Forks, Noodles and the Burau representation for $n=4$

TL;DR

This paper investigates the faithfulness of the Burau representation for four-strand braids using homological techniques, proposing a conjecture and providing computational and geometric evidence.

Contribution

It introduces a new approach using forks and noodles homological techniques to analyze the Burau representation at n=4, including a conjecture on its faithfulness.

Findings

Proposes a conjecture implying faithfulness for n=4

Provides geometric examples and computational data

Offers arguments supporting the conjecture's validity

Abstract

\begin{abstract} The reduced Burau representation is a natural action of the braid group $B_n$ on the first homology group $H_1({\tilde{D}}_n;\mathbb{Z})$ of a suitable infinite cyclic covering space ${\tilde{D}}_n$ of the $n$--punctured disc $D_n$. It is known that the Burau representation is faithful for $n\le 3$ and that it is not faithful for $n\ge 5$. We use forks and noodles homological techniques and Bokut--Vesnin generators to analyze the problem for $n=4$. We present a Conjecture implying faithfulness and a Lemma explaining the implication. We give some arguments suggesting why we expect the Conjecture to be true. Also, we give some geometrically calculated examples and information about data gathered using a C\texttt{++} program.

Equations

$$\rho :B_4\to Aut\left(H_1\left({\tilde{D}}_4;\mathbb{Z}\right)\right)$$

$$\rho \left(\sigma \right)={\tilde{\phi }}_{\ast },\forall \sigma \in B_4.$$

$$⟨-,-⟩:H_1\left({\tilde{D}}_4;\mathbb{Z}\right)\times H_1\left({\tilde{D}}_4,\partial {\tilde{D}}_4;\mathbb{Z}\right)\to \mathbb{Z}\left[t,t^{-1}\right].$$

$$⟨-,-⟩:H_1\left({\tilde{D}}_4;\mathbb{Z}\right)\times H_1\left({\tilde{D}}_4,\partial {\tilde{D}}_4;\mathbb{Z}\right)\to \mathbb{Z}\left[t,t^{-1}\right]$$

$$⟨F,N⟩=\sum _{1\le i\le n}{\epsilon }_i{t}^{e_i}\in \mathbb{Z}\left[t,t^{-1}\right].$$

$${\rho }_{ij}\left(\sigma \right)=⟨F_i\sigma ,N_j⟩.$$

$$\rho\left({\sigma}_{1}\right)=\left(\begin{array}[]{ccc}-t^{-1}&0&0\\ t^{-1}&1&0\\ 0&0&1\end{array}\right),\rho\left({\sigma}_{2}\right)=\left(\begin{array}[]{ccc}1&1&0\\ 0&-t^{-1}&0\\ 0&t^{-1}&1\end{array}\right)$$

$$\rho\left({\sigma}_{3}\right)=\left(\begin{array}[]{ccc}1&0&0\\ 0&1&1\\ 0&0&-t^{-1}\end{array}\right).$$

$${\rho }_{11}({\sigma }_1)=⟨F_1{\sigma }_1,N_1⟩=-t^{-1}$$

$$B_4=⟨{\sigma }_1,{\sigma }_2,{\sigma }_3|{\sigma }_1{\sigma }_2{\sigma }_1={\sigma }_2{\sigma }_1{\sigma }_2,{\sigma }_3{\sigma }_2{\sigma }_3={\sigma }_2{\sigma }_3{\sigma }_2,{\sigma }_3{\sigma }_1={\sigma }_1{\sigma }_3⟩,$$

$$B_3=⟨{\sigma }_1,{\sigma }_2|{\sigma }_1{\sigma }_2{\sigma }_1={\sigma }_2{\sigma }_1{\sigma }_2⟩.$$

$$\phi \left({\sigma }_1\right)={\sigma }_1,\phi \left({\sigma }_2\right)={\sigma }_2,\phi \left({\sigma }_3\right)={\sigma }_1.$$

$$a={\sigma }_1{\sigma }_2{\sigma }_1^{-1}{\sigma }_3{\sigma }_2^{-1}{\sigma }_1^{-1},b={\sigma }_3{\sigma }_1^{-1}.$$

$$\begin{gathered} U_1^1\to U_1^1, \\ {U}_2^2\to U_2^2, \\ {U}_3^3\to U_1^1. \end{gathered}$$

$$\rho\left(a\right)=\left(\begin{array}[]{ccc}-t^{-1}+1&-t^{-1}+t&-t^{-1}\\ 0&-t&0\\ -1&0&0\end{array}\right),$$

$$\rho\left(b\right)=\left(\begin{array}[]{ccc}-t&0&0\\ 1&1&1\\ 0&0&-t^{-1}\end{array}\right).$$

$$\rho\left(a^{-1}\right)=\left(\begin{array}[]{ccc}0&0&-1\\ 0&-t^{-1}&0\\ -t&t^{-1}-t&1-t\end{array}\right),$$

$$\rho\left(b^{-1}\right)=\left(\begin{array}[]{ccc}-t^{-1}&0&0\\ t^{-1}&1&t\\ 0&0&-t\end{array}\right).$$

$${\rho }_{11}\left(a^n\sigma \right)=P\left(t,t^{-1}\right)\left(1-t^{-1}\right)+Q\left(t,t^{-1}\right)+R\left(t,t^{-1}\right)(1-t),$$

