Braid groups in handlebodies and corresponding Hecke algebras

TL;DR

This paper investigates the structure of braid groups in handlebodies, showing their kernels form semi-direct products of free groups, and introduces a new algebra analogous to Hecke algebras associated with these braid groups.

Contribution

It characterizes the kernel of the handlebody braid group homomorphism and introduces a new algebraic structure related to these braid groups.

Findings

Kernel of the handlebody braid group homomorphism is a semi-direct product of free groups

Introduces an algebra $H_{g,n}(q)$ analogous to the Hecke algebra

Provides structural insights into handlebody braid groups

Abstract

In this paper we study the kernel of the homomorphism $B_{g,n} \to B_n$ of the braid group $B_{g,n}$ in the handlebody $\mathcal{H}_g$ to the braid group $B_n$. We prove that this kernel is a semi-direct product of free groups. Also, we introduce an algebra $H_{g,n}(q)$, which is some analog of the Hecke algebra $H_n(q)$, constructed by the braid group $B_n$.

Equations

$${\sigma }_1,{\sigma }_2,\dots ,{\sigma }_{m-1},$$

$$\begin{array}[]{ll}\sigma_{i}\sigma_{j}=\sigma_{j}\sigma_{i},&~{}\mbox{for}~{}|i-j|>1,\\ \sigma_{i}\sigma_{i+1}\sigma_{i}=\sigma_{i+1}\sigma_{i}\sigma_{i+1},&~{}\mbox{for}~{}i=1,2,\ldots,m-2.\end{array}$$

$$a_{ij}={\sigma }_{j-1}{\sigma }_{j-2}\dots {\sigma }_{i+1}{\sigma }_i^2{\sigma }_{i+1}^{-1}\dots {\sigma }_{j-2}^{-1}{\sigma }_{j-1}^{-1},1\le i<j\le m,$$

$$a_{ik}{a}_{ij}{a}_{kj}=a_{kj}{a}_{ik}{a}_{ij},$$

$$a_{nj}{a}_{kn}{a}_{kj}=a_{kj}{a}_{nj}{a}_{kn},\text{mbox}forn<j,$$

$$(a_{kn}{a}_{kj}{a}_{kn}^{-1})a_{in}=a_{in}(a_{kn}{a}_{kj}{a}_{kn}^{-1}),\text{mbox}fori<k<n<j,$$

$$a_{kj}{a}_{in}=a_{in}{a}_{kj},\text{mbox}fork<i<n<j\text{mbox}orn<k.$$

$${\sigma }_k^{-1}{a}_{ij}{\sigma }_k=a_{ij},\text{mbox}fork\ne i-1,i,j-1,j,$$

$${\sigma }_i^{-1}{a}_{i,i+1}{\sigma }_i=a_{i,i+1},$$

$${\sigma }_{i-1}^{-1}{a}_{ij}{\sigma }_{i-1}=a_{i-1,j},$$

$${\sigma }_i^{-1}{a}_{ij}{\sigma }_i=a_{i+1,j}[a_{i,i+1}^{-1},a_{ij}^{-1}],\text{mbox}forj\ne i+1$$

$${\sigma }_{j-1}^{-1}{a}_{ij}{\sigma }_{j-1}=a_{i,j-1},$$

$${\sigma }_j^{-1}{a}_{ij}{\sigma }_j=a_{ij}{a}_{i,j+1}{a}_{ij}^{-1},$$

$$U_i=⟨a_{1i},a_{2i},\dots ,a_{i-1,i}⟩,i=2,\dots ,m,$$

$$a_{ik}^{-\epsilon }{a}_{kj}{a}_{ik}^{\epsilon }=(a_{ij}{a}_{kj}{)}^{\epsilon }{a}_{kj}(a_{ij}{a}_{kj}{)}^{-\epsilon },$$

$$a_{kn}^{-\epsilon }{a}_{kj}{a}_{kn}^{\epsilon }=(a_{kj}{a}_{nj}{)}^{\epsilon }{a}_{kj}(a_{kj}{a}_{nj}{)}^{-\epsilon },\text{mbox}forn<j,$$

$$a_{in}^{-\epsilon }{a}_{kj}{a}_{in}^{\epsilon }=[a_{ij}^{-\epsilon },a_{nj}^{-\epsilon }{]}^{\epsilon }{a}_{kj}[a_{ij}^{-\epsilon },a_{nj}^{-\epsilon }{]}^{-\epsilon },\text{mbox}fori<k<n,$$

$$a_{in}^{-\epsilon }{a}_{kj}{a}_{in}^{\epsilon }=a_{kj},\text{mbox}fork<i<n<j\text{mbox}orn<k.$$

