Another Proof of the Faithfulness of the Lawrence-Krammer Representation of the Braid Group 𝐵_𝟑

Abstract

The Lawrence-Krammer representation of the braid group 𝐵_𝑛 was proved to be faithful for 𝑛≥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 𝐵₃ 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 𝐺𝐿_𝑛(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 𝑛≥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 𝑡 is a real number with 0<𝑡<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 𝑞 and 𝑡 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 𝑞 and 𝑡 in [2] and assumed that 𝑞 is a real number with 0<𝑞<1.

Our work uses a different method that basically depends on matrix computations to show that the Lawrence-Krammer representation of 𝐵₃ is faithful. In our argument, we specialize the indeterminates 𝑞 and 𝑡 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 𝐵₃ 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 𝑞 and 𝑡 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 𝑞 and 𝑡 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 𝐵₃ 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, 𝐵_𝑛, is the abstract group with presentation 𝐵_𝑛=⟨𝜎₁,…,𝜎_𝑛−1∣𝜎_𝑖𝜎_𝑖+1𝜎_𝑖=𝜎_𝑖+1𝜎_𝑖𝜎_𝑖+1 for 𝑖=1,2,…,𝑛−2, and 𝜎_𝑖𝜎_𝑗=𝜎_𝑗𝜎_𝑖 if |𝑖−𝑗|≥2⟩. The generators 𝜎₁,…,𝜎_𝑛−1 are called the standard generators of 𝐵_𝑛.

The Lawrence-Krammer representation of braid groups is a representation of the braid group 𝐵_𝑛 in 𝐺𝐿_𝑚(ℤ[𝑞^±1,𝑡^±1])=Aut(𝑉₀), where 𝑚=𝑛(𝑛−1)/2 and 𝑉₀ is the free module of rank 𝑚 over ℤ[𝑞^±1,𝑡^±1]. The representation is denoted by 𝐾(𝑞,𝑡). For simplicity, we write 𝐾 instead of 𝐾(𝑞,𝑡).

Definition 2.2. With respect to {𝑥_𝑖,𝑗}_{1≤𝑖<𝑗≤𝑛}, the free basis of 𝑉₀, the image of each Artin generator under Krammer's representation is written as 𝐾𝜎_𝑘𝑥_𝑖,𝑗=⎧⎪⎪⎪⎨⎪⎪⎪⎩𝑡𝑞²(1−𝑥𝑞)𝑥_𝑘,𝑘+1,_𝑖,𝑘𝑖=𝑘,𝑗=𝑘+1,𝑥+𝑞𝑥_𝑖,𝑘+1,𝑗=𝑘,𝑖<𝑘,_𝑖,𝑘𝑡𝑞(𝑞+−𝑡𝑞1)𝑥^{𝑘−𝑖+1}(𝑞_𝑘,𝑘+1−1)𝑥_𝑘,𝑘+1,𝑗=𝑘+1,𝑖<𝑘,𝑥+𝑞𝑥_𝑘+1,𝑗,𝑖=𝑘,𝑘+1<𝑗,_𝑘,𝑗𝑥+(1−𝑞)𝑥_𝑘+1,𝑗,𝑖=𝑘+1,𝑘+1<𝑗,_𝑖,𝑗𝑥,𝑖<𝑗<𝑘or𝑘+1<𝑖<𝑗,_𝑖,𝑗+𝑡𝑞^𝑘−𝑖(𝑞−1)²𝑥_𝑘,𝑘+1,𝑖<𝑘<𝑘+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(𝑞,𝑡)∶𝐵₃→𝐺𝐿₃(ℤ[𝑞^±1,𝑡^±1]), where 𝐾𝜎₁=⎛⎜⎜⎜⎜⎝𝑡𝑞²𝑡𝑞(𝑞0−1)1001−0𝑞𝑞⎞⎟⎟⎟⎟⎠,𝐾𝜎₂=⎛⎜⎜⎜⎜⎝1−10𝑞00𝑞𝑡𝑞²(𝑡𝑞𝑞0₂−1)⎞⎟⎟⎟⎟⎠.(2.2)

Here, ℤ[𝑞^±1,𝑡^±1] is the ring of Laurent polynomials on two variables. For simplicity, we denote 𝐾(𝜎_𝑖) by 𝑋_𝑖, where 𝑖=1,2. Here, 𝑋^∗𝑖 is the conjugate transpose of 𝑋, where 𝑡=𝑡⁻¹ and 𝑞=𝑞⁻¹.

3. The Lawrence-Krammer Representation of 𝐵₃ Is Equivalent to a Unitary Representation

Budney has proved that the L-K representation of the braid group 𝐵_𝑛 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 𝑞 and 𝑡 to nonzero complex numbers on the unit circle. We consider the complex representation 𝐾∶𝐵₃→𝐺𝐿₃(ℂ) and prove that it is unitary relative to a positive definite Hermitian form.

Theorem 3.1. The images of the generators of 𝐵₃ under the Lawrence-Krammer representation are unitary relative to a Hermitian matrix 𝑀 given by 𝑀=⎛⎜⎜⎜⎜⎜⎜⎝1+𝑞²(1+𝑡𝑞)𝑡𝑞²−1𝑡𝑞²(𝑞−1)1+𝑞²(1−𝑡𝑞(𝑞−1))𝑡𝑞²1+𝑞²1+𝑞²𝑡𝑞²+𝑞−1𝑞²1+𝑞²−1+𝑡1−(𝑞−2)²𝑞+𝑡𝑞³𝑡𝑞²(𝑞−1)1+𝑞²𝑡𝑞²+𝑞−1𝑞₂𝑞+1𝑞²1+𝑞²(1−𝑡𝑞(𝑞−1))𝑡𝑞³1+𝑞²(1+𝑡𝑞)𝑡𝑞²−1𝑡𝑞²(𝑞−1)⎞⎟⎟⎟⎟⎟⎟⎠(3..1)
That is, 𝑋_𝑖𝑀𝑋^∗𝑖=𝑀 for 𝑖=1,2, where 𝑀^∗=𝑀.

