TetsuyaIto (OsakaUniv.), ToshioSaito (JoetsuUniv.Edu.)E-KOOKSeminar2017.8.30,OsakaInst.Tech. Basedonajointworkwith ChirallycosmeticsurgeriesandCassoninvariantsKazuhiroIchihara

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Chirally cosmetic surgeries and Casson invariants

Kazuhiro Ichihara

Nihon University, College of Humanities and Sciences

Based on a joint work with

Tetsuya Ito

(Osaka Univ.),

Toshio Saito

(Joetsu Univ. Edu.) E-KOOK Seminar

2017.8.30, Osaka Inst.Tech.

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Cosmetic surgery Casson invariant SL(2,C)Casson invariant Applications

Dehn surgery on a knot

K

: a knot (i.e., embedded circle) in a 3-manifold

M

Dehn surgery on K (operation to produce a “NEW” 3-mfd)

1) remove the open neighborhood ofKfromM

(to obtain theexterior E(K)ofK) 2) glue a solid torus back (along a slopeγ)

γ m

f

We denote the obtained manifold by

M_K(γ),

or, by

K(γ)

if K is a knot in S

³

.

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Cosmetic surgery conjecture

It is natural to ask:

Can a pair of distinct Dehn surgeries give the same manifold?

Two surgeries on inequivalent slopes are never purely cosmetic.

•

Two slopes for a knot K are called equivalent

if

^∃

homeo. of the exterior of K taking one slope to the other.

•

Two Dehn surgeries on K are called purely cosmetic if

^∃

orientation preserving homeo. between the manifolds obtained by the surgeries.

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Cosmetic surgery conjecture

It is natural to ask:

Can a pair of distinct Dehn surgeries give the same manifold?

Conjecture. (Problem 1.81(A) in Kirby’s list)

Two surgeries on inequivalent slopes are never purely cosmetic.

•

Two slopes for a knot K are called equivalent

if

^∃

homeo. of the exterior of K taking one slope to the other.

•

Two Dehn surgeries on K are called purely cosmetic if

^∃

orientation preserving homeo. between the manifolds obtained by the surgeries.

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Chirally cosmetic case

For “Orientation reversing” case, there exist (counter-)examples.

[Mathieu, 1992]

There exist some knots admitting

“chirally” cosmetic surgeries along inequivalent slopes.

In fact, (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.

Further examples were obtained by [Rong], [Matignon], [Bleiler-Hodgson-Weeks], [Hoffman-Matignon], [I.-Jong].

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On amphicheiral knots

When K is amphicheiral, for

^∀

slope r 6∈ {0, 1/0},

r- & (−r)-surgeries are chirally cosmetic along equivalent slopes.

6 ∃ other cosmetic surgeries on an amphicheiral knot in S

³

:

If K is amphicheiral and K(r) ∼ = −K(r

⁰

), then

K(−r

⁰

) ∼ = −K(r

⁰

) ∼ = K(r). By [Ni-Wu], this implies r = ±r

⁰

. Ni-Wu (2011)

If the surgeries along distinct slopes r

1

and r

2

are purely cosmetic, then r

₁

, r

₂

satisfy that

(a)

r1=−r2

,

(b) q

²

≡ −1 mod p for r

1

= p/q,

(c) τ (K) = 0 (the invariant defined by Ozsv´ ath-Szab´ o).

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On amphicheiral knots

When K is amphicheiral, for

^∀

slope r 6∈ {0, 1/0},

r- & (−r)-surgeries are chirally cosmetic along equivalent slopes.

6 ∃ other cosmetic surgeries on an amphicheiral knot in S

³

: If K is amphicheiral and K(r) ∼ = −K (r

⁰

), then

K(−r

⁰

) ∼ = −K(r

⁰

) ∼ = K(r). By [Ni-Wu], this implies r = ±r

⁰

. Ni-Wu (2011)

If the surgeries along distinct slopes r

1

and r

2

are purely cosmetic, then r

₁

, r

₂

satisfy that

(a)

r1=−r2

,

(b) q

²

≡ −1 mod p for r

1

= p/q,

(c) τ (K) = 0 (the invariant defined by Ozsv´ ath-Szab´ o).

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Casson-Walker invariant and Casson-Gordon invariant Let K be a knot in an integral homology 3-sphere Σ.

The Casson-Walker invariant λ and the Casson-Gordon invariant τ satisfy the following surgery formulae [Walker, Boyer-Lines]:

λ(ΣK(p/q)) = q

pa2(K)−1

2s(q, p), τ(ΣK(p/q)) =−4p·s(q, p)+σ(K, p).