$$P_m=U_m⋋(U_{m-1}⋋(\dots ⋋(U_3⋋U_2)\dots )).$$

$$1=U_m^{(m)}\le U_m^{(m-1)}\le \dots \le U_m^{(2)}\le U_m^{(1)}=P_m,$$

$$U_m^{(r)}/U_m^{(r+1)}\cong F_r,r=1,2,\dots ,m-1.$$

$$V_k=⟨a_{k,k+1},a_{k,k+2},\dots ,a_{k,m}⟩,k=1,2,\dots ,m-1.$$

$$P_m=V_1⋋(V_2⋋(\dots ⋋(V_{m-2}⋋V_{m-1})\dots )).$$

$$1=V_m^{(1)}\le V_m^{(2)}\le \dots \le V_m^{(m-1)}\le V_m^{(m)}=P_m,$$

$$V_m^{(r)}/V_m^{(r-1)}\cong F_{m-r+1},r=2,3,\dots ,m.$$

$$\begin{array}[]{ccccc}a_{12},&a_{13},&\ldots&a_{1,m-1},&a_{1,m},\\ &a_{23},&\ldots&a_{2,m-1},&a_{2,m},\\ \ldots&\ldots&\ldots&\ldots&\ldots\\ &&&a_{m-2,m-1},&a_{m-2,m},\\ &&&&a_{m-1,m},\end{array}$$

$${\tau }_1,{\tau }_2,\dots {\tau }_g,{\sigma }_{g+1},{\sigma }_{g+2},\dots ,{\sigma }_{g+n-1},$$

$$\begin{array}[]{ll}\sigma_{i}\sigma_{j}=\sigma_{j}\sigma_{i},&~{}\mbox{for}~{}|i-j|>1,\\ \sigma_{i}\sigma_{i+1}\sigma_{i}=\sigma_{i+1}\sigma_{i}\sigma_{i+1},&~{}\mbox{for}~{}i=g+1,\ldots,g+n-2,\\ \tau_{k}\sigma_{i}=\sigma_{i}\tau_{k},&~{}\mbox{for}~{}k\geq 1,~{}~{}i\geq g+2,\\ \tau_{k}(\sigma_{g+1}\tau_{k}\sigma_{g+1})=(\sigma_{g+1}\tau_{k}\sigma_{g+1})\tau_{k},&~{}\mbox{for}~{}k=1,\ldots,g,\\ \tau_{k}(\sigma_{g+1}^{-1}\tau_{k+l}\sigma_{g+1})=(\sigma_{g+1}^{-1}\tau_{k+l}\sigma_{g+1})\tau_{k},&~{}\mbox{for}~{}k=1,\ldots,g-1,~{}~{}l=1,\ldots,g-k.\end{array}$$

$${\tau }_k={\sigma }_g{\sigma }_{g-1}\dots {\sigma }_{k+1}{\sigma }_k^2{\sigma }_{k+1}^{-1}\dots {\sigma }_{g-1}^{-1}{\sigma }_g^{-1},k=1,2,\dots ,g.$$

$$\begin{array}[]{ccccc}a_{1,g+1},&a_{1,g+2},&\ldots&a_{1,g+n-1},&a_{1,g+n},\\ a_{2,g+1},&a_{2,g+2},&\ldots&a_{2,g+n-1},&a_{2,g+n},\\ \ldots&\ldots&\ldots&\ldots&\ldots\\ a_{g-1,g+1},&a_{g-1,g+2},&\ldots&a_{g-1,g+n-1},&a_{g-1,g+n},\\ a_{g,g+1},&a_{g,g+2},&\ldots&a_{g,g+n-1},&a_{g,g+n},\\ \hline\cr&a_{g+1,g+2},&\ldots&a_{g+1,g+n-1},&a_{g+1,g+n},\\ &\ldots&\ldots&\ldots&\ldots\\ &&&a_{g+n-2,g+n-1},&a_{g+n-2,g+n},\\ &&&&a_{g+n-1,g+n}.\\ \end{array}$$

$${\tilde{P}}_n={\tilde{U}}_n⋋({\tilde{U}}_{n-1}⋋(\dots ⋋({\tilde{U}}_3⋋{\tilde{U}}_2)\dots )),$$

$${\tilde{U}}_i=⟨a_{g+1,g+i},a_{g+2,g+i},\dots ,a_{g+i-1,g+i}⟩,i=2,3,\dots ,n.$$

$${\psi }_n:B_{g,n}⟶S_n,$$

$${\psi }_n({\tau }_k)=1,k=1,2,\dots ,g,{\psi }_n({\sigma }_i)=(i,i+1),i=g+1,g+2,\dots ,g+n-1.$$

$$1⟶P_{g,n}⟶B_{g,n}⟶S_n⟶1,$$

$$P_{g,n}=U_{g+n}⋋(U_{g+n-1}⋋(\dots ⋋(U_{g+2}⋋U_{g+1})\dots )).$$

Full text

1 Introduction

Let $\mathcal{H}_{g}$ be a handlebody of genus $g$. The braid group $B_{g,n}$ on $n$ stings in the handlebody $\mathcal{H}_{g}$ was introduced by A. B. Sossinsky S and independently S. Lambropoulou L1 . Properties of this group are studied by V. V. Vershinin V ; V1 and by S. Lambropolou L1 ; L2 .

