Krammer's Representation of the Pure Braid Group, 𝑃₃

Abstract

We consider Krammer's representation of the pure braid group on three strings: 𝑃₃→𝐺𝐿(3,𝑍[𝑡^±1,𝑞^±1]), where 𝑡 and 𝑞 are indeterminates. As it was done in the case of the braid group, 𝐵₃, we specialize the indeterminates 𝑡 and 𝑞 to nonzero complex numbers. Then we present our main theorem that gives us a necessary and sufficient condition that guarantees the irreducibility of the complex specialization of Krammer's representation of the pure braid group, 𝑃₃.

1. Introduction

Let 𝐵_𝑛 be the braid group on 𝑛 strings. There are a lot of linear representations of 𝐵_𝑛. The earliest was the Artin representation, which is an embedding 𝐵_𝑛→𝐴𝑢𝑡(𝐹_𝑛), the automorphism group of a free group on 𝑛 generators. Applying the free differential calculus to elements of 𝐴𝑢𝑡(𝐹_𝑛) sometimes gives rise to linear representations of 𝐵_𝑛 and its normal subgroup, the pure braid group denoted by 𝑃_𝑛 [1]. The Burau, Gassner, and Krammer's representations arise this way. In a previous paper, we considered Krammer's representation of the braid group on three strings and we specialized the indeterminates to nonzero complex numbers. We then found a necessary and sufficient condition that guarantees the irreducibility of such a representation. For more details, see [2].

In Section 2, we introduce some definitions of the pure braid group and Krammer's representation. In Sections 3 and 4, we present our work that leads to our main theorem, Theorem 4.2, which gives a necessary and sufficient condition for the specialization of Krammer's representation of 𝑃₃ to be irreducible.

2. Definitions

Definition 2.1 (see [1]). The braid group on 𝑛 strings, 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,𝜎_𝑖𝜎_𝑗=𝜎_𝑗𝜎_𝑖 if |𝑖−𝑗|>1}.

The generators 𝜎₁,…,𝜎_𝑛−1 are called the standard generators of 𝐵_𝑛.

Definition 2.2. The kernel of the group homomorphism 𝐵_𝑛→𝑆_𝑛 is called the pure braid group on 𝑛 strands and is denoted by 𝑃_𝑛. It consists of those braids which connect the 𝑖th item of the left set to the 𝑖th item of the right set, for all 𝑖. The generators of 𝑃_𝑛 are 𝐴_𝑖,𝑗, 1≤𝑖<𝑗≤𝑛, where 𝐴_𝑖,𝑗=𝜎_𝑗−1𝜎_𝑗−2⋯𝜎_𝑖+1𝜎²_𝑖𝜎⁻¹_𝑖+1⋯𝜎⁻¹_𝑗−2𝜎⁻¹_𝑗−1.

Let us recall the Lawrence-Krammer representation of braid groups. This is a representation of 𝐵_𝑛 in 𝐺𝐿(𝑚,𝑍[𝑡^±1,𝑞^±1])=𝐴𝑢𝑡(𝑉₀), where 𝑚=𝑛(𝑛−1)/2 and 𝑉₀ is the free module of rank 𝑚 over 𝑍[𝑡^±1,𝑞^±1]. The representation is denoted by 𝐾(𝑞,𝑡). For simplicity we write 𝐾 instead of 𝐾(𝑞,𝑡). What distinguishes this representation from others is that Krammer's representation defined on the braid group, 𝐵_𝑛, is a faithful representation for all 𝑛≥3 [3]. The question of whether or not a specific linear representation of an abstract group is irreducible has always been a significant question to answer, especially those representations of the braid group and its normal subgroups. In a previous result, we determined a necessary and sufficient condition for the specialization of Krammer's representation of 𝐵₃ to be irreducible [2]. In our current work, we apply Krammer's representation on the normal subgroup of 𝐵₃, namely, the pure braid group, 𝑃₃. Having done some computations, we succeed in establishing a necessary and sufficient condition for the complex specialization of Krammer's representation of 𝑃₃ to be irreducible.

