New presentations of surface braid groups

arXiv:math/9910020v1 [math.GT] 5 Oct 1999 New presentations of surface braid groups Juan Gonz´alez-Meneses October, 1999 Abstract In this paper we give new presentations of the braid groups and the pure braid groups of a closed surface. We also give an algorithm to solve the word problem in these groups, using the given presentations. 1 Introduction Let Mbe a closed surface, not necessarily orientable, and let P={P1,...,Pn}be a set of ndistinct points of M. A geometric braid over Mbased at Pis an n-tuple Γ = (γ1,...,γn) of paths, γi: [0,1] −→ M, such that (1) γi(0) = Pifor all i= 1,...,n, (2) γi(1) ∈ P for all i= 1,...,n, (3) {γ1(t),...,γn(t)}are ndistinct points of Mfor all t∈[0,1]. For all i= 1,...,n, we will call γithe i-th string of Γ. Two geometric braids based at Pare said to be equivalent if there exists a homotopy which deforms one of them into the other, provided that at any time we always have a geometric braid based at P. We can naturally define the product of two braids as induced by the usual product of paths: for every i= 1,...,n, we compose the string of the first braid which ends at Pi, with the i-th string of the second braid. This product is clearly well defined, and it endows the set of equivalence classes of braids with a group structure. This group is called the braid group on nstrings over Mbased at P, and is denoted by Bn(M, P). This group does not depend, up to isomorphism, on the choice of P, but only on the number of strings, so we may write Bn(M) instead of Bn(M, P). A braid Γ = (γ1,...,γn) is said to be pure if γi(1) = Pifor all i= 1,...,n, that is, if all its strings are loops. The set of equivalence classes of pure braids forms a subgroup of Keywords: Braid - Surface - Presentation - Word Problem. Mathematics Subject Classification: Primary: 20F36. Secondary: 57N05. Partially supported by DGES-PB97-0723 and by the european network TMR Sing. Eq. Diff. et Feuill. 1 Bn(M, P) called pure braid group on nstrings over Mbased at P, and denoted P Bn(M, P). Again, we may write PBn(M) since it does not depend on the choice of P. Note that if n= 1, then B1(M) = PB1(M) = π1(M), the fundamental group of M. There exists an interpretation of braid groups as fundamental groups of some spaces, called configuration spaces. Let FnMdenote the space of n-tuples of distinct points of M, that is, FnM=Mn\∆, where ∆ = {(x1,...,xn)∈Mn/ xi=xjfor some i6=j}. It is clear that P Bn(M)≃π1(FnM). Now consider the symmetric group on nelements, Σn. This group acts naturally on FnMby permuting coordinates, so we can consider the configuration space: b FnM=FnM/Σn, which can be seen as the space of embeddings of npoints in M. We clearly have Bn(M)≃ π1(b FnM). This way to look at braids provides some useful exact sequences, derived from fibrations. The first one comes from the covering space map FnM−→ b FnM, with fiber Σn. It induces the following exact sequence: 1−→ P Bn(M)e −→ Bn(M)f −→ Σn−→ 1.(1) The homomorphism eis the natural inclusion, and fmaps a given braid to the permutation that it induces on P. Now we consider the Fadell-Neuwirth fibration ([FN]): given 1 ≤m < n, the map p:FnM−→ FmM (x1,...,xn)7−→ (xn−m+1,...,xn) is a locally trivial fibration with fiber Fn−m(M\{Q1,...,Qm}), for any choice of the points {Q1,...,Qm}. Set P′={P2,...,Pn}, take m=n−1, and consider Mdifferent from the sphere and from the projective plane (so π2(M) = 1). By the long exact sequence of homotopy groups of this fibration, we obtain 1−→ π1(M\P′, P1)u −→ P Bn(M, P)v −→ P Bn−1(M, P′)−→ 1.