DETERMINING THE HURWITZ ORBIT OF THE STANDARD GENERATORS OF A BRAID GROUP

Yaguchi, Y. Osaka J. Math. 52 (2015), 59 70 DETERMINING THE HURWITZ ORBIT OF THE STANDARD GENERATORS OF A BRAID GROUP YOSHIRO YAGUCHI (Received January 16, 2012, revised June 18, 2013) Abstract The Hurwitz action of the n-braid group B n on the n-fold product (B m ) n of the m-braid group B m is studied. Using a natural action of B n on trees with n labeled edges and n 1 labeled vertices, we determine all elements of the orbit of every n-tuple of the n distinct standard generators of B n 1 under the Hurwitz action of B n. 1. Introduction Let B n denote the n-braid group, which has the following presentation [1, 4]., 1,, n 1 i j i j i j ( i j 1), i j j i ( i j 1) where i is the i-th standard generator represented by a geometric n-braid depicted in Fig. 1.1. Throughout this paper, n is an integer with n 2. Let G be a group and let G n be the n-fold direct product of G. For elements g and h of G, let g h denote h 1 gh and let g h denote g (h 1 ) hgh 1. DEFINITION 1.1. The Hurwitz action of B n on G n is the right action defined by and (g 1,, g i 1, g i, g i 1, g i 2,, g n ) i (g 1,, g i 1, g i 1, g i g i 1, g i 2,, g n ) (g 1,, g i 1, g i, g i 1, g i 2,, g n ) 1 i (g 1,, g i 1, g i 1 g i, g i, g i 2,, g n ), where 1,, n 1 are the standard generators of B n. We call the orbit of (g 1,, g n ) ¾ G n under the Hurwitz action of B n the Hurwitz orbit of (g 1,, g n ) and denote it by (g 1,, g n ) B n. We say two elements of G n are Hurwitz equivalent if they belong to the same Hurwitz orbit. 2010 Mathematics Subject Classification. Primary 20F36; Secondary 20F34.

60 Y. YAGUCHI Fig. 1.1. When G is a braid group B m and each g j ( j 1,, n) is a positive simple braid (that means a conjugate of 1 ), the Hurwitz orbit of an n-tuple (g 1,, g n ) corresponds to an equivalence class of an algebraic braided surface in a bidisk D 2 D 2 with n branch points [11, 13, 14, 15, 17]. Here, we say that two braided surfaces S and S ¼ are equivalent if there is a fiber-preserving diffeomorphism fï D 2 D 2 D 2 D 2 (that is, f (D 2 {x}) D 2 {g(x)} for some diffeomorphism gï D 2 D 2 ) carrying S to S ¼ rel D 2 D 2. It is natural to ask if two given braided surfaces are equivalent or not. This question is equivalent to asking if two given n-tuples (g 1,, g n ) and (g ¼ 1,, g¼ n ) of positive simple braids determined by their monodromies are Hurwitz equivalent or not. This is a very hard problem and no algorithm to solve it is known. However, we can determine all elements of the Hurwitz orbits of some n-tuples of positive simple braids. Throughout this paper, we use the symbol s i to denote the i-th standard generator of B n 1, and i to denote that of B n. We prove that for every permutation ³ of {1,, n}, there is a natural bijection from the Hurwitz orbit of (s ³(1),, s ³(n) ) to a set consisting of all trees satisfying the conditions of Definition 2.2. As a result, we determine all elements in the Hurwitz orbit of (s ³(1),, s ³(n) ) for every permutation ³ of {1,, n} (see Theorem 4.3). We will also prove Theorem 1.2 by using Theorem 4.3. Let X n 1 be the set of the integers {2, 3,, n}. For