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

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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

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1. Main Theorem.

K : a prime knot G(K) : the knot group of K i.e. G(K) = π1(E(K)) = π1(S3 − K)

31
41

× 31♯41

2

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1. Main Theorem.

K : a prime knot G(K) : the knot group of K i.e. G(K) = π1(E(K)) = π1(S3 − K)

31
41

× 31♯41

Defintion.

✓ ✏

K, K′ : two prime knots

∃ϕ : G(K) −

→ − → G(K′) ⇐ ⇒ K ≥ K′

✒ ✑

3

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∃ϕ : G(K) −

→ − → G(K′) ⇐ ⇒ K ≥ K′ Lemma.

✓ ✏

The above ≥ is a partial order on the set of prime knots.

K ≥ K
K1 ≥ K2, K2 ≥ K1 ⇒ K1 = K2
K1 ≥ K2, K2 ≥ K3 ⇒ K1 ≥ K3

✒ ✑

4

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∃ϕ : G(K) −

→ − → G(K′) ⇐ ⇒ K ≥ K′ Lemma.

✓ ✏

The above ≥ is a partial order on the set of prime knots.

K ≥ K
K1 ≥ K2, K2 ≥ K1 ⇒ K1 = K2
K1 ≥ K2, K2 ≥ K3 ⇒ K1 ≥ K3

✒ ✑

Problem.

✓ ✏

Determine this partial order ≥ on the set of knots in the Rolfsen’s table.

✒ ✑

5

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85, 810, 815, 818, 819, 820, 821, 91, 96, 916, 923, 924, 928, 940, 105, 109, 1032, 1040, 1061, 1062, 1063, 1064, 1065, 1066, 1076, 1077, 1078, 1082, 1084, 1085, 1087, 1098, 1099, 10103, 10106, 10112, 10114, 10139, 10140, 10141, 10142, 10143, 10144, 10159, 10164 ≥ 31 818, 937, 940, 1058, 1059, 1060, 10122, 10136, 10137, 10138 ≥ 41 1074, 10120, 10122 ≥ 52

6

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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′, then there exists no surjective homomorphism ϕ : G(K) − → − → G(K′).

✒ ✑

7

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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′, then there exists no surjective homomorphism ϕ : G(K) − → − → G(K′).

✒ ✑

(2) By the twisted Alexander invariant

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Twisted Alexander invariant

✓ ✏

G : a finitely presentable group α : G − → − → Zl : a surjective homomorphism to a free abelian group ρ : G − → GL(n; R) : a representation of G R : UFD = ⇒ ∆G,ρ : twisted Alexander invariant

✒ ✑

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Twisted Alexander invariant

✓ ✏

G : a finitely presentable group α : G − → − → Zl : a surjective homomorphism to a free abelian group 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 α : G(K) − → − → Z : the abelianization ρ : G(K) − → SL(2; R) : a representation = ⇒ ∆K,ρ : twisted Alexander invariant

✒ ✑

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G(K) = x1, . . . , xu | r1, . . . , ru−1 Wirtinger presentation Fu = x1, . . . , xu : the free group of rank u α : G(K) − → − → Z ≃ t : the abelianization (generated by the meridian) ρ : G(K) − → SL(2; R) a representation of G(K), ρ(xi) = Xi Fu − → Z[Fu] − → Z[G(K)] − → M

2 ; R[t±1]
∈

∈ ∈ ∈ ri − →

∂ri ∂xj

− →

∂ri ∂xj

− → Φ ∂ri

∂xj

↑

Replacing xk in ∂ri

∂xj by tXk

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G(K) = x1, . . . , xu | r1, . . . , ru−1 Wirtinger presentation Fu = x1, . . . , xu : the free group of rank u α : G(K) − → − → Z ≃ t : the abelianization (generated by the meridian) ρ : G(K) − → SL(2; R) a representation of G(K), ρ(xi) = Xi Fu − → Z[Fu] − → Z[G(K)] − → M

2 ; R[t±1]
∈

∈ ∈ ∈ ri − →

∂ri ∂xj

− →

∂ri ∂xj

− → Φ ∂ri

∂xj

↑

Replacing xk in ∂ri

∂xj by tXk

M =

Φ

∂ri ∂xj

i,j

∈ M

2(u − 1), 2u ; R[t±1]
12

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Fu − → Z[Fu] − → Z[G(K)] − → M

