power of $q$ depending whether we
have $so_{N}$ or $sp_{N}$). For details about that along the same lines as
above, we refer for example to [48, §8.6]. This concludes our discussion on
the interpretation of Hecke and BMW algebras as centralisers of
representations of quantum groups.

##  5 Fusion Procedure for the Yang–Baxter Equation

The fusion procedure was designed very early in the history of the YB equation
[37, 38, 52]. It consists in a general procedure to construct new solutions
starting from a known one. This procedure can be split in two steps and
presented schematically like in the following table:

Vector spaces | Matrices | Algebras
---|---|---
$V$ | basic solution on $V\otimes V$ |  $\hookleftarrow\ $ Hecke algebra $H_{n}(q)$
(generic fusion) $\downarrow$ | $\downarrow$ |
$V^{\otimes k}$ | solution on $V^{\otimes k}\otimes V^{\otimes k}$ |  $\hookleftarrow\ $ (bigger) Hecke algebra $H_{kn}(q)$
(projection) $\downarrow$ | $\downarrow$ |
$Pr(V^{\otimes k})$ |  solution on $Pr(V^{\otimes k})\otimes Pr(V^{\otimes k})$ |  $\hookleftarrow\ $ _?? fused Hecke algebra ??_

To give a rough idea of what is going on, the first two columns describe the
fusion procedure that was originally designed, for matrix solutions of the YB
equation. Starting from a known solution on $V$, an explicit formula builds a
new solution on the much larger vector space $V^{\otimes k}$. This is the
first step. The second step identifies invariant subspaces in $V^{\otimes k}$
for this new solution. This is quite general and valid for any solution $R(u)$
of the YB equation.

It is important to indicate that we have some free parameters in the fusion
procedure, and it turns out that for some specific choices of these free
parameters, the invariant subspaces can be especially interesting. To give at
once the explicit example we will be interested in, say the solution on $V$
comes from the Hecke algebra and look at the tensor product $V^{\otimes n}$ as
a representation of $sl_{N}$ (or $U_{q}(sl_{N})$), where $N=\dim(V)$. Then it
decomposes as:

$V^{\otimes k}=S^{k}(V)\oplus\dots\,,$

where $S^{k}(V)$ is the $k$-th symmetric power of $V$ (or its analogue for
$U_{q}(sl_{N})$). If we choose wisely and precisely the parameters in the
fusion procedure, then the big solution $R^{(k)}(u)$ restricts to a solution
on the subspace $S^{k}(V)$. This is for the matrix side of the picture.

The third column intends to indicate what should be the “algebra” counterpart
of this. Well it turns out that the first step of the fusion procedure can be
carried out directly in the algebra. So if we started in $H_{n}(q)$, we end up
in the larger Hecke algebra $H_{kn}(q)$. The number of strands has been
multiplied by $k$, this is called cabling from the point of view of braids.

It also turns out that the projector, denoted here $Pr$, on the subspace
$S^{k}(V)$, can be seen directly in the Hecke algebra $H_{k}(q)$. In fact this
is part of the statement of the Schur–Weyl duality, and it turns out that in
this case $Pr$ is simply the (image of the) $q$-symmetriser of $H_{k}(q)$.
With a little algebraic thinking on how to project a representation, we
realise that the algebra controlling the new fused solution on $S^{k}(V)$
should be:

$\text{``fused Hecke algebra''}\ =\ P_{k,n}.H_{kn}(q).P_{k,n}\ ,$

where the idempotent $P_{k,n}$ in $H_{kn}(q)$ is made up by plugging the
$q$-symmetriser on the first $k$ strands, and also on the next $k$ strands and
so on up to the last $k$ strands. If we want to conclude this section now, we
can sum up the algebraic procedure like this:

######  Conclusion 4.

The fused Hecke algebra is obtained with the following two steps:

$H_{n}(q)\ \ \stackrel{{\scriptstyle\text{``cabling''}}}{{\longrightarrow}}\ \
H_{kn}(q)\ \
\stackrel{{\scriptstyle\text{``projecting''}}}{{\longrightarrow}}\ \
P_{k,n}.H_{kn}(q).P_{k,n}\ \ \text{($=$ fused Hecke algebra)}$

If one is satisfied enough with this brief description, one can proceed
directly to Section 6 for a description of the fused Hecke algebra (and should
not be surprised to find a solution of the YB equation in there). Otherwise,
in the present section, we shall present more precisely the machinery of the
fusion procedure. More details can be found in [62].

