Braids and homeomorphisms

Braids and Braid Groups

1.6 Braids and homeomorphisms

1.6.2 Half-twists

LetM be an oriented surface (possibly with boundary) and letQbe a finite subset ofM^◦. By aspanning arc on (M, Q), we mean a subset ofM homeo- morphic toI= [0,1] and disjoint fromQ∪∂M except at its two endpoints, which should lie inQ. Let us stress that all arcs considered here are simple, i.e., have no self-intersections.

Letα⊂M be a spanning arc on (M, Q). Thehalf-twist τα: (M, Q)→(M, Q)

is obtained as the result of the isotopy of the identity map id : M → M rotatingαin M about its midpoint by the angleπin the direction provided by the orientation of M. The half-twist τα is the identity outside a small neighborhood ofα in M. Clearly, τα(α) = α, τα(Q) = Q, andτα induces a transposition on Q permuting the endpoints of α. Note that rotating α as above but in the opposite direction, we obtainτ_α⁻¹.

For completeness, we give a more formal definition ofτα. Let us identify a small neighborhoodU of αwith the open unit disk {z∈C| |z|<1} so that α= [−1/2,1/2] and the orientation inM corresponds to the counterclockwise orientation inC. The homeomorphismτ_α:M →M is the identity outsideU, sends anyz∈Cwith|z| ≤1/2 to−z, and sends anyz∈Cwith 1/2≤ |z|<1 to exp(−2πi|z|)z. Clearly,τα∈M(M, Q) does not depend on the choice ofU. The action ofταon a curve onM transversely meetingαin one point is shown in Figure 1.14.

α −→

Fig. 1.14.The action ofταon a transversal curve

We state a few properties of half-twists.

(i) If f : (M, Q) →(M, Q) is an orientation-preserving homeomorphism of two pairs as above andαis a spanning arc on (M, Q), thenf(α) is a spanning arc on (M, Q) andτ_f(α)=f ταf⁻¹∈M(M, Q).

This property is obvious. Informally speaking, it says that applying the construction of a half-twist on two copies of the same surface, we obtain two copies of the same homeomorphism.

(ii) If two spanning arcsα, αon (M, Q) are isotopic in the class of spanning arcs on (M, Q) (in particular they must have the same endpoints), then τα=τα inM(M, Q).

Indeed, if α, α are isotopic, then there is a self-homeomorphism f of (M, Q) that is the identity onQ, is isotopic to the identity, and sendsα ontoα. By (i),

τ_α =τ_f(α)=f τ_αf⁻¹. Sincef is isotopic to the identity,f ταf⁻¹=τα.

(iii) A self-homeomorphism of (M, Q) induces a self-homeomorphism ofM by forgetting Q. The resulting group homomorphism M(M, Q) →M(M) sendsτα to 1. This is clear from the definitions.

(iv) Ifα, βare disjoint spanning arcs on (M, Q), then

τατβ =τβτα∈M(M, Q). (1.12) This is obtained by using disjoint neighborhoods ofα, βin the construc- tion ofτα, τβ.

(v) For any two spanning arcsα, βon (M, Q) that share one common end- point and are disjoint otherwise,

τ_ατ_βτ_α=τ_βτ_ατ_β∈M(M, Q). (1.13) To prove this fundamental formula, we begin with the equality

τα(β) =τ_β⁻¹(α),

which can be verified by drawing the arcsτ_α(β) andτ_β⁻¹(α). The equality here is understood as isotopy in the class of spanning arcs on (M, Q).

By (ii),

τ_τα(β)=τ_τ−1 β (α).

By (i), this impliesτ_ατ_βτ_α⁻¹=τ_β⁻¹τ_ατ_β. This is equivalent to (1.13).

1.6.3 The isomorphismBn∼=M(D, Qn)

Forn ≥ 1, let Q_n ⊂R² be the n-point set {(1,0),(2,0), . . . ,(n,0)}. Let D be a closed Euclidean disk in R² containing the set Q_n in its interior. We orientD counterclockwise. For every i= 1,2, . . . , n−1, consider the arc

αi = [i, i+ 1]× {0} ⊂D .

This arc meetsQn only at its endpoints and hence gives rise to a half-twist τα_i ∈M(D, Qn).

Formulas (1.12) and (1.13) imply thatτα₁, . . . , τα_n−1satisfy the braid relations of Section 1.1. By Lemma 1.2, there is a group homomorphism

η:Bn→M(D, Qn) such thatη(σi) =ταi for alli= 1, . . . , n−1.

Recall the group of braid automorphisms Bn defined in Section 1.5.1.

