Seifert fibered surgeries on Montesinos knots

日大トポロジーセミナー

, 2008/12/6, 13:30–15:00

Seifert fibered surgeries on Montesinos knots

In Dae Jong (

ちょん

鄭 仁大 )

Osaka City University ( _{大阪市立大学} )

joint work with

Kazuhiro Ichihara (Nara University of Education) and

Shigeru Mizushima (Tokyo Institute of Technology)

An aim of today’s talk

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A study on Dehn surgery by using knot invariants.

In particular, we focus on

• alt( K ) : the alternation number of a knot K ,

• s ( K ) : the Rasmussen invariant of a knot K ,

• σ ( K ) : the signature of a knot K .

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These invariants are closely related in terms of

Gordian distance on knots.

§ 1. Introduction

Dehn surgery on a knot K in S ³

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1) Remove a neighborhood of K from S ³

2) Gluing a solid torus back (along slope γ )

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f

γ m _{γ = [ f ( m )]}

By using a preferred meridian-longitude system, one can parameterize slopes by irreducible

fractions.

i.e., { slopes } ←→ ^1:1 Q ^{∪ { 1 / 0 }}

For r ∈ ^Q ,

r -surgery: surgery along a slope parameterized by r , K ( r ): the manifold obtained by r -surgery

along a knot K .

Today, we assume that all surgeries are non-trivial.

Namely, we set aside the trivial(1 / 0-) surgery.

Problem

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On hyperbolic knots in S ³ , determine all non- trivial Dehn surgeries producing non-hyperbolic 3- mfds. (Determine exceptional surgeries.)

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Such surgeries are only finitely many . [Thurston]

Types of exceptional surgeries

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• Reducible surgery

• Toroidal surgery

• Seifert fibered (SF) surgery

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(as a consequence of Geometrization conjecture.)

Today’s problem

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Determine all exceptional surgeries on hyperbolic Montesinos knots.

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Montesinos link M ( R ₁ , . . . , R _l )

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R R R ₁

2 l

l : length. ( R _i ∈ ^Q ↔ a rational tangle)

If R _i = 1 /a _i for all i ( a _i ∈ ^Z \{ 0 } ), then we denote it by P ( a ₁ , . . . , a _l ) a pretzel knot of type ( a ₁ , . . . , a _l ).

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Review

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For a hyperbolic Montesinos knot with length l ,

• l ≤ 2 ⇒ K is 2-bridge knot. All exceptional surgeries are classified [Brittenham-Wu ’95].

• l ≥ 4 ⇒ ̸ ∃ exceptional surgery [Wu ’96].

• ̸ ∃ reducible surgery [Wu ’96].

• All toroidal surgeries are classified [Wu ’06].

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Remains are SF surgeries with l = 3.

Target: K = M ( R ₁ , R ₂ , R ₃ ), K ( r ) : SF.

r -surgery on K is cyclic (resp. finite)

= π ₁ ( K ( r )) is cyclic (resp. finite).

Theorem 1 [Ichihara-J.] (arXiv:0807.0905v3)

K : a hyperbolic Montesinos knot.

(i) If r -surgery on K is cyclic,

then K = P ( − 2 , 3 , 7) and r = 18 or 19.

(ii) If r -surgery on K is acyclic finite,

then K = P ( − 2 , 3 , 7) and r = 17, or

K = P ( − 2 , 3 , 9) and r = 22 or 23.

Remains are SF surgeries with l = 3

and with | π ₁ ( K ( r )) | = ∞ .

Together with the result by [Wu], we have;

Corollary

Among hyperbolic arborescent knots,

only the knots in Thm.1 admit such surgeries.

[Watson]

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For p ∈ { 5 , 7 , · · · , 25 } , the ( − 2 , p, p )-pretzel knot admit no finite surgeries.

(by using Khovanov homology) ( arXiv:0807.1341v2)

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[Futer-Ishikawa-Kabaya-Mattman-Shimokawa]

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A complete classification of finite surgeries on ( − 2 , p, q )-pretzel knots with p, q : odd positive.

( arXiv:0809.4278)

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Result in the case of | π ₁ ( K ( r )) | = ∞

Theorem 2 [Ichihara-J.-Mizushima].

K : alternating hyperbolic Montesinos knot.

If r -surgery on K is SF, then K = P ( p, q, r )

with p, q, r ≥ 3: mutually inequivalent odds.

In _§ 2, we show an outline of the proof of Thm.2.

In _§ 3, we show Thm.1 (if we have enough time).

§ 2. Proof of Thm.2

Let K be an alternating hyperbolic Montesinos knot.