$${\rho }_{31}\left(a^n\sigma \right)=-P\left(t,t^{-1}\right)-Q\left(t,t^{-1}\right)-R\left(t,t^{-1}\right).$$

$${\rho }_{11}\left(a^{m+n}\sigma \right)=P\left(t,t^{-1}\right)\left(S_{m+1}(t^{-1})\right)+Q\left(t,t^{-1}\right)+R\left(t,t^{-1}\right)\left(S_{m+1}(t)\right),$$

$${\rho }_{31}\left(a^{m+n}\sigma \right)=-P\left(t,t^{-1}\right)\left(S_m(t^{-1})\right)-Q\left(t,t^{-1}\right)-R\left(t,t^{-1}\right)\left(S_m(t)\right).$$

$${\rho }_{11}\left(a^n\sigma \right)=⟨F_1{a}^n,N_1{\sigma }^{-1}⟩,$$

$${\rho }_{31}\left(a^n\sigma \right)=⟨F_3{a}^n,N_1{\sigma }^{-1}⟩.$$

$${\rho }_{11}\left(a^n\sigma \right)=P\left(t,t^{-1}\right)\left(1-t^{-1}\right)+Q\left(t,t^{-1}\right)+R\left(t,t^{-1}\right)(1-t).$$

$${\rho }_{31}\left(a^n\sigma \right)=-P\left(t,t^{-1}\right)-Q\left(t,t^{-1}\right)-R\left(t,t^{-1}\right).$$

$${\rho }_{11}\left(a^{n+1}\sigma \right)=P\left(t,t^{-1}\right)\left(1-t^{-1}+t^{-2}\right)+$$

$$Q\left(t,t^{-1}\right)+R\left(t,t^{-1}\right)\left(1-t^1+t^2\right),$$

$${\rho }_{31}\left(a^{n+1}\sigma \right)=-P\left(t,t^{-1}\right)\left(1-t^{-1}\right)-Q\left(t,t^{-1}\right)-R\left(t,t^{-1}\right)\left(1-t^1\right).$$

$${\rho }_{11}\left(a^2{b}^{-1}{a}^{-1}b\right)=-t^{-3}+t^{-2}-1+2t-t^2-t^3+2t^4-t^5=$$

Full text

1 Introduction

Let us recall the definition of the reduced Burau representation in terms of the first homology group $H_{1}({\tilde{D}}_{4};\mathbb{Z})$ of a suitable infinite cyclic covering space ${\tilde{D}}_{4}$ of the $4$-punctured disc $D_{4}$. Let $D_{4}$ be the unit closed disc on the plane with center $(0,0)$ and four punctures at: $p_{1}=\left(-\frac{1}{2},\frac{1}{2}\right)$, $\ p_{2}=\left(\frac{1}{2},\frac{1}{2}\right)$, $p_{3}=\left(\frac{1}{2},-\frac{1}{2}\right)$, $\ p_{4}=\left(-\frac{1}{2},-\frac{1}{2}\right)$ (see Figure 1).

The braid group $B_{4}$ is the group of all equivalence classes of orientation preserving homeomorphisms $\varphi:D_{4}\to D_{4}$ which fix the boundary $\partial D_{4}$ pointwise, where equivalence relation is isotopy relative to $\partial D_{4}$. Let ${\pi}_{1}\left(D_{4}\right)$ be the fundamental group of the 4–punctured disc $D_{4}$ with respect to the basepoint $p_{0}=(-1,0)$. Consider the map $\varepsilon:{\pi}_{1}\left(D_{4}\right)\to\left\langle t\right\rangle$ which sends a loop $\gamma\in{\pi}_{1}\left(D_{4}\right)$ to $t^{\left[\gamma\right]}$, where $\left[\gamma\right]$ is the winding number of $\gamma$ around punctured points $p_{1},p_{2},p_{3},p_{4}$ (meaning: the sum of the four winding numbers for individual points). Let $\pi:{\tilde{D}}_{4}\to D_{4}$ be the infinite cyclic covering space corresponding to the kernel $\ker\left(\varepsilon\right)$ of the map $\varepsilon:{\pi}_{1}\left(D_{4}\right)\to\left\langle t\right\rangle$. Let ${\tilde{p}}_{0}$ be any fixed basepoint which is a lift of the basepoint $p_{0}$. In this case $H_{1}({\tilde{D}}_{4};\mathbb{Z})$ is free $\mathbb{Z}\left[t,t^{-1}\right]$–module of rank 3 (see [3]). Let $\varphi:D_{4}\to D_{4}$ be a homeomorphism representing of an element $\sigma\in B_{4}$. It can be lifted to a map $\widetilde{\varphi}:{\tilde{D}}_{4}\to{\tilde{D}}_{4}$ which fixes the fiber over $p_{0}.$ Therefore it induces a $\mathbb{Z}\left[t,t^{-1}\right]$–module automorphism ${\widetilde{\varphi}}_{*}:H_{1}\left({\tilde{D}}_{4};\mathbb{Z}\right)\to H_{1}\left({\tilde{D}}_{4};\mathbb{Z}\right)$. Consequently, the reduced Burau representation

[TABLE]

is given [3] by

[TABLE]

