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ISRN Algebra
VolumeΒ 2012Β (2012), Article IDΒ 858959, 13 pages
http://dx.doi.org/10.5402/2012/858959
Research Article

Another Proof of the Faithfulness of the Lawrence-Krammer Representation of the Braid Group π΅πŸ‘

Department of Mathematics, Beirut Arab University, P.O. Box 11-5020, Beirut 11072809, Lebanon

Received 16 March 2012; Accepted 6 May 2012

Academic Editors: P.Β Koshlukov, H.Β Li, S.Β Yang, and Y.Β Zhou

Copyright Β© 2012 Mohammad N. Abdulrahim and Mariam Hariri. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

The Lawrence-Krammer representation of the braid group 𝐡𝑛 was proved to be faithful for 𝑛β‰₯3 by Bigelow and Krammer. In our paper, we give a new proof in the case 𝑛=3 by using matrix computations. First, we prove that the representation of the braid group 𝐡3 is unitary relative to a positive definite Hermitian form. Then we show the faithfulness of the representation by specializing the indeterminates q and t to complex numbers on the unit circle rather than specializing them to real numbers as what was done by Krammer.

1. Introduction

A group is said to be linear if it admits a faithful representation into 𝐺𝐿𝑛(𝐾) for some natural number 𝑛 and some field 𝐾. The question of faithfulness of the braid group has been the subject of research for a long time. It has been shown that the Burau representation is faithful for 𝑛≀3 and is not faithful for 𝑛β‰₯5 [1]. However, the faithfulness of the Burau representation in the case 𝑛=4 is still unknown. In [2], Krammer proved that the Lawrence-Krammer representation is faithful in the case 𝑛=4, where he assumed that 𝑑 is a real number with 0<𝑑<1. After that, Bigelow [3] used topological methods that are very different from the algebraic methods used by Krammer to prove that the Lawrence-Krammer representation is faithful for all 𝑛. In his argument, he assumed that π‘ž and 𝑑 are variables. Shortly after, Krammer [4] found a new proof that shows that this representation is faithful for all 𝑛 using algebraic methods, in which he interchanged the roles of π‘ž and 𝑑 in [2] and assumed that π‘ž is a real number with 0<π‘ž<1.

Our work uses a different method that basically depends on matrix computations to show that the Lawrence-Krammer representation of 𝐡3 is faithful. In our argument, we specialize the indeterminates π‘ž and 𝑑 to nonzero complex numbers on the unit circle rather than specializing them to real numbers. For a long time, it is known that the braid group 𝐡3 is linear (see [1]). What makes our approach different from the previous methods is the fact that our work is computational, which depends only on simple rules in linear algebra. Also, the difference between our work and the previous ones is in specializing π‘ž and 𝑑 to nonzero complex numbers on the unit circle rather than to real numbers in the interval [0,1] as in [2] or [4].

In Section 3, We show that if π‘ž and 𝑑 are chosen to be appropriate algebraically independent complex numbers on the unit circle, then this representation is equivalent to a unitary representation. This will be a tool to find a necessary and sufficient condition for an element of 𝐡3 possibly to belong to the kernel of the Lawrence-Krammer representation.

In Section 4, we prove the faithfulness of this representation by a purely computational method.

2. Definitions

Definition 2.1. The braid group on 𝑛 strands, 𝐡𝑛, is the abstract group with presentation 𝐡𝑛=⟨𝜎1,…,πœŽπ‘›βˆ’1βˆ£πœŽπ‘–πœŽπ‘–+1πœŽπ‘–=πœŽπ‘–+1πœŽπ‘–πœŽπ‘–+1 for 𝑖=1,2,…,π‘›βˆ’2, and πœŽπ‘–πœŽπ‘—=πœŽπ‘—πœŽπ‘– if |π‘–βˆ’π‘—|β‰₯2⟩. The generators 𝜎1,…,πœŽπ‘›βˆ’1 are called the standard generators of 𝐡𝑛.

The Lawrence-Krammer representation of braid groups is a representation of the braid group 𝐡𝑛 in πΊπΏπ‘š(β„€[π‘žΒ±1,𝑑±1])=Aut(𝑉0), where π‘š=𝑛(π‘›βˆ’1)/2 and 𝑉0 is the free module of rank π‘š over β„€[π‘žΒ±1,𝑑±1]. The representation is denoted by 𝐾(π‘ž,𝑑). For simplicity, we write 𝐾 instead of 𝐾(π‘ž,𝑑).

Definition 2.2. With respect to {π‘₯𝑖,𝑗}1≀𝑖<𝑗≀𝑛, the free basis of 𝑉0, the image of each Artin generator under Krammer's representation is written as πΎξ€·πœŽπ‘˜π‘₯𝑖,𝑗=⎧βŽͺβŽͺβŽͺ⎨βŽͺβŽͺβŽͺβŽ©π‘‘π‘ž2π‘₯π‘˜,π‘˜+1,𝑖=π‘˜,𝑗=π‘˜+1,(1βˆ’π‘ž)π‘₯𝑖,π‘˜+π‘žπ‘₯𝑖,π‘˜+1π‘₯,𝑗=π‘˜,𝑖<π‘˜,𝑖,π‘˜+π‘‘π‘žπ‘˜βˆ’π‘–+1(π‘žβˆ’1)π‘₯π‘˜,π‘˜+1,𝑗=π‘˜+1,𝑖<π‘˜,π‘‘π‘ž(π‘žβˆ’1)π‘₯π‘˜,π‘˜+1+π‘žπ‘₯π‘˜+1,𝑗π‘₯,𝑖=π‘˜,π‘˜+1<𝑗,π‘˜,𝑗+(1βˆ’π‘ž)π‘₯π‘˜+1,𝑗π‘₯,𝑖=π‘˜+1,π‘˜+1<𝑗,𝑖,𝑗π‘₯,𝑖<𝑗<π‘˜orπ‘˜+1<𝑖<𝑗,𝑖,𝑗+π‘‘π‘žπ‘˜βˆ’π‘–(π‘žβˆ’1)2π‘₯π‘˜,π‘˜+1,𝑖<π‘˜<π‘˜+1<𝑗.(2.1)

Using the Magnus representation of subgroups of the automorphism group of a free group with three generators, we determine Krammer's representation 𝐾(π‘ž,𝑑)∢𝐡3→𝐺𝐿3(β„€[π‘žΒ±1,𝑑±1]), where πΎξ€·πœŽ1ξ€Έ=βŽ›βŽœβŽœβŽœβŽœβŽπ‘‘π‘ž2βŽžβŽŸβŽŸβŽŸβŽŸβŽ ξ€·πœŽ00π‘‘π‘ž(π‘žβˆ’1)0π‘ž011βˆ’π‘ž,𝐾2ξ€Έ=βŽ›βŽœβŽœβŽœβŽœβŽ1βˆ’π‘žπ‘ž010π‘‘π‘ž2(π‘žβˆ’1)00π‘‘π‘ž2⎞⎟⎟⎟⎟⎠.(2.2)

Here, β„€[π‘žΒ±1,𝑑±1] is the ring of Laurent polynomials on two variables. For simplicity, we denote 𝐾(πœŽπ‘–) by 𝑋𝑖, where 𝑖=1,2. Here, π‘‹βˆ—π‘– is the conjugate transpose of 𝑋, where 𝑑=π‘‘βˆ’1 and π‘ž=π‘žβˆ’1.

