Twisted Alexander invariant and a partial order in the knot table II - PowerPoint PPT Presentation

Twisted Alexander invariant and a partial order in the knot table II Masaaki Suzuki (University of Tokyo) Joint work with T. Kitano (Tokyo Inst. Tech.) Contents. 1. Main Theorem. 2. Sketch of Proof (twisted Alexander invariant). 3. Problems. 1

1 . Main Theorem. K : a prime knot i.e. G ( K ) = π 1 ( E ( K )) = π 1 ( S 3 − K ) G ( K ) : the knot group of K ◦ 3 1 ◦ 4 1 × 3 1 ♯ 4 1 2

1 . Main Theorem. K : a prime knot i.e. G ( K ) = π 1 ( E ( K )) = π 1 ( S 3 − K ) G ( K ) : the knot group of K ◦ 3 1 ◦ 4 1 × 3 1 ♯ 4 1 ✓ Defintion. ✏ K, K ′ : two prime knots ∃ ϕ : G ( K ) − → G ( K ′ ) ⇐ ⇒ K ≥ K ′ − → ✒ ✑ 3

∃ ϕ : G ( K ) − → G ( K ′ ) ⇐ ⇒ K ≥ K ′ − → ✓ Lemma. ✏ The above ≥ is a partial order on the set of prime knots. • K ≥ K • K 1 ≥ K 2 , K 2 ≥ K 1 ⇒ K 1 = K 2 • K 1 ≥ K 2 , K 2 ≥ K 3 ⇒ K 1 ≥ K 3 ✒ ✑ 4

∃ ϕ : G ( K ) − → G ( K ′ ) ⇐ ⇒ K ≥ K ′ − → ✓ Lemma. ✏ The above ≥ is a partial order on the set of prime knots. • K ≥ K • K 1 ≥ K 2 , K 2 ≥ K 1 ⇒ K 1 = K 2 • K 1 ≥ K 2 , K 2 ≥ K 3 ⇒ K 1 ≥ K 3 ✒ ✑ ✓ ✏ Problem. Determine this partial order ≥ on the set of knots in the Rolfsen’s table. ✒ ✑ 5

8 5 , 8 10 , 8 15 , 8 18 , 8 19 , 8 20 , 8 21 , 9 1 , 9 6 , 9 16 , 9 23 , 9 24 , 9 28 , 9 40 , 10 5 , 10 9 , 10 32 , 10 40 , 10 61 , 10 62 , 10 63 , 10 64 , 10 65 , 10 66 , 10 76 , ≥ 3 1 10 77 , 10 78 , 10 82 , 10 84 , 10 85 , 10 87 , 10 98 , 10 99 , 10 103 , 10 106 , 10 112 , 10 114 , 10 139 , 10 140 , 10 141 , 10 142 , 10 143 , 10 144 , 10 159 , 10 164 8 18 , 9 37 , 9 40 , 10 58 , 10 59 , 10 60 , 10 122 , 10 136 , 10 137 , 10 138 ≥ 4 1 10 74 , 10 120 , 10 122 ≥ 5 2 6

2 . Sketch of Proof. To prove the non-existence of a surjective homomorphism (1) By the (classical) Alexander polynomial K : a knot ∆ K : the Alexander polynomial of K ✓ Fact. ✏ If ∆ K can not be divided by ∆ K ′ , → G ( K ′ ). then there exists no surjective homomorphism ϕ : G ( K ) − − → ✒ ✑ 7

2 . Sketch of Proof. To prove the non-existence of a surjective homomorphism (1) By the (classical) Alexander polynomial K : a knot ∆ K : the Alexander polynomial of K ✓ Fact. ✏ If ∆ K can not be divided by ∆ K ′ , → G ( K ′ ). then there exists no surjective homomorphism ϕ : G ( K ) − − → ✒ ✑ (2) By the twisted Alexander invariant 8

✓ ✏ Twisted Alexander invariant G : a finitely presentable group → Z l : a surjective homomorphism to a free abelian group α : G − − → ρ : G − → GL ( n ; R ) : a representation of G R : UFD = ⇒ ∆ G,ρ : twisted Alexander invariant ✒ ✑ 9

✓ ✏ Twisted Alexander invariant G : a finitely presentable group → Z l : a surjective homomorphism to a free abelian group α : G − − → ρ : G − → GL ( n ; R ) : a representation of G R : UFD = ⇒ ∆ G,ρ : twisted Alexander invariant ✒ ✑ ✓ ✏ K : a knot G ( K ) : the knot group of a knot K → Z : the abelianization α : G ( K ) − − → ρ : G ( K ) − → SL (2; R ) : a representation = ⇒ ∆ K,ρ : twisted Alexander invariant ✒ ✑ 10

