Vassiliev invariants for braids on surfaces

arXiv:math/0006014v2 [math.GT] 9 Jun 2000 Vassiliev invariants for braids on surfaces Juan Gonz´alez-Meneses Luis Paris May, 2000 Abstract We show that Vassiliev invariants separate braids on a closed oriented surface, and we exhibit an universal Vassiliev invariant for these braids in terms of chord diagrams labeled by elements of the fundamental group of the considered surface. 1 Definitions and statements 1.1 Introduction Vassiliev knot invariants were introduced by V. A. Vassiliev ([V1], [V2]; see also [B2], [B-N1]), and they have been generalized to several other knot-like objects, such as links, braids, tangles, string links, knotted graphs, etc. The purpose of this paper is to consider Vassiliev invariants of braids on surfaces, and to extend some well known results of Vassiliev invariants of Artin braids to the case of braids on surfaces. Our study of Vassiliev invariants is inspired by Papadima’s work [P] on Vassiliev invariants for Artin braids with values in Z. However, the presence of the fundamental group of the surface varies substantially the analysis. Anyway, the Vassiliev theory for braids on surfaces, exposed in this paper, appears to be a natural generalization of the corresponding theory for Artin braids. Aknowledgement. We are grateful to S¸tefan Papadima for stimulating conversations and suggestions which were the starting point of this work. 1.2 Braids and singular braids on surfaces Throughout this paper Mwill denote a closed, orientable surface of genus g≥1, and P= {P1,...,Pn}a set of ndistinct points in M. Define a n-braid based at Pto be a collection b= (b1,...,bn) of disjoint smooth paths in M×[0,1], called strings of b, such that the i-th string biruns monotonically in t∈[0,1] from the point (Pi,0) to some point (Pj,1), Pj∈ P. An isotopy in this context is a deformation through braids (which fixes the ends). Multiplication of braids is defined by concatenation, generalizing the construction of the fundamental group. The isotopy classes of braids with this multiplication form the group Bn(M, P), called braid group with nstrings on Mbased at P. Note that the group Bn(M, P) does not depend, up to isomorphism, on the set Pof points but only on the cardinality n=|P|. So we may write Bn(M) in place of Bn(M, P). In the same way as Artin braid groups have been extended to singular braid monoids ([B2],[Ba]), one can extend the braid group Bn(M) to SBn(M), the monoid of singular braids with nstrings on M. The strings of a singular braid are now allowed to intersect transversely, but only in finitely many double points, called singular points. As with braids, isotopy is a deformation through singular braids (which fixes the ends), and multiplication is by concatenation. Note that the isotopy classes of singular braids form a monoid and not a group: the singular braids with one or more singular points being non-invertible. Keywords: Braid - Surface - Vassiliev Invariant - Finite Type Invariant. Mathematics Subject Classification: Primary: 20F36. Secondary: 57M27, 57N05. First author partially supported by DGESIC-PB97-0723 and by the european network TMR Sing. Eq. Diff. et Feuill. 1 1.3 Vassiliev invariants and Vassiliev filtration An invariant of braids on Mwith values in an abelian group A, is a set-mapping v:Bn(M)→A. Like for knots and Artin braids, one can extend vto singular braids by using the recursive rule v−v=v The picture on the left hand side represents a small neighborhood of a singular point in a singular braid. Those on the right hand side represent the braids which are obtained from the previous one by resolution of that singular point. That is, we modify the first braid inside the neighborhood of the singular point, in a positive and a negative way, to obtain two singular braids having one less singular point. Remark that this is well defined since Mis orientable. In a non-orientable case, one also has two different modifications, but it would not be possible to differentiate the positive one from the negative one. Let dbe an integer. A Vassiliev invariant of type dis an invariant vsuch that v(b) = 0 for every singular braid bwith more than dsingular points. There is an equivalent definition of a Vassiliev invariant, in terms of the so-called Vassiliev filtration. First, consider the group ring Z[Bn(M)]. We can define a map η:SBn(M)−→ Z[Bn(M)] which “resolves” all the singular points of a given