(2) If γ∈π1(M\P′, P1), then u(γ) = (γ, eP2,...,ePn), where ePidenotes the constant path on Pi, and, for Γ = (γ1,...,γn)∈P Bn(M, P), one has v(Γ) = (γ2,...,γn). The goal of this paper is to determine new presentations of the braid groups of closed surfaces different from the sphere and from the projective plane. These presentations are 2 much simpler than those which were known before ([S]). Moreover, the generators and the relations have an easy geometric interpretation. We also show that these presentations furnish an algorithm to solve the word problem for surface braid groups. Notice that similar presentations of the braid groups of the sphere and of the projective plane can be found in [FvB] and in [vB], respectively. Our work is organized as follows. In Section 2 we state the results, introducing the generators and relations of our new presentations. Then we explain in Section 3 the method followed in the proofs, which we apply throughout Sections 4 and 5, for orientable and non-orientable surfaces, respectively. Finally, we describe in Section 6 an algorithm to solve the word problem in surface braid groups. I would like to thank Luis Paris for giving me the idea of applying Lemma 3.1 to surface braid groups, and also for its valuable help in the writing of this paper. 2 Statements The aim of this section is to state our presentations of surface braid groups, defining the generators and showing that the proposed relations are satisfied. We start with the case of an oriented surface different from the sphere. Let Mbe a closed, orientable surface of genus g≥1. The first thing we want is to have a geometrical representation of a braid over M. We represent Mas a polygon Lof 4gsides, identified in the way of Figure 1 (See [M], page 34, ex. 8.9). α1 α2g α2g−1 α1 α2 α2g−1 α2g α2 Figure 1: The polygon Lrepresenting M. We could now take the cylinder L×I(I= [0,1]), and represent a braid Γ over Mas it is usually done for the open disc, that is, in L× {t}we draw the npoints γ1(t),...,γn(t). But in this case a string could “go through a wall” of the cylinder and appear from the other side. Hence, if we look at the cylinder from the usual viewpoint, it would not be clear which are the “crossed walls” (see the left hand side of Figure 2). 3 L I Figure 2: A braid over a surface of genus 2: two different viewpoints. The solution we propose is to look at the cylinder from above, as in the right hand side of Figure 2. In this way, we get rid of the ambiguity, and moreover we see the strings again as paths in the surface. When two strings cross, we see passing above the one that reaches before the crossing point. Anyway, it is good to keep in mind the idea that we are looking to a cylinder, and to consider the paths as strings: in this manner, one can see more easily when two geometric braids are equivalent. Now we can define the generators of Bn(M). We choose the nbase points along the horizontal diameter of L, as in Figure 3. Now given r, 1 ≤r≤2gwe define the braid ar as follows: its only nontrivial string is the first one, which goes through the r-th wall, in the way of Figure 3. That is, the first string will go upwards if ris odd, and downwards otherwise. We also define, for all i= 1,...,n−1, the braid σias in Figure 3. Note that σ1,...,σn−1 are the classical generators of the braid group Bnof the disc. a2k+1 α2k+1 α2k+1 Pn P1P1 α2k α2k Pn a2kσi Pn Pi+1 P1Pi Figure 3: The generators of BnM. 4 We will see later that the set {a1,...,a2g, σ1,...,σn−1}is