a permutation ³ of {1,, n}, let A ³ {i ¾ X n 1 ³ 1 (i 1) ³ 1 (i)}. Theorem 1.2. For permutations ³ and of {1,, n}, the following conditions are mutually equivalent. (1) (s ³(1),, s ³(n) ) and (s (1),, s (n) ) are Hurwitz equivalent. (2) The products s ³(1) s ³(n) and s (1) s (n) are equal in B n 1. (3) The sets A ³ and A are equal. REMARK 1.3. Let B n 1 be the semi-group which has the generators s 1,, s n and the relations s i s j s j s i if i j 2 and s i s j s i s j s i s j if i j 1. Let i Ï B n 1 B n 1 be the natural semi-group homomorphism, i.e., i(s i ) s i for each i ¾ {1,,n}. In [7], Garside proved that i is injective. (We call an element of i(b n 1 ) a positive element of B n 1.) Using the Garside s theorem, T. Ben-Itzhak and M. Teicher showed the equivalence of the conditions (1) and (2) of Theorem 1.2 in [2]. This is

DETERMINING THE HURWITZ ORBIT 61 explained as follows. If i j 2, then (s i, s j ) and (s j, s i ) are Hurwitz equivalent. If i j 1, then (s i, s j, s i ) and (s j, s i, s j ) are Hurwitz equivalent. (See [2].) Thus, for permutations ³ and of {1,, n}, (s ³(1),, s ³(n) ) and (s (1),, s (n) ) are Hurwitz are equal. equivalent if and only if the products s ³(1) s ³(n) and (1) s s (n) in B n 1 By the Garside s theorem, this condition is equal to the condition that the products s ³(1) s ³(n) and s (1) s (n) in B n 1 are equal. Thus, we have the equivalence of (1) and (2) of Theorem 1.2. In this paper, we will prove the equivalence of (1) and (3) of Theorem 1.2 by using Theorem 4.3. 2. Main result Let S n 1 be the symmetric group of degree n 1 and let T n 1 be the set of elements ( 1,, n ) of the n-fold direct product (S n 1 ) n of S n 1 such that 1,, n are transpositions which generate S n 1. For any element ( 1,, n ) of T n 1, the Hurwitz orbit ( 1,, n ) B n is contained in T n 1 [10, 12]. Thus, we obtain an action of B n on T n 1 as a restriction of the Hurwitz action on (S n 1 ) n. Let G n 1 be the set of trees with n edges labeled {e 1,, e n } and n 1 vertices labeled {Ú 1,, Ú n 1 }. A natural action of B n on G n 1 has been observed [5, 6, 8]. It is explained as follows. Take an element ( 1,, n ) of T n 1 and put i (k i l i ) (k i l i ) for i ¾ {1,, n}. We define ¼( 1,, n ) as a graph with n 1 vertices labeled {Ú 1,, Ú n 1 } and n edges labeled {e 1,, e n } such that the edge labeled e i connects with two vertices labeled Ú ki and Ú li for i ¾ {1,, n}. Since 1,, n generate S n 1, the graph ¼( 1,, n ) must be a tree [5, 6]. Hence, ¼( 1,, n )¾G n 1. We call the induced map ¼Ï T n 1 G n 1 the graphic map. The map ¼ is bijective [5, 6]. Thus, we obtain the action of B n on G n 1 defined by ¼((¼ 1 ( )) ) for ¾ G n 1 and ¾ B n [5, 6, 8]. We call this action the Hurwitz action of B n on G n 1. For 1 i n, let t i be the transposition (i i 1) of S n 1. Take a permutation ³ of {1,, n}. Let pï B n 1 S n 1 be the canonical projection and let P p n Ï (B n 1 ) n (S n 1 ) n be the map defined by (b 1,,b n ) (p(b 1 ),, p(b n )). Then, for any ¾ B n, it holds P((b 1,,b n ) ) (P(b 1,,b n )). Thus, for every permutation ³ of {1,,n}, we have P((s ³(1),, s ³(n) ) B n ) (t ³(1),, t ³(n) ) B n. Since (t ³(1),, t ³(n) ) ¾T n 1 (and hence, ¼(t ³(1),, t ³(n) ) ¾ G n 1 ), we have ¼(t ³(1),, t ³(n) ) B n G n 1. In [16], it is proved that the map P P (s³(1),,s ³(n) ) B n Ï (s ³(1),, s ³(n) ) B n (t ³(1),, t ³(n) ) B n is bijective. Then, ¼Æ P((s ³(1),,s ³(n) ) B n ) ¼((t ³(1), t ³(n) ) B n ) ¼(t ³(1),,t ³(n) ) B n. Since ¼ and P are bijective, the map ¼Æ PÏ (s ³(1),,s ³(n) ) B n ¼(t ³(1),,t ³(n) ) B n is bijective. REMARK 2.1. Take an element ( 1,, n ) of T n 1. Let T n 1 ( 1,, n ) be the subset of T n 1 defined by {( ¼ 1,, ¼ n )¾T n 1 ¼ 1 ¼ n 1 n }. In [10], A. Hurwitz proved that there exists (n 1) n 1 elements of T n 1 ( 1,, n ). In [12], P. Kluitmann proved that the Hurwitz orbit ( 1,, n ) B n is equal to the set T n 1 ( 1,, n ) by an induction on n and using combinatorially calculations of S n 1. Hence, there exists

62 Y. YAGUCHI (n 1) n 1 elements in the Hurwitz orbit ( 1,, n ) B n. In [9], S.P. Humphries proved that the Hurwitz orbit (s 1,,s n ) B n consists of (n 1) n 1 elements. In [16], the author proved that for any permutation ³ of {1,, n}, the Hurwitz orbit (s ³(1),, s ³(n) ) B n consists of (n 1) n 1 elements. Let C be a circle in Ê 2. Let P n 1 (C) C C denote the product space of n 1 copies of C. Let Q n 1 (C) {(q 1,, q n 1 ) ¾ P n 1 (C) q i q j for i j}. We call Q n 1 (C) the configuration space of ordered n 1 distinct points of C. Fix an element q (q 1,, q n 1 ) ¾ Q n 1 (C) and let ¼ q n 1 be the set of segments {q iq j 1 i j n 1} in Ê 2. Take elements e and e ¼ of ¼ q n 1. If e {q i, q i ¼}, e ¼ {q i, q i ¼¼} and q i ¼ q i ¼¼, i.e., e and e ¼ share a common end point q i, then we say that e and e ¼ are adjacent (at q i ). Moreover, if the end points q i ¼, q i and q i ¼¼ appear on C counterclockwise in this order, then we say that e ¼ is a right adjacent to e (at q i ). DEFINITION 2.2. An n-tuple (e 1,,e n ) of elements of ¼ q n 1 is good if it satisfies the conditions (i) (iii). (i) If k l, then e k and e l are disjoint or adjacent. (ii) If k l and e k and e l are adjacent, then e l is a right adjacent to e k. (iii) The union e 1 e n is contractible. Let G(¼ q n 1 ) be the set of n-tuples of elements of ¼q n 1 which are good. When (e 1,, e n ) is an element of G(¼ q n 1 ), the union e 1 e n is regarded as a graph with n 1 vertices q 1,, q n 1 and n edges e 1,, e n. Note that the graph e 1 e n is a tree. Putting labels Ú i on the points q i for 1 i n 1, we regard G(¼ q n 1 ) as a subset of G n 1. The following is our main theorem. Theorem 2.3. For a permutation ³ of {1,, n}, there exists an element q of Q n 1 (C) such that the Hurwitz orbit ¼(t ³(1),, t ³(n) ) B n is equal to the set G(¼ q n 1 ( ) G n 1 ). Hence, the map ¼ Æ P gives a bijection from the Hurwitz orbit (s ³(1),, s ³(n) ) B n to the set G(¼ q n 1 ). REMARK 2.4. In [16], the author found an element q of Q n 1 (C) such that the Hurwitz orbit ¼(t ³(1),, t ³(n) ) B n is contained in G(¼ q n 1 ). Thus, we obtained an action of B n on G(¼ q n 1 ) as the restriction of the Hurwitz action of B n on G n 1. In this paper, we will show the transitively of this action. 3. Some notions Throughout this section, X n 1 is the set of the integers {2, 3,, n} and A is a fixed subset of X n 1.

