The Characterization of a Line Arrangement Whose Fundamental Group of the Complement Is a Direct Sum of Free Groups

Algebraic & Geometric Topology 10 (2010) 1285–1304 1285

The characterization of a line arrangement whose fundamental group of the complement is a direct sum of free groups

MEITAL ELIYAHU ERAN LIBERMAN MALKA SCHAPS MINA TEICHER

Kwai Man Fan proved that if the intersection lattice of a line arrangement does not contain a cycle, then the fundamental group of its complement is a direct sum of inﬁnite and cyclic free groups. He also conjectured that the converse is true as well. The main purpose of this paper is to prove this conjecture.

14F35, 32S22; 14N30, 52C30

1 Introduction

An arrangement of lines is a finite collection of C–affine subspaces of dimension 1. For such an arrangement † C2 , there is a natural projective arrangement † of Â lines in CP 2 associated to it, obtained by adding to each line its corresponding point at infinity. The problem of finding connections between the topology of C2 † and the combinatorial theory of † is one of the main problems in the theory of line arrangements (see for example Cohen and Suciu [3]). The main motivations for studying the topology of C2 † are derived from the areas of hypergeometric functions, singularity theory and algebraic geometry. Given an arrangement †, we define the graph G.†/ in the way it is defined by Fan [7]. We first connect all higher multiple points of † which lie on a line Li by an arbitrary simple arc ˛i Li which does not go through any double point of †. Taking the union of the set of all higher multiple points and all of this simple arcs, we obtain G.†/. Note that if 3 points are on the same line we do not consider it as a cycle. In 1994, Jiang and Yau [11] defined the concept of a “nice” arrangement. For †, they define a graph G.V; E/: The vertices are the multiple points of †, and vertices u; v are connected by an edge if there exists l † such that u; v l . For v V define a 2 2 2

Published: 4 June 2010 DOI: 10.2140/agt.2010.10.1285 1286 Meital Eliyahu, Eran Liberman, Malka Schaps and Mina Teicher

subgraph G†.v/: The vertex set is v and all his neighbors from † (note this deﬁnition differs from the deﬁnition of Fan).

An arrangement † is nice if there is V 0 V such that E†.v/ E†.u/ ∅ for all \ D u; v V , and if we delete the vertex v and the edges of its subgraph G†.v/ from G , 2 0 for all v V we get a forest (ie a graph without cycles). 2 0 Jiang and Yau have proven several properties of nice arrangements:

(a) If A1; A2 are nice arrangements and their lattices are isomorphic, then their 2 2 complements C A1 and C A2 are diffeomorphic. This property naturally 2 2 implies that 1.C A1/ 1.C A2/. Š (b) As a consequence of (a), they showed that the presentation of the fundamental group of the complement can be written explicitly and depends only on the lattice of the line arrangement.

In 2005, Wang and Yau [20] continued in this direction and proved that the results of Jiang and Yau hold for a much larger family of arrangements, which they call simple arrangements. Falk [4] finds several examples of line arrangements with the same homotopy types but with different lattices. Fan [6] proved that for up to 6 lines, the fundamental group of a real line arrangement is determined by the lattice. Garber, Teicher and Vishne [9] proved this for a real line arrangement with up to 8 lines. Let G; H be groups whose abelianizations are free abelian groups of finite rank. Choudary, Dimca and Papadima [2] defined a set ˆ of natural group isomorphisms 2 G H (which they call a 1–marking). In the case of G 1.C †/ (where † is W ! D an affine arrangement), takes the topological structure of C2 † into consideration. They prove that if A; B are line arrangements and B is nice in the sense of Jiang and Yau, then the lattices are isomorphic if and only if there is an isomorphism ˆ 2 2 2 where 1.C A/ 1.C B/. W ! Fan [7] showed that if the graph G.†/ is a forest (ie a graph without cycles), then the fundamental group is a direct sum of free groups. He also conjectured that the converse of his theorem is true. In [5] Fan proved that if the fundamental group of the complement is a direct sum of free groups, then the arrangement is composed of parallel lines. In this paper, we prove his conjecture. Our theorem will state that if the fundamental group is isomorphic to a direct sum of free groups, then the graph has no cycles. We would like to emphasize that we make no restrictions on our isomorphisms.

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The structure of the paper is as follows. In Section 2, we give basic definitions related to groups. In Section 3 we give basic definitions related to line arrangements and the fundamental group of the complement of a line arrangement. In Section 4, we define a function and a special set induced by it. Section 5 deals with some special properties of fundamental groups of the complements of line arrangements which are direct sum of free groups. In Section 6, we prove the main result of the paper.

2 Deﬁnitions and notation

This section presents the needed deﬁnitions for the paper.

2.1 Lower central series

We start by deﬁning the lower central series of a group G which will be used throughout the paper.

