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Subgroups and Quotients of Fundamental Groups By Samuel Mark Corson Dissertation Submitted

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Topology and its Applications 130 (2003) 159–173 www.elsevier.com/locate/topol

Fundamental groups of some quadric-line arrangements Meirav Amram a,∗ , Mina Teicher b , A. Muhammed Uludag c a Meirav Amram, Mathematisches Institut, Bismarck Strasse 1 1/2, Erlangen, Germany b Mathematics Department, Bar-Ilan University, Ramat-Gan, Israel c Mathematics Department, Galatasaray University, Ortaköy, Istanbul, Turkey

Received 15 June 2002; received in revised form 29 July 2002

Abstract In this paper we obtain presentations of fundamental groups of the complements of three quadricline arrangements in P2 . The first arrangement is a smooth quadric Q with n tangent lines to Q, and the second one is a quadric Q with n lines passing through a point p ∈ / Q. The last arrangement consists of a quadric Q with n lines passing through a point p ∈ Q. 2002 Elsevier Science B.V. All rights reserved. Keywords: Fundamental groups; Complements of curve; Conic-line arrangements

1. Introduction This is the first of a series of articles in which we shall study the fundamental groups of complements of some quadric-line arrangements. In contrast with the extensive literature on line arrangements and the fundamental groups of their complements, (see, e.g., [14,7, 15]), only a little known about the quadric-line arrangements (see [12,1,2]). The present article is dedicated to the computation of the fundamental groups of the complements of three infinite families of such arrangements. A similar analysis for the quadric-line arrangements up to degree six will be done in our next paper. Let C ⊂ P2 be a plane curve and ∗ ∈ P2 \ C a base point. By abuse of language we will call the group π1 (P2 \ C, ∗) the fundamental group of C, and we shall frequently omit base * Corresponding author.

E-mail addresses: [email protected], [email protected] (M. Amram), [email protected] (M. Teicher), [email protected] (A.M. Uludag). 0166-8641/02/$ – see front matter 2002 Elsevier Science B.V. All rights reserved. doi:10.1016/S0166-8641(02)00218-3

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Fig. 1. The arrangement A3 .

points and write π1 (P2 \ C). One is interested in the group π1 (P2 \ C) mainly for the study of the Galois coverings X → P2 branched along C. Many interested surfaces have been constructed as branched Galois coverings of the plane, for example for the arrangement A3 in Fig. 1, there are Galois coverings X → P2 branched along A3 such that X P1 × P1 , or X is an abelian surface, a K3 surface, or a quotient of the two-ball B2 (see [9,8,17]). Moreover, some line arrangements defined by unitary reflection groups studied in [13] are related to A3 via orbifold coverings. For example, if L is the line arrangement given by the equation xyz(x + y + z)(x + y − z)(x − y + z)(x − y − z) = 0 then the image of L under the branched covering map [x : y : z] ∈ P2 → [x 2 : y 2 : z2 ] ∈ P2 is the arrangement A3 , see [17] for details. The standard tool for fundamental group computations is the Zariski–van Kampen algorithm [19,18], see [3] for a modern approach. We use a variation of this algorithm developed in [16] for computing the fundamental groups of real line arrangements and avoids lengthy monodromy computations. The arrangements Bn and Cn discussed below are of fiber type, so presentations of their fundamental groups could be easily found as an extension of a free group by a free group. However, our approach has the advantage that it permits to capture the local fundamental groups around the singular points of these arrangements. The local fundamental groups are needed for the study of the singularities of branched of P2 branched along these arrangements. In Section 2 below, we give fundamental group presentations and prove some immediate corollaries. In Section 3 we deal with the computations of fundamental group presentations given in Section 2.

2. Results Let C ⊂ P2 be a plane curve and B an irreducible component of C. Recall that a meridian µ of B in P2 \ C with the base point ∗ ∈ P2 is a loop in P2 \ C obtained by following a path ω with ω(0) = ∗ and ω(1) belonging to a small neighborhood of a smooth point p ∈ B \ C, turning around C in the positive sense along the boundary of a small disc ∆ intersecting B transversally at p, and then turning back to ∗ along ω. The meridian µ represents a homotopy class in π1 (P2 \ C, ∗), which we also call a meridian of B. Any

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two meridians of B in P2 \ C are conjugate elements of π1 (P2 \ C) (see, e.g., [10, 7.5]), hence the meridians of irreducible components of C are supplementary invariants of the pair (P2 , C). These meridians are specified in presentations of the fundamental group given below, they will be used in orbifold-fundamental group computations in [17]. 2.1. The arrangement An Theorem 1. Let An := Q ∪ T1 ∪ · · · ∪ Tn be an arrangement consisting of a smooth quadric Q with n distinct tangent lines T1 , . . . , Tn . Then κi = τi κi−1 τ −1 , 2 i n i 2 τ1 , . . . , τn , (κi τi )2 = (τi κi )2 , 1 i n (1) π1 P \ An κ1 , . . . , κn κi−1 τi κi , τj = 1, 1 i < j n τ · · · τ κ2 = 1 n 1 1 where κi are meridians of Q and τi is a meridian of Ti for 1 i n. Local fundamental groups around the singular points of An are generated by κi−1 τi κi , τj for the nodes Ti ∩ Tj and by κi , τi for the tangent points Ti ∩ Q. Part (i) of the corollary below is almost trivial. Part (ii) appears in [6], and part (iii) was given in [4]. Corollary 2. (i) One has: π1 (P2 \ A1 ) Z. (ii) The group π1 (P2 \ A2 ) admits the presentation π1 P2 \ A2 τ, κ | (τ κ)2 = (κτ )2 ,

(2)

where κ is a meridian of Q and τ is a meridian of T1 . A meridian of T2 is given by κ −2 τ −1 . (iii) The group π1 (P2 \ A3 ) admits the presentation π1 P2 \ A3 τ, σ, κ | (τ κ)2 = (κτ )2 , (σ κ)2 = (κσ )2 , [σ, τ ] = 1 (3) where σ , τ are meridians of T1 and T3 respectively, and κ is a meridian of Q. A meridian of T2 is given by (κτ κσ )−1 . A group G is said to be big if it contains a non-abelian free subgroup, and small if G is almost solvable. In [6], it was proved by V. Lin that the group (2) is big. Below we give an alternative proof: Proposition 3. For n > 1, the group π1 (P2 \ An ) is big. Proof. A group with a big quotient is big. Since τn+1 is a meridian of Tn+1 in π1 (P2 \ An+1 ), one has π1 P2 \ An π1 P2 \ An+1 /τn+1 ,

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and it suffices to show that the group π1 (P2 \ A2 ) is big. In the presentation (2), applying the change of generators α := τ κ, β := τ gives π1 P2 \ A2 α, β | α 2 , β = 1 . Adding the relations α 2 = β 3 = 1 to the latter presentation gives a surjection π1 (P2 \ A2 ) Z/(2) ∗ Z/(3). Since the commutator subgroup of Z/(2) ∗ Z/(3) is the free group on two generators (see [5]), we get the desired result. ✷ 2.2. The arrangement Bn Theorem 4. Let Bn := Q ∪ T1 ∪ T2 ∪ L1 ∪ · · · ∪ Ln be an arrangement consisting of a smooth quadric Q with n + 2 distinct lines T1 , T2 , L1 , . . . , Ln all passing through a point p∈ / Q such that T1 , T2 are tangent to Q. Then one has (κτ )2 = (τ κ)2 2 λ ] = 1, 1 i n π1 P \ Bn τ, κ, λ1 , . . . , λn [κ, (4) i τ −1 κτ, λi = 1, 1 i n where τ is a meridian of T1 , λi is a meridian of Li for 1 i n, and κ is a meridian of Q. A meridian σ of T2 is given by σ := (λn . . . λ1 κ 2 τ )−1 . Local fundamental groups around the singular points of Bn are generated by κ, λi and τ −1 κτ, λi for the nodes Li ∩ Q, by κ, τ for the tangent point T1 ∩ Q, and by κ, σ for the tangent point T2 ∩ Q. Corollary 5. (i) Put Bn := Bn \ T1 and Bn := Bn \ T2 . Then π1 P2 \ Bn π1 P2 \ Bn+1 κ, λ1 , . . . , λn | [κ, λi ] = 1, 1 i n .