3. The Lawrence-Krammer Representation of 𝐡3 Is Equivalent to a Unitary Representation

Budney has proved that the L-K representation of the braid group 𝐡𝑛 is unitary relative to a hermitian form. His hermitian form turns out to be negative-definite (see [5]). In our work, we specialize the indeterminates π‘ž and 𝑑 to nonzero complex numbers on the unit circle. We consider the complex representation 𝐾∢𝐡3→𝐺𝐿3(β„‚) and prove that it is unitary relative to a positive definite Hermitian form.

Theorem 3.1. The images of the generators of 𝐡3 under the Lawrence-Krammer representation are unitary relative to a Hermitian matrix 𝑀 given by βŽ›βŽœβŽœβŽœβŽœβŽœβŽœβŽξ€·π‘€=1+π‘ž2ξ€Έξ€·(1+π‘‘π‘ž)π‘‘π‘ž2ξ€Έβˆ’1π‘‘π‘ž2ξ€·(π‘žβˆ’1)1+π‘ž2ξ€Έ(1βˆ’π‘‘π‘ž(π‘žβˆ’1))π‘‘π‘ž21+π‘ž2ξ€·1+π‘ž2ξ€Έξ€·π‘‘π‘ž2ξ€Έ+π‘žβˆ’1π‘ž2ξ€·1+π‘ž2ξ€·ξ€Έξ€·βˆ’1+𝑑1βˆ’(π‘žβˆ’2)2π‘ž+π‘‘π‘ž3ξ€Έξ€Έπ‘‘π‘ž2ξ€·(π‘žβˆ’1)1+π‘ž2ξ€Έξ€·π‘‘π‘ž2ξ€Έ+π‘žβˆ’1π‘žπ‘ž2+1π‘ž2ξ€·1+π‘ž2ξ€Έ(1βˆ’π‘‘π‘ž(π‘žβˆ’1))π‘‘π‘ž3ξ€·1+π‘ž2ξ€Έξ€·(1+π‘‘π‘ž)π‘‘π‘ž2ξ€Έβˆ’1π‘‘π‘ž2⎞⎟⎟⎟⎟⎟⎟⎠.(π‘žβˆ’1)(3.1)
That is, π‘‹π‘–π‘€π‘‹βˆ—π‘–=𝑀 for 𝑖=1,2, where π‘€βˆ—=𝑀.

The choice of 𝑀 is unique only if we specialize the indeterminates π‘ž and 𝑑 in a way that Krammer's representation becomes irreducible. A necessary and sufficient condition for irreducibility is given in [6].

Our objective is to show that a certain specialization π‘€ξ…ž of 𝑀 is equivalent to the identity matrix, that is, π‘ˆπ‘€ξ…žπ‘ˆβˆ—=𝐼 for some matrix π‘ˆ. In other words, we need to show that π‘€ξ…ž=π‘‰π‘‰βˆ— for some matrix 𝑉; or equivalently, π‘€ξ…ž is hermitian.

Theorem 3.2. Let π‘ž and 𝑑 be nonzero complex numbers on the unit circle such that π‘ž=π‘’π‘–πœƒ and 𝑑=𝑒𝑖𝛼. The matrix 𝑀 is positive definite when πœƒ and 𝛼 are chosen in either one of the following intervals:(i)βˆ’πœ‹/2<πœƒ<0 and βˆ’πœ‹βˆ’(3πœƒ/2)<𝛼<βˆ’(3πœƒ/2), (ii)0<πœƒ<πœ‹/2 and βˆ’(3πœƒ/2)<𝛼<πœ‹βˆ’(3πœƒ/2).

Proof . We denote the principal minors of 𝑀 by 𝑑𝑖, where 𝑖=1,2,3. We have 𝑑1=ξ€·1+π‘ž2ξ€Έξ€·(1+π‘‘π‘ž)π‘‘π‘ž2ξ€Έβˆ’1π‘‘π‘ž2,𝑑(π‘žβˆ’1)2=ξ€·1+π‘ž2ξ€Έ2𝑑2π‘ž3ξ€Έβˆ’1(βˆ’1+π‘‘π‘ž(βˆ’2+π‘ž(2+π‘‘π‘ž)))𝑑2π‘ž4(π‘žβˆ’1)2,𝑑3=ξ€·1+π‘ž2ξ€Έ3ξ€·(1+𝑑)π‘‘π‘ž3π‘‘βˆ’1ξ€Έξ€·2π‘ž3ξ€Έβˆ’12𝑑3π‘ž6(π‘žβˆ’1)3.(3.2)
We find the intervals of π‘ž and 𝑑 for which 𝑑1,𝑑2, and 𝑑3 are strictly positive. Let π‘ž=π‘’π‘–πœƒ and 𝑑=𝑒𝑖𝛼, where βˆ’πœ‹β‰€πœƒβ‰€πœ‹ and βˆ’πœ‹β‰€π›Όβ‰€πœ‹. Then we have 𝑑1ξ‚ƒπœƒ=2cosπœƒ1+csc2ξ‚€sin3πœƒ2,𝑑+𝛼2=4cos2πœƒπœƒcsc2ξ‚€sin3πœƒ2πœƒ+𝛼2+csc2ξ‚€sin3πœƒ2,𝑑+𝛼3=8cos3πœƒcsc3πœƒ2sin2ξ‚€3πœƒ2+𝛼sin3πœƒ2ξ‚€+sin3πœƒ2.+𝛼(3.3)
In order to have 𝑑1,𝑑2, and 𝑑3 strictly positive real numbers, we consider the intervals given by the hypothesis and make the following observations.
In the interval (i), it is easy to see that csc(πœƒ/2)<0, sin(3πœƒ/2+𝛼)<0, cosπœƒ>0, and sin(3πœƒ/2)+sin(3πœƒ/2+𝛼)<0.
In the interval (ii), we see that csc(πœƒ/2)>0, sin(3πœƒ/2+𝛼)>0, cosπœƒ>0, and sin(3πœƒ/2)+sin(3πœƒ/2+𝛼)>0.

We construct a homomorphism that specializes the indeterminates π‘ž and 𝑑 to nonzero complex numbers, on the unit circle, which are transcendentally independent over β„€.

