# Virtual and universal braid groups, their quotients and representations

V. Bardakov, I. Emel’yanenkov, M. Ivanov, T. Kozlovskaya, T. Nasybullov, A.
Vesnin

###### Abstract.

In the present paper we study structural aspects of certain quotients of braid
groups and virtual braid groups. In particular, we construct and study linear
representations $B_{n}\to{\rm GL}_{n(n-1)/2}\left(\mathbb{Z}[t^{\pm
1}]\right)$, $VB_{n}\to{\rm GL}_{n(n-1)/2}\left(\mathbb{Z}[t^{\pm
1},t_{1}^{\pm 1},t_{2}^{\pm 1},\ldots,t_{n-1}^{\pm 1}]\right)$ which are
connected with the famous Lawrence-Bigelow-Krammer representation. It turns
out that these representations are faithful representations of
crystallographic groups $B_{n}/P_{n}^{\prime}$, $VB_{n}/VP_{n}^{\prime}$,
respectively. Using these representations we study certain properties of the
groups $B_{n}/P_{n}^{\prime}$, $VB_{n}/VP_{n}^{\prime}$. Moreover, we
construct new representations and decompositions of universal braid groups
$UB_{n}$.

Keywords: virtual braid, universal braid, crystallographic group,
representations by automorphisms, linear group.

Mathematics Subject Classification 2010: 20F36, 20F29, 57M27, 57M25.

###### Contents

  1. 1 Preliminaries
    1. 1.1 Braid groups and conjugating automorphism groups
    2. 1.2 The singular braid monoid
    3. 1.3 The virtual braid group and the welded braid group
    4. 1.4 Braid groups and crystallographic groups
  2. 2 Linear representation of $B_{n}/P_{n}^{\prime}$
  3. 3 Linear representation of $VB_{n}/VP_{n}^{\prime}$
  4. 4 Decomposition and linearity of $VB_{3}$
  5. 5 Universal braid group
    1. 5.1 Representations by automorphisms of a free group
    2. 5.2 Linear representations
    3. 5.3 Universal pure braid group

## Introduction

Braid groups were introduced by E. Artin as a tool for working with classical
knots and links. The famous Alexander theorem says that every classical link
is equivalent to the closure of some braid, and the well-known Markov theorem
describes braids which have equivalent closures.

In 1999 L. Kauffman introduced the virtual knot theory which generalizes the
classical knot theory [33]. Virtual braid groups were introduced in the same
paper as a tool for working with virtual knots and links. The analogues of the
Alexander and the Markov theorems for virtual braids and links were formulated
and proved by S. Kamada in [31]. Another form of the Markov theorem for
virtual braids and links was introduced by Kauffman and Lambropoulou in [35].

Welded braid groups, unrestricted virtual braid groups, flat virtual braid
groups, flat welded braid groups, Gauss virtual braid groups are quotients of
virtual braid groups. These groups have various applications in algebra and
topology, in particular, each of these groups corresponds to some knot theory,
which has a geometric interpretation that is often very useful when studying
applications. Welded and unrestricted virtual braid groups are used for
working with welded links [25] and fused links [2, 41], respectively. Flat
virtual braid groups were introduced in [34] for working with flat virtual
knots and links, which were introduced in [33], and which model curves on
surfaces. The notion of flat virtual links is the same to the notion of
virtual strings introduced by V. Turaev in [48]. In the papers [30, 36]
invariants for flat virtual knots and links were used for constructing
invariants for virtual knots and links. Gauss virtual braid groups are used
for studying Gauss virtual links, which are called “homotopy classes of Gauss
words” by Turaev in [49], and called “free knots” by Manturov in [38]. All
together classical braid groups, virtual braid groups, welded braid groups,
unrestricted virtual braid groups, flat virtual braid groups, flat welded
braid groups and Gauss virtual braid groups are called braid-like groups.
Universal braid groups were introduced in [4]. The name “universal” in this
paper is motivated by the fact that universal braid groups contain classical
braid groups, and moreover these groups have singular braid groups and all
braid-like groups as quotients. Hence the structure of these groups play an
important role in the study of all braid-like groups.

One way of studying braid-like groups is to study representations of these
groups by automorphisms of different algebraic systems. Using the
representations one can prove that various algorithmic problems (the word
problem, the conjugacy problem etc) can be solved in these groups. Also using
representations of braid-like groups one can construct invariants for the
corresponding knot theories [12, 13, 14].

One of the first representations of braid-like groups is the Artin’s
representation $\varphi:B_{n}\to{\rm Aut}(F_{n})$ of the braid group $B_{n}$
by automorphisms of the free group $F_{n}$ with $n$ generators. The Artin’s
representation is faithful. During the years a lot of representations of
braid-like groups were investigated. For example, there are the Silver-
Williams representation $\varphi_{SW}:VB_{n}\to{\rm
Aut}(F_{n}*\mathbb{Z}^{n+1})$ (see [46]), the Boden-Dies-Gaudreau-Gerlings-
Harper-Nicas representation $\varphi_{BD}:VB_{n}\to{\rm
Aut}(F_{n}*\mathbb{Z}^{2})$ (see [21]), the Kamada representation
$\varphi_{K}:VB_{n}\to{\rm Aut}(F_{n}*\mathbb{Z}^{2n})$ (see [15]), the
representations $\varphi_{M}:VB_{n}\to{\rm Aut}(F_{n}*\mathbb{Z}^{2n+1})$,
$\tilde{\varphi}_{M}:VB_{n}\to{\rm Aut}(F_{n}*\mathbb{Z}^{n})$ of Bardakov-
Mikhalchishina-Neshchadim (see [10, 11]). In the paper [8] there are
representations of the flat virtual braid groups by automorphisms of the free
group, which do not preserve forbidden relations.

The study of linear representations of braid groups was started when Burau
constructed his famous representation $\varphi_{B}:B_{n}\to{\rm
GL}_{n}(\mathbb{Z}[t,t^{-1}])$ (see, for example, [20, Section 3]). For
$n=2,3$ the Burau representation is known to be faithful, for $n\geq 5$ the
Burau representation is known to have non-trivial kernel [18], for $n=4$ the
question on faithfulness of the Burau representation is still open (see some
positive results, for example, in [17]). The Burau representation was
generalized in 1961 by Gassner [27] to a related representation of the pure
braid group $\varphi_{G}:P_{n}\to{\rm GL}_{n}(\mathbb{Z}[t_{1}^{\pm
1},t_{2}^{\pm 1},\dots,t_{n}^{\pm 1}])$. It is not known whether the Gassner
representation is faithful for $n>3$. Some new results about representations
of virtual braid groups can be found, for example, in [12, 14]. Certain linear
representations of flat virtual braid groups and Gauss virtual braid groups
are studied in [8].

Another way to study braid-like groups is to explore normal forms of elements
in braid-like groups which usually (but not always) arise when a braid-like
group can be written as a product (for example, direct product, semiderict
product, free product etc). There are several normal forms of elements in
classical braid groups which can be found, for example, in [32]. Attempts to
construct a normal form for elements of the virtual braid groups were made in
the paper [4]. Paper [7] contains results about the normal form of elements in
unrestricted virtual braid groups and flat welded braid groups.

In the present paper we study structural aspects of certain quotients of braid
groups and virtual braid groups using both representations and decompositions.
In particular, we construct and study linear representations

$\displaystyle B_{n}\to{\rm GL}_{n(n-1)/2}\left(\mathbb{Z}[t^{\pm 1}]\right),$
$\displaystyle VB_{n}\to{\rm GL}_{n(n-1)/2}\left(\mathbb{Z}[t^{\pm
1},t_{1}^{\pm 1},t_{2}^{\pm 1},\ldots,t_{n-1}^{\pm 1}]\right)$

which are connected with the famous Lawrence-Bigelow-Krammer representation.
It turns out that these representations are faithful representations of
crystallographic groups $B_{n}/P_{n}^{\prime}$, $VB_{n}/VP_{n}^{\prime}$,
respectively. Using these representations we study certain properties of the
groups $B_{n}/P_{n}^{\prime}$, $VB_{n}/VP_{n}^{\prime}$. Moreover, we
construct new representations and decompositioms of universal braid groups.

The paper is organized as follows. In Section 1 we recall definitions and
basic properties of braid groups, virtual braid groups, welded braid groups
and pure subgroups of these groups. In Section 2 we construct a faithful
linear representation of the crystallographic group $B_{n}/P_{n}^{\prime}$
that is the quotient of the braid group $B_{n}$ by the commutator subgroup
$P_{n}^{\prime}=[P_{n},P_{n}]$ of the pure braid group $P_{n}$ (Theorem 2.1).
The torsion in this group is studied in [28]. Using the constructed
representation we describe all elements of order three in
$B_{3}/P_{3}^{\prime}$ (Proposition 2.1). Note that this group has only three-
torsion. In Section 3 we construct a faithful linear representation of another
crystallographic group $VB_{n}/VP_{n}^{\prime}$ which contains
$B_{n}/P_{n}^{\prime}$ (Theorem 3.1). In Proposition 3.2 we demonstrate that
$VB_{3}/VP_{3}^{\prime}$ contains infinitely many elements of order two. Note
that $B_{n}/P_{n}^{\prime}$ has no two-torsion. In Section 4 we show that
$VP_{3}$ can be decomposed into the free product of infinite cyclic group and
group that is a semi-direct product of free group of rank three and free group
of rank two (Proposition 4.1). In Theorem 4.1 we prove that $VB_{3}$ is
linear. In Section 5 we discuss the universal braid group $UB_{n}$ introduced
in [4], and its pure subgroup $UP_{n}$. In particular we construct several
representations of $UP_{n}$ by automorphisms of free groups which act
faithfully on a subgroup $T_{n}$ of $UP_{n}$. Also in Theorem 5.2 we
demonstrate that the groups $UP_{2}$ and $UP_{3}$ admit decompositions into
free products.

##  1\. Preliminaries

In this section we give necessary preliminaries. We use classical notations.
If $G$ is a group, and $a,b\in G$, then we denote by $a^{b}=b^{-1}ab$ the
conjugate of $a$ by $b$, and by $[a,b]=a^{-1}b^{-1}ab$ the commutator of the
elements $a,b$. If $X$ is a set, then we denote by ${\rm Sym}(X)$ the set of
all bijections from $X$ to $X$. All actions are supposed to be right, i. e. if
$f,g\in{\rm Sym}(X)$, then for $x\in X$ we denote $fg(x)=g(f(x))$.

###  1.1. Braid groups and conjugating automorphism groups

For $n\geq 2$ the braid group $B_{n}$ on $n$ strands is defined as the group
generated by $(n-1)$ elements $\sigma_{1},\sigma_{2},\ldots,\sigma_{n-1}$ with
defining relations

(1) $\displaystyle\sigma_{i}\sigma_{i+1}\sigma_{i}$
$\displaystyle=\sigma_{i+1}\sigma_{i}\sigma_{i+1},$ $\displaystyle
i=1,2,\ldots,n-2,$ (2) $\displaystyle\sigma_{i}\sigma_{j}$
$\displaystyle=\sigma_{j}\sigma_{i},$ $\displaystyle|i-j|\geq 2.$

Relations (1) and (2) are called _braid relations_. There is a homomorphism
$\varphi$ from $B_{n}$ to the symmetric group $S_{n}$ on $n$ symbols
$\\{1,2,\ldots,n\\}$, which maps the generator $\sigma_{i}$ to the
transposition $(i,i+1)$ for $i=1,2,\ldots,n-1$. The kernel
$\operatorname{Ker}(\varphi)$ is called the _pure braid group_ on $n$ strands
and is denoted by $P_{n}$. From the definition it is clear that we have the
following short exact sequence:

$1\to P_{n}\to B_{n}\stackrel{{\scriptstyle\varphi}}{{\to}}S_{n}\to 1.$

It is known that the pure braid group $P_{n}$ can be generated by the elements
$a_{i,j}$ for $1\leq i<j\leq n$, where

$\displaystyle a_{i,i+1}$ $\displaystyle=\sigma_{i}^{2},$ $\displaystyle
i=1,2,\dots,n-1,$ $\displaystyle a_{i,j}$
$\displaystyle=\sigma_{j-1}\,\sigma_{j-2}\ldots\sigma_{i+1}\,\sigma_{i}^{2}\,\sigma_{i+1}^{-1}\ldots\sigma_{j-2}^{-1}\,\sigma_{j-1}^{-1},$
$\displaystyle i+1<j\leq n.$

Also it is know that the pure braid group $P_{n}$ can be decomposed into a
semi-direct product $P_{n}=U_{n}\rtimes P_{n-1}$, where $U_{n}$ is the free
group with free generators $a_{1,n},a_{2,n},\ldots,a_{n-1,n}$, and $P_{n-1}$
is the pure braid group on $(n-1)$ strands. Here $U_{n}$ is a normal subgroup
of $P_{n}$. Similar, $P_{n-1}$ is a semi-direct product of the free group
$U_{n-1}$ with the free generators $a_{1,n-1},a_{2,n-1}\ldots,a_{n-2,n-1}$ and
$P_{n-2}$, and so on. Therefore $P_{n}$ has the following decomposition into
the iterated semi-direct product (see [39]):

$P_{n}=U_{n}\rtimes(U_{n-1}\rtimes(\ldots\rtimes(U_{3}\rtimes U_{2}))\ldots),$

where $U_{i}\simeq F_{i-1}$ is the free group on $(i-1)$ generators for
$i=2,3,\ldots,n$.