It is known that the Burau representation is faithful for $n\leq 3$ [1], [2] and it is not faithful for $n\geq 5$ [2], [6], [7]. Therefore, the problem is open for $n=4$. In this paper, we use the Bokut-Vesnin generators $a,a^{-1},b,b^{-1}$ of a certain free subgroup of $B_{4}$ (see [4]) and a technique developed in [3], to prove the crucial lemma, which gives the opportunity to decompose entries $\rho_{11}(a^{n}\sigma)$ and $\rho_{13}(a^{n}\sigma)$ of the Burau matrix $\rho(a^{n}\sigma)$ as a sum of three uniquely determined polynomials and the formula to calculate $\rho_{13}(a^{n+m}\sigma)$ and $\rho(a^{n+m}\sigma)$ polynomials using the given decomposition. Besides, we formulate Conjecture 4.2, which implies that if a non–trivial braid $\sigma\in\ker\rho$ has a certain additional property, then there exists a sufficiently large $l_{0}$ with respect to the length of $\sigma$ (to be explained in Section 3, Corollary 3.2) and a sufficiently large $m_{0}$ such that for each $m>m_{0}$ and $l>l_{0}$ the difference of lowest degrees of polynomials $\rho_{13}(a^{n+m}\sigma)$ and $\rho(a^{n+m}\sigma)$ is $-1$. We will present arguments and experimental data showing why we expect the conjecture to be true. Also, we will consider several examples calculated geometrically. We will show that the conjecture implies faithfulness of the Burau representation for $n=4$.

2 The Burau representation, Forks and Noodles

The Burau representation for $n=4$ was defined by (1.1) and (1.2). On the other hand $H_{1}({\tilde{D}}_{4};\mathbb{Z})$ is a free $\mathbb{Z}\left[t,t^{-1}\right]$–module of rank 3 and if we take a basis of it, then $Aut\left(H_{1}\left({\tilde{D}}_{4};\mathbb{Z}\right)\right)$ can be identified with $GL\left(3,\ \mathbb{Z}\left[t,t^{-1}\right]\right)$. For this reason we will review the definition of the forks.

Definition 2.1**.**

A fork is an embedded oriented tree $F$ in the disc $D$ with four vertices $p_{0},p_{i},p_{j}$ and $z$, where $i\neq j,i,j\in\{1,2,3,4\}$ such that (see [3]):

$F$* meets the puncture points only at $p_{i}$ and $p_{j}$;* 2. 2.

$F$* meets the boundary $\partial D_{4}$ only at $p_{0}$;* 3. 3.

All three edges of $F$ have $z$ as a common vertex.

The edge of $F$ which contains $p_{0}$ is called the handle. The union of the other two edges is denoted by $T\left(F\right)$ and it is called tine of $F$. Orient $T\left(F\right)$ so that the handle of $F$ lies to the right of $T\left(F\right)$ (see Figure 2) [3].

For a given fork $F$, let $h:I\to D_{4}$ be the handle of $F$, viewed as a path in $D_{4}$ and take a lift $\tilde{h}:I\to{\tilde{D}}_{4}$ of $h$ so that $\tilde{h}\left(0\right)={\tilde{p}}_{0}$. Let $\tilde{T}\left(F\right)$ be the connected component of ${\pi}^{-1}\left(T\left(F\right)\right)$ which contains the point $\tilde{h}\left(1\right)$. In this case any element of $H_{1}\left({\tilde{D}}_{4};\mathbb{Z}\right)$ can be viewed as a homology class of $\tilde{T}\left(F\right)$ and it is denoted by $F$ [3].

Standard fork F_{i},\ \ i=1,2,3\ is the fork whose tine edge is the straight arc connecting the i-th and the (i+1)-st punctured points and whose handle has the form as in Figure 3. It is known that if $F_{1},{\ F}_{2}$ and $F_{3}$ are the corresponding homology classes, then they form a basis of $H_{1}\left({\tilde{D}}_{4};\mathbb{Z}\right)$ (see [3]).

Using the basis derived from $F_{1},F_{2},F_{3}$, any automorphism ${\widetilde{\varphi}}_{*}:H_{1}\left({\tilde{D}}_{4};\mathbb{Z}\right)\to H_{1}\left({\tilde{D}}_{4};\mathbb{Z}\right)$ can be viewed as a $3\times 3$ matrix with elements in the free $\mathbb{Z}\left[t,t^{-1}\right]$–module [3]. If $\varphi:D_{4}\to D_{4}$ is representing an element $\sigma\in B_{4}$, then we need to write the matrix $\rho\left(\sigma\right)={\widetilde{\varphi}}_{*}$ in terms of homology (algebraic) intersection pairing

[TABLE]

For this aim we need to define the noodles which represent relative homology classes in $H_{1}\left({\tilde{D}}_{4},\partial{\tilde{D}}_{4};\mathbb{Z}\right)$.

Definition 2.2**.**

A noodle is an embedded oriented arc in $D_{4}$, which begins at the base point $p_{0}$ and ends at some point of the boundary $\partial D_{4}$ [3].