Let 𝑓𝑀 be a homomorphism π‘“π‘€βˆΆβ„€[π‘žΒ±1,𝑑±1]β†’β„‚ defined as follows: 𝑓𝑀(π‘ž)=𝑀1,𝑓𝑀(𝑑)=𝑀2,𝑓𝑀(𝑧)=𝑧 for π‘§βˆˆβ„€, where 𝑀=(𝑀1,𝑀2) and 𝑀1,𝑀2 are complex numbers on the unit circle. Let 𝑓𝑀 also denote the group homomorphism 𝐺𝐿3(β„€[π‘žΒ±1,𝑑±1])→𝐺𝐿3(β„‚). We choose 𝑀 such that 𝑓𝑀(𝑀) is positive definite; that is, 𝑓𝑀(𝑀)=π‘‰π‘‰βˆ—, for some π‘‰βˆˆπΊπΏ3(β„‚).

Consider now the composition map π‘“π‘€βˆ˜πΎβˆΆπ΅3→𝐺𝐿3(β„‚). It was proved in [6] that the specialization of Krammer's representation 𝐡3→𝐺𝐿3(β„‚) is irreducible if and only if π‘‘β‰ βˆ’1,π‘‘π‘ž3β‰ 1, and 𝑑2π‘ž3β‰ 1. It is easy to see that the conditions for irreducibility are achieved once we choose π‘ž and 𝑑 in either one of the intervals (i) or (ii) in the hypothesis of Theorem 3.2. The uniqueness of 𝑓𝑀(𝑀) up to scalar multiplication follows from Shur's Lemma and the fact that the specialization of Krammer's representation of 𝐡3 is irreducible.

Theorem 3.3. The complex representation of 𝐡3, π‘“π‘€βˆ˜πΎ, is conjugate to a unitary representation.

Proof. Since π‘‹π‘–π‘€π‘‹βˆ—π‘–=𝑀 for 𝑖=1,2, it follows that 𝑓𝑀𝑋𝑖𝑓𝑀(𝑀)π‘“π‘€ξ€·π‘‹βˆ—π‘–ξ€Έ=𝑓𝑀(𝑀).(3.4) But 𝑓𝑀(𝑀)=π‘‰π‘‰βˆ—, then ξ€·π‘‰βˆ’1π‘“π‘€ξ€·π‘‹π‘–ξ€Έπ‘‰π‘‰ξ€Έξ€·βˆ’1π‘“π‘€ξ€·π‘‹π‘–ξ€Έπ‘‰ξ€Έβˆ—=𝐼.(3.5) Now, let π‘ˆ=π‘‰βˆ’1𝑓𝑀𝑋𝑖𝑉,(3.6) then π‘ˆπ‘ˆβˆ—=𝐼.(3.7) Hence π‘ˆβˆ—π‘ˆ=𝐼 and so π‘ˆ is unitary.

We now find a necessary and sufficient condition for an element of 𝐡3 possibly to belong to the kernel of the Lawrence-Krammer representation.

Theorem 3.4. An element of 𝐡3 lies in the kernel of the Lawrence-Krammer representation of 𝐡3 if and only if the trace of its image is equal to three.

Proof . If tr(𝑓𝑀(𝑋𝑖))=3, then tr(π‘ˆ)=3, where π‘ˆ=π‘‰βˆ’1𝑓𝑀(𝑋𝑖)𝑉. Since π‘ˆ is unitary, it follows that π‘ˆ is diagonalizable, that is, there exists a matrix 𝑃 such that π‘ƒβˆ’1π‘ˆπ‘ƒ=𝐷, where 𝐷 is a diagonal matrix with eigenvalues of π‘ˆ as the diagonal entries. Hence πœ†1+πœ†2+πœ†3=3, where the πœ†ξ…žπ‘–π‘  are the eigenvalues of π‘ˆ. Being unitary, it has its eigenvalues on the unit circle. Therefore, we get πœ†1=πœ†2=πœ†3=1. Thus, 𝐷 is the identity matrix and so is π‘ˆ. This implies that 𝑓𝑀(𝑋𝑖)=𝐼3.

Therefore, we conclude that if there exists a nontrivial element in 𝐡3 such that the trace of its image under π‘“π‘€βˆ˜πΎβˆΆπ΅3→𝐺𝐿3(β„‚) is three then, the element lies in the kernel of this representation.

4. The Faithfulness of the L-K Representation for 𝑛=3

In this section, we prove our main theorem, which shows the faithfulness of our representation for certain values of π‘ž and 𝑑. Let π‘ž and 𝑑 be nonzero complex numbers on the unit circle such that π‘ž=π‘’π‘–πœƒ and 𝑑=𝑒𝑖𝛼.

Let βŽ›βŽœβŽœβŽœβŽπ›½=arg1+π‘‘π‘ž2+𝑑2π‘ž4βˆ’π‘ž(1+𝑑)βˆ’π‘ž3𝑑(1+𝑑)+βˆ’4π‘ž4𝑑2+ξ€·ξ€·1+(βˆ’1+π‘ž(π‘žβˆ’1)𝑑)1+π‘‘π‘ž2ξ€Έξ€Έ22π‘‘π‘ž2⎞⎟⎟⎟⎠.(4.1)

Consider the values of π‘ž and 𝑑 whose arguments belong to the set 𝐺 defined as follows: ξ‚»πœ‹πΊ=(πœƒ,𝛼)∈𝐴βˆͺπ΅βˆ£πœƒβ‰ Β±3,𝛼+3πœƒπœ‹π›½βˆ‰β„š,πœ‹ξ‚Όβˆ‰β„š,(4.2) where ξ‚†πœ‹π΄=(πœƒ,𝛼)βˆ£βˆ’2πœ‹<πœƒ<0,βˆ’3βˆ’2πœƒ<𝛼<βˆ’3πœƒ2,ξ‚†πœ‹π΅=(πœƒ,𝛼)∣0<πœƒ<2,βˆ’3πœƒ2πœ‹<𝛼<3.βˆ’2πœƒ(4.3)

The set 𝐺 is a subset of the union of the intervals given in Theorem 3.2. We choose 𝑀 such that (πœƒ,𝛼)∈𝐺. Therefore, for values of π‘ž and 𝑑 such that (πœƒ,𝛼)∈𝐺, we have 𝑓𝑀(𝑀) being positive definite and π‘“π‘€βˆ˜πΎβˆΆπ΅3→𝐺𝐿3(β„‚) irreducible. For simplicity, we still write 𝐾 instead of π‘“π‘€βˆ˜πΎ.

We now present our second main theorem.

Theorem 4.1. The complex specialization of the Lawrence-Krammer representation 𝐾∢𝐡3→𝐺𝐿3(β„‚) is faithful for a dense subset of the set 𝐺 defined above.