The motivation for studying braids from $B_{g,n}$ comes from studying oriented knots and links in knot complements in compact connected oriented 3-manifolds and in handlebodies, since these spaces may be represented by a fixed braid or a fixed integer-framed braid in $\mathbb{S}^{3}$ LR1 ; L1 ; HL ; LR2 . Then knots and links in these spaces may be represented by elements of the braid group $B_{g,n}$ or an appropriate cosets of these group L1 . In particular, if $M$ denotes the complements of the $g$-unlink or a connected sum of $g$ lens spaces of type $L(p,1)$ or a handlebody of genus $g$, then knots and links in these spaces may be represented precisely by the braids in $B_{g,n}$ for $n\in\mathbb{N}$. In the case $g=1$, $B_{1,n}$ is the Artin group of type $\mathcal{B}$.

The group $B_{g,n}$ can be considered as a subgroup of the braid group $B_{g+n}$ on $g+n$ strings such that the braids in $B_{g,n}$ leave the first $g$ strings identically fixed. Using this fact V. V. Vershinin V and S. Lambropoulou L1 defined a epimorphism of $B_{g,n}$ onto the symmetric group $S_{n}$ and prove that the kernel of this epimorphism: $P_{g,n}$ is a subgroup of the pure braid group $P_{g+n}$ and is semi-direct products of free groups. On the other side, V. V. Vershinin V noted that there is a decomposition $B_{g,n}=R_{g,n}\leftthreetimes B_{n}$ for some group $R_{g,n}$ and the braid group $B_{n}$. S. Lambropoulou proved in L2 that $R_{1,n}$ is isomorphic to a free group $F_{n}$ of rank $n$.

The Hecke algebra of type $\mathrm{A}$, $H_{n}(q)$ was used by V. F. R. Jones J for the construction of polynomial invariant for classical links, the well known HOMFLYPT polynomial. S. Lambropoulou L2 defined a generalization of the Hecke algebra of type $\mathrm{B}$ to construct a HOMFLYPT-type polynomial invariant for links in the solid torus, which was then used in DiLaPr for extending the study to the lens spaces $L(p,1)$.

In the present paper we define some decomposition of $P_{g,n}$ as semi-direct product of free groups, which is different from the decomposition defined in L1 ; V . Also, we study the group $R_{g,n}$ and prove that this group is semi-direct products of free groups. Using this decomposition and the decomposition $B_{g,n}=R_{g,n}\leftthreetimes B_{n}$ we define some algebra, which is a generalization of the Hecke algebra $H_{n}(q)$.

The paper is organized as follows. In Section 2, we remind some facts on the braid group $B_{n}$. In particular, we define vertical and horizontal decompositions of the pure braid group $P_{n}$. In Section 3, we recall some facts on the group $B_{g,n}$, we describe the vertical decomposition of $P_{g,n}$ and we define the horizontal decomposition of $P_{g,n}$. In Section 4, we study $R_{g,n}$ and we construct vertical and horizontal decompositions for this group. In Section 5 we introduce some algebra $H_{g,n}(q)$ which is a generalization of the Hecke algebra $H_{n}(q)$ and we find the quotient of $B_{g,n}$ by the relations $\sigma_{i}^{2}=1$.

Acknowledgements.

The author gratefully acknowledges Prof. Lambropoulou and her students and colleagues: Neslihan Gügümcü, Stathis Antoniou, Dimos Goundaroulis, Ioannis Diamantis, Dimitrios Kodokostas for the kind invitation to the Athens, where this paper was written, for conversation and interesting discussions.

This work was supported by the Russian Foundation for Basic Research (project 16-01-00414).

2 Braid group

In this section we recall some facts on the braid groups (see Bir ; M ).

The braid group $B_{m}$, $m\geq 2$, on $m$ strings is generated by elements

[TABLE]

and is defined by relations

[TABLE]

A subgroup of $B_{m}$ which is generated by elements

[TABLE]

is called the pure braid group and is denoted $P_{m}$. This group is defined by the relations

[TABLE]

The subgroup $P_{m}$ is normal in $B_{m}$, and the quotient $B_{m}/P_{m}$ is the symmetric group $S_{m}$. The generators of $B_{m}$ act on the generator $a_{ij}\in P_{m}$ by the rules:

[TABLE]

where $[a,b]=a^{-1}b^{-1}ab$.