DETERMINING THE HURWITZ ORBIT 63 For integers i and j with 1 i j n 1, we define s A i j ¾ B n 1 and s A ji ¾ B n 1 by si A j s A ji s i j 1 k i 1 where k 1 if k ¾ A and k 1 if k A. We call si A j associated with A. Note that a standard generator s i of B n 1 is a band generator sii 1 A. Let n 1 A be the set of band generators {sa i j ¾ B n 1 1 i j n 1} associated with A. Let T n 1 be the set of transpositions of S n 1. The natural projection pï B n 1 S n 1 gives the bijection p p A n 1 Ï n 1 A T n 1 which satisfies p(si A j ) (i j) for 1 i j n 1. Let P k (k, 0) ¾ Ê 2 for an integer with 1 k n 1. Let C 1 be the circle in Ê 2 whose diameter is the segment P 1 P n 1. Take the points Q k ¾ C 1 for 1 k n 1 such that Q 1 P 1, Q n 1 P n 1 and Q k (k, y k ) for each 2 k n, where y k 0 if k ¾ A and y k 0 if k A. We call the points Q 1,, Q n 1 on the circle C 1 the points associated with A. We call the element Q (Q 1,, Q n 1 ) of the configuration space Q n 1 (C 1 ) the ordered n 1 points associated with A. Let ¼n 1 A denote the set of segments ¼n 1 A ¼ Q n 1 {Q i Q j 1 i j n 1} in Ê 2. We have a bijection from A n 1 to ¼A n 1 defined by sa i j Q i Q j segment Q i Q j the segment corresponding to s A i j. s k k, a band generator of B n 1 for 1 i j n 1, and we call the REMARK 3.1. In [16], the reason why we call Q i Q j the segment corresponding to si A j is explained as follows. Let P 0 Q 0 (0, 0)¾Ê 2 and P n 2 Q n 2 (n 2, 0)¾Ê 2. Let C 2 be the circle in Ê 2 whose diameter is the segment P 0 P n 2. Let D be the disk in Ê 2, with D C 2. Take an isotopy {h u } u¾[0,1] of D such that for each u ¾ [0, 1], h 0 id, h u D id, and for each u ¾ [0, 1] and each (x, y) ¾ Ë n 1 i 0 Q iq i 1, h u (x, y) (x, (1 u)y). Then h 1 (Q i ) P i for any i. For 1 i j n 1, we define «A i j by the arc h 1 (Q i Q j ) in D. Note that «A i j {P i, P j }, «A i j is upper than P k if k ¾ A and «A i j is lower than P k if k A. The braid group B n 1 is isomorphic to the mapping class group of (D, {P 1,, P n 1 }) relative to the boundary (cf. [3]). The band generator s A i j corresponds to the isotopy class of a homeomorphism from (D,{P 1,,P n 1 }) to itself which twists a sufficiently small disk neighborhood of the arc «i A j by 180 Æ -rotation clockwise using its collar neighborhood. By the homeomorphism h 1 Ï (D, {Q 1,, Q n 1 }) (D,{P 1,, P n 1 }), we identify the mapping class group of (D,{Q 1,, Q n 1 }) and that of (D, {P 1,, P n 1 }). Then the band generator si A j corresponds to the isotopy class of a homeomorphism from (D, {Q 1,, Q n 1 }) to itself which twists a sufficiently small disk neighborhood of the segment Q i Q j by 180 Æ -rotation clockwise. Thus, we say that the segment Q i Q j corresponds to the band generator si A j ¾ A.