We need to state some lemmas that help us prove Theorem 4.1. We already know that 𝐡3=⟨𝜎1,𝜎2∣𝜎1𝜎2𝜎1=𝜎2𝜎1𝜎2⟩, where 𝜎1 and 𝜎2 are called the standard generators of 𝐡3. Consider the product of generators of 𝐡3, namely, 𝐽=𝜎1𝜎2. We also let 𝑆=𝜎1𝐽. It was proved in [7] that 𝐡3 is generated by the two elements 𝑆 and 𝐽 and the relation 𝑆2=𝐽3. That is, we have that 𝐡3=βŸ¨π‘†,π½βˆ£π‘†2=𝐽3⟩. Also, it was proved in [8] that the center 𝑍(𝐡3) of 𝐡3 is generated by the one generator 𝐽3, that is, 𝑍(𝐡3)=⟨𝐽3⟩. Thus, the elements of 𝐡3 are of the form: 𝑆𝑙, π½π‘š, π½π‘š1𝑆𝑙1π½π‘š2𝑆𝑙2,…, and so forth, where all the exponents are integers.

Consider the element 𝑆𝑙. If 𝑙 is odd, then 𝑆𝑙=𝑆2π‘˜+1=𝑆⋅𝑆2π‘˜=𝑆𝐽3π‘˜, where π‘˜βˆˆβ„€. If 𝑙 is even then 𝑆𝑙=𝑆2π‘˜=𝐽3π‘˜, where π‘˜βˆˆβ„€. Therefore, any element of 𝐡3 is of one of the following forms: 𝑆, π½π‘š, π‘†π½π‘š1π‘†π½π‘š2…, or π½π‘š1π‘†π½π‘š2𝑆…, where π‘šπ‘–'s βˆˆβ„€.

Lemma 4.2. The possible nontrivial elements in the kernel of the representation 𝐾∢𝐡3→𝐺𝐿3(β„‚) are(i)π‘†π½π‘Ÿ1π‘†π½π‘Ÿ2β‹―π‘†π½π‘Ÿ2π‘›βˆ’1π‘†π½βˆ‘βˆ’3π‘›βˆ’2π‘›βˆ’1𝑖=1π‘Ÿπ‘–, where the number of 𝑆's is 2𝑛 and π‘Ÿπ‘–=1 or 2 for 1≀𝑖<2𝑛,(ii)elements obtained from (𝑖) by permuting 𝑆 and 𝐽.

Proof. Recall that π‘ž=π‘’π‘–πœƒ,𝑑=𝑒𝑖𝛼, and 𝐾(πœŽπ‘–)=𝑋𝑖, where 𝑖=1,2. For simplicity, we use the symbol |𝑋| instead of det(𝑋). We have that 𝐽=𝜎1𝜎2, 𝑆=𝜎1𝐽, and |𝑋1|=|𝑋2|=βˆ’π‘‘π‘ž3. In general, elements of 𝐡3 are of the form: 𝑆, π½π‘š, π‘†π½π‘š1π‘†π½π‘š2β‹―π‘†π½π‘šπ‘˜, or elements obtained by permuting 𝑆 and 𝐽 in the element π‘†π½π‘š1π‘†π½π‘š2β‹―π‘†π½π‘šπ‘˜. We deal with each of these forms separately.
Consider the element 𝑆. We have |𝐾(𝑆)|=|𝑋1|3=(βˆ’π‘‘π‘ž3)3. Assume that |𝐾(𝑆)|=1. Then (π‘‘π‘ž3)6=1 and so 6(𝛼+3πœƒ)=2π‘πœ‹,π‘βˆˆβ„€. This implies that (𝛼+3πœƒ)/πœ‹=(𝑐/3)βˆˆβ„š, which is a contradiction when (πœƒ,𝛼)∈𝐺. Therefore, |𝐾(𝑆)|β‰ 1 and so π‘†βˆ‰Ker𝐾.
Consider the element π½π‘š, where π‘šβ‰ 0. We have |𝐾(π½π‘š)|=|𝑋1|2π‘š=(π‘‘π‘ž3)2π‘š. Assume that |𝐾(π½π‘š)|=1. Then (π‘‘π‘ž3)2π‘š=1 and so 2π‘š(𝛼+3πœƒ)=2π‘πœ‹, π‘βˆˆβ„€. This implies that (𝛼+3πœƒ)/πœ‹=(𝑐/π‘š)βˆˆβ„š, which is a contradiction when (πœƒ,𝛼)∈𝐺. Therefore, |𝐾(π½π‘š)|β‰ 1 and so π½π‘šβˆ‰Ker𝐾.
Consider the element 𝑒=π‘†π½π‘š1π‘†π½π‘š2β‹―π‘†π½π‘š2𝑛+1, where the number of 𝑆's is 2𝑛+1. We have |𝐾(𝑒)|=|𝑋1|β„Ž, where βˆ‘β„Ž=3(2𝑛+1)+22𝑛+1𝑖=1π‘šπ‘–βˆˆβ„€. It is clear that we have β„Žβ‰ 0. Assume that |𝐾(𝑒)|=1. Then |𝑋1|β„Ž=(βˆ’π‘‘π‘ž3)β„Ž=1 and so (π‘‘π‘ž3)2β„Ž=1. Then 2β„Ž(𝛼+3πœƒ)=2π‘πœ‹,π‘βˆˆβ„€. This implies that (𝛼+3πœƒ)/πœ‹=(𝑐/β„Ž)βˆˆβ„š, which is a contradiction when (πœƒ,𝛼)∈𝐺. Therefore, |𝐾(𝑒)|β‰ 1, that is, π‘’βˆ‰Ker𝐾. It follows, by Theorem 3.4, that tr(𝐾(𝑒))β‰ 3. Now, consider an element 𝑣 obtained by permuting 𝑆 and 𝐽 in 𝑒. Then 𝑣 is an element with odd number of 𝑆's. The trace of the image of 𝑉 is equal to that of the image of an element having the same form as 𝑒. This implies that tr(𝐾(𝑣))β‰ 3 and thus π‘£βˆ‰Ker𝐾.
Consider the element 𝑒=π‘†π½π‘š1π‘†π½π‘š2β‹―π‘†π½π‘š2𝑛, where the number of 𝑆's is 2𝑛. Then |𝐾(𝑒)|=|𝑋1|β„Ž, where βˆ‘β„Ž=2(3𝑛+2𝑛𝑖=1π‘šπ‘–)βˆˆβ„€. For π‘’βˆˆKer𝐾, we have |𝐾(𝑒)|=1, that is, (βˆ’π‘‘π‘ž3)β„Ž=1. If β„Žβ‰ 0, then (π‘‘π‘ž3)2β„Ž=1 and so 2β„Ž(𝛼+3πœƒ)=2π‘πœ‹, π‘βˆˆβ„€. This implies that (𝛼+3πœƒ)/πœ‹=(𝑐/β„Ž)βˆˆβ„š, which contradicts the fact that (πœƒ,𝛼)∈𝐺. Thus β„Ž=0, that is, βˆ‘2𝑛𝑖=1π‘šπ‘–=βˆ’3𝑛. We write π‘š2π‘›βˆ‘=βˆ’3π‘›βˆ’2π‘›βˆ’1𝑖=1π‘šπ‘–. Then 𝑒=π‘†π½π‘š1π‘†π½π‘š2β‹―π‘†π½π‘š2π‘›βˆ’1π‘†π½βˆ‘βˆ’3π‘›βˆ’2π‘›βˆ’1𝑖=1π‘šπ‘–.
If π‘šπ‘–=3π‘˜π‘– for some 1≀𝑖<2𝑛, then π½π‘šπ‘–βˆˆπ‘(𝐡3). It follows that π‘†π½π‘šπ‘–=π½π‘šπ‘–π‘†. Using the relation 𝑆2=𝐽3, 𝑒 will be reduced to an element with less number of 𝑆's and π‘šπ‘–β‰ 3π‘˜π‘– for all 𝑖.
In the case π‘šπ‘–β‰ 3π‘˜π‘– for every 1≀𝑖<2𝑛, we have π‘šπ‘–=3π‘˜π‘–+π‘Ÿπ‘–, where π‘Ÿπ‘–=1 or 2. Then 𝑒=𝑆𝐽3π‘˜1+π‘Ÿ1⋯𝑆𝐽3π‘˜2π‘›βˆ’1+π‘Ÿ2π‘›βˆ’1π‘†π½βˆ’3π‘›βˆ’(3π‘˜1+π‘Ÿ1+3π‘˜2+π‘Ÿ2+β‹―+3π‘˜2π‘›βˆ’1+π‘Ÿ2π‘›βˆ’1). Since 𝐽3π‘˜π‘–βˆˆπ‘(𝐡3), it follows that 𝑒=π‘†π½π‘Ÿ1π‘†π½π‘Ÿ2β‹―π‘†π½π‘Ÿ2π‘›βˆ’1π‘†π½βˆ‘βˆ’3π‘›βˆ’2π‘›βˆ’1𝑖=1π‘Ÿπ‘–, where π‘Ÿπ‘–=1 or 2. Since π‘’βˆˆKer𝐾, it follows that the trace of the image of 𝑒 is equal to 3. Consider an element 𝑣 obtained by permuting 𝑆 and 𝐽 in the element 𝑒. Then 𝑣 is an element that has an even number of 𝑆's. The trace of the image of 𝑉 is equal to that of an element having the same form as 𝑒. This implies that tr(𝐾(𝑣))=3. It follows, by Theorem 3.4, that π‘£βˆˆKer𝐾.
Therefore, the possible nontrivial elements in the kernel of 𝐾∢𝐡3→𝐺𝐿3(β„‚) are π‘†π½π‘Ÿ1π‘†π½π‘Ÿ2β‹―π‘†π½π‘Ÿ2π‘›βˆ’1π‘†π½βˆ‘βˆ’3π‘›βˆ’2π‘›βˆ’1𝑖=1π‘Ÿπ‘–, and other elements that are obtained by permuting 𝑆 and 𝐽. Here, the number of 𝑆's is 2𝑛 and π‘Ÿπ‘–=1 or 2 for 1≀𝑖<2𝑛.