Denote by

[TABLE]

a subgroup of $P_{m}$. It is known that $U_{i}$ is a free group of rank $i-1$. One can rewrite the relations of $P_{m}$ as the following conjugation rules (for $\varepsilon=\pm 1$):

[TABLE]

From these rules it follows that $U_{m}$ is normal in $P_{m}$ and hence $P_{m}$ has the following decomposition: $P_{m}=U_{m}\leftthreetimes P_{m-1}$, where the action of $P_{m-1}$ on $U_{m}$ is define by the rules (11-14). By induction on $m$, $P_{m}$ is the semi-direct product of free groups:

[TABLE]

We will call this decomposition vertical decomposition.

Let $U_{m}^{(k)}$, $k=1,2,\ldots,m$, be the subgroup of $P_{m}$ which is generated by $a_{ij}$, where $k<j$. Then $U_{m}^{(k)}=U_{m}\leftthreetimes(U_{m-1}\leftthreetimes(\ldots\leftthreetimes(U_{k+2}\leftthreetimes U_{k+1})\ldots))$. By definition $U_{m}^{(1)}=P_{m}$ and these groups form the normal series

[TABLE]

where

[TABLE]

On the other side, define the following subgroups of $P_{m}$:

[TABLE]

This group is free of rank $m-k$. Using the defining relation of $P_{m}$, it is not difficult to prove the following:

Lemma 1

In $P_{n}$ hold the following conjugation rules (for $\varepsilon=\pm 1$):

1) $a_{kj}^{-\varepsilon}a_{ik}a_{kj}^{\varepsilon}=(a_{ik}a_{ij})^{\varepsilon}a_{ik}(a_{ik}a_{ij})^{-\varepsilon},$ where $i<k<j$;

2) $a_{jk}^{-\varepsilon}a_{ik}a_{jk}^{\varepsilon}=(a_{ij}a_{ik})^{\varepsilon}a_{ik}(a_{ij}a_{ik})^{-\varepsilon},$ where $i<j<k$;

3) $a_{kn}^{-\varepsilon}a_{ij}a_{kn}^{\varepsilon}=[a_{ik}^{-\varepsilon},a_{in}^{-\varepsilon}]^{\varepsilon}a_{ij}[a_{ik}^{-\varepsilon},a_{in}^{-\varepsilon}]^{-\varepsilon},$ where $i<k<j<n$;

4) $a_{in}^{-\varepsilon}a_{kj}a_{in}^{\varepsilon}=a_{kj},$ where $k<i$ and $n<j$;

5) $a_{kj}^{-\varepsilon}a_{in}a_{kj}^{\varepsilon}=a_{in},$ where $n<k$.

From this lemma it follows that $V_{1}$ is normal in $P_{m}$ and we have decomposition $P_{m}=V_{1}\leftthreetimes P_{m-1}$. By induction on $m$, $P_{m}$ is the semi-direct products of free groups:

[TABLE]

We will call this decomposition horizontal decomposition. Let $V_{m}^{(k)}$ be a subgroup of $P_{m}$ which is generated by $a_{ij}$ for $i<k$. Then we have the normal series

[TABLE]

where

[TABLE]

A motivation for the terms vertical and horizontal is as follows. If we put the generators of $P_{m}$ in the following table

[TABLE]

then the generators from the $k$th row generate $V_{k}$ and the generators from the $k$th column generate $U_{k+1}$. The group $U_{m}^{(r)}$ is generated by the last $m-r$ columns of this table and the group $V_{m}^{(r)}$ is generated by the first $r-1$ rows of this table.

3 Braid groups in handlebodies

Recall some facts on the braid group $B_{g,n}$ on $n$ strings in the handlebody $\mathcal{H}_{g}$ (see L1 ; S ; V ). The group $B_{g,n}$ is generated by elements

[TABLE]

and is defined by the following list of relations

[TABLE]

The group $B_{g,n}$ can be considered as a subgroup of the classical braid group $B_{g+n}$ on $n+g$ strings such that the braids from $B_{g,n}$ leave the first $g$ strings unbraided. Then $\tau_{k}=a_{k,g+1}$, in $B_{g+n}$, i. e.

[TABLE]

The elements $\tau_{k}$, $k=1,2,\ldots,g$, generate a free group of rank $g$ which is isomorphic to $U_{g+1}=\langle a_{1,g+1},a_{2,g+1},\ldots,a_{g,g+1}\rangle$ in $B_{g+n}$. Also, we see that some other generators of $P_{g+n}$ lie in $B_{g,n}$. Put them in the table bellow:

[TABLE]

We will denote by $\widetilde{B}_{n}$ the subgroup of $B_{g,n}$ which is generated by $\sigma_{g+1},$ $\sigma_{g+2},$ $\ldots,\sigma_{g+n-1}$. It is evident that $\widetilde{B}_{n}$ is isomorphic to $B_{n}$. The corresponding pure braid group of $\widetilde{B}_{n}$ will be denote $\widetilde{P}_{n}$. This group is isomorphic to $P_{n}$ and is generated by elements from the above table which lie under the horizontal line. The group $\widetilde{P}_{n}$ has the following vertical decomposition

[TABLE]

where

[TABLE]

There is an epimorphism:

[TABLE]

which is defined by the rule

[TABLE]

This epimorphism is induced by the standard epimorphism $B_{g+n}\longrightarrow S_{g+n}$. Let $P_{g,n}=\mathrm{Ker}(\psi_{n})$. Then $P_{g,n}$ is generated by the element from the table above. In L1 ; V it was proved that there exists the following short exact sequence:

[TABLE]

and was found the vertical decomposition of $P_{g,n}$:

[TABLE]

If we let $U_{g+n}^{(g+n-i)}=U_{g+n}\leftthreetimes(U_{g+n-1}\leftthreetimes(\ldots\leftthreetimes(U_{g+n-i+2}\leftthreetimes U_{g+n-i+1})\ldots))$, then we get the normal series:

[TABLE]

where

[TABLE]

The homomorphism which is induced by the embedding $B_{g,n}\longrightarrow B_{g+n}$ sends this normal series to the corresponding normal series for $P_{g+n}$.

Let us construct the horizontal decomposition for $P_{g,n}$. To do this, define the following subgroups in $P_{g,n}$:

[TABLE]

We see that $V_{g,g+i}=V_{g+i}$ for all $i=1,2,\ldots,n-1$, these subgroups lie in $\widetilde{P}_{n}$ and as we know

[TABLE]

is the horizontal decomposition of $\widetilde{P}_{n}$.

We see that the vertical decomposition of $P_{g,n}$ is a part of the vertical decomposition for $P_{g+n}$. For the horizontal decomposition situation is more complicated.

Example 1

The horizontal decomposition of $P_{4}$ has the form

[TABLE]

where

[TABLE]

and $V_{1}$ is normal in $P_{4}$, $V_{2}$ is normal in $V_{2}\leftthreetimes V_{3}$. The group $P_{2,2}$ contains subgroups

[TABLE]

but in this case $V_{2,1}$ is not normal in $P_{2,2}$, Indeed, from Lemma 1 we have the following relations

[TABLE]

Hence, $P_{2,2}$ contains not only $V_{2,1}$ but its normal closure in $P_{2,2}$ and we get the horizontal decomposition

[TABLE]

where $\overline{V}_{2,1}=\langle V_{2,1}\rangle^{P_{2,2}}$ is the normal closure of $V_{2,1}$ in $P_{2,2}$.

In the general case, let $\overline{V}_{g,i}$ be the normal closure of $V_{g,i}$ in the subgroup $\langle V_{g,i},V_{g,i+1},\ldots,V_{g,g},\widetilde{P}_{n}\rangle$, i.e.

[TABLE]

Lemma 2

1) $\overline{V}_{g,g}=V_{g,g}\cong F_{n}.$

2) $\overline{V}_{g,i}$ is a subgroup of $V_{i}$ for every $i=1,2,\ldots,g$ and, in particular, is a free group.

Proof

1. We see that $V_{g,g}=V_{g}$ and from Lemma 1 it follows that $V_{g}$ is normal in $\langle V_{g,g},\widetilde{P}_{n}\rangle$.

2. The fact that $\overline{V}_{g,i}$ is a subgroup of $V_{i}$ follows from the conjugation rules of Lemma 1. The fact that $\overline{V}_{g,i}$ is free follows from the fact that $V_{i}$ is free.

Since $\widetilde{P}_{n}\cong P_{n}$, it has the horizontal decomposition:

[TABLE]

Using Lemma 2, one can construct the horizontal decomposition of $P_{g,n}$.

Theorem 3.1

The group $P_{g,n}$ is the semi-direct products of groups:

[TABLE]

Proof

Use induction on $g$. If $g=1$, then $P_{1,n}=P_{n+1}$ and the horizontal decomposition for $P_{n+1}$ gives the horizontal decomposition: $P_{1,n}=V_{g}\leftthreetimes\widetilde{P}_{n}$. As follows from Lemma 2, $V_{g}=\overline{V}_{g,g}$.

Let $g>1$. Define a homomorphism of $P_{g,n}$ onto the group $P_{g-1,n}$ which sends all generators of $V_{g,1}$ to the unit and keeps all other generators. The kernel of this homomorphism is the normal closure of $V_{g,1}$ in $P_{g,n}$. Denote this kernel by $\overline{V}_{g,1}=\langle V_{g,1}\rangle^{P_{g,n}}$. Since, $V_{g,1}$ is a subgroup in $V_{1}$ and $V_{1}$ is normal in $P_{g+n}$ we get that $\overline{V}_{g,1}$ lies in $V_{1}$ and hence is a free. We have the decomposition $P_{g,n}=\overline{V}_{g,1}\leftthreetimes P_{g-1,n}$. Using the induction hypothesis we get the required decomposition.