Definition 2.3 (see [3]). 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)=𝑡𝑞²𝑥_𝑘,𝑘+1,𝑖=𝑘,𝑗=𝑘+1;(1−𝑞)𝑥_𝑖,𝑘+𝑞𝑥_𝑖,𝑘+1𝑥,𝑗=𝑘,𝑖<𝑘;_𝑖,𝑘+𝑡𝑞^{𝑘−𝑖+1}(𝑞−1)𝑥_𝑘,𝑘+1,𝑗=𝑘+1,𝑖<𝑘;𝑡𝑞(𝑞−1)𝑥_𝑘,𝑘+1+𝑞𝑥_𝑘+1,𝑗𝑥,𝑖=𝑘,𝑘+1<𝑗;_𝑘,𝑗+(1−𝑞)𝑥_𝑘+1,𝑗𝑥,𝑖=𝑘+1,𝑘+1<𝑗;_𝑖,𝑗,𝑖<𝑗<𝑘or𝑥𝑘+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 automorphisms group of free group with 𝑛(𝑛−1)/2 generators, we determine Krammer's representation 𝐾(𝑞,𝑡)∶𝑃₃→𝐺𝐿(3,𝑍[𝑡^±1,𝑞^±1]). Here 𝑍[𝑡^±1,𝑞^±1] is the ring of Laurent polynomials on two variables. The images of the generators under Krammer's representation are given by𝐾(𝐴_1,2𝑡)=²𝑞⁴𝑡00₂𝑞³(𝑞−1)𝑞𝑞(1−𝑞)𝑡𝑞(𝑞−1)1−𝑞1−𝑞+𝑞²,𝐾(𝐴_2,3)=1−𝑞+𝑞²𝑞(1−𝑞)𝑡𝑞³(𝑞−1)1−𝑞𝑞𝑡²𝑞⁴(𝑞−1)00𝑡²𝑞⁴,𝐾(𝐴_1,3)=𝑞𝑞(𝑞−1)1−𝑞−𝑡𝑞(𝑞−1)²𝑡−𝑡𝑞(𝑞−1)²𝑡𝑞[𝑡𝑞²(𝑞²−𝑞+1)−(𝑞−1)³]𝑚𝑡𝑞(1−𝑞)𝑡𝑞(𝑞−1)(1−𝑞+𝑡𝑞²,)𝑛 (2.2) where𝑚=−1+𝑞[2−2𝑞+𝑞²+𝑡(𝑞−1)⁴+𝑞²(1−𝑞)(1+𝑞(𝑞−1))𝑡²],𝑛=1+𝑞(𝑞−1)[1+𝑡(𝑞−1)(−1+𝑞−𝑡𝑞²)]. (2.3)

Specializing 𝑡 and 𝑞 to non zero complex numbers, we consider the complex linear representation 𝐾(𝑞,𝑡)∶𝑃₃→𝐺𝐿(3,𝐶). We show that the only non zero invariant subspace under the action of specialization of Krammer's representation of 𝑃₃ coincides with the vectorspace 𝐶³. Here, we regard 𝑀₃(𝐶) as acting from the left on column vectors so that eigenvectors and invariant subspaces lie in 𝐶³.

3. Sufficient Condition for Irreducibility

In this section, we find a sufficient condition for the irreducibility of Krammer's representation of the pure braid group on three strings 𝑃₃.

Theorem 3.1. For (𝑞,𝑡)∈(𝐶^∗)², Krammer's representation 𝐾(𝑞,𝑡)∶𝑃₃→𝐺𝐿(3,𝐶) is irreducible if 𝑡²𝑞³≠1,𝑡𝑞³≠1,𝑡≠−1,𝑞≠1,𝑡𝑞≠1, and 𝑡𝑞²≠−1.