Now, we present a lemma that will be used in the proof of Theorem 4.1.

Lemma 4.3. 𝐾(π½π‘›βŽ§βŽͺ⎨βŽͺβŽ©ξ€·π‘‘)=2π‘ž6ξ€Έπ‘˜ξ€·π‘‘πΌ,𝑛=3π‘˜,2π‘ž6ξ€Έπ‘˜ξ€·π‘‘πΎ(𝐽),𝑛=3π‘˜+1,2π‘ž6ξ€Έπ‘˜πΎξ€·π½2ξ€Έ,𝑛=3π‘˜+2.(4.4) Here π‘˜βˆˆβ„• and 𝐼 is the identity matrix.

Proof. Direct computations show that βŽ›βŽœβŽœβŽœβŽœβŽπΎ(𝐽)=βˆ’π‘‘π‘ž2(π‘žβˆ’1)π‘‘π‘ž30βˆ’π‘‘π‘ž(π‘žβˆ’1)2π‘‘π‘ž2(π‘žβˆ’1)π‘‘π‘ž3⎞⎟⎟⎟⎟⎠,𝐾𝐽1002ξ€Έ=βŽ›βŽœβŽœβŽœβŽœβŽ00𝑑2π‘ž6π‘‘π‘ž30𝑑2π‘ž5(π‘žβˆ’1)βˆ’π‘‘π‘ž2(π‘žβˆ’1)π‘‘π‘ž30⎞⎟⎟⎟⎟⎠,𝐾𝐽3ξ€Έ=𝑑2π‘ž6𝐽𝐼,𝐾4ξ€Έ=𝑑2π‘ž6𝐽𝐾(𝐽),𝐾5ξ€Έ=𝑑2π‘ž6𝐾𝐽2ξ€Έ.(4.5)
Using mathematical induction on the integer π‘˜, we prove the lemma.

5. Proof of Theorem 4.1

In order to prove that the representation 𝐾∢𝐡3→𝐺𝐿3(β„‚) is faithful, we have to show that the elements in Lemma 4.2 do not belong to Ker𝐾 if we choose π‘ž and 𝑑 in a way that their arguments πœƒ and 𝛼 belong to the previously defined set 𝐺.

That is, we need to show that the element 𝑒=π‘†π½π‘Ÿ1π‘†π½π‘Ÿ2β‹―π‘†π½π‘Ÿ2π‘›βˆ’1π‘†π½βˆ‘βˆ’3π‘›βˆ’2π‘›βˆ’1𝑖=1π‘Ÿπ‘– and others obtained from 𝑒 by permuting 𝑆 and 𝐽 do not belong to the kernel of this representation if (πœƒ,𝛼)∈𝐺.

Consider the element 𝑒=π‘†π½π‘Ÿ1π‘†π½π‘Ÿ2β‹―π‘†π½π‘Ÿ2π‘›βˆ’1π‘†π½βˆ‘βˆ’3π‘›βˆ’2π‘›βˆ’1𝑖=1π‘Ÿπ‘–, where the number of 𝑆's is 2𝑛 and π‘Ÿπ‘–=1 or 2 for 1≀𝑖<2𝑛. We consider five cases and we prove that the element 𝑒, in each of the cases, does not belong to the kernel of the complex specialization 𝐾∢𝐡3→𝐺𝐿3(β„‚).

