Acta Mathematica Sinica, English Series Jun., 2009, Vol. 25, No. 1, pp. 1 16 Published online: July 1, 2009 DOI: 10.1007/s10114-009-6473-8 Http://www.ActaMath.com Acta Mathematica Sinica, English Series The Editorial Office of AMS & Springer-Verlag 2009 The Fundamental Group of the Complement of the Branch Curve of CP 1 T Abstract Meirav AMRAM Michael FRIEDMAN Mina TEICHER Department of Mathematics, Bar-Ilan University, 52900 Ramat Gan, Israel E-mail : meirav@macs.biu.ac.il fridmam@macs.biu.ac.il teicher@macs.biu.ac.il Denoting by T the comple projective torus, we can embed the surface CP 1 T in CP 5. In this paper we compute the fundamental group of the complement of the branch curve of this surface. Since the embedding is not ample enough, the embedded surface does not belong to the classes of surfaces where the fundamental group is virtually solvable: a property which holds for these groups for ample enough embeddings. On the other hand, as it is the first eample of this computation for non simply-connected surfaces, the structure of this group (as shown in this paper) give rise to the etension of the conjecture regarding the structure of those fundamental groups of any surface. Keywords fundamental group, generic projection, curves and singularities, branch curve MR(2000) Subject Classification 14F35, 14H30, 14J99, 14Q05, 14Q10 1 Introduction Let X be a projective algebraic surface embedded in a projective space in CP n. Considering S CP 2, the branch locus of a generic projection X CP 2 (or S a generic affine piece of S), the fundamental groups G = π 1 (CP 2 S) and G = π 1 (C 2 S) have always been important topological invariants of X. It was proven in [1] that these fundamental groups G and G, derived from the braid monodromy factorizations can distinguish between surfaces that are not deformation of one another. That is, if two surfaces have non-isomorphic fundamental groups, then they are not deformation equivalent. In this paper, we compute the groups G and G for the surface CP 1 T, where T is a comple projective torus, T CP 2. We embed CP 1 T by the Segre map into CP 5 (see [2 3]). We also give the results when degenerating T into a cycle of n projective lines, when the embedding of CP 1 T is into CP 2n 1. Till now, all of the known eamples of those groups were computed for surfaces which were simply-connected: X p,q the embedding of CP 1 CP 1 into CP N w.r.t. a linear system pl 1 +ql 2 (see [4]); V n the Veronese surface of order n 3 (see [5]); F 1,(a,b) the embedding of a Hirzebruch surface F 1 w.r.t. the linear system ac + be 0 (see [6]); the Hirzebruch surface Received September 13, 2006, Accepted May 21, 2008 The third author is partially supported by DAAD and EU-network HPRN-CT-2009-00099(EAGER); The Emmy Noether Research Institute for Mathematics and the Minerva Foundation of Germany; The Israel Science Foundation grant # 8008/02-3 (Ecellency Center Group Theoretic Methods in the Study of Algebraic Varieties ).
2 Amram M. et al. F 2,(2,2) (see [7]); and complete intersections eamples (see [8]). These eamples show us, on the one hand, that when the embedding is ample enough (or sufficiently ample, as in [9]), those fundamental groups satisfy the conjecture posed in [10]: they are almost solvable (they contain a subgroup of finite inde which is solvable), i.e., they are etensions of a solvable group by the symmetric group (for eample, this happens when p, q > 1 for X p,q ;when a, b > 1 for F 1,(a,b), or when n 3 for V n ). On the other hand, when the embedding is not ample enough, the fundamental groups are big they contain a free noncommutative subgroup with 2 elements; see, for