# Krammer's Representation of the Pure Braid Group,P3

Authors: Mohammad N. Abdulrahim; Madline Al-Tahan
Journal: International Journal of Mathematics and Mathematical Sciences
 (2010)

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## 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.


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## Body

## 1. Introduction

LetBn be the braid group on n strings. There are a lot of linear representations of Bn. The earliest was the Artin representation, which is an embedding Bn→Aut(Fn), the automorphism group of a free group on n generators. Applying the free differential calculus to elements of Aut(Fn) sometimes gives rise to linear representations of Bn and its normal subgroup, the pure braid group denoted by Pn [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 Section2, 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 onn strings, Bn, is the abstract group with presentation Bn={σ1,…,σn-1/σiσi+1σi=σi+1σiσi+1 for i=1,2,…n-2,σiσj=σjσi if |i-j|>1}.The generatorsσ1,…,σn-1 are called the standard generators of Bn.Definition 2.2.
The kernel of the group homomorphismBn→Sn is called the pure braid group on n strands and is denoted by Pn. 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 Pn are Ai,j, 1≤i<j≤n, where Ai,j=σj-1σj-2⋯σi+1σi2σi+1-1⋯σj-2-1σj-1-1.Let us recall the Lawrence-Krammer representation of braid groups. This is a representation ofBn 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, Bn, 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{xi,j}1≤i<j≤n, the free basis of V0, the image of each Artin generator under Krammer's representation is written as
(2.1)K(σk)(xi,j)={tq2xk,k+1,i=k,j=k+1;(1-q)xi,k+qxi,k+1,j=k,i<k;xi,k+tqk-i+1(q-1)xk,k+1,j=k+1,i<k;tq(q-1)xk,k+1+qxk+1,j,i=k,k+1<j;xk,j+(1-q)xk+1,j,i=k+1,k+1<j;xi,j,i<j<kork+1<i<j;xi,j+tqk-i(q-1)2xk,k+1,i<k<k+1<j.Using the Magnus representation of subgroups of the automorphisms group of free group withn(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 by(2.2)K(A1,2)=(t2q400t2q3(q-1)qq(1-q)tq(q-1)1-q1-q+q2),K(A2,3)=(1-q+q2q(1-q)tq3(q-1)1-qqt2q4(q-1)00t2q4),K(A1,3)=(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),
where(2.3)m=-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)].Specializingt 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 stringsP3.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 writeK(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 ofS is 1:
The subspaceS 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〉).
Sincee1∈S, it follows that A1,2(e1)∈S which implies that t2q3(q-1)=0, a contradiction.Case 2 (S=〈e2〉).
Sincee2∈S,it follows that A1,2(e2)∈S which implies that 1-q=0, a contradiction.Case 3 (S=〈e3〉).
Sincee3∈S,  it follows that A1,2(e3)∈S which implies that q(1-q)=0, a contradiction.Case 4 (S=〈e1+ue2〉,u≠0).
Sincee1+ue2∈S, it follows that A1,2(e1+ue2)∈S. This implies that
(3.1)(t2q4t2q3(q-1)+qutq(q-1)+(1-q)u)=m(1u0),
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).
Sincee2+ue3∈S, it follows that A2,3(e2+ue3)∈S. This implies that
(3.2)(q(1-q)+tq3(q-1)uq+t2q4(q-1)ut2q4u)=m(01u),
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).
Sincee1+ue3∈S, it follows that A1,2(e1+ue3)∈S. This implies that
(3.3)(t2q4t2q3(q-1)+q(1-q)utq(q-1)+(1-q+q2)u)=m(10u),
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 A2,3(e1+ue3)∈S. This implies that
(3.4)(1-q+q2+tq3(q-1)u1-q+t2q4(q-1)ut2q4u)=n(10u),
where n is a complex number. By solving this system of equations, we get that t2q3=-1. This means that
(3.5)t2q3=tq(tq2)=tq(1-q)=tq-tq2=tq-1+q.
This implies that q(t+1)=0, which contradicts the hypothesis.Case 7 (S=〈e1+ue2+ve3〉,u,v≠0).
Sincee1+ue2+ve3∈S, it follows that A1,2(e1+ue2+ve3)∈S. This implies that
(3.6)(t2q4t2q3(q-1)+qu+q(1-q)vtq(q-1)+(1-q)u+(1-q+q2)v)=m(1uv),
where m is a complex number. Since A2,3(e1+ue2+ve3)∈S, it follows that
(3.7)(1-q+q2+q(1-q)u+tq3(q-1)v1-q+qu+t2q4(q-1)vt2q4v)=n(1uv),
where n is a complex number. Solving these two system of equations, we get that m=n=t2q4. Also, we have that
(3.8)q(t2q3-1)u+q(q-1)v=t2q3(q-1),(3.9)(q-1)u+(t2q4-q2+q-1)v=tq(q-1),(3.10)q(1-q)u+tq3(q-1)v=t2q4-q2+q-1,(3.11)q(t2q3-1)u-t2q4(q-1)v=1-q.
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.
If1+t2q3=0, then we rewrite (3.8), (3.9), (3.10), and (3.11) to become as follows:
(3.12)2qu-q(q-1)v=q-1,(3.13)(q-1)u-(q2+1)v=tq(q-1),(3.14)q(1-q)u+tq3(q-1)v=-(q2+1).
Multiplying (3.13) by q and adding it to (3.14) we get that
(3.15)q(tq3-tq2-q2-1)v=tq3-tq2-q2-1.
A simple computation shows thattq3-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 that1+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 ofS is 2:
Easy computations show that the subspaceS cannot be in the form S=〈ei,ej〉 or S=〈ei+uej,ek〉 for i≠j≠k.
It suffices to consider only the caseS=〈e1+ue2,e1+ve3〉, where u,v≠0.
Sincee1+ue2∈S, it follows that A1,2(e1+ue2)∈S and so
(3.16)(t2q4t2q3(q-1)+qutq(q-1)+(1-q)u)∈S.
Also, we have thate1+ve3∈S,then A1,2(e1+ve3)∈S, and so
(3.17)(t2q4t2q3(q-1)+q(1-q)vtq(q-1)+(1-q+q2)v)∈S.
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 thatue2-ve3∈S, we get that {u((1-q+q2)v+(q-1)u)+v((q-q2)v-qu)}e3∈S and so
(3.18)(u+qv)(u-v)e3∈S.
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:
(3.19)letu=-qv. Since(0q1)∈S,it follows that(0t2q3t2q2)∈S.
On the other hand, we have that(3.20)q-2A2,3(0q1)=((q-1)(tq-1)1+t2q2(q-1)t2q2)∈S.
Substracting (3.19) from (3.20) we get that ((q-1)(tq-1)1-t2q20)∈S.
This means that(3.21)(q-1)(tq-1)e1+(1-t2q2)e2∈S.
We also have that(3.22)e1-qve2∈S.
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
(3.23)(1u0)∈Sthen(1-q1+tq0)∈S.
Also, we have that(3.24)A2,3(1-q1+tq0)=((1-q)(1+q2+tq2)1+q2+tq2-q0)∈S.
Solving (3.23) and (3.24) implies that q(1+t)(1+tq2)e2∈S and thus e2∈S. Hence S=C3.
Letu=v. Since e2-e3∈S, it follows that A2,3(e2-e3)∈S. That is, we have that
(3.25)((q-1)(-1-tq2)1-t2q3(q-1)-t2q3)∈S.
We also have that(3.26)(0t2q3-t2q3)∈S.
Substracting (3.26) from (3.25), we get that
(3.27)(q-1)(-1-tq2)e1+(1-t2q4)e2∈S.
Also we have that(3.28)e1+ve2∈S.
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 thate1+ue2∈S and so
(3.29)((q-1)(1+q2-tq3)(tq2-1)(1+q2-tq3)0)∈S.
We also have that(3.30)A2,3(q-1tq2-10)=((q-1)(1+q2-tq3)-q2+q-1+tq30)∈S.
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 ofP3.