(1) If π‘Ÿπ‘–=1 for all 1≀𝑖<2𝑛, then 𝑒=𝑆𝐽1ξ€Έ2π‘›βˆ’1π‘†π½βˆ’3π‘›βˆ’(2π‘›βˆ’1)=𝑆𝐽1ξ€Έ2π‘›βˆ’1π‘†π½βˆ’3(2π‘›βˆ’1)+π‘›βˆ’2.(5.1) Since π½βˆ’3βˆˆπ‘(𝐡3), it follows that 𝑒=(π‘†π½βˆ’2)2π‘›βˆ’1π‘†π½π‘›βˆ’2 and so 𝑒=(π‘†π½βˆ’2)2𝑛𝐽𝑛. The eigenvalues of 𝐾(π‘†π½βˆ’2) are 1,βˆ’1/π‘ž and 1/π‘‘π‘ž2, which are easily shown to be distinct for (πœƒ,𝛼)∈𝐺. We diagonalize 𝐾(π‘†π½βˆ’2) by a matrix 𝑇 given by βŽ›βŽœβŽœβŽœβŽœβŽœβŽπ‘‡=π‘‘π‘ž2βˆ’π‘‘π‘ž3𝑑2π‘ž4π‘‘π‘ž(π‘žβˆ’1)βˆ’π‘‘π‘ž2(π‘žβˆ’1)π‘ž(βˆ’1+π‘‘π‘ž(βˆ’1+π‘ž+π‘ž(1+(π‘žβˆ’1)π‘ž)𝑑))βŽžβŽŸβŽŸβŽŸβŽŸβŽŸβŽ π‘žβˆ’1111.(5.2) It is easy to see that det(𝑇)=βˆ’(π‘ž3(1+π‘ž)𝑑(1+π‘‘π‘ž)(π‘‘π‘ž2βˆ’1))/(π‘žβˆ’1)β‰ 0 for (πœƒ,𝛼)∈𝐺.

Then π‘‡βˆ’1πΎξ€·π‘†π½βˆ’2ξ€ΈβŽ›βŽœβŽœβŽœβŽœβŽœβŽ1𝑇=1000βˆ’π‘ž0100π‘‘π‘ž2⎞⎟⎟⎟⎟⎟⎠,π‘‡βˆ’1βŽ›βŽœβŽœβŽœβŽœβŽœβŽ01𝐾(𝑒)𝑇=100π‘ž2𝑛0100ξ€·π‘‘π‘ž2ξ€Έ2π‘›βŽžβŽŸβŽŸβŽŸβŽŸβŽŸβŽ π‘‡βˆ’1𝐾(𝐽𝑛)𝑇.(5.3)

We take the following possibilities regarding the value of 𝑛.

(a) For 𝑛=3π‘˜, we have, from Lemma 4.3, that π‘‡βˆ’1𝐾(𝐽𝑛)𝑇=(𝑑2π‘ž6)π‘˜πΌ. Then we get π‘‡βˆ’1βŽ›βŽœβŽœβŽœβŽœβŽœβŽξ€·π‘‘πΎ(𝑒)𝑇=2π‘ž6ξ€Έπ‘˜000𝑑2π‘˜0100𝑑4π‘ž6ξ€Έπ‘˜βŽžβŽŸβŽŸβŽŸβŽŸβŽŸβŽ .(5.4) The entry (𝑑2π‘ž6)π‘˜β‰ 1; otherwise, π‘’π‘–π‘˜(2𝛼+6πœƒ)=1. That is, (𝛼+3πœƒ)/πœ‹βˆˆβ„š, which contradicts the fact that (πœƒ,𝛼)∈𝐺.

(b) For 𝑛=3π‘˜+1, we have, from Lemma 4.3, that π‘‡βˆ’1𝐾(𝐽𝑛)𝑇=(𝑑2π‘ž6)π‘˜π‘‡βˆ’1𝐾(𝐽)𝑇. Direct computations give π‘‡βˆ’1βŽ›βŽœβŽœβŽœβŽœβŽœβŽβˆ—πΎ(𝑒)𝑇=π‘‘π‘ž3𝑑2π‘ž6ξ€Έπ‘˜ξ€·1βˆ’π‘‘π‘ž3ξ€Έξ€·(1+π‘ž)π‘‘π‘ž2ξ€Έβˆ—βŽžβŽŸβŽŸβŽŸβŽŸβŽŸβŽ βˆ’1βˆ—βˆ—βˆ—βˆ—βˆ—βˆ—.(5.5) It is easy to show that (π‘‘π‘ž3(𝑑2π‘ž6)π‘˜(1βˆ’π‘‘π‘ž3))/((1+π‘ž)(π‘‘π‘ž2βˆ’1))β‰ 0 for (πœƒ,𝛼)∈𝐺.

(c) For 𝑛=3π‘˜+2, we have, from Lemma 4.3, that π‘‡βˆ’1𝐾(𝐽𝑛)𝑇=(𝑑2π‘ž6)π‘˜π‘‡βˆ’1𝐾(𝐽2)𝑇. We get, by direct computations, π‘‡βˆ’1βŽ›βŽœβŽœβŽœβŽœβŽœβŽβˆ—πΎ(𝑒)𝑇=π‘‘π‘ž4𝑑2π‘ž6ξ€Έπ‘˜ξ€·π‘‘π‘ž3ξ€Έβˆ’1ξ€·(1+π‘ž)π‘‘π‘ž2ξ€Έβˆ—βŽžβŽŸβŽŸβŽŸβŽŸβŽŸβŽ βˆ’1βˆ—βˆ—βˆ—βˆ—βˆ—βˆ—.(5.6) It is easy to show that (π‘‘π‘ž4(𝑑2π‘ž6)π‘˜(π‘‘π‘ž3βˆ’1))/((1+π‘ž)(π‘‘π‘ž2βˆ’1))β‰ 0 for (πœƒ,𝛼)∈𝐺.

(2) If π‘Ÿπ‘–=2 for all 1≀𝑖<2𝑛, then 𝑒=𝑆𝐽2ξ€Έ2π‘›βˆ’1π‘†π½βˆ’3π‘›βˆ’2(2π‘›βˆ’1)=𝑆𝐽2ξ€Έ2π‘›βˆ’1π‘†π½βˆ’3(2π‘›βˆ’1)βˆ’π‘›βˆ’1.(5.7) Since π½βˆ’3βˆˆπ‘(𝐡3), it follows that 𝑒=(π‘†π½βˆ’1)2π‘›βˆ’1π‘†π½βˆ’π‘›βˆ’1 and so 𝑒=(π‘†π½βˆ’1)2π‘›π½βˆ’π‘›. The eigenvalues of 𝐾(π‘†π½βˆ’1) are 1,βˆ’π‘ž, and π‘‘π‘ž2. It is easy to show that these eigenvalues are distinct when (πœƒ,𝛼)∈𝐺. We diagonalize 𝐾(π‘†π½βˆ’1) by a matrix 𝐻 given by βŽ›βŽœβŽœβŽœβŽœβŽœβŽξ€·π»=00(1+π‘‘π‘ž)π‘‘π‘ž2ξ€Έβˆ’1𝑑(π‘žβˆ’1)π‘žβˆ’1βˆ’1+π‘ž+π‘‘π‘ž2⎞⎟⎟⎟⎟⎟⎠111.(5.8)

It is clear that det(𝐻)=((1+π‘ž)(1+π‘‘π‘ž)(π‘‘π‘ž2βˆ’1))/(𝑑(π‘žβˆ’1))β‰ 0 for (πœƒ,𝛼)∈𝐺.