a set of generators of Bn(M). There are two relations between these generators that we can deduce as follows. Consider the interior of L. It is a subsurface Dof Mhomeomorphic to a disc, so clearly every relation satisfied in the braid group Bn=Bn(D) will be satisfied as well in Bn(M) (the same homotopy can be used in both cases). In fact, since g≥1, it is known that Bnis a subgroup of Bn(M) (see [PR]). Hence, from the classical presentation of Bn, we obtain two relations in Bn(M): σiσj=σjσi(|i−j| ≥ 2), σiσi+1σi=σi+1σiσi+1 (1 ≤i≤n−2). Note also that if i∈ {2,...,n−1}and r∈ {1,...,2g}, then the non-trivial strings of σiand the one of armay be taken to be disjoint. This clearly implies that these two braids commute. Hence we have arσi=σiar(1 ≤r≤2g;i≥2). Now, in order to find more relations between the set of generators, we do the following construction. Denote by srthe first string of ar, for all r= 1,...,2g, and consider all the paths s1,...,s2g. We can “cut” the polygon Lalong them, and “glue” the pieces along the paths α1,...,α2g. We obtain another polygon of 4gsides which are labeled by s1,...,s2g (see in Figure 4 the case of a surface of genus 2; the general case is analogous). We will call this new polygon the P1-polygon of M, since all of its vertices are identified to P1, while L will be called the initial polygon. We obtain in this way a new representation of the surface M. P1 α3 α3α2 α2 α4 α1α4 α1 Pn s4 s2 s3 s1 P1 s2 s3 s1 s4 α1α4 P1 Pn P1 α2α3 s2 s1 s4 s3 P1 P1 P1 P1P1 Figure 4: The initial and the P1-polygon of a surface of genus 2. We will use the P1-polygon to show three more relations in Bn(M). For instance, consider the braid a1···a2ga−1 1···a−1 2g. If we look at it in the P1-polygon, it is clear that it 5 is equivalent to the braid of Figure 5. But this one can be seen into the initial polygon as a braid that does not go through the walls, namely, an element of Bn, the braid group of the disc. Then we can easily show that it is equivalent to the braid σ1···σn−2σ2 n−1σn−2···σ1. So we have: a1···a2ga−1 1···a−1 2g=σ1···σn−2σ2 n−1σn−2···σ1. Pn s1 s2g P1 P1 s2g s1 Figure 5: The braid a1···a2ga−1 1···a−1 2g. Now we define, for each r= 1,...,2g, the braid A2,r =σ−1 1a1···ar−1a−1 r+1 ···a−1 2gσ−1 1. We will use the P1-polygon to see how it looks like. In the left hand side of Figure 6, we can see a braid which is clearly equivalent to A2,r (if ris odd, the other case being analogous). If we “cut” and “glue” to see this braid in the P1-polygon, we obtain the situation of the right hand side of Figure 6. That is, A2,r can be seen as a braid whose only nontrivial string is the second one, which goes upwards and crosses once the r-th wall sr. Note that, unlike the case of ar,A2,r always points upwards in the P1-polygon, no matter the parity of r. Therefore we have seen that the braid A2,r can be represented by a geometric braid, whose only non trivial string can be taken disjoint from all the paths st,t6=r. This clearly implies that atA2,r =A2,rat(1 ≤t, r ≤2g;t6=r). Now we finish our set of relations by considering the commutator of the braids (a1···ar) and A2,r, for all r= 1,...,2g. In Figure 7 we can see a sketch of the homotopy which starts with this commutator and deforms it to a braid clearly equivalent to σ2 1. Therefore, we obtain the relation: (a1···ar)A2,r =σ2 1A2,r (a1···ar) (1 ≤r≤2g). 