eample, V 2 or a surface of deg = n in CP 3 (see [11 12]). There are two motivations for the computations in this paper. First of all, the results of this paper are the first computations for a non-simply connected surface. They indicate that these groups have, in some sense, a similar structure to the braid group B n (or to the cylindrical braid group), but they are highly different in other respects. Note that this similarity to B n is derived from the fact that the degeneration of CP 1 T can be obtained from the degeneration of X 1,3 simply by identifying the etreme edges; One can epect the mentioned similarity since for the surface X 1,3, the corresponding groups G and Ḡ are a quotient of B 6. This similarity suggest that the embedding of CP 1 T,in this case, is not ample enough. The second important aspect is the fact that the structure of G and G indicates what will happen when one would try to compute the fundamental group of the complement of the branch curve of a surface X, when X is not simply-connected, as in our article. Note that since CP 1 can be embedded in CP m (m 2), and T can be embedded in CP n 1 (n 3), CP 1 T can be embedded into a larger projective space by the Segre embedding of CP m CP n 1. In the end of our paper we propose a conjecture about the structure of the fundamental group of the complement of the branch curve of these surfaces. This paper is organized as follows. In Section 2, we define the doubled cylindrical braid group DΩ n ; in Section 3, we prove that G (and G) is a quotient of this group by a single relation, and present our conjecture. 2 B n, Ω n and DΩ n Groups This section describes a certain modification of the braid groups such that eventually the fundamental groups G and G will be a quotient of this group. Definition 2.1 Let G be a group, a, b G. Define [a, b] = ab(ba) 1, a, b = aba(bab) 1, a b = b 1 ab. A relation in G of the form a, b = 1 is called a triple relation. Denote by B n the braid group. We can consider B n as the group generated by (n 1) generators 1,..., n 1, with the relations: [ i, j ] = e for i j > 1, i, j = e for i j = 1, where e is the unit element. vertices and (n 1) edges: Note that we can represent the generators in a graph with n
Fundamental Group of the Complement of the Branch Curve of CP 1 T 3 1 2 3 n 1 2 n 1 Figure 1 The generators of B n Remark 2.2 We can say that the relations in B n are induced from the graph, such that every two consecutive edges, i, i+1, maintain a triple relations: i, i+1 = e; and every two disjoint edges, i, j, commute: [ i, j ] = e. Recall that every edge that connects 2 vertices is called a half-twist. We know that every braid can be also represented as an equivalence class of isotopic motions of n points on a plane, embedded in R 3, for eample see Figure 2. Figure 2 We now recall another definition of B n. A braid in the plane Let D be a closed disk in R 2, K Int (D), K finite, n = #K. Recall that the braid group B n [D, K] can be defined as the group of all equivalent diffeomorphisms β of D such that β(k) = K, β D = Id D. Definition 2.3 H(σ), half-twist defined by σ Let a, b K, and let σ be a smooth simple path in Int (D) connecting a with b s.t. σ K = {a, b}. Choose a small regular neighborhood U of σ contained in Int (D), s.t. U K = {a, b}. Denote by H(σ) the diffeomorphism of D which switches a and b by a counterclockwise 180 degree rotation and is the identity on D \ U. Thus it defines an element of B n [D, K], called the half-twist defined by σ. Assume that all of the points of K are on the -ais (when considering D in R 2 ). In this situation, if a, b K, and z a,b is a path that connects them, then we denote it by Z a,b = H(z a,b ). If z a,b is a path that goes below the -ais, then we denote it by Z a,b, or just Z a,b. If z a,b is a path that goes above the -ais, then we denote it by Z a,b. See [13] for additional notations. Definition 2.4 Ω n Take the graph that represents B n+1, and identify the beginning and the end vertices in the graph, as in Figure 3. 1 2 3 n+1 1 2 n 2 1 n 1 Figure 3
