Homology classes associated with spanning arcs

Homological Representations of the Braid Groups

3.7 Proof of Theorem 3.15

3.7.1 Homology classes associated with spanning arcs

Lemma 3.22.We have (q−1)²(qt+ 1) [∂S] = 0in H1(U;Z).

Proof. Let (p1,0), (p2,0) be the endpoints ofα, wherep1, p2∈ {1,2, . . . , n}. For brevity, we shall denote the point (pi,0) simply bypi, wherei= 1,2. For i= 1,2, pick a pointui∈Upi lying in the strip betweenα⁻ andα. Consider the pointsA, A, B, B∈Σ and the eight paths

α1, α2, α3, β1, β2, β3, γ1, γ2

inΣdrawn in Figure 3.14. The pathsα1, α2, α3, β1, β2, β3are embedded arcs, while γ_i is a loop in U_pi encircling p_i and based at u_i for i = 1,2. It is understood that αgoes along a radius of U_p1 from p₁ to A, then alongα₂ from A to A, and then along a radius of Up₂ from A to p2 (the radii in question are not drawn in Figure 3.14). The arcα⁻goes along a radius ofUp₁

fromp1toB, then along the pathβ⁻₂¹inverse toβ2, and then along a radius ofUp₂ from B top2. One should think ofα2 (resp. ofβ2) as being long and almost entirely exhaustingα(resp.α⁻), while the radii ofUp₁, Up₂ and the arcsα1, β3⊂Up₁, α3, β1⊂Up₂ are short.

α1

α2

α3

γ₂ u2

β1

β2

β3

u₁ γ1

A

B

A B

θ1 p1 θ2 p2

Up₁ Up₂

Fig. 3.14.The arcsα1, α2, α3, β1, β2, β3, γ1, γ2

Consider the following loops inU based ate={u1, u2}={u2, u1} ∈U: a₁={γ₁, u₂}, a₂={u₁, γ₂},

b1={α1, β1β2β3}{α2α3, u1}, b2={α1α2α3, β1}{u2, β2β3}, whereu₁, u₂stand for the constant paths in the pointsu₁, u₂. Note that both loopsb1, b2 are homotopic inC to the loop

{α1α2α3, β1β2β3}.

(This certainly does not imply thatb₁, b₂ are homotopic in U.) The homo- topy classes of the loopsa1, a2, b1, b2 in the fundamental groupπ=π1(U, e) will be denoted by the same symbolsa1, a2, b1, b2. The symbol∼will denote homotopy inU for loops inU based ate. For anyx, y ∈π, set

x^y =y⁻¹xy∈π and [x, y] =x⁻¹x^y =x⁻¹y⁻¹xy∈π .

Observe the following relations inπ:

[a1, a2] = 1, [a1, b1a1b1] = 1, [a2, b2a2b2] = 1. (3.24) The first relation is obvious, since

a1a2∼ {γ1, γ2} ∼a2a1.

The relations [a1, b1a1b1] = 1 and [a2, b2a2b2] = 1 are proven similarly, and we shall prove only the first one. Consider the oriented arcs θ1, θ2 on ∂Up₁

as shown in Figure 3.14. These arcs lead from B to A and from A to B respectively, and their productθ1θ2 is a loop parametrizing∂Up₁. We claim that

b1a1b1∼ {u1, β1β2θ1α2α3}. (3.25) This will imply that

a₁b₁a₁b₁∼ {γ₁, u₂}{u₁, β₁β₂θ₁α₂α₃}

∼ {γ1, β1β2θ1α2α3}

∼ {u1, β1β2θ1α2α3}{γ1, u2}

∼b₁a₁b₁a₁.

Hence [a1, b1a1b1] = 1. We now prove (3.25). Observe first that b₁a₁∼ {α₁, β₁β₂β₃}{α₂α₃, γ₁}

and

b1∼ {u1, β1β2}{α1α2α3, β3}={β1β2, u1}{β3, α1α2α3}. Therefore,

b1a1b1∼ {α1, β1β2β3}{α2α3β1β2, γ1}{β3, α1α2α3}.

The pathα₂α₃β₁β₂is homotopic inΣ−γ₁toθ₂. (By a homotopy of paths we always mean a homotopy keeping the endpoints of the paths fixed.) Hence,

b1a1b1∼ {α1, β1β2β3}{θ2, γ1}{β3, α1α2α3}

∼ {α1, β1β2}{A, β3}{θ2, γ1}{B, α1}{β3, α2α3}

∼ {α1, β1β2}{θ2, β3γ1α1}{β3, α2α3}.