###  5.1 Fusion procedure for matrix solutions

In what follows, we make the choice to work with multiplicative spectral
parameters in the YB equation. This is more adapted to the Hecke algebra
situation. Of course, everything is also valid in the additive case with
suitable modifications.

##### First step.

We start with an arbitrary solution of the YB equation on a vector space $V$.
Here we use the following form of the YB equation:

$R_{12}(u)R_{13}(uv)R_{23}(v)=R_{23}(v)R_{13}(uv)R_{12}(u)\ \quad\ \ \ \
\text{on $V\otimes V\otimes V$.}$

Let $\textbf{c}:=(c_{1},\dots,c_{k})$ be a $k$-tuple of non-zero complex
parameters. We consider the space $V^{\otimes k}\otimes V^{\otimes k}$ and we
find it convenient to label the copies of $V$ by
$1,\dots,k,\underline{1},\dots,\underline{k}$ (from left to right). For
example, the operator $R_{a,\underline{b}}(u)$ stands for the operator
$R_{a,k+b}(u)$ with the notation used before.

Then we define the following operator in $\text{End}(V^{\otimes k}\otimes
V^{\otimes k})$.

$R^{(\textbf{c})}(u):=\prod_{i=1,\dots,k}^{\leftarrow}R_{1,\underline{i}}(u\frac{c_{1}}{c_{i}})R_{2,\underline{i}}(u\frac{c_{2}}{c_{i}})\dots\dots
R_{k,\underline{i}}(u\frac{c_{k}}{c_{i}})\ ,$ (27)

where the arrow means that the product is taken from left to right in
decreasing order in the index $i$ (the first factor is
$R_{1,\underline{n}}(u\frac{c_{1}}{c_{k}})$ and the last is
$R_{n,\underline{1}}(u\frac{c_{k}}{c_{1}})$).

By a direct calculation using repeatedly the YB equation for $R(u)$, one can
check that this new operator also satisfies the YB equation:

$R^{(\textbf{c})}_{12}(u)R^{(\textbf{c})}_{13}(uv)R^{(\textbf{c})}_{23}(v)=R^{(\textbf{c})}_{23}(v)R^{(\textbf{c})}_{13}(uv)R^{(\textbf{c})}_{12}(u)\
\quad\ \ \ \text{on $V^{\otimes k}\otimes V^{\otimes k}\otimes V^{\otimes
k}$,}$

where now the indices $1,2,3$ refer to the new vector space $V^{\otimes k}$.

######  Example 5.1.

Within the “space-time trajectories” interpretation of the YB equation in (7),
this first step of the fusion procedure is easy to understand. The new
solution, given by Formula (27), corresponds to interactions of multiplets of
$k$ particles. We have a “beam” of $k$ particles crossing another beam, and
the operator $R^{(k)}(u)$ describes this interaction, factorised as a product
of $2$-particles interaction. For example, for $k=2$, the operator
$R^{(2)}(u)$ corresponds to:

$\scriptstyle{1}$$\scriptstyle{2}$$\scriptstyle{\underline{1}}$$\scriptstyle{\underline{2}}$$\scriptstyle{u_{2\underline{1}}}$$\scriptstyle{u_{1\underline{1}}}$$\scriptstyle{u_{2\underline{2}}}$$\scriptstyle{u_{1\underline{2}}}$$:$$\
\ \
R^{(2)}(u)=R_{1\underline{2}}(u_{1\underline{2}})R_{2\underline{2}}(u_{2\underline{2}})R_{1\underline{1}}(u_{1\underline{1}})R_{2\underline{1}}(u_{2\underline{1}})\
,$

where $u_{i\underline{j}}:=u\displaystyle\frac{c_{i}}{c_{j}}$. The parameters
$\textbf{c}=(c_{1},\dots,c_{k})$ accounts for the different rapidities of the
particles. Note that we should have taken in general a different set
$\underline{\textbf{c}}=(c_{\underline{1}},\dots,c_{\underline{k}})$ of
parameters for the particles $\underline{1},\dots,\underline{k}$. For
simplicity, and as it is enough for what follows, we took the same set c
twice.