4 Amram M. et al. We look at the Artin group of this graph. This group, denoted by Ω n, has n generators: 1,..., n with the following relations: [ i, j ] = e for (i j) (mod n) > 1, i, j = e for (i j) (mod n) = 1. These relations can also be induced from the graph in Figure 3. Ω n can be considered, though we will not prove it, as the group of equivalence classes of isotopic motions of n points on a cylinder, embedded in R 3 (see Figure 4). We can also see that Ω n = π 1 (C(n, S 1 )) the fundamental group of the configuration space of S 1 (this is similar to the fact that B n = π 1 (C(n, R 2 ))). Figure 4 A braid on the cylinder As an eample, the following two motions in the figure below represent the braid 1 2 1 3 in B 4 and in Ω 3. 1 2 3 B 4 Ω 3 1 2 3 Figure 5 The braid 1 2 1 3 Remark 2.5 Note that it is obvious that one can embed B n in Ω n 1. Since B n is big (for n > 2; that is - it contains a free commutative subgroup with 2 elements), so does Ω n, for n > 3. The net step would be to check what happens to Ω n if we double one of the edges, say 1, in Figure 3. 3 2 1 1 n Figure 6 A planar graph with 1 and 1
Fundamental Group of the Complement of the Branch Curve of CP 1 T 5 We can look at a planar graph; in this case, as in Figure 6. However, due to the fact that the degeneration of CP 1 T is not planar, we look at another graph, which is not planar, as depicted in Figure 7. 1 3 2 1 n Figure 7 A non-planar graph with 1 and 1 Note that the edge 1 goes below the edge n and doesn t intersect it, while all the edges 1,..., n are in the same plane. We now define the relations involving 1, which derived from Figure 7. 2 (1) 1, 2 = e; induced from 1 1 (2) 1, n = e; from n 2 1 (3) ( 1 ) 1, 1 = e; from 2 ( 1 ) 2 1 (4) ( 1 ) 1, 1 = e; from n Definition 2.6 1 ( 1 ) n 1 n DΩ n : the doubled cylindrical braid group. Denote by i the generator i or i. We define the doubled cylindrical braid group [ i, j ] = e for (i j) (mod n) > 1 DΩ n = 1, 1, 2,..., n i, j = e for (i j) (mod n) = 1 and relations (3), (4) Remark 2.7 (i) Relations (1), (2) are already included from the triple relation. (ii) DΩ n is big, for n > 3.. In the following section, we will see that the fundamental groups G, G will be a quotient of DΩ n by a single relation. 3 Eplicit Computation of G and G In this section, we recall how to degenerate the surface CP 1 T and how to derive the braid monodromy related to the branch curve. Let T be a comple torus in CP 2. We define the surface X = CP 1 T and embed it into a projective space CP 5, using the Segre map CP 1 CP 2 CP 5. Projecting X onto CP 2 by a generic projection, we get the branch curve S. We refer the reader to [2] for a complete introduction of the braid monodromy factorization of CP 1 T. In order to compute the fundamental group G, we use the degeneration technique (see [14]