Proof. For simplicity, we write 𝐾(𝛼) instead of 𝐾(𝑞,𝑡)(𝛼), where 𝛼∈𝑃₃. Suppose, to get contradiction, that 𝐾(𝑞,𝑡)∶𝑃₃→𝐺𝐿(3,𝐶) is reducible; then there exists a proper nonzero invariant subspace 𝑆, where the dimension of 𝑆 is either 1 or 2. We will show that a contradiction is obtained in each of these cases.
Assume that dimension of 𝑆 is 1:
The subspace 𝑆 has to be one of the following subspaces: ⟨𝑒₁⟩, ⟨𝑒₂⟩, ⟨𝑒₃⟩, ⟨𝑒₁+𝑢𝑒₂⟩, ⟨𝑒₂+𝑢𝑒₃⟩, ⟨𝑒₁+𝑢𝑒₃⟩, ⟨𝑒₁+𝑢𝑒₂+𝑣𝑒₃⟩, where 𝑢,𝑣 are non zero complex numbers.Case 1 (𝑆=⟨𝑒₁⟩). Since 𝑒₁∈𝑆, it follows that 𝐴_1,2(𝑒₁)∈𝑆 which implies that 𝑡²𝑞³(𝑞−1)=0, a contradiction.Case 2 (𝑆=⟨𝑒₂⟩). Since 𝑒₂∈𝑆,it follows that 𝐴_1,2(𝑒₂)∈𝑆 which implies that 1−𝑞=0, a contradiction.Case 3 (𝑆=⟨𝑒₃⟩). Since 𝑒₃∈𝑆, it follows that 𝐴_1,2(𝑒₃)∈𝑆 which implies that 𝑞(1−𝑞)=0, a contradiction.Case 4 (𝑆=⟨𝑒₁+𝑢𝑒₂⟩,𝑢≠0). Since 𝑒₁+𝑢𝑒₂∈𝑆, it follows that 𝐴_1,2(𝑒₁+𝑢𝑒₂)∈𝑆. This implies that 𝑡²𝑞⁴t²𝑞³1𝑢0(𝑞−1)+𝑞𝑢,𝑡𝑞(𝑞−1)+(1−𝑞)𝑢=𝑚 (3.1) where 𝑚 is a complex number. Solving this system of equations implies that (𝑡𝑞−1)(𝑡𝑞²+1)=0, which is a contradiction to the hypothesis.Case 5 (𝑆=⟨𝑒₂+𝑢𝑒₃⟩,𝑢≠0). Since 𝑒₂+𝑢𝑒₃∈𝑆, it follows that 𝐴_2,3(𝑒₂+𝑢𝑒₃)∈𝑆. This implies that 𝑞(1−𝑞)+𝑡𝑞³(𝑞−1)𝑢𝑞+𝑡²𝑞⁴𝑡(𝑞−1)𝑢₂𝑞⁴𝑢01𝑢=𝑚, (3.2) where 𝑚 is a complex number. By solving this system of equations, we get that (𝑡𝑞−1)(𝑡𝑞²+1)=0, which is a contradiction.Case 6 (𝑆=⟨𝑒₁+𝑢𝑒₃⟩,𝑢≠0). Since 𝑒₁+𝑢𝑒₃∈𝑆, it follows that 𝐴_1,2(𝑒₁+𝑢𝑒₃)∈𝑆. This implies that 𝑡²𝑞⁴𝑡²𝑞³(𝑞−1)+𝑞(1−𝑞)𝑢𝑡𝑞(𝑞−1)+(1−𝑞+𝑞²10𝑢,)𝑢=𝑚 (3.3) where 𝑚 is a complex number. By solving this system of equations, we get that (𝑡𝑞−1)(𝑡𝑞²+1)(𝑡𝑞²+𝑞−1)=0.