4 The kernel of the epimorphism $B_{g,n}\to\widetilde{B}_{n}$

As was noted in V there is an epimorphism

[TABLE]

where the subgroup $\widetilde{B}_{n}=\langle\sigma_{g+1},\sigma_{g+2},\ldots,\sigma_{g+n-1}\rangle$ is isomorphic to $B_{n}$. This endomorphism is defined by the rule

[TABLE]

If we denote $R_{g,n}=\mathrm{Ker}(\varphi_{n})$, then $B_{g,n}=R_{g,n}\leftthreetimes\widetilde{B}_{n}$. The purpose of this section is the description of the group $R_{g,n}$.

Considering the table with the generators of $P_{g,n}$ (see Section 3), we see that all generators, which lie in the fist $g$ rows of this table, are elements of $R_{g,n}$. Denote by $Q_{g,n}$ the subgroup of $R_{g,n}$ that is generated by these elements, i.e. $Q_{g,n}=\langle V_{g,1},V_{g,2},\ldots,V_{g,g}\rangle$. Then $R_{g,n}=\langle Q_{g,n}\rangle^{B_{g,n}}$ is the normal closure of $Q_{g,n}$ in $B_{g,n}$.

On the other side, if we denote

[TABLE]

then $Q_{g,n}=\langle U_{g,g+1},U_{g,g+2},\ldots,U_{g,g+n}\rangle$. Note that $U_{g,g+1}=U_{g+1}$. Let $\overline{U}_{g,g+i}$ be the normal closure of $U_{g,g+i}$ in $U_{g+i}$, $i=1,2,\ldots,n$. In this notations it holds:

Theorem 4.1

The group $R_{g,n}$ has the following decompositions:

*1)  $R_{g,n}=\overline{U}_{g,g+n}\leftthreetimes(\overline{U}_{g,g+n-1}\leftthreetimes(\ldots\leftthreetimes(\overline{U}_{g,g+2}\leftthreetimes\overline{U}_{g,g+1})\ldots)),$

where $\overline{U}_{g,g+1}=U_{g+1}$.*

*2)  $R_{g,n}=\overline{V}_{g,1}\leftthreetimes(\overline{V}_{g,2}\leftthreetimes(\ldots\leftthreetimes(\overline{V}_{g,g-1}\leftthreetimes\overline{V}_{g,g})\ldots)),$

where $\overline{V}_{g,g}=V_{g}$.*

Proof

As we know, $R_{g,n}$ is the normal closure of $Q_{g,n}$ in $B_{g,n}$. To find this closure, at first, consider conjugations of the generators of $Q_{g,n}$ by $\sigma_{g+k}$, $k=1,2,\ldots,n-1$.

Conjugating generators from the $k$th column of our table by $\sigma_{g+k}$, we have

[TABLE]

Conjugating generators from the $(k+1)$st column of our table by $\sigma_{g+k}$, we have

[TABLE]

Generators from all other columns commute with $\sigma_{g+k}$.

Hence, for every generator $a_{ij}$ of $Q_{g,n}$ and every $\sigma_{k}$, $k=g+1,g+2,\ldots,g+n-1$, the element $\sigma_{k}^{-1}a_{ij}\sigma_{k}$ lies in $Q_{g,n}$.

For the group $\widetilde{B}_{n}$ there exists the following exact sequence

[TABLE]

Let $m_{kl}=\sigma_{k-1}\,\sigma_{k-2}\ldots\sigma_{l}$ for $l<k$ and $m_{kl}=1$ in other cases. Then the set

[TABLE]

is a Schreier set of coset representatives of $\widetilde{P}_{n}$ in $\widetilde{B}_{n}$.

From the previous observations it follows that if $\alpha\in\Lambda$, then for every generator $a_{ij}$ of $Q_{g,n}$ the element $\alpha^{-1}a_{ij}\alpha$ lies in $Q_{g,n}$.

Now we will consider the conjugations of the generators of $Q_{g,n}$ by the generators of $P_{g,n}$.

1. To prove the first decomposition, take some element $h\in B_{g,n}$ and using the vertical decomposition, write it in the normal form:

[TABLE]

In every group $U_{k}$, $k=g+1,g+2,\ldots,g+n$, we have two subgroups:

[TABLE]

We see that $U_{g,k}$ is a subgroup of $P_{g,n}$, $\widetilde{U}_{k}$ is a subgroup of $\widetilde{P}_{n}$ and $U_{k}=\langle U_{g,k},\widetilde{U}_{k}\rangle$ is a free group. Define the projection $\pi_{k}:U_{k}\longrightarrow\widetilde{U}_{k}$ by the rules

[TABLE]

[TABLE]

The kernel $\mathrm{Ker}(\pi_{k})=\overline{U}_{g,k}$ is the normal closure $\langle U_{g,k}\rangle^{U_{k}}$ of $U_{g,k}$ into $U_{k}$. Hence, element $u_{k}$ can be written in the form $u_{k}=\overline{u}_{k}\pi_{k}(u_{k})$ for some $\overline{u}_{k}\in\overline{U}_{g,k}$. Denote for simplicity $w_{g+i}=\pi_{g+1}(u_{g+i})$, we can rewrite $h$ in the form

[TABLE]

Shifting all $w_{g+i}$ to the right we get

[TABLE]

The image $\varphi_{n}(h)=w_{g+1}w_{g+2}\ldots w_{g+n}\alpha$. Hence, the element

[TABLE]

lies in the kernel $\mathrm{Ker}(\varphi_{n})=R_{g,n}$. We proved that any element from $R_{g,n}$ lies in the product

[TABLE]

On the other side, this product contains $Q_{g,n}$ and is normal in $B_{g,n}$. Hence, $R_{g,n}$ has the required decomposition.