6 P2α1 α2 αr+1 αr−1 αr+1 αr−1 α2g α2g−2 α3 P1 α1 α2 α3 α2g α2g−1 α2g−1 α2g−2 αr αr P2 sr s2g s2g−1 s1 s2 s2 s1 s2g s2g−1 sr−1 sr+1 sr P1 sr−1sr+1 Figure 6: The braid A2,r: In the P1-polygon and in the initial one. Now we claim that the six relations that we have considered form a complete set of defining relations of Bn(M). In other words, we have the following result. Theorem 2.1. If Mis a closed, orientable surface of genus g≥1, then Bn(M)admits the following presentation: •Generators: σ1,...,σn−1, a1,...,a2g. •Relations: (R1) σiσj=σjσi(|i−j| ≥ 2) (R2) σiσi+1σi=σi+1σiσi+1 (1 ≤i≤n−2) (R3) a1···a2ga−1 1···a−1 2g=σ1···σn−2σ2 n−1σn−2···σ1 (R4) arA2,s =A2,sar(1 ≤r, s ≤2g;r6=s) (R5) (a1···ar)A2,r =σ2 1A2,r (a1···ar) (1 ≤r≤2g) (R6) arσi=σiar(1 ≤r≤2g;i≥2) where A2,r =σ−1 1a1···ar−1a−1 r+1 ···a−1 2gσ−1 1. Now we turn to the non-orientable case. Let Mbe a closed non-orientable surface of genus g≥2. To represent a braid in Mwe will also present the surface as a polygon, this time of 2gsides, as in Figure 8, and we make an additional cut: define the path eas in 7 P1P2 sr sr s1s1 P2 P1 sr srsr P1P2 s1 sr Figure 7: The braid [a1···ar, A2,r]. the left hand side of Figure 8, and cut the polygon along it. We get Mrepresented as in the right hand side of the same figure, where we can also see how we choose the points P1,...,Pn. α2 α1 e α1 αg−1 αg−1 αg αg α2α1 ee Pn P1 αg αg α1 Figure 8: Representation of a non-orientable surface M. We define now the generators of Bn(M). They will be similar to those of the orientable surface braid groups. For all i∈ {1,...n−1}, the braid σiwill be the same as in the orientable case. For all r∈ {1,...,g}, the braid arconsists on the first string passing through the r-th wall, in the way of Figure 9, while the other strings are trivial paths. There are six relations in the braid group of Mthat are analogous to those considered for an orientable surface. They can be shown to hold in the same way as in the orientable case; the only difference is the construction of the P1-polygon. We denote by s1,...,sgthe first string of a1,...,ag, respectively, and in this case we define another path, e1, which goes from P1to the final point of e(see Figure 9). Then we cut along the paths s1,...,sg, e1and glue along α1,...,αg, e. The result is the P1-polygon of Mwhose sides, reading clockwise, 8 ar ee αrαr Pn e1 P1Pn Pi+1 P1Pi σi ee αg αg α1 α1 Figure 9: The generators of Bn(M). are labeled by s1, s1, s2, s2,...,sg, sg, e1, e−1 1. We claim that the six mentioned relations form a set of defining relations of Bn(M). To be more precise, we claim the following. Theorem 2.2. If Mis a closed, non-orientable surface of genus g≥2, then Bn(M)admits the following presentation: •Generators: σ1,...,σn−1, a1,...,ag. •Relations: (r1) σiσj=σjσi(|i−j| ≥ 2) (r2) σiσi+1σi=σi+1σiσi+1 (1 ≤i≤n−2) (r3) a2 1···a2 g=σ1···σn−2σ2 n−1σn−2···σ1 (r4) arA2,s =A2,sar(1 ≤r, s ≤g;r6=s) (r5) a2 1···a2 r−1arA2,r =σ2 1A2,r a2 1···a2 r−1ar(1 ≤r≤g) (r6) arσj=σjar(1 ≤r≤g;j≥2) where A2,r =σ−1 1a2 1···a2 r−1a−1 ra−2 r−1···a−2 1σ1. 9 Note that the left hand side maps by ϕto ˜r, where ris a relator of PBn−1(M) corresponding to (PR1), while the right hand side maps by ϕto a word over GA. Hence, Uis equal to the right hand side of the equation, and this yields the first relation of Type 2. The remaining relations of Type 2 are also images by ϕof relations in Presentation 1; namely (PR2), (PR3), (PR4) and (PR5) when i≥2, (PR6) when i≥2 and j≥2, and (PR7), (PR8) when j≥2. For all these relations, the word Uis just the trivial word, except for (PR8), for which U=a−1 1,2g···a−1 1,1T1,j−1T−1 1,j a1,1···a1,2g. Finally, we find the relations of Type 3. For i= 1, (PR2) becomes Aj,sa1,rA−1 j,s =a1,r (r6=s), so V=a1,r. Next, using (PR2), Relation (PR3) turns to be equivalent to Aj,ra1,rA−1 j,r =a−1 1,r−1···a−1 1,1T1,j−1T−1 1,j (a1,1···a1,r), so Vequals the right hand side of this equation. Relations of the form Tk,lT1,jT−1 k,l =V, where Vis a word over GA, follow from (PR4)-(PR5), while those of the form Tk,la1,rT−1 k,l = Vfollow from (PR6), when i= 1. Also, if j > k, we obtain from (PR6) the relations Aj,rT1,kA−1 j,r =V, where Vis a word over GA. The only remaining relations are those of the form Aj,rT1,kA−1 j,r =V, when 1 < j ≤k, which are deduced as follows: By (PR7), we know that aj,s commutes with the element a−1 1,2g···a−1 1,1T1,ka1,2g···a1,1 for s= 1,...,2g. This implies that Aj,r commutes with the same element, so a−1 1,2g···a−1 1,1T1,ka1,2g···a1,1=Aj,r a−1 1,2g···a−1 1,1T1,ka1,2g···a1,1A−1 j,r =Aj,ra−1 1,2gA−1 j,r ···Aj,ra−1 1,1A−1 j,r Aj,rT1,kA−1 j,r Aj,ra1,2gA−1 j,r ···Aj,ra1,1A−1 j,r . But using (PR2) and (PR3) we know how to write all the terms in the above product (except the middle one) as words over GA, so we are done. Hence, we have shown that PBn(M)ϕ ≃PBn(M) and therefore, we have proved: Theorem 4.2. If Mis a closed, orientable surface of genus g≥1, then PBn(M)admits Presentation 1 (and also Presentation 2) as presentation. Step 3. Now we want to find a presentation of Bn(M), for g≥1. We define then the group Bn(M), given by the presentation in Theorem 2.1. This is the most reduced presentation we have found. But to show its validity we will modify it, obtaining a new one with more generators and relations, but equivalent to the first one. 16 First, we change our notation, and call a1,r the generators ar, for r= 1,...,2g. Then we must simply add to the given presentation the generators −ai,r i= 2,...,n;r= 1,...,2g, −Tj,k 1≤j < k ≤n, and the relations (R7) aj+1,r =σjaj,rσj( 1 ≤j≤n−1; 1 ≤r≤2g;reven). (R8) aj+1,r =σ−1 jaj,rσ−1 j(1 ≤j≤n−1; 1 ≤r≤2g;rodd). (R9) Tj,k =σjσj+1 ···σk−2σ2 k−1σk−2···σj(1 ≤j < k ≤n). Clearly, both presentations define the same group, that is, Bn(M). Now we define ψ:Bn(M)→Bn(M) in the natural way. It is an easy exercise to show, using the same methods as before, that Relations (R7), (R8) and (R9) map to relations in Bn(M). Therefore, ψis a well defined homomorphism. Recall now the exact sequence (1): 1−→ P Bn(M)e −→ Bn(M)f −→ Σn−→ 1. We know by Theorem 4.2 a presentation of P Bn(M) (say Presentation 1), and it is also known that a presentation of Σnis •Generators: δ1,...,δn−1. •Relations: –δiδj=δjδi|i−j| ≥ 2, –δiδi+1δi=δi+1δiδi+1 1≤i≤n−2, –δ2 i= 1 1 ≤i≤n−1, where δiis the permutation (i, i + 1), for any i. Now σiis clearly a pre-image by fof δi, so by Lemma 3.1 Bn(M) and Bn(M) have the same generators, and ψis surjective. Similarly to what we did in Step 2, we show now that ψis an isomorphism by the following procedure. First, we denote by GAthe set of generators of PBn(M), and by Gthe set of generators of Bn(M). For each relation in the presentation of PBn(M), we consider it via eas a relation in Bn(M), and we show that it also holds in Bn(M). 17 Next, for each relator rof Σn, we consider its canonical pre-image by f, denoted by ˜r. Then we find a word Uover Gsuch that the equality ˜r=Uholds in Bn(M), and such that ψ(U) is a word over GA. Finally, for each x∈GAand each generator δiof Σn, we find a word Vover Gsuch that the equality σix σ−1 i=Vholds in Bn(M), and such that ψ(V) is a word over GA. This gives us the relations of Types 1, 2, and 3 of Lemma 3.1 and, therefore, a presentation of Bn(M), and, at the same time, this shows that ψis injective, and, consequently, that ψis an isomorphism. Let us then verify in Bn(M) the relations of Type 1. In the case of (PR1), we start with (R3): a1,1···a1,2ga−1 1,1···a−1 1,2g=σ1···σn−2σ2 n−1σn−2· · ·σ1.