Recall:

a

₂

(K) :=

¹₂

∆

⁰⁰_K

(1),

(∆_K(t): normalized Alexander polynomial ofK)

In the case Σ = S

³

,

a

2

(K) is the 2nd coeff. of the Conway poly. ∇

_K

(z) of K.

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Casson-Walker invariant and Casson-Gordon invariant Let K be a knot in an integral homology 3-sphere Σ.

The Casson-Walker invariant λ and the Casson-Gordon invariant τ satisfy the following surgery formulae [Walker, Boyer-Lines]:

λ(ΣK(p/q)) = q

pa2(K)−1

2s(q, p), τ(ΣK(p/q)) =−4p·s(q, p)+σ(K, p).

Recall:

a

₂

(K) :=

¹₂

∆

⁰⁰_K

(1),

(∆_K(t): normalized Alexander polynomial ofK)

In the case Σ = S

³

,

a

2

(K) is the 2nd coeff. of the Conway poly. ∇

_K

(z) of K.

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Dedekind sum and total p-signature

Dedekind sum s(q, p)

For coprime integers p, q with p > 0,

s(q, p) :=

p−1

X

k=1

k p

kq p

with ((x)) = x − bxc −

¹₂

and the floor function bxc for x ∈

Q

. the total p-signature σ(K, p)

σ(K, p) :=

X

ω^p=1

σ

ω

(K) for ω ∈ {z ∈

C

| |z| = 1}

where σ

_ω

(K) denotes

the Levine-Tristram signature,

i.e., the signature of(1−ω)S+ (1−ω)S^T for a Seifert matrixS ofK.

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Result (1)

Theorem 1.

Let K be a knot in an integral homology 3-sphere Σ.

If Σ

_K

(p/q) ∼ = −Σ

_K

(p/q

⁰

), then we have the following.

4(q + q

⁰

)a

2

(K ) = 2p(s(q, p) + s(q

⁰

, p)) = σ(K, p)

Proof

Suppose that Σ

_K

(p/q) ∼ = −Σ

_K

(p/q

⁰

).

Recall: λ(−M) = −λ(M ) and τ (−M ) = −τ (M).

Then we have the following by the surgery formulae above.

(λ(ΣK(p/q))−(−λ(ΣK(p/q⁰))) = ^q+q_p⁰a2(K)− ¹₂s(q, p) + ¹₂s(q⁰, p)

= 0, τ(ΣK(p/q))−(−τ(ΣK(p/q⁰))) =−4p(s(q, p) +s(q⁰, p)) + 2σ(K, p) = 0.

These imply the equalities which we want.

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Let K be a knot in ZHS Σ, and assume: Σ

_K

(p/q) ∼ = −Σ

_K

(p/q

⁰

).

Corollary

If 1 ≤ p ≤ 10, then one of the following holds.

1. a2(K) = 0orq=−q⁰, andσ(K, p) = 0.

2. p= 7,q= 7s+ 1,q⁰=−7s−2 (s∈Z),a2(K) =−1andσ(K,7) = 4.

3. p= 7,q= 7s+ 2,q⁰=−7s−1 (s∈Z),a2(K) =−1andσ(K,9) =−4.

4. p= 9,q= 1 + 9s,q⁰= 2−9s(s∈Z),a2(K) = 1andσ(K,9) = 12.

5. p= 9,q=−1 + 9s,q⁰=−2−9s(s∈Z),a2(K) = 1andσ(K,9) =−12.

6. p= 9,q= 1 + 9s,q⁰=−4−9s(s∈Z),a2(K) =−1andσ(K,9) = 12.

7. p= 9,q=−1 + 9s,q⁰= 4−9s(s∈Z),a2(K) =−1andσ(K,9) =−12.

Corollary

If ∆

_K

(ζ) 6= 0 for any p-th root of unity ζ and σ(K) ≡ 0 (mod 4), then p must be odd.

In fact, (chirally) cosmetic surgeries with surgery slopes of even numerators seems difficult to find (c.f. [Ichihara]).

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Result (2) :by SL(2, C ) Casson invariant

(Very Rough) Definition. [Curtis, 2001]

For a closed orientable 3-manifold Σ = W

₁

∪

_F

W

₂

, the

SL(2,C)

Casson invariant λ

_SL(2,_C₎

(Σ) is defined as an oriented intersection number of X

^∗

(W

1

) and X

^∗

(W

2

) in X

^∗

(F ) which counts only compact, zero-dimensional components of the intersection.

Theorem 2.