We now define a certain group homomorphism ρ : M(D, Qn) → Bn. Pick a base pointd∈∂Das in Figure 1.15. It is clear that the fundamental group π₁(D−Q_n, d) is a free group F_n of rank n with generators x₁, x₂, . . . , x_n represented by the loops X1, X2, . . . , Xn shown in Figure 1.15. Every self- homeomorphismf of (D, Qn) can be restricted toD−Qn and yields in this way a self-homeomorphism ofD−Qn. The latter sendsd∈∂D to itself and induces a group automorphismρ(f) of Fn =π1(D−Qn, d). This automor- phism depends only on the isotopy class of f: if two self-homeomorphisms of (D, Qn) are isotopic, then their restrictions to D−Qnare isotopic relative to∂D, and therefore they induce the same automorphism ofFn.

d

X₁ X_i X_n

· · · · · ·

Fig. 1.15.The loops X1, . . . , Xn onD−Qn

Let us verify that ρ(f) is a braid automorphism of Fn. The loop Xk in Figure 1.15 can be deformed inD−Qn into a small loop encircling clockwise the point (k,0). The homeomorphismf maps the latter loop onto a small loop encircling clockwise the point (μ(k),0) for some μ(k) ∈ {1,2, . . . , n}. This small loop can be deformed into the loopX_μ(k) in D−Q_n. Hence, the loop f(X_k) can be deformed intoX_μ(k)inD−Q_n. (Under the deformation, the base pointf(d) =dmay move inD−Q_n.) This implies that the homotopy classes of these two loops ρ(f)(x_k) and x_μ(k) are conjugate in π₁(D−Q_n, d). This verifies Condition (i) in the definition of a braid automorphism. Condition (ii) follows from the fact that the productx1x2· · ·xnis represented by the loop∂D based atd. This loop is preserved byf pointwise, and therefore its homotopy class inπ1(D−Qn, d) is invariant underρ(f).

We conclude that the formulaf →ρ(f) defines a mapρfrom M(D, Qn) toBn. This map is a group homomorphism, since

ρ(f g) =ρ(f ◦g) =ρ(f)◦ρ(g) =ρ(f)ρ(g), for anyf, g∈M(D, Qn).

We can now state the main theorem relating braids to homeomorphisms.

Theorem 1.33.For any n ≥ 1, the homomorphisms η and ρ are isomor- phisms. The following diagram is commutative:

Bn η

β→β

$$I

II II II II I

M(D, Q_n) ^ρ //Bn

(1.14)

whereβ →β:B_n→B_n is the isomorphism defined in Section 1.5.

This fundamental theorem allows us to identifyB_nwith the mapping class groupM(D, Q_n). We have by now three different geometric interpretations ofB_n: via geometric braids onnstrings, via the configuration space ofnpoints in the plane, and via the group of homeomorphisms of a 2-disk withn dis- tinguished points. It is this variety of geometric facets ofBn that makes this group so appealing.

The commutativity of the diagram (1.14) means thatβ=ρ(η(β)) for any β∈Bn. This can be verified at once. Sinceρ, η, andβ→βare group homo- morphisms, it suffices to verify this equality for the generatorsσ1, . . . , σn−1. We need to check that ρ(ταi) = σi for i = 1,2, . . . , n−1. The formulas ρ(ταi)(xk) = xk for k = i, i+ 1 and ρ(ταi)(xi) = xi+1 follow directly from the definition ofταi. The equalityρ(ταi)(xi+1) =x⁻_i+1¹xixi+1 can be verified directly or deduced from the formulaρ(ταi)(x1· · ·xn) =x1· · ·xn. Hence, we haveρ(ταi) =σi. In view of the commutativity of the diagram (1.14) and The- orem 1.31, to prove Theorem 1.33 we need only show thatηis an isomorphism.

This will be done in Section 1.7.

It is clear that for all i = 1, . . . , n−1, the half-twist τ_αi : D → D is a diffeomorphism with respect to the standard smooth structure onD induced by that on R². Integral powers of diffeomorphisms and their products are also diffeomorphisms. Therefore the surjectivity of η implies the following assertion.

Corollary 1.34.An arbitrary self-homeomorphism of the pair(D, Qn)is iso- topic in the class of self-homeomorphisms of this pair to a diffeomorphism (D, Qn)→(D, Qn).

Exercise 1.6.1.LetM, Qbe as in Section 1.6.2.

(a) Consider an embeddedr-gonP ⊂M (withr≥3) meetingQprecisely in its vertices. Moving along∂P in the direction provided by the orientation of M, we meet consecutivelyr edges, say α1, α2, . . . , αr, of P. Each αi is a spanning arc on (M, Q). Prove that

τα₁τα₂· · ·τα_r−1 =τα₂τα₃· · ·τα_r.

(Hint: Forr= 3 rewrite asτ_α⁻₂¹τα1τα2 =τα3; forr≥4 use induction.)

(b) Consider r≥2 spanning arcs on (M, Q) with one common endpoint a∈Qand disjoint otherwise. Moving around ain the direction given by the orientation ofM, denote these arcs byα₁, α₂, . . . , α_r. Prove that

τ_α⁻¹

1τ_α2τ_α1 =τ_α2τ_α1τ_α⁻¹

2

commutes withτ_αi for 3≤i≤r. Deduce that τα1β τα2τα1 =τα2τα1β τα2

for any elementβ of the group generated byτα3, τα4, . . . , ταr.

Exercise 1.6.2.Prove that M(S¹) = {1}. (Hint: Composing an arbitrary self-homeomorphismf ofS¹ with a rotation of S¹ into itself, we can assume that f has a fixed point. Cutting out S¹ at a fixed point of f, we obtain a self-homeomorphism of a closed interval, which, as we know, is isotopic to the identity.)