64 Y. YAGUCHI Fig. 3.1. Let (g 1,, g n ) be an element of the n-fold product ( n 1 A )n of n 1 A and let (a 1,, a n ) be the n-tuple of the segments a i corresponding to g i. Then, we call (a 1,, a n ) the segment system corresponding to (g 1,, g n ). DEFINITION 3.2. An element of (¼n 1 A )n is A-good if it is good. An element of ( n 1 A )n is A-good if the corresponding segment system is A-good. We denote the set of elements of ( n 1 A )n (resp. (¼n 1 A )n ) which are A-good by G( n 1 A ) (resp. G(¼A n 1 )). EXAMPLE 3.3. Let n 4 and A {2}. Then, the element (s A 23, sa 24, sa 13, sa 25 ) of ( A 5 )4 is an A-good element. The segments a 1,, a 4 corresponding to s A 23, sa 24, sa 13, sa 25 are depicted in Fig. 3.1. Let Ë A {2,,n} ( A n 1 )n and ¼ Ë A {2,,n} (¼A n 1 )n. Let Ï ¼ be the map which sends an element of ( n 1 A )n to the segment system corresponding to it for each subset A of {2,,n}. It is obvious that the map is bijective and (( n 1 ) A )n (¼n 1 A )n. By Definition 3.2, (G( n 1 A )) G(¼A n 1 ). For each subset A of {2,, n}, we simply write for the bijective map G( A n 1 )Ï G( A n 1 ) G(¼A n 1 ). Proposition 3.4. Let A and B be subsets of {2,, n}. Then, the conditions ) and A B are mutually equivalent. G( A n 1 ) G( B n 1 ), G(¼A n 1 ) G(¼B n 1

Proof. DETERMINING THE HURWITZ ORBIT 65 If A B, then it is obvious that the other two conditions hold. By virtue of the bijective map Ï ¼, the conditions G( A n 1 ) G( B n 1 ) and G(¼A n 1 ) G(¼n 1 B ) are equivalent. We prove that the condition G(¼A n 1 ) G(¼B n 1 ) implies A B. Let Q X k denote the point (k, y k), on the circle C 1, associated with a subset X {2,,n}. Take elements (a 1,,a n )¾G(¼n 1 A ) and (b 1,,b n )¾G(¼n 1 B ). Then, by the conditions of Definition 2.2, a 1 a n {Q A 1,,QA n 1 } and b 1 b n {Q B 1,,QB n 1 }. Hence, the condition G(¼n 1 A ) G(¼B n 1 ) implies QA k Q B k for 1 k n 1, and this implies A B. This completes the proof of Proposition 3.4. Let A be a subset of {2,,n} and let Q 1,,Q n 1 be the points, on the circle C 1, associated with A. We regard G(¼ A n 1 ) as a subset of G n 1 putting labels Ú 1,, Ú n 1 on the points Q 1,, Q n 1, resp. Recall that the Hurwitz action of B n on G n 1 is defined by ¼((¼ 1 ( )) ) for ¾ G n 1 and ¾ B n, where ¼ is the graphic map ¼Ï T n 1 G n 1. By Proposition 3.5 (2), we obtain an action of B n on G(¼ A n 1 ) as a restriction of the Hurwitz action of B n on G n 1. We call it the Hurwitz action of B n on G(¼ A n 1 ). Proposition 3.5 ([16]). (1) If (g 1,, g n ) ¾ G( n 1 A ), then, for any ¾ B n, we have (g 1,, g n ) ¾ G( n 1 A ). (2) If (a 1,, a n )¾G(¼n 1 A ), then, for any ¾ B n, we have (a 1,, a n ) ¾G(¼n 1 A ). Take an element (g 1,, g n ) (si A 1 j 1,, si A n j n ) ¾ G( n 1 A ). Then, (g 1,, g n ) (Q i1 Q j1,, Q in Q jn ) ¼((i 1 j 1 ),, (i n j n )) ¼Æ P(g 1,, g n ) (¾ G(¼n 1 A )). Moreover, for any element ¾ B n, (g 1,, g n ) ¾ G( n 1 A ) by Proposition 3.5 (1), and we have ((g 1,,g n ) ) ¼Æ P((g 1,,g n ) ) ¼(P(g 1,,g n ) ) ¼(((i 1 j 1 ),,(i n j n )) ) ¼((¼ 1 (Q i1 Q j1,, Q in Q jn )) ) (Q i1 Q j1,, Q in Q jn ) (g 1,, g n ). Thus, we have the following proposition. Proposition 3.6. For an element (g 1,, g n ) of G( n 1 A ), we have )) and (1) (g 1,, g n ) ¼ Æ P(g 1,, g n ) (¾ G(¼n 1 A (2) for any element ¾ B n, ((g 1,, g n ) ) (g 1,, g n ). By Proposition 3.6 (2), the bijective map Ï G( n 1 A ) G(¼A n 1 ) implies that two statements of (1) and (2) of Proposition 3.5 are equivalent. 4. Proof of Theorem 1.2 and Theorem 2.3. Throughout this section, n is a fixed integer n with n 2 and X n 1 is the set of the integers {2,, n}. For a permutation ³ of {1,, n}, let A ³ {i ¾ X n 1 ³ 1 (i 1) ³ 1 (i)}. For the first step of the proof of Theorem 1.2 and Theorem 2.3, we prepare the following lemma which is proved in [16]:

66 Y. YAGUCHI Lemma 4.1 ([16]). (s ³(1),, s ³(n) ) is an element of G( A³ n 1 ). Let A be a subset of X n 1. By virtue of the bijective map Ï G( A n 1 ) G(¼A n 1 ), two statements of (1) and (2) of Theorem 4.2 are equivalent. Theorem 4.2 is also our main result in this paper. Theorem 4.2. (1) For each element (a 1,, a n ) of G(¼n 1 A ), the Hurwitz orbit (a 1,, a n ) B n is equal to the set G(¼ A n 1 ). (2) For each element (g 1,, g n ) of G( A n 1 ), the Hurwitz orbit (g 1,, g n ) B n is equal to the set G( A n 1 ). The following theorem is directly obtained from Lemma 4.1 and Theorem 4.2 (2). Theorem 4.3. For a permutation ³ of {1,, n}, the Hurwitz orbit (s ³(1),, s ³(n) ) B n is equal to the set G( A³ n 1 ). We can prove Theorem 1.2 and Theorem 2.3 by using Theorem 4.3. Proof of Theorem 1.2. We prove that the conditions (1) and (3) are equivalent. Take permutations ³ and of {1,, n}. The condition (1) is equivalent to (s ³(1),, s ³(n) ) B n (s (1),, s (n) ) B n. By Theorem 4.3, this is equivalent to G( A³ G( A n 1 ). By Proposition 3.4, this is equivalent to A³ A, and we have the result. Proof of Theorem 2.3. n 1 ) Fix a permutation ³ of {1,,n} and let A A ³. By Theorem 4.3, (s ³(1),,s ³(n) ) B n G( A n 1 ). Then, ((s ³(1),,s ³(n) ) B n ) (G( A n 1 )) G(¼ A n 1 ). Let Q (Q 1,, Q n 1 ) be the ordered n 1 points of associated with A, that is an element of the configuration space Q n 1 (C 1 ) of the circle C 1. Then, G(¼n 1 A ) G(¼ Q n 1 ) and we regard G(¼Q n 1 ) as a subset of G n 1 putting labels Ú 1,, Ú n 1 on the points Q 1,, Q n 1, resp. By Proposition 3.6 (2), ((s ³(1),, s ³(n) ) B n ) (s ³(1),, s ³(n) ) B n. By Proposition 3.6 (1), (s ³(1),, s ³(n) ) B n ¼Æ P((s ³(1),, s ³(n) ) B n ) ¼((P(s ³(1),, s ³(n) )) B n ) ¼((t ³(1),, t ³(n) ) B n ) ¼(t ³(1),, t ³(n) ) B n, where ¼Ï T n 1 G n 1 is the graphic map. Thus, ¼(t ³(1),, t ³(n) ) B n G(¼ Q n 1 ), and we have the result. Let C be a circle in Ê 2. Fix an element q (q 1,, q n 1 ) of the configuration space Q n 1 (C). Let ¼ n 1 denote the set of segments ¼ q n 1 {q i q j 1 i j n 1} in Ê 2. Let G(¼ n 1 ) be the set of good elements of (¼ n 1 ) n. Let r 1,, r n 1 be n 1 points with {r 1,, r n 1 } {q 1,, q n 1 } such that r 1,, r n 1 stand on C counterclockwise in this order. Let A be an subset of X n 1 and let Q 1,, Q n 1 be the n 1 points, on the circle the C 1, associated with A. Let R 1,, R n 1 be n 1 points with {R 1,, R n 1 }

DETERMINING THE HURWITZ ORBIT 67 {Q 1,, Q n 1 } such that R 1,, R n 1 stand on the circle C 1 counterclockwise in this order. Let hï Ê 2 Ê 2 be a self-homeomorphism of Ê 2 such that h(c 1 ) C, h(r i ) r i for 1 i n 1 and h(r i R j ) r i r j for 1 i j n 1. Then, h induces the bijection hï ¼n 1 A ¼ n 1. Let f Ï (¼n 1 A )n (¼ n 1 ) n be the n-fold product of h, i.e., for each element (a 1,,a n ) of (¼n 1 A )n, f (a 1,,a n ) (h(a 1 ),,h(a n )). Then, it is obvious that f (G(¼n 1 A )) G(¼ n 1) and the map F f G(¼ A n 1 )Ï G(¼n 1 A ) G(¼ n 1) is bijective. If (e 1,,e n )¾G(¼ n 1 ), then, for any ¾ B n, we easily see