The group $B_{n}$ has a faithful representation by automorphisms of the free
group $F_{n}=\langle x_{1},x_{2},\ldots,x_{n}\rangle$, where the generator
$\sigma_{i}$, $i=1,2,\ldots,n-1$, defines the automorphism (which we denote by
the same symbol $\sigma_{i}\in\operatorname{Aut}(F_{n})$) which acts on the
generators $x_{1},x_{2},\dots,x_{n}$ of $F_{n}$ by the following rule

$\sigma_{i}:\begin{cases}x_{i}\mapsto x_{i}x_{i+1}x_{i}^{-1},&\\\
x_{i+1}\mapsto x_{i},&\\\ x_{l}\mapsto x_{l},&l\neq i,i+1.\end{cases}$

By the Artin’s theorem [20, Theorem 1.9] an automorphism $\beta$ from
$\operatorname{Aut}(F_{n})$ belongs to $B_{n}<\operatorname{Aut}(F_{n})$ if
and only if $\beta$ satisfies the following two conditions:

  1. 1)

$\beta(x_{i})=a_{i}^{-1}\,x_{\pi(i)}\,a_{i}$ for $1\leq i\leq n$,

  2. 2)

$\beta(x_{1}x_{2}\cdots x_{n})=x_{1}x_{2}\cdots x_{n}$,

where $\pi$ is a permutation from $S_{n}$ and $a_{i}\in F_{n}$ for
$i=1,2,\dots,n$.

The automorphism $\beta$ of the free group $F_{n}$ is called a _conjugating
automorphism_ (or permutation-conjugating by terminology from [25]) if it
satisfies condition 1) from the list above. It is clear that every
automorphism from $B_{n}<{\rm Aut}(F_{n})$ is a conjugating automorphism. The
group of all conjugating automorphisms of $F_{n}$ is denoted by $C_{n}$. It is
known that this group is generated by the automorphisms $\sigma_{i}$ for
$i=1,2,\ldots,n-1$ and automorphisms $\alpha_{i}$ for $i=1,2,\ldots,$ $n-1$,
where $\alpha_{i}$ acts on the generators $x_{1},x_{2},\dots,x_{n}$ of $F_{n}$
by the following rule

$\alpha_{i}:\begin{cases}x_{i}\mapsto x_{i+1},&\\\ x_{i+1}\mapsto x_{i},&\\\
x_{l}\mapsto x_{l},&l\neq i,i+1.\end{cases}$

Generators $\sigma_{1},\sigma_{2},\ldots,\sigma_{n-1}$ of $C_{n}$ satisfy the
braid relations (1) and (2) of $B_{n}$. Also it is easy to see that the
automorphisms $\alpha_{1},\alpha_{2},\dots,\alpha_{n-1}$ of $C_{n}$ generate
the symmetric group $S_{n}$ in $C_{n}$ which has the following defining
relations

(3) $\displaystyle\alpha_{i}\alpha_{i+1}\alpha_{i}$
$\displaystyle=\alpha_{i+1}\alpha_{i}\alpha_{i+1},$ $\displaystyle
i=1,2,\ldots,n-2,$ (4) $\displaystyle\alpha_{i}\alpha_{j}$
$\displaystyle=\alpha_{j}\alpha_{i},$ $\displaystyle|i-j|\geq 2,$ (5)
$\displaystyle\alpha_{i}^{2}$ $\displaystyle=1,$ $\displaystyle
i=1,2,\ldots,n-1.$

Moreover, the generators
$\sigma_{1},\sigma_{2},\ldots,\sigma_{n-1},\alpha_{1},\alpha_{2},\dots,\alpha_{n-1}$
of $C_{n}$ satisfy the following mixed relations (see [25, 44]):

(6) $\displaystyle\alpha_{i}\sigma_{j}$ $\displaystyle=\sigma_{j}\alpha_{i},$
$\displaystyle|i-j|\geq 2,$ (7)
$\displaystyle\sigma_{i}\alpha_{i+1}\alpha_{i}$
$\displaystyle=\alpha_{i+1}\alpha_{i}\sigma_{i+1},$ $\displaystyle
i=1,2,\ldots,n-2,$ (8) $\displaystyle\sigma_{i+1}\sigma_{i}\alpha_{i+1}$
$\displaystyle=\alpha_{i}\sigma_{i+1}\sigma_{i},$ $\displaystyle
i=1,2,\ldots,n-2.$

Summarizing, the group $C_{n}$ has generators
$\sigma_{1},\sigma_{2},\ldots,\sigma_{n-1},\alpha_{1},\alpha_{2},\dots,\alpha_{n-1}$
with the defining relations (1)-(8).

For $1\leq i\neq j\leq n$ denote by $\varepsilon_{i,j}$ the automorphism of
$F_{n}$ which acts on the generators of $F_{n}$ by the rule

$\varepsilon_{i,j}:\begin{cases}x_{i}\mapsto x_{j}^{-1}x_{i}x_{j},&i\neq j,\\\
x_{l}\mapsto x_{l},&l\neq i.\end{cases}$

The subgroup of ${\rm Aut}(F_{n})$ generated by $\varepsilon_{i,j}$ for $1\leq
i\neq j\leq n$ is called the _group of basis-conjugating automorphisms_ and is
denoted by $Cb_{n}$. It is clear that every generator $\varepsilon_{i,j}$
belongs to $C_{n}$, therefore the group $Cb_{n}$ of basis-conjugating
automorphisms is a subgroup of the group $C_{n}$ of conjugating automorphisms.
In [40] it is proved that $Cb_{n}$ has the following defining relations:

(9) $\displaystyle\varepsilon_{i,j}\varepsilon_{k,l}$
$\displaystyle=\varepsilon_{k,l}\varepsilon_{i,j},$ (10)
$\displaystyle\varepsilon_{i,j}\varepsilon_{k,j}$
$\displaystyle=\varepsilon_{k,j}\varepsilon_{i,j},$ (11)
$\displaystyle(\varepsilon_{i,j}\varepsilon_{k,j})\varepsilon_{i,k}$
$\displaystyle=\varepsilon_{i,k}(\varepsilon_{i,j}\varepsilon_{k,j}),$

where different letters denote different indices.

It is easy to see that the group $C_{n}$ of conjugating automorphisms can be
decomposed into a semi-direct product $C_{n}=Cb_{n}\rtimes S_{n}$, where the
permutation group $S_{n}$ is generated by the automorphisms
$\alpha_{1},\alpha_{2},\ldots,\alpha_{n-1}$. In [44] it is prove that the
following relations hold:

$\displaystyle\varepsilon_{i,i+1}$ $\displaystyle=\alpha_{i}\sigma^{-1}_{i},$
$\displaystyle\varepsilon_{i+1,i}$ $\displaystyle=\sigma^{-1}_{i}\alpha_{i},$
$\displaystyle\varepsilon_{i,j}$
$\displaystyle=\alpha_{j-1}\alpha_{j-2}\ldots\alpha_{i+1}\varepsilon_{i,i+1}\alpha_{i+1}\ldots\alpha_{j-2}\alpha_{j-1},$
$\displaystyle\text{if}~{}i<j,$ $\displaystyle\varepsilon_{j,i}$
$\displaystyle=\alpha_{j-1}\alpha_{j-2}\ldots\alpha_{i+1}\alpha_{i}\varepsilon_{i,i+1}\alpha_{i}\alpha_{i+1}\ldots\alpha_{j-2}\alpha_{j-1},$
$\displaystyle\text{if}~{}i<j.$

The structure of $Cb_{n}$ was studied in [4, 5], where it was shown that
$Cb_{n}$, $n\geq 2$, has a decomposition into an iterated semi-direct product

$Cb_{n}=D_{n-1}\rtimes(D_{n-2}\rtimes(\ldots.\rtimes(D_{2}\rtimes
D_{1}))\ldots),$

where $D_{i}$ denotes a subgroup of $Cb_{n}$ generated by the elements

$\\{\varepsilon_{i+1,1},\varepsilon_{i+1,2},\ldots,\varepsilon_{i+1,i},\varepsilon_{1,i+1},\varepsilon_{2,i+1},\ldots\varepsilon_{i,i+1}\\},$

where $i=1,2,\ldots,n-1,$ The elements
$\\{\varepsilon_{i+1,1},\varepsilon_{i+1,2},\ldots,\varepsilon_{i+1,i}\\}$
generate the free group of rank $i$ in $D_{i}$, while the elements
$\\{\varepsilon_{1,i+1},\varepsilon_{2,i+1},\ldots,\varepsilon_{i,i+1}\\}$
generate the free abelian group of rank $i$ in $D_{i}$.

The pure braid group $P_{n}$ is a subgroup of $Cb_{n}$. The generators
$a_{i,j}$ (for $1\leq i<j\leq n$) of $P_{n}$ can be written in $Cb_{n}$ in the
following form

$\displaystyle a_{i,i+1}$
$\displaystyle=\varepsilon_{i,i+1}^{-1}\varepsilon_{i+1,i}^{-1}$

for $i=1,2,\ldots,n-1$, and

$\displaystyle a_{i,j}$
$\displaystyle=\varepsilon_{j-1,i}\varepsilon_{j-2,i}\ldots\varepsilon_{i+1,i}(\varepsilon_{i,j}^{-1}\varepsilon_{j,i}^{-1})\varepsilon_{i+1,i}^{-1}\ldots\varepsilon_{j-2,i}^{-1}\varepsilon_{j-1,i}^{-1}$
$\displaystyle=\varepsilon_{j-1,j}^{-1}\varepsilon_{j-2,j}^{-1}\ldots\varepsilon_{i+1,j}^{-1}(\varepsilon_{i,j}^{-1}\varepsilon_{j,i}^{-1})\varepsilon_{i+1,j}\ldots\varepsilon_{j-2,j}\varepsilon_{j-1,j},$

for $2\leq i+1<j\leq n$.

###  1.2. The singular braid monoid

The Baez-Birman monoid [3, 19] or the singular braid monoid $SB_{n}$ is
defined as the monoind with the generators
$\sigma_{1},\sigma_{2},\dots,\sigma_{n-1}$,
$\sigma_{1}^{-1},\sigma_{2}^{-1},\dots,\sigma_{n-1}^{-1}$,
$\tau_{1},\tau_{2},\dots,\tau_{n-1}$, and definig relations (1), (2) and
addinitonal relations

(12) $\displaystyle\sigma_{i}\,\sigma_{i}^{-1}$
$\displaystyle=\sigma_{i}^{-1}\sigma_{i}=1,$ $\displaystyle i=1,2,\dots,n-1,$
(13) $\displaystyle\tau_{i}\,\tau_{j}$ $\displaystyle=\tau_{j}\,\tau_{i},$
$\displaystyle|i-j|\geq 1,$ (14) $\displaystyle\tau_{i}\,\sigma_{j}$
$\displaystyle=\sigma_{j}\,\tau_{i},$ $\displaystyle|i-j|\geq 1,$ (15)
$\displaystyle\tau_{i}\,\sigma_{i}$ $\displaystyle=\sigma_{i}\,\tau_{i},$
$\displaystyle i=1,2,\ldots,n-1,$ (16)
$\displaystyle\sigma_{i}\,\sigma_{i+1}\,\tau_{i}$
$\displaystyle=\tau_{i+1}\,\sigma_{i}\,\sigma_{i+1},$ $\displaystyle
i=1,2,\ldots,n-2,$ (17) $\displaystyle\sigma_{i+1}\,\sigma_{i}\,\tau_{i+1}$
$\displaystyle=\tau_{i}\,\sigma_{i+1}\,\sigma_{i},$ $\displaystyle
i=1,2,\ldots,n-2.$

In the work [26] it was proved that the singular braid monoid $SB_{n}$ is
embedded into the group $SG_{n}$, which is called the singular braid group and
has the same defining relations as $SB_{n}$.

###  1.3. The virtual braid group and the welded braid group

The _virtual braid group_ $VB_{n}$ on $n$ strands, introduced in [33], is the
group with the generators $\sigma_{1},\sigma_{2},\dots,\sigma_{n-1}$,
$\rho_{1},\rho_{2},\dots,\rho_{n-1}$, which is defined by the relations
(1)-(2) and additional relations

(18) $\displaystyle\rho_{i}\rho_{i+1}\rho_{i}$
$\displaystyle=\rho_{i+1}\rho_{i}\rho_{i+1},$ $\displaystyle
i=1,2,\ldots,n-2,$ (19) $\displaystyle\rho_{i}\rho_{j}$
$\displaystyle=\rho_{j}\rho_{i},$ $\displaystyle|i-j|\geq 2,$ (20)
$\displaystyle\rho_{i}^{2}$ $\displaystyle=1,$ $\displaystyle
i=1,2,\ldots,n-1,$ (21) $\displaystyle\sigma_{i}\rho_{j}$
$\displaystyle=\rho_{j}\sigma_{i},$ $\displaystyle|i-j|\geq 1,$ (22)
$\displaystyle\rho_{i}\rho_{i+1}\sigma_{i}$
$\displaystyle=\sigma_{i+1}\rho_{i}\rho_{i+1},$ $\displaystyle
i=1,2,\ldots,n-2.$

The elements $\sigma_{1},\sigma_{2},\dots,\sigma_{n-1}$ generate the braid
group $B_{n}$ with the braid relations (1)-(2) in $VB_{n}$. Relations
(18)-(20) are defining relations of the permutation group $S_{n}$, where the
elements $\rho_{1},\rho_{2},\dots,\rho_{n-1}$ generate the permutation group
$S_{n}$ in $VB_{n}$, hence the virtual braid group $VB_{n}$ is generated by
the braid group $B_{n}$ and the permutation group $S_{n}$. Relations (21)-(22)
are called the mixed relations of the virtual braid group. Note that relations
(22) are equivalent to the relations

$\rho_{i+1}\,\rho_{i}\,\sigma_{i+1}=\sigma_{i}\,\rho_{i+1}\,\rho_{i}$

for $i=1,2,\dots,n-1$. It was shown in [29] that the relations

(23)
$\displaystyle\rho_{i}\,\sigma_{i+1}\,\sigma_{i}=\sigma_{i+1}\,\sigma_{i}\,\rho_{i+1},$
$\displaystyle\rho_{i+1}\,\sigma_{i}\,\sigma_{i+1}=\rho_{i}\,\sigma_{i+1}\,\sigma_{i}.$

donot hold in $VB_{n}$. Relations (23) are usually called forbidden relations.