braid, with the corresponding signs. That is, − η This map is a well defined multiplicative morphism. Remark that a singular braid with d singular points is mapped to an alternate sum of 2dnon-singular braids, each one having coefficient +1 or −1 depending on the sign of its corresponding resolutions. Let SdBn(M) denote the set of isotopy classes of singular braids with dsingular points. We denote by Vdthe Z-module generated by η(SdBn(M)). One can easily verify that Vdis a (twosided) ideal of Z[Bn(M)] and that we have the inclusions Vd+1 ⊂Vdand Vd1Vd2⊂Vd1+d2, for all d1, d2, d ∈N. We have then obtained a filtration Z[Bn(M)] = V0⊃V1⊃V2⊃ · · · , which is called the Vassiliev filtration of Z[Bn(M)]. The definition of a Vassiliev invariant in terms of the Vassiliev filtration is as follows. One can extend any invariant v:Bn(M)→Aby linearity to a morphism of Z-modules v:Z[Bn(M)] →A. Note that the previous extension of vto singular braids can also be expressed by v(b) = v(η(b)), for b∈SBn(M). Then, vis a Vassiliev invariant of type dif and only if it vanishes on Vd+1. Therefore, the set of Vassiliev invariants of type dwith values in Ais equal to HomZ(Z[Bn(M)]/Vd+1 , A). 1.4 Statements We have two goals in this paper. The first one is to show that Vassiliev invariants separate braids on surfaces, that is, to prove the following. 2 Theorem 1.1. Given two non-equivalent braids band con M, there exists an integer N≥1and a Vassiliev invariant vNof type Nsuch that vN(b)6=vN(c). Moreover, vNcan be chosen to take values in Z. This result is known to hold for Artin braids ([B-N2], [K], [P]), but it is still a conjecture for knots. Actually, Theorem 1.1 is a corollary of the following theorem. Theorem 1.2. Let {Vd}∞ d=1 be the Vassiliev filtration of Z[Bn(M)]. One has: 1. T∞ d=0 Vd={0}, 2. Vd/Vd+1 is a free Z-module for all d≥0. Indeed, if Theorem 1.2 holds, then given two non-equivalent braids b, c ∈Bn(M), there exists an integer Nsuch that b−c /∈VN+1. Then we can take vNto be the canonical projection from Z[Bn(M)] to Z[Bn(M)]/VN+1. In addition, if Vd/Vd+1 is a free Z-module for all d, then Z[Bn(M)]/VN+1 ≃(Z[Bn(M)]/V1)⊕(V1/V2)⊕ · · · ⊕ (VN/VN+1) is also a free Z-module, so we can obviously compose the above projection with a map from Z[Bn(M)]/VN+1 to Z, in such a way that the image of b−cis non-zero. Therefore, our first goal will be achieved by proving Theorem 1.2. Our second goal is to define a universal Vassiliev invariant for Bn(M) which generalizes the notion of chord diagrams for Artin braids. Recall that a chord diagram is a diagram made of n vertical lines and of a finite number of horizontal segments, called chords, connecting the lines. A M-labeled chord diagram is a chord diagram such that each chord is labeled by an element of π1(M) (see Figure 1). Note that the set of M-labeled chord diagrams is equipped with a multiplication defined by concatenation. The free Z-module generated by the chord diagrams is a Z-algebra which can be identified with Z[ti,j,γ ], the free non-commutative Z-algebra freely generated by the ti,j,γ , where i, j ∈ {1,...,n},i6=j,γ∈π1(M), and where ti,j,γ =tj,i,γ−1(see Figure 1). γ i j γ ti,j,γ α β Figure 1: A M-labeled chord diagram and the generator ti,j,γ . We denote by Anthe quotient Z-algebra obtained from Z[ti,j,γ ] by imposing the relations •[ti,j,γ , tk,l,δ] = 0,for all distinct i, j, k, l ∈ {1,...,n}and all γ, δ ∈π1(M), •[ti,j,γ , tj,k,δ +ti,k,(γδ)] = 0, for all distinct i, j, k ∈ {1,...,n}and all γ, δ ∈π1(M), and we denote by b Anits natural completion. Note that the symmetric group Σnacts on π1(M)nby permuting coordinates, so we can consider the induced semi-direct product Hn=π1(M)n⋊Σn. In addition, it is straightforward to show that Hnacts on b An, defining the semi-direct product b An⋊ Z[Hn]. The action is defined by the following relations. •σ ti,j,γ σ−1=tσ(i),σ(j),γ , for all σ∈Σn, •µ(k)ti,j,γ µ(k)−1=ti,j,γ , for all µ∈π1(M) and all k6=i, j, 3 •µ(i)ti,j,γ µ(i)−1=ti,j,(µγ), for all µ∈π1(M), where, µ(i) = (1,...,1,(i) µ,1,...,1) ∈π1(M)n. Note that one also has the following relation: µ(j)ti,j,γ µ(j)−1=µ(j)tj,i,γ−1µ(j)−1=tj,i,(µγ−1)=ti,j,(γµ−1). The Z-algebra b An⋊Z[Hn] carries the filtration induced by that of b An, so its associated graded algebra is An⋊ Z[Hn]. We also have grVZ[Bn(M)] = L∞ d=0(Vd/Vd+1). Our second main result will be: Theorem 1.3. There exists a homomorphism of Z-modules u:Z[Bn(M)] →b An⋊ Z[Hn]such that the corresponding