For each $a\in H_{1}\left({\tilde{D}}_{4};\mathbb{Z}\right)$ and $b\in H_{1}\left({\tilde{D}}_{4},\partial{\tilde{D}}_{4};\mathbb{Z}\right)$ we should take the corresponding fork $F$ and noodle $N$ and define the polynomial $\left\langle F,N\right\rangle\in\mathbb{Z}\left[t,t^{-1}\right]$. It does not depend on the choice of representatives of homology classes and so

[TABLE]

is well-defined [3]. The map defined by the above formula is called the noodle–fork paring. Note that geometrically it can be computed in the following way: Let $F$ be a fork and $N$ be a noodle, such that $T\left(F\right)$ intersects $N$ transversely. Let $z_{1},z_{2},\ \dots,z_{n}$ be the intersection points. For each point $z_{i}$ let ${\varepsilon}_{i}$ be the sign of the intersection between $T\left(F\right)$ and $N$ at $z_{i}$ (the intersection is positive if going from tine to noodle according to the chosen directions means turning left) and $e_{i}=\left[{\gamma}_{i}\right]$ be the winding number of the loop ${\gamma}_{i}$ around the puncture points $p_{1},p_{2},p_{3},p_{4}$, where ${\gamma}_{i}$ is the composition of three paths $h$, $t_{i}$ and $n_{i}$:

$h$ is a path from $p_{0}$ to $z$ along the handle of $F$ (see Figure 4a); 2. 2.

$t_{i}$ is a path from $z$ to $z_{i}$ along the tine $T(F)$ (see Figure 4b); 3. 3.

$n_{i}$ is a path from $z_{i}$ to $p_{0}$ along the noodle $N$ (see Figure 4c).

In such case the noodle-fork pairing of $F$ and $N$ is given by (see [3]):

[TABLE]

Let $N_{1}$, $N_{2}$ and $N_{3}$ be the noodles given in Figure 5. These are called standard noodles. For each braid $\sigma\in B_{4}$, the corresponding Burau matrix $\rho\left(\sigma\right)$ can be computed using noodle-fork pairing of standard noodles and standard forks. In particular the following is true.

Lemma 2.3**.**

(see [5]). Let $\sigma\in B_{n}$. Then for $1\leq i,j\leq n-1$, the entry ${\rho}_{ij}\left(\sigma\right)$ of its Burau matrix $\rho\left(\sigma\right)$ is given by

[TABLE]

Note that under the convention adopted here we have

[TABLE]

and

[TABLE]

For example, to calculate $\rho_{1,1}(\sigma_{1})$ entry of the matrix $\rho(\sigma_{1})$ see the corresponding Figure 6. Note that intersection of the tine $T\left(F_{1}\sigma_{1}\right)$ of the fork $F_{1}\sigma_{1}$ and the noodle $N_{1}$ at point $z_{1}$ is negative which means that ${\varepsilon}_{i}=-1$ (see Figure 6a). On the other hand the winding number $e_{1}$ of the loop ${\gamma}_{1}$ (see Figure 6b) around puncture points equals $-1$ because the considered loop misses $p_{1},p_{3}$ and $p_{4}$ and it goes around $p_{2}$ once in anti–clockwise direction. Therefore

[TABLE]

3 The Bokut-Vesnin generators and kernel elements of the Burau representation

The braid groups $B_{4}$ and $B_{3}$ are defined by the following standard presentations [1]:

[TABLE]

[TABLE]

Let $\varphi:B_{4}\to B_{3}$ be the homomorphism defined by

[TABLE]

The kernel of $\varphi$ is known to be a free group $F\left(a,b\right)$ of two generators [4];

[TABLE]

This was proved by L. Bokut and A. Vesnin [4]. We will refer to $a$ and $b$ as the Bokut–Vesnin generators. The generators $a$ and $b$ are in fact much more similar than they look at the first glance. This becomes obvious when we interpret $B_{4}$ as the mapping class group of the $4$–punctured disc. In this well–known approach a braid is an isotopy class of homeomorphisms of the punctured disc fixing the boundary. Figure 7 shows $a$ and $b$ as homeomorphisms of the punctured disc. The punctures are arranged to make the similarity more visible. Another advantage of this approach is that it gives natural interpretation to various actions of $B_{4}$ to be considered later in this paper.

The following Proposition is crucial to our considerations.

Proposition 3.1**.**

$\ker\rho_{4}\subset\ker\varphi$**

Proof.

Let us make a slight detour into the realm of the Temperley-Lieb algebras $TL_{3}$ and $TL_{4}$. The Temperley-Lieb algebra $TL_{n}$ is defined as an algebra over $\mathbb{Z}[t,t^{-1}]$. It has $n-1$ generators $\{U^{i}_{i}\}_{i=1}^{n-1}$, and the following relations:

(TL1) $U^{i}_{i}U^{i}_{i}=(-t^{-2}-t^{2})U_{i}^{i},$

(TL2) $U^{i}_{i}U^{j}_{j}U^{i}_{i}=U_{i}^{i},$ for $|i-j|=1,$

(TL3) $U^{i}_{i}U^{j}_{j}=U^{j}_{j}U^{i}_{i},$ for $|i-j|>1.$

Let us consider the homomorphism $\psi:TL_{4}\to TL_{3}$ defined by

[TABLE]

Also, we need to use the Jones’ representation $\theta:B_{n}\to TL_{n}$ defined by sending $\sigma_{i}$ to $A+A^{-1}U^{i}_{i}$. It is known (see [3], Proposition 1.5) that for $n=3,4$ we have $\ker\theta_{n}=\ker\rho_{n}$. Moreover, the following diagram is obviously commutative:

B_{4}$$TL_{4}$$B_{3}$$TL_{3}$$\theta_{4}$$\varphi$$\theta_{3}$$\psi

On the other hand the representation $\theta_{3}$ is faithful and therefore $\ker\rho_{4}=\ker\theta_{4}\subset\ker\varphi$. ∎

Corollary 3.2**.**

All kernel elements of the Burau representation may be written as words in the Bokut–Vesnin generators $a$, $b$, $a^{-1}$, $b^{-1}$. Moreover, all possible $nontrivial$ elements in the kernel may be written as reduced words of positive length.