So this first step amounts to the remarkable property that if the one-particle
interaction satisfies the YB equation then so does the $k$-particles
interaction.

######  Remark 5.2.

The extension from one particle to $k$ particles is reminiscent in braid
theory of a particular case of what is called _cabling_. Start from a braid,
and duplicate all strands, replacing a strand by $k$ new strands. This gives a
new braid. For example, on the elementary braiding for $k=2$:

$\rightsquigarrow$$\rightsquigarrow$

It is immediate that the new braids obtained from the elementary braidings
also satisfy the braid relations. This is the analogue of the first step of
the fusion procedure. ∎

##### Second step.

Now we consider the following operator on $V^{\otimes k}$:

$F(\textbf{c}):=\prod_{1\leq i<j\leq
k}^{\rightarrow}R_{j,i}(\frac{c_{j}}{c_{i}})\ ,$ (28)

where the arrow indicates that the product is taken from left to right in
increasing order for the lexicographic order:
$F(\textbf{c})=R_{21}(\frac{c_{2}}{c_{1}})R_{31}(\frac{c_{3}}{c_{1}})\dots
R_{k1}(\frac{c_{k}}{c_{1}})\cdot
R_{32}(\frac{c_{3}}{c_{2}})......R_{k,k-1}(\frac{c_{k}}{c_{k-1}})$ . The
subspace we are interested in is the image of this operator $F(\textbf{c})$:

$W_{\textbf{c}}:=\textrm{Im}(F(\textbf{c}))\subset V^{\otimes k}\ .$ (29)

We note that $F(\textbf{c})$ may not be defined for any value of the
parameters c, since the function $R(u)$ can have singularities (for example,
the solution coming from the Hecke algebra can not be evaluated at $u=1$).
Nevertheless, we assume in our discussion that the set of parameters c is such
that the operator $F(\textbf{c})$ is a well-defined element of
$\text{End}(V^{\otimes k})$.

So now we have a subspace $W_{\textbf{c}}\otimes W_{\textbf{c}}$ of
$V^{\otimes k}\otimes V^{\otimes k}$. The main fact about it is that it is an
invariant subspace for the operator $R^{(\textbf{c})}(u)$. In other words, the
operator $R^{(\textbf{c})}(u)$ restricts to $W_{\textbf{c}}\otimes
W_{\textbf{c}}$ and provides a solution of the YB equation on this subspace.

######  Conclusion 5.

For any set of parameters $\textbf{c}=(c_{1},\dots,c_{k})$ such that the
subspace $W_{\textbf{c}}\subset V^{\otimes n}$ is defined, we have a solution
$R^{fus}(u)$ of the YB equation on $W_{\textbf{c}}\otimes W_{\textbf{c}}$.

We denoted by $R^{fus}(u)$ the restriction to $W_{\textbf{c}}\otimes
W_{\textbf{c}}$ of $R^{(\textbf{c})}(u)$, and we call this element
$R^{fus}(u)$ a fused solution.

We should note that for parameters c such that the operator $F(\textbf{c})$ is
invertible then the subspace $W_{\textbf{c}}$ is in fact $V^{\otimes k}$ and
the second step has given us nothing new. Of course, if we went into all this
trouble, it is because we hope to find more interesting subspaces than that.
We will show convincing examples below, in the Hecke algebra situation.

###  5.2 Fusion procedure, the braided version

Recall from Section 1.2 that the YB equation for $R(u)$ is equivalent to the
braided YB equation for $\check{R}(u)$, where both are related by the
permutation operator $P$ of $V\otimes V$:

$R(u)=P\check{R}(u)\ .$

In our setting, it is the braided version $\check{R}(u)$ which can be seen as
coming from a Baxterization formula of an algebra. While $R(u)$ a priori can
not be expressed directly in the algebra, because the permutation operator $P$
may have no meaning in the algebra. So our goal should be first to express the
fusion procedure using only the “braided” operator $\check{R}(u)$. It turns
out to be possible.