Then π»βˆ’1πΎξ€·π‘†π½βˆ’1ξ€ΈβŽ›βŽœβŽœβŽœβŽœβŽπ»=1000βˆ’π‘ž000π‘‘π‘ž2⎞⎟⎟⎟⎟⎠,π»βˆ’1βŽ›βŽœβŽœβŽœβŽœβŽπΎ(𝑒)𝐻=1000π‘ž2𝑛0ξ€·00π‘‘π‘ž2ξ€Έ2π‘›βŽžβŽŸβŽŸβŽŸβŽŸβŽ π»βˆ’1𝐾(π½βˆ’π‘›)𝐻.(5.9)

Using Lemma 4.3, we can easily see that 𝐾(π½βˆ’π‘›βŽ§βŽͺβŽͺβŽͺ⎨βŽͺβŽͺβŽͺ⎩1)=𝑑2π‘ž6ξ€Έπ‘˜1𝐼,𝑛=3π‘˜,𝑑2π‘ž6ξ€Έπ‘˜πΎξ€·π½βˆ’1ξ€Έ1,𝑛=3π‘˜+1,𝑑2π‘ž6ξ€Έπ‘˜πΎξ€·π½βˆ’2ξ€Έ,𝑛=3π‘˜+2.(5.10)

Here π‘˜ is a positive integer. We deal with the following possibilities regarding the value of 𝑛.

(a) For 𝑛=3π‘˜, a comuptation shows that π»βˆ’1πΎβŽ›βŽœβŽœβŽœβŽœβŽœβŽ1(𝑒)𝐻=𝑑2π‘ž6ξ€Έπ‘˜0100𝑑2π‘˜0𝑑004π‘ž6ξ€Έπ‘˜βŽžβŽŸβŽŸβŽŸβŽŸβŽŸβŽ .(5.11)

The entry 1/(𝑑2π‘ž6)π‘˜β‰ 1; otherwise, π‘’π‘–π‘˜(2𝛼+6πœƒ)=1, which implies that (𝛼+3πœƒ)/πœ‹βˆˆβ„š. This contradicts the fact that (πœƒ,𝛼)∈𝐺.

(b) For 𝑛=3π‘˜+1, a computation shows that π»βˆ’1βŽ›βŽœβŽœβŽœβŽœβŽœβŽβˆ—πΎ(𝑒)𝐻=1βˆ’π‘‘π‘ž3π‘‘π‘ž3𝑑2π‘ž6ξ€Έπ‘˜ξ€·(1+π‘ž)π‘‘π‘ž2ξ€Έβˆ—βŽžβŽŸβŽŸβŽŸβŽŸβŽŸβŽ βˆ’1βˆ—βˆ—βˆ—βˆ—βˆ—βˆ—.(5.12)

It is easy to show that (1βˆ’π‘‘π‘ž3)/(π‘‘π‘ž3(𝑑2π‘ž6)π‘˜(1+π‘ž)(π‘‘π‘ž2βˆ’1))β‰ 0 for (πœƒ,𝛼)∈𝐺.

(c) For 𝑛=3π‘˜+2, a computation shows that π»βˆ’1βŽ›βŽœβŽœβŽœβŽœβŽœβŽβˆ—πΎ(𝑒)𝐻=π‘‘π‘ž3βˆ’1π‘‘π‘ž4𝑑2π‘ž6ξ€Έπ‘˜ξ€·(1+π‘ž)π‘‘π‘ž2ξ€Έβˆ—βŽžβŽŸβŽŸβŽŸβŽŸβŽŸβŽ βˆ’1βˆ—βˆ—βˆ—βˆ—βˆ—βˆ—.(5.13)

We can also see that (π‘‘π‘ž3βˆ’1)/(π‘‘π‘ž4(𝑑2π‘ž6)π‘˜(1+π‘ž)(π‘‘π‘ž2βˆ’1))β‰ 0 for (πœƒ,𝛼)∈𝐺.

(3) If for 1≀𝑖<2𝑛, we have that π‘Ÿπ‘–=ξ‚»1,𝑖isodd,2,𝑖iseven,(5.14) then 𝑒=𝑆𝐽1𝑆𝐽2ξ€Έπ‘›βˆ’1𝑆𝐽1π‘†π½βˆ’3π‘›βˆ’[3(π‘›βˆ’1)+1]=𝑆𝐽1𝑆𝐽2ξ€Έπ‘›βˆ’1𝑆𝐽1π‘†π½βˆ’3(2π‘›βˆ’1)βˆ’1.(5.15)

Since π½βˆ’3βˆˆπ‘(𝐡3), it follows that 𝑒=(π‘†π½βˆ’2π‘†π½βˆ’1)π‘›βˆ’1π‘†π½βˆ’2π‘†π½βˆ’1 and so 𝑒=(π‘†π½βˆ’2π‘†π½βˆ’1)𝑛. The eigenvalues of 𝐾(π‘†π½βˆ’2π‘†π½βˆ’1) are 𝑒1=1,𝑒2√=(π‘Žβˆ’π‘)/2π‘‘π‘ž2,𝑒3√=(π‘Ž+𝑏)/2π‘‘π‘ž2, where π‘Ž=1+π‘‘π‘ž2+𝑑2π‘ž4βˆ’π‘ž(1+𝑑)βˆ’π‘‘π‘ž3(1+𝑑) and 𝑏=βˆ’4𝑑2π‘ž4+(1+π‘ž(βˆ’1+(π‘žβˆ’1)𝑑)(1+π‘‘π‘ž2))2. These eigenvalues are distinct for (πœƒ,𝛼)∈𝐺. Now, we diagonalize 𝐾(π‘†π½βˆ’2π‘†π½βˆ’1) by a matrix 𝐿 given by βŽ›βŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽπ‘žπ‘žξ‚€βˆšπΏ=βˆ’π‘βˆ’π‘ξ‚βˆšβˆ’π‘+π‘βˆ’π‘žξ‚€βˆšβˆ’π‘+π‘ξ‚βˆšπ‘+𝑏1βˆ’1+π‘ž+π‘‘π‘žβˆ’2π‘ž+2(π‘žβˆ’1)2𝑑2π‘ž3βˆšβˆ’π‘+π‘βˆ’βˆ’2π‘ž+2(π‘žβˆ’1)2𝑑2π‘ž3βˆšπ‘+π‘βŽžβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽ 111,(5.16) where 𝑏 is defined above and 𝑐=1+π‘‘π‘ž2+π‘‘π‘ž3(π‘‘βˆ’1)βˆ’π‘‘2π‘ž4βˆ’π‘ž(1+𝑑).