The choice of 𝑀 is unique only if we specialize the indeterminates 𝑞 and 𝑡 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 𝑀^ of 𝑀 is equivalent to the identity matrix, that is, 𝑈𝑀^𝑈^∗=𝐼 for some matrix 𝑈. In other words, we need to show that 𝑀^=𝑉𝑉^∗ for some matrix 𝑉; or equivalently, 𝑀^ is hermitian.

Theorem 3.2. Let 𝑞 and 𝑡 be nonzero complex numbers on the unit circle such that 𝑞=𝑒^𝑖𝜃 and 𝑡=𝑒^𝑖𝛼. The matrix 𝑀 is positive definite when 𝜃 and 𝛼 are chosen in either one of the following intervals:(i)−𝜋/2<𝜃<0 and −𝜋−(3𝜃/2)<𝛼<−(3𝜃/2), (ii)0<𝜃<𝜋/2 and −(3𝜃/2)<𝛼<𝜋−(3𝜃/2).

Proof . We denote the principal minors of 𝑀 by 𝑑_𝑖, where 𝑖=1,2,3. We have 𝑑₁=1+𝑞²(1+𝑡𝑞)𝑡𝑞²−1𝑑𝑡𝑞²(𝑞−1)₂,=1+𝑞²²𝑡²𝑞³−1(−1+𝑡𝑞(−2+𝑞(2+𝑡𝑞)))𝑡²𝑞⁴(𝑞−1)²𝑑₃,=1+𝑞²³(1+𝑡)𝑡𝑞³−1𝑡²𝑞³−1²𝑡³𝑞⁶(𝑞−1)³.(3.2)
We find the intervals of 𝑞 and 𝑡 for which 𝑑₁,𝑑₂, and 𝑑₃ are strictly positive. Let 𝑞=𝑒^𝑖𝜃 and 𝑡=𝑒^𝑖𝛼, where −𝜋≤𝜃≤𝜋 and −𝜋≤𝛼≤𝜋. Then we have 𝑑₁=2cos𝜃1+csc𝜃𝑑2sin32𝜃+𝛼₂,=4cos²𝜃csc𝜃𝑑2sin32𝜃+𝛼2+csc𝜃2sin32𝜃+𝛼₃,=8cos³𝜃csc³𝜃2sin²32𝜃+𝛼sin32𝜃+sin32𝜃+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(𝜃/2)<0, sin(3𝜃/2+𝛼)<0, cos𝜃>0, and sin(3𝜃/2)+sin(3𝜃/2+𝛼)<0.
In the interval (ii), we see that csc(𝜃/2)>0, sin(3𝜃/2+𝛼)>0, cos𝜃>0, and sin(3𝜃/2)+sin(3𝜃/2+𝛼)>0.

We construct a homomorphism that specializes the indeterminates 𝑞 and 𝑡 to nonzero complex numbers, on the unit circle, which are transcendentally independent over ℤ.

Let 𝑓_𝑤 be a homomorphism 𝑓_𝑤∶ℤ[𝑞^±1,𝑡^±1]→ℂ defined as follows: 𝑓_𝑤(𝑞)=𝑤₁,𝑓_𝑤(𝑡)=𝑤₂,𝑓_𝑤(𝑧)=𝑧 for 𝑧∈ℤ, where 𝑤=(𝑤₁,𝑤₂) and 𝑤₁,𝑤₂ are complex numbers on the unit circle. Let 𝑓_𝑤 also denote the group homomorphism 𝐺𝐿₃(ℤ[𝑞^±1,𝑡^±1])→𝐺𝐿₃(ℂ). We choose 𝑤 such that 𝑓_𝑤(𝑀) is positive definite; that is, 𝑓_𝑤(𝑀)=𝑉𝑉^∗, for some 𝑉∈𝐺𝐿₃(ℂ).

Consider now the composition map 𝑓_𝑤∘𝐾∶𝐵₃→𝐺𝐿₃(ℂ). It was proved in [6] that the specialization of Krammer's representation 𝐵₃→𝐺𝐿₃(ℂ) is irreducible if and only if 𝑡≠−1,𝑡𝑞³≠1, and 𝑡²𝑞³≠1. It is easy to see that the conditions for irreducibility are achieved once we choose 𝑞 and 𝑡 in either one of the intervals (i) or (ii) in the hypothesis of Theorem 3.2. The uniqueness of 𝑓_𝑤(𝑀) up to scalar multiplication follows from Shur's Lemma and the fact that the specialization of Krammer's representation of 𝐵₃ is irreducible.

Theorem 3.3. The complex representation of 𝐵₃, 𝑓_𝑤∘𝐾, is conjugate to a unitary representation.

Proof. Since 𝑋_𝑖𝑀𝑋^∗𝑖=𝑀 for 𝑖=1,2, it follows that 𝑓_𝑤𝑋_𝑖𝑓_𝑤(𝑀)𝑓_𝑤𝑋^∗𝑖=𝑓_𝑤(𝑀).(3.4) But 𝑓_𝑤(𝑀)=𝑉𝑉^∗, then 𝑉⁻¹𝑓_𝑤𝑋_𝑖𝑉𝑉⁻¹𝑓_𝑤𝑋_𝑖𝑉^∗=𝐼.(3.5) Now, let 𝑈=𝑉⁻¹𝑓_𝑤𝑋_𝑖𝑉,(3.6) then 𝑈𝑈^∗=𝐼.(3.7) Hence 𝑈^∗𝑈=𝐼 and so 𝑈 is unitary.