By our hypothesis, (𝑡𝑞−1)(𝑡𝑞²+1)≠0. This implies that 𝑡𝑞²+𝑞−1=0. That is, 𝑡𝑞²=1−𝑞. Also, we have that 𝐴_2,3(𝑒₁+𝑢𝑒₃)∈𝑆. This implies that 1−𝑞+𝑞²+𝑡𝑞³(𝑞−1)𝑢1−𝑞+𝑡²𝑞⁴𝑡(𝑞−1)𝑢₂𝑞⁴𝑢10𝑢=𝑛, (3.4) where 𝑛 is a complex number. By solving this system of equations, we get that 𝑡²𝑞³=−1. This means that 𝑡²𝑞³=𝑡𝑞(𝑡𝑞²)=𝑡𝑞(1−𝑞)=𝑡𝑞−𝑡𝑞²=𝑡𝑞−1+𝑞. (3.5) This implies that 𝑞(𝑡+1)=0, which contradicts the hypothesis.Case 7 (𝑆=⟨𝑒₁+𝑢𝑒₂+𝑣𝑒₃⟩,𝑢,𝑣≠0). Since 𝑒₁+𝑢𝑒₂+𝑣𝑒₃∈𝑆, it follows that 𝐴_1,2(𝑒₁+𝑢𝑒₂+𝑣𝑒₃)∈𝑆. This implies that 𝑡²𝑞⁴𝑡²𝑞³(𝑞−1)+𝑞𝑢+𝑞(1−𝑞)𝑣𝑡𝑞(𝑞−1)+(1−𝑞)𝑢+(1−𝑞+𝑞²1𝑢𝑣,)𝑣=𝑚 (3.6) where 𝑚 is a complex number. Since 𝐴_2,3(𝑒₁+𝑢𝑒₂+𝑣𝑒₃)∈𝑆, it follows that 1−𝑞+𝑞²+𝑞(1−𝑞)𝑢+𝑡𝑞³(𝑞−1)𝑣1−𝑞+𝑞𝑢+𝑡²𝑞⁴𝑡(𝑞−1)𝑣₂𝑞⁴𝑣1𝑢𝑣=𝑛, (3.7) where 𝑛 is a complex number. Solving these two system of equations, we get that 𝑚=𝑛=𝑡²𝑞⁴. Also, we have that 𝑞(𝑡²𝑞³−1)𝑢+𝑞(𝑞−1)𝑣=𝑡²𝑞³(𝑞−1),(𝑞−1)𝑢+(𝑡²𝑞⁴−𝑞²+𝑞−1)𝑣=𝑡𝑞(𝑞−1),𝑞(1−𝑞)𝑢+𝑡𝑞³(𝑞−1)𝑣=𝑡²𝑞⁴−𝑞²+𝑞−1,𝑞(𝑡²𝑞³−1)𝑢−𝑡²𝑞⁴(𝑞−1)𝑣=1−𝑞. (3.8)(3.9)(3.10)(3.11)
Substracting (3.11) from (3.8), we get that 𝑞(1+𝑡²𝑞³)𝑣=1+𝑡²𝑞³. Here, we have 2 cases whether or not (1+𝑡²𝑞³) is zero.
If 1+𝑡²𝑞³=0, then we rewrite (3.8), (3.9), (3.10), and (3.11) to become as follows: 2𝑞𝑢−𝑞(𝑞−1)𝑣=𝑞−1,(𝑞−1)𝑢−(𝑞²+1)𝑣=𝑡𝑞(𝑞−1),𝑞(1−𝑞)𝑢+𝑡𝑞³(𝑞−1)𝑣=−(𝑞²+1). (3.12)(3.13)(3.14)
Multiplying (3.13) by 𝑞 and adding it to (3.14) we get that 𝑞(𝑡𝑞³−𝑡𝑞²−𝑞²−1)𝑣=𝑡𝑞³−𝑡𝑞²−𝑞²−1. (3.15)
A simple computation shows that 𝑡𝑞³−𝑡𝑞²−𝑞²−1≠0. Thus 𝑣=1/𝑞. Substituting 𝑣=1/𝑞 in (3.12), we get that 𝑢=(𝑞−1)/𝑞. Substituting 𝑢 and 𝑣 in (3.14), we get that 𝑡𝑞²=𝑡𝑞−2. Having that 𝑡²𝑞³=−1 implies that 𝑡²𝑞³=𝑡𝑞(𝑡𝑞²)=𝑡𝑞(𝑡𝑞−2). This implies that (𝑡𝑞−1)²=0 which contradicts the hypothesis.
This means that 1+𝑡²𝑞³≠0. Then 𝑣=1/𝑞 and 𝑢=(𝑞−1)/𝑞 by (3.8). Substituting 𝑢 and 𝑣 in (3.9), we get that (𝑡𝑞−1)(𝑡𝑞²+1)=0, which contradicts the hypothesis.Assume that dimension of 𝑆 is 2:
Easy computations show that the subspace 𝑆 cannot be in the form 𝑆=⟨𝑒_𝑖,𝑒_𝑗⟩ or 𝑆=⟨𝑒_𝑖+𝑢𝑒_𝑗,𝑒_𝑘⟩ for 𝑖≠𝑗≠𝑘.
It suffices to consider only the case 𝑆=⟨𝑒₁+𝑢𝑒₂,𝑒₁+𝑣𝑒₃⟩, where 𝑢,𝑣≠0.
Since 𝑒₁+𝑢𝑒₂∈𝑆, it follows that 𝐴_1,2(𝑒₁+𝑢𝑒₂)∈𝑆 and so 𝑡²𝑞⁴𝑡²𝑞³(𝑞−1)+𝑞𝑢𝑡𝑞(𝑞−1)+(1−𝑞)𝑢∈𝑆. (3.16)
