Homological Representations of the Braid Groups
3.7 Proof of Theorem 3.15
3.7.2 Surfaces associated with noodles For a noodle N on D, the set
3.7.2 Surfaces associated with noodles
The rest of the definition is quite standard. To define the intersection number F·v∈Zforv∈ H, we pick a 2-cycleV inC◦representingv. By the remarks above,V meetsF≈R2 along a compact subset, which necessarily lies inside a closed 2-disk inF. We can slightly deformV inC◦to make it transversal to this disk, keepingV disjoint from the rest ofF. The setF∩V is then discrete and compact. It is therefore finite, so that one can count its points with signs±determined by the orientation of C, F, and V. A standard argument from the theory of homological intersections shows that the resulting integer F·v =F·V depends only onv. Specifically, any two 2-cyclesV1, V2 in C◦ representingv differ by the boundary of a 3-chain inC◦; such a chain can be made transversal toFand then its intersection withFis a compact oriented 1-manifold. The fact that this 1-manifold has the same numbers of inputs and outputs implies thatF·V1=F·V2.
In analogy with formula (3.17), we set for anyv∈ H, F , v =
k,∈Z
(qktF·v)qkt. (3.30)
HereqktF is the image ofF under the covering transformationsqkt of the coveringC → C . Note that whenk, run overZ, the surfaceqktF runs over all possible lifts ofF to C. A priori, the sum on the right-hand side of (3.30) may be infinite; Lemma 3.25 below shows that it is finite.
The same computations as in the proof of Lemma 3.18 show that under a different choice of the liftF ofF, the expression F , v is multiplied by a monomial inq±1, t±1.
Lemma 3.25.Let r→r∗ be the involution of the ring R=Z[q±1, t±1]send- ing q to q and t to −t. Let N be a noodle on D and let α be an oriented spanning arc on(D, Q). Then for anyα-classv∈ H,
FN, v=−(q−1)2(qt+ 1)N, α∗, (3.31) whereN, α ∈R is the algebraic intersection defined in Section 3.6.2.
Proof. Note that the left-hand side of (3.31) is defined up to multiplication by monomials inq±1,t±1, while the right-hand side is defined up to multiplication by monomials inq±1, t±2. The equality is understood in the sense that the sides have a common representative. Then all representatives of the right-hand side represent also the left-hand side.
Pushing the endpoints ofN along∂D, we can deformN into a noodleN with starting pointd1and terminal pointd2, whered1, d2∈∂Dare the points used in the construction ofC. The surfacesFN andFN differ only in a subset of a cylinder neighborhood of∂CinC. We can choose the liftsFN andFN so that they differ only in a subset of a cylinder neighborhood of∂CinC. Sincev can be represented by a 2-cycle in the complement of such a neighborhood, FN, v=FN, v. It follows from the definitions thatN, α=N, α. Thus,
without loss of generality we can assume that the starting point of N is d1
and the terminal point ofN isd2.
It is enough to prove (3.31) for a specific choice ofF=FN ⊂C. Fix a lift
c∈Cofc={d1, d2} ∈ C. ForF, we take the lift ofF=FN containingc.
We need to specify a liftSα⊂Cof the surfaceSαdefined in Section 3.7.1.
To this end, fix pointsz−∈α−, z∈αand fix pathsθ−, θinΣ=D−Qhaving disjoint images and leading from d1 to z− and from d2 to z, respectively.
Consider the path{θ−, θ}inC leading fromc={d1, d2}to{z−, z}. LetΘbe the lift of this path toCstarting atc. The pathΘ terminates at a pointΘ(1) lying over{z−, z} ∈Sα. We choose forSα⊂Cthe lift ofSαcontainingΘ(1).
The surfacesSαandSαare oriented as in Section 3.7.1.
Assume thatN intersectsα(resp.α−) transversely inmpointsz1, . . . , zm
(resp.z1−, . . . , zm−) as in Section 3.6.2. ThenFintersectsSαtransversely in the points{z−i , zj}, wherei, j = 1, . . . , m. Therefore for anyk, ∈Z, the image ofFunder the covering transformationqktmeetsSαtransversely in at most m2points. Adding the corresponding intersection signs, we obtain an integer, denoted byqkt(F)·Sα∈Z. Set
σ=
k,∈Z
(qktF·Sα)qkt∈R .