We now find a necessary and sufficient condition for an element of 𝐵₃ possibly to belong to the kernel of the Lawrence-Krammer representation.

Theorem 3.4. An element of 𝐵₃ lies in the kernel of the Lawrence-Krammer representation of 𝐵₃ if and only if the trace of its image is equal to three.

Proof . If tr(𝑓_𝑤(𝑋_𝑖))=3, then tr(𝑈)=3, where 𝑈=𝑉⁻¹𝑓_𝑤(𝑋_𝑖)𝑉. Since 𝑈 is unitary, it follows that 𝑈 is diagonalizable, that is, there exists a matrix 𝑃 such that 𝑃⁻¹𝑈𝑃=𝐷, where 𝐷 is a diagonal matrix with eigenvalues of 𝑈 as the diagonal entries. Hence 𝜆₁+𝜆₂+𝜆₃=3, where the 𝜆^𝑖𝑠 are the eigenvalues of 𝑈. Being unitary, it has its eigenvalues on the unit circle. Therefore, we get 𝜆₁=𝜆₂=𝜆₃=1. Thus, 𝐷 is the identity matrix and so is 𝑈. This implies that 𝑓_𝑤(𝑋_𝑖)=𝐼₃.

Therefore, we conclude that if there exists a nontrivial element in 𝐵₃ such that the trace of its image under 𝑓_𝑤∘𝐾∶𝐵₃→𝐺𝐿₃(ℂ) 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 𝑞 and 𝑡. Let 𝑞 and 𝑡 be nonzero complex numbers on the unit circle such that 𝑞=𝑒^𝑖𝜃 and 𝑡=𝑒^𝑖𝛼.

Let 𝛽=arg⎛⎜⎜⎜⎝1+𝑡𝑞²+𝑡²𝑞⁴−𝑞(1+𝑡)−𝑞³𝑡(1+𝑡)+−4𝑞⁴𝑡²+1+(−1+𝑞(𝑞−1)𝑡)1+𝑡𝑞²²2𝑡𝑞²⎞⎟⎟⎟⎠.(4.1)

Consider the values of 𝑞 and 𝑡 whose arguments belong to the set 𝐺 defined as follows: 𝐺=(𝜃,𝛼)∈𝐴∪𝐵∣𝜃≠±𝜋3,𝛼+𝜋3𝜃∉ℚ,𝜋𝛽∉ℚ,(4.2) where 𝐴=(𝜃,𝛼)∣−𝜋𝐵=(𝜃,𝛼)∣02<<𝜃𝜃<<𝜋0,−𝜋3−2𝜃<𝛼<−32𝜃,2,−32𝜃<𝛼<𝜋3−2𝜃.(4.3)

The set 𝐺 is a subset of the union of the intervals given in Theorem 3.2. We choose 𝑤 such that (𝜃,𝛼)∈𝐺. Therefore, for values of 𝑞 and 𝑡 such that (𝜃,𝛼)∈𝐺, we have 𝑓_𝑤(𝑀) being positive definite and 𝑓_𝑤∘𝐾∶𝐵₃→𝐺𝐿₃(ℂ) irreducible. For simplicity, we still write 𝐾 instead of 𝑓_𝑤∘𝐾.

We now present our second main theorem.

Theorem 4.1. The complex specialization of the Lawrence-Krammer representation 𝐾∶𝐵₃→𝐺𝐿₃(ℂ) 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 𝐵₃=⟨𝜎₁,𝜎₂∣𝜎₁𝜎₂𝜎₁=𝜎₂𝜎₁𝜎₂⟩, where 𝜎₁ and 𝜎₂ are called the standard generators of 𝐵₃. Consider the product of generators of 𝐵₃, namely, 𝐽=𝜎₁𝜎₂. We also let 𝑆=𝜎₁𝐽. It was proved in [7] that 𝐵₃ is generated by the two elements 𝑆 and 𝐽 and the relation 𝑆²=𝐽³. That is, we have that 𝐵₃=⟨𝑆,𝐽∣𝑆²=𝐽³⟩. Also, it was proved in [8] that the center 𝑍(𝐵₃) of 𝐵₃ is generated by the one generator 𝐽³, that is, 𝑍(𝐵₃)=⟨𝐽³⟩. Thus, the elements of 𝐵₃ are of the form: 𝑆^𝑙, 𝐽^𝑚, 𝐽^𝑚1𝑆^𝑙1𝐽^𝑚2𝑆^𝑙2,…, and so forth, where all the exponents are integers.

Consider the element 𝑆^𝑙. If 𝑙 is odd, then 𝑆^𝑙=𝑆^2𝑘+1=𝑆⋅𝑆^2𝑘=𝑆𝐽^3𝑘, where 𝑘∈ℤ. If 𝑙 is even then 𝑆^𝑙=𝑆^2𝑘=𝐽^3𝑘, where 𝑘∈ℤ. Therefore, any element of 𝐵₃ is of one of the following forms: 𝑆, 𝐽^𝑚, 𝑆𝐽^𝑚1𝑆𝐽^𝑚2…, or 𝐽^𝑚1𝑆𝐽^𝑚2𝑆…, where 𝑚_𝑖's ∈ℤ.

Lemma 4.2. The possible nontrivial elements in the kernel of the representation 𝐾∶𝐵₃→𝐺𝐿₃(ℂ) are(i)𝑆𝐽^𝑟1𝑆𝐽^𝑟2⋯𝑆𝐽^{𝑟2𝑛−1}𝑆𝐽^{∑−3𝑛−2𝑛−1𝑖=1𝑟𝑖}, where the number of 𝑆's is 2𝑛 and 𝑟_𝑖=1 or 2 for 1≤𝑖<2𝑛,(ii)elements obtained from (𝑖) by permuting 𝑆 and 𝐽.