Observe that the pathβ3γ1α1 is homotopic inUp₁ to θ1. Therefore b₁a₁b₁∼ {α₁, β₁β₂}{θ₂, θ₁}{β₃, α₂α₃}.

Since the productα1θ2β3 is homotopic to the constant pathu1, we obtain b₁a₁b₁∼ {u₁, β₁β₂θ₁α₂α₃},

which proves (3.25).

We define the following four elements ofπ:

a=a⁻₂¹a1, b=b⁻₂¹b1, c1= [a1, b1], c2= [a2, b2]. Then

a^a1 =a , c^b₁^1a1c1= 1 =c^b₂^2a2c2, c2ba^b1 =ab^a1c1. (3.26) To see this, rewrite all four relations viaa1, a2, b1, b2. The first three relations are consequences of (3.24); in the last one, both sides are equal toa⁻₂¹b⁻₂¹a1b1. Pick a lifte∈Cofe={u1, u2}. The groupπ=π1(U , e) is the subgroup of π = π1(U, e) formed by the homotopy classes of loops ξ in U such that w(ξ) =u(ξ) = 0. We claim thata, b, c1, c2 ∈π. Indeed, fori= 1,2, we have w(ai) =w(γi) = 1 and

w(bi) =w(α1α2α3β1β2β3) = 0.

It follows from the definitions thatu(a1) =u(a2) = 0 andu(b1) =u(b2) = 1.

Hencew(a) =u(a) = 0 andw(b) =u(b) = 0, so thata, b∈π. The commutator of any two elements ofπbelongs toπ, so that c1, c2∈π.

The image of anyx∈π under the natural projectionπ→H1(U;Z) will be denoted by [x]. It is clear that if x∈πand y∈π, then x^y ∈π. We claim that for allx∈πand y∈π,

[x^y] =q^−w(y)t^−u(y)[x], (3.27) where we use theR-module structure onH1(U;Z). To see this, present x, y by loopsξ, η in U, based at e. Thenx^y∈πis represented by the loop η⁻¹ξη inU. This loop lifts to a pathμ1μ2μ3inU, where the pathμ1is the lift ofη⁻¹ beginning ateand ending at the point

e=q^w(y−1)t^u(y−1)e=q^−w(y)t^−u(y)e ,

the path μ₂ is the lift ofξ beginning ate, and μ₃ is the lift of η beginning at the terminal endpoint of μ₂. Since ξ represents x ∈ π, the path μ₂ is a loop beginning and ending ate. The pathμ3, being the lift of η beginning ate, must be the inverse ofμ1. Therefore the pathμ1μ2μ3 is a loop and its homology class in H1(U;Z) is equal to the homology class of μ2. The latter is equal toq^−w(y)t^−u(y)[x].

Applying (3.27), we obtain [a^a1] =q⁻¹[a] and

[c^b₁^1a1c1] =q⁻¹t⁻¹[c1] + [c1], [c^b₂^2a2c2] =q⁻¹t⁻¹[c2] + [c2], [c2ba^b1] = [c2] + [b] +t⁻¹[a], [ab^a1c1] = [a] +q⁻¹[b] + [c1]. Together with (3.26), this gives the following relations inH1(U;Z):

(q−1)[a] = 0, (qt+ 1)[c1] = 0 = (qt+ 1)[c2], (q⁻¹−1)[b] = (t⁻¹−1)[a] + [c2]−[c1].

Combining these relations, we obtain

(q−1)²(qt+ 1)[b] = 0. (3.28) To compute the homology class [S] ∈ H₂(C,U;Z), we need to choose the arcs s ⊂ α and s⁻ ⊂ α⁻ used in the definition of S. We take s = α2

ands⁻ =β2. The endpoints of these arcs and their complements inα⁻,αlie in Up₁ ∪Up₂ ⊂n

i=1Ui, as required. The circle∂S ⊂U is parametrized by a loop inU that is a lift of the following loopb⊂U based at{A, B}:

b={A, β₂}{α₂, B}{A, β₂}⁻¹{α₂, B}⁻¹.