Proposition 1

If K admits SF surgery, then either

(i) K = P ( p, q, r ) with p, q, r ≥ 3: odd, (ii) K = P (3 , 3 , 2 n ) with n ≥ 1, or

(iii) K = M ( ¹ ₃ , ¹ ₃ , ^{2 m} _{2 m} ^{− 1} ) with m ≥ 2.

This is proved based on Delman’s work:

C. Delman, Preprint (unpublished, 1995).

“Constructing essential laminations and

taut foliations which survive all Dehn surgeries”

[Delman] showed that K admits

an essential lamination (introduced by [Gabai-Oertel]) in its exterior surviving essential after all non-trivial

surgeries (called persistent lamination).

Also [Brittenham] showed that

“certain” (called genuine) essential lamination cannot exist in a SF manifold.

Outline of Proof (Proposition 1)

1) Check whether the Delman’s lami. is genuine, 2) if not, construct “new” one, which is genuine.

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Proposition 2

(1) K ̸ = P ( p, q, q ) with odd p, q ≥ 3.

(2) K ̸ = P (2 n, q, q ) with n ≥ 1 and odd q ≥ 3.

(3) K ̸ = M ( ¹ ₃ , ¹ ₃ , ^{2 m} _{2 m} ^{− 1} ) with m ≥ 2.

In the following, we prove Proposition 2 (1) . Fact 1 [Ichihara]

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All exceptional surgeries on alternating knots are integral surgeries.

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Since K is alternating, we may assume that r ∈ ^Z .

Set K = P ( p, q, q ). Suppose that K ( r ) is SF.

Then, by so-called Montesinos trick,

K ( r ) = 2-fold branched cover of ∼ S ³ along L ( p, q, r ).

α

… ……

q p

r+2(p+q)

… …

…

quotient

q p { q

L (p,q,r) K(r)

{ { { {

{

Remark

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L ( p, q, r ) is a knot ( ⇔ r : odd)

or a 2-component link ( ⇔ r : even).

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Fact 2

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K ( r ): SF ⇒

L ( p, q, r ) must be a Montesinos or Seifert link.

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Seifert link = the exterior admits a Seifert fibration.

Lemma A.

L ( p, q, r ) is not a Montesinos link.

Lemma B.

L ( p, q, r ) is not a Seifert link.

These give a contradiction.

Then we complete the proof of Proposition 2 (1).

From now on, we will show Lemmas A and B.

[Proof of Lem. A ( r : even, i.e., L ( p, q, r ): 2-comp.)]

Fact 3 [Motegi]

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L ( p, q, r ) ̸ = Montesinos link of l > 3.

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Fact 4

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L : Montesinos link with length l ⇒ b ( L ) ≤ l .

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Claim 1.

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r : even ⇒ L ( p, q, r ) ̸ = Montesinos link.

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[Proof ] Suppose that r is even and L ( p, q, r ) is a Montesinos link. By Facts 3 & 4, b ( L ( p, q, r )) ≤ 3.

On the other hand, L ( p, q, r ) = T _{2 ,q} ∪ T _{2 ,p + q}

( T _x,y : ( x, y )-torus knot).

Then we have b ( L ( p, q, r )) ≥ 4. ⇒ contradiction. _¤

[Proof of Lem. A ( r : odd, i.e., L ( p, q, r ): knot)]

Here we introduce the alternation number.

alternation number [Kawauchi]

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A = _{ alternating links _} ( ∋ trivial links).

d _G ( · , · ) : the Gordian distance.

alt( L ) = min

L ^′ ∈A d _G ( L, L ^′ ) .

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Remark

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• alt( K ) ≤ u ( K )

• ∀ n ∈ ^N , ∃ K s.t. alt( K ) = n . [Abe], [Kawauchi]

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Fact 5. [Abe-J.-Kishimoto]

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alt( L ) ≤ 1 for any Montesinos link L .

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Fact 6. [Abe]

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For a knot K , alt( K ) ≥ ^{| s ( K ) −} ₂ ^{σ ( K ) |}

where s ( K ): the Rasmussen invariant of K , and σ ( K ): the signature of K

with σ (right-handed trefoil) = 2.

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Claim 2.

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alt( L ( p, q, r )) ≥ 2.

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Claim 2 guarantees L ( p, q, r ) ̸ = Montesinos knot.

Namely, Claim 2 is true ⇒ Lemma A is true .

[Proof of Claim 2.]

Fact 7 [Ichihara]

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K : hyperbolic knot. r, r ^′ ∈ ^Z .

r, r ^′ -surgereis: exceptional ⇒ ∆( r, r ^′ ) ≤ 8.

