KazuhiroIchihara Cosmeticsurgeryonknots

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Introduction Known Facts Results (2-bridge knots) Results (New Example)

Cosmetic surgery on knots

Kazuhiro Ichihara

Nihon University

College of Humanities and Sciences

Joint works with

Toshio Saito

(Joetsu Univ. of Edu.)

In Dae Jong

(Kinki Univ.)

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Complement determines the knot type

Knot Complement Conjecture

A pair of knots in S

³

have homeomorphic complements if and only if they are equivalent.

(i.e.,

^∃

h : S

³ →

S

³

, homeo. which takes one knot to the other) This was conjectured by Tietze (1908),

They actually showed the following. (The KCC is a corollary to it) Theorem (Gordon-Luecke, 1989)

On a nontrivial knot in S

³

, nontrivial Dehn surgery never yields S

³

.

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Complement determines the knot type

Knot Complement Conjecture

A pair of knots in S

³

have homeomorphic complements if and only if they are equivalent.

(i.e.,

^∃

h : S

³ →

S

³

, homeo. which takes one knot to the other) This was conjectured by Tietze (1908),

and was proved by Gordon-Luecke (1989).

They actually showed the following. (The KCC is a corollary to it)

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Dehn surgery

E(K): the exterior of a

knot K in a 3-mfd M

(i.e., M

−

(open tubular nbd of K)) Glue solid torus to E(K) along slope γ;

γ m

f

We denote the obtained mfd by

K(γ).

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Generalization

The Knot Complement Conjecture can be generalized as follows.

Oriented Knot Complement Conjecture (Bleiler (Kirby’s list Problem 1.81(D)))

Let K

₁

and K

₂

be knots in a closed, oriented 3-manifold M. If the complements of K

1

and K

2

are homeomorphic

via an orientation-preserving homeomorphism,

then there exists an orientation-preserving homeomorphism of M

taking K

1

to K

2

.

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Generalization

The Knot Complement Conjecture can be generalized as follows.

Oriented Knot Complement Conjecture (Bleiler (Kirby’s list Problem 1.81(D)))

Let K

₁

and K

₂

be knots in a closed, oriented 3-manifold M . If the complements of K

1

and K

2

are homeomorphic

via an orientation-preserving homeomorphism,

then there exists an orientation-preserving homeomorphism of M

taking K

1

to K

2

.

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Cosmetic surgery

The OKCC follows from

Cosmetic Surgery Conjecture

(Bleiler (Kirby’s list Problem 1.81(A)))

Two surgeries on inequivalent slopes are never purely cosmetic. i.e., if K(r

1

)

∼

= K(r

2

) for inequivalent slopes r

1

, r

2

,

then the homeomorphism is orientation reversing.

•

Two slopes are equivalent if there exists

a homeo. of E(K) taking one slope to the other.

•

Two surgeries on K along slopes r

₁

, r

₂

are purely cosmetic

if there exists an ori.-pres. homeo. between K(r

1

) & K(r

2

),

and chirally cosmetic if the homeo. is orientation-reversing.

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Cosmetic surgery

The OKCC follows from Cosmetic Surgery Conjecture

(Bleiler (Kirby’s list Problem 1.81(A)))

Two surgeries on inequivalent slopes are never purely cosmetic.

i.e., if K(r

1

)

∼

= K(r

2

) for inequivalent slopes r

1

, r

2

, then the homeomorphism is orientation reversing.

•

Two slopes are equivalent if there exists

a homeo. of E(K ) taking one slope to the other.

•

Two surgeries on K along slopes r

₁

, r

₂

are purely cosmetic

if there exists an ori.-pres. homeo. between K(r

1

) & K (r

2

),

and chirally cosmetic if the homeo. is orientation-reversing.

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Remark

Remark: (Mathieu, 1992)

For “Orientation reversing” case, there exist examples.

Actually (18k + 9)/(3k + 1)- and (18k + 9)/(3k + 2)-surgeries on the trefoil knot T

2,3

in S

³

yield

orientation-reversingly homeomorphic pairs for any k

≥

0.