Also, we have that 𝑒₁+𝑣𝑒₃∈𝑆,then 𝐴_1,2(𝑒₁+𝑣𝑒₃)∈𝑆, and so 𝑡²𝑞⁴𝑡²𝑞³(𝑞−1)+𝑞(1−𝑞)𝑣𝑡𝑞(𝑞−1)+(1−𝑞+𝑞²)𝑣∈𝑆. (3.17)
This implies that ((𝑞−𝑞²)𝑣−𝑞𝑢)𝑒₂+((1−𝑞+𝑞²)𝑣+(𝑞−1)𝑢)𝑒₃∈𝑆. Note that ((𝑞−𝑞²)𝑣−𝑞𝑢) and ((1−𝑞+𝑞²)𝑣+(𝑞−1)𝑢) cannot be both zeros. Assume then that (𝑞−𝑞²)𝑣−𝑞𝑢≠0.
Having that 𝑢𝑒₂−𝑣𝑒₃∈𝑆, we get that {𝑢((1−𝑞+𝑞²)𝑣+(𝑞−1)𝑢)+𝑣((𝑞−𝑞²)𝑣−𝑞𝑢)}𝑒₃∈𝑆 and so (𝑢+𝑞𝑣)(𝑢−𝑣)𝑒₃∈𝑆. (3.18) If (𝑢+𝑞𝑣)(𝑢−𝑣)≠0, then 𝑒₃∈𝑆 and thus 𝑆 is the whole space. Now if (𝑢+𝑞𝑣)(𝑢−𝑣)=0, then we have 2 cases: 𝑢=−𝑞𝑣 and 𝑢=𝑣: let𝐮=−𝐪𝐯.Since0𝑞1∈𝑆,itfollowsthat0𝑡²𝑞³𝑡²𝑞²∈𝑆. (3.19)
On the other hand, we have that 𝑞⁻²𝐴_2,30𝑞1=(𝑞−1)(𝑡𝑞−1)1+𝑡²𝑞²𝑡(𝑞−1)₂𝑞²∈𝑆. (3.20)
Substracting (3.19) from (3.20) we get that (^{(𝑞−1)(𝑡𝑞−1)}_1−𝑡²_𝑞²₀)∈𝑆.
This means that (𝑞−1)(𝑡𝑞−1)𝑒₁+(1−𝑡²𝑞²)𝑒₂∈𝑆. (3.21)
We also have that 𝑒₁−𝑞𝑣𝑒₂∈𝑆. (3.22)
Solving (3.21) and (3.22), we get that ((1+𝑡𝑞)+𝑞(1−𝑞)𝑣)𝑒₂∈𝑆.
If (1+𝑡𝑞)+𝑞(1−𝑞)𝑣≠0, we are done. Otherwise, we have that 𝑣=(𝑡𝑞+1)/𝑞(𝑞−1) and 𝑢=−𝑞𝑣=(𝑡𝑞+1)/(1−𝑞). On the other hand, we have that 1𝑢0∈𝑆then01−𝑞1+𝑡𝑞∈𝑆. (3.23)
Also, we have that 𝐴_2,30=1−𝑞1+𝑡𝑞(1−𝑞)(1+𝑞²+𝑡𝑞²)1+𝑞²+𝑡𝑞²0−𝑞∈𝑆. (3.24)
Solving (3.23) and (3.24) implies that 𝑞(1+𝑡)(1+𝑡𝑞²)𝑒₂∈𝑆 and thus 𝑒₂∈𝑆. Hence 𝑆=𝐶³.
Let 𝐮=𝐯. Since 𝑒₂−𝑒₃∈𝑆, it follows that 𝐴_2,3(𝑒₂−𝑒₃)∈𝑆. That is, we have that (𝑞−1)(−1−𝑡𝑞²)1−𝑡²𝑞³(𝑞−1)−𝑡²𝑞³∈𝑆. (3.25)
We also have that 0𝑡²𝑞³−𝑡²𝑞³∈𝑆. (3.26)
Substracting (3.26) from (3.25), we get that (𝑞−1)(−1−𝑡𝑞²)𝑒₁+(1−𝑡²𝑞⁴)𝑒₂∈S. (3.27)
Also we have that 𝑒₁+𝑣𝑒₂∈𝑆. (3.28)
Solving (3.27) and (3.28), we get that {(1+𝑡𝑞²)[(1−𝑡𝑞²)+𝑣(𝑞−1)]}𝑒₂∈𝑆. If [(1−𝑡𝑞²)+𝑣(𝑞−1)]=0, then we get that 𝑢=𝑣=(𝑡𝑞²−1)/(𝑞−1).
Now we have that 𝑒₁+𝑢𝑒₂∈𝑆 and so (𝑞−1)(1+𝑞²−𝑡𝑞³)(𝑡𝑞²−1)(1+𝑞²−𝑡𝑞³)0∈𝑆. (3.29)
We also have that 𝐴_2,3𝑞−1𝑡𝑞²0−1=(𝑞−1)(1+𝑞²−𝑡𝑞³)−𝑞²+𝑞−1+𝑡𝑞³0∈𝑆. (3.30)
Substracting (3.30) from (3.29), we get that 𝑞(1−𝑡𝑞)(𝑡𝑞³−1)𝑒₂∈𝑆 and so 𝑒₂∈𝑆. Thus 𝑆=𝐶³.