Deﬁnition 2.1 (Commutator group and lower central series) Let G be a group. The commutator group of G is 1 1 G G2 ŒG; G aba b a; b G 0 D D D h j 2 i The subgroup G0 is normal in G with an abelian quotient. We can deﬁne the lower central series of G recursively:

G1 G D G2 ŒG; G D G3 ŒG; G2 D: :

Gn ŒG; Gn 1: D Since Gn 1 contains the commutators of Gn , we have Gn 1 Gn and the quotient C C G Gn=Gn 1 is abelian for all n N . C 2 To understand these groups, the following identities are needed.

Proposition 2.2 (Witt–Hall identities [15]) (1) Œa; bŒb; a e. D (2) Œa; bc Œa; bŒa; cŒŒc; a; b. D (3) Œab; c Œa;Œb; cŒb; cŒa; c. D

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From the second and third identities we get:

Lemma 2.3 Let G be a group and let x1;:::; x be the generators of G . Then f k g G2=G3 Œxi; xj i j ; 1 i; j k : D h j ¤ Ä Ä i

Proof If Œxy; z G2 , then 2 Œxy; z xyz.xy/ 1z 1 xyzy 1x 1z 1 xŒy; zx 1Œx; z: D D D 1 1 1 Over G3 , we have Œx;Œy; z xŒy; zx Œy; z e Œy; z xŒy; zx . We get D D ) D Œxy; z Œy; zŒx; z over G3 . D This result means that G2=G3 is ﬁnitely generated by the set

Œxi; xj i j ; 1 i; j k ; f j ¤ Ä Ä g where x1;:::; xk are the generators of G .

Proposition 2.4 Let G x1;:::; x R , where the relations R are commutator D h k j i type relations (ie every relation r R can be written as r Œw1; w2; w1; w2 G ). 2 D 2 Then * ˇ 1 + ˇ Œxk ; xl Œxl ; xk ; ˇ D G2=G3 Œxi; xj ˇ all generators commute; D ˇ R where R is set of the relations R written by means of the generators of G2 , taken modulo G3 .

The next Deﬁnition and theorem will give us a better understanding of G2=G3 and help us in the future.

Deﬁnition 2.5 [10, page 165] Let G be a group generated by x1;:::; xr . We consider formal words or strings b1 b2 bn where each b is one of the generators. We also introduce formal commutators cj and weights !.cj / by the rules:

(1) ci xi; i 1;:::; r are the commutators of weight 1; ie !.xi/ 1. D D D (2) If ci and cj are commutators, then c Œci; cj and !.c / !.ci/ !.cj /. k D k D C Theorem 2.6 (Basis theorem [10]) If F is the free group with free generators y1;:::; yr and if in a sequence of basic commutators c1;:::; ct are those of weights 1; 2;:::; n then an arbitrary element f F has a unique representation f ce1ce2 cet 2 D 1 2 t mod Fn 1 . C The basic commutators of weight n form a basis for the free abelian group Fn=Fn 1 . C

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2.2 Line arrangements

2 Deﬁnition 2.7 (Line arrangement) A line arrangement † L1;:::; Ls C is a D f g Â union of copies of C1 .

Remark 2.8 Each time we are mentioning a line arrangement † C2 , we assume Â that there are no parallel lines in †. Pay attention we can project every line arrangement in CP 2 to C2 such that the new arrangement does not contain parallel lines.

Deﬁnition 2.9 Let † be a line arrangement. An intersection point in † is called simple if there are precisely two lines which meet at that point. Otherwise, we call it multiple.

Definition 2.10 (Cycle) A cycle is a nonempty ordered set of multiple intersection points p1;:::; pk , such that any pair of adjacent points pj ; pj 1 and the points f g C p1; p are connected by lines of the arrangement. Moreover, if 1 j k 2, then k Ä Ä the line connecting pj to pj 1 and the line connecting pj 1 to pj 2 are different. C C C Also, the line connecting pk 1 to pk is different from the line connecting pk to p1 . Let † CP 2 be a line arrangement. The invariant ˇ.†/ which is defined in [7], Â counts the number of independent cycles in the graph. Fan [7] showed that ˇ.†/ 0 if and only if G.†/ has no cycles. D In an analogous way for † C2 , we define ˇ.†/ ˇ.† L / where L is Â WD [ 1 1 the projective line at infinity of C2 and G.†/ G.† L /. Note that the new WD [ 1 definitions are well defined. One of the most important invariants of a line arrangement is the fundamental group of 2 its complement, denoted by 1.C †/. The next lemma presents its computation.

Lemma 2.11 (Constructing the fundamental group [1; 18; 19; 3, page 304]) Let 2 † L1;:::; Ln C be a line arrangement that is enumerated as in [3]. D f g Â We associate a generator i to each line Li with base point as in [3] such that 2 G 1.C †/ 1; : : : ; n R ; D D h j i where R is a set of relations generated as follows.

Every intersection point of lines Li1 ;:::; Lim creates a set of relations

x1 x2 xm xm x1 xm 1 x2 xm x1 i1 i2 im im i1 im 1 i2 im i1 D D xi 1 where xi G and xi ixi . 2 i D

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It is easy to see that this set is equivalent to the following set: x Œ j ; x1 xm e; 1 j m: ij i1 im D Ä Ä

We get a similar presentation for a real arrangement by using the Moishezon–Teicher algorithm [16] and the van Kampen Theorem [13].