2. Prove the second decomposition. Denote $G=\overline{V}_{g,1}\leftthreetimes(\overline{V}_{g,2}\leftthreetimes(\ldots\leftthreetimes(\overline{V}_{g,g-1}\leftthreetimes\overline{V}_{g,g})\ldots))$ the group from the right side of the decomposition of $R_{g,n}$. Take an arbitrary element $h\in B_{g,n}$. By the theorem on the horizontal decomposition of $B_{g,n}$ it has the following normal form:

[TABLE]

and $v_{1}v_{2}\ldots v_{g}$ lies in $R_{g,n}$. Under the endomorphism $\varphi_{n}:B_{g,n}\longrightarrow\widetilde{B}_{n}$ element $h$ goes to the element $b$. Hence, $G$ lies in $R_{g,n}$.

On the other side we must prove that $R_{g,n}$ lies in $G$. We know that $R_{g,n}$ is the normal closure of $Q_{g,n}$ in $B_{g,n}$. As was shown before, if $\alpha\in\Lambda_{n}$ is a coset representative of $\widetilde{P}_{n}$ into $\widetilde{B}_{n}$ and $a_{ij}$ is some generator of $Q_{g,n}$, then $a_{ij}^{\alpha}$ lies in $Q_{g,n}$. Considering the conjugation of $a_{ij}\in Q_{g,n}$ by generators of $P_{g,n}$ and using the Lemma 1 we get an element which lies in $G$.

In the case $g=1$ we have $P_{1,n}=P_{n+1}$ and $R_{1,n}=\langle a_{12},a_{13},$ $\ldots,$ $a_{1,n+1}\rangle\cong F_{n}$ and we get the decomposition which was found in L2 .

Corollary 1

$B_{1,n}\cong F_{n}\leftthreetimes B_{n}$.

5 Some analog of the Hecke algebra for the braid group in the handlebody

Let $q$ be some complex number. Recall that the Hecke algebra $H_{n}(q)$ is an associative $\mathbb{C}$-algebra with unit, which is generated by elements

[TABLE]

and is defined by the relations

[TABLE]

The algebra $H_{n}(q)$ has the following linear basis:

[TABLE]

for $1\leq i_{1}<\ldots<i_{p}\leq n-1$. The basis $S$ is used in the construction of the Markov trace, leading to the HOMFLYPT or $2$-variable Jones polynomial (see J ).

The braid group in the solid torus $B_{1,n}$ is the Artin group of the Coxeter group of type $\mathcal{B}$, which is related to the Hecke algebra of type $\mathcal{B}$. The generalized Hecke algebra of type $\mathcal{B}$, $H_{1,n}(q)$ is defined by S. Lambropoulou in L . $H_{1,n}(q)$ is isomorphic to the affine Hecke algebra of type $\mathcal{A}$, $\widetilde{H}_{n}(q)$. A unique Markov trace is constructed on the algebras $H_{1,n}(q)$ that leads to an invariant for links in the solid torus, the universal analogue of the HOMFLYPT polynomial for the solid torus.

The algebra $H_{1,n}(q)$ is generated by elements

[TABLE]

and is defined by the relations

[TABLE]

Hence,

[TABLE]

Note that in $H_{1,n}(q)$ the generator $t$ satisfies no polynomial relation, making the algebra $H_{1,n}(q)$ infinite dimensional. If we set $t=0$ in $H_{1,n}(q)$, we obtain the Hecke algebra $H_{n}(q)$.

In $H_{1,n}(q)$ are defined in L2 the elements

[TABLE]

It was then proved that the following sets form linear bases for $H_{1,n}(q)$:

[TABLE]

[TABLE]

where $k_{1},k_{2},\ldots,k_{r}\in\mathbb{Z}$ and $\sigma$ a basis element in $H_{n}(q)$. The basis $\Sigma^{\prime}_{n}$ is used in L ; L2 for constructing a Markov trace on $\bigcup_{n=1}^{\infty}H_{1,n}(q)$ and a universal HOMFLYPT-type invariant for oriented links in the solid torus.