Proof. Recall that 𝑞=𝑒^𝑖𝜃,𝑡=𝑒^𝑖𝛼, and 𝐾(𝜎_𝑖)=𝑋_𝑖, 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|=−𝑡𝑞³. In general, elements of 𝐵₃ are of the form: 𝑆, 𝐽^𝑚, 𝑆𝐽^𝑚1𝑆𝐽^𝑚2⋯𝑆𝐽^𝑚𝑘, or elements obtained by permuting 𝑆 and 𝐽 in the element 𝑆𝐽^𝑚1𝑆𝐽^𝑚2⋯𝑆𝐽^𝑚𝑘. We deal with each of these forms separately.
Consider the element 𝑆. We have |𝐾(𝑆)|=|𝑋₁|³=(−𝑡𝑞³)³. Assume that |𝐾(𝑆)|=1. Then (𝑡𝑞³)⁶=1 and so 6(𝛼+3𝜃)=2𝑐𝜋,𝑐∈ℤ. This implies that (𝛼+3𝜃)/𝜋=(𝑐/3)∈ℚ, which is a contradiction when (𝜃,𝛼)∈𝐺. Therefore, |𝐾(𝑆)|≠1 and so 𝑆∉Ker𝐾.
Consider the element 𝐽^𝑚, where 𝑚≠0. We have |𝐾(𝐽^𝑚)|=|𝑋₁|^2𝑚=(𝑡𝑞³)^2𝑚. Assume that |𝐾(𝐽^𝑚)|=1. Then (𝑡𝑞³)^2𝑚=1 and so 2𝑚(𝛼+3𝜃)=2𝑐𝜋, 𝑐∈ℤ. This implies that (𝛼+3𝜃)/𝜋=(𝑐/𝑚)∈ℚ, which is a contradiction when (𝜃,𝛼)∈𝐺. Therefore, |𝐾(𝐽^𝑚)|≠1 and so 𝐽^𝑚∉Ker𝐾.
Consider the element 𝑢=𝑆𝐽^𝑚1𝑆𝐽^𝑚2⋯𝑆𝐽^𝑚2𝑛+1, where the number of 𝑆's is 2𝑛+1. We have |𝐾(𝑢)|=|𝑋₁|^ℎ, where ℎ=3(2𝑛+1)+2∑^{2𝑛+1𝑖=1}𝑚_𝑖∈ℤ. It is clear that we have ℎ≠0. Assume that |𝐾(𝑢)|=1. Then |𝑋₁|^ℎ=(−𝑡𝑞³)^ℎ=1 and so (𝑡𝑞³)^2ℎ=1. Then 2ℎ(𝛼+3𝜃)=2𝑐𝜋,𝑐∈ℤ. This implies that (𝛼+3𝜃)/𝜋=(𝑐/ℎ)∈ℚ, which is a contradiction when (𝜃,𝛼)∈𝐺. Therefore, |𝐾(𝑢)|≠1, that is, 𝑢∉Ker𝐾. It follows, by Theorem 3.4, that tr(𝐾(𝑢))≠3. Now, consider an element 𝑣 obtained by permuting 𝑆 and 𝐽 in 𝑢. 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 𝑢. This implies that tr(𝐾(𝑣))≠3 and thus 𝑣∉Ker𝐾.
Consider the element 𝑢=𝑆𝐽^𝑚1𝑆𝐽^𝑚2⋯𝑆𝐽^𝑚2𝑛, where the number of 𝑆's is 2𝑛. Then |𝐾(𝑢)|=|𝑋₁|^ℎ, where ℎ=2(3𝑛+∑^2𝑛𝑖=1𝑚_𝑖)∈ℤ. For 𝑢∈Ker𝐾, we have |𝐾(𝑢)|=1, that is, (−𝑡𝑞³)^ℎ=1. If ℎ≠0, then (𝑡𝑞³)^2ℎ=1 and so 2ℎ(𝛼+3𝜃)=2𝑐𝜋, 𝑐∈ℤ. This implies that (𝛼+3𝜃)/𝜋=(𝑐/ℎ)∈ℚ, which contradicts the fact that (𝜃,𝛼)∈𝐺. Thus ℎ=0, that is, ∑^2𝑛𝑖=1𝑚_𝑖=−3𝑛. We write 𝑚_2𝑛=−3𝑛−∑^{2𝑛−1𝑖=1}𝑚_𝑖. Then 𝑢=𝑆𝐽^𝑚1𝑆𝐽^𝑚2⋯𝑆𝐽^{𝑚2𝑛−1}𝑆𝐽^{∑−3𝑛−2𝑛−1𝑖=1𝑚𝑖}.
If 𝑚_𝑖=3𝑘_𝑖 for some 1≤𝑖<2𝑛, then 𝐽^𝑚𝑖∈𝑍(𝐵₃). It follows that 𝑆𝐽^𝑚𝑖=𝐽^𝑚𝑖𝑆. Using the relation 𝑆²=𝐽³, 𝑢 will be reduced to an element with less number of 𝑆's and 𝑚_𝑖≠3𝑘_𝑖 for all 𝑖.
In the case 𝑚_𝑖≠3𝑘_𝑖 for every 1≤𝑖<2𝑛, we have 𝑚_𝑖=3𝑘_𝑖+𝑟_𝑖, where 𝑟_𝑖=1 or 2. Then 𝑢=𝑆𝐽^3𝑘1+𝑟1⋯𝑆𝐽^{3𝑘2𝑛−1+𝑟2𝑛−1}𝑆𝐽^{−3𝑛−(3𝑘1+𝑟1+3𝑘2+𝑟2+⋯+3𝑘2𝑛−1+𝑟2𝑛−1)}. Since 𝐽^3𝑘𝑖∈𝑍(𝐵₃), it follows that 𝑢=𝑆𝐽^𝑟1𝑆𝐽^𝑟2⋯𝑆𝐽^{𝑟2𝑛−1}𝑆𝐽^{∑−3𝑛−2𝑛−1𝑖=1𝑟𝑖}, where 𝑟_𝑖=1 or 2. Since 𝑢∈Ker𝐾, it follows that the trace of the image of 𝑢 is equal to 3. Consider an element 𝑣 obtained by permuting 𝑆 and 𝐽 in the element 𝑢. 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 𝑢. This implies that tr(𝐾(𝑣))=3. It follows, by Theorem 3.4, that 𝑣∈Ker𝐾.
Therefore, the possible nontrivial elements in the kernel of 𝐾∶𝐵₃→𝐺𝐿₃(ℂ) are 𝑆𝐽^𝑟1𝑆𝐽^𝑟2⋯𝑆𝐽^{𝑟2𝑛−1}𝑆𝐽^{∑−3𝑛−2𝑛−1𝑖=1𝑟𝑖}, and other elements that are obtained by permuting 𝑆 and 𝐽. Here, the number of 𝑆's is 2𝑛 and 𝑟_𝑖=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(𝐽^𝑛)=⎧⎪⎨⎪⎩𝑡²𝑡𝑞⁶^𝑘𝐼,𝑛=3𝑘,₂𝑡𝑞⁶^𝑘𝐾(𝐽),𝑛=3𝑘+1,₂𝑞⁶^𝑘𝐾𝐽²,𝑛=3𝑘+2.(4.4) Here 𝑘∈ℕ and 𝐼 is the identity matrix.