The virtual pure braid group introduced in [4] is the kernel
$VP_{n}=\operatorname{Ker}(\nu)$ of the homomorphism $\nu:VB_{n}\to S_{n}$
from the virtual braid group $VB_{n}$ onto the symmetric group $S_{n}$ given
on generators
$\sigma_{1},\sigma_{2},\dots,\sigma_{n-1},\rho_{1},\rho_{2},\dots,\rho_{n-1}$
of $VB_{n}$ by the following rule

$\nu(\sigma_{i})=\nu(\rho_{i})=\rho_{i},\qquad\text{for}~{}i=1,2,\ldots,n-1.$

From the definition it is clear that the following short exact sequence holds:

$1\to VP_{n}\to VB_{n}\stackrel{{\scriptstyle\nu}}{{\to}}S_{n}\to 1.$

Following [4] for $1\leq i\neq j\leq n$ let us definte the elements
$\lambda_{i,j}$ by the following formulas

$\displaystyle\lambda_{i,i+1}$ $\displaystyle=\rho_{i}\,\sigma_{i}^{-1},$
$\displaystyle\lambda_{i+1,i}$
$\displaystyle=\rho_{i}\,\lambda_{i,i+1}\,\rho_{i}=\sigma_{i}^{-1}\,\rho_{i}$

for $i=1,2,\ldots,n-1$, and

$\displaystyle\lambda_{i,j}$
$\displaystyle=\rho_{j-1}\,\rho_{j-2}\ldots\rho_{i+1}\,\lambda_{i,i+1}\,\rho_{i+1}\ldots\rho_{j-2}\,\rho_{j-1},$
$\displaystyle\lambda_{j,i}$
$\displaystyle=\rho_{j-1}\,\rho_{j-2}\ldots\rho_{i+1}\,\lambda_{i+1,i}\,\rho_{i+1}\ldots\rho_{j-2}\,\rho_{j-1}$

for $1\leq i<j-1\leq n-1$. The following theorem is proved in [4].

######  Theorem 1.1.

The group $VP_{n}$ admits a presentation with the generators $\lambda_{k,l}$
for $1\leq k\neq l\leq n$ and the defining relations

(24) $\displaystyle\lambda_{i,j}\,\lambda_{k,\,l}$
$\displaystyle=\lambda_{k,\,l}\,\lambda_{i,j},$ (25)
$\displaystyle\lambda_{k,i}\,\lambda_{k,j}\,\lambda_{i,j}$
$\displaystyle=\lambda_{i,j}\,\lambda_{k,j}\,\lambda_{k,i},$

where different letters denote different indices.

The _welded braid group_ $WB_{n}$ on $n$ strands introduced in [25] is the
group with the generators $\sigma_{1},\sigma_{2},\dots,\sigma_{n-1}$,
$\alpha_{1},\alpha_{2},\dots,\alpha_{n-1}$, where the elements
$\sigma_{1},\sigma_{2},\dots,\sigma_{n-1}$ generate the braid group $B_{n}$ on
$n$ strands (i. e. satisfy relations (1)-(2)), the elements
$\alpha_{1},\alpha_{2},\dots,\alpha_{n-1}$ generate the symmetric group
$S_{n}$ (i. e. satisfy relations (18)-(20) where we change the letter $\rho$
by the letter $\alpha$), and moreover, the following mixed relations hold:

(26) $\displaystyle\alpha_{i}\,\sigma_{j}$
$\displaystyle=\sigma_{j}\,\alpha_{i},$ $\displaystyle|i-j|\geq 2,$ (27)
$\displaystyle\sigma_{i}\,\alpha_{i+1}\,\alpha_{i}$
$\displaystyle=\alpha_{i+1}\,\alpha_{i}\,\sigma_{i+1},$ $\displaystyle
i=1,2,\ldots,n-2,$ (28) $\displaystyle\sigma_{i+1}\,\sigma_{i}\,\alpha_{i+1}$
$\displaystyle=\alpha_{i}\,\sigma_{i+1}\,\sigma_{i},$ $\displaystyle
i=1,2,\ldots,n-2.$

It was observed in [25] that $WB_{n}$ is isomorphic to the group of
conjugating automorphisms $C_{n}$. It was proved in [25] that the following
relations symmetric to (27)

$\sigma_{i+1}\,\alpha_{i}\,\alpha_{i+1}=\alpha_{i}\,\alpha_{i+1}\,\sigma_{i},$

hold in $WB_{n}$ for $i=1,2,\dots,n-2$, while the relations

$\alpha_{i+1}\,\sigma_{i}\,\sigma_{i+1}=\sigma_{i}\,\sigma_{i+1}\,\alpha_{i}$

do not hold.

Comparing defining relations (21)–(22) of $VB_{n}$ with defining relations
(26)–(28) of $WB_{n}$, one can see that the difference between $WB_{n}$ and
$VB_{n}$ is given by relations (28). Therefore, there exists a homomorphism

$\varphi_{VW}:VB_{n}\longrightarrow WB_{n},$

which maps $\sigma_{i}$ to $\sigma_{i}$, and $\rho_{i}$ to $\alpha_{i}$ for
$i=1,2\dots,n-1$. Hence, $WB_{n}$ is the homomorphic image of $VB_{n}$.

Recall that a group $G$ is said to be linear if there exists a faithful
representation of $G$ into the general linear group ${\rm GL}_{m}(k)$ for some
integer $m\geq 2$ and a field $k$. It was shown in [18, 37] that the braid
group $B_{n}$ is linear for every $n\geq 2$. In [50] there were constructed
linear representations of the virtual braid groups $VB_{n}$ and the welded
braid groups $WB_{n}$ by matrices from $\mbox{GL}_{n}(\mathbb{Z}[t,t^{-1}])$.
These representations extend the well-known Burau representation. In [5] there
was constructed a linear representation of the group $C_{n}\simeq WB_{n}$
which is an extension (with additional conditions on parameters) of the well-
known Lawrence-Bigelow-Krammer representation.

###  1.4. Braid groups and crystallographic groups

An interesting connection between braid groups and crystallographic groups was
observed in [28]. Recall that a discrete and uniform subgroup $G$ of
$\mathbb{R}^{n}\rtimes
O_{n}(\mathbb{R})\subseteq\operatorname{Aff}(\mathbb{R}^{n})$ is said to be a
crystallographic group of dimension $n$. If in addition $G$ is torsion free,
then $G$ is called a Bieberbach group of dimension $n$. The correspondence
between Bieberbach groups and fundamental groups of closed flat Riemannian
manifold is presented in [51].

The following characterization of crystallographic groups was used in [28,
Lemma 8] in order to demonstrate the relation between braid groups and
crystallographic groups.

######  Lemma 1.1.

Let $G$ be a group. Then $G$ is a crystallographic group if and only if there
exist an integer $n\in\mathbb{N}$ and a short exact sequence

$0\to\mathbb{Z}^{n}\to G\to\Phi\to 1$

such that the following conditions hold:

  1. (a)

$\Phi$ is a finite group,

  2. (b)

the integral representation
$\theta:\Phi\to\operatorname{Aut}(\mathbb{Z}^{n})$, induced by conjugation on
$\mathbb{Z}^{n}$ and defined by $\theta(\varphi)(x)=\pi x\pi^{-1}$, where
$x\in\mathbb{Z}^{n}$, $\varphi\in\Phi$ and $\pi\in G$ is such that
$\zeta(\pi)=\varphi$, is faithful.

If $G$ is a crystallographic group, then the integer $n$ from Lemma 1.1 is
called the dimension of $G$, the finite group $\Phi$ is called the holonomy
group of $G$, and the integral representation
$\theta:\Phi\to\operatorname{Aut}(\mathbb{Z}^{n})$ is called the holonomy
representation of $G$.

Let $P^{\prime}_{n}=[P_{n},P_{n}]$ be the commutator subgroup of the pure
braid group $P_{n}$. It was shown in [28] that the quotient
$B_{n}/P_{n}^{\prime}$ is a crystallographic group for all $n\geq 3$.
Moreover, it was proved that $B_{n}/P_{n}^{\prime}$ possesses torsion, and
there is a one-to-one correspondence between the conjugacy classes of the
finite-order elements of $B_{n}/P_{n}^{\prime}$ and the conjugacy classes of
the elements of odd order of the symmetric group $S_{n}$.

The groups $B_{n}/P_{n}^{\prime}$ also arise in the study of pseudo-symmetric
braided categories [42]. In this paper the authors consider the quotient,
denoted by $PS_{n}$, of $B_{n}$ by the normal subgroup generated by the
relations

$s_{i}s^{-1}_{i+1}s_{i}=s_{i+1}s^{-1}_{i}s_{i+1},\qquad i=1,\ldots,n-2,$

and show that $PS_{n}$ is isomorphic to $B_{n}/P_{n}^{\prime}$.

##  2\. Linear representation of $B_{n}/P_{n}^{\prime}$

Let $R=\mathbb{Z}[q^{\pm 1},t^{\pm 1}]$ be the ring of Laurent polynomials
with two variables $q,t$, and $V$ be the free $R$-module with the basis
$e_{i,j}$ for $1\leq i<j\leq n$. The Lawrence-Bigelow-Krammer representation
(shortly LBKR) is the linear representation of the braid group $B_{n}$ by the
automorphisms of $V$

$\psi=\psi_{q,t}:B_{n}\to\operatorname{Aut}(V)$

which is given on the generators $\sigma_{1},\sigma_{2},\dots,\sigma_{n-1}$ of
$B_{n}$ by the following rule

(29) $\psi(\sigma_{k})(e_{i,j})=\begin{cases}e_{i,j},&k<i-1\text{ or }k>j,\\\
e_{i-1,j}+(1-q)e_{i,j},&k=i-1,\\\ tq(q-1)e_{i,i+1}+qe_{i+1,j},&k=i<j-1,\\\
tq^{2}e_{i,j},&k=i=j-1,\\\ e_{i,j}+tq^{k-i}(q-1)^{2}e_{k,k+1},&i<k<j-1,\\\
e_{i,j-1}+tq^{j-i}(q-1)e_{j-1,i},&i<k=j-1,\\\ (1-q)e_{i,j}+qe_{i,j+1},&k=j.\\\
\end{cases}$

######  Example 2.1.

If $n=3$, then the braid group $B_{3}$ is generated by
$\sigma_{1},\sigma_{2}$, and the module $V$ has the basis
$\\{e_{1,2},e_{1,3},e_{2,3}\\}$. From formula (29) it follows that the linear
maps $\psi(\sigma_{1})$, $\psi(\sigma_{2})$, $\psi(\sigma_{1})^{-1}$,
$\psi(\sigma_{2})^{-1}$ can be presented by the following matrices, which are
written in the basis $e_{1,2},e_{1,3},e_{2,3}$ of $V$

$\displaystyle[\psi(\sigma_{1})]$ $\displaystyle=\begin{pmatrix}tq^{2}&0&0\\\
tq(q-1)&0&q\\\ 0&1&1-q\end{pmatrix},$ $\displaystyle[\psi(\sigma_{2})]$
$\displaystyle=\begin{pmatrix}1-q&q&0\\\ 1&0&tq^{2}(q-1)\\\
0&0&tq^{2}\end{pmatrix},$ $\displaystyle[\psi(\sigma_{1})]^{-1}$
$\displaystyle=\begin{pmatrix}t^{-1}q^{-2}&0&0\\\
-1+2q^{-1}-q^{-2}&1-q^{-1}&1\\\ -q^{-1}+q^{-2}&q^{-1}&0\end{pmatrix},$
$\displaystyle[\psi(\sigma_{2})]^{-1}$
$\displaystyle=\begin{pmatrix}0&1&-q+1\\\ q^{-1}&1-q^{-1}&-q+2-q^{-1}\\\
0&0&t^{-1}q^{-2}\end{pmatrix}.$

For two complex numbers $q_{0},t_{0}\in\mathbb{C}$ denote by $V_{q_{0},t_{0}}$
the free module over the ring $\mathbb{Z}[q_{0},t_{0}]$ with the same basis as
$V$, and denote by $\psi_{q_{0},t_{0}}$ the specialization of $\psi_{q,t}$
obtained by putting $q=q_{0}$, $t=t_{0}$. The above specialization gives the
linear representation
$\psi_{q_{0},t_{0}}:B_{n}\to\operatorname{Aut}(V_{q_{0},t_{0}})$. In
particular, the map $\psi_{1,t}:B_{n}\to\operatorname{Aut}(V_{1,t})$ is
defined on the generators $\sigma_{1},\sigma_{2},\dots,\sigma_{n-1}$ of
$B_{n}$ by the following rules

$\psi_{1,t}(\sigma_{k}):\begin{cases}e_{k,k+1}\mapsto te_{k,k+1},&\\\
e_{i,j}\mapsto
e_{s_{k}(i),s_{k}(j)},&\text{if}~{}(i,j)\not=(k,k+1)~{}\text{and}~{}1\leq
i<j\leq n,\\\ \end{cases}$

where $s_{k}=(k,k+1)$ is the transposition from $S_{n}$. In this formula we
assume that $e_{j,i}=e_{i,j}$ if $i<j$.

######  Lemma 2.1.

If $q=1$ and $t=\pm 1$, then the image of $B_{n}$ under the specialization
$\psi_{1,\pm 1}$ of LBKR is isomorphic to the symmetric group $S_{n}$.