In what follows, we use indices $1,\dots,2k$ to label the factors in
$V^{\otimes k}\otimes V^{\otimes k}=V^{\otimes 2k}$. For the first step of the
fusion procedure, a direct calculation shows that

$R^{(\textbf{c})}(u)=P_{V^{\otimes k},V^{\otimes
k}}\cdot\prod_{i=1,\dots,k}^{\leftarrow}\check{R}_{i}(u\frac{c_{1}}{c_{i}})\check{R}_{i+1}(u\frac{c_{2}}{c_{i}})\dots\dots\check{R}_{i+k-1}(u\frac{c_{k}}{c_{i}})\
,$

where $P_{V^{\otimes k},V^{\otimes k}}$ is the permutation operator on
$V^{\otimes k}\otimes V^{\otimes k}$. But of course, multiplication on the
left by $P_{V^{\otimes k},V^{\otimes k}}$ is just a way to go the braided YB
equation. So we conclude that we could just have started with the following
definition666However, the nice interpretation with interacting particles would
have been less apparent.:

$\check{R}^{(\textbf{c})}(u)=\prod_{i=1,\dots,k}^{\leftarrow}\check{R}_{i}(u\frac{c_{1}}{c_{i}})\check{R}_{i+1}(u\frac{c_{2}}{c_{i}})\dots\dots\check{R}_{i+k-1}(u\frac{c_{k}}{c_{i}})\
.$ (30)

And the first step of the fusion procedure asserts that the operator
$\check{R}^{(\textbf{c})}(u)$ satisfies the braided YB equation:

$\check{R}^{(\textbf{c})}_{1}(u)\check{R}^{(\textbf{c})}_{2}(uv)\check{R}^{(\textbf{c})}_{1}(v)=\check{R}^{(\textbf{c})}_{2}(v)\check{R}^{(\textbf{c})}_{1}(uv)\check{R}^{(\textbf{c})}_{2}(u)\quad\
\ \ \ \ \text{on $V^{\otimes k}\otimes V^{\otimes k}\otimes V^{\otimes k}$.}$

Now for the second step, we should try to express the operator $F(\textbf{c})$
also in terms of the braided element $\check{R}(u)$. This is indeed possible,
a direct calculation showing that:

$F(\textbf{c})=\Bigl{(}\prod_{i=1,\dots,k-1}^{\rightarrow}\check{R}_{i}(\frac{c_{i+1}}{c_{1}})\check{R}_{i-1}(\frac{c_{i+1}}{c_{2}})\dots\dots\check{R}_{1}(\frac{c_{i+1}}{c_{i}})\Bigr{)}\cdot
P_{w_{0}}\ \ \ \ \quad\ \text{on $V^{\otimes k}$,}$

where $P_{w_{0}}$ is the operator on $V^{\otimes k}$ permuting the factors
according to the permutation completely reversing the indices: $(1,\dots,k)$
goes to $(k,\dots,1)$. The map $P_{w_{0}}$ is invertible so it sends
$V^{\otimes k}$ ot itself, and therefore the image of $F(\textbf{c})$ is the
same as the image of the element in the big parenthesis just above. Thus we
conclude that we could just have started with the following definition:

$\check{F}(\textbf{c})=\prod_{i=1,\dots,k-1}^{\rightarrow}\check{R}_{i}(\frac{c_{i+1}}{c_{1}})\check{R}_{i-1}(\frac{c_{i+1}}{c_{2}})\dots\dots\check{R}_{1}(\frac{c_{i+1}}{c_{i}})\
,$ (31)

and define

$W_{\textbf{c}}:=\textrm{Im}(\check{F}(\textbf{c}))\subset V^{\otimes k}\ .$
(32)

