Homological representations

Homological Representations of the Braid Groups

3.2 Nonfaithfulness of the Burau representation

3.2.1 Homological representations

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.)

Recall theintersection formH×H→Z. This is a skew-symmetric bilinear form whose value [α]·[β]∈Zon the homology classes [α],[β]∈H represented by oriented loopsα, βonΣis the algebraic intersection number of these loops computed as follows. Deforming slightly αand β, we can assume that they meet transversely in a finite set of points that are not self-crossings ofαor ofβ. Then

[α]·[β] =

p∈α∩β

ε_p,

whereεp = +1 if the tangent vectors ofα, β at pform a positively oriented basis andεp =−1 otherwise. This sum does not depend on the choice of the loopsα,β in their homology classes and defines a bilinear formH×H →Z.

The identity [α]·[β] = −[β]·[α] shows that this intersection form is skew- symmetric. The action ofM(Σ) onH preserves the intersection form.

The homological representation has a more general “twisted” version, which comes up in the following setting. Suppose for concreteness that

∂Σ=∅and fix a base pointd∈∂Σ. Consider a surjective homomorphismϕ from π1(Σ, d) onto a group G. Let Σ → Σ be the covering corresponding to the kernel ofϕ. The group of covering transformations of Σ is identified withG. Pick an arbitrary pointd∈∂Σ lying overdand consider the relative homology groupH =H1(Σ, G d;Z), whereGdis theG-orbit ofd, i.e., the set of all points ofΣlying overd. The action ofGonΣinduces a left action ofG onH and turnsH into a left module over the group ringZ[G]. This module is free of rankn= rkH₁(Σ;Z). This follows from the fact thatΣ deformation retracts onto a union ofnsimple closed loops onΣmeeting only at their com- mon origind(here we crucially use the assumption ∂Σ =∅; cf. Figure 1.15, whereΣ is the complement ofn points in a disk). Let Aut(H) be the group ofZ[G]-linear automorphisms ofH. Clearly, Aut( H)∼= GLn(Z[G]).

Any self-homeomorphism f of Σ fixes the boundary∂Σ pointwise and, in particular, fixes d. It induces therefore an automorphism f# of the fun- damental group π1(Σ, d). Let Mϕ(Σ, d) be the group of isotopy classes of self-homeomorphismsf of Σ such that ϕ◦f# = ϕ. We construct a homo- morphism Mϕ(Σ, d) → Aut(H) called the twisted homological representa- tion ofMϕ(Σ, d). Every self-homeomorphismf ofΣrepresenting an element of Mϕ(Σ, d) lifts uniquely to a homeomorphism f: Σ → Σ fixing d. The equality ϕ◦f# = ϕ ensures that fcommutes with the action of G on Σ.

Therefore ffixes the set Gdpointwise: f(gd) = gf(d) = gdfor all g ∈ G.

Letf_∗be the automorphism ofH =H1(Σ, G d;Z) induced byf. Since fcom- mutes with the action ofG, this automorphism isZ[G]-linear. The mapf →f_∗ defines a group homomorphismMϕ(Σ, d)→Aut(H), which is the homolog- ical representation in question. The group H carries a natural intersection form preserved byMϕ(Σ, d) but we shall not need it.

3.2.2 The homomorphismΨn

We apply the general scheme of twisted homological representations to punc- tured disks. Fix n≥1. Let Qbe the set{(1,0),(2,0), . . . ,(n,0)} ⊂R² and letD be a closed Euclidean disk inR² containingQin its interior. We pro- videD with the counterclockwise orientation as in Figure 1.15. Observe that for any pointpin the interior ofD, the group

H1(D− {p};Z)∼=Z

is generated by the homology class of a small loop encirclingpcounterclock- wise. Each loopγinD− {p}representsktimes this generator, wherekis the winding number ofγaroundp. Set

Σ=D−Q

and fix a base point d ∈ ∂Σ = ∂D. Consider the group homomorphism ϕ fromπ1(Σ, d) to the infinite cyclic group{t^k}k∈Z sending the homotopy class of a loopγ to t^−w(γ), wherew(γ) is the total winding number of γ defined as the sum of its winding numbers around the points (1,0),(2,0), . . . ,(n,0).

The kernel ofϕdetermines an infinite cyclic coveringΣ→Σ. We identify its group of covering transformations with the infinite cyclic group{t^k}k∈Z. Pick a pointd∈∂Σ overdand set

H =H1 Σ,

k∈Z

t^kd;Z .

Observe that any self-homeomorphism ofD permuting the points of Q pre- serves the total winding number of loops inΣ. This is obvious for the small loops encircling the points of Q and holds for arbitrary loops, since their total winding numbers depend only on their homology classes in the group H1(Σ;Z)∼=Zⁿ, which is generated by the homology classes of the small loops.

Therefore the restriction toΣdefines a group homomorphism M(D, Q)→Mϕ(Σ, d).

(It is actually an isomorphism but we do not need this.) Composing this homo- morphism with the twisted homological representationMϕ(Σ, d)→Aut(H) defined in Section 3.2.1, we obtain a group homomorphism

Ψn :M(D, Q)→Aut(H).

The image off ∈M(D, Q) underΨn is the automorphism f_∗ of H induced by the liftf:Σ →Σ off|Σ:Σ→Σ fixingd.

In the next two subsections we show that KerΨn ={1} for n≥6. After that we show thatΨn is equivalent to the Burau representationψn for alln.

This will imply the nonfaithfulness of the latter forn≥6.