Notation 2.12 We denote by P the set of intersection points. For p P , we denote 2 by .p/ the set of generators attached to the lines passing through the point p and by .p/c the set of generators attached to the lines not passing through the point p.

3 Decomposition of G2=G3

Let G be a group. The abelianization of G , denoted by G , is G G=G2 . If g G , x x D 2 we denote g g G2 . x D Remark 3.1 Let † C2 be a line arrangement, and let p P . Then, G 2 2 c 2 D 1.C †/ .p/; .p/ R , D h j i c c G=G2 G Ab.G/ .p/; .p/ R;Œx; y e; x; y .p/ .p/ : D x D D h j D 2 [ i

This Lemma is an immediate implementation of the last section.

Lemma 3.2 (An implementation for line arrangements) Let † be a line arrangement, then the abelian group G2=G3 can be written as 1 * ˇ Œi; j Œj ; i ; + ˇ D G2=G3 Œi; j ˇ Œi; j Œ ; Œ ; Œi; j ; : D ˇ k l D k l ˇ Q P x .p/ Œx; y; p ; y .p/ 2 2 2 Proof A simple implementation of Proposition 2.4 on the presentation of G from Lemma 2.11.

Remark 3.3 We can see that if 1 and 2 are associated with lines meeting in one point and 3 and 4 are associated with lines meeting in a different point, there is no relation combining Œ1; 2 and Œ3; 4. Therefore, M G2=G3 Cp D p P 2

Algebraic & Geometric Topology, Volume 10 (2010) Characterization of line arrangements 1291 where * ˇ Œ ; Œ ; 1; + ˇ i j j i ˇ D Cp Œi;j ;i;j .p/ˇ Œi;j Œk ;l Œk ;l Œi;j ;i;j ;k ;l .p/; : D 2 ˇ Q D 2 x .p/ Œx;y;y .p/ 2 2 We can see that the generators of the different groups Cp in the direct sum are disjoint. L Consequently, let x G2=G3 , then x p P cp , where cp Cp . 2 D 2 2 For each r P , consider the projection 2 r G2=G3 Cr W ! M given by r .x/ r cp cr : D D p P 2 c If i .r/ , then for all j ; r .Œi; j / r .Œj ; i/ e. If i; j .r/, then 2 D D 2 r .Œi; j / Œi; j . D

4 The stabilizer of an intersection point

2 Let G 1.C †/. Deﬁne D f G=G2 G=G2 G2=G3 W ! by f .a; b/ Œa; b=G3: x x D

This function is well-deﬁned: If a1 a2 and b1 b2 , then a2 a1x where x G2 , S DS S D S D 2 b2 b1y where y G2 . Then, by Proposition 2.2, D 2 Œa2; b2 Œa1x; b1y Œa1x; b1Œa1x; y Œa1x; b1 Œa1; b1Œx; b1 Œa1; b1; D D D D D so f .a1; b1/ f .a2; b2/. S S D S S The following lemma presents some properties of f :

Lemma 4.1 Let a; b; c G=G2 . Then: 2 (1) f .a b; c/ f .a; c/ f .b; c/. D (2) f .a; b c/ f .a; b/ f .a; c/. D (3) f .an; bm/ f .a; b/nm for m; n Z. D 2 (4) f .b; a/ .f .a; b// 1 . D

Algebraic & Geometric Topology, Volume 10 (2010) 1292 Meital Eliyahu, Eran Liberman, Malka Schaps and Mina Teicher

Proof (1) Let A; B; C G such that A a, B b, and C c. This means that 2 x D x D x D AB ab, so by definition, f .ab; c/ ŒAB; C =G3 .ŒA; C ŒB; C /=G3 which by D D D definition is equal to f .a; c/f .b; c/. (2) The proof is the same as .1/. (3) Simple induction on .1/ and .2/. (4) Let A; B G such that A a, B b. By definition, 2 x D x D 1 1 f .b; a/ ŒB; A=G3 .ŒA; B=G3/ .f .a; b// : D D D

Now for any x G=G2 we define x 2 S.x/ y G=G2 f .y; x/ eG : x D f 2 j x x D 3 g By Lemma 4.1, S.x/ is a subgroup of G=G2 . x From now on, we talk about a specific intersection point Q. The lines passing through this point are L ;:::; L . Define M x1 x2 xm . Then, as noted in Lemma i1 im i1 i2 im f g D x1 2.11, the relations induced from the point Q can be translated to Œi ; M x 1 D D Œ m ; M e. im D Let M i i i . S D 1 2 m xj Since for each j , i we get that j D ij xj (1) f .i ; M / Œ ; M =G3 eG : j S D ij D 3

Theorem 4.2 Let Q P be an intersection point. Let .Q/ i1 ; : : : ; im and x x 2 D f g M 1 m . Then D i1 im \ S.M / .Q/ S./ : S D [ x .Q/ 2 We call S.M / the stabilizer of the intersection point Q. S