(5)

Proof. One has π1 (P2 \ Bn ) π1 (P2 \ Bn )/τ . Setting τ = 1 in presentation (4) gives π1 P2 \ Bn κ, λ1 , . . . , λn | [κ, λi ] = 1, 1 i n . Setting τ = 1 in the expression for a meridian σ of T2 given in Theorem 4 shows that (λn . . . λ1 κ 2 )−1 is a meridian of T2 in π1 (P2 \ Bn ). In order to find π1 (P2 \ Bn ), it suffices to set λn . . . λ1 κ 2 = 1 in the presentation of π1 (P2 \ Bn ). Eliminating λn by this relation yields the presentation π1 P2 \ Bn κ, λ1 , . . . , λn−1 | [λi , κ] = λn−1 . . . λ1 κ 2 , κ = 1 .

Fig. 2. Arrangements B2 and B2 .

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Fig. 3. Arrangements C3 and C3 .

Since the last relation above is redundant, we get the desired isomorphism π1 (P2 \ Bn ) π1 (P2 \ Bn+1 ). ✷ Note that the groups π1 (P2 \ Bi ) are abelian for i = 0, 1, 2. Hence, the groups π1 (P2 \ Bi ) are abelian for i = 0, 1. Otherwise, setting κ = 1 in presentation (5) gives the free group on n − 1 generators, which shows that these groups are big. The groups π1 (P2 \ Bn ) are always big, since the arrangement B0 is same as A2 , and π1 (P2 \ A2 ) is big by Proposition 3. 2.3. The arrangement Cn Theorem 6. Let Cn := Q ∪ T ∪ L1 ∪ · · · ∪ Ln be an arrangement consisting of a smooth quadric Q with n + 1 distinct lines T , L1 , . . . , Ln , all passing through a point p ∈ Q such that T is tangent to Q. Then one has π1 P2 \ Cn κ, λ1 , . . . , λn | [κ, λi ] = 1, 1 i n , (6) where κ is a meridian of Q and λi is a meridian of Li for 1 i n. A meridian τ of T is given by τ := (λn . . . λ1 κ 2 )−1 . Local fundamental groups around the singular points of Cn are generated by κ, λi for the nodes Li ∩ Q, and by τ, λ1 , . . . , λn , κ for the point p. Note that the arrangement Cn is a degeneration (in the sense of Zariski) of the arrangement Bn as the point p approaches to Q. By Zariski’s “semicontinuity” theorem of the fundamental group [19] (see also [5]), there is a surjection π1 (P2 \ Cn ) π1 (P2 \ Bn ). In our case, this is also an injection: Corollary 7. (i) π1 (P2 \ Bn ) π1 (P2 \ Cn ). (ii) Put Cn := Cn \ T . Then π1 (P2 \ Cn ) π1 (P2 \ Cn+1 ). Proof. Part (i) is obvious. The proof of part (ii) is same as the proof of Corollary 5, (ii).

✷

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3. The arrangement An It is easily seen that any two arrangements An with fixed n are isotopic. In particular, the groups π1 (P2 \ An ) are isomorphic. Hence one can take as a model of the arrangements An the quadric Q defined by x 2 + y 2 = z2 , where [x : y : z] ∈ P2 is a fixed coordinate system in P2 . Pass to the affine coordinates in C2 P2 \ {z = 0}. Choose real numbers x1 , . . . , xn such that −1 < x1 < x2 < · · · < xn < 0, and define yi to be the positive solution of xi2 + yi2 = 1 for 1 i n. Put ti := (xi , yi ) ∈ Q, and take Ti to be the tangent line to Q at the point ti (see Fig. 4). Let pr1 : C2 \ An → C be the first projection. The base of this projection will be denoted by B. Put Fx := pr−1 1 (x), and denote by S the set of singular fibers of pr1 . It is clear that if Fx ∈ S, then x ∈ [−1, 1]. There are three types of singular fibers: (i) The fibers F1 and F−1 , corresponding to the ‘branch points’ (−1, 0) and (1, 0). (ii) The fibers Fxi (1 i n) corresponding to the ‘tangent points’ ti = (xi , yi ) = Ti ∩Q. (iii) The fibers Fai,j (1 i = j n) corresponding to the nodes ni,j = (ai,j , bi,j ) := Ti ∩ Tj . One can arrange the lines Ti such that −1 < x1 < a1,2 < a1,3 < · · · < a1,n < x2 < a2,3 < · · · < xn < 1. Identify the base B of the projection pr1 with the line y = −2 ⊂ C2 . Let N be the number of singular fibers and let −1 = s1 < s2 < · · · < sN−1 < sN = 1 be the elements of S ∩ B (so that s2 = x1 , s3 = a1,2 , s4 = a1,3 , and so on). In B, take small discs ∆i around the points si , and denote by ci , di (ci < di ) the points ∂∆i ∩ R for 1 i N (see Fig. 5). Put B1 := [c1 , c2 ] ∪ ∆1 and for 2 i N let Bi := [c1 , ci+1 ] ∪ ∆1 ∪ · · · ∪ ∆i . Let Xi := pr−1 (Bi ) be the restriction of the fibration pr to Bi . Let Ai := ∆i ∪ ∂ (z) 0, c2 (z) ci \ (∆2 ∪ ∆3 ∪ · · · ∪ ∆i−1 ) and let Yi := pr−1 (Ai ) be the restriction of the fibration pr to Ai (see Fig. 6). Clearly, Xi = Xi−1 ∪ Yi for 2 i N . We will use this fact to compute the groups π1 (Xi , ∗) recursively, where ∗ := (c2 , −2) is the base point. For details of the algorithm we apply below, see [16].

Fig. 4.

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Fig. 5. The base B.

Fig. 6. The space Ai .

Fig. 7.

Identify the fibers of pr1 with F0 via the second projection pr2 : (x, y) ∈ C2 → y ∈ C. In each one of the fibers Fci (respectively Fdi ) take a basis for π1 (Fci , −2) (respectively for π1 (Fdi , −2)) as in Fig. 7 (for Fdi , just replace γ ’s by θ ’s in Fig. 7). We shall denote (i) (i) these basis by the vectors Γi := [γ1(i) , . . . , γn+2 ] (respectively Θi := [θ1(i) , . . . , θn+2 ]). Let νi ⊂ Bi ⊂ B be a path starting at νi (0) = c2 , ending at νi (1) = ci and such that νi [0, 1] = ∂ (z) 0, c2 (z) ci \ (∆2 ∪ ∆3 ∪ · · · ∪ ∆i−1 ) . Similarly, let ηi ⊂ Bi ⊂ B be a path starting at η(0) = c2 , ending at η(0) = di and such that ηi [0, 1] = ∂ (z) 0, c2 (z) di \ (∆2 ∪ ∆3 ∪ · · · ∪ ∆i ) . For 2 i N and 1 j n + 2 each loop γ˜j := νi · γj · νi−1 represents a homotopy (i) class in π1 (Xi , ∗), where ∗ := (c2 , −2) is the base point. Similarly, each loop θ˜ := (i)

(i)

j

ηi · θi · ηi−1 represents a homotopy class in π1 (Xi , ∗). Denote Γi := [γ˜1 , . . . , γ˜n+2 ], and i := [θ˜ (i) , . . . , θ˜ (i) ]. Θ (i)

1

n+2

(i)

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It is well known that the group π1 (Yi , ∗) has the presentation

(i) γ˜1 , . . . , γ˜n+2 | γ˜j(i) = Mi γ˜j(i) , 1 j n + 2

(7)

where Mi : π1 (Fci , −2) → π1 (Fci , −2) is the monodromy operator around the singular fiber above si . It is also well known that if it is the branches of An corresponding to (i) the loops γ˜k(i) and γ˜k+1 that meet above si , then the only non-trivial relation in (7) is (i)

(i)

(i)

(i)

(i) (i)

γ˜k = γ˜k+1 in case of a branch point, [γ˜k , γ˜k+1 ] = 1 in case of a node, and (γ˜k γ˜k+1 )2 = (i) (i)