Proof. Direct computations show that 𝐾(𝐽)=⎛⎜⎜⎜⎜⎝−𝑡𝑞²−𝑡𝑞(𝑞(𝑞−−11)₂)𝑡𝑞³0𝑡𝑞²(𝑞−1)𝑡𝑞³𝐾𝐽100₂⎞⎟⎟⎟⎟⎠,=⎛⎜⎜⎜⎜⎝00𝑡²𝑞⁶−𝑡𝑞₂𝑡𝑞³0𝑡²𝑞⁵(𝑞−1)𝐾𝐽₃(𝑞−1)𝑡𝑞³0⎞⎟⎟⎟⎟⎠,=𝑡²𝑞⁶𝐼,𝐾𝐽⁴=𝑡²𝑞⁶𝐾(𝐽),𝐾𝐽⁵=𝑡²𝑞⁶𝐾𝐽².(4.5)
Using mathematical induction on the integer 𝑘, we prove the lemma.

5. Proof of Theorem 4.1

In order to prove that the representation 𝐾∶𝐵₃→𝐺𝐿₃(ℂ) is faithful, we have to show that the elements in Lemma 4.2 do not belong to Ker𝐾 if we choose 𝑞 and 𝑡 in a way that their arguments 𝜃 and 𝛼 belong to the previously defined set 𝐺.

That is, we need to show that the element 𝑢=𝑆𝐽^𝑟1𝑆𝐽^𝑟2⋯𝑆𝐽^{𝑟2𝑛−1}𝑆𝐽^{∑−3𝑛−2𝑛−1𝑖=1𝑟𝑖} and others obtained from 𝑢 by permuting 𝑆 and 𝐽 do not belong to the kernel of this representation if (𝜃,𝛼)∈𝐺.

Consider the element 𝑢=𝑆𝐽^𝑟1𝑆𝐽^𝑟2⋯𝑆𝐽^{𝑟2𝑛−1}𝑆𝐽^{∑−3𝑛−2𝑛−1𝑖=1𝑟𝑖}, where the number of 𝑆's is 2𝑛 and 𝑟_𝑖=1 or 2 for 1≤𝑖<2𝑛. We consider five cases and we prove that the element 𝑢, in each of the cases, does not belong to the kernel of the complex specialization 𝐾∶𝐵₃→𝐺𝐿₃(ℂ).

(1) If 𝑟_𝑖=1 for all 1≤𝑖<2𝑛, then 𝑢=𝑆𝐽¹^2𝑛−1𝑆𝐽^{−3𝑛−(2𝑛−1)}=𝑆𝐽¹^2𝑛−1𝑆𝐽^{−3(2𝑛−1)+𝑛−2}.(5.1) Since 𝐽⁻³∈𝑍(𝐵₃), it follows that 𝑢=(𝑆𝐽⁻²)^2𝑛−1𝑆𝐽^𝑛−2 and so 𝑢=(𝑆𝐽⁻²)^2𝑛𝐽^𝑛. The eigenvalues of 𝐾(𝑆𝐽⁻²) are 1,−1/𝑞 and 1/𝑡𝑞², which are easily shown to be distinct for (𝜃,𝛼)∈𝐺. We diagonalize 𝐾(𝑆𝐽⁻²) by a matrix 𝑇 given by 𝑇=⎛⎜⎜⎜⎜⎜⎝𝑡𝑞²−𝑡𝑞³𝑡²𝑞⁴𝑡𝑞(𝑞−1)−𝑡𝑞²(𝑞−1)𝑞(−1+𝑡𝑞(−1+𝑞+𝑞(1+(𝑞−1)𝑞)𝑡))11𝑞−11⎞⎟⎟⎟⎟⎟⎠.(5.2) It is easy to see that det(𝑇)=−(𝑞³(1+𝑞)𝑡(1+𝑡𝑞)(𝑡𝑞²−1))/(𝑞−1)≠0 for (𝜃,𝛼)∈𝐺.

