Attacks on the Faithfulness of the Burau Representation of the Braid Group B 4

The faithfulness of the Burau representation of the 4-strand braid group, B4, remains an open question. In this work, there are two main results. First, we specialize the indeterminate t to a complex number on the unit circle, and we find a necessary condition for a word of B4 to belong to the kernel of the representation. Second, by using a simple algorithm, we will be able to exclude a family of words in the generators from belonging to the kernel of the reduced Burau representation.

1. Introduction Magnus and Peluso (1969) showed that the Burau representation is faithful for n ≤ 3. Moody (1991) showed that it is not faithful for n ≥ 9; this result was improved to n ≥ 6 by Long and Paton (1992).The non-faithfulness for n = 5 was shown by Bigelow (1999).The question of whether or not the Burau representation for n = 4 is faithful is still open.
In our work, we attack the question of faithfulness of the Burau representation of B 4 .In section 3, we specialize the indeterminate t to a complex number e iα , where α ∈ R. Then we show that if α π Q and 4ϵ + 3m 0, then the word b ϵ 1 a m 1 b ϵ 2 a m 2 .......b ϵ n a m n does not belong to the kernel of the representation .Here ϵ i = 0, 1 or 2, m i ∈ Z, ϵ = n ∑ i=1 ϵ i and m = n ∑ i=1 m i , for 1 ≤ i ≤ n.In section 4, we let a = σ 1 σ 2 σ 3 and b = aσ 1 , where σ 1 , σ 2 , σ 3 are generators of B 4 .Then we find the general form of the words a n and b n and we prove that they are not in the kernel of the representation for any non-zero natural number n.In section 5, we introduce a simple algorithm which computes all words of the form a i b j and a i b j a k for integers i, j and k.We then conclude that there is no word of such forms in the kernel of the representation.

Preliminaries
Definition 1. (Artin, 1965) The braid group, B n , is an abstract group generated by σ 1 , σ 2 , ..., σ n−1 with the following relations Definition 2. (Burau, 1936) The reduced Burau representation of B n is defined by In particular, setting n = 4, we have where t is an indeterminate.
Let a = σ 1 σ 2 σ 3 and b = aσ 1 .Then we get Theorem 1. (Holtzma, 2008) Let B 4 be the braid group of order 4 then Using Theorem 1, we can show that the elements of B 4 are either of the form b ϵ 1 a m 1 b ϵ 2 a m 2 .......b ϵ n a m n or of the form obtained by permuting a m i and b ϵ i .Here

Necessary Condition for Elements in the Kernel of the Reduced Burau Representation
Let t be a non-zero complex number on the unit circle, t = e iα , where α is a non-zero real number.
Likewise for a word obtained from u by permuting b ϵ i and a m i .
Corollary 1.For a word u of the form b ϵ 1 a m 1 b ϵ 2 a m 2 .......b ϵ n a m n to belong to the kernel of the representation, m has to belong to 4Z and ϵ has to belong to 3Z.

The Words a n and b n
In this section, we find the general form of the words a n and b n , for any integer n.Denote by I 3 the identity matrix.We recall, from section 2, that a = σ 1 σ 2 σ 3 and b = aσ Proof.We prove this proposition using mathematical induction principle.For k = 0 and k = 1, direct computations give us that J 4 = 1, and Suppose that the proposition is true for all integers less than or equal to k.We then show it is still true for k + 1.We show (1): We show (2): Therefore the proposition is true for all k ∈ N.
Proposition 2. For all k ∈ N, we have Proof.We prove the proposition using mathematical induction principle.Direct computations show Suppose that it is true for all integers less than or equal to k.We now show it is still true for k + 1.We show (1).
As for (2): Lemma 1.For any non-zero integer n, a n and b n do not belong to the kernel of the reduced burau representation B 4 → GL(3, Z[t, t −1 ]). Proof.
This implies, by Proposition 1, that a n has to be one of the following words: We denote by J k ii the diagonal entry of the matrix J k , which lies in the ith row and in the ith column (1 11 and J 3 22 are zeros, it follows that a n is not the empty word for any integer n.On the other hand, n is written in either one of the following forms: ±3k, ±(3k + 1), ±(3k + 2), where k ∈ Z This implies, by Proposition 2, that b n has to be one of the following words: The 1-1 entries in both of the matrices b and b 2 are equal to zeros , it follows that b n is not the empty word for any integer n.
5. Words of The Form a i b j And a i b j a k In this section, we use Mathematica to excute a program that computes words of the form a i b j and a i b j a n for all non zero integers i, j and n.In order to excute this program, we consider the following notations: Then we excute the following codes: These codes compute all words of the form a i b j and a i b j a k for all non zero integers i, j and k.Among these words, the ones that might possibly belong to the kernel of the representation are those which have the form t α I 3 .More precisely, these words are It is clear that each of these words is the empty word.For example, c 1 d 1 z 5 = t α I 3 , where α = 4k 1 + 4s 1 − 4x 5 .If α = 0, then x 5 = k 1 + s 1 .
Therefore we get the following theorem.Theorem 3.For integers i, j, k, there are no words of the form a i b j a k which lies in the kernel of the Burau representation.