Let K be a hyperbolic small knot in ZHS Σ. Assume that two slopes p/q and p/q

⁰

are admissible with Σ

_K

(p/q) ∼ = ±Σ

_K

(p/q

⁰

).

If all the boundary slopes are non-negative (resp. non-positive), then

^q+q_p ⁰

> 0 (resp.

^q+q_p ⁰

< 0).

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Result (3)

Theorem 3.

Let K be a

two-bridge knot of genus one.

If the r- and r

⁰

-surgeries on K are chirally cosmetic, then either (i) K is amphicheiral and r = −r

⁰

, or

(ii) K is the positive or the negative trefoil, and

{r, r⁰}=

18k+ 9

3k+ 1 ,18k+ 9 3k+ 2

,

−18k+ 9

3k+ 1 ,−18k+ 9 3k+ 2

(k∈Z).

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Double twist knots Let K be a two-bridge knot of genus one.

the double twist knot J (`, m) with even `, m & ` > 0.

|{z}

` half twists

| {z }

m half twists

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Double twist knots Let K be a two-bridge knot of genus one.

From [Hatcher-Thurston], K must be represented as the double twist knot J (`, m) with even `, m & ` > 0.

|{z}

` half twists

| {z }

m half twists

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The case of `, m > 0 (1)

Assume: K is not trefoil, i.e., a hyperbolic small knot.

Suppose that K(p/q) ∼ = −K(p/q

⁰

)

•

K is trefoil, i.e., ` = m = 2

⇔ a

2

(K) = 1

⇔ ∆

K

(t) has root of unity as root.

•

By [Boden-Curtis], all the slopes for 2-br. knots are regular.

⇒ a slope is admissible iff it is NOT a

∂-slope.

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The case of `, m > 0 (1)

Assume: K is not trefoil, i.e., a hyperbolic small knot.

Suppose that K(p/q) ∼ = −K(p/q

⁰

)

To apply Theorem 2, we need to check the

admissibility

of slopes.

•

K is trefoil, i.e., ` = m = 2

⇔ a

2

(K) = 1

⇔ ∆

K

(t) has root of unity as root.

•

By [Boden-Curtis], all the slopes for 2-br. knots are regular.

⇒ a slope is admissible iff it is NOT a

∂-slope.

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The case of `, m > 0 (2)

Suppose: p/q and p/q

⁰

are both admissible

The mirror image K! of K = J (`, m) is a positive knot.

Then, by [Mattman-Maybrun-Robinson], K! enjoys

Property (+): all the boundary slopes are non-negative

By Property (+) and Theorem 2, we have

^q+q_p ⁰

> 0.

i.e., σ(K!, p) < 0 [Przytycki-Taniyama].

(b)

All the coeff. of the Conway poly. of K! are non-negative. In particular, a

₂

(K!) ≥ 0 [Cromwell].

By (a), (b) and Theorem 1, we have

^q+q_p ⁰

< 0. A contradiction occurs.

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The case of `, m > 0 (2)

Suppose: p/q and p/q

⁰

are both admissible

The mirror image K! of K = J (`, m) is a positive knot.

Then, by [Mattman-Maybrun-Robinson], K! enjoys

Property (+): all the boundary slopes are non-negative

By Property (+) and Theorem 2, we have

^q+q_p ⁰

> 0.

(a)

The total p-signature of K ! is always negative:

i.e., σ(K!, p) < 0 [Przytycki-Taniyama].

(b)

All the coeff. of the Conway poly. of K! are non-negative.

In particular, a

₂

(K!) ≥ 0 [Cromwell].

By (a), (b) and Theorem 1, we have

^q+q_p ⁰

< 0.

A contradiction occurs.

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The case of m < 0 < ` (1)

Claim

The knot J(`, m) is amphicheiral if and only if m = −`.

Proof

From the diagram, J(`, m) is amphicheiral if m = −`.

For K = J(`, m), v

3

(K) is computed as v

3

(J(`, m)) =

^m`₆₄

(m + `). If K is amphicheiral knot K, v

3

(K) = 0 must hold

(since v

₃

(K) = −v

₃

(K!) for the mirror image K! of K). Thus, if J (`, m) is amphicheiral, then m = −` holds.

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The case of m < 0 < ` (1)

Claim

The knot J(`, m) is amphicheiral if and only if m = −`.

Proof

From the diagram, J(`, m) is amphicheiral if m = −`.

v3(K): the primitive finite type inv. of deg. 3 of a knot

K in S

³

For K = J(`, m), v

3

(K) is computed as v

3

(J (`, m)) =

^m`₆₄

(m + `).