G ( K ) = � x 1 , . . . , x u | r 1 , . . . , r u − 1 � Wirtinger presentation F u = � x 1 , . . . , x u � : the free group of rank u → Z ≃ � t � : the abelianization (generated by the meridian) α : G ( K ) − − → ρ : G ( K ) − → SL (2; R ) a representation of G ( K ), ρ ( x i ) = X i 2 ; R [ t ± 1 ] � � Z [ F u ] Z [ G ( K )] F u − → − → − → M ∈ ∈ ∈ ∈ � ∂r i � ∂r i ∂r i r i �− → �− → �− → Φ ∂x j ∂x j ∂x j ↑ Replacing x k in ∂r i ∂x j by tX k 11

G ( K ) = � x 1 , . . . , x u | r 1 , . . . , r u − 1 � Wirtinger presentation F u = � x 1 , . . . , x u � : the free group of rank u → Z ≃ � t � : the abelianization (generated by the meridian) α : G ( K ) − − → ρ : G ( K ) − → SL (2; R ) a representation of G ( K ), ρ ( x i ) = X i 2 ; R [ t ± 1 ] � � Z [ F u ] Z [ G ( K )] F u − → − → − → M ∈ ∈ ∈ ∈ � ∂r i � ∂r i ∂r i r i �− → �− → �− → Φ ∂x j ∂x j ∂x j ↑ Replacing x k in ∂r i ∂x j by tX k � ∂r i � �� � � 2( u − 1) , 2 u ; R [ t ± 1 ] M = Φ ∈ M ∂x j i,j 12

2 ; R [ t ± 1 ] � � Z [ F u ] Z [ G ( K )] F u − → − → − → M ∈ ∈ ∈ ∈ � ∂r i � ∂r i ∂r i r i �− → �− → �− → Φ ∂x j ∂x j ∂x j � ∂r i � �� M = Φ : 2( u − 1) × 2 u -matrix ∂x j i,j Remove the 2 k − 1 , 2 k -th column = ⇒ M k : 2( u − 1) × 2( u − 1)-matrix 13

2 ; R [ t ± 1 ] � � Z [ F u ] Z [ G ( K )] F u − → − → − → M ∈ ∈ ∈ ∈ � ∂r i � ∂r i ∂r i r i �− → �− → �− → Φ ∂x j ∂x j ∂x j � ∂r i � �� M = Φ : 2( u − 1) × 2 u -matrix ∂x j i,j Remove the 2 k − 1 , 2 k -th column = ⇒ M k : 2( u − 1) × 2( u − 1)-matrix Definition. (Twisted Alexander invariant) ✓ ✏ det M k ∆ K,ρ = det ( tX k − I 2 ) where I 2 is the identity matrix. ✒ ✑ 14

To prove the non-existence of a surjective homomorphism (1) By the (classical) Alexander polynomial 15

To prove the non-existence of a surjective homomorphism (1) By the (classical) Alexander polynomial (2) By the twisted Alexander invariant ∆ K,ρ : the twisted Alexander invariant of K ∆ N K,ρ , ∆ D K,ρ : the numerator and denominator of ∆ K,ρ Theorem. (Kitano-S.-Wada) ✓ ✏ If there exists a representation ρ ′ : G ( K ′ ) → SL (2; Z /p Z ) such that for any representation ρ : G ( K ) → SL (2; Z /p Z ), ∆ N K,ρ is not divisible by ∆ N K ′ ,ρ ′ or ∆ D K,ρ � = ∆ D K ′ ,ρ ′ , → G ( K ′ ). then there exists no surjective homomorphism ϕ : G ( K ) − − → ✒ ✑ 16

i.e. ? ∃ ϕ : G (8 21 ) − Example. 8 21 ≥ 4 1 ? − → → G (4 1 ) 8 21 = , 4 1 = 17

i.e. ? ∃ ϕ : G (8 21 ) − Example. 8 21 ≥ 4 1 ? − → → G (4 1 ) 8 21 = , 4 1 = ∆ 8 21 = t 4 − 4 t 3 + 5 t 2 − 4 t + 1 , ∆ 4 1 = t 2 − 3 t + 1 ∆ 8 21 = t 2 − t + 1 ∆ 4 1 Then ∆ 4 1 divides ∆ 8 21 , 18

i.e. ? ∃ ϕ : G (8 21 ) − Example. 8 21 ≥ 4 1 ? − → → G (4 1 ) 8 21 = , 4 1 = ∆ 8 21 = t 4 − 4 t 3 + 5 t 2 − 4 t + 1 , ∆ 4 1 = t 2 − 3 t + 1 ∆ 8 21 = t 2 − t + 1 ∆ 4 1 Then ∆ 4 1 divides ∆ 8 21 , we cannot determine the existence of a surjective homomorphism from G (8 21 ) to G (4 1 ) by the Alexander polynomial. 19

For a certain representation ρ ′ : G (4 1 ) − → SL (2; Z / 3 Z ), 4 1 ,ρ ′ = t 4 + t 2 + 1 , 4 1 ,ρ ′ = t 2 + t + 1 ∆ N ∆ D 20