We have explained that this definition of $W_{\textbf{c}}$ coincides with the
original one in (29). The second step of the fusion procedure asserts that the
operator $\check{R}^{(\textbf{c})}(u)$ leaves the subspace
$W_{\textbf{c}}\otimes W_{\textbf{c}}$ invariant. In fact, we can show that:

$\check{R}^{(\textbf{c})}(u)\ \text{commutes with
}\check{F}(\textbf{c})\otimes\check{F}(\textbf{c})\,.$

We denote by $\check{R}^{fus}(u)$ the restriction to $W_{\textbf{c}}\otimes
W_{\textbf{c}}$ of $\check{R}^{(\textbf{c})}(u)$, and we also call this
element $\check{R}^{fus}(u)$ a fused solution. At the end of the day, of
course we have:

$\check{R}^{fus}(u)=P_{W_{\textbf{c}}\otimes W_{\textbf{c}}}R^{fus}(u)\ \ \ \
\ \ \ \ \text{on $W_{\textbf{c}}\otimes W_{\textbf{c}}$\ ,}$

but our point in this subsection was mostly to express $\check{R}^{fus}(u)$
and $\check{F}(\textbf{c})$ in terms of $\check{R}(u)$.

###  5.3 Fusion procedure in the Hecke algebra

We will now describe the procedure directly in the Hecke algebra. We deal with
the example of the Hecke algebra to simplify the discussion, but part of it,
especially the beginning, can be formulated for an arbitrary algebra with a
Baxterization formula and a local representation.

So we assume now that the solution $\check{R}(u)$ comes from the Hecke
algebra. It means that $\check{R}_{i}(u)$ is the image of
$\sigma_{i}(u)=\sigma_{i}+(q-q^{-1})\frac{1}{u-1}$ in the local representation
of the Hecke algebra $H_{n}(q)$ on $V^{\otimes n}$, see Section 2.1.

The first step of the fusion procedure is easy to formulate in the algebra.
From its definition in (30), it is immediate that the new solution
$\check{R}^{(\textbf{c})}(u)$ in $\text{End}(V^{\otimes k}\otimes V^{\otimes
k})$ is the image of the following element of $H_{2k}(q)$:

$\sigma_{1}^{(\textbf{c})}(u)=\prod_{i=1,\dots,k}^{\leftarrow}\sigma_{i}(u\frac{c_{1}}{c_{i}})\sigma_{i+1}(u\frac{c_{2}}{c_{i}})\dots\dots\sigma_{i+k-1}(u\frac{c_{k}}{c_{i}})\
.$ (33)

By shifting the indices by $k$, we can easily define
$\sigma_{2}^{(\textbf{c})}(u)$, and in turn $\sigma_{3}^{(\textbf{c})}(u)$ and
so on. The elements
$\sigma_{1}^{(\textbf{c})}(u),\dots,\sigma_{n-1}^{(\textbf{c})}(u)$ satisfy
the braided Yang–Baxter equation

$\sigma^{(\textbf{c})}_{i}(u)\sigma^{(\textbf{c})}_{i+1}(uv)\sigma^{(\textbf{c})}_{i}(v)=\sigma^{(\textbf{c})}_{i+1}(v)\sigma^{(\textbf{c})}_{i}(uv)\sigma^{(\textbf{c})}_{i+1}(u)\
,$

directly in the algebra, which is here $H_{kn}(q)$ (the number of strands has
simply been multiplied by $k$). The local representation of the algebra
$H_{kn}(q)$ is naturally on $(V^{\otimes k})^{\otimes n}$, so we have
interpreted the solution $\check{R}^{(\textbf{c})}(u)$ as coming from an
algebra with a local representation.