(3) Using (R7) and (R8), we see that the left hand side of Equation (3) becomes σ1···σn−1(an,1···an,2g)σ−1 n−1···σ−2 1···σ−1 n−1a−1 n,1···a−1 n,2gσn−1···σ1. On the other hand, from (R1), (R2) (braid relations) and (R9), we get T−1 i,n−1Ti,n =σ−1 iσ−1 i+1 ···σ−1 n−2σ2 n−1σn−2···σi=σn−1···σi+1σ2 iσ−1 i+1 ···σ−1 n−1, so n−1 Y i=1 T−1 i,n−1Ti,n =σn−1···σ2 1···σn−1. Therefore, Equation (3) becomes an,1···an,2g n−1 Y i=1 T−1 i,n−1Ti,n!−1 a−1 n,1···a−1 n,2g= 1, which is clearly equivalent to (PR1). We will use in what follows some relations of Bn(M) easily deduced from (R1)-(R9). From (R7) and (R8), we get ai,r =σ−1 i−1···σ−1 1a1,r σ−1 1···σ−1 i−1if ris odd. (4) ai,r = (σi−1···σ1)a1,r (σ1···σi−1) if ris even. (5) Aj,s =σ−1 j−1···σ−1 2A2,s σ−1 2···σ−1 j−1=σ−1 j−1···σ−1 1A1,s σ−1 1···σ−1 j−1.(6) Also, from (R1) and (R2), we obtain σj(σkσk−1···σi) = (σkσk−1···σi)σj+1 (i≤j < k).(7) σjσ−1 kσ−1 k−1···σ−1 i=σ−1 kσ−1 k−1···σ−1 iσj+1 (i≤j < k).(8) 18 σi···σk−1σ2 kσ−1 k−1···σ−1 i=σ−1 k···σ−1 i+1σ2 iσi+1 ···σk.(9) Now using (6), (7), (8) and (R6), we see that if 1 ≤k≤j−2; σkAj,s =σkσ−1 j−1···σ−1 1A1,s σ−1 1···σ−1 j−1 =σ−1 j−1···σ−1 1σk+1A1,s σ−1 1···σ−1 j−1 =σ−1 j−1···σ−1 1A1,sσk+1 σ−1 1···σ−1 j−1=Aj,sσk.(10) In the same way, using (6), (R6) and (R4), we get a1,rAj,s =a1,r σ−1 j−1···σ−1 2A2,s σ−1 2···σ−1 j−1=Aj,sa1,r, if r6=sand 1 < j. Therefore, if i < j and r6=s, by (4) and (5) ai,r is a product of elements which commute with Aj,s, so we obtain ai,rAj,s =Aj,sai,r, which shows that (PR2) holds in Bn(M). Now we verify Relation (PR3). We will do the case when ris odd, the other case being analogous. It is clear which of the known relations of Bn(M) we are using at each of the following equalities: (ai,1. . . ai,r)Aj,r =σ−1 i−1···σ−1 1(a1,1. . . a1,r)σ−1 1···σ−1 i−1Aj,r =σ−1 i−1···σ−1 1(a1,1. . . a1,r)Aj,r σ−1 1···σ−1 i−1 =σ−1 i−1···σ−1 1(a1,1. . . a1,r)σ−1 j−1···σ−1 2A2,r σ−1 2···σ−1 j−1σ−1 1···σ−1 i−1 =σ−1 i−1···σ−1 1σ−1 j−1···σ−1 2(a1,1. . . a1,r)A2,r σ−1 2···σ−1 j−1σ−1 1···σ−1 i−1 =σ−1 i−1···σ−1 1σ−1 j−1···σ−1 2σ2 1A2,r (a1,1. . . a1,r)σ−1 2···σ−1 j−1σ−1 1···σ−1 i−1 =σi···σj−2σ2 j−1σ−1 j−2···σ−1 1σ−1 j−1···σ−1 2A2,rσ−1 2···σ−1 j−1(a1,1. . . a1,r)σ−1 1···σ−1 i−1 =σi···σj−2σ2 j−1σ−1 j−2···σ−1 1Aj,r (a1,1. . . a1,r)σ−1 1···σ−1 i−1 =σi···σj−2σ2 j−1σ−1 j−2···σ−1 iAj,r (ai,1. . . ai,r) =Ti,jT−1 i,j−1Aj,r (ai,1. . . ai,r). 19 This shows the case of (PR3). Relations (PR4) and (PR5) are actually relations in the braid group of the disc, so they are a consequence of (R1) and (R2). (PR6) is obtained easily from (R9), (4), (5) and the braid relations (R1) and (R2). So we may turn to (PR7): It is clear that it suffices to show that in Bn(M), Ai,r commutes with a−1 j,2g···a−1 j,1Tj,kaj,2g···aj,1for 1 ≤j < i ≤k < n. This is shown as follows (remember that we can already use (PB1)-(PB6)): Ai,r a−1 j,2g···a−1 j,1Tj,kaj,2g···aj,1 =a−1 j,2g···a−1 j,r+1Ai,r a−1 j,r ···a−1 j,1Tj,kaj,2g···aj,1 =a−1 j,2g···a−1 j,1Tj,iT−1 j,i−1Ai,rTj,kaj,2g···aj,1 =a−1 j,2g···a−1 j,1Tj,iT−1 j,i−1Ai,r σj···σ2 k−1···σjaj,2g···aj,1 =a−1 j,2g···a−1 j,1Tj,iT−1 j,i−1Ai,r σj···σ2 k−1···σ1a1,2g···a1,1σ−1 1· · · σ−1 j−1 (using (R3)) =a−1 j,2g···a−1 j,1Tj,iT−1 j,i−1Ai,r σj···σk−1σ−1 k···σ−2 n−1···σ−1 1a1,1···a1,2gσ−1 1···σ−1 j−1 (by (R9) and (10)) =a−1 j,2g···a−1 j,1σj···σ2 i−1Ai,rσi−1···σk−1σ−1 k···σ−2 n−1···σ−1 