# Another Proof of the Faithfulness of the Lawrence-Krammer Representation of the Braid GroupB3

Authors: Mohammad N. Abdulrahim; Mariam Hariri
Journal: ISRN Algebra
 (2012)

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## Abstract

The Lawrence-Krammer representation of the braid groupBn 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.


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## Body

## 1. Introduction

A group is said to be linear if it admits a faithful representation intoGLn(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 ofB3 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 Section3, 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 Section4, we prove the faithfulness of this representation by a purely computational method.


## 2.   Definitions

Definition 2.1.
The braid group onn strands, Bn, is the abstract group with presentation Bn=〈σ1,…,σn-1∣σiσi+1σi=σi+1σiσi+1 for i=1,2,…,n-2, and σiσj=σjσi if |i-j|≥2〉. The generators σ1,…,σn-1 are called the standard generators of Bn.The Lawrence-Krammer representation of braid groups is a representation of the braid groupBn in GLm(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{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 automorphism group of a free group with three generators, we determine Krammer's representationK(q,t):B3→GL3(ℤ[q±1,t±1]), where
(2.2)K(σ1)=(tq200tq(q-1)0q011-q),K(σ2)=(1-qq010tq2(q-1)00tq2).Here,ℤ[q±1,t±1] is the ring of Laurent polynomials on two variables. For simplicity, we denote K(σi) by Xi, where i=1,2. Here, Xi* is the conjugate transpose of X, where t¯=t-1 and q¯=q-1.


## 3.   The Lawrence-Krammer Representation ofB3 Is Equivalent to a Unitary Representation

Budney has proved that the L-K representation of the braid groupBn 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 ofB3 under the Lawrence-Krammer representation are unitary relative to a Hermitian matrix M given by
(3.1)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)qq2+1q2(1+q2)(1-tq(q-1))tq3(1+q2)(1+tq)(tq2-1)tq2(q-1)).
That is,XiMXi*=M for i=1,2, where M*=M.The choice ofM 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 specializationM′ 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.
Letq and t be nonzero complex numbers on the unit circle such that q=eiθ and t=eiα. 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 ofM by di, where i=1,2,3. We have
(3.2)d1=(1+q2)(1+tq)(tq2-1)tq2(q-1),d2=(1+q2)2(t2q3-1)(-1+tq(-2+q(2+tq)))t2q4(q-1)2,d3=(1+q2)3(1+t)(tq3-1)(t2q3-1)2t3q6(q-1)3.
We find the intervals ofq and t for which d1,d2, and d3 are strictly positive. Let q=eiθ and t=eiα, where -π≤θ≤π and -π≤α≤π. Then we have
(3.3)d1=2cosθ[1+cscθ2sin(3θ2+α)],d2=4cos⁡2θcscθ2sin⁡(3θ2+α)[2+cscθ2sin⁡(3θ2+α)],d3=8cos3θcsc3θ2sin2(3θ2+α)[sin3θ2+sin(3θ2+α)].
In order to haved1,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 thatcsc(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 thatcsc(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 indeterminatesq and t to nonzero complex numbers, on the unit circle, which are transcendentally independent over ℤ.Letfw be a homomorphism fw:ℤ[q±1,t±1]→ℂ defined as follows: fw(q)=w1,fw(t)=w2,fw(z)=z for z∈ℤ, where w=(w1,w2) and w1,w2 are complex numbers on the unit circle. Let fw also denote the group homomorphism GL3(ℤ[q±1,t±1])→GL3(ℂ). We choose w such that fw(M) is positive definite; that is, fw(M)=VV*, for some V∈GL3(ℂ).Consider now the composition mapfw∘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 fw(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 ofB3, fw∘K, is conjugate to a unitary representation.Proof.
SinceXiMXi*=M for i=1,2, it follows that
(3.4)fw(Xi)fw(M)fw(Xi*)=fw(M).
But fw(M)=VV*, then
(3.5)(V-1fw(Xi)V)(V-1fw(Xi)V)*=I.
Now, let
(3.6)U=V-1fw(Xi)V,
then
(3.7)UU*=I.
Hence U*U=I and so U is unitary.We now find a necessary and sufficient condition for an element ofB3 possibly to belong to the kernel of the Lawrence-Krammer representation.Theorem 3.4.
An element ofB3 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 .
Iftr(fw(Xi))=3, then tr(U)=3, where U=V-1fw(Xi)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 λi′s 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 fw(Xi)=I3.Therefore, we conclude that if there exists a nontrivial element inB3 such that the trace of its image under fw∘K:B3→GL3(ℂ) is three then, the element lies in the kernel of this representation.