6 Amram M. et al. and [2]). This technique enables us to degenerate X into a union of 6 planes, as depicted in Figure 8. Figure 8 3 5 2 1 4 6 1 The degeneration of CP 1 T We numerate the intersection lines of the planes as 1 to 6, and their intersection points by V 1 to V 6. Note that the etreme edges of the degeneration are identified. The branch curve S 0 of the degeneration in CP 2 is a line arrangement of 6 lines (the projections of the lines 1,..., 6 in Figure 8). We numerate the projection of V 1 to V 6 as V 1 to V 6 as well (see Figure 9). V V V V 4 5 6 4 Figure 9 V V V V 1 2 3 1 the numeration of the vertices We recall now the definition of braid monodromy related to a curve S. For a full definition, see [2] and [15]. Definition 3.1 The braid monodromy w.r.t. S, π, u Let S be a curve, S C 2 Let π : S C be defined by π(, y) =. We denote deg π by m. Let N = { C #π 1 () < m}. Take u / N. Let C 1 u = {(u, y)}. There is a natural defined homomorphism π 1 (C 1 N, u) ϕ B m [C 1 u, C 1 u S] which is called the braid monodromy w.r.t. S, π, u, where B m is the braid group. We sometimes denote ϕ by ϕ u. Note that for computational reasons we demand that u is real, s.t. Real() u N. Denote the generator of the center of B n as 2. We recall Artin s theorem on the presentation of 2 as a product of braid monodromy elements of a geometric-base (an ordered base of π 1 = π 1 (C 1 N, u) with certain properties; see [15] for definitions). Theorem 3.2 Let S be a curve transversal to the line in infinity, and ϕ is a braid monodromy of S, ϕ : π 1 B m. Let δ i be a geometric (free) base (g-base) of π 1. Then 2 = ϕ(δ i ). This product is also defined as the braid monodromy factorization (BMF) related to a curve S. Since S 0 is a line arrangement, we can compute the braid monodromy factorization as in [15]. In order to compute the braid monodromy factorization of S, we use the regeneration rules ([13, 16]). The regeneration methods are actually, locally, the reverse process of the degeneration method. When regenerating a singular configuration consisting of lines and conics, the final stage in the regeneration process involves doubling each line, so that each point of K
Fundamental Group of the Complement of the Branch Curve of CP 1 T 7 (which is the set of points in the disk, that is C 1 u S 0 ) corresponding to a line labelled i is replaced by a pair of points, labelled i and i. The purpose of the regeneration rules is to eplain how the braid monodromy behaves when lines are doubled in this manner. The braid monodromy factorization of S is of the form 6 j=1 C jh Vj. The braids C j correspond to intersection of lines in CP 2, which do not intersect in the degeneration in CP 5 (they are known as parasitic intersection, see [17]). The braids H Vj are the local braid monodromy obtained by regeneration around the verte V j. The monodromies H Vj, 1 j 6, were computed in [2]. We list them here H V1 = Z 3 11,3 Z 33 (1), H V2 = Z 3 44,5 Z 55 (4), H V3 = Z 3 2,66 Z 22 (6), H V4 = Z 3 11,2 Z 22 (1), H V5 = Z 3 3,44 Z 33 (4), H V6 = Z 3 5,66 Z 55 (6). By [15] and [13], it is known that each braid is a power of some half-twist, which can be represented as a path; recall from Definition 2.2 that Z = H(z), when z is a path connecting points in C 1 u S. The braids appearing in the H Vj s now can be described: (1) The product of braids Zαα 3,β = Z3 α,β Z3 Z α,β 3 α,β corresponds to the three paths α α β β and Z 3 α,ββ = Z3 α,β Z3 α,β Z 3 α,β to α α β β (2) The braid Z αα (β) corresponds to the path α α β β and Z ββ (α) to α α β β