Also, we have that det(𝐿)=((βˆ’1+π‘ž+π‘ž(1+π‘ž+π‘ž2)π‘‘βˆ’(π‘žβˆ’1)𝑑2π‘ž3)βˆšπ‘)/(π‘ž3(1+(π‘žβˆ’1)π‘ž)𝑑3)β‰ 0 when πœ‹πœƒβ‰ Β±3,𝑏≠0,1+2cosπœƒ+2cos(πœƒ+𝛼)βˆ’2cos(2πœƒ+𝛼)β‰ 0.(5.17) Then πΏβˆ’1πΎξ€·ξ€·π‘†π½βˆ’2π‘†π½βˆ’1ξ€Έπ‘›ξ€ΈβŽ›βŽœβŽœβŽœβŽœβŽπΏ=1000𝑒𝑛2000𝑒𝑛3⎞⎟⎟⎟⎟⎠.(5.18)

If 𝑒𝑛3=1, then 𝑒𝑖𝛽𝑛=1, where 𝛽=arg(𝑒3). This implies that 𝛽/πœ‹βˆˆβ„š, which contradicts the fact that (πœƒ,𝛼)∈𝐺. Therefore, πΏβˆ’1𝐾((π‘†π½βˆ’2π‘†π½βˆ’1)𝑛)𝐿≠𝐼 and so 𝐾((π‘†π½βˆ’2π‘†π½βˆ’1)𝑛)≠𝐼. We conclude that 𝑒=(π‘†π½βˆ’2π‘†π½βˆ’1)π‘›βˆ‰Ker𝐾.

(4) If for 1≀𝑖<2𝑛, we have that π‘Ÿπ‘–=ξ‚»2,𝑖isodd,1,𝑖iseven,(5.19) then 𝑒=𝑆𝐽2𝑆𝐽1ξ€Έπ‘›βˆ’1𝑆𝐽2π‘†π½βˆ’3π‘›βˆ’[3(π‘›βˆ’1)+2]=𝑆𝐽2𝑆𝐽1ξ€Έπ‘›βˆ’1𝑆𝐽2𝑆𝐽(βˆ’3(2π‘›βˆ’1)βˆ’2).(5.20) Since π½βˆ’3βˆˆπ‘(𝐡3), it follows that 𝑒=(π‘†π½βˆ’1π‘†π½βˆ’2)π‘›βˆ’1π‘†π½βˆ’1π‘†π½βˆ’2 and so 𝑒=(π‘†π½βˆ’1π‘†π½βˆ’2)𝑛. Using the properties of the trace, we have that tr(𝐾(𝑒))=tr(𝐾((π‘†π½βˆ’2π‘†π½βˆ’1)𝑛)). By our result in case (3) and Theorem 3.4, we have that tr(𝐾(𝑒))β‰ 3 and so 𝑒=(π‘†π½βˆ’1π‘†π½βˆ’2)π‘›βˆ‰Ker𝐾.

(5) Let 𝑒=π‘†π½π‘Ÿ1π‘†π½π‘Ÿ2β‹―π‘†π½π‘Ÿ2π‘›βˆ’1π‘†π½βˆ‘βˆ’3π‘›βˆ’2π‘›βˆ’1𝑖=1π‘Ÿπ‘– such that not all π‘Ÿπ‘–'s are 1 and not all π‘Ÿπ‘–'s are 2. We also assume that there exists an integer 𝑗 such that π‘Ÿπ‘—=π‘Ÿπ‘—+1, where 1≀𝑗<2𝑛. We write 𝑒=π΄π½βˆ’π‘, where 𝐴=π‘†π½π‘Ÿ1π‘†π½π‘Ÿ2β‹―π‘†π½π‘Ÿ2π‘›βˆ’1𝑆 and βˆ‘π‘=3𝑛+2π‘›βˆ’1𝑖=1π‘Ÿπ‘–. Without loss of generality, we assume that 𝐾(𝐴) and 𝐾(π½βˆ’π‘) do not commute for some values of π‘ž and 𝑑 in 𝐺. For this reason, we require that 𝑝≑1 or 2 (mod 3). If 𝐾(𝐴) and 𝐾(π½βˆ’π‘) happened to commute for some values of π‘ž and 𝑑 then we change the values of π‘ž and 𝑑, slightly in a way that (πœƒ,𝛼) is still in 𝐺 and the matrices 𝐾(𝐴),𝐾(π½βˆ’π‘) do not commute anymore. This is due to the fact that (π‘ž,𝑑) belongs to a dense subset of 𝐺 by the hypothesis of the theorem.

The eigenvalues of π½βˆ’1 are 1/(π‘ž2𝑑2/3),βˆ’(βˆ’1)1/3/(π‘ž2𝑑2/3) and (βˆ’1)2/3/(π‘ž2𝑑2/3). It is clear that these eigenvalues are distinct. We diagonalize 𝐾(π½βˆ’1) by a matrix π‘Š given by βŽ›βŽœβŽœβŽœβŽœβŽπ‘ž2𝑑2/3(βˆ’1)2/3π‘ž2𝑑2/3(βˆ’1)4/3π‘ž2𝑑2/3π‘žξ€·1+(π‘žβˆ’1)𝑑1/3𝑑1/3(βˆ’π‘‘)1/3π‘žξ€·βˆ’1+(βˆ’1)1/3(π‘žβˆ’1)𝑑1/3𝑑1/3π‘žξ€·1+(βˆ’1)2/3(ξ€Έπ‘‘π‘žβˆ’1)1/3⎞⎟⎟⎟⎟⎠111.(5.21)

Under direct computations, we see that √det(π‘Š)=βˆ’3𝑖3π‘ž3𝑑≠0. Here βˆšπ‘–=βˆ’1. Then we get that π‘Šβˆ’1πΎξ€·π½βˆ’1ξ€ΈβŽ›βŽœβŽœβŽœβŽœβŽœβŽœβŽ1π‘Š=π‘ž2𝑑2/3(000βˆ’βˆ’1)1/3π‘ž2𝑑2/3000(βˆ’1)2/3π‘ž2𝑑2/3⎞⎟⎟⎟⎟⎟⎟⎠.(5.22) We have that π‘Šβˆ’1ξ€·π‘ŠπΎ(𝑒)π‘Š=βˆ’1π‘ŠπΎ(𝐴)π‘Šξ€Έξ€·βˆ’1𝐾(π½βˆ’π‘ξ€Έ.)π‘Š(5.23) Assume, to get contradiction, that π‘’βˆˆKer(𝐾). Then π‘Šβˆ’1ξ€·π‘ŠπΎ(𝐴)π‘Š=βˆ’1π‘ŠπΎ(𝑒)π‘Šξ€Έξ€·βˆ’1𝐾(π½βˆ’π‘ξ€Έ)π‘Šβˆ’1=ξ€·π‘Šβˆ’1𝐾(π½βˆ’π‘ξ€Έ)π‘Šβˆ’1.(5.24)

It follows that π‘Šβˆ’1𝐾(𝐴)π‘Š is a diagonal matrix. Therefore, π‘Šβˆ’1𝐾(𝐴)π‘Š and π‘Šβˆ’1𝐾(π½βˆ’π‘)π‘Š commute, that is, π‘Šβˆ’1𝐾(𝑒)π‘Š=π‘Šβˆ’1𝐾(π½βˆ’π‘)𝐾(𝐴)π‘Š.(5.25)

Then 𝐾(𝑒)=𝐾(π½βˆ’π‘)𝐾(𝐴).(5.26)

Therefore, 𝐾(𝐴) and 𝐾(π½βˆ’π‘) commute, which is a contradiction.

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