graded map gru: grVZ[Bn(M)] −→ An⋊ Z[Hn] is an isomorphism of graded Z-algebras. We end this section by showing why uis called a universal Vassiliev invariant for Bn(M). Corollary 1.4. Every Vassiliev invariant of Bn(M)factors through uin a unique way. Proof: By Theorem 1.2, we know that Z[Bn(M)]/VN+1 is a free Z-module for all N≥0, hence, Z[Bn(M)] ≃(Z[Bn(M)]/VN+1)⊕VN+1. Recall that grVZ[Bn(M)] = ∞ M d=0 (Vd/Vd+1)≃(Z[Bn(M)]/VN+1)⊕ M d>N (Vd/Vd+1)!. Now, since gruis an isomorphism, we conclude that, for all N≥0, A(≤N) n⋊ Z[Hn] is also a free Z-module. Therefore, one has: b An⋊ Z[Hn]≃(A(≤N) n⋊ Z[Hn]) ⊕(b A(>N) n⋊ Z[Hn]), and An⋊ Z[Hn]≃(A(≤N) n⋊ Z[Hn]) ⊕(A(>N) n⋊ Z[Hn]). Every Vassiliev invariant v∈HomZ(Z[Bn(M)]/VN+1 , A) can then be seen as a linear map from grVZ[Bn(M)] to Awhich vanishes on Ld>N (Vd/Vd+1). Via gru, this means that vis a linear map from An⋊ Z[Hn] to A, which vanishes on A(>N) n⋊Z[Hn]. Therefore, if vis a Vassiliev invariant of type N, it can be lifted in a unique way to a linear map bv:b An⋊ Z[Hn]→A, which verifies v=bv◦u. 2 Vassiliev invariants separate braids Our strategy for proving Theorem 1.2 is the following. In a first subsection, we introduce some ideal Jof Z[Bn(M)] given by its generators and we prove that Vd=Jdfor all d≥0. In a second subsection, we consider an exact sequence 1 →Kn→Bn(M)→Hn→1, and we prove that Jd is equal in some sense to I(Kn)d⊗Z[Hn], where I(Kn) denotes the augmentation ideal of Kn. In a third subsection, we prove that Kncan be expressed as an iterated semi-direct product of free groups (of infinite rank). Finally, in the fourth subsection, we use the results of the previous ones to prove Theorem 1.2. 4 2.1 The Vassiliev filtration coincides with the J-adic filtration The aim of this subsection is to introduce an ideal Jof Z[Bn(M)] defined by its generators, and to show that the Vassiliev filtration coincides with the J-adic filtration (i.e. Vd=Jdfor all d∈N). We begin by explaining our “visualization” of (singular) braids, and by exhibiting generators for SBn(M). We represent the surface Mas a polygon of 4gsides which are identified in the way of Figure 2. α1 α2g α2g−1 α1 α2 α2g−1 α2g α2 Figure 2: A representation of the surface M. We draw braids over Min this polygon, as if we looked at the cylinder M×[0,1] from above, that is, we project the braid over M× {0}. Like for the planar representations of knots, we see over and under-crossings, and we can always move our braid via a convenient isotopy to avoid triple crossing points in the projection. See Figure 3 for an example. L I Figure 3: A braid with 3 strings on a surface of genus 2: two different viewpoints. Now, for every i∈ {1, . . . , n}and every r∈ {1,...,2g}, we define the braid ai,r as follows. All the strings of ai,r are trivial except the i-th one which goes through the r-th wall in the way of Figure 4. It goes upwards if ris odd and it goes downwards if ris even. We also define, for all j= 1,...,n−1, the braid σjas follows. All the strings of σjare trivial except the j-th one and the (j+ 1)-th one. The j-th string goes from (Pj,0) to (Pj+1,1) and the (j+ 1)-th string goes from (Pj+1,0) to (Pj,1) according to Figure 4. Note that σ1,...,σn−1are the classical generators of the braid group Bnof the disc. It is easy to show that {ai,r;i= 1,...,n, r = 1,...,2g} ∪ {σ1,...,σn−1}is a generator set for Bn(M). Actually, there is no need to include ai,r if i≥2, but it is better for our purposes. One can find in [G-M] a presentation for Bn(M) which involves these generators. For every i= 1,...,n−1, we define the singular braid τi∈S1Bn(M) as in Figure 5. This singular braid has a unique singular point on which intersect the i-th string and the (i+ 1)-th string. The i-th string goes from (Pi,0) to (Pi+1,1) and the (i+ 1)-th string goes from (Pi+1,0) to (Pi,1). The other strings are trivial. By a suitable isotopy, any singular braid b∈SkBn(M) can be written in the form b=c1τj1c2τj2···ckτjkck+1, 5 α2k+1 Pi P1Pn α2k+1 ai,2k+1 ai,2k P1Pn Pi α2k α2k Pj+1 Pn P1 σj Pj Figure 4: Generators for BnM. Pn τi P1PiPi+1 Figure 5: The singular braid τi. where ci∈Bn(M). So the following set generates SBn(M) (as a monoid). {a±1 i,r ;i= 1,...,n, r= 1,...,2g} ∪ {σ±1 1,...,σ±1 n−1} ∪ {τ1, . . . , τn−1}. Now, the morphism η:SBn(M)→Z[Bn(M)] sends σ±1 ito σ±1 i,a±1 i,r to a±1 i,r and τito σi−σ−1 i. Recall that Vddenotes the Z-module of Z[Bn(M)] generated by η(SdBn(M)). From the above considerations, it