We will use this fact in the next section.

We present for future use the images of $a$, $b$, $a^{-1}$ and $b^{-1}$ under the Burau representation:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

4 Faithfulness Problem of the Burau representation

Let us outline the strategy for analyzing $\ker\rho_{4}$ in general terms. Consider a braid $\sigma$ that is a candidate for a non–trivial kernel element of the Burau representation. Of course we can exclude from our considerations all those non–trivial braids for which we know for whatever reason that they do not belong to the kernel. Also, we can adjust the remaining candidates in some ways — like replacing $\sigma$ with a suitably chosen conjugate of $\sigma$. For such a suitably chosen braid $\sigma$ we need to give some argument which shows that $\rho_{11}(\sigma)$ and $\rho_{31}(\sigma)$ should be non–zero and that $\deg_{min}(\rho_{11})-\deg_{min}(\rho_{31})=-1$, where $\deg_{min}$ denotes the exponent of the lowest degree term in the considered Laurent polynomial.

To simplify notation we will denote by $S_{i}(t^{\pm 1})$ the $i^{th}$ partial sum of the geometric series with initial term $1$ and quotient $-t$ or $-t^{-1}$ (e.g. $S_{2}(t^{-1})=1-t^{-1}+t^{-2}$).

Lemma 4.1**.**

*For each braid $\sigma\in B_{4}$ there exists $n\in\mathbb{N}$, such that

(1) the ${\rho}_{11}(a^{n}\sigma)$ and ${\rho}_{31}(a^{n}\sigma)$ entries of the Burau matrix $\rho\left(a^{n}\sigma\right)$ can be decomposed as a sum of three uniquely determined polynomials*

[TABLE]

[TABLE]

*such that

(2) for each $m\in\mathbb{N}$ we have*

[TABLE]

[TABLE]

(3) Moreover, if $\sigma$ is a pure braid, then the polynomial $P$ is non–zero.

Proof.

First of all let us observe that uniqueness of $P,Q$ and $R$ follows from properties (1) and (2) and general algebra. This means that we only need to prove existence and property (3). While it is possible to give specific algebraic formulas for $P,Q$ and $R$ we prefer to prove existence using forks and noodles. We will always assume that the fork/noodle configuration considered is irreducible.

Let $\sigma\in B_{4}$ be any braid. By Lemma 2.3 ${\rho}_{11}\left(a^{n}\sigma\right)=\left\langle F_{1}a^{n}\sigma,N_{1}\right\rangle$ and ${\rho}_{31}\left(a^{n}\sigma\right)=\left\langle F_{3}a^{n}\sigma,N_{1}\right\rangle$. On the other hand $\left\langle-,-\right\rangle$ is a bilinear form, so $\left\langle F_{1}a^{n}\sigma,N_{1}\right\rangle=\left\langle F_{1}a^{n},N_{1}{\sigma}^{-1}\right\rangle$ and $\left\langle F_{3}a^{n}\sigma,N_{1}\right\rangle=\left\langle F_{3}a^{n},N_{1}{\sigma}^{-1}\right\rangle$. It follows that

[TABLE]

[TABLE]

Let us consider $N_{1}{\sigma}^{-1}$, the image of the standard noodle $N_{1}$ under the action of $\sigma^{-1}$. $N_{1}{\sigma}^{-1}$ is a path in $D_{4}$ that begins at the base point $p_{0}$ and ends at the point $\left(0,1\right)\in\partial D_{4}$. By the definition of the standard noodle $N_{1}$ it is clear that $N_{1}{\sigma}^{-1}$ divides $D_{4}$ into two components, such that there is one puncture point in one component and three puncture points in the other. Let us assume that the single point is $p_{1}$. For example see Figure 8.

We intend to define $P$ and $R$ by grouping some terms in the sum originally used to define the representation in terms of fork/noodle pairing. The pairing is defined as a certain sum (2.1) of terms corresponding to crossings between forks and noodles. We will choose some of the crossings to define $P$ and some other to define $R$. In order to do this we will need some preparations.

Let $T$ be the boundary of the square whose vertices are the puncture points. We denote the sides with $T_{1},\ldots,T_{4}$, where $T_{i}$ connects $P_{i}$ with the next crossing (clockwise). We would like to work with a fork/noodle arrangement that has certain special properties. We need the pair (of a fork and a noodle) to be irreducible. We need the fork to be drawn in the standard way. We need the noodle to intersect $T$ transversally with minimum possible number of intersection points. And finally we need the tine of the fork to intersect all segments at which the noodle intersects $T_{4}$ and $T_{2}$. While general position arguments show that we can take care of the first three conditions, there is no possibility of the fourth being satisfied without some further adjustments. Figure 9 shows an example.

However it is automatically corrected if we increase the exponent $n$. The effect is just that we add a number of turns around two pairs of punctures. They do not affect the three properties already dealt with and with sufficient increase of $n$ we obtain the fourth property. So we are interested in strings between puncture points which has transversal intersection with $T$ (see Figure 10).

Note that it is possible to be no such string between $p_{2}$ and $p_{3}$ or $p_{3}$ and $p_{4}$, but by our assumption ($p_{1}$ is in the first component) there is an odd number of strings between $p_{1}$ and $p_{2}$ and an odd number between $p_{1}$ and $p_{4}$ (this guaranties that $P$ and $Q$ are not zero).