Then 𝑇⁻¹𝐾𝑆𝐽⁻²𝑇=⎛⎜⎜⎜⎜⎜⎝10−01000𝑞01𝑡𝑞²𝑇₋₁⎞⎟⎟⎟⎟⎟⎠,𝐾(𝑢)𝑇=⎛⎜⎜⎜⎜⎜⎝001𝑞001_2𝑛001𝑡𝑞²^2𝑛⎞⎟⎟⎟⎟⎟⎠𝑇₋₁𝐾(𝐽^𝑛)𝑇.(5.3)

We take the following possibilities regarding the value of 𝑛.

(a) For 𝑛=3𝑘, we have, from Lemma 4.3, that 𝑇⁻¹𝐾(𝐽^𝑛)𝑇=(𝑡²𝑞⁶)^𝑘𝐼. Then we get 𝑇⁻¹𝐾(𝑢)𝑇=⎛⎜⎜⎜⎜⎜⎝𝑡²𝑞0⁶^𝑘𝑡0_2𝑘00001𝑡⁴𝑞⁶^𝑘⎞⎟⎟⎟⎟⎟⎠.(5.4) The entry (𝑡²𝑞⁶)^𝑘≠1; otherwise, 𝑒^{𝑖𝑘(2𝛼+6𝜃)}=1. That is, (𝛼+3𝜃)/𝜋∈ℚ, which contradicts the fact that (𝜃,𝛼)∈𝐺.

(b) For 𝑛=3𝑘+1, we have, from Lemma 4.3, that 𝑇⁻¹𝐾(𝐽^𝑛)𝑇=(𝑡²𝑞⁶)^𝑘𝑇⁻¹𝐾(𝐽)𝑇. Direct computations give 𝑇⁻¹𝐾(𝑢)𝑇=⎛⎜⎜⎜⎜⎜⎝∗𝑡𝑞³𝑡²𝑞⁶^𝑘1−𝑡𝑞³∗∗(1+𝑞)∗∗𝑡𝑞²−1∗∗∗⎞⎟⎟⎟⎟⎟⎠.(5.5) It is easy to show that (𝑡𝑞³(𝑡²𝑞⁶)^𝑘(1−𝑡𝑞³))/((1+𝑞)(𝑡𝑞²−1))≠0 for (𝜃,𝛼)∈𝐺.

(c) For 𝑛=3𝑘+2, we have, from Lemma 4.3, that 𝑇⁻¹𝐾(𝐽^𝑛)𝑇=(𝑡²𝑞⁶)^𝑘𝑇⁻¹𝐾(𝐽²)𝑇. We get, by direct computations, 𝑇⁻¹𝐾(𝑢)𝑇=⎛⎜⎜⎜⎜⎜⎝∗𝑡𝑞⁴𝑡²𝑞⁶^𝑘𝑡𝑞³−1∗∗(1+𝑞)∗∗𝑡𝑞²−1∗∗∗⎞⎟⎟⎟⎟⎟⎠.(5.6) It is easy to show that (𝑡𝑞⁴(𝑡²𝑞⁶)^𝑘(𝑡𝑞³−1))/((1+𝑞)(𝑡𝑞²−1))≠0 for (𝜃,𝛼)∈𝐺.

(2) If 𝑟_𝑖=2 for all 1≤𝑖<2𝑛, then 𝑢=𝑆𝐽²^2𝑛−1𝑆𝐽^{−3𝑛−2(2𝑛−1)}=𝑆𝐽²^2𝑛−1𝑆𝐽^{−3(2𝑛−1)−𝑛−1}.(5.7) Since 𝐽⁻³∈𝑍(𝐵₃), it follows that 𝑢=(𝑆𝐽⁻¹)^2𝑛−1𝑆𝐽^−𝑛−1 and so 𝑢=(𝑆𝐽⁻¹)^2𝑛𝐽^−𝑛. The eigenvalues of 𝐾(𝑆𝐽⁻¹) are 1,−𝑞, and 𝑡𝑞². It is easy to show that these eigenvalues are distinct when (𝜃,𝛼)∈𝐺. We diagonalize 𝐾(𝑆𝐽⁻¹) by a matrix 𝐻 given by 𝐻=⎛⎜⎜⎜⎜⎜⎝00(1+𝑡𝑞)𝑡𝑞²−1𝑞−1−1𝑡+(𝑞𝑞−+1)𝑡𝑞₂111⎞⎟⎟⎟⎟⎟⎠.(5.8)

It is clear that det(𝐻)=((1+𝑞)(1+𝑡𝑞)(𝑡𝑞²−1))/(𝑡(𝑞−1))≠0 for (𝜃,𝛼)∈𝐺.

Then 𝐻⁻¹𝐾𝑆𝐽⁻¹𝐻=⎛⎜⎜⎜⎜⎝100−00𝑞𝑡𝑞00₂𝐻₋₁⎞⎟⎟⎟⎟⎠,𝐾(𝑢)𝐻=⎛⎜⎜⎜⎜⎝100𝑞00^2𝑛𝑡𝑞00₂^2𝑛⎞⎟⎟⎟⎟⎠𝐻₋₁𝐾(𝐽^−𝑛)𝐻.(5.9)

Using Lemma 4.3, we can easily see that 𝐾(𝐽^−𝑛)=⎧⎪⎪⎪⎨⎪⎪⎪⎩1𝑡²𝑞⁶^𝑘1𝐼,𝑛=3𝑘,𝑡²𝑞⁶^𝑘1𝐾𝐽⁻¹,𝑛=3𝑘+1,𝑡²𝑞⁶^𝑘𝐾𝐽⁻²,𝑛=3𝑘+2.(5.10)

Here 𝑘 is a positive integer. We deal with the following possibilities regarding the value of 𝑛.