The sum on the right-hand side is finite (it has at mostm2terms).
We compute σ as follows. Observe that for every pair i, j ∈ {1, . . . , m}, there are unique integerski,j, i,j∈Zsuch that qki,jti,jF intersects Sα at a point lying over{zi−, zj} ∈ C. Letεi,j=±1 be the corresponding intersection sign. Then
σ= m i=1
m j=1
εi,jqki,jti,j.
We now express the right-hand side in terms of the loopsξi,j and other data introduced in Section 3.6.2 (whered−=d1 andd=d2). We claim that
qki,jti,j =ϕ(ξi,j),
or in other words, thatki,j = w(ξi,j) and i,j = u(ξi,j) for all i, j. Indeed, we can lift ξi,j to a path Θβγ in Cbeginning at c, where Θ, β, γ are lifts of {θ−, θ},{βi−, βj},{γi,j−, γi,j}, respectively. By the choice of Sα, the point Θ(1) =β(0) lies on Sα. Then the pathβ lies entirely onSα. The pathΘβγ, being a lift of the loopξi,j, ends at
γ(1) =ϕ(ξi,j)(c)∈ϕ(ξi,j)FN.
Hence, the liftγ of{γi,j−, γi,j}lies onϕ(ξi,j)F and the pointγ(0) =β(1) lies over{zi−, zj}and belongs toϕ(ξi,j)F∩Sα. This proves our claim.
We now claim that for alli, j,
εi,j=−(−1)u(ξi,j)εiεj,
where εi (resp. εj) is the intersection sign of N and α at zi (resp. at zj).
Observe first thatεi,jis the intersection sign of the surfacesFN andSαat the point{zi−, zj} ∈ C. Let x− (resp.x) be a positive tangent vector ofN atzi− (resp. atzj). Lety− (resp.y) be a positive tangent vector ofα− atz−i (resp.
ofαatzj). Assume for concreteness that the pointzi−lies closer tod1alongN thanzj. Then the orientation of FN at the point {zi−, zj} is determined by the pair of vectors (x−, x). The orientation of Sα at {z−i , zj} is determined by the pair of vectors (y−, y). The distinguished orientation ofC at {zi−, zj} is equal toεiεj times the orientation of C determined by the following tuple of four tangent vectors:
(x−, y−, x, y). Then
εi,j=−εiεj =−(−1)u(ξi,j)εiεj,
since in the case at hand the pathsγi,j− andγi,j end atd1andd2, respectively, and the integeru(ξi,j) is even. The case in whichzj lies closer tod1alongN thanz−i is treated similarly.
To sum up, σ=
m i=1
m j=1
−(−1)u(ξi,j)εiεjqw(ξi,j)tu(ξi,j)=−N, α∗.
We can now prove (3.31). Let U1, . . . , Un and U be as in Section 3.7.1.
Choosing the disksU1, . . . , Un small enough, we can assume that they do not meetN. Then
qktF∩U =∅ (3.32)
for allk, ∈Z. Recall that theα-classv is represented by a sum of a 2-chain in U and a 2-chain (q−1)2(qt+ 1)S. By (3.32), the 2-chain in U does not contribute toF , v , so that we can safely replacev by (q−1)2(qt+ 1)S. By definition,S ⊂Sα is a subsurface ofSαsuch thatSα−S⊂U. Therefore, a similar argument shows that in the computation ofF , v , we can replace S bySα. Using the same computations as in the proof of Lemma 3.18, we obtain the equalities
F , v = (q−1)2(qt+ 1)
k,l∈Z
(qktlF·Sα)qktl
= (q−1)2(qt+ 1)σ
=−(q−1)2(qt+ 1)N, α∗.
Lemma 3.26.If a self-homeomorphism f of(D, Q)represents an element of the kernelKer (Bn →AutR(H)), then N, f(α)= N, α for any noodle N and any oriented spanning arcαon (D, Q).
Proof. As was already observed above, the homomorphismf∗:H → Htrans- forms anyα-classv∈ Hinto anf(α)-class. Formula (3.31) and the assumption f∗= id imply that
−(q−1)2(qt+ 1)N, f(α)∗=F , f∗(v)
=F , v
=−(q−1)2(qt+ 1)N, α∗.
Therefore,N, f(α)=N, α.