The pictures of $T\left(F_{1}\right)a^{n}$ and $T\left(F_{3}\right)a^{n}$ are as given in Figure 11. Therefore, they differ from each other by just one string and by the direction. The number of strings around $p_{1}$, $p_{4}$ and $p_{2}$, $p_{3}$ is $n$ for $T\left(F_{1}\right)a^{n}$ and $n-1$ for $T\left(F_{3}\right)a^{n}$.

If we take curves $T\left(F_{1}\right)a$ and $N_{1}{\sigma}^{-1}$ in the same $D_{4}$ and assume that their intersection is transversal, then it is possible that $T\left(F_{1}\right)a$ does not intersect all strings between $p_{1}$ and $p_{4}$ or $p_{2}$ and $p_{3}$. For example see Figure 9.

In this case we must take more numbers of $a$’s and finally we will obtain the curves $T\left(F_{1}\right)a^{n}$ and $N_{1}{\sigma}^{-1}$ such that we can find neighborhoods $U_{1}$ and $U_{2}$ of $T_{4}$ and $T_{2}$ respectively, with the following picture, illustrated in Figure 12a.

In this case for Figure 12a the polynomial corresponding to intersections inside $U_{1}$ and $U_{2}$ can be written as $P(t,t^{-1})(1-t^{-1})$ and $R(t,t^{-1})(1-t)$ respectively. Let $Q\left(t,t^{-1}\right)={\rho}_{11}\left(a^{n}\sigma\right)-P\left(t,t^{-1}\right)\left(1-t^{-1}\right)-R(t,t^{-1})(1-t)$, then we have

[TABLE]

On the other hand if we look at Figure 12b and keep in mind that directions of $T\left(F_{1}\right)a^{n}$ and $T\left(F_{3}\right)a^{n}$ are different we can say that

[TABLE]

After that if we multiply the braid $a^{n}\sigma$ by $a$ on the left side then we obtain the following picture, illustrated in Figure 13.

Therefore we will have

[TABLE]

[TABLE]

[TABLE]

Note that the same argument it sufficient to complete the proof. ∎

Example 1. Let $\sigma=b^{-1}a^{-1}b$, then for $n=2$ we have

[TABLE]

[TABLE]

[TABLE]

See Figure 14. Let $m=3$, then we can see that

[TABLE]

[TABLE]

[TABLE]

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[TABLE]

[TABLE]

[TABLE]

[TABLE]

Example 2. Let $\sigma=ba^{-2}b^{-1}$, then for $n=2$ we have

[TABLE]

[TABLE]

[TABLE]

See Figure 15 Let $m=4$ then we can see that

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Example 3. Let $\sigma=ab^{2}ab^{-1}$, then for $n=2$ we have

[TABLE]

[TABLE]

[TABLE]

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See Figure 16 Let $m=2$ then we can see that

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

We formulate a conjecture that describes a certain regularity, experimentally observed for images (matrices) of braids of a special form. For future reference let us state clearly that what we mean by regularity is that the $(1,1)$ and $(3,1)$ entries in the considered matrix are non–zero Laurent polynomials and that the difference of the degrees of lowest degree terms is equal to $-1$.

Conjecture 4.2**.**

Let $\ \sigma\in B_{4}$ be any non–trivial pure braid which is not equivalent to $\Delta^{m},$ for some $m\in\mathbb{N}.$ We assume that $\sigma$ acts non–trivially on $T_{4}.$ Then there exists a sufficiently large $l_{0}\in\mathbb{N}$ with respect to the length of $\sigma$ and a sufficiently large $m_{0}\in\mathbb{N}$ such that for each $m>m_{0},l\geq l_{0}$ the difference of the lowest degrees of the polynomials ${\rho}_{11}(a^{m}\sigma a^{-l})$ and ${\rho}_{31}(a^{m}\sigma a^{-l})$ is equal to $-1$ and the polynomials are non-zero.

While experimental data suggest that the Conjecture is true as formulated, we are really interested in the situation when $\sigma$ is a product of Bokut–Vesnin generators. Therefore we may refer to the length of $\sigma$, meaning the length of $\sigma$ as a reduced word in $a,b,a^{-1},b^{-1}$. Now, we give some arguments showing why we expect the Conjecture to be true. Take a sufficiently large $l_{0}\in\mathbb{N}$ with respect to the length of $\sigma$. Consider the curves $N_{1}a^{l_{0}}$ and neighborhood $U_{1}$ of $T_{4}$ inside of which it looks as in Figure 17:

Apply to the curve $N_{1}a^{l_{0}}$ the transformation corresponding to the braid $\sigma^{-1}$. Note that $l_{0}$ is sufficiently large with respect to the length of $\sigma^{-1}$ and so in the curve $N_{1}a^{l_{0}}\sigma^{-1}$ almost all parallel lines to line $T_{4}$ are followed by the curve $T_{4}\sigma^{-1}$. On the other hand the transformation corresponding to the braid $\sigma^{-1}$ acts non–trivially on the line $T_{4}$. Therefore the final image $N_{1}a^{l_{0}}\sigma^{-1}$ does not have problematic strings around $T_{4}$, as in Figure 18:

In general, if we take any braid $\sigma$, then $F_{1}\sigma$ may have the strings in the form illustrated in Figure 18. For example if $\sigma=b^{-1}ab^{-1}$ or $\sigma=a^{3}$ then the corresponding curves are shown in Figure 19.