Proof We start by proving that \ S.M / .Q/ S./ : S Ã [ x .Q/ 2 .Q/ S.M /: We have already shown that if .Q/, then f .; M / e and Â S 2 x S D x hence f .M ; / f ..; M // 1 e. S x D x S D x

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T T .Q/ S./ S.M /: If x .Q/ S./, then for any .Q/, we have 2 x Â S 2 2 Qx 2 f .; x/ e. This means that also the product .Q/ f .; x/ e, so by Lemma 4.1, x D 2 x D x Y Y e f .; x/ f ; x f .M ; x/: x D x D x D S .Q/ .Q/ 2 2 Therefore x S.M /. 2 S Since S.M / is a subgroup and we have shown that it contains the union of .Q/ and T S .Q/ S./, it clearly contains the subgroup generated by the union. 2 x To complete the proof, we prove the opposite inclusion: \ S.M / .Q/ S./ : S Â [ x .Q/ 2 c Let x S.M / G=G2 .Q/; .Q/ R . Then x can be written as x z y x 2 S Â D h c j xi Tx x D x x where z .Q/ ; y .Q/ . We will prove y .Q/ S./. x 2 h i x 2 h i x 2 2 x Since z 1 .Q/ S.M /, so y z 1 z y z 1x. Hence we get y S.M /. 2 h i Â S x D x x D x x 2 S Let lx be a line passing through Q and x be the generator associated with it. Recall that y S./ if f .; y/ e. We need to prove that f .x; y/ e in G2=G3 . From x 2 x x x D x x D x Remark 3.3 we know that G2=G3 is a direct sum of groups M G2=G3 Cp; D p P 2 so it remains to prove that f .Œx; y/ e in Cp , for all p P . Let p P . We have x D x 2 2 to show that the projection of the coset Œx; y=G3 on Cp is trivial. x We separate our proof into three cases. Case 1 p Q. D c c Since y .Q/ .p/ then the projection is trivial by the deﬁnition of Cp . x 2 h i D h i Case 2 lx does not pass through p.

c In this case, x .p/ and therefore the projection is trivial, by the deﬁnition 2 h i of Cp .

Case 3 p Q and lx does pass through p. ¤ Q By deﬁnition of S, f . .Q/ ; y/ e. This means by the properties of f (Lemma Q 2 x x D x 4.1) that .Q/ f .; y/ e. Since lx passes through p, all the other lines passing 2 x x D x

Algebraic & Geometric Topology, Volume 10 (2010) 1294 Meital Eliyahu, Eran Liberman, Malka Schaps and Mina Teicher

through Q do not cross p. So if we project any f .z; y/ where z is any other x generator from .Q/, we get the identity. So in Cp , Y e f .; y/ f .x; y/; x D x x D x .Q/ 2 and thus f .x; y/ e. x D x

5 Fundamental groups which are semidirect product

Ln l Theorem 5.1 Let G i 1 Ai Z where Ai are free groups and let G G D D ˚ W ! be a projection where im. / y1;:::; y F . Then, there exists r; 1 r n, D h k i Š k Ä Ä such that for all g G; pr .g/ e implies that .g/ e and pr , where 2 r D D D ı r pr G G is the projection onto the subgroup of G naturally isomorphic to Ar . r W !

l Proof Let us denote the Z component as A0 . Since is a projection, .y1/ y1 D and .y2/ y2 . Then .Œy1; y2/ Œy1; y2 e. Note D D ¤ n M Œy1; y2 pr .Œy1; y2/ pr .Œy1; y2/: D i C 0 i 1 D Since A0 is abelian, pr .Œy1; y2/ e, and therefore 0 D n M Œy1; y2 pr .Œy1; y2/: D i i 1 D Since is a homomorphism,

.Œy1; y2/ .pr .Œy1; y2/ pr .Œy1; y2// D 1 ˚ ˚ n .pr .Œy1; y2// .pr .Œy1; y2// e: D 1 ˚ ˚ n ¤ This means that there is r; 1 r n, such that .pr Œy1; y2/ e, so if we denote Ä Ä r ¤ a pr .y1/ and b pr .y2/ we get Œ .a/; .b/ e. WD r WD r ¤ Since a; b Ar , if x Aj where j r , then Œx; a Œx; b e. We get that 2 2 ¤ D D Œ .x/; .a/ Œ .x/; .b/ e: D D So .x/ commutes with two noncommutative elements in a free group and hence .x/ e, for all x Aj , j r . This means that if g G and pr .g/ e, then D 2 ¤ 2 r D .g/ e. D

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l For each g G , g v1 vn w; vj Aj ; w Z ; we have that .g/ 2 D ˚ ˚ ˚ 2 2 D .v1 vn w/ .v1/ .vn/ .w/ .vr / .pr .g//. Therefore ˚ ˚ ˚ D ˚ ˚ ˚ D D r pr . D ı r For proving the next result, we use the following theorem from Lyndon and Schupp [14]:

Theorem 5.2 Let be an epimorphism from a ﬁnitely generated free group F to a free group G . Then F has a basis Z Z1 Z2 such that maps Z1 isomorphically D [ h i onto G and maps Z2 to e. h i Ln l Corollary 5.3 Let G i 1 Ai Z where Ai are free groups and let G G D D ˚ W ! be a projection where im. / y1;:::; y F . For the same r as in Theorem 5.1, D h k i Š k we can ﬁnd elements z1;:::; zm m 0 Ar such that f j g Â pr .y1/; : : : ; pr .y /; z1;:::; zm m 0 f r r k j g are generators of Ar and .zi/ e. D Proof From Theorem 5.1 and the fact that is a projection ( ), we get that ı D pr pr pr . Therefore, pr is a projection. r ı ı r ı D r ı r ı Now since .pr /.Ar / Ar , r ı Â .pr / A .pr / A ..pr / .pr // A .pr / A : r ı j r ı r ı j r D r ı ı r ı j r D r ı j r By assumption,

( ) im. / y1;:::; y is a free group. D h k i Now we claim pr .y1/; : : : ; pr .y / are the generators of free group F . If not, there f r r k g k is a nontrivial word w W.y1;:::; y / so that prr.w/ W.pr .y1/; : : : ; pr .y // e. D k D r r k D Therefore . pr /.w/ .pr .w// .e/ e: ı r D r D D By Theorem 5.1, pr , therefore .w/ . pr /.w/ e which means that D ı r D ı r D w e, a contradiction to ( ). D In conclusion, we get that .pr / A Ar Ar is a projection such that im.pr / r ı j r W ! r ı D pr .y1/; : : : ; pr .y / F (pr .y1/; : : : ; pr .y / are the generators of F ). h r r k i Š k r r k k By Theorem 5.2, Ai has a basis B B1 B2 such that im.pr / B1 D [ r ı D h i and .pr /.B2/ e. Since pr is a projection, we can assume that B1 r ı D r ı D pr .y1/; : : : ; pr .y / . Let B2 z1;:::; zm pr .zi/ e , which means that h r r k i D f j r ı D g . pr /.zi/ e. As we proved earlier, pr , so . /.zi/ e. But ı r ı D ı r D ı D , therefore .zi/ e. ı D D

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This implies that Ar has generators

pr .y1/; : : : ; pr .y /; z1;:::; zm f r r k g which satisfy the needed properties.

Without loss of generality, we temporarily change to a local numeration. 2 Let G 1.C †/ 1; : : : ; n; n 1; : : : ; l R , where .A/ 1; : : : ; n D D h C j i D f g are the generators of the lines which participate in a speciﬁc multiple intersection point called A.

Definition 5.4 [12] Let G be a group with a given presentation G X R . D h j i A Tietze transformation Ti .1 i 4/ is a transformation of X R to a presentation Ä Ä h j i G X R of one of the following types: D h 0 j 0i (T1 ) If r is a word in X and r e is a relation that holds in G , then let X h i D 0 D X; R R r : 0 D [ f g (T2 ) If r R is such that the relation r e holds in the group X R r , then let 2 D h j i X X; R R r . 0 D 0 D n f g 1 (T3 ) If w is a word in X and z X , put X X z ; R R wz . h i … 0 D [ f g 0 D [ f g 1 (T4 ) If z X and w is a word in the other elements of X such that wz R, 2 2 then substitute w for z in every other element of R to get R and take X z 0 D X z ; R R. f g 0 D z Before introducing the notion of a braid monodromy [16], we have to make some constructions. From now, we will work in C2 . Let E (resp. D) be a closed disk on x–axis (resp. y –axis), and let C be a part of an algebraic curve in C2 located in E D. Let 1 E D E and 2 E D D be the canonical projections, W ! W ! and let 1 C C E. Assume is a proper map, and deg n. Let N D 1j W ! D D x E # .x/ < n , and assume N @E ∅. Now choose M @E and let f 2 j g \ D 2 K K.M / 1.M /. By the assumption that deg n . #K n/, we can D D D ) D write K a1; a2;:::; an . Under these constructions, from each loop in E N , we D f g can define a braid in BnŒM D; K in the following way: (1) Since deg n, we can lift any loop in E N with a base point M to a D system of n paths in .E N / D which start and finish at a1; a2;:::; an . f g (2) Project this system into D (by 2 ), to get n paths in D which start and end at the image of K in D (under 2 ). These paths actually form a motion. (3) Induce a braid from this motion; see Garber [8].

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To conclude, we can match a braid to each loop. So we get a map ˛ 1.E N; M / W ! BnŒM D; K, which is also a group homomorphism. This map is called the braid monodromy of C with respect to E D; 1; M [16; 8]. The following remark demonstrates the necessity of the condition that there are no parallel lines in the afﬁne plane:

Remark 5.5 Let † be a line arrangement with no parallel line in the afﬁne plane. Let 2 1; : : : ; be the generators of 1.C †/. Then l 2 1 Z.1.C †//: l 2

Proof Let ˛ 1.C †/ Bn be the braid monodromy of †. By Remark 4.7 W ! in [3], the closed braid determined by the product ˛.1/ ˛.l / is actually a link 2 of the curve at inﬁnity. In a generic curve, we have ˛.1/ ˛. / , where l D is the braid which rotates by 180ı counterclockwise all the strands together. One of the presentations of the fundamental group is 1; : : : ; ˛.si/.j / j ; i; j , h l j D 8 8 i where si is the loop created in the x–axis as a result of the projection of the line arrangement on the plane [13]. Therefore, ˛.si/ is a braid acting on the generators of the fundamental group. As a result ˛.s1 sn/.j / j , for all j , which means 2 2 D1 l 1 l .j / j , for all j . It is known that .x/ x and thus j j . D 2 D D Hence, 1 Z.1.C †//. l 2

2 Let G 1.C †/ 1; : : : n; n 1; : : : ; l R , where .A/ 1; : : : ; n D D h C j i D f g are the generators of the lines which pass through in a speciﬁc multiple intersection point called A.

