ISRN AlgebraVolume 2012 (2012), Article ID 858959, 13 pagesdoi:10.5402/2012/858959

Research Article

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

Mohammad N. Abdulrahim and Mariam Hariri

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. Koshlukov, H. Li, S. Yang, and Y. Zhou

Copyright © 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 was proved to be faithful for by Bigelow and Krammer. In our paper, we give a new proof in the case by using matrix computations. First, we prove that the representation of the braid group 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 for some natural number and some field . 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 and is not faithful for [1]. However, the faithfulness of the Burau representation in the case is still unknown. In [2], Krammer proved that the Lawrence-Krammer representation is faithful in the case , where he assumed that is a real number with . 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 . In his argument, he assumed that and are variables. Shortly after, Krammer [4] found a new proof that shows that this representation is faithful for all using algebraic methods, in which he interchanged the roles of and in [2] and assumed that is a real number with .

Our work uses a different method that basically depends on matrix computations to show that the Lawrence-Krammer representation of is faithful. In our argument, we specialize the indeterminates and 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 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 and to nonzero complex numbers on the unit circle rather than to real numbers in the interval as in [2] or [4].

In Section 3, We show that if and 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 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 strands, , is the abstract group with presentation for , and if . The generators are called the standard generators of .

The Lawrence-Krammer representation of braid groups is a representation of the braid group in , where and is the free module of rank over . The representation is denoted by . For simplicity, we write instead of .

Definition 2.2. With respect to , the free basis of , the image of each Artin generator under Krammer's representation is written as

Using the Magnus representation of subgroups of the automorphism group of a free group with three generators, we determine Krammer's representation , where

Here, is the ring of Laurent polynomials on two variables. For simplicity, we denote by , where . Here, is the conjugate transpose of , where and .

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

Budney has proved that the L-K representation of the braid group 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 and to nonzero complex numbers on the unit circle. We consider the complex representation and prove that it is unitary relative to a positive definite Hermitian form.

Theorem 3.1. The images of the generators of under the Lawrence-Krammer representation are unitary relative to a Hermitian matrix given by That is, for , where .

The choice of is unique only if we specialize the indeterminates and 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 of is equivalent to the identity matrix, that is, for some matrix . In other words, we need to show that for some matrix ; or equivalently, is hermitian.

Theorem 3.2. Let and be nonzero complex numbers on the unit circle such that and . The matrix is positive definite when and are chosen in either one of the following intervals: (i) and , (ii) and .

Proof . We denote the principal minors of by , where . We have We find the intervals of and for which , and are strictly positive. Let and , where and . Then we have In order to have , and 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 , , , and .In the interval (ii), we see that , , , and .

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

Let be a homomorphism defined as follows: for , where and are complex numbers on the unit circle. Let also denote the group homomorphism . We choose such that is positive definite; that is, , for some .

Consider now the composition map . It was proved in [6] that the specialization of Krammer's representation is irreducible if and only if , and . It is easy to see that the conditions for irreducibility are achieved once we choose and in either one of the intervals (i) or (ii) in the hypothesis of Theorem 3.2. The uniqueness of up to scalar multiplication follows from Shur's Lemma and the fact that the specialization of Krammer's representation of is irreducible.

Theorem 3.3. The complex representation of , , is conjugate to a unitary representation.

Proof. Since for , it follows that But , then Now, let then Hence and so is unitary.

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

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

Proof . If , then , where . Since is unitary, it follows that is diagonalizable, that is, there exists a matrix such that , where is a diagonal matrix with eigenvalues of as the diagonal entries. Hence , where the are the eigenvalues of . Being unitary, it has its eigenvalues on the unit circle. Therefore, we get . Thus, is the identity matrix and so is . This implies that .

Therefore, we conclude that if there exists a nontrivial element in such that the trace of its image under is three then, the element lies in the kernel of this representation.

4. The Faithfulness of the L-K Representation for

In this section, we prove our main theorem, which shows the faithfulness of our representation for certain values of and . Let and be nonzero complex numbers on the unit circle such that and .

Let

Consider the values of and whose arguments belong to the set defined as follows: where

The set is a subset of the union of the intervals given in Theorem 3.2. We choose such that . Therefore, for values of and such that , we have being positive definite and irreducible. For simplicity, we still write instead of .

We now present our second main theorem.

Theorem 4.1. The complex specialization of the Lawrence-Krammer representation is faithful for a dense subset of the set defined above.

We need to state some lemmas that help us prove Theorem 4.1. We already know that , where and are called the standard generators of . Consider the product of generators of , namely, . We also let . It was proved in [7] that is generated by the two elements and and the relation . That is, we have that . Also, it was proved in [8] that the center of is generated by the one generator , that is, . Thus, the elements of are of the form: , , , and so forth, where all the exponents are integers.