###### Proof.

It is enough to observe that $\left(\psi_{1,\pm 1}(\sigma_{i})\right)^{2}$ is
the identity automorphism. This property immidiately follows from the
definition of $\psi_{1,\pm 1}$. ∎

From the expressions of the generators $a_{i,j}~{}(1\leq i<j\leq n)$ of
$P_{n}$ introduced in Section 1.1 and direct calculations we have the
following result.

######  Lemma 2.2.

The representation $\psi_{1,t}:B_{n}\to{\rm Aut}(V_{1,t})$ acts on the
generators $a_{i,j}~{}(1\leq i<j\leq n)$ of $P_{n}$ by the following rules

$\psi_{1,t}(a_{ij}):\begin{cases}e_{i,j}\mapsto t^{2}e_{i,j},&\\\
e_{k,l}\mapsto e_{k,l},&~{}\text{for}~{}(i,j)\not=(k,l).\\\ \end{cases}$

The following theorem is the main results of the present section.

######  Theorem 2.1.

The image of $B_{n}$ under $\psi_{1,t}$ is isomorphic to
$B_{n}/P_{n}^{\prime}$.

###### Proof.

In Section 1.1 we have noticed that the following short exact sequence holds

$1\to P_{n}\to B_{n}\to S_{n}\stackrel{{\scriptstyle\pi}}{{\to}}1.$

From this short exact sequence and the isomorphism

$(B_{n}/P_{n}^{\prime})/(P_{n}/P_{n}^{\prime})=B_{n}/P_{n}$

we have the following short exact sequence

$1\to P_{n}/P_{n}^{\prime}\to
B_{n}/P_{n}^{\prime}\stackrel{{\scriptstyle\pi}}{{\to}}S_{n}\to 1.$

From this short exact sequence it follows that every element $b\in
B_{n}/P_{n}^{\prime}$ can be expressend in the form $b=p\lambda$, where $p\in
P_{n}/P_{n}^{\prime}$ and $\lambda$ is a coset representative of
$P_{n}/P_{n}^{\prime}$ in $B_{n}/P_{n}^{\prime}$. In order to prove the
statement of the theorem we have to prove that if $b$ is non-trivial, then
$\psi_{1,t}(b)$ in non-identity automorphism.

Let $b=p\lambda\neq 1$. If $\lambda\not=1$, then $\pi(b)\not=1$, and hence the
composition of $\psi_{1,t}$ and specialization $t=1$ send $b$ to non-trivial
automorphism. Therefore, $\psi_{1,t}(b)$ is non-trivial. If $\lambda=1$ and
$b\not=1$, then $b$ can be presented in the form

$b=a_{1,2}^{\alpha_{1,2}}\left(a_{1,3}^{\alpha_{1,3}}a_{2,3}^{\alpha_{2,3}}\right)\ldots\left(a_{1,n}^{\alpha_{1,n}}\ldots
a_{n-1,n}^{\alpha_{n-1,n}}\right)P_{n}^{\prime}$

for some integers $\alpha_{i,j}\in\mathbb{Z}$. From Lemma 2.2 it follows that
the matrix of the automorphism $\psi_{1,t}(b)$ is the following diagonal
matrix

$[\psi_{1,t}(b)]=\operatorname{diag}(t^{2\alpha_{1,2}},t^{2\alpha_{1,3}},t^{2\alpha_{2,3}}\ldots
t^{2\alpha_{1,n}}\ldots t^{2\alpha_{n-1,n}}),$

which is not the identity matrix. Hence, we have proved that if $b$ is non-
trivial, then its image $\psi_{1,t}$ is non-trivial. ∎

It was shown in [28] that the group $B_{n}/P_{n}^{\prime}$ possesses torsion,
and there is a one-to-one correspondence between the conjugacy classes of the
finite-order elements of $B_{n}/P_{n}^{\prime}$ and the conjugacy classes of
the elements of odd order of the symmetric group $S_{n}$. The group
$B_{3}/P_{3}^{\prime}$ was considered separately in the paper [28]. Using the
linear representation $\psi_{1,t}$ we obtain the following statement.

######  Proposition 2.1.

The group $B_{3}/P_{3}^{\prime}$ has only $3$-torsion. Every element which has
one of the following forms

$\displaystyle
a_{1,2}^{\alpha}a_{1,3}^{\beta}a_{2,3}^{\gamma}\sigma_{1}\sigma_{2},$
$\displaystyle
a_{1,2}^{\alpha}a_{1,3}^{\beta}a_{2,3}^{\gamma}\sigma_{2}\sigma_{1},$

where $\alpha,\beta,\gamma\in\mathbb{Z}$ are such that
$\alpha+\beta+\gamma=-1$, has order $3$. Any other non-trivial element has
infinite order.

###### Proof.

Since $\psi_{1,t}(B_{n})\cong B_{3}/P_{3}^{\prime}$, we can present elements
of $B_{3}/P_{3}^{\prime}$ by matrices in
$\operatorname{GL}_{3}(\mathbb{Z}[t^{\pm 1}])$. For $n=3$ any element $b\in
B_{3}/P_{3}^{\prime}$ can be presented in the form $b=p\lambda$, where $p\in
P_{3}/P_{3}^{\prime}$ has the following form

$p=\operatorname{diag}(t^{2\alpha},t^{2\beta},t^{2\gamma}),\qquad~{}\text{for}~{}\alpha,\beta,\gamma\in\mathbb{Z},$

and $\lambda$ is one of the following matrices

$\displaystyle\lambda_{0}$ $\displaystyle=\begin{pmatrix}1&0&0\\\ 0&1&0\\\
0&0&1\end{pmatrix},$ $\displaystyle\lambda_{1}=\begin{pmatrix}t&0&0\\\
0&0&1\\\ 0&1&0\end{pmatrix},$
$\displaystyle\lambda_{2}=\begin{pmatrix}0&1&0\\\ 1&0&0\\\
0&0&t\end{pmatrix},$ $\displaystyle\lambda_{3}$
$\displaystyle=\begin{pmatrix}0&t&0\\\ 0&0&t\\\ 1&0&0\end{pmatrix},$
$\displaystyle\lambda_{4}=\begin{pmatrix}0&0&1\\\ t&0&0\\\
0&t&0\end{pmatrix},$ $\displaystyle\lambda_{5}=\begin{pmatrix}0&0&t\\\
0&t&0\\\ t&0&0\end{pmatrix}.$

Let us consider each $\lambda_{i}$ separately. First, it is obvious that the
element $p\lambda_{0}$ has infinite order. If the element $p\lambda_{1}$ has a
finite order, then from the matrix form of $\lambda_{1}$ it is clear that this
order has to be even. Since for an arbitrary integer $k$ we have

$(p\lambda_{1})^{2k}=\operatorname{diag}(t^{2k(2\alpha+1)},t^{2k(\beta+\gamma)},t^{2k(\beta+\gamma)}),$

the element $p\lambda_{1}$ has infinite order. Further if the element
$p\lambda_{2}$ has a finite order, then from the matrix form of $\lambda_{2}$
we see that this order is even. Since for an arbitrary integer $k$ we have

$(p\lambda_{2})^{2k}=\operatorname{diag}(t^{2k(\alpha+\beta)},t^{2k(\alpha+\beta)},t^{2k(2\gamma+1)}),$

the element $p\lambda_{2}$ has infinite order. From the matrix representations
it is clear that the order of the element $p\lambda_{3}$ is divisible by $3$.
For an arbitrary $k$ we have the equality

$(p\lambda_{3})^{3k}=\operatorname{diag}(t^{2k(\alpha+\beta+\gamma+1)},t^{2k(\alpha+\beta+\gamma+1)},t^{2k(\alpha+\beta+\gamma+1)}),$

i. e. if $\alpha+\beta+\gamma+1=0$, then $p\lambda_{3}$ has order 3, otherwise
$p\lambda_{3}$ has infinite order. Using the same argument, the order of the
element $p\lambda_{4}$ is divisible by $3$, and for an arbitrary $k$ we have

$(p\lambda_{4})^{3k}=\operatorname{diag}(t^{2k(\alpha+\beta+\gamma+1)},t^{2k(\alpha+\beta+\gamma+1)},t^{2k(\alpha+\beta+\gamma+1)}),$

i. e. if $\alpha+\beta+\gamma+1=0$, then $p\lambda_{4}$ has order 3, otherwise
$p\lambda_{4}$ has infinite order. It is easy to check that elements of the
form $p\lambda_{5}$ have infinite order ∎

In general case it is not difficult to prove the following statement.

######  Proposition 2.2.

If $b$ is a non-trivial element in $B_{n}/P_{n}^{\prime}$ such that the
permutation $\pi(b)$ is non-identity but fixes some points, then $b$ has
infinite order.

##  3\. Linear representation of $VB_{n}/VP_{n}^{\prime}$

In Section 1.3 we have noticed that the following short exact sequence holds

$1\to VP_{n}\to VB_{n}\to S_{n}\to 1.$

From this short exact sequence and the isomorphism

$(VB_{n}/VP_{n}^{\prime})/(VP_{n}/VP_{n}^{\prime})=VB_{n}/VP_{n}$

we have the following short exact sequence

$1\to VP_{n}/VP_{n}^{\prime}\to VB_{n}/VP_{n}^{\prime}\to S_{n}\to 1,$

where $VP_{n}/VP_{n}^{\prime}=\mathbb{Z}^{n(n-1)}$ (since $VP_{n}$ is
generated by the elements $\lambda_{i,j}$ for $1\leq i\neq j\leq n$),
$S_{n}=\langle\rho_{1},\rho_{2},\ldots,\rho_{n-1}\rangle$, and the group
$S_{n}$ acts on the generators of $VP_{n}/VP_{n}^{\prime}$ by the permutation
of indices. From the equality $VB_{n}=VP_{n}\rtimes S_{n}$ it follows that
$VB_{n}/VP_{n}^{\prime}=(VP_{n}/VP_{n}^{\prime})\rtimes S_{n}$, and hence,
$VB_{n}/VP_{n}^{\prime}$ is a crystallographic group.

In this section we construct an extension of the representation $\psi_{1,t}$
from the braid group $B_{n}$ to the virtual braid group $VB_{n}$. Let $U$ be
the free module with the basis $e_{i,j}$, $1\leq i<j\leq n$ over the ring
$\mathbb{Z}[t^{\pm 1},t_{1}^{\pm 1},t_{2}^{\pm 1},\ldots,t_{n-1}^{\pm 1}]$ of
Laurent polinomials with $n$ variables $t,t_{1},\dots,t_{n-1}$. Let us define
the map $\Psi:VB_{n}\to\operatorname{Aut}(U)$ on the generators
$\sigma_{1},\sigma_{2},\dots,\sigma_{n-1}$,
$\rho_{1},\rho_{2},\dots,\rho_{n-1}$ of $VB_{n}$ by the following rules:

(30) $\displaystyle\Psi(\sigma_{i})$ $\displaystyle=\psi_{1,t}(\sigma_{i}),$
(31) $\displaystyle\Psi(\rho_{i})$
$\displaystyle:\begin{cases}e_{i,i+1}\mapsto e_{i,i+1},&\\\ e_{k,l}\mapsto
e_{k,l},&k\not=i,i+1,l\not=i,i+1,\\\ e_{i,j}\mapsto t_{i}e_{i+1,j},&i<j-1,\\\
e_{i+1,j}\mapsto t_{i}^{-1}e_{i,j},&\\\ e_{k,i}\mapsto t_{i}e_{k,i+1},&\\\
e_{k,i+1}\mapsto t_{i}^{-1}e_{k,i}.&\end{cases}$

The following property can be easy verified using direct calculations.

######  Proposition 3.1.

The map $\Psi:VB_{n}\to\operatorname{Aut}(U)$, defined by (30) and (31) is a
representation. The images $\Psi(\sigma_{i})$ and $\Psi(\rho_{i})$ satisfy
both forbidden relations (23) for $VB_{n}$.

The following statement describes the image of $\Psi$.

######  Theorem 3.1.

The image $\Psi(VB_{n})$ is isomorphic to $VB_{n}/VP_{n}^{\prime}$.

###### Proof.

Using direct calculations it is easy to see that the image $\Psi(b)$ of every
element $b\in VP_{n}^{\prime}$ is the trivial automorphism. Hence, in order to
prove the theorem we have to prove that the image of any elements $b\in
VB_{n}\setminus VP_{n}^{\prime}$ is non-trivial. If $b$ does not belong to
$VP_{n}$, then the image of $b$ under $\Psi$ is clearly non-trivial
automorphism of $U$. If $b$ is a non-trivial element of $VP_{n}$, then looking
to the representative of $b$ in $bVP_{n}^{\prime}$ it is easy to see that the
matrix of $\Psi(b)$ is a non-identity diagonal matrix. ∎

The following questions arise naturally from the paper [28] where the same
questions are studied for classical linear groups.

######  Question 3.1.

What are the possible finite orders of elements from $VB_{n}/VP_{n}^{\prime}$
which do not belong to $S_{n}$?

######  Question 3.2.

Which Bieberbach subgroups belong to $VB_{n}/VP_{n}^{\prime}$?