If K is amphicheiral knot K, v

3

(K) = 0 must hold (since v

₃

(K) = −v

₃

(K!) for the mirror image K! of K).

Thus, if J (`, m) is amphicheiral, then m = −` holds.

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The case of m < 0 < ` (2) Suppose that K(p/q) ∼ = −K(p/q

⁰

)

⇒ a

2

(K ) =

^`m₄

< 0.

⇒ ∆

_K

(t) has no roots on the unit circle {z ∈

C

| |z| = 1}.

⇒ σ

_ω

(K) = 0 for all ω ∈ {z ∈

C

| |z| = 1}. In particular, σ(K, p) = 0 for all p.

By Theorem 1, we obtain that q + q

⁰

= 0, i.e., q = −q

⁰

. However, v

₃

(K) =

^`m₆₄

(` + m) 6= 0, contradicting to: Theorem [Ito, Corollary 1.3]

Assume thatp/q- andp/q⁰-surgeries on a knotKinS³are chirally cosmetic (i) Ifq=−q⁰, thenv3(K) = 0.

(ii) Ifq6=−q⁰, thenv3(K)6= 0and p

q+q⁰ =7a2(K)²−a2(K)−10a4(K)

8v3(K) .

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The case of m < 0 < ` (2)

Suppose that K(p/q) ∼ = −K(p/q

⁰

)

Suppose for a contrary that K is NOT amphicheiral, i.e., m 6= −`

⇒ a

2

(K ) =

^`m₄

< 0.

⇒ ∆

_K

(t) has no roots on the unit circle {z ∈

C

| |z| = 1}.

⇒ σ

_ω

(K) = 0 for all ω ∈ {z ∈

C

| |z| = 1}.

In particular, σ(K, p) = 0 for all p.

By Theorem 1, we obtain that q + q

⁰

= 0, i.e., q = −q

⁰

. However, v

₃

(K) =

^`m₆₄

(` + m) 6= 0, contradicting to: Theorem [Ito, Corollary 1.3]

q+q⁰ =7a2(K)²−a2(K)−10a4(K)

8v3(K) .

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The case of m < 0 < ` (2)

Suppose that K(p/q) ∼ = −K(p/q

⁰

)

Suppose for a contrary that K is NOT amphicheiral, i.e., m 6= −`

⇒ a

2

(K ) =

^`m₄

< 0.

⇒ ∆

_K

(t) has no roots on the unit circle {z ∈

C

| |z| = 1}.

⇒ σ

_ω

(K) = 0 for all ω ∈ {z ∈

C

| |z| = 1}.

In particular, σ(K, p) = 0 for all p.

By Theorem 1, we obtain that q + q

⁰

= 0, i.e., q = −q

⁰

.

Theorem [Ito, Corollary 1.3]

q+q⁰ =7a2(K)²−a2(K)−10a4(K)

8v3(K) .

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The case of m < 0 < ` (2)

Suppose that K(p/q) ∼ = −K(p/q

⁰

)

Suppose for a contrary that K is NOT amphicheiral, i.e., m 6= −`

⇒ a

2

(K ) =

^`m₄

< 0.

⇒ ∆

_K

(t) has no roots on the unit circle {z ∈

C

| |z| = 1}.

⇒ σ

_ω

(K) = 0 for all ω ∈ {z ∈

C

| |z| = 1}.

In particular, σ(K, p) = 0 for all p.

By Theorem 1, we obtain that q + q

⁰

= 0, i.e., q = −q

⁰

. However, v

₃

(K) =

^`m₆₄

(` + m) 6= 0, contradicting to:

Theorem [Ito, Corollary 1.3]

q+q⁰ =7a2(K)²−a2(K)−10a4(K)

8v3(K) .

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Trefoil

We can give a complete list of cosmetic surgeries on torus knots, based on [Rong].

Theorem

Let T

r,s

be the (r, s)-torus knot. If T

r,s

(p/q) ∼ = ±T

r,s

(p/q

⁰

), then T

r,s

(p/q) ∼ = −T

r,s

(p/q

⁰

), s = 2 and

p/q = 2r

²

(2m + 1)

r(2m + 1) + 1 , p/q

⁰

= 2r

²

(2m + 1) r(2m + 1) − 1 for a positive integer m. Conversely, if p/q and p/q

⁰

are such rational numbers, then T

r,2

(p/q) ∼ = −T

r,2

(p/q

⁰

) for any odd integer r ≥ 3.

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Acknowledgements

Thank you for your attention.

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