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In general, ∆( r, r ^′ ) ≤ 8. [Lackenby-Meyerhoff ] (arXiv:0808.1176)

SubClaim 1

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− 8 ≤ r ≤ 8.

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[Proof ] K (0) contains essential torus ⇒

0-surgery is exceptional. By Fact 7, − 8 ≤ r ≤ 8. _¤ Remark

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L ( p, q, r ) is a positive knot for − 8 ≤ r ≤ 8.

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SubClaim 2

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s ( L ( p, q, r )) = s ( L ( p, q − 2 , r )) + 8.

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[Proof ] Direct calculation by using Fact 8 [Rasmussen]

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For a positive knot K with a positive diagram D , s ( K ) = c ( D ) − o ( D ) + 1

where c ( D ): the number of crossings of D ,

o ( D ): the number of Seifert circles of D . In particular, σ ( L ( p, q, r )) ≤ s ( L ( p, q, r )) = 2 g _∗ ( K ).

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SubClaim 3

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σ ( L ( p, q, r )) ≤ σ ( L ( p, q − 2 , r )) + 4.

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[Proof ] L ( p, q − 2 , r ) is obtained from L ( p, q, r ) by a single ♯ -unknotting operation.

Applying Fact 9,

we complete the proof of SubClaim 3. _¤

Fact 9 [H. Murakami]

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K K’

a #-unknotting operation

K

K ₀ : 2-component link ⇒ σ ( K ) ≤ σ ( K ^′ ) + 4 .

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By Facts 6, 8 and SubClaims 1, 2, 3,

2 alt( L ( p, q, r )) ≥ s ( L ( p, q, r )) − σ ( L ( p, q, r ))

≥ ( s ( L ( p, q − 2 , r )) + 8) − ( σ ( L ( p, q − 2 , r )) + 4)

= s ( L ( p, q − 2 , r )) − σ ( L ( p, q − 2 , r )) + 4 ≥ 4 .

This completes [Proof of Claim 2] alt( L ( p, q, r )) ≥ 2

By Fact 5, L ( p, q, r ) ̸ = Montesinos knot

( p, q ≥ 3 , − 8 ≤ r ≤ 8).

Consequently, we completes [Proof of Lemma A].

Namely, L ( p, q, r ) ̸ = Montesinos.

[Proof of Lem B.] L ( p, q, r ) ̸ = Seifert link Fact 10

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A Seifert link with (# of comp.) ≤ 2 is either (i) a torus knot, (ii) a torus link, or

(iii) (a torus knot) ∪ (the core of the torus).

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Claim 3.

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L ( p, q, r ) ̸ = a link of (ii) and of (iii).

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[Proof ] r :even ⇒ L ( p, q, r ) = T _{2 ,q} ∪ T _{2 ,p + q} .

A torus link has parallel components.

A link of (iii) has a trivial component. _¤

⋆ braid index of L ( p, q, r ) ≤ 4 ⇒ Check T _{3 ,x} , T _{4 ,x} .

Claim 4.

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L ( p, q, r ) ̸ = T _{4 ,x} with odd x ≥ 5.

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[Proof ] Consider det( K )(= ∆ _K ( − 1)) of a knot K . Note 1 : det( L ( p, q, r )) = | H ₁ ( K ( r )) | = | r | .

Note 2 : det( T _{4 ,x} ) = x with x ≥ 5. ⇒ | r | = x . Recall − 8 ≤ r ≤ 8 . (see Proof of SubClaim 1.)

Then we have r = ± 5, x = 5, or r = ± 7, x = 7 .

⋆ g ( L ( p, q, r )) = ^{c ( D ) − o} ₂ ^{( D )+1} _∵ L ( p, q, r ): positive Note 3 : g ( T _{4 ,x} ) = ³

2 ( x − 1) ≤ ³ ₂ (7 − 1) = 9.

Note 4 : g ( L ( p, q, r )) = 3( p + q ) + ¹

2 ( r − 3)

≥ 3(3 + 3) + ¹

2 ( − 7 − 3) = 13.

This completes Proof of Claim 4. _¤

Claim 5.

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L ( p, q, r ) ̸ = T _{3 ,x}

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[Proof ] Comparing the determinants, we have

r =











± 1 if x ≡ ± 1 mod 6

± 3 if x ≡ ± 2 mod 6 . (i) r = 1 and x ≡ ± 1 mod 6

g ( L ( p, q, 1)) = 3( p + q ) − 1 ≡ − 1 mod 6.

g ( T _{3 ,x} ) = x − 1 ̸≡ − 1. ⇒ L ( p, q, 1) ̸ = T _{3 ,x} .