(a) For 𝑛=3𝑘, a comuptation shows that 𝐻⁻¹𝐾(𝑢)𝐻=⎛⎜⎜⎜⎜⎜⎝1𝑡²𝑞⁶^𝑘0100𝑡^2𝑘00𝑡₄0𝑞⁶^𝑘⎞⎟⎟⎟⎟⎟⎠.(5.11)

The entry 1/(𝑡²𝑞⁶)^𝑘≠1; otherwise, 𝑒^{𝑖𝑘(2𝛼+6𝜃)}=1, which implies that (𝛼+3𝜃)/𝜋∈ℚ. This contradicts the fact that (𝜃,𝛼)∈𝐺.

(b) For 𝑛=3𝑘+1, a computation shows that 𝐻⁻¹𝐾(𝑢)𝐻=⎛⎜⎜⎜⎜⎜⎝∗1−𝑡𝑞³∗∗𝑡𝑞³𝑡²𝑞⁶^𝑘(1∗∗+𝑞)𝑡𝑞²−1∗∗∗⎞⎟⎟⎟⎟⎟⎠.(5.12)

It is easy to show that (1−𝑡𝑞³)/(𝑡𝑞³(𝑡²𝑞⁶)^𝑘(1+𝑞)(𝑡𝑞²−1))≠0 for (𝜃,𝛼)∈𝐺.

(c) For 𝑛=3𝑘+2, a computation shows that 𝐻⁻¹𝐾(𝑢)𝐻=⎛⎜⎜⎜⎜⎜⎝∗𝑡𝑞³−1∗∗𝑡𝑞⁴𝑡²𝑞⁶^𝑘(1∗∗+𝑞)𝑡𝑞²−1∗∗∗⎞⎟⎟⎟⎟⎟⎠.(5.13)

We can also see that (𝑡𝑞³−1)/(𝑡𝑞⁴(𝑡²𝑞⁶)^𝑘(1+𝑞)(𝑡𝑞²−1))≠0 for (𝜃,𝛼)∈𝐺.

(3) If for 1≤𝑖<2𝑛, we have that 𝑟_𝑖=12,,𝑖𝑖iissoevdedn,,(5.14) then 𝑢=𝑆𝐽¹𝑆𝐽²^𝑛−1𝑆𝐽¹𝑆𝐽^{−3𝑛−[3(𝑛−1)+1]}=𝑆𝐽¹𝑆𝐽²^𝑛−1𝑆𝐽¹𝑆𝐽^{−3(2𝑛−1)−1}.(5.15)

Since 𝐽⁻³∈𝑍(𝐵₃), it follows that 𝑢=(𝑆𝐽⁻²𝑆𝐽⁻¹)^𝑛−1𝑆𝐽⁻²𝑆𝐽⁻¹ and so 𝑢=(𝑆𝐽⁻²𝑆𝐽⁻¹)^𝑛. The eigenvalues of 𝐾(𝑆𝐽⁻²𝑆𝐽⁻¹) are 𝑒₁=1,𝑒₂=(𝑎−√𝑏)/2𝑡𝑞²,𝑒₃=(𝑎+√𝑏)/2𝑡𝑞², where 𝑎=1+𝑡𝑞²+𝑡²𝑞⁴−𝑞(1+𝑡)−𝑡𝑞³(1+𝑡) and 𝑏=−4𝑡²𝑞⁴+(1+𝑞(−1+(𝑞−1)𝑡)(1+𝑡𝑞²))². These eigenvalues are distinct for (𝜃,𝛼)∈𝐺. Now, we diagonalize 𝐾(𝑆𝐽⁻²𝑆𝐽⁻¹) by a matrix 𝐿 given by 𝐿=⎛⎜⎜⎜⎜⎜⎜⎜⎝𝑞𝑞−𝑐−√𝑏−𝑐+√𝑏−𝑞−𝑐+√𝑏−1+𝑞+1𝑐+√𝑏𝑡𝑞−2𝑞+2(𝑞−1)²𝑡²𝑞³−𝑐+√𝑏−−2𝑞+2(𝑞−1)²𝑡²𝑞³11𝑐+1√𝑏⎞⎟⎟⎟⎟⎟⎟⎟⎠,(5.16) where 𝑏 is defined above and 𝑐=1+𝑡𝑞²+𝑡𝑞³(𝑡−1)−𝑡²𝑞⁴−𝑞(1+𝑡).

Also, we have that det(𝐿)=((−1+𝑞+𝑞(1+𝑞+𝑞²)𝑡−(𝑞−1)𝑡²𝑞³)√𝑏)/(𝑞³(1+(𝑞−1)𝑞)𝑡³)≠0 when 𝜃≠±𝜋3,𝑏≠0,1+2cos𝜃+2cos(𝜃+𝛼)−2cos(2𝜃+𝛼)≠0.(5.17) Then 𝐿⁻¹𝐾𝑆𝐽⁻²𝑆𝐽⁻¹^𝑛𝐿=⎛⎜⎜⎜⎜⎝010𝑒00^𝑛2𝑒00_𝑛3⎞⎟⎟⎟⎟⎠.(5.18)

If 𝑒^𝑛3=1, then 𝑒^𝑖𝛽𝑛=1, where 𝛽=arg(𝑒₃). This implies that 𝛽/𝜋∈ℚ, which contradicts the fact that (𝜃,𝛼)∈𝐺. Therefore, 𝐿⁻¹𝐾((𝑆𝐽⁻²𝑆𝐽⁻¹)^𝑛)𝐿≠𝐼 and so 𝐾((𝑆𝐽⁻²𝑆𝐽⁻¹)^𝑛)≠𝐼. We conclude that 𝑢=(𝑆𝐽⁻²𝑆𝐽⁻¹)^𝑛∉Ker𝐾.