8 Amram M. et al. The factors in the braids C j correspond to the following paths (each figure below represents four paths; Note that each C j is a product of few braids. See [2] for eplicit computation): 2 2 3 3 1 1 2 2 3 3 4 4 2 2 3 3 4 4 1 1 5 5 2 2 5 5 3 3 4 4 5 5 1 1 5 5 6 6 3 3 4 4 5 5 6 6 4 4 5 5 6 6 Figure 10 Let π : C 2 C be the projection to the first coordinate. M S is the set of points where π is not etale. Let M = π(m ) and u C M a generic point. Choose a point u 0 π 1 (u), u 0 S. Choose a good system of generators for π 1 (C 2 S, u 0 ), denoted as {Γ j }, encircling the point on π 1 (u 0 ). In our case, we have the generators {Γ j, Γ j } 6 j=1. The reason is that through the regeneration, each point in π 1 (u) is replaced by two close points. We eplain how to associate a relation in G to a braid. Take a braid which is a half-twist that corresponds to a path σ from k to l via u 0. Let δ 1 (resp. δ 2 ) be the path from u 0 to k (resp. l) along σ, going around k (resp. l) and coming back to u 0 along the same path, respectively. A and B are the homotopy classes of a loop around k (resp. l) along δ 1 (resp. δ 2 ). A (resp. B) is a conjugation of Γ k (resp. Γ l ). The following figure illustrates how to find A, B from the half-twist V = H(σ):
Fundamental Group of the Complement of the Branch Curve of CP 1 T 9 σ σ k u 0 l B A u u 0 0 Figure 11 We cite now the Van Kampen Theorem [18], from which we induce the relations in G. Theorem 3.3 (van Kampen Theorem) Let S CP 2 be a projective curve, which is transversal to the line at infinity. Let S = S C 2. Let ϕ : π 1 (C N, u) B m [C 1 u, C 1 u S] be the braid monodromy w.r.t. S, π, u. Then (a) π 1 (C 2 S, ) = π 1 (C u S u, )/{β(v ) = V β Imϕ, V π 1 (C 1 u S)}. (b) π 1 (CP 2 S) π 1 (C 2 S)/ Γ where Γ is a simple loop in C 1 u S around S C 1 u = {q 1,..., q m }. Thus, we have one of the following relations in G (fied according to the type of singularity, from which we have the path σ): 1. A = B, if the singularity is a branch point, 2. [A, B] = ABA 1 B 1 = e if it is a node, 3. A, B = ABAB 1 A 1 B 1 = e if it is a cusp. The loops A and B are determined as follows. Each loop is a conjugation of Γ j, when j is the number of the point which is circled, with a sequence of Γ i s. The conjugation is determined as follows: whenever we follow from the starting point u 0 to the circled loop; when the loop goes below a point i 0, no conjugation happens; when the loop goes above a point i 0 clockwise, we conjugate by i 0 ; when the loop goes above a point i 0 counterclockwise, we conjugate by Γ i0. Notation Γ jj stands for Γ j or Γ j. Theorem 3.4 relations: The group G is generated by Γ 1, Γ 1, Γ 2, Γ 3, Γ 4, Γ 5, Γ 6 and admit the following [Γ 11, Γ 4 ] = [Γ 11, Γ 5 ] = [Γ 11, Γ 6 ] = e (begin1) [Γ 2, Γ 3 ] = [Γ 2, Γ 4 ] = [Γ 2, Γ 5 ] = e [Γ 3, Γ 5 ] = [Γ 3, Γ 6 ] = e [Γ 4, Γ 6 ] = e Γ 2, Γ 6 = Γ 3, Γ 4 = Γ 4, Γ 5 = Γ 5, Γ 6 = e Γ 2, Γ 1 = Γ 2, Γ 1 = Γ 2, Γ 1 Γ 1 Γ 1 1 = e Γ 3, Γ 1 = Γ 3, Γ 1 = Γ 3, Γ 1 Γ 1 Γ 1 1 = e (begin2)
10 Amram M. et al. [Γ 1, Γ 3 1 4 Γ 1 5 Γ 1 6 Γ 1 2 Γ 3 Γ 4 Γ 5 Γ 6 1 Γ 2] = e. (begin3) Proof We list now the relations in G, using the Van Kampen Theorem. We start with the relations derived from the branch point braids Z αα (β) and Z ββ (α) : Γ 3 Γ 5 Γ 2 Γ 6 Γ 4 Γ 6 = 1 Γ 1 1 Γ 3Γ 1 Γ 1 (br1) = 4 Γ 1 4 Γ 5Γ 4 Γ 4 (br2) = 1 Γ 1 1 Γ 2Γ 1 Γ 1 (br3) = 2 Γ 1 2 Γ 6Γ 2 Γ 2 (br4) = 3 Γ 1 3 Γ 4Γ 3 Γ 3 (br5) = 5 Γ 1 5 Γ 6Γ 5 Γ 5. (br6) The triple relations derived from the cuspidal braids Z 3 αα,β and Z3 α,ββ are Γ 5, Γ 6 = Γ 5 Γ 6 = Γ 5 Γ 6 Γ 6 6 = e (tr1) Γ 3, Γ 4 = Γ 3, Γ 4 = Γ 3, Γ 4 Γ 4 4 = e (tr2) Γ 2, Γ 6 = Γ 2, Γ 6 = Γ 2, Γ 6 Γ 6 6 = e (tr3) Γ 2, Γ 1 = Γ 2, Γ 1 = Γ 2 Γ 1 Γ 1 1 = e (tr4) Γ 5, Γ 4 = Γ 5, Γ 4 = Γ 5, Γ 4 Γ 4 4 = e (tr5) Γ 3, Γ 1 = Γ 3, Γ 1 = Γ 3, Γ 1 Γ 1 1 = e. (tr6) The relations derived from the parasitic intersections are [Γ 22, Γ 33 ] = [Γ 22, Γ 44 ] = [Γ 22, Γ 55 ] = e (C 1 ) [Γ 11, Γ 44 ] = [Γ 11, Γ 55 ] = [Γ 11, Γ 66 ] = e (C 2 ) [Γ 33, Γ 55 ] = [Γ 33, Γ 66 ] = e (C 3 ) [Γ 44, Γ 66 ] = e (C 4 ) [Γ 11, 3 Γ 1 3 Γ 44 Γ 3 Γ 3] = e (C 5 ) [Γ 55, Γ 4 Γ 4 Γ 33 Γ 3 4 Γ 1 4 ] = e (C 6) [Γ 66, Γ 2 Γ 2 Γ 11 2 Γ 1 2 ] = e (C 7) [Γ 66, Γ 5 Γ 5 Γ 44 5 Γ 1 5 ] = e. (C 8) First, by (br1) (br6), Γ 5 is replaced by Γ 5 in (tr1) and (tr5); Γ 3 is replaced by Γ 3 in (tr2) and (tr6); and Γ 2 is replaced by Γ 2 in (tr3) and (tr4). We omit the generators Γ 2, Γ 3, Γ 4, Γ 5 and Γ 6. By (br4) and (br6), we can omit first Γ 6 : [Γ 44, Γ 6 ] (br4) = [Γ 44, 2 Γ 1 2 Γ 6Γ 2 Γ 2 ] (C 1 ) = [Γ 44, Γ 6 ] = e, [Γ 33, Γ 6 ] (br4) = [Γ 33, 2 Γ 1 2 Γ 6Γ 2 Γ 2 ] (C 1 ) = [Γ 33, Γ 6 ] = e, [Γ 11, Γ 6 ] (br6) = [Γ 11, 5 Γ 1 5 Γ 6Γ 5 Γ 5 ] (C 2 ) = [Γ 11, Γ 6 ] = e,
