The Burau representation

Homological Representations of the Braid Groups

3.1 The Burau representation

Homological Representations

whereIk denotes the unitk×k matrix. Wheni= 1, there is no unit matrix in the upper left corner ofUi. Wheni=n−1, there is no unit matrix in the lower right corner ofU_i. Substitutingt= 1 in the definition ofU₁, . . . , U_n−1, we obtain permutationn×nmatrices. One can therefore view U₁, . . . , U_n−1 as one-parameter deformations of permutation matrices.

Each matrixUi has a block-diagonal form with blocks being the unit ma- trices and the 2×2 matrix

U =

1−t t

1 0

. (3.1)

By the Cayley–Hamilton theorem, any 2×2 matrixM over the ringΛsatisfies M²−tr(M)M+ det(M)I2= 0, where tr(M) is the trace ofM and det(M) is the determinant ofM. ForM =U, this givesU²−(1−t)U−tI₂= 0. Since the unit matrices also satisfy this equation,

U_i²−(1−t)U_i−tI_n = 0

for all i. This can be rewritten as Ui(Ui−(1−t)In) = tIn. Hence, Ui is invertible overΛand its inverse is computed by

U_i⁻¹=t⁻¹(Ui−(1−t)In) =

⎛

⎜⎜

⎝

Ii−1 0 0 0

0 0 1 0

0 t⁻¹ 1−t⁻¹ 0

0 0 0 In−i−1

⎞

⎟⎟

⎠.

The block form of the matricesU1, . . . , Un−1implies thatUiUj=UjUifor alli, j with|i−j| ≥2. We also have

UiUi+1Ui=Ui+1UiUi+1

fori= 1, . . . , n−2. To check this, it is enough to verify the equality

⎛

⎝1−t t 0

1 0 0

0 0 1

⎞

⎠

⎛

⎝1 0 0 0 1−t t

0 1 0

⎞

⎠

⎛

⎝1−t t 0

1 0 0

0 0 1

⎞

⎠

=

⎛

⎝1 0 0 0 1−t t

0 1 0

⎞

⎠

⎛

⎝1−t t 0

1 0 0

0 0 1

⎞

⎠

⎛

⎝1 0 0 0 1−t t

0 1 0

⎞

⎠.

This equality is an exercise in matrix multiplication.

By Lemma 1.2, the formula ψn(σi) = Ui with i = 1, . . . , n−1 defines a group homomorphism ψn from the braid group Bn with n ≥ 2 to the group GLn(Λ) of invertiblen×nmatrices over Λ. This is the Burau repre- sentationofBn. In particular, forn= 2, this representation is the homomor- phismB₂→GL₂(Λ), sending the generatorσ₁ofB₂∼=Zto the matrix (3.1).

By convention, the Burau representationψ1 of the (trivial) groupB1 is the trivial homomorphismB1→GL1(Λ).

Observe that detU_i=−t for alli. This implies that for anyβ∈B_n, detψn(β) = (−t)^β,

whereβ ∈Zis the image of β under the homomorphism B_n →Zsending the generatorsσ₁, . . . , σ_n−1 to 1.

The Burau representations{ψn}n≥1 are compatible with the natural in- clusionsι:Bn→Bn+1: for anyn≥1 andβ∈Bn,

ψ_n+1(ι(β)) =

ψ_n(β) 0

0 1

. (3.2)

3.1.2 Unitarity

The study of the Burau representationψn : Bn → GLn(Λ) has to a great extent been focused on its kernel and image. We establish here a simple pro- perty of the image showing that it is contained in a rather narrow subgroup of GLn(Λ). This property will not be used in the sequel.

Consider the involutive automorphism of the ring Λ, λ → λ for λ ∈ Λ, sending t to t⁻¹. For a matrix A = (λi,j) over Λ, set A = (λi,j) and let A^T = (λj,i) be the transpose of A. Let Ωn be the lower triangular n×n matrix overΛ with all diagonal terms equal to 1 and all subdiagonal terms equal to 1−t:

Ωn=

⎛

⎜⎜

⎜⎜

⎜⎝

1 0 0 · · · 0 1−t 1 0 · · · 0 1−t 1−t 1 · · · 0 ... ... ... . .. ... 1−t 1−t 1−t · · · 1

⎞

⎟⎟

⎟⎟

⎟⎠ .

Theorem 3.1.For anyn≥1 andA∈ψ_n(B_n)⊂GL_n(Λ),

A ΩnA^T =Ωn. (3.3)

Proof. If (3.3) holds for a matrix A, then it holds for its inverse: multiply- ing (3.3) on the left byA⁻¹ and on the right by (A^T)⁻¹, we obtain the same formula withAreplaced byA⁻¹. If (3.3) holds for two matricesA1, A2, then it holds for their product:

A1A2Ωn(A1A2)^T =A1A2ΩnA^T₂A^T₁ =A1ΩnA^T₁ =Ωn.

Now, since the matricesU1, . . . , Un−1generate the groupψn(Bn), it is enough to prove (3.3) forA=Ui withi= 1, . . . , n−1. PresentA=UiandΩn in the block form

A=

⎛

⎝I_i−1 0 0

0 U 0

0 0 In−i−1

⎞

⎠, Ω_n=

⎛

⎝ Ω_i−1 0 0

K2,i−1 Ω2 0

Kn−i−1,i−1 Kn−i−1,2 Ωn−i−1

⎞

⎠,

where

U =

1−t t

1 0

, Ω₂=

1 0 1−t 1

,

andK_p,qis thep×qmatrix with all entries equal to 1−t. A direct computation gives

A ΩnA^T =

⎛

⎝ Ω_i−1 0 0 U K2,i−1 U Ω2U^T 0 Kn−i−1,i−1 Kn−i−1,2U^T Ωn−i−1

⎞

⎠.