We claim that b is homotopic to the following loop b in U also based at{A, B}:

b={A, β2β3}{α2α3, u1}{u2, β2β3}⁻¹{α2α3, B}⁻¹. (3.29) To see this, observe the obvious equalities of paths (up to homotopy inU)

{A, β3}{α2α3, u1}={α2α3, β3}={α2, B}{α3, β3}. Therefore

{α2α3, u1}={A, β3}⁻¹{α2, B}{α3, β3}. A similar argument shows that

{u₂, β₂β₃}⁻¹={α₃, β₃}⁻¹{A, β₂}⁻¹{α₃, B}. Substituting these expressions in (3.29) and observing that

{A, β2β3}={A, β2}{A, β3} and {α2α3, B}⁻¹={α3, B}⁻¹{α2, B}⁻¹, we conclude thatb is homotopic tob. Observe now that

b1={α1, β1β2β3}{α2α3, u1} ∼ {α1, β1}{A, β2β3}{α2α3, u1}, b₂={α₁α₂α₃, β₁}{u₂, β₂β₃} ∼ {α₁, β₁}{α₂α₃, B}{u₂, β₂β₃}. Therefore the loopb is homotopic to the loop

{α1, β1}⁻¹b1b⁻₂¹{α1, β1}

inU. The latter loop is freely homotopic inUtob1b⁻₂¹. Sinceb1b⁻₂¹is conjugate tob=b⁻₂¹b1 inπ, the loopsbandbare freely homotopic inU. We conclude that b is freely homotopic to b in U. Since b lifts to a loop ∂S in U, any homotopy ofb lifts to a homotopy of∂SinU. Hence,∂S is freely homotopic to a lift ofbtoU. Now the claim of the lemma directly follows from (3.28).

Lemma 3.22 and the exact homology sequence of the pair (C,U),

· · · →H2(U;Z)→ H →H2(C,U;Z)→H1(U;Z)→ · · ·,

imply that the homology class (q−1)²(qt+ 1)[S]∈H2(C,U;Z) is the image of a certainv∈ Hunder the inclusion homomorphismH →H2(C,U;Z). Any suchv∈ His called anα-class with respect to the disks U1, . . . , Unor, shorter, anα-class. Anα-class can be represented by a 2-cycle inCobtained by gluing the 2-chain (q−1)²(qt+1)Swith a 2-chain inUbounded by (q−1)²(qt+1)∂S.

It is clear that the α-class is determined by α only up to addition of elements of the image of the homomorphismH₂(U;Z) → Hinduced by the inclusionU →Cand up to multiplication by monomials inq, t (the latter is due to the indeterminacy in the choice ofSα). This describes completely the indeterminacy in the construction of anα-class. Indeed, it is easy to check that the set ofα-classes does not depend on the choice of the arcss⊂α−∂αand s⁻⊂α⁻−∂α⁻ used in the definition of the surfaceS. (To see this, observe that the surfacesSdetermined bys, s⁻ and by a pair of bigger arcs differ by an annulus inU.) We show now that the set ofα-classes is independent of the choice of the disksU1, . . . , Un.

Lemma 3.23.The set ofα-classes inHdoes not depend on the choice of the disksU₁, . . . , U_n.

Proof. Let{Ui}ⁿi=1 and{U_i}ⁿi=1be two systems of closed disk neighborhoods of the points ofQ={(1,0),(2,0), . . . ,(n,0)}inD^◦ as at the beginning of this subsection. LetU andUbe the subsets ofCassociated with these systems of disks as above. Suppose first thatU_i⊂Ui for alli. We can viewUiandU_ias concentric disks with center (i,0). By the assumptions, the arcαeither does not meet the disk Ui or meets it along a radius whose intersection with U_i is the radius of the latter. Contracting eachUi into U_i along the radii, we obtain an isotopy{Fs:D→D}s∈I ofDinto itself such thatF0= id,Fsfixes

∂D∪Qpointwise and fixesαsetwise for all s∈I, andF₁(U_i) =U_i for alli.

The induced homeomorphisms {F_s : C → C} s∈I form an isotopy of Cinto itself such thatF1(U) =U.

Observe now that any self-homeomorphismf of (D, Q) transformsαinto a spanning arcf(α) on (D, Q), and the orientation ofαinduces an orientation off(α) viaf. It is clear from the definitions that the induced homomorphism f_∗:H → Hsends the set ofα-classes with respect to the disks{Ui}ionto the set off(α)-classes with respect to the disks{f(Ui)}i. Applying this tof =F1

and observing that f(α) = α, f(Ui) = U_i for all i, and f_∗ = id (because f=F1 is isotopic—and hence homotopic—toF0= id), we conclude that the set of α-classes with respect to the disks {Ui}ⁿi=1 coincides with the set of α-classes with respect to the disks {U_i}ⁿi=1. The general case is obtained by transitivity using a third system of disks{U_i}ⁿi=1such thatU_i⊂Ui∩U_i for

alli= 1, . . . , n.

3.7.2 Surfaces associated with noodles