1a1,1···a1,2gσ−1 1···σ−1 j−1 =a−1 j,2g···a−1 j,1σj···σi−1Ai−1,rσi···σk−1σ−1 k···σ−2 n−1···σ−1 1a1,1···a1,2gσ−1 1···σ−1 j−1 =a−1 j,2g···a−1 j,1σj···σk−1σ−1 k···σ−2 n−1···σ−1 iσi−1σ−1 i−2···σ−1 1Ai,ra1,1···a1,2gσ−1 1···σ−1 j−1 =a−1 j,2g···a−1 j,1σj···σk−1σ−1 k···σ−2 n−1···σ−1 1a1,1···a1,2gσ−1 1···σ−1 j−1Ai,r (by (R3) again) =a−1 j,2g···a−1 j,1Tj,k σj−1···σ1a1,2g···a1,1σ−1 1···σ−1 j−1Ai,r =a−1 j,2g···a−1 j,1Tj,kaj,2g···aj,1Ai,r. Finally, Relation (PR8) is verified using some intermediary results. The first is evident: by (R4) we see that in Bn(M), A1,2gA2,2g=A2,2gA1,2g, and moreover this braid commutes 20 with σ1, since A1,2gA2,2gσ1=A1,2gσ−1 1A1,2g=σ1A2,2gA1,2g=σ1A1,2gA2,2g. Analogously, one shows that (a1,2ga2,2g) commutes with σ1. The following result is a consequence of the previous ones and of (R5): a1,2gA2,2ga−1 1,2g=a−1 1,2g−1···a−1 1,1σ2 1A2,2g(a1,1···a1,2g−1) =A−1 1,2gσ2 1A2,2gA1,2g=A−1 1,2gA2,2gA1,2gσ2 1=A2,2gσ2 1, so we obtain a−1 1,2gA2,2g=A2,2ga−1 1,2gσ−2 1.(11) Now we consider the factors in the right hand side of (PR8), and we see that a−1 i,2g···a−1 i,1Ti,j−1T−1 i,j (ai,1···ai,2g) =a−1 i,2g···a−1 i,1σi···σj−2σ2 j−1σ−1 j−2···σ−1 i(ai,1···ai,2g) =σ−1 i−1···σ−1 1a−1 1,2g···a−1 1,1σ1···σj−2σ2 j−1σ−1 j−2···σ−1 1(a1,1· · ·a1,2g)σ1···σi−1 (by (9)) =σ−1 i−1···σ−1 1a−1 1,2g···a−1 1,1σ−1 j−1···σ−1 2σ−2 1σ2···σj−1(a1,1···a1,2g)σ1···σi−1 =σ−1 i−1···σ−1 1σ−1 j−1···σ−1 2a−1 1,2g···a−1 1,1σ−2 1(a1,1···a1,2g)σ2···σj−1σ1···σi−1 =σ−1 i−1···σ−1 1σ−1 j−1···σ−1 2a−1 1,2gA−1 1,2gσ−2 1A1,2ga1,2g(σ2···σj−1σ1···σi−1) (since (A1,2gA2,2g) commutes with σ1) =σ−1 i−1···σ−1 1σ−1 j−1···σ−1 2a−1 1,2gA2,2gσ−2 1A−1 2,2ga1,2g(σ2···σj−1σ1···σi−1) (by (11)) =σ−1 i−1···σ−1 1σ−1 j−1···σ−1 2A2,2ga−1 1,2gσ−2 1a1,2gA−1 2,2g(σ2···σj−1σ1···σi−1) (since, (a1,2ga2,2g) commutes with σ1) =σ−1 i−1···σ−1 1σ−1 j−1···σ−1 2A2,2ga2,2gσ−2 1a−1 2,2gA−1 2,2g(σ2···σj−1σ1···σi−1) 21 =σ−1 i−1···σ−1 1Aj,2gaj,2gσ−1 j−1···σ−1 2σ−2 1σ2···σj−1a−1 j,2gA−1 j,2g(σ1···σi−1) (by (10)) =Aj,2gaj,2gσ−1 i−1···σ−1 1σ−1 j−1···σ−1 2σ−2 1σ2···σj−1(σ1· · · σi−1)a−1 j,2gA−1 j,2g (by (9)) =aj,1···aj,2gσ−1 j−1···σ−1 i+1σ−2 iσi+1 ···σj−1a−1 j,2g···a−1 j,1. And this clearly yields (PR8): Qj−1 i=1 a−1 i,2g···a−1 i,1Ti,j−1T−1 i,j ai,1···ai,2gaj,1···aj,2ga−1 j,1···a−1 j,2g =Qj−1 i=1 aj,1···aj,2gσ−1 j−1···σ−1 i+1σ−2 iσi+1 ···σj−1a−1 j,2g···a−1 j,1aj,1···aj,2ga−1 j,1···a−1 j,2g =aj,1···aj,2gσ−1 j−1···σ−1 2σ−2 1σ−1 2···σ−1 j−1a−1 j,1···a−1 j,2g = (σj···σn−1)an,1···an,2gσ−1 n−1···σ−2 1···σ−1 n−1a−1 n,1···a−1 n,2g(σn−1···σj) (by (R9) and (PR1)) = (σj···σn−1) (σn−1···σj) = Tj,n. We have thus finished with relations of Type 1. Consider now those of Type 2. For each relator in the presentation of Σn, we must find the word Umentioned above. The first relator is δiδjδ−1 iδ−1 j, when |i−j| ≥ 2 which, by (R1), yields in Bn(M) the relation σiσjσ−1 iσ−1 j= 1 (|i−j| ≥ 2). Clearly, Uis the trivial word. The second relator, δiδi+1δiδ−1 i+1δ−1 iδ−1 i+1 gives, by (R2), σiσi+1σiσ−1 i+1σ−1 iσ−1 i+1 = 1 (i= 1,···, n −2), so in this case Uis also the trivial word. Finally, by the third relator δ2 i, we obtain, using (R9), σ2 i=Ti,i+1 (i= 1,···, n −1), 22 hence U=Ti,i+1. So we have obtained the relations in Bn(M) mapped by ψto the relations of Type 2. We finish the proof of Theorem 2.1 obtaining the relations of Type 3. They are very easy to deduce, using (10), (R1), (R2), (R7), (R8) and (R9). They are the following: σiaj,rσ−1 i=aj,r (j6=i, i + 1), σiai,rσ−1 i=ai+1,rT−1 i,i+1 if ris even, σiai,rσ−1 i=Ti,i+1ai+1,r if ris odd, σiai+1,rσ−1 i=Ti,i+1ai,r if ris even, σiai+1,rσ−1 i=ai,rT−1 i,i+1 if ris odd, σiTj,kσ−1 i=Tj,k (i6=j−1, j, k), σiTi+1,kσ−1 i=Ti,kT−1 i,i+1, σiTi,kσ−1 i=Ti,i+1Ti+1,k, σiTj,iσ−1 i=Tj,i−1T−1 j,i Tj,i+1. 