Consider the element . If is odd, then , where . If is even then , where . Therefore, any element of is of one of the following forms: , , , or , where 's .

Lemma 4.2. The possible nontrivial elements in the kernel of the representation are (i), where the number of 's is and or for ,(ii)elements obtained from by permuting and .

Proof. Recall that , and , where . For simplicity, we use the symbol instead of . We have that , , and . In general, elements of are of the form: , , , or elements obtained by permuting and in the element . We deal with each of these forms separately.Consider the element . We have . Assume that . Then and so . This implies that , which is a contradiction when . Therefore, and so .Consider the element , where . We have . Assume that . Then and so , . This implies that , which is a contradiction when . Therefore, and so .Consider the element , where the number of 's is . We have , where . It is clear that we have . Assume that . Then and so . Then . This implies that , which is a contradiction when . Therefore, , that is, . It follows, by Theorem 3.4, that . Now, consider an element obtained by permuting and in . Then is an element with odd number of 's. The trace of the image of is equal to that of the image of an element having the same form as . This implies that and thus .Consider the element , where the number of 's is . Then , where . For , we have , that is, . If , then and so , . This implies that , which contradicts the fact that . Thus , that is, . We write . Then .If for some , then . It follows that . Using the relation , will be reduced to an element with less number of 's and for all .In the case for every , we have , where or . Then . Since , it follows that , where or . Since , it follows that the trace of the image of is equal to . Consider an element obtained by permuting and in the element . Then is an element that has an even number of 's. The trace of the image of is equal to that of an element having the same form as . This implies that . It follows, by Theorem 3.4, that . Therefore, the possible nontrivial elements in the kernel of are , and other elements that are obtained by permuting and . Here, the number of 's is and or for .

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

Lemma 4.3. Here and is the identity matrix.

Proof. Direct computations show that Using mathematical induction on the integer , we prove the lemma.

5. Proof of Theorem 4.1

In order to prove that the representation is faithful, we have to show that the elements in Lemma 4.2 do not belong to if we choose and in a way that their arguments and belong to the previously defined set .

That is, we need to show that the element and others obtained from by permuting and do not belong to the kernel of this representation if .

Consider the element , where the number of 's is and or for . We consider five cases and we prove that the element , in each of the cases, does not belong to the kernel of the complex specialization .

(1) If for all , then Since , it follows that and so . The eigenvalues of are and , which are easily shown to be distinct for . We diagonalize by a matrix given by It is easy to see that for .

Then

We take the following possibilities regarding the value of .

(a) For , we have, from Lemma 4.3, that . Then we get The entry ; otherwise, . That is, , which contradicts the fact that .

(b) For , we have, from Lemma 4.3, that . Direct computations give It is easy to show that for .

(c) For , we have, from Lemma 4.3, that . We get, by direct computations, It is easy to show that for .

(2) If for all , then Since , it follows that and so . The eigenvalues of are , and . It is easy to show that these eigenvalues are distinct when . We diagonalize by a matrix given by

It is clear that for .

Then

Using Lemma 4.3, we can easily see that

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

(a) For , a comuptation shows that

The entry ; otherwise, , which implies that . This contradicts the fact that .

(b) For , a computation shows that

It is easy to show that for .

(c) For , a computation shows that

We can also see that for .

(3) If for , we have that then

Since , it follows that and so . The eigenvalues of are , where and . These eigenvalues are distinct for . Now, we diagonalize by a matrix given by where is defined above and .

Also, we have that when Then

If , then , where . This implies that , which contradicts the fact that . Therefore, and so . We conclude that .

(4) If for , we have that then Since , it follows that and so . Using the properties of the trace, we have that . By our result in case (3) and Theorem 3.4, we have that and so .

(5) Let such that not all 's are and not all 's are . We also assume that there exists an integer such that , where . We write , where and . Without loss of generality, we assume that and do not commute for some values of and in . For this reason, we require that or 2 (mod 3). If and happened to commute for some values of and then we change the values of and , slightly in a way that is still in and the matrices do not commute anymore. This is due to the fact that belongs to a dense subset of by the hypothesis of the theorem.

The eigenvalues of are and . It is clear that these eigenvalues are distinct. We diagonalize by a matrix given by

Under direct computations, we see that . Here . Then we get that We have that Assume, to get contradiction, that . Then

It follows that is a diagonal matrix. Therefore, and commute, that is,

Then

Therefore, and commute, which is a contradiction.

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