(4) If for 1≤𝑖<2𝑛, we have that 𝑟_𝑖=21,,𝑖𝑖iissoevdedn,,(5.19) then 𝑢=𝑆𝐽²𝑆𝐽¹^𝑛−1𝑆𝐽²𝑆𝐽^{−3𝑛−[3(𝑛−1)+2]}=𝑆𝐽²𝑆𝐽¹^𝑛−1𝑆𝐽²𝑆𝐽(−3(2𝑛−1)−2).(5.20) Since 𝐽⁻³∈𝑍(𝐵₃), it follows that 𝑢=(𝑆𝐽⁻¹𝑆𝐽⁻²)^𝑛−1𝑆𝐽⁻¹𝑆𝐽⁻² and so 𝑢=(𝑆𝐽⁻¹𝑆𝐽⁻²)^𝑛. Using the properties of the trace, we have that tr(𝐾(𝑢))=tr(𝐾((𝑆𝐽⁻²𝑆𝐽⁻¹)^𝑛)). By our result in case (3) and Theorem 3.4, we have that tr(𝐾(𝑢))≠3 and so 𝑢=(𝑆𝐽⁻¹𝑆𝐽⁻²)^𝑛∉Ker𝐾.

(5) Let 𝑢=𝑆𝐽^𝑟1𝑆𝐽^𝑟2⋯𝑆𝐽^{𝑟2𝑛−1}𝑆𝐽^{∑−3𝑛−2𝑛−1𝑖=1𝑟𝑖} such that not all 𝑟_𝑖's are 1 and not all 𝑟_𝑖's are 2. We also assume that there exists an integer 𝑗 such that 𝑟_𝑗=𝑟_𝑗+1, where 1≤𝑗<2𝑛. We write 𝑢=𝐴𝐽^−𝑝, where 𝐴=𝑆𝐽^𝑟1𝑆𝐽^𝑟2⋯𝑆𝐽^{𝑟2𝑛−1}𝑆 and 𝑝=3𝑛+∑^{2𝑛−1𝑖=1}𝑟_𝑖. Without loss of generality, we assume that 𝐾(𝐴) and 𝐾(𝐽^−𝑝) do not commute for some values of 𝑞 and 𝑡 in 𝐺. For this reason, we require that 𝑝≡1 or 2 (mod 3). If 𝐾(𝐴) and 𝐾(𝐽^−𝑝) happened to commute for some values of 𝑞 and 𝑡 then we change the values of 𝑞 and 𝑡, slightly in a way that (𝜃,𝛼) is still in 𝐺 and the matrices 𝐾(𝐴),𝐾(𝐽^−𝑝) do not commute anymore. This is due to the fact that (𝑞,𝑡) belongs to a dense subset of 𝐺 by the hypothesis of the theorem.

The eigenvalues of 𝐽⁻¹ are 1/(𝑞²𝑡^2/3),−(−1)^1/3/(𝑞²𝑡^2/3) and (−1)^2/3/(𝑞²𝑡^2/3). It is clear that these eigenvalues are distinct. We diagonalize 𝐾(𝐽⁻¹) by a matrix 𝑊 given by ⎛⎜⎜⎜⎜⎝𝑞²𝑡^2/3(−1)^2/3𝑞²𝑡^2/3(−1)^4/3𝑞²𝑡^2/3𝑞1+(𝑞−1)𝑡^1/3𝑡^1/3(−𝑡)^1/3𝑞−1+(−1)^1/3(𝑞−1)𝑡^1/3𝑡^1/3𝑞1+(−1)^2/3(𝑞−1)𝑡^1/3111⎞⎟⎟⎟⎟⎠.(5.21)

Under direct computations, we see that det(𝑊)=−3𝑖√3𝑞³𝑡≠0. Here 𝑖=√−1. Then we get that 𝑊⁻¹𝐾𝐽⁻¹𝑊=⎛⎜⎜⎜⎜⎜⎜⎝1𝑞²𝑡^2/30−(−01)_1/30𝑞²𝑡^2/300(−10)_2/3𝑞²𝑡^2/3⎞⎟⎟⎟⎟⎟⎟⎠.(5.22) We have that 𝑊⁻¹𝐾(𝑢)𝑊=𝑊⁻¹𝐾(𝐴)𝑊𝑊⁻¹𝐾(𝐽^−𝑝)𝑊.(5.23) Assume, to get contradiction, that 𝑢∈Ker(𝐾). Then 𝑊⁻¹𝐾(𝐴)𝑊=𝑊⁻¹𝐾(𝑢)𝑊𝑊⁻¹𝐾(𝐽^−𝑝)𝑊⁻¹=𝑊⁻¹𝐾(𝐽^−𝑝)𝑊⁻¹.(5.24)

It follows that 𝑊⁻¹𝐾(𝐴)𝑊 is a diagonal matrix. Therefore, 𝑊⁻¹𝐾(𝐴)𝑊 and 𝑊⁻¹𝐾(𝐽^−𝑝)𝑊 commute, that is, 𝑊⁻¹𝐾(𝑢)𝑊=𝑊⁻¹𝐾(𝐽^−𝑝)𝐾(𝐴)𝑊.(5.25)

Then 𝐾(𝑢)=𝐾(𝐽^−𝑝)𝐾(𝐴).(5.26)

Therefore, 𝐾(𝐴) and 𝐾(𝐽^−𝑝) commute, which is a contradiction.