(γk γk+1 )2 in case of a tangent point. Now suppose that the group π1 (Xi−1 , ∗) is known, with generators Γ2 . Recall that Xi = Xi−1 ∪ Yi . In order to find the group π1 (Xi , ∗), one has to express the base Γi in terms of the base Γi . Adding to the presentation of π1 (Xi−1 ) the relation obtained by writing the relation of π1 (Yi ) in the new base then yields a presentation of π1 (Xi ). Note that, since the space Yi is eventually glued to Xi−1 , it suffices to find an expression of Γi in terms of the base Γ2 in the group π1 (Xi−1 , ∗). Since all the points of An above the interval [di−1 , ci ] are smooth and real, one has Fact. The loops θ˜j(i−1) and γ˜j(i) are homotopic in Xi (or in Yi ) for 2 i N and i−1 and Γi are homotopic. 1 j n + 2. In other words, the bases Θ i in terms of the base Γi the following lemma will be In order to express the base Θ helpful. Lemma 8. Let Ck : x 2 − y k+1 = 0 be an Ak singularity, where k = 1 or k = 3. Put D := {(x, y): |x| 1, |y| 1} and let pr1 := (x, y) ∈ D \ Ck → (x, −1) be the first projection. Denote by Fx the fiber of pr1 above (x, −1). Identify the fibers of pr1 via the second projection. Let −1 < c < 0 be a real number, and put d := −c. In Fc (respectively in Fd ) take a basis Γ := [γ1 , γ2 ] for π1 (Fc , −1) (respectively a basis Θ := [θ1, θ2 ] for π1 (Fd , −1)) as in Fig. 8. Let η be the path η(t) := ceπit , and put θ˜i := η · θ · η−1 for i = 1, 2. Then γi , θ˜i are loops in D \ Ck based at ∗ := (c, −1), and one has (i) If k = 1, then θ˜1 is homotopic to γ2 , and θ˜2 is homotopic to γ1 , in other words, = [γ2 , γ1 ]. Θ = [γ2 γ1 γ −1 , γ −1 γ2 γ1 ]. (ii) If k = 3, then Θ 2 1

Fig. 8.

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Fig. 9.

Proof. Since π1 (D \ C2 ) is abelian, part (i) is obvious. For part (ii), note that the points of intersection Fη(t ) ∩ C4 are y1 := c2 e2πit and y2 := −c2 e2πit . Hence, when we move the fiber Fc over Fd along the path η, y1 and y2 make one complete turn around the origin in the positive sense. The loops γ1 , γ2 are transformed to loops γ 1 , γ 2 ⊂ Fd as in Fig. 9. It follows that the loop η · γ i · η−1 is homotopic to γi for i = 1, 2. This homotopy can be constructed explicitly as follows: Let Φη(t ) : Fc → Fη(t ) be the corresponding Leftschez homeomorphism (see [11]). Then 0 s t/3, η(3s), Φ γ 3(s − t/3)/(3 − 2t) , t/3 s 1 − t/3, H (s, t) := η(t ) i η3(1 − s), 1 − t/3 s 1 gives a homotopy between γi and γ i . Expressing θ˜i in terms of γ i , we get −1 −1 −1 θ˜1 = γ −1 1 γ 2 γ 1 γ 2 γ 1 = γ1 γ2 γ1 γ2 γ1 , −1 θ˜2 = γ −1 1 γ 2 γ 1 = γ1 γ2 γ1 .

Since from the monodromy one has the relation (γ1 γ2 )2 = (γ2 γ1 )2 , the expression for θ˜1 can be simplified to get θ˜1 = γ2 γ1 γ2−1 . ✷ Now we proceed with the computation of the groups π1 (Xi ). Clearly, the group π1 (X2 ) is generated by the base (2) Γ2 = γ1(2), γ2(2) , . . . , γn+2 with the only relations γ1(2) = γ2(2) and

(2) (2) 2

γ2 γ3

(8) (2) (2) 2 = γ3 γ2 .

(9)

Put [κ1 , κ1 , τ1 , . . . , τn ] := Γ2 . Then relation (9) becomes (κ1 τ1 )2 = (τ1 κ1 )2 .

(10)

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By Lemma 8 and the above Fact, one has 2 = κ1 , τ1 κ1 τ −1 , κ −1 τ1 κ1 , τ2 , . . . , τn . Γ3 = Θ 1 1 Since s3 corresponds to the node T1 ∩ T2 , the next relation is −1 κ1 τ1 κ1 , t2 = 1.

(11)

Hence,

π1 (X3 , ∗) κ1 , τ1 , . . . , τn | (κ1 τ1 )2 = (τ1 κ1 )2 , κ1−1 τ1 κ1 , τ2 = 1 . By Lemma 8, one has 3 = κ1 , τ1 κ1 τ −1 , τ2 , κ −1 τ1 κ1 , τ3 , . . . , τn . Γ4 = Θ 1 1 Since s4 corresponds to the node T1 ∩ T3 , one has the relation −1 κ1 τ1 κ1 , τ3 = 1. Hence,

π1 (X4 , ∗) κ1 , τ1 , . . . , τn | (κ1 τ1 )2 = (τ1 κ1 )2 , −1 κ1 τ1 κ1 , τ2 = κ1−1 τ1 κ1 , τ3 = 1 . By Lemma 8, one has 4 = κ1 , τ1 κ1 τ −1 , τ2 , τ3 , κ −1 τ1 κ1 , τ4 , . . . , τn . Γ5 = Θ 1 1 Since sk corresponds to the node T1 ∩ Tk−1 for 2 k n + 1, repeating the above procedure gives the presentation

π1 (Xn+1 , ∗) κ1 , τ1 , . . . , τn | (κ1 τ1 )2 = (τ1 κ1 )2 , κ1−1 τ1 κ1 , τk = 1, 2 k n and

n+1 = κ1 , κ2 , τ2 , τ3 , . . . , τn , κ −1 τ1 κ1 , Γn+2 = Θ 1

where we put κi+1 := τi κi τi−1 for 1 i n − 1. The next point sn+2 corresponds to the tangent point T2 ∩ Q. This gives the relation (κ2 τ2 )2 = (τ2 κ2 )2 and

(12)

n+1 = κ1 , τ2 κ2 τ −1 , κ −1 τ2 κ2 , τ3 , . . . , τn , κ −1 τ1 κ1 . Γn+2 = Θ 2 2 1

Now comes the n − 2 points sk corresponding to the nodes T2 ∩ Tk−n for n + 3 2n + 1. These give the relations −1 κ2 τ2 κ2 , τk = 1, 3 k n. Hence, one has

π1 (Xn+1 , ∗) κ1 , κ2 , τ1 , . . . , τn | κ2 = τ1 κ1 τ1−1 , (κi τi )2 = (τi κi )2 , −1 κi τi κi , τk = 1, i < k n, i = 1, 2 .

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We proceed in this manner until the last singular fiber sN . Since this is a branch point, the final relation is κn = κ1 . This gives the presentation π1 (XN , ∗) τi , κi , 1 i n

(13) κi = τi κi−1 τ −1 , 2 i n i (κi τi )2 = (τi κi )2 , 1 i n κ −1 τ κ , τ = 1, 1 i < j n i i i j κ =κ 1 n

.

(14)

Adding to this presentation of π1 (XN , ∗) the projective relation τn · · · τ1 κ12 = 1 gives the presentation κi = τi κi−1 τ −1 , 2 i n i (κi τi )2 = (τi κi )2 , 1 i n 2 π1 P \ An τi , κi , 1 i n −1 . (15) κi τi κi , τj = 1, 1 i < j n τ . . . τ κ 2 = 1, κ = κ n 1 1 1 n Note that the relation κ1 = κn is redundant. Indeed, since κi = τi κi−1 τi−1 , one has κn = (τn . . . τ1 )κ1 (τn . . . τ1 )−1 .

(16)

But τn . . . τ1 = κ −2 by the projective relation. Substituting this in (16) yields the relation κ1 = κn . This finally gives the presentation (1) and proves Theorem 1. Claims regarding the local fundamental groups around the singular points of An are direct consequences of the above algorithm. 3.1. Proof of Corollary 2 (i) The arrangement A1 . Writing down the presentation (1) explicitly for n = 1 gives (κ1 τ1 )2 = (κ1 τ1 )2 2 π1 P \ A1 κ1 , τ1 2 . t1 κ1 = 1 Eliminating τ1 from the last relation shows that π1 (P2 \ A1 ) Z. (ii) The arrangement A2 . Writing down the presentation (1) explicitly for n = 2 gives (1) κ2 = τ1 κ1 τ −1 1 (2) (κ1 τ1 )2 = (τ1 κ1 )2 2 2 2 . π1 P \ A2 κ1 , κ2 , τ1 , τ2 (3) (κ2 τ2 ) = (τ2 κ2 ) (4) κ −1 t κ , t = 1 1 1 1 2 (5) τ τ κ 2 = 1 2 1 1 Eliminating κ2 by (1) and τ2 by (5) one easily shows that the relations (3) and (4) are redundant. This leaves (2) and gives the desired presentation.

