Bridge numbers of links and minimal numbers of meridian generators - PowerPoint PPT Presentation

Bridge numbers of links and minimal numbers of meridian generators of link groups Yeonhee JANG Hiroshima University, JSPS research fellow ———– CONTENTS ———– 1. Introduction. 2. Result. 3. Outline of Proof. 4. Open Problems.

1. Introduction. • n -bridge presentation • n -bridge link : link which admits an n -bridge presentation but does not admit an ( n − 1)-bridge presentation Example) 2-bridge link

� � � � � � � � � � � � � � 1. Introduction. Fact ( n -bridge links) (1) The only 1-bridge link is the unknot. (2) (Schubert) completely classified 2-bridge links. (3) (Boileau-Zieschang) determined the bridge numbers of (gener- alized) Montesinos links. (4) (J.) A 3-bridge “arborescent” link is equivalent to one of the following links. 3 β α ′ 1 3 β α ′ 1 2 β α ′ β α 2 1 β α ′ 2 2 1 -b 1 β α α 0 1 β α β α β α 1 β α 1 2 3 1 2 3 1 2 β α β α 1 ′ 2 1 ′ ′ β α β α ′ 2 2 2 2 n

� � 1. Introduction. • rational tangle : ( B 3 , T ) 1 = 12 3 + − 2 + 1 5 = 3 • arborescent tangle : a tangle obtained from rational tangles by the following operations T 1 T 2 T 1 T 2 T 1 T 2 T 1 T 2 T 1 T 2 • arborescent link : a link obtained from two arborescent tangles by gluing their boundaries

1. Introduction. • If a link L ⊂ S 3 admits an n -bridge presentation, then its group π 1 ( S 3 \ L ) is generated by n meridians. ∵ Note that π 1 ( B 1 \ L ) = � m 1 , m 2 , . . . , m n � , π 1 ( B 2 \ L ) = � m ′ 1 , m ′ 2 , . . . , m ′ n � and π 1 ( S \ L ) = � n 1 , n 2 , . . . , n 2 n − 1 � . By van Kampen’s theorem, we have π 1 ( S 3 \ L ) = � m 1 , m 2 , . . . , m n , m ′ 1 , m ′ 2 , . . . , m ′ n | m ′ 1 = w 1 ( m 1 , m 2 , . . . , m n ) , m ′ 1 = w 2 ( m 1 , m 2 , . . . , m n ) , . . . � = � m 1 , m 2 , . . . , m n | w 1 ( m 1 , . . . , m n ) = w 2 ( m 1 , . . . , m n ) , . . . � .

� � 1. Introduction. Example m 1 m 2 n 3 n 4 n 1 n 2 m 1 m m 1 m 2 m 1 m 2 m 1 2 n 1 ❀ r 1 : m ′ 1 = m 1 m 1 m m 1 m 2 m 1 2 n 2 ❀ r 2 : m ′ 1 = m 1 m 2 m 1 m 2 ¯ m 1 ¯ m 2 ¯ m 1 m 1 m 2 m 1 n 3 ❀ r 3 : m ′ 2 = m 1 m 2 m 1 ¯ m 2 ¯ m 1 m 1 m 2 π 1 ( S 3 \ L ) = � m 1 , m 2 , m ′ 1 , m ′ 2 | r 1 , r 2 , r 3 � = � m 1 , m 2 | m 1 m 2 m 1 = m 2 m 1 m 2 � .

1. Introduction. • b ( L ) : the bridge number of L , • w ( L ) : the minimal number of meridians needed to generate the group π 1 ( S 3 \ L ) ❀ w ( L ) ≤ b ( L ) Question w ( L ) = b ( L )? Fact True when w ( L ) = 2 (Boileau-Zimmermann), (i.e., w ( L ) = 2 ⇐ ⇒ b ( L ) = 2), for generalized Montesinos links (Boileau-Zieschang), for torus links (Rost-Zieschang).

2. Result. Theorem (Boileau-J.) Let L be an arborescent link. Then w ( L ) = 3 ⇐ ⇒ b ( L ) = 3 .

3. Outline of Proof. Case 1 : M 2 ( L ) is a Seifert fibered space ❀ By [Dunbar] and [Burde-Murasugi], ❀ [Boileau-Zieschang] L : generalized Montesinos link L : tor u s link T ❀ [Rost-Zieschang] 3 ,n L : S ei f ert link ❀ 3-bridge!! L : T its c ore 2 ,n Case 2 : M 2 ( L ) is a graph manifold with a nontrivial JSJ decomposition ✬ ✩ Proposition 1 Assume that w ( L ) = 3 ⇒ τ L induces an “inversion” of π 1 ( M 2 ( L )). In particular, if M 2 ( L ) is a graph manifold, then g ( M 2 ( L )) = 2. ✫ ✪ ❀ It suffices to show that τ L is the “hyper-elliptic” involution asso- ciated with a genus-2 Heegaard splitting of M 2 ( L ).

3. Outline of Proof. Case 2 (continued) : M 2 ( L ) is a graph manifold with a nontrivial JSJ decomposition ✬ ✩ Proposition 2 Let L be an arborescent link such that w ( L ) = 3. ⇒ L is either hyperbolic or equivalent to the following link. n ✫ ✪ ✬ ✩ Proposition 3 M : closed ori. 3-mfd with nontrivial JSJ decomp, τ : involution on M s.t. M → M/τ ( ∼ = S 3 ) : double branched cover / hyperbolic link L . ⇒ (1) τ is hyper-elliptic on each JSJ torus. (2) τ preserves each JSJ piece and each singular fiber. ✫ ✪

4. Open Problems. Q1 : w ( L ) = 3 ⇒ b ( L ) = 3 ? Q1-1 : Is Q1 true if M 2 ( L ) is a graph manifold? ( ♣ links containing essential tori in their exterior) Q1-2 : Is Q1 true if M 2 ( L ) admits a nontrivial JSJ decomp.? ( ♣ hyperbolic 1-bridge knots in lens spaces) Q2 : w ( L ) = b ( L ) ? Q2-1 : When K ⊂ S 3 satisfies w ( K ) = b ( K ), does C p,q ( K ) have the same property? ( ♣ true when K : 2-bridge knot and ( p, q ) = (1 , 2)) Q2-2 : When K i ⊂ S 3 satisfies w ( K i ) = b ( K i ) ( i = 1 , 2), w ( K 1 ♯K 2 ) = b ( K 1 ♯K 2 )(= b ( K 1 ) + b ( K 2 ) − 1)?