Let us recall that the lines passing through Q are Li1 ;:::; Lim . Recall also M x x x f g D 1 2 m . i1 i2 im Theorem 5.6 With the same assumptions of Theorem 5.1 and Remark 5.5, let H c WD i1 ; : : : ; im 1 . Let N be the normal closure of .Q/ ; M . Let f1 H G be h i f g W ! the natural embedding and f2 G G=N the natural homomorphism. Then: W !

(1) G=N Fm 1 which is generated by f2.ij /, 1 j m 1. Š Ä Ä (2) H Fm 1 . Š (3) G N Ì Fm 1 . D (4) There exists a projection h G G .h2 h/, such that im.h/ H and W ! D D ker.h/ N . D

Algebraic & Geometric Topology, Volume 10 (2010) 1298 Meital Eliyahu, Eran Liberman, Malka Schaps and Mina Teicher

Proof (1) We can present G=N .Q/; .Q/c R; M; .Q/c . Denote .Q/c D h j i D z1;:::; z . By applying iteratively Tietze’s transformation (T4 ) for every zi f k g 2 .Q/c , we obtain G=N .Q/ R; M , where R; M are obtained from R; M by D h j y i y substituting e for zi , 1 i k , in every other element of R and M , respectively. Ä Ä Let v P with v Q be an intersection point, and let r be the relations induced by v. 2 ¤ Now we can get the following:

(a) If there is a line L that goes through both points v and Q and the generator attached to L is i for some j , 1 j m, then the relations r become j Ä Ä Œx ; 1;:::; 1 for some x. These relations are trivial. ij (b) If the points v and Q do not share any line, then the relations r become Œ1;:::; 1, which are also trivial.

Let us denote by W the word W after rewriting. Due to the above observations, using the presentation G=N .Q/ R; M , then D h j z i x fi1 xim G=N i1 ; : : : ; im Œ ; : : : ; ; M : D h j i1 im i This is equal to

x fi1 xim i1 ; : : : ; im Œi ; M; : : : ; Œi ; M; M Fm 1: h j 1 m e i D (2) Assume otherwise. Then there exists a nontrivial word w.i1 ; : : : ; im 1 / e. D So applying f2 , we get a nontrivial word w.f2.i1 /; : : : ; f2.im 1 // which is not possible by (1).

(3) From the ﬁrst paragraph, we get that f2 fe1 is a surjective function from Fm 1 ı to Fm 1 and therefore it is an isomorphism. (4) Derived directly from (2) and (3).

By the last theorem: Let Q P , .Q/ i ; : : : ; i . Then there is a projection 2 D f 1 m g h G G such that im.h/ i ; : : : ; i and ker.h/ is the normal closure of W ! D h 1 m i .Q/c; M . h i By Theorem 5.1 and Corollary 5.3 we get that there is r , 1 r n, such that Ä Ä Ar w1; : : : ; wn 1; z1;:::; zk k 0 and h h prr , prr h is a projection, D f j 1 g D ı ı i h.wi/; h.zi/ e; bi iwi ; h.bi/ e (note that h has the role of ). D D D D

Claim 5.7 Ar prr .S.M //. D x S

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Proof We start by proving that pr .M / e: for all x; y and j j . Since r 1 2 ij1 D x y ¤ and ij are different generators of the free group im.h/, h.Œi i / e. We know 2 j1 j2 ¤ that h h pr , so we get D ı r x y x y .h prr /.Œi i / h.Œi i / e; ı j1 j2 D j1 j2 ¤

x y xj1 xj2 so prr .Œ / e. Hence, for x xj1 ; y xj2 , prr .Œ ; / e. We know ij1 ij2 ij1 ij2 xj ¤ D xj D ¤ that Œi 1 ; M e, so the elements prr .i /; j 1; 2, commute with M but do not j1 D j D commute with each other in the free group Ar , therefore pr .M / e. r D The next step is to prove that Ar S.M /. Let a Ar . Then there exists a Ar Â S 2 2 such that a a. Since pr .M / e and a Ar , by the properties of direct sum D r D 2 Œa ; M e. Therefore Œa ; M =G3 eG . By deﬁnition f .a; M / eG which D D 3 S D 3 implies that a S.M /. 2 S Hence, we have Ar prr .S.M //. The opposite inclusion is trivial by deﬁnition, and Â x S thus Ar prr .S.M //. D x S ˝ ˚ Tm ˛ By Theorem 4.2, S.M / i1 ;:::; im k 1 S.ik / , therefore S D f g [ D m \ Ar prr i1 ; : : : ; im S.ik / D x f g [ k 1 D m \ prr .i1 /; : : : ; prr .im / prr S.ik / : D f x x g [ x k 1 D We claim that the right part of this generating set is trivial:

Tm Claim 5.8 prr k 1 S.ik / e . x D D f g Proof It is known that for any two sets A; B and a function F that F.A B/ Tm Tm \ Â F.A/ F.B/. Therefore prr k 1 S.ik / k 1 prr .S.ik // . \ x D Â D x As we mentioned Ar pr .1/; : : : ; pr .m/; z1;:::; zq q 0 . D h r r j i In other words, Ar ˇ1; : : : ; ˇm q where D h C i ( 6 prr .ik / k m; ˇk D zk m m 1 6 k 6 m q: C C t1 tm q Let g prr .S.ik // Ar , therefore g ˇ1 ˇq m C . Without loss of gener- x 2 x Â t1 xtmDq C ality, we can choose that g ˇ1 ˇm Cq . By Theorem 5.1 we know that D C ` G A1 Ar An Z : D ˚ ˚ ˚ ˚ ˚

Algebraic & Geometric Topology, Volume 10 (2010) 1300 Meital Eliyahu, Eran Liberman, Malka Schaps and Mina Teicher

` L We define D A1 Acr An Z , then G Ar D. D ˚ ˚ ˚ ˚ ˚ D Hence, there exists g2 D such that g g2 S.i /. 2 x ˚ 2 k Let ' G D be the natural projection, and define '.i /. Hence, it implies W ! k WD k that i pr .i / . k D r k ˚ k eG f .i ; g g2/ f .pr .i / ; g g2/ 3 D k ˚ D r k ˚ k ˚ f .pr .i /; g/f .pr .i /; g2/f . ; g/f . ; g2/: D r k x r k k x k By the definition of f , this implies that

Œpr .i /; gŒpr .i /; g2Œ ; gŒ ; g2 G3: r k x r k k x k 2 Since pr .i /; g Ar , and g2; D, then Œpr .i /; g2 Œ ; g eG . Therefore r k 2 k 2 r k D k D 3 Œpr .i /; gŒ ; g2 G3 .Ar /3 D3 , which means Œpr .i /; g .Ar /3 and r k k 2 D ˚ r k 2 Œ ; g2 D3 . k 2 The fact that .Ar /2=.Ar /3 Œˇi; ˇj 1 i < j m q is derived directly from D h j Ä Ä C i t1 tm q the deﬁnition of Ar . Hence, .Ar /3 Œprr .ik /; g Œˇk ; g Œˇk ; ˇ1 ˇm Cq t1 tk 3tm q D D C D Œˇk ; ˇ1 Œˇk ; ˇk Œˇk ; ˇm q C . C Note that Œˇ ; ˇ e. By [10], Œˇ ; ˇj j k are generators (or inverses) of k k D f k j ¤ g .Ar /2=.Ar /3 which is a free abelian group. Since every generator appears at most once, we get tj 0 for all j k . D ¤ tk In conclusion, g ˇ ˇ pr .i / . So g pr .i / . x D k 2 h k i D h r k i x 2 h r k i x To summarize, we have shown that if g prr S.ik / then g prr ik . There- x 2 h x i x 2 h x i fore, prr S.ik / prr .ik / . h x i Â h x i Tm Tm Tm In conclusion, pr .S.i // pr .i / e, so pr S.i / k 1h r k i Â k 1h r k i D r k 1 k D e as claimed. D x D x x D f g

Since i1 im e, Ar i1 ;:::; im 1 . As a result, Ar prr i1 ; : : : ; im 1 . D D h i D xh i Consequently, rank.Ar / m 1. Ä We know that Ar w1; : : : ; wm 1; z1;:::; zt t 0 . Combining these two facts to- D h j i xj gether and by Theorem 5.2, we get Ar w1; : : : ; wm 1 , where yi .prr h/.i /. D h i D ı j From Theorem 5.1, h pr h. Since h is a projection h2 h. Hence, we get ı r D D pr h pr h pr h h pr h; r ı ı r ı D r ı ı D r ı which means that pr h is a projection too. Recall that r ı

Ar .prr h/.i1 /; : : : ; .prr h/.im 1 / : D h ı ı i

Algebraic & Geometric Topology, Volume 10 (2010) Characterization of line arrangements 1301

Therefore .pr h/.wj / .pr h/ .pr h/.i / .pr h/.i / wj . Hence we r ı D r ı ı r ı j D r ı j D get that .pr h/.Ar / Ar . r ı D Now if li is a line that does not pass through the point Q and i is the generator of li , then h.i/ e and therefore .pr h/.i/ pr .e/ e. D r ı D r D From this investigation, we get the following theorem.