2 ; R[t±1]
∈

∈ ∈ ∈ ri − →

∂ri ∂xj

− →

∂ri ∂xj

− → Φ ∂ri

∂xj

M =
Φ

∂ri ∂xj

i,j

: 2(u − 1) × 2u-matrix Remove the 2k − 1, 2k-th column = ⇒ Mk : 2(u − 1) × 2(u − 1)-matrix

13

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Fu − → Z[Fu] − → Z[G(K)] − → M

2 ; R[t±1]
∈

∈ ∈ ∈ ri − →

∂ri ∂xj

− →

∂ri ∂xj

− → Φ ∂ri

∂xj

M =
Φ

∂ri ∂xj

i,j

: 2(u − 1) × 2u-matrix Remove the 2k − 1, 2k-th column = ⇒ Mk : 2(u − 1) × 2(u − 1)-matrix

Definition. (Twisted Alexander invariant)

✓ ✏

∆K,ρ = det Mk det (tXk − I2) where I2 is the identity matrix.

✒ ✑

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

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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/pZ) such that for any representation ρ : G(K) → SL(2; Z/pZ), ∆N

K,ρ is not divisible by ∆N K′,ρ′ or ∆D K,ρ = ∆D K′,ρ′,

then there exists no surjective homomorphism ϕ : G(K) − → − → G(K′).

✒ ✑

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Example. 821 ≥ 41 ? i.e. ?∃ϕ : G(821) − → − → G(41) 821 = , 41 =

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Example. 821 ≥ 41 ? i.e. ?∃ϕ : G(821) − → − → G(41) 821 = , 41 = ∆821 = t4 − 4t3 + 5t2 − 4t + 1, ∆41 = t2 − 3t + 1 ∆821 ∆41 = t2 − t + 1 Then ∆41 divides ∆821,

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Example. 821 ≥ 41 ? i.e. ?∃ϕ : G(821) − → − → G(41) 821 = , 41 = ∆821 = t4 − 4t3 + 5t2 − 4t + 1, ∆41 = t2 − 3t + 1 ∆821 ∆41 = t2 − t + 1 Then ∆41 divides ∆821, we cannot determine the existence of a surjective homomorphism from G(821) to G(41) by the Alexander polynomial.

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For a certain representation ρ′ : G(41) − → SL(2; Z/3Z), ∆N

41,ρ′ = t4 + t2 + 1,

∆D

41,ρ′ = t2 + t + 1

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For a certain representation ρ′ : G(41) − → SL(2; Z/3Z), ∆N

41,ρ′ = t4 + t2 + 1,

∆D

41,ρ′ = t2 + t + 1

The following is the table of the twisted Alexnader invariants of G(821) for all representations ρ : G(821) − → SL(2; Z/3Z) ∆N

821,ρi

∆D

821,ρi

ρ1 t8 + t4 + 1 t2 + 1 ρ2 t8 + t7 + 2t6 + 2t4 + 2t2 + t + 1 t2 + t + 1 ρ3 t8 + t7 + 2t6 + 2t4 + 2t2 + t + 1 t2 + 2t + 1 ρ4 t8 + 2t7 + 2t6 + 2t4 + 2t2 + 2t + 1 t2 + t + 1 ρ5 t8 + 2t7 + 2t6 + 2t4 + 2t2 + 2t + 1 t2 + 2t + 1

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For a certain representation ρ′ : G(41) − → SL(2; Z/3Z), ∆N

41,ρ′ = t4 + t2 + 1,

∆D

41,ρ′ = t2 + t + 1

The following is the table of the twisted Alexnader invariants of G(821) for all representations ρ : G(821) − → SL(2; Z/3Z) ∆N

821,ρi

∆D

821,ρi

ρ1 t8 + t4 + 1 t2 + 1 ρ2 t8 + t7 + 2t6 + 2t4 + 2t2 + t + 1 t2 + t + 1 ρ3 t8 + t7 + 2t6 + 2t4 + 2t2 + t + 1 t2 + 2t + 1 ρ4 t8 + 2t7 + 2t6 + 2t4 + 2t2 + 2t + 1 t2 + t + 1 ρ5 t8 + 2t7 + 2t6 + 2t4 + 2t2 + 2t + 1 t2 + 2t + 1 821 41