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Known facts (Examples)

Rong (1995)

Seifert knots in closed 3-manifolds (except lens spaces) admitting cosmetic surgeries are classified.

Matignion (2010)

Non-hyperbolic knots in lens spaces admitting cosmetic surgeries are classified.

The surgeries are all chirally cosmetic.

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Known facts (Examples)

Theorem [Bleiler-Hodgson-Weeks (1999)]

There exists a hyperbolic knot which admits a pair of surgeries along inequivalent slopes yielding oppositely oriented lens spaces.

The lens spaces are: L(49,

−

19)

↔

L(49,

−

18) (mirror image)

It was announced by Matignion that there are infinite family

of such examples extended above.

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Known facts (Examples)

Theorem [Bleiler-Hodgson-Weeks (1999)]

There exists a hyperbolic knot which admits a pair of surgeries along inequivalent slopes yielding oppositely oriented lens spaces.

The lens spaces are: L(49,

−

19)

↔

L(49,

−

18) (mirror image)

It was announced by Matignion that there are infinite family

of such examples extended above.

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Known facts (Criteria)

Let

∆_K(t)

denote the Alexander polynomial of a knot K in S

³

normalized to be symmetric and satisfy ∆

K

(1) = 1.

Boyer-Lines (1990)

A knot K satisfying

∆⁰⁰_K(1)6= 0

has no cosmetic surgeries.

They use the (original) Casson invariant

(i.e., defined by using SU (2)-representations).

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Recent Progress

Theorem [Ni-Wu (2011)]

Suppose K is a nontrivial knot in S

³

& r

₁

, r

₂

are distinct slopes such that K(r

1

)

∼

= K(r

2

) as oriented manifolds.

Then r

1

, r

2

satisfy that (a) r

₁

=

−

r

₂

(b) suppose r

₁

= p/q, then q

² ≡ −

1 (mod p)

(c)

τ(K) = 0, where

τ is the invariant defined by Ozsv´ ath-Szab´ o.

based on Heegaard Floer Homology (d-invariant and τ -invariant).

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2-bridge knots with at most 9 crossings

Proposition.

All the two-bridge knots of at most 9 crossings other than 9

27

admits no cosmetic surgery pairs.

9

27

= S(49, 18) = C[2, 2,

−

2, 2, 2,

−

2]

[I., 2013]

For K = 9

27

, K(10/3)

6∼

= K(

−

10/3).

(Tohoku Knot Semi 2011@Sendai)

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2-bridge knots with at most 9 crossings

Proposition.

All the two-bridge knots of at most 9 crossings other than 9

27

admits no cosmetic surgery pairs.

9

27

= S(49, 18) = C[2, 2,

−

2, 2, 2,

−

2]

[I., 2013]

For K = 9

27

, K(10/3)

6∼

= K(

−

10/3).

(Tohoku Knot Semi 2011@Sendai)

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Table : 2-bridge knots of at most 9 crossings withτ = 0

Name Schubert Form Alexander Polynomial ∆⁰⁰_K(1)

41 S(5,2) t⁻¹−3 +t 2

61 S(9,7) 2t⁻¹−5 + 2t 4

63 S(13,5) t⁻²−3t⁻¹+ 5−3t+t² 2

77 S(21,8) t⁻²−5t⁻¹+ 9−5t+t² -2

81 S(13,11) 3t⁻¹−7 + 3t 6

83 S(17,4) 4t⁻¹−9 + 4t 8

88 S(25,9) 2t⁻²−6t⁻¹+ 9−6t+ 2t² 4

89 S(25,7) t⁻³−3t⁻²+ 5t⁻¹−7 + 5t−3t²+t³ 4

812 S(29,12) t⁻²−7t⁻¹+ 13−7t+t² -6

813 S(29,11) 2t⁻²−7t⁻¹+ 11−7t+ 2t² 2 914 S(37,14) 2t⁻²−9t⁻¹+ 15−9t+ 2t² -2 919 S(41,16) 2t⁻²−10t⁻¹+ 17−10t+ 2t² -4 927 S(49,19) t⁻³−5t⁻²+ 11t⁻¹−15 + 11t−5t²+t³ 0

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A family including 9

₂₇

Theorem [I.-Saito].

Let K

x

be a 2-bridge knot C[2x, 2,

−

2x, 2x, 2,

−

2x] with x

≥

1. Then K

x

admits no cosmetic surgery pairs yielding homology 3-spheres.

i.e., any

_n¹

- and

_m¹

-surgeries are not purely cosmetic for K

x

.

Remark:

For K

_x

, ∆

⁰⁰_K

x

(1) = 0 and τ (K

_x

) = 0 hold. In particular, K

₁

= 9

₂₇

.

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Key Ingredients

C. L. Curtis, An intersection theory count of the SL2(C)- representations of the fundamental group of a 3-manifold, Topology 40(2001), no. 4, 773–787.

H. U. Boden and C. L. Curtis,TheSL(2,C)Casson invariantfor Dehn surgeries on two-bridge knots, Algebr. Geom. Topol. 12 (2012), no. 4, 2095–2126.

T. Ohtsuki, Ideal points and incompressible surfaces in two-bridge knot complements, J. Math. Soc. Japan46(1994), no. 1, 51–87.

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Montesinos trick & Banding

Montesinos trick [Montesinos, 1975]

Let M be a 3-mfd with a double br.cover M ¯ along a link L

⊂

M. For a knot K in M, which is strongly invertible w.r.t. the axis ¯ L, ¯ and for a slope r

∈Z, the surgered mfd.

K(r) is homeomorphic to the double br.cover along the link obtained from L by a banding.

q ] ^q ] ^q ]

]

2q−4±1

Kq±

··· ··· ···

· · ·

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Bleiler-Hodgson-Weeks’s Example

Bleiler-Hodgson-Weeks’s example (1999);

yieldingL(49,−19).

↓

(Cosmetic banding on the knot 9

27

)

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Bleiler-Hodgson-Weeks’s Example

Bleiler-Hodgson-Weeks’s example (1999);

yieldingL(49,−19).

↓

(Cosmetic banding on the knot 9

27

)

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Bleiler-Hodgson-Weeks’s Example

mirroring

& 2π/3-rot.

banding of slope 0

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New Example

a (chirally)

cosmetic banding.

•

The double branched cover M along K is hyperbolic.

•

The knot K ¯ in M corresponding to the banding is hyperbolic.

•

The knot K ¯ is not

amphicheiral.

This gives a counter-example to: Conjecture (BHW)

Cusped hyperbolic manifolds admit no

cosmetic fillings, true or reflective,

yielding hyperbolic manifolds.

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New Example

•

This knot K admits

a (chirally)

cosmetic banding.

•

The double branched cover M along K is hyperbolic.

•

The knot K ¯ in M corresponding to the banding is hyperbolic.

•

The knot K ¯ is not

amphicheiral.

This gives a counter-example to: Conjecture (BHW)

Cusped hyperbolic manifolds admit no

cosmetic fillings, true or reflective,

yielding hyperbolic manifolds.

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New Example

•

This knot K admits

a (chirally)

cosmetic banding.

•

The double branched cover M along K is hyperbolic.

•

The knot K ¯ in M corresponding to the banding is hyperbolic.

•

The knot K ¯ is not

amphicheiral.

This gives a counter-example to:

Conjecture (BHW)

Cusped hyperbolic manifolds admit no

cosmetic fillings, true or reflective,

yielding hyperbolic manifolds.

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How to check?

Theorem [I.-Jong].

The knot K ¯ admits a pair of chirally cosmetic surgeries yielding hyperbolic manifolds.

•

The double branched cover M along K is hyperbolic.

•

The knot K ¯ in M corresponding to the banding is hyperbolic.

by HIKMOT [To appear in Exper.Math., arXiv:1310.3410.]

•

The knot K ¯ is not

amphicheiral.