Definition 1

Let $q\in\mathbb{C}$. The algebra $H_{g,n}(q)$ is an associative algebra over $\mathbb{C}$ with unit that is generated by

[TABLE]

and is defined by the following relations

[TABLE]

Remark 1

More natural to consider the generators $s_{1},s_{2},\ldots s_{n-1}$ instead $s_{g+1},s_{g+2},\ldots s_{g+n-1}$, but for technical reasons we will use our notation.

We see that

[TABLE]

If we consider the algebra $H_{n}(q)$ as a vector space over $\mathbb{C}$, then it is isomorphic to the vector space $\mathbb{C}[S_{n}]$. Thus to study $H_{g,n}(q)$, we define a group $G_{g,n}$ as the quotient of $B_{g,n}$ by the relations

[TABLE]

Denote the natural homomorphism $B_{g,n}\longrightarrow G_{g,n}$ by $\psi$ and let $\psi(a_{ij})=b_{ij}$.

Theorem 5.1

The group $G_{g,n}$ is the semi-direct product $G_{g,n}=F_{g}^{n}\leftthreetimes S_{n}$ of the direct product of $n$ copies of free group $F_{g}$ of rank $g$ and the symmetric group $S_{n}$.

Proof

We know that $B_{g,n}=R_{g,n}\leftthreetimes\widetilde{B}_{n}$ and

[TABLE]

On the other side, $U_{k}$, $k=g+1,g+2,\ldots,g+n$, is generated by two subgroups:

[TABLE]

Note that under the homomorphism $\psi$ the subgroups $\widetilde{U}_{k}$ go to 1. Hence, $\psi(\overline{U}_{g,g+k})=\psi(U_{g,g+k})\cong F_{g}$ and

[TABLE]

is the semi-direct product of $n$ copies of free group $F_{g}$. Also, $G_{g,n}=\psi(R_{g,n})\leftthreetimes\psi(\widetilde{B}_{n})\cong\psi(R_{g,n})\leftthreetimes S_{n}$.

Let us show that in fact $\psi(R_{g,n})$ is the direct product of $n$ copies of free group $F_{g}$. To do it it is enough to prove that

[TABLE]

Denote $b_{ij}=\psi(a_{ij})$, consider the relations (11)-(14) and find their images under the homomorphism $\psi$. The relation (11) goes to the relation

[TABLE]

but $b_{g+j,g+n}=1$ and we have the trivial relations.

The relation (12) goes to the relation

[TABLE]

but $b_{g+j,g+n}=1$ and we have the relation

[TABLE]

The relation (13) goes to the relation

[TABLE]

but $b_{g+j,g+n}=1$ and we have the relation

[TABLE]

The relation (14) goes to the relation

[TABLE]

The image of $\widetilde{B}_{n}\leq B_{g,n}$ is isomorphic to $S_{n}$. Thus from the decomposition $B_{g,n}=R_{g,n}\leftthreetimes\widetilde{B}_{n}$ we get the required decomposition for $G_{g,n}$.

From this theorem we have

Corollary 2

Every element $h\in G_{g,n}$ has the unique normal form

[TABLE]

where $h_{i}$ is a reduced word in the free group

[TABLE]

$\alpha\in\Lambda_{n}$* is a coset representative of $\psi(R_{g,n})$ in $G_{g,n}$.*

Using this normal form and the fact that these normal forms form a linear basis of $\mathbb{C}[G_{g,n}]$, we can try to define a basis of $H_{g,n}(q)$. In the algebra $H_{g,n}(q)$ define the following elements

[TABLE]

where

[TABLE]

These elements correspond to the elements $b_{ij}$ from $G_{g,n}$, which were defined in 5.1 as the images of the elements $a_{ij}$ under the map $\psi$. It is not difficult to see that the elements

[TABLE]

generate a free group of rank $g$.

The algebra $H_{g,n}(q)$ contains the subalgebra with the set of generators $s_{g+1},s_{g+2},\ldots s_{g+n-1}$, which is isomorphic to $H_{n}(q)$. Let $\Sigma_{g,n}$ be the following set in $H_{g,n}(q)$:

[TABLE]

where $u_{i}$ is a reduced word in the free group

[TABLE]

$\sigma$ is a basis element in $H_{n}(q)$.

Conjecture 1

There is an isomorphism $H_{g,n}(q)\cong\mathbb{C}[G_{g,n}]$ as $\mathbb{C}$-modules. In particular, the set $\Sigma_{g,n}$ is a basis of the algebra $H_{g,n}(q)$.

The algebra $H_{2,n}(q)$ is the subject of study in KL , where one set of elements in $H_{2,n}(q)$, different from our $\Sigma_{2,n}$, is proved to be a spanning set for the algebra.

At the end, we formulate the following question for further investigation.

Question 1

Is it possible to define a Markov trace on the algebra $\bigcup_{n=1}^{\infty}H_{g,n}(q)$ and construct some analogue of the HOMLYPT polynomial that is an invariant of links in the handlebody $\mathcal{H}_{g}$?