that (e 1,,e n ) F((F 1 (e 1,, e n )) ). By virtue of the map F and the Hurwitz action of B n on G(¼ n 1 ), Proposition 3.5 is equivalent to Proposition 4.4, and Theorem 4.2 is equivalent to Theorem 4.5. The rest of this section is devoted to proving Theorem 4.5 (and hence Theorem 4.2). Proposition 4.4. If (e 1,, e n ) ¾ G(¼ n 1 ), then, for any k ¾ {1,, n 1} and any ¾ { 1, 1}, we have (e 1,, e n ) k ¾ G(¼ n 1 ). Theorem 4.5. For each element (e 1,, e n ) of G(¼ n 1 ), the Hurwitz orbit (e 1,, e n ) B n is equal to the set G(¼ n 1 ). For elements a and b of ¼ n 1 satisfying that a and b are disjoint or b is a right adjacent to a, we define the elements a b and b a of ¼ n 1 as follows. (1) If a and b are disjoint, then, let a b a and let b a b. (2) If b is a right adjacent to a and a q x q y, b q y q z, then, let a b b a q x q z. Let (e 1,, e n ) be an element of G(¼ n 1 ). By the conditions (i) and (ii) of Definition 2.2, for any k ¾ {1,, n 1}, e k and e k 1 are disjoint or e k 1 is a right adjacent to e k. This implies that the elements e k e k 1 and e k 1 e k of ¼ n 1 are defined. Then, we have the following proposition. Proposition 4.6 ([16]). The Hurwitz action of B n on G(¼ n 1 ) is given by and (e 1,, e k 1, e k, e k 1, e k 2,, e n ) k (e 1,, e k 1, e k 1, e k e k 1, e k 2,, e n ) (e 1,, e k 1, e k, e k 1, e k 2,, e n ) 1 k (e 1,, e k 1, e k 1 e k, e k, e k 2,, e n ). For a power «(1 2 n 1) t (t ¾ ) of the n 1-cyclic permutation (1 2 n 1), let c k r «(k) r «(k 1) for 1 k n. Then, we see that (c 1,, c n ) is an element of G(¼ n 1 ) and we call it cyclic. Lemma 4.7. Any two cyclic elements of G(¼ n 1 ) are Hurwitz equivalent.

68 Y. YAGUCHI Proof. Let c0 (r 1 r 2,,r n r n 1 ), that is a cyclic element of G(¼ n 1 ). It is enough to prove that any cyclic element c is Hurwitz equivalent to c0. Put c (c 1,,c n ) (r «(1) r «(2),, r «(n) r «(n 1) ) for a power «of the cyclic permutation (1 2 n 1). By Proposition 4.6, c ( 1 2 n 1 ) (c 2, c 3,, c n, ( ((c 1 c 2 ) c 3 ) ) c n ). Since the points r 1,,r n 1 stand on C 1 counterclockwise in this order, so do the points r «(1), r «(i) and r «(i 1) for 2 i n. Hence, r «(1) r «(i) is a right adjacent to c i r «(i) r «(i 1). Since c 1 r «(1) r «(2) and r «(1) r «(i) c i r «(1) r «(i 1), we have ( ((c 1 c 2 ) c 3 ) ) c n r «(1) r «(n 1). Thus, c ( 1 n 1 ) (r «(2) r «(3),, r «(n) r «(n 1), r «(n 1) r «(1) ). If «(t) n 1, then c ( 1 n 1 ) t c0 and we have the result. Let C¼ n 1 {r 1 r 2,, r n r n 1, r n 1 r 1 }. Then, the following lemma holds. Lemma 4.8. k ¾ {1,, n} such that e k ¾ C¼ n 1. Proof. For any element (e 1,, e n ) of G(¼ n 1 ), there exists an element Suppose that there exists an element (e 1,, e n ) of G(¼ n 1 ) such that e k ¾ ¼ n 1 ÒC¼ n 1 for each k ¾ {1,, n}. Since the n 1-gon r 1 r 2 r n 1 is convex, there exist elements i, j ¾ {1,, n} (i j) such that e i and e j intersect in their interior. This contradicts the condition (i) of Definition 2.2. Thus, we have the result. For a power «of the n 1-cyclic permutation (1 2 n 1), let G «(¼ n 1 ) be the set defined by {(e 1,, e n ) ¾ G(¼ n 1 ) e n r «(n) r «(n 1) }. Let ¼ n,«be the subset of ¼ n 1 defined by {r i r j ¾ ¼ n 1 i, j «(n 1)} and