## 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 onB3. 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(A1,2), K(A2,3), and K(A1,3) that will be used in the proof are those given in Definition 2.3.Proof.
We have that(4.1)K(A1,2)=(t200010001),K(A2,3)=(10001000t2),K(A1,3)=(1000t20001).
We take the invariant subspace as the one generated bye1=(1,0,0).Proof.
We have that(4.2)K(A1,2)=(q200q(q-1)qq(1-q)q-11-q1-q+q2),K(A2,3)=(1-q+q2q(1-q)q2(q-1)1-qqq2(q-1)00q2),K(A1,3)=(qq(q-1)(1-q)q2-(q-1)21+2q(q-1)-q(q-1)21-qq-1q).
We take the invariant subspace as the one generated bym=(0,q,1)T. More precisely, we have that
(4.3)K(A1,2)(m)=m,K(A2,3)(m)=q2m,K(A1,3)(m)=q2m.Proof.
We have that(4.4)K(A1,2)=(1001+tqqq(1-q)-1-tq1-q1-q+q2),K(A2,3)=(1-q+q2q(1-q)q(1-q)1-qqq-1001),K(A1,3)=(qq(q-1)q(q-1)q-2-tqq2-2q+2(q-1)2tq+1q-1q).
We take the invariant subspace as the one generated bym=(-q,1,0)T. More precisely, we have that
(4.5)K(A1,2)(m)=m,K(A2,3)(m)=q2m,K(A1,3)(m)=m.Combining Theorems3.1 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.



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