Fundamental Group of the Complement of the Branch Curve of CP 1 T 11 [Γ 6, Γ 2 Γ 2 Γ 11 2 Γ 1 2 ] (br6) = [ 5, Γ 1 5 Γ 6Γ 5 Γ 5, Γ 2 Γ 2 Γ 11 2 Γ 1 2 ] (C 1 ) = (C 2 ) [Γ 6, Γ 2 Γ 2 Γ 11 2 Γ 1 2 ] = e, [Γ 6, Γ 5 Γ 5 Γ 44 5 Γ 1 5 ] (br4) = [ 2, Γ 1 2 Γ 6Γ 2 Γ 2, Γ 5 Γ 5 Γ 44 5 Γ 1 5 ] (C 1 ) = [Γ 6, Γ 5 Γ 5 Γ 44 5 Γ 1 5 ] = e, Γ 5, Γ 6 (br4) = Γ 5, 2 Γ 1 2 Γ 6Γ 2 Γ 2 (C 1 ) = Γ 5, Γ 6, Γ 5, Γ 6 Γ 6 6 = Γ 1 6 Γ 5Γ 6, Γ 6 (tr1) = Γ 5 Γ 6 5, Γ 6 = Γ 6, 5 Γ 6Γ 5 (br6) = 5 Γ 1 5 Γ 6Γ 5 Γ 5, 5 Γ 6Γ 5 (tr1) = Γ 6 Γ 5 6, Γ 6 = Γ 5, Γ 6 (br2) = 4 Γ 1 4 Γ 5Γ 4 Γ 4, Γ 6 (C 4 ) = Γ 5, Γ 6 = e, Γ 2, Γ 6 (br6) = Γ 2, 5 Γ 1 5 Γ 6Γ 5 Γ 5 (C 1 ) = Γ 2, Γ 6 = e, Γ 2, Γ 6 Γ 6 6 = Γ 1 6 Γ 2Γ 6, Γ 6 (tr3) = Γ 2 Γ 6 2, Γ 6 = Γ 6, 2 Γ 6Γ 2 (br4) = 2 Γ 1 2 Γ 6Γ 2 Γ 2, 2 Γ 6Γ 2 (tr3) = Γ 6 Γ 2 6, Γ 6 = Γ 2, Γ 6 (br3) = 1 Γ 1 1 Γ 2Γ 1 Γ 1, Γ 6 (C 2 ) = Γ 2, Γ 6 = e. In order to omit Γ 5, we use (br2) and (br6). [Γ 11, Γ 5 ] = e and [Γ 22, Γ 5 ] = e are translated to [Γ 11, Γ 5 ] = e and [Γ 22, Γ 5 ] = e, respectively. [Γ 33, Γ 5 ] = e gets the form [Γ 33, Γ 6 ] = e which already appears. The relation [Γ 4 Γ 4 Γ 33, 4 Γ 1 4, Γ 5 ] = e is omitted by substituting Γ (br6) 5 = 6 Γ 5Γ 6 5 Γ 6 (br4) = 6 Γ 5( 2 Γ 1 2 Γ 6Γ 2 Γ 2 ) 5 Γ 6, and the relation [Γ 5 Γ 5 Γ 44 5 Γ 1 5, Γ 6] = e gets the form [Γ 44, Γ 6 ] = e, which appears in our list. At this stage we are left with {Γ j, Γ j } 4 j=1 and {Γ 5, Γ 6 }. We have the following relations: Γ 5, Γ 6 = e, (t1) Γ 3, Γ 4 = Γ 3, Γ 4 = Γ 3, Γ 4 Γ 4 4 = e, (t2) Γ 2, Γ 6 = e, (t3) Γ 2, Γ 1 = Γ 2, Γ 1 = Γ 2, Γ 1 Γ 1 1 = e, (t4) Γ 5, Γ 4 = Γ 5, Γ 4 = Γ 5, Γ 4 Γ 4 4 = e, (t5) Γ 3, Γ 1 = Γ 3, Γ 1 = Γ 3, Γ 1 Γ 1 1 = e, (t6) [Γ 22, Γ 33 ] = [Γ 22, Γ 44 ] = [Γ 22, Γ 5 ] = e, (C 1) [Γ 11, Γ 44 ] = [Γ 11, Γ 5 ] = [Γ 11, Γ 6 ] = e, (C 2) [Γ 33, Γ 5 ] = [Γ 33, Γ 6 ] = e, (C 3) [Γ 44, Γ 6 ] = e, (C 4) [Γ 11, 3 Γ 1 3 Γ 44 Γ 3 Γ 3] == e, (C 5) [Γ 5, Γ 4 Γ 4 Γ 33 4 Γ 1 4 ] = e, (C 6) [Γ 6, Γ 2 Γ 2 Γ 11 2 Γ 1 2 ] = e. (C 7) We omit now Γ 4. Substituting (br5) in [Γ 4, Γ 6 ] = e and [Γ 22, Γ 4 ] = e, we get [Γ 4, Γ 6 ] = e
12 Amram M. et al. and [Γ 22, Γ 4 ] = e, respectively. Substituting (br2) in [Γ 11, Γ 4 ] = e, we get [Γ 11, Γ 5 ] = e which appear in (C 2). We epress Γ 4 as follows Γ 4 (br2) = Γ 5 Γ 4 Γ 5 4 Γ (br6) 5 = 5 Γ 4( 6 Γ 5Γ 6 5 Γ 6) 4 Γ 5 (br4 ) (br4) = 5 Γ 4( 6 Γ 5( 2 Γ 1 2 Γ 6Γ 2 Γ 2 ) 5 Γ 6) 4 Γ 5. Substituting (br4 ) in (C 5) gives us the relation [Γ 2 Γ 2 Γ 11 2 Γ 1 2, Γ 6], which is (C 7). And in the same manner, (C 6) gets the form [Γ 33, Γ 6 ] = e, which appears in (C 3). Substituting (br4 ) in Γ 3, Γ 4 = e and in Γ 3, Γ 4 Γ 4 4 = e gives Γ 4, Γ 5 = e (which appears in (t5)) and Γ 2, Γ 6 = e (which appears in (t3)), respectively. Again by (br4 ), Γ 5, Γ 4 = e and Γ 5, Γ 4 Γ 4 4 = e both get the form Γ 4, Γ 5 = e. In order to omit Γ 3, we have to consider the relations [Γ 22, Γ 3 ] = e, [Γ 3, Γ 5 ] = e, [Γ 3, Γ 6 ] = e. By (br1), [Γ 3, Γ 6 ] = e and [Γ 3, Γ 5 ] = e are translated to [Γ 3, Γ 6 ] = e and [Γ 3, Γ 5 ] = e, respectively. By (br5), Γ 3 = 4 Γ 3Γ 4 3 Γ 4, and substituting (br4 ), we have Γ 3 = 4 Γ 3( 5 Γ 4( 6 Γ 5 2 Γ 1 2 Γ 6Γ 2 Γ 2 5 Γ 6) 4 Γ 5) 3 