Because the curve $N_{1}a^{l_{0}}\sigma^{-1}$ does not have any problematic strings for $n=3$ the intersection of curves $F_{1}a^{3}$ and $N_{1}a^{l_{0}}\sigma^{-1}$ inside the neighbourhoods $U_{1}$ and $U_{2}$ of $T_{4}$ and $T_{2}$ looks as in Figure 13. So the ${\rho}_{11}(a^{3}\sigma a^{-{l_{o}}})$ and ${\rho}_{31}(a^{3}\sigma a^{-{l_{0}}})$ entries of the Burau matrix $\rho\left(a^{n}\sigma a^{-{l_{0}}}\right)$ can be written as

[TABLE]

[TABLE]

where polynomial $P\left(t,t^{-1}\right)$ is not zero and for each $m^{\prime}\in\mathbb{N}$ we have

\begin{array}[]{lclll}{\rho}_{11}\left(a^{3+m^{\prime}}\sigma a^{-{l_{0}}}\right)&=&\hphantom{-}P\left(t,t^{-1}\right)\left(S_{m^{\prime}+1}(t^{-1})\right)&+Q\left(t,t^{-1}\right)\cr&&+R\left(t,t^{-1}\right)\left(S_{m^{\prime}+1}(t)\right),\cr{\rho}_{31}\left(a^{3+m^{\prime}}\sigma a^{-l_{0}}\right)&=&-P\left(t,t^{-1}\right)\left(S_{m^{\prime}}(t^{-1})\right)&-Q\left(t,t^{-1}\right)\cr&&-R\left(t,t^{-1}\right)\left(S_{m^{\prime}}(t)\right).\cr\end{array}

On the other hand if we compare the curves $N_{1}a^{{l_{0}}+1}$ and $N_{1}a^{l_{0}}$ (see Figure 20), it is clear that they differ only by strings around $T_{4}$.

Moreover, if we consider the intersections of curves $N_{1}a^{l_{0}+1}$ and $N_{1}a^{l_{0}}$ with the strings between the puncture points $p_{1}$ and $p_{4}$ as in Figure 21, then corresponding polynomials up to sign $\epsilon$ and multiplications $t^{\alpha}$ have the forms:

[TABLE]

[TABLE]

Therefore their lowest degrees are equal. Note that, $l_{0}$ is sufficiently large with respect to the length of $\sigma$ and so the pictures of curves $N_{1}a^{l_{0}+1}\sigma^{-1}$ and $N_{1}a^{l_{0}}\sigma^{-1}$ ‘globally’ are the same. That means that in some ’local’ pictures there are just different numbers of strings. Now we must look at pictures $N_{1}a^{l_{0}+1}\sigma^{-1}$ and $N_{1}a^{l_{0}}\sigma^{-1}$ inside a neighborhood of $T$ as it was done in the previous proof (see Figure 10). Note that by the arguments in the proof of Lemma 4.1 and same ‘global’ picture of the curves $N_{1}a^{l_{0}+1}\sigma^{-1}$ and $N_{1}a^{l_{0}}\sigma^{-1}$ we have

[TABLE]

[TABLE]

and for each $m^{\prime}\in\mathbb{N}$ we have:

\begin{array}[]{lclll}{\rho}_{11}\left(a^{3+m^{\prime}}\sigma a^{-l_{0}-1}\right)&=&\hphantom{-}P^{\prime}\left(t,t^{-1}\right)\left(S_{m^{\prime}+1}(t^{-1})\right)&+Q^{\prime}\left(t,t^{-1}\right)\cr&&+R^{\prime}\left(t,t^{-1}\right)\left(S_{m^{\prime}+}(t)\right),\cr{\rho}_{31}\left(a^{3+m^{\prime}}\sigma a^{-l_{0}-1}\right)&=&-P^{\prime}\left(t,t^{-1}\right)\left(S_{m^{\prime}}(t^{-1})\right)&-Q^{\prime}\left(t,t^{-1}\right)\cr&&-R^{\prime}\left(t,t^{-1}\right)\left(S_{m^{\prime}+}(t)\right).\end{array}

Take a large $m_{0}=3+m^{\prime}$ such that the difference of lowest degrees of polynomials ${\rho}_{11}\left(a^{m_{0}}\sigma a^{-{l_{0}}}\right)$ and ${\rho}_{31}\left(a^{m_{0}}\sigma a^{-{l_{0}}}\right)$ is equal to $-1$ and these lowest degrees come from the lowest degree of the polynomial $P(t,t^{-1})$. By (*) and (**) the polynomials $P\left(t,t^{-1}\right)$ and $P^{\prime}\left(t,t^{-1}\right)$, $Q\left(t,t^{-1}\right)$ and $Q^{\prime}\left(t,t^{-1}\right)$ and also $R\left(t,t^{-1}\right)$ and $R^{\prime}\left(t,t^{-1}\right)$ have the same lowest degrees and so the same regularity will be true for the polynomials ${\rho}_{11}\left(a^{m}\sigma ba^{-l_{0}-1}\right)$ and ${\rho}_{31}\left(a^{m}\sigma a^{-l_{0}-1}\right)$. By induction on the length of $\sigma$ it will be true for the braid $a^{m}\sigma a^{-l},l>l_{0}$ as well.