Note thatU K2,i−1=K2,i−1, since U

1−t 1−t

=

1−t⁻¹ t⁻¹

1 0

1−t 1−t

= 1−t

1−t

. Similarly,Kn−i−1,2U^T =Kn−i−1,2, since

(1−t,1−t)U^T = (1−t,1−t)

1−t 1

t 0

= (1−t,1−t).

A direct computation givesU Ω2U^T =Ω2. Substituting these formulas in the expression forA ΩnA^T, we conclude that A ΩnA^T =Ωn. Applying the involutionA→Aand the transposition to (3.3), we obtain A Ω^T_nA^T =Ω^T_n. Therefore for anyA∈ψn(Bn) andλ, μ∈Λ,

A(λΩn+μΩ^T_n)A^T =λΩn+μΩ^T_n. In particular, settingλ=μ= 1, we obtain

A Θ_nA^T =Θ_n, (3.4)

whereΘ_n =Ω_n+Ω^T_n is the followingn×nmatrix:

Θn=

⎛

⎜⎜

⎜⎜

⎜⎝

2 1−t⁻¹ 1−t⁻¹ · · · 1−t⁻¹ 1−t 2 1−t⁻¹ · · · 1−t⁻¹ 1−t 1−t 2 · · · 1−t⁻¹

... ... ... . .. ... 1−t 1−t 1−t · · · 2

⎞

⎟⎟

⎟⎟

⎟⎠ .

The matrixΘn is “Hermitian” in the sense thatΘ^T_n =Θn.

Remark 3.2.Sending t∈Λ to a complex numberζ of absolute value 1, we obtain a ring homomorphismpζ :Λ→C. The involutionλ→λon the ringΛ corresponds under p_ζ to complex conjugation. Applyingp_ζ to the entries of n×nmatrices overΛ, we obtain a group homomorphism GL_n(Λ)→GL_n(C), also denoted bypζ. This gives a representation

P_ζ =p_ζψ_n :B_n →GL_n(C). Formula (3.4) implies that

Pζ(β)pζ(Θn)Pζ(β)^T =pζ(Θn)

for allβ ∈ Bn. For ζ = 1, we have pζ(Θn) = 2In. Therefore the Hermitian matrixpζ(Θn) is positive definite for allζsufficiently close to 1. For suchζ, the matrices inPζ(Bn)⊂GLn(C) are obtained by transposition and conjugation from unitary matrices.

3.1.3 The kernel of ψn

A homomorphism from a group to a group of matrices is said to befaithful if its kernel is trivial. The homomorphismψ1 is faithful, sinceB1={1}. The homomorphism ψ2 is also faithful. Indeed, the matrix U = U1 ∈ GL2(Λ), which is the image of the generatorσ1 ofB2∼=Z, satisfies

(1,−1)U = (−t, t) =−t(1,−1).

Hence, (1,−1)U^k= (−t)^k(1,−1) for allk∈Zand we can conclude thatU is of infinite order in GL₂(Λ). In Section 3.3.2 we shall show that Kerψ₃={1}. Forn ≥ 4, the question whether ψ_n is faithful, i.e., whether Kerψ_n ={1}, remained open for a long time. Note that Kerψ_n ⊂Kerψ_n+1under the inclu- sionBn ⊂Bn+1. Therefore, if Kerψn={1}, then we have also Kerψm={1} for allm≥n.

Theorem 3.3.Kerψn={1} for n≥5.

At the moment of writing (2007), it is unknown whether Kerψ4={1}. We point out explicit braids on five and six strings annihilated by the Burau representation. Set

γ=σ4σ⁻₃¹σ⁻₂¹σ²₁σ⁻₂¹σ⁻₁²σ⁻₂²σ₁⁻¹σ₄⁻⁵σ2σ3σ³₄σ2σ²₁σ2σ⁻₃¹ ∈B5. Then the commutator

ρ= [γσ4γ⁻¹, σ4σ3σ2σ²₁σ2σ3σ4]

is a nontrivial element of Kerψ5⊂B5. Here for elementsa, bof a group, [a, b] =a⁻¹b⁻¹ab .

The braid ρ is represented by a word of length 120 in the generators σ₁^±1, σ^±₂¹, σ^±₃¹, σ₄^±1(observe thatγhas length 26, whileσ⁻₄¹γandγ⁻¹σ4have length 25). Forn= 6 we can produce a shorter word representing an element of the kernel. Set

γ=σ4σ⁻₅²σ⁻₂¹σ³₁σ⁻₂¹σ⁻₅¹σ4 ∈B6. The commutator

ρ= [γσ3γ⁻¹, σ3]

is a nontrivial element of Kerψ6⊂B6. The braidρ is represented by a word of length 44 in the generators. Thatρ, ρ lie in the kernel of the Burau repre- sentation can in principle be verified by a direct computation. That they are nontrivial braids can be obtained using the solution of the word problem inB_n given in Section 1.5.1, or the normal form of braids discussed in Section 6.5.4, or the prime handle reduction of Section 7.5. These computations, however, shed no light on the geometric reasons forcingρ, ρ to lie in the kernel. These reasons will be discussed in Section 3.2.

Exercise 3.1.1.Show that Kerψn ⊂ Bn is invariant under the involutive anti-automorphismh : Bn → Bn sending σi to itself for i = 1, . . . , n−1.

(Hint: Verify thatU_i^T =DUiD⁻¹, where i= 1, . . . , n−1 andD=Dn is the diagonaln×nmatrix with diagonal terms 1, t, t², . . . , tⁿ⁻¹. Deduce that

ψn(h(β)) =D⁻¹ψn(β)^TD for allβ ∈Bn.)