Next, we find a necessary condition that guarantees the irreducibility of the complex specialization of Krammer's representation of 𝑃₃.

4. Necessary Condition for Irreducibility

We present the following theorem.

Theorem 4.1. For (𝑞,𝑡)∈(𝐶^∗)², Krammer's representation 𝐾(𝑞,𝑡)∶𝑃₃→𝐺𝐿(3,𝐶) is reducible if one of the following conditions is satisfied:(1)𝑡²𝑞³=1,(2)𝑡𝑞³=1,(3)𝑡=−1,(4)𝑞=1,(5)𝑡𝑞=1,(6)𝑡𝑞²=−1.

Proof. Notice that the first three conditions followed from the reducibility on 𝐵₃. Under each of the last three conditions of our hypothesis, we find a proper nonzero invariant subspace under the action of complex specialization of Krammer’s representation of 𝑃₃. Recall that the matrices 𝐾(𝐴_1,2), 𝐾(𝐴_2,3), and 𝐾(𝐴_1,3) that will be used in the proof are those given in Definition 2.3.

Proof. We have that 𝐾(𝐴_1,2𝑡)=²00010001,𝐾(𝐴_2,3)=10001000𝑡²,𝐾(𝐴_1,3)=1000𝑡²0.001 (4.1)
We take the invariant subspace as the one generated by 𝑒₁=(1,0,0).

Proof. We have that 𝐾(𝐴_1,2𝑞)=²00𝑞(𝑞−1)𝑞𝑞(1−𝑞)𝑞−11−𝑞1−𝑞+𝑞²,𝐾(𝐴_2,3)=1−𝑞+𝑞²𝑞(1−𝑞)𝑞²(𝑞−1)1−𝑞𝑞𝑞²(𝑞−1)00𝑞²,𝐾(𝐴_1,3)=𝑞𝑞(𝑞−1)(1−𝑞)𝑞²−(𝑞−1)²1+2𝑞(𝑞−1)−𝑞(𝑞−1)².1−𝑞𝑞−1𝑞 (4.2)
We take the invariant subspace as the one generated by 𝑚=(0,𝑞,1)^𝑇. More precisely, we have that 𝐾(𝐴_1,2)(𝑚)=𝑚,𝐾(𝐴_2,3)(𝑚)=𝑞²𝑚,𝐾(𝐴_1,3)(𝑚)=𝑞²𝑚. (4.3)

Proof. We have that 𝐾(𝐴_1,2)=1001+𝑡𝑞𝑞𝑞(1−𝑞)−1−𝑡𝑞1−𝑞1−𝑞+𝑞²,𝐾(𝐴_2,3)=1−𝑞+𝑞²,𝑞(1−𝑞)𝑞(1−𝑞)1−𝑞𝑞𝑞−1001𝐾(𝐴_1,3)=𝑞𝑞(𝑞−1)𝑞(𝑞−1)𝑞−2−𝑡𝑞𝑞²−2𝑞+2(𝑞−1)².𝑡𝑞+1𝑞−1𝑞 (4.4)
We take the invariant subspace as the one generated by 𝑚=(−𝑞,1,0)^𝑇. More precisely, we have that 𝐾(𝐴_1,2)(𝑚)=𝑚,𝐾(𝐴_2,3)(𝑚)=𝑞²𝑚,𝐾(𝐴_1,3)(𝑚)=𝑚. (4.5)

Combining Theorems 3.1 and 4.1, we obtain our main theorem.

Theorem 4.2. For (𝑞,𝑡)∈(𝐶^∗)², Krammer's representation 𝐾(𝑞,𝑡)∶𝑃₃→𝐺𝐿(3,𝐶) is irreducible if and only if  𝑡²𝑞³≠1,  𝑡𝑞³≠1,𝑡≠−1,𝑞≠1,𝑡𝑞≠1,  and  𝑡𝑞²≠−1.