3.2.3 The kernel of Ψn

We give a construction of elements in KerΨnusing half-twists about spanning arcs as introduced in Section 1.6.2.

We say that two spanning arcsα, βon (D, Q) aretransversal if they have no common endpoints and meet transversely at a finite number of points.

For any transversal spanning arcs α, β on (D, Q), we define their algebraic intersection α, β ∈ Λ =Z[t, t⁻¹]. Consider the open arcs α∩Σ =α−∂α andβ ∩Σ =β−∂β onΣ =D−Q. Orient these arcs in an arbitrary way and pick arbitrary lifts α, β⊂Σ of α, β with induced orientations. Now we can set

α, β=

k∈Z

(t^kα·β) t^k ∈Λ , (3.5) wheret^kα·β∈Zis the algebraic intersection number of the oriented arcst^kα and β on Σ. Note that although the arcs t^kα and β are not compact, they have only a finite number of intersections, and moreover, the sum on the right- hand side of (3.5) is finite. This is so because the covering projectionΣ→Σ mapsβbijectively onto β and maps the set (

k∈Zt^kα) ∩βbijectively onto the finite setα∩β. This shows also that every point p∈α∩β ⊂Σ lifts to an intersection point oft^kα withβfor exactly onek=kp∈Z. Therefore,

α, β=

p∈α∩β

ε_pt^kp, (3.6)

where εp = ±1 is the intersection sign of αand β at p. As an exercise, the reader may verify that for anyp, q∈α∩β, the differencekp−kq is the total winding number of the loop inΣ going frompto qalongαand then fromq topalongβ. The expressionα, βis defined only up to multiplication by±1 and a power oftdepending on the choice of orientations onα, βand the choice of their lifts α, β. This will not be important for us, since we are interested only in whetherα, β= 0. Note that

β, α=

k∈Z

(t^kβ·α) t^k =

k∈Z

(β·t^−kα) t^k

=−

k∈Z

(t^−kα·β) t^k=−

k∈Z

(t^kα·β) t^−k

=−α, β,

where the overbar denotes the ring involution onΛ sending tto t⁻¹. Hence, α, β= 0⇒ β, α= 0.

As we know, every spanning arc α on (D, Q) gives rise to a half-twist τα: (D, Q)→(D, Q) acting as the identity outside a disk neighborhood ofα and mappingαonto itself via an orientation-reversing involution. Restricting ταtoΣ=D−Q, we obtain a self-homeomorphism ofΣ, denoted again byτα.

Lemma 3.4.Let α, β be transversal spanning arcs on (D, Q). If α, β= 0, thenΨn(τατβ) =Ψn(τβτα).

Proof. To prove the lemma we compute the homological action of the half- twists. As a warmup, we compute the action ofταonH =H1(Σ;Z). Consider the loopα on D drawn in Figure 3.1. This loop has a “figure-eight” shape and its only self-crossing lies onα. We orientαandα so that [α]·[α] =−2, where [α] ∈ H1(D, Q;Z) is the relative homology class of α and [α] ∈ H is the homology class ofα. The dot· denotes the bilinear intersection form H1(D, Q;Z)×H →Zdetermined by the counterclockwise orientation ofD.

The effect of the half-twistτ_α on an oriented curve transversal toαis to insert (α)^±1 at each crossing ofαwith this curve; see Figure 1.14. It is easy to check that for anyh∈H,

(τα)_∗(h) =h+ ([α]·h) [α].

α α

Fig. 3.1.The loopαassociated with a spanning arcα

The automorphism Ψn(τα) of H = H1(Σ,

k∈Z t^kd;Z) is defined by Ψ_n(τ_α) = (τ_α)_∗, where τ_α : Σ → Σ is the lift of τ_α : Σ → Σ fixing d.

Observe that the loopα onΣassociated toαhas zero total winding number and therefore lifts to a loopα on Σ. Consider an arbitrary oriented path γ in Σ with endpoints in

k∈Z t^kd. The effect of τ_α on γ is to insert a lift of (α)^±1at each crossing of γ with the preimage ofαinΣ. Thus ( τ_α)_∗ acts on the relative homology class [γ]∈H by

(τα)_∗([γ]) = [γ] +λγ[α],

where λγ ∈ Λ is a Laurent polynomial whose coefficients are the algebraic intersection numbers ofγ with lifts ofαtoΣ. Since α, β= 0, any lift of α has algebraic intersection number zero with any lift ofβ toΣ and hence with any lift β of β to Σ. Therefore, λ_β = 0 and (τα)_∗([β]) = [β]. Similarly, (τβ)_∗([γ]) = [γ] +μγ[β] for all γ as above and some μγ ∈ Λ. The equality β, α= 0 implies that (τβ)_∗([α]) = [α]. We conclude that for allγ,

(τατβ)_∗([γ]) = [γ] +λγ[α] +μγ[β] = (τβτα)_∗([γ]).

Therefore (τατβ)_∗= (τβτα)_∗.

To prove that KerΨn={1}, it remains to construct two spanning arcsα,β satisfying the conditions of Lemma 3.4 and such thatτατβ=τβταinM(D, Q).

For n = 6, such spanning arcs α, β are drawn in Figure 3.2. To check the equality α, β = 0, one applies (3.6) and the computations after it (this is left as an exercise for the reader). To prove thatτα and τβ do not commute in M(D, Q), one can use a brute-force computation using, for instance, the action of the mapping class group onπ1(Σ, d). We give a geometric argument in the next subsection.

α β

Fig. 3.2. Spanning arcsα, βforn= 6