Then we note that from its definition (31), the operator
$\check{F}(\textbf{c})$ in $\text{End}(V^{\otimes k})$ is the image of the
following element of $H_{k}(q)$:

$\Phi(\textbf{c})=\prod_{i=1,\dots,k-1}^{\rightarrow}\sigma_{i}(\frac{c_{i+1}}{c_{1}})\sigma_{i-1}(\frac{c_{i+1}}{c_{2}})\dots\dots\sigma_{1}(\frac{c_{i+1}}{c_{i}})\
.$ (34)

What does it mean in the Hecke algebra that $\check{R}^{(\textbf{c})}(u)$
leaves invariant the subspace $W_{\textbf{c}}\otimes W_{\textbf{c}}$? Well,
first we have to note that $H_{k}(q)\otimes H_{k}(q)$ is included in
$H_{2k}(q)$. The first factor is generated by $\sigma_{1},\dots,\sigma_{k-1}$,
the second factor is generated by $\sigma_{k+1},\dots,\sigma_{2k-1}$. So we
have an element $\Phi(\textbf{c})\otimes\Phi(\textbf{c})$ in $H_{2k}(q)$. Then
the second step of the fusion procedure, directly inside the Hecke algebra is
that:

$[\sigma_{1}^{(\textbf{c})}(u),\Phi(\textbf{c})\otimes\Phi(\textbf{c})]=0\ \ \
\ \ \text{in $H_{2k}(q)$.}$ (35)

This formula can be obtained directly from a straightforward calculation using
repeatedly the Yang–Baxter equation for $\sigma_{i}(u)$.

To conclude and summarize the description of the fusion procedure inside the
algebra:

| Algebra | Representation
---|---|---
1st step |  $\sigma_{i}^{(\textbf{c})}(u)$: satisfies YB in $H_{kn}(q)$ |  $\check{R}_{i}^{(\textbf{c})}(u)$: satisfies YB on $(V^{\otimes k})^{\otimes n}$
2nd step |  $\sigma_{1}^{(\textbf{c})}(u)$ commutes with $\Phi(\textbf{c})\otimes\Phi(\textbf{c})$ |  $\check{R}^{(\textbf{c})}(u)$ commutes with $\check{F}(\textbf{c})\otimes\check{F}(\textbf{c})$
|  | $\Downarrow$
|  |  $\check{R}_{i}^{(\textbf{c})}(u)$ leaves invariant $(W_{\textbf{c}})^{\otimes n}$

##### Interesting subspaces $W_{\textbf{c}}$.

Finally, we are ready to study explicit examples of interesting subspaces
$W_{\textbf{c}}$. To do so, we have to study the function $\Phi(\textbf{c})$
and to find particular specialisations of the parameters c which makes it
remarkable.

First, a tiny bit of combinatorics notations. A pair $(x,y)\in\mathbb{Z}^{2}$
is called a node. The $q$-content of the node $\theta=(x,y)$ is
$\mathrm{c}(\theta)=q^{2(y-x)}$.

We have already defined the notion of a partition $\lambda\vdash k$. The Young
diagram of $\lambda=(\lambda_{1},\dots,\lambda_{l})$ is the set of nodes
$(x,y)$ such that $x\in\\{1,\dots,l\\}$ and $y\in\\{1,\dots,\lambda_{x}\\}$.
The Young diagram of $\lambda$ will be seen as a left-justified array of $l$
rows such that the $j$-th row contains $\lambda_{j}$ nodes for all
$j=1,\dots,l$ (a node will be pictured by an empty box).

A standard Young tableau of shape $\lambda\vdash k$ is an assignment of a
number from $\\{1,\dots,k\\}$ to each node of the Young diagram of $\lambda$,
such that the numbers are strictly ascending along rows and down columns of
the Young diagram. It is represented by filling the nodes of the Young diagram
of $\lambda$ with the numbers from $\\{1,\dots,k\\}$. Here is an example

$\boldsymbol{t}=\begin{array}[]{ccc}\framebox{\scriptsize{$1$}}&\framebox{\scriptsize{$2$}}&\framebox{\scriptsize{$4$}}\\\\[-1.99997pt]
\framebox{\scriptsize{$3$}}&&\end{array}\
:\qquad\mathrm{c}_{1}(\boldsymbol{t})=1\,,\ \ \
\mathrm{c}_{2}(\boldsymbol{t})=q^{2}\,,\ \ \
\mathrm{c}_{3}(\boldsymbol{t})=q^{-2}\,,\ \ \
\mathrm{c}_{4}(\boldsymbol{t})=q^{4}\,.$

The partition is $\lambda=(3,1)$. Next to the Young tableau $\boldsymbol{t}$
is its sequence of $q$-contents, where $\mathrm{c}_{i}(\boldsymbol{t})$ is the
$q$-content of the node of $\boldsymbol{t}$ with number $i$.