## 4.   The Faithfulness of the L-K Representation forn=3

In this section, we prove our main theorem, which shows the faithfulness of our representation for certain values ofq and t. Let q and t be nonzero complex numbers on the unit circle such that q=eiθ and t=eiα.Let(4.1)β=arg(1+tq2+t2q4-q(1+t)-q3t(1+t)+-4q4t2+(1+(-1+q(q-1)t)(1+tq2))22tq2).Consider the values ofq and t whose arguments belong to the set G defined as follows:
(4.2)G={(θ,α)∈A∪B∣θ≠±π3,α+3θπ∉Q,βπ∉Q},
where
(4.3)A={(θ,α)∣-π2<θ<0,-π3-2θ<α<-3θ2},B={(θ,α)∣0<θ<π2,-3θ2<α<π3-2θ}.The setG 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 fw(M) being positive definite and fw∘K:B3→GL3(ℂ) irreducible. For simplicity, we still write K instead of fw∘K.We now present our second main theorem.Theorem 4.1.
The complex specialization of the Lawrence-Krammer representationK: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 Theorem4.1. 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: Sl, Jm, Jm1Sl1Jm2Sl2,…, and so forth, where all the exponents are integers.Consider the elementSl. If l is odd, then Sl=S2k+1=S·S2k=SJ3k, where k∈ℤ. If l is even then Sl=S2k=J3k, where k∈ℤ. Therefore, any element of B3 is of one of the following forms: S, Jm, SJm1SJm2…, or Jm1SJm2S…, where mi's ∈ℤ.Lemma 4.2.
The possible nontrivial elements in the kernel of the representationK:B3→GL3(ℂ) are(i)
SJr1SJr2⋯SJr2n-1SJ-3n-∑i=12n-1‍ri, where the number of S's is 2n and ri=1 or 2 for 1≤i<2n,(ii)
elements obtained from(i) by permuting S and J.Proof.
Recall thatq=eiθ,t=eiα, and K(σi)=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|=-tq3. In general, elements of B3 are of the form: S, Jm, SJm1SJm2⋯SJmk, or elements obtained by permuting S and J in the element SJm1SJm2⋯SJmk. We deal with each of these forms separately.
Consider the elementS. 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 elementJm, where m≠0. We have |K(Jm)|=|X1|2m=(tq3)2m. Assume that |K(Jm)|=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(Jm)|≠1 and so Jm∉KerK.
Consider the elementu=SJm1SJm2⋯SJm2n+1, where the number of S's is 2n+1. We have |K(u)|=|X1|h, where h=3(2n+1)+2∑i=12n+1‍mi∈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 elementu=SJm1SJm2⋯SJm2n, where the number of S's is 2n. Then |K(u)|=|X1|h, where h=2(3n+∑i=12n‍mi)∈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, ∑i=12n‍mi=-3n. We write m2n=-3n-∑i=12n-1‍mi. Then u=SJm1SJm2⋯SJm2n-1SJ-3n-∑i=12n-1‍mi.
Ifmi=3ki for some 1≤i<2n, then Jmi∈Z(B3). It follows that SJmi=JmiS. Using the relation S2=J3, u will be reduced to an element with less number of S's and mi≠3ki for all i.
In the casemi≠3ki for every 1≤i<2n, we have mi=3ki+ri, where ri=1 or 2. Then u=SJ3k1+r1⋯SJ3k2n-1+r2n-1SJ-3n-(3k1+r1+3k2+r2+⋯+3k2n-1+r2n-1). Since J3ki∈Z(B3), it follows that u=SJr1SJr2⋯SJr2n-1SJ-3n-∑i=12n-1‍ri, where ri=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 ofK:B3→GL3(ℂ) are SJr1SJr2⋯SJr2n-1SJ-3n-∑i=12n-1‍ri, and other elements that are obtained by permuting S and J. Here, the number of S's is 2n and ri=1 or 2 for 1≤i<2n.Now, we present a lemma that will be used in the proof of Theorem4.1.Lemma 4.3.
(4.4)
K
(
J
n
)
=
{
(
t
2
q
6
)
k
I
,
n
=
3
k
,
(
t
2
q
6
)
k
K
(
J
)
,
n
=
3
k
+
1
,
(
t
2
q
6
)
k
K
(
J
2
)
,
n
=
3
k
+
2
.
Here k∈ℕ and I is the identity matrix.Proof.
Direct computations show that(4.5)K(J)=(-tq2(q-1)tq30-tq(q-1)2tq2(q-1)tq3100),K(J2)=(00t2q6tq30t2q5(q-1)-tq2(q-1)tq30),K(J3)=(t2q6)I,K(J4)=(t2q6)K(J),K(J5)=(t2q6)K(J2).
Using mathematical induction on the integerk, we prove the lemma.