let G(¼ n,«) be the set of good elements of (¼ n,«) n 1. Then, we have the following. Lemma 4.9. Let (e 1,, e n ) be an element of G «(¼ n 1 ) for a power «of the (n 1)-cyclic permutation (1 2 n 1). Then, we have: (1) the degree of the vertex r «(n 1) of the graph e 1 e n is 1, and (2) (e 1,, e n 1 ) is an element of G(¼ n,«). Proof. First, we prove (1). Suppose that the degree of the vertex r «(n 1) is greater than 1. Then, there exists r «(n 1) r x ¾ {e 1,, e n 1 } for some x ¾ {1,, n 1} Ò {«(n),«(n 1)}. Since the points r «(1),,r «(n) and r «(n 1) stand on C counterclockwise in this order, so do the points r «(n), r «(n 1) and r x in this order. Thus, r «(n 1) r x is right adjacent to r «(n) r «(n 1) e n. This contradicts the condition (ii) of Definition 2.2. Thus, the degree of the vertex r «(n 1) must be 1. By (1), the union e 1 e n 1 is contractible, hence (e 1,, e n 1 ) satisfies the condition (iii) of Definition 2.2. It is obvious that (e 1,, e n 1 ) satisfies (i) and (ii) of Definition 2.2. Thus, we obtain (2). Theorem 4.5 is proved by Proposition 4.4, Lemma 4.7 and the following lemma.

DETERMINING THE HURWITZ ORBIT 69 Lemma 4.10. of G(¼ n 1 ). Any element of G(¼ n 1 ) is Hurwitz equivalent to a cyclic element Proof. Let (e 1,, e n ) be an element of G(¼ n 1 ). We prove this by induction on n. First, consider a case where n 2. Then, (e 1, e 2 ) (r 1 r 2, r 2 r 3 ), (r 2 r 3, r 3 r 1 ) or (r 3 r 1, r 1 r 2 ). They are cyclic and we have the result when n 2. Next, consider a case where n 2. By Lemma 4.8, we can take an element k ¾ {1,, n} such that e k ¾ C¼ n 1. Let (e ¼ 1,, e¼ n ) (e 1,, e n ) ( n 1 n 2 k ) 1. Using Proposition 4.6, by direct calculations (e ¼ 1,, e¼ n ) (e 1, e 2,, e k 1, e k 1 e k, e k 2 e k,, e n e k, e k ). By Proposition 4.4, (e ¼ 1,, e¼ n ) is an element of G(¼ n 1). Since e ¼ n e k ¾ C¼ n 1, e ¼ n r «(n) r «(n 1) for a power «of the n 1-cyclic permutation (1 2 n 1). Then, (e ¼ 1,, e¼ n ) is an element of G «(¼ n 1 ). By Lemma 4.9 (2), (e ¼ 1,, e¼ n 1 ) is an element of G(¼ n,«). By the assumption of the induction, (e ¼ 1,, e¼ n 1 ) is Hurwitz equivalent to a cyclic element (r («(1)) r («(2)),, r («(n 1)) r («(n)) ) of G(¼ n,«), where is a power of the n-cyclic permutation («(1) «(2) «(n)). By Lemma 4.7, it is Hurwitz equivalent to (r «(1) r «(2),, r «(n 1) r «(n) ). Then, (e ¼ 1,, e¼ n 1, e¼ n ) is Hurwitz equivalent to (r «(1) r «(2),, r «(n 1) r «(n), e ¼ n ) (r «(1)r «(2),, r «(n) r «(n 1) ), that is a cyclic element of G(¼ n 1 ), and we have the result. Proof of Theorem 4.5. By Proposition 4.4, it is sufficient to prove that any two elements of G(¼ n 1 ) are Hurwitz equivalent. Take an element g of G(¼n 1 ). By Lemma 4.10, g is Hurwitz equivalent to a cyclic element c of G(¼n 1 ). By Lemma 4.7, c is Hurwitz equivalent to a special cyclic element c0. Hence, any element of G(¼ n 1 ) is Hurwitz equivalent to c0. This completes the proof of Theorem 4.5. ACKNOWLEDGMENTS. The author would like to thank Professor Takao Matumoto, Professor Seiichi Kamada and Professor Makoto Sakuma for their help. This work was supported by JSPS KAKENHI Grant Number 23840026.

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