Example 4. Let $\sigma=\ b^{6}a\ b^{-1}a^{-1}b^{-6}a^{-6}$. Our aim is to find $n$ which satisfies the conditions of Lemma 4.1 and to calculate the corresponding polynomials $P$, $Q$ and $R$. Then we will take any $l>6$ (In our case we consider $l=9$) and will show that lowest degrees of the corresponding polynomials do not change. For the given braid it is difficult to see the picture and write down the polynomials $P$, $Q$ and $R$. Therefore we will use the following method: If $n$ (in our case $n=2$) is a number as in Lemma 4.1, then we have

[TABLE]

[TABLE]

Therefore

[TABLE]

[TABLE]

In this way we can see that for the braid $\sigma=a^{2}b^{6}ab^{-1}a^{-1}b^{-6}a^{-6}$ we have

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Similarly, for the braid $\sigma^{\prime}=a^{2}b^{6}ab^{-1}a^{-1}b^{-6}a^{-9}$ we obtain

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Therefore the lowest degrees of polynomials $P\left(t,t^{-1}\right)$ and $P^{\prime}\left(t,t^{-1}\right)$ (same situation is with the polynomials $Q\left(t,t^{-1}\right)$ and $Q^{\prime}\left(t,t^{-1}\right)$ or $R\left(t,t^{-1}\right)$ and $R^{\prime}\left(t,t^{-1}\right)$) are equal.

Theorem 4.3**.**

Conjecture 4.2 implies faithfulness of the Burau representation for $n=4$ .

Proof.

Let us consider a nontrivial braid $\sigma\in B_{4}$ written as a reduced word in the Bokut–Vesnin generators. We may assume that it begins and ends with $a$ or $a^{-1}$ (otherwise we will conjugate by a suitable power of $a$).

If we interpret the braid group as the mapping class group, then there is a natural induced action on the set of isotopy classes of forks. Let $\sigma$ act non–trivially on $T_{4}$. Then by Lemma 4.1 it is possible to find sufficiently large $l_{0}$ with respect to the length of $\sigma$ and sufficiently large $m_{0}$, such that for each $m>m_{0}$ and $l>l_{0}$ the difference of lowest degrees of the polynomials ${\rho}_{11}(a^{m}\sigma a^{-l})$ and ${\rho}_{31}(a^{m}\sigma a^{-l})$ is equal to $-1$ and the polynomials are both non-zero. In particular, we can assume that $m=l$ and so ${\rho}_{11}(a^{m}\sigma a^{-m})$ and ${\rho}_{31}(a^{m}\sigma a^{-m})$ are both non-zero which contradicts the assumption that $a^{m}\sigma a^{-m}\in\ker\rho$.

The general case (when we do not assume that $\sigma$ acts non–trivially on $T_{4}$) is easily reduced to the one discussed above. The reason is that if $\sigma$ acts trivially on all four segments, then $\sigma$ is a power of $\Delta$ which is not possible if $\sigma$ is a product of the Bokut–Vesnin generators. And if $\sigma$ acts non–trivially on at least one of the four segments, then we can rotate the whole disc to make the action non–trivial for $T_{4}$.

∎

Remark 4.4**.**

We have a C++ program checking whether our regularity works or not for randomly generated examples. We calculated millions of examples and the regularity was always confirmed. In fact we considered examples of type $a^{3}b^{3}wb^{-3}a^{-3}$, where $a^{3}b^{3}wb^{-3}a^{-3}$ is a reduced word in the Bokut–Vesnin generators. Such a version of Proposition 3.1 is sufficient for the Burau representation faithfulness problem.

Acknowledgements:

Most of this research was conducted while the first author was a postdoc at the University of Warsaw during the Spring 2014 semester, with support of Erasmus Mundus Project (WEBB).

REFERENCES

Joan S Birman. Braids, links, and mapping class groups. Annals of Mathematics Studies, No. 82, Princeton University Press, Princeton, NJ (1974) 2. 2.

Stephen Bigelow. The Burau representation is not faithful for $n=5$. Geom. Topol. 3 (1999), 397-404 3. 3.

Stephen Bigelow. Does the Jones polynomial detect the unknot? J. Knot Theory and Ramifications (4) 11 (2002), 493-505 4. 4.

Leonid Bokut and Andrei Vesnin. New rewriting system for the braid group $B_{4}$. * in: Proceedings of Symposium in honor of Bruno Buchberger’s 60th birthday ”Logic, Mathematics and Computer Sciences: Intersections”, Research Institute for Symbolic Computations, Linz, Austia, 2002, Report Series No. 02-60, 48-60 * 5. 5.

Matthieu Calvez and Tetsuya Ito. Garside-theoretic analysis of Burau representations. * arXiv:1401.2677v2 * 6. 6.

D. D.Long and M. Paton. The Burau representation is not faithful for $n\geq 6$. Topology **32 **(1993), no. 2, 439—447 7. 7.

John Atwell Moody. The Burau representation of the braid group $B_{n}$ is unfaithful for large $n$. Bull. Amer. Math. Soc. (N.S.) **25 **(1991) no. 2, 379–384

Authors’ addresses:

Anzor Beridze,

Department of Mathematics,

Batumi ShotaRustaveli State University,

35, Ninoshvili St., Batumi 6010,

Georgia

a.beridze@bsu.edu.ge;

Pawel Traczyk,

Institute of Mathematics,

University of Warsaw,

Banacha 2, 02-097 Warszawa,

Poland

traczyk@mimuw.edu.pl