Γ 4. (br3 ) We substitute (br3 ) in [Γ 22, Γ 3 ] = e to get [ 22 Γ 6Γ 22, 2 Γ 1 2 Γ 5Γ 6 5 Γ 2 Γ 2] = e. We check the two cases separately. The first relation is [ 2 Γ 6Γ 2, 2 Γ 1 2 Γ 5Γ 6 5 Γ 2 Γ 2] = [Γ 2 Γ 6 2, Γ 5Γ 6 5 ] which appears in (C 1). The second relation is [ 2 Γ 6Γ 2, 2 Γ 1 2 Γ 5Γ 6 5 Γ 2 Γ 2] (t3) = [ 6 Γ 2 Γ 6, 6 Γ 5Γ 6 ] = [Γ 2, Γ 5 ] = e, = [Γ 2 2 Γ 6Γ 2 2, Γ 5 2 Γ 6Γ 2 5 ] (tr3) = [Γ 2 Γ 6 Γ 2 6 Γ 1 2, Γ 5Γ 6 Γ 2 6 Γ 1 5 ] (br4) = [Γ 2 Γ 6 ( 6 Γ 2Γ 6 2 Γ 6) 6 Γ 1 2, Γ 5Γ 6 ( 6 Γ 2Γ 6 2 Γ 6) 6 Γ 1 5 ] = [Γ 2 2Γ 6 Γ 2 2, Γ 2Γ 5 Γ 6 5 Γ 1 2 ] = [Γ 2Γ 6 2, Γ 5Γ 6 which also appears in (C 1). 5 ] (tr3) = (tr1) [Γ 2, Γ 5 ] = e, The last generator which we omit is Γ 2. We have to consider [Γ 2, Γ 5 ] = e, [Γ 2, Γ 4 ] = e, [Γ 2, Γ 3 ] = e and (C 7). By (br3), the relations [Γ 2, Γ 5 ] = e and [Γ 2, Γ 4 ] = e get the forms [Γ 2, Γ 5 ] = e and [Γ 2, Γ 4 ] = e, respectively. By (br4), the relation [Γ 2, Γ 3 ] = e is transferred to [Γ 3, Γ 6 ] = e, which appears in (C 3). Now, relation (C 7) gets a new form as follows: [ 2 Γ 6Γ 2, Γ 2 Γ 11 2 ] (br4) = [Γ 6 Γ 2 6, Γ 2Γ 11 2 ] (br4) = [Γ 6 ( 6 Γ 2Γ 6 2 Γ 6)G 1 6, 2Γ 11 ]G 1 2 ] = [Γ 6, Γ 11 ] = e. This relation already appears in (C 2).
Fundamental Group of the Complement of the Branch Curve of CP 1 T 13 Therefore, we are left with Γ 1 and {Γ j } 6 j=1. The relations in π 1(C 2 S) are Γ 5, Γ 6 = Γ 3, Γ 4 = Γ 2, Γ 6 = Γ 4, Γ 5 = e, Γ 2, Γ 1 = Γ 2, Γ 1 = Γ 2, Γ 1 Γ 1 1 = e, Γ 3, Γ 1 = Γ 3, Γ 1 = Γ 3, Γ 1 Γ 1 1 = e, [Γ 11, Γ 4 ] = [Γ 11, Γ 5 ] = [Γ 11, Γ 6 ] = e, [Γ 2, Γ 3 ] = [Γ 2, Γ 4 ] = [Γ 2, Γ 5 ] = e, [Γ 3, Γ 5 ] = [Γ 3, Γ 6 ] = e, [Γ 4, Γ 6 ] = e. These are eactly the relations (begin1) (begin2) mentioned in the theorem. The relation (begin3) can be obtained from (br1) (br6) as follows. We equate first (br4) and (br6): 2 Γ 1 2 Γ 6Γ 2 Γ 2 = 5 Γ 1 5 Γ 6Γ 5 Γ 5. We use (tr1) and (tr3) to get 2 Γ 6Γ 2 6 Γ 2 = 5 Γ 6Γ 5 6 Γ 5. Substituting (br3) and (br2) in these relations, we have which by (tr5) get the form We substitute (br5) to get 2 Γ 6( 1 Γ 1 1 Γ 2Γ 1 Γ 1 ) 6 Γ 2 = 5 Γ 6( 4 Γ 1 4 Γ 5Γ 4 Γ 4 ) 6 Γ 5, 2 Γ 6( 1 Γ 1 1 Γ 2Γ 1 Γ 1 ) 6 Γ 2 = 4 Γ 5Γ 4 5 Γ 4 6 Γ 5. 2 Γ 6 1 Γ 1 1 Γ 2Γ 1 Γ 1 6 Γ 2 = 4 Γ 5( 3 Γ 4Γ 3 4 Γ 3) 5 Γ 4 6 Γ 5. And by (br1), we get 2 Γ 6 1 Γ 1 1 Γ 2Γ 1 Γ 1 6 Γ 2 = 4 Γ 5 3 Γ 4( 1 Γ 1 1 Γ 3Γ 1 Γ 1 ) 4 Γ 3 5 Γ 4 6 Γ 5. By (tr4) and (tr6), 2 Γ 6 1 Γ 2Γ 1 2 Γ 1 6 Γ 2 = 4 Γ 5 3 Γ 4 1 Γ 3Γ 1 3 Γ 1 4 Γ 3 5 Γ 4 6 Γ 5. (C*) By the commutations, we rewrite the equations as follows: Γ 1 = (Γ 1 ) 4 Γ 1 5 Γ 1 6 Γ1Γ 1 2 Γ3Γ4Γ5Γ6Γ 1 1 Γ2. This means [Γ 1, 3 Γ 1 4 Γ 1 5 Γ 1 6 Γ 1 2 Γ 3 Γ 4 Γ 5 Γ 6 1 Γ 2] = e. The proof of this theorem supports our claim at the end of Section 2. If we denote then G = π 1 (C 2 S) = DΩ 6 / ρ = e. Theorem 3.5 ρ = [Γ 1, 3 Γ 1 4 Γ 1 5 Γ 1 6 Γ 1 2 Γ 3 Γ 4 Γ 5 Γ 6 1 Γ 2], The group G is isomorphic to the quotient of G by the relation 3 Γ 1Γ 3 Γ 4 Γ 5 Γ 6 Γ 2 (Γ 1 Γ 1 )Γ 2 Γ 6 Γ 5 Γ 4 1 Γ 3 = 1. (pr)
14 Amram M. et al. Proof In order to find G, we have to add the projective relation Γ 6 Γ 6 Γ 1 Γ 1 = e to the presentation G. We compute first each factor Γ j Γ j separately. Denote the left hand-side of (C*) as a. By (br1) (br6) and (C*), we have Γ 6 Γ 6 = a Γ 6 Γ 5 Γ 5 = 6 Γ 5a 5 Γ 6 Γ 5 Γ 4 Γ 4 = 5 Γ 4 6 Γ 5a 4 Γ 5 Γ 4 Γ 3 Γ 3 = 4 Γ 3 5 Γ 4 6 Γ 5a 4 Γ 5 3 Γ 4 Γ 3 Γ 2 Γ 2 = 6 Γ 2a 2 Γ 6 Γ 2. By relations (begin1) (begin2), the projective relation is transformed to (pr) through the following steps. aγ 6 6 Γ 5a 5 Γ 6Γ 5 5 Γ 4 6 Γ 5a 4 Γ 5Γ 4 4 Γ 3 5 Γ 4 6 Γ 5a 4 Γ 5 3 Γ 4Γ 3 6 Γ 2a 2 Γ 6 Γ 2 Γ 1 Γ 1 = e, aγ 5 a 5 Γ 4Γ 5 a 5 Γ 1 4 Γ 3Γ 4 Γ 5 a 4 Γ 5 3 Γ 4Γ 3 6 Γ 1 2 Γ 1 Γ 1 = e, 2 (Γ 1 Γ 1) 2 (Γ 1 Γ 1)Γ 2 Γ 5 2 (Γ 1 Γ 1) 1 2 Γ 6Γ 2 (Γ 1 Γ 1 )Γ 2 5 Γ 4Γ 5 2 (Γ 1 Γ 1) 1 2 Γ 6Γ 2 (Γ 1 Γ 1 )Γ 2 5 Γ 1 4 Γ 3Γ 4 Γ 5 2 (Γ 1 Γ 1) 1 2 Γ 6Γ 2 (Γ 1 Γ 1 )Γ 2 4 Γ 5 3 Γ 4Γ 3 6 Γ 2 2 (Γ 1 Γ 1) 1 2 Γ 6Γ 2 (Γ 1 Γ 1 )Γ 2 2 Γ 6Γ 2 (Γ 1 Γ 1 ) = e, Γ 6 Γ 5 Γ 6 5 Γ 4Γ 5 Γ 6 Γ 2 (Γ 1 Γ 1 ) 5 Γ 1 4 Γ 3Γ 4 Γ 5 (Γ 1 Γ 1 ) 1 2 Γ 6Γ 2 (Γ 1 Γ 1 )Γ 2 4 Γ 5 3 Γ 4Γ 3 (Γ 1 Γ 1 ) 1 Γ 2 (Γ 1 Γ 1 )Γ 2 (Γ 1 Γ 1 ) 2 (Γ 1 Γ 1) 1 2 = e Γ 5 Γ 6 Γ 4 Γ 5 Γ 6 5 Γ 2(Γ 1 Γ 1 ) 4 Γ 3Γ 4 (Γ 1 Γ 1 ) 1 2 Γ 5Γ 6 5 Γ 2(Γ 1 Γ 1 )Γ 2 Γ 6 4 Γ 5 3 Γ 4Γ 3 (Γ 1 Γ 1 ) 1 Γ 2 (Γ 1 Γ 1 )Γ 2 (Γ 1 Γ 1 ) 2 Γ 1 2 = e, Γ 5 Γ 4 Γ 5 Γ 6 Γ 2 (Γ 1 Γ 1 ) 4 Γ 3Γ 4 (Γ 1 Γ 1 ) 1 2 Γ 5Γ 6 5 Γ 2(Γ 1 Γ 1 )Γ 2 Γ 6 4 Γ 5 3 Γ 4Γ 3 (Γ 1 Γ 1 ) 1 Γ 2 (Γ 1 Γ 1 )Γ 2 (Γ 1 Γ 1 ) 2 (Γ 1 Γ 1) 1 2 = e, Γ 6 Γ 2 (Γ 1 Γ 1 )Γ 3 (Γ 1 Γ 1 ) 1 2 Γ 4 6 Γ 5Γ 6 Γ 2 (Γ 1 Γ 1 )Γ 2 Γ 6 Γ 5 Γ 4 Γ 3 = Γ 2 (Γ 1 Γ 1 )Γ 2 (Γ 1 Γ 1 ) 1 2 (Γ 1 Γ 1) 1 2 (Γ 1 Γ 1). The right hand-side of this equation gets the form Γ 2 Γ 1 Γ 1 Γ 2 1 Γ 1 1 Γ 1 2 Γ 1 1 Γ 1 1 Γ 1 2 Γ 1 Γ 1 = Γ 2 Γ 1 2 Γ 1Γ 2 1 Γ 1 2 Γ 1 1 Γ 1 1 Γ 1 2 Γ 1Γ 1 Therefore, we are left with = Γ 2 Γ 1 2 Γ 1 1 Γ 1 2 Γ 1 Γ 1 1 Γ 1 1 Γ 1 2 Γ 1 Γ 1 = Γ 2 Γ 1 2 Γ 1 1 Γ 1 Γ 1 1 Γ 1 1 Γ 1 1 Γ 1 2 Γ 1 Γ 1 1 Γ 1 1 Γ 1 = Γ 2 Γ 1 2 Γ 1 1 Γ 1 2 Γ 1 = e. Γ 6 Γ 2 (Γ 1 Γ 1 )Γ 3 (Γ 1 Γ 1 ) 1 2 Γ 4Γ 5 Γ 6 5 Γ 2(Γ 1 Γ 1 )Γ 2 Γ 6 Γ 5 Γ 4 Γ 4 Γ 3 = e.
Fundamental Group of the Complement of the Branch Curve of CP 1 T 15 This relation gets the form Γ 2 Γ 1 Γ 3 Γ 3 1 Γ 1 1 Γ 1 2 Γ 4Γ 5 Γ 6 5 Γ 2Γ 1 Γ 1 Γ 2 Γ 6 Γ 5 Γ 6 Γ 4 Γ 3 = e, Γ 1 Γ 1 Γ 3 1 Γ 1 1 Γ 4Γ 5 2 Γ 6Γ 2 }{{} Γ 6Γ 2 6 Γ 1 Γ 1 Γ 2 Γ 6 Γ 2 }{{} Γ 6Γ 2Γ 6 Γ 5 Γ 4 Γ 3 = e, Γ 1 Γ 1 Γ 3 (Γ 1 Γ 1 ) 1 Γ 4 Γ 5 Γ 6 Γ 2 (Γ 1 Γ 1 )Γ 2 Γ 6 Γ 5 Γ 4 Γ 3 = e, Γ 3 Γ 4 Γ 5 Γ 6 Γ 2 (Γ 1 Γ 1 )Γ 2 Γ 6 Γ 5 Γ 4 Γ 3 (Γ 1 Γ 1 )Γ 3 (Γ 1 Γ 1 ) 3 = e. The factor Γ 3 Γ 1 Γ 1 Γ 3 1 Γ 1 1 Γ 1 3 equals 1 Γ 3Γ 1 3 Γ 1. Therefore, we get 3 Γ 1Γ 3 Γ 4 Γ 5 Γ 6 Γ 2 Γ 1 Γ 1 Γ 2 Γ 6 Γ 5 Γ 4 1 Γ 3 = 1. Therefore, by substituting the projective relation in the commutative we get totally a new relation: Γ 6 Γ 5 Γ 4 Γ 3 (Γ 1 Γ 1 )Γ 3 Γ 4 Γ 5 Γ 6 ( 1 Γ 2 3 Γ 1) = (Γ 1 2 Γ 3 1 )Γ 4Γ 5 Γ 6 Γ 2 (Γ 1 Γ 1 )Γ 2 Γ 6 Γ 5 Γ 4. That means that the projective group G is a quotient of DΩ 6 as G is, but under the above relation, instead ρ. Remark 3.6 When embedding T in CP n 1, then CP 1 T can be embedded in CP 2n 1. By abuse of notation, we denote by G and by G the corresponding fundamental groups in this case. Therefore, by the same methods as above, G and G are quotients of DΩ 2n by a single relation. We now propose our conjecture about the structure of G and G in the sufficiently-ample embedding case. Conjecture 3.7 Embed CP 1 in CP m (m 2), and T in CP n 1 (n 3); CP 1 T is then embedded by the Segre embedding CP 1 CP n 1 CP mn+n 1. Denote G m,n = π 1 (CP 2 S m,n ), where S m,n is the branch curve of a generic projection of the embedded surface, and G m,n = π 1 (C 2 S m,n ) where S m,n is a generic affine piece of S m,n. We present in Figure 12 the degeneration of the embedded surface, and a graph T m,n which is related to it. Denote by A Tn,m the Artin group corresponding to T n,m. Notice that after numeration of the 2nm planes, there eist an epimorphism to the symmetric group A Tn,m Sym 2nm, sending each generator to the transposition (i j), s.t. the generator connects the planes i and j. Denote also à Tn,m = A Tn,m / [X, Y ], where X, Y are two transversal braids; and by  Tn,m,0 = ker (ÃT n,m Sym 2mn ) ker (Ab(ÃT n,m )). Then G = (ÃT n,m ÂT n,m,0)/n, where N is a normal subgroup of the above semi-direct product and gcd (m, n) = 1; and when gcd (m, n) > 1, replace, in G, the factor ÂT n,m,0 by ÂT n,m,0 ÂT n,m,0.
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