2 Ln l Theorem 5.9 Let † be a line arrangement and 1.C †/ . i 1 Ai/ Z where ' D ˚ Ai is a free group. Then for any multiple point Q in the arrangement with k lines, namely, l1;:::; l , there exists r; 1 r n,and a projection onto Ar , 'Q G G f k g Ä Ä W ! such that Ar 'Q.1/; : : : ; 'Q.k / Fk 1 . If lj is a line which does not pass D h i Š through the point, then 'Q.j / e. D Moreover, if p1;:::; pm are the multiple points of † and ni is the number of lines Lm pass through the point pi , then G i 1 Ci B , where Ci Fni 1 . If l is a Š D ˚ Š line which does not pass through pi and let be its corresponding generator, then pr ./ e ( where pr is the projection onto Ci ). i D i

6 Main theorems

Theorem 6.1 Let † C2 be a line arrangement which has no pair of parallel lines. Â Then if r 2 M l 1.C †/ Ai Z ; D ˚ i 1 D where Ai are free groups. Then ˇ.†/ 0. D Proof Assume by negation that ˇ.†/ 0 which means that there is at least one cycle ¤ in the graph of †. Let us pick a minimal cycle in the following sense. The cycle contains r points, namely p1;:::; pr and pi is connected in the cycle only to pi 1 f g and pi 1 (the indices are taken modulo r ). In other words, a cycle with no subcycles. C Lr By Theorem 5.9, we can write G . Ci/ B1 where B1 is not necessar- D i 1 ˚ ily abelian. If l is a line that does not passD through the points in the cycle and let be its corresponding generator, then pr ./ e; 1 i r . Deﬁne N i D Ä Ä D X ; : : : ; X ; 1 n , where X ; : : : ; X are the generators of lines that do not h 1 t i 1 t participate in an intersection point pi; 1 i r . Denote Z 1 n . Let Ä Ä D H G=N . WD Let ni be the number of lines passing through pi .

Algebraic & Geometric Topology, Volume 10 (2010) 1302 Meital Eliyahu, Eran Liberman, Malka Schaps and Mina Teicher

Pr On one hand, since we have b i 1 .ni 1/ lines participating in the cycle, then WD D if we denote by 1; : : : ; b the generators associated with the lines participating in the cycle, then H 1; : : : ; R; Z where R is the relations which are derived from Š h b j z zi z the original relations. Therefore, it is easy to see that rank.H / b 1. x Ä

On the other hand, for 1 i r and 1 j t , pr .X / e and also 1 n Z.G/ Ä Ä Ä Ä i j D 2 so pri.1 n/ e. Thus pri.N / e and hence Ci=N Ci . This implies that rD D Š H G=N L C B =N . Since C F and C Zni 1 , we have H i 1 i 1 i ni 1 i x LDr Š D ˚ ŠLr ni 1 Š b D . Ci/ B1=N which implies that H . Z / B1=N Z B1=N . i 1 ˚ x D i 1 ˚ Š ˚ ThusD rank.H / b, a contradiction. D x

We are now ready to prove the converse of Fan’s theorem.

Theorem 6.2 Let † CP 2 be a line arrangement. If Â r 2 M l 1.CP †/ Ai Z ; D ˚ i 1 D where Ai are free groups, then ˇ.†/ 0. D

Proof Suppose that † CP 2 is an arrangement of complex projective lines such 2 2 that 1.CP †/ is isomorphic to a direct product of free groups. Let L0 CP be n a complex projective line which is in general position to †. Then L0 and † intersect at double points. Choose an arbitrary complex projective line of † and denote this line 2 2 2 by L1 . Note that C CP L0 CP L1 . By applying the product theorem of Š n Š n Oka and Sakamoto [17], we have

2 2 1.CP .† L0// 1..CP L1/ ..† L1/ L0// n [ D n n n [ 2 2 1..CP L1/ .† L1// 1..CP L1/ L0/ Š n n n ˚ n n 2 1.CP †/ Z: Š n ˚ 2 This calculation shows that 1.CP .† L0// is a product of free group. Note that n [ 2 2 1.CP .† L0// 1..CP L0/ .† L0//: n [ D n n n 2 Note that † L0 is a union of complex lines in the complex plane CP L0 such that n n no two lines of this arrangement are parallel.

By Theorem 6.1 ˇ.† L0/ 0, therefore ˇ.†/ ˇ.† L0/ 0. n D D n D

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Acknowledgements We wish to thank David Garber for the experience he shared with us and the time he invested in improving the paper. We are extremely grateful for his great efforts in order to make this paper worthy. We would also like to thank the anonymous referee for his fruitful advice. In an earlier version submitted to the journal, the paper was written under the implicit assumption that no two lines of the arrangement are parallel to each other. Under this assumption we give the proof of Theorem 6.1. The referee suggested us to make this assumption explicit. The proof of Theorem 6.2 is given by the referee in earlier referee report of this paper.

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Department of Mathematics, Bar Ilan University, 52900 Ramat Gan, Israel [email protected], [email protected], [email protected], [email protected] http://www.cs.biu.ac.il/~teicher/

Received: 26 February 2009 Revised: 14 October 2009

Algebraic & Geometric Topology, Volume 10 (2010)