Let us consider the particular case $n=3$. The virtual braid group $VB_{3}$ is
generated by four elements $\sigma_{1},\sigma_{2},\rho_{1},\rho_{2}$ and it
has the following defining relations

$\displaystyle\sigma_{1}\sigma_{2}\sigma_{1}=\sigma_{2}\sigma_{1}\sigma_{2},$
$\displaystyle\rho_{1}\rho_{2}\rho_{1}=\rho_{2}\rho_{1}\rho_{2},$
$\displaystyle\rho_{1}^{2}=\rho_{2}^{2}=1,$
$\displaystyle\rho_{1}\rho_{2}\sigma_{1}=\sigma_{2}\rho_{1}\rho_{2}.$

The images of the generators $\sigma_{1},\sigma_{2},\rho_{1},\rho_{2}$ of
$VB_{n}$ under the representation
$\Psi:VB_{3}\to\operatorname{GL}_{3}(\mathbb{Z}[t^{\pm 1},t_{1}^{\pm
1},t_{2}^{\pm 1}]$ are the following matrices

$\displaystyle\Psi(\sigma_{1})$ $\displaystyle=\begin{pmatrix}t&0&0\\\
0&0&1\\\ 0&1&0\end{pmatrix},$
$\displaystyle\Psi(\sigma_{2})=\begin{pmatrix}0&1&0\\\ 1&0&0\\\
0&0&t\end{pmatrix},$ $\displaystyle\Psi(\rho_{1})$
$\displaystyle=\begin{pmatrix}1&0&0\\\ 0&0&t_{1}\\\
0&t_{1}^{-1}&0\end{pmatrix},$
$\displaystyle\Psi(\rho_{2})=\begin{pmatrix}0&t_{2}&0\\\ t_{2}^{-1}&0&0\\\
0&0&1\end{pmatrix}.$

The group $VP_{3}$ is generated by the elements

$\displaystyle\lambda_{12}=\rho_{1}\sigma_{1}^{-1},$
$\displaystyle\lambda_{21}=\sigma_{1}^{-1}\rho_{1},$
$\displaystyle\lambda_{13}=\rho_{2}\lambda_{12}\rho_{2},$
$\displaystyle\lambda_{31}=\rho_{2}\lambda_{21}\rho_{2},$
$\displaystyle\lambda_{23}=\rho_{2}\sigma_{2}^{-1},$
$\displaystyle\lambda_{32}=\sigma_{2}^{-1}\rho_{2}.$

Using direct calculations it is easy to see that the images of the elements
$\lambda_{12}$, $\lambda_{21}$, $\lambda_{13}$, $\lambda_{31}$,
$\lambda_{23}$, $\lambda_{32}$ under the representations $\Psi$ are the
following diagonal matrices

$\displaystyle\Psi(\lambda_{12})=\operatorname{diag}(t^{-1},t_{1},t_{1}^{-1}),$
$\displaystyle\Psi(\lambda_{21})=\operatorname{diag}(t^{-1},t_{1}^{-1},t_{1}),$
$\displaystyle\Psi(\lambda_{13})=\operatorname{diag}(t_{1},t,t_{1}^{-1}),$
$\displaystyle\Psi(\lambda_{31})=\operatorname{diag}(t_{1}^{-1},t^{-1},t_{1}),$
$\displaystyle\Psi(\lambda_{23})=\operatorname{diag}(t_{2},t_{2}^{-1},t^{-1}),$
$\displaystyle\Psi(\lambda_{32})=\operatorname{diag}(t_{2}^{-1},t_{2},t^{-1}),$

in particular, $\Psi(\lambda_{12})\not=\Psi(\lambda_{21})$. Let us demonstrate
that there are torsion elements in $VB_{3}/VP_{3}^{\prime}$ which do not
belong to $S_{3}$. Consider the element

$p=\lambda_{12}^{a}\lambda_{21}^{\alpha}\lambda_{13}^{b}\lambda_{23}^{c}\lambda_{31}^{\beta}\lambda_{32}^{\gamma}d\rho_{1},$

where $\alpha,\beta,\gamma,a,b,c\in\mathbb{Z}$ and $d\in VP_{3}^{\prime}$.
Then

$\Psi(p)^{2}=\operatorname{diag}\left(t^{-2(a+\alpha)}t_{1}^{2(b-\beta)}t_{2}^{-2(c-\gamma)},t^{b-\beta-\gamma-c}t_{1}^{\beta-b}t_{2}^{\gamma-c},t^{b-\beta-\gamma-c}t_{1}^{\beta-b}t_{2}^{2(\gamma-c)}\right).$

From this formula it is clear that if $\alpha=-a$, $\beta=b$, and
$\gamma=c=0$, then $\Psi(p)^{2}=E$. The same arguments lead to the following
statement.

######  Proposition 3.2.

The group $VB_{3}/VP_{3}^{\prime}$ has infinitely many elements of order 2.

##  4\. Decomposition and linearity of $VB_{3}$

Properties of the group $VP_{3}$ were studied in [9]. In particular, the
certain decomposition of this group was found in [9]. In the present section
we construct a simpler decomposition and prove that $VP_{3}$, and hence
$VB_{3}$, is linear. The following statement is the first main result of this
section.

######  Proposition 4.1.

$VP_{3}=(F_{3}\rtimes F_{2})*\mathbb{Z}$.

###### Proof.

By Theorem 1.1, the group $VP_{3}$ has a presentation with six generators
$\lambda_{12}$, $\lambda_{21}$, $\lambda_{13}$, $\lambda_{23}$,
$\lambda_{32}$, $\lambda_{31}$, and six relations which can be write as the
following conjugations:

(32)
$\displaystyle\lambda_{12}(\lambda_{13}\lambda_{23})=(\lambda_{23}\lambda_{13})\lambda_{12}$
$\displaystyle\Leftrightarrow$
$\displaystyle(\lambda_{13}\lambda_{23})^{\lambda_{12}^{-1}}=\lambda_{23}\lambda_{13},$
(33)
$\displaystyle\lambda_{13}(\lambda_{12}\lambda_{32})=(\lambda_{32}\lambda_{12})\lambda_{13}$
$\displaystyle\Leftrightarrow$
$\displaystyle(\lambda_{32}\lambda_{13}^{-1})^{\lambda_{12}^{-1}}=\lambda_{13}^{-1}\lambda_{32},$
(34)
$\displaystyle\lambda_{31}(\lambda_{32}\lambda_{12})=(\lambda_{12}\lambda_{32})\lambda_{31}$
$\displaystyle\Leftrightarrow$
$\displaystyle(\lambda_{32}\lambda_{31}^{-1})^{\lambda_{12}^{-1}}=\lambda_{31}^{-1}\lambda_{32},$
(35)
$\displaystyle\lambda_{21}(\lambda_{23}\lambda_{13})=(\lambda_{13}\lambda_{23})\lambda_{21}$
$\displaystyle\Leftrightarrow$
$\displaystyle(\lambda_{23}\lambda_{13})^{\lambda_{21}^{-1}}=\lambda_{13}\lambda_{23},$
(36)
$\displaystyle\lambda_{23}(\lambda_{21}\lambda_{31})=(\lambda_{31}\lambda_{21})\lambda_{23}$
$\displaystyle\Leftrightarrow$
$\displaystyle(\lambda_{31}\lambda_{23}^{-1})^{\lambda_{21}^{-1}}=\lambda_{23}^{-1}\lambda_{31},$
(37)
$\displaystyle\lambda_{32}(\lambda_{31}\lambda_{21})=(\lambda_{21}\lambda_{31})\lambda_{32}$
$\displaystyle\Leftrightarrow$
$\displaystyle(\lambda_{31}\lambda_{32})^{\lambda_{21}^{-1}}=\lambda_{32}\lambda_{31}.$

Note that equations (32) and (33) are also equivalent to the following
commutation relations

$\displaystyle(\lambda_{13}\lambda_{23})^{\lambda_{12}^{-1}\lambda_{13}^{-1}}=\lambda_{13}\lambda_{23},$
$\displaystyle(\lambda_{32}\lambda_{13}^{-1})^{\lambda_{12}^{-1}\lambda_{13}^{-1}}=\lambda_{32}\lambda_{13}^{-1},$

respectively. Let us define the following elements of $VP_{3}$:

$\displaystyle b_{11}=\lambda_{21}\lambda_{12},$ $\displaystyle
b_{12}=\lambda_{13}\lambda_{12},$ $\displaystyle
b_{21}=\lambda_{13}\lambda_{23},$ $\displaystyle
b_{22}=\lambda_{32}\lambda_{31},$ $\displaystyle
b_{23}=\lambda_{32}\lambda_{23},$

and rewrite all relations (32)-(37) in order to express how the group $\langle
b_{11},b_{12}\rangle$ acts on the group $\langle b_{21},b_{22},b_{23}\rangle$
by conjugations:

$\displaystyle b_{21}^{b_{12}^{-1}}$
$\displaystyle=(\lambda_{13}\lambda_{23})^{\lambda_{12}^{-1}\lambda_{13}^{-1}}\stackrel{{\scriptstyle(\ref{reln1})}}{{=}}\lambda_{13}\lambda_{23}=b_{21},$
$\displaystyle b_{22}^{b_{12}^{-1}}$
$\displaystyle=\lambda_{13}\lambda_{12}\cdot\lambda_{32}\lambda_{31}\cdot\lambda_{12}^{-1}\lambda_{13}^{-1}\stackrel{{\scriptstyle(\ref{reln3})}}{{=}}\lambda_{13}\cdot\lambda_{31}\lambda_{32}\cdot\lambda_{13}^{-1}$
$\displaystyle=\lambda_{13}\lambda_{23}\cdot\lambda_{23}^{-1}\lambda_{32}^{-1}\cdot\lambda_{32}\lambda_{31}\cdot\lambda_{32}\lambda_{23}\cdot\lambda_{23}^{-1}\lambda_{13}^{-1}=b_{22}^{b_{23}b_{21}^{-1}},$
$\displaystyle b_{23}^{b_{12}^{-1}}$
$\displaystyle=(\lambda_{32}\lambda_{23})^{\lambda_{12}^{-1}\lambda_{13}^{-1}}=(\lambda_{32}\lambda_{13}^{-1}\cdot\lambda_{13}\lambda_{23})^{\lambda_{12}^{-1}\lambda_{13}^{-1}}$
$\displaystyle\stackrel{{\scriptstyle(\ref{reln1}),(\ref{reln2})}}{{=}}\lambda_{32}\lambda_{13}^{-1}\cdot\lambda_{13}\lambda_{23}=b_{23},$
$\displaystyle b_{21}^{b_{11}^{-1}}$
$\displaystyle=(\lambda_{13}\lambda_{23})^{\lambda_{12}^{-1}\lambda_{21}^{-1}}\stackrel{{\scriptstyle(\ref{reln1}),(\ref{reln4})}}{{=}}\lambda_{13}\lambda_{23}=b_{21},$
$\displaystyle b_{22}^{b_{11}^{-1}}$
$\displaystyle=(\lambda_{32}\lambda_{31})^{\lambda_{12}^{-1}\lambda_{21}^{-1}}\stackrel{{\scriptstyle(\ref{reln3}),(\ref{reln6})}}{{=}}\lambda_{32}\lambda_{31}=b_{22},$
$\displaystyle b_{23}^{b_{11}^{-1}}$
$\displaystyle=(\lambda_{32}\lambda_{13}^{-1}\cdot\lambda_{13}\lambda_{23})^{\lambda_{12}^{-1}\lambda_{21}^{-1}}\stackrel{{\scriptstyle(\ref{reln2}),(\ref{reln1})}}{{=}}(\lambda_{13}^{-1}\lambda_{32}\cdot\lambda_{23}\lambda_{13})^{\lambda_{21}^{-1}}$
$\displaystyle=(\lambda_{13}^{-1}\lambda_{23}^{-1}\cdot\lambda_{23}\lambda_{31}^{-1}\cdot\lambda_{31}\lambda_{32}\cdot\lambda_{23}\lambda_{13})^{\lambda_{21}^{-1}}$
$\displaystyle\stackrel{{\scriptstyle(\ref{reln4}),(\ref{reln5}),(\ref{reln6}),(\ref{reln4})}}{{=}}\lambda_{23}^{-1}\lambda_{13}^{-1}\cdot\lambda_{31}^{-1}\lambda_{23}\cdot\lambda_{32}\lambda_{31}\cdot\lambda_{13}\lambda_{23}$
$\displaystyle=\lambda_{23}^{-1}\lambda_{13}^{-1}\cdot\lambda_{31}^{-1}\lambda_{32}^{-1}\cdot\lambda_{32}\lambda_{23}\cdot\lambda_{32}\lambda_{31}\cdot\lambda_{13}\lambda_{23}=b_{23}^{b_{22}b_{21}}.$

To summarize, we have six calculations which show the following six relations:

(38) $\displaystyle b_{21}^{b_{12}^{-1}}=b_{21},$ $\displaystyle
b_{22}^{b_{12}^{-1}}=b_{22}^{b_{23}b_{21}^{-1}},$ $\displaystyle
b_{23}^{b_{12}^{-1}}=b_{23},$ (39) $\displaystyle
b_{21}^{b_{11}^{-1}}=b_{21},$ $\displaystyle b_{22}^{b_{11}^{-1}}=b_{22},$
$\displaystyle b_{23}^{b_{11}^{-1}}=b_{23}^{b_{22}b_{21}}.$

Moreover, in the sequence of six calculations we made, we used exactly one of
the relations (32)-(37) which was not yet used in the preceding ones. It means
that the set of relations (38)-(39) is equivalent to the set of relations
(32)-(37). In particular, the groups $F_{2}=\langle b_{11},b_{12}\rangle$,
$F_{3}=\langle b_{21},b_{22},b_{23}\rangle$ are free, and the five elements
$b_{11},b_{12},b_{21},b_{22},b_{23}$ generate the semidirect product
$F_{3}\rtimes F_{2}$, where $F_{2}$ acts on $F_{3}$ by the way described in
(32)-(37). Finally, the five generators $b_{11},b_{12},b_{21},b_{22},b_{23}$
together with any one of the generators $\lambda_{ij}$, for instance with
$\lambda_{13}$, generate the full group $VP_{3}$. Since the six relations
describing the semidirect product structure form a complete set of relations
of $VP_{3}$, the group $\langle\lambda_{13}\rangle$ forms a free factor in
$VP_{3}$ and we have the decomposition