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(iii) The arrangement A3 . Writing down the presentation (1) explicitly for n = 3 gives (1) κ2 = τ1 κ1 τ1−1 (2) κ3 = (τ2 τ1 )κ1 (τ2 τ1 )−1 (3) (κ1 τ1 )2 = (τ1 κ1 )2 (4) (κ2 τ2 )2 = (τ2 κ2 )2 2 κ1 , κ2 , κ3 , −1 2 2 π1 P \ A3 (5) (κ3 τ3 ) = (τ3 κ3 ) (6) κ1 τ1 κ1 , τ2 = 1 . τ1 , τ2 , τ3 (7) κ −1 τ1 κ1 , τ3 = 1 (8) κ −1 τ2 κ2 , τ3 = 1 1 2 (9) τ τ τ κ 2 = 1 3 2 1 1

Eliminate κ2 by (1), κ3 by (2), and τ2 by (9). It can be shown that the relations (4), (6) and (8) are consequences of the remaining relations. The relation (5) becomes (κ1 τ3 )2 = (τ3 κ1 )2 . This gives the presentation

π1 P2 \ A3 κ1 , τ1 , τ3 | (κ1 τ1 )2 = (τ1 κ1 )2 ,

(κ1 τ3 )2 = (τ3 κ1 )2 , κ1−1 τ1 κ1 , τ3 = 1 .

Finally, put κ := κ1 , τ := κ1−1 τ1 κ1 and σ := τ3 . Then τ1 = κτ κ −1 , and the first relation in the above presentation becomes (κ 2 τ κ −1 )2 = (κτ )2 ⇒ (κτ )2 = (τ κ)2 . This gives the desired presentation.

4. The arrangement Bn As in the case of the arrangements An , it is readily seen that arrangements Bn are all isotopic to each other for fixed n, so one can compute π1 (P2 \ Bn ) from the following model for Bn ’s (see Fig. 10): The quadric Q is given by the equation x 2 + y 2 = 1, and p is the point (2, 0). The lines Li intersect Q above the x-axis. The projection to the x-axis has four types of singular fibers:

Fig. 10.

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(i) The fibers F1 and F−1 , corresponding to ‘branch points’. (ii) The fiber Fx corresponding to the ‘tangent points’ (x, y) = t1 := T1 ∩ Q and (x, −y) = t2 := T2 ∩ Q. (iii) The fibers Fa1 , . . . , Fan (−1 < an < · · · < a1 < x) corresponding to the nodes Li ∩ Q lying on the right of the tangent points and the fibers Fb1 , . . . , Fbn (x < b1 < · · · < bn < 1) corresponding to the nodes Li ∩ Q lying on the left of the tangent points. (iv) The fiber F2 , corresponding to the point p. In order to find the group π1 (P2 \ Bn ), we shall apply the same procedure as in the computation of π1 (P2 \ An ). Let y ∈ R be such that −1 < y < an , and take Fy to be the base fiber. Let s1 := −1, si+1 := bi for 1 i n, sn+2 := x, sn+2+i := an+1−i for 1 i n, and s2n+3 := 1, and s2n+4 = 2. Take a basis Γ2 := [τ, κ1 , κ2 , λ1 , . . . , λn , σ ] for Fy as in Fig. 7. Since s1 corresponds to a branch point, one has the relation κ1 = κ2 . Put κ := κ1 = κ2 . The point s2 is a node, and yields the relation [κ, λ1 ] = 1, and one has Γ3 = [τ, κ, λ1 , κ, λ2 , . . . , λn , σ ]. Repeating this for the nodes s3 , . . . , sn+1 gives the relations [κ, λi ] = 1 for 1 i n, and one has Γn+2 = [τ, κ, λ1 , . . . , λn , κ, σ ]. The monodromy around the fiber Fx gives the relations (τ κ)2 = (κτ )2 and (σ κ)2 = (κσ )2 . One has Γn+3 = κτ κ −1 , τ −1 κτ, λ1 , . . . , λn , σ κσ −1 , κ −1 σ κ . Since the points sn+3 , . . . , s2n+2 corresponds to nodes, one has the relations [λi , σ κσ −1 ] = 1, and Γ2n+3 = κτ κ −1 , τ −1 κτ, σ κσ −1 , λ1 , . . . , λn , κ −1 σ κ . The branch point corresponding to s2n+3 yields the relation τ −1 κτ = σ κσ −1 . Together with the projective relation σ λn . . . λ1 κ 2 τ = 1 these relations already gives a presentation of π1 (P2 \ Bn ), since one can always ignore one of the singular fibers when computing the monodromy (see [16]). We obtained the presentation (1) (κτ )2 = (τ κ)2 (2) (κσ )2 = (σ κ)2 −1 −1 (3) τ κτ = σ κσ 2 . π1 P \ Bn Λ, τ, κ, λ1 , . . . , λn , σ =1 1i n (4) [κ, λi ]−1 (5) σ κσ , λi = 1 1 i n (6) σ λ . . . λ κ 2 τ = 1 n

1

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Put Λ := λn . . . λ1 . Eliminating σ by (7), it is easily seen that (3) is redundant. Relation (2) becomes 2 −1 2 −1 2 2 = κ Λκ τ ⇒ τ −1 κτ, Λ = 1. Λκ τ κ But this relation is a consequence of (4), so that (2) is also redundant. Since τ −1 κτ = σ κσ −1 by (3), the relation (5) can be written as [τ −1 κτ, λi ] = 1. This gives the presentation (4) and proves Theorem 4. 5. The arrangement Cn In order to compute the group, consider the model of Cn shown in Fig. 11, where Q is given by x 2 + y 2 = 1. Suppose that the second points of intersection of the lines Li with Q lie above the x-axis. As in the previous cases, take an initial base Γ2 := [κ1 , κ2 , λ1 , . . . , λn , τ ]. The relation induced by the branch point is κ1 = κ2 =: κ. The nodes of Cn will give the relations [κ, λi ] = 1 for 1 i n, and one has Γn+2 = [κ, λ1 , . . . , λn , κ, τ ]. One can simplify the computation of the monodromy around the complicated singular fiber as follows: Put Λ := λn . . . λ1 . By the projective relation one has τ Λκ 2 = 1 ⇒ τ = κ −2 Λ−1 . Hence, [κ, τ ] = 1. Since we also have [κ, λi ] = 1 for 1 i n, this means that when computing the monodromy around this fiber, one can ignore the branch Q. This leaves n + 1 branches intersecting transversally, and the induced relation is (see [16]) [τ Λ, λi ] = [τ Λ, τ ] = 1,

(17)

and one has Γn+3 = [κ, κ, . . .]. The last relation induced by the branch point yields the trivial relation κ = κ, as expected.

Fig. 11.

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Eliminating τ shows that the relations (17) are redundant and gives the presentation π1 P2 \ Cn κ, λ1 , . . . , λn | [λi , κ] = 1 . ✷

Acknowledgements This work was partially supported by the Emmy Noether Research Institute for Mathematics (center of the Minerva Foundation of Germany), the Excellency Center “Group Theoretic Methods in the Study of Algebraic Varieties” of the Israel Science Foundation, and EAGER (EU network, HPRN-CT-2009-00099).