## 5. Proof of Theorem4.1

In order to prove that the representationK: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 elementu=SJr1SJr2⋯SJr2n-1SJ-3n-∑i=12n-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 elementu=SJr1SJr2⋯SJr2n-1SJ-3n-∑i=12n-1‍ri, where the number of S's is 2n and ri=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) Ifri=1 for all 1≤i<2n, then
(5.1)u=(SJ1)2n-1SJ-3n-(2n-1)=(SJ1)2n-1SJ-3(2n-1)+n-2.
Since  J-3∈Z(B3), it follows that u=(SJ-2)2n-1SJn-2 and so u=(SJ-2)2nJn. 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
(5.2)T=(tq2-tq3t2q4tq(q-1)-tq2(q-1)q(-1+tq(-1+q+q(1+(q-1)q)t))q-1111).
It is easy to see that det(T)=-(q3(1+q)t(1+tq)(tq2-1))/(q-1)≠0 for (θ,α)∈G.Then(5.3)T-1K(SJ-2)T=(1000-1q0001tq2),T-1K(u)T=(10001q2n0001(tq2)2n)T-1K(Jn)T.We take the following possibilities regarding the value ofn.(a) Forn=3k, we have, from Lemma 4.3, that T-1K(Jn)T=(t2q6)kI. Then we get
(5.4)T-1K(u)T=((t2q6)k000t2k0001(t4q6)k).
The entry (t2q6)k≠1; otherwise, eik(2α+6θ)=1. That is, (α+3θ)/π∈ℚ, which contradicts the fact that (θ,α)∈G.(b) Forn=3k+1, we have, from Lemma 4.3, that T-1K(Jn)T=(t2q6)kT-1K(J)T. Direct computations give
(5.5)T-1K(u)T=(*tq3(t2q6)k(1-tq3)(1+q)(tq2-1)*******).
It is easy to show that (tq3(t2q6)k(1-tq3))/((1+q)(tq2-1))≠0 for (θ,α)∈G.(c) Forn=3k+2, we have, from Lemma 4.3, that T-1K(Jn)T=(t2q6)kT-1K(J2)T. We get, by direct computations,
(5.6)T-1K(u)T=(*tq4(t2q6)k(tq3-1)(1+q)(tq2-1)*******).
It is easy to show that (tq4(t2q6)k(tq3-1))/((1+q)(tq2-1))≠0 for (θ,α)∈G.(2) Ifri=2 for all 1≤i<2n, then
(5.7)u=(SJ2)2n-1SJ-3n-2(2n-1)=(SJ2)2n-1SJ-3(2n-1)-n-1.
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
(5.8)H=(00(1+tq)(tq2-1)t(q-1)q-1-1+q+tq2111).It is clear thatdet(H)=((1+q)(1+tq)(tq2-1))/(t(q-1))≠0 for (θ,α)∈G.Then(5.9)H-1K(SJ-1)H=(1000-q000tq2),H-1K(u)H=(1000q2n000(tq2)2n)H-1K(J-n)H.Using Lemma4.3, we can easily see that
(5.10)K(J-n)={1(t2q6)kI,n=3k,1(t2q6)kK(J-1),n=3k+1,1(t2q6)kK(J-2),n=3k+2.Herek is a positive integer. We deal with the following possibilities regarding the value of n.(a) Forn=3k, a comuptation shows that
(5.11)H-1K(u)H=(1(t2q6)k0001t2k000(t4q6)k).The entry1/(t2q6)k≠1; otherwise, eik(2α+6θ)=1, which implies that (α+3θ)/π∈ℚ. This contradicts the fact that (θ,α)∈G.(b) Forn=3k+1, a computation shows that