$VP_{3}=(\langle b_{21},b_{22},b_{23}\rangle\rtimes\langle
b_{11},b_{12}\rangle)*\langle\lambda_{13}\rangle.$

The proposition is proved. ∎

Let us now prove that the group $VP_{3}$ is linear. Denote by $\mu:VB_{n}\to
S_{n}$ the homomorphism defined on the generators
$\sigma_{1},\sigma_{2},\dots,\sigma_{n-1},\rho_{1},\rho_{2},\dots,\rho_{n-1}$
of $VB_{n}$ as follows:

$\displaystyle\mu(\sigma_{i})=1,$ $\displaystyle\mu(\rho_{i})=\rho_{i},$
$\displaystyle i=1,2,\dots,n-1\,,$

where $S_{n}=\langle\rho_{1},\rho_{2},\ldots,\rho_{n-1}\rangle$. Denote by
$H_{n}$ the normal closure of $B_{n}$ in $VB_{n}$. It is obvious that
$\operatorname{Ker}(\mu)$ coincides with $H_{n}$. Now, define the following
elements:

$\displaystyle x_{i,i+1}=\sigma_{i},$ $\displaystyle
x_{i+1,i}=\rho_{i}\sigma_{i}\rho_{i},$

and

$\displaystyle x_{i,j}$
$\displaystyle=\rho_{j-1}\cdots\rho_{i+1}\sigma_{i}\rho_{i+1}\cdots\rho_{j-1}$
$\displaystyle\mbox{for}\quad 1\leq i<j-1\leq n-1,$ $\displaystyle x_{j,i}$
$\displaystyle=\rho_{j-1}\cdots\rho_{i+1}\rho_{i}\sigma_{i}\rho_{i}\rho_{i+1}\cdots\rho_{j-1}$
$\displaystyle\mbox{for}\quad 1\leq i<j-1\leq n-1.$

The symmetric group $S_{n}=\langle\rho_{1},\rho_{2},\dots,\rho_{n-1}\rangle$
acts on the elements $x_{i,j}$ for $1\leq i\not=j\leq n$ by permutation of
indices, i. e.,

$\rho x_{i,j}\rho^{-1}=x_{\rho(i),\rho(j)},\qquad\rho\in S_{n}.$

The following proposition is proved in [43] (see also [6]).

######  Proposition 4.2.

1) The group $H_{n}$ admits a presentation with the generators $x_{k,l}$ for
$1\leq k\neq l\leq n$, and the defining relations:

(40) $\displaystyle x_{i,j}\,x_{k,\,l}$ $\displaystyle=x_{k,\,l}\,x_{i,j},$
(41) $\displaystyle x_{i,k}\,x_{k,j}\,x_{i,k}$
$\displaystyle=x_{k,j}\,x_{i,k}\,x_{k,j},$

where distinct letters stand for distinct indices.

2) The group $VB_{n}$ is isomorphic to $H_{n}\rtimes S_{n}$ where $S_{n}$ acts
on the generators $x_{i,j}$ of $H_{n}$ by permutation of indices.

######  Corollary 4.1.

The group $H_{2}$ is isomorphic to $\mathbb{Z}*\mathbb{Z}$ which is residually
nilpotent.

Now we are ready prove the second main result of this section.

######  Theorem 4.1.

The group $VB_{3}$ is linear.

###### Proof.

Since $VB_{3}=H_{3}\rtimes S_{3}$, the group $VB_{3}$ is linear if and only if
$H_{3}$ is linear. The group $H_{3}$ decomposes into a free product
$G_{1}*G_{2}$, where $G_{1}=\langle x_{1,2},x_{2,3},x_{3,1}\rangle$ and
$G_{2}=\langle x_{1,3},x_{3,2},x_{2,1}\rangle$. The group $G_{i},$ $i=1,2$ is
isomorphic to the $2$-nd affine Artin-Tits group of type $\mathcal{A}$, also
called circular braid group on $3$ strands (see [1]). Since $G_{i}$ can be
embedded in $B_{4}$ (see [47]), it is linear. Since the free product of linear
groups is linear, $H_{3}$ is linear. ∎

######  Question 4.1.

What is the minimal dimension of the faithful linear representation of
$VP_{3}$?

######  Question 4.2.

What is the image of $H_{n}$ under the representation $\Psi$?

##  5\. Universal braid group

The universal braid group $UB_{n}$ was defined in [4] as a group with
generators $\sigma_{1},\sigma_{2},\ldots,\sigma_{n-1},$
$\tau_{1},\tau_{2},\ldots,\tau_{n-1}$ which satisfy braid relations (1)-(2),
commutativity relations

$\tau_{i}\,\tau_{j}=\tau_{j}\,\tau_{i},\qquad|i-j|\geq 2,$

and mixed relations

$\tau_{i}\,\sigma_{j}=\sigma_{j}\,\tau_{i}\qquad|i-j|\geq 2.$

Recall [22] that the Artin’s group of type $I$, denoted by $A_{I}$, is the
group with generators $a_{i}$, $i\in I$ and defining relations

$a_{i}\,a_{j}\,a_{i}\ldots=a_{j}\,a_{i}\,a_{j}\ldots,\qquad i,j\in I,$

where the words from the left and from the right hand side of the equality
above consist of $m_{ij}$ alternating letters $a_{i}$ and $a_{j}$. The
following statement is proved in [4].

######  Proposition 5.1.

1) The elements $\sigma_{1},\sigma_{2},\dots,\sigma_{n-1}$ generate the braid
group $B_{n}$ in the universal braid group $UB_{n}$.

2) There are homomorphisms $\varphi_{US}:UB_{n}\to SG_{n}$,
$\varphi_{UV}:UB_{n}\to VB_{n}$ and $\varphi_{UB}:UB_{n}\to B_{n}$.

3) The group $UB_{n}$ is the Artin’s group.

Statement 2) of Proposition 5.1 is the reason why the group $UB_{n}$ is called
the universal braid group. The commutator subgroup of $UB_{n}$ is studied in
[23].

Denote by $T_{n}$ the group generated by $\tau_{1}$, $\tau_{2}$, $\ldots$,
$\tau_{n-1}$ with the following defining commutativity relations

$\tau_{i}\tau_{j}=\tau_{j}\tau_{i},\qquad|i-j|>1.$

It is clear that $T_{n}$ forms a subgroup in the singular braid group $SG_{n}$
and a subgroup in the universal braid group $UB_{n}$. By adding to $T_{n}$ the
relations $\tau_{i}^{2}=1$ for $i=1,2,\ldots,n-1$ we get the twin group or the
flat braid group (see [16]).

Since the elements $\sigma_{1},\sigma_{2},\dots,\sigma_{n-1}$ generate a
subgroup of $UB_{n}$ which is isomorphic to the braid group $B_{n}$, the
elements $\tau_{1},\tau_{2},\dots,\tau_{n-1}$ generate a subgroup of $UB_{n}$
which is isomorphic to $T_{n}$, and the group $UB_{n}$ is generated by the
elements $\sigma_{1},\sigma_{2},\dots,\sigma_{n-1}$,
$\tau_{1},\tau_{2},\dots,\tau_{n-1}$, in order to define some representation
of $UB_{n}$ we have to define a representation of $B_{n}<UB_{n}$, a
representation of $T_{n}<UB_{n}$, and then match these two representations.

###  5.1. Representations by automorphisms of a free group

Further we consider several representations of $T_{n}$ by automorphisms of
free groups. The first one-parametric family of representations consists of
the representations $\phi_{k}:T_{n}\to\operatorname{Aut}(F_{n})$ for
$k\in\mathbb{Z}$, where $\phi_{k}$ acts on the generators
$\tau_{1},\tau_{2},\dots,\tau_{n-1}$ of $T_{n}$ by the following rule

$\phi_{k}(\tau_{i}):\begin{cases}x_{i}\to x_{i+1}^{-k}x_{i}x_{i+1}^{k},&\\\
x_{j}\to x_{j},&\mbox{if}\quad j\neq i.\end{cases}$

The second family of representations is the two-parametric family of
representations $\psi_{k,l}:T_{n}\to\operatorname{Aut}(F_{n})$ for
$k,l\in\mathbb{Z}\setminus\\{0\\}$, where $\phi_{k,l}$ acts on the generators
$\tau_{1},\tau_{2},\dots,\tau_{n-1}$ of $T_{n}$ by the following rule

$\psi_{k,l}(\tau_{i}):\begin{cases}x_{i}\to
x_{i}^{-l}x_{i+1}^{-k}x_{i}x_{i+1}^{k}x_{i}^{l},&\\\ x_{i+1}\to
x_{i}^{-l}x_{i+1}x_{i}^{l},&\\\ x_{j}\to x_{j}&\mbox{if}\quad j\neq
i,i+1.\end{cases}$

It is evidently that $\psi_{k,0}=\phi_{k}$.

Finally, the representation $\xi:T_{n}\to\operatorname{Aut}(F_{3n-2})$, where
the free group $F_{3n-2}$ has the set of free generators

$\\{x_{1},\ldots,x_{n},y_{1,1},\ldots,y_{1,n-1},y_{2,1},\ldots,y_{2,n-1}\\},$

acts on the generators $\tau_{1},\tau_{2},\dots,\tau_{n-1}$ of $T_{n}$ by the
following rule

$\xi(\tau_{i}):\begin{cases}x_{i}\to x_{i}y_{i,1},&\\\ x_{i+1}\to
x_{i+1}y_{i,2},&\\\ x_{j}\to x_{j},&\mbox{if}\quad j\neq i\\\ y_{i,m}\to
y_{i,m},&\mbox{if}\quad i=1,2\quad\mbox{and}\quad m=1,\ldots,n-1.\end{cases}$

From the definition it is clear that
$\phi_{k}(\tau_{i})=\varepsilon_{i,i+1}^{k}$ and
$\psi_{k,l}(\tau_{i})=\varepsilon_{i,i+1}^{k}\varepsilon_{i+1,i}^{l}$,
therefore, representations $\phi_{k}$ and $\psi_{k,l}$ act from $T_{n}$ to the
group of basis conjugating automorphisms $Cb_{n}$.

######  Lemma 5.1.

Let $w$ be an element in $T_{n}$, and $s\in\\{1,2,\dots,n-1\\}$ be a minimal
number such that $w$ cannot be written in terms of generators
$\tau_{1},\tau_{2},\dots,\tau_{n-1}$ without $\tau_{s}$. If $k\neq 0$, then
$\phi_{k}(w)(x_{s})\neq x_{s}$.

###### Proof.

Every word $w$ in the group $T_{n}$ can be written in the normal form

$w=\tau_{s_{1}}^{\alpha_{1}}\tau_{s_{2}}^{\alpha_{2}}\dots\tau_{s_{r}}^{\alpha_{r}},$

where $\alpha_{1},\alpha_{2},\dots,\alpha_{r}$ are non-zero integers, and for
all $i=1,2,\dots,r-1$ we have either $s_{i}<s_{i+1}$ or $s_{i}=s_{i+1}+1$. If
the word $w$ is written in the normal form using generators
$\tau_{s_{1}},\tau_{s_{2}},\dots,\tau_{s_{r}}$, then $w$ cannot be written
(even not in normal form) withour using these generators. From this fact it
follows that the word $w$ from the formulation of the lemma can be written in
terms of $\tau_{s},\tau_{s+1},\dots,\tau_{n-1}$, and the generator $\tau_{s}$
is used in the expression of $w$.

We will prove the lemma using specific induction on $n$. First, we will prove
that the lemma is correct in the case when $s=n-1$, then we will assume that
the lemma is correct for all numbers $s+1,s+2,\dots,n-1$, and using this
conjecture we will prove that the lemma is correct for $s$.

The basis of the induction is obvious: if $s=n-1$, then
$w=\tau_{n-1}^{\alpha}$ for $\alpha\neq 0$, and from the definition of
$\phi_{k}$ it is clear that

$\phi_{k}(w)(x_{n-1})=\phi_{k}(\tau_{n-1}^{\alpha})(x_{n-1})\neq x_{n-1}.$

Let us assume that the lemma is correct for all numbers $s+1,s+2,\dots,n-1$,
and let us prove the lemma for $s$.

By contrary, suppose that $w$ is a counter example with the minimal value
$|w|$, where $|w|$ denotes the length of $w$, when it is written in terms of
generators $\tau_{1},\tau_{2},\dots,\tau_{n-1}$. This assumption means that
$w$ cannot be written in terms of generators
$\tau_{1},\tau_{2},\dots,\tau_{n-1}$ without $\tau_{s}$ and
$\phi_{k}(w)(x_{s})=x_{s}$. Since $s\in\\{1,2,\dots,n-1\\}$ is a minimal
number such that $w$ cannot be written in terms of generators
$\tau_{1},\tau_{2},\dots,\tau_{n-1}$ without $\tau_{s}$, the element $w$ can
be written in the following form

(42) $w=w_{1}\tau_{s}^{\alpha_{1}}w_{2}\tau_{s}^{\alpha_{2}}\dots
w_{m}\tau_{s}^{\alpha_{m}}w_{m+1},$

where $\alpha_{1},\alpha_{2},\dots,\alpha_{m}$ are non-zero integers, and
$w_{1},w_{2},\dots,w_{m+1}$ are elements from $T_{n}$ which can be written in
terms of $\tau_{s+1},\tau_{s+2},\dots,\tau_{n-1}$.