References [1] M. Amram, Braid group and braid monodromy, M.Sc. Thesis, 1995. [2] M. Amram, M. Teicher, Braid monodromy of special curves, J. Knot Theory Ramifications 10 (2) (2001) 171–212. [3] D. Cheniot, Topologie du complémentaire d’un ensemble algébrique projectif, Enseign. Math., II. Ser. 37 (3/4) (1991) 293–402. [4] A. Degtyarev, Quintics in CP 2 with non-abelian fundamental group, St. Petersburg Math. J. 11 (5) (2000) 809–826. [5] A. Dimca, Singularities and the Topology of Hypersurface Complements, in: Universitext, Springer-Verlag, New York, 1992. [6] G. Detloff, S. Orevkov, M. Zaidenberg, Plane curves with a big fundamental group of the complement, Amer. Math. Soc. Transl. Ser. 2 184 (1998). [7] D. Graber, M. Teicher, The fundamental group’s structure of the complement of some configurations of real line arrangements, in: Complex Analysis and Algebraic Geometry. A Volume in Memory of Michael Schneider, de Gruyter, Berlin, 2000, pp. 173–223. [8] R.P. Holzapfel, N. Vladov, Quadric-line configurations degenerating plane Picard–Einstein metrics I–II, in: Proceedings to 60th Birthday of H. Kurke, Math. Ges., Berlin, 2000, to appear. [9] J. Kaneko, S. Tokunaga, M. Yoshida, Complex crystallographic groups II, J. Math. Soc. Japan 34 (4) (1982). [10] K. Lamotke, The topology of complex projective varieties after S. Leftschez, Topology 20 (1979) 15–51. [11] B. Moishezon, M. Teicher, Braid group technique in complex geometry I, Line arrangements in CP2 , Contemp. Math. 78 (1988) 425–555. [12] B. Moishezon, M. Teicher, Braid group technique in complex geometry II, From arrangements of lines and conics to cuspidal curves, in: Algebraic Geometry, in: Lecture Notes in Math., Vol. 1479, Springer-Verlag, Berlin, 1990. [13] P. Orlik, L. Salomon, Arrangements defined by unitary reflection groups, Math. Ann. 261 (1982) 339–357. [14] P. Orlik, H. Terao, Arrangements of Hyperplanes, in: Grundlagen Mat. Wiss., Vol. 300, Springer-Verlag, Berlin, 1992. [15] A. Suciu, Fundamental groups of line arrangements: Enumerative aspects, Contemp. Math. 276 (2001) 43– 79. [16] A.M. Uluda˘g, On finite smooth uniformizations of the plane along line arrangements, to be published. [17] A.M. Uluda˘g, Covering relations between ball quotient orbifolds, to be published. [18] E.R. van Kampen, On the fundamental group of an algebraic curve, Amer. J. Math. 55 (1933) 255–260. [19] O. Zariski, On the Poincare group of rational plane curves, Amer. J. Math. 58 (1936) 607–619.

Fundamental groups of some quadric-line arrangements Meirav Amram a,∗ , Mina Teicher b , A. Muhammed Uludag c a Meirav Amram, Mathematisches Institut, Bismarck Strasse 1 1/2, Erlangen, Germany b Mathematics Department, Bar-Ilan University, Ramat-Gan, Israel c Mathematics Department, Galatasaray University, Ortaköy, Istanbul, Turkey

Received 15 June 2002; received in revised form 29 July 2002

Abstract In this paper we obtain presentations of fundamental groups of the complements of three quadricline arrangements in P2 . The first arrangement is a smooth quadric Q with n tangent lines to Q, and the second one is a quadric Q with n lines passing through a point p ∈ / Q. The last arrangement consists of a quadric Q with n lines passing through a point p ∈ Q. 2002 Elsevier Science B.V. All rights reserved. Keywords: Fundamental groups; Complements of curve; Conic-line arrangements

1. Introduction This is the first of a series of articles in which we shall study the fundamental groups of complements of some quadric-line arrangements. In contrast with the extensive literature on line arrangements and the fundamental groups of their complements, (see, e.g., [14,7, 15]), only a little known about the quadric-line arrangements (see [12,1,2]). The present article is dedicated to the computation of the fundamental groups of the complements of three infinite families of such arrangements. A similar analysis for the quadric-line arrangements up to degree six will be done in our next paper. Let C ⊂ P2 be a plane curve and ∗ ∈ P2 \ C a base point. By abuse of language we will call the group π1 (P2 \ C, ∗) the fundamental group of C, and we shall frequently omit base * Corresponding author.

E-mail addresses: [email protected], [email protected] (M. Amram), [email protected] (M. Teicher), [email protected] (A.M. Uludag). 0166-8641/02/$ – see front matter 2002 Elsevier Science B.V. All rights reserved. doi:10.1016/S0166-8641(02)00218-3

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Fig. 1. The arrangement A3 .

points and write π1 (P2 \ C). One is interested in the group π1 (P2 \ C) mainly for the study of the Galois coverings X → P2 branched along C. Many interested surfaces have been constructed as branched Galois coverings of the plane, for example for the arrangement A3 in Fig. 1, there are Galois coverings X → P2 branched along A3 such that X P1 × P1 , or X is an abelian surface, a K3 surface, or a quotient of the two-ball B2 (see [9,8,17]). Moreover, some line arrangements defined by unitary reflection groups studied in [13] are related to A3 via orbifold coverings. For example, if L is the line arrangement given by the equation xyz(x + y + z)(x + y − z)(x − y + z)(x − y − z) = 0 then the image of L under the branched covering map [x : y : z] ∈ P2 → [x 2 : y 2 : z2 ] ∈ P2 is the arrangement A3 , see [17] for details. The standard tool for fundamental group computations is the Zariski–van Kampen algorithm [19,18], see [3] for a modern approach. We use a variation of this algorithm developed in [16] for computing the fundamental groups of real line arrangements and avoids lengthy monodromy computations. The arrangements Bn and Cn discussed below are of fiber type, so presentations of their fundamental groups could be easily found as an extension of a free group by a free group. However, our approach has the advantage that it permits to capture the local fundamental groups around the singular points of these arrangements. The local fundamental groups are needed for the study of the singularities of branched of P2 branched along these arrangements. In Section 2 below, we give fundamental group presentations and prove some immediate corollaries. In Section 3 we deal with the computations of fundamental group presentations given in Section 2.

2. Results Let C ⊂ P2 be a plane curve and B an irreducible component of C. Recall that a meridian µ of B in P2 \ C with the base point ∗ ∈ P2 is a loop in P2 \ C obtained by following a path ω with ω(0) = ∗ and ω(1) belonging to a small neighborhood of a smooth point p ∈ B \ C, turning around C in the positive sense along the boundary of a small disc ∆ intersecting B transversally at p, and then turning back to ∗ along ω. The meridian µ represents a homotopy class in π1 (P2 \ C, ∗), which we also call a meridian of B. Any

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two meridians of B in P2 \ C are conjugate elements of π1 (P2 \ C) (see, e.g., [10, 7.5]), hence the meridians of irreducible components of C are supplementary invariants of the pair (P2 , C). These meridians are specified in presentations of the fundamental group given below, they will be used in orbifold-fundamental group computations in [17]. 2.1. The arrangement An Theorem 1. Let An := Q ∪ T1 ∪ · · · ∪ Tn be an arrangement consisting of a smooth quadric Q with n distinct tangent lines T1 , . . . , Tn . Then κi = τi κi−1 τ −1 , 2 i n i 2 τ1 , . . . , τn , (κi τi )2 = (τi κi )2 , 1 i n (1) π1 P \ An κ1 , . . . , κn κi−1 τi κi , τj = 1, 1 i < j n τ · · · τ κ2 = 1 n 1 1 where κi are meridians of Q and τi is a meridian of Ti for 1 i n. Local fundamental groups around the singular points of An are generated by κi−1 τi κi , τj for the nodes Ti ∩ Tj and by κi , τi for the tangent points Ti ∩ Q. Part (i) of the corollary below is almost trivial. Part (ii) appears in [6], and part (iii) was given in [4]. Corollary 2. (i) One has: π1 (P2 \ A1 ) Z. (ii) The group π1 (P2 \ A2 ) admits the presentation π1 P2 \ A2 τ, κ | (τ κ)2 = (κτ )2 ,

(2)

where κ is a meridian of Q and τ is a meridian of T1 . A meridian of T2 is given by κ −2 τ −1 . (iii) The group π1 (P2 \ A3 ) admits the presentation π1 P2 \ A3 τ, σ, κ | (τ κ)2 = (κτ )2 , (σ κ)2 = (κσ )2 , [σ, τ ] = 1 (3) where σ , τ are meridians of T1 and T3 respectively, and κ is a meridian of Q. A meridian of T2 is given by (κτ κσ )−1 . A group G is said to be big if it contains a non-abelian free subgroup, and small if G is almost solvable. In [6], it was proved by V. Lin that the group (2) is big. Below we give an alternative proof: Proposition 3. For n > 1, the group π1 (P2 \ An ) is big. Proof. A group with a big quotient is big. Since τn+1 is a meridian of Tn+1 in π1 (P2 \ An+1 ), one has π1 P2 \ An π1 P2 \ An+1 /τn+1 ,

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and it suffices to show that the group π1 (P2 \ A2 ) is big. In the presentation (2), applying the change of generators α := τ κ, β := τ gives π1 P2 \ A2 α, β | α 2 , β = 1 . Adding the relations α 2 = β 3 = 1 to the latter presentation gives a surjection π1 (P2 \ A2 ) Z/(2) ∗ Z/(3). Since the commutator subgroup of Z/(2) ∗ Z/(3) is the free group on two generators (see [5]), we get the desired result. ✷ 2.2. The arrangement Bn Theorem 4. Let Bn := Q ∪ T1 ∪ T2 ∪ L1 ∪ · · · ∪ Ln be an arrangement consisting of a smooth quadric Q with n + 2 distinct lines T1 , T2 , L1 , . . . , Ln all passing through a point p∈ / Q such that T1 , T2 are tangent to Q. Then one has (κτ )2 = (τ κ)2 2 λ ] = 1, 1 i n π1 P \ Bn τ, κ, λ1 , . . . , λn [κ, (4) i τ −1 κτ, λi = 1, 1 i n where τ is a meridian of T1 , λi is a meridian of Li for 1 i n, and κ is a meridian of Q. A meridian σ of T2 is given by σ := (λn . . . λ1 κ 2 τ )−1 . Local fundamental groups around the singular points of Bn are generated by κ, λi and τ −1 κτ, λi for the nodes Li ∩ Q, by κ, τ for the tangent point T1 ∩ Q, and by κ, σ for the tangent point T2 ∩ Q. Corollary 5. (i) Put Bn := Bn \ T1 and Bn := Bn \ T2 . Then π1 P2 \ Bn π1 P2 \ Bn+1 κ, λ1 , . . . , λn | [κ, λi ] = 1, 1 i n .