immediately follows: Proposition 2.1. Let Jbe the two-sided ideal of Z[Bn(M)] generated by {σi−σ−1 i;i= 1,...,n−1}. Then Vd=Jdfor all d∈N. 2.2 From Jdto I(Kn)d Recall that Hndenotes the semi-direct product π1(M)n⋊Σn. We define a homomorphism ϕ: Bn(M)→Hnas follows. We fix a disc Dembedded in Mwhich contains Pand, for all i, j ∈ {1,...,n}, a path αi,j in Dgoing from Pito Pj. Pick a braid b= (b1,...,bn), bi: [0,1] →M×[0,1], based at P. Let s∈Σnbe the permutation induced by b. Let bi: [0,1] →Mbe the projection of bion the first coordinate, and let µibe the loop based at Pidefined by µi=biαs(i),i. Then we set ϕ(b) = (µ1,...,µn)s∈π1(M)n⋊Σn=Hn. One can easily verify that ϕ:Bn(M)→Hnis a well defined homomorphism, and that its definition depends on the choice of Dbut not on the choice of the paths αi,j. Let Kndenote the kernel of ϕ. It is a classical matter that a set-section σ:Hn→Bn(M) of ϕdetermines a Z-isomorphism Φ : Z[Bn(M)] →Z[Kn]⊗Z[Hn] defined by Φ(b) = b(σ◦ϕ)(b)−1⊗ϕ(b). 6 Let us fix such a set-section. Recall that the augmentation ideal of a group Gis defined to be the two-sided ideal I(G) of Z[G] generated by the set {1−g;g∈G}. In this subsection, we prove the following. Proposition 2.2. The isomorphism Φ : Z[Bn(M)] →Z[Kn]⊗Z[Hn]sends isomorphically Jdto I(Kn)d⊗Z[Hn]for all d∈N. Note that Proposition 2.2 implies that, in order to prove Theorem 1.2, it will suffice to prove the following two conditions. 1. T∞ d=0 I(Kn)d={0}, 2. I(Kn)d/I(Kn)d+1 is a free Z-module for all d≥1. To prove Proposition 2.2, we will make use of some classical exact sequences involving braid groups (see [B1]). The first one comes from the homomorphism πwhich maps a given braid to the permutation that it induces on P. The kernel of this (clearly well defined) homomorphism is a subgroup of Bn(M) denoted by PBn(M), whose elements are called pure braids. Then one has: 1−→ P Bn(M)−→ Bn(M)π −→ Σn−→ 1. On the other hand, there is a homomorphism :PBn(M)→PBn−1(M) which sends (b1,...,bn) to (b2,...,bn). If we set Pn−1={P2,...,Pn}, then the kernel of can be seen as the group π1(M\Pn−1). This gives: 1−→ π1(M\Pn−1)−→ PBn(M) −→ PBn−1(M)−→ 1. Finally, if bis a pure braid, the projection of each string bi(i∈ {1, . . . , n}) over M, denoted by bi, is a loop in Mbased at Pi, which determines an element µi∈π1(M). This gives a homomorphism θ:PBn(M)→π1(M)n, which sends (b1,...,bn) to (µ1, . . . , µn). One can easily verify that Kn= ker θ, and that the exact sequence 1−→ Kn−→ PBn(M)θ −→ π1(M)n−→ 1 extends to the exact sequence 1−→ Kn−→ Bn(M)ϕ −→ Hn−→ 1. Moreover, Knis the normal closure in P Bn(M) of the subgroup P Bn(D), where Dis a disc in M which contains P(see [B1]). In what follows, we write I=I(Kn) and we consider Z[Kn] as a subring of Z[Bn(M)]. The next lemma is a preliminary result to the proof of Proposition 2.2. Lemma 2.3. Let B=Z[Bn(M)]. For every d≥1, one has Jd=B IdB=BId=IdB. Proof: Since Knis a normal subgroup of Bn(M), it is straightforward to prove that B IdB= BId=IdB. So, it suffices to prove that J=B I B. The inclusion J⊂B I B is obvious, once we notice that σ2 i∈Knand that σi−σ−1 i= σ−1 i(σ2 i−1) ∈B I B. For the other inclusion, we must prove that for all p∈Knone has p−1∈J. Suppose that p=p1p2, with p1, p2∈Kn; then p−1 = p1(p2−1)+(p1−1), so it suffices to show it for a set of generators of Kn. As we said before, Knis the normal closure of P Bn(D) in PBn(M), 7 so a set of generators of Knconsists on the elements of the form α b α−1, where α∈PBn(M) and b∈PBn(D). Take an element α b α−1as above. One has: α b α−1−1 = α(b−1) α−1, so we only have to show that b−1∈Jfor b∈P Bn(D). It is known ([St], Lemma 1.2) that b−1 belongs to the ideal of Z[Bn(D)] generated by {σi−σ−1 i;i= 1,...,n−1}. But the extension of this ideal to Z[Bn(M)] ⊃Z[Bn(D)] is precisely J, so b−1∈J. Proof of Proposition 2.2. First, we show that Φ(Jd)⊂Id⊗Z[Hn] for all d≥1. By Lemma 2.3, we know that Jd=IdB, thus Jdis generated as a Z-module by the elements of the form (k1−1) ···(kd−1)b, where b∈Bn(M) and ki∈Knfor i= 1,...,d. Now, the image of such an element by Φ is (k1−1) ···(kd−1) b′⊗ϕ(b), where b′=b(σ◦ϕ)(b)−1, which clearly belongs to Id⊗Z[Hn]. The inclusion Id⊗Z[Hn]⊂Φ(Jd) follows from the facts that Id⊗Z[Hn] is generated as a Z-module by the elements of the form (k1−1) ···(kd−1)k⊗β, where k1,...,kd, k ∈Knand β∈Hn, and that such an element is the image by Φ of (k1−1) ···(kd−1)k σ(β)∈IdB=Jd. 2.3 The structure of Kn The goal of this