(5.12)H-1K(u)H=(*1-tq3tq3(t2q6)k(1+q)(tq2-1)*******).It is easy to show that(1-tq3)/(tq3(t2q6)k(1+q)(tq2-1))≠0 for (θ,α)∈G.(c) Forn=3k+2, a computation shows that
(5.13)H-1K(u)H=(*tq3-1tq4(t2q6)k(1+q)(tq2-1)*******).We can also see that(tq3-1)/(tq4(t2q6)k(1+q)(tq2-1))≠0 for (θ,α)∈G.(3) If for1≤i<2n, we have that
(5.14)ri={1,iis  odd,2,iis  even,
then
(5.15)u=(SJ1SJ2)n-1SJ1SJ-3n-[3(n-1)+1]=(SJ1SJ2)n-1SJ1SJ-3(2n-1)-1.SinceJ-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
(5.16)L=(qq(-c-b)-c+b-q(-c+b)c+b-1+q+1tq-2q+2(q-1)2t2q3-c+b--2q+2(q-1)2t2q3c+b111),
where b is defined above and c=1+tq2+tq3(t-1)-t2q4-q(1+t).Also, we have thatdet(L)=((-1+q+q(1+q+q2)t-(q-1)t2q3)b)/(q3(1+(q-1)q)t3)≠0 when
(5.17)θ≠±π3,b≠0,1+2cos⁡θ+2cos⁡(θ+α)-2cos⁡(2θ+α)≠0.
Then
(5.18)L-1K((SJ-2SJ-1)n)L=(1000e2n000e3n).Ife3n=1, then eiβn=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 for1≤i<2n, we have that
(5.19)ri={2,iis  odd,1,iis  even,
then
(5.20)u=(SJ2SJ1)n-1SJ2SJ-3n-[3(n-1)+2]=(SJ2SJ1)n-1SJ2SJ(-3(2n-1)-2).
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) Letu=SJr1SJr2⋯SJr2n-1SJ-3n-∑i=12n-1‍ri such that not all ri's are 1 and not all ri's are 2. We also assume that there exists an integer j such that rj=rj+1, where 1≤j<2n. We write u=AJ-p, where A=SJr1SJr2⋯SJr2n-1S and p=3n+∑i=12n-1‍ri. 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 ofJ-1 are 1/(q2t2/3),-(-1)1/3/(q2t2/3) and (-1)2/3/(q2t2/3). It is clear that these eigenvalues are distinct. We diagonalize K(J-1) by a matrix W given by
(5.21)(q2t2/3(-1)2/3q2t2/3(-1)4/3q2t2/3q(1+(q-1)t1/3)t1/3(-t)1/3q(-1+(-1)1/3(q-1)t1/3)t1/3q(1+(-1)2/3(q-1))t1/3111).Under direct computations, we see thatdet(W)=-3i3q3t≠0. Here i=-1. Then we get that
(5.22)W-1K(J-1)W=(1q2t2/3000-(-1)1/3q2t2/3000(-1)2/3q2t2/3).
We have that
(5.23)W-1K(u)W=(W-1K(A)W)(W-1K(J-p)W).
Assume, to get contradiction, that u∈Ker(K). Then
(5.24)W-1K(A)W=(W-1K(u)W)(W-1K(J-p)W)-1=(W-1K(J-p)W)-1.It follows thatW-1K(A)W is a diagonal matrix. Therefore, W-1K(A)W and W-1K(J-p)W commute, that is,
(5.25)W-1K(u)W=W-1K(J-p)K(A)W.Then(5.26)K(u)=K(J-p)K(A).Therefore,K(A) and K(J-p) commute, which is a contradiction.



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*Source: 858959-2012-06-25.xml*