For a certain representation ρ ′ : G (4 1 ) − → SL (2; Z / 3 Z ), 4 1 ,ρ ′ = t 4 + t 2 + 1 , 4 1 ,ρ ′ = t 2 + t + 1 ∆ N ∆ D The following is the table of the twisted Alexnader invariants of G (8 21 ) → SL (2; Z / 3 Z ) for all representations ρ : G (8 21 ) − ∆ N ∆ D 8 21 ,ρ i 8 21 ,ρ i ρ 1 t 8 + t 4 + 1 t 2 + 1 ρ 2 t 8 + t 7 + 2 t 6 + 2 t 4 + 2 t 2 + t + 1 t 2 + t + 1 ρ 3 t 8 + t 7 + 2 t 6 + 2 t 4 + 2 t 2 + t + 1 t 2 + 2 t + 1 ρ 4 t 8 + 2 t 7 + 2 t 6 + 2 t 4 + 2 t 2 + 2 t + 1 t 2 + t + 1 ρ 5 t 8 + 2 t 7 + 2 t 6 + 2 t 4 + 2 t 2 + 2 t + 1 t 2 + 2 t + 1 21

For a certain representation ρ ′ : G (4 1 ) − → SL (2; Z / 3 Z ), 4 1 ,ρ ′ = t 4 + t 2 + 1 , 4 1 ,ρ ′ = t 2 + t + 1 ∆ N ∆ D The following is the table of the twisted Alexnader invariants of G (8 21 ) → SL (2; Z / 3 Z ) for all representations ρ : G (8 21 ) − ∆ N ∆ D 8 21 ,ρ i 8 21 ,ρ i ρ 1 t 8 + t 4 + 1 t 2 + 1 ρ 2 t 8 + t 7 + 2 t 6 + 2 t 4 + 2 t 2 + t + 1 t 2 + t + 1 ρ 3 t 8 + t 7 + 2 t 6 + 2 t 4 + 2 t 2 + t + 1 t 2 + 2 t + 1 ρ 4 t 8 + 2 t 7 + 2 t 6 + 2 t 4 + 2 t 2 + 2 t + 1 t 2 + t + 1 ρ 5 t 8 + 2 t 7 + 2 t 6 + 2 t 4 + 2 t 2 + 2 t + 1 t 2 + 2 t + 1 8 21 � 4 1 22

To prove the existence of a surjective homomorphism By constructing explicitly a surjective homomorphism between knot groups. 23

To prove the existence of a surjective homomorphism By constructing explicitly a surjective homomorphism between knot groups. i.e. ? ∃ ϕ : G (8 18 ) − Example. 8 18 ≥ 3 1 ? − → → G (3 1 ) 8 18 = , 3 1 = � � � x 1 , x 2 , x 3 , x 4 , x 4 x 1 ¯ x 4 ¯ x 2 , x 5 x 3 ¯ x 5 ¯ x 2 , x 6 x 3 ¯ x 6 ¯ x 4 , x 7 x 5 ¯ x 7 ¯ x 4 , � G (8 18 ) = � � x 5 , x 6 , x 7 , x 8 x 8 x 5 ¯ x 8 ¯ x 6 , x 1 x 7 ¯ x 1 ¯ x 6 , x 2 x 7 ¯ x 2 ¯ x 8 � G (3 1 ) = � y 1 , y 2 , y 3 | y 3 y 1 ¯ y 3 ¯ y 2 , y 1 y 2 ¯ y 1 ¯ y 3 � 24

To prove the existence of a surjective homomorphism By constructing explicitly a surjective homomorphism between knot groups. i.e. ? ∃ ϕ : G (8 18 ) − Example. 8 18 ≥ 3 1 ? − → → G (3 1 ) 8 18 = , 3 1 = � � � x 1 , x 2 , x 3 , x 4 , x 4 x 1 ¯ x 4 ¯ x 2 , x 5 x 3 ¯ x 5 ¯ x 2 , x 6 x 3 ¯ x 6 ¯ x 4 , x 7 x 5 ¯ x 7 ¯ x 4 , � G (8 18 ) = � � x 5 , x 6 , x 7 , x 8 x 8 x 5 ¯ x 8 ¯ x 6 , x 1 x 7 ¯ x 1 ¯ x 6 , x 2 x 7 ¯ x 2 ¯ x 8 � G (3 1 ) = � y 1 , y 2 , y 3 | y 3 y 1 ¯ y 3 ¯ y 2 , y 1 y 2 ¯ y 1 ¯ y 3 � ϕ ( x 1 ) = y 1 , ϕ ( x 2 ) = y 2 , ϕ ( x 3 ) = y 1 , ϕ ( x 4 ) = y 3 , ϕ ( x 5 ) = y 3 , ϕ ( x 6 ) = y 1 y 3 ¯ y 1 , ϕ ( x 7 ) = y 3 , ϕ ( x 8 ) = y 1 8 18 ≥ 3 1 25