subsection is to prove the following. Proposition 2.4. For n≥2, there exists a free group Fnsuch that Kn=Fn⋊Kn−1. Moreover, the action of Kn−1on the abelianization of Fnis trivial. Remarks. (i) The notation Fnmay lead to some confusion; indeed, here Fnis not a free group of rank n. It is actually of infinite rank. (ii) A direct consequence of Proposition 2.4 is that Kncan be expressed as an iterated semi-direct product of (infinitely generated) free groups Kn=Fn⋊(Fn−1⋊(···⋊(F3⋊F2)···)). Recall the exact sequences defined in the previous subsection. Since Knis a subgroup of PBn(M), we can consider the image by of Kn. By definition, it is equal to Kn−1. If we denote Fn= ker ∩Kn, we obtain the following commutative diagram, where all rows and columns are exact: 1 1 1 ↑ ↑ ↑ 1→π1(M, P1)→π1(M)n→π1(M)n−1→1 ↑ ↑ θ↑θ 1→π1(M\Pn−1)→PBn(M) →PBn−1(M)→1 ↑ ↑ ↑ 1→Fn→Kn  →Kn−1→1 ↑ ↑ ↑ 1 1 1 Notice that Fnis a free group, since it is a subgroup of π1(M\Pn−1), which is a free group. We are specially interested in the lowest row of the diagram. In particular, in order to show Proposition 2.4, we will show that there exists a homomorphism s:Kn−1→Knwhich is a section of , and that Kn−1acts trivially on the abelianization of Fn. We turn first to find a free set of generators for Fn. Let Ω = {ω1,...,ω2g}be a set of 2gletters. It is well known that a presentation for π1(M) is as follows. π1(M) = Ω; (ω1ω2···ω2gω−1 1ω−1 2···ω−1 2g) = 1. 8 For every element γ∈π1(M) we choose a unique word eγover Ω ∪Ω−1which represents γ. We call this word the normal form of γ. Normal forms are chosen in such a way that they are prefix-closed (namely, if ω1ω2is a normal form, then ω1is also a normal form). For every word ω over Ω ∪Ω−1, we will denote by ω(i)the word over {a±1 i,1,...,a±1 i,2g}obtained from ωby replacing ω±1 jby a±1 i,j , for all j= 1, . . . , 2g. Let us consider, for 1 ≤i < j ≤n, the braid Ti,j drawn in Figure 6. All its strings are trivial except the i-th one which goes around the points Pi+1,...,Pjand turns back to Pi. Ti,j P1Pn PiPj Figure 6: The path (or braid) Ti,j. Notice that in π1(M\Pn−1), viewed as a subgroup of P Bn(M), one has T1,n =a1,1···a1,2ga−1 1,1···a−1 1,2g. Lemma 2.5. The following set is a free system of generators for Fn. B={eγ(1) T1,j eγ−1 (1) ; 2 ≤j≤nand γ∈π1(M)}. Proof: Consider the Cayley graph of π1(M), which is defined as follows. Its vertices are the elements of π1(M), and its edges are labeled by Ω. For every vertex γ∈π1(M) and for every i∈1,...,2g, there is exactly one edge labeled by ωi, with source γand target γωi. In this graph, the normal form of an element γ∈π1(M) corresponds to a unique path going from 1 to γ. By the prefix-closed condition mentioned above, the set of normal forms of π1(M) defines a maximal tree Tof the Cayley graph. The Cayley graph of π1(M) can be seen as the one-skeleton of a tiling of the plane. For every vertex γ, the path which starts at γand which is labeled by ω1...ω2gω−1 1. . . ω−1 2gbounds a fundamental region Rγof this tiling, and all fundamental regions are obtained in this way. Hence, there is a one to one correspondence between the vertices of the Cayley graph and its fundamental regions. Therefore, the fundamental group of the Cayley graph of π1(M) is the free group with free system of generators {eγ(ω1. . . ω2gω−1 1. . . ω−1 2g)eγ−1;γ∈π1(M, P1)}. We now define a graph Γ as follows. Take the Cayley graph of π1(M) and replace the labels ωiby a1,i. Then, for every vertex γ, add n−2 edges with source and target γ, labeled by T1,2,...,T1,n−1, respectively. Notice that the fundamental group of Γ is the free group with free system of generators B, where T1,n =a1,1···a1,2ga−1 1,1···a−1 1,2g. Recall the exact sequence 1−→ Fn−→ π1(M\Pn−1)θ −→ π1(M, P1)−→ 1. 9 Proof: This lemma follows from the well known congruences [ti,j, tk,l]≡0 (mod (PBn(D))3), [ti,j, ti,k]≡[tj,k, ti,j] (mod (PBn(D))3), (see, for example, [B-N3]), together with the inclusion (PBn(D))3⊂(Kn)3. Lemma 3.7. There is a well defined Lie algebra homomorphism ψn:Ln−→ gr(Kn)which sends ti,j,γ to fi,j,γ for all i, j ∈ {1,...,n},i6=j, and all γ∈π1(M). Proof: We have to show that the following congruences hold: (R1) fi,j,γ ≡fj,i,γ−1(mod (Kn)2), for all i, j ∈ {1,...,n},i6=j, and all γ∈π1(M); (R2) [fi,j,γ , fk,l,δ]≡0 (mod (Kn)3), for all distinct i, j, k, l ∈ {1,...,n} and all γ, δ ∈π1(M); (R3) [fi,j,γ , fj,k,δ]≡[fi,k,(γδ), fi,j,γ ] (mod (Kn)3), for all distinct i, j, k ∈ {1,...,n} and all