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

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To prove the existence of a surjective homomorphism By constructing explicitly a surjective homomorphism between knot groups. Example. 818 ≥ 31 ? i.e. ?∃ϕ : G(818) − → − → G(31) 818 = , 31 = G(818) =

x1, x2, x3, x4,

x5, x6, x7, x8

x4x1¯

x4¯ x2, x5x3¯ x5¯ x2, x6x3¯ x6¯ x4, x7x5¯ x7¯ x4, x8x5¯ x8¯ x6, x1x7¯ x1¯ x6, x2x7¯ x2¯ x8

G(31) = y1, y2, y3 | y3y1¯

y3¯ y2, y1y2¯ y1¯ y3

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To prove the existence of a surjective homomorphism By constructing explicitly a surjective homomorphism between knot groups. Example. 818 ≥ 31 ? i.e. ?∃ϕ : G(818) − → − → G(31) 818 = , 31 = G(818) =

x1, x2, x3, x4,

x5, x6, x7, x8

x4x1¯

x4¯ x2, x5x3¯ x5¯ x2, x6x3¯ x6¯ x4, x7x5¯ x7¯ x4, x8x5¯ x8¯ x6, x1x7¯ x1¯ x6, x2x7¯ x2¯ x8

G(31) = y1, y2, y3 | y3y1¯

y3¯ y2, y1y2¯ y1¯ y3 ϕ(x1) = y1, ϕ(x2) = y2, ϕ(x3) = y1, ϕ(x4) = y3, ϕ(x5) = y3, ϕ(x6) = y1y3¯ y1, ϕ(x7) = y3, ϕ(x8) = y1 818 ≥ 31

25

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Problem.

✓ ✏

Determine this partial order ≥ on the set of knots in the Rolfsen’s table.

✒ ✑

26

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Problem.

✓ ✏

Determine this partial order ≥ on the set of knots in the Rolfsen’s table.

✒ ✑

The number of prime knots with up to 10 crossings is 249. Then the number of cases to consider is 249P2 = 61752. 61508 cases : non-existence of a surjection by the Alexander polynomial

27

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Problem.

✓ ✏

Determine this partial order ≥ on the set of knots in the Rolfsen’s table.

✒ ✑

The number of prime knots with up to 10 crossings is 249. Then the number of cases to consider is 249P2 = 61752. 61508 cases : non-existence of a surjection by the Alexander polynomial The rest are 244 cases. 186 cases : non-existence of a surjection by the twisted Alexander invariant. 58 cases : existence of a surjection by constructing explicitly.

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85, 810, 815, 818, 819, 820, 821, 91, 96, 916, 923, 924, 928, 940, 105, 109, 1032, 1040, 1061, 1062, 1063, 1064, 1065, 1066, 1076, 1077, 1078, 1082, 1084, 1085, 1087, 1098, 1099, 10103, 10106, 10112, 10114, 10139, 10140, 10141, 10142, 10143, 10144, 10159, 10164 ≥ 31 818, 937, 940, 1058, 1059, 1060, 10122, 10136, 10137, 10138 ≥ 41 1074, 10120, 10122 ≥ 52

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3. Problems
1. Are three knots 31, 41 and 52 minimal elements?
2. What is the geometric interpretation of each surjective homomorphism?

30

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3. Problems
1. Are three knots 31, 41 and 52 minimal elements?
2. What is the geometric interpretation of each surjective homomorphism?

Theorem.

✓ ✏

Two knots 31 and 41 are minimal elements under the partial order ≥.

✒ ✑

Sketch of Proof.

K: a fibered knot of genus g

ϕ : G(K) − → − → G(K′) : a surjective homomorphism = ⇒ K′ : a fibered knot of genus g′, g ≥ g′

All fibered knots of genus 1 are just 31 and 41.