(5)

Proof. One has π1 (P2 \ Bn ) π1 (P2 \ Bn )/τ . Setting τ = 1 in presentation (4) gives π1 P2 \ Bn κ, λ1 , . . . , λn | [κ, λi ] = 1, 1 i n . Setting τ = 1 in the expression for a meridian σ of T2 given in Theorem 4 shows that (λn . . . λ1 κ 2 )−1 is a meridian of T2 in π1 (P2 \ Bn ). In order to find π1 (P2 \ Bn ), it suffices to set λn . . . λ1 κ 2 = 1 in the presentation of π1 (P2 \ Bn ). Eliminating λn by this relation yields the presentation π1 P2 \ Bn κ, λ1 , . . . , λn−1 | [λi , κ] = λn−1 . . . λ1 κ 2 , κ = 1 .

Fig. 2. Arrangements B2 and B2 .

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Fig. 3. Arrangements C3 and C3 .

Since the last relation above is redundant, we get the desired isomorphism π1 (P2 \ Bn ) π1 (P2 \ Bn+1 ). ✷ Note that the groups π1 (P2 \ Bi ) are abelian for i = 0, 1, 2. Hence, the groups π1 (P2 \ Bi ) are abelian for i = 0, 1. Otherwise, setting κ = 1 in presentation (5) gives the free group on n − 1 generators, which shows that these groups are big. The groups π1 (P2 \ Bn ) are always big, since the arrangement B0 is same as A2 , and π1 (P2 \ A2 ) is big by Proposition 3. 2.3. The arrangement Cn Theorem 6. Let Cn := Q ∪ T ∪ L1 ∪ · · · ∪ Ln be an arrangement consisting of a smooth quadric Q with n + 1 distinct lines T , L1 , . . . , Ln , all passing through a point p ∈ Q such that T is tangent to Q. Then one has π1 P2 \ Cn κ, λ1 , . . . , λn | [κ, λi ] = 1, 1 i n , (6) where κ is a meridian of Q and λi is a meridian of Li for 1 i n. A meridian τ of T is given by τ := (λn . . . λ1 κ 2 )−1 . Local fundamental groups around the singular points of Cn are generated by κ, λi for the nodes Li ∩ Q, and by τ, λ1 , . . . , λn , κ for the point p. Note that the arrangement Cn is a degeneration (in the sense of Zariski) of the arrangement Bn as the point p approaches to Q. By Zariski’s “semicontinuity” theorem of the fundamental group [19] (see also [5]), there is a surjection π1 (P2 \ Cn ) π1 (P2 \ Bn ). In our case, this is also an injection: Corollary 7. (i) π1 (P2 \ Bn ) π1 (P2 \ Cn ). (ii) Put Cn := Cn \ T . Then π1 (P2 \ Cn ) π1 (P2 \ Cn+1 ). Proof. Part (i) is obvious. The proof of part (ii) is same as the proof of Corollary 5, (ii).

✷

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3. The arrangement An It is easily seen that any two arrangements An with fixed n are isotopic. In particular, the groups π1 (P2 \ An ) are isomorphic. Hence one can take as a model of the arrangements An the quadric Q defined by x 2 + y 2 = z2 , where [x : y : z] ∈ P2 is a fixed coordinate system in P2 . Pass to the affine coordinates in C2 P2 \ {z = 0}. Choose real numbers x1 , . . . , xn such that −1 < x1 < x2 < · · · < xn < 0, and define yi to be the positive solution of xi2 + yi2 = 1 for 1 i n. Put ti := (xi , yi ) ∈ Q, and take Ti to be the tangent line to Q at the point ti (see Fig. 4). Let pr1 : C2 \ An → C be the first projection. The base of this projection will be denoted by B. Put Fx := pr−1 1 (x), and denote by S the set of singular fibers of pr1 . It is clear that if Fx ∈ S, then x ∈ [−1, 1]. There are three types of singular fibers: (i) The fibers F1 and F−1 , corresponding to the ‘branch points’ (−1, 0) and (1, 0). (ii) The fibers Fxi (1 i n) corresponding to the ‘tangent points’ ti = (xi , yi ) = Ti ∩Q. (iii) The fibers Fai,j (1 i = j n) corresponding to the nodes ni,j = (ai,j , bi,j ) := Ti ∩ Tj . One can arrange the lines Ti such that −1 < x1 < a1,2 < a1,3 < · · · < a1,n < x2 < a2,3 < · · · < xn < 1. Identify the base B of the projection pr1 with the line y = −2 ⊂ C2 . Let N be the number of singular fibers and let −1 = s1 < s2 < · · · < sN−1 < sN = 1 be the elements of S ∩ B (so that s2 = x1 , s3 = a1,2 , s4 = a1,3 , and so on). In B, take small discs ∆i around the points si , and denote by ci , di (ci < di ) the points ∂∆i ∩ R for 1 i N (see Fig. 5). Put B1 := [c1 , c2 ] ∪ ∆1 and for 2 i N let Bi := [c1 , ci+1 ] ∪ ∆1 ∪ · · · ∪ ∆i . Let Xi := pr−1 (Bi ) be the restriction of the fibration pr to Bi . Let Ai := ∆i ∪ ∂ (z) 0, c2 (z) ci \ (∆2 ∪ ∆3 ∪ · · · ∪ ∆i−1 ) and let Yi := pr−1 (Ai ) be the restriction of the fibration pr to Ai (see Fig. 6). Clearly, Xi = Xi−1 ∪ Yi for 2 i N . We will use this fact to compute the groups π1 (Xi , ∗) recursively, where ∗ := (c2 , −2) is the base point. For details of the algorithm we apply below, see [16].

Fig. 4.

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Fig. 5. The base B.

Fig. 6. The space Ai .

Fig. 7.

Identify the fibers of pr1 with F0 via the second projection pr2 : (x, y) ∈ C2 → y ∈ C. In each one of the fibers Fci (respectively Fdi ) take a basis for π1 (Fci , −2) (respectively for π1 (Fdi , −2)) as in Fig. 7 (for Fdi , just replace γ ’s by θ ’s in Fig. 7). We shall denote (i) (i) these basis by the vectors Γi := [γ1(i) , . . . , γn+2 ] (respectively Θi := [θ1(i) , . . . , θn+2 ]). Let νi ⊂ Bi ⊂ B be a path starting at νi (0) = c2 , ending at νi (1) = ci and such that νi [0, 1] = ∂ (z) 0, c2 (z) ci \ (∆2 ∪ ∆3 ∪ · · · ∪ ∆i−1 ) . Similarly, let ηi ⊂ Bi ⊂ B be a path starting at η(0) = c2 , ending at η(0) = di and such that ηi [0, 1] = ∂ (z) 0, c2 (z) di \ (∆2 ∪ ∆3 ∪ · · · ∪ ∆i ) . For 2 i N and 1 j n + 2 each loop γ˜j := νi · γj · νi−1 represents a homotopy (i) class in π1 (Xi , ∗), where ∗ := (c2 , −2) is the base point. Similarly, each loop θ˜ := (i)

(i)

j

ηi · θi · ηi−1 represents a homotopy class in π1 (Xi , ∗). Denote Γi := [γ˜1 , . . . , γ˜n+2 ], and i := [θ˜ (i) , . . . , θ˜ (i) ]. Θ (i)

1

n+2

(i)