(ii) r = 3 and x ≡ ± 2 mod 6

g ( L ( p, q, 3)) = 3( p + q ) ≡ 0 mod 6.

g ( T _{3 ,x} ) = x − 1 ̸≡ 0. ⇒ L ( p, q, 3) ̸ = T _{3 ,x} .

(iii) r = − 1 and x ≡ ± 1 mod 6

g ( L ( p, q, − 1)) = 3( p + q ) − 2 ≡ − 2 mod 6.

g ( T _{3 ,x} ) = x − 1 ≡ 0 or − 2 mod 6 .

⇒ x = 3( p + q ) − 1. Thus, L ( p, q, − 1) = ^? T _{3 , 3( p + q ) − 1} . (iv) r = − 3 and x ≡ ± 2 mod 6

g ( L ( p, q, − 3)) = 3( p + q ) − 3 ≡ 3 mod 6.

g ( T _{3 ,x} ) = x − 1 ≡ 1 or 3 mod 6 .

⇒ x = 3( p + q ) − 2. Thus, L ( p, q, − 3) = ^? T _{3 , 3( p + q ) − 2} . Calculating the Alexander polynomials, we can show that these are inequivalent. _¤ (Claim 5) This completes the proof

[Lemma B] ⇒ [Proposition 2 (1)] _¤

Next, we prove Proposition 2 (2) . Proposition 2

(2) K ̸ = P (2 n, q, q ) with n ≥ 1 and odd q ≥ 3.

Key tools of Proof of Prop. 2 (2) & (3)

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• Symmetries (strong inversion & cyclic period)

• alternation number

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Fact 11 [Motegi, Miyazaki-Motegi]

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K : periodic with period 2 and strongly invertible.

K ^′ : factor knot on the period.

Suppose that K ( r ): SF and K ^′ : non-trivial

⇒ K ^′ ( r/ 2): lens space ( ̸ = S ³ , S ² × S ¹ )

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Set K = P (2 n, q, q ). Suppose that K ( r ) is SF.

Note that K ^′ = T _{2 ,q} . By Fact 11, K ^′ ( ^r

2 ): lens.

Fact 12 [Moser]

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T _x,y ( ^m

n ): lens space ⇔ ^m _n = xy ± _n ¹ .

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Then, by Fact 12, we have ^r

2 = 2 q ± ¹ ₂ ⇔ r = 4 q ± 1.

Then, by Montesinos trick again,

K ( r ) = 2-fold branched cover of ∼ S ³ along K ( n, q, r ).

r+2(2n

q)

K (n,q,r)

… ……

{ {

{

…

q {

q { ²ⁿ {

K(r)

q 2n

We can show that K ( n, q, r ) is neither Montesinos knot nor torus knot by the same way of (1) ^(with

some calculations) .

Note: K ( n, q, r ) is positive knot ( r = 4 q ± 1).

¤ (Proof of Proposition 2 (2))

Next, we prove Proposition 2 (3) . Proposition 2

(3) K ̸ = M ( ¹

3 , ¹ ₃ , ^{2 m} _{2 m} ^{− 1} ) with m ≥ 2.

M (3, 3/4, 3) ( m =2)

factor knot

=

Set K = M ( ¹

3 , ¹ ₃ , ^{2 m} _{2 m} ^{− 1} ). Suppose that K ( r ) is SF.

By Facts 11 and 12, we have r = 11 or 13.

Claim 6 [Ichihara]

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K : hyperbolic knot.

F : essential surface in E ( K ) with ∂ -slope β ∈ ^Z . Suppose that K ( r ) is non-hyperbolic for r ∈ ^Z .

⇒ | r − β | < 5 . 05( ⁻ _| ^{χ ( F )}

∂F | + 1).

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β = 14 − 4 m χ ( F ) = − 2

| ∂F | = 1

| r − (14 − 4 m ) | < 5 . 05(2 + 1)

⇔ − 15 . 15 < r + 4 m − 14 < 15 . 15

⇔ − 4 m − 1 . 15 < r < − 4 m + 29 . 15

⇒ − 4 m − 1 ≤ r ≤ − 4 m + 29.

Since r = 11 or 13, m = 2 , 3 , 4.

Then, by Montesinos trick again,

K ( r ) = 2-fold branched cover of ∼ S ³ along a certain knot K ( m, r ).

We omit the details.

Actually, we can apply parallel argument to show that K ( m, r ) is neither Montesinos knot nor torus knot by the same way of (1) & (2).