From the definition of the representation $\phi_{k}$ it is clear that since
$w_{m+1}$ can be written in terms of generators
$\tau_{s+1},\tau_{s+2},\dots,\tau_{n-1}$, and $\phi_{k}(w)(x_{s})=x_{s}$ then
for the element

$\widetilde{w}=w_{m+1}ww_{m+1}^{-1}=w_{m+1}w_{1}\tau_{s}^{\alpha_{1}}w_{2}\tau_{s}^{\alpha_{2}}\dots
w_{m}\tau_{s}^{\alpha_{m}}$

we have the equality $\phi_{k}(\widetilde{w})(x_{s})=x_{s}$. Since
$|\widetilde{w}|\leq|w|$, considering the element $\widetilde{w}$ instead of
$w$ we can assume that the equality $w_{m+1}=1$ holds in (42), i. e.

(43) $w=w_{1}\tau_{s}^{\alpha_{1}}w_{2}\tau_{s}^{\alpha_{2}}\dots
w_{m}\tau_{s}^{\alpha_{m}}.$

Since the element $w_{1}$ can be written in terms of generators
$\tau_{s+1},\tau_{s+2},\dots,\tau_{n-1}$ from the definition of the
representation $\phi_{k}$ it is clear that

$\phi_{k}(w_{1}\tau_{s}^{\alpha_{1}}w_{2}\tau_{s}^{\alpha_{2}}\dots
w_{m}\tau_{s}^{\alpha_{m}})(x_{s})=\phi_{k}(\tau_{s}^{\alpha_{1}}w_{2}\tau_{s}^{\alpha_{2}}\dots
w_{m}\tau_{s}^{\alpha_{m}})(x_{s}).$

Since $|\tau_{s}^{\alpha_{1}}w_{2}\tau_{s}^{\alpha_{2}}\dots
w_{m}\tau_{s}^{\alpha_{m}}|\leq|w_{1}\tau_{s}^{\alpha_{1}}w_{2}\tau_{s}^{\alpha_{2}}\dots
w_{m}\tau_{s}^{\alpha_{m}}|=|w|$, considering the element
$\tau_{s}^{\alpha_{1}}w_{2}\tau_{s}^{\alpha_{2}}\dots
w_{m}\tau_{s}^{\alpha_{m}}$ instead of $w$ we can assume that the equality
$w_{1}=1$ holds in (42), i. e.

(44) $w=\tau_{s}^{\alpha_{1}}w_{2}\tau_{s}^{\alpha_{2}}w_{3}\dots
w_{m}\tau_{s}^{\alpha_{m}}.$

Consider the following elements in $T_{n}$:

$\displaystyle u_{1}$ $\displaystyle=w_{2}w_{3}w_{4}\dots w_{m},$
$\displaystyle u_{2}$ $\displaystyle=w_{3}w_{4}\dots w_{m},$ $\displaystyle
u_{3}$ $\displaystyle=w_{4}\dots w_{m},$ $\displaystyle~{}~{}\vdots$
$\displaystyle u_{m-1}$ $\displaystyle=w_{m},$ $\displaystyle u_{m}$
$\displaystyle=1.$

From the definition of the representation $\phi_{k}$ and direct calculations
it follows that

(45)
$\phi_{k}(w)(x_{s})=x_{s}^{\left(\phi_{k}(u_{m})(x_{s+1}^{\beta_{m}})\right)\left(\phi_{k}(u_{m-1})(x_{s+1}^{\beta_{m-1}})\right)\dots\left(\phi_{k}(u_{2})(x_{s+1}^{\beta_{2}})\right)\left(\phi_{k}(u_{1})(x_{s+1}^{\beta_{1}})\right)},$

where $\beta_{i}=k\alpha_{i}$ for $i=1,2,\dots,m$. Since
$\phi_{k}(w)(x_{s})=x_{s}$ and (from the definition of $\phi_{k}$) the element

$\left(\phi_{k}(u_{m})(x_{s+1}^{\beta_{m}})\right)\left(\phi_{k}(u_{m-1})(x_{s+1}^{\beta_{m-1}})\right)\dots\left(\phi_{k}(u_{2})(x_{s+1}^{\beta_{2}})\right)\left(\phi_{k}(u_{1})(x_{s+1}^{\beta_{1}})\right)$

can be written in terms of $x_{s+1},x_{s+2},\dots,x_{n}$, from equality (45)
we conclude the equality

(46)
$\left(\phi_{k}(u_{m})(x_{s+1}^{\beta_{m}})\right)\left(\phi_{k}(u_{m-1})(x_{s+1}^{\beta_{m-1}})\right)\dots\left(\phi_{k}(u_{2})(x_{s+1}^{\beta_{2}})\right)\left(\phi_{k}(u_{1})(x_{s+1}^{\beta_{1}})\right)=1.$

From the definition of $\phi_{k}$ it is easy to see that for $i=1,2,\dots,m$
we have the equality

(47)
$\left(\phi_{k}(u_{i})(x_{s+1}^{\beta_{i}})\right)=\left(x_{s+1}^{\beta_{i}}\right)^{v_{i}},$

where $v_{1},v_{2},\dots,v_{m}$ are elements in $F_{n}$ which can be written
in terms of $x_{s+2},x_{s+3},\dots,x_{n}$. From the equality $u_{m}=1$ we have

$\left(\phi_{k}(u_{m})(x_{s+1}^{\beta_{m}})\right)=x_{s+1}^{\beta_{m}}$

and taking into account equality (47) we can rewrite equality (46) in the
following form

(48)
$x_{s+1}^{\beta_{m}}\left(x_{s+1}^{\beta_{m-1}}\right)^{v_{m-1}}\dots\left(x_{s+1}^{\beta_{2}}\right)^{v_{2}}\left(x_{s+1}^{\beta_{1}}\right)^{v_{1}}=1.$

Due to the fact that for $i=1,2,\dots,m$ the elements
$v_{2},v_{3},\dots,v_{m}\in F_{n}$ can be written in terms of
$x_{s+2},x_{s+3},\dots,x_{n}$, in order for the element $x_{s+1}^{\beta_{m}}$
in equality (48) to be reduced with something, there must be a number $t$ such
that $\left(x_{s+1}^{\beta_{t}}\right)^{v_{t}}=x_{s+1}^{\beta_{t}}$ and

(49)
$x_{s+1}^{\beta_{m}}\left(x_{s+1}^{\beta_{m-1}}\right)^{v_{m-1}}\dots\left(x_{s+1}^{\beta_{t+1}}\right)^{v_{t+1}}\left(x_{s+1}^{\beta_{t}}\right)^{v_{t}}=1.$

From equality (49) it follows that

$\displaystyle x_{s}$
$\displaystyle=x_{s}^{x_{s+1}^{\beta_{m}}\left(x_{s+1}^{\beta_{m-1}}\right)^{v_{m-1}}\dots\left(x_{s+1}^{\beta_{t+1}}\right)^{v_{t+1}}\left(x_{s+1}^{\beta_{t}}\right)^{v_{t}}}$
$\displaystyle=x_{s}^{\left(\phi_{k}(u_{m})(x_{s+1}^{\beta_{m}})\right)\left(\phi_{k}(u_{m-1})(x_{s+1}^{\beta_{m-1}})\right)\dots\left(\phi_{k}(u_{t})(x_{s+1}^{\beta_{t}})\right)}$
(50)
$\displaystyle=\phi_{k}(\tau_{s}^{\alpha_{t}}w_{t+1}\tau_{s}^{\alpha_{t+1}}w_{t+2}\dots\tau_{s}^{\alpha_{m-1}}w_{m}\tau_{s}^{\alpha_{m}})(x_{s})$

If $t\neq 1$, then

$|\tau_{s}^{\alpha_{t}}w_{t+1}\tau_{s}^{\alpha_{t+1}}w_{t+2}\dots\tau_{s}^{\alpha_{m-1}}w_{m}\tau_{s}^{\alpha_{m}}|<|\tau_{s}^{\alpha_{1}}w_{2}\tau_{s}^{\alpha_{2}}w_{3}\dots
w_{m}\tau_{s}^{\alpha_{m}}|=|w|$

and equality (50) contradicts the fact that $w$ is a counter example to the
lemma with the minimal value $|w|$, hence we have $t=1$, what means that

$\phi_{k}(u_{1})(x_{s+1}^{\beta_{1}})=\left(x_{s+1}^{\beta_{1}}\right)^{v_{1}}=x_{s+1}^{\beta_{1}}.$

This equality together with the definition of $\phi_{k}$ implies that the
eqiality $\phi_{k}(u_{1})(x_{s+1})=x_{s+1}$ holds.

Let us calculate $\phi_{k}(w)(x_{s+1})$. From the definition of $\phi_{k}$ we
have the following equality

$\phi_{k}(w)(x_{s+1})=\phi_{k}(\tau_{s}^{\alpha_{1}}w_{2}\tau_{s}^{\alpha_{2}}w_{3}\dots
w_{m}\tau_{s}^{\alpha_{m}})(x_{s+1})=\\\ =\phi_{k}(w_{2}w_{3}\dots
w_{m})(x_{s+1})=\phi_{k}(u_{1})(x_{s+1})=x_{s+1}.$

So, now we have the following equalities

(51) $\displaystyle w$
$\displaystyle=\tau_{s}^{\alpha_{1}}w_{2}\tau_{s}^{\alpha_{2}}w_{3}\dots
w_{m}\tau_{s}^{\alpha_{m}},$ (52) $\displaystyle\phi_{k}(w)(x_{s})$
$\displaystyle=x_{s},$ (53) $\displaystyle\phi_{k}(w)(x_{s+1})$
$\displaystyle=x_{s+1}.$

Note that in equality (51) the number $m$ can not be equal to $1$, since if
$m=1$, then $w=\tau_{s}^{\alpha_{1}}$, and it is clear that
$\phi_{k}(\tau_{s}^{\alpha_{1}})(x_{s})\neq x_{s}$ for $\alpha_{1}\neq 0$.
Therefore $m>1$.

Consider the element

$\widetilde{w}=\tau_{s}^{-\alpha_{1}}w\tau_{s}^{\alpha_{1}}=w_{2}\tau_{s}^{\alpha_{2}}w_{3}\dots
w_{m}\tau_{s}^{\alpha_{m}+\alpha_{1}}.$

From equalities (52), (53) we have the equality

$\phi_{k}(\widetilde{w})(x_{s})=x_{s}.$

Since the element $w_{2}$ can be written in terms of generators
$\tau_{s+1},\tau_{s+2},\dots,\tau_{n-1}$ from the definition of the
representation $\phi_{k}$ it is clear that

$x_{s}=\phi_{k}(w_{2}\tau_{s}^{\alpha_{2}}w_{3}\dots
w_{m}\tau_{s}^{\alpha_{m}+\alpha_{1}})(x_{s})=\phi_{k}(\tau_{s}^{\alpha_{2}}w_{3}\dots
w_{m}\tau_{s}^{\alpha_{m}+\alpha_{1}})(x_{s})$

and since $|\tau_{s}^{\alpha_{2}}w_{3}\dots
w_{m}\tau_{s}^{\alpha_{m}+\alpha_{1}}|<|w_{2}\tau_{s}^{\alpha_{2}}w_{3}\dots
w_{m}\tau_{s}^{\alpha_{m}+\alpha_{1}}|$ we must have that the element
$\tau_{s}^{\alpha_{2}}w_{3}\dots w_{m}\tau_{s}^{\alpha_{m}+\alpha_{1}}$
doesn’t depent on $\tau_{s}$. It means that $m=2$ and
$\alpha_{1}+\alpha_{m}=0$, and hence

$w=\tau_{s}^{\alpha_{1}}w_{2}\tau_{s}^{-\alpha_{1}},$

where the word $w_{2}$ can be written in terms of generators
$\tau_{s+1},\tau_{s+2},\dots,\tau_{n}$ and $\phi_{k}(w_{2})(x_{s+1})=x_{s+1}$.
By the unduction conjecture the word $w$ can be written in terms of
$\tau_{s+2},\tau_{s+3},\dots,\tau_{n-1}$, therefore from the definint
relations of $T_{n}$ we have
$w=\tau_{s}^{\alpha_{1}}w_{2}\tau_{s}^{-\alpha_{1}}=w_{2}$, i. e. $w$ can be
written without using $\tau_{s}$, what contradicts conditions of the lemma. ∎

The following theorem is the main result of this section.

######  Theorem 5.1.

The following statements hold.

  1. (1)

If $k\neq 0$, then the representation $\phi_{k}:T_{n}\to Cb_{n}$ is faithful.

  2. (2)

If $k\neq 0$, then the representation $\psi_{k,k}:T_{n}\to Cb_{n}$ is
faithful.

  3. (3)

The representation $\xi:T_{n}\to\operatorname{Aut}(F_{3n-2})$ is faithful for
$n=3$ and has non-trivial kernel for $n\geqslant 4$.

###### Proof.

(1) Let $w$ be an element from ${\rm Ker}(\phi_{k})$. Since for all
$s\in\\{1,2,\dots,n\\}$ we have $\phi_{k}(w)(x_{s})=x_{s}$, from Lemma 5.1 it
follows that in the representation of $w$ in terms of
$\tau_{1},\tau_{2},\dots,\tau_{n-1}$ there are no generators
$\tau_{1},\tau_{2},\dots,\tau_{n-1}$, hence, $w=1$.