γ, δ ∈π1(M). Notice that (R1) follows from Lemma 3.3. So, it remains to prove (R2) and (R3). We argue by induction on n. The conditions (R2) and (R3) being empty if n= 2, we may assume that n > 2, that (R2) holds if i, j, k, l ∈ {2,...,n}(by induction), and that (R3) holds if i, j, k ∈ {2,...,n} (by induction). We turn now to prove (R2) for k= 1. By Lemma 3.4, there exists W1∈(Kn)2such that fi,j,γ e δ(1) f−1 i,j,γ =W1e δ(1). Also, by Lemma 3.5, there exists W2∈(Kn)2such that eγ−1 (i)t1,l eγ(i)=W2t1,l. Then fi,j,γ f1,l,δ f−1 i,j,γ =fi,j,γ e δ(1) f−1 i,j,γfi,j,γ t1,l f−1 i,j,γ fi,j,γ e δ−1 (1) f−1 i,j,γ =W1e δ(1) fi,j,γ t1,l f−1 i,j,γ e δ−1 (1)W−1 1 ≡e δ(1) fi,j,γ t1,l f−1 i,j,γ e δ−1 (1) (mod (Kn)3) =e δ(1) eγ(i)ti,j eγ−1 (i)t1,l eγ(i)t−1 i,j eγ−1 (i)e δ−1 (1) =e δ(1) eγ(i)ti,j W2t1,l t−1 i,j eγ−1 (i)e δ−1 (1) ≡e δ(1) eγ(i)W2ti,j t1,l t−1 i,j eγ−1 (i)e δ−1 (1) (mod (Kn)3) ≡e δ(1) eγ(i)W2t1,l eγ−1 (i)e δ−1 (1) (mod (Kn)3) (by Lemma 3.6) =e δ(1) t1,l e δ−1 (1) =f1,l,δ. Therefore, (R2) holds for k= 1. The congruence (R2) holds for either i= 1, or j= 1, or l= 1, because of the above case and of the relation (R1). We turn now to prove (R3) for i= 1. By Lemma 3.4, there exists W1∈(Kn)2such that fj,k,δ eγ(1) f−1 j,k,δ =W1eγ(1). 16 Also, by Lemma 3.3, there exists W2∈(Kn)2such that e δ−1 (j)t1,j e δ(j)=e δ(1) t1,j e δ−1 (1) W2. Then fj,k,δ f1,j,γ f−1 j,k,δ =fj,k,δ eγ(1) f−1 j,k,δfj,k,δ t1,j f−1 j,k,δfj,k,δ eγ−1 (1) f−1 j,k,δ =W1eγ(1) fj,k,δ t1,j f−1 j,k,δ eγ−1 (1)W−1 1 ≡eγ(1) fj,k,δ t1,j f−1 j,k,δ eγ−1 (1) (mod (Kn)3) =eγ(1) e δ(j)tj,k e δ−1 (j)t1,j e δ(j)t−1 j,k e δ−1 (j)eγ−1 (1) =eγ(1) e δ(j)tj,k e δ(1) t1,j e δ−1 (1) W2t−1 j,k e δ−1 (j)eγ−1 (1) ≡eγ(1) e δ(j)tj,k e δ(1) t1,j e δ−1 (1) t−1 j,kW2e δ−1 (j)eγ−1 (1) (mod (Kn)3) =eγ(1) e δ(j)htj,k ,e δ(1) t1,j e δ−1 (1)ie δ(1) t1,j e δ−1 (1)W2e δ−1 (j)eγ−1 (1) =eγ(1) e δ(j)htj,k ,e δ(1) t1,j e δ−1 (1)ie δ−1 (j)t1,j eγ−1 (1). Since e δ(1) commutes with tj,k, it follows that: fj,k,δ f1,j,γ f−1 j,k,δ ≡eγ(1) e δ(j)e δ(1) [tj,k , t1,j]e δ−1 (1) e δ−1 (j)t1,j eγ−1 (1) (mod (Kn)3) ≡eγ(1) e δ(j)e δ(1) [t1,j , t1,k]e δ−1 (1) e δ−1 (j)t1,j eγ−1 (1) (mod (Kn)3) (by Lemma 3.6) =eγ(1) [e δ(j),e δ(1)]e δ(1)e δ(j)[t1,j, t1,k]e δ−1 (j)e δ−1 (1)[e δ(j),e δ(1)]−1t1,j eγ−1 (1) . Notice that [e δ(j),e δ(1)]∈Knand [t1,j, t1,k]∈(Kn)2, thus fj,k,δ f1,j,γ f−1 j,k,δ ≡eγ(1) e δ(1)e δ(j)[t1,j, t1,k]e δ−1 (j)e δ−1 (1)t1,j eγ−1 (1) (mod (Kn)3) ≡eγ(1) ht1,j ,e δ(1) t1,k e δ−1 (1)it1,j eγ−1 (1) (mod (Kn)3) (by Lemma 3.3 and Lemma 3.5) =heγ(1) t1,j eγ−1 (1) ,eγ(1) e δ(1) t1,k e δ−1 (1) eγ−1 (1) ieγ(1) t1,j eγ−1 (1) . Let k=eγ(1) e δ(1) f γδ−1 (1). One has k∈Kn, thus fj,k,δ f1,j,γ f−1 j,k,δ ≡eγ(1) t1,j eγ−1 (1) , k g (γδ)(1) t1,k g (γδ)−1 (1) k−1eγ(1) t1,j eγ−1 (1)  (mod (Kn)3) =f1,j,γ , k f1,k,(γδ)k−1f1,j,γ ≡f1,j,γ , f1,k,(γδ)f1,j,γ (mod (Kn)3). This proves that (R3) holds for i= 1. For j= 1, (R3) holds because of the above case and of (R1), since one has: [fi,1,γ , f1,k,δ]≡[f1,i,γ−1, f1,k,δ]≡[fi,k,(γδ), f1,i,γ−1]≡[fi,k,(γδ), fi,1,γ ]. Finally, (R3) also holds for k= 1, since in this case: [fi,j,γ , fj,1,δ]≡[fj,i,γ−1, fj,1,δ]≡[fi,1,(γδ), fj,i,γ−1]≡[fi,1,(γδ), fi,j,γ ]. 17 Proof of Proposition 3.2: It suffices to prove that the homomorphism ψn:Ln→gr(Kn) of Lemma 3.7 is an isomorphism. We argue by induction on n. For n= 2, K2=F2is a free group freely generated by F1,2={f1,2,γ;γ∈π1(M)}, so gr(K2) is the free Lie algebra generated by F1,2. On the other hand, L2is by definition the free Lie algebra generated by {t1,2,γ;γ∈π1(M)}. Therefore, ψ2is a Lie algebra isomorphism. Suppose now that ψmis an isomorphism for m < n. Recall that Kn=Fn⋊Kn−1, and that we have the exact sequence 1−→ Fn−→ Kn  −→ Kn−1−→ 1 f1,j,γ 7−→ f1,j,γ 7−→ 1 fi+1,j+1,γ 7−→ fi,j,γ . Since Kn−1acts trivially on the abelianization of Fn, we can apply the result in [FR] which claims that the associated graded sequence of Lie algebras is exact, that is, 1−→ gr(Fn)i −→ gr(Kn)gr −→ gr(Kn−1)−→ 1, where iis the natural inclusion. Besides, since gr(Fm) is a free Lie algebra for all m≥2, the above sequence shows, by induction, that gr(Kn) is a free Z-module. This fact will be used later on. Let us now define the following Lie algebra homomorphism: en:Ln−→ Ln−1 t1,j,γ 7−→ 0 ti+1,j+1,γ 7−→ ti,j,γ . Looking at the relations (L1), (L2) and (L3), we see that enis a well defined epimorphism of Lie algebras. We will denote Qn= ker en. In this way, we obtain the following commutative diagram: 1−→ gr(Fn)i −→ gr(Kn)gr() −→ gr(Kn−1)−→ 1 ↑ηn↑ψn↑ψn−1 1−→ Qnei −→ Lnen −→ Ln−1−→ 1, where ηnis the restriction of ψnto Qn. Notice that t1,j,γ ∈Qnfor all j= 2,...,n and all γ∈π1(M). Notice as well that ηn(t1,j,γ ) = f1,j,γ, and that gr(Fn) is the free Lie algebra generated by F1,n. Therefore, if we show that Qn is generated (as a Lie algebra) by B1,n ={t1,j,γ;j= 2,...,n;γ∈π1(M)}, then Qnwill be the free Lie algebra generated by B1,n, and ηnwill be an isomorphism. In this case, since ψn−1is an isomorphism by the induction hypothesis, ψnwill also be a Lie algebra isomorphism, as we want to show. Let l=Pk i=1 libe an element of Qn, where each liis a Lie bracket over the generators of Ln. We can decompose l= (Pr i=1 li) + (Pk i=r+1 li), where {l1,...,lr}are the Lie brackets in which some t1,j,γ appears, and {lr+1,...,lk}are Lie brackets over {ti,j,γ; 2 ≤i < j ≤n, γ ∈π1(M)}. For all i= 1,...,r,li∈Qn, hence Pk i=r+1 li∈Qn. But if en(Pk i=r+1 li) = 0 in Ln−1, then Pk i=r+1 li= 0 in Ln, since the relations in Ln−1are the images by enof the same relations in Ln which involve no t1,j,γ . Therefore, l=Pr i=1 li, where each licontains some t1,j,γ . We must then show that each limay be written as a sum of brackets over B1,n. If liis a bracket of length 2, the result is a direct consequence of (L1), (L2) and (L3). Suppose that the result is true for brackets of length d−1, and consider li= [a, b], a bracket of length d > 2. We can suppose that acontains some t1,j,γ, and by induction, that it is a bracket over B1,n. If length(a)≥2, then a= [a1, a2], where a1,a2are brackets over B1,n. By the Jacoby identity, li= [[a1, a2], b] = −[[a2, b], a1]−[[b, a1], a2], 18 where [a2, b] and [b, a1] can be written, by the induction hypothesis, as a sum of brackets over B1,n, so the result follows. If length(a) = 1, then length(b)≥2, so b= [b1, b2]. Hence, [a, [b1, b2]] = [b1,[b2, a]] −[b2,[a, b1]], and we reduce to the previous case. Therefore, Qnis generated by B1,n, and hence ψnis a Lie algebra isomorphism. 3.3 The isomorphism χ2:Ugr(Kn)→grIZ[Kn] We start this subsection stating a result due to Quillen. Theorem 3.8. (Quillen [Q]). Let Gbe a group. Let I=I(G)be the augmentation ideal of G, let grIZ[G]be the graded ring associated with the I-adic filtration, and let G=G1⊃G2⊃ · · · ⊃ Gi⊃ ··· be the lower central series of G. Then the maps κi:Gi→Ii,g7→ g−1induce a surjective homomorphism κ:Ugr(G)→grIZ[G]of Z-algebras. Moreover, κ⊗Qis an isomorphism of Q-algebras. Notice that, if gr(G) is a free Z-module, then Ugr(G) is also a free Z-module. Thus: Corollary 3.9. If gr(G)is a free Z-module, then the maps κi:Gi→Ii,g7→ g−1induce an isomorphism κ:Ugr(G)→grIZ[G]of Z-algebras. Now, it is shown in the proof of Proposition 3.2 that gr(Kn) is a free Z-module. So: Proposition 3.10. There is a well defined isomorphism χ2:Ugr(Kn)→grIZ[Kn]which sends fi,j,γ to fi,j,γ −1, for all i, j ∈ {1,...,n},i6=j, and all γ∈π1(M). 3.4 grvis the inverse of χ=χ2◦χ1 We have shown in Subsections 3.2 and 3.3 that there is a well defined isomorphism χ=χ2◦χ1: An−→ grIZ[Kn] which sends ti,j,γ to fi,j,γ −1 for all i, j ∈ {1,...,n},i6=j, and all γ∈π1(M). We turn now to prove the following. Proposition 3.11. The homomorphism grvis the inverse of χ, hence it is an isomorphism of graded Z-algebras. Proof: We only need to prove that grvis the inverse of χas a homomorphism of Z-modules. For d≥1, let A(d) n=b A(≥d) n/b A(≥d+1) nbe the submodule of Anconsisting of the homogeneous polynomials of degree d. Consider as well i, j, k, l ∈ {1,...,n}, where i < j,k < l and i < k. By Relations (L1), (L2) and (L3), seen as relations in the enveloping algebra Anof Ln, one has: tk,l,δ ti,j,γ =   ti,j,γ tk,l,δ if i, j, k, l are all distinct. ti,j,γ tk,l,δ +ti,j,γ ti,l,(γδ)−ti,l,(γδ)ti,j,γ if j=k ti,j,γ tk,l,δ +ti,j,γ ti,k,(γδ−1)−ti,k,(γδ−1)ti,j,γ if j=l Therefore, a set of generators for A(d) nas a Z-module consists on the elements of the form R=ti1,j1,γ1ti2,j2,γ2···tid,jd,γd, where i1≤i2≤ · · · ≤ idand ik< jkfor all k= 1,...,d. But χ(R) = (fi1,j1,γ1−1)(fi2,j2,γ2−1) ···(fid,jd,γd−1), 19 so, by definition of grv, and since i1≤i2≤ · · · ≤ id, one has grv(χ(R)) = R. This is true for all d≥1, so it follows that grv◦χ= idAn. Hence, since χis an isomorphism, grvis its inverse, as we wanted to show. This result implies the following. Theorem 3.12. gruis an isomorphism of Z-modules. Proof: Recall that, by Proposition 2.2, the ideal Vd=Jdof Z[Bn(M)] is isomorphic to Id n⊗Z[Hn] via Φ, for all d≥0. Moreover, since Z[Hn] is a free Z-module, one has: Vd/Vd+1 ≃Id n/Id+1 n⊗Z[Hn]. Hence, grVZ[Bn(M)] ≃(grIZ[Kn]) ⊗Z[Hn] via grΦ. Now, gru= (grv⊗id) ◦grΦ and both grΦ and grv⊗id are isomorphisms of Z-modules, thus gruis an isomorphism of Z-modules. 