5 The braid groups of a non-orientable surface This section is devoted to prove Theorem 2.2, using the same method as before. Thus, let Mbe a closed, non-orientable surface of genus g≥2. Step 1. Denote by PBn(M) the group defined by the following presentation. Presentation 3 •Generators: {ai,r; 1 ≤i≤n, 1≤r≤g} ∪ {Tj,k; 1 ≤j < k ≤n}. •Relations: (Pr1) a2 n,1···a2 n,g =Qn−1 i=1 T−1 i,n−1Ti,n. (Pr2) ai,rAj,s =Aj,sai,r (1 ≤i < j ≤n; 1 ≤r, s ≤g;r6=s). (Pr3) a2 i,1···a2 i,r−1ai,rAj,r a−1 i,r a−2 i,r−1···a−2 i,1A−1 j,r =Ti,j T−1 i,j−1 (1 ≤i < j ≤n; 1 ≤r≤g). (Pr4) Ti,jTk,l =Tk,lTi,j (1 ≤i < j < k < l ≤nor 1 ≤i < k < l ≤j≤n). (Pr5) Tk,lTi,jT−1 k,l =Ti,k−1T−1 i,k Ti,jT−1 i,l Ti,kT−1 i,k−1Ti,l (1 ≤i < k ≤j < l ≤n). 23 (Pr6) ai,rTj,k =Tj,kai,r (1 ≤i < j < k ≤nor 1 ≤j < k < i ≤n),(1 ≤r≤g). (Pr7) ai,r a−2 j,g ···a−2 j,1Tj,k=a−2 j,g ···a−2 j,1Tj,kai,r (1 ≤j < i ≤k≤n). (Pr8) Tj,n =a2 j,1···a2 j,g Qj−1 i=1 T−1 j−i,jTj−i,j−1. Where Aj,r =a2 j,1···a2 j,r−1a−1 j,r a−2 j,r−1···a−2 j,1. We shall need, as in the orientable case, another presentation of PBn(M), which is the following one. Presentation 4 •Generators: {Ai,r; 1 ≤i≤n, 1≤r≤g} ∪ {Tj,k; 1 ≤j < k ≤n}. •Relations: the same as in Presentation 3, where ai,r =A2 i,1···A2 i,r−1A−1 i,r A−2 i,r−1A−2 i,1. It is clear that Presentation 3 and Presentation 4 are equivalent, in the same way as they were Presentation 1 and Presentation 2. We must now define the homomorphism PBn(M)ϕ −→ P Bn(M), by giving the image of the generators. They will be similar to those of the orientable surface. For all iand jsuch that 1 ≤i≤j≤n, the braid Ti,j will be the same as in Section 4. For all i,r, such that 1 ≤i≤nand 1 ≤r≤g, the braid ai,r will represent the i-th string passing through the r-th wall, in the way of Figure 13. We define as well the path ei(i= 1,...,n), which goes from Pito the final point of e, as in Figure 13. Given i∈ {1,...,n}, denote by si,r the i-th string of ai,r. We can proceed as we did for the P1-polygon in Section 2 to get the Pi-polygon: Cut along the paths eiand si,1,...,si,g, and glue along eand α1, . . . , αg. The resulting Pi-polygon is labeled by the paths si,1, si,1, si,2, si,2,...,si,g, si,g, ei, e−1 i, reading clockwise. Now we can repeat the process of Section 2 to see that for 1 ≤i < j, the braid Aj,r =a2 j,1···a2 j,r−1a−1 j,r a−2 j,r−1···a−2 j,1 can be represented in the Pi-polygon in the way of Figure 14. The remainder of Step 1, that is to show that ϕis a well defined homomorphism, is analogous to the orientable case. That is, Relations (Pr4), (Pr5) and (Pr6) are obvious; Relations (Pr1), (Pr2) and (Pr3) are analogous to Relations (r3), (r4) and (r5) of Theorem 2.2; and we can easily check Relations (Pr7) and (Pr8) in the Pj-polygon. 24 ai,r ee αrαr ei Pi Pn P1 P1Pn Pi Ti,j Pj Figure 13: The generators of PBn(M). si,r ei Pi si,r Pn Pj P1 si,1si,g ei Figure 14: The braid Aj,r in the Pi-polygon (i < j). Step 2. This step parallels, up to evident substitutions, the corresponding one in Section 4, showing the following theorem: Theorem 5.1. If Mis a closed, non-orientable surface of genus g≥2, then P Bn(M) admits Presentation 3(and also Presentation 4) as presentation. Step 3. Denote by Bn(M) the group defined by the presentation of Theorem 2.2. Call a1,r the elements arfor r= 1,...,g, and then add the generators −ai,r i= 2,...,n;r= 1,...,g, −Tj,k 1≤j < k ≤n, and the relations 25