Here is the result providing nice specialisations of $\Phi(\textbf{c})$.
Below, $\sigma_{w_{0}}^{-1}$ is the element of the standard basis of the Hecke
algebra corresponding to the longest element $w_{0}$ of the symmetrix group
$\mathfrak{S}_{k}$. Its presence here is harmless to us since, being an
invertible element, in a representation it will not change the subspace
obtained as the image of $\Phi(\textbf{c})$. Classical versions for the
symmetric group can be found in [15, 56, 60].

######  Theorem 5.3 ([34]).

Let $\boldsymbol{t}$ be a standard Young tableau of shape $\lambda\vdash k$.
The element obtained by the following consecutive evaluations

$\Phi(\textbf{c})\sigma_{w_{0}}^{-1}\Bigr{|}_{c_{1}=\mathrm{c}_{1}(\boldsymbol{t})}\Bigr{|}_{c_{2}=\mathrm{c}_{2}(\boldsymbol{t})}\dots\Bigr{|}_{c_{k}=\mathrm{c}_{k}(\boldsymbol{t})}$
(36)

is non-zero and proportional to an idempotent $E_{\boldsymbol{t}}$ of
$H_{k}(q)$.

In fact, we can say more about $E_{\boldsymbol{t}}$. The idempotent
$E_{\boldsymbol{t}}$ is a primitive idempotent corresponding to the
irreducible representation $S_{\lambda}$ of $H_{k}(q)$. That is equivalent to
say that the action of $E_{\boldsymbol{t}}$ in an irreducible representation
of $H_{k}(q)$ is as follows:

$E_{\boldsymbol{t}}(S_{\lambda})\ \text{is a one-dimensional subspace}\quad\ \
\ \text{and}\quad\ \ \ \ E_{\boldsymbol{t}}(S_{\mu})=\\{0\\}\ \ \text{if
$\mu\neq\lambda$}.$

So finally, take the parameters c as the sequence of contents of a standard
Young tableau of some shape $\lambda\vdash k$. Recall from the Schur–Weyl
duality that the tensor product $V^{\otimes k}$ decomposes as follows:

$V^{\otimes k}=\bigoplus_{\begin{subarray}{c}\mu\vdash k\\\\[1.39998pt]
\ell(\mu)\leq N\end{subarray}}L_{\mu}^{N}\otimes S_{\mu}\,,$

as a representation of $U_{q}(sl_{N})\otimes H_{k}(q)$. To determine the
subspace $W_{\textbf{c}}$, we need to calcuate the image in this
representation of $\Phi(\textbf{c})$, which is the same as the image of the
idempotent $E_{\boldsymbol{t}}$. From what we have said above, we can
conclude:

$E_{\boldsymbol{t}}(V^{\otimes k})=\bigoplus_{\begin{subarray}{c}\mu\vdash
k\\\\[1.39998pt] \ell(\mu)\leq N\end{subarray}}L_{\mu}^{N}\otimes
E_{\boldsymbol{t}}(S_{\mu})=L_{\lambda}^{N}\ .$

######  Conclusion 6.

If the parameters c correspond to a standard Young tableau of shape
$\lambda\vdash k$, then the subspace $W_{\textbf{c}}$ coincides with the
irreducible representation $L^{N}_{\lambda}$ of $U_{q}(sl_{N})$.

$\leadsto$ The fused solution $\check{R}^{fus}(u)$ is thus a solution on
$L_{\lambda}^{N}\otimes L_{\lambda}^{N}$.

##### What is the algebra for $\check{R}^{fus}(u)$?

We still have to understand what could be the algebra controlling the fused
solution $\check{R}^{fus}(u)$ on $L_{\lambda}^{N}\otimes L_{\lambda}^{N}$. So
we need to find an algebra with a solution of the YB equation and a
representation on $(L_{\lambda}^{N})^{\otimes n}$ which recovers