3.5 gruis a homomorphism In this subsection, we finish the proof of Theorem 1.3 by showing that gruis a homomorphism. We start by defining an algebra structure on grIZ[Kn]⊗Z[Hn]. Consider the action of Bn(M) on Knby conjugation: an element b∈Bn(M) sends k∈Knto bkb−1∈Kn. This action extends naturally to Z[Kn] and preserves the I-adic filtration, so it defines an action of Bn(M) on grIZ[Kn]. This action restricted to Knbecomes trivial, since if k, k′∈Kn, k(k′−1)k−1=k k′k−1−1 = [k, k′]k′−1, so, in grIZ[Kn], k(k′−1)k−1≡([k, k′]−1)k′+ (k′−1) ≡(k′−1). Therefore, the action induced on grIZ[Kn] by an element b∈Bn(M) depends only on ϕ(b)∈Hn. Recall the set map section σ:Hn→Bn(M). Now, define the product in grIZ[Kn]⊗Z[Hn] by (k1⊗β1)(k2⊗β2) = (k1σ(β1)k2σ(β1)−1)⊗β1β2. By the above discussion, this product does not depend on σ, and it endows grIZ[Kn]⊗Z[Hn] with aZ-algebra structure. Now, in order to prove that gru= (grv⊗id) ◦grΦ is a homomorphism of graded Z-algebras, we turn to prove that both grΦ and (grv⊗id) are homomorphisms of graded Z-algebras. Lemma 3.13. grΦ : grVZ[Bn(M)] →grIZ[Kn]⊗Z[Hn]is a homomorphism of graded Z-algebras. Proof: Let b1, b2∈Bn(M). Write βi=ϕ(bi) and ki=bi(σ◦ϕ)(bi)−1for i= 1,2. Then grΦ(b1) grΦ(b2) = (k1⊗β1)(k2⊗β2) = (k1σ(β1)k2σ(β1)−1)⊗β1β2, grΦ(b1b2) = (k1σ(β1)k2σ(β2)σ(β1β2)−1)⊗β1β2. So, in order to prove that grΦ(b1b2) = grΦ(b1) grΦ(b2), it suffices to show that σ(β1)σ(β2)≡σ(β1β2) (mod V1). But ϕ(σ(β1)σ(β2)) = β1β2=ϕ(σ(β1β2)), thus there exists k∈Knsuch that σ(β1)σ(β2) = k σ(β1β2) with k∈Kn, hence, in Z[Bn(M)], σ(β1)σ(β2)−σ(β1β2) = (k−1) σ(β1β2)∈V1, since k−1∈V1, as we wanted to show. 20 Lemma 3.14. grv⊗id : grIZ[Kn]⊗Z[Hn]→ An⋊ Z[Hn]is a homomorphism of graded Z-algebras. Proof: Write g= grvand g′= grv⊗id = g⊗id, to simplify notation. Write as well β′ 1=σ(β1). We know that gis a Z-algebra isomorphism, so g′((k1⊗β1)(k2⊗β2)) = g′((k1β′ 1k2β′−1 1)⊗β1β2) =g(k1β′ 1k2β′−1 1)⊗β1β2 =g(k1)g(β′ 1k2β′−1 1)⊗β1β2. On the other hand: g′(k1⊗β1)g′(k2⊗β2) = (g(k1)⊗β1) (g(k2)⊗β2) =g(k1) (β1g(k2)β−1 1)⊗β1β2. Therefore, we need to show that, in An, g(σ(β1)k2σ(β1)−1) = β1g(k2)β−1 1. Since the action by conjugation does not depend on σ, we only need to verify the above formula when β1is a generator of Hn. In addition, since gis a homomorphism of Z-algebras, it suffices to verify it when k2is a generator of grIZ[Kn] as a Z-algebra, that is, when k2=fi,j,γ −1, i < j. Hence, it suffices to prove Lemma 3.15 below. Lemma 3.15. In grIZ[Kn]one has the following relations, for all i, j, k ∈ {1,...,n}and all γ∈π1(M). •σkfi,j,γ σ−1 k=fsk(i),sk(j),γ , where skis the transposition (k k + 1), •ak,r fi,j,γ a−1 k,r =fi,j,γ, if k6=i, j, •ai,r fi,j,γ a−1 i,r =fi,j,(ωrγ), where {σ1,...,σn−1}and {ai,r; 1 ≤i≤nand 1≤r≤2g}are the braids described in Subsection 2.1. Proof: The first equation is a consequence of the following relations in Bn(M), which are easily verified. σkai,r σ−1 k=           ai,r if k6=i−1, i. ai+1,r t−1 i,i+1 if k=iand ris even. ti,i+1 ai+1,r if k=iand ris odd. ti−1,i ai−1,r if k=i−1 and ris even. ai−1,r t−1 i−1,i if k=i−1 and ris odd. σkti,j σ−1 k=           ti−1,j if k=i−1. ti,i+1 ti+1,j t−1 i,i+1 if k=i. ti,j−1if k=j−1. t−1 i,j ti,j+1 ti,j if k=j. ti,j otherwise. The second equation comes from Lemma 3.5, and from the following relations, where i6=k and we have denoted bl,m =al,m if mis odd, and bl,m =a−1 l,m if mis even. bk,r bi,s b−1 k,r =           t−1 i,k bi,s if s < r and i < k. bi,s (b−1 i,r ti,k bi,r) if s > r and i < k. bi,s (b−1 i,r t−1 k,i bi,r) if s < r and i > k. tk,i bi,s if s > r and i > k. bi,s if s=r. 21 Indeed, in this case, bk,r fi,j,γ b−1 k,r ≡bk,r eγ(i)ti,j eγ−1 (i)b−1 k,r ≡eγ(i)bk,r ti,j b−1 k,r eγ−1 (i), and by Lemma 3.5, this is equivalent to fi,j,γ . Finally, the third equation is verified as follows. ai,r fi,j,γ a−1 i,r ≡ai,r eγ(i)ti,j eγ−1 (i)a−1 i,r ≡k^ (ωrγ)(i)ti,j ^ (ωrγ) −1 (i)k−1, where k∈Kn, so this is equivalent to fi,j,(ωrγ). References [Ba] J. C. BAEZ, Link invariants of finite type and perturbation theory, Lett. Math. Phys. 26 (1992), no. 1, 43-51. [B-N1] D. BAR-NATAN, On the Vassiliev knot invariants, Topology 34 (1995), no. 2, 423-472. [B-N2] D. 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