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It is well known that the group π1 (Yi , ∗) has the presentation

(i) γ˜1 , . . . , γ˜n+2 | γ˜j(i) = Mi γ˜j(i) , 1 j n + 2

(7)

where Mi : π1 (Fci , −2) → π1 (Fci , −2) is the monodromy operator around the singular fiber above si . It is also well known that if it is the branches of An corresponding to (i) the loops γ˜k(i) and γ˜k+1 that meet above si , then the only non-trivial relation in (7) is (i)

(i)

(i)

(i)

(i) (i)

γ˜k = γ˜k+1 in case of a branch point, [γ˜k , γ˜k+1 ] = 1 in case of a node, and (γ˜k γ˜k+1 )2 = (i) (i)

(γk γk+1 )2 in case of a tangent point. Now suppose that the group π1 (Xi−1 , ∗) is known, with generators Γ2 . Recall that Xi = Xi−1 ∪ Yi . In order to find the group π1 (Xi , ∗), one has to express the base Γi in terms of the base Γi . Adding to the presentation of π1 (Xi−1 ) the relation obtained by writing the relation of π1 (Yi ) in the new base then yields a presentation of π1 (Xi ). Note that, since the space Yi is eventually glued to Xi−1 , it suffices to find an expression of Γi in terms of the base Γ2 in the group π1 (Xi−1 , ∗). Since all the points of An above the interval [di−1 , ci ] are smooth and real, one has Fact. The loops θ˜j(i−1) and γ˜j(i) are homotopic in Xi (or in Yi ) for 2 i N and i−1 and Γi are homotopic. 1 j n + 2. In other words, the bases Θ i in terms of the base Γi the following lemma will be In order to express the base Θ helpful. Lemma 8. Let Ck : x 2 − y k+1 = 0 be an Ak singularity, where k = 1 or k = 3. Put D := {(x, y): |x| 1, |y| 1} and let pr1 := (x, y) ∈ D \ Ck → (x, −1) be the first projection. Denote by Fx the fiber of pr1 above (x, −1). Identify the fibers of pr1 via the second projection. Let −1 < c < 0 be a real number, and put d := −c. In Fc (respectively in Fd ) take a basis Γ := [γ1 , γ2 ] for π1 (Fc , −1) (respectively a basis Θ := [θ1, θ2 ] for π1 (Fd , −1)) as in Fig. 8. Let η be the path η(t) := ceπit , and put θ˜i := η · θ · η−1 for i = 1, 2. Then γi , θ˜i are loops in D \ Ck based at ∗ := (c, −1), and one has (i) If k = 1, then θ˜1 is homotopic to γ2 , and θ˜2 is homotopic to γ1 , in other words, = [γ2 , γ1 ]. Θ = [γ2 γ1 γ −1 , γ −1 γ2 γ1 ]. (ii) If k = 3, then Θ 2 1

Fig. 8.

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Fig. 9.

Proof. Since π1 (D \ C2 ) is abelian, part (i) is obvious. For part (ii), note that the points of intersection Fη(t ) ∩ C4 are y1 := c2 e2πit and y2 := −c2 e2πit . Hence, when we move the fiber Fc over Fd along the path η, y1 and y2 make one complete turn around the origin in the positive sense. The loops γ1 , γ2 are transformed to loops γ 1 , γ 2 ⊂ Fd as in Fig. 9. It follows that the loop η · γ i · η−1 is homotopic to γi for i = 1, 2. This homotopy can be constructed explicitly as follows: Let Φη(t ) : Fc → Fη(t ) be the corresponding Leftschez homeomorphism (see [11]). Then 0 s t/3, η(3s), Φ γ 3(s − t/3)/(3 − 2t) , t/3 s 1 − t/3, H (s, t) := η(t ) i η3(1 − s), 1 − t/3 s 1 gives a homotopy between γi and γ i . Expressing θ˜i in terms of γ i , we get −1 −1 −1 θ˜1 = γ −1 1 γ 2 γ 1 γ 2 γ 1 = γ1 γ2 γ1 γ2 γ1 , −1 θ˜2 = γ −1 1 γ 2 γ 1 = γ1 γ2 γ1 .

Since from the monodromy one has the relation (γ1 γ2 )2 = (γ2 γ1 )2 , the expression for θ˜1 can be simplified to get θ˜1 = γ2 γ1 γ2−1 . ✷ Now we proceed with the computation of the groups π1 (Xi ). Clearly, the group π1 (X2 ) is generated by the base (2) Γ2 = γ1(2), γ2(2) , . . . , γn+2 with the only relations γ1(2) = γ2(2) and

(2) (2) 2

γ2 γ3

(8) (2) (2) 2 = γ3 γ2 .

(9)

Put [κ1 , κ1 , τ1 , . . . , τn ] := Γ2 . Then relation (9) becomes (κ1 τ1 )2 = (τ1 κ1 )2 .

(10)

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By Lemma 8 and the above Fact, one has 2 = κ1 , τ1 κ1 τ −1 , κ −1 τ1 κ1 , τ2 , . . . , τn . Γ3 = Θ 1 1 Since s3 corresponds to the node T1 ∩ T2 , the next relation is −1 κ1 τ1 κ1 , t2 = 1.

(11)

Hence,

π1 (X3 , ∗) κ1 , τ1 , . . . , τn | (κ1 τ1 )2 = (τ1 κ1 )2 , κ1−1 τ1 κ1 , τ2 = 1 . By Lemma 8, one has 3 = κ1 , τ1 κ1 τ −1 , τ2 , κ −1 τ1 κ1 , τ3 , . . . , τn . Γ4 = Θ 1 1 Since s4 corresponds to the node T1 ∩ T3 , one has the relation −1 κ1 τ1 κ1 , τ3 = 1. Hence,

π1 (X4 , ∗) κ1 , τ1 , . . . , τn | (κ1 τ1 )2 = (τ1 κ1 )2 , −1 κ1 τ1 κ1 , τ2 = κ1−1 τ1 κ1 , τ3 = 1 . By Lemma 8, one has 4 = κ1 , τ1 κ1 τ −1 , τ2 , τ3 , κ −1 τ1 κ1 , τ4 , . . . , τn . Γ5 = Θ 1 1 Since sk corresponds to the node T1 ∩ Tk−1 for 2 k n + 1, repeating the above procedure gives the presentation

π1 (Xn+1 , ∗) κ1 , τ1 , . . . , τn | (κ1 τ1 )2 = (τ1 κ1 )2 , κ1−1 τ1 κ1 , τk = 1, 2 k n and

n+1 = κ1 , κ2 , τ2 , τ3 , . . . , τn , κ −1 τ1 κ1 , Γn+2 = Θ 1

where we put κi+1 := τi κi τi−1 for 1 i n − 1. The next point sn+2 corresponds to the tangent point T2 ∩ Q. This gives the relation (κ2 τ2 )2 = (τ2 κ2 )2 and

(12)

n+1 = κ1 , τ2 κ2 τ −1 , κ −1 τ2 κ2 , τ3 , . . . , τn , κ −1 τ1 κ1 . Γn+2 = Θ 2 2 1

Now comes the n − 2 points sk corresponding to the nodes T2 ∩ Tk−n for n + 3 2n + 1. These give the relations −1 κ2 τ2 κ2 , τk = 1, 3 k n. Hence, one has

π1 (Xn+1 , ∗) κ1 , κ2 , τ1 , . . . , τn | κ2 = τ1 κ1 τ1−1 , (κi τi )2 = (τi κi )2 , −1 κi τi κi , τk = 1, i < k n, i = 1, 2 .

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We proceed in this manner until the last singular fiber sN . Since this is a branch point, the final relation is κn = κ1 . This gives the presentation π1 (XN , ∗) τi , κi , 1 i n

(13) κi = τi κi−1 τ −1 , 2 i n i (κi τi )2 = (τi κi )2 , 1 i n κ −1 τ κ , τ = 1, 1 i < j n i i i j κ =κ 1 n

.

(14)

Adding to this presentation of π1 (XN , ∗) the projective relation τn · · · τ1 κ12 = 1 gives the presentation κi = τi κi−1 τ −1 , 2 i n i (κi τi )2 = (τi κi )2 , 1 i n 2 π1 P \ An τi , κi , 1 i n −1 . (15) κi τi κi , τj = 1, 1 i < j n τ . . . τ κ 2 = 1, κ = κ n 1 1 1 n Note that the relation κ1 = κn is redundant. Indeed, since κi = τi κi−1 τi−1 , one has κn = (τn . . . τ1 )κ1 (τn . . . τ1 )−1 .