(2) From result of Collins [24] follows that the subgroup
$\langle\sigma_{1}^{2},\sigma_{2}^{2},\dots,\sigma_{n-1}^{2}\rangle$ of
$B_{n}$ has the presentation

$\langle\sigma_{1}^{2},\sigma_{2}^{2},\dots,\sigma_{n-1}^{2}~{}|~{}\sigma_{i}^{2}\sigma_{j}^{2}=\sigma_{j}^{2}\sigma_{i}^{2}\text{
for }|i-j|\geqslant 2\rangle,$

hence the map $\tau_{i}\to\sigma_{i}^{2}$ for $i=1,2,\dots,n-1$ gives an
isomorphism

$T_{n}\to\langle\sigma_{1}^{2},\sigma_{2}^{2},\dots,\sigma_{n-1}^{2}\rangle<P_{n}.$

From the main result of the paper [45] it follows that if $k\neq 0$, then the
map
$\sigma_{i}^{2}\mapsto\varepsilon_{i,i+1}^{k}\varepsilon_{i+1,i}^{k}=\psi_{k,k}(\tau_{i})$
for $i=1,2,\dots,n-1$ gives a faithful representation of $P_{n}$ to ${\rm
Aut}(F_{n})$. Since $\psi_{k,k}$ is the superposition of these two maps

$T_{n}\to\langle\sigma_{1}^{2},\sigma_{2}^{2},\dots,\sigma_{n-1}^{2}\rangle\to{\rm
Aut}(F_{n}),$

the representation $\psi_{k,k}$ is faithful.

(3) For $n=3$ the group $\xi(T_{3})$ acts on the generator $x_{2}$ as the free
group $F_{2}$ which is isomorphic to $T_{3}$, therefore the representation
$\xi:T_{n}\to\operatorname{Aut}(F_{3n-2})$ is faithful for $n=3$. If
$n\geqslant 4$, then from direct calculations it is easy to see that the
element

$\tau_{1}\tau_{2}\tau_{3}\tau_{2}^{-1}\tau_{1}^{-1}\tau_{2}\tau_{3}^{-1}\tau_{2}^{-1}$

belongs to the kernel of $\xi$. ∎

Let us now consider extensions of the representations $\phi_{k},\psi_{k,l}$
from the group $T_{n}$ to the group $UB_{n}$,

$\displaystyle\widetilde{\phi}_{k}$
$\displaystyle:UB_{n}\to\operatorname{Aut}(F_{n}),$ $\displaystyle
k\in\mathbb{Z},$ $\displaystyle\widetilde{\psi}_{k,l}$
$\displaystyle:UB_{n}\to\operatorname{Aut}(F_{n}),$ $\displaystyle
k,l\in\mathbb{Z}$

which act on the generators $\sigma_{1},\sigma_{2},\dots,\sigma_{n-1}$ of
$UB_{n}$ as Artin representation presented in Section 1.1. The following
statement says that despite the fact that according to Theorem 5.1 the
representation $\phi_{k}:T_{3}\to{\rm Aut}(F_{3})$ is faithful for $k\neq 0$,
the extension $\widetilde{\phi}_{k}:UB_{3}\to\operatorname{Aut}(F_{3})$ is not
faithful.

######  Proposition 5.2.

The representation $\widetilde{\phi}_{k}:UB_{3}\to\operatorname{Aut}(F_{3})$
has non-trivial kernel for any $k\in\mathbb{Z}$.

###### Proof.

Using direct calculations it is easy to see that the element

$\sigma_{1}\tau_{1}^{k}\sigma_{2}\tau_{2}^{k}\sigma_{1}\tau_{1}^{k}\tau_{2}^{-k}\sigma_{2}^{-1}\tau_{1}^{-k}\sigma_{1}^{-1}\tau_{2}^{-k}\sigma_{2}^{-1}$

belongs to the kernel of $\widetilde{\phi}_{k}$. ∎

######  Problem 5.1.

Is it true that $\widetilde{\psi}_{k,k}$ is faithful for $k\not=0$?

At the end of this subsection we introduce a representation

$\Psi_{k,l}:UB_{n}\to{\rm Aut}(F_{n+1}),$

where $F_{n+1}=\langle x_{1},x_{2},\dots,x_{n},y\rangle$ which generalizes the
representation $\widetilde{\psi}_{k,l}$, and which acts on the generators
$\sigma_{1},\sigma_{2},\dots,\sigma_{n-1},\tau_{1},\tau_{2},\dots,\tau_{n-1}$
of $UB_{n}$ by the rules

$\displaystyle\Psi_{k,l}(\sigma_{i})$ $\displaystyle:\begin{cases}x_{i}\mapsto
x_{i}x_{i+1}x_{i}^{-1},&\\\ x_{i+1}\mapsto x_{i},&\\\ x_{j}\mapsto
x_{j},&j\neq i,i+1,\\\ y\mapsto y,&\end{cases}$
$\displaystyle\Psi_{k,l}(\tau_{i})$ $\displaystyle:\begin{cases}x_{i}\to
y^{-1}x_{i}^{-l}x_{i+1}^{-k}x_{i}x_{i+1}^{k}x_{i}^{l}y,&\\\ x_{i+1}\to
y^{-1}x_{i}^{-l}x_{i+1}x_{i}^{l}y,&\\\ x_{j}\to x_{j}&\mbox{if}\quad j\neq
i,i+1,\\\ y\mapsto y,&\end{cases}$

for $i=1,2,\ldots,n-1$. It is easy to see that $\Psi_{k,l}$ generalizes the
representation $\widetilde{\psi}_{k,l}$: if we put $y=1$, then $\Psi_{k,l}$
goes to $\widetilde{\psi}_{k,l}$.

######  Problem 5.2.

Do there exist some integers $k,l$ such that the representation
$\Psi_{k,l}:UB_{n}\to{\rm Aut}(F_{n-1})$ is faithful for all $n>2$?

###  5.2. Linear representations

As a corollary from the previous section, in the present section we construct
a linear representations of the groups $T_{n}$. In order to do it we use the
Magnus approach for construction of linear representations from
representations by automorphisms of free groups. This approach is described,
for example, in [8, Section 4.1]. In the present section we will just
introduce the final representation we got.

Let $M$ be the free module with the basis $\\{e_{1},e_{2},\ldots,e_{n-1}\\}$
over the ring $R=\mathbb{Z}[t_{1}^{\pm 1},t_{2}^{\pm 1},\ldots,t_{n}^{\pm 1}]$
of Laurent polynomials with variables $t_{1},t_{2},\dots,t_{n}$. Define the
map $\Theta:T_{n}\to\mathrm{Aut}(M)$ which acts on the generators
$\tau_{1},\tau_{2},\dots,\tau_{n-1}$ by the following rule

$\Theta(\tau_{i}):\begin{cases}e_{i}\longmapsto
t_{i+1}^{-1}e_{i}+t_{i+1}^{-1}(t_{i}-1)e_{i+1},&\\\ e_{j}\longmapsto
e_{j},&j\not=i.\end{cases}$

The following statement obviously follows from the fact that the matrices of
the elements $\Theta(\tau_{i})$ in the basis $e_{1},e_{2},\dots,e_{n}$ are
triangular.

######  Proposition 5.3.

The image $\operatorname{Im}\Theta$ is a solvable group. Hence, for $n>2$ the
representation $\Theta$ has a non-trivial kernel.

###  5.3. Universal pure braid group

In this section we study universal pure braid groups on small amount of
strands. Consider the homomorphism $\pi:UB_{n}\to S_{n}$ given on generators
of $UB_{n}$ by the rule $\pi(\sigma_{i})=\pi(\tau_{i})=(i,i+1)$ for
$i=1,2,\ldots,n-1$. The kernel $\operatorname{Ker}(\pi)$ of this homomorphism
is known as the _universal pure braid group_ , denoted by $U{P_{n}}$. The
following short exact sequence holds:

$1\to UP_{n}\to UB_{n}\stackrel{{\scriptstyle\pi}}{{\to}}S_{n}\to 1.$

It is obvious that $P_{n}$ is a subgroup of $U{P_{n}}$. Hence, the last group
contains all generators and relations of $P_{n}$.

######  Theorem 5.2.

The following properties hold:

1) $UB_{2}=B_{2}*T_{2}\cong F_{2}$, $UB_{3}=B_{3}*T_{3}=B_{3}*F_{2}$,

2) $U{P_{2}}=F_{3}$, $UP_{3}={P_{3}}*{F_{12}}$.

###### Proof.

1) Follows from the presentations of $UB_{2}$ and $UB_{3}$.

2) We will find the presentation of $UP_{3}=\operatorname{Ker}(\pi)$ by the
Reidemeister-Shreier method using the Shreier set
$\\{1,\sigma_{1},\sigma_{2},\sigma_{1}\sigma_{2},\sigma_{2}\sigma_{1},\sigma_{1}\sigma_{2}\sigma_{1},\sigma_{2}\sigma_{1}\sigma_{2}\\}$.
Since the restriction of homomorphism $\pi$ on $B_{3}\subset UB_{3}$ has
$P_{3}$ as a kernel, it is enough to consider only generators
$\tau_{1},\tau_{2}$ of $UB_{3}$. Hence we get

$\displaystyle
S_{1,\tau_{1}}=\tau_{1}\cdot(\overline{\tau_{1}})^{-1}=\tau_{1}\cdot\sigma_{1}^{-1},$
$\displaystyle
S_{1,\tau_{2}}=\tau_{2}\cdot(\overline{\tau_{2}})^{-1}=\tau_{2}\cdot\sigma_{2}^{-1},$
$\displaystyle
S_{\sigma_{1},\tau_{1}}=\sigma_{1}\tau_{1}\cdot(\overline{\sigma_{1}\tau_{1}})^{-1}=\sigma_{1}\tau_{1},$
$\displaystyle
S_{\sigma_{1},\tau_{2}}=\sigma_{1}\tau_{2}\cdot(\overline{\sigma_{1}\tau_{2}})^{-1}=\sigma_{1}\tau_{2}\sigma_{2}^{-1}\sigma_{1}^{-1},$
$\displaystyle
S_{\sigma_{2},\tau_{1}}=\sigma_{2}\tau_{1}\cdot(\overline{\sigma_{2}\sigma_{1}})^{-1}=\sigma_{2}\tau_{1}\sigma_{1}^{-1}\sigma_{2}^{-1},$
$\displaystyle S_{\sigma_{2},\tau_{2}}=\sigma_{2}\tau_{2},$ $\displaystyle
S_{\sigma_{1}\sigma_{2},\tau_{1}}=\sigma_{1}\sigma_{2}\tau_{1}\sigma_{1}^{-1}\sigma_{2}^{-1}\sigma_{1}^{-1},$
$\displaystyle
S_{\sigma_{1}\sigma_{2},\tau_{2}}=\sigma_{1}\sigma_{2}\tau_{2}\sigma_{1}^{-1},$
$\displaystyle
S_{\sigma_{2}\sigma_{1},\tau_{1}}=\sigma_{2}\sigma_{1}\tau_{1}\sigma_{2}^{-1},$
$\displaystyle
S_{\sigma_{2}\sigma_{1},\tau_{2}}=\sigma_{2}\sigma_{1}\tau_{2}\sigma_{2}^{-1}\sigma_{1}^{-1}\sigma_{2}^{-1},$
$\displaystyle
S_{\sigma_{1}\sigma_{2}\sigma_{1},\tau_{1}}=\sigma_{1}\sigma_{2}\sigma_{1}\tau_{1}\sigma_{2}^{-1}\sigma_{1}^{-1},$
$\displaystyle
S_{\sigma_{1}\sigma_{2}\sigma_{1},\tau_{2}}=\sigma_{1}\sigma_{2}\sigma_{1}\tau_{2}\sigma_{2}^{-1}\sigma_{1}^{-1}\sigma_{2}^{-1}\sigma_{1}^{-1}.$

Since, $T_{3}$ is free, all of its subgroups are also free. Hence, we have the
decomposition for $UP_{3}$. The group $UP_{2}$ is its subgroup and it is
generated by

$a_{12}=\sigma_{1}^{2},~{}~{}S_{1,\tau_{1}}=\tau_{1}\sigma_{1}^{-1},~{}~{}~{}S_{\sigma_{1},\tau_{1}}=\sigma_{1}\tau_{1}.$

We do not have relations between them. Hence, $UP_{2}$ is free of rank $3$.∎

######  Corollary 5.1.

The following properties hold:

  * 1)

$UB_{3}$ is linear;

  * 2)

$UP_{3}$ is residually nilpotent.

###### Proof.

1) Follows from the fact that the free product of two linear groups is linear.

2) Follows from the fact that the free product of two residually nilpotent
groups that do not have torsion is residually nilpotent. ∎

######  Remark 1.

If $n>3$, then we can not say that $UB_{n}$ is the free product of $B_{n}$ and
$T_{n}$. In this case we have mixed relations of the form

$[\sigma_{i},\tau_{j}]=1,~{}~{}|i-j|>1.$

Also, if $n>3$ we can not say that $UP_{n}$ is the free product of $P_{n}$ and
a free group. In this case we have relations of the form

$[\tau_{i}^{2},\tau_{j}^{2}]=1,~{}~{}|i-j|>1.$

######  Question 5.1.

For $n>3$ find a set of generators and defining relations of the group
$UP_{n}$.

## Acknowledgements

The results were obtained during the program on knots and braid groups advised
by V. Bardakov, T. Nasybullov and A. Vesnin in the frame of The First Workshop
at the Mathematical Center in Akademgorodok supported by the Ministry of
Science and Higher Education of the Russian Federation (agreement no.
075–2019–1675).

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Valeriy Bardakov∗,†,‡ (bardakov@math.nsc.ru),

Ivan Emel’yanenkov∗ (i.emelianenkov@g.nsu.ru),

Maxim Ivanov∗ (m.ivanov2@g.nsu.ru),

Tatyana Kozlovskaya† (t.kozlovskaya@math.tsu.ru),

Timur Nasybullov∗,†,‡ (timur.nasybullov@mail.ru),

Andrei Vesnin∗,†,‡<EMAIL_ADDRESS>
∗ Novosibirsk State University, Pirogova 1, 630090 Novosibirsk, Russia,
† Tomsk State University, Lenin Ave. 36, 634050 Tomsk, Russia,
‡ Sobolev Institute of Mathematics, Acad. Koptyug Ave. 4, 630090 Novosibirsk,
Russia,