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Geometric settings of a surjective homomorhism between knot groups

From any knot to the trivial knot

G( 31 ) − → − → G( 01 )

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Geometric settings of a surjective homomorhism between knot groups

From any knot to the trivial knot

G( 31 ) − → − → G( 01 )

From a connected sum of knots to each knot

G( 31♯41 ) − → − → G( 31 )

− → − →

G( 41 )

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Geometric settings of a surjective homomorhism between knot groups

Degree one map

f : (E(K), ∂E(K)) − → (E(K′), ∂E(K′)) : a degree one map i.e. f∗ : H3(E(K), ∂E(K), Z) − → H3(E(K′), ∂E(K′), Z) ∈ ∈ [E(K), ∂E(K)] − → [E(K′), ∂E(K′)] = ⇒ f∗ : G(K) − → − → G(K′)

34

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Geometric settings of a surjective homomorhism between knot groups

Degree one map

f : (E(K), ∂E(K)) − → (E(K′), ∂E(K′)) : a degree one map i.e. f∗ : H3(E(K), ∂E(K), Z) − → H3(E(K′), ∂E(K′), Z) ∈ ∈ [E(K), ∂E(K)] − → [E(K′), ∂E(K′)] = ⇒ f∗ : G(K) − → − → G(K′)

From a periodic knot to the factor knot

G( 818 ) − → − → G( 41 )

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Which partial order can be realized by degree one maps? Proposition.

✓ ✏

ϕ : G(K) − → − → G(K′) ϕ(Meridian of K) = Meridian of K′, ϕ(Longitude of K) = Longitude of K′ = ⇒ The surjective homomorhpism is realized by a degree one map.

✒ ✑

36

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Which partial order can be realized by degree one maps? Proposition.

✓ ✏

ϕ : G(K) − → − → G(K′) ϕ(Meridian of K) = Meridian of K′, ϕ(Longitude of K) = Longitude of K′ = ⇒ The surjective homomorhpism is realized by a degree one map.

✒ ✑

Example. 818 ≥ 31 818 = , 31 = We constructed explicitly a surjective homomorphism from G(818) to G(31).

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G(818) =

x1, x2, x3, x4,

x5, x6, x7, x8

x4x1¯

x4¯ x2, x5x3¯ x5¯ x2, x6x3¯ x6¯ x4, x7x5¯ x7¯ x4, x8x5¯ x8¯ x6, x1x7¯ x1¯ x6, x2x7¯ x2¯ x8

G(31) = y1, y2, y3 | y3y1¯

y3¯ y2, y1y2¯ y1¯ y3 Meridian of 818 = x1, Longitude of 818 = ¯ x4x5¯ x6x7¯ x8x1¯ x2x3 Meridian of 31 = y1, Longitude of 31 = ¯ y3¯ y1¯ y2y1y1y1

38

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G(818) =

x1, x2, x3, x4,

x5, x6, x7, x8

x4x1¯

x4¯ x2, x5x3¯ x5¯ x2, x6x3¯ x6¯ x4, x7x5¯ x7¯ x4, x8x5¯ x8¯ x6, x1x7¯ x1¯ x6, x2x7¯ x2¯ x8

G(31) = y1, y2, y3 | y3y1¯

y3¯ y2, y1y2¯ y1¯ y3 Meridian of 818 = x1, Longitude of 818 = ¯ x4x5¯ x6x7¯ x8x1¯ x2x3 Meridian of 31 = y1, Longitude of 31 = ¯ y3¯ y1¯ y2y1y1y1 ϕ(x1) = y1, ϕ(x2) = y2, ϕ(x3) = y1, ϕ(x4) = y3, ϕ(x5) = y3, ϕ(x6) = y1y3¯ y1, ϕ(x7) = y3, ϕ(x8) = y1 ϕ(Meridian of 818) = ϕ(x1) = y1 = Meridian of 31 ϕ(Longitude of 818) = ϕ(¯ x4x5¯ x6x7¯ x8x1¯ x2x3) = ¯ y3¯ y1¯ y2y1y1y1 = Longitude of 31

39

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G(818) =

x1, x2, x3, x4,

x5, x6, x7, x8

x4x1¯

x4¯ x2, x5x3¯ x5¯ x2, x6x3¯ x6¯ x4, x7x5¯ x7¯ x4, x8x5¯ x8¯ x6, x1x7¯ x1¯ x6, x2x7¯ x2¯ x8

G(31) = y1, y2, y3 | y3y1¯

y3¯ y2, y1y2¯ y1¯ y3 Meridian of 818 = x1, Longitude of 818 = ¯ x4x5¯ x6x7¯ x8x1¯ x2x3 Meridian of 31 = y1, Longitude of 31 = ¯ y3¯ y1¯ y2y1y1y1 ϕ(x1) = y1, ϕ(x2) = y2, ϕ(x3) = y1, ϕ(x4) = y3, ϕ(x5) = y3, ϕ(x6) = y1y3¯ y1, ϕ(x7) = y3, ϕ(x8) = y1 ϕ(Meridian of 818) = ϕ(x1) = y1 = Meridian of 31 ϕ(Longitude of 818) = ϕ(¯ x4x5¯ x6x7¯ x8x1¯ x2x3) = ¯ y3¯ y1¯ y2y1y1y1 = Longitude of 31 = ⇒ The surjective homomorhpism ϕ is realized by a degree one map.

40

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Which partial order can be realized by periods of knots? Example. 818 ≥ 41, 01 818

✲

period 2 41

❍❍❍❍❍❍❍❍ ❍ ❥

period 4 01

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85, 810, 815, 818, 819, 820, 821, 91, 96, 916, 923, 924, 928, 940, 105, 109, 1032, 1040, 1061, 1062, 1063, 1064, 1065, 1066, 1076, 1077, 1078, 1082, 1084, 1085, 1087, 1098, 1099, 10103, 10106, 10112, 10114, 10139, 10140, 10141, 10142, 10143, 10144, 10159, 10164 ≥ 31 818, 937, 940, 1058, 1059, 1060, 10122, 10136, 10137, 10138 ≥ 41 1074, 10120, 10122 ≥ 52

42

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period degree one map 85, 810, 815, 818, 819, 820, 821, 91, 96, 916, 923, 924, 928, 940, 105, 109, 1032, 1040, 1061, 1062, 1063, 1064, 1065, 1066, 1076, 1077, 1078, 1082, 1084, 1085, 1087, 1098, 1099, 10103, 10106, 10112, 10114, 10139, 10140, 10141, 10142, 10143, 10144, 10159, 10164 ≥ 31 818, 937, 940, 1058, 1059, 1060, 10122, 10136, 10137, 10138 ≥ 41 1074, 10120, 10122 ≥ 52

43

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1084, 1085, 1087, 1099, 10140, 10143, 10144 ≥ 31

1059, 10137 ≥ 41

44

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1084, 1085, 1087, 1099, 10140, 10143, 10144 ≥ 31

1059, 10137 ≥ 41
2. ∃ϕ : G(K) −

→ − → G(K′) = ⇒ ϕ(Meridian of K) = Meridian of K′ ?

45

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1084, 1085, 1087, 1099, 10140, 10143, 10144 ≥ 31

1059, 10137 ≥ 41
2. ∃ϕ : G(K) −

→ − → G(K′) = ⇒ ϕ(Meridian of K) = Meridian of K′ ?

In the knot table =

⇒ O.K.

46

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→ − → G(K′) = ⇒ c(K) ≥ c(K′) ?

47

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→ − → G(K′) = ⇒ c(K) ≥ c(K′) ?

K : up to 10 crossings =

⇒ O.K.

48

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→ − → G(K′) = ⇒ c(K) ≥ c(K′) ?

K : up to 10 crossings =

⇒ O.K.

K : alternating knots with up to 11 crossings =

⇒ O.K.

49

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→ − → G(K′) = ⇒ c(K) ≥ c(K′) ?

K : up to 10 crossings =

⇒ O.K.

K : alternating knots with up to 11 crossings =

⇒ O.K.

4. Relation of “≥” and Taniyama partial order “≥T”.

50

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→ − → G(K′) = ⇒ c(K) ≥ c(K′) ?

K : up to 10 crossings =

⇒ O.K.

K : alternating knots with up to 11 crossings =

⇒ O.K.

4. Relation of “≥” and Taniyama partial order “≥T”.
In the knot table, K ≥ K′ =

⇒ K ≥T K′

51

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→ − → G(K′) = ⇒ c(K) ≥ c(K′) ?

K : up to 10 crossings =

⇒ O.K.

K : alternating knots with up to 11 crossings =

⇒ O.K.

4. Relation of “≥” and Taniyama partial order “≥T”.
In the knot table, K ≥ K′ =

⇒ K ≥T K′

K ≥T K′ =

⇒ c(K) ≥ c(K′)

52