(16)

But τn . . . τ1 = κ −2 by the projective relation. Substituting this in (16) yields the relation κ1 = κn . This finally gives the presentation (1) and proves Theorem 1. Claims regarding the local fundamental groups around the singular points of An are direct consequences of the above algorithm. 3.1. Proof of Corollary 2 (i) The arrangement A1 . Writing down the presentation (1) explicitly for n = 1 gives (κ1 τ1 )2 = (κ1 τ1 )2 2 π1 P \ A1 κ1 , τ1 2 . t1 κ1 = 1 Eliminating τ1 from the last relation shows that π1 (P2 \ A1 ) Z. (ii) The arrangement A2 . Writing down the presentation (1) explicitly for n = 2 gives (1) κ2 = τ1 κ1 τ −1 1 (2) (κ1 τ1 )2 = (τ1 κ1 )2 2 2 2 . π1 P \ A2 κ1 , κ2 , τ1 , τ2 (3) (κ2 τ2 ) = (τ2 κ2 ) (4) κ −1 t κ , t = 1 1 1 1 2 (5) τ τ κ 2 = 1 2 1 1 Eliminating κ2 by (1) and τ2 by (5) one easily shows that the relations (3) and (4) are redundant. This leaves (2) and gives the desired presentation.

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(iii) The arrangement A3 . Writing down the presentation (1) explicitly for n = 3 gives (1) κ2 = τ1 κ1 τ1−1 (2) κ3 = (τ2 τ1 )κ1 (τ2 τ1 )−1 (3) (κ1 τ1 )2 = (τ1 κ1 )2 (4) (κ2 τ2 )2 = (τ2 κ2 )2 2 κ1 , κ2 , κ3 , −1 2 2 π1 P \ A3 (5) (κ3 τ3 ) = (τ3 κ3 ) (6) κ1 τ1 κ1 , τ2 = 1 . τ1 , τ2 , τ3 (7) κ −1 τ1 κ1 , τ3 = 1 (8) κ −1 τ2 κ2 , τ3 = 1 1 2 (9) τ τ τ κ 2 = 1 3 2 1 1

Eliminate κ2 by (1), κ3 by (2), and τ2 by (9). It can be shown that the relations (4), (6) and (8) are consequences of the remaining relations. The relation (5) becomes (κ1 τ3 )2 = (τ3 κ1 )2 . This gives the presentation

π1 P2 \ A3 κ1 , τ1 , τ3 | (κ1 τ1 )2 = (τ1 κ1 )2 ,

(κ1 τ3 )2 = (τ3 κ1 )2 , κ1−1 τ1 κ1 , τ3 = 1 .

Finally, put κ := κ1 , τ := κ1−1 τ1 κ1 and σ := τ3 . Then τ1 = κτ κ −1 , and the first relation in the above presentation becomes (κ 2 τ κ −1 )2 = (κτ )2 ⇒ (κτ )2 = (τ κ)2 . This gives the desired presentation.

4. The arrangement Bn As in the case of the arrangements An , it is readily seen that arrangements Bn are all isotopic to each other for fixed n, so one can compute π1 (P2 \ Bn ) from the following model for Bn ’s (see Fig. 10): The quadric Q is given by the equation x 2 + y 2 = 1, and p is the point (2, 0). The lines Li intersect Q above the x-axis. The projection to the x-axis has four types of singular fibers:

Fig. 10.

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171

(i) The fibers F1 and F−1 , corresponding to ‘branch points’. (ii) The fiber Fx corresponding to the ‘tangent points’ (x, y) = t1 := T1 ∩ Q and (x, −y) = t2 := T2 ∩ Q. (iii) The fibers Fa1 , . . . , Fan (−1 < an < · · · < a1 < x) corresponding to the nodes Li ∩ Q lying on the right of the tangent points and the fibers Fb1 , . . . , Fbn (x < b1 < · · · < bn < 1) corresponding to the nodes Li ∩ Q lying on the left of the tangent points. (iv) The fiber F2 , corresponding to the point p. In order to find the group π1 (P2 \ Bn ), we shall apply the same procedure as in the computation of π1 (P2 \ An ). Let y ∈ R be such that −1 < y < an , and take Fy to be the base fiber. Let s1 := −1, si+1 := bi for 1 i n, sn+2 := x, sn+2+i := an+1−i for 1 i n, and s2n+3 := 1, and s2n+4 = 2. Take a basis Γ2 := [τ, κ1 , κ2 , λ1 , . . . , λn , σ ] for Fy as in Fig. 7. Since s1 corresponds to a branch point, one has the relation κ1 = κ2 . Put κ := κ1 = κ2 . The point s2 is a node, and yields the relation [κ, λ1 ] = 1, and one has Γ3 = [τ, κ, λ1 , κ, λ2 , . . . , λn , σ ]. Repeating this for the nodes s3 , . . . , sn+1 gives the relations [κ, λi ] = 1 for 1 i n, and one has Γn+2 = [τ, κ, λ1 , . . . , λn , κ, σ ]. The monodromy around the fiber Fx gives the relations (τ κ)2 = (κτ )2 and (σ κ)2 = (κσ )2 . One has Γn+3 = κτ κ −1 , τ −1 κτ, λ1 , . . . , λn , σ κσ −1 , κ −1 σ κ . Since the points sn+3 , . . . , s2n+2 corresponds to nodes, one has the relations [λi , σ κσ −1 ] = 1, and Γ2n+3 = κτ κ −1 , τ −1 κτ, σ κσ −1 , λ1 , . . . , λn , κ −1 σ κ . The branch point corresponding to s2n+3 yields the relation τ −1 κτ = σ κσ −1 . Together with the projective relation σ λn . . . λ1 κ 2 τ = 1 these relations already gives a presentation of π1 (P2 \ Bn ), since one can always ignore one of the singular fibers when computing the monodromy (see [16]). We obtained the presentation (1) (κτ )2 = (τ κ)2 (2) (κσ )2 = (σ κ)2 −1 −1 (3) τ κτ = σ κσ 2 . π1 P \ Bn Λ, τ, κ, λ1 , . . . , λn , σ =1 1i n (4) [κ, λi ]−1 (5) σ κσ , λi = 1 1 i n (6) σ λ . . . λ κ 2 τ = 1 n

1

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Put Λ := λn . . . λ1 . Eliminating σ by (7), it is easily seen that (3) is redundant. Relation (2) becomes 2 −1 2 −1 2 2 = κ Λκ τ ⇒ τ −1 κτ, Λ = 1. Λκ τ κ But this relation is a consequence of (4), so that (2) is also redundant. Since τ −1 κτ = σ κσ −1 by (3), the relation (5) can be written as [τ −1 κτ, λi ] = 1. This gives the presentation (4) and proves Theorem 4. 5. The arrangement Cn In order to compute the group, consider the model of Cn shown in Fig. 11, where Q is given by x 2 + y 2 = 1. Suppose that the second points of intersection of the lines Li with Q lie above the x-axis. As in the previous cases, take an initial base Γ2 := [κ1 , κ2 , λ1 , . . . , λn , τ ]. The relation induced by the branch point is κ1 = κ2 =: κ. The nodes of Cn will give the relations [κ, λi ] = 1 for 1 i n, and one has Γn+2 = [κ, λ1 , . . . , λn , κ, τ ]. One can simplify the computation of the monodromy around the complicated singular fiber as follows: Put Λ := λn . . . λ1 . By the projective relation one has τ Λκ 2 = 1 ⇒ τ = κ −2 Λ−1 . Hence, [κ, τ ] = 1. Since we also have [κ, λi ] = 1 for 1 i n, this means that when computing the monodromy around this fiber, one can ignore the branch Q. This leaves n + 1 branches intersecting transversally, and the induced relation is (see [16]) [τ Λ, λi ] = [τ Λ, τ ] = 1,

(17)

and one has Γn+3 = [κ, κ, . . .]. The last relation induced by the branch point yields the trivial relation κ = κ, as expected.

Fig. 11.

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Eliminating τ shows that the relations (17) are redundant and gives the presentation π1 P2 \ Cn κ, λ1 , . . . , λn | [λi , κ] = 1 . ✷

Acknowledgements This work was partially supported by the Emmy Noether Research Institute for Mathematics (center of the Minerva Foundation of Germany), the Excellency Center “Group Theoretic Methods in the Study of Algebraic Varieties” of the Israel Science Foundation, and EAGER (EU network, HPRN-CT-2009-00099).

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