¤ (Proof of Proposition 2 (3))

§ 3 Proof of Theorem 1

Theorem 1 [Ichihara-J.] (arXiv:0807.0905v3)

K : a hyperbolic Montesinos knot.

(i) If r -surgery on K is a non-trivial cyclic, then K = P ( − 2 , 3 , 7) and r = 18 or 19.

(ii) If r -surgery on K is a non-trivial acyclic finite, then K = P ( − 2 , 3 , 7) and r = 17, or

K = P ( − 2 , 3 , 9) and r = 22 or 23.

Key ingredients

• Essential lamination

• Heegaard Floer homology

[Outline of Proof of Thm 1.]

K : a hyperbolic Montesinos knot.

Fact 14. [Delman]

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If K admits a cyclic / finite surgery, then K is equivalent to either

(i) P ( − 2 ℓ, p, q ), (ii) P ( − 1 , 2 n, p, q ), or (iii) P ( − 1 , − 1 , 2 m, p, q ).

Here ℓ > 1, n ̸ = 0, m > 1, and 3 ≤ p ≤ q : odd.

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[Delman] (unpublished, 1995).

“Constructing essential laminations and taut foliations which

survive all Dehn surgeries”

[Delman]

Every hyperbolic Montesinos knot except for the three families

admits an essential lamination in its exterior surviving after all non-trivial Dehn surgeries.

Essential lamination (introduced by [Gabai-Oertel]) Actually they showed that

if a 3-mfd. M contains an essential lamination, then its universal cover must be _R ³ .

In particular, π ₁ ( M ) is never cyclic/finite.

Case 1: P ( − 2 ℓ, p, q )

Fact 15. [Mattman]

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If K admits a cyclic / finite surgery, then

K ̸∼ = P ( − 2 ℓ, p, q ) with ℓ > 1 & 3 ≤ p ≤ q : odd.

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[Mattman],

“Cyclic and finite surgeries on pretzel knots”,

J. Knot Theory Ramifications, 11(6):891–902, 2002.

Case 2: P ( − 1 , 2 n, p, q )

L-space

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A rational homology sphere Y is an L-space if the rank of HF

( Y ) is equal to | H ₁ ( Y ; _Z ) | .

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In fact, Ozsv´ ath-Szab´ o showed that

M has π ₁ ( M ); cyclic / finite ⇒ M is an L-space .

Fact 16. [Ozsv´ ath-Szab´ o]

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If a knot K in S ³ admits an integral surgery yielding an L-space,

then every non-zero coefficient of ∆ _K ( t ) is ± 1.

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Lemma C.

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If P ( − 1 , 2 n, p, q ) with n ̸ = 0 & 3 ≤ p ≤ q : odd

admits a cyclic/finite surgery, then ( n, p ) = (1 , 3).

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Fact 17.

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For a knot K ⊂ S ³ ,

K ( p/q ) is an L-space ⇒ K ( p ) is also an L-space.

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[Proof of Lem. C] Suppose that K = P ( − 1 , 2 n, p, q ) admits a cyclic / finite surgery.

By Facts 16 & 17,

every non-zero coefficient of ∆ _K ( t ) must be ± 1.

Notation & Normalization

Set ∆ _K ( t ) = a ₀ + a ₁ t + · · · + a _k t ^k with a ₀ ̸ = 0.

Claim 6.

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• If n ≤ − 1, then a ₁ =







− 4 if n = − 1

− 3 if n ≤ − 2

• If n ≥ 2, then a ₃ = 2.

• If n = 1 & 5 ≤ p ≤ q : odd, then a ₄ = − 2.

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¤ (Lemma C)

By this Lemma C,

if K admits a cyclic/finite surgery, then

K = P ( − 1 , 2 , 3 , q ) = ∼ P ( − 2 , 3 , q ) with q ≥ 3:odd.

Then [Mattman] already showed:

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Among such knots, only P ( − 2 , 3 , 7) & P ( − 2 , 3 , 9) can have cyclic / finite surgeries,

and the surgery slopes are the ones in Thm. 1.

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Case 3: P ( − 1 , − 1 , 2 m, p, q )

Fact 18 [Ni]

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If a knot in S ³ admits a cyclic / finite surgery, then it must be a fibered knot.

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Claim 6.

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P ( − 1 , − 1 , 2 m, p, q ) is not fibered

with m > 1 and 3 ≤ p ≤ q : odd.

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Claim 6 is shown by using an algorithm due